Problem Either solve y^{\prime \prime }+4y=\cos x with y^{\prime }\left ( 0\right ) =0,y^{\prime }\left ( \pi \right ) =0 or show it has no solution.
Solution The homogeneous solution y_{h} can be easily found to be y_{h}=c_{1}\cos \left ( 2x\right ) +c_{2}\sin \left ( 2x\right ) Therefore the basis solutions are \begin{align*} y_{1} & =\cos 2x\\ y_{2} & =\sin 2x \end{align*}
And \begin{align*} y_{1}^{\prime } & =-2\sin 2x\\ y_{2}^{\prime } & =2\cos 2x \end{align*}
Hence y_{h}^{\prime }\left ( x\right ) =-2c_{1}\sin 2x+2c_{2}\cos 2x To find particular solution, let y_{p}=A\cos x The original ODE becomes\begin{align*} -A\cos x+4A\cos x & =\cos x\\ 3A\cos x & =\cos x\\ A & =\frac{1}{3} \end{align*}
Hence the full solution is\begin{align*} y\left ( x\right ) & =y_{h}+y_{p}\\ & =-2c_{1}\sin 2x+2c_{2}\cos 2x+\frac{1}{3}\cos x \end{align*}
Therefore y^{\prime }\left ( x\right ) =-4c_{1}\cos 2x-4c_{2}\sin 2x-\frac{1}{3}\sin x First B.C. gives\begin{align*} y^{\prime }\left ( 0\right ) & =0=-4c_{1}\\ c_{1} & =0 \end{align*}
Therefore the solution now becomes y\left ( x\right ) =2c_{2}\cos 2x+\frac{1}{3}\cos x and y^{\prime }\left ( x\right ) =-4c_{2}\sin 2x-\frac{1}{3}\sin x. The second B.C. gives\begin{align*} y^{\prime }\left ( \pi \right ) & =0=-4c_{2}\left ( 0\right ) \\ 0 & =-4c_{2}\left ( 0\right ) \end{align*}
Hence c_{2} can be any value. Therefore, there is no unique solution. There are infinite number of solutions.
Final solution is y\left ( x\right ) =2c_{2}\cos 2x+\frac{1}{3}\cos x Since 2c_{2} is constant, we can rename it to A and write the above as \fbox{$y\left ( x\right ) =A\cos 2x+\frac{1}{3}\cos x$} To verify that there is no unique solution, we set up W where y_{1}=\cos 2x,y_{2}=\sin 2x, and y_{1},y_{2} as found above. These are the two basis solutions for the homogeneous ODE. W=\begin{vmatrix} y_{1}^{\prime }\left ( 0\right ) & y_{2}^{\prime }\left ( 0\right ) \\ y_{1}^{\prime }\left ( \pi \right ) & y_{2}^{\prime }\left ( \pi \right ) \end{vmatrix} =\begin{vmatrix} 0 & 2\\ 0 & 2 \end{vmatrix} =0 Since W=0, this implies there is no unique solution. Therefore the ODE can has no solution, or it can have an infinite number of solutions. In this case, as shown above, it has infinite number of solutions.
Problem Either solve x^{2}y^{\prime \prime }+3xy^{\prime }+y=x^{2} with y\left ( 1\right ) =0,y\left ( e\right ) =0 or show it has no solution.
Solution The homogeneous solution is first found. This is a Euler ODE. Let y_{h}=x^{r}, then y_{h}^{\prime }=rx^{r-1},y_{h}^{\prime \prime }=r\left ( r-1\right ) x^{r-2} and the homogeneous ODE becomes
\begin{align*} r\left ( r-1\right ) x^{r}+3rx^{r}+x^{r} & =0\\ r\left ( r-1\right ) +3r+1 & =0\\ r^{2}-r+3r+1 & =0\\ r^{2}+2r+1 & =0\\ \left ( r+1\right ) \left ( r+1\right ) & =0 \end{align*}
Hence double roots. Therefore the solution is y_{h}=c_{1}\frac{1}{x}+c_{2}\frac{1}{x}\ln x To find particular solution, let y_{p}=c_{1}+c_{2}x+c_{3}x^{2}. Plugging this in original ODE gives\begin{align*} x^{2}\left ( 2c_{3}\right ) +3x\left ( c_{2}+2c_{3}x\right ) +\left ( c_{1}+c_{2}x+c_{3}x^{2}\right ) & =x^{2}\\ x^{2}\left ( 2c_{3}\right ) +c_{1}+x\left ( 3c_{2}+c_{2}\right ) +x^{2}\left ( 6c_{3}+c_{3}\right ) & =x^{2} \end{align*}
Comparing coefficients gives\begin{align*} c_{1} & =0\\ 4c_{2} & =0\\ 9c_{3} & =1 \end{align*}
Hence solution is c_{2}=0,c_{1}=0,c_{3}=\frac{1}{9}. Therefore y_{p}=\frac{1}{9}x^{2} and the full solution is\begin{equation} \fbox{$y\left ( x\right ) =c_1\frac{1}{x}+c_2\frac{1}{x}\ln x+\frac{1}{9}x^2$} \tag{1} \end{equation} Boundary conditions are now applied to find c_{1},c_{2}. First BC gives\begin{align*} 0 & =c_{1}+c_{2}\ln 1+\frac{1}{9}\\ 0 & =c_{1}+\frac{1}{9}\\ c_{1} & =-\frac{1}{9} \end{align*}
Second BC y\left ( e\right ) =0 gives\begin{align*} 0 & =c_{1}\frac{1}{e}+c_{2}\frac{1}{e}\ln e+\frac{1}{9}e^{2}\\ 0 & =-\frac{1}{9e}+c_{2}\frac{1}{e}+\frac{1}{9}e^{2}\\ c_{2} & =\frac{1}{9}-\frac{1}{9}e^{3}\\ & =\frac{1-e^{3}}{9} \end{align*}
Therefore the solution (1) becomes \fbox{$y\left ( x\right ) =-\frac{1}{9x}+\frac{x^2}{9}+\left ( \frac{1-e^3}{9}\right ) \frac{1}{x}\ln x$} Therefore solution exist and is unique. This is verified using W where now y_{1}=\frac{1}{x},y_{2}=\frac{1}{x}\ln x. These are found above as the bases solutions for the homogeneous ODE. W=\begin{vmatrix} y_{1}\left ( 1\right ) & y_{2}\left ( 1\right ) \\ y_{1}\left ( e\right ) & y_{2}\left ( e\right ) \end{vmatrix} =\begin{vmatrix} 1 & 0\\ \frac{1}{e} & \frac{1}{e}\end{vmatrix} =\frac{1}{e}\neq 0 This confirms that a unique solution exists.
Problem Find eigenvalue and eigenfunction of y^{\prime \prime }+\lambda y=0 with y\left ( 0\right ) =0,y^{\prime }\left ( \pi \right ) =0.
Solution
Assuming the solution is y=Ae^{rx}, then the characteristic equation is\begin{align*} r^{2}+\lambda & =0\\ r & =\pm \sqrt{-\lambda } \end{align*}
Case \lambda <0
In this case -\lambda is positive and hence \sqrt{-\lambda } is also positive. Let \sqrt{-\lambda }=\mu where \mu >0. Hence the roots are \pm \mu . This gives the solution y=c_{1}\cosh \left ( \mu x\right ) +c_{2}\sinh \left ( \mu x\right ) First BC gives 0=c_{1} Hence solution becomes y\left ( x\right ) =c_{2}\sinh \left ( \mu x\right ) Second BC gives\begin{align*} y^{\prime }\left ( x\right ) & =\mu c_{2}\cosh \left ( \mu x\right ) \\ 0 & =\mu c_{2}\cosh \left ( \mu \pi \right ) \end{align*}
But \cosh \mu \pi \neq 0, hence only other choice is c_{2}=0, leading to trivial solution. Therefore \lambda <0 is not eigenvalue.
Case \lambda =0, then the homogenous solution is y\left ( x\right ) =c_{1}+c_{2}x First BC gives 0=c_{1} Hence solution becomes y\left ( x\right ) =c_{2}x Second BC gives\begin{align*} y^{\prime }\left ( x\right ) & =c_{2}\\ 0 & =c_{2} \end{align*}
Leading to trivial solution. Therefore \lambda =0 is not eigenvalue.
Case \lambda >0, then the homogenous solution is y\left ( x\right ) =c_{1}\cos \left ( \sqrt{\lambda }x\right ) +c_{2}\sin \left ( \sqrt{\lambda }x\right ) First BC gives 0=c_{1} Hence solution becomes y\left ( x\right ) =c_{2}\sin \left ( \sqrt{\lambda }x\right ) Second BC gives\begin{align*} y^{\prime }\left ( x\right ) & =\sqrt{\lambda }c_{2}\cos \left ( \sqrt{\lambda }x\right ) \\ 0 & =\sqrt{\lambda }c_{2}\cos \left ( \sqrt{\lambda }\pi \right ) \end{align*}
Non-trivial solution requires \cos \left ( \sqrt{\lambda }\pi \right ) =0 or \sqrt{\lambda }\pi =\frac{n\pi }{2} for n=1,3,5,\cdots . Therefore\begin{align*} \sqrt{\lambda _{n}}\pi & =\frac{n\pi }{2}\\ \sqrt{\lambda _{n}} & =\frac{n}{2}\qquad n=1,3,5,\cdots \end{align*}
Hence the eigenvalues are \lambda _{n}=\left ( \frac{n}{2}\right ) ^{2}\qquad n=1,3,5,\cdots And the corresponding eigenfunction is \sin \left ( \frac{n}{2}x\right ) for n=1,3,5,\cdots . The solution is y\left ( x\right ) =\sum _{n=1,3,5,\cdots }^{\infty }c_{n}\sin \left ( \frac{n}{2}x\right )
Problem Find eigenvalue and eigenfunction of x^{2}y^{\prime \prime }-xy^{\prime }+\lambda y=0 with y\left ( 1\right ) =0,y\left ( L\right ) =0,L>1
Solution
This is Euler type ODE. Using standard substitution, ley y=x^{r}. The ODE now becomes\begin{align*} x^{2}r\left ( r-1\right ) x^{r-2}-xrx^{r-1}+\lambda x^{r} & =0\\ r\left ( r-1\right ) -r+\lambda & =0\\ r^{2}-2r+\lambda & =0 \end{align*}
The above is called the characteristic equations. Its roots give the solution. The roots are r=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{2\pm \sqrt{4-4\lambda }}{2}=1\pm \sqrt{1-\lambda } case 1-\lambda >0
Let 1-\lambda =\mu ^{2} for some real \mu . Then the roots are 1\pm \mu and hence the solution is\begin{align*} y & =c_{1}x^{r_{1}}+c_{2}x^{r_{2}}\\ & =c_{1}x^{1+\mu }+c_{2}x^{1-\mu }\\ & =x\left ( c_{1}x^{\mu }+c_{2}\frac{1}{x^{\mu }}\right ) \end{align*}
At first BC y\left ( 1\right ) =0 the above gives 0=c_{1}+c_{2} At second BC y\left ( L\right ) =0\begin{align*} 0 & =L\left ( c_{1}L^{\mu }+c_{2}\frac{1}{L^{\mu }}\right ) \\ 0 & =c_{1}L^{\mu }+c_{2}\frac{1}{L^{\mu }}\\ 0 & =\frac{c_{1}L^{2\mu }+c_{2}}{L^{\mu }} \end{align*}
Hence c_{1}L^{2\mu }+c_{2}=0 But c_{2}=-c_{1}, therefore\begin{align*} c_{1}L^{2\mu }-c_{1} & =0\\ c_{1}\left ( L^{2\mu }-1\right ) & =0 \end{align*}
For arbitrary L>0 the above can only be satisfied if c_{1}=0. This means both c_{1},c_{2} are zero. Hence 1-\lambda >0 is not possible.
case 1-\lambda =0
Hence the roots now are r=1. Double root. We now in the case of double root the solution can be written as\begin{align*} y & =c_{1}x^{r_{1}}+c_{2}x^{r_{1}}\ln x\\ & =c_{1}x+c_{2}x\ln x \end{align*}
At first BC y\left ( 1\right ) =0 the above gives 0=c_{1} Therefore the solution now becomes y=c_{2}x\ln x. At second BC y\left ( L\right ) =0\begin{align*} 0 & =c_{2}L\ln L\\ 0 & =c_{2}\ln L \end{align*}
Since L>0 then only possibility is that c_{2}=0.This means both c_{1},c_{2} are zero. Hence 1-\lambda =0 is not possible.
case 1-\lambda <0
Let 1-\lambda =-\mu ^{2} for some real \mu . Then the roots are 1\pm i\mu and hence the solution is\begin{align*} y & =c_{1}x^{r_{1}}+c_{2}x^{r_{2}}\\ & =c_{1}x^{1+i\mu }+c_{2}x^{1-i\mu }\\ & =x\left ( c_{1}x^{i\mu }+c_{2}x^{-iu}\right ) \end{align*}
The above can be written as \begin{align*} y & =x\left ( c_{1}e^{\ln x^{i\mu }}+c_{2}e^{\ln x^{-i\mu }}\right ) \\ & =x\left ( c_{1}e^{i\mu \ln x}+c_{2}e^{-i\mu \ln x}\right ) \end{align*}
Hence c_{1}e^{i\mu \ln x}+c_{2}e^{-i\mu \ln x} can be written as C_{1}\cos \left ( \mu \ln x\right ) +C_{2}\sin \left ( \mu \ln x\right ) . This is done using Euler relation and the new constants C_{1},C_{2} are not the same as c_{1},c_{2}. The solution becomes y=x\left ( C_{1}\cos \left ( \mu \ln x\right ) +C_{2}\sin \left ( \mu \ln x\right ) \right ) First BC y\left ( 1\right ) =0 the above becomes\begin{align*} 0 & =C_{1}\cos \left ( \mu \ln 1\right ) +C_{2}\sin \left ( \mu \ln 1\right ) \\ & =C_{1} \end{align*}
Therefore the solution is\begin{equation} y=xC_{2}\sin \left ( \mu \ln x\right ) \tag{1} \end{equation} For second BC y\left ( L\right ) =0 the above becomes\begin{align*} 0 & =LC_{2}\sin \left ( \mu \ln L\right ) \\ 0 & =C_{2}\sin \left ( \mu \ln L\right ) \end{align*}
Non-trivial solution requires\sin \left ( \mu \ln L\right ) =0 or \mu \ln L=n\pi for n=1,2,3,\cdots . This means \mu =\frac{n\pi }{\ln L}\qquad n=1,2,3,\cdots But 1-\lambda =-\mu ^{2}, or \lambda =1+\mu ^{2}, therefore\begin{equation} \lambda _{n}=1+\left ( \frac{n\pi }{\ln L}\right ) ^{2}\qquad n=1,2,3,\cdots \tag{2} \end{equation} These are the eigenvalues. The corresponding eigenfunctions are from (1)\begin{align*} y_{n}\left ( x\right ) & =c_{n}x\sin \left ( \mu _{n}\ln x\right ) \\ & =c_{n}x\sin \left ( \sqrt{\lambda _{n}-1}\ln x\right ) \\ & =c_{n}x\sin \left ( \sqrt{1+\left ( \frac{n\pi }{\ln L}\right ) ^{2}-1}\ln x\right ) \\ & =c_{n}x\sin \left ( \sqrt{\left ( \frac{n\pi }{\ln L}\right ) ^{2}}\ln x\right ) \\ & =c_{n}x\sin \left ( \frac{n\pi }{\ln L}\ln x\right ) \qquad n=1,2,3,\cdots \end{align*}
Hence the solution is y\left ( x\right ) =x\sum _{n=1}^{\infty }c_{n}\sin \left ( \frac{n\pi }{\ln L}\ln x\right )
This is standard ODE with constant coefficients. Just integrating and substitutions.
Problem Sketch the graph of f\left ( x\right ) =-x,-L\leq x<L where f\left ( x+2L\right ) =f\left ( x\right ) and find the Fourier series of the function
Solution
This is an odd function. Only b_{n} needs to be evaluated. b_{n}=\frac{1}{T/2}\int _{-T/2}^{T/2}f\left ( x\right ) \sin \left ( n\frac{2\pi }{T}x\right ) T is the period of f\left ( x\right ) which is 2L. The above becomes b_{n}=\frac{1}{L}\int _{-L}^{L}-x\sin \left ( n\frac{\pi }{L}x\right ) Since x is odd and \sin is odd then the product is even and the above simplifies to\begin{equation} b_{n}=\frac{-2}{L}\int _{0}^{L}x\sin \left ( n\frac{\pi }{L}x\right ) \tag{1} \end{equation} Using integration by parts \int udv=uv-\int vdu where u=x,dv=\sin \left ( n\frac{\pi }{L}x\right ) , therefore du=1 and v=-\frac{\cos \left ( n\frac{\pi }{L}x\right ) }{n\frac{\pi }{L}}=\frac{-L}{n\pi }\cos \left ( n\frac{\pi }{L}x\right ) Integral (1) becomes\begin{align*} b_{n} & =\frac{-2}{L}\left ( \left [ \frac{-L}{n\pi }x\cos \left ( n\frac{\pi }{L}x\right ) \right ] _{0}^{L}-\int _{0}^{L}\frac{-L}{n\pi }\cos \left ( n\frac{\pi }{L}x\right ) dx\right ) \\ & =\frac{-2}{L}\left ( \left [ \frac{-L^{2}}{n\pi }\cos \left ( n\pi \right ) -0\right ] +\frac{-L}{n\pi }\int _{0}^{L}\cos \left ( n\frac{\pi }{L}x\right ) dx\right ) \\ & =\frac{-2}{L}\left ( \frac{-L^{2}}{n\pi }\cos \left ( n\pi \right ) +\frac{-L}{n\pi }\frac{1}{n\frac{\pi }{L}}\left [ \sin \left ( n\frac{\pi }{L}x\right ) \right ] _{0}^{L}\right ) \\ & =\frac{-2}{L}\left ( \frac{-L^{2}}{n\pi }\cos \left ( n\pi \right ) +\frac{-L^{2}}{n^{2}\pi ^{2}}\left [ \sin \left ( n\pi \right ) -0\right ] \right ) \\ & =\frac{-2}{L}\frac{-L^{2}}{n\pi }\cos \left ( n\pi \right ) \\ & =\frac{2L}{n\pi }\cos \left ( n\pi \right ) \end{align*}
For n=1,2,3,\cdots . Looking at few n values gives\begin{align*} b_{n} & =\frac{2L}{\pi }\left ( -1\right ) ,\frac{2L}{2\pi },\frac{2L}{3\pi }\left ( -1\right ) ,\cdots \\ & =\frac{2L}{n\pi }\left ( -1\right ) ^{n} \end{align*}
Therefore the Fourier series is\begin{align*} f\left ( x\right ) & =\sum _{n=1}^{\infty }\frac{2L}{n\pi }\left ( -1\right ) ^{n}\sin \left ( \frac{n\pi }{L}x\right ) \\ & =\frac{2L}{\pi }\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}}{n}\sin \left ( \frac{n\pi }{L}x\right ) \end{align*}
Problem Sketch the graph and find the Fourier series of the function f\left ( x\right ) =\left \{ \begin{array} [c]{cc}0 & -2\leq x\leq -1\\ x & -1<x<1\\ 0 & 1\leq x<2 \end{array} \right . And f\left ( x+4\right ) =f\left ( x\right )
Solution
f\left ( x\right ) is an odd function. Therefore only b_{n} needs to be evaluated. b_{n}=\frac{1}{L}\int _{-L}^{L}f\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) 2L is the period of f\left ( x\right ) which is 4. Hence L=2. The above becomes\begin{align*} b_{n} & =\frac{1}{2}\int _{-2}^{2}f\left ( x\right ) \sin \left ( \frac{n\pi }{2}x\right ) \\ & =\frac{1}{2}\left ( \int _{-2}^{-1}f\left ( x\right ) \sin \left ( \frac{n\pi }{2}x\right ) +\int _{-1}^{1}f\left ( x\right ) \sin \left ( \frac{n\pi }{2}x\right ) +\int _{1}^{2}f\left ( x\right ) \sin \left ( \frac{n\pi }{2}x\right ) \right ) \\ & =\frac{1}{2}\int _{-1}^{1}f\left ( x\right ) \sin \left ( \frac{n\pi }{2}x\right ) \\ & =\frac{1}{2}\int _{-1}^{1}x\sin \left ( \frac{n\pi }{2}x\right ) \end{align*}
Since x is odd and \sin is odd then the product is even and the above simplifies to\begin{equation} b_{n}=\int _{0}^{1}x\sin \left ( \frac{n\pi }{2}x\right ) \tag{1} \end{equation} Using integration by parts \int udv=uv-\int vdu where u=x,dv=\sin \left ( \frac{n\pi }{2}x\right ) , therefore du=1 and v=-\frac{\cos \left ( \frac{n\pi }{2}x\right ) }{\frac{n\pi }{2}}=\frac{-2}{n\pi }\cos \left ( \frac{n\pi }{2}x\right ) Integral (1) becomes\begin{align*} b_{n} & =\frac{-2}{n\pi }\left [ x\cos \left ( n\frac{\pi }{2}x\right ) \right ] _{0}^{1}-\int _{0}^{1}\frac{-2}{n\pi }\cos \left ( n\frac{\pi }{2}x\right ) dx\\ & =\frac{-2}{n\pi }\left [ \cos \left ( \frac{n}{2}\pi \right ) \right ] +\frac{2}{n\pi }\int _{0}^{1}\cos \left ( n\frac{\pi }{2}x\right ) dx\\ & =\left ( \frac{-2}{n\pi }\cos \left ( \frac{n}{2}\pi \right ) +\frac{2}{n\pi }\frac{1}{n\frac{\pi }{2}}\left [ \sin \left ( n\frac{\pi }{2}x\right ) \right ] _{0}^{1}\right ) \\ & =\left ( \frac{-2}{n\pi }\cos \left ( \frac{n}{2}\pi \right ) +\frac{4}{n^{2}\pi ^{2}}\sin \left ( \frac{n}{2}\pi \right ) \right ) \end{align*}
Therefore b_{n}=\left ( \frac{2}{n\pi }\right ) ^{2}\sin \left ( \frac{n}{2}\pi \right ) -\frac{2}{n\pi }\cos \left ( \frac{n}{2}\pi \right ) \qquad n=1,2,3,\cdots The Fourier series is f\left ( x\right ) =\sum _{n=1}^{\infty }\left [ \left ( \frac{2}{n\pi }\right ) ^{2}\sin \left ( \frac{n}{2}\pi \right ) -\frac{2}{n\pi }\cos \left ( \frac{n}{2}\pi \right ) \right ] \sin \left ( \frac{n\pi }{2}x\right )
Problem Assume the function is periodically extended outside the original interval. (a) Find the Fourier series of the extended function. (b) Sketch the graph of the function to which the series converges for three periods. f\left ( x\right ) =\left \{ \begin{array} [c]{ccc}0 & & -\pi \leq x<0\\ x & & 0\leq x<\pi \end{array} \right . Solution
This is plot of the above function for one period, and then for 3 periods
The calculation of the Fourier series will have a_{n},b_{n} and will follow same methods as before. The period here is 2\pi .
Since both f\left ( x\right ) and f^{\prime }\left ( x\right ) are piecewise continuous, then the Fourier series will converge to the function f\left ( x\right ) . But at the points where f\left ( x\right ) has jumps (such as at x=\pm \pi ) the Fourier series will converge to the average value of f\left ( x\right ) at these points.
Problem Assume the function is periodically extended outside the original interval. (a) Find the Fourier series of the extended function. (b) Sketch the graph of the function to which the series converges for three periods.
f\left ( x\right ) =1-x^{2}\qquad -1\leq x<1
Solution
This is plot of the above function for one period, and then for 3 periods
The calculation of the Fourier series will have only a_{n} since f\left ( x\right ) is even, and will follow same methods as before. The period here is 2.
Since both f\left ( x\right ) and f^{\prime }\left ( x\right ) are piecewise continuous, then the Fourier series will converge to the function f\left ( x\right ) for all x.
Problem (a) Find the Fourier series of the given function (b) Sketch the graph of the function to which the series converges for three periods. f\left ( x\right ) =1\qquad 0\leq x\leq \pi Use cosine series, with period 2\pi .
Solution
Extending this as even function gives f_{e}\left ( x\right ) =1\qquad -\pi <x\leq \pi Hence, since period is 2\pi , then L=\pi now and a_{0}=\frac{1}{L}\int _{-L}^{L}f_{e}\left ( x\right ) dx=\frac{1}{\pi }\int _{-\pi }^{\pi }dx=\frac{2}{\pi }\int _{0}^{\pi }dx=2 And a_{n}=\frac{1}{L}\int _{-L}^{L}f_{e}\left ( x\right ) \cos \left ( \frac{n\pi }{L}x\right ) dx=\frac{2}{\pi }\int _{0}^{\pi }\cos \left ( nx\right ) dx=\frac{2}{n\pi }\left ( -\sin \left ( nx\right ) \right ) _{0}^{\pi }=0 Therefore the cosine extension Fourier series is\begin{align*} f\left ( x\right ) & =\frac{a_{0}}{2}+\sum _{n=1}^{\infty }a_{n}\cos \left ( nx\right ) \\ & =\frac{a_{0}}{2}\\ & =1 \end{align*}
Problem (a) Find the Fourier series of the given function (b) Sketch the graph of the function to which the series converges for three periods.
f\left ( x\right ) =1\qquad 0<x<\pi Use sin series, with period 2\pi .
Solution
Extending this as odd function gives f_{o}\left ( x\right ) =\left \{ \begin{array} [c]{ccc}1 & & 0<x<\pi \\ -1 & & -\pi <x\leq 0 \end{array} \right . Hence, since period is 2\pi , then L=\pi now and, since this is an odd function, only b_{n} terms will show up\begin{align*} b_{n} & =\frac{1}{L}\int _{-L}^{L}f_{o}\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx\\ & =\frac{1}{\pi }\int _{-\pi }^{\pi }f_{o}\left ( x\right ) \sin \left ( nx\right ) dx \end{align*}
But now f_{o}\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) is even, therefore the above simplifies to\begin{align*} b_{n} & =\frac{2}{\pi }\int _{0}^{\pi }f_{o}\left ( x\right ) \sin \left ( nx\right ) dx\\ & =\frac{2}{\pi }\int _{0}^{\pi }\sin \left ( nx\right ) dx\\ & =\frac{-2}{\pi }\left ( \frac{\cos \left ( nx\right ) }{n}\right ) _{0}^{\pi }\\ & =\frac{-2}{n\pi }\left ( \cos \left ( n\pi \right ) -1\right ) \\ & =\frac{-2}{n\pi }\left ( -1^{n}-1\right ) \end{align*}
Therefore the sine extension Fourier series is\begin{align*} f\left ( x\right ) & =\sum _{n=1}^{\infty }b_{n}\sin \left ( nx\right ) \\ & =\frac{-2}{\pi }\sum _{n=1}^{\infty }\frac{1}{n}\left ( -1^{n}-1\right ) \sin \left ( nx\right ) \end{align*}
Problem Find solution to u_{t}=100u_{xx} with 0<x<1,t>0 and boundary conditions u\left ( 0,t\right ) =u\left ( 1,t\right ) =0 and initial conditions u\left ( x,0\right ) =\sin \left ( 2\pi x\right ) -\sin \left ( 5\pi x\right )
Solution
The fundamental solution for this problem with homogenous B.C. was derived in earlier problem and it is given as u\left ( x,t\right ) =\sum _{n=1}^{\infty }c_{n}e^{-\lambda _{n}kt}\sin \left ( \sqrt{\lambda _{n}}x\right ) Where in this problem k=100 and \lambda _{n}=\left ( n\pi \right ) ^{2},n=1,2,3,\ldots . The c_{n} terms is the Fourier sine coefficients of the initial conditions. But the initial conditions is already expressed as sum of sine terms. Therefore the c_{n} coefficient can be read directly from f\left ( x\right ) , giving c_{2}=1,c_{5}=-1. Therefore only two terms exist in the sum above, leading to the solution\begin{align*} u\left ( x,t\right ) & =c_{2}e^{-\left ( 2\pi \right ) ^{2}\left ( 100\right ) t}\sin \left ( 2\pi x\right ) +c_{5}e-^{\left ( 5\pi \right ) ^{2}\left ( 100\right ) t}\sin \left ( 5\pi x\right ) \\ & =e^{-400\pi ^{2}t}\sin \left ( 2\pi x\right ) -e^{-25000\pi t}\sin \left ( 5\pi x\right ) \end{align*}
Problem Solve u_{t}=u_{xx}, with 0<x<L and L=40cm and boundary conditions u\left ( 0,t\right ) =u\left ( L,t\right ) =0^{0} with initial conditions u\left ( x,0\right ) =\left \{ \begin{array} [c]{ccc}x & & 0\leq x<20\\ 40-x & & 20\leq x\leq 40 \end{array} \right . Solution
The fundamental solution for this problem with homogenous B.C. was derived in earlier problem and it is given as u\left ( x,t\right ) =\sum _{n=1}^{\infty }c_{n}e^{-\lambda _{n}kt}\sin \left ( \sqrt{\lambda _{n}}x\right ) Where in this problem k=1 and \lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2},n=1,2,3,\ldots .and L=40 cm. To find c_{n}, initial conditions are used. At t=0 f\left ( x\right ) =\sum _{n=1}^{\infty }c_{n}\sin \left ( \sqrt{\lambda _{n}}x\right ) Applying orthogonality result in\begin{align*} c_{n} & =\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) dx\\ & =\frac{2}{40}\left ( \int _{0}^{20}x\sin \left ( \sqrt{\lambda _{n}}x\right ) dx+\int _{20}^{40}\left ( 40-x\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) dx\right ) \\ & =\frac{2}{40}\left ( \frac{3200}{n^{2}\pi ^{2}}\sin \left ( \frac{n\pi }{2}\right ) \right ) \\ & =\frac{160}{n^{2}\pi ^{2}}\sin \left ( \frac{n\pi }{2}\right ) \end{align*}
Hence the solution is u\left ( x,t\right ) =\frac{160}{\pi ^{2}}\sum _{n=1}^{\infty }\frac{1}{n^{2}}\sin \left ( \frac{n\pi }{2}\right ) e^{-\left ( \frac{n\pi }{40}\right ) ^{2}t}\sin \left ( \frac{n\pi }{40}x\right )
Problem
Solve u_{t}=u_{xx}, with 0<x<L and L=40cm and boundary conditions u\left ( 0,t\right ) =u\left ( L,t\right ) =0^{0} with initial conditions u\left ( x,0\right ) =\left \{ \begin{array} [c]{ccc}0 & & 0\leq x<10\\ 50 & & 10\leq x\leq 30\\ 0 & & 30\leq x\leq 40 \end{array} \right . Solution
The fundamental solution for this problem with homogenous B.C. was derived in earlier problem and it is given as u\left ( x,t\right ) =\sum _{n=1}^{\infty }c_{n}e^{-\lambda _{n}kt}\sin \left ( \sqrt{\lambda _{n}}x\right ) Where in this problem k=1 and \lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2},n=1,2,3,\ldots .and L=40 cm. To find c_{n}, initial conditions are used. At t=0 f\left ( x\right ) =\sum _{n=1}^{\infty }c_{n}\sin \left ( \sqrt{\lambda _{n}}x\right ) Applying orthogonality result in\begin{align*} c_{n} & =\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) dx\\ & =\frac{2}{40}\left ( \int _{0}^{10}0\sin \left ( \sqrt{\lambda _{n}}x\right ) dx+\int _{10}^{30}50\sin \left ( \sqrt{\lambda _{n}}x\right ) dx+\int _{30}^{40}0\sin \left ( \sqrt{\lambda _{n}}x\right ) dx\right ) \\ & =\frac{2}{40}\int _{10}^{30}50\sin \left ( \sqrt{\lambda _{n}}x\right ) dx\\ & =\frac{200}{n\pi }\sin \frac{n\pi }{4}\sin \frac{n\pi }{2} \end{align*}
Hence the solution is u\left ( x,t\right ) =\frac{200}{\pi }\sum _{n=1}^{\infty }\frac{1}{n}\sin \left ( \frac{n\pi }{4}\right ) \sin \left ( \frac{n\pi }{2}\right ) e^{-\left ( \frac{n\pi }{40}\right ) ^{2}t}\sin \left ( \frac{n\pi }{40}x\right )
Problem Find steady state solution that satisfies the given boundary conditions u_{t}=\alpha ^{2}u_{xx} with u\left ( 0,t\right ) =0,u_{x}\left ( L,t\right ) =0
solution at steady state\begin{align*} v^{\prime \prime }\left ( x\right ) & =0\\ v\left ( 0\right ) & =0\\ v^{\prime }\left ( L\right ) & =0 \end{align*}
Solution to the above ODE is v\left ( x\right ) =c_{1}x+c_{2}. At x=0, this leads to c_{2}=0. Therefore the solution now becomes v\left ( x\right ) =c_{1}x and v^{\prime }\left ( x\right ) =c_{1}. Second boundary condition implies c_{1}=0 as well. Therefore v\left ( x\right ) =0 is the steady state solution.
Problem Find steady state solution that satisfies the given boundary conditions u_{t}=\alpha ^{2}u_{xx} with u_{x}\left ( 0,t\right ) -u\left ( 0,t\right ) =0,u\left ( L,t\right ) =T
solution at steady state\begin{align*} v^{\prime \prime }\left ( x\right ) & =0\\ v^{\prime }\left ( 0\right ) -v\left ( 0\right ) & =0\\ v\left ( L\right ) & =T \end{align*}
Solution to the above ODE is v\left ( x\right ) =c_{1}x+c_{2}. At x=0, this leads to c_{1}-c_{2}=0. Second boundary condition implies c_{1}L+c_{2}=0. Two equations in 2 unknowns\begin{align*} c_{1}-c_{2} & =0\\ c_{1}L+c_{2} & =T \end{align*}
From first equation, c_{1}=c_{2}. Second equation becomes c_{2}\left ( 1+L\right ) =T or c_{2}=\frac{T}{1+L}. Therefore the steady state solution\begin{align*} v\left ( x\right ) & =\frac{T}{1+L}x+\frac{T}{1+L}\\ & =\frac{T}{1+L}\left ( 1+x\right ) \end{align*}
Problem Let L=20 cm, with initial temperature 25^{0}C, an initial conditions u\left ( 0,x\right ) =0,u\left ( L,0\right ) =60^{0}C. (a) Find u\left ( x,t\right ) . (b) Plot initial temperature distribution, final steady state solution and solution are two intermediate times on same axes. (c) Plot u vs. t for x=5,10,15. (d) determine how much time has elapsed before the temperature at x=5 cm comes and remains with 1\% of the steady state value. Use \alpha ^{2}=0.86
solution
\begin{align*} u_{t} & =\alpha ^{2}u_{xx}\\ u\left ( 0,x\right ) & =0\\ u\left ( L,0\right ) & =60 \end{align*}
Let solution be u\left ( x,t\right ) =w\left ( x,t\right ) +v\left ( x\right ) where v\left ( x\right ) is solution to v^{\prime \prime }\left ( x\right ) =0 with boundary conditions v\left ( 0\right ) =0,v\left ( L\right ) =60. Hence the solution is v\left ( x\right ) =c_{1}x+c_{2} At x=0, this leads to c_{2}=0. Therefore solution is v\left ( x\right ) =c_{1}x. At x=L, 60=c_{1}L or c_{1}=\frac{60}{L}=\frac{60}{20}=3. Therefore v\left ( x\right ) =3x Hence the complete solution is u\left ( x,t\right ) =\left ( \sum _{n=1}^{\infty }c_{n}e^{-\alpha ^{2}\lambda _{n}t}\sin \left ( \sqrt{\lambda _{n}}x\right ) \right ) +3x Where \lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2} for n=1,2,3,\cdots . c_{n} is now found from initial conditions. At t=0\begin{align*} 25 & =\left ( \sum _{n=1}^{\infty }c_{n}\sin \left ( \sqrt{\lambda _{n}}x\right ) \right ) +3x\\ 25-3x & =\sum _{n=1}^{\infty }c_{n}\sin \left ( \sqrt{\lambda _{n}}x\right ) \end{align*}
Applying orthogonality gives\begin{align*} \int _{0}^{L}\left ( 25-3x\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) dx & =c_{n}\frac{L}{2}\\ c_{n} & =\frac{2}{L}\int _{0}^{L}\left ( 25-3x\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) dx\\ & =\frac{2}{20}\int _{0}^{L}\left ( 25-3x\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) dx \end{align*}
Integrating gives c_{n}=\frac{50+70\left ( -1\right ) ^{n}}{n\pi }. Therefore the solution is u\left ( x,t\right ) =\left ( \sum _{n=1}^{\infty }\frac{50+70\left ( -1\right ) ^{n}}{n\pi }e^{-\alpha ^{2}\lambda _{n}t}\sin \left ( \sqrt{\lambda _{n}}x\right ) \right ) +3x
solution
To do.
Problem Consider elastic string of length L with ends held fixed. Let initial position u\left ( x,0\right ) =f\left ( x\right ) and u_{t}\left ( x,0\right ) =0. Let L=10,a=1. (a) Find u\left ( x,t\right ) . (b) Plot u\left ( x,t\right ) vs x for 0\leq x\leq 10 and for several values of time between t=0 and t=20 (c) Plot u\left ( x,t\right ) vs. t for 0\leq t\leq 20 and for several values of x (d) Construct an animation of the solution for at least one period. (e) Describe the motion of the string. Let f\left ( x\right ) =\frac{8x\left ( L-x\right ) ^{2}}{L^{3}}
solution Since domain is finite, it is easier to use the series solution for wave equation than D’Alembert solution. This is given by u\left ( x,t\right ) =\sum _{n=1}^{\infty }c_{n}\cos \left ( \sqrt{\lambda _{n}}at\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) Where \lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2},n=1,2,3,\cdots and c_{n}=\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) dx. Hence, since a=1 and L=10, the solution becomes\begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }c_{n}\cos \left ( \frac{n\pi }{10}t\right ) \sin \left ( \frac{n\pi }{10}x\right ) \\ c_{n} & =\frac{2}{10}\int _{0}^{10}\frac{8x\left ( L-x\right ) ^{2}}{L^{3}}\sin \left ( \frac{n\pi }{10}x\right ) dx\\ & =\frac{2}{10}\int _{0}^{10}\frac{8x\left ( 10-x\right ) ^{2}}{10^{3}}\sin \left ( \frac{n\pi }{10}x\right ) dx \end{align*}
Integrating gives c_{n}=\frac{32\left ( 2+\left ( -1\right ) ^{n}\right ) }{n^{3}\pi ^{3}} Hence solution is u\left ( x,t\right ) =\frac{32}{\pi ^{3}}\sum _{n=1}^{\infty }\frac{2+\left ( -1\right ) ^{n}}{n^{3}}\cos \left ( \frac{n\pi }{10}t\right ) \sin \left ( \frac{n\pi }{10}x\right )
Problem Consider elastic string of length L with ends held fixed. Let initial position u\left ( x,0\right ) =0 and u_{t}\left ( x,0\right ) =g\left ( x\right ) . Let L=10,a=1. (a) Find u\left ( x,t\right ) . (b) Plot u\left ( x,t\right ) vs x for 0\leq x\leq 10 and for several values of time between t=0 and t=20 (c) Plot u\left ( x,t\right ) vs. t for 0\leq t\leq 20 and for several values of x (d) Construct an animation of the solution for at least one period. (e) Describe the motion of the string. Let g\left ( x\right ) =\frac{8x\left ( L-x\right ) ^{2}}{L^{3}}
solution Since domain is finite, it is easier to use the series solution for wave equation than D’Alembert solution. The eigenvalue ODE is gives solution for \lambda >0 as X_{n}\left ( x\right ) =c_{n}\sin \left ( \sqrt{\lambda _{n}}x\right ) Where \lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2},n=1,2,3,\cdots \ The time solution is T_{n}\left ( t\right ) =A_{n}\cos \left ( \sqrt{\lambda _{n}}at\right ) +B_{n}\sin \left ( \sqrt{\lambda _{n}}at\right ) . At t=0, this gives 0=A_{n}. Therefore T_{n}\left ( t\right ) =B_{n}\sin \left ( \sqrt{\lambda _{n}}at\right ) . Hence the complete solution is\begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }c_{n}T_{n}\left ( t\right ) X_{n}\left ( x\right ) \\ & =\sum _{n=1}^{\infty }c_{n}\sin \left ( \sqrt{\lambda _{n}}at\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) \end{align*}
To find c_{n}, time derivative of the above is taken giving \frac{\partial }{\partial t}u\left ( x,t\right ) =\sum _{n=1}^{\infty }c_{n}\sqrt{\lambda _{n}}\cos \left ( \sqrt{\lambda _{n}}at\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) At t=0 the above becomes g\left ( x\right ) =\sum _{n=1}^{\infty }c_{n}\sqrt{\lambda _{n}}\sin \left ( \sqrt{\lambda _{n}}x\right ) Applying orthogonality\begin{align*} \int _{0}^{L}g\left ( x\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) dx & =\sqrt{\lambda _{n}}c_{n}\frac{L}{2}\\ c_{n} & =\frac{2}{L\sqrt{\lambda _{n}}}\int _{0}^{L}g\left ( x\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) dx \end{align*}
Hence since g\left ( x\right ) =\frac{8x\left ( L-x\right ) ^{2}}{L^{3}}, L=10,a=1 the above becomes c_{n}=\frac{2}{10\left ( \frac{n\pi }{10}\right ) }\int _{0}^{10}\frac{8x\left ( 10-x\right ) ^{2}}{10^{3}}\sin \left ( \frac{n\pi }{10}x\right ) dx Integrating the above gives c_{n}=\frac{320\left ( 2+\left ( -1\right ) ^{n}\right ) }{n^{4}\pi ^{4}} Therefore the solution is\begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }\frac{320\left ( 2+\left ( -1\right ) ^{n}\right ) }{n^{4}\pi ^{4}}T_{n}\left ( t\right ) X_{n}\left ( x\right ) \\ & =\frac{320}{\pi ^{4}}\sum _{n=1}^{\infty }\frac{2+\left ( -1\right ) ^{n}}{n^{4}}\sin \left ( \sqrt{\lambda _{n}}at\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) \end{align*}
Where \lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2},n=1,2,3,\cdots
solution
The eigenvalue ODE is X\left ( x\right ) =A\cos \left ( \sqrt{\lambda }x\right ) +B\sin \left ( \sqrt{\lambda }x\right ) Boundary condition at x=0 gives 0=A Therefore the solution becomes X\left ( x\right ) =B\sin \left ( \sqrt{\lambda }x\right ) . And X^{\prime }\left ( x\right ) =B\sqrt{\lambda }\cos \left ( \sqrt{\lambda }x\right ) . Applying boundary conditions at x=L gives 0=B\sqrt{\lambda }\cos \left ( \sqrt{\lambda }L\right ) Therefore \sqrt{\lambda }L=\left \{ \frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\cdots \right \} Hence\begin{align*} \sqrt{\lambda _{n}} & =\frac{n\pi }{2L}\qquad n=1,3,5,\cdots \\ \sqrt{\lambda _{n}} & =\frac{\left ( 2n-1\right ) \pi }{2L}\qquad n=1,2,3,\cdots \end{align*}
Therefore X_{n}\left ( x\right ) =c_{n}\sin \left ( \frac{\left ( 2n-1\right ) \pi }{2L}x\right ) And \begin{align*} T_{n}\left ( t\right ) & =A_{n}\cos \left ( \sqrt{\lambda _{n}}at\right ) +B_{n}\sin \left ( \sqrt{\lambda _{n}}at\right ) \\ T_{n}^{\prime }\left ( t\right ) & =-A_{n}a\sqrt{\lambda _{n}}\sin \left ( \sqrt{\lambda _{n}}at\right ) +B_{n}a\sqrt{\lambda _{n}}\cos \left ( \sqrt{\lambda _{n}}at\right ) \end{align*}
Since initial velocity is zero, the above gives 0=B_{n}a\sqrt{\lambda _{n}} Which means B_{n}=0. Hence T_{n}\left ( t\right ) =A_{n}\cos \left ( \sqrt{\lambda _{n}}at\right ) Therefore the complete solution becomes u\left ( x,t\right ) =\sum _{n=1}^{\infty }c_{n}\cos \left ( \frac{\left ( 2n-1\right ) \pi }{2L}at\right ) \sin \left ( \frac{\left ( 2n-1\right ) \pi }{2L}x\right ) c_{n} is found from initial position by applying orthogonality.
Solution
Straight forward.
Solution
To do.
Convert to form \left ( py^{\prime }\right ) ^{\prime }+q\left ( x\right ) y=0 y^{\prime \prime }-2xy^{\prime }+\lambda y=0 Solution
Writing the ODE as p\left ( x\right ) y^{\prime \prime }+Q\left ( x\right ) y^{\prime }+R\left ( x\right ) y=0, hence\begin{align*} p\left ( x\right ) & =1\\ Q\left ( x\right ) & =-2x\\ R\left ( x\right ) & =\lambda \end{align*}
Then the new form is \left ( \mu \left ( x\right ) p\left ( x\right ) y^{\prime }\right ) ^{\prime }+\mu \left ( x\right ) R\left ( x\right ) y=0, where \begin{align*} \mu \left ( x\right ) & =\frac{1}{p\left ( x\right ) }e^{\int ^{x}\frac{Q\left ( s\right ) }{P\left ( s\right ) }ds}\\ & =e^{\int ^{x}-2sds}\\ & =e^{-x^{2}} \end{align*}
Therefore the new form is \left ( e^{-x^{2}}y^{\prime }\right ) ^{\prime }+e^{-x^{2}}\lambda y=0
Convert to form \left ( py^{\prime }\right ) ^{\prime }+q\left ( x\right ) y=0 x^{2}y^{\prime \prime }+xy^{\prime }+\left ( x^{2}-v^{2}\right ) y=0 Solution
Writing the ODE as p\left ( x\right ) y^{\prime \prime }+Q\left ( x\right ) y^{\prime }+R\left ( x\right ) y=0, hence\begin{align*} p\left ( x\right ) & =x^{2}\\ Q\left ( x\right ) & =x\\ R\left ( x\right ) & =\left ( x^{2}-v^{2}\right ) \end{align*}
The new form is \left ( \mu \left ( x\right ) p\left ( x\right ) y^{\prime }\right ) ^{\prime }+\mu \left ( x\right ) R\left ( x\right ) y=0, where \begin{align*} \mu \left ( x\right ) & =\frac{1}{p\left ( x\right ) }e^{\int ^{x}\frac{Q\left ( s\right ) }{P\left ( s\right ) }ds}\\ & =\frac{1}{x^{2}}e^{\int ^{x}\frac{1}{s}ds}\\ & =\frac{1}{x^{2}}e^{\left \vert \ln x\right \vert }\\ & =\frac{1}{x^{2}}x\\ & =\frac{1}{x} \end{align*}
Therefore the new form is \left ( \frac{1}{x}y^{\prime }\right ) ^{\prime }+\frac{1}{x}\left ( x^{2}-v^{2}\right ) y=0
Solution
Let y\left ( x\right ) =s\left ( x\right ) u\left ( x\right ) . Then y^{\prime }=s^{\prime }u+su^{\prime } and y^{\prime \prime }=s^{\prime \prime }u+s^{\prime }u^{\prime }+s^{\prime }u^{\prime }+su^{\prime \prime }=s^{\prime \prime }u+2\left ( s^{\prime }u^{\prime }\right ) +su^{\prime \prime }. Therefore the original ODE becomes s^{\prime \prime }u+2\left ( s^{\prime }u^{\prime }\right ) +su^{\prime \prime }+4\left ( s^{\prime }u+su^{\prime }\right ) +\left ( 4+9\lambda \right ) su=0 Collecting terms in u gives su^{\prime \prime }+u^{\prime }\left ( 2s^{\prime }+4s\right ) +\left ( s^{\prime \prime }+4s^{\prime }+\left ( 4+9\lambda \right ) s\right ) u=0 Making u^{\prime } term vanish requires that 2s^{\prime }+4s or s^{\prime }+2s=0. Hence \frac{d}{dx}\left ( se^{2x}\right ) =0 or s=e^{-2x}. Hence s^{\prime }=-2e^{-2x},s^{\prime \prime }=4e^{-2x}. Substituting these into the above gives\begin{align*} e^{-2x}u^{\prime \prime }+\left ( 4e^{-2x}+4\left ( -2e^{-2x}\right ) +\left ( 4+9\lambda \right ) e^{-2x}\right ) u & =0\\ u^{\prime \prime }+\left ( 4+4\left ( -2\right ) +\left ( 4+9\lambda \right ) \right ) u & =0\\ u^{\prime \prime }+\left ( 4-8+4+9\lambda \right ) u & =0\\ u^{\prime \prime }+9\lambda u & =0 \end{align*}
Let 9\lambda =\hat{\lambda } so the above becomes u^{\prime \prime }+\hat{\lambda }u=0 With boundary conditions u\left ( 0\right ) =\frac{y\left ( 0\right ) }{s\left ( 0\right ) }=0 and u^{\prime }\left ( L\right ) =\frac{y^{\prime }\left ( L\right ) }{s^{\prime }\left ( L\right ) }=0. This was solved before, the eigenfunctions of the above are\begin{align*} \Phi _{n}\left ( x\right ) & =\sin \left ( \sqrt{\hat{\lambda }_{n}}x\right ) \\ \hat{\lambda }_{n} & =\left ( \frac{n\pi }{2L}\right ) ^{2}\qquad n=1,3,5,\cdots \end{align*}
But \hat{\lambda }_{n}=9\lambda _{n}, therefore the above becomes\begin{align*} \Phi _{n}\left ( x\right ) & =\sin \left ( 3\sqrt{\lambda _{n}}x\right ) \\ \lambda _{n} & =\frac{1}{9}\left ( \frac{n\pi }{2L}\right ) ^{2}\qquad n=1,2,3,\cdots \end{align*}
Or \Phi _{n}\left ( x\right ) =\sin \left ( \frac{n\pi }{2L}x\right ) Now the eigenfunction is normalized\begin{align*} \int _{0}^{1}\left ( k_{n}\Phi _{n}\left ( x\right ) \right ) ^{2}dx & =1\\ k_{n}^{2}\int _{0}^{1}\Phi _{n}\left ( x\right ) ^{2}dx & =1\\ k_{n}^{2}\int _{0}^{1}\sin ^{2}\left ( \frac{n\pi }{2L}x\right ) dx & =1\\ k_{n}^{2}\frac{L}{2} & =1\\ k_{n} & =\sqrt{\frac{2}{L}} \end{align*}
Hence k_{n}=\sqrt{\frac{2}{L}} And \hat{\Phi }_{n}\left ( x\right ) =\sqrt{\frac{2}{L}}\sin \left ( \frac{n\pi }{2L}x\right ) Mapping back to y\left ( x\right ) =s\left ( x\right ) u\left ( s\right ) , and since s\left ( x\right ) =e^{-2x} then the eigenfunction in y space is \Phi _{n}\left ( x\right ) =e^{-2x}\sqrt{\frac{2}{L}}\sin \left ( \frac{n\pi }{2L}x\right ) \qquad n=1,3,5,\cdots
Now the ODE is solved directly. y^{\prime \prime }+4y^{\prime }+\left ( 4+9\lambda \right ) y=0. The characteristic equation is r^{2}+4r+\left ( 4+9\lambda \right ) =0 Hence roots are \begin{align*} r & =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{-4\pm \sqrt{16-4\left ( 4+9\lambda \right ) }}{2}\\ & =\frac{-4\pm \sqrt{16-16-36\lambda }}{2}=-2\pm 3\sqrt{-\lambda } \end{align*}
We know that \lambda >0. So the roots are r=-2\pm i\sqrt{\lambda } and the solution is y\left ( x\right ) =e^{-2x}\left ( A\cos \left ( \sqrt{\lambda }x\right ) +B\sin \left ( \sqrt{\lambda }x\right ) \right ) Applying boundary conditions y\left ( 0\right ) =0 leads to A=0. So the solution becomes y\left ( x\right ) =e^{-2x}B\sin \left ( \sqrt{\lambda }x\right ) Hence y^{\prime }\left ( x\right ) =-2e^{-2x}B\sin \left ( \sqrt{\lambda }x\right ) +e^{-2x}B\sqrt{\lambda }\cos \left ( \sqrt{\lambda }x\right ) Applying second B.C. y^{\prime }\left ( L\right ) =0 the above becomes\begin{align*} 0 & =-2e^{-2L}B\sin \left ( \sqrt{\lambda }L\right ) +e^{-2L}B\sqrt{\lambda }\cos \left ( \sqrt{\lambda }L\right ) \\ & =B\left ( -2\sin \left ( \sqrt{\lambda }L\right ) +\sqrt{\lambda }\cos \left ( \sqrt{\lambda }L\right ) \right ) \end{align*}
Non-trivial solution requires that \begin{align*} -2\sin \left ( \sqrt{\lambda }L\right ) +\sqrt{\lambda }\cos \left ( \sqrt{\lambda }L\right ) & =0\\ -2\tan \sqrt{\lambda }L+\sqrt{\lambda } & =0\\ \tan \sqrt{\lambda }L & =\frac{1}{2}\sqrt{\lambda } \end{align*}
Hence the direct method finds that the eigenvalues \lambda _{n} are the solutions to the above nonlinear equation and the corresponding eigenfunctions are e^{-2x}\sin \left ( \sqrt{\lambda _{n}}x\right ) .
Determine the real eigenvalues and eigenfunctions.\begin{align*} y^{\prime \prime }+y^{\prime }+\lambda \left ( y^{\prime }+y\right ) & =0\\ y^{\prime }\left ( 0\right ) & =0\\ y\left ( 1\right ) & =0 \end{align*}
Solution
Writing the ODE as y^{\prime \prime }+\left ( 1+\lambda \right ) y^{\prime }+\lambda y=0 Case \lambda =0 y^{\prime \prime }+y^{\prime }=0 The characteristic equation is \begin{align*} r^{2}+r & =0\\ r\left ( r+1\right ) & =0 \end{align*}
The roots are r=0,-1. Hence the solution is y=c_{1}+c_{2}e^{-x}. Hence y^{\prime }=-c_{2}e^{-x}. First BC gives y^{\prime }\left ( 0\right ) =0\rightarrow 0=-c_{2}. Therefore the solution becomes y=c_{1}. Second BC gives y\left ( 1\right ) =0\rightarrow 0=c_{1}. Therefore trivial solution and \lambda =0 is not eigenvalue.
Case \lambda <0 Let \lambda =-m^{2} for some real m. The ODE becomes y^{\prime \prime }+\left ( 1-m^{2}\right ) y^{\prime }-m^{2}y=0 The characteristic equation is r^{2}+\left ( 1-m^{2}\right ) r-m^{2}=0 The roots are\begin{align*} r & =\frac{-\left ( 1-m^{2}\right ) }{2}\pm \frac{1}{2}\sqrt{\left ( 1-m^{2}\right ) ^{2}+4m^{2}}\\ & =\frac{-\left ( 1-m^{2}\right ) }{2}\pm \frac{1}{2}\sqrt{1+m^{4}-2m^{2}+4m^{2}}\\ & =\frac{-\left ( 1-m^{2}\right ) }{2}\pm \frac{1}{2}\sqrt{\left ( 1+m^{2}\right ) ^{2}}\\ & =\frac{-\left ( 1-m^{2}\right ) }{2}\pm \frac{1}{2}\left ( 1+m^{2}\right ) \end{align*}
Hence roots are r_{1}=\frac{-\left ( 1-m^{2}\right ) }{2}+\frac{1}{2}\left ( 1+m^{2}\right ) =m^{2} and r_{2}=\frac{-\left ( 1-m^{2}\right ) }{2}-\frac{1}{2}\left ( 1+m^{2}\right ) =-1. Therefore the solution is y=c_{1}e^{m^{2}x}+c_{2}e^{-x} Hence y^{\prime }=m^{2}c_{1}e^{m^{2}x}-c_{2}e^{-x}. First BC gives y^{\prime }\left ( 0\right ) =0\rightarrow 0=m^{2}c_{1}-c_{2} or c_{2}=m^{2}c_{1}. Therefore the solution becomes\begin{align*} y & =c_{1}e^{m^{2}x}+m^{2}c_{1}e^{-x}\\ & =c_{1}\left ( e^{m^{2}x}+m^{2}e^{-x}\right ) \end{align*}
Second BC gives y\left ( 1\right ) =0\rightarrow 0=c_{1}\left ( e^{m^{2}}+m^{2}e^{-1}\right ) therefore c_{1}=0 and trivial solution. Hence \lambda <0 is not eigenvalue.
Case \lambda >0 The characteristic equation is r^{2}+\left ( 1+\lambda \right ) r+\lambda =0 The roots are\begin{align*} r & =\frac{-\left ( 1+\lambda \right ) }{2}\pm \frac{1}{2}\sqrt{\left ( 1+\lambda \right ) ^{2}-4\lambda }\\ & =\frac{-\left ( 1+\lambda \right ) }{2}\pm \frac{1}{2}\sqrt{1+\lambda ^{2}+2\lambda -4\lambda }\\ & =\frac{-\left ( 1+\lambda \right ) }{2}\pm \frac{1}{2}\sqrt{\left ( 1-\lambda \right ) ^{2}}\\ & =\frac{-\left ( 1+\lambda \right ) }{2}\pm \frac{1}{2}\left ( 1-\lambda \right ) \end{align*}
Hence roots are r_{1}=\frac{-1}{2}\left ( 1+\lambda \right ) +\frac{1}{2}\left ( 1-\lambda \right ) =-\lambda and r_{2}=\frac{-1}{2}\left ( 1+\lambda \right ) -\frac{1}{2}\left ( 1-\lambda \right ) =-1. Therefore the solution is y=c_{1}e^{-\lambda x}+c_{2}e^{-x} Hence y^{\prime }=-\lambda c_{1}e^{\lambda x}-c_{2}e^{-x}. First BC gives y^{\prime }\left ( 0\right ) =0\rightarrow 0=-\lambda c_{1}-c_{2} or c_{2}=-\lambda c_{1}. Therefore the solution becomes\begin{align*} y & =c_{1}e^{-\lambda x}-\lambda c_{1}e^{-x}\\ & =c_{1}\left ( e^{-\lambda x}-\lambda e^{-x}\right ) \end{align*}
Second BC gives y\left ( 1\right ) =0\rightarrow 0=c_{1}\left ( e^{-\lambda }-\lambda e^{-1}\right ) For non-trivial solution, we need e^{-\lambda }-\lambda e^{-1}=0. The solution to this is \lambda =1.
When \lambda =1 the eigenfunction is y\left ( x\right ) =c_{1}\left ( e^{-x}-e^{-x}\right ) =0 But eigenfunction can not be zero. Therefore there is eigenvalue when \lambda >0. Hence for all cases, there is no eigenvalue with corresponding nonzero eigenfunction.
Determine the real eigenvalues and eigenfunctions.\begin{align*} x^{2}y^{\prime \prime }-\lambda \left ( xy^{\prime }-y\right ) & =0\\ y\left ( 1\right ) & =0\\ y\left ( 2\right ) -y^{\prime }\left ( 2\right ) & =0 \end{align*}
Solution
This is a Euler ODE. x^{2}y^{\prime \prime }-\lambda xy^{\prime }+\lambda y=0. Let y=x^{r}, then y^{\prime }=rx^{r-1},y^{\prime \prime }=r\left ( r-1\right ) x^{r-2}. The ODE becomes\begin{align*} x^{2}r\left ( r-1\right ) x^{r-2}-\lambda xrx^{r-1}+\lambda x^{r} & =0\\ r\left ( r-1\right ) x^{r}-\lambda rx^{r}+\lambda x^{r} & =0\\ r\left ( r-1\right ) -\lambda r+\lambda & =0 \end{align*}
Case \lambda =0
The characteristic equation becomes r\left ( r-1\right ) =0 The roots are r=0,r=1, hence the solution is y=c_{1}+c_{2}x At BC y\left ( 1\right ) =0\rightarrow 0=c_{1}+c_{2}. Hence c_{1}=-c_{2} and the solution becomes\ y=c_{1}-c_{1}x=c_{1}\left ( 1-x\right ) . Hence y^{\prime }=-c_{1}. Second BC y\left ( 2\right ) -y^{\prime }\left ( 2\right ) =0 gives\begin{align*} 0 & =c_{1}\left ( 1-2\right ) +c_{1}\\ 0 & =-c_{1}+c_{1}\\ 0 & =0 \end{align*}
Therefore any c_{1} will work. Giving a solution y=c_{1}\left ( 1-x\right ) Therefore \lambda =0 is an eigenvalue with eigenfunction \Phi _{0}\left ( x\right ) =1-x.
Case \lambda <0 Let \lambda =-m^{2}. The characteristic equation becomes \begin{align*} r\left ( r-1\right ) +m^{2}r-m^{2} & =0\\ r^{2}-r+m^{2}r-m^{2} & =0\\ r^{2}+r\left ( m^{2}-1\right ) -m^{2} & =0 \end{align*}
The roots are\begin{align*} r & =\frac{-\left ( m^{2}-1\right ) }{2}\pm \frac{1}{2}\sqrt{\left ( m^{2}-1\right ) ^{2}+4m^{2}}\\ & =\frac{-\left ( m^{2}-1\right ) }{2}\pm \frac{1}{2}\sqrt{m^{4}-2m^{2}+1+4m^{2}}\\ & =\frac{-\left ( m^{2}-1\right ) }{2}\pm \frac{1}{2}\sqrt{\left ( 1+m^{2}\right ) ^{2}}\\ & =-\frac{1}{2}\left ( m^{2}-1\right ) \pm \frac{1}{2}\left ( 1+m^{2}\right ) \end{align*}
Roots are r=-\frac{1}{2}\left ( m^{2}-1\right ) +\frac{1}{2}\left ( 1+m^{2}\right ) =1 or r=-\frac{1}{2}\left ( m^{2}-1\right ) -\frac{1}{2}\left ( 1+m^{2}\right ) =-m^{2}. Hence solution is y=c_{1}x+c_{2}x^{-m^{2}} At BC y\left ( 1\right ) =0\rightarrow 0=c_{2}. Therefore the solution is y=c_{1}x and y^{\prime }=c_{1}. Second BC gives y\left ( 2\right ) -y^{\prime }\left ( 2\right ) =0 or \begin{align*} 0 & =2c_{1}-c_{1}\\ 0 & =c_{1} \end{align*}
Hence trivial solution. So \lambda <0 is not an eigenvalue.
Case \lambda >0
The characteristic equation becomes \begin{align*} r^{2}-r-\lambda r+\lambda & =0\\ r^{2}-r\left ( 1+\lambda \right ) +\lambda & =0 \end{align*}
The roots are\begin{align*} r & =\frac{1+\lambda }{2}\pm \frac{1}{2}\sqrt{\left ( 1+\lambda \right ) ^{2}-4\lambda }\\ & =\frac{1+\lambda }{2}\pm \frac{1}{2}\sqrt{1+\lambda ^{2}-2\lambda }\\ & =\frac{1+\lambda }{2}\pm \frac{1}{2}\sqrt{\left ( 1-\lambda \right ) ^{2}}\\ & =\frac{1+\lambda }{2}\pm \frac{1}{2}\left ( 1-\lambda \right ) \end{align*}
Roots are r=\frac{1}{2}\left ( 1+\lambda \right ) +\frac{1}{2}\left ( 1-\lambda \right ) =1 or r=\frac{1}{2}\left ( 1+\lambda \right ) -\frac{1}{2}\left ( 1-\lambda \right ) =\lambda . Hence solution is y=c_{1}x+c_{2}x^{\lambda } This is similar to the case above for \lambda <0. Hence there is no eigenvalue for \lambda >0.
Determine the normalized eigenfunction for\begin{align} y^{\prime \prime }+\lambda y & =0\tag{1}\\ y\left ( 0\right ) & =0\nonumber \\ y^{\prime }\left ( 1\right ) & =0\nonumber \end{align}
Solution
The eigenfunction for the above problem can be easily found using chapter 10 methods to be \Phi _{n}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{n}}x\right ) \qquad n=1,3,5,\cdots Where \lambda _{n}=\frac{n\pi }{2L}=\frac{n\pi }{2} The normalized \hat{\Phi }_{n}\left ( x\right ) =k_{n}\Phi _{n}\left ( x\right ) . Where \int _{0}^{1}\hat{\Phi }_{n}^{2}\left ( x\right ) dx=1 Hence solving the above for k_{n} gives\begin{align*} \int _{0}^{1}\left ( k_{n}\Phi _{n}\left ( x\right ) \right ) ^{2}dx & =1\\ k_{n}^{2}\int _{0}^{1}{}\Phi _{n}^{2}\left ( x\right ) dx & =1 \end{align*}
But \int _{0}^{1}{}\Phi _{n}^{2}\left ( x\right ) dx=\int _{0}^{1}\sin ^{2}\left ( \sqrt{\lambda _{n}}x\right ) dx=\int _{0}^{1}\sin ^{2}\left ( \frac{n\pi }{2}x\right ) dx=\frac{1}{2}. Hence the above becomes\begin{align*} k_{n}^{2}\frac{1}{2} & =1\\ k_{n} & =\sqrt{2} \end{align*}
Therefore\begin{align*} \hat{\Phi }_{n}\left ( x\right ) & =\sqrt{2}\Phi _{n}\left ( x\right ) \\ & =\sqrt{2}\sin \left ( \frac{n\pi }{2}x\right ) \qquad n=1,3,5,\cdots \\ & =\left \{ \sqrt{2}\sin \left ( \frac{\pi }{2}x\right ) ,\sqrt{2}\sin \left ( \frac{3\pi }{2}x\right ) ,\sqrt{2}\sin \left ( \frac{5\pi }{2}x\right ) ,\cdots \right \} \end{align*}
Determine the normalized eigenfunction for
\begin{align} y^{\prime \prime }+\lambda y & =0\tag{1}\\ y^{\prime }\left ( 0\right ) & =0\nonumber \\ y\left ( 1\right ) & =0\nonumber \end{align}
Solution
The eigenfunction for the above problem can be found using chapter 10 methods to be \Phi _{n}\left ( x\right ) =\cos \left ( \sqrt{\lambda _{n}}x\right ) \qquad n=1,3,5,\cdots Where \lambda _{n}=\frac{n\pi }{2L}=\frac{n\pi }{2} The normalized \hat{\Phi }_{n}\left ( x\right ) =k_{n}\Phi _{n}\left ( x\right ) . Where \int _{0}^{1}\hat{\Phi }_{n}^{2}\left ( x\right ) dx=1 Hence solving the above for k_{n} gives\begin{align*} \int _{0}^{1}{}\left ( k_{n}\Phi _{n}\left ( x\right ) \right ) ^{2}dx & =1\\ k_{n}^{2}\int _{0}^{1}{}\Phi _{n}^{2}\left ( x\right ) dx & =1 \end{align*}
But \int _{0}^{1}{}\Phi _{n}^{2}\left ( x\right ) dx=\int _{0}^{1}\cos ^{2}\left ( \sqrt{\lambda _{n}}x\right ) dx=\int _{0}^{1}\cos ^{2}\left ( \frac{n\pi }{2}x\right ) dx=\frac{1}{2}. Hence the above becomes\begin{align*} k_{n}^{2}\frac{1}{2} & =1\\ k_{n} & =\sqrt{2} \end{align*}
Therefore\begin{align*} \hat{\Phi }_{n}\left ( x\right ) & =\sqrt{2}\Phi _{n}\left ( x\right ) \\ & =\sqrt{2}\cos \left ( \frac{n\pi }{2}x\right ) \qquad n=1,3,5,\cdots \\ & =\left \{ \sqrt{2}\cos \left ( \frac{\pi }{2}x\right ) ,\sqrt{2}\cos \left ( \frac{3\pi }{2}x\right ) ,\sqrt{2}\cos \left ( \frac{5\pi }{2}x\right ) ,\cdots \right \} \end{align*}
Determine the normalized eigenfunction for\begin{align} y^{\prime \prime }+\lambda y & =0\tag{1}\\ y^{\prime }\left ( 0\right ) & =0\nonumber \\ y^{\prime }\left ( 1\right ) & =0\nonumber \end{align}
Solution
The eigenfunctions are first found. Let the solution be y=Ae^{rx}. This leads to the characteristic equation\begin{align*} r^{2}+\lambda & =0\\ r & =\pm \sqrt{-\lambda } \end{align*}
Case \lambda <0
In this case -\lambda is positive and hence \sqrt{-\lambda } is also positive. Let \sqrt{-\lambda }=\mu where \mu >0. Hence the roots are \pm \mu . This gives the solution\begin{align*} y & =c_{1}\cosh \left ( \mu x\right ) +c_{2}\sinh \left ( \mu x\right ) \\ y^{\prime } & =c_{1}\mu \sinh \left ( \mu x\right ) +c_{2}\mu \cosh \left ( \mu x\right ) \end{align*}
First B.C. y^{\prime }\left ( 0\right ) =0 gives\begin{align*} 0 & =c_{2}\mu \\ c_{2} & =0 \end{align*}
Hence solution becomes y\left ( x\right ) =c_{1}\cosh \left ( \mu x\right ) Second B.C. y^{\prime }\left ( 1\right ) =0 gives 0=c_{1}\mu \sinh \left ( \mu \right ) But \sinh \left ( \mu \right ) can not be zero since \mu \neq 0, hence c_{1}=0, Leading to trivial solution. Therefore \lambda <0 is not eigenvalue.
Let \lambda =0, The solution is
y\left ( x\right ) =c_{1}+c_{2}x First B.C. y^{\prime }\left ( 0\right ) =0 gives 0=c_{2} The solution becomes y\left ( x\right ) =c_{1} Second B.C. y^{\prime }\left ( 1\right ) =0 gives 0=0 Therefore c_{1} can be any value. Therefore \lambda =0 is an eigenvalue and the corresponding eigenfunction is any constant, say 1.
Case \lambda >0, The solution is\begin{align*} y\left ( x\right ) & =c_{1}\cos \left ( \sqrt{\lambda }x\right ) +c_{2}\sin \left ( \sqrt{\lambda }x\right ) \\ y^{\prime }\left ( x\right ) & =-c_{1}\sqrt{\lambda }\sin \left ( \sqrt{\lambda }x\right ) +c_{2}\sqrt{\lambda }\cos \left ( \sqrt{\lambda }x\right ) \end{align*}
First B.C. y^{\prime }\left ( 0\right ) =0 gives\begin{align*} 0 & =c_{2}\sqrt{\lambda }\\ c_{2} & =0 \end{align*}
The solution becomes y\left ( x\right ) =c_{1}\cos \left ( \sqrt{\lambda }x\right ) Second B.C. y^{\prime }\left ( 1\right ) =0 gives 0=-c_{1}\sqrt{\lambda }\sin \left ( \sqrt{\lambda }\right ) For non-trivial solution, we want \sin \left ( \sqrt{\lambda }\right ) =0 or \sqrt{\lambda }=n\pi for n=1,2,3,\cdots Therefore \lambda _{n}=\left ( n\pi \right ) ^{2}\qquad n=1,2,3,\cdots And the corresponding eigenfunctions are \Phi _{n}\left ( x\right ) =\cos \left ( \sqrt{\lambda }x\right ) \qquad n=1,2,3,\cdots Hence\begin{align*} \Phi _{0}\left ( x\right ) & =1\\ \Phi _{n}\left ( x\right ) & =\cos \left ( \sqrt{\lambda _{n}}x\right ) \qquad n=1,2,3,\cdots \end{align*}
The normalized \hat{\Phi }_{0}\left ( x\right ) =k_{0}\Phi _{0}\left ( x\right ) . Where \int _{0}^{1}r\left ( x\right ) \hat{\Phi }_{0}^{2}\left ( x\right ) dx=1 But r\left ( x\right ) =1. Therefore solving the above for k_{0} gives\begin{align*} \int _{0}^{1}{}\left ( k_{0}\Phi _{0}\left ( x\right ) \right ) ^{2}dx & =1\\ k_{0}^{2}\int _{0}^{1}{}dx & =1\\ k_{0} & =1 \end{align*}
And for n=1,2,3,\cdots we obtain\begin{align*} \int _{0}^{1}\hat{\Phi }_{n}^{2}\left ( x\right ) dx & =1\\ \int _{0}^{1}{}\left ( k_{n}\Phi _{n}\left ( x\right ) \right ) ^{2}dx & =1\\ k_{n}^{2}\int _{0}^{1}{}\Phi _{n}^{2}\left ( x\right ) dx & =1\\ k_{n}^{2}\int _{0}^{1}{}\cos ^{2}\left ( \sqrt{n\pi }x\right ) dx & =1 \end{align*}
But \int _{0}^{1}{}\cos ^{2}\left ( \sqrt{n\pi }x\right ) =\frac{1}{2}. Hence the above becomes\begin{align*} k_{n}^{2}\frac{1}{2} & =1\\ k_{n} & =\sqrt{2} \end{align*}
Therefore \hat{\Phi }_{0}\left ( x\right ) =1 And for n=1,2,3,\cdots \begin{align*} \hat{\Phi }_{n}\left ( x\right ) & =\sqrt{2}\Phi _{n}\left ( x\right ) \\ & =\sqrt{2}\cos \left ( n\pi x\right ) \\ & =\left \{ \sqrt{2}\cos \left ( \pi x\right ) ,\sqrt{2}\cos \left ( 2\pi x\right ) ,\sqrt{2}\cos \left ( 3\pi x\right ) ,\cdots \right \} \end{align*}
Determine the normalized eigenfunction for\begin{align} y^{\prime \prime }+\lambda y & =0\tag{1}\\ y^{\prime }\left ( 0\right ) & =0\nonumber \\ y^{\prime }\left ( 1\right ) +y\left ( 1\right ) & =0\nonumber \end{align}
Solution
The eigenfunctions for the above problem are first found. Let the solution be y=Ae^{rx}. This leads to the characteristic equation\begin{align*} r^{2}+\lambda & =0\\ r & =\pm \sqrt{-\lambda } \end{align*}
Case \lambda <0
In this case -\lambda is positive and hence \sqrt{-\lambda } is also positive. Let \sqrt{-\lambda }=\mu where \mu >0. Hence the roots are \pm \mu . This gives the solution\begin{align*} y & =c_{1}\cosh \left ( \mu x\right ) +c_{2}\sinh \left ( \mu x\right ) \\ y^{\prime } & =c_{1}\mu \sinh \left ( \mu x\right ) +c_{2}\mu \cosh \left ( \mu x\right ) \end{align*}
First B.C. y^{\prime }\left ( 0\right ) =0 gives\begin{align*} 0 & =c_{2}\mu \\ c_{2} & =0 \end{align*}
Hence solution becomes y\left ( x\right ) =c_{1}\cosh \left ( \mu x\right ) Second B.C. y\left ( 1\right ) +y^{\prime }\left ( 1\right ) =0 gives 0=c_{1}\left ( \cosh \left ( \mu \right ) +\mu \sinh \left ( \mu \right ) \right ) But \sinh \left ( \mu \right ) can not be negative since its argument is positive here. And \cosh \mu is always positive. In addition \cosh \left ( \mu \right ) +\mu \sinh \left ( \mu \right ) can not be zero since \sinh \left ( \mu \right ) can not be zero as \mu \neq 0 and \cosh \left ( \mu \right ) is not zero. Therefore c_{1}=0, Leading to trivial solution. Therefore \lambda <0 is not eigenvalue.
Case \lambda =0, The solution is y\left ( x\right ) =c_{1}+c_{2}x First B.C. y^{\prime }\left ( 0\right ) =0 gives 0=c_{2} The solution becomes y\left ( x\right ) =c_{1} Second B.C. y\left ( 1\right ) +y^{\prime }\left ( 1\right ) =0 gives 0=c_{1} This gives trivial solution. Therefore \lambda =0 is not eigenvalue.
Case \lambda >0, The solution is\begin{align*} y\left ( x\right ) & =c_{1}\cos \left ( \sqrt{\lambda }x\right ) +c_{2}\sin \left ( \sqrt{\lambda }x\right ) \\ y^{\prime }\left ( x\right ) & =-c_{1}\sqrt{\lambda }\sin \left ( \sqrt{\lambda }x\right ) +c_{2}\sqrt{\lambda }\cos \left ( \sqrt{\lambda }x\right ) \end{align*}
First B.C. y^{\prime }\left ( 0\right ) =0 gives\begin{align*} 0 & =c_{2}\sqrt{\lambda }\\ c_{2} & =0 \end{align*}
The solution becomes y\left ( x\right ) =c_{1}\cos \left ( \sqrt{\lambda }x\right ) Second B.C. y\left ( 1\right ) +y^{\prime }\left ( 1\right ) =0 gives\begin{align*} 0 & =c_{1}\cos \left ( \sqrt{\lambda }\right ) -c_{1}\sqrt{\lambda }\sin \left ( \sqrt{\lambda }\right ) \\ & =c_{1}\left ( \cos \left ( \sqrt{\lambda }\right ) -\sqrt{\lambda }\sin \left ( \sqrt{\lambda }\right ) \right ) \end{align*}
For non-trivial solution the above implies \begin{equation} \cos \left ( \sqrt{\lambda }\right ) -\sqrt{\lambda }\sin \left ( \sqrt{\lambda }\right ) =0 \tag{1} \end{equation} Therefore the eigenvalues are the solution to the above nonlinear equation. And the corresponding eigenfunctions are \Phi _{n}=\cos \left ( \sqrt{\lambda _{n}}x\right ) \qquad n=1,2,3,\cdots Where \lambda _{n} are the roots of equation (1).
The normalized \hat{\Phi }_{n}=k_{n}\Phi _{n} eigenfunctions are now found. \int _{0}^{1}r\left ( x\right ) \hat{\Phi }_{n}^{2}dx=1 Since the weight function is r\left ( x\right ) =1, then\begin{align*} \int _{0}^{1}\hat{\Phi }_{n}^{2}dx & =1\\ \int _{0}^{1}k_{n}^{2}\Phi _{n}^{2}dx & =1\\ k_{n}^{2}\int _{0}^{1}\Phi _{n}^{2}dx & =1\\ k_{n}^{2}\int _{0}^{1}\cos ^{2}\left ( \sqrt{\lambda _{n}}x\right ) dx & =1 \end{align*}
But \int _{0}^{1}\cos ^{2}\left ( ax\right ) dx=\left ( \frac{x}{2}+\frac{\sin 2ax}{4a}\right ) _{0}^{1}=\left ( \frac{x}{2}+\frac{\sin \left ( 2\sqrt{\lambda _{n}}x\right ) }{4\sqrt{\lambda _{n}}}\right ) _{0}^{1}=\left ( \frac{1}{2}+\frac{\sin \left ( 2\sqrt{\lambda _{n}}\right ) }{4\sqrt{\lambda _{n}}}\right ) =\left ( \frac{2\sqrt{\lambda _{n}}+\sin \left ( 2\sqrt{\lambda _{n}}\right ) }{4\sqrt{\lambda _{n}}}\right ) . Hence the above becomes\begin{align*} k_{n}^{2} & =\frac{1}{\frac{2\sqrt{\lambda _{n}}+\sin \left ( 2\sqrt{\lambda _{n}}\right ) }{4\sqrt{\lambda _{n}}}}\\ & =\frac{4\sqrt{\lambda _{n}}}{2\sqrt{\lambda _{n}}+\sin \left ( 2\sqrt{\lambda _{n}}\right ) } \end{align*}
But \sin \left ( 2a\right ) =2\sin a\cos a and the above can be written as k_{n}^{2}=\frac{4\sqrt{\lambda _{n}}}{2\sqrt{\lambda _{n}}+2\sin \left ( \sqrt{\lambda _{n}}\right ) \cos \sqrt{\lambda _{n}}} But from (1) earlier, we found \cos \left ( \sqrt{\lambda }\right ) -\sqrt{\lambda }\sin \left ( \sqrt{\lambda }\right ) =0 or \cos \left ( \sqrt{\lambda }\right ) =\sqrt{\lambda }\sin \left ( \sqrt{\lambda }\right ) . Substituting this into the above gives k_{n}^{2}=\frac{4\sqrt{\lambda _{n}}}{2\sqrt{\lambda _{n}}+2\sqrt{\lambda _{n}}\sin ^{2}\left ( \sqrt{\lambda _{n}}\right ) } And since \lambda _{n}\neq 0 the above simplifies to \begin{align*} k_{n}^{2} & =\frac{2}{1+\sin ^{2}\left ( \sqrt{\lambda _{n}}\right ) }\\ & =\frac{4}{4+\sin ^{2}\left ( \sqrt{\lambda _{n}}\right ) } \end{align*}
Therefore k_{n}=\sqrt{\frac{2}{1+\sin ^{2}\left ( \sqrt{\lambda _{n}}\right ) }} Since there is no closed form solution to \lambda _{n} as it is a root of nonlinear equation \sqrt{\lambda _{n}}\tan \left ( \sqrt{\lambda _{n}}L\right ) =1.
Hence the normalized eigenfunctions are\begin{align*} \hat{\Phi }_{n} & =k_{n}\Phi _{n}\\ & =\frac{\sqrt{2}}{\sqrt{1+\sin ^{2}\left ( \sqrt{\lambda _{n}}\right ) }}\cos \left ( \sqrt{\lambda _{n}}x\right ) \end{align*}
Determine the normalized eigenfunction for\begin{align} y^{\prime \prime }-2y^{\prime }+\left ( 1+\lambda \right ) y & =0\tag{1}\\ y\left ( 0\right ) & =0\nonumber \\ y\left ( 1\right ) & =0\nonumber \end{align}
Solution
Let y\left ( x\right ) =s\left ( x\right ) u\left ( x\right ) . Then y^{\prime }=s^{\prime }u+su^{\prime } and y^{\prime \prime }=s^{\prime \prime }u+s^{\prime }u^{\prime }+s^{\prime }u^{\prime }+su^{\prime \prime }=s^{\prime \prime }u+2\left ( s^{\prime }u^{\prime }\right ) +su^{\prime \prime }. Therefore the original ODE becomes s^{\prime \prime }u+2\left ( s^{\prime }u^{\prime }\right ) +su^{\prime \prime }-2\left ( s^{\prime }u+su^{\prime }\right ) +\left ( 1+\lambda \right ) su=0 Collecting terms in u the above becomes su^{\prime \prime }+u^{\prime }\left ( 2s^{\prime }-2s\right ) +u\left ( \left ( 1+\lambda \right ) s+s^{\prime \prime }-2s^{\prime }\right ) =0 To get rid of u^{\prime } we therefore want 2s^{\prime }-2s=0 or s^{\prime }-s=0. Hence the integrating factor is I=e^{-x} and the solution is obtained from \frac{d}{dx}\left ( se^{-x}\right ) =0 or s=e^{x}. Therefore, if s=e^{x} then the original ODE becomes\begin{align*} e^{x}u^{\prime \prime }+u\left ( \left ( 1+\lambda \right ) e^{x}+e^{x}-2e^{x}\right ) & =0\\ u^{\prime \prime }+u\left ( \left ( 1+\lambda \right ) +1-2\right ) & =0\\ u^{\prime \prime }+u\left ( \left ( 1+\lambda \right ) -1\right ) & =0\\ u^{\prime \prime }+\lambda u & =0 \end{align*}
With the boundary conditions u\left ( 0\right ) =\frac{y\left ( 0\right ) }{s\left ( 0\right ) }=\frac{y\left ( 0\right ) }{e^{0}}=0 and u\left ( 1\right ) =\frac{y\left ( 1\right ) }{s\left ( 1\right ) }=0. Hence we need to find the eigenfunctions for\begin{align*} u^{\prime \prime }+\lambda u & =0\\ u\left ( 0\right ) & =0\\ u\left ( 1\right ) & =0 \end{align*}
But this we did before. It has \Phi _{n}\left ( x\right ) =\sin \left ( n\pi x\right ) for n=1,2,\cdots . And the normalized \hat{\Phi }_{n}\left ( x\right ) =\sqrt{2}\sin \left ( n\pi x\right ) . Mapping this normalized eigenfunction back to y\left ( x\right ) using the transformation y\left ( x\right ) =s\left ( x\right ) u\left ( x\right ) gives the normalized eigenfunction in y space as \hat{\Phi }_{n}\left ( x\right ) =e^{x}\sqrt{2}\sin \left ( n\pi x\right ) \qquad n=1,2,3,\cdots
Here, example 1 is solved again, but without using normalization. Showing that one does not need to normalize the eigenfunctions as the book shows and will get same answer. Solve \begin{equation} y^{\prime \prime }+2y=-x \tag{1} \end{equation} With boundary conditions y\left ( 0\right ) =0,y\left ( 1\right ) +y^{\prime }\left ( 1\right ) =0. Using the method of eigenfunction expansion without normalization.
Solution
The idea behind solving using eigenfunction expansion, is that\begin{equation} -\left ( py^{\prime }\right ) ^{\prime }+q\left ( x\right ) y\left ( x\right ) =\mu r\left ( x\right ) y\left ( x\right ) +f\left ( x\right ) \tag{1A} \end{equation} Is solved using the eigenfunctions of the corresponding homogeneous eigenvalue ODE\begin{equation} -\left ( py^{\prime }\right ) ^{\prime }+q\left ( x\right ) y\left ( x\right ) =\lambda r\left ( x\right ) y\left ( x\right ) \tag{2A} \end{equation} Where in (1A) \mu is just a constant. And in (2A), \lambda is an eigenvalue. Writing (1) in same form as (1A) leads to\begin{align} -\left ( y^{\prime }\right ) ^{\prime }-2y & =x\nonumber \\ -\left ( y^{\prime }\right ) ^{\prime } & =2y+x \tag{3A} \end{align}
Therefore \mu =2 and r\left ( x\right ) =1. The corresponding homogeneous eigenvalue problem is -\left ( y^{\prime }\right ) ^{\prime }=\lambda y\left ( x\right ) Or y^{\prime \prime }+\lambda y\left ( x\right ) =0 With boundary conditions y\left ( 0\right ) =0,y\left ( 1\right ) +y^{\prime }\left ( 1\right ) =0. The solution of the above is used to solve (3A), which is the original ODE. The solution to the above eigenvalue problem was done before. The result is that \lambda _{n} is the solution of nonlinear equation \sin \left ( \sqrt{\lambda }x\right ) +\sqrt{\lambda }\cos \left ( \sqrt{\lambda }x\right ) =0 Solving this numerically for the first 10 eigenvalues gives \lambda _{n}=\left \{ 4.116,24.139,63.659,122.889,201.851,300.55,418.987,557.162,715.077,892.73\right \} And the eigenfunctions are \Phi _{n}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{n}}x\right ) \qquad n=1,2,3,\cdots Notice that the eigenfunction above is not normalized as in the text book. Now assuming that the solution of the original nonhomogeneous ODE (3A) is given by y\left ( x\right ) =\sum _{n=1}^{\infty }b_{n}\Phi _{n}\left ( x\right ) Where b_{n} is unknown as of now and substituting the above into (3A) gives -\frac{d^{2}}{dx^{2}}\sum _{n=1}^{\infty }b_{n}\Phi _{n}\left ( x\right ) =2\sum _{n=1}^{\infty }b_{n}\Phi _{n}\left ( x\right ) +\sum _{n=1}^{\infty }q_{n}\Phi _{n}\left ( x\right ) Where \sum _{n=1}^{\infty }q_{n}\Phi _{n}\left ( x\right ) is the eigenfunction expansion of the forcing terms -x. In this expression q_{n} is still not known. Now assuming that differentiation can be moved inside the summation above (this needs conditions which assumed valid here). The above equation now becomes\begin{equation} -\sum _{n=1}^{\infty }b_{n}\Phi _{n}^{\prime \prime }\left ( x\right ) -2\sum _{n=1}^{\infty }b_{n}\Phi _{n}\left ( x\right ) =\sum _{n=1}^{\infty }q_{n}\Phi _{n}\left ( x\right ) \tag{1A} \end{equation} q_{n} is now found. This is done by applying orthogonality as follows. Let x=\sum _{n=1}^{\infty }q_{n}\Phi _{n}\left ( x\right ) . Multiplying both sides by \Phi _{m}\left ( x\right ) and integrating over the domain gives\begin{align} \int _{0}^{1}x\Phi _{m}\left ( x\right ) dx & =\sum _{n=1}^{\infty }q_{n}\int _{0}^{1}\Phi _{n}\left ( x\right ) \Phi _{m}\left ( x\right ) dx\nonumber \\ \int _{0}^{1}x\Phi _{m}\left ( x\right ) dx & =q_{n}\int _{0}^{1}\Phi _{m}^{2}\left ( x\right ) dx \tag{2} \end{align}
Since \Phi _{n}\left ( x\right ) is not normalized, one can not replace the integral by 1 as in the book. But since \Phi _{n}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{n}}x\right ) , the integrals can be evaluated as follows. The right side of (2) is\begin{equation} \int _{0}^{1}\sin ^{2}\left ( \sqrt{\lambda _{n}}x\right ) dx=\frac{1}{2}-\frac{\sin \left ( 2\sqrt{\lambda _{n}}\right ) }{4\sqrt{\lambda _{n}}} \tag{3} \end{equation} And the left side of (2) is found by integration by parts\begin{align} \int _{0}^{1}x\Phi _{m}\left ( x\right ) dx & =\int _{0}^{1}x\sin \left ( \sqrt{\lambda _{n}}x\right ) dx\nonumber \\ & =\frac{\sin \sqrt{\lambda _{n}}-\sqrt{\lambda _{n}}\cos \sqrt{\lambda _{n}}}{\lambda _{n}} \tag{4} \end{align}
Using (3) and (4) in (2) q_{n} is solved for giving\begin{align} \frac{\sin \sqrt{\lambda _{n}}-\sqrt{\lambda _{n}}\cos \sqrt{\lambda _{n}}}{\lambda _{n}} & =q_{n}\left ( \frac{1}{2}-\frac{\sin \left ( 2\sqrt{\lambda _{n}}\right ) }{4\sqrt{\lambda _{n}}}\right ) \nonumber \\ q_{n} & =\frac{\sin \sqrt{\lambda _{n}}-\sqrt{\lambda _{n}}\cos \sqrt{\lambda _{n}}}{\lambda _{n}\left ( \frac{1}{2}-\frac{\sin \left ( 2\sqrt{\lambda _{n}}\right ) }{4\sqrt{\lambda _{n}}}\right ) } \tag{5} \end{align}
Now that q_{n} is known, b_{n} is found from (1A) -\sum _{n=1}^{\infty }b_{n}\Phi _{n}^{\prime \prime }\left ( x\right ) -2\sum _{n=1}^{\infty }b_{n}\Phi _{n}\left ( x\right ) =\sum _{n=1}^{\infty }q_{n}\Phi _{n}\left ( x\right ) Since \Phi _{n}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{n}}x\right ) then \Phi _{n}^{\prime }\left ( x\right ) =\sqrt{\lambda _{n}}\cos \left ( \sqrt{\lambda _{n}}x\right ) ,\Phi _{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}\sin \left ( \sqrt{\lambda _{n}}x\right ) = -\lambda _{n}\Phi _{n}\left ( x\right ) and the above simplifies to \sum _{n=1}^{\infty }b_{n}\lambda _{n}\Phi _{n}\left ( x\right ) -2\sum _{n=1}^{\infty }b_{n}\Phi _{n}\left ( x\right ) =\sum _{n=1}^{\infty }q_{n}\Phi _{n}\left ( x\right ) Canceling summations and also \Phi _{n}\left ( x\right ) since \Phi _{n}\left ( x\right ) \neq 0 the above simplifies to\begin{align*} b_{n}\lambda _{n}-2b_{n} & =q_{n}\\ b_{n} & =\frac{q_{n}}{\lambda _{n}-2} \end{align*}
Hence the solution to the original ODE is\begin{align*} y\left ( x\right ) & =\sum _{n=1}^{\infty }b_{n}\Phi _{n}\left ( x\right ) \\ & =\sum _{n=1}^{\infty }\left ( \frac{q_{n}}{\lambda _{n}-2}\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) \end{align*}
Using the value found for q_{n} in (5), the above becomes\begin{equation} y\left ( x\right ) =\sum _{n=1}^{\infty }\frac{1}{\lambda _{n}\left ( \lambda _{n}-2\right ) }\frac{\sin \sqrt{\lambda _{n}}-\sqrt{\lambda _{n}}\cos \sqrt{\lambda _{n}}}{\frac{1}{2}-\frac{\sin \left ( 2\sqrt{\lambda _{n}}\right ) }{4\sqrt{\lambda _{n}}}}\sin \left ( \sqrt{\lambda _{n}}x\right ) \tag{6} \end{equation} The above is the solution, found without normalization. The book solution is\begin{equation} y\left ( x\right ) =4\sum _{n=1}^{\infty }\frac{1}{\lambda _{n}\left ( \lambda _{n}-2\right ) }\frac{1}{\left ( 1+\cos ^{2}\left ( \sqrt{\lambda _{n}}\right ) \right ) }\sin \left ( \sqrt{\lambda _{n}}x\right ) \tag{7} \end{equation} To show that (6) and (7) are actually the same, they are plotted against each others, using 10 terms in the sum, which is more than enough. The result shows identical plots.
They also plotted against the solution found using standard methods, which is y=\frac{\sin \left ( \sqrt{2}x\right ) }{\sin \left ( \sqrt{2}\right ) +\sqrt{2}\cos \left ( \sqrt{2}\right ) }-\frac{x}{2} And both (6,7) matched exactly the above solution.
Determine if the given boundary value problem is self-adjoint\begin{align*} y^{\prime \prime }+y^{\prime }+2y & =0\\ y\left ( 0\right ) & =0\\ y\left ( 1\right ) & =0 \end{align*}
Solution
The ODE can be written as \left ( y^{\prime }+y\right ) ^{\prime }+2y=0. Hence the operator is L\left [ y\right ] =\left ( y^{\prime }+y\right ) ^{\prime }+2y The ODE is self-adjoint if \left \langle L\left [ u\right ] ,v\right \rangle =\left \langle u,L\left [ v\right ] \right \rangle For any two functions u,v that satisfy the ODE. One way to proceed, is to start from the left side of the above equation and see if the right side can be arrived at. By definition \begin{align} \left \langle L\left [ u\right ] ,v\right \rangle & =\int _{0}^{1}L\left [ u\right ] vdx\nonumber \\ & =\int _{0}^{1}\left [ \left ( u^{\prime }+u\right ) ^{\prime }+2u\right ] vdx\nonumber \\ & =\int _{0}^{1}\left ( u^{\prime }+u\right ) ^{\prime }v+uvdx\nonumber \\ & =\int _{0}^{1}\overset{dv}{\overbrace{\left ( u^{\prime }+u\right ) ^{\prime }}}\overset{u}{\overbrace{v}}dx+\int _{0}^{1}uvdx \tag{1} \end{align}
integration by parts of the above gives\begin{align*} \left \langle L\left [ u\right ] ,v\right \rangle & =\left [ \left ( u^{\prime }+u\right ) v\right ] _{0}^{1}-\int _{0}^{1}\left ( u^{\prime }+u\right ) v^{\prime }dx+\int _{0}^{1}uvdx\\ & =\left [ \left ( u^{\prime }+u\right ) v\right ] _{0}^{1}-\int _{0}^{1}\left ( u^{\prime }v^{\prime }+uv^{\prime }\right ) dx+\int _{0}^{1}uvdx\\ & =\left [ \left ( u^{\prime }+u\right ) v\right ] _{0}^{1}-\left ( \int _{0}^{1}u^{\prime }v^{\prime }dx+\int _{0}^{1}uv^{\prime }dx\right ) +\int _{0}^{1}uvdx \end{align*}
Integrating by parts the term \int _{0}^{1}u^{\prime }v^{\prime }dx=\left [ uv^{\prime }\right ] _{0}^{1}-\int _{0}^{1}uv^{\prime \prime }dx the above becomes\begin{align*} \left \langle L\left [ u\right ] ,v\right \rangle & =\left [ \left ( u^{\prime }+u\right ) v\right ] _{0}^{1}-\left ( \left [ uv^{\prime }\right ] _{0}^{1}-\int _{0}^{1}uv^{\prime \prime }dx+\int _{0}^{1}uv^{\prime }dx\right ) +\int _{0}^{1}uvdx\\ & =\left [ \left ( u^{\prime }+u\right ) v-uv^{\prime }\right ] _{0}^{1}-\left ( -\int _{0}^{1}uv^{\prime \prime }dx+\int _{0}^{1}uv^{\prime }dx\right ) +\int _{0}^{1}uvdx\\ & =\left [ \left ( u^{\prime }+u\right ) v-uv^{\prime }\right ] _{0}^{1}+\int _{0}^{1}uv^{\prime \prime }dx-\int _{0}^{1}uv^{\prime }dx+\int _{0}^{1}uvdx\\ & =\left [ \left ( u^{\prime }+u\right ) v-uv^{\prime }\right ] _{0}^{1}+\int _{0}^{1}\left ( v^{\prime \prime }-v^{\prime }+v\right ) udx \end{align*}
The above can never be \left \langle u,L\left [ v\right ] \right \rangle even if the boundary terms vanish, since \int _{0}^{1}\left ( v^{\prime \prime }-v^{\prime }+v\right ) udx\neq \int _{0}^{1}\left ( v^{\prime \prime }+v^{\prime }+v\right ) udx. There is a different sign in the operator obtained. Hence the ode is not self adjoint.
Determine if the given boundary value problem is self-adjoint\begin{align*} \left ( 1+x^{2}\right ) y^{\prime \prime }+2xy^{\prime }+y & =0\\ y^{\prime }\left ( 0\right ) & =0\\ y\left ( 1\right ) +2y^{\prime }\left ( 1\right ) & =0 \end{align*}
Solution
The ODE can be written as \left ( \left ( 1+x^{2}\right ) y^{\prime }\right ) ^{\prime }+y=0 The operator L\left [ y\right ] =\left ( \left ( 1+x^{2}\right ) y^{\prime }\right ) ^{\prime }+y The ODE is self-adjoint if \left \langle L\left [ u\right ] ,v\right \rangle =\left \langle u,L\left [ v\right ] \right \rangle For any two functions u,v that satisfy the ODE. One way to proceed, is to start from the left side of the above equation and see if the right side can be arrived at. By definition \begin{align} \left \langle L\left [ u\right ] ,v\right \rangle & =\int _{0}^{1}L\left [ u\right ] vdx\nonumber \\ & =\int _{0}^{1}\left [ \left ( \left ( 1+x^{2}\right ) u^{\prime }\right ) ^{\prime }+u\right ] vdx\nonumber \\ & =\int _{0}^{1}\left ( \left ( 1+x^{2}\right ) u^{\prime }\right ) ^{\prime }v+uvdx\nonumber \\ & =\int _{0}^{1}\left ( \left ( 1+x^{2}\right ) u^{\prime }\right ) ^{\prime }vdx+\int _{0}^{1}uvdx \tag{1} \end{align}
Starting with the first integral in (1) and using integration by parts \int _{0}^{1}\left ( \left ( 1+x^{2}\right ) u^{\prime }\right ) ^{\prime }vdx=\int _{0}^{1}\overset{dv}{\overbrace{\left ( \left ( 1+x^{2}\right ) u^{\prime }\right ) ^{\prime }}}\overset{u}{\overbrace{v}}dx By integration by parts, where \int udv=\left \vert uv\right \vert -\int vdu, the above becomes\begin{align*} \int _{0}^{1}\left ( \left ( 1+x^{2}\right ) u^{\prime }\right ) ^{\prime }vdx & =\left [ \left ( 1+x^{2}\right ) u^{\prime }v\right ] _{0}^{1}-\int _{0}^{1}\left ( 1+x^{2}\right ) u^{\prime }v^{\prime }dx\\ & =\left [ \left ( 1+x^{2}\right ) u^{\prime }v\right ] _{0}^{1}-\int _{0}^{1}\overset{u}{\overbrace{\left ( 1+x^{2}\right ) v^{\prime }}}\overset{dv}{\overbrace{u^{\prime }}}dx \end{align*}
Doing integration by parts again. But notice the choice of u and dv made above. This is important in order to get to the form needed. The above becomes\begin{align*} \int _{0}^{1}\left ( \left ( 1+x^{2}\right ) u^{\prime }\right ) ^{\prime }vdx & =\left [ \left ( 1+x^{2}\right ) u^{\prime }v\right ] _{0}^{1}-\left ( \left [ u\left ( 1+x^{2}\right ) v^{\prime }\right ] _{0}^{1}-\int _{0}^{1}\left ( \left ( 1+x^{2}\right ) v^{\prime }\right ) ^{\prime }udx\right ) \\ & =\left [ \left ( 1+x^{2}\right ) u^{\prime }v-u\left ( 1+x^{2}\right ) v^{\prime }\right ] _{0}^{1}+\int _{0}^{1}\left ( \left ( 1+x^{2}\right ) v^{\prime }\right ) ^{\prime }udx \end{align*}
Going back to (1) and adding the second integral which is left there gives\begin{align*} \left \langle L\left [ u\right ] ,v\right \rangle & =\left [ \left ( 1+x^{2}\right ) u^{\prime }v-u\left ( 1+x^{2}\right ) v^{\prime }\right ] _{0}^{1}+\int _{0}^{1}\left ( \left ( 1+x^{2}\right ) v^{\prime }\right ) ^{\prime }udx+\int _{0}^{1}uvdx\\ & =\left [ \left ( 1+x^{2}\right ) u^{\prime }v-u\left ( 1+x^{2}\right ) v^{\prime }\right ] _{0}^{1}+\int _{0}^{1}\left [ \left ( \left ( 1+x^{2}\right ) v^{\prime }\right ) ^{\prime }+v\right ] udx \end{align*}
But \int _{0}^{1}\left [ \left ( \left ( 1+x^{2}\right ) v^{\prime }\right ) ^{\prime }+v\right ] udx=\left \langle u,L\left [ v\right ] \right \rangle , hence the above becomes\begin{equation} \left \langle L\left [ u\right ] ,v\right \rangle =\left [ \left ( 1+x^{2}\right ) u^{\prime }v-u\left ( 1+x^{2}\right ) v^{\prime }\right ] _{0}^{1}+\left \langle u,L\left [ v\right ] \right \rangle \tag{2} \end{equation} We are almost there. If the boundary terms above all go to zero, then it is self-adjoint. If the boundary terms do not vanish, then the problem is not self adjoint. Evaluating the boundary terms in (2)\begin{align*} \Delta & =\left [ \left ( 1+x^{2}\right ) u^{\prime }v-u\left ( 1+x^{2}\right ) v^{\prime }\right ] _{0}^{1}\\ & =\left [ 2u^{\prime }\left ( 1\right ) v\left ( 1\right ) -2u\left ( 1\right ) v^{\prime }\left ( 1\right ) \right ] -\left [ u^{\prime }\left ( 0\right ) v\left ( 0\right ) -u\left ( 0\right ) v^{\prime }\left ( 0\right ) \right ] \end{align*}
Since u^{\prime }\left ( 0\right ) =0 and v^{\prime }\left ( 0\right ) =0, from the given boundary conditions, then above simplifies to \Delta =2\left ( u^{\prime }\left ( 1\right ) v\left ( 1\right ) -u\left ( 1\right ) v^{\prime }\left ( 1\right ) \right ) But u\left ( 1\right ) =-2u^{\prime }\left ( 1\right ) and v\left ( 1\right ) =-2v^{\prime }\left ( 1\right ) , hence the above becomes\begin{align*} \Delta & =2\left ( u^{\prime }\left ( 1\right ) \left ( -2v^{\prime }\left ( 1\right ) \right ) -\left ( -2u^{\prime }\left ( 1\right ) \right ) v^{\prime }\left ( 1\right ) \right ) \\ & =4\left ( -u^{\prime }\left ( 1\right ) v^{\prime }\left ( 1\right ) +u^{\prime }\left ( 1\right ) v^{\prime }\left ( 1\right ) \right ) \\ & =0 \end{align*}
Since the boundary terms \Delta vanish, then from (2) \begin{equation} \left \langle L\left [ u\right ] ,v\right \rangle =\left \langle u,L\left [ v\right ] \right \rangle \tag{3} \end{equation} Hence the ODE is self-adjoint.
Determine if the given boundary value problem is self-adjoint\begin{align*} y^{\prime \prime }+y & =\lambda y\\ y\left ( 0\right ) -y^{\prime }\left ( 1\right ) & =0\\ y^{\prime }\left ( 0\right ) -y\left ( 1\right ) & =0 \end{align*}
Solution
The operator is L\left [ y\right ] =y^{\prime \prime }+y The ODE is self-adjoint if \left \langle L\left [ u\right ] ,v\right \rangle =\left \langle u,L\left [ v\right ] \right \rangle For any two functions u,v that satisfy the ODE. One way to proceed, is to start from the left side of the above equation and see if the right side can be arrived at. By definition \begin{align} \left \langle L\left [ u\right ] ,v\right \rangle & =\int _{0}^{1}L\left [ u\right ] vdx\nonumber \\ & =\int _{0}^{1}\left ( u^{\prime \prime }+u\right ) vdx\nonumber \\ & =\int _{0}^{1}u^{\prime \prime }vdx+\int _{0}^{1}uvdx\nonumber \\ & =\int _{0}^{1}\overset{dv}{\overbrace{u^{\prime \prime }}}\overset{u}{\overbrace{v}}dx+\int _{0}^{1}uvdx \tag{1} \end{align}
Integrating by parts \left \langle L\left [ u\right ] ,v\right \rangle =\left [ u^{\prime }v\right ] _{0}^{1}-\int _{0}^{1}\overset{dv}{\overbrace{u^{\prime }}}\overset{u}{\overbrace{v^{\prime }}}dx+\int _{0}^{1}uvdx Integrating by parts again\begin{align} \left \langle L\left [ u\right ] ,v\right \rangle & =\left [ u^{\prime }v\right ] _{0}^{1}-\left ( \left [ uv^{\prime }\right ] _{0}^{1}-\int _{0}^{1}uv^{\prime \prime }dx\right ) +\int _{0}^{1}uvdx\nonumber \\ & =\left [ u^{\prime }v-uv^{\prime }\right ] _{0}^{1}+\int _{0}^{1}uv^{\prime \prime }dx+\int _{0}^{1}uvdx\nonumber \\ & =\left [ u^{\prime }v-uv^{\prime }\right ] _{0}^{1}+\int _{0}^{1}\left ( v^{\prime \prime }+v\right ) udx\nonumber \\ & =\left [ u^{\prime }v-uv^{\prime }\right ] _{0}^{1}+\left \langle u,L\left [ v\right ] \right \rangle \tag{2} \end{align}
Hence if the boundary terms vanish, then it is self adjoint else it is not. Evaluating the boundary terms in (2)\begin{align*} \Delta & =\left [ u^{\prime }v-uv^{\prime }\right ] _{0}^{1}\\ & =\left [ u^{\prime }\left ( 1\right ) v\left ( 1\right ) -u\left ( 1\right ) v^{\prime }\left ( 1\right ) \right ] -\left [ u^{\prime }\left ( 0\right ) v\left ( 0\right ) -u\left ( 0\right ) v^{\prime }\left ( 0\right ) \right ] \end{align*}
But u^{\prime }\left ( 1\right ) =u\left ( 0\right ) and v^{\prime }\left ( 1\right ) =v\left ( 0\right ) and u^{\prime }\left ( 0\right ) =u\left ( 1\right ) and v^{\prime }\left ( 0\right ) =v\left ( 1\right ) from the given boundary conditions. Substituting these into the above gives\begin{align*} \Delta & =\left [ u\left ( 0\right ) v\left ( 1\right ) -u\left ( 1\right ) v\left ( 0\right ) \right ] -\left [ u\left ( 1\right ) v\left ( 0\right ) -u\left ( 0\right ) v\left ( 1\right ) \right ] \\ & =2u\left ( 1\right ) v\left ( 0\right ) \\ & \neq 0 \end{align*}
Since the boundary terms \Delta do not vanish, then from (2) \left \langle L\left [ u\right ] ,v\right \rangle \neq \left \langle u,L\left [ v\right ] \right \rangle Hence the ODE is not self-adjoint.
Determine if the given boundary value problem is self-adjoint\begin{align*} \left ( 1+x^{2}\right ) y^{\prime \prime }+2xy^{\prime }+y & =\lambda \left ( 1+x^{2}\right ) y\\ y\left ( 0\right ) -y^{\prime }\left ( 1\right ) & =0\\ y^{\prime }\left ( 0\right ) +2y\left ( 1\right ) & =0 \end{align*}
Solution
The ode can be written as \left ( \left ( 1+x^{2}\right ) y^{\prime }\right ) ^{\prime }+y=\lambda \left ( 1+x^{2}\right ) y Hence the operator is L\left [ y\right ] =\left ( \left ( 1+x^{2}\right ) y^{\prime }\right ) ^{\prime }+y The ODE is self-adjoint if \left \langle L\left [ u\right ] ,v\right \rangle =\left \langle u,L\left [ v\right ] \right \rangle For any two functions u,v that satisfy the ODE. One way to proceed, is to start from the left side of the above equation and see if the right side can be arrived at. By definition \begin{align} \left \langle L\left [ u\right ] ,v\right \rangle & =\int _{0}^{1}L\left [ u\right ] vdx\nonumber \\ & =\int _{0}^{1}\left ( \left ( \left ( 1+x^{2}\right ) u^{\prime }\right ) ^{\prime }+u\right ) vdx\nonumber \\ & =\int _{0}^{1}\overset{dv}{\overbrace{\left ( \left ( 1+x^{2}\right ) u^{\prime }\right ) ^{\prime }}}\overset{u}{\overbrace{v}}dx+\int _{0}^{1}uvdx\nonumber \end{align}
Integrating by parts\begin{align*} \left \langle L\left [ u\right ] ,v\right \rangle & =\left [ \left ( 1+x^{2}\right ) u^{\prime }v\right ] _{0}^{1}-\int _{0}^{1}\left ( 1+x^{2}\right ) u^{\prime }v^{\prime }dx+\int _{0}^{1}uvdx\\ & =\left [ \left ( 1+x^{2}\right ) u^{\prime }v\right ] _{0}^{1}-\int _{0}^{1}\overset{u}{\overbrace{\left ( 1+x^{2}\right ) v^{\prime }}}\overset{dv}{\overbrace{u^{\prime }}}dx+\int _{0}^{1}uvdx \end{align*}
Integrating by parts\begin{align*} \left \langle L\left [ u\right ] ,v\right \rangle & =\left [ \left ( 1+x^{2}\right ) u^{\prime }v\right ] _{0}^{1}-\left ( \left [ u\left ( 1+x^{2}\right ) v^{\prime }\right ] _{0}^{1}-\int _{0}^{1}\left ( \left ( 1+x^{2}\right ) v^{\prime }\right ) ^{\prime }udx\right ) +\int _{0}^{1}uvdx\\ & =\left [ \left ( 1+x^{2}\right ) u^{\prime }v-u\left ( 1+x^{2}\right ) v^{\prime }\right ] _{0}^{1}+\int _{0}^{1}\left ( \left ( 1+x^{2}\right ) v^{\prime }\right ) ^{\prime }udx+\int _{0}^{1}uvdx\\ & =\left [ \left ( 1+x^{2}\right ) u^{\prime }v-u\left ( 1+x^{2}\right ) v^{\prime }\right ] _{0}^{1}+\int _{0}^{1}\left [ \left ( \left ( 1+x^{2}\right ) v^{\prime }\right ) ^{\prime }+v\right ] udx\\ & =\left [ \left ( 1+x^{2}\right ) u^{\prime }v-u\left ( 1+x^{2}\right ) v^{\prime }\right ] _{0}^{1}+\left \langle u,L\left [ v\right ] \right \rangle \end{align*}
Therefore, if the boundary terms vanish, then the ODE is self adjoint. \Delta =\left [ 2u^{\prime }\left ( 1\right ) v\left ( 1\right ) -2u\left ( 1\right ) v^{\prime }\left ( 1\right ) \right ] -\left [ u^{\prime }\left ( 0\right ) v\left ( 0\right ) -u\left ( 0\right ) v^{\prime }\left ( 0\right ) \right ] But u^{\prime }\left ( 1\right ) =u\left ( 0\right ) and v^{\prime }\left ( 1\right ) =v\left ( 0\right ) and u^{\prime }\left ( 0\right ) =2u\left ( 1\right ) and v^{\prime }\left ( 0\right ) =2v\left ( 1\right ) , from the given boundary conditions. Substituting these in the above gives\begin{align*} \Delta & =\left [ 2u\left ( 0\right ) v\left ( 1\right ) -2u\left ( 1\right ) v\left ( 0\right ) \right ] -\left [ 2u\left ( 1\right ) v\left ( 0\right ) -u\left ( 0\right ) 2v\left ( 1\right ) \right ] \\ & =2u\left ( 0\right ) v\left ( 1\right ) -2u\left ( 1\right ) v\left ( 0\right ) -2u\left ( 1\right ) v\left ( 0\right ) +u\left ( 0\right ) 2v\left ( 1\right ) \\ & =4u\left ( 0\right ) v\left ( 1\right ) -4u\left ( 1\right ) v\left ( 0\right ) \\ & =0 \end{align*}
Hence \left \langle L\left [ u\right ] ,v\right \rangle =\left \langle u,L\left [ v\right ] \right \rangle , therefore the ODE is self-adjoint.
Determine if the given boundary value problem is self-adjoint\begin{align*} y^{\prime \prime }+\lambda y & =0\\ y\left ( 0\right ) & =0\\ y\left ( \pi \right ) +y^{\prime }\left ( \pi \right ) & =0 \end{align*}
Solution
The ode can be written as y^{\prime \prime }=-\lambda y Hence L\left [ y\right ] =y^{\prime \prime }. The ODE is self-adjoint if \left \langle L\left [ u\right ] ,v\right \rangle =\left \langle u,L\left [ v\right ] \right \rangle For any two functions u,v that satisfy the ODE. One way to proceed, is to start from the left side of the above equation and see if the right side can be arrived at. By definition \begin{align} \left \langle L\left [ u\right ] ,v\right \rangle & =\int _{0}^{\pi }L\left [ u\right ] vdx\nonumber \\ & =\int _{0}^{\pi }u^{\prime \prime }vdx\nonumber \end{align}
Integrating by parts once \left \langle L\left [ u\right ] ,v\right \rangle =\left [ u^{\prime }v\right ] _{0}^{\pi }-\int _{0}^{\pi }u^{\prime }v^{\prime }dx Integrating by parts again\begin{align*} \left \langle L\left [ u\right ] ,v\right \rangle & =\left [ u^{\prime }v\right ] _{0}^{\pi }-\left ( \left [ uv^{\prime }\right ] _{0}^{\pi }-\int _{0}^{\pi }uv^{\prime \prime }dx\right ) \\ & =\left [ u^{\prime }v-uv^{\prime }\right ] _{0}^{\pi }+\int _{0}^{\pi }uv^{\prime \prime }dx\\ & =\left [ u^{\prime }v-uv^{\prime }\right ] _{0}^{\pi }+\left \langle u,L\left [ v\right ] \right \rangle \end{align*}
Now we will check if the boundary terms vanish or not. \begin{align*} \Delta & =\left [ u^{\prime }v-uv^{\prime }\right ] _{0}^{\pi }\\ & =\left [ u^{\prime }\left ( \pi \right ) v\left ( \pi \right ) -u\left ( \pi \right ) v^{\prime }\left ( \pi \right ) \right ] -\left [ u^{\prime }\left ( 0\right ) v\left ( 0\right ) -u\left ( 0\right ) v^{\prime }\left ( 0\right ) \right ] \end{align*}
Since u\left ( 0\right ) =0,v\left ( 0\right ) =0 then the above simplifies to \Delta =u^{\prime }\left ( \pi \right ) v\left ( \pi \right ) -u\left ( \pi \right ) v^{\prime }\left ( \pi \right ) But u^{\prime }\left ( \pi \right ) =-u\left ( \pi \right ) and v^{\prime }\left ( \pi \right ) =-v\left ( \pi \right ) the above becomes\begin{align*} \Delta & =-u\left ( \pi \right ) v\left ( \pi \right ) +u\left ( \pi \right ) v\left ( \pi \right ) \\ & =0 \end{align*}
Hence \left \langle L\left [ u\right ] ,v\right \rangle =\left \langle u,L\left [ v\right ] \right \rangle and the ODE is self adjoint.
Solve by method of eigenfunction expansion\begin{align*} y^{\prime \prime }+2y & =-x\\ y\left ( 0\right ) & =0\\ y\left ( 1\right ) & =0 \end{align*}
Solution
The corresponding homogeneous eigenvalue ODE is y^{\prime \prime }+\lambda y=0 with y\left ( 0\right ) =0,y\left ( 1\right ) =0. This was solved before. \begin{align*} \hat{\Phi }_{n}\left ( x\right ) & =\sqrt{2}\sin \left ( \sqrt{\lambda _{n}}x\right ) \\ \lambda _{n} & =\left ( n\pi \right ) ^{2}\qquad n=1,2,3,\cdots \end{align*}
Hence eigenvalues are \lambda _{n}=\left \{ \pi ^{2},4\pi ^{2},9\pi ^{2},\cdots \right \} . None of the eigenvalues is 2. Therefore the solution to the original ODE can be assumed to be\begin{equation} y=\sum _{n=1}^{\infty }b_{n}\hat{\Phi }_{n}\left ( x\right ) \tag{1} \end{equation} Substituting this into the original ODE gives \sum _{n=1}^{\infty }b_{n}\hat{\Phi }_{n}^{\prime \prime }\left ( x\right ) +2\sum _{n=1}^{\infty }b_{n}\hat{\Phi }_{n}\left ( x\right ) =-x Expanding -x using same basis function as the solution gives\begin{equation} \sum _{n=1}^{\infty }b_{n}\hat{\Phi }_{n}^{\prime \prime }\left ( x\right ) +2\sum _{n=1}^{\infty }b_{n}\hat{\Phi }_{n}\left ( x\right ) =\sum _{n=1}^{\infty }q_{n}\hat{\Phi }_{n}\left ( x\right ) \tag{2} \end{equation} Where q_{n} is found by applying orthogonality on\begin{align*} -x & =\sum _{n=1}^{\infty }q_{n}\hat{\Phi }_{n}\left ( x\right ) \\ -\int _{0}^{1}x\hat{\Phi }_{m}\left ( x\right ) dx & =\sum _{n=1}^{\infty }q_{n}\int _{0}^{1}\hat{\Phi }_{n}\left ( x\right ) \hat{\Phi }_{m}\left ( x\right ) dx\\ & =q_{m}\int _{0}^{1}\hat{\Phi }_{m}^{2}\left ( x\right ) dx \end{align*}
Since normalized, \int _{0}^{1}\hat{\Phi }_{m}^{2}\left ( x\right ) dx=1 and the above simplifies to -\int _{0}^{1}x\hat{\Phi }_{m}\left ( x\right ) dx=q_{m} But \hat{\Phi }_{m}\left ( x\right ) =\sqrt{2}\sin \left ( n\pi x\right ) and the above becomes -\sqrt{2}\int _{0}^{1}x\sin \left ( n\pi x\right ) dx=q_{n} Using \int x\sin \left ( ax\right ) dx=\frac{\sin ax}{a^{2}}-\frac{x\cos ax}{a} the above gives\begin{align*} -\sqrt{2}\left ( \frac{\sin \left ( n\pi x\right ) }{\left ( n\pi \right ) ^{2}}-\frac{x\cos \left ( n\pi x\right ) }{n\pi }\right ) _{0}^{1} & =q_{n}\\ -\sqrt{2}\left ( \frac{\sin \left ( n\pi \right ) }{\left ( n\pi \right ) ^{2}}-\frac{\cos \left ( n\pi \right ) }{n\pi }\right ) & =q_{n}\\ \sqrt{2}\left ( \frac{\cos \left ( n\pi \right ) }{n\pi }\right ) & =q_{n}\\ \sqrt{2}\left ( \frac{-1^{n}}{n\pi }\right ) & =q_{n} \end{align*}
Now that q_{n} is found, then b_{n} can be solved for form (2) above giving\begin{equation} \sum _{n=1}^{\infty }b_{n}\hat{\Phi }_{n}^{\prime \prime }\left ( x\right ) +2\sum _{n=1}^{\infty }b_{n}\hat{\Phi }_{n}\left ( x\right ) =\sum _{n=1}^{\infty }\sqrt{2}\left ( \frac{-1^{n}}{4n\pi }\right ) \hat{\Phi }_{n}\left ( x\right ) \tag{2A} \end{equation} But \hat{\Phi }_{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}\hat{\Phi }_{n}\left ( x\right ) since the eigenfunction satisfy the ode y^{\prime \prime }=-\lambda y and the above simplifies to -\sum _{n=1}^{\infty }b_{n}\lambda _{n}\hat{\Phi }_{n}\left ( x\right ) +2\sum _{n=1}^{\infty }b_{n}\hat{\Phi }_{n}\left ( x\right ) =\sum _{n=1}^{\infty }\sqrt{2}\left ( \frac{-1^{n}}{4n\pi }\right ) \hat{\Phi }_{n}\left ( x\right ) Since \hat{\Phi }_{n}\left ( x\right ) \neq 0 the above simplifies to -b_{n}\lambda _{n}+2b_{n}=\sqrt{2}\left ( \frac{-1^{n}}{n\pi }\right ) Therefore \begin{align*} b_{n} & =\frac{\sqrt{2}\left ( \frac{-1^{n}}{n\pi }\right ) }{2-\lambda _{n}}\\ & =\frac{\sqrt{2}\left ( -1\right ) ^{n}}{\left ( 2-\left ( n\pi \right ) ^{2}\right ) n\pi } \end{align*}
Therefore the solution from (1) is y=\sum _{n=1}^{\infty }\frac{\sqrt{2}\left ( -1\right ) ^{n}}{\left ( 2-\left ( n\pi \right ) ^{2}\right ) n\pi }\hat{\Phi }_{n}\left ( x\right ) But \hat{\Phi }_{n}\left ( x\right ) =\sqrt{2}\Phi _{n}\left ( x\right ) =\sqrt{2}\sin \left ( n\pi x\right ) and the above becomes y=2\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}}{\left ( 2-\left ( n\pi \right ) ^{2}\right ) n\pi }\sin \left ( n\pi x\right ) Or y=2\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n+1}}{\left ( \left ( n\pi \right ) ^{2}-2\right ) n\pi }\sin \left ( n\pi x\right )
Solve by method of eigenfunction expansion\begin{align*} y^{\prime \prime }+2y & =-x\\ y\left ( 0\right ) & =0\\ y^{\prime }\left ( 1\right ) & =0 \end{align*}
Solution
The corresponding homogeneous eigenvalue ODE is y^{\prime \prime }+\lambda y=0 with y\left ( 0\right ) =0,y^{\prime }\left ( 1\right ) =0. This was solved before. \begin{align*} \Phi _{n}\left ( x\right ) & =\sin \left ( \sqrt{\lambda _{n}}x\right ) \\ \lambda _{n} & =\left ( \frac{n\pi }{2}\right ) ^{2}\qquad n=1,3,5,\cdots \end{align*}
Or, to keep the sum continuous, it can be written as \lambda _{n}=\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}\qquad n=1,2,3,\cdots The normalized eigenfunctions weight k_{n} is found from solving \int _{0}^{1}k_{n}^{2}\sin ^{2}\left ( \frac{n\pi }{2}x\right ) dx=1 which results in k_{n}=\sqrt{2}
Hence \hat{\Phi }_{n}\left ( x\right ) =\sqrt{2}\sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) \qquad n=1,2,3,\cdots
The eigenvalues are \lambda _{n}=\left \{ \left ( \frac{\pi }{2}\right ) ^{2},9\left ( \frac{\pi }{2}\right ) ^{2},25\left ( \frac{\pi }{2}\right ) ^{2},\cdots \right \} . None of the eigenvalues is 2. Therefore the solution to the original ODE can be assumed to be\begin{equation} y=\sum _{n=1}^{\infty }b_{n}\hat{\Phi }_{n}\left ( x\right ) \tag{1} \end{equation} Substituting this into the original ODE gives \sum _{n=1}^{\infty }b_{n}\hat{\Phi }_{n}^{\prime \prime }\left ( x\right ) +2\sum _{n=1}^{\infty }b_{n}\hat{\Phi }_{n}\left ( x\right ) =-x Expanding -x using same basis function as the solution gives\begin{equation} \sum _{n=1}^{\infty }b_{n}\hat{\Phi }_{n}^{\prime \prime }\left ( x\right ) +2\sum _{n=1}^{\infty }b_{n}\hat{\Phi }_{n}\left ( x\right ) =\sum _{n=1}^{\infty }q_{n}\hat{\Phi }_{n}\left ( x\right ) \tag{2} \end{equation} Where q_{n} is found by applying orthogonality on\begin{align*} -x & =\sum _{n=1}^{\infty }q_{n}\hat{\Phi }_{n}\left ( x\right ) \\ -\int _{0}^{1}x\hat{\Phi }_{m}\left ( x\right ) dx & =\sum _{n=1}^{\infty }q_{n}\int _{0}^{1}\hat{\Phi }_{n}\left ( x\right ) \hat{\Phi }_{m}\left ( x\right ) dx\\ & =q_{m}\int _{0}^{1}\hat{\Phi }_{m}^{2}\left ( x\right ) dx \end{align*}
Since normalized, \int _{0}^{1}\hat{\Phi }_{m}^{2}\left ( x\right ) dx=1 and the above simplifies to -\int _{0}^{1}x\hat{\Phi }_{m}\left ( x\right ) dx=q_{m} But \hat{\Phi }_{m}\left ( x\right ) =\sqrt{2}\sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) and the above becomes -\sqrt{2}\int _{0}^{1}x\sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) dx=q_{n} Using \int x\sin \left ( ax\right ) dx=\frac{\sin ax}{a^{2}}-\frac{x\cos ax}{a} the above gives\begin{align*} -\sqrt{2}\left ( \frac{\sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) }{\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}}-\frac{x\cos \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) }{\left ( 2n-1\right ) \frac{\pi }{2}}\right ) _{0}^{1} & =q_{n}\\ -\sqrt{2}\left ( \frac{\sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) }{\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}}-\frac{\cos \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) }{\left ( 2n-1\right ) \frac{\pi }{2}}\right ) _{0}^{1} & =q_{n}\\ -\sqrt{2}\left ( \frac{\sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) }{\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}}\right ) & =q_{n} \end{align*}
Using \sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) =-\cos \left ( n\pi \right ) which for n=1,2,3,\cdots can be written as -\left ( -1\right ) ^{n} or \left ( -1\right ) ^{n+1}. The above simplifies to\begin{align*} -\sqrt{2}\left ( \frac{\left ( -1\right ) ^{n+1}}{\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}}\right ) & =q_{n}\\ \sqrt{2}\left ( \frac{\left ( -1\right ) ^{n}}{\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}}\right ) & =q_{n} \end{align*}
Now that q_{n} is found, then b_{n} can be solved for form (2) above giving\begin{equation} \sum _{n=1}^{\infty }b_{n}\hat{\Phi }_{n}^{\prime \prime }\left ( x\right ) +2\sum _{n=1}^{\infty }b_{n}\hat{\Phi }_{n}\left ( x\right ) =\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}\sqrt{2}}{\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}}\hat{\Phi }_{n}\left ( x\right ) \tag{2A} \end{equation} But \hat{\Phi }_{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}\hat{\Phi }_{n}\left ( x\right ) since the eigenfunction satisfy the ode y^{\prime \prime }=-\lambda y and the above simplifies to -\sum _{n=1}^{\infty }b_{n}\lambda _{n}\hat{\Phi }_{n}\left ( x\right ) +2\sum _{n=1}^{\infty }b_{n}\hat{\Phi }_{n}\left ( x\right ) =\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}\sqrt{2}}{\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}}\hat{\Phi }_{n}\left ( x\right ) Since \hat{\Phi }_{n}\left ( x\right ) \neq 0 the above simplifies to -b_{n}\lambda _{n}+2b_{n}=\frac{\left ( -1\right ) ^{n}\sqrt{2}}{\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}} Therefore \begin{align*} b_{n} & =\frac{\frac{\left ( -1\right ) ^{n}\sqrt{2}}{\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}}}{2-\lambda _{n}}\\ & =\frac{\frac{\left ( -1\right ) ^{n}\sqrt{2}}{\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}}}{\left ( 2-\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}\right ) }\\ & =\frac{\left ( -1\right ) ^{n}\sqrt{2}}{\left ( 2-\left ( 2n-1\right ) ^{2}\left ( \frac{\pi }{2}\right ) ^{2}\right ) \left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}} \end{align*}
Therefore the solution from (1) is y=\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}\sqrt{2}}{\left ( 2-\left ( 2n-1\right ) ^{2}\left ( \frac{\pi }{2}\right ) ^{2}\right ) \left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}}\hat{\Phi }_{n}\left ( x\right ) But \hat{\Phi }_{n}\left ( x\right ) =\sqrt{2}\Phi _{n}\left ( x\right ) =\sqrt{2}\sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) and the above becomes y=2\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n+1}}{\left ( \left ( 2n-1\right ) ^{2}\left ( \frac{\pi }{2}\right ) ^{2}-2\right ) \left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}}\sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) Since \ \left ( 2n-1\right ) \frac{\pi }{2}=\left ( n-\frac{1}{2}\right ) \pi , the above can also be written as (to match back of book solution) y=2\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n+1}}{\left ( \left ( n-\frac{1}{2}\right ) ^{2}\pi ^{2}-2\right ) \left ( \left ( n-\frac{1}{2}\right ) \pi \right ) ^{2}}\sin \left ( \left ( n-\frac{1}{2}\right ) \pi x\right )
Solve by method of eigenfunction expansion\begin{align*} y^{\prime \prime }+2y & =-x\\ y^{\prime }\left ( 0\right ) & =0\\ y^{\prime }\left ( 1\right ) & =0 \end{align*}
Solution
The corresponding homogeneous eigenvalue ODE is y^{\prime \prime }+\lambda y=0 with y^{\prime }\left ( 0\right ) =0,y^{\prime }\left ( 1\right ) =0. This was solved above in Chapter 11.2, problem 3. The eigenvalues are \begin{align*} \lambda _{n} & =\left \{ 0,\pi ^{2},\left ( 2\pi \right ) ^{2},\left ( 3\pi \right ) ^{2},\cdots \right \} \\ & =\left ( n\pi \right ) ^{2}\qquad n=0,1,2,\cdots \end{align*}
The normalized eigenfunctions are \hat{\Phi }_{0}\left ( x\right ) =1 And for n=1,2,3,\cdots \begin{align*} \hat{\Phi }_{n}\left ( x\right ) & =\sqrt{2}\Phi _{n}\left ( x\right ) \\ & =\sqrt{2}\cos \left ( n\pi x\right ) \\ & =\left \{ \sqrt{2}\cos \left ( \pi x\right ) ,\sqrt{2}\cos \left ( 2\pi x\right ) ,\sqrt{2}\cos \left ( 3\pi x\right ) ,\cdots \right \} \end{align*}
Since none of the eigenvalues is 2, the solution to the original ODE can be assumed to be\begin{equation} y=\sum _{n=0}^{\infty }b_{n}\hat{\Phi }_{n}\left ( x\right ) \tag{1} \end{equation} Substituting this into the original ODE gives \sum _{n=0}^{\infty }b_{n}\hat{\Phi }_{n}^{\prime \prime }\left ( x\right ) +2\sum _{n=0}^{\infty }b_{n}\hat{\Phi }_{n}\left ( x\right ) =-x Expanding -x using same basis function as the solution gives\begin{equation} \sum _{n=0}^{\infty }b_{n}\hat{\Phi }_{n}^{\prime \prime }\left ( x\right ) +2\sum _{n=0}^{\infty }b_{n}\hat{\Phi }_{n}\left ( x\right ) =\sum _{n=0}^{\infty }c_{n}\hat{\Phi }_{n}\left ( x\right ) \tag{2} \end{equation} Where c_{n} is found by applying orthogonality on\begin{align*} -x & =\sum _{n=0}^{\infty }c_{n}\hat{\Phi }_{n}\left ( x\right ) \\ -\int _{0}^{1}x\hat{\Phi }_{m}\left ( x\right ) dx & =\sum _{n=0}^{\infty }c_{n}\int _{0}^{1}\hat{\Phi }_{n}\left ( x\right ) \hat{\Phi }_{m}\left ( x\right ) dx\\ & =c_{m}\int _{0}^{1}\hat{\Phi }_{m}^{2}\left ( x\right ) dx \end{align*}
Since normalized then \int _{0}^{1}\hat{\Phi }_{m}^{2}\left ( x\right ) dx=1 and the above simplifies to -\int _{0}^{1}x\hat{\Phi }_{n}\left ( x\right ) dx=c_{n} For n=0 the eigenfunction is \hat{\Phi }_{0}\left ( x\right ) =1 and the above gives c_{0}=-\frac{1}{2}\left [ x^{2}\right ] _{0}^{1}=-\frac{1}{2} and for n>0 the eigenfunction is \hat{\Phi }_{n}\left ( x\right ) =\sqrt{2}\cos \left ( n\pi x\right ) and the integrals becomes -\sqrt{2}\int _{0}^{1}x\cos \left ( n\pi x\right ) dx=c_{n} Using \int x\cos \left ( ax\right ) dx=\frac{\cos ax}{a^{2}}+\frac{x\sin ax}{a} the above gives\begin{align*} c_{n} & =-\sqrt{2}\left ( \frac{\cos \left ( n\pi x\right ) }{\left ( n\pi \right ) ^{2}}+\frac{x\sin \left ( n\pi x\right ) }{n\pi }\right ) _{0}^{1}\\ & =-\sqrt{2}\left ( \frac{\cos \left ( n\pi \right ) }{\left ( n\pi \right ) ^{2}}+\frac{\sin \left ( n\pi \right ) }{n\pi }-\frac{1}{\left ( n\pi \right ) ^{2}}\right ) \\ & =-\sqrt{2}\left ( \frac{\cos \left ( n\pi \right ) }{\left ( n\pi \right ) ^{2}}-\frac{1}{\left ( n\pi \right ) ^{2}}\right ) \\ & =\frac{-\sqrt{2}}{\left ( n\pi \right ) ^{2}}\left ( \cos \left ( n\pi \right ) -1\right ) \qquad \qquad n=1,2,\cdots \end{align*}
When n is odd then c_{n}=\frac{2\sqrt{2}}{\left ( n\pi \right ) ^{2}} and when n is even it is zero. Now that q_{n} is found, then b_{n} can be solved for form (2) above giving\begin{equation} \sum _{n=0}^{\infty }b_{n}\hat{\Phi }_{n}^{\prime \prime }\left ( x\right ) +2\sum _{n=0}^{\infty }b_{n}\hat{\Phi }_{n}\left ( x\right ) =\sum _{n=0}^{\infty }c_{n}\hat{\Phi }_{n}\left ( x\right ) \tag{2A} \end{equation} But \hat{\Phi }_{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}\hat{\Phi }_{n}\left ( x\right ) since the eigenfunction satisfies the ode y^{\prime \prime }=-\lambda y and the above simplifies to -\sum _{n=0}^{\infty }b_{n}\lambda _{n}\hat{\Phi }_{n}\left ( x\right ) +2\sum _{n=0}^{\infty }b_{n}\hat{\Phi }_{n}\left ( x\right ) =\sum _{n=0}^{\infty }c_{n}\hat{\Phi }_{n}\left ( x\right ) Since \hat{\Phi }_{n}\left ( x\right ) \neq 0 the above simplifies to\begin{align*} -b_{n}\lambda _{n}+2b_{n} & =c_{n}\\ b_{n} & =\frac{c_{n}}{2-\lambda _{n}} \end{align*}
Therefore the solution from (1) is\begin{align*} y & =\sum _{n=0}^{\infty }\frac{c_{n}}{2-\lambda _{n}}\hat{\Phi }_{n}\left ( x\right ) \\ & =\frac{c_{0}}{2-\lambda _{0}}\hat{\Phi }_{0}\left ( x\right ) +\sum _{n=1,3,5,\cdots }^{\infty }\frac{c_{n}}{2-\lambda _{n}}\hat{\Phi }_{n}\left ( x\right ) \end{align*}
But \lambda _{0}=0,c_{0}=-\frac{1}{2}and \hat{\Phi }_{0}\left ( x\right ) =1, therefore the above becomes\begin{align*} y\left ( x\right ) & =-\frac{1}{4}+\sum _{n=1,3,5,\cdots }^{\infty }\frac{\frac{2\sqrt{2}}{\left ( n\pi \right ) ^{2}}}{2-\left ( n\pi \right ) ^{2}}\sqrt{2}\cos \left ( n\pi x\right ) \\ & =-\frac{1}{4}+\sum _{n=1,3,5,\cdots }^{\infty }\frac{2\sqrt{2}}{\left ( 2-\left ( n\pi \right ) ^{2}\right ) \left ( n\pi \right ) ^{2}}\sqrt{2}\cos \left ( n\pi x\right ) \\ & =-\frac{1}{4}-4\sum _{n=1,3,5,\cdots }^{\infty }\frac{1}{\left ( \left ( n\pi \right ) ^{2}-2\right ) \left ( n\pi \right ) ^{2}}\cos \left ( n\pi x\right ) \end{align*}
To make the sum continuous, let m=\left ( 2n-1\right ) and now m runs from 1,2,3,\cdots and above becomes y\left ( x\right ) =-\frac{1}{4}-4\sum _{n=1,3,5,\cdots }^{\infty }\frac{\cos \left ( \left ( 2n-1\right ) \pi x\right ) }{\left ( \left ( \left ( 2n-1\right ) \pi \right ) ^{2}-2\right ) \left ( \left ( 2n-1\right ) \pi \right ) ^{2}}
Determine if there is any value of the constant a for which the ODE has a solution. Find the solution for each such value\begin{align*} y^{\prime \prime }+\pi ^{2}y & =a+x\\ y\left ( 0\right ) & =0\\ y\left ( 1\right ) & =0 \end{align*}
Solution
The eigenvalues of the corresponding homogenous eigenvalue ODE y^{\prime \prime }+\lambda y=0 with same homogenous boundary conditions are \lambda _{n}=\left ( n\pi \right ) ^{2}\, for n=1,2,\cdots . Therefore one can see that \lambda _{1} is eigenvalue in the original ODE y^{\prime \prime }+\pi ^{2}y=a+x. This means there is a solution (which will be non unique) only if the forcing function is orthogonal to the specific eigenfunction \Phi _{1}\left ( x\right ) . Therefore the condition is\begin{align*} \int _{0}^{1}f\left ( x\right ) \Phi _{1}\left ( x\right ) dx & =0\\ \int _{0}^{1}\left ( a+x\right ) \sin \left ( \pi x\right ) dx & =0\\ \int _{0}^{1}a\sin \left ( \pi x\right ) dx+\int _{0}^{1}x\sin \left ( \pi x\right ) dx & =0\\ a\left ( -\frac{\cos \pi x}{\pi }\right ) _{0}^{1}+\left [ \frac{\sin \pi x}{\pi ^{2}}-\frac{x\cos \pi x}{\pi }\right ] _{0}^{1} & =0\\ -\frac{a}{\pi }\left ( \cos \pi -1\right ) +\left [ \frac{\sin \pi }{\pi ^{2}}-\frac{\cos \pi }{\pi }\right ] & =0\\ -\frac{a}{\pi }\left ( -1-1\right ) +\left [ -\frac{-1}{\pi }\right ] & =0\\ \frac{2a}{\pi }+\frac{1}{\pi } & =0 \end{align*}
Hence a=\frac{-1}{2} Only when a is the above value, is there a solution. The original ODE is now solved using the direct method (meaning, not eigenfunction expansion) when a=\frac{-1}{2} as follows. Solve\begin{align*} y^{\prime \prime }+\pi ^{2}y & =-\frac{1}{2}+x\\ y\left ( 0\right ) & =0\\ y\left ( 1\right ) & =0 \end{align*}
The homogeneous solution is easily found to be y_{h}=A\cos \left ( \pi x\right ) +B\sin \left ( \pi x\right ) . Since the RHS is a polynomial, let the particular solution be y_{p}=c_{1}+c_{2}x. Then y_{p}^{\prime }=c_{2} and y_{p}^{\prime \prime }=0. Then \begin{align*} \pi ^{2}\left ( c_{1}+c_{2}x\right ) & =-\frac{1}{2}+x\\ c_{1}\pi ^{2}+c_{2}\pi ^{2}x & =-\frac{1}{2}+x \end{align*}
Therefore c_{2}\pi ^{2}=1 or c_{2}=\frac{1}{\pi ^{2}} and c_{1}\pi ^{2}=-\frac{1}{2} or c_{1}=-\frac{1}{2\pi ^{2}}. Hence y_{p}=-\frac{1}{2\pi ^{2}}+\frac{1}{\pi ^{2}}x. The solution is\begin{align*} y & =y_{h}+y_{p}\\ & =A\cos \left ( \pi x\right ) +B\sin \left ( \pi x\right ) -\frac{1}{2\pi ^{2}}+\frac{1}{\pi ^{2}}x \end{align*}
Applying boundary conditions, at y\left ( 0\right ) =0 the above becomes\begin{align*} 0 & =A-\frac{1}{2\pi ^{2}}\\ A & =\frac{1}{2\pi ^{2}} \end{align*}
Hence the solution becomes y\left ( x\right ) =\frac{1}{2\pi ^{2}}\cos \left ( \pi x\right ) +B\sin \left ( \pi x\right ) -\frac{1}{2\pi ^{2}}+\frac{1}{\pi ^{2}}x At y\left ( 1\right ) =0 the above gives\begin{align*} 0 & =\frac{1}{2\pi ^{2}}\cos \left ( \pi \right ) +B\sin \left ( \pi \right ) -\frac{1}{2\pi ^{2}}+\frac{1}{\pi ^{2}}\\ 0 & =\frac{-1}{2\pi ^{2}}-\frac{1}{2\pi ^{2}}+\frac{1}{\pi ^{2}}\\ 0 & =0 \end{align*}
Therefore B can be any value. Hence the final solution is \fbox{$y\left ( x\right ) =\frac{1}{2\pi ^2}\cos \left ( \pi x\right ) +B\sin \left ( \pi x\right ) +\frac{1}{\pi ^2}\left ( x-\frac{1}{2}\right ) $} The solution is not unique as expected. Any arbitrary value of B gives a solution.
Determine if there is any value of the constant a for which the ODE has a solution. Find the solution for each such value\begin{align*} y^{\prime \prime }+4\pi ^{2}y & =a+x\\ y\left ( 0\right ) & =0\\ y\left ( 1\right ) & =0 \end{align*}
Solution
The eigenvalues of the corresponding homogenous eigenvalue ODE y^{\prime \prime }+\lambda y=0 with same homogenous boundary conditions are \lambda _{n}=\left ( n\pi \right ) ^{2}\, for n=1,2,\cdots . Therefore \lambda _{2}=4\pi ^{2} is eigenvalue in the original ODE y^{\prime \prime }+4\pi ^{2}y=a+x. This means there is a solution (which will be non unique) only if the forcing function is orthogonal to the eigenfunction \Phi _{2}\left ( x\right ) . Therefore the condition is\begin{align*} \int _{0}^{1}f\left ( x\right ) \Phi _{2}\left ( x\right ) dx & =0\\ \int _{0}^{1}\left ( a+x\right ) \sin \left ( 2\pi x\right ) dx & =0\\ \int _{0}^{1}a\sin \left ( 2\pi x\right ) dx+\int _{0}^{1}x\sin \left ( 2\pi x\right ) dx & =0\\ a\left ( -\frac{\cos 2\pi x}{2\pi }\right ) _{0}^{1}+\left [ \frac{\sin \left ( 2\pi x\right ) }{4\pi ^{2}}-\frac{x\cos \left ( 2\pi x\right ) }{2\pi }\right ] _{0}^{1} & =0\\ -\frac{a}{2\pi }\left ( \cos 2\pi -1\right ) +\left [ \frac{\sin 2\pi }{4\pi ^{2}}-\frac{\cos 2\pi }{2\pi }\right ] & =0\\ -\frac{a}{2\pi }\left ( 1-1\right ) +\left [ -\frac{1}{2\pi }\right ] & =0\\ -\frac{1}{2\pi } & =0 \end{align*}
But this is not possible. Hence there is no a which makes \int _{0}^{1}\left ( a+x\right ) \sin \left ( 2\pi x\right ) dx=0. This means there is no solution for any a.
Determine if there is any value of the constant a for which the ODE has a solution. Find the solution for each such value\begin{align*} y^{\prime \prime }+\pi ^{2}y & =a\\ y^{\prime }\left ( 0\right ) & =0\\ y^{\prime }\left ( 1\right ) & =0 \end{align*}
Solution
The eigenvalues of the corresponding homogenous eigenvalue ODE y^{\prime \prime }+\lambda y=0 with same homogenous boundary conditions are \lambda _{0}=0 and \lambda _{n}=\left ( n\pi \right ) ^{2}\, for n=1,2,\cdots . Therefore \lambda _{1}=\pi ^{2} is eigenvalue in the original ODE y^{\prime \prime }+\pi ^{2}y=a+x. This means there is a solution (which will be non unique) only if the forcing function is orthogonal to \Phi _{1}\left ( x\right ) . The eigenfunctions in this case are \Phi _{n}\left ( x\right ) =\cos \left ( n\pi x\right ) . Therefore the condition is\begin{align*} \int _{0}^{1}f\left ( x\right ) \Phi _{1}\left ( x\right ) dx & =0\\ \int _{0}^{1}a\cos \left ( \pi x\right ) dx & =0\\ a\left ( \frac{\sin \pi x}{\pi }\right ) _{0}^{1} & =0\\ \frac{a}{\pi }\left ( 0\right ) & =0 \end{align*}
Hence any a will satisfy this. Therefore there is a solution for any a. The solution is y=A\cos \left ( \pi x\right ) +B\sin \left ( \pi x\right ) +y_{p} Since the RHS is a constant, let y_{p}=k. This leads to \pi ^{2}k=a or k=\frac{a}{\pi ^{2}}. Hence the solution is y=A\cos \left ( \pi x\right ) +B\sin \left ( \pi x\right ) +\frac{a}{\pi ^{2}} Or y^{\prime }\left ( x\right ) =-\pi A\sin \left ( \pi x\right ) +B\pi \cos \left ( \pi x\right ) At y^{\prime }\left ( 0\right ) =0 the above becomes 0=B\pi Hence B=0 and the solution now becomes\begin{align*} y & =A\cos \left ( \pi x\right ) +\frac{a}{\pi ^{2}}\\ y^{\prime } & =-A\pi \sin \left ( \pi x\right ) \end{align*}
At y\left ( 1\right ) =0 the above becomes\begin{align*} 0 & =-A\pi \sin \pi \\ & =-A\left ( 0\right ) \end{align*}
Therefore A is arbitrary. Any A will give a solution. Hence the final solution is \fbox{$y=A\cos \left ( \pi x\right ) +\frac{a}{\pi ^2}$} For any A and where a is the given a in the original ODE which can take in any value.
Determine if there is any value of the constant a for which the ODE has a solution. Find the solution for each such value\begin{align*} y^{\prime \prime }+\pi ^{2}y & =a-\cos \pi x\\ y\left ( 0\right ) & =0\\ y\left ( 1\right ) & =0 \end{align*}
Solution
The eigenvalues of the corresponding homogenous eigenvalue ODE y^{\prime \prime }+\lambda y=0 with same homogenous boundary conditions are \lambda _{0}=0 and \lambda _{n}=\left ( n\pi \right ) ^{2}\, for n=1,2,\cdots . Therefore \lambda _{1}=\pi ^{2} is eigenvalue in the original ODE y^{\prime \prime }+\pi ^{2}y=a-\cos \pi x. This means there is a solution (which will be non unique) only if the forcing function is orthogonal to \Phi _{1}\left ( x\right ) . The eigenfunctions in this case are \Phi _{n}\left ( x\right ) =\sin \left ( n\pi x\right ) . Therefore the condition is\begin{align*} \int _{0}^{1}f\left ( x\right ) \Phi _{1}\left ( x\right ) dx & =0\\ \int _{0}^{1}\left ( a-\cos \pi x\right ) \sin \left ( \pi x\right ) dx & =0\\ \int _{0}^{1}a\sin \left ( \pi x\right ) dx-\int _{0}^{1}\cos \left ( \pi x\right ) \sin \left ( \pi x\right ) dx & =0 \end{align*}
Using \sin A\cos B=\frac{1}{2}\left ( \sin \left ( A-B\right ) +\sin \left ( A+B\right ) \right ) then \sin \left ( \pi x\right ) \cos \left ( \pi x\right ) =\frac{1}{2}\left ( \sin \left ( 0\right ) +\sin \left ( 2\pi x\right ) \right ) =\frac{1}{2}\sin \left ( 2\pi x\right ) and the above becomes\begin{align*} \int _{0}^{1}a\sin \left ( \pi x\right ) dx-\frac{1}{2}\int _{0}^{1}\sin \left ( 2\pi x\right ) dx & =0\\ -\frac{a}{\pi }\left [ \cos \pi x\right ] _{0}^{1}+\frac{1}{4\pi }\left [ \cos \left ( 2\pi x\right ) \right ] _{0}^{1} & =0\\ -\frac{a}{\pi }\left ( \cos \pi -1\right ) +\frac{1}{4\pi }\left ( \cos \left ( 2\pi \right ) -1\right ) & =0\\ \frac{2a}{\pi } & =0 \end{align*}
Hence a=0. Therefore there is a solution only when a=0\,. The original ODE then becomes y^{\prime \prime }+\pi ^{2}y=-\cos \pi x The homogenous solution is y_{h}=A\cos \left ( \pi x\right ) +B\sin \left ( \pi x\right ) Since the forcing function matches one of the basis solution, then the particular solution guess is multiplied by extra x. Therefore \begin{align*} y_{p} & =x\left ( c_{1}\cos \left ( \pi x\right ) +c_{2}\sin \left ( \pi x\right ) \right ) \\ y_{p}^{\prime } & =c_{1}\cos \left ( \pi x\right ) +c_{2}\sin \left ( \pi x\right ) +x\left ( -c_{1}\pi \sin \left ( \pi x\right ) +c_{2}\pi \cos \left ( \pi x\right ) \right ) \\ y_{p}^{\prime \prime } & =-c_{1}\pi \sin \left ( \pi x\right ) +c_{2}\pi \cos \left ( \pi x\right ) +\left ( -c_{1}\pi \sin \left ( \pi x\right ) +c_{2}\pi \cos \left ( \pi x\right ) \right ) +x\left ( -c_{1}\pi ^{2}\cos \left ( \pi x\right ) -c_{2}\pi ^{2}\sin \left ( \pi x\right ) \right ) \\ & =\sin \left ( \pi x\right ) \left ( -2c_{1}\pi -c_{2}\pi ^{2}x\right ) +\cos \left ( \pi x\right ) \left ( 2c_{2}\pi -c_{1}x\pi ^{2}\right ) \end{align*}
Substituting back into the ODE gives\begin{align*} \sin \left ( \pi x\right ) \left ( -2c_{1}\pi -c_{2}\pi ^{2}x\right ) +\cos \left ( \pi x\right ) \left ( 2c_{2}\pi -c_{1}x\pi ^{2}\right ) +\pi ^{2}\left ( x\left ( c_{1}\cos \left ( \pi x\right ) +c_{2}\sin \left ( \pi x\right ) \right ) \right ) & =-\cos \pi x\\ \sin \left ( \pi x\right ) \left ( -2c_{1}\pi -c_{2}\pi ^{2}x+\pi ^{2}xc_{2}\right ) +\cos \left ( \pi x\right ) \left ( 2c_{2}\pi -c_{1}x\pi ^{2}+\pi ^{2}xc_{1}\right ) & =-\cos \pi x\\ -2c_{1}\pi \sin \left ( \pi x\right ) +2c_{2}\pi \cos \left ( \pi x\right ) & =-\cos \pi x \end{align*}
Hence\begin{align*} -2c_{1}\pi & =0\\ 2c_{2}\pi & =-1 \end{align*}
Or\begin{align*} c_{1} & =0\\ c_{2} & =-\frac{1}{2\pi } \end{align*}
Therefore y_{p}=-\frac{1}{2\pi }x\sin \left ( \pi x\right ) And the general solution is y\left ( x\right ) =A\cos \left ( \pi x\right ) +B\sin \left ( \pi x\right ) -\frac{1}{2\pi }x\sin \left ( \pi x\right ) At y\left ( 0\right ) =0 the above becomes 0=A\cos \left ( \pi x\right ) Hence A=0 and the solution now becomes \fbox{$y\left ( x\right ) =B\sin \left ( \pi x\right ) -\frac{1}{2\pi }x\sin \left ( \pi x\right ) $} One can stop here, since it is known that the solution is not unique and must contain an arbitrary constant. It is not possible to solve for B using the second boundary conditions.
Show that the problem y^{\prime \prime }+\pi ^{2}y=\pi ^{2}x,y\left ( 0\right ) =1,y\left ( 1\right ) =0 has solution y=c_{1}\sin \pi x+c_{2}\cos \pi x+x also show that the solution can not be obtained by splitting the problem as suggested in problem 15 since neither of the two subsidiary problems can be solve in this case.
Solution
To attempt to solve the problem by splitting, the solution is first assumed to be y=u+v where u is the solution to u^{\prime \prime }+\pi ^{2}u=0,u\left ( 0\right ) =1,u\left ( 1\right ) =0 and v is the solution to v^{\prime \prime }+\pi ^{2}v=\pi ^{2}x,v\left ( 0\right ) =0,v\left ( 1\right ) =0. Let us now try to solve the u ODE. The solution is u\left ( x\right ) =A\cos \pi x+B\sin \pi x Applying first BC u\left ( 0\right ) =1 gives A=1. Hence the solution becomes u=\cos \pi x+B\sin \pi x. Applying second BC u\left ( 1\right ) =0 gives\begin{align*} 0 & =\cos \pi +B\sin \pi \\ 0 & =1+B\tan \pi \\ B & =\frac{-1}{\tan \pi }=\frac{-1}{0} \end{align*}
Therefore there is no solution for u. Hence no solution is possible by splitting it was suggested in problem 15 for this problem. Now the problem is solved using the direct method. The homogeneous solution is y_{h}=A\cos \pi x+B\sin \pi x Since the forcing function \pi ^{2}x is a polynomial, let y_{p} guess be y_{p}=kxsubstituting this back into the ODE gives k=1. Hence the solution becomes\begin{align*} y & =y_{h}+y_{p}\\ & =A\cos \pi x+B\sin \pi x+x \end{align*}
Applying first BC y\left ( 0\right ) =1 gives 1=A. Hence the solution now becomes y=\cos \pi x+B\sin \pi x+x. Applying second BC y\left ( 1\right ) =0 gives \begin{align*} 0 & =\cos \pi +B\sin \pi +1\\ 0 & =-1+B\tan \pi +1\\ 0 & =B\tan \pi \\ 0 & =B\left ( 0\right ) \end{align*}
Therefore, any B will work. Hence the solution is not unique. Let B=1. Therefore the final solution is y=\cos \pi x+\sin \pi x+x This is solution is not unique. This is also a solution y=\cos \pi x+3\sin \pi x+x and also this y=\cos \pi x+100\sin \pi x+x and also y=\cos \pi x+x and so on.
Use eigenfunction expansion to solve
u_{t}=u_{xx}-x
With initial condition u\left ( x,0\right ) =\sin \left ( \frac{\pi x}{2}\right ) and boundary conditions u\left ( 0,t\right ) =0,u_{x}\left ( 1,t\right ) =0
Solution
The homogenous PDE is first solved to find the eigenfunctions, and these are used to expand the non-homogenous term -x in the PDE. By separation of variables, the spatial eigenvalue ODE is\begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X^{\prime }\left ( 1\right ) & =0 \end{align*}
The eigenfunctions for this ODE are \Phi _{n}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{n}}x\right ) with \lambda _{n}=\left ( \frac{n\pi }{2}\right ) ^{2} for n=1,3,5,\cdots . or \lambda _{n}=\left ( 2n-1\right ) ^{2}\left ( \frac{\pi }{2}\right ) ^{2} for n=1,2,3,\cdots . with now \Phi _{n}\left ( x\right ) =\sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) .
The normalized eigenfunctions are \hat{\Phi }_{n}\left ( x\right ) =\sqrt{2}\sin \left ( \sqrt{\lambda _{n}}x\right ) . Using these, the original PDE is now solved by assuming the solution is u\left ( x,t\right ) =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) The coefficient b_{n}\left ( t\right ) must be a function of time, since it includes all time contributions to the solution. Substituting the above back into the original PDE gives \sum _{n=1}^{\infty }b_{n}^{\prime }\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) =\frac{d^{2}}{dx^{2}}\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) +\sum _{n=1}^{\infty }c_{n}\hat{\Phi }_{n}\left ( x\right ) Where \sum _{n=1}^{\infty }c_{n}\hat{\Phi }_{n}\left ( x\right ) is the eigenfunction expansion of -x. Assuming term by term differentiation is allowed (can be shown to be justified here), the above becomes \sum _{n=1}^{\infty }b_{n}^{\prime }\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \hat{\Phi }_{n}^{\prime \prime }\left ( x\right ) +\sum _{n=1}^{\infty }c_{n}\hat{\Phi }_{n}\left ( x\right ) But \hat{\Phi }_{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}\hat{\Phi }_{n}\left ( x\right ) then the above becomes\begin{equation} \sum _{n=1}^{\infty }\left ( b_{n}^{\prime }\left ( t\right ) +\lambda _{n}b_{n}\left ( t\right ) \right ) \hat{\Phi }_{n}\left ( x\right ) =\sum _{n=1}^{\infty }c_{n}\hat{\Phi }_{n}\left ( x\right ) \tag{1} \end{equation} Now c_{n} is found. Since -x=\sum _{n=1}^{\infty }c_{n}\hat{\Phi }_{n}\left ( x\right ) , then applying orthogonality gives -\int _{0}^{1}r\left ( x\right ) x\hat{\Phi }_{m}\left ( x\right ) dx=\sum _{n=1}^{\infty }c_{n}\int _{0}^{1}r\left ( x\right ) \hat{\Phi }_{n}\left ( x\right ) \hat{\Phi }_{m}dx But the weight r\left ( x\right ) =1, hence the above simplifies to -\int _{0}^{1}x\hat{\Phi }_{m}\left ( x\right ) dx=c_{n}\int _{0}^{1}\hat{\Phi }_{m}^{2}\left ( x\right ) dx Since eigenfunctions are normalized, then \int _{0}^{1}r\left ( x\right ) \hat{\Phi }_{m}^{2}\left ( x\right ) dx=1 and the above reduces to\begin{align*} c_{n} & =-\int _{0}^{1}x\hat{\Phi }_{n}\left ( x\right ) dx\\ & =-\int _{0}^{1}x\sqrt{2}\sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) dx\\ & =-\sqrt{2}\left [ \frac{\sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) }{\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}}-\frac{x\cos \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) }{\left ( 2n-1\right ) \frac{\pi }{2}}\right ] _{0}^{1}\\ & =-\sqrt{2}\left [ \frac{\sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) }{\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}}-\frac{\cos \left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) }{\left ( 2n-1\right ) \frac{\pi }{2}}\right ] \end{align*}
But \cos \left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) =0 for all n, and the above now simplifies to\begin{align*} c_{n} & =-\sqrt{2}\frac{\sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) }{\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}}\\ & =-4\sqrt{2}\frac{\sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) }{\left ( \left ( 2n-1\right ) \pi \right ) ^{2}} \end{align*}
But \sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) =\left ( -1\right ) ^{n-1} for n=1,2,3,\cdots , hence the above becomes\begin{align*} c_{n} & =-4\sqrt{2}\frac{\left ( -1\right ) ^{n-1}}{\left ( \left ( 2n-1\right ) \pi \right ) ^{2}}\\ & =4\sqrt{2}\frac{\left ( -1\right ) ^{n}}{\left ( \left ( 2n-1\right ) \pi \right ) ^{2}} \end{align*}
Now that c_{n} is found, (1) is used to solve for b_{n}\left ( t\right ) \sum _{n=1}^{\infty }\left ( b_{n}^{\prime }\left ( t\right ) +\lambda _{n}b_{n}\left ( t\right ) \right ) \hat{\Phi }_{n}\left ( x\right ) =\sum _{n=1}^{\infty }c_{n}\hat{\Phi }_{n}\left ( x\right ) The above simplifies to b_{n}^{\prime }\left ( t\right ) +\lambda _{n}b_{n}\left ( t\right ) =c_{n} The integrating factor is e^{\int \lambda _{n}dt}=e^{\lambda _{n}t}, therefore \frac{d}{dt}\left ( b_{n}\left ( t\right ) e^{\lambda _{n}t}\right ) =c_{n}e^{\lambda _{n}t}. Integrating gives\begin{align*} b_{n}\left ( t\right ) e^{\lambda _{n}t} & =b\left ( 0\right ) +c_{n}\int _{0}^{t}e^{\lambda _{n}s}ds\\ b_{n}\left ( t\right ) & =b\left ( 0\right ) e^{-\lambda _{n}t}+c_{n}e^{-\lambda _{n}t}\int _{0}^{t}e^{\lambda _{n}s}ds\\ & =b\left ( 0\right ) e^{-\lambda _{n}t}+c_{n}e^{-\lambda _{n}t}\frac{\left ( e^{\lambda _{n}t}-1\right ) }{\lambda _{n}}\\ & =b\left ( 0\right ) e^{-\lambda _{n}t}+\frac{c_{n}}{\lambda _{n}}\left ( 1-e^{-\lambda _{n}t}\right ) \end{align*}
Therefore the solution becomes\begin{align} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) \nonumber \\ & =\sum _{n=1}^{\infty }\left ( b\left ( 0\right ) e^{-\lambda _{n}t}+\frac{c_{n}}{\lambda _{n}}\left ( 1-e^{-\lambda _{n}t}\right ) \right ) \hat{\Phi }_{n}\left ( x\right ) \tag{2} \end{align}
At t=0, the initial conditions is u\left ( x,0\right ) =\sin \left ( \frac{\pi x}{2}\right ) , therefore the above becomes\begin{align*} \sin \left ( \frac{\pi x}{2}\right ) & =\sum _{n=1}^{\infty }\left ( b\left ( 0\right ) +\frac{c_{n}}{\lambda _{n}}\left ( 1-1\right ) \right ) \hat{\Phi }_{n}\left ( x\right ) \\ & =\sum _{n=1}^{\infty }b\left ( 0\right ) \hat{\Phi }_{n}\left ( x\right ) \\ & =\sum _{n=1}^{\infty }b\left ( 0\right ) \sqrt{2}\sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) \end{align*}
Hence only n=1 gives a solution for b\left ( 0\right ) , and therefore the above becomes \sin \left ( \frac{\pi x}{2}\right ) =b\left ( 0\right ) \sqrt{2}\sin \left ( \frac{\pi }{2}x\right ) Or \fbox{$b\left ( 0\right ) =\frac{1}{\sqrt{2}}$} Therefore the solution (2) now becomes\begin{equation} u\left ( x,t\right ) =\left ( b\left ( 0\right ) e^{-\lambda _{1}t}+c_{1}e^{-\lambda _{1}t}\frac{\left ( e^{\lambda _{1}t}-1\right ) }{\lambda _{1}}\right ) \hat{\Phi }_{1}\left ( x\right ) +\sum _{n=2}^{\infty }\frac{c_{n}}{\lambda _{n}}\left ( 1-e^{-\lambda _{n}t}\right ) \hat{\Phi }_{n}\left ( x\right ) \tag{3} \end{equation} Where \begin{align*} c_{n} & =4\sqrt{2}\frac{\left ( -1\right ) ^{n}}{\left ( \left ( 2n-1\right ) \pi \right ) ^{2}}\\ b\left ( 0\right ) & =\frac{1}{\sqrt{2}}\\ \lambda _{n} & =\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}\qquad n=1,2,3,\cdots \\ \hat{\Phi }_{n}\left ( x\right ) & =\sqrt{2}\sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) \end{align*}
Hence the solution (3) becomes\begin{align*} u\left ( x,t\right ) & =\left ( \frac{1}{\sqrt{2}}e^{-\frac{\pi ^{2}}{4}t}+c_{1}e^{-\frac{\pi ^{2}}{4}t}\frac{\left ( e^{\frac{\pi ^{2}}{4}t}-1\right ) }{\frac{\pi ^{2}}{4}}\right ) \sqrt{2}\sin \left ( \frac{\pi }{2}x\right ) \\ & +\sum _{n=2}^{\infty }\frac{c_{n}}{\left ( 2n-1\right ) ^{2}\frac{\pi ^{2}}{4}}\left ( 1-e^{-\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}t}\right ) \sqrt{2}\sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) \end{align*}
To make it the same as back of the book solution, some more manipulation is needed.\begin{align*} u\left ( x,t\right ) & =e^{-\frac{\pi ^{2}}{4}t}\sin \left ( \frac{\pi }{2}x\right ) +4\sqrt{2}\frac{c_{1}}{\pi ^{2}}e^{-\frac{\pi ^{2}}{4}t}\left ( e^{\frac{\pi ^{2}}{4}t}-1\right ) \sin \left ( \frac{\pi }{2}x\right ) \\ & +\sqrt{2}\sum _{n=2}^{\infty }\frac{4c_{n}}{\left ( 2n-1\right ) ^{2}\pi ^{2}}e^{-\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}t}\left ( e^{\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}t}-1\right ) \sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) \end{align*}
Or\begin{align*} u\left ( x,t\right ) & =e^{-\frac{\pi ^{2}}{4}t}\sin \left ( \frac{\pi }{2}x\right ) +4\sqrt{2}\frac{c_{1}}{\pi ^{2}}\left ( 1-e^{-\frac{\pi ^{2}}{4}t}\right ) \sin \left ( \frac{\pi }{2}x\right ) \\ & +\sqrt{2}\sum _{n=2}^{\infty }\frac{4c_{n}}{\left ( 2n-1\right ) ^{2}\pi ^{2}}\left ( 1-e^{-\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}t}\right ) \sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) \end{align*}
Or\begin{align*} u\left ( x,t\right ) & =e^{-\frac{\pi ^{2}}{4}t}\sin \left ( \frac{\pi }{2}x\right ) +4\sqrt{2}\frac{c_{1}}{\pi ^{2}}\sin \left ( \frac{\pi }{2}x\right ) -4\sqrt{2}\frac{c_{1}}{\pi ^{2}}e^{-\frac{\pi ^{2}}{4}t}\sin \left ( \frac{\pi }{2}x\right ) \\ & +\sqrt{2}\sum _{n=2}^{\infty }\frac{4c_{n}}{\left ( 2n-1\right ) ^{2}\pi ^{2}}\left ( 1-e^{-\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}t}\right ) \sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) \end{align*}
Or\begin{align*} u\left ( x,t\right ) & =\sqrt{2}\left [ 4\frac{c_{1}}{\pi ^{2}}+\left ( \frac{1}{\sqrt{2}}-4\frac{c_{1}}{\pi ^{2}}\right ) e^{-\frac{\pi ^{2}}{4}t}\right ] \sin \left ( \frac{\pi }{2}x\right ) \\ & +\sqrt{2}\sum _{n=2}^{\infty }\frac{4c_{n}}{\left ( 2n-1\right ) ^{2}\pi ^{2}}\left ( 1-e^{-\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}t}\right ) \sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) \end{align*}
The back of the book uses c_{n}=4\sqrt{2}\frac{\left ( -1\right ) ^{n+1}}{\left ( \left ( 2n-1\right ) \pi \right ) ^{2}} instead of c_{n}=4\sqrt{2}\frac{\left ( -1\right ) ^{n}}{\left ( \left ( 2n-1\right ) \pi \right ) ^{2}} as was done in this solution. Therefore, changing c_{n} to be as the back of the book means flipping the sign of each c_{n}. (or multiplying by -1). Hence the solution becomes now the same as the back of the book\begin{align*} u\left ( x,t\right ) & =\sqrt{2}\left [ -4\frac{c_{1}}{\pi ^{2}}+\left ( \frac{1}{\sqrt{2}}+4\frac{c_{1}}{\pi ^{2}}\right ) e^{-\frac{\pi ^{2}}{4}t}\right ] \sin \left ( \frac{\pi }{2}x\right ) \\ & -\sqrt{2}\sum _{n=2}^{\infty }\frac{4c_{n}}{\left ( 2n-1\right ) ^{2}\pi ^{2}}\left ( 1-e^{-\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}t}\right ) \sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}x\right ) \end{align*}
Where in the above, c_{n}=4\sqrt{2}\frac{\left ( -1\right ) ^{n+1}}{\left ( \left ( 2n-1\right ) \pi \right ) ^{2}} Both solutions are the same. The sign is either added to c_{n} or left outside. This completes the solution.
Use eigenfunction expansion to solve u_{t}=u_{xx}+e^{-t} With initial condition u\left ( x,0\right ) =1-x and boundary conditions u_{x}\left ( 0,t\right ) =0,u_{x}\left ( 1,t\right ) +u\left ( 1,t\right ) =0
Solution
The homogenous PDE is solved first to obtain the eigenfunctions. These are then used to expand the non-homogenous term e^{-t} in the PDE. By separation of variables, the spatial eigenvalue ODE is\begin{align*} X^{\prime \prime }+\lambda X & =0\\ X^{\prime }\left ( 0\right ) & =0\\ X^{\prime }\left ( 1\right ) +X\left ( 1\right ) & =0 \end{align*}
The eigenfunctions for this ODE were found earlier in problem 4, Chapter 11.2. They are \begin{align*} \hat{\Phi }_{n} & =k_{n}\Phi _{n}\\ & =\frac{\sqrt{2}}{\sqrt{1+\sin ^{2}\left ( \sqrt{\lambda _{n}}\right ) }}\cos \left ( \sqrt{\lambda _{n}}x\right ) \end{align*}
Where \lambda _{n} are the roots of \cos \left ( \sqrt{\lambda _{n}}\right ) -\sqrt{\lambda _{n}}\sin \left ( \sqrt{\lambda _{n}}\right ) =0 For n=1,2,3,\cdots . Using these, the original PDE is now solved by assuming the solution is u\left ( x,t\right ) =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) The coefficient b_{n}\left ( t\right ) must be a function of time, since it includes all time contributions to the solution. Substituting the above back into the original PDE gives \sum _{n=1}^{\infty }b_{n}^{\prime }\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) =\frac{d^{2}}{dx^{2}}\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) +\sum _{n=1}^{\infty }c_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) Where \sum _{n=1}^{\infty }c_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) is the eigenfunction expansion of e^{-t}. In the above c_{n}\left ( t\right ) is now a function of time, since the forcing function depends on time in this problem. Assuming term by term differentiation is allowed the above becomes \sum _{n=1}^{\infty }b_{n}^{\prime }\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \hat{\Phi }_{n}^{\prime \prime }\left ( x\right ) +\sum _{n=1}^{\infty }c_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) But \hat{\Phi }_{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}\hat{\Phi }_{n}\left ( x\right ) therefore\begin{equation} \sum _{n=1}^{\infty }\left ( b_{n}^{\prime }\left ( t\right ) +\lambda _{n}b_{n}\left ( t\right ) \right ) \hat{\Phi }_{n}\left ( x\right ) =\sum _{n=1}^{\infty }c_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) \tag{1} \end{equation} Now c_{n}\left ( t\right ) is found. Since e^{-t}=\sum _{n=1}^{\infty }c_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) , then applying orthogonality gives \int _{0}^{1}r\left ( x\right ) e^{-t}\hat{\Phi }_{m}\left ( x\right ) dx=\sum _{n=1}^{\infty }c_{n}\left ( t\right ) \int _{0}^{1}r\left ( x\right ) \hat{\Phi }_{n}\left ( x\right ) \hat{\Phi }_{m}dx But the weight r\left ( x\right ) =1, hence the above simplifies to e^{-t}\int _{0}^{1}\hat{\Phi }_{m}\left ( x\right ) dx=c_{n}\left ( t\right ) \int _{0}^{1}\hat{\Phi }_{m}^{2}\left ( x\right ) dx Since eigenfunctions are normalized, then \int _{0}^{1}r\left ( x\right ) \hat{\Phi }_{m}^{2}\left ( x\right ) dx=1 and the above reduces to e^{-t}\int _{0}^{1}\hat{\Phi }_{m}\left ( x\right ) dx=c_{n}\left ( t\right ) Hence\begin{align} c_{n}\left ( t\right ) & =e^{-t}\int _{0}^{1}k_{n}\cos \left ( \sqrt{\lambda _{n}}x\right ) dx\nonumber \\ & =e^{-t}\frac{k_{n}}{\sqrt{\lambda _{n}}}\left [ \sin \left ( \sqrt{\lambda _{n}}x\right ) \right ] _{0}^{1}\nonumber \\ & =e^{-t}\frac{k_{n}}{\sqrt{\lambda _{n}}}\sin \left ( \sqrt{\lambda _{n}}\right ) \tag{2} \end{align}
To make it match the way the back of the book expressed the above, let us write c_{n}\left ( t\right ) =e^{-t}c_{n} Where now c_{n}=\frac{k_{n}}{\sqrt{\lambda _{n}}}\sin \left ( \sqrt{\lambda _{n}}\right ) This makes it easier to verify the final solution found here is the same as the back of the book.
Now that c_{n}\left ( t\right ) is found, (1) is used to solve for b_{n}\left ( t\right ) \sum _{n=1}^{\infty }\left ( b_{n}^{\prime }\left ( t\right ) +\lambda _{n}b_{n}\left ( t\right ) \right ) \hat{\Phi }_{n}\left ( x\right ) =\sum _{n=1}^{\infty }e^{-t}c_{n}\hat{\Phi }_{n}\left ( x\right ) The above simplifies to b_{n}^{\prime }\left ( t\right ) +\lambda _{n}b_{n}\left ( t\right ) =e^{-t}c_{n} The integrating factor is e^{\int \lambda _{n}dt}=e^{\lambda _{n}t}, therefore \frac{d}{dt}\left ( b_{n}\left ( t\right ) e^{\lambda _{n}t}\right ) =e^{-t}c_{n}e^{\lambda _{n}t}. Integrating gives\begin{align} b_{n}\left ( t\right ) e^{\lambda _{n}t} & =b\left ( 0\right ) +c_{n}\int _{0}^{t}e^{-s}e^{\lambda _{n}s}ds\nonumber \\ b_{n}\left ( t\right ) & =b\left ( 0\right ) e^{-\lambda _{n}t}+c_{n}e^{-\lambda _{n}t}\int _{0}^{t}e^{\left ( \lambda _{n}-1\right ) s}ds\nonumber \\ & =b\left ( 0\right ) e^{-\lambda _{n}t}+c_{n}e^{-\lambda _{n}t}\frac{\left [ e^{\left ( \lambda _{n}-1\right ) s}\right ] _{0}^{t}}{\lambda _{n}-1}\nonumber \\ & =b\left ( 0\right ) e^{-\lambda _{n}t}+c_{n}e^{-\lambda _{n}t}\frac{e^{\left ( \lambda _{n}-1\right ) t}-1}{\lambda _{n}-1} \tag{3} \end{align}
Using the above in u\left ( x,t\right ) =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) gives the solution as \begin{equation} u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( b\left ( 0\right ) e^{-\lambda _{n}t}+c_{n}e^{-\lambda _{n}t}\frac{e^{\left ( \lambda _{n}-1\right ) t}-1}{\lambda _{n}-1}\right ) \hat{\Phi }_{n}\left ( x\right ) \tag{4} \end{equation} At t=0, the above simplifies to 1-x=\sum _{n=1}^{\infty }b\left ( 0\right ) \hat{\Phi }_{n}\left ( x\right ) Applying orthogonality gives\begin{align*} \int _{0}^{1}r\left ( x\right ) \left ( 1-x\right ) \hat{\Phi }_{m}\left ( x\right ) dx & =\sum _{n=1}^{\infty }b\left ( 0\right ) \int _{0}^{1}r\left ( x\right ) \hat{\Phi }_{n}\left ( x\right ) \hat{\Phi }_{m}\left ( x\right ) dx\\ \int _{0}^{1}r\left ( x\right ) \left ( 1-x\right ) \hat{\Phi }_{m}\left ( x\right ) dx & =b\left ( 0\right ) \int _{0}^{1}r\left ( x\right ) \hat{\Phi }_{m}^{2}\left ( x\right ) dx \end{align*}
But r\left ( x\right ) =1 and \int _{0}^{1}r\left ( x\right ) \hat{\Phi }_{m}^{2}\left ( x\right ) dx=1 therefore\begin{align*} b\left ( 0\right ) & =\int _{0}^{1}\left ( 1-x\right ) \hat{\Phi }_{n}\left ( x\right ) dx\\ & =\int _{0}^{1}\hat{\Phi }_{n}\left ( x\right ) dx-\int _{0}^{1}x\hat{\Phi }_{n}\left ( x\right ) dx\\ & =k_{n}\left ( \int _{0}^{1}\Phi _{n}\left ( x\right ) dx-\int _{0}^{1}x\Phi _{n}\left ( x\right ) dx\right ) \end{align*}
But \Phi _{n}\left ( x\right ) =\cos \left ( \sqrt{\lambda _{n}}x\right ) , hence the above becomes\begin{align*} b\left ( 0\right ) & =k_{n}\left ( \int _{0}^{1}\cos \left ( \sqrt{\lambda _{n}}x\right ) dx-\int _{0}^{1}x\cos \left ( \sqrt{\lambda _{n}}x\right ) dx\right ) \\ & =k_{n}\left ( \left [ \frac{\sin \left ( \sqrt{\lambda _{n}}x\right ) }{\sqrt{\lambda _{n}}}\right ] _{0}^{1}-\left [ \frac{\cos \left ( \sqrt{\lambda _{n}}x\right ) }{\lambda _{n}}+\frac{x\sin \left ( \sqrt{\lambda _{n}}x\right ) }{\sqrt{\lambda _{n}}}\right ] _{0}^{1}\right ) \\ & =k_{n}\left ( \left [ \frac{\sin \left ( \sqrt{\lambda _{n}}\right ) }{\sqrt{\lambda _{n}}}\right ] -\left [ \frac{\cos \left ( \sqrt{\lambda _{n}}\right ) }{\lambda _{n}}+\frac{\sin \left ( \sqrt{\lambda _{n}}\right ) }{\sqrt{\lambda _{n}}}-\frac{1}{\lambda _{n}}\right ] \right ) \\ & =k_{n}\left ( \frac{\sin \left ( \sqrt{\lambda _{n}}\right ) }{\sqrt{\lambda _{n}}}-\frac{\cos \left ( \sqrt{\lambda _{n}}\right ) }{\lambda _{n}}-\frac{\sin \left ( \sqrt{\lambda _{n}}\right ) }{\sqrt{\lambda _{n}}}+\frac{1}{\lambda _{n}}\right ) \\ & =\frac{k_{n}}{\lambda _{n}}\left ( 1-\cos \left ( \sqrt{\lambda _{n}}\right ) \right ) \end{align*}
Now that b\left ( 0\right ) is found, then the solution (4) becomes \begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }\left ( \frac{k_{n}}{\lambda _{n}}\left ( 1-\cos \left ( \sqrt{\lambda _{n}}\right ) \right ) e^{-\lambda _{n}t}+c_{n}e^{-\lambda _{n}t}\frac{e^{\left ( \lambda _{n}-1\right ) t}-1}{\lambda _{n}-1}\right ) \hat{\Phi }_{n}\left ( x\right ) \\ & =\sum _{n=1}^{\infty }\left ( \frac{k_{n}}{\lambda _{n}}\left ( 1-\cos \left ( \sqrt{\lambda _{n}}\right ) \right ) e^{-\lambda _{n}t}+\frac{c_{n}}{\lambda _{n}-1}\left ( e^{-t}-e^{-\lambda _{n}t}\right ) \right ) k_{n}\cos \left ( \sqrt{\lambda _{n}}x\right ) \end{align*}
But k_{n}=\frac{\sqrt{2}}{\sqrt{1+\sin ^{2}\left ( \sqrt{\lambda _{n}}\right ) }}, hence the above becomes u\left ( x,t\right ) =\sqrt{2}\sum _{n=1}^{\infty }\left ( \alpha _{n}e^{-\lambda _{n}t}+\frac{c_{n}}{\lambda _{n}-1}\left ( e^{-t}-e^{-\lambda _{n}t}\right ) \right ) \frac{\cos \left ( \sqrt{\lambda _{n}}x\right ) }{\sqrt{1+\sin ^{2}\left ( \sqrt{\lambda _{n}}\right ) }} Where \begin{align*} \alpha _{n} & =\frac{k_{n}}{\lambda _{n}}\left ( 1-\cos \left ( \sqrt{\lambda _{n}}\right ) \right ) \\ & =\frac{\sqrt{2}\left ( 1-\cos \left ( \sqrt{\lambda _{n}}\right ) \right ) }{\lambda _{n}\sqrt{1+\sin ^{2}\left ( \sqrt{\lambda _{n}}\right ) }} \end{align*}
And
\begin{align*} c_{n} & =\frac{k_{n}}{\sqrt{\lambda _{n}}}\sin \left ( \sqrt{\lambda _{n}}\right ) \\ & =\frac{\sqrt{2}}{\sqrt{\lambda _{n}}}\frac{\sin \left ( \sqrt{\lambda _{n}}\right ) }{\sqrt{1+\sin ^{2}\left ( \sqrt{\lambda _{n}}\right ) }} \end{align*}
Use eigenfunction expansion to solve u_{t}=u_{xx}+e^{-t}\left ( 1-x\right ) With initial condition u\left ( x,0\right ) =0 and boundary conditions u\left ( 0,t\right ) =0,u_{x}\left ( 1,t\right ) =0
Solution
The homogenous PDE is solved first to obtain the eigenfunctions. These are then used to expand the non-homogenous term e^{-t}\left ( 1-x\right ) in the PDE. By separation of variables, the spatial eigenvalue ODE is\begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X^{\prime }\left ( 1\right ) & =0 \end{align*}
The eigenfunctions for this ODE were found earlier. They are \begin{align*} \hat{\Phi }_{n} & =k_{n}\Phi _{n}\\ & =\sqrt{2}\sin \left ( \sqrt{\lambda _{n}}x\right ) \end{align*}
Where \lambda _{n}=\left ( \frac{n\pi }{2}\right ) ^{2} for n=1,3,5,\cdots . Or \begin{align*} \hat{\Phi }_{n} & =\sqrt{2}\sin \left ( \sqrt{\lambda _{n}}x\right ) \\ \lambda _{n} & =\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}\qquad n=1,2,3,\cdots \end{align*}
The original PDE is now solved by assuming the solution is u\left ( x,t\right ) =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) The coefficient b_{n}\left ( t\right ) must be a function of time, since it includes all time contributions to the solution. Substituting the above back into the original PDE gives \sum _{n=1}^{\infty }b_{n}^{\prime }\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) =\frac{d^{2}}{dx^{2}}\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) +\sum _{n=1}^{\infty }c_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) Where \sum _{n=1}^{\infty }c_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) is the eigenfunction expansion of e^{-t}\left ( 1-x\right ) . In the above c_{n}\left ( t\right ) is now a function of time, since the forcing function depends on time in this problem. Assuming term by term differentiation is allowed the above becomes \sum _{n=1}^{\infty }b_{n}^{\prime }\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \hat{\Phi }_{n}^{\prime \prime }\left ( x\right ) +\sum _{n=1}^{\infty }c_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) But \hat{\Phi }_{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}\hat{\Phi }_{n}\left ( x\right ) therefore\begin{equation} \sum _{n=1}^{\infty }\left ( b_{n}^{\prime }\left ( t\right ) +\lambda _{n}b_{n}\left ( t\right ) \right ) \hat{\Phi }_{n}\left ( x\right ) =\sum _{n=1}^{\infty }c_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) \tag{1} \end{equation} Now c_{n}\left ( t\right ) is found. Since e^{-t}\left ( 1-x\right ) =\sum _{n=1}^{\infty }c_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) , then applying orthogonality gives \int _{0}^{1}r\left ( x\right ) e^{-t}\left ( 1-x\right ) \hat{\Phi }_{m}\left ( x\right ) dx=\sum _{n=1}^{\infty }c_{n}\left ( t\right ) \int _{0}^{1}r\left ( x\right ) \hat{\Phi }_{n}\left ( x\right ) \hat{\Phi }_{m}dx But the weight r\left ( x\right ) =1, hence the above simplifies to e^{-t}\int _{0}^{1}\left ( 1-x\right ) \hat{\Phi }_{m}\left ( x\right ) dx=c_{n}\left ( t\right ) \int _{0}^{1}\hat{\Phi }_{m}^{2}\left ( x\right ) dx Since eigenfunctions are normalized, then \int _{0}^{1}r\left ( x\right ) \hat{\Phi }_{m}^{2}\left ( x\right ) dx=1 and the above reduces to e^{-t}\int _{0}^{1}\left ( 1-x\right ) \hat{\Phi }_{m}\left ( x\right ) dx=c_{n}\left ( t\right ) Hence\begin{align} c_{n}\left ( t\right ) & =e^{-t}\int _{0}^{1}\left ( 1-x\right ) k_{n}\sin \left ( \sqrt{\lambda _{n}}x\right ) dx\nonumber \\ & =e^{-t}\sqrt{2}\left ( \int _{0}^{1}\sin \left ( \sqrt{\lambda _{n}}x\right ) dx-\int _{0}^{1}x\sin \left ( \sqrt{\lambda _{n}}x\right ) dx\right ) \nonumber \\ & =e^{-t}\sqrt{2}\left ( \left [ \frac{-\cos \left ( \sqrt{\lambda _{n}}x\right ) }{\sqrt{\lambda _{n}}}\right ] _{0}^{1}-\left [ \frac{\sin \sqrt{\lambda _{n}}x}{\lambda _{n}}-\frac{x\cos \sqrt{\lambda _{n}}x}{\sqrt{\lambda _{n}}}\right ] _{0}^{1}\right ) \nonumber \\ & =e^{-t}\sqrt{2}\left ( \left [ \frac{-\cos \left ( \sqrt{\lambda _{n}}\right ) }{\sqrt{\lambda _{n}}}+\frac{1}{\sqrt{\lambda _{n}}}\right ] -\left [ \frac{\sin \sqrt{\lambda _{n}}}{\lambda _{n}}-\frac{\cos \sqrt{\lambda _{n}}}{\sqrt{\lambda _{n}}}\right ] \right ) \nonumber \\ & =e^{-t}\sqrt{2}\left ( \frac{-\cos \left ( \sqrt{\lambda _{n}}\right ) }{\sqrt{\lambda _{n}}}+\frac{1}{\sqrt{\lambda _{n}}}-\frac{\sin \sqrt{\lambda _{n}}}{\lambda _{n}}+\frac{\cos \sqrt{\lambda _{n}}}{\sqrt{\lambda _{n}}}\right ) \nonumber \\ & =e^{-t}\sqrt{2}\left ( \frac{1}{\sqrt{\lambda _{n}}}-\frac{\sin \sqrt{\lambda _{n}}}{\lambda _{n}}\right ) \nonumber \\ & =e^{-t}\frac{\sqrt{2}}{\lambda _{n}}\left ( \sqrt{\lambda _{n}}-\sin \sqrt{\lambda _{n}}\right ) \tag{2} \end{align}
But \lambda _{n}=\left ( 2n-1\right ) \frac{\pi }{2}, therefore \sin \left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) =\left \{ 1,-1,1,-1,\cdots \right \} or \left ( -1\right ) ^{n-1} and the above becomes\begin{align*} c_{n}\left ( t\right ) & =e^{-t}\frac{\sqrt{2}}{\lambda _{n}}\left ( \sqrt{\lambda _{n}}-\left ( -1\right ) ^{n-1}\right ) \\ & =e^{-t}\frac{\sqrt{2}}{\lambda _{n}}\left ( \sqrt{\lambda _{n}}+\left ( -1\right ) ^{n}\right ) \\ & =e^{-t}\frac{\sqrt{2}}{\lambda _{n}}\left ( \sqrt{\lambda _{n}}+\left ( -1\right ) ^{n}\right ) \end{align*}
To make it match the way the back of the book expressed the above, let us write c_{n}\left ( t\right ) =e^{-t}c_{n} Where \begin{equation} c_{n}=\frac{\sqrt{2}}{\lambda _{n}}\left ( \sqrt{\lambda _{n}}+\left ( -1\right ) ^{n}\right ) \tag{2A} \end{equation} Now that c_{n}\left ( t\right ) is found, (1) is used to solve for b_{n}\left ( t\right ) \sum _{n=1}^{\infty }\left ( b_{n}^{\prime }\left ( t\right ) +\lambda _{n}b_{n}\left ( t\right ) \right ) \hat{\Phi }_{n}\left ( x\right ) =\sum _{n=1}^{\infty }e^{-t}c_{n}\hat{\Phi }_{n}\left ( x\right ) The above simplifies to b_{n}^{\prime }\left ( t\right ) +\lambda _{n}b_{n}\left ( t\right ) =e^{-t}c_{n} The integrating factor is e^{\int \lambda _{n}dt}=e^{\lambda _{n}t}, therefore \frac{d}{dt}\left ( b_{n}\left ( t\right ) e^{\lambda _{n}t}\right ) =e^{-t}c_{n}e^{\lambda _{n}t}. Integrating gives\begin{align} b_{n}\left ( t\right ) e^{\lambda _{n}t} & =b\left ( 0\right ) +c_{n}\int _{0}^{t}e^{-s}e^{\lambda _{n}s}ds\nonumber \\ b_{n}\left ( t\right ) & =b\left ( 0\right ) e^{-\lambda _{n}t}+c_{n}e^{-\lambda _{n}t}\int _{0}^{t}e^{\left ( \lambda _{n}-1\right ) s}ds\nonumber \\ & =b\left ( 0\right ) e^{-\lambda _{n}t}+c_{n}e^{-\lambda _{n}t}\frac{\left [ e^{\left ( \lambda _{n}-1\right ) s}\right ] _{0}^{t}}{\lambda _{n}-1}\nonumber \\ & =b\left ( 0\right ) e^{-\lambda _{n}t}+c_{n}e^{-\lambda _{n}t}\frac{e^{\left ( \lambda _{n}-1\right ) t}-1}{\lambda _{n}-1} \tag{3} \end{align}
Using the above in u\left ( x,t\right ) =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \hat{\Phi }_{n}\left ( x\right ) gives the solution as \begin{equation} u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( b\left ( 0\right ) e^{-\lambda _{n}t}+c_{n}e^{-\lambda _{n}t}\frac{e^{\left ( \lambda _{n}-1\right ) t}-1}{\lambda _{n}-1}\right ) \hat{\Phi }_{n}\left ( x\right ) \tag{4} \end{equation} At t=0, the initial conditions are zero, and above simplifies to 0=\sum _{n=1}^{\infty }b\left ( 0\right ) \hat{\Phi }_{n}\left ( x\right ) Which implies b\left ( 0\right ) =0. Now that b\left ( 0\right ) is found, then the solution (4) becomes \begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }c_{n}e^{-\lambda _{n}t}\frac{e^{\left ( \lambda _{n}-1\right ) t}-1}{\lambda _{n}-1}\hat{\Phi }_{n}\left ( x\right ) \\ & =\sqrt{2}\sum _{n=1}^{\infty }c_{n}\left ( \frac{e^{-t}-e^{-\lambda _{n}t}}{\lambda _{n}-1}\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) \end{align*}
Where c_{n}=\frac{\sqrt{2}}{\lambda _{n}}\left ( \sqrt{\lambda _{n}}+\left ( -1\right ) ^{n}\right ) and \lambda _{n}=\left ( \left ( 2n-1\right ) \frac{\pi }{2}\right ) ^{2}. This completes the solution.
The solution was animated and verified it is correct against a numerical solution.
Solve u_{t}=u_{xx}-2 With initial condition u\left ( x,0\right ) =x^{2}-2x+2 and boundary conditions u\left ( 0,t\right ) =1,u\left ( 1,t\right ) =0
Solution
Let u\left ( x,t\right ) =w\left ( x,t\right ) +v\left ( x\right ) where v\left ( x\right ) is steady state solution which only needs to satisfy the non-homogenous boundary conditions and w\left ( x,t\right ) is the transient solution which needs to satisfy the homogeneous boundary conditions.
At steady state, the PDE becomes an ODE 0=v^{\prime \prime }\left ( x\right ) -2 This has the solution v\left ( x\right ) =c_{1}+c_{2}x+x^{2} Where x^{2} is the particular solution. From boundary conditions v\left ( 0\right ) =1,v\left ( 1\right ) =0, the solution becomes v\left ( x\right ) =1-2x+x^{2} Hence u\left ( x,t\right ) =w\left ( x,t\right ) +1-2x+x^{2}. Substituting this into the PDE u_{t}=u_{xx}-2 results in\begin{align*} w_{t} & =w_{xx}+v^{\prime \prime }\left ( x\right ) -2\\ & =w_{xx}+2-2\\ & =w_{xx} \end{align*}
Hence the PDE to solve is w_{t}=w_{xx} with w\left ( 0,t\right ) =0,w\left ( 1,t\right ) =0. This heat PDE was solved before. Its solution is\begin{equation} w\left ( x,t\right ) =\sum _{n=1}^{\infty }c_{n}e^{-\lambda _{n}t}\sin \left ( \sqrt{\lambda _{n}}x\right ) \tag{1} \end{equation} Where \lambda _{n}=\left ( n\pi \right ) ^{2} for n=1,2,3,\cdots . At t=0, since u\left ( x,0\right ) =w\left ( x,0\right ) +v\left ( x\right ) then w\left ( x,0\right ) =u\left ( x,0\right ) -v\left ( x\right ) which gives \begin{align*} w\left ( x,0\right ) & =\left ( x^{2}-2x+2\right ) -\left ( 1-2x+x^{2}\right ) \\ & =1 \end{align*}
Hence at t=0, (1) becomes \begin{equation} 1=\sum _{n=1}^{\infty }c_{n}\sin \left ( \sqrt{\lambda _{n}}x\right ) \tag{1A} \end{equation} Applying orthogonality gives\begin{align*} \int _{0}^{1}\sin \left ( \sqrt{\lambda _{n}}x\right ) dx & =\frac{1}{2}c_{n}\\ c_{n} & =2\int _{0}^{1}\sin \left ( \sqrt{\lambda _{n}}x\right ) dx\\ & =-2\left [ \frac{\cos \left ( \sqrt{\lambda _{n}}x\right ) }{\sqrt{\lambda _{n}}}\right ] _{0}^{1}\\ & =\frac{-2}{\sqrt{\lambda _{n}}}\left [ \cos \left ( \sqrt{\lambda _{n}}\right ) -1\right ] \\ & =\frac{-2}{n\pi }\left [ \cos \left ( n\pi \right ) -1\right ] \end{align*}
For even n the above is zero. And for odd n the above becomes c_{n}=\frac{4}{n\pi }\qquad n=1,3,5,\cdots Therefore from (1) the solution to w\left ( x,t\right ) is w\left ( x,t\right ) =\frac{4}{\pi }\sum _{n=1,3,5,\cdots }^{\infty }\frac{1}{n}e^{-\lambda _{n}t}\sin \left ( \sqrt{\lambda _{n}}x\right ) The above can also be written as w\left ( x,t\right ) =\frac{4}{\pi }\sum _{n=1}^{\infty }\frac{1}{\left ( 2n-1\right ) }e^{-\left ( 2n-1\right ) ^{2}\pi ^{2}t}\sin \left ( \left ( 2n-1\right ) \pi x\right ) Now, since u\left ( x,t\right ) =w\left ( x,t\right ) +v\left ( x\right ) , then the final solution is u\left ( x,t\right ) =x^{2}-2x+1+\frac{4}{\pi }\sum _{n=1}^{\infty }\frac{1}{\left ( 2n-1\right ) }e^{-\left ( 2n-1\right ) ^{2}\pi ^{2}t}\sin \left ( \left ( 2n-1\right ) \pi x\right )
Solve u_{t}=u_{xx}-\pi ^{2}\cos \pi x With initial condition u\left ( x,0\right ) =\cos \left ( \frac{3\pi }{2}x\right ) -\cos \left ( \pi x\right ) and boundary conditions u_{x}\left ( 0,t\right ) =0,u\left ( 1,t\right ) =1
Solution
Let u\left ( x,t\right ) =w\left ( x,t\right ) +v\left ( x\right ) where v\left ( x\right ) is steady state solution which only needs to satisfy the non-homogenous boundary conditions and w\left ( x,t\right ) is the transient solution which needs to satisfy the homogeneous version of boundary conditions.
At steady state, the PDE becomes an ODE 0=v^{\prime \prime }\left ( x\right ) -\pi ^{2}\cos \pi x This ODE can be easily solved giving v\left ( x\right ) =-\cos \left ( \pi x\right ) Hence u\left ( x,t\right ) =w\left ( x,t\right ) -\cos \left ( \pi x\right ) . Substituting this into the PDE u_{t}=u_{xx}-\pi ^{2}\cos \pi x results in w_{t}=w_{xx}+v^{\prime \prime }\left ( x\right ) -\pi ^{2}\cos \pi x But v^{\prime }\left ( x\right ) =\pi \sin \left ( \pi x\right ) and v^{\prime \prime }\left ( x\right ) =\pi ^{2}\cos \left ( \pi x\right ) . The above becomes w_{t}=w_{xx} With boundary conditions w_{x}\left ( 0,t\right ) =0,w\left ( 1,t\right ) =0. This was solved before. It has the solution\begin{equation} w\left ( x,t\right ) =\sum _{n=1,3,5,\cdots }^{\infty }c_{n}e^{-\lambda _{n}t}\cos \left ( \sqrt{\lambda _{n}}x\right ) \tag{1} \end{equation} Where \lambda _{n}=\left ( \frac{n\pi }{2}\right ) ^{2} with n=1,3,5,\cdots . At t=0, from u\left ( x,0\right ) =w\left ( x,0\right ) +v\left ( x\right ) , then w\left ( x,0\right ) =u\left ( x,0\right ) -v\left ( x\right ) or\begin{align*} w\left ( x,0\right ) & =\cos \left ( \frac{3\pi }{2}x\right ) -\cos \left ( \pi x\right ) +\cos \left ( \pi x\right ) \\ & =\cos \left ( \frac{3\pi }{2}x\right ) \end{align*}
Therefore, from (1) and at t=0 we obtain\begin{align*} w\left ( x,0\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }c_{n}\cos \left ( \sqrt{\lambda _{n}}x\right ) \\ \cos \left ( \frac{3\pi }{2}x\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }c_{n}\cos \left ( \frac{n\pi }{2}x\right ) \end{align*}
Therefore, only for n=3 is there a solution. Therefore c_{3}=1. Hence (1) becomes\begin{align*} w\left ( x,t\right ) & =e^{-\lambda _{3}t}\cos \left ( \sqrt{\lambda _{3}}x\right ) \\ & =e^{-\left ( \frac{3\pi }{2}\right ) ^{2}t}\cos \left ( \frac{3\pi }{2}x\right ) \end{align*}
Therefore the final solution is\begin{align*} u\left ( x,t\right ) & =w\left ( x,t\right ) +v\left ( x\right ) \\ & =-\cos \left ( \pi x\right ) +e^{-\frac{9\pi ^{2}}{4}t}\cos \left ( \frac{3\pi }{2}x\right ) \end{align*}
Part (a) Show that by method of variation of parameters that general solution to y^{\prime \prime }\left ( x\right ) =-f\left ( x\right ) can be written as
y=c_{1}+c_{2}x-\int _{0}^{x}\left ( x-s\right ) f\left ( s\right ) ds part (b). Let the solution required to satisfy boundary conditions y\left ( 0\right ) =0,y\left ( 1\right ) =0. Show that c_{1}=0,c_{2}=\int _{0}^{1}\left ( 1-x\right ) f\left ( s\right ) ds
part (c). Defining G\left ( x,s\right ) =\left \{ \begin{array} [c]{ccc}s\left ( 1-x\right ) & & 0\leq s\leq x\\ x\left ( 1-s\right ) & & x\leq s\leq 1 \end{array} \right . \ show that the solution can be written as y\left ( x\right ) =\int _{0}^{1}G\left ( x,s\right ) f\left ( s\right ) ds
Solution
The solution is y=y_{h}+y_{p}. Where y_{h}^{\prime \prime }=0. This has the solution y_{h}=c_{1}+c_{2}x In this expression, the basis solutions are \begin{align*} y_{1} & =1\\ y_{2} & =x. \end{align*}
The particular solution is now found using variation of parameters, where it is assumed that \begin{equation} y_{p}=y_{1}u_{1}+y_{2}u_{2} \tag{1} \end{equation} And u_{1},u_{2} are two functions to be determined. Using the standard formulas for finding u_{1},u_{2} gives\begin{equation} u_{1}=\int _{0}^{x}\frac{-y_{2}F\left ( s\right ) }{W\left ( s\right ) }ds \tag{2} \end{equation} Where in the above, F\left ( s\right ) is the forcing function in the RHS\ of the original ODE which is -f\left ( x\right ) here, and W is the Wronskian. The Wronskian is found as follows W=\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} Substituting y_{1}=1,y_{2}=x in the above gives W=\begin{vmatrix} 1 & x\\ 0 & 1 \end{vmatrix} =1 Therefore (2) becomes\begin{align} u_{1} & =\int _{0}^{x}-s\left ( -f\left ( s\right ) \right ) ds\nonumber \\ & =\int _{0}^{x}sf\left ( s\right ) ds \tag{3} \end{align}
Similarly, u_{2} is found using\begin{align} u_{2} & =\int _{0}^{x}\frac{y_{1}F\left ( s\right ) }{W\left ( s\right ) }ds\nonumber \\ & =\int _{0}^{x}-f\left ( s\right ) ds \tag{4} \end{align}
Using (3,4) in (1) gives the particular solution as\begin{align*} y_{p} & =y_{1}\int _{0}^{x}sf\left ( s\right ) ds-y_{2}\int _{0}^{x}f\left ( s\right ) ds\\ & =\int _{0}^{x}sf\left ( s\right ) ds-x\int _{0}^{x}f\left ( s\right ) ds\\ & =\int _{0}^{x}sf\left ( s\right ) ds-\int _{0}^{x}xf\left ( s\right ) ds\\ & =\int _{0}^{x}\left ( s-x\right ) f\left ( s\right ) ds\\ & =-\int _{0}^{x}\left ( x-s\right ) f\left ( s\right ) ds \end{align*}
Now that particular solution is found, the complete solution is found from y=y_{h}+y_{p} giving\begin{equation} y=c_{1}+c_{2}x-\int _{0}^{x}\left ( x-s\right ) f\left ( s\right ) ds \tag{5} \end{equation}
Using the BC y\left ( 0\right ) =0 on (5) gives\begin{align*} 0 & =c_{1}-\int _{0}^{0}-sf\left ( s\right ) ds\\ c_{1} & =0 \end{align*}
Hence c_{1}=0 and the solution (5) now becomes\begin{equation} y=c_{2}x-\int _{0}^{x}\left ( x-s\right ) f\left ( s\right ) ds \tag{6} \end{equation} Using the second BC y\left ( 1\right ) =0 the above becomes\begin{align*} 0 & =c_{2}-\int _{0}^{1}\left ( 1-s\right ) f\left ( s\right ) ds\\ c_{2} & =\int _{0}^{1}\left ( 1-s\right ) f\left ( s\right ) ds \end{align*}
Hence the solution (6) now becomes\begin{align*} y & =x\int _{0}^{1}\left ( 1-s\right ) f\left ( s\right ) ds-\int _{0}^{x}\left ( x-s\right ) f\left ( s\right ) ds\\ & =\int _{0}^{1}x\left ( 1-s\right ) f\left ( s\right ) ds-\int _{0}^{x}\left ( x-s\right ) f\left ( s\right ) ds \end{align*}
Writing \int _{0}^{1}x\left ( 1-s\right ) f\left ( s\right ) ds=\int _{0}^{x}x\left ( 1-s\right ) f\left ( s\right ) ds+\int _{x}^{1}x\left ( 1-s\right ) f\left ( s\right ) ds then the above becomes y=\int _{0}^{x}x\left ( 1-s\right ) f\left ( s\right ) ds+\int _{x}^{1}x\left ( 1-s\right ) f\left ( s\right ) ds-\int _{0}^{x}\left ( x-s\right ) f\left ( s\right ) ds Combining the first and third integrals gives\begin{align} y & =\int _{0}^{x}\left [ x\left ( 1-s\right ) -\left ( x-s\right ) \right ] f\left ( s\right ) ds+\int _{x}^{1}x\left ( 1-s\right ) f\left ( s\right ) ds\nonumber \\ & =\int _{0}^{x}\left [ x-xs-x+s\right ] f\left ( s\right ) ds+\int _{x}^{1}x\left ( 1-s\right ) f\left ( s\right ) ds\nonumber \\ & =\int _{0}^{x}\left ( -xs+s\right ) f\left ( s\right ) ds+\int _{x}^{1}x\left ( 1-s\right ) f\left ( s\right ) ds\nonumber \\ & =\int _{0}^{x}s\left ( 1-x\right ) f\left ( s\right ) ds+\int _{x}^{1}x\left ( 1-s\right ) f\left ( s\right ) ds \tag{7} \end{align}
Which is the result required to show.
From part (b) above, the solution in (7) can be written as\begin{equation} y=\int _{0}^{x}G_{L}\left ( x,s\right ) f\left ( s\right ) ds+\int _{x}^{1}G_{R}\left ( x,s\right ) f\left ( s\right ) ds \tag{8} \end{equation} Where G\left ( x,s\right ) =\left \{ \begin{array} [c]{c}G_{L}\left ( x,x\right ) \\ G_{L}\left ( x,s\right ) \end{array} \right . =\left \{ \begin{array} [c]{ccc}s\left ( 1-x\right ) & & 0\leq s\leq x\\ x\left ( 1-s\right ) & & x\leq s\leq 1 \end{array} \right . Hence (8) can be combined into one integral y=\int _{0}^{1}G\left ( x,s\right ) f\left ( s\right ) ds
By using procedure in problem 28 show that solution to y^{\prime \prime }+y=-f\left ( x\right ) ,y\left ( 0\right ) =0,y\left ( 1\right ) =0 is y=\int _{0}^{1}G\left ( x,s\right ) f\left ( s\right ) Where G\left ( x,s\right ) =\left \{ \begin{array} [c]{ccc}\frac{\sin \left ( s\right ) \sin \left ( 1-x\right ) }{\sin \left ( 1\right ) } & & 0\leq s\leq x\\ \frac{\sin \left ( x\right ) \sin \left ( 1-s\right ) }{\sin \left ( 1\right ) } & & x\leq s\leq 1 \end{array} \right . Solution
Let y=y_{h}+y_{p}. Where y_{h} is solution to y_{h}^{\prime \prime }+y_{h}=0. This has the solution y_{h}=c_{1}\cos x+c_{2}\sin x. Hence the bases solutions are \begin{align*} y_{1} & =\cos x\\ y_{2} & =\sin x \end{align*}
And therefore the Wronskian is W=\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} \cos x & \sin x\\ -\sin x & \cos x \end{vmatrix} =\cos ^{2}x+\sin ^{2}x=1 Hence u_{1}=\int _{0}^{x}\frac{-y_{2}F\left ( s\right ) }{W\left ( s\right ) }ds Where in the above, F\left ( s\right ) is the forcing function in the RHS\ of the original ODE which is -f\left ( x\right ) here, and W is the Wronskian. Therefore\begin{align*} u_{1} & =\int _{0}^{x}-\sin \left ( s\right ) \left ( -f\left ( s\right ) \right ) ds\\ & =\int _{0}^{x}\sin \left ( s\right ) f\left ( s\right ) ds \end{align*}
Similarly, u_{2} is found using\begin{align*} u_{2} & =\int _{0}^{x}\frac{y_{1}F\left ( s\right ) }{W\left ( s\right ) }ds\\ & =\int _{0}^{x}\cos \left ( s\right ) \left ( -f\left ( s\right ) \right ) ds \end{align*}
Hence the particular solution is\begin{align*} y_{p} & =y_{1}u_{1}+y_{2}u_{2}\\ & =\cos \left ( x\right ) \int _{0}^{x}\sin \left ( s\right ) f\left ( s\right ) ds-\sin \left ( x\right ) \int _{0}^{x}\cos \left ( s\right ) f\left ( s\right ) ds\\ & =\int _{0}^{x}\cos \left ( x\right ) \sin \left ( s\right ) f\left ( s\right ) ds-\int _{0}^{x}\sin \left ( x\right ) \cos \left ( s\right ) f\left ( s\right ) ds\\ & =\int _{0}^{x}\left ( \cos \left ( x\right ) \sin \left ( s\right ) -\sin \left ( x\right ) \cos \left ( s\right ) \right ) f\left ( s\right ) ds \end{align*}
Applying \left ( \sin A\cos B-\cos A\sin B\right ) =\sin \left ( A-B\right ) to the integrand above, where A=x,B=s gives y_{p}=-\int _{0}^{x}\sin \left ( x-s\right ) f\left ( s\right ) ds Therefore the solution is\begin{align} y & =y_{h}+y_{p}\nonumber \\ & =\left ( c_{1}\cos x+c_{2}\sin x\right ) -\int _{0}^{x}\sin \left ( x-s\right ) f\left ( s\right ) ds \tag{1} \end{align}
Applying BC y\left ( 0\right ) =0 the above becomes\begin{align*} 0 & =c_{1}-\int _{0}^{0}\sin \left ( -s\right ) f\left ( s\right ) ds\\ c_{1} & =0 \end{align*}
And the solution (1) simplifies to\begin{equation} y\left ( x\right ) =c_{2}\sin x-\int _{0}^{x}\sin \left ( x-s\right ) f\left ( s\right ) ds \tag{2} \end{equation} Applying BC y\left ( 1\right ) =0 the above becomes y\left ( x\right ) =c_{2}\sin 1-\int _{0}^{1}\sin \left ( 1-s\right ) f\left ( s\right ) ds Hence c_{2}=\frac{1}{\sin 1}\int _{0}^{1}\sin \left ( 1-s\right ) f\left ( s\right ) ds The solution in (2) now becomes\begin{align*} y\left ( x\right ) & =\frac{\sin x}{\sin 1}\int _{0}^{1}\sin \left ( 1-s\right ) f\left ( s\right ) ds-\int _{0}^{x}\sin \left ( x-s\right ) f\left ( s\right ) ds\\ & =\frac{1}{\sin 1}\int _{0}^{1}\sin x\sin \left ( 1-s\right ) f\left ( s\right ) ds-\int _{0}^{x}\sin \left ( x-s\right ) f\left ( s\right ) ds \end{align*}
Writing \int _{0}^{1}\sin x\sin \left ( 1-s\right ) f\left ( s\right ) ds=\int _{0}^{x}\sin x\sin \left ( 1-s\right ) f\left ( s\right ) ds+ \int _{x}^{1}\sin x\sin \left ( 1-s\right ) f\left ( s\right ) ds then the above becomes\begin{align} y\left ( x\right ) & =\frac{1}{\sin 1}\left ( \int _{0}^{x}\sin x\sin \left ( 1-s\right ) f\left ( s\right ) ds+\int _{x}^{1}\sin x\sin \left ( 1-s\right ) f\left ( s\right ) ds\right ) -\int _{0}^{x}\sin \left ( x-s\right ) f\left ( s\right ) ds\nonumber \\ & =\int _{0}^{x}\frac{\sin x\sin \left ( 1-s\right ) }{\sin \left ( 1\right ) }f\left ( s\right ) -\int _{0}^{x}\sin \left ( x-s\right ) f\left ( s\right ) ds+\int _{x}^{1}\frac{\sin x\sin \left ( 1-s\right ) }{\sin \left ( 1\right ) }f\left ( s\right ) ds\nonumber \\ & =\int _{0}^{x}\left [ \frac{\sin x\sin \left ( 1-s\right ) }{\sin \left ( 1\right ) }-\sin \left ( x-s\right ) \right ] f\left ( s\right ) ds+\int _{x}^{1}\frac{\sin x\sin \left ( 1-s\right ) }{\sin \left ( 1\right ) }f\left ( s\right ) ds\nonumber \\ & =\frac{1}{\sin \left ( 1\right ) }\int _{0}^{x}\left ( \sin x\sin \left ( 1-s\right ) -\sin \left ( 1\right ) \sin \left ( x-s\right ) \right ) f\left ( s\right ) ds+\int _{x}^{1}\frac{\sin x\sin \left ( 1-s\right ) }{\sin \left ( 1\right ) }f\left ( s\right ) ds \tag{3} \end{align}
Using \sin \left ( A-B\right ) =\sin A\cos B-\cos A\sin B, where now A=1,B=s, then \sin \left ( 1-s\right ) =\sin 1\cos s-\cos 1\sin s And also \sin \left ( x-s\right ) =\sin x\cos s-\cos x\sin s Using the above two relations in first integral of (3) which is I=\int _{0}^{x}\left ( \sin x\sin \left ( 1-s\right ) -\sin \left ( 1\right ) \sin \left ( x-s\right ) \right ) f\left ( s\right ) ds gives\begin{align*} I & =\int _{0}^{x}\left ( \sin x\left ( \sin 1\cos s-\cos 1\sin s\right ) -\sin 1\left ( \sin x\cos s-\cos x\sin s\right ) \right ) f\left ( s\right ) ds\\ & =\int _{0}^{x}\left ( \sin x\sin 1\cos s-\sin x\cos 1\sin s-\sin 1\sin x\cos s+\sin 1\cos x\sin s\right ) f\left ( s\right ) ds\\ & =\int _{0}^{x}\left ( -\sin x\cos 1\sin s+\sin 1\cos x\sin s\right ) f\left ( s\right ) ds\\ & =\int _{0}^{x}\left ( \sin s\left ( \sin 1\cos x-\sin x\cos 1\right ) \right ) f\left ( s\right ) ds\\ & =\int _{0}^{x}\left ( \sin s\sin \left ( 1-x\right ) \right ) f\left ( s\right ) ds \end{align*}
Substituting the above result in (3) results in\begin{equation} y\left ( x\right ) =\int _{0}^{x}\frac{\sin s\sin \left ( 1-x\right ) }{\sin 1}f\left ( s\right ) ds+\int _{x}^{1}\frac{\sin x\sin \left ( 1-s\right ) }{\sin \left ( 1\right ) }f\left ( s\right ) ds \tag{4} \end{equation} Let G\left ( x,s\right ) =\left \{ \begin{array} [c]{ccc}\frac{\sin \left ( s\right ) \sin \left ( 1-x\right ) }{\sin \left ( 1\right ) } & & 0\leq s\leq x\\ \frac{\sin \left ( x\right ) \sin \left ( 1-s\right ) }{\sin \left ( 1\right ) } & & x\leq s\leq 1 \end{array} \right . Then the solution (4) can be written as y\left ( x\right ) =\int _{0}^{1}G\left ( x,s\right ) f\left ( s\right ) ds
By using procedure in problem 30 find Green function and express solution as definite integral for\begin{align*} -y^{\prime \prime } & =f\left ( x\right ) \\ y^{\prime }\left ( 0\right ) & =0\\ y\left ( 1\right ) & =0 \end{align*}
Solution
The first step is to determine y_{1}\left ( x\right ) ,y_{2}\left ( x\right ) . These are the two fundamental solutions of y^{\prime \prime }=0. As the book says, to simplify the derivation, y_{1}\left ( x\right ) is selected to be the solution that satisfies the boundary conditions at the left end of domain (x=0 in this problem) and y_{2}\left ( x\right ) satisfies the boundary condition on the right end (x=1).
The homogeneous solution to y^{\prime \prime }=0 is y_{h}\left ( x\right ) =c_{1}+c_{2}x Therefore y_{1}^{\prime }\left ( 0\right ) =0. This gives c_{2}=0. Hence y_{1}\left ( x\right ) =1 The second boundary conditions y_{2}\left ( 1\right ) =0 gives 0=c_{1}+c_{2}, or c_{1}=-c_{2} and this leads to y_{2}\left ( x\right ) =c_{2}\left ( -1+x\right ) . Or y_{2}\left ( x\right ) =x-1 Given y_{1},y_{2} found above, the next step is to determine the Wronskian as follows W\left ( x\right ) =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} 1 & x-1\\ 0 & 1 \end{vmatrix} =1 Therefore, Green function is now computed using equation (iv) on page 701 of text book giving G\left ( x,s\right ) =\frac{-1}{p\left ( x\right ) W\left ( x\right ) }\left \{ \begin{array} [c]{ccc}y_{1}\left ( s\right ) y_{2}\left ( x\right ) & & 0\leq s\leq x\\ y_{1}\left ( x\right ) y_{2}\left ( s\right ) & & x\leq s\leq 1 \end{array} \right . But p\left ( x\right ) =1 and W\left ( x\right ) =1, and using values found earlier for y_{1},y_{2}, the above becomes\begin{align*} G\left ( x,s\right ) & =-1\left \{ \begin{array} [c]{ccc}\left ( x-1\right ) & & 0\leq s\leq x\\ \left ( s-1\right ) & & x\leq s\leq 1 \end{array} \right . \\ & =\left \{ \begin{array} [c]{ccc}x-1 & & 0\leq s\leq x\\ s-1 & & x\leq s\leq 1 \end{array} \right . \end{align*}
Hence the solution is\begin{equation} y\left ( x,s\right ) =\int _{0}^{1}G\left ( x,s\right ) f\left ( s\right ) ds \tag{1} \end{equation}
To verify this solution, it is compared to solution to same ODE using the direct method. Let f\left ( x\right ) =x. Hence the ODE is\begin{align*} -y^{\prime \prime } & =x\\ y^{\prime }\left ( 0\right ) & =0\\ y\left ( 1\right ) & =0 \end{align*}
The solution found above in (1) can now be found as\begin{align} y\left ( x\right ) & =\int _{0}^{x}G\left ( x,s\right ) sds+\int _{x}^{1}G\left ( x,s\right ) sds\nonumber \\ & =\int _{0}^{x}\left ( 1-x\right ) sds+\int _{x}^{1}\left ( 1-s\right ) sds\nonumber \\ & =\left ( \frac{s^{2}}{2}-x\frac{s^{2}}{2}\right ) _{0}^{x}+\left ( \frac{s^{2}}{2}-\frac{s^{3}}{3}\right ) _{x}^{1}\nonumber \\ & =\left ( \frac{x^{2}}{2}-\frac{x^{3}}{2}\right ) +\left ( \left ( \frac{1}{2}-\frac{1}{3}\right ) -\left ( \frac{x^{2}}{2}-\frac{x^{3}}{3}\right ) \right ) \nonumber \\ & =\frac{1}{6}-\frac{1}{6}x^{3} \tag{2} \end{align}
Verification The solution is verified by solving the same problem using the direct method. The homogenous solution is y_{h}=c_{1}+c_{2}x. Since the forcing function is -x, let the particular solution be y_{p}=kx^{3},y_{p}^{\prime }=3kx^{2},y^{\prime \prime }=6kx. Therefore 6kx=-x or k=\frac{-1}{6}. Therefore the particular solution is y_{p}=\frac{-1}{6}x^{3} and the general solution is y\left ( x\right ) =c_{1}+c_{2}x-\frac{1}{6}x^{3} Applying BC y^{\prime }\left ( 0\right ) =0 gives c_{2}=0 Hence the solution becomes y\left ( x\right ) =c_{1}-\frac{1}{6}x^{3}. Applying BC y\left ( 1\right ) =0 gives 0=c_{1}-\frac{1}{6} or c_{1}=\frac{1}{6}. Therefore the solution is\begin{equation} y\left ( x\right ) =\frac{1}{6}-\frac{1}{6}x^{3} \tag{3} \end{equation} Which is the same answer found using Green function method. Of course in this case the direct method is much simpler and easier to find. The advantage of Green method, is that once the G\left ( x,s\right ) is found, then for any new f\left ( x\right ) only integration is needed to find the new solution, since G\left ( x,s\right ) does not change when f\left ( x\right ) changes. The direct method requires one to find the particular solution each time, and to determine the constants c_{1},c_{2} again from boundary conditions each time f\left ( x\right ) changes since the particular solution changes when f\left ( x\right ) changes. With Green function method, all the work in using G\left ( x,y\right ) is done in the integration step only. The solution found using Green function already incorporated the boundary conditions in it.
By using procedure in problem 30 find Green function and express solution as definite integral for\begin{align*} -y^{\prime \prime } & =f\left ( x\right ) \\ y\left ( 0\right ) & =0\\ y\left ( 1\right ) +y^{\prime }\left ( 1\right ) & =0 \end{align*}
Solution
The first step is to determine y_{1}\left ( x\right ) ,y_{2}\left ( x\right ) , where these are the fundamental solutions of y^{\prime \prime }=0 where y_{1}\left ( x\right ) satisfies the boundary conditions at the left end of domain (x=0) and y_{2}\left ( x\right ) satisfies the boundary condition on the right end (x=1).
Since the homogeneous solution to y^{\prime \prime }=0 is y_{h}\left ( x\right ) =c_{1}+c_{2}x Then y_{1}\left ( 0\right ) =0 gives c_{1}=0. Therefore y_{1}\left ( x\right ) =x And to satisfy y_{2}\left ( 1\right ) +y_{2}^{\prime }\left ( 1\right ) =0 then \begin{align*} 0 & =\left ( c_{1}+c_{2}\right ) +c_{2}\\ c_{1} & =-2c_{2} \end{align*}
Therefore \begin{align*} y_{2}\left ( x\right ) & =-2c_{2}+c_{2}x\\ & =c_{2}\left ( x-2\right ) \end{align*}
Hence y_{2}\left ( x\right ) =x-2 Now that y_{1},y_{2} are found, the next step is to find the Wronskian. W\left ( x\right ) =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} x & x-2\\ 1 & 1 \end{vmatrix} =x-\left ( x-2\right ) =2 Therefore, Green function is, using equation (iv) on page 701 of text book G\left ( x,s\right ) =\frac{-1}{p\left ( x\right ) W\left ( x\right ) }\left \{ \begin{array} [c]{ccc}y_{1}\left ( s\right ) y_{2}\left ( x\right ) & & 0\leq s\leq x\\ y_{1}\left ( x\right ) y_{2}\left ( s\right ) & & x\leq s\leq 1 \end{array} \right . But p\left ( x\right ) =1 and W\left ( x\right ) =1, and using values found earlier for y_{1},y_{2}, then the above becomes\begin{align*} G\left ( x,s\right ) & =\frac{-1}{2}\left \{ \begin{array} [c]{ccc}s\left ( x-2\right ) & & 0\leq s\leq x\\ x\left ( s-2\right ) & & x\leq s\leq 1 \end{array} \right . \\ & =\left \{ \begin{array} [c]{ccc}\frac{s\left ( 2-x\right ) }{2} & & 0\leq s\leq x\\ \frac{x\left ( 2-s\right ) }{2} & & x\leq s\leq 1 \end{array} \right . \end{align*}
And the solution is\begin{equation} y\left ( x,s\right ) =\int _{0}^{1}G\left ( x,s\right ) f\left ( s\right ) ds \tag{1} \end{equation} To verify this solution, it is compared to solution to same ODE using the direct method. Let f\left ( x\right ) =x. Hence the ODE is\begin{align*} -y^{\prime \prime } & =x\\ y^{\prime }\left ( 0\right ) & =0\\ y\left ( 1\right ) & =0 \end{align*}
The solution found above in (1) is now found as\begin{align} y\left ( x\right ) & =\int _{0}^{x}G\left ( x,s\right ) sds+\int _{x}^{1}G\left ( x,s\right ) sds\nonumber \\ & =\int _{0}^{x}\frac{s\left ( 2-x\right ) }{2}sds+\int _{x}^{1}\frac{x\left ( 2-s\right ) }{2}sds\nonumber \\ & =\frac{1}{2}\int _{0}^{x}\left ( 2s^{2}-xs^{2}\right ) ds+\frac{1}{2}\int _{x}^{1}\left ( 2xs-xs^{2}\right ) ds\nonumber \\ & =\frac{1}{2}\left ( \frac{2s^{3}}{3}-x\frac{s^{3}}{3}\right ) _{0}^{x}+\frac{1}{2}\left ( xs^{2}-x\frac{s^{3}}{3}\right ) _{x}^{1}\nonumber \\ & =\frac{1}{6}\left ( 2x^{3}-x^{4}\right ) +\frac{1}{2}\left ( \left ( x-\frac{x}{3}\right ) -\left ( x^{3}-\frac{x^{4}}{3}\right ) \right ) \nonumber \\ & =\frac{1}{6}\left ( 2x-x^{3}\right ) \tag{2} \end{align}
Verification The solution is now verified by solving the same problem using the direct method. The homogenous solution is y_{h}=c_{1}+c_{2}x. Since the forcing function is -x, let the particular solution be y_{p}=kx^{3},y_{p}^{\prime }=3kx^{2},y^{\prime \prime }=6kx. Therefore 6kx=-x or k=\frac{-1}{6}. Therefore the particular solution is y_{p}=\frac{-1}{6}x^{3} and the general solution is y\left ( x\right ) =c_{1}+c_{2}x-\frac{1}{6}x^{3} Applying BC y\left ( 0\right ) =0 gives c_{1}=0 Hence the solution becomes \begin{align*} y\left ( x\right ) & =c_{2}x-\frac{1}{6}x^{3}\\ y^{\prime }\left ( x\right ) & =c_{2}-\frac{1}{2}x^{2} \end{align*}
Applying BC y\left ( 1\right ) +y^{\prime }\left ( 1\right ) =0 gives \begin{align*} 0 & =\left ( c_{2}-\frac{1}{6}\right ) +\left ( c_{2}-\frac{1}{2}\right ) \\ 0 & =2c_{2}-\frac{2}{3}\\ c_{2} & =\frac{1}{3} \end{align*}
Therefore the solution is\begin{align} y\left ( x\right ) & =\frac{1}{3}x-\frac{1}{6}x^{3}\nonumber \\ & =\frac{1}{6}\left ( 2x-x^{3}\right ) \tag{3} \end{align}
Which is the same as (2) using Green function.
By using procedure in problem 30 find Green function and express solution as definite integral for\begin{align*} -\left ( y^{\prime \prime }+y\right ) & =f\left ( x\right ) \\ y^{\prime }\left ( 0\right ) & =0\\ y\left ( 1\right ) & =0 \end{align*}
Solution
The first step is to determine y_{1}\left ( x\right ) ,y_{2}\left ( x\right ) , where these are the fundamental solutions of y^{\prime \prime }+y=0 where y_{1}\left ( x\right ) satisfies the boundary conditions at the left end of domain (x=0) and y_{2}\left ( x\right ) satisfies the boundary condition on the right end (x=1).
Since the homogeneous solution to y^{\prime \prime }+y=0 is y_{h}\left ( x\right ) =c_{1}\cos x+c_{2}\sin x Then y_{1}^{\prime }= -c_{1}\sin x+c_{2}\cos x and y_{1}^{\prime }\left ( 0\right ) =0 leads to c_{2}=0, therefore y_{1}\left ( x\right ) =\cos x And to satisfy y_{2}\left ( 1\right ) =0 then 0=c_{1}\cos 1+c_{2}\sin 1, hence c_{2}=-c_{1}\frac{\cos \left ( 1\right ) }{\sin \left ( 1\right ) }. therefore\begin{align*} y_{2}\left ( x\right ) & =c_{1}\cos x-c_{1}\frac{\cos \left ( 1\right ) }{\sin \left ( 1\right ) }\sin x\\ & =c_{1}\left ( \cos x-\frac{\cos \left ( 1\right ) }{\sin \left ( 1\right ) }\sin x\right ) \end{align*}
Hence y_{2}\left ( x\right ) =\cos x-\frac{\cos \left ( 1\right ) }{\sin \left ( 1\right ) }\sin x Now that y_{1},y_{2} are found, the next step is to determine the Wronskian. \begin{align*} W\left ( x\right ) & =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} \\ & =\begin{vmatrix} \cos x & \left ( \cos x-\frac{\cos \left ( 1\right ) }{\sin \left ( 1\right ) }\sin x\right ) \\ -\sin x & -\left ( \sin x+\frac{\cos \left ( 1\right ) }{\sin \left ( 1\right ) }\cos x\right ) \end{vmatrix} \\ & =-\cos x\left ( \sin x+\frac{\cos \left ( 1\right ) }{\sin \left ( 1\right ) }\cos x\right ) +\sin x\left ( \cos x-\frac{\cos \left ( 1\right ) }{\sin \left ( 1\right ) }\sin x\right ) \\ & =-\cos x\sin x-\frac{\cos \left ( 1\right ) }{\sin \left ( 1\right ) }\cos ^{2}x+\sin x\cos x-\frac{\cos \left ( 1\right ) }{\sin \left ( 1\right ) }\sin ^{2}x\\ & =-\frac{\cos \left ( 1\right ) }{\sin \left ( 1\right ) }\left ( \cos ^{2}x+\sin ^{2}x\right ) \\ & =-\frac{\cos \left ( 1\right ) }{\sin \left ( 1\right ) } \end{align*}
Therefore, Green function is, using equation (iv) on page 701 of text book G\left ( x,s\right ) =\frac{-1}{p\left ( x\right ) W\left ( x\right ) }\left \{ \begin{array} [c]{ccc}y_{1}\left ( s\right ) y_{2}\left ( x\right ) & & 0\leq s\leq x\\ y_{1}\left ( x\right ) y_{2}\left ( s\right ) & & x\leq s\leq 1 \end{array} \right . But p\left ( x\right ) =1 and W\left ( x\right ) =1, and using values found earlier for y_{1},y_{2}, then the above becomes (using p\left ( x\right ) =1)\begin{align*} G\left ( x,s\right ) & =\frac{-1}{-\frac{\cos \left ( 1\right ) }{\sin \left ( 1\right ) }}\left \{ \begin{array} [c]{ccc}\cos s\left ( \cos x-\frac{\cos \left ( 1\right ) }{\sin \left ( 1\right ) }\sin x\right ) & & 0\leq s\leq x\\ \cos x\left ( \cos s-\frac{\cos \left ( 1\right ) }{\sin \left ( 1\right ) }\sin s\right ) & & x\leq s\leq 1 \end{array} \right . \\ & =\frac{\sin \left ( 1\right ) }{\cos \left ( 1\right ) }\left \{ \begin{array} [c]{ccc}\cos s\left ( \cos x-\frac{\cos \left ( 1\right ) }{\sin \left ( 1\right ) }\sin x\right ) & & 0\leq s\leq x\\ \cos x\left ( \cos s-\frac{\cos \left ( 1\right ) }{\sin \left ( 1\right ) }\sin s\right ) & & x\leq s\leq 1 \end{array} \right . \\ & =\left \{ \begin{array} [c]{ccc}\frac{\cos s}{\cos \left ( 1\right ) }\left ( \sin \left ( 1\right ) \cos x-\cos \left ( 1\right ) \sin x\right ) & & 0\leq s\leq x\\ \frac{\cos x}{\cos \left ( 1\right ) }\left ( \sin \left ( 1\right ) \cos s-\cos \left ( 1\right ) \sin s\right ) & & x\leq s\leq 1 \end{array} \right . \end{align*}
Using \sin A\cos B-\cos A\sin B=\sin \left ( A-B\right ) then \sin \left ( 1\right ) \cos x-\cos \left ( 1\right ) \sin x=\sin \left ( 1-x\right ) and \sin \left ( 1\right ) \cos s-\cos \left ( 1\right ) \sin s=\sin \left ( 1-s\right ) and the above becomes G\left ( x,s\right ) =\left \{ \begin{array} [c]{ccc}\frac{\cos s}{\cos \left ( 1\right ) }\sin \left ( 1-x\right ) & & 0\leq s\leq x\\ \frac{\cos x}{\cos \left ( 1\right ) }\sin \left ( 1-s\right ) & & x\leq s\leq 1 \end{array} \right . And the solution is y\left ( x,s\right ) =\int _{0}^{1}G\left ( x,s\right ) f\left ( s\right ) ds To verify this solution, it is compared to the solution to same ODE using the direct method. Let f\left ( x\right ) =x. Hence the ODE is\begin{align*} -\left ( y^{\prime \prime }+y\right ) & =x\\ y^{\prime }\left ( 0\right ) & =0\\ y\left ( 1\right ) & =0 \end{align*}
The solution found above in (1) is now computed as\begin{align} y\left ( x\right ) & =\int _{0}^{x}G\left ( x,s\right ) sds+\int _{x}^{1}G\left ( x,s\right ) sds\nonumber \\ & =\int _{0}^{x}\frac{\cos s}{\cos \left ( 1\right ) }\sin \left ( 1-x\right ) sds+\int _{x}^{1}\frac{\cos x}{\cos \left ( 1\right ) }\sin \left ( 1-s\right ) sds\nonumber \\ & =I_{1}+I_{2} \tag{1} \end{align}
The first integral is\begin{align*} I_{1} & =\frac{\sin \left ( 1-x\right ) }{\cos \left ( 1\right ) }\int _{0}^{x}s\cos sds\\ & =\frac{\sin \left ( 1-x\right ) }{\cos \left ( 1\right ) }\left ( \cos s+s\sin s\right ) _{0}^{x}\\ & =\frac{\sin \left ( 1-x\right ) }{\cos \left ( 1\right ) }\left ( \cos x+x\sin x-1\right ) \end{align*}
The second integral is\begin{align*} I_{2} & =\frac{\cos x}{\cos \left ( 1\right ) }\int _{x}^{1}s\sin \left ( 1-s\right ) ds\\ & =\frac{\cos x}{\cos \left ( 1\right ) }\left ( s\cos \left ( s-1\right ) -\sin \left ( s-1\right ) \right ) _{x}^{1}\\ & =\frac{\cos x}{\cos \left ( 1\right ) }\left ( \left ( \cos \left ( 1-1\right ) -\sin \left ( 1-1\right ) \right ) -\left ( x\cos \left ( x-1\right ) -\sin \left ( x-1\right ) \right ) \right ) \\ & =\frac{\cos x}{\cos \left ( 1\right ) }\left ( 1-\left ( x\cos \left ( x-1\right ) -\sin \left ( x-1\right ) \right ) \right ) \end{align*}
Hence (1) becomes\begin{align*} y\left ( x\right ) & =\frac{\sin \left ( 1-x\right ) }{\cos \left ( 1\right ) }\left ( \cos x+x\sin x-1\right ) +\frac{\cos x}{\cos \left ( 1\right ) }\left ( 1-\left ( x\cos \left ( x-1\right ) -\sin \left ( x-1\right ) \right ) \right ) \\ & =\frac{1}{\cos \left ( 1\right ) }\left ( \cos x\sin \left ( 1-x\right ) +x\sin x\sin \left ( 1-x\right ) -\sin \left ( 1-x\right ) +\cos x-x\cos x\cos \left ( x-1\right ) -\cos x\sin \left ( x-1\right ) \right ) \\ & =\frac{1}{\cos \left ( 1\right ) }\left ( x\sin x\sin \left ( 1-x\right ) -\sin \left ( 1-x\right ) +\cos x-x\cos x\cos \left ( x-1\right ) \right ) \\ & =\frac{1}{\cos 1}\left ( x\left ( \sin x\sin \left ( 1-x\right ) -\cos x\cos \left ( x-1\right ) \right ) -\sin \left ( 1-x\right ) +\cos x\right ) \end{align*}
But \sin A\sin B-\cos A\cos B=-\cos \left ( A+B\right ) , using this in the above, where now x=A,B=\left ( 1-x\right ) gives\begin{align} y\left ( x\right ) & =\frac{1}{\cos 1}\left ( x\left ( -\cos \left ( x+1-x\right ) \right ) -\sin \left ( 1-x\right ) +\cos x\right ) \nonumber \\ & =\frac{1}{\cos 1}\left ( -x\cos \left ( 1\right ) -\sin \left ( 1-x\right ) +\cos x\right ) \nonumber \\ & =\frac{\cos x}{\cos \left ( 1\right ) }-\frac{\sin \left ( 1-x\right ) }{\cos \left ( 1\right ) }-x \tag{2} \end{align}
Verification The solution is now verified by solving the same problem using the direct method. The homogenous solution to y^{\prime \prime }+y=0 is y_{h}=c_{1}\cos x+c_{2}\sin x. Since the forcing function is -x, let the particular solution be y_{p}=k_{1}x,y_{p}^{\prime }=k_{1},y^{\prime \prime }=0. Therefore k_{1}x=-x or k=-1. Therefore the particular solution is y_{p}=-x and the general solution is y\left ( x\right ) =c_{1}\cos x+c_{2}\sin x-x Now BC y^{\prime }\left ( 0\right ) =0 is applied. y^{\prime }\left ( x\right ) =-c_{1}\sin x+c_{2}\cos x-1, therefore\begin{align*} 0 & =c_{2}-1\\ c_{2} & =1 \end{align*}
Hence the solution becomes y\left ( x\right ) =c_{1}\cos x+\sin x-x Applying BC y\left ( 1\right ) =0 gives \begin{align*} 0 & =c_{1}\cos \left ( 1\right ) +\sin \left ( 1\right ) -1\\ c_{1} & =\frac{1-\sin \left ( 1\right ) }{\cos \left ( 1\right ) } \end{align*}
Therefore the solution is\begin{align*} y\left ( x\right ) & =\frac{\left ( 1-\sin \left ( 1\right ) \right ) }{\cos \left ( 1\right ) }\cos x+\sin \left ( x\right ) -x\\ & =\frac{\cos x}{\cos \left ( 1\right ) }+\frac{-\cos x\sin \left ( 1\right ) }{\cos \left ( 1\right ) }+\sin \left ( x\right ) -x\\ & =\frac{\cos x}{\cos \left ( 1\right ) }+\frac{\sin \left ( x\right ) \cos \left ( 1\right ) -\cos x\sin \left ( 1\right ) }{\cos \left ( 1\right ) }-x \end{align*}
But \sin \left ( x\right ) \cos \left ( 1\right ) -\cos x\sin \left ( 1\right ) =\sin \left ( x-1\right ) =-\sin \left ( 1-x\right ) , hence the above becomes\begin{equation} \fbox{$y\left ( x\right ) =\frac{\cos x}{\cos \left ( 1\right ) }-\frac{\sin \left ( 1-x\right ) }{\cos \left ( 1\right ) }-x$} \tag{3} \end{equation} Which is the same solution in (2) found using Green function.
By using procedure in problem 30 find Green function and express solution as definite integral for\begin{align*} -y^{\prime \prime } & =f\left ( x\right ) \\ y\left ( 0\right ) & =0\\ y^{\prime }\left ( 1\right ) & =0 \end{align*}
Solution
The first step is to determine y_{1}\left ( x\right ) ,y_{2}\left ( x\right ) , where these are the fundamental solutions of y^{\prime \prime }=0 where y_{1}\left ( x\right ) satisfies the boundary conditions at the left end of domain (x=0) and y_{2}\left ( x\right ) satisfies the boundary condition on the right end (x=1).
Since the homogeneous solution to y^{\prime \prime }=0 is y_{h}\left ( x\right ) =c_{1}+c_{2}x Then y_{1}\left ( 0\right ) =0 gives c_{1}=0. Therefore y_{1}\left ( x\right ) =x And to satisfy y_{2}^{\prime }\left ( 1\right ) =0 then 0=c_{2}. and this leads to y_{2}\left ( x\right ) =1 Now that y_{1},y_{2} are found, the next step is to find the Wronskian. W\left ( x\right ) =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =\begin{vmatrix} x & 1\\ 1 & 0 \end{vmatrix} =-1 Therefore, Green function is, using equation (iv) on page 701 of text book G\left ( x,s\right ) =\frac{-1}{p\left ( x\right ) W\left ( x\right ) }\left \{ \begin{array} [c]{ccc}y_{1}\left ( s\right ) y_{2}\left ( s\right ) & & 0\leq s\leq x\\ y_{1}\left ( x\right ) y_{2}\left ( x\right ) & & x\leq s\leq 1 \end{array} \right . But p\left ( x\right ) =1 and W\left ( x\right ) =-1, and using values found earlier for y_{1},y_{2}, then the above becomes G\left ( x,s\right ) =\left \{ \begin{array} [c]{ccc}s & & 0\leq s\leq x\\ x & & x\leq s\leq 1 \end{array} \right . And the solution is\begin{equation} y\left ( x,s\right ) =\int _{0}^{1}G\left ( x,s\right ) f\left ( s\right ) ds \tag{1} \end{equation} To verify this solution, it is now compared to the solution to same ODE using the direct method. Let f\left ( x\right ) =x. Hence the ODE now is\begin{align*} -y^{\prime \prime } & =x\\ y\left ( 0\right ) & =0\\ y^{\prime }\left ( 1\right ) & =0 \end{align*}
The solution found above in (1) is now computed as\begin{align} y\left ( x\right ) & =\int _{0}^{x}G\left ( x,s\right ) sds+\int _{x}^{1}G\left ( x,s\right ) sds\nonumber \\ & =\int _{0}^{x}\left ( s\right ) sds+\int _{x}^{1}\left ( x\right ) sds\nonumber \\ & =\left ( \frac{s^{3}}{3}\right ) _{0}^{x}+x\left ( \frac{s^{2}}{2}\right ) _{x}^{1}\nonumber \\ & =\frac{1}{3}x^{3}+\frac{x}{2}\left ( 1-x^{2}\right ) \nonumber \\ & =\frac{1}{2}x-\frac{1}{6}x^{3} \tag{2} \end{align}
Verification The above solution is now verified by solving the same problem using the direct method. The homogenous solution to y^{\prime \prime }=0 is y_{h}=c_{1}+c_{2}x. Since the forcing function is -x, the particular solution is y_{p}=\frac{-1}{6}x^{3} and the general solution is y\left ( x\right ) =c_{1}+c_{2}x-\frac{1}{6}x^{3} BC y\left ( 0\right ) =0 gives c_{1}=0. The solution becomes y\left ( x\right ) =c_{2}x-\frac{1}{6}x^{3} and y^{\prime }\left ( x\right ) =c_{2}-\frac{1}{2}x^{2}. BC y^{\prime }\left ( 1\right ) =0 gives
\begin{align*} 0 & =c_{2}-\frac{1}{2}\\ c_{2} & =\frac{1}{2} \end{align*}
Hence the solution becomes y\left ( x\right ) =\frac{1}{2}x-\frac{1}{6}x^{3} Which is the same solution in (2) found using Green function.
Find formal solution to -\left ( xy^{\prime }\right ) ^{\prime }=\mu xy+f\left ( x\right ) where y,y^{\prime } bounded as x\rightarrow 0 and y\left ( 1\right ) =0
Solution
The given ODE can be written as \begin{equation} -\frac{1}{x}\left ( xy^{\prime }\right ) ^{\prime }=\mu y+\frac{f\left ( x\right ) }{x} \tag{1} \end{equation} The corresponding homogeneous ODE \begin{equation} -\frac{1}{x}\left ( xy^{\prime }\right ) ^{\prime }=\lambda y \tag{2} \end{equation} Where p=x,q=0,r=x. This was solved in the textbook at page 707. The fundamental solution is given by y_{n}=\Phi _{n}\left ( x\right ) =J_{0}\left ( \sqrt{\lambda _{n}}x\right ) where the eigenvalues \lambda _{n} are the roots of J_{o}\left ( \sqrt{\lambda _{n}}\right ) =0. These eigenfunctions are not normalized. Therefore, the solution of the inhomogeneous ODE (1) can be now written as y\left ( x\right ) =\sum _{n=1}^{\infty }b_{n}\Phi _{n}\left ( x\right ) Using this in (1) gives -\frac{1}{x}\left ( xy^{\prime }\right ) ^{\prime }=\mu \sum _{n=1}^{\infty }b_{n}\Phi _{n}\left ( x\right ) +\sum _{n=1}^{\infty }c_{n}\Phi _{n}\left ( x\right ) But from (2), -\frac{1}{x}\left ( xy^{\prime }\right ) ^{\prime } can be replaced by \lambda y, so the above becomes\begin{equation} \sum _{n=1}^{\infty }\lambda _{n}b_{n}\Phi _{n}\left ( x\right ) =\mu \sum _{n=1}^{\infty }b_{n}\Phi _{n}\left ( x\right ) +\sum _{n=1}^{\infty }c_{n}\Phi _{n}\left ( x\right ) \tag{3} \end{equation} Where \sum _{n=1}^{\infty }c_{n}\Phi _{n}\left ( x\right ) =\frac{f\left ( x\right ) }{x} c_{n} is now found by orthogonality. Multiplying both sides of the above by r\left ( x\right ) \Phi _{m}\left ( x\right ) , where the weight r\left ( x\right ) =x, and integrating gives \begin{align*} \int _{0}^{1}x\frac{f\left ( x\right ) }{x}\Phi _{m}\left ( x\right ) dx & =\sum _{n=1}^{\infty }c_{n}\int _{0}^{1}x\Phi _{n}\left ( x\right ) \Phi _{m}\left ( x\right ) dx\\ \int _{0}^{1}f\left ( x\right ) \Phi _{m}\left ( x\right ) dx & =\sum _{n=1}^{\infty }c_{n}\int _{0}^{1}x\Phi _{n}\left ( x\right ) \Phi _{m}\left ( x\right ) dx \end{align*}
Due to orthogonality of the eigenfunctions, the above simplifies to\begin{equation} c_{n}=\frac{\int _{0}^{1}f\left ( x\right ) \Phi _{n}\left ( x\right ) dx}{\int _{0}^{1}x\Phi _{n}^{2}\left ( x\right ) dx} \tag{4} \end{equation} Since \Phi _{n}\left ( x\right ) is not normalized, \int _{0}^{1}x\Phi _{n}^{2}\left ( x\right ) dx can not be replaced by 1. The above is left as is. Substituting (4) in (3) and simplifying gives\begin{align*} \lambda _{n}b_{n} & =\mu b_{n}+c_{n}\\ b_{n} & =\frac{c_{n}}{\left ( \lambda _{n}-\mu \right ) } \end{align*}
Where \lambda _{n}\neq \mu . Hence the formal solution y=\sum _{n=1}^{\infty }b_{n}\Phi _{n}\left ( x\right ) can be written as y\left ( x\right ) =\sum _{n=1}^{\infty }\frac{c_{n}}{\left ( \lambda _{n}-\mu \right ) }J_{o}\left ( \sqrt{\lambda _{n}}x\right ) Using (4) in the above gives y\left ( x\right ) =\sum _{n=1}^{\infty }\left ( \frac{\int _{0}^{1}f\left ( x\right ) \Phi _{n}\left ( x\right ) dx}{\int _{0}^{1}x\Phi _{n}^{2}\left ( x\right ) dx}\right ) \frac{J_{o}\left ( \sqrt{\lambda _{n}}x\right ) }{\left ( \lambda _{n}-\mu \right ) }
Consider BVP -\left ( xy^{\prime }\right ) ^{\prime }=\lambda xy where y,y^{\prime } bounded as x\rightarrow 0 and y^{\prime }\left ( 1\right ) =0. (a) Show that \lambda _{0}=0 is eigenvalue corresponding to \Phi _{0}=1. If \lambda >0 show formally that the eigenfunctions are given by \Phi _{n}=J_{0}\left ( \sqrt{\lambda _{n}}x\right ) where \sqrt{\lambda _{n}} is the n^{th} positive root in increasing order of J_{0}^{\prime }\left ( \sqrt{\lambda _{n}}\right ) =0. It is possible to show there are infinite sequence of such roots.
(b) Show that if m=0,1,2,\cdots then \int _{0}^{1}x\Phi _{m}\left ( x\right ) \Phi _{n}\left ( x\right ) dx=0,m\neq n.
(c) Find formal solution to nonhomogeneous problem -\left ( xy^{\prime }\right ) ^{\prime }=\mu xy+f\left ( x\right ) , where y,y^{\prime } bounded as x\rightarrow 0 and y^{\prime }\left ( 1\right ) =0, where f is given continuous function on 0\leq x\leq 1 and \mu is not eigenvalue of the corresponding homogeneous ODE.
Solution
The given ODE can be written as \begin{equation} xy^{\prime \prime }+y^{\prime }+\lambda xy=0 \tag{1} \end{equation} Let t=\sqrt{\lambda }x, then \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{dy}{dt}\sqrt{\lambda } and \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left ( \frac{dy}{dt}\sqrt{\lambda }\right ) =\sqrt{\lambda }\frac{d^{2}y}{dt^{2}}\frac{dt}{dx}=\sqrt{\lambda }\frac{d^{2}y}{dt^{2}}\sqrt{\lambda }=\lambda \frac{d^{2}y}{dt^{2}}. Hence (1) becomes\begin{align*} \frac{t}{\sqrt{\lambda }}\lambda y^{\prime \prime }\left ( t\right ) +\sqrt{\lambda }y^{\prime }\left ( t\right ) +\lambda \frac{t}{\sqrt{\lambda }}y\left ( t\right ) & =0\\ t\sqrt{\lambda }y^{\prime \prime }\left ( t\right ) +\sqrt{\lambda }y^{\prime }\left ( t\right ) +\sqrt{\lambda }ty\left ( t\right ) & =0 \end{align*}
Since problem says that \lambda >0, then dividing by \sqrt{\lambda } the above simplifies to ty^{\prime \prime }\left ( t\right ) +y^{\prime }\left ( t\right ) +ty\left ( t\right ) =0 This is Bessel ODE of zero order. Its solution is y\left ( t\right ) =c_{1}J_{0}\left ( t\right ) +c_{2}Y_{0}\left ( t\right ) . Where J_{0}\left ( 0\right ) =0 and \lim _{t\rightarrow 0}Y_{0}\left ( t\right ) \rightarrow \infty . Hence a bounded solution requires that c_{2}=0. Therefore the solution becomes y\left ( t\right ) =c_{1}J_{0}\left ( t\right ) or in terms of x y\left ( x\right ) =c_{1}J_{0}\left ( \sqrt{\lambda }x\right ) To satisfy the second boundary condition, since y^{\prime }\left ( x\right ) =c_{1}J_{0}^{\prime }\left ( \sqrt{\lambda }x\right ) =-c_{1}J_{1}\left ( \sqrt{\lambda }x\right ) . Therefore the eigenvalues are roots of J_{1}\left ( \sqrt{\lambda }x\right ) =0 Plotting J_{1}\left ( \sqrt{\lambda }x\right ) shows that the first roots are \lambda =0. Numerically, the first few eigenvalues are\begin{equation} \lambda =\{0,14.682,49.2185,103,499,177.532,\cdots \} \tag{2} \end{equation} Hence the fundamental solution is y\left ( x\right ) =J_{0}\left ( \sqrt{\lambda _{n}}x\right ) where \lambda _{n} is given by above. When \lambda =0,J_{0}\left ( 0\right ) =1. Therefore the eigenfunction associated with \lambda =0 is \Phi _{0}\left ( x\right ) =1. Since there are infinite eigenvalues (2), there are infinite eigenfunctions \Phi _{n}\left ( x\right ) =J_{0}\left ( \sqrt{\lambda _{n}}x\right ) where n=0,1,2,3,\cdots
Let \Phi _{n}\left ( x\right ) ,\Phi _{m}\left ( x\right ) be any two eigenfunctions of \left ( xy^{\prime }\right ) ^{\prime }+\lambda xy=0. Therefore each satisfies the ODE. Hence\begin{align} \left ( x\Phi _{n}^{\prime }\right ) ^{\prime }+\lambda _{n}x\Phi _{n}\left ( x\right ) & =0\tag{3A}\\ \left ( x\Phi _{m}^{\prime }\right ) ^{\prime }+\lambda _{m}x\Phi _{m}\left ( x\right ) & =0 \tag{3B} \end{align}
Multiplying (3A) by \Phi _{m} and (3B) by \Phi _{n} and subtracting gives\begin{align*} \Phi _{m}\left ( x\Phi _{n}^{\prime }\right ) ^{\prime }+\lambda _{n}x\Phi _{m}\Phi _{n}\left ( x\right ) -\Phi _{n}\left ( x\Phi _{m}^{\prime }\right ) ^{\prime }-\lambda _{m}x\Phi _{n}\Phi _{m}\left ( x\right ) & =0\\ \Phi _{m}\left ( x\Phi _{n}^{\prime }\right ) ^{\prime }-\Phi _{n}\left ( x\Phi _{m}^{\prime }\right ) ^{\prime }+\left ( \lambda _{n}-\lambda _{m}\right ) x\Phi _{n}\Phi _{m}\left ( x\right ) & =0 \end{align*}
Integrating from 0\cdots 1 gives\begin{equation} \int _{0}^{1}\Phi _{m}\left ( x\Phi _{n}^{\prime }\right ) ^{\prime }dx-\int _{0}^{1}\Phi _{n}\left ( x\Phi _{m}^{\prime }\right ) ^{\prime }dx+\left ( \lambda _{n}-\lambda _{m}\right ) \int _{0}^{1}x\Phi _{n}\Phi _{m}\left ( x\right ) dx=0 \tag{4} \end{equation} Integrating \int _{0}^{1}\Phi _{m}\left ( x\Phi _{n}^{\prime }\right ) ^{\prime }dx by parts gives\begin{equation} \int _{0}^{1}\overset{u}{\overbrace{\Phi _{m}}}\overset{dv}{\overbrace{\left ( x\Phi _{n}^{\prime }\right ) ^{\prime }}}dx=\left [ \Phi _{m}x\Phi _{n}^{\prime }\right ] _{0}^{1}-\int _{0}^{1}\Phi _{m}^{\prime }\left ( x\Phi _{n}^{\prime }\right ) dx \tag{5A} \end{equation} And similarly, Integrating \int _{0}^{1}\Phi _{n}\left ( x\Phi _{m}^{\prime }\right ) ^{\prime }dx by parts gives\begin{equation} \int _{0}^{1}\overset{u}{\overbrace{\Phi _{n}}}\overset{dv}{\overbrace{\left ( x\Phi _{m}^{\prime }\right ) ^{\prime }}}dx=\left [ \Phi _{n}x\Phi _{m}^{\prime }\right ] _{0}^{1}-\int _{0}^{1}\Phi _{n}^{\prime }\left ( x\Phi _{m}^{\prime }\right ) dx \tag{5B} \end{equation} Substituting (5A,5B) back in (4) gives \left [ \Phi _{m}x\Phi _{n}^{\prime }\right ] _{0}^{1}-\int _{0}^{1}\Phi _{m}^{\prime }\left ( x\Phi _{n}^{\prime }\right ) dx-\left [ \Phi _{n}x\Phi _{m}^{\prime }\right ] _{0}^{1}+\int _{0}^{1}\Phi _{n}^{\prime }\left ( x\Phi _{m}^{\prime }\right ) dx+\left ( \lambda _{n}-\lambda _{m}\right ) \int _{0}^{1}x\Phi _{n}\Phi _{m}\left ( x\right ) dx=0 The above simplifies to\begin{equation} \left [ \Phi _{m}x\Phi _{n}^{\prime }-\Phi _{n}x\Phi _{m}^{\prime }\right ] _{0}^{1}+\left ( \lambda _{n}-\lambda _{m}\right ) \int _{0}^{1}x\Phi _{n}\Phi _{m}\left ( x\right ) dx=0 \tag{6} \end{equation} The boundary terms above simplifies to \left [ \Phi _{m}x\Phi _{n}^{\prime }-\Phi _{n}x\Phi _{m}^{\prime }\right ] _{0}^{1}=\left [ \Phi _{m}\left ( 1\right ) \Phi _{n}^{\prime }\left ( 1\right ) -\Phi _{n}\left ( 1\right ) \Phi _{m}^{\prime }\left ( 1\right ) \right ] But \Phi _{n}^{\prime }\left ( 1\right ) and \Phi _{m}^{\prime }\left ( 1\right ) are zero. This is because of the given boundary conditions y^{\prime }\left ( 1\right ) =0. Hence \left [ \Phi _{m}x\Phi _{n}^{\prime }-\Phi _{n}x\Phi _{m}^{\prime }\right ] _{0}^{1}=0. Therefore (6) now simplifies to \left ( \lambda _{n}-\lambda _{m}\right ) \int _{0}^{1}x\Phi _{n}\Phi _{m}\left ( x\right ) dx=0 But since \lambda _{n}-\lambda _{m}\neq 0, since these are different eigenvalues, then one concludes that \int _{0}^{1}x\Phi _{n}\Phi _{m}\left ( x\right ) dx=0 Which is the result asked to show.
The problem to solve is written as\begin{equation} -\frac{1}{x}\left ( xy^{\prime }\right ) ^{\prime }=\mu y+\frac{f\left ( x\right ) }{x} \tag{A} \end{equation} The solution to the corresponding homogeneous ODE -\frac{1}{x}\left ( xy^{\prime }\right ) ^{\prime }=\lambda y was found in part (a). Using eigenfunction expansion, the solution of the nonhomogeneous ODE (A) can then be written as \begin{equation} y\left ( x\right ) =\sum _{n=0}^{\infty }b_{n}\Phi _{n}\left ( x\right ) \tag{7} \end{equation} Where \Phi _{n}\left ( x\right ) =J_{0}\left ( \sqrt{\lambda _{n}}x\right ) ,n=0,1,2,\cdots and \lambda _{n} are roots of -J_{1}\left ( \sqrt{\lambda }\right ) =0. Using (7) in -\left ( xy^{\prime }\right ) ^{\prime }=\mu xy+f\left ( x\right ) gives -\frac{1}{x}\left ( xy^{\prime }\right ) ^{\prime }=x\sum _{n=0}^{\infty }b_{n}\Phi _{n}\left ( x\right ) +\sum _{n=0}^{\infty }c_{n}\Phi _{n}\left ( x\right ) But since -\frac{1}{x}\left ( xy^{\prime }\right ) ^{\prime }=\lambda y from part (a), then the above becomes \begin{equation} \sum _{n=0}^{\infty }\lambda _{n}b_{n}\Phi _{n}\left ( x\right ) =\mu \sum _{n=0}^{\infty }b_{n}\Phi _{n}\left ( x\right ) +\sum _{n=0}^{\infty }c_{n}\Phi _{n}\left ( x\right ) \tag{8} \end{equation} Where \sum _{n=0}^{\infty }c_{n}\Phi _{n}\left ( x\right ) =\frac{f\left ( x\right ) }{x} c_{n} is now found by orthogonality. Multiplying both sides of the above by r\left ( x\right ) \Phi _{m}\left ( x\right ) , where the weight r\left ( x\right ) =x, and integrating gives \begin{align} \int _{0}^{1}x\frac{f\left ( x\right ) }{x}\Phi _{m}\left ( x\right ) dx & =c_{0}\int _{0}^{1}x\Phi _{0}\left ( x\right ) \Phi _{m}\left ( x\right ) dx+\sum _{n=1}^{\infty }c_{n}\int _{0}^{1}x\Phi _{n}\left ( x\right ) \Phi _{m}\left ( x\right ) dx\nonumber \\ \int _{0}^{1}f\left ( x\right ) \Phi _{m}\left ( x\right ) dx & =c_{0}\int _{0}^{1}x\Phi _{0}\left ( x\right ) \Phi _{m}\left ( x\right ) dx+\sum _{n=1}^{\infty }c_{n}\int _{0}^{1}x\Phi _{n}\left ( x\right ) \Phi _{m}\left ( x\right ) dx \tag{9} \end{align}
For m=0, the eigenfunction is \Phi _{0}\left ( x\right ) =1, and the above becomes\begin{align*} \int _{0}^{1}f\left ( x\right ) dx & =c_{0}\int _{0}^{1}xdx\\ & =c_{0}\left [ \frac{x^{2}}{2}\right ] _{0}^{1}=\frac{c_{0}}{2} \end{align*}
Therefore \begin{equation} c_{0}=2\int _{0}^{1}f\left ( x\right ) dx \tag{10} \end{equation} For m>0, (9) becomes \int _{0}^{1}f\left ( x\right ) \Phi _{m}\left ( x\right ) dx=\sum _{n=1}^{\infty }c_{n}\int _{0}^{1}x\Phi _{n}\left ( x\right ) \Phi _{m}\left ( x\right ) dx Due to orthogonality of the eigenfunctions from part (b) \int _{0}^{1}x\Phi _{n}\left ( x\right ) \Phi _{m}\left ( x\right ) dx=0 for m\neq n\,, and the above simplifies to\begin{equation} c_{n}=\frac{\int _{0}^{1}f\left ( x\right ) \Phi _{n}\left ( x\right ) dx}{\int _{0}^{1}x\Phi _{n}^{2}\left ( x\right ) dx} \tag{11} \end{equation} Since \Phi _{n}\left ( x\right ) is not normalized, \int _{0}^{1}x\Phi _{n}^{2}\left ( x\right ) dx can not be replaced by 1. The above is left as is. Substituting (10,11) in (8) and simplifying gives\begin{equation} \sum _{n=0}^{\infty }\lambda _{n}b_{n}\Phi _{n}\left ( x\right ) =\mu \sum _{n=0}^{\infty }b_{n}\Phi _{n}\left ( x\right ) +\sum _{n=0}^{\infty }c_{n}\Phi _{n}\left ( x\right ) \tag{12} \end{equation} For n=0 only, and since \lambda _{n}=0 then (12) gives 0=\mu b_{0}\Phi _{0}\left ( x\right ) +c_{0}\Phi _{0}\left ( x\right ) But \Phi _{0}\left ( x\right ) =1, hence\begin{align*} 0 & =\mu b_{0}+c_{0}\\ b_{0} & =-\frac{c_{0}}{\mu } \end{align*}
For n>0, then (12) gives\begin{align*} \sum _{n=1}^{\infty }\lambda _{n}b_{n}\Phi _{n}\left ( x\right ) & =\mu \sum _{n=1}^{\infty }b_{n}\Phi _{n}\left ( x\right ) +\sum _{n=1}^{\infty }c_{n}\Phi _{n}\left ( x\right ) \\ \lambda _{n}b_{n} & =\mu b_{n}+c_{n}\\ b_{n} & =\frac{c_{n}}{\left ( \lambda _{n}-\mu \right ) } \end{align*}
Where \lambda _{n}\neq \mu . Hence the formal solution y=\sum _{n=0}^{\infty }b_{n}\Phi _{n}\left ( x\right ) can be written as\begin{align*} y\left ( x\right ) & =b_{0}\Phi _{0}\left ( x\right ) +\sum _{n=1}^{\infty }b_{n}\Phi _{0}\left ( x\right ) \\ & =-\frac{c_{0}}{\mu }+\sum _{n=1}^{\infty }\frac{c_{n}}{\left ( \lambda _{n}-\mu \right ) }J_{o}\left ( \sqrt{\lambda _{n}}x\right ) \\ & =-\frac{2}{\mu }\int _{0}^{1}f\left ( x\right ) dx+\sum _{n=1}^{\infty }\frac{1}{\left ( \lambda _{n}-\mu \right ) }\frac{\int _{0}^{1}f\left ( x\right ) J_{o}\left ( \sqrt{\lambda _{n}}x\right ) dx}{\int _{0}^{1}xJ_{o}^{2}\left ( \sqrt{\lambda _{n}}x\right ) dx}J_{o}\left ( \sqrt{\lambda _{n}}x\right ) \end{align*}
But \int _{0}^{1}xJ_{o}^{2}\left ( \sqrt{\lambda _{n}}x\right ) dx=\frac{1}{2}\left ( J_{o}^{2}\left ( \sqrt{\lambda _{n}}\right ) +J_{1}^{2}\left ( \sqrt{\lambda _{n}}\right ) \right ) , hence the above becomes y\left ( x\right ) =-\frac{2}{\mu }\int _{0}^{1}f\left ( x\right ) dx+2\sum _{n=1}^{\infty }\frac{1}{\left ( \lambda _{n}-\mu \right ) }\frac{\int _{0}^{1}f\left ( x\right ) J_{o}\left ( \sqrt{\lambda _{n}}x\right ) dx}{J_{o}^{2}\left ( \sqrt{\lambda _{n}}\right ) +J_{1}^{2}\left ( \sqrt{\lambda _{n}}\right ) }J_{o}\left ( \sqrt{\lambda _{n}}x\right )
Consider -\left ( xy^{\prime }\right ) ^{\prime }+\frac{k^{2}}{x}y=\lambda xy. with y,y^{\prime } bounded as x\rightarrow 0 and y\left ( 1\right ) =0, where k is positive integer. (a) using t=\sqrt{\lambda }x show the ODE reduces to Bessel of order k. (b) show formally that the eigenvalues \lambda _{1},\lambda _{2},\cdots of the given differential equation are the squares of positive zeros of J_{k}\left ( \sqrt{\lambda }\right ) and that the corresponding eigenfunctions are \Phi _{n}\left ( x\right ) =J_{k}\left ( \sqrt{\lambda }x\right ) . It is possible to show there as infinite sequence of such zeros. (c) Show that the eigenfunctions \Phi _{n}\left ( x\right ) satisfy the orthogonality relation
\int _{0}^{1}x\Phi _{m}\left ( x\right ) \Phi _{n}\left ( x\right ) dx=0\qquad m\neq n (d) Determine the coefficients of the formal series expansion f\left ( x\right ) =\sum _{n=1}^{\infty }a_{n}\Phi _{n}\left ( x\right ) . (e) Final formal solution of the nonhomogeneous problem -\left ( xy^{\prime }\right ) ^{\prime }+\frac{k^{2}}{x}y=\mu xy+f\left ( x\right ) With y,y^{\prime } bounded as x\rightarrow 0 and y\left ( 1\right ) =0, where f is given continuous function on 0\leq x\leq 1 and \mu is eigenvalue of the corresponding homogeneous problem.
Solution
The ODE to solve is -\left ( xy^{\prime }\right ) ^{\prime }+\frac{k^{2}}{x}y-\lambda xy=0 Note: The problem seems to not have mentioned that \lambda >0 here as well, as in the problem above it. This condition is needed to fully solve this problem with y,y^{\prime } bounded as x\rightarrow 0 and y\left ( 1\right ) =0. The ODE can be written as\begin{align} -xy^{\prime \prime }-y^{\prime }+y\left ( \frac{k^{2}}{x}-\lambda x\right ) & =0\nonumber \\ xy^{\prime \prime }+y^{\prime }+y\left ( \lambda x-\frac{k^{2}}{x}\right ) & =0 \tag{1} \end{align}
Let t=\sqrt{\lambda }x, then \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{dy}{dt}\sqrt{\lambda } and \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left ( \frac{dy}{dt}\sqrt{\lambda }\right ) =\sqrt{\lambda }\frac{d^{2}y}{dt^{2}}\frac{dt}{dx}=\sqrt{\lambda }\frac{d^{2}y}{dt^{2}}\sqrt{\lambda }=\lambda \frac{d^{2}y}{dt^{2}}. Hence (1) becomes\begin{align*} \frac{t}{\sqrt{\lambda }}\lambda y^{\prime \prime }\left ( t\right ) +\sqrt{\lambda }y^{\prime }\left ( t\right ) +y\left ( t\right ) \left ( \lambda \frac{t}{\sqrt{\lambda }}-\frac{k^{2}}{t}\sqrt{\lambda }\right ) & =0\\ t\sqrt{\lambda }y^{\prime \prime }\left ( t\right ) +\sqrt{\lambda }y^{\prime }\left ( t\right ) +\sqrt{\lambda }y\left ( t\right ) \left ( t-\frac{k^{2}}{t}\right ) & =0\\ t^{2}y^{\prime \prime }+ty^{\prime }+\left ( t^{2}-k^{2}\right ) y & =0 \end{align*}
This is Bessel ODE of k order.
The solution to the above ODE is known to be y\left ( t\right ) =c_{1}J_{k}\left ( t\right ) +c_{2}Y_{k}\left ( t\right ) Where J_{k}\left ( 0\right ) =0 and \lim _{t\rightarrow 0}Y_{k}\left ( t\right ) \rightarrow \infty . Hence a bounded solution requires that c_{2}=0. Therefore the solution becomes y\left ( t\right ) =c_{1}J_{k}\left ( t\right ) Or in terms of x y\left ( x\right ) =c_{1}J_{k}\left ( \sqrt{\lambda }x\right ) To satisfy the second boundary condition y\left ( 1\right ) =0 gives c_{1}J_{k}\left ( \sqrt{\lambda }\right ) =0 Non-trivial solution implies J_{k}\left ( \sqrt{\lambda }\right ) =0. Therefore the eigenvalues are the square of positive roots of this equation. Even though there are negative and positive roots for J_{k}\left ( \sqrt{\lambda }\right ) =0 but for real root, \lambda must be non-negative. It assumed \lambda >0. There are infinite number of positive roots for J_{k}\left ( \sqrt{\lambda }\right ) =0. Hence the eigenfunctions are \Phi _{n}\left ( x\right ) =J_{k}\left ( \sqrt{\lambda _{n}}x\right ) \qquad n=1,2,3,\cdots Where \lambda _{n} are square of the all positive zeros of J_{k}\left ( \sqrt{\lambda }\right ) =0.
Show that the eigenfunctions \Phi _{n}\left ( x\right ) satisfy the orthogonality relation \int _{0}^{1}x\Phi _{m}\left ( x\right ) \Phi _{n}\left ( x\right ) dx=0\qquad m\neq n Let \Phi _{n}\left ( x\right ) ,\Phi _{m}\left ( x\right ) be any two eigenfunctions of -\left ( xy^{\prime }\right ) ^{\prime }+\frac{k^{2}}{x}y=\lambda xy where now \Phi _{n}\left ( x\right ) =J_{k}\left ( \sqrt{\lambda _{n}}x\right ) and \Phi _{m}\left ( x\right ) =J_{k}\left ( \sqrt{\lambda _{m}}x\right ) . Therefore each satisfies the ODE. Hence\begin{align} -\left ( x\Phi _{n}^{\prime }\right ) ^{\prime }+\frac{k^{2}}{x}\Phi _{n}\left ( x\right ) -\lambda _{n}x\Phi _{n}\left ( x\right ) & =0\tag{3A}\\ -\left ( x\Phi _{m}^{\prime }\right ) ^{\prime }+\frac{k^{2}}{x}\Phi _{m}\left ( x\right ) -\lambda _{m}x\Phi _{m}\left ( x\right ) & =0 \tag{3B} \end{align}
Multiplying 3A by \Phi _{m} and 3B by \Phi _{n} and subtracting gives\begin{align*} -\Phi _{m}\left ( x\Phi _{n}^{\prime }\right ) ^{\prime }+\Phi _{m}\frac{k^{2}}{x}\Phi _{n}\left ( x\right ) -\lambda _{n}x\Phi _{m}\Phi _{n}\left ( x\right ) -\left ( -\left ( \Phi _{n}x\Phi _{m}^{\prime }\right ) ^{\prime }+\frac{k^{2}}{x}\Phi _{n}\Phi _{m}\left ( x\right ) -\lambda _{m}x\Phi _{n}\Phi _{m}\left ( x\right ) \right ) & =0\\ -\Phi _{m}\left ( x\Phi _{n}^{\prime }\right ) ^{\prime }+\frac{k^{2}}{x}\Phi _{m}\Phi _{n}\left ( x\right ) -\lambda _{n}x\Phi _{m}\Phi _{n}\left ( x\right ) +\Phi _{n}\left ( x\Phi _{m}^{\prime }\right ) ^{\prime }-\frac{k^{2}}{x}\Phi _{n}\Phi _{m}\left ( x\right ) +\lambda _{m}x\Phi _{n}\Phi _{m}\left ( x\right ) & =0\\ -\left ( x\Phi _{n}^{\prime }\right ) ^{\prime }+\left ( x\Phi _{m}^{\prime }\right ) ^{\prime }+\left ( \lambda _{m}-\lambda _{n}\right ) x\Phi _{n}\Phi _{m}\left ( x\right ) & =0 \end{align*}
Integrating from 0\cdots 1 gives\begin{equation} \int _{0}^{1}\Phi _{m}\left ( x\Phi _{n}^{\prime }\right ) ^{\prime }dx-\int _{0}^{1}\Phi _{n}\left ( x\Phi _{m}^{\prime }\right ) ^{\prime }dx+\left ( \lambda _{n}-\lambda _{m}\right ) \int _{0}^{1}x\Phi _{n}\Phi _{m}\left ( x\right ) dx=0 \tag{4} \end{equation} Integrating \int _{0}^{1}\Phi _{m}\left ( x\Phi _{n}^{\prime }\right ) ^{\prime }dx by parts gives\begin{equation} \int _{0}^{1}\overset{u}{\overbrace{\Phi _{m}}}\overset{dv}{\overbrace{\left ( x\Phi _{n}^{\prime }\right ) ^{\prime }}}dx=\left [ \Phi _{m}x\Phi _{n}^{\prime }\right ] _{0}^{1}-\int _{0}^{1}\Phi _{m}^{\prime }\left ( x\Phi _{n}^{\prime }\right ) dx \tag{5A} \end{equation} And similarly, Integrating \int _{0}^{1}\Phi _{n}\left ( x\Phi _{m}^{\prime }\right ) ^{\prime }dx by parts gives\begin{equation} \int _{0}^{1}\overset{u}{\overbrace{\Phi _{n}}}\overset{dv}{\overbrace{\left ( x\Phi _{m}^{\prime }\right ) ^{\prime }}}dx=\left [ \Phi _{n}x\Phi _{m}^{\prime }\right ] _{0}^{1}-\int _{0}^{1}\Phi _{n}^{\prime }\left ( x\Phi _{m}^{\prime }\right ) dx \tag{5B} \end{equation} Substituting (5A,5B) back in (4) gives \left [ \Phi _{m}x\Phi _{n}^{\prime }\right ] _{0}^{1}-\int _{0}^{1}\Phi _{m}^{\prime }\left ( x\Phi _{n}^{\prime }\right ) dx-\left [ \Phi _{n}x\Phi _{m}^{\prime }\right ] _{0}^{1}+\int _{0}^{1}\Phi _{n}^{\prime }\left ( x\Phi _{m}^{\prime }\right ) dx+\left ( \lambda _{n}-\lambda _{m}\right ) \int _{0}^{1}x\Phi _{n}\Phi _{m}\left ( x\right ) dx=0 The above simplifies to\begin{equation} \left [ \Phi _{m}x\Phi _{n}^{\prime }-\Phi _{n}x\Phi _{m}^{\prime }\right ] _{0}^{1}+\left ( \lambda _{n}-\lambda _{m}\right ) \int _{0}^{1}x\Phi _{n}\Phi _{m}\left ( x\right ) dx=0 \tag{6} \end{equation} Let \Delta =\left [ \Phi _{m}x\Phi _{n}^{\prime }-\Phi _{n}x\Phi _{m}^{\prime }\right ] _{0}^{1}, then the boundary terms above simplifies to \Delta =\left [ \Phi _{m}\left ( 1\right ) \Phi _{n}^{\prime }\left ( 1\right ) -\Phi _{n}\left ( 1\right ) \Phi _{m}^{\prime }\left ( 1\right ) \right ] -\lim _{x\rightarrow 0}\left [ x\Phi _{m}\left ( x\right ) \Phi _{n}^{\prime }\left ( x\right ) -x\Phi _{n}\left ( x\right ) \Phi _{m}^{\prime }\left ( x\right ) \right ] But \Phi _{n}\left ( 1\right ) and \Phi _{m}\left ( 1\right ) are zero. This is because of the given boundary conditions. Hence the above simplifies to \left [ \Phi _{m}x\Phi _{n}^{\prime }-\Phi _{n}x\Phi _{m}^{\prime }\right ] _{0}^{1}=-\lim _{x\rightarrow 0}\left ( x\left ( \Phi _{m}\left ( x\right ) \Phi _{n}^{\prime }\left ( x\right ) -\Phi _{n}\left ( x\right ) \Phi _{m}^{\prime }\left ( x\right ) \right ) \right ) But since both \Phi _{m}\left ( x\right ) ,\Phi _{n}\left ( x\right ) ,\Phi _{n}^{\prime }\left ( x\right ) ,\Phi _{m}^{\prime }\left ( x\right ) are bounded as x\rightarrow 0 then the above vanishes. This means the all the boundary terms are zero and (6) simplifies to \left ( \lambda _{n}-\lambda _{m}\right ) \int _{0}^{1}x\Phi _{n}\Phi _{m}\left ( x\right ) dx=0 But since \lambda _{n}-\lambda _{m}\neq 0, since these are different eigenvalues, therefore \int _{0}^{1}x\Phi _{n}\Phi _{m}\left ( x\right ) dx=0 Which is the result asked to show.
This is both parts combined. To solve -\left ( xy^{\prime }\right ) ^{\prime }+\frac{k^{2}}{x}y=\mu xy+f\left ( x\right ) , we start with dividing by x to get the ODE to the form\begin{equation} -\frac{1}{x}\left ( xy^{\prime }\right ) ^{\prime }+\frac{k^{2}}{x^{2}}y=\mu y+\frac{f\left ( x\right ) }{x} \tag{1} \end{equation} The homogeneous ode -\frac{1}{x}\left ( xy^{\prime }\right ) ^{\prime }+\frac{k^{2}}{x^{2}}y=\lambda y was solved in part (a,b). And since the problem says that \lambda \neq \mu , then the solution to the above nonhomogeneous ODE is\begin{equation} y\left ( x\right ) =\sum _{n=1}^{\infty }b_{n}\Phi _{n}\left ( x\right ) \tag{1} \end{equation} Where \Phi _{n}\left ( x\right ) are eigenfunctions of the homogeneous ODE found above to be \Phi _{n}\left ( x\right ) =J_{k}\left ( \sqrt{\lambda _{n}}x\right ) \qquad n=1,2,3,\cdots Substituting (2) in RHS of (1) gives -\frac{1}{x}\left ( xy^{\prime }\right ) ^{\prime }+\frac{k^{2}}{x^{2}}y=\mu \sum _{n=1}^{\infty }b_{n}\Phi _{n}\left ( x\right ) +\sum _{n=1}^{\infty }c_{n}\Phi _{n}\left ( x\right ) Where \sum _{n=1}^{\infty }c_{n}\Phi _{n}\left ( x\right ) =\frac{f\left ( x\right ) }{x}. But -\frac{1}{x}\left ( xy^{\prime }\right ) ^{\prime }+\frac{k^{2}}{x^{2}}y=\lambda y from part (a,b). Therefore the above becomes \sum _{n=1}^{\infty }\lambda _{n}b_{n}\Phi _{n}\left ( x\right ) =\mu \sum _{n=1}^{\infty }b_{n}\Phi _{n}\left ( x\right ) +\sum _{n=1}^{\infty }c_{n}\Phi _{n}\left ( x\right ) Or\begin{align*} \lambda _{n}b_{n} & =\mu b_{n}+c_{n}\\ b_{n} & =\frac{c_{n}}{\lambda _{n}-\mu } \end{align*}
What is left is to to find c_{n} (called a_{n} in this problem). Since \sum _{n=1}^{\infty }c_{n}\Phi _{n}\left ( x\right ) =\frac{f\left ( x\right ) }{x}, then applying orthogonality gives c_{n}\int _{0}^{1}r\left ( x\right ) \Phi _{n}^{2}\left ( x\right ) dx=\int _{0}^{1}r\left ( x\right ) \frac{f\left ( x\right ) }{x}\Phi _{n}\left ( x\right ) dx But r\left ( x\right ) =x, and the above becomes\begin{align*} c_{n}\int _{0}^{1}xJ_{k}^{2}\left ( \sqrt{\lambda _{n}}x\right ) dx & =\int _{0}^{1}f\left ( x\right ) J_{k}\left ( \sqrt{\lambda _{n}}x\right ) dx\\ c_{n} & =\frac{\int _{0}^{1}f\left ( x\right ) J_{k}\left ( \sqrt{\lambda _{n}}x\right ) dx}{\int _{0}^{1}xJ_{k}^{2}\left ( \sqrt{\lambda _{n}}x\right ) dx} \end{align*}
This complete the solution.\begin{align*} y\left ( x\right ) & =\sum _{n=1}^{\infty }b_{n}J_{k}\left ( \sqrt{\lambda _{n}}x\right ) \\ & =\sum _{n=1}^{\infty }\frac{c_{n}}{\lambda _{n}-\mu }J_{k}\left ( \sqrt{\lambda _{n}}x\right ) \\ & =\sum _{n=1}^{\infty }\frac{\int _{0}^{1}f\left ( x\right ) J_{k}\left ( \sqrt{\lambda _{n}}x\right ) dx}{\int _{0}^{1}xJ_{k}^{2}\left ( \sqrt{\lambda _{n}}x\right ) dx}\frac{J_{k}\left ( \sqrt{\lambda _{n}}x\right ) }{\lambda _{n}-\mu } \end{align*}
Consider Legendre equation -\left ( \left ( 1-x^{2}\right ) y^{\prime }\right ) ^{\prime }=\lambda y subject to boundary conditions y\left ( 0\right ) =0 with y,y^{\prime } bounded as x\rightarrow 1 and \Phi _{1}\left ( x\right ) =P_{1}\left ( x\right ) , \Phi _{2}\left ( x\right ) =P_{3}\left ( x\right ) , \Phi _{n}\left ( x\right ) =P_{2n-1}\left ( x\right ) corresponding to eigenvalues \lambda _{1}=2,\lambda _{2}=4\cdot 3,\cdots ,\lambda _{n}=2n\left ( 2n-1\right ) . (a) Show that the eigenfunctions \Phi _{n}\left ( x\right ) satisfy the orthogonality relation \int _{0}^{1}\Phi _{m}\left ( x\right ) \Phi _{n}\left ( x\right ) dx=0\qquad m\neq n (b) Final formal solution of the nonhomogeneous problem -\left ( \left ( 1-x^{2}\right ) y^{\prime }\right ) ^{\prime }=\mu y+f\left ( x\right ) where y\left ( 0\right ) =0 with y,y^{\prime } bounded as x\rightarrow 1 where f\left ( x\right ) is continuous function on 0\leq x\leq 1 and \mu is not eigenvalue of -\left ( \left ( 1-x^{2}\right ) y^{\prime }\right ) ^{\prime }=\lambda y
Solution
Let \Phi _{n}\left ( x\right ) ,\Phi _{m}\left ( x\right ) be any two eigenfunctions of -\left ( \left ( 1-x^{2}\right ) y^{\prime }\right ) ^{\prime }=\lambda y associated with eigenvalues \lambda _{n},\lambda _{m}, where \Phi _{n}\left ( x\right ) =P_{n}\left ( x\right ) and \Phi _{m}\left ( x\right ) =P_{m}\left ( x\right ) . Therefore each satisfies the ODE. Hence\begin{align} \left ( \left ( 1-x^{2}\right ) \Phi _{n}^{\prime }\left ( x\right ) \right ) ^{\prime }+\lambda _{n}\Phi _{n} & =0\tag{3A}\\ \left ( \left ( 1-x^{2}\right ) \Phi _{m}^{\prime }\left ( x\right ) \right ) ^{\prime }+\lambda _{m}\Phi _{m} & =0 \tag{3B} \end{align}
Multiplying 3A by \Phi _{m} and 3B by \Phi _{n} and subtracting gives\begin{align*} \Phi _{m}\left ( \left ( 1-x^{2}\right ) \Phi _{n}^{\prime }\left ( x\right ) \right ) ^{\prime }+\lambda _{n}\Phi _{m}\Phi _{n}-\left ( \Phi _{n}\left ( \left ( 1-x^{2}\right ) \Phi _{m}^{\prime }\left ( x\right ) \right ) ^{\prime }+\lambda _{m}\Phi _{n}\Phi _{m}\right ) & =0\\ \Phi _{m}\left ( \left ( 1-x^{2}\right ) \Phi _{n}^{\prime }\left ( x\right ) \right ) ^{\prime }-\Phi _{n}\left ( \left ( 1-x^{2}\right ) \Phi _{m}^{\prime }\left ( x\right ) \right ) ^{\prime }+\left ( \lambda _{n}-\lambda _{m}\right ) \Phi _{n}\Phi _{m} & =0 \end{align*}
Integrating from 0\cdots 1 gives (all upper limits below show be \lim _{\varepsilon \rightarrow 0^{-}}\int _{0}^{1-\varepsilon } instead of \int _{0}^{1} but to simplify notation, the latter is used and at the end, it is switched back to former. \begin{equation} \int _{0}^{1}\Phi _{m}\left ( \left ( 1-x^{2}\right ) \Phi _{n}^{\prime }\left ( x\right ) \right ) ^{\prime }dx-\int _{0}^{1}\Phi _{n}\left ( \left ( 1-x^{2}\right ) \Phi _{m}^{\prime }\left ( x\right ) \right ) ^{\prime }dx+\left ( \lambda _{n}-\lambda _{m}\right ) \int _{0}^{1}\Phi _{n}\Phi _{m}\left ( x\right ) dx=0 \tag{4} \end{equation} The first integral in (4) \int _{0}^{1}\overset{u}{\overbrace{\Phi _{m}}}\overset{dv}{\overbrace{\left ( \left ( 1-x^{2}\right ) \Phi _{n}^{\prime }\left ( x\right ) \right ) ^{\prime }}}dx is integrated by parts, giving
\begin{align} \int _{0}^{1}\Phi _{m}\left ( \left ( 1-x^{2}\right ) \Phi _{n}^{\prime }\left ( x\right ) \right ) ^{\prime }dx & =\left [ \Phi _{m}\left ( 1-x^{2}\right ) \Phi _{n}^{\prime }\left ( x\right ) \right ] _{0}^{1}-\int _{0}^{1}\Phi _{m}^{\prime }\left ( \left ( 1-x^{2}\right ) \Phi _{n}^{\prime }\left ( x\right ) \right ) dx\nonumber \\ & =\left [ \Phi _{m}\left ( 1-x^{2}\right ) \Phi _{n}^{\prime }\left ( x\right ) \right ] _{0}^{1}-\int _{0}^{1}\Phi _{n}^{\prime }\left ( \left ( 1-x^{2}\right ) \Phi _{m}^{\prime }\left ( x\right ) \right ) dx \tag{4A} \end{align}
Similarly, the second integral in (4) \int _{0}^{1}\overset{u}{\overbrace{\Phi _{n}}}\overset{dv}{\overbrace{\left ( \left ( 1-x^{2}\right ) \Phi _{m}^{\prime }\left ( x\right ) \right ) ^{\prime }}}dx is integrated by parts, giving
\begin{align} \int _{0}^{1}\Phi _{n}\left ( \left ( 1-x^{2}\right ) \Phi _{m}^{\prime }\left ( x\right ) \right ) ^{\prime }dx & =\left [ \Phi _{n}\left ( 1-x^{2}\right ) \Phi _{m}^{\prime }\left ( x\right ) \right ] _{0}^{1}-\int _{0}^{1}\Phi _{n}^{\prime }\left ( \left ( 1-x^{2}\right ) \Phi _{m}^{\prime }\left ( x\right ) \right ) dx\nonumber \\ & =\left [ \Phi _{m}\left ( 1-x^{2}\right ) \Phi _{n}^{\prime }\left ( x\right ) \right ] _{0}^{1}-\int _{0}^{1}\Phi _{n}^{\prime }\left ( \left ( 1-x^{2}\right ) \Phi _{m}^{\prime }\left ( x\right ) \right ) dx \tag{4B} \end{align}
Substituting (4A) and (4B) back into (4) gives
 |
Let \Delta =\left [ \Phi _{m}\left ( 1-x^{2}\right ) \Phi _{n}^{\prime }\left ( x\right ) -\Phi _{m}\left ( 1-x^{2}\right ) \Phi _{n}^{\prime }\left ( x\right ) \right ] _{0}^{1}. The boundary terms above are evaluated as follows \Delta =\lim _{x\rightarrow 1}\left [ \Phi _{m}\left ( x\right ) \left ( 1-x^{2}\right ) \Phi _{n}^{\prime }\left ( x\right ) -\Phi _{m}\left ( x\right ) \left ( 1-x^{2}\right ) \Phi _{n}^{\prime }\left ( x\right ) \right ] -\left ( \Phi _{m}\left ( 0\right ) \Phi _{n}^{\prime }\left ( 0\right ) -\Phi _{m}\left ( 0\right ) \Phi _{n}^{\prime }\left ( 0\right ) \right ) Since \Phi _{m}\left ( 0\right ) =0,\Phi _{m}\left ( 0\right ) =0, the above simplifies to\begin{align*} \Delta & =\lim _{x\rightarrow 1}\left [ \Phi _{m}\left ( x\right ) \left ( 1-x^{2}\right ) \Phi _{n}^{\prime }\left ( x\right ) -\Phi _{m}\left ( x\right ) \left ( 1-x^{2}\right ) \Phi _{n}^{\prime }\left ( x\right ) \right ] \\ & =\lim _{x\rightarrow 1}\left ( 1-x^{2}\right ) \left [ \Phi _{m}\left ( x\right ) \Phi _{n}^{\prime }\left ( x\right ) -\Phi _{m}\left ( x\right ) \Phi _{n}^{\prime }\left ( x\right ) \right ] \end{align*}
Since \Phi _{m}\left ( x\right ) ,\Phi _{n}^{\prime }\left ( x\right ) ,\Phi _{m}\left ( x\right ) \Phi _{n}^{\prime }\left ( x\right ) are all bounded as x\rightarrow 1 then the above goes to zero in the limit. Which means all boundary conditions term vanish. Hence (5) reduces to \left ( \lambda _{n}-\lambda _{m}\right ) \int _{0}^{1}\Phi _{n}\Phi _{m}\left ( x\right ) dx=0 But since \lambda _{n}-\lambda _{m}\neq 0, since these are different eigenvalues, therefore \int _{0}^{1}\Phi _{n}\Phi _{m}\left ( x\right ) dx=0 Which is the result asked to show.
Since \lambda \neq \mu , then the the solution to nonhomogeneous ODE is\begin{equation} y\left ( x\right ) =\sum _{n=1}^{\infty }b_{n}\Phi _{n}\left ( x\right ) \tag{1} \end{equation} Where \Phi _{n}\left ( x\right ) are eigenfunctions \Phi _{n}\left ( x\right ) =P_{\left ( 2n-1\right ) }\left ( x\right ) . Substituting (1) in -\left ( \left ( 1-x^{2}\right ) y^{\prime }\right ) ^{\prime }=\mu y+f\left ( x\right ) gives -\left ( \left ( 1-x^{2}\right ) y^{\prime }\right ) ^{\prime }=\mu \sum _{n=1}^{\infty }b_{n}\Phi _{n}\left ( x\right ) +\sum _{n=1}^{\infty }c_{n}\Phi _{n}\left ( x\right ) Where \sum _{n=1}^{\infty }c_{n}\Phi _{n}\left ( x\right ) =f\left ( x\right ) . But -\left ( \left ( 1-x^{2}\right ) y^{\prime }\right ) ^{\prime }=\lambda y, therefore the above becomes \sum _{n=1}^{\infty }\lambda _{n}b_{n}\Phi _{n}\left ( x\right ) =\mu \sum _{n=1}^{\infty }b_{n}\Phi _{n}\left ( x\right ) +\sum _{n=1}^{\infty }c_{n}\Phi _{n}\left ( x\right ) Or\begin{align*} \lambda _{n}b_{n} & =\mu b_{n}+c_{n}\\ b_{n} & =\frac{c_{n}}{\lambda _{n}-\mu } \end{align*}
What is left is to to find c_{n}. Since \sum _{n=1}^{\infty }c_{n}\Phi _{n}\left ( x\right ) =f\left ( x\right ) , then applying orthogonality gives c_{n}\int _{0}^{1}r\left ( x\right ) \Phi _{n}^{2}\left ( x\right ) dx=\int _{0}^{1}r\left ( x\right ) f\left ( x\right ) \Phi _{n}\left ( x\right ) dx But r\left ( x\right ) =1, and the above becomes\begin{align*} c_{n}\int _{0}^{1}P_{\left ( 2n-1\right ) }^{2}\left ( x\right ) dx & =\int _{0}^{1}f\left ( x\right ) P_{\left ( 2n-1\right ) }\left ( x\right ) dx\\ c_{n} & =\frac{\int _{0}^{1}f\left ( x\right ) P_{\left ( 2n-1\right ) }\left ( x\right ) dx}{\int _{0}^{1}P_{\left ( 2n-1\right ) }^{2}\left ( x\right ) dx} \end{align*}
This complete the solution.\begin{align*} y\left ( x\right ) & =\sum _{n=1}^{\infty }b_{n}P_{\left ( 2n-1\right ) }\left ( x\right ) \\ & =\sum _{n=1}^{\infty }\frac{c_{n}}{\lambda _{n}-\mu }P_{\left ( 2n-1\right ) }\left ( x\right ) \\ & =\sum _{n=1}^{\infty }\frac{\int _{0}^{1}f\left ( x\right ) P_{\left ( 2n-1\right ) }\left ( x\right ) dx}{\int _{0}^{1}P_{\left ( 2n-1\right ) }^{2}\left ( x\right ) dx}\frac{P_{\left ( 2n-1\right ) }\left ( x\right ) }{\lambda _{n}-\mu } \end{align*}
Equation \left ( 1-x^{2}\right ) y^{\prime \prime }-xy^{\prime }+\lambda y=0 is Chebyshev’s equation. (a) show it can be written as -\left ( \sqrt{1-x^{2}}y^{\prime }\right ) ^{\prime }=\frac{\lambda }{\sqrt{1-x^{2}}}y\qquad -1<x<1 (b) consider boundary conditions y,y^{\prime } bounded as x\rightarrow -1 and x\rightarrow +1. Show that the problem is self adjoint. (c) Show that \int _{-1}^{1}\frac{T_{m}\left ( x\right ) T_{n}\left ( x\right ) }{\sqrt{1-x^{2}}}dx=0 Where T_{n}\left ( x\right ) are the eigenfunctions :T_{0}\left ( x\right ) =1,T_{1}\left ( x\right ) =x,T_{2}\left ( x\right ) =1-2x^{2},\cdots and eigenvalues are \lambda _{n}=n^{2} for n=0,1,2,\cdots
Solution
Writing the ODE \left ( 1-x^{2}\right ) y^{\prime \prime }-xy^{\prime }+\lambda y=0 as P\left ( x\right ) y^{\prime \prime }+Q\left ( x\right ) y^{\prime }+R\left ( x\right ) y=0 Where P\left ( x\right ) =\left ( 1-x^{2}\right ) ,Q\left ( x\right ) =-x,R\left ( x\right ) =\lambda , then the integrating factor is \begin{align*} \mu & =\frac{1}{P}e^{\int \frac{Q\left ( x\right ) }{P\left ( x\right ) }dx}\\ & =\frac{1}{\left ( 1-x^{2}\right ) }e^{\int \frac{-x}{\left ( 1-x^{2}\right ) }dx} \end{align*}
But \int \frac{x}{\left ( 1-x^{2}\right ) }dx=\frac{1}{2}\ln \left \vert 1-x^{2}\right \vert , therefore e^{\frac{1}{2}\ln \left \vert 1-x^{2}\right \vert }=\sqrt{1-x^{2}} and the above becomes \mu =\frac{1}{\sqrt{1-x^{2}}}. Hence the SL form is\begin{align*} \left ( \mu Py^{\prime }\right ) ^{\prime }+\mu R\left ( x\right ) y & =0\\ \left ( \frac{1}{\sqrt{1-x^{2}}}\left ( 1-x^{2}\right ) y^{\prime }\right ) ^{\prime }+\frac{1}{\sqrt{1-x^{2}}}\lambda y & =0\\ -\left ( \sqrt{1-x^{2}}y^{\prime }\right ) ^{\prime } & =\frac{1}{\sqrt{1-x^{2}}}\lambda y \end{align*}
A problem is self adjoint if \left \langle L\left [ u\right ] ,v\right \rangle =\left \langle u,L\left [ v\right ] \right \rangle Where u,v are any two arbitrary eigenfunctions of the ODE which therefore by definition satisfy the ODE and the boundary conditions as given. Starting with \left \langle L\left [ u\right ] ,v\right \rangle and it is evaluated to see if it leads to \left \langle u,L\left [ v\right ] \right \rangle . The operator is defined as (from part (a)) as L\left [ y\right ] =-\left ( \sqrt{1-x^{2}}y^{\prime }\right ) ^{\prime }=\frac{1}{\sqrt{1-x^{2}}}\lambda y Therefore \left \langle L\left [ u\right ] ,v\right \rangle =\int _{-1}^{1}\overset{dv}{\overbrace{-\left ( \sqrt{1-x^{2}}u^{\prime }\right ) ^{\prime }}}\overset{u}{\overbrace{v}}dx Integrating by parts gives\begin{align*} \left \langle L\left [ u\right ] ,v\right \rangle & =\left [ -\left ( \sqrt{1-x^{2}}u^{\prime }\right ) v\right ] _{-1}^{1}-\int _{-1}^{1}-\left ( \sqrt{1-x^{2}}u^{\prime }\right ) v^{\prime }dx\\ & =\left [ -\left ( \sqrt{1-x^{2}}u^{\prime }\right ) v\right ] _{-1}^{1}-\int _{-1}^{1}\overset{u}{\overbrace{-\left ( \sqrt{1-x^{2}}v^{\prime }\right ) }}\overset{dv}{\overbrace{u^{\prime }}}dx \end{align*}
Integrating by parts again gives\begin{align*} \left \langle L\left [ u\right ] ,v\right \rangle & =\left [ -\left ( \sqrt{1-x^{2}}u^{\prime }\right ) v\right ] _{-1}^{1}-\left ( \left [ -\left ( \sqrt{1-x^{2}}v^{\prime }\right ) u\right ] _{-1}^{1}-\int _{-1}^{1}-\left ( \sqrt{1-x^{2}}v^{\prime }\right ) ^{\prime }udx\right ) \\ & =\left [ -\sqrt{1-x^{2}}u^{\prime }v+\sqrt{1-x^{2}}v^{\prime }u\right ] _{-1}^{1}+\int _{-1}^{1}-\left ( \sqrt{1-x^{2}}v^{\prime }\right ) ^{\prime }udx\\ & =\left [ \sqrt{1-x^{2}}\left ( v^{\prime }u-u^{\prime }v\right ) \right ] _{-1}^{1}+\left \langle u,L\left [ v\right ] \right \rangle \end{align*}
Therefore the ODE is self adjoint if the boundary terms vanish. Let \Delta =\left [ \sqrt{1-x^{2}}\left ( v^{\prime }u-u^{\prime }v\right ) \right ] _{-1}^{1}. Evaluating this gives \Delta =\lim _{x\rightarrow 1}\sqrt{1-x^{2}}\left ( v^{\prime }\left ( x\right ) u\left ( x\right ) -u^{\prime }\left ( x\right ) v\left ( x\right ) \right ) -\lim _{x\rightarrow -1}\sqrt{1-x^{2}}\left ( v^{\prime }\left ( x\right ) u\left ( x\right ) -u^{\prime }\left ( x\right ) v\left ( x\right ) \right ) But since u,u^{\prime } are bounded as x\rightarrow -1 and x\rightarrow +1 and also v,v^{\prime } are bounded as x\rightarrow -1 and x\rightarrow +1, then this shows that \Delta \rightarrow 0. Therefore \left \langle L\left [ u\right ] ,v\right \rangle =\left \langle u,L\left [ v\right ] \right \rangle Hence the ODE is self adjoint.
Since T_{n}\left ( x\right ) ,T_{m}\left ( x\right ) are two eigenfunctions of -\left ( \sqrt{1-x^{2}}y^{\prime }\right ) ^{\prime }=\frac{1}{\sqrt{1-x^{2}}}\lambda y then each satisfies the ODE. Hence\begin{align} \left ( \sqrt{1-x^{2}}T_{n}^{\prime }\right ) ^{\prime }+\frac{1}{\sqrt{1-x^{2}}}\lambda _{n}T_{n} & =0\tag{3A}\\ \left ( \sqrt{1-x^{2}}T_{m}^{\prime }\right ) ^{\prime }+\frac{1}{\sqrt{1-x^{2}}}\lambda _{m}T_{m} & =0 \tag{3B} \end{align}
Multiplying 3A by T_{m} and 3B by T_{n} and subtracting gives\begin{align*} T_{m}\left ( \sqrt{1-x^{2}}T_{n}^{\prime }\right ) ^{\prime }+\frac{1}{\sqrt{1-x^{2}}}\lambda _{n}T_{m}T_{n}-\left ( T_{n}\left ( \sqrt{1-x^{2}}T_{m}^{\prime }\right ) ^{\prime }+\frac{1}{\sqrt{1-x^{2}}}\lambda _{m}T_{n}T_{m}\right ) & =0\\ T_{m}\left ( \sqrt{1-x^{2}}T_{n}^{\prime }\right ) ^{\prime }-T_{n}\left ( \sqrt{1-x^{2}}T_{m}^{\prime }\right ) ^{\prime }+\left ( \lambda _{n}-\lambda _{m}\right ) \frac{1}{\sqrt{1-x^{2}}}T_{m}T_{n} & =0 \end{align*}
Integrating from -1\cdots 1 gives\begin{equation} \int _{-1}^{1}T_{m}\left ( \sqrt{1-x^{2}}T_{n}^{\prime }\right ) ^{\prime }dx-\int _{-1}^{1}T_{n}\left ( \sqrt{1-x^{2}}T_{m}^{\prime }\right ) ^{\prime }dx+\left ( \lambda _{n}-\lambda _{m}\right ) \int _{-1}^{1}\frac{T_{m}T_{n}}{\sqrt{1-x^{2}}}dx=0 \tag{1} \end{equation} Integrating by parts the first integral in (1) above gives\begin{equation} \int _{-1}^{1}T_{m}\left ( \sqrt{1-x^{2}}T_{n}^{\prime }\right ) ^{\prime }dx=\left [ T_{m}\sqrt{1-x^{2}}T_{n}^{\prime }\right ] _{-1}^{1}-\int _{-1}^{1}T_{m}^{\prime }\left ( \sqrt{1-x^{2}}T_{n}^{\prime }\right ) dx \tag{1A} \end{equation} Integrating by parts the second integral in (1) gives\begin{equation} \int _{-1}^{1}T_{n}\left ( \sqrt{1-x^{2}}T_{m}^{\prime }\right ) ^{\prime }dx=\left [ T_{n}\sqrt{1-x^{2}}T_{m}^{\prime }\right ] _{-1}^{1}-\int _{-1}^{1}T_{n}^{\prime }\left ( \sqrt{1-x^{2}}T_{m}^{\prime }\right ) dx \tag{1B} \end{equation} Substituting (1A) and (1B) back into (1) and simplifying gives \begin{align} \left [ T_{m}\sqrt{1-x^{2}}T_{n}^{\prime }\right ] _{-1}^{1}-\left [ T_{n}\sqrt{1-x^{2}}T_{m}^{\prime }\right ] _{-1}^{1}+\left ( \lambda _{n}-\lambda _{m}\right ) \int _{-1}^{1}\frac{T_{m}T_{n}}{\sqrt{1-x^{2}}}dx & =0\nonumber \\ \left [ T_{m}\sqrt{1-x^{2}}T_{n}^{\prime }-T_{n}\sqrt{1-x^{2}}T_{m}^{\prime }\right ] _{-1}^{1}+\left ( \lambda _{n}-\lambda _{m}\right ) \int _{-1}^{1}\frac{T_{m}T_{n}}{\sqrt{1-x^{2}}}dx & =0\nonumber \\ \left [ \sqrt{1-x^{2}}\left ( T_{m}T_{n}^{\prime }-T_{n}T_{m}^{\prime }\right ) \right ] _{-1}^{1}+\left ( \lambda _{n}-\lambda _{m}\right ) \int _{-1}^{1}\frac{T_{m}T_{n}}{\sqrt{1-x^{2}}}dx & =0 \tag{1C} \end{align}
Let \Delta =\left [ \sqrt{1-x^{2}}\left ( T_{m}T_{n}^{\prime }-T_{n}T_{m}^{\prime }\right ) \right ] _{-1}^{1}, then \Delta =\lim _{x\rightarrow 1}\sqrt{1-x^{2}}\left ( T_{m}\left ( x\right ) T_{n}^{\prime }\left ( x\right ) -T_{n}\left ( x\right ) T_{m}^{\prime }\left ( x\right ) \right ) -\lim _{x\rightarrow -1}\sqrt{1-x^{2}}\left ( T_{m}\left ( x\right ) T_{n}^{\prime }\left ( x\right ) -T_{n}\left ( x\right ) T_{m}^{\prime }\left ( x\right ) \right ) But since T_{n}\left ( x\right ) ,T_{m}\left ( x\right ) ,T_{n}^{\prime }\left ( x\right ) ,T_{m}^{\prime }\left ( x\right ) are all bounded as x\rightarrow -1 and as x\rightarrow +1, then \Delta \rightarrow 0. Therefore (1C) becomes \left ( \lambda _{n}-\lambda _{m}\right ) \int _{-1}^{1}\frac{T_{m}T_{n}}{\sqrt{1-x^{2}}}dx=0 But since \lambda _{n}\neq \lambda _{m}, since m\neq n, then \int _{-1}^{1}\frac{T_{m}T_{n}}{\sqrt{1-x^{2}}}dx=0 Which is what we are asked to show.
Find displacement u\left ( r,t\right ) in vibrating circular elastic membrane of radius 1 that satisfies the boundary conditions u\left ( 1,t\right ) =0\qquad t\geq 0 And initial conditions\begin{align*} u\left ( r,0\right ) & =0\\ u_{t}\left ( r,0\right ) & =g\left ( r\right ) \end{align*}
For 0\leq r\leq 1, where g\left ( 1\right ) =0.
Solution
The wave equation is u_{tt}=a^{2}\left ( u_{xx}+u_{yy}\right ) . In polar coordinates this becomes \frac{1}{a^{2}}u_{tt}=u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta } Due to circular symmetry, the above simplifies to \frac{1}{a^{2}}u_{tt}=u_{rr}+\frac{1}{r}u_{r} Applying separation of variables. Let u=T\left ( t\right ) R\left ( r\right ) . Substituting this in the above PDE gives \frac{1}{a^{2}}T^{\prime \prime }R=R^{\prime \prime }T+\frac{1}{r}R^{\prime }T Dividing by RT results in \frac{1}{a^{2}}\frac{T^{\prime \prime }}{T}=\frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}=-\lambda ^{2} Where \lambda is the sepration constant. For \lambda >0 (it is known \lambda =0 is not eigenvalue, as well as there are no negative eigenvalues.) The above gives two ODE T^{\prime \prime }+\lambda ^{2}a^{2}T=0 And\begin{equation} rR^{\prime \prime }\left ( r\right ) +R^{\prime }\left ( r\right ) +\lambda ^{2}rR\left ( r\right ) =0 \tag{1} \end{equation} With the boundary conditions R\left ( 1\right ) =0 and to R\left ( 0\right ) is bounded. This comes from physics, since one expects the vibration not to blow up in the center of the membrane. The ODE (1) is now transformed to Bessel ODE using \xi =\lambda r Hence \frac{dR}{dr}=\frac{dR}{d\xi }\frac{d\xi }{dr}=\lambda \frac{dR}{d\xi } and \frac{d^{2}R}{dr^{2}}=\lambda ^{2}\frac{d^{2}R}{d\xi ^{2}}. Therefore (1) becomes \frac{\xi }{\lambda }\lambda ^{2}R^{\prime \prime }\left ( \xi \right ) +\lambda R^{\prime }\left ( \xi \right ) +\lambda ^{2}\frac{\xi }{\lambda }R\left ( \xi \right ) =0 The above simplifies to \xi R^{\prime \prime }\left ( \xi \right ) +R^{\prime }\left ( \xi \right ) +\xi R\left ( \xi \right ) =0 The above is Bessel ODE of order zero. Its solution is R\left ( \xi \right ) =c_{1}J_{0}\left ( \xi \right ) +c_{2}Y_{0}\left ( \xi \right ) Converting back to r the above becomes R\left ( r\right ) =c_{1}J_{0}\left ( r\lambda \right ) +c_{2}Y_{0}\left ( r\lambda \right ) Since R\left ( r\right ) is bounded as r\rightarrow 0, then c_{2}=0 as Y_{0}\left ( r\lambda \right ) blows up at r=0. Therefore the radial solution becomes R\left ( r\right ) =c_{1}J_{0}\left ( r\lambda \right ) At boundary conditions R\left ( 1\right ) =0 the above becomes 0=c_{1}J_{0}\left ( \lambda \right ) Non trivial solution requires J_{0}\left ( \lambda \right ) =0. Therefore the eigenvalues are the the positive roots of J_{0}\left ( \lambda \right ) =0. The first few eigenvalues are \lambda _{1}=5.78319,\lambda _{2}=30.4713,\lambda _{3}=74.887,\cdots . Hence R_{n}\left ( r\right ) =c_{n}J_{0}\left ( \lambda _{n}r\right ) \qquad n=1,2,3,\cdots Now the time ODE is T^{\prime \prime }+\lambda ^{2}a^{2}T=0 Since \lambda >0 then the solution is T_{n}\left ( t\right ) =A_{n}\cos \left ( \lambda _{n}at\right ) +B_{n}\sin \left ( \lambda _{n}at\right ) Therefore the fundamental solution is u_{n}\left ( r,t\right ) =T_{n}\left ( t\right ) R_{n}\left ( r\right ) And by superposition, the general solution is\begin{equation} u\left ( r,t\right ) =\sum _{n=1}^{\infty }\left ( A_{n}\cos \left ( \lambda _{n}at\right ) +B_{n}\sin \left ( \lambda _{n}at\right ) \right ) J_{0}\left ( \lambda _{n}r\right ) \tag{1A} \end{equation}
Where the c_{n} is merged into A_{n},B_{n} due to the product. At t=0 and since u\left ( r,0\right ) =0, the above becomes 0=\sum _{n=1}^{\infty }A_{n}J_{0}\left ( \lambda _{n}r\right ) Hence A_{n}=0. The solution simplifies to u\left ( r,t\right ) =\sum _{n=1}^{\infty }B_{n}\sin \left ( \lambda _{n}at\right ) J_{0}\left ( \lambda _{n}r\right ) Taking time derivative gives u_{t}\left ( r,t\right ) =\sum _{n=1}^{\infty }B_{n}\lambda _{n}a\cos \left ( \lambda _{n}at\right ) J_{0}\left ( \lambda _{n}r\right ) At t=0, and from initial conditions, the above becomes g\left ( r\right ) =\sum _{n=1}^{\infty }B_{n}\lambda _{n}aJ_{0}\left ( \lambda _{n}r\right ) Applying orthogonality, and since the weight is r, therefore\begin{align} \int _{0}^{1}rg\left ( r\right ) J_{0}\left ( \lambda _{n}r\right ) dr & =B_{n}\lambda _{n}a\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{n}r\right ) dr\nonumber \\ B_{n} & =\frac{1}{\lambda _{n}a}\frac{\int _{0}^{1}rg\left ( r\right ) J_{0}\left ( \lambda _{n}r\right ) dr}{\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{n}r\right ) dr} \tag{2} \end{align}
Therefore the final solution is u\left ( r,t\right ) =\sum _{n=1}^{\infty }B_{n}\sin \left ( \lambda _{n}at\right ) J_{0}\left ( \lambda _{n}r\right ) With B_{n} given by (2).
Find displacement u\left ( r,t\right ) in vibrating circular elastic membrane of radius 1 that satisfies the boundary conditions u\left ( 1,t\right ) =0\qquad t\geq 0 And initial conditions\begin{align*} u\left ( r,0\right ) & =f\left ( r\right ) \\ u_{t}\left ( r,0\right ) & =g\left ( r\right ) \end{align*}
For 0\leq r\leq 1, where g\left ( 1\right ) =0.
Solution
The same steps are used to reach the general solution as was done in the above problem. The difference is when initial conditions are used to determine the coefficients.
The general solution from the above problem was found to be\begin{equation} u\left ( r,t\right ) =\sum _{n=1}^{\infty }\left ( A_{n}\cos \left ( \lambda _{n}at\right ) +B_{n}\sin \left ( \lambda _{n}at\right ) \right ) J_{0}\left ( \lambda _{n}r\right ) \tag{1A} \end{equation} At t=0 f\left ( r\right ) =\sum _{n=1}^{\infty }A_{n}J_{0}\left ( \lambda _{n}r\right ) Applying orthogonality, and since the weight is r results in\begin{align} \int _{0}^{1}rf\left ( r\right ) J_{0}\left ( \lambda _{n}r\right ) dr & =A_{n}\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{n}r\right ) dr\nonumber \\ A_{n} & =\frac{\int _{0}^{1}rf\left ( r\right ) J_{0}\left ( \lambda _{n}r\right ) dr}{\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{n}r\right ) dr} \tag{2} \end{align}
Taking time derivative of the solution (1A) u_{t}\left ( r,t\right ) =\sum _{n=1}^{\infty }-A_{n}\sqrt{\lambda _{n}}a\sin \left ( \lambda _{n}at\right ) +B_{n}\lambda _{n}a\cos \left ( \lambda _{n}at\right ) J_{0}\left ( \lambda _{n}r\right ) At t=0, and from initial conditions, the above becomes g\left ( r\right ) =\sum _{n=1}^{\infty }B_{n}\lambda _{n}aJ_{0}\left ( \lambda _{n}r\right ) Applying orthogonality, and since the weight is r, therefore\begin{align} \int _{0}^{1}rg\left ( r\right ) J_{0}\left ( \lambda _{n}r\right ) dr & =B_{n}\lambda _{n}a\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{n}r\right ) dr\nonumber \\ B_{n} & =\frac{1}{\lambda _{n}a}\frac{\int _{0}^{1}rg\left ( r\right ) J_{0}\left ( \lambda _{n}r\right ) dr}{\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{n}r\right ) dr} \tag{3} \end{align}
The two coefficients A_{n},B_{n} are now found. Therefore the final solution is u\left ( r,t\right ) =\sum _{n=1}^{\infty }\left ( A_{n}\cos \left ( \sqrt{\lambda _{n}}at\right ) +B_{n}\sin \left ( \sqrt{\lambda _{n}}at\right ) \right ) J_{0}\left ( \sqrt{\lambda _{n}}r\right ) With A_{n} given by (2) and B_{n} given by (3)
The wave equation in polar coordinates is \frac{1}{a^{2}}u_{tt}=u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta } Show that if u\left ( r,\theta ,t\right ) =R\left ( r\right ) \Theta \left ( \theta \right ) T\left ( t\right ) then R,\Theta ,T\, satisfy the ODE’s\begin{align*} r^{2}R^{\prime \prime }+rR^{\prime }+\left ( \lambda ^{2}r^{2}-n^{2}\right ) R & =0\\ \Theta ^{\prime \prime }+n^{2}\Theta & =0\\ T^{\prime \prime }+\lambda ^{2}a^{2}T & =0 \end{align*}
Solution
Let u\left ( r,\theta ,t\right ) =R\left ( r\right ) \Theta \left ( \theta \right ) T\left ( t\right ) . Substituting in the wave PDE gives \frac{1}{a^{2}}T^{\prime \prime }R\Theta =R^{\prime \prime }T\Theta +\frac{1}{r}R^{\prime }T\Theta +\frac{1}{r^{2}}\Theta ^{\prime \prime }RT dividing by R\Theta T gives \frac{1}{a^{2}}\frac{T^{\prime \prime }}{T}=\frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}+\frac{1}{r^{2}}\frac{\Theta ^{\prime \prime }}{\Theta }=-\lambda ^{2} Where \lambda is separation constant. The above now become\begin{align} \frac{1}{a^{2}}\frac{T^{\prime \prime }}{T} & =-\lambda ^{2}\tag{1}\\ \frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}+\frac{1}{r^{2}}\frac{\Theta ^{\prime \prime }}{\Theta } & =-\lambda ^{2}\nonumber \end{align}
The second ODE above can now be written as\begin{align*} r^{2}\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R}+\frac{\Theta ^{\prime \prime }}{\Theta } & =-r^{2}\lambda ^{2}\\ r^{2}\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R}+r^{2}\lambda ^{2} & =-\frac{\Theta ^{\prime \prime }}{\Theta }=n^{2} \end{align*}
Where n is the new separation constant (I do not like using n for this, but this is what the book did). The above now gives the ODE’s\begin{align} -\frac{\Theta ^{\prime \prime }}{\Theta } & =n^{2}\tag{2}\\ r^{2}\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R}+r^{2}\lambda ^{2} & =n^{2} \tag{3} \end{align}
Therefore (1,2,3) becomes\begin{align} T^{\prime \prime }+a^{2}\lambda ^{2}T & =0\tag{1A}\\ \Theta ^{\prime \prime }+n^{2}\Theta & =0\tag{2A}\\ r^{2}R^{\prime \prime }+rR^{\prime }+\left ( r^{2}\lambda ^{2}-n^{2}\right ) R & =0 \tag{3A} \end{align}
Which is what the problem asked to show.
In the circular cylindrical coordinates r,\theta ,z defined by \begin{align*} x & =r\cos \theta \\ y & =r\sin \theta \\ z & =z \end{align*}
Laplace equation is u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta }+u_{zz}=0 (a) Show that if u\left ( r,\theta ,t\right ) =R\left ( r\right ) \Theta \left ( \theta \right ) Z\left ( z\right ) then R,\Theta ,Z\, satisfy the ODE’s\begin{align*} r^{2}R^{\prime \prime }+rR^{\prime }+\left ( \lambda ^{2}r^{2}-n^{2}\right ) R & =0\\ \Theta ^{\prime \prime }+n^{2}\Theta & =0\\ Z^{\prime \prime }-\lambda ^{2}Z & =0 \end{align*}
(b) Show that if u\left ( r,\theta ,z\right ) is independent of \theta then the first equation in (a) becomes r^{2}R^{\prime \prime }+rR^{\prime }+\lambda ^{2}r^{2}R=0 The second is omitted altogether and the third is unchanged.
Solution
Let u\left ( r,\theta ,z\right ) =R\left ( r\right ) \Theta \left ( \theta \right ) Z\left ( z\right ) . Substituting in the wave PDE u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta }+u_{zz}=0 gives R^{\prime \prime }\Theta Z+\frac{1}{r}R^{\prime }\Theta Z+\frac{1}{r^{2}}\Theta ^{\prime \prime }RZ+Z^{\prime \prime }R\Theta =0 dividing by R\Theta Z gives \frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}+\frac{1}{r^{2}}\frac{\Theta ^{\prime \prime }}{\Theta }=-\frac{Z^{\prime \prime }}{Z}=-\lambda ^{2} Where \lambda is separation constant. The above now become\begin{align} Z^{\prime \prime }-\lambda ^{2}Z & =0\tag{1}\\ \frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}+\frac{1}{r^{2}}\frac{\Theta ^{\prime \prime }}{\Theta } & =-\lambda ^{2}\nonumber \end{align}
The second ODE above can now be written as\begin{align*} r^{2}\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R}+\frac{\Theta ^{\prime \prime }}{\Theta } & =-r^{2}\lambda ^{2}\\ r^{2}\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R}+r^{2}\lambda ^{2} & =-\frac{\Theta ^{\prime \prime }}{\Theta }=n^{2} \end{align*}
Where n is the new separation constant. The above now gives the ODE’s\begin{align} -\frac{\Theta ^{\prime \prime }}{\Theta } & =n^{2}\tag{2}\\ r^{2}\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R}+r^{2}\lambda ^{2} & =n^{2} \tag{3} \end{align}
Therefore (1,2,3) becomes\begin{align} Z^{\prime \prime }-\lambda ^{2}Z & =0\tag{1A}\\ \Theta ^{\prime \prime }+n^{2}\Theta & =0\tag{2A}\\ r^{2}R^{\prime \prime }+rR^{\prime }+\left ( r^{2}\lambda ^{2}-n^{2}\right ) R & =0 \tag{3A} \end{align}
When no dependency on \theta then the ODE becomes u_{rr}+\frac{1}{r}u_{r}+u_{zz}=0. Let u\left ( r,z\right ) =R\left ( r\right ) Z\left ( z\right ) . Substituting into the wave PDE R^{\prime \prime }Z+\frac{1}{r}R^{\prime }Z+Z^{\prime \prime }R=0 dividing by RZ gives \frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}=-\frac{Z^{\prime \prime }}{Z}=-\lambda ^{2} The above gives\begin{align*} \frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R} & =-\lambda ^{2}\\ -\frac{Z^{\prime \prime }}{Z} & =-\lambda ^{2} \end{align*}
Or\begin{align*} R^{\prime \prime }+\frac{1}{r}R^{\prime }+\lambda ^{2}R & =0\\ Z^{\prime \prime }-\lambda ^{2}Z & =0 \end{align*}
Find steady state solution in semi-infinite rod 0<z<\infty ,0\leq r\leq 1 if the temperature is independent of \theta and approaches zero as z\rightarrow \infty . Assume u\left ( r,z\right ) satisfies boundary conditions\begin{align*} u\left ( 1,z\right ) & =0\qquad z>0\\ u\left ( r,0\right ) & =f\left ( r\right ) \qquad 0\leq r\leq 1 \end{align*}
Solution
The PDE is u_{rr}+\frac{1}{r}u_{r}+u_{zz}=0 By separation of variables, as was done in problem 5 above, this gives\begin{align} R^{\prime \prime }+\frac{1}{r}R^{\prime }+\lambda ^{2}R & =0\tag{1}\\ R\left ( 1\right ) & =0\nonumber \\ \lim _{r\rightarrow 0}R\left ( 0\right ) & \rightarrow \text{bounded}\nonumber \end{align}
And\begin{align} Z^{\prime \prime }-\lambda ^{2}Z & =0\tag{2}\\ Z\left ( 0\right ) & =f\left ( r\right ) \nonumber \\ \lim _{z\rightarrow \infty }Z\left ( r\right ) & \rightarrow 0\nonumber \end{align}
The solution to (2) is known to be R\left ( r\right ) =c_{n}J_{0}\left ( \lambda _{n}r\right ) Where \lambda _{n} are the positive roots of J_{0}\left ( \lambda _{n}\right ) =0. The solution to (2) is Z\left ( z\right ) =A_{n}e^{\lambda _{n}z}+B_{n}e^{-\lambda _{n}z} Since u goes to zero as z\rightarrow \infty , then this implies A_{n}=0. Hence Z\left ( z\right ) =B_{n}e^{-\lambda _{n}z} Hence the overall solution becomes u\left ( r,z\right ) =\sum _{n=1}^{\infty }B_{n}e^{-\lambda _{n}z}J_{0}\left ( \lambda _{n}r\right ) Where c_{n} is combined with B_{n}. To find B_{n}, using the final boundary condition u\left ( r,0\right ) =f\left ( r\right ) gives f\left ( r\right ) =\sum _{n=1}^{\infty }B_{n}J_{0}\left ( \lambda _{n}r\right ) Applying orthogonality and using the weight of r gives
\begin{align*} \int _{0}^{1}rf\left ( r\right ) J_{0}\left ( \lambda _{n}r\right ) dr & =B_{n}\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{n}r\right ) dr\\ B_{n} & =\frac{\int _{0}^{1}rf\left ( r\right ) J_{0}\left ( \lambda _{n}r\right ) dr}{\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{n}r\right ) dr} \end{align*}
Hence the solution is now complete. It is given by u\left ( r,z\right ) =\sum _{n=1}^{\infty }\frac{\int _{0}^{1}rf\left ( r\right ) J_{0}\left ( \lambda _{n}r\right ) dr}{\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{n}r\right ) dr}e^{-\lambda _{n}z}J_{0}\left ( \lambda _{n}r\right )
Solution
Substituting v\left ( r,\theta \right ) =R\left ( r\right ) \Theta \left ( \theta \right ) into the PDE gives R^{\prime \prime }\Theta +\frac{1}{r}R^{\prime }\Theta +\frac{1}{r^{2}}\Theta ^{\prime \prime }R+k^{2}R\Theta =0 Dividing by R\Theta gives\begin{align*} \frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}+\frac{1}{r^{2}}\frac{\Theta ^{\prime \prime }}{\Theta }+k^{2} & =0\\ r^{2}\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R}+r^{2}k^{2} & =-\frac{\Theta ^{\prime \prime }}{\Theta }=\lambda ^{2} \end{align*}
Where \lambda is the separation constant. This gives r^{2}\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R}+r^{2}k^{2}-\lambda ^{2}=0 And -\frac{\Theta ^{\prime \prime }}{\Theta }=\lambda ^{2} Hence\begin{align*} r^{2}R^{\prime \prime }+rR^{\prime }+R\left ( r^{2}k^{2}-n^{2}\right ) & =0\\ \Theta ^{\prime \prime }+\lambda ^{2}\Theta & =0 \end{align*}
Starting with \Theta ^{\prime \prime }+\lambda ^{2}\Theta =0. The eigenvalue \lambda can not be negative. The following two cases are considered.
Case \lambda =0
Solution is \Theta \left ( \theta \right ) =c_{1}\theta +c_{2} The boundary conditions are periodic with period 2\pi , meaning\begin{align*} \Theta \left ( 0\right ) & =\Theta \left ( 2\pi \right ) \\ \Theta ^{\prime }\left ( 0\right ) & =\Theta ^{\prime }\left ( 2\pi \right ) \end{align*}
Applying first BC gives\begin{equation} c_{2}=c_{1}2\pi +c_{2} \tag{1} \end{equation} Applying second BC gives\begin{equation} c_{1}=c_{1} \tag{2} \end{equation} So c_{1} can be any value. But to solve (1) c_{1} must be zero. Hence first BC now gives c_{2}=c_{2} Which means c_{2} can be any value, say 1. Therefore \lambda =0 is an eigenvalue with eigenfunction \Phi _{0}\left ( \theta \right ) =1
Case \lambda >0
The solution now is \Theta \left ( \theta \right ) =A\cos \left ( \lambda \theta \right ) +B\sin \left ( \lambda \theta \right ) The boundary conditions are periodic with period 2\pi , meaning\begin{align*} \Theta \left ( 0\right ) & =\Theta \left ( 2\pi \right ) \\ \Theta ^{\prime }\left ( 0\right ) & =\Theta ^{\prime }\left ( 2\pi \right ) \end{align*}
Applying the above boundary conditions gives\begin{align*} A & =A\cos \left ( \lambda 2\pi \right ) +B\sin \left ( \lambda 2\pi \right ) \\ B\lambda & =A\lambda \sin \left ( \lambda 2\pi \right ) +B\lambda \cos \left ( \lambda 2\pi \right ) \end{align*}
This means \lambda must be an integer n=1,2,\cdots for the above relations be satisfied. Since only when n is an integer, the above gives A=A and B\lambda =B\lambda . Hence the eigenfunction in this case is \Phi _{n}\left ( \theta \right ) =A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \qquad n=1,2,\cdots Now that the eigenvalues are found, the solution to the R ODE is found. Summary of the above result: The eigenvalues are n=0 with eigenfunction \Phi _{0}\left ( \theta \right ) =1 and n=1,2,3,\cdots with eigenfunction \Phi _{n}\left ( \theta \right ) =A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) .
Case \lambda =n=0
In this case, the R ODE above r^{2}R^{\prime \prime }+rR^{\prime }+R\left ( r^{2}k^{2}-\lambda ^{2}\right ) =0 reduces to r^{2}R^{\prime \prime }+rR^{\prime }+Rr^{2}k^{2}=0 let t=rk Therefore R^{\prime }\left ( r\right ) =R^{\prime }\left ( t\right ) k and R^{\prime \prime }\left ( r\right ) =R^{\prime \prime }\left ( t\right ) k^{2}. Substituting these in the above ODE gives\begin{align*} \frac{t^{2}}{k^{2}}k^{2}R^{\prime \prime }\left ( t\right ) +\frac{t}{k}kR^{\prime }\left ( r\right ) +R\frac{t^{2}}{k^{2}}k^{2} & =0\\ t^{2}R^{\prime \prime }\left ( t\right ) +tR^{\prime }\left ( t\right ) +t^{2}R\left ( t\right ) & =0 \end{align*}
This is now Bessel ODE of order zero. Its solution is R_{0}\left ( t\right ) =A_{0}J_{0}\left ( t\right ) +B_{0}Y_{0}\left ( t\right ) Converting back to r, the above becomes R_{0}\left ( r\right ) =A_{0}J_{0}\left ( rk\right ) +B_{0}Y_{0}\left ( rk\right ) Since R is bounded at r=0, this implies B_{0}=0, since Y_{0}\left ( rk\right ) blows up at r=0. Hence R_{0}\left ( r\right ) =A_{0}J_{0}\left ( rk\right ) This is the solution for eigenvalue n=0.
Case \lambda =n>0
The Bessel PDE now has the form r^{2}R^{\prime \prime }\left ( r\right ) +rR^{\prime }\left ( r\right ) +\left ( r^{2}k^{2}-n^{2}\right ) R\left ( r\right ) =0. To convert the ODE to standard Bessel form let t=rk Therefore R^{\prime }\left ( r\right ) =R^{\prime }\left ( t\right ) k and R^{\prime \prime }\left ( r\right ) =R^{\prime \prime }\left ( t\right ) k^{2}. Substituting these in the above ODE gives\begin{align*} \frac{t^{2}}{k^{2}}k^{2}R^{\prime \prime }\left ( t\right ) +\frac{t}{k}kR^{\prime }\left ( r\right ) +R\left ( \frac{t^{2}}{k^{2}}k^{2}-n^{2}\right ) & =0\\ t^{2}R^{\prime \prime }\left ( t\right ) +tR^{\prime }\left ( t\right ) +R\left ( t\right ) \left ( t^{2}-n^{2}\right ) & =0 \end{align*}
This is now Bessel ODE of order n. Its solution is R_{n}\left ( t\right ) =A_{n}J_{n}\left ( t\right ) +B_{n}Y_{n}\left ( t\right ) Converting back to r, the above becomes R_{n}\left ( r\right ) =A_{n}J_{n}\left ( rk\right ) +B_{n}Y_{n}\left ( rk\right ) Since R is bounded at r=0, this implies B_{n}=0, since Y_{n}\left ( rk\right ) blows up at r=0. Hence R\left ( r\right ) =A_{n}J_{n}\left ( rk\right ) . This is the solution for eigenvalue n>0.
Hence the fundamental solution is\begin{align*} v_{0}\left ( r,\theta \right ) & =\Phi _{0}\left ( \theta \right ) R_{0}\left ( r\right ) \\ & =A_{0}J_{0}\left ( rk\right ) \end{align*}
Since \Phi _{0}\left ( \theta \right ) =1 and\begin{align*} v_{n}\left ( r,\theta \right ) & =\Phi _{n}\left ( \theta \right ) R_{n}\left ( r\right ) \\ & =\left ( A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \right ) J_{n}\left ( rk\right ) \end{align*}
Where the constants are combined. Therefore the general solution becomes\begin{equation} v\left ( r,\theta \right ) =A_{0}J_{0}\left ( rk\right ) +\sum _{n=1}^{\infty }\left ( A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \right ) J_{n}\left ( rk\right ) \tag{3} \end{equation} Constants A_{0},A_{n},B_{n} are found from boundary conditions. At r=c, u\left ( c,\theta \right ) =f\left ( \theta \right ) and the above becomes f\left ( \theta \right ) =A_{0}J_{0}\left ( ck\right ) +\sum _{n=1}^{\infty }\left ( A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \right ) J_{n}\left ( ck\right ) For n=0 only and applying orthogonality\begin{align*} \int _{0}^{2\pi }f\left ( \theta \right ) d\theta & =\int _{0}^{2\pi }A_{0}J_{0}\left ( ck\right ) d\theta \\ \int _{0}^{2\pi }f\left ( \theta \right ) d\theta & =A_{0}J_{0}\left ( ck\right ) \int _{0}^{2\pi }d\theta \\ & =2\pi A_{0}J_{0}\left ( ck\right ) \end{align*}
Hence A_{0}=\frac{\int _{0}^{2\pi }f\left ( \theta \right ) d\theta }{2\pi J_{0}\left ( ck\right ) } And for n>0\begin{align*} \int _{0}^{2\pi }f\left ( \theta \right ) \sin \left ( m\theta \right ) d\theta & =\sum _{n=1}^{\infty }\int _{0}^{2\pi }\left ( A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \right ) \sin \left ( m\theta \right ) J_{n}\left ( ck\right ) d\theta \\ & =\sum _{n=1}^{\infty }J_{n}\left ( ck\right ) A_{n}\int _{0}^{2\pi }\cos \left ( n\theta \right ) \sin \left ( m\theta \right ) d\theta +B_{n}\sum _{n=1}^{\infty }J_{n}\left ( ck\right ) \int _{0}^{2\pi }\sin \left ( n\theta \right ) \sin \left ( m\theta \right ) d\theta \end{align*}
But \int _{0}^{2\pi }\cos \left ( n\theta \right ) \sin \left ( m\theta \right ) d\theta =0 for all n,m and the above now is solved for B_{n}\begin{align*} \int _{0}^{2\pi }f\left ( \theta \right ) \sin \left ( m\theta \right ) d\theta & =B_{n}\sum _{n=1}^{\infty }J_{n}\left ( ck\right ) \int _{0}^{2\pi }\sin \left ( n\theta \right ) \sin \left ( m\theta \right ) d\theta \\ & =B_{m}J_{m}\left ( ck\right ) \int _{0}^{2\pi }\sin ^{2}\left ( m\theta \right ) d\theta \\ & =B_{m}J_{m}\left ( ck\right ) \pi \end{align*}
Hence B_{n}=\frac{\int _{0}^{2\pi }f\left ( \theta \right ) \sin \left ( n\theta \right ) d\theta }{\pi J_{n}\left ( ck\right ) } Similarly, to find A_{n}\begin{align*} \int _{0}^{2\pi }f\left ( \theta \right ) \cos \left ( m\theta \right ) d\theta & =\sum _{n=1}^{\infty }\int _{0}^{2\pi }\left ( A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \right ) \cos \left ( m\theta \right ) J_{n}\left ( ck\right ) d\theta \\ & =\sum _{n=1}^{\infty }J_{n}\left ( ck\right ) A_{n}\int _{0}^{2\pi }\cos \left ( n\theta \right ) \cos \left ( m\theta \right ) d\theta +B_{n}\sum _{n=1}^{\infty }J_{n}\left ( ck\right ) \int _{0}^{2\pi }\sin \left ( n\theta \right ) \cos \left ( m\theta \right ) d\theta \end{align*}
But \int _{0}^{2\pi }\sin \left ( n\theta \right ) \cos \left ( m\theta \right ) d\theta =0 for all n,m and the above now is solved for A_{n}\begin{align*} \int _{0}^{2\pi }f\left ( \theta \right ) \sin \left ( m\theta \right ) d\theta & =A_{n}\sum _{n=1}^{\infty }J_{n}\left ( ck\right ) \int _{0}^{2\pi }\cos \left ( n\theta \right ) \cos \left ( m\theta \right ) d\theta \\ & =A_{m}J_{m}\left ( ck\right ) \int _{0}^{2\pi }\cos ^{2}\left ( m\theta \right ) d\theta \\ & =A_{m}J_{m}\left ( ck\right ) \pi \end{align*}
Hence A_{n}=\frac{\int _{0}^{2\pi }f\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta }{\pi J_{n}\left ( ck\right ) } The complete solution from (3) becomes\begin{align*} v\left ( r,\theta \right ) & =A_{0}J_{0}\left ( rk\right ) +\sum _{n=1}^{\infty }\left ( A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \right ) J_{n}\left ( rk\right ) \\ A_{0} & =\frac{\int _{0}^{2\pi }f\left ( \theta \right ) d\theta }{2\pi J_{0}\left ( ck\right ) }\\ B_{n} & =\frac{\int _{0}^{2\pi }f\left ( \theta \right ) \sin \left ( n\theta \right ) d\theta }{\pi J_{n}\left ( ck\right ) }\\ A_{n} & =\frac{\int _{0}^{2\pi }f\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta }{\pi J_{n}\left ( ck\right ) } \end{align*}
Solution
Let u\left ( r,t\right ) =R\left ( r\right ) T\left ( t\right ) . Substituting into the PDE gives \frac{1}{\alpha ^{2}}T^{\prime }R=R^{\prime \prime }T+\frac{1}{r}R^{\prime }T Dividing by RT gives \frac{1}{\alpha ^{2}}\frac{T^{\prime }}{T}=\frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}=-\lambda ^{2} Where \lambda is the separation constant. This gives the ODE\begin{align} R^{\prime \prime }+\frac{1}{r}R+\lambda ^{2}R & =0\nonumber \\ rR^{\prime \prime }+R+\lambda ^{2}rR & =0\nonumber \\ \left ( rR^{\prime }\right ) ^{\prime }+\lambda ^{2}rR & =0 \tag{1} \end{align}
With BC\begin{align*} R\left ( 1\right ) & =0\\ \lim _{r\rightarrow 0}R\left ( r\right ) & \rightarrow \text{bounded} \end{align*}
And\begin{equation} T^{\prime }+\alpha ^{2}\lambda ^{2}R=0 \tag{2} \end{equation} ODE (1) is Sturm Liouville ODE where p=r,q=0 and the weight is r. The eigenvalue can not be negative. Two cases to consider.
Case \lambda =0
The ODE becomes \left ( rR^{\prime }\right ) ^{\prime }=0 which has solution rR^{\prime }=c_{1} or r\frac{dR}{dr}=c_{1} or dR=\frac{c_{1}}{r}dr. Integrating gives R\left ( r\right ) =c_{1}\ln r+c_{2} Since R is bounded at r=0, then c_{1}=0. The solution becomes R\left ( r\right ) =c_{2}. Since R\left ( 1\right ) =0 then c_{2}=0. Hence trivial solution. Therefore \lambda =0 is not an eigenvalue.
Case \lambda >0
The ODE now becomes rR^{\prime \prime }\left ( r\right ) +R\left ( r\right ) +\lambda ^{2}rR\left ( r\right ) =0. Let t=\lambda r. Hence R^{\prime }\left ( r\right ) =\lambda R^{\prime }\left ( t\right ) and R^{\prime \prime }\left ( r\right ) =\lambda ^{2}R^{\prime \prime }\left ( t\right ) and the ODE becomes\begin{align*} \frac{t}{\lambda }\lambda ^{2}R^{\prime \prime }\left ( t\right ) +\lambda R^{\prime }\left ( t\right ) +\lambda ^{2}\frac{t}{\lambda }R\left ( t\right ) & =0\\ t\lambda R^{\prime \prime }\left ( t\right ) +\lambda R^{\prime }\left ( t\right ) +\lambda tR\left ( t\right ) & =0\\ tR^{\prime \prime }\left ( t\right ) +R^{\prime }\left ( t\right ) +tR\left ( t\right ) & =0 \end{align*}
This is Bessel ODE of order zero. Its solution is R\left ( t\right ) =c_{1}J_{o}\left ( t\right ) +c_{2}Y_{o}\left ( t\right ) Converting back to r R\left ( r\right ) =c_{1}J_{o}\left ( \lambda r\right ) +c_{2}Y_{o}\left ( \lambda r\right ) Since R is bounded at r=0 then c_{2}=0 and the solution becomes R\left ( r\right ) =c_{1}J_{o}\left ( \lambda r\right ) Since R\left ( 1\right ) =0 then 0=c_{1}J_{o}\left ( \lambda \right ) For nontrivial solution, J_{o}\left ( \lambda \right ) =0. This gives the eigenvalues as the positive roots of J_{o}\left ( \lambda \right ) =0. Hence the solution is R_{n}\left ( r\right ) =c_{n}J_{o}\left ( \lambda _{n}r\right ) Where \lambda _{n} are roots of J_{o}\left ( \lambda \right ) =0 for n=1,2,3,\cdots . The Time ODE (2) has solution T_{n}\left ( t\right ) =A_{n}e^{-\lambda _{n}^{2}\alpha ^{2}t} Hence the final solution is u\left ( r,t\right ) =\sum _{n=1}^{\infty }c_{n}e^{-\lambda _{n}^{2}\alpha ^{2}t}J_{0}\left ( \lambda _{n}r\right ) Where constants A_{n},c_{n} are combined into c_{n}. c_{n} is now found from initial conditions. At t=0 the above becomes u\left ( r,0\right ) =f\left ( r\right ) =\sum _{n=1}^{\infty }c_{n}J_{0}\left ( \lambda _{n}r\right ) The weight is r, since the R ODE in S.L. form is \left ( rR^{\prime }\right ) ^{\prime }+\lambda ^{2}rR=0. Therefore, applying orthogonality gives\begin{align*} \int _{0}^{1}rf\left ( r\right ) J_{0}\left ( \lambda _{n}r\right ) dr & =c_{n}\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{n}r\right ) dr\\ c_{n} & =\frac{\int _{0}^{1}rf\left ( r\right ) J_{0}\left ( \lambda _{n}r\right ) dr}{\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{n}r\right ) dr} \end{align*}
This completes the solution. u\left ( r,t\right ) =\sum _{n=1}^{\infty }\frac{\int _{0}^{1}rf\left ( r\right ) J_{0}\left ( \lambda _{n}r\right ) dr}{\int _{0}^{1}rJ_{0}^{2}\left ( \lambda _{n}r\right ) dr}e^{-\lambda _{n}^{2}\alpha ^{2}t}J_{0}\left ( \lambda _{n}r\right )
Solution
Let u\left ( \rho ,\theta ,\phi \right ) =P\left ( \rho \right ) \Theta \left ( \theta \right ) \Phi \left ( \phi \right ) Substituting the above in the Laplace PDE given results in \rho ^{2}P^{\prime \prime }\Theta \Phi +2\rho P^{\prime }\Theta \Phi +\left ( \csc ^{2}\phi \right ) \Theta ^{\prime \prime }P\Phi +\Phi ^{\prime \prime }P\Theta +\cot \left ( \phi \right ) \Phi ^{\prime }P\Theta =0 Dividing by P\Theta \Phi gives\begin{align*} \rho ^{2}\frac{P^{\prime \prime }}{P}+2\rho \frac{P^{\prime }}{P}+\left ( \csc ^{2}\phi \right ) \frac{\Theta ^{\prime \prime }}{\Theta }+\frac{\Phi ^{\prime \prime }}{\Phi }+\cot \left ( \phi \right ) \frac{\Phi ^{\prime }}{\Phi } & =0\\ \rho ^{2}\frac{P^{\prime \prime }}{P}+2\rho \frac{P^{\prime }}{P} & =-\left ( \csc ^{2}\phi \right ) \frac{\Theta ^{\prime \prime }}{\Theta }-\frac{\Phi ^{\prime \prime }}{\Phi }-\cot \left ( \phi \right ) \frac{\Phi ^{\prime }}{\Phi }=\mu ^{2} \end{align*}
Where \mu is the first separation constant. The above gives\begin{align*} \rho ^{2}\frac{P^{\prime \prime }}{P}+2\rho \frac{P^{\prime }}{P} & =\mu ^{2}\\ -\left ( \csc ^{2}\phi \right ) \frac{\Theta ^{\prime \prime }}{\Theta }-\frac{\Phi ^{\prime \prime }}{\Phi }-\cot \left ( \phi \right ) \frac{\Phi ^{\prime }}{\Phi }-\mu ^{2} & =0 \end{align*}
The first ODE above becomes \rho ^{2}P^{\prime \prime }+2\rho P^{\prime }-P\mu ^{2}=0 And the second equation is now separated again into two additional ODE’s as follows\begin{align*} -\frac{\Theta ^{\prime \prime }}{\Theta }-\frac{1}{\csc ^{2}\phi }\frac{\Phi ^{\prime \prime }}{\Phi }-\frac{\cot \phi }{\csc ^{2}\phi }\frac{\Phi ^{\prime }}{\Phi }-\frac{\mu ^{2}}{\csc ^{2}\phi } & =0\\ \frac{1}{\csc ^{2}\phi }\frac{\Phi ^{\prime \prime }}{\Phi }+\frac{\cot \left ( \phi \right ) }{\csc ^{2}\phi }\frac{\Phi ^{\prime }}{\Phi }+\frac{\mu ^{2}}{\csc ^{2}\phi } & =-\frac{\Theta ^{\prime \prime }}{\Theta }=\lambda ^{2} \end{align*}
Where \lambda is the second separation constant. The above gives the following two ODE’s \Theta ^{\prime \prime }+\lambda ^{2}\Theta =0 And, since \csc ^{2}\phi =\frac{1}{\sin ^{2}\phi } and \cot \left ( \phi \right ) =\frac{1}{\tan \phi }, the third ODE is\begin{align*} \sin ^{2}\phi \frac{\Phi ^{\prime \prime }}{\Phi }+\frac{\sin ^{2}\phi }{\tan \phi }\frac{\Phi ^{\prime }}{\Phi }+\mu ^{2}\sin ^{2}\phi & =\lambda ^{2}\\ \sin ^{2}\phi \frac{\Phi ^{\prime \prime }}{\Phi }+\sin \phi \cos \phi \frac{\Phi ^{\prime }}{\Phi }+\mu ^{2}\sin ^{2}\phi & =\lambda ^{2}\\ \left ( \sin ^{2}\phi \right ) \Phi ^{\prime \prime }+\left ( \sin \phi \cos \phi \right ) \Phi ^{\prime }+\left ( \mu ^{2}\sin ^{2}\phi -\lambda ^{2}\right ) \Phi & =0 \end{align*}
If u is independent of \theta then the PDE simplifies to\begin{equation} \rho ^{2}u_{\rho \rho }+2\rho u_{\rho }+u_{\phi \phi }+\cot \phi u_{\phi }=0 \tag{1} \end{equation} Let u\left ( \rho ,\phi \right ) =P\left ( \rho \right ) \Phi \left ( \phi \right ) Substituting the above in the Laplace PDE (1) results in \rho ^{2}P^{\prime \prime }\Phi +2\rho P^{\prime }\Phi +\Phi ^{\prime \prime }P+\cot \left ( \phi \right ) \Phi ^{\prime }P=0 Dividing by P\Phi gives\begin{align*} \rho ^{2}\frac{P^{\prime \prime }}{P}+2\rho \frac{P^{\prime }}{P}+\frac{\Phi ^{\prime \prime }}{\Phi }+\cot \left ( \phi \right ) \frac{\Phi ^{\prime }}{\Phi } & =0\\ \rho ^{2}\frac{P^{\prime \prime }}{P}+2\rho \frac{P^{\prime }}{P} & =-\frac{\Phi ^{\prime \prime }}{\Phi }-\cot \left ( \phi \right ) \frac{\Phi ^{\prime }}{\Phi }=\mu ^{2} \end{align*}
Where \mu is the first separation constant. The above gives\begin{align*} \rho ^{2}\frac{P^{\prime \prime }}{P}+2\rho \frac{P^{\prime }}{P} & =\mu ^{2}\\ -\frac{\Phi ^{\prime \prime }}{\Phi }-\cot \left ( \phi \right ) \frac{\Phi ^{\prime }}{\Phi }-\mu ^{2} & =0 \end{align*}
The first ODE above becomes \rho ^{2}P^{\prime \prime }+2\rho P^{\prime }-\mu ^{2}P=0 And the second ODE becomes\begin{align*} -\Phi ^{\prime \prime }-\cot \left ( \phi \right ) \Phi ^{\prime }-\mu ^{2}\Phi & =0\\ \Phi ^{\prime \prime }+\frac{1}{\tan \phi }\Phi ^{\prime }+\mu ^{2}\Phi & =0\\ \Phi ^{\prime \prime }+\frac{\cos \phi }{\sin \phi }\Phi ^{\prime }+\mu ^{2}\Phi & =0\\ \left ( \sin \phi \right ) \Phi ^{\prime \prime }+\left ( \cos \phi \right ) \Phi ^{\prime }+\left ( \mu ^{2}\sin \phi \right ) \Phi & =0 \end{align*}
Multiplying again by \sin \phi to get it to the form needed gives\begin{equation} \sin ^{2}\phi \Phi ^{\prime \prime }+\left ( \sin \phi \cos \phi \right ) \Phi ^{\prime }+\left ( \mu ^{2}\sin ^{2}\phi \right ) \Phi =0 \tag{2} \end{equation} Therefore the first PDE in P\left ( \rho \right ) , the second ODE in \Theta \left ( \theta \right ) is now eliminated, and the third ODE changes to the above.
The equation for \Phi found in part (b) is\begin{equation} \sin ^{2}\phi \frac{d^{2}\Phi }{d\phi ^{2}}+\left ( \sin \phi \cos \phi \right ) \frac{d\Phi }{d\phi }+\left ( \mu ^{2}\sin ^{2}\phi \right ) \Phi =0 \tag{1} \end{equation} Let s=\cos \phi , then\begin{align} \frac{d\Phi }{d\phi } & =\frac{d\Phi }{ds}\frac{ds}{d\phi }\nonumber \\ & =\frac{d\Phi }{ds}\left ( -\sin \phi \right ) \tag{2} \end{align}
And\begin{align} \frac{d^{2}\Phi }{d\phi ^{2}} & =\frac{d}{d\phi }\left ( \frac{d\Phi }{d\phi }\right ) \nonumber \\ & =\frac{d}{d\phi }\left ( \frac{d\Phi }{ds}\left ( -\sin \phi \right ) \right ) \nonumber \\ & =\frac{d^{2}\Phi }{ds^{2}}\left ( \sin ^{2}\phi \right ) -\frac{d\Phi }{ds}\left ( \cos \phi \right ) \tag{3} \end{align}
Substituting (2,3) into (1) gives \sin ^{2}\phi \left ( \frac{d^{2}\Phi }{ds^{2}}\left ( \sin ^{2}\phi \right ) -\frac{d\Phi }{ds}\left ( \cos \phi \right ) \right ) +\left ( \sin \phi \cos \phi \right ) \left ( \frac{d\Phi }{ds}\left ( -\sin \phi \right ) \right ) +\left ( \mu ^{2}\sin ^{2}\phi \right ) \Phi =0 Dividing by \sin ^{2}\phi gives\begin{align*} \frac{d^{2}\Phi }{ds^{2}}\sin ^{2}\phi -\frac{d\Phi }{ds}\cos \phi -\cos \phi \frac{d\Phi }{ds}+\mu ^{2}\Phi & =0\\ \frac{d^{2}\Phi }{ds^{2}}\sin ^{2}\phi -2\frac{d\Phi }{ds}\cos \phi +\mu ^{2}\Phi & =0 \end{align*}
But \cos \phi =s and \sin ^{2}\phi =1-\cos ^{2}\phi =1-s^{2}, therefore the above reduces to \left ( 1-s^{2}\right ) \frac{d^{2}\Phi }{ds^{2}}-2s\frac{d\Phi }{ds}+\mu ^{2}\Phi =0
Which is Legendre’s equation.
Solution
TO DO