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\begin{document}
\begin{center}
Laplace Equation-Separation of Variables
\end{center}
\bigskip Consider
\begin{align*}
\Delta u &= \frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u} {%
\partial y^{2}}+\frac{\partial ^{2}u}{\partial z^{2}}= 0\qquad 0\leq x\leq
a\quad 0\leq y\leq b \\
\frac{\partial u}{\partial y}+u &= 0\qquad y=0 \\
u(0,y) &= 0 \\
\frac{\partial u}{\partial x} &= 0\qquad x=a \\
u(x,b) &=g(x) \qquad y=b
\end{align*}
\bigskip
\noindent Assume (to be justified later)%
\begin{equation*}
u=X(x)Y(y)
\end{equation*}
\noindent Substituting into the Laplace equation we get%
\begin{align*}
X^{\prime \prime}Y+Y^{\prime \prime}X &= 0 \\
\frac{X^{\prime \prime}}{X}+\frac{Y^{\prime \prime}}{Y} &= 0
\end{align*}
\noindent Hence,%
\begin{align*}
X^{\prime \prime }+\lambda ^{2}X& =0\qquad X(0)=0\qquad X^{\prime }(a)=0 \\
Y^{\prime \prime }-\lambda ^{2}Y& =0\qquad Y^{\prime }(0)+Y(0)=0
\end{align*}
\noindent Solving we get%
\begin{align*}
X_{n}(x)& =\sin (\lambda _{n}x)\qquad \lambda _{n}=\frac{(n+\frac{1}{2})\pi
}{a} \\
Y_{n}(y)& =A\cosh (\lambda _{n}y)+B\sinh (\lambda _{n}y)
\end{align*}
\noindent Using the boundary conditions%
\begin{equation*}
0=Y^{\prime}_{n}(0)+Y_{n}(0)=B\beta_n+A
\end{equation*}
\noindent So%
\begin{equation*}
B=-1\qquad A=\beta _{n}
\end{equation*}%
\noindent and%
\begin{equation*}
Y_{n}(y)=\lambda _{n}\cosh (\lambda _{n}y)-\sinh (\lambda _{n}y)
\end{equation*}%
\noindent and%
\begin{equation*}
u(x,y)=\tsum\limits_{n=0}^{\infty }A_{n}\sin (\lambda _{n}x)\left[ \lambda
_{n}\cosh (\lambda _{n}y)-\sinh (\lambda _{n}y)\right]
\end{equation*}%
\noindent At $y=b$ we have%
\begin{align*}
u(x,b)=g(x)& =\tsum\limits_{n=0}^{\infty }A_{n}\sin (\lambda _{n}x)\left[
\lambda _{n}\cosh (\lambda _{n}b)-\sinh (\lambda _{n}b)\right] \\
g(x)& =\tsum\limits_{n=0}^{\infty }\widetilde{A}_{n}\sin (\lambda
_{n}x)\quad \widetilde{A}_{n}=A_{n}\left[ \lambda _{n}\cosh (\lambda
_{n}b)-\sinh (\lambda _{n}b)\right]
\end{align*}%
So%
\begin{eqnarray*}
\widetilde{A}_{n} &=&\frac{2}{a}\tint\limits_{0}^{a}g(x)\sin (\beta _{n}x)dx
\\
A_{n} &=&\frac{2}{a}\frac{1}{\lambda _{n}\cosh (\lambda _{n}b)-\sinh
(\lambda _{n}b)}\tint\limits_{0}^{a}g(x)\sin (\lambda _{n}x)dx
\end{eqnarray*}
\bigskip
\begin{center}
Theory
\end{center}
\bigskip
We consider the easier case
\begin{align*}
\Delta u& =0\qquad 0\leq x\leq L\quad 0\leq y\leq L \\
u(x,0)& =0 \\
u(0,y)& =u(L,y)=0 \\
u(x,L)& =g(x)
\end{align*}%
with $|g(x)|\leq L$.
Usingseparation of variables we get%
\begin{equation*}
u(x,y)=\tsum\limits_{n=0}^{\infty }A_{n}X_{n}(x)\sinh \left( \frac{n\pi }{L}%
y\right) =\tsum\limits_{n=0}^{\infty }A_{n}\sqrt{\frac{2}{L}}\sin \left(
\frac{n\pi x}{L}\right) \sinh \left( \frac{n\pi y}{L}\right)
\end{equation*}
\noindent So%
\begin{equation*}
|X_{n}(x)|\leq \sqrt{\frac{2}{L}}
\end{equation*}
\noindent
\begin{equation*}
A_{n}=\frac{1}{\sinh (n\pi )}\tint\limits_{0}^{L}X_{n}(x)g(x)dx
\end{equation*}
\noindent So%
\begin{align*}
|A_{n}|\; &\leq \frac{1}{|\sinh (n\pi )|}\tint\limits_{0}^{L}|X_{n}(x)|\cdot
|g(x)|dx \\
&\leq \frac{K\sqrt{\frac{2}{L}}L}{|\sinh (n\pi )|}=\frac{K\sqrt{2L}}{|\sinh
(n\pi )|}
\end{align*}
\noindent So for $y\leq y_{0}