The Fourier sin series of \(f\left ( x\right ) =1\) on \(0\leq x\leq L\) is given by\begin{equation} f\left ( x\right ) \sim \sum _{n=1}^{\infty }b_{n}\sin \left ( \frac{n\pi }{L}x\right ) \tag{1} \end{equation} Where \begin{align*} b_{n} & =\frac{1}{L}\int _{-L}^{L}f\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx\\ & =\frac{1}{L}\left ( \int _{-L}^{0}\left ( -1\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx+\int _{0}^{L}\left ( +1\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx\right ) \\ & =\frac{1}{L}\left ( -\left [ -\frac{L}{n\pi }\cos \left ( \frac{n\pi }{L}x\right ) \right ] _{-L}^{0}+\left [ -\frac{L}{n\pi }\cos \left ( \frac{n\pi }{L}x\right ) \right ] _{0}^{L}\right ) \\ & =\frac{1}{L}\left ( \frac{L}{n\pi }\left [ \cos \left ( \frac{n\pi }{L}x\right ) \right ] _{-L}^{0}-\frac{L}{n\pi }\left [ \cos \left ( \frac{n\pi }{L}x\right ) \right ] _{0}^{L}\right ) \\ & =\frac{1}{L}\left ( \frac{L}{n\pi }\left [ \cos \left ( 0\right ) -\cos \left ( n\pi \right ) \right ] -\frac{L}{n\pi }\left [ \cos \left ( n\pi \right ) -\cos \left ( 0\right ) \right ] \right ) \\ & =\frac{1}{L}\left ( \frac{L}{n\pi }\left [ 1-\cos \left ( n\pi \right ) \right ] -\frac{L}{n\pi }\left [ \cos \left ( n\pi \right ) -1\right ] \right ) \end{align*}
We see that \(b_{n}=0\) for \(n=2,4,6,\cdots ,\) and \(b_{n}\) odd for \(n=1,3,5,\cdots \) so we can simplify the above to be\begin{align*} b_{n} & =\frac{1}{L}\left ( \frac{L}{n\pi }\left [ 1-\left ( -1\right ) \right ] -\frac{L}{n\pi }\left [ -1-1\right ] \right ) \\ & =\frac{1}{L}\left ( \frac{L}{n\pi }\left [ 2\right ] -\frac{L}{n\pi }\left [ -2\right ] \right ) \\ & =\frac{1}{L}\left ( \frac{4L}{n\pi }\right ) \\ & =\frac{4}{n\pi } \end{align*}
Equation (1) becomes\begin{equation} f\left ( x\right ) \sim \sum _{n=1,3,5,\cdots }^{\infty }\frac{4}{n\pi }\sin \left ( \frac{n\pi }{L}x\right ) \tag{2} \end{equation} mean-square error is, from textbook, page 213, is given by equation 5.10.11\begin{equation} E=\int _{0}^{L}f^{2}\left ( x\right ) \sigma \left ( x\right ) dx-\sum _{n=1,3,5,\cdots }^{\infty }\alpha _{n}^{2}\int _{0}^{L}\phi _{n}^{2}\sigma \left ( x\right ) dx \tag{5.10.11} \end{equation} In this problem, \(\phi _{n}=\sin \left ( \frac{n\pi }{L}x\right ) \) and \(\alpha _{n}=a_{n}=\frac{4}{n\pi }.\) The above equation becomes\begin{align*} E & =\int _{0}^{L}f^{2}\left ( x\right ) \sigma \left ( x\right ) dx-\sum _{n=1,3,5,\cdots }^{\infty }\left ( \frac{4}{n\pi }\right ) ^{2}\int _{0}^{L}\sin ^{2}\left ( \frac{n\pi }{L}x\right ) \sigma \left ( x\right ) dx\\ & =\int _{0}^{L}f^{2}\left ( x\right ) \sigma dx-\sum _{n=1,3,5,\cdots }^{\infty }\frac{16}{n^{2}\pi ^{2}}\int _{0}^{L}\sin ^{2}\left ( \frac{n\pi }{L}x\right ) \sigma dx \end{align*}
For \(\sigma =1\) we know that\[ \int _{0}^{L}\sin ^{2}\left ( \frac{n\pi }{L}x\right ) \sigma dx=\frac{L}{2}\] Hence \(E\) becomes\[ E=\int _{0}^{L}f^{2}\left ( x\right ) \sigma dx-\sum _{n=1,3,5,\cdots }^{\infty }\frac{16}{n^{2}\pi ^{2}}\frac{L}{2}\] But \(\int _{0}^{L}f^{2}\left ( x\right ) \sigma dx\) for \(\sigma =1\) is just \(\int _{0}^{L}1^{2}dx=L\), and the above becomes\begin{align*} E & =L-\frac{L}{2}\frac{16}{\pi ^{2}}\sum _{n=1,3,5,\cdots }^{\infty }\frac{1}{n^{2}}\\ & =L-\frac{8L}{\pi ^{2}}\sum _{n=1,3,5,\cdots }^{\infty }\frac{1}{n^{2}} \end{align*}
We need to find \(N\) so that \(E=0.01L\). The above becomes\[ 0.01L=L-\frac{8L}{\pi ^{2}}\sum _{n=1,3,5,\cdots }^{N}\frac{1}{n^{2}}\] We need now to solve for \(N\) in the above\begin{align*} 0.01L-L & =-\frac{8L}{\pi ^{2}}\sum _{n=1,3,5,\cdots }^{N}\frac{1}{n^{2}}\\ 0.99L\left ( \frac{\pi ^{2}}{8L}\right ) & =\sum _{n=1,3,5,\cdots }^{N}\frac{1}{n^{2}}\\ 1.2214 & =\sum _{n=1,3,5,\cdots }^{N}\frac{1}{n^{2}} \end{align*}
A small Mathematica program written which prints the RHS sum for each \(n\), and was visually checked when it reached \(1.2214\), here is the result
Counting the number of terms needed to reach \(1.2214\), we see there are \(21\) terms (21 rows in the table, since only odd entries are counted, the table above skips the even \(n\) values in the sum since these are all zero).
Parseval’s equality is given by equation 5.10.14, page 214 in textbook\begin{equation} \int _{a}^{b}f^{2}\sigma dx=\sum _{n=1}^{\infty }a_{n}^{2}\int _{a}^{b}\phi _{n}^{2}\sigma dx \tag{5.10.14} \end{equation} The books uses \(\alpha _{n}\) instead of \(a_{n}\), but it is the same, these are the coefficients in the Fourier series for \(f\left ( x\right ) \). We now need to find the cosine Fourier series for \(f\left ( x\right ) =x\). This is given by\[ f\left ( x\right ) =a_{0}+\sum _{n=1}^{\infty }a_{n}\cos \left ( \frac{n\pi }{L}x\right ) \] Where\begin{align*} a_{n} & =\frac{1}{L}\int _{-L}^{L}f\left ( x\right ) \cos \left ( \frac{n\pi }{L}x\right ) dx\\ & =\frac{1}{L}\left [ \int _{-L}^{0}\left ( -x\right ) \cos \left ( \frac{n\pi }{L}x\right ) dx+\int _{0}^{L}\left ( +x\right ) \cos \left ( \frac{n\pi }{L}x\right ) dx\right ] \\ & =\frac{1}{L}\left [ -\int _{-L}^{0}x\cos \left ( \frac{n\pi }{L}x\right ) dx+\int _{0}^{L}x\cos \left ( \frac{n\pi }{L}x\right ) dx\right ] \end{align*}
But \begin{align*} \int _{-L}^{0}x\cos \left ( \frac{n\pi }{L}x\right ) dx & =-\frac{-1+\left ( -1\right ) ^{n}}{n^{2}\pi ^{2}}L^{2}\\ \int _{0}^{L}x\cos \left ( \frac{n\pi }{L}x\right ) dx & =\frac{-1+\left ( -1\right ) ^{n}}{n^{2}\pi ^{2}}L^{2} \end{align*}
Hence\begin{align*} a_{n} & =\frac{1}{L}\left [ 2\frac{-1+\left ( -1\right ) ^{n}}{n^{2}\pi ^{2}}L^{2}\right ] \\ & =\frac{2L}{\pi ^{2}}\frac{\left ( -1+\left ( -1\right ) ^{n}\right ) }{n^{2}} \end{align*}
Looking at few terms to see the pattern\[ a_{n}=\frac{2L}{\pi ^{2}}\left \{ \frac{-2}{1},0,\frac{-2}{3^{2}},0,\frac{-2}{5^{2}},\cdots \right \} \] Therefore, we can write \(a_{n}\) as\[ a_{n}=\frac{-4L}{\pi ^{2}n^{2}}\qquad n=1,3,5,\cdots \] And \begin{align*} a_{0} & =\frac{1}{2L}\int _{-L}^{L}f\left ( x\right ) dx\\ & =\frac{1}{2L}\left [ \int _{-L}^{0}\left ( -x\right ) dx+\int _{0}^{L}\left ( +x\right ) dx\right ] \\ & =\frac{1}{2L}\left [ -\left [ \frac{x^{2}}{2}\right ] _{-L}^{0}+\left [ \frac{x^{2}}{2}\right ] _{0}^{L}\right ] \\ & =\frac{1}{2L}\left [ -\left [ 0-\frac{L^{2}}{2}\right ] +\left [ \frac{L^{2}}{2}-0\right ] \right ] \\ & =\frac{1}{2L}\left [ \frac{L^{2}}{2}+\frac{L^{2}}{2}\right ] \\ & =\frac{L}{2} \end{align*}
Hence the Fourier series is\[ f\left ( x\right ) =\frac{L}{2}+\sum _{n=1,3,5,\cdots }^{\infty }\frac{-4L}{\pi ^{2}n^{2}}\cos \left ( \frac{n\pi }{L}x\right ) \] We now go back to equation 5.10.14 (but need to add \(a_{0}\) to it, since there is this extra term with cosine Fourier series)\begin{align*} \int _{a}^{b}f^{2}\sigma dx & =a_{0}^{2}\int _{a}^{b}1^{2}dx+\sum _{n=1}^{\infty }a_{n}^{2}\int _{a}^{b}\phi _{n}^{2}\sigma dx\\ \int _{0}^{L}x^{2}dx & =\left ( \frac{L}{2}\right ) ^{2}\int _{0}^{L}dx+\sum _{n=1,3,5,\cdots }^{\infty }\left ( \frac{-4L}{\pi ^{2}n^{2}}\right ) ^{2}\int _{0}^{L}\cos ^{2}\left ( \frac{n\pi }{L}x\right ) dx\\ \left [ \frac{x^{3}}{3}\right ] _{0}^{L} & =\left ( \frac{L^{2}}{4}\right ) L+\sum _{n=1,3,5,\cdots }^{\infty }\frac{16L^{2}}{\pi ^{4}n^{4}}\int _{0}^{L}\cos ^{2}\left ( \frac{n\pi }{L}x\right ) dx \end{align*}
Since \(\int _{0}^{L}\cos ^{2}\left ( \frac{n\pi }{L}x\right ) dx=\frac{L}{2}\) the above becomes\begin{align*} \frac{L^{3}}{3} & =\left ( \frac{L^{2}}{4}\right ) L+\sum _{n=1,3,5,\cdots }^{\infty }\frac{16L^{2}}{\pi ^{4}n^{4}}\frac{L}{2}\\ \frac{L^{3}}{3} & =\left ( \frac{L^{2}}{4}\right ) L+\sum _{n=1,3,5,\cdots }^{\infty }\frac{8L^{3}}{\pi ^{4}n^{4}}\\ \frac{L^{3}}{3} & =\frac{L^{3}}{4}+\frac{8L^{3}}{\pi ^{4}}\sum _{n=1,3,5,\cdots }^{\infty }\frac{1}{n^{4}} \end{align*}
Simplifying\begin{align*} \frac{1}{3} & =\frac{1}{4}+\frac{8}{\pi ^{4}}\sum _{n=1,3,5,\cdots }^{\infty }\frac{1}{n^{4}}\\ \sum _{n=1,3,5,\cdots }^{\infty }\frac{1}{n^{4}} & =\left ( \frac{1}{3}-\frac{1}{4}\right ) \frac{\pi ^{4}}{8}\\ \sum _{n=1,3,5,\cdots }^{\infty }\frac{1}{n^{4}} & =\frac{\pi ^{4}}{96} \end{align*}
Hence \[ \frac{\pi ^{4}}{96}=1+\frac{1}{3^{4}}+\frac{1}{5^{4}}+\frac{1}{7^{4}}+\cdots \] Which agrees with the book solution given in back of book.
\begin{align} \int _{a}^{b}fg\sigma dx & =\int _{a}^{b}\left ( \sum _{n=1}^{\infty }\alpha _{n}\phi _{n}\right ) \left ( \sum _{n=1}^{\infty }\beta _{n}\phi _{n}\right ) \sigma dx\nonumber \\ & =\int _{a}^{b}\left ( \alpha _{1}\phi _{1}+\alpha _{2}\phi _{2}+\cdots \right ) \left ( \beta _{1}\phi _{1}+\beta _{2}\phi _{2}+\cdots \right ) \sigma dx \tag{1} \end{align}
But \begin{align*} \left ( \alpha _{1}\phi _{1}+\alpha _{2}\phi _{2}+\cdots \right ) \left ( \beta _{1}\phi _{1}+\beta _{2}\phi _{2}+\cdots \right ) & =\alpha _{1}\beta _{1}\phi _{1}^{2}+\alpha _{1}\beta _{2}\phi _{1}\phi _{2}+\alpha _{1}\beta _{3}\phi _{1}\phi _{3}+\cdots \\ & +\alpha _{2}\beta _{1}\phi _{2}\phi _{1}+\alpha _{2}\beta _{2}\phi _{2}^{2}+\alpha _{2}\beta _{3}\phi _{1}\phi _{3}+\cdots \\ & +\alpha _{3}\beta _{1}\phi _{3}\phi _{1}+\alpha _{3}\beta _{2}\phi _{3}\phi _{2}+\alpha _{3}\beta _{3}\phi _{3}^{2}+\cdots \\ & \vdots \end{align*}
Which means when expanding the product of the two series, only the terms on the diagonal (the terms with \(\alpha _{i}\beta _{j}\phi _{i}\phi _{j}\) with \(i=j\)) will survive. This due to orthogonality. To show this more clearly, we put the above expansion back into the integral (1) and break up the integral into sum of integrals\begin{align*} \int _{a}^{b}fg\sigma dx & =\int _{a}^{b}\alpha _{1}\beta _{1}\phi _{1}^{2}\sigma dx+\int _{a}^{b}\alpha _{1}\beta _{2}\phi _{1}\phi _{2}\sigma dx+\int _{a}^{b}\alpha _{1}\beta _{3}\phi _{1}\phi _{3}\sigma dx+\cdots \\ & +\int _{a}^{b}\alpha _{2}\beta _{1}\phi _{2}\phi _{1}\sigma dx+\int _{a}^{b}\alpha _{2}\beta _{2}\phi _{2}^{2}\sigma dx+\int _{a}^{b}\alpha _{2}\beta _{3}\phi _{1}\phi _{3}\sigma dx+\cdots \\ & +\int _{a}^{b}\alpha _{3}\beta _{1}\phi _{3}\phi _{1}\sigma dx+\int _{a}^{b}\alpha _{3}\beta _{2}\phi _{3}\phi _{2}\sigma dx+\int _{a}^{b}\alpha _{3}\beta _{3}\phi _{3}^{2}\sigma dx+\cdots \\ & \vdots \end{align*}
The above simplifies to\[ \int _{a}^{b}fg\sigma dx=\int _{a}^{b}\alpha _{1}\beta _{1}\phi _{1}^{2}\sigma dx+\int _{a}^{b}\alpha _{2}\beta _{2}\phi _{2}^{2}\sigma dx+\int _{a}^{b}\alpha _{3}\beta _{3}\phi _{3}^{2}\sigma dx+\cdots \int _{a}^{b}\alpha _{n}\beta _{n}\phi _{n}^{2}\sigma dx+\cdots \] Since all other terms vanish due to orthogonality of eigenfunctions. The above simplifies to\begin{align*} \int _{a}^{b}fg\sigma dx & =\sum _{n=1}^{\infty }\int _{a}^{b}\alpha _{n}\beta _{n}\phi _{n}^{2}\sigma dx\\ & =\sum _{n=1}^{\infty }\left ( \alpha _{n}\beta _{n}\int _{a}^{b}\phi _{n}^{2}\sigma dx\right ) \end{align*}
Because the eigenfunctions are normalized, then \(\int _{a}^{b}\phi _{n}^{2}\sigma dx=1\) and the above reduces to the result needed\[ \int _{a}^{b}fg\sigma \ dx=\sum _{n=1}^{\infty }\alpha _{n}\beta _{n}\]
Let \(u=X\left ( x\right ) Y\left ( y\right ) T\left ( t\right ) \). Substituting this back into the PDE gives\[ T^{\prime \prime }XY=c^{2}\left ( X^{\prime \prime }YT+Y^{\prime \prime }XT\right ) \] Dividing by \(XYT\neq 0\) gives\[ \frac{1}{c^{2}}\frac{T^{\prime \prime }}{T}=\frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y}\] Since left side depends on \(t\) only and right side depends on \(\left ( x,y\right ) \) only, then both must be equal to some constant, say \(-\lambda \)\[ \frac{1}{c^{2}}\frac{T^{\prime \prime }}{T}=\frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y}=-\lambda \] We obtain the following\begin{align*} T^{\prime \prime }+c^{2}\lambda T & =0\\ \frac{X^{\prime \prime }}{X} & =-\lambda -\frac{Y^{\prime \prime }}{Y} \end{align*}
Again, looking at the second ODE above, we see that the left side depends on \(x\) only, and the right side on \(y\) only. Then they must be equal to some constant, say \(-\mu \) and we obtain\[ \frac{X^{\prime \prime }}{X}=\left ( -\lambda -\frac{Y^{\prime \prime }}{Y}\right ) =-\mu \] Which results in two ODE’s. The first is \begin{align*} X^{\prime \prime }+\mu X & =0\\ X\left ( 0\right ) & =0\\ X\left ( L\right ) & =0 \end{align*}
And the second is\begin{align*} \lambda +\frac{Y^{\prime \prime }}{Y} & =\mu \\ Y^{\prime \prime } & =Y\mu -\lambda Y\\ Y^{\prime \prime }+Y\left ( \lambda -\mu \right ) & =0 \end{align*}
With B.C.\begin{align*} Y^{\prime }\left ( 0\right ) & =0\\ Y^{\prime }\left ( H\right ) & =0 \end{align*}
Starting with the \(X\) ODE since it is simpler, the solution is\[ X=c_{1}\cos \left ( \sqrt{\mu }x\right ) +c_{2}\sin \left ( \sqrt{\mu }x\right ) \] Applying \(X\left ( 0\right ) =0\) gives\[ 0=c_{1}\] Hence solution is\[ X=c_{2}\sin \left ( \sqrt{\mu }x\right ) \] Applying \(X\left ( L\right ) =0\) gives\[ 0=c_{2}\sin \left ( \sqrt{\mu }L\right ) \] For non-trivial solution\[ \mu _{n}=\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \] And the eigenfunctions are \[ X_{n}\left ( x\right ) =\sin \left ( \frac{n\pi }{L}x\right ) \] We now solve the \(Y\) ODE. \[ Y^{\prime \prime }+\left ( \lambda -\mu _{n}\right ) Y=0 \] Assuming that \(\left ( \lambda -\mu \right ) >0\) for all \(\lambda ,\mu \), (we know this is the only case, since only positive \(\lambda -\mu _{n}\) will be possible when B.C. are homogeneous Dirichlet). Then, for \(\left ( \lambda -\mu \right ) >0,\) the solution is \begin{align*} Y\left ( y\right ) & =c_{1}\cos \left ( \sqrt{\lambda -\mu _{n}}y\right ) +c_{2}\sin \left ( \sqrt{\lambda -\mu _{n}}y\right ) \\ Y^{\prime } & =-c_{1}\sqrt{\lambda -\mu _{n}}\sin \left ( \sqrt{\lambda -\mu _{n}}y\right ) +c_{2}\sqrt{\lambda -\mu _{n}}\cos \left ( \sqrt{\lambda -\mu _{n}}y\right ) \end{align*}
Applying B.C. \(Y^{\prime }\left ( 0\right ) =0\) the above becomes\[ 0=c_{2}\sqrt{\lambda -\mu _{n}}\] Hence \(c_{2}=0\) and the solution becomes\begin{align*} Y & =c_{1}\cos \left ( \sqrt{\lambda -\mu _{n}}y\right ) \\ Y^{\prime } & =-c_{1}\sqrt{\lambda -\mu _{n}}\sin \left ( \sqrt{\lambda -\mu _{n}}y\right ) \end{align*}
Applying second B.C. \(Y^{\prime }\left ( H\right ) =0\) gives\[ 0=-c_{1}\sqrt{\lambda -\mu _{n}}\sin \left ( \sqrt{\lambda -\mu _{n}}H\right ) \] For non-trivial solution we want \begin{align*} \sin \left ( \sqrt{\lambda -\mu _{n}}H\right ) & =0\\ \sqrt{\lambda _{nm}-\mu _{n}} & =m\frac{\pi }{H}\qquad \\ \lambda _{nm}-\mu _{n} & =\left ( m\frac{\pi }{H}\right ) ^{2}\\ \lambda _{nm} & =\left ( m\frac{\pi }{H}\right ) ^{2}+\mu _{n}\qquad m=0,1,2,\cdots \end{align*}
Hence the eigenfunctions are \[ Y_{nm}=\cos \left ( m\frac{\pi }{H}y\right ) \qquad m=0,1,2,\cdots ,n=1,2,3,\cdots \] For each \(n,m,\) we find solution of \(T^{\prime \prime }+c^{2}\lambda _{nm}T=0.\)The solution is \[ T_{nm}\left ( t\right ) =A_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) +B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \] Putting all these results together gives\begin{align*} u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }T_{nm}\left ( t\right ) X_{nm}\left ( x\right ) Y_{nm}\left ( y\right ) \\ & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }\left [ A_{nm}\cos \left ( c\sqrt{\lambda _{n,m}}t\right ) +B_{nm}\sin \left ( c\sqrt{\lambda _{n,m}}t\right ) \right ] \sin \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) \\ & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) \end{align*}
We now apply initial conditions to find \(A_{nm},B_{nm}\). At \(t=0\)\begin{align*} u\left ( x,y,0\right ) & =0\\ & =\sum _{n=1}^{\infty }\sum _{m=0}^{\infty }A_{nm}\sin \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) \end{align*}
Hence \[ A_{nm}=0 \] And the solution becomes\[ u\left ( x,y,t\right ) =\sum _{n=1}^{\infty }\sum _{m=0}^{\infty }B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) \] Taking derivative of the solution w.r.t. time \(t\) gives\[ \frac{\partial }{\partial t}u\left ( x,y,t\right ) =\sum _{n=1}^{\infty }\sum _{m=0}^{\infty }c\sqrt{\lambda _{nm}}B_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) \] At \(t=0\) the above becomes\[ \alpha \left ( x,y\right ) =\sum _{n=1}^{\infty }\sum _{m=0}^{\infty }c\sqrt{\lambda _{nm}}B_{nm}\sin \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) \] Multiplying both sides by \(\sin \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) \) and integrating gives\begin{align*} \int _{0}^{L}\int _{0}^{H}\alpha \left ( x,y\right ) \sin \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) \ dxdy & =c\sqrt{\lambda _{nm}}B_{nm}\sum _{n=1}^{\infty }\sum _{m=0}^{\infty }\int _{0}^{L}\int _{0}^{H}\sin ^{2}\left ( \frac{n\pi }{L}x\right ) \cos ^{2}\left ( m\frac{\pi }{H}y\right ) \ dxdy\\ & =c\sqrt{\lambda _{nm}}B_{nm}\int _{0}^{L}\int _{0}^{H}\sin ^{2}\left ( \frac{n\pi }{L}x\right ) \cos ^{2}\left ( m\frac{\pi }{H}y\right ) \ dxdy\\ & =c\sqrt{\lambda _{nm}}B_{nm}\left ( \frac{L}{2}\right ) \left ( \frac{H}{2}\right ) \end{align*}
Hence\[ B_{nm}=\frac{4}{LHc\sqrt{\lambda _{nm}}}\int _{0}^{L}\int _{0}^{H}\alpha \left ( x,y\right ) \sin \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) dxdy \] Summary of solution\begin{align*} X_{n}\left ( x\right ) & =\sin \left ( \frac{n\pi }{L}x\right ) \qquad n=1,2,3,\cdots \\ \mu _{n} & =\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \\ Y_{nm}\left ( y\right ) & =\cos \left ( m\frac{\pi }{H}y\right ) \qquad m=0,1,2,\cdots \\ \lambda _{nm}-\mu _{n} & =\left ( m\frac{\pi }{H}\right ) ^{2}\qquad m=0,1,2,\cdots ,n=1,2,3,\cdots \\ T_{nm}\left ( t\right ) & =B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \\ u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=0}^{\infty }B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) \\ B_{nm} & =\frac{4}{LHc\sqrt{\lambda _{nm}}}\int _{0}^{L}\int _{0}^{H}\alpha \left ( x,y\right ) \sin \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) dxdy \end{align*}
In this case we have\begin{align*} X^{\prime \prime }+\mu X & =0\\ X^{\prime }\left ( 0\right ) & =0\\ X^{\prime }\left ( L\right ) & =0 \end{align*}
And the second spatial ODE is\begin{align*} \lambda +\frac{Y^{\prime \prime }}{Y} & =\mu \\ Y^{\prime \prime } & =Y\mu -\lambda Y\\ Y^{\prime \prime }+Y\left ( \lambda -\mu \right ) & =0 \end{align*}
With B.C.\begin{align*} Y^{\prime }\left ( 0\right ) & =0\\ Y^{\prime }\left ( H\right ) & =0 \end{align*}
Starting with the \(X\) ODE. The solution is\begin{align*} X & =c_{1}\cos \left ( \sqrt{\mu }x\right ) +c_{2}\sin \left ( \sqrt{\mu }x\right ) \\ X^{\prime } & =-c_{1}\sqrt{\mu }\sin \left ( \sqrt{\mu }x\right ) +c_{2}\sqrt{\mu }\cos \left ( \sqrt{\mu }x\right ) \end{align*}
First B.C. gives\[ 0=c_{2}\sqrt{\mu }\] Hence \(c_{2}=0\) and the solution becomes\begin{align*} X & =c_{1}\cos \left ( \sqrt{\mu }x\right ) \\ X^{\prime } & =-c_{1}\sqrt{\mu }\sin \left ( \sqrt{\mu }x\right ) \end{align*}
Second B.C. gives\[ 0=-c_{1}\sqrt{\mu }\sin \left ( \sqrt{\mu }L\right ) \] Hence \begin{align*} \sqrt{\mu }L & =n\pi \\ \mu & =\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=0,1,2,\cdots \end{align*}
Now for the \(Y\) solution. This is the same as part (a).\begin{align*} Y_{nm}\left ( y\right ) & =\cos \left ( m\frac{\pi }{H}y\right ) \\ \lambda _{nm}-\mu _{n} & =\left ( m\frac{\pi }{H}\right ) ^{2}\\ \lambda _{nm} & =\left ( m\frac{\pi }{H}\right ) ^{2}+\mu _{n}\\ & =\left ( m\frac{\pi }{H}\right ) ^{2}+\left ( \frac{n\pi }{L}\right ) ^{2}\qquad m=0,1,2,\cdots ,n=0,1,2,\cdots \end{align*}
For each \(n,m,\) we find solution of \(T^{\prime \prime }+c^{2}\lambda _{nm}T=0.\)When \(n=0,m=0\), \(\lambda _{nm}=0\) and the ODE becomes\[ T^{\prime \prime }=0 \] With solution \[ T=At+B \] And total solution is\begin{align*} u\left ( x,y,t\right ) & =T_{nm}\left ( t\right ) X_{nm}\left ( x\right ) Y_{nm}\left ( y\right ) \\ & =T_{00}\left ( t\right ) X_{00}\left ( x\right ) Y_{00}\left ( y\right ) \\ & =\left ( At+B\right ) \end{align*}
Since \(X_{00}\left ( x\right ) =1\) and \(Y_{00}\left ( y\right ) =1\). Applying initial conditions gives \[ u\left ( x,y,0\right ) =0=B \] Therefore the solution is \(u\left ( x,y,t\right ) =At\). Applying second initial conditions gives\[ A=\alpha \left ( x,y\right ) \] Hence the time solution for \(n=m=0\) is \[ T_{00}=t\alpha \left ( x,y\right ) \] For each \(n,m,\) other than \(n=m=0\), the time solution of \(T^{\prime \prime }+c^{2}\lambda _{nm}T=0\) is\[ T_{nm}\left ( t\right ) =A_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) +B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \] Putting all these results together, we obtain\begin{align*} u\left ( x,y,t\right ) & =\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }T_{nm}\left ( t\right ) X_{nm}\left ( x\right ) Y_{nm}\left ( y\right ) \\ & =\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }\left [ A_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) +B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \right ] \cos \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) \\ & =\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }A_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) \cos \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) \\ & +\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \cos \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) \end{align*}
The difference in part(b) from part(a), is that the space solutions eigenfunctions are now all cosine instead of cosine and sine. When the eigenfunction is \(\cos \) the sum starts from zero. When eigenfunction is \(\sin \) the sum starts from \(1\). Now initial conditions are applied as in part (a). \[ u\left ( x,y,0\right ) =0=\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }A_{nm}\cos \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) \] Hence \(A_{nm}=0\). And the solution becomes\[ u\left ( x,y,t\right ) =\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \cos \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) \] Taking derivative of the solution w.r.t. time\[ \frac{\partial }{\partial t}u\left ( x,y,t\right ) =\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }c\sqrt{\lambda _{nm}}B_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) \cos \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) \] At \(t=0\) the above becomes\[ \alpha \left ( x,y\right ) =\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }c\sqrt{\lambda _{nm}}B_{nm}\cos \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) \] Multiplying both sides by \(\cos \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) \) and integrating gives\begin{align*} \int _{0}^{L}\int _{0}^{H}\alpha \left ( x,y\right ) \cos \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) dxdy & =c\sqrt{\lambda _{nm}}B_{nm}\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }\int _{0}^{L}\int _{0}^{H}\cos ^{2}\left ( \frac{n\pi }{L}x\right ) \cos ^{2}\left ( m\frac{\pi }{H}y\right ) dxdy\\ & =c\sqrt{\lambda _{nm}}B_{nm}\left ( \frac{L}{2}\right ) \left ( \frac{H}{2}\right ) \end{align*}
Hence\[ B_{nm}=\frac{4}{LHc\sqrt{\lambda _{nm}}}\int _{0}^{L}\int _{0}^{H}\alpha \left ( x,y\right ) \cos \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) dxdy \] Summary of solution\begin{align*} X_{n}\left ( x\right ) & =\cos \left ( \frac{n\pi }{L}x\right ) \qquad n=0,1,2,\cdots \\ \mu _{n} & =\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=0,1,2,\cdots \\ Y_{nm}\left ( y\right ) & =\cos \left ( m\frac{\pi }{H}y\right ) \qquad m=0,1,2,\cdots \\ \lambda _{nm}-\mu _{n} & =\left ( m\frac{\pi }{H}\right ) ^{2}\qquad m=0,1,2,\cdots ,n=0,1,2,\cdots \\ T_{nm}\left ( t\right ) & =\left \{ \begin{array} [c]{ccc}t\alpha \left ( x,y\right ) & & n=m=0\\ B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) & & \text{otherwise}\end{array} \right . \\ u\left ( x,y,t\right ) & =\left \{ \begin{array} [c]{ccc}t\alpha \left ( x,y\right ) & & n=m=0\\ \sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \cos \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) & & \text{otherwise}\end{array} \right . \\ B_{nm} & =\frac{4}{LHc\sqrt{\lambda _{nm}}}\int _{0}^{L}\int _{0}^{H}\alpha \left ( x,y\right ) \cos \left ( \frac{n\pi }{L}x\right ) \cos \left ( m\frac{\pi }{H}y\right ) dxdy \end{align*}
Same problem, but using the following boundary conditions\begin{align*} u\left ( 0,y,t\right ) & =0\\ u\left ( L,y,t\right ) & =0\\ u\left ( x,0,t\right ) & =0\\ u\left ( x,H,t\right ) & =0 \end{align*}
Since the boundary conditions are homogeneous Dirichlet then the \(X\left ( x\right ) \) ODE solution is\begin{align*} X_{n} & =\sin \left ( \frac{n\pi }{L}x\right ) \\ \mu & =\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \end{align*}
And \(Y\left ( y\right ) \) ODE solution is\begin{align*} Y_{nm}\left ( y\right ) & =\sin \left ( \frac{m\pi }{H}y\right ) \\ \lambda _{nm} & =\left ( m\frac{\pi }{H}\right ) ^{2}+\mu _{n}\\ & =\left ( m\frac{\pi }{H}\right ) ^{2}+\left ( \frac{n\pi }{L}\right ) ^{2}\qquad m=1,2,3,\cdots ,n=1,2,3,\cdots \end{align*}
And the time solution is\[ T_{nm}\left ( t\right ) =A_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) +B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \qquad m=1,2,3,\cdots ,n=1,2,3,\cdots \] Hence the total solution is\begin{align*} u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }T_{nm}\left ( t\right ) X_{nm}\left ( x\right ) Y_{nm}\left ( y\right ) \\ & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{H}y\right ) \\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{H}y\right ) \end{align*}
At \(t=0\)\[ 0=\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{H}y\right ) \] Hence \(A_{nm}=0\) and the solution becomes\[ u\left ( x,y,t\right ) =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{H}y\right ) \] Taking derivative\[ \frac{\partial }{\partial t}u\left ( x,y,t\right ) =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{nm}c\sqrt{\lambda _{nm}}\cos \left ( c\sqrt{\lambda _{nm}}t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{H}y\right ) \] At \(t=0\)\[ \alpha \left ( x,y\right ) =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{nm}c\sqrt{\lambda _{nm}}\sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{H}y\right ) \] Therefore, using orthogonality in 2D, we find\[ B_{nm}=\frac{4}{LHc\sqrt{\lambda _{nm}}}\int _{0}^{L}\int _{0}^{H}\alpha \left ( x,y\right ) \sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{H}y\right ) dxdy \] Summary of solution\begin{align*} X_{n}\left ( x\right ) & =\sin \left ( \frac{n\pi }{L}x\right ) \qquad n=1,2,3,\cdots \\ \mu _{n} & =\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \\ Y_{nm}\left ( y\right ) & =\cos \left ( m\frac{\pi }{H}y\right ) \qquad m=1,2,3,\cdots \\ \lambda _{nm}-\mu _{n} & =\left ( m\frac{\pi }{H}\right ) ^{2}\qquad m=1,2,3,\cdots ,n=1,2,3,\cdots \\ T_{nm}\left ( t\right ) & =B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \\ u\left ( x,y,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{nm}\sin \left ( c\sqrt{\lambda _{nm}}t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( m\frac{\pi }{H}y\right ) \\ B_{nm} & =\frac{4}{LHc\sqrt{\lambda _{nm}}}\int _{0}^{L}\int _{0}^{H}\alpha \left ( x,y\right ) \sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{H}y\right ) dxdy \end{align*}
\[ \nabla ^{2}u=\frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{\partial y^{2}}+\frac{\partial ^{2}u}{\partial z^{2}}=0 \] Let \(u=XYZ\) where \(X\equiv X\left ( x\right ) ,Y\equiv Y\left ( y\right ) ,Z\equiv Z\left ( z\right ) \). Substituting this back in the above gives\[ X^{\prime \prime }YZ+Y^{\prime \prime }XZ+Z^{\prime \prime }XY=0 \] Dividing by \(XYZ\neq 0\) gives\begin{align*} \frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y}+\frac{Z^{\prime \prime }}{Z} & =0\\ \frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y} & =-\frac{Z^{\prime \prime }}{Z} \end{align*}
Since the left side depends on \(x,y\) only and the right side depends on \(z\) only and they are equal, they must both be the same constant. Say \(-\lambda \), and we write \begin{align} \frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y} & =-\lambda \tag{1}\\ \frac{Z^{\prime \prime }}{Z} & =\lambda \nonumber \end{align}
The problem asks to separate the \(z\) variable, then the ODE for this variable is\begin{equation} Z^{\prime \prime }-\lambda Z=0 \tag{2} \end{equation} With boundary conditions\begin{align*} Z^{\prime }\left ( 0\right ) & =0\\ Z\left ( H\right ) & =f\left ( x,y\right ) \end{align*}
We will continue separation from part(a). From (1) in part (a)\[ \frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y}=-\lambda \] We now need to separate \(X,Y\). Therefore\[ \frac{X^{\prime \prime }}{X}=-\lambda -\frac{Y^{\prime \prime }}{Y}\] As the left side depends on \(x\) only and right side depends on \(y\) only and both are equal, then they are equal to some constant, say \(-\mu \)\begin{align*} \frac{X^{\prime \prime }}{X} & =-\mu \\ -\lambda -\frac{Y^{\prime \prime }}{Y} & =-\mu \end{align*}
The \(x\) ODE becomes\begin{align} X^{\prime \prime }+\mu X & =0\tag{1}\\ X\left ( 0\right ) & =0\nonumber \\ X\left ( L\right ) & =0\nonumber \end{align}
And the \(y\) ODE becomes\begin{align} -\frac{Y^{\prime \prime }}{Y} & =-\mu +\lambda \nonumber \\ Y^{\prime \prime }+\left ( \lambda -\mu \right ) Y & =0\tag{2} \end{align}
With B.C.\begin{align*} Y\left ( 0\right ) & =0\\ Y\left ( W\right ) & =0 \end{align*}
Now that we have the three ODE’s we start solving them. Starting with the \(x\) ODE (1). The solution is\begin{align*} X_{n} & =\sin \left ( \frac{n\pi }{L}x\right ) \\ \mu & =\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \end{align*}
For each \(n\) there is solution for the \(y\) ODE\begin{align*} Y_{nm} & =\sin \left ( \frac{m\pi }{W}y\right ) \\ \lambda _{nm}-\mu _{n} & =\left ( \frac{m\pi }{W}\right ) ^{2}\qquad m=1,2,3,\cdots \end{align*}
Or\[ \lambda _{nm}=\left ( \frac{m\pi }{W}\right ) ^{2}+\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots ,m=1,2,3,\cdots \] And for each \(n\) and for each \(m\) there is a solution for the \(z\) ODE we found in part (a), which is\begin{align*} Z^{\prime \prime }-\lambda _{nm}Z & =0\\ Z^{\prime }\left ( 0\right ) & =0 \end{align*}
The solution is, since \(\lambda _{nm}>0\) is\begin{align*} Z & =c_{1}\cosh \left ( \sqrt{\lambda _{nm}}z\right ) +c_{2}\sinh \left ( \sqrt{\lambda _{nm}}z\right ) \\ Z^{\prime } & =c_{1}\sqrt{\lambda _{nm}}\sinh \left ( \sqrt{\lambda _{nm}}z\right ) +c_{2}\sqrt{\lambda _{nm}}\cosh \left ( \sqrt{\lambda _{nm}}z\right ) \end{align*}
Applying B.C. \(Z^{\prime }\left ( 0\right ) =0\) gives\[ 0=c_{2}\sqrt{\lambda _{nm}}\] Hence \(c_{2}=0\) and the solution becomes\[ Z=c_{nm}\cosh \left ( \sqrt{\lambda _{nm}}z\right ) \] Putting all these solutions together, we obtain\[ u\left ( x,y,z\right ) =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }c_{nm}\sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{W}y\right ) \cosh \left ( \sqrt{\lambda _{nm}}z\right ) \] Only now we apply the last boundary condition \(u\left ( x,y,H\right ) =f\left ( x,y\right ) \) to find \(c_{nm}\).\[ f\left ( x,y\right ) =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }c_{nm}\sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{W}y\right ) \cosh \left ( \sqrt{\lambda _{nm}}H\right ) \] Applying 2D orthogonality gives\begin{align*} \int _{0}^{L}\int _{0}^{W}f\left ( x,y\right ) \sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{W}y\right ) dxdy & =c_{nm}\cosh \left ( \sqrt{\lambda _{nm}}H\right ) \int _{0}^{L}\int _{0}^{W}\sin ^{2}\left ( \frac{n\pi }{L}x\right ) \sin ^{2}\left ( \frac{m\pi }{W}y\right ) dxdy\\ & =c_{nm}\cosh \left ( \sqrt{\lambda _{nm}}H\right ) \left ( \frac{L}{2}\right ) \left ( \frac{W}{2}\right ) \end{align*}
Hence\begin{align*} c_{nm} & =\frac{\int _{0}^{L}\int _{0}^{W}f\left ( x,y\right ) \sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{W}y\right ) dxdy}{\cosh \left ( \sqrt{\lambda _{nm}}H\right ) \left ( \frac{L}{2}\right ) \left ( \frac{W}{2}\right ) }\\ & =\frac{4}{LW\cosh \left ( \sqrt{\lambda _{nm}}H\right ) }\int _{0}^{L}\int _{0}^{W}f\left ( x,y\right ) \sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{W}y\right ) dxdy \end{align*}
Summary of solution
\begin{align*} X_{n} & =\sin \left ( \frac{n\pi }{L}x\right ) \\ Y_{nm} & =\sin \left ( \frac{m\pi }{W}y\right ) \\ \lambda _{nm} & =\left ( \frac{m\pi }{W}\right ) ^{2}+\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots ,m=1,2,3,\cdots \\ u\left ( x,y,z\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }c_{nm}\sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{W}y\right ) \cosh \left ( \sqrt{\lambda _{nm}}z\right ) \\ c_{nm} & =\frac{4}{LW\cosh \left ( \sqrt{\lambda _{nm}}H\right ) }\int _{0}^{L}\int _{0}^{W}f\left ( x,y\right ) \sin \left ( \frac{n\pi }{L}x\right ) \sin \left ( \frac{m\pi }{W}y\right ) dxdy \end{align*}
Equation 7.4.7 is\begin{align*} \nabla ^{2}\phi +\lambda \phi & =0\\ \phi \left ( 0,y\right ) & =0\\ \phi \left ( L,y\right ) & =0\\ \phi \left ( x,0\right ) & =0\\ \phi \left ( x,H\right ) & =0 \end{align*}
And 7.4.6 is\[ \lambda =\frac{-{\displaystyle \oint } \phi \nabla \phi \cdot \hat{n}ds+{\displaystyle \iint \limits _{R}} \left \vert \nabla \phi \right \vert ^{2}dxdy}{{\displaystyle \iint \limits _{R}} \phi ^{2}dxdy}\] \({\displaystyle \oint } \phi \nabla \phi \cdot \hat{n}ds=0\) as we are told \(\phi =0\) on the boundary and this integration is for the boundary only. Hence \(\lambda \) simplifies to \[ \lambda =\frac{{\displaystyle \iint \limits _{R}} \left \vert \nabla \phi \right \vert ^{2}dxdy}{{\displaystyle \iint \limits _{R}} \phi ^{2}dxdy}\] The numerator can not be negative, since the integrand \(\left \vert \nabla \phi \right \vert ^{2}\) is not negative. Similarly, the denominator has positive integrand, because \(\phi \) can not be identically zero, as it is an eigenfunction. Hence we conclude that \(\lambda \geq 0.\)
\begin{align} \nabla \cdot \left ( u\nabla v\right ) & =u\nabla ^{2}v+\nabla u\cdot \nabla v\tag{1}\\ \nabla \cdot \left ( v\nabla u\right ) & =v\nabla ^{2}u+\nabla v\cdot \nabla u \tag{2} \end{align}
Equation (1)-(2) leads to\begin{align*} \nabla \cdot \left ( u\nabla v\right ) -\nabla \cdot \left ( v\nabla u\right ) & =\left ( u\nabla ^{2}v+\nabla u\cdot \nabla v\right ) -\left ( v\nabla ^{2}u+\nabla v\cdot \nabla u\right ) \\ \nabla \cdot \left ( u\nabla v-v\nabla u\right ) & =u\nabla ^{2}v-v\nabla ^{2}u+\nabla u\cdot \nabla v-\nabla v\cdot \nabla u \end{align*}
But \(\nabla u\cdot \nabla v=\nabla v\cdot \nabla u\) so the above reduces to\[ \nabla \cdot \left ( u\nabla v-v\nabla u\right ) =u\nabla ^{2}v-v\nabla ^{2}u \] Therefore \begin{equation}{\displaystyle \iint } \left ( u\nabla ^{2}v-v\nabla ^{2}u\right ) dxdy={\displaystyle \iint } \nabla \cdot \left ( u\nabla v-v\nabla u\right ) dxdy \tag{3} \end{equation} But the RHS of the above is of the form \({\displaystyle \iint } \left ( \nabla \cdot A\right ) dxdy\) where \(A=\left ( u\nabla v-v\nabla u\right ) \) here. Which we can apply divergence theorem on it and obtain \({\displaystyle \oint } \left ( A\cdot \hat{n}\right ) ds\). Therefore, using divergence theorem on the RHS of (3), then (3) can be written as\[{\displaystyle \iint } \left ( u\nabla ^{2}v-v\nabla ^{2}u\right ) dxdy={\displaystyle \oint } \left ( u\nabla v-v\nabla u\right ) \cdot \hat{n}ds \] Which is what is required to show.
Equation 7.4.5 is\begin{equation}{\displaystyle \iint \limits _{R}} \phi _{\lambda _{1}}\phi _{\lambda _{2}}dxdy=0\qquad \text{if }\lambda _{1}\neq \lambda _{2} \tag{7.4.5} \end{equation} From part (a), we found\begin{equation}{\displaystyle \iint } \left ( u\nabla ^{2}v-v\nabla ^{2}u\right ) dxdy={\displaystyle \oint } \left ( u\nabla v-v\nabla u\right ) \cdot \hat{n}ds \tag{1} \end{equation} But we know that, since both \(u,v\) satisfy the multidimensional eigenvalue problem on same domain, then\begin{align} \nabla ^{2}v+\lambda _{v}v & =0\tag{2}\\ \beta _{1}v+\beta _{2}\left ( \nabla v\cdot \hat{n}\right ) & =0 \tag{3} \end{align}
And similarly \begin{align} \nabla ^{2}u+\lambda _{u}u & =0\tag{4}\\ \beta _{1}u+\beta _{2}\left ( \nabla u\cdot \hat{n}\right ) & =0 \tag{5} \end{align}
Now we will use (2,3,4,5) into (1) to obtain 7.4.5. From (2), we see that \(\nabla ^{2}v=-\lambda _{v}v\) and from (4) \(\nabla ^{2}u=-\lambda _{u}u\) and from (3) \(\nabla v\cdot \hat{n}=-\frac{\beta _{1}}{\beta _{2}}v\) and from (5) \(\nabla u\cdot \hat{n}=-\frac{\beta _{1}}{\beta _{2}}u\). Substituting all of these back into (1) gives\begin{align}{\displaystyle \iint } \left ( u\left ( -\lambda _{v}v\right ) -v\left ( -\lambda _{u}u\right ) \right ) \ dxdy & ={\displaystyle \oint } u\left ( \nabla v\cdot \hat{n}\right ) -v\left ( \nabla u\cdot \hat{n}\right ) ds\nonumber \\{\displaystyle \iint } \left ( -\lambda _{v}uv+\lambda _{u}vu\right ) \ dxdy & ={\displaystyle \oint } u\left ( -\frac{\beta _{1}}{\beta _{2}}v\right ) -v\left ( -\frac{\beta _{1}}{\beta _{2}}u\right ) ds\nonumber \\{\displaystyle \iint } \left ( \lambda _{u}-\lambda _{v}\right ) uv\ dxdy & ={\displaystyle \oint } \frac{\beta _{1}}{\beta _{2}}\left [ -uv+uv\right ] ds\nonumber \\ \left ( \lambda _{u}-\lambda _{v}\right ){\displaystyle \iint } uv\ dxdy & =0 \tag{6} \end{align}
We now use (6) the above to show that 7.4.5 is correct. In (6), if we replace \(u=\phi _{\lambda _{1}},v=\phi _{\lambda 1}\) and \(\lambda _{u}=\lambda _{1},\lambda _{v}=\lambda _{2}\) then (6) becomes\[ \left ( \lambda _{1}-\lambda _{2}\right ){\displaystyle \iint } \left ( \phi _{\lambda _{1}}\phi _{\lambda 1}\right ) \ dxdy=0 \] We see now that for \(\lambda _{1}\neq \lambda _{2}\), then \({\displaystyle \iint } \left ( \phi _{\lambda _{1}}\phi _{\lambda 1}\right ) \ dxdy=0\). Which is what we asked to show.