Solution
\begin{align*} c_{n} & =\left \langle f\left ( x\right ) ,\phi _{n}\left ( x\right ) \right \rangle \\ & =\int _{1}^{b}p\left ( x\right ) f\left ( x\right ) \phi _{n}\left ( x\right ) dx \end{align*}
But \(p\left ( x\right ) =\frac{1}{x}\) and \(\phi _{n}\left ( x\right ) =\sqrt{\frac{2}{\ln b}}\sin \left ( \alpha _{n}\ln x\right ) \) and \(f\left ( x\right ) =x\) therefore the above becomes\begin{align*} c_{n} & =\int _{1}^{b}\frac{1}{x}x\sqrt{\frac{2}{\ln b}}\sin \left ( \alpha _{n}\ln x\right ) dx\\ & =\sqrt{\frac{2}{\ln b}}\int _{1}^{b}\sin \left ( \alpha _{n}\ln x\right ) dx \end{align*}
But \(\alpha _{n}=\frac{n\pi }{\ln b}\), therefore\[ c_{n}=\sqrt{\frac{2}{\ln b}}\int _{1}^{b}\sin \left ( \frac{n\pi }{\ln b}\ln x\right ) dx \] Let \(s=\pi \frac{\ln x}{\ln b}\ \), hence \(\frac{ds}{dx}=\frac{\pi }{\ln b}\frac{1}{x}\). When \(x=1\rightarrow s=0\) and when \(x=b\rightarrow s=\pi \). The above becomes\[ c_{n}=\sqrt{\frac{2}{\ln b}}\int _{0}^{\pi }\sin \left ( ns\right ) \frac{\ln \left ( b\right ) }{\pi }xds \] But \(\ln x=\frac{s}{\pi }\ln b\), hence \(x=e^{s\frac{\ln b}{\pi }}\), and the above becomes\begin{equation} c_{n}=\frac{\sqrt{2\ln \left ( b\right ) }}{\pi }\int _{0}^{\pi }e^{s\frac{\ln b}{\pi }}\sin \left ( ns\right ) ds\tag{1} \end{equation} Using\[ \int e^{ax}\sin \left ( bx\right ) ds=\frac{e^{ax}}{a^{2}+b^{2}}\left ( a\sin bx-b\cos bx\right ) \] Where in our case \(a=\frac{\ln b}{\pi }\) and \(b=n\). Applying the above gives \begin{align*} \int _{0}^{\pi }e^{s\frac{\ln b}{\pi }}\sin \left ( ns\right ) ds & =\left [ \frac{e^{\frac{\ln b}{\pi }x}}{\left ( \frac{\ln b}{\pi }\right ) ^{2}+n^{2}}\left ( \frac{\ln b}{\pi }\sin nx-n\cos nx\right ) \right ] _{0}^{\pi }\\ & =\frac{1}{\left ( \frac{\ln b}{\pi }\right ) ^{2}+n^{2}}\left [ e^{\frac{\ln b}{\pi }\pi }\left ( \frac{\ln b}{\pi }\sin n\pi -n\cos n\pi \right ) -\left ( 0-n\right ) \right ] \end{align*}
But \(\sin n\pi =0\) since \(n\) integer, giving\begin{align*} \int _{0}^{\pi }e^{s\frac{\ln b}{\pi }}\sin \left ( ns\right ) ds & =\frac{1}{\left ( \frac{\ln b}{\pi }\right ) ^{2}+n^{2}}\left [ -bn\cos n\pi +n\right ] \\ & =\frac{\pi ^{2}}{\left ( \ln b\right ) ^{2}+\pi ^{2}n^{2}}\left [ -bn\left ( -1\right ) ^{n}+n\right ] \\ & =\frac{\pi ^{2}\left ( bn\left ( -1\right ) ^{n+1}+n\right ) }{\left ( \ln b\right ) ^{2}+\pi ^{2}n^{2}} \end{align*}
Hence (1) becomes\begin{align*} c_{n} & =\frac{\sqrt{2\ln \left ( b\right ) }}{\pi }\frac{n\pi ^{2}\left ( 1+\left ( -1\right ) ^{n+1}b\right ) }{\left ( \ln b\right ) ^{2}+\left ( \pi n\right ) ^{2}}\\ & =\sqrt{2\ln \left ( b\right ) }\frac{n\pi \left ( 1+\left ( -1\right ) ^{n+1}b\right ) }{\left ( \ln b\right ) ^{2}+\left ( \pi n\right ) ^{2}} \end{align*}
Where \(n=1,2,3,\cdots \), which is the result required to show.
Solution
\begin{align*} \phi _{n}\left ( x\right ) & =\sqrt{\frac{2}{c}}\sin \left ( \alpha _{n}x\right ) \qquad n=1,2,3,\cdots \\ \alpha _{n} & =\pi \frac{2n-1}{2x} \end{align*}
Since \(\phi _{n}\left ( x\right ) \) are complete, then we can represent \(f\left ( x\right ) \) using \(\phi _{n}\left ( x\right ) \) as generalized Fourier series using\[ f\left ( x\right ) =\sum _{n=1}^{\infty }B_{n}\phi _{n}\left ( x\right ) \qquad 0<x<c \] To find \(B_{n}\), since \(\phi _{n}\left ( x\right ) \) are orthonormal eigenfunctions then\begin{align*} B_{n} & =\left \langle f\left ( x\right ) ,\phi _{n}\left ( x\right ) \right \rangle \\ & =\int _{0}^{c}p\left ( x\right ) f\left ( x\right ) \phi _{n}dx \end{align*}
But problem (7) section 72 is \(X^{\prime \prime }+\lambda X=0\) which implies that \(p\left ( x\right ) =1\). Hence the above becomes\begin{align*} B_{n} & =\int _{0}^{c}f\left ( x\right ) \sqrt{\frac{2}{c}}\sin \left ( \alpha _{n}x\right ) dx\\ & =\sqrt{\frac{2}{c}}\int _{0}^{c}f\left ( x\right ) \sin \left ( \alpha _{n}x\right ) dx \end{align*}
Which is the result required to show.
Theorem 2 section 15 gives the conditions on \(f\left ( x\right ) \) for it to have a Fourier sine series which converges to \(f\left ( x\right ) \) where \(f\left ( x\right ) \) is continuous and converges to mean value of \(f\left ( x\right ) \) where \(f\left ( x\right ) \) have a jump discontinuity.
Since \(f\left ( x\right ) \) is piecewise continuous in this problem, then for those regions where \(f\left ( x\right ) \) is continuous between \(0<x<c\), the series found in part(a) converges to \(f\left ( x\right ) \) and is valid Fourier sine series representation of \(f\left ( x\right ) \) there.
Solution
Solution (6) is given by\begin{equation} u\left ( x,t\right ) =\sum _{n=1}^{\infty }A_{n}\exp \left ( -\alpha _{n}^{2}kt\right ) \cos \left ( \alpha _{n}x\right ) \tag{6} \end{equation} Where\[ A_{n}=\frac{2h}{h+\sin ^{2}\alpha _{n}}\int _{0}^{1}f\left ( x\right ) \cos \left ( \alpha _{n}x\right ) dx \] But \(f\left ( x\right ) =1\) which reduces the above to\begin{align*} A_{n} & =\frac{2h}{h+\sin ^{2}\alpha _{n}}\int _{0}^{1}\cos \left ( \alpha _{n}x\right ) dx\\ & =\frac{2h}{h+\sin ^{2}\alpha _{n}}\left [ \sin \left ( \alpha _{n}x\right ) \right ] _{0}^{1}\\ & =\frac{2h}{h+\sin ^{2}\alpha _{n}}\sin \left ( \alpha _{n}\right ) \end{align*}
Hence (6) becomes\[ u\left ( x,t\right ) =2h\sum _{n=1}^{\infty }\frac{\sin \left ( \alpha _{n}\right ) }{h+\sin ^{2}\alpha _{n}}\exp \left ( -\alpha _{n}^{2}kt\right ) \cos \left ( \alpha _{n}x\right ) \] But from example 1, section 72 we are given that \(\tan \left ( \alpha _{n}c\right ) =\frac{h}{\alpha _{n}}\). But \(c=1\) in this problem, hence\[ \tan \left ( \alpha _{n}\right ) =\frac{h}{\alpha _{n}}\] Which is what required to show.
Solution
\(u\left ( 0,t\right ) =0\) means that the left surface is kept at fixed temperature which is zero. And \(u_{x}\left ( 1,t\right ) +hu\left ( 1,t\right ) =0\) means that the surface heat transfer takes place at face \(x=1\) into the medium at temperature zero. To solve the PDE, we first check the boundary conditions by writing them as\begin{align*} a_{1}u\left ( 0,t\right ) +a_{2}u_{x}\left ( 0,t\right ) & =0\\ b_{1}u\left ( 1,t\right ) +b_{2}u_{x}\left ( 1,t\right ) & =0 \end{align*}
Then \(a_{1}=0,a_{2}=0\). Hence \(a_{1}a_{2}=0\). And \(b_{1}=1,b_{2}=h\). Then since it is assumed that \(h>0\) per section 26, then \(b_{1}b_{2}\geq 0\). And since \(q\left ( x\right ) =0\) from the PDE itself, then we know that eigenvalues are \(\lambda \geq 0\).
Let \(u=X\left ( x\right ) T\left ( t\right ) \) then the PDE becomes\begin{align*} T^{\prime }X & =X^{\prime \prime }T\\ \frac{T^{\prime }}{T} & =\frac{X^{\prime \prime }}{X}=-\lambda \end{align*}
Hence the Sturm Liouville problem is \begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X^{\prime }\left ( 1\right ) +hX\left ( 1\right ) & =0 \end{align*}
Where \(p\left ( x\right ) =1\).
Case \(\lambda =0\)
Solution is \[ X\left ( x\right ) =Ax+B \] At \(x=0\)\[ 0=B \] Hence solution becomes\[ X\left ( x\right ) =Ax \] At \(x=1\) the second boundary conditions gives\begin{align*} A+hA & =0\\ A\left ( 1+h\right ) & =0 \end{align*}
For non trivial solution \(1+h=0\) or \(h=-1\). But we assumed that \(h>0\). Therefore \(\lambda =0\) is not eigenvalue.
Case \(\lambda >0\)
Let \(\lambda =\alpha ^{2},\alpha >0\). Hence solution is \[ X\left ( x\right ) =A\cos \left ( \alpha x\right ) +B\sin \left ( \alpha x\right ) \] At \(X\left ( 0\right ) =0\)\[ 0=A \] The solution becomes\[ X\left ( x\right ) =B\sin \left ( \alpha x\right ) \] At \(x=1\) the second boundary conditions gives\begin{align*} B\alpha \cos \left ( \alpha \right ) +hB\sin \left ( \alpha \right ) & =0\\ \alpha \cos \left ( \alpha \right ) +h\sin \left ( \alpha \right ) & =0\\ \tan \left ( \alpha \right ) & =-\frac{\alpha }{h} \end{align*}
Therefore the eigenvalues are given by solution to \[ \tan \left ( \alpha _{n}\right ) =-\frac{\alpha _{n}}{h}\qquad n=1,2,3,\cdots \] And eigenfunctions are\[ X_{n}\left ( x\right ) =\sin \left ( \alpha _{n}x\right ) \] The normalized eigenfunctions are\[ \phi _{n}\left ( x\right ) =\frac{X_{n}\left ( x\right ) }{\left \Vert X_{n}\left ( x\right ) \right \Vert }\] But \begin{align*} \left \Vert X_{n}\left ( x\right ) \right \Vert ^{2} & =\int _{0}^{1}p\left ( x\right ) X_{n}^{2}\left ( x\right ) dx\\ & =\int _{0}^{1}\sin ^{2}\left ( \alpha _{n}x\right ) dx\\ & =\frac{1}{2}\int _{0}^{1}1-\cos \left ( 2\alpha _{n}x\right ) dx\\ & =\frac{1}{2}\left ( 1-\left [ \frac{\sin \left ( 2\alpha _{n}x\right ) }{2\alpha _{n}}\right ] _{0}^{1}\right ) \\ & =\frac{1}{2}\left ( 1-\frac{1}{2\alpha _{n}}\left [ \sin \left ( 2\alpha _{n}x\right ) \right ] _{0}^{1}\right ) \\ & =\frac{1}{2}\left ( 1-\frac{\sin \left ( 2\alpha _{n}\right ) }{2\alpha _{n}}\right ) \\ & =\frac{1}{2}-\frac{\sin \left ( 2\alpha _{n}\right ) }{4\alpha _{n}} \end{align*}
But \(\sin \left ( 2\alpha _{n}\right ) =2\sin \alpha _{n}\cos \alpha _{n}\) and \(\alpha _{n}=-h\frac{\sin \left ( \alpha _{n}\right ) }{\cos \left ( \alpha _{n}\right ) }\), therefore the above becomes\begin{align*} \left \Vert X_{n}\left ( x\right ) \right \Vert ^{2} & =\frac{1}{2}+\frac{2\sin \alpha _{n}\cos \alpha _{n}}{4h\frac{\sin \left ( \alpha _{n}\right ) }{\cos \left ( \alpha _{n}\right ) }}\\ & =\frac{1}{2}+\frac{\cos ^{2}\alpha _{n}}{2h}\\ & =\frac{h+\cos ^{2}\alpha _{n}}{2h} \end{align*}
Hence\begin{align*} \phi _{n}\left ( x\right ) & =\frac{X_{n}\left ( x\right ) }{\sqrt{\frac{h+\cos ^{2}\alpha _{n}}{2h}}}\\ & =\sqrt{\frac{2h}{h+\cos ^{2}\alpha _{n}}}\sin \left ( \alpha _{n}x\right ) \end{align*}
Now we use generalized Fourier series to find the solution. Let\begin{equation} u\left ( x,t\right ) =\sum _{n=1}^{\infty }B_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \tag{1} \end{equation} Substituting this back into the PDE gives\[ \sum _{n=1}^{\infty }B_{n}^{\prime }\left ( t\right ) \phi _{n}\left ( x\right ) =k\sum _{n=1}^{\infty }B_{n}\left ( t\right ) \phi _{n}^{\prime \prime }\left ( x\right ) \] But \(\phi _{n}^{\prime \prime }\left ( x\right ) =-\lambda _{n}\phi _{n}\left ( x\right ) =-\alpha _{n}^{2}\phi _{n}\left ( x\right ) \). The above becomes\begin{align*} \sum _{n=1}^{\infty }B_{n}^{\prime }\left ( t\right ) \phi _{n}\left ( x\right ) & =-k\sum _{n=1}^{\infty }B_{n}\left ( t\right ) \alpha _{n}^{2}\phi _{n}\left ( x\right ) \\ B_{n}^{\prime }\left ( t\right ) +k\alpha _{n}^{2}B_{n}\left ( t\right ) & =0 \end{align*}
The solution is \[ B_{n}\left ( t\right ) =B_{n}\left ( 0\right ) e^{-k\alpha _{n}^{2}t}\] Hence (1) becomes\[ u\left ( x,t\right ) =\sum _{n=1}^{\infty }B_{n}\left ( 0\right ) e^{-k\alpha _{n}^{2}t}\phi _{n}\left ( x\right ) \] At \(t=0\) the above becomes\[ f\left ( x\right ) =\sum _{n=1}^{\infty }B_{n}\left ( 0\right ) \phi _{n}\left ( x\right ) \] Therefore\begin{align*} B_{n}\left ( 0\right ) & =\left \langle f\left ( x\right ) ,\phi _{n}\left ( x\right ) \right \rangle \\ & =\int _{0}^{1}p\left ( x\right ) f\left ( x\right ) \phi _{n}\left ( x\right ) dx\\ & =\sqrt{\frac{2h}{h+\cos ^{2}\alpha _{n}}}\int _{0}^{1}f\left ( x\right ) \sin \left ( \alpha _{n}x\right ) dx \end{align*}
Therefore \begin{align*} B_{n}\left ( t\right ) & =B_{n}\left ( 0\right ) e^{-k\alpha _{n}^{2}t}\\ & =\left ( \sqrt{\frac{2h}{h+\cos ^{2}\alpha _{n}}}\int _{0}^{1}f\left ( x\right ) \sin \left ( \alpha _{n}x\right ) dx\right ) e^{-k\alpha _{n}^{2}t} \end{align*}
and solution (1) becomes\begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }\sqrt{\frac{2h}{h+\cos ^{2}\alpha _{n}}}\left ( \int _{0}^{1}f\left ( x\right ) \sin \left ( \alpha _{n}x\right ) dx\right ) e^{-k\alpha _{n}^{2}t}\sqrt{\frac{2h}{h+\cos ^{2}\alpha _{n}}}\sin \left ( \alpha _{n}x\right ) \\ & =\frac{2h}{h+\cos ^{2}\alpha _{n}}\sum _{n=1}^{\infty }\left ( \int _{0}^{1}f\left ( x\right ) \sin \left ( \alpha _{n}x\right ) dx\right ) e^{-k\alpha _{n}^{2}t}\sin \left ( \alpha _{n}x\right ) \end{align*}
Which is what required to show.
We need to show that the solution found in part (a) also satisfies the PDE when \(-1<x<1\)\[ u_{t}=ku_{xx}\qquad -1<x<1,t>0 \] With boundary conditions (9)\begin{align*} u_{x}\left ( -1,t\right ) & =hu\left ( -1,t\right ) \\ u_{x}\left ( 1,t\right ) & =-hu\left ( 1,t\right ) \end{align*}
And initial conditions (10)\[ u\left ( x,0\right ) =f\left ( x\right ) \] When \(f\left ( x\right ) \) is odd.
The solution found in \(a\) already satisfies the above PDE with the second boundary conditions in (9). Since sine is odd then the solution in part(a) is also odd. Then its partial derivative is even in \(x\), hence the first boundary conditions in (9) is also satisfied \[ u_{x}\left ( -1,t\right ) =hu\left ( -1,t\right ) =-u_{x}\left ( 1,t\right ) =hu\left ( 1,t\right ) \] Finally we know that \(u\left ( x,0\right ) =f\left ( x\right ) \) for \(0<x<1\). Furthermore when \(-1<x<0\) the fact that \(u\) and \(f\left ( x\right ) \) are odd enables us to write\[ u\left ( -x,0\right ) =-u\left ( x,0\right ) =f\left ( -x\right ) =-f\left ( x\right ) \]
Solution
Solve
\[ u_{t}=u_{xx}\qquad 0<x<1,t>0 \] With boundary conditions\begin{align*} u_{x}\left ( 0,t\right ) -hu\left ( 0,t\right ) & =0\\ u\left ( 1,t\right ) & =1 \end{align*}
With \(h>0\). And initial conditions \(u\left ( x,0\right ) =f\left ( x\right ) \).
Because the second B.C. is not zero, we need to introduce a reference function \(r\left ( x\right ) \) which satisfies the nonhomogeneous boundary conditions.
Let \(r\left ( x\right ) =Ax+B\). When \(x=0\) then the first BC gives\[ A-hB=0 \] And the second BC gives\[ A+B=1 \] From the first equation \(A=hB\). Substituting in the second equation give \(hB+B=1\) or \(B\left ( 1+h\right ) =1\) or \(B=\frac{1}{1+h}\). Hence \(A=\frac{h}{1+h}\). Therefore\begin{align} r\left ( x\right ) & =Ax+B\nonumber \\ & =\frac{h}{1+h}x+\frac{1}{1+h}\nonumber \\ & =\frac{hx+1}{1+h} \tag{1} \end{align}
To verify. \(r_{x}=\frac{h}{1+h}\). When \(x=0\) then \(r\left ( 0\right ) =\frac{1}{1+h}\). Hence \(r_{x}\left ( 0\right ) -hr\left ( 0\right ) =\frac{h}{1+h}-h\frac{1}{1+h}=0\) as expected. And when \(x=1\) then \(r\left ( 1\right ) =1\) as expected. Now that we found \(r\left ( x\right ) \) then we write\[ u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x\right ) \] Where \(v\left ( x,t\right ) \) is the solution to the homogenous PDE\[ v_{t}=v_{xx}\qquad 0<x<1,t>0 \] With boundary conditions\begin{align*} v_{x}\left ( 0,t\right ) -hv\left ( 0,t\right ) & =0\\ v\left ( 1,t\right ) & =0 \end{align*}
We can now solve for \(v\left ( x,t\right ) \,\) using separation of variables since boundary conditions are homogenous. Separation of variables gives\begin{align*} X^{\prime \prime }+\lambda X & =0\\ X^{\prime }\left ( 0\right ) -hX\left ( 0\right ) & =0\\ X\left ( 1\right ) & =0 \end{align*}
Using problem 5 section 72, the eigenfunctions and eigenvalues for the above are\begin{align*} \phi _{n}\left ( x\right ) & =\sqrt{\frac{2h}{h+\cos ^{2}\alpha _{n}}}\sin \left ( \alpha _{n}\left ( 1-x\right ) \right ) \qquad n=1,2,\cdots \\ \tan \left ( \alpha _{n}\right ) & =\frac{-\alpha _{n}}{h} \end{align*}
With \(\alpha _{n}>0\). Hence the solution \(v\left ( x,t\right ) \) using generalized Fourier series is\begin{equation} v\left ( x,t\right ) =\sum _{n=1}^{\infty }B_{n}\left ( t\right ) \phi _{n}\left ( x\right ) \tag{2} \end{equation} Substituting into the PDE \(v_{t}=v_{xx}\) gives\begin{align*} \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 }B_{n}\left ( t\right ) \alpha _{n}^{2}\phi _{n}\left ( x\right ) \end{align*}
Therefore the ODE is\[ B_{n}^{\prime }\left ( t\right ) +\alpha _{n}^{2}B_{n}\left ( t\right ) =0 \] The solution is \[ B_{n}\left ( t\right ) =B_{n}\left ( 0\right ) e^{-\alpha _{n}^{2}t}\] Hence (2) becomes\[ v\left ( x,t\right ) =\sum _{n=1}^{\infty }B_{n}\left ( 0\right ) e^{-\alpha _{n}^{2}t}\phi _{n}\left ( x\right ) \] And since \(u\left ( x,t\right ) =v\left ( x,t\right ) +r\left ( x\right ) \) then\[ u\left ( x,t\right ) =\sum _{n=1}^{\infty }B_{n}\left ( 0\right ) e^{-\alpha _{n}^{2}t}\phi _{n}\left ( x\right ) +\frac{hx+1}{1+h}\] Now we find \(B_{n}\left ( 0\right ) \) from initial conditions. At \(t=0\) the above becomes\begin{align*} 0 & =\sum _{n=1}^{\infty }B_{n}\left ( 0\right ) \phi _{n}\left ( x\right ) +\frac{hx+1}{1+h}\\ -\frac{hx+1}{1+h} & =\sum _{n=1}^{\infty }B_{n}\left ( 0\right ) \phi _{n}\left ( x\right ) \end{align*}
Hence \begin{align} B_{n}\left ( 0\right ) & =\left \langle -\frac{hx+1}{1+h},\phi _{n}\left ( x\right ) \right \rangle \nonumber \\ & =-\int _{0}^{1}p\left ( x\right ) \frac{hx+1}{1+h}\phi _{n}\left ( x\right ) dx\nonumber \\ & =-\int _{0}^{1}\frac{hx+1}{1+h}\sqrt{\frac{2h}{h+\cos ^{2}\alpha _{n}}}\sin \left ( \alpha _{n}\left ( 1-x\right ) \right ) dx\nonumber \\ & =-\frac{1}{1+h}\sqrt{\frac{2h}{h+\cos ^{2}\alpha _{n}}}\int _{0}^{1}\left ( hx+1\right ) \sin \left ( \alpha _{n}\left ( 1-x\right ) \right ) dx \tag{3} \end{align}
But\begin{align*} \int _{0}^{1}\left ( hx+1\right ) \sin \left ( \alpha _{n}\left ( 1-x\right ) \right ) dx & =\int _{0}^{1}\sin \left ( \alpha _{n}\left ( 1-x\right ) \right ) dx+h\int _{0}^{1}x\sin \left ( \alpha _{n}\left ( 1-x\right ) \right ) dx\\ & =\left [ \frac{\cos \left ( \alpha _{n}\left ( 1-x\right ) \right ) }{\alpha _{n}}\right ] _{0}^{1}+h\left [ \frac{\alpha _{n}x\cos \left ( \alpha _{n}\left ( 1-x\right ) \right ) +\sin \left ( \alpha _{n}\left ( 1-x\right ) \right ) }{\alpha _{n}^{2}}\right ] _{0}^{1}\\ & =\frac{1-\cos \left ( \alpha _{n}\right ) }{\alpha _{n}}+\frac{h}{\alpha _{n}^{2}}\left [ \alpha _{n}x\cos \left ( \alpha _{n}\left ( 1-x\right ) \right ) +\sin \left ( \alpha _{n}\left ( 1-x\right ) \right ) \right ] _{0}^{1}\\ & =\frac{1-\cos \left ( \alpha _{n}\right ) }{\alpha _{n}}+\frac{h}{\alpha _{n}^{2}}\left [ \alpha _{n}-\sin \alpha _{n}\right ] \\ & =\frac{\alpha _{n}-\alpha _{n}\cos \left ( \alpha _{n}\right ) +h\alpha _{n}-h\sin \alpha _{n}}{\alpha _{n}^{2}} \end{align*}
But \(\frac{\sin \left ( \alpha _{n}\right ) }{\cos \left ( \alpha _{n}\right ) }=-\frac{\alpha _{n}}{h}\) or \(h\sin \left ( \alpha _{n}\right ) =-\alpha _{n}\cos \left ( \alpha _{n}\right ) \) or \(-h\sin \alpha _{n}=\alpha _{n}\cos \left ( \alpha _{n}\right ) \), hence the above simplifies to \begin{align*} \int _{0}^{1}\left ( hx+1\right ) \sin \left ( \alpha _{n}\left ( 1-x\right ) \right ) dx & =\frac{\alpha _{n}+h\alpha _{n}}{\alpha _{n}^{2}}\\ & =\frac{1+h}{\alpha _{n}} \end{align*}
Therefore (3) becomes\begin{align*} B_{n}\left ( 0\right ) & =\frac{-1}{1+h}\sqrt{\frac{2h}{h+\cos ^{2}\alpha _{n}}}\left ( \frac{1+h}{\alpha _{n}}\right ) \\ & =-\frac{1}{\alpha _{n}}\sqrt{\frac{2h}{h+\cos ^{2}\alpha _{n}}} \end{align*}
Hence final solution becomes\begin{align*} u\left ( x,t\right ) & =\frac{hx+1}{1+h}+\sum _{n=1}^{\infty }B_{n}\left ( 0\right ) e^{-\alpha _{n}^{2}t}\phi _{n}\left ( x\right ) \\ & =\frac{hx+1}{1+h}+\sum _{n=1}^{\infty }B_{n}\left ( 0\right ) \exp \left ( -\alpha _{n}^{2}t\right ) \sqrt{\frac{2h}{h+\cos ^{2}\alpha _{n}}}\sin \left ( \alpha _{n}\left ( 1-x\right ) \right ) \\ & =\frac{hx+1}{1+h}+\sum _{n=1}^{\infty }-\frac{1}{\alpha _{n}}\sqrt{\frac{2h}{h+\cos ^{2}\alpha _{n}}}\exp \left ( -\alpha _{n}^{2}t\right ) \sqrt{\frac{2h}{h+\cos ^{2}\alpha _{n}}}\sin \left ( \alpha _{n}\left ( 1-x\right ) \right ) \\ & =\frac{hx+1}{1+h}-2h\sum _{n=1}^{\infty }\frac{\sin \left ( \alpha _{n}\left ( 1-x\right ) \right ) }{\alpha _{n}\left ( h+\cos ^{2}\alpha _{n}\right ) }\exp \left ( -\alpha _{n}^{2}t\right ) \end{align*}
Which is what required to show.