Problem Solve the PDE
\begin{equation} u_{t}=u_{xx}+xt\qquad 0\leq x\leq 1,t\geq 0 \tag{1} \end{equation} With boundary conditions\begin{align*} u\left ( 0,t\right ) & =0\\ u\left ( 1,t\right ) & =0 \end{align*}
And initial condition\[ u\left ( x,0\right ) =\sin \left ( \pi x\right ) \] Solution
The corresponding homogeneous PDE \(u_{t}=u_{xx}\) with the same homogeneous boundary conditions was solved before. It was found to have eigenfunctions\[ \Phi _{n}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{n}}x\right ) \] With corresponding eigenvalues\[ \lambda _{n}=n^{2}\pi ^{2}\qquad n=1,2,3,\cdots \] Using eigenfunction expansion, it is now assumed that the solution to the given inhomogeneous PDE is given by\[ u\left ( x,t\right ) =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \] Substituting the above into the original PDE (1), and since term by term differentiation is justified (eigenfunctions are continuous) results in\begin{equation} \sum _{n=1}^{\infty }b_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \Phi _{n}^{\prime \prime }\left ( x\right ) +\sum _{n=1}^{\infty }\gamma _{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \tag{1A} \end{equation} Where \(\sum _{n=1}^{\infty }\gamma _{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \) is the expansion of the forcing function \(xt\) using same eigenfunctions\begin{equation} xt=\sum _{n=1}^{\infty }\gamma _{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \tag{1B} \end{equation} But \(\Phi _{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}\Phi _{n}\left ( x\right ) \) since the eigenfunctions satisfy the eigenvalue ODE \(X^{\prime \prime }=-\lambda _{n}X\). Therefore (1A) simplifies to\begin{align} \sum _{n=1}^{\infty }b_{n}^{\prime }\left ( t\right ) \Phi _{n}\left ( x\right ) & =\sum _{n=1}^{\infty }-\lambda _{n}b_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) +\sum _{n=1}^{\infty }\gamma _{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \nonumber \\ b_{n}^{\prime }\left ( t\right ) +\lambda _{n}b_{n}\left ( t\right ) & =\gamma _{n}\left ( t\right ) \tag{2} \end{align}
\(\gamma _{n}\left ( t\right ) \) is now found by applying orthogonality to (1B), and using the weight \(r\left ( x\right ) =1\) gives\[ t\int _{0}^{1}x\Phi _{n}\left ( x\right ) dx=\gamma _{n}\left ( t\right ) \int _{0}^{1}\Phi _{n}^{2}\left ( x\right ) dx \] Using \(\Phi _{n}\left ( x\right ) =\sin \left ( \sqrt{\lambda _{n}}x\right ) =\sin \left ( n\pi x\right ) \) and \(\int _{0}^{1}\sin ^{2}\left ( n\pi x\right ) dx\) \(=\frac{1}{2}\), the above simplifies to\begin{align} t\int _{0}^{1}x\sin \left ( n\pi x\right ) dx & =\gamma _{n}\left ( t\right ) \frac{1}{2}\nonumber \\ \gamma _{n}\left ( t\right ) & =2t\int _{0}^{1}x\sin \left ( n\pi x\right ) dx \tag{3} \end{align}
The integral on the right side above is found using\(\ \int x\sin \left ( ax\right ) dx=\frac{\sin ax}{a^{2}}-\frac{x\cos ax}{a}\), therefore\begin{align*} \int _{0}^{1}x\sin \left ( n\pi x\right ) dx & =\left ( \frac{\sin n\pi x}{n^{2}\pi ^{2}}-\frac{x\cos n\pi x}{n\pi }\right ) _{0}^{1}\\ & =\left ( \frac{\sin n\pi }{n^{2}\pi ^{2}}-\frac{\cos n\pi }{n\pi }\right ) \\ & =-\frac{\cos n\pi }{n\pi }\\ & =\frac{-\left ( -1\right ) ^{n}}{n\pi }\\ & =\frac{\left ( -1\right ) ^{n+1}}{n\pi } \end{align*}
Hence equation (3) now can be written as\[ \gamma _{n}\left ( t\right ) =\frac{2\left ( -1\right ) ^{n+1}}{n\pi }t \] Substituting the above in (2) gives the first order ODE to solve for \(b_{n}\left ( t\right ) \)\[ b_{n}^{\prime }\left ( t\right ) +\left ( n\pi \right ) ^{2}b_{n}\left ( t\right ) =\frac{2\left ( -1\right ) ^{n+1}}{n\pi }t \] The integrating factor is \(I=e^{n^{2}\pi ^{2}t}\). Hence the above becomes, after multiplying both sides by \(I\)\[ \frac{d}{dt}\left ( e^{n^{2}\pi ^{2}t}b_{n}\left ( t\right ) \right ) =\frac{2\left ( -1\right ) ^{n+1}}{n\pi }te^{n^{2}\pi ^{2}t}\] Integrating both sides gives\begin{equation} e^{n^{2}\pi ^{2}t}b_{n}\left ( t\right ) =\frac{2\left ( -1\right ) ^{n+1}}{n\pi }\int _{0}^{t}se^{n^{2}\pi ^{2}s}ds+b_{n}\left ( 0\right ) \tag{4} \end{equation} Where \(b_{n}\left ( 0\right ) \) is the constant of integration. Dividing both sides by \(e^{n^{2}\pi ^{2}t}\) gives \[ b_{n}\left ( t\right ) =\frac{2\left ( -1\right ) ^{n+1}}{n\pi }\int _{0}^{t}se^{n^{2}\pi ^{2}\left ( s-t\right ) }ds+b_{n}\left ( 0\right ) e^{-n^{2}\pi ^{2}t}\] But \(\int _{0}^{t}se^{n^{2}\pi ^{2}\left ( s-t\right ) }ds=\frac{n^{2}\pi ^{2}t-1+e^{-n^{2}\pi ^{2}t}}{n^{4}\pi ^{4}}\) by integration by parts. The above now becomes\[ b_{n}\left ( t\right ) =2\left ( -1\right ) ^{n+1}\left ( \frac{n^{2}\pi ^{2}t-1+e^{-n^{2}\pi ^{2}t}}{n^{5}\pi ^{5}}\right ) +b_{n}\left ( 0\right ) e^{-n^{2}\pi ^{2}t}\] Now that \(b_{n}\left ( t\right ) \) is found, the final solution is\begin{align} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \Phi _{n}\left ( x\right ) \nonumber \\ & =\sum _{n=1}^{\infty }\left ( 2\left ( -1\right ) ^{n+1}\left ( \frac{n^{2}\pi ^{2}t-1+e^{-n^{2}\pi ^{2}t}}{n^{5}\pi ^{5}}\right ) +b_{n}\left ( 0\right ) e^{-n^{2}\pi ^{2}t}\right ) \sin \left ( n\pi x\right ) \tag{5} \end{align}
\(b_{n}\left ( 0\right ) \) is determined from the given initial conditions \(u\left ( x,0\right ) =\sin \pi x\). The above becomes at \(t=0\)\begin{align*} \sin \pi x & =\sum _{n=1}^{\infty }\left ( 2\left ( -1\right ) ^{n+1}\left ( \frac{-1+1}{n^{5}\pi ^{5}}\right ) +b_{n}\left ( 0\right ) \right ) \sin \left ( n\pi x\right ) \\ & =\sum _{n=1}^{\infty }b_{n}\left ( 0\right ) \sin \left ( n\pi x\right ) \end{align*}
Therefore when \(n=1\) (since LHS is \(\sin \pi x\,\)) the above gives\[ b_{1}\left ( 0\right ) =1 \] And \(b_{n}\left ( 0\right ) =0\) for all other \(n\). Equation (5) now simplifies to \[ u\left ( x,t\right ) =\overset{n=1\text{ term }}{\overbrace{\left ( 2\left ( \frac{\pi ^{2}t-1+e^{-\pi ^{2}t}}{\pi ^{5}}\right ) +e^{-\pi ^{2}t}\right ) \sin \left ( \pi x\right ) }}+\frac{1}{\pi ^{5}}\sum _{n=2}^{\infty }\frac{2}{n^{5}}\left ( -1\right ) ^{n+1}\left ( n^{2}\pi ^{2}t+e^{-n^{2}\pi ^{2}t}-1\right ) \sin \left ( n\pi x\right ) \] To verify the above solution, it was plotted against numerical solution for different instances of time and also animated. It gave an exact match. A small number of terms was needed in the summation since convergence was fast and is of order \(O\left ( \frac{1}{n^{3}}\right ) \). The following is a plot of the above solution for different instances of times using \(5\) terms.
Problem Show that \[ \left ( \lambda -\mu \right ) \int _{0}^{1}xJ_{o}\left ( \sqrt{\lambda }x\right ) J_{o}\left ( \sqrt{\mu }x\right ) dx=\sqrt{\mu }J_{o}^{\prime }\left ( \sqrt{\mu }\right ) J_{o}\left ( \sqrt{\lambda }\right ) -\sqrt{\lambda }J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) J_{o}\left ( \sqrt{\mu }\right ) \] Hint: Use the same method that proves orthogonality of eigenfunctions in 11.4
Solution
In the above, \(\lambda \) and \(\mu \) are the eigenvalues, with the corresponding eigenfunctions \begin{align} \Phi _{\lambda }\left ( x\right ) & =J_{o}\left ( \sqrt{\lambda }x\right ) \tag{1}\\ \Phi _{\mu }\left ( x\right ) & =J_{o}\left ( \sqrt{\mu }x\right ) \tag{2} \end{align}
These come from the Sturm Liouville equation\begin{equation} -\left ( xy^{\prime }\right ) ^{\prime }=\lambda xy\tag{3} \end{equation} Where \begin{align*} p\left ( x\right ) & =x\\ q\left ( x\right ) & =0\\ r\left ( x\right ) & =x \end{align*}
In operator form\begin{equation} L\left [ \Phi _{\lambda }\right ] =-\left ( \Phi _{\lambda }^{\prime }\right ) ^{\prime }=\lambda x\Phi _{\lambda }\tag{4} \end{equation} Similarly for any other eigenvalue such as \(\mu \). Multiplying both sides of (4) by \(\Phi _{\mu }\left ( x\right ) \) and integrating gives\[ \int _{0}^{1}L\left [ \Phi _{\lambda }\right ] \Phi _{\mu }dx=\int _{0}^{1}\overset{dv}{\overbrace{-\left ( \Phi _{\lambda }^{\prime }\right ) ^{\prime }}}\overset{u}{\overbrace{\Phi _{\mu }}}dx \] Integrating by part the right side results in\[ \int _{0}^{1}L\left [ \Phi _{\lambda }\right ] \Phi _{\mu }dx=\left [ -\Phi _{\lambda }^{\prime }\Phi _{\mu }\right ] _{0}^{1}-\int _{0}^{1}-\Phi _{\lambda }^{\prime }\Phi _{\mu }^{\prime }dx \] Integrating by parts again the second integral above, where now \(dv=-\Phi _{\lambda }^{\prime },u=\Phi _{\mu }^{\prime }\) gives\begin{align*} \int _{0}^{1}L\left [ \Phi _{\lambda }\right ] \Phi _{\mu }dx & =\left [ -\Phi _{\lambda }^{\prime }\Phi _{\mu }\right ] _{0}^{1}-\left ( \left [ -\Phi _{\lambda }\Phi _{\mu }^{\prime }\right ] _{0}^{1}-\int _{0}^{1}-\Phi _{\lambda }\Phi _{\mu }^{\prime \prime }dx\right ) \\ & =\left [ -\Phi _{\lambda }^{\prime }\Phi _{\mu }\right ] _{0}^{1}-\left [ -\Phi _{\lambda }\Phi _{\mu }^{\prime }\right ] _{0}^{1}+\int _{0}^{1}-\Phi _{\lambda }\Phi _{\mu }^{\prime \prime }dx\\ & =\left [ -\Phi _{\lambda }^{\prime }\Phi _{\mu }+\Phi _{\lambda }\Phi _{\mu }^{\prime }\right ] _{0}^{1}+\int _{0}^{1}\Phi _{\lambda }\left ( -\Phi _{\mu }^{\prime }\right ) ^{\prime }dx \end{align*}
But \(\left ( -\Phi _{\mu }^{\prime }\right ) ^{\prime }=L\left [ \Phi _{\mu }\right ] \). Hence the above can be written as\begin{align*} \int _{0}^{1}L\left [ \Phi _{\lambda }\right ] \Phi _{\mu }dx & =\left [ -\Phi _{\lambda }^{\prime }\Phi _{\mu }+\Phi _{\lambda }\Phi _{\mu }^{\prime }\right ] _{0}^{1}+\int _{0}^{1}L\left [ \Phi _{\mu }\right ] \Phi _{\lambda }dx\\ \int _{0}^{1}L\left [ \Phi _{\lambda }\right ] \Phi _{\mu }dx-\int _{0}^{1}L\left [ \Phi _{\mu }\right ] \Phi _{\lambda }dx & =\left [ -\Phi _{\lambda }^{\prime }\Phi _{\mu }+\Phi _{\lambda }\Phi _{\mu }^{\prime }\right ] _{0}^{1}\\ \int _{0}^{1}\left ( L\left [ \Phi _{\lambda }\right ] \Phi _{\mu }-L\left [ \Phi _{\mu }\right ] \Phi _{\lambda }\right ) dx & =\left [ -\Phi _{\lambda }^{\prime }\Phi _{\mu }+\Phi _{\lambda }\Phi _{\mu }^{\prime }\right ] _{0}^{1} \end{align*}
But \(L\left [ \Phi _{\lambda }\right ] =\lambda x\Phi _{\lambda }\) and \(L\left [ \Phi _{\mu }\right ] =\mu x\Phi _{\mu },\) therefore the above can be written as\begin{align} \int _{0}^{1}\left ( \lambda x\Phi _{\lambda }\Phi _{\mu }-\mu x\Phi _{\mu }\Phi _{\lambda }\right ) dx & =\left [ -\Phi _{\lambda }^{\prime }\Phi _{\mu }+\Phi _{\lambda }\Phi _{\mu }^{\prime }\right ] _{0}^{1}\nonumber \\ \int _{0}^{1}\left ( \lambda -\mu \right ) \left ( x\Phi _{\lambda }\Phi _{\mu }\right ) dx & =\left [ -\Phi _{\lambda }^{\prime }\Phi _{\mu }+\Phi _{\lambda }\Phi _{\mu }^{\prime }\right ] _{0}^{1}\nonumber \\ \left ( \lambda -\mu \right ) \int _{0}^{1}x\Phi _{\lambda }\Phi _{\mu }dx & =\left [ -\Phi _{\lambda }^{\prime }\Phi _{\mu }+\Phi _{\lambda }\Phi _{\mu }^{\prime }\right ] _{0}^{1}\tag{5} \end{align}
Since \(\Phi _{\lambda }\left ( x\right ) =J_{o}\left ( \sqrt{\lambda }x\right ) ,\Phi _{\lambda }^{\prime }\left ( x\right ) =\sqrt{\lambda }J_{o}^{\prime }\left ( \sqrt{\lambda }x\right ) \) and \(\Phi _{\mu }\left ( x\right ) =J_{o}\left ( \sqrt{\mu }x\right ) ,\Phi _{\mu }^{\prime }\left ( x\right ) =\sqrt{\mu }J_{o}^{\prime }\left ( \sqrt{\mu }x\right ) \), then the above simplifies to\[ \left ( \lambda -\mu \right ) \int _{0}^{1}xJ_{o}\left ( \sqrt{\lambda }x\right ) J_{o}\left ( \sqrt{\mu }x\right ) dx=\left [ -\sqrt{\lambda }J_{o}^{\prime }\left ( \sqrt{\lambda }x\right ) J_{o}\left ( \sqrt{\mu }x\right ) +J_{o}\left ( \sqrt{\lambda }x\right ) \sqrt{\mu }J_{o}^{\prime }\left ( \sqrt{\mu }x\right ) \right ] _{0}^{1}\] What is left is to evaluate the boundary terms \(\Delta =\left [ -\sqrt{\lambda }J_{o}^{\prime }\left ( \sqrt{\lambda }x\right ) J_{o}\left ( \sqrt{\mu }x\right ) +J_{o}\left ( \sqrt{\lambda }x\right ) \sqrt{\mu }J_{o}^{\prime }\left ( \sqrt{\mu }x\right ) \right ] _{0}^{1}.\) This gives\[ \Delta =\left [ -\sqrt{\lambda }J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) J_{o}\left ( \sqrt{\mu }\right ) +J_{o}\left ( \sqrt{\lambda }\right ) \sqrt{\mu }J_{o}^{\prime }\left ( \sqrt{\mu }\right ) \right ] -\left [ -\sqrt{\lambda }J_{o}^{\prime }\left ( 0\right ) J_{o}\left ( 0\right ) +J_{o}\left ( 0\right ) \sqrt{\mu }J_{o}^{\prime }\left ( 0\right ) \right ] \] But \(J_{o}^{\prime }\left ( 0\right ) =0\) (since \(J_{o}^{\prime }\left ( x\right ) =-J_{1}\left ( x\right ) \) and \(J_{1}\left ( 0\right ) =0\) ). Therefore the boundary terms reduces to\[ \Delta =J_{o}\left ( \sqrt{\lambda }\right ) \sqrt{\mu }J_{o}^{\prime }\left ( \sqrt{\mu }\right ) -\sqrt{\lambda }J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) J_{o}\left ( \sqrt{\mu }\right ) \] Substituting this back in (5) gives the desired result\[ \left ( \lambda -\mu \right ) \int _{0}^{1}xJ_{o}\left ( \sqrt{\lambda }x\right ) J_{o}\left ( \sqrt{\mu }x\right ) dx=\sqrt{\mu }J_{o}^{\prime }\left ( \sqrt{\mu }\right ) J_{o}\left ( \sqrt{\lambda }\right ) -\sqrt{\lambda }J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) J_{o}\left ( \sqrt{\mu }\right ) \]
Problem By letting \(\mu \rightarrow \lambda \) in the formula of problem 2, derive a formula for \(\int _{0}^{1}xJ_{0}^{2}\left ( \sqrt{\lambda }x\right ) dx\). Then show that the normalized eigenfunctions of the eigenvalue problem in section 11.4 is\[ \hat{\Phi }_{n}\left ( x\right ) =\frac{\sqrt{2}J_{0}\left ( j_{n}x\right ) }{\left \vert J_{0}^{\prime }\left ( j_{n}\right ) \right \vert }\] where \(0<j_{1}<j_{2}<j_{3}<\cdots \) denote the positive zeros of \(J_{0}\)
Solution
From problem 3, the formula obtained is\[ \left ( \lambda -\mu \right ) \int _{0}^{1}xJ_{o}\left ( \sqrt{\lambda }x\right ) J_{o}\left ( \sqrt{\mu }x\right ) dx=\sqrt{\mu }J_{o}^{\prime }\left ( \sqrt{\mu }\right ) J_{o}\left ( \sqrt{\lambda }\right ) -\sqrt{\lambda }J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) J_{o}\left ( \sqrt{\mu }\right ) \] Moving \(\left ( \lambda -\mu \right ) \) to the right side gives \[ \int _{0}^{1}xJ_{o}\left ( \sqrt{\lambda }x\right ) J_{o}\left ( \sqrt{\mu }x\right ) dx=\frac{\sqrt{\mu }J_{o}^{\prime }\left ( \sqrt{\mu }\right ) J_{o}\left ( \sqrt{\lambda }\right ) -\sqrt{\lambda }J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) J_{o}\left ( \sqrt{\mu }\right ) }{\left ( \lambda -\mu \right ) }\] Taking the limit \(\lim \mu \rightarrow \lambda \) then the integral on the left becomes \(\int _{0}^{1}x\Phi _{\lambda }^{2}dx\) resulting in\begin{equation} \int _{0}^{1}xJ_{o}^{2}\left ( \sqrt{\lambda }x\right ) dx=\lim _{\mu \rightarrow \lambda }\frac{\sqrt{\mu }J_{o}^{\prime }\left ( \sqrt{\mu }\right ) J_{o}\left ( \sqrt{\lambda }\right ) -\sqrt{\lambda }J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) J_{o}\left ( \sqrt{\mu }\right ) }{\left ( \lambda -\mu \right ) } \tag{1} \end{equation} When \(\mu \rightarrow \lambda \) the right side becomes indeterminate form \(\frac{0}{0}\). Therefore L’hospital rule is used, which says that\[ \lim _{x\rightarrow a}\frac{f\left ( x\right ) }{g\left ( x\right ) }=\lim _{x\rightarrow a}\frac{f^{\prime }\left ( x\right ) }{g^{\prime }\left ( x\right ) }\] Comparing the above to (1) shows that \(\mu \) is now like \(x\) and \(\lambda \) is like \(a\). Therefore \(f^{\prime }\left ( x\right ) \) is like \begin{align*} f^{\prime }\left ( x\right ) & \equiv \frac{d}{d\mu }\left ( \sqrt{\mu }J_{o}^{\prime }\left ( \sqrt{\mu }\right ) J_{o}\left ( \sqrt{\lambda }\right ) -\sqrt{\lambda }J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) J_{o}\left ( \sqrt{\mu }\right ) \right ) \\ & \equiv \frac{d}{d\mu }\sqrt{\mu }J_{o}^{\prime }\left ( \sqrt{\mu }\right ) J_{o}\left ( \sqrt{\lambda }\right ) -\frac{d}{d\mu }\sqrt{\lambda }J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) J_{o}\left ( \sqrt{\mu }\right ) \\ & \equiv \frac{1}{2}\frac{1}{\sqrt{\mu }}J_{o}^{\prime }\left ( \sqrt{\mu }\right ) J_{o}\left ( \sqrt{\lambda }\right ) +\sqrt{\mu }\frac{1}{2\sqrt{\mu }}J_{o}^{\prime \prime }\left ( \sqrt{\mu }\right ) J_{o}\left ( \sqrt{\lambda }\right ) -\frac{1}{2\sqrt{\mu }}\sqrt{\lambda }J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) J_{o}^{\prime }\left ( \sqrt{\mu }\right ) \end{align*}
And \(g^{\prime }\left ( x\right ) \) is like \(\frac{d}{d\mu }\left ( \lambda -\mu \right ) =-1\). Using the above result back in (1) gives\begin{align*} \int _{0}^{1}xJ_{o}^{2}\left ( \sqrt{\lambda }\right ) dx & \equiv \lim _{x\rightarrow a}\frac{f^{\prime }\left ( x\right ) }{g^{\prime }\left ( x\right ) }\\ & =\lim _{\mu \rightarrow \lambda }\left ( -\frac{1}{2}\frac{1}{\sqrt{\mu }}J_{o}^{\prime }\left ( \sqrt{\mu }\right ) J_{o}\left ( \sqrt{\lambda }\right ) -\sqrt{\mu }\frac{1}{2\sqrt{\mu }}J_{o}^{\prime \prime }\left ( \sqrt{\mu }\right ) J_{o}\left ( \sqrt{\lambda }\right ) +\frac{1}{2\sqrt{\mu }}\sqrt{\lambda }J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) J_{o}^{\prime }\left ( \sqrt{\mu }\right ) \right ) \\ & =\lim _{\mu \rightarrow \lambda }\left ( -\frac{1}{2}\frac{1}{\sqrt{\mu }}J_{o}^{\prime }\left ( \sqrt{\mu }\right ) J_{o}\left ( \sqrt{\lambda }\right ) -\frac{1}{2}J_{o}^{\prime \prime }\left ( \sqrt{\mu }\right ) J_{o}\left ( \sqrt{\lambda }\right ) +\frac{1}{2}J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) J_{o}^{\prime }\left ( \sqrt{\mu }\right ) \right ) \end{align*}
Now the limit is taken, since there is no indeterminate form. The above becomes\begin{align} \int _{0}^{1}xJ_{o}^{2}\left ( \sqrt{\lambda }x\right ) dx & =-\frac{1}{2}\frac{1}{\sqrt{\lambda }}J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) J_{o}\left ( \sqrt{\lambda }\right ) -\frac{1}{2}J_{o}^{\prime \prime }\left ( \sqrt{\lambda }\right ) J_{o}\left ( \sqrt{\lambda }\right ) +\frac{1}{2}J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) \nonumber \\ & =\frac{1}{2}\left ( \left [ J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) \right ] ^{2}-\frac{1}{\sqrt{\lambda }}J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) J_{o}\left ( \sqrt{\lambda }\right ) -J_{o}^{\prime \prime }\left ( \sqrt{\lambda }\right ) J_{o}\left ( \sqrt{\lambda }\right ) \right ) \tag{2} \end{align}
To simplify the above, the following relations were obtained from dlmf.NIST.gov to simplify the above\begin{align*} J_{n}^{\prime }\left ( x\right ) & =J_{n-1}\left ( x\right ) -\frac{\left ( n+1\right ) }{x}J_{n}\left ( x\right ) \\ J_{n}^{\prime }\left ( x\right ) & =-J_{n+1}\left ( x\right ) +\frac{n}{x}J_{n}\left ( x\right ) \end{align*}
Using these, then \(J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) =-J_{1}\left ( \sqrt{\lambda }\right ) \) and \(J_{0}^{\prime \prime }\left ( \sqrt{\lambda }\right ) =-J_{0}\left ( \sqrt{\lambda }\right ) +\frac{1}{\sqrt{\lambda }}J_{1}\left ( \sqrt{\lambda }\right ) \). Equation (2) now simplifies to\begin{align*} \int _{0}^{1}xJ_{o}^{2}\left ( \sqrt{\lambda }x\right ) dx & =\frac{1}{2}\left ( \left [ J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) \right ] ^{2}-\frac{1}{\sqrt{\lambda }}\left ( -J_{1}\left ( \sqrt{\lambda }\right ) \right ) J_{o}\left ( \sqrt{\lambda }\right ) -\left ( -J_{0}\left ( \sqrt{\lambda }\right ) +\frac{1}{\sqrt{\lambda }}J_{1}\left ( \sqrt{\lambda }\right ) \right ) J_{o}\left ( \sqrt{\lambda }\right ) \right ) \\ & =\frac{1}{2}\left ( \left [ J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) \right ] ^{2}+\frac{1}{\sqrt{\lambda }}J_{1}\left ( \sqrt{\lambda }\right ) J_{o}\left ( \sqrt{\lambda }\right ) +J_{0}\left ( \sqrt{\lambda }\right ) J_{o}\left ( \sqrt{\lambda }\right ) -\frac{1}{\sqrt{\lambda }}J_{1}\left ( \sqrt{\lambda }\right ) J_{o}\left ( \sqrt{\lambda }\right ) \right ) \end{align*}
The second term cancels with the last term above giving the final result\begin{equation} \int _{0}^{1}xJ_{o}^{2}\left ( \sqrt{\lambda }x\right ) dx=\frac{1}{2}\left ( \left [ J_{o}^{\prime }\left ( \sqrt{\lambda }\right ) \right ] ^{2}+J_{0}^{2}\left ( \sqrt{\lambda }\right ) \right ) \tag{3} \end{equation}
\(\sqrt{\lambda _{n}}\) are the positive zeros of \(J_{0}\left ( \sqrt{\lambda _{n}}\right ) =0\). Below, \(\sqrt{\lambda _{n}}\) is replaced by \(j_{n}\) where now \(j_{n}\) are the zeros of \(J_{0}\left ( j_{n}\right ) \). One way to find the normalized eigenfunction \(\hat{J}_{0}\left ( j_{n}x\right ) \) is by dividing \(J_{0}\left ( j_{n}x\right ) \) by its norm. In other words, \begin{equation} \hat{J}_{0}\left ( j_{n}x\right ) =\frac{J_{0}\left ( j_{n}x\right ) }{\left \Vert J_{0}\left ( j_{n}x\right ) \right \Vert } \tag{1A} \end{equation} But\[ \left \Vert J_{0}\left ( j_{n}x\right ) \right \Vert =\sqrt{\int _{0}^{1}r\left ( x\right ) J_{0}^{2}\left ( j_{n}x\right ) dx}\] Which is by the definition of the norm of a function with the corresponding weight \(r\left ( x\right ) \). But from part(a) \(\left \Vert J_{0}\left ( j_{n}x\right ) \right \Vert =\int _{0}^{1}r\left ( x\right ) J_{0}^{2}\left ( j_{n}x\right ) dx\) was found to be \(\frac{1}{2}\left ( \left [ J_{o}^{\prime }\left ( j_{n}\right ) \right ] ^{2}+J_{0}^{2}\left ( j_{n}\right ) \right ) \). Therefore (1A) becomes\begin{align*} \hat{J}_{0}\left ( j_{n}x\right ) & =\frac{J_{0}\left ( j_{n}x\right ) }{\sqrt{\frac{1}{2}\left ( \left [ J_{o}^{\prime }\left ( j_{n}\right ) \right ] ^{2}+J_{0}^{2}\left ( j_{n}\right ) \right ) }}\\ & =\frac{\sqrt{2}J_{0}\left ( j_{n}x\right ) }{\sqrt{\left [ J_{o}^{\prime }\left ( j_{n}\right ) \right ] ^{2}+J_{0}^{2}\left ( j_{n}\right ) }} \end{align*}
But since \(j_{n}\) are the zeros of \(J_{0}\left ( j_{n}\right ) \), then all the \(J_{0}\left ( j_{n}\right ) \) terms above vanish giving\begin{align} \hat{J}_{0}\left ( j_{n}x\right ) & =\frac{\sqrt{2}J_{0}\left ( j_{n}x\right ) }{\sqrt{\left [ J_{o}^{\prime }\left ( j_{n}\right ) \right ] ^{2}}}\nonumber \\ & =\frac{\sqrt{2}J_{0}\left ( j_{n}x\right ) }{\left \vert J_{o}^{\prime }\left ( j_{n}\right ) \right \vert } \tag{1} \end{align}
Another way to find the normalized eigenfunctions \(\hat{J}_{0}\left ( j_{n}x\right ) \) is as was done in the text book, which is to first determine \(k_{n}\) as follows. Let \(\hat{J}_{0}\left ( j_{n}x\right ) =k_{n}J_{0}\left ( j_{n}x\right ) \), then the following equation is solved for \(k_{n}\)\begin{equation} \int _{0}^{1}r\left ( x\right ) \left [ \hat{J}_{0}\left ( j_{n}x\right ) \right ] ^{2}dx=1 \tag{2} \end{equation} But the weight \(r\left ( x\right ) =x,\) equation (2) becomes\[ k_{n}^{2}\int _{0}^{1}xJ_{0}^{2}\left ( j_{n}x\right ) dx=1 \] But from part(a), \(\int _{0}^{1}xJ_{0}^{2}\left ( j_{n}x\right ) dx=\frac{1}{2}\left ( \left [ J_{o}^{\prime }\left ( j_{n}\right ) \right ] ^{2}+J_{0}^{2}\left ( j_{n}\right ) \right ) \). Hence the above becomes\begin{align*} k_{n}^{2} & =\frac{1}{\frac{1}{2}\left ( \left [ J_{o}^{\prime }\left ( j_{n}\right ) \right ] ^{2}+J_{0}^{2}\left ( j_{n}\right ) \right ) }\\ k_{n} & =\frac{\sqrt{2}}{\sqrt{\left [ J_{o}^{\prime }\left ( j_{n}\right ) \right ] ^{2}+J_{0}^{2}\left ( j_{n}\right ) }} \end{align*}
As above, since all \(J_{0}\left ( j_{n}\right ) =0\) then \[ k_{n}=\frac{\sqrt{2}}{\sqrt{\left [ J_{o}^{\prime }\left ( j_{n}\right ) \right ] ^{2}}}\] And the normalized eigenfunction become \begin{align*} \hat{J}_{0}\left ( j_{n}x\right ) & =k_{n}J_{0}\left ( j_{n}x\right ) \\ & =\frac{\sqrt{2}J_{0}\left ( j_{n}x\right ) }{\sqrt{\left [ J_{o}^{\prime }\left ( j_{n}\right ) \right ] ^{2}}}\\ & =\frac{\sqrt{2}J_{0}\left ( j_{n}x\right ) }{\left \vert J_{o}^{\prime }\left ( j_{n}\right ) \right \vert } \end{align*}
Which is the same result as (1).
Problem Solve the inhomogeneous differential equation\[ -\left ( \left ( 1-x^{2}\right ) y^{\prime }\right ) ^{\prime }=y+x^{3}\qquad -1<x<1 \] With boundary conditions \(y\left ( x\right ) ,y^{\prime }\left ( x\right ) \) bounded as \(x\rightarrow -1^{+}\) and \(x\rightarrow 1^{-}\).
Solution
This problem is solved using 11.3 method (Eigenfunction expansion). The ODE is written as\begin{equation} -\left ( \left ( 1-x^{2}\right ) y^{\prime }\right ) ^{\prime }=\mu y+x^{3} \tag{1} \end{equation} Where \(\mu =1\) in this case. The corresponding homogeneous eigenvalue ODE to solve is then\begin{equation} -\left ( \left ( 1-x^{2}\right ) y^{\prime }\right ) ^{\prime }=\lambda y \tag{2} \end{equation} Comparing to Sturm-Liouville form \(-\left ( py^{\prime }\right ) ^{\prime }+qy=r\lambda y\), then \(p\left ( x\right ) =\left ( 1-x^{2}\right ) ,q=0,r=1\). Since \(p\left ( x\right ) \) must be positive over all points in the domain, and since in this problem \(p\left ( -1\right ) =0\) and \(p\left ( 1\right ) =0\), then both \(x=-1,+1\) are singular points. They can be shown to be regular singular points.
Equation (2), where \(\lambda \) is now is an eigenvalue, is the Legendre equation\[ \left ( 1-x^{2}\right ) y^{\prime \prime }-2xy^{\prime }+\lambda y=0 \] Comparing to the standard Legendre equation form in chapter 5\begin{equation} \left ( 1-x^{2}\right ) y^{\prime \prime }-2xy^{\prime }+n\left ( n+1\right ) y=0 \tag{3} \end{equation} There are two cases to consider. \(n\) is integer and \(n\) is not an integer.
Case \(n\) is not an integer. It is know that now the solution to (3) is\[ y\left ( x\right ) =c_{1}\bar{P}_{n}\left ( x\right ) +c_{2}\bar{Q}_{n}\left ( x\right ) \] Where \(\bar{P}_{n}\left ( x\right ) \) is called the Legendre function of order \(n\,\) and \(\bar{Q}_{n}\left ( x\right ) \) is called the Legendre function of the second kind of order \(n\). These solutions are valid for \(\left \vert x\right \vert <1\) since series expansion was about point \(x=0\). But both of these functions are unbounded at the end points (\(\bar{Q}_{n}\left ( x\right ) \) blows up at \(x=\pm 1\) and \(\bar{P}_{n}\left ( x\right ) \) blows up at \(x=-1\)) leading to trivial solution.
This means \(n\) must be an integer. When \(n\) is an integer, then \(\lambda _{n}=n\left ( n+1\right ) \). It is known (from chapter 5), that in this case the solution to (3) becomes a terminating power series (a polynomial), which is called the Legendre polynomial \(P_{n}\left ( x\right ) .\)These polynomials are there bounded everywhere, including at the end points \(x=\pm 1\), and therefore these solutions satisfy the boundary conditions. Hence the Legendre \(P_{n}\left ( x\right ) \) are the eigenfunctions to (3). This table summaries the result found
| \(n\) | eigenvalue | eigenfunctions |
| \(0\) | \(\lambda _{0}=0\) | \(P_{0}\left ( x\right ) =1\) |
| \(1\) | \(\lambda _{1}=2\) | \(P_{1}\left ( x\right ) =x\) |
| \(2\) | \(\lambda _{2}=6\) | \(P_{2}\left ( x\right ) =\frac{1}{2}\left ( 3x^{2}-1\right ) \) |
| \(3\) | \(\lambda _{3}=12\) | \(P_{3}\left ( x\right ) =\frac{1}{2}\left ( 5x^{3}-3x\right ) \) |
| \(\vdots \) | \(\vdots \) | \(\vdots \) |
| \(n\) | \(\lambda _{n}=n\left ( n+1\right ) \) | \(P_{n}\left ( x\right ) =\frac{1}{2^{n}n!}\frac{d^{n}}{dx^{2}}\left ( x^{2}-1\right ) ^{n}\) |
What the above says, is that the solution to \[ \left ( 1-x^{2}\right ) P_{n}^{\prime \prime }\left ( x\right ) -2xP_{n}^{\prime }\left ( x\right ) +\lambda _{n}P_{n}\left ( x\right ) =0 \] Is \(P_{n}\left ( x\right ) \) with the corresponding eigenvalue \(\lambda _{n}=n\left ( n+1\right ) \) as given by the above table. Now that the eigenfunctions of the corresponding homogeneous eigenvalue ODE are found, they are used to solve the given inhomogeneous ODE\begin{equation} -\left ( \left ( 1-x^{2}\right ) y^{\prime }\right ) ^{\prime }=\mu y+x^{3} \tag{4} \end{equation} Using eigenfunction expansion method. Since \(\mu =1\) and since there is no eigenvalue which is also \(1\), then a solution exists. Let the solution be\[ y\left ( x\right ) =\sum _{n=0}^{\infty }c_{n}P_{n}\left ( x\right ) \] Substituting this solution into (4), and noting that \(L\left [ y\right ] =-\left ( \left ( 1-x^{2}\right ) y^{\prime }\right ) ^{\prime }=\lambda _{n}y\) gives\[ \lambda _{n}\sum _{n=0}^{\infty }c_{n}P_{n}\left ( x\right ) =\mu \sum _{n=0}^{\infty }c_{n}P_{n}\left ( x\right ) +x^{3}\] Expanding \(x^{3}\) using the same eigenfunctions (this can be done, since \(x^{3}\) is continuous function and the eigenfunctions are complete), then the above becomes\begin{align*} \lambda _{n}\sum _{n=0}^{\infty }c_{n}P_{n}\left ( x\right ) & =\mu \sum _{n=0}^{\infty }c_{n}P_{n}\left ( x\right ) +\sum _{n=0}^{\infty }d_{n}P_{n}\left ( x\right ) \\ \lambda _{n}c_{n} & =\mu c_{n}+d_{n}\\ c_{n} & =\frac{d_{n}}{\lambda _{n}-\mu } \end{align*}
What is left is to determine \(d_{n}\) from\[ x^{3}=\sum _{n=0}^{\infty }d_{n}P_{n}\left ( x\right ) \] The above can be solved for \(d_{n}\) using orthogonality, or by direct expansion (otherwise called undetermined coefficients method). Since the force \(x^{3}\) is already a polynomial in \(x\) and of a small order, then direct expansion is simpler. The above then becomes\[ x^{3}=d_{0}P_{0}\left ( x\right ) +d_{1}P_{1}\left ( x\right ) +d_{2}P_{2}\left ( x\right ) +d_{3}P_{3}\left ( x\right ) \] There is no need to expand for more than \(n=3\), since the LHS polynomial is of order \(3\). Substituting the known \(P_{n}\left ( x\right ) \) expressions into the above equation gives\begin{align*} x^{3} & =d_{0}+d_{1}x+d_{2}\frac{1}{2}\left ( 3x^{2}-1\right ) +d_{3}\frac{1}{2}\left ( 5x^{3}-3x\right ) \\ & =d_{0}+d_{1}x+d_{2}\left ( \frac{3}{2}x^{2}-\frac{1}{2}\right ) +d_{3}\left ( \frac{5}{2}x^{3}-\frac{3}{2}x\right ) \end{align*}
Collecting terms of equal powers in \(x\) results in\[ x^{3}=x^{0}\left ( d_{0}-\frac{1}{2}d_{2}\right ) +x\left ( d_{1}-\frac{3}{2}d_{3}\right ) +x^{2}\left ( \frac{3}{2}d_{2}\right ) +x^{3}\left ( \frac{5}{2}d_{3}\right ) \] Or\begin{align*} d_{0}-\frac{1}{2}d_{2} & =0\\ d_{1}-\frac{3}{2}d_{3} & =0\\ \frac{3}{2}d_{2} & =0\\ \frac{5}{2}d_{3} & =1 \end{align*}
From third equation, \(d_{2}=0\). From first equation \(d_{0}=0\), and substituting last equation in the second equation give \(d_{1}=\frac{3}{2}\). Therefore\begin{align*} d_{1} & =\frac{3}{5}\\ d_{3} & =\frac{2}{5} \end{align*}
And all other \(d_{n}\) are zero. Now the \(c_{n}\) are found using \(c_{n}=\frac{d_{n}}{\lambda _{n}-\mu }.\) For \(n=1\)\[ c_{1}=\frac{d_{1}}{\lambda _{1}-\mu }=\frac{\frac{3}{5}}{2-1}=\frac{3}{5}\] And for \(n=3\)\[ c_{3}=\frac{d_{3}}{\lambda _{3}-\mu }=\frac{\frac{2}{5}}{12-1}=\frac{2}{55}\] And all other \(c_{n}\) are zero. Hence the final solution from \(y\left ( x\right ) =\sum _{n=0}^{\infty }c_{n}P_{n}\left ( x\right ) \) reduces to only two terms in the sum\begin{align*} y\left ( x\right ) & =c_{1}P_{1}\left ( x\right ) +c_{3}P_{3}\left ( x\right ) \\ & =\frac{3}{5}x+\frac{2}{55}\left ( \frac{1}{2}\left ( 5x^{3}-3x\right ) \right ) \end{align*}
Giving the final solution as\[ \fbox{$y\left ( x\right ) =\frac{1}{11}x\left ( x^2+6\right ) $}\] This is a plot of the solution
Initially I did not know we had to use eigenfunction expansion, so solved it directly as follows. Let the solution to \[ \left ( 1-x^{2}\right ) y^{\prime \prime }-2xy^{\prime }+y=x^{3}\] Be \[ y\left ( x\right ) =y_{h}\left ( x\right ) +y_{p}\left ( x\right ) \] Where \(y_{h}\left ( x\right ) \) is the homogeneous solution to \(\left ( 1-x^{2}\right ) y^{\prime \prime }-2xy^{\prime }+y=0\) and \(y_{p}\left ( x\right ) \) is a particular solution. Now, since \(\left ( 1-x^{2}\right ) y^{\prime \prime }-2xy^{\prime }+y=0\) is a Legendre ODE but with a non-integer order, then its solution is not a terminating polynomials, but instead is given by\[ y_{h}\left ( x\right ) =c_{1}\bar{P}_{n}\left ( x\right ) +c_{2}\bar{Q}_{n}\left ( x\right ) \] Where \(\bar{P}_{n}\left ( x\right ) \) is called the Legendre function of order \(n\,\) and \(\bar{Q}_{n}\left ( x\right ) \) is called the Legendre function of the second kind of order \(n\), and \(y_{p}\left ( x\right ) \) is a particular solution. The particular solution can be found, using method of undetermined coefficients to be \(y_{p}\left ( x\right ) =\frac{1}{11}x^{3}+\frac{6}{11}x\). Hence the general solution becomes\[ y\left ( x\right ) =c_{1}\bar{P}_{n}\left ( x\right ) +c_{2}\bar{Q}_{n}\left ( x\right ) +\frac{1}{11}x\left ( x^{2}+6\right ) \] Now since the solution must be bounded as \(x\rightarrow \pm 1\), then we must set \(c_{1}=0\) and \(c_{2}=0\), because both \(\bar{P}_{n}\left ( x\right ) \) and \(\bar{Q}_{n}\left ( x\right ) \) are unbounded at the end points (\(\bar{Q}_{n}\left ( x\right ) \) blows up at \(x=\pm 1\) and \(\bar{P}_{n}\left ( x\right ) \) blows up at only \(x=-1\)), therefore the final solution contains only the particular solution\[ y\left ( x\right ) =\frac{1}{11}x\left ( x^{2}+6\right ) \] Which is the same solution found using eigenfunction expansion. At first I thought I made an error somewhere, since I did not think all of the homogenous solution basis could vanish leaving only a particular solution.