<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}

NOTE REFERENCE TO SHEET 5, QUESTION 8

{\bf Question}

Reconsider the conformal map of Q8, exercises 5 and the example in
the lecture notes. Find a solution $\phi$ of Laplace's equation
$\nabla ^2 \phi=0$ inside the ellipse
$\left(\ds\frac{x}{5}\right)^2+\left(\ds\frac{y}{3}\right)^2=\ds\frac{1}{4}$,
if $\phi=0$ on th ellipse boundary and $\phi=V$ when $y=0,\
|x|&lt;2$. This situation models the electric potential inside an
ellipsoidal capacitor.


\medskip

{\bf Answer}

From Q8, Ex 5 we have:

PICTURE \vspace{2in}

Now solution of $\nabla^2\phi(u,v)=0$ in concentric circles (see
lecture notes) is

\begin{eqnarray*} \phi_1(u,v) &amp; = &amp; A\log|w|+B\\ &amp; = &amp;
A\log\sqrt{u^2+v^2}+b\\ &amp; = &amp; Re[A\log w +B],\ \ A,B, \in {\bf{R}}
\end{eqnarray*}

Boundary conditions on $|w|=r$,

On $r=1,\ V=A\log 1+B \Rightarrow B=V$

On $r=2,\ 0=A\log 2+B \Rightarrow A=-\ds\frac{V}{\log 2}$

Therefore

\begin{eqnarray*} \phi_1(u,v) &amp; = &amp; \left[V-\ds\frac{V}{\log
2}\log |w|\right]\\ &amp; = &amp; \ds\frac{V}{\log 2}\log
\left(\ds\frac{2}{|w|}\right) \end{eqnarray*}

Therefore $\Phi_1(w)=\ds\frac{V}{\log
2}\log\left(\ds\frac{2}{w}\right)$

So $\Phi(z)=\Phi_1(w)=\Phi_1\left(z+\ds\frac{1}{z}\right)$

Hence back in the $z$-plane we have,

$$\Phi(z)=\ds\frac{V}{\log
2}\log\left(\ds\frac{2}{z+\frac{1}{z}}\right)$$

and so the harmonic solution $\nabla^2\phi(x,y)=0$ which satisfies
the boundary conditions is:

\begin{eqnarray*} \phi(x,y) &amp; = &amp; Re[\Phi(z)]\\ &amp; = &amp;
\ds\frac{V}{\log 2}\log \left|\ds\frac{2}{z+\frac{1}{z}}\right|\\
&amp; = &amp; \ds\frac{V}{\log 2}\log \left|\ds\frac{2z}{z^2+1}\right|\\ &amp;
= &amp; -\ds\frac{V}{\log 2}\log
\left|\ds\frac{1}{2}\left(z+\ds\frac{1}{z}\right)\right|
\end{eqnarray*}

So setting $z=x+iy$, we get

\begin{eqnarray*}
\left|\ds\frac{1}{2}\left(z+\ds\frac{1}{z}\right)\right| &amp; = &amp;
\ds\frac{1}{2}\left|x+iy+\ds\frac{x-iy}{x^2+y^2}\right|\\ &amp; = &amp;
\ds\frac{1}{2}\left|x\ds\frac{(x^2+y^2+1)}{x^2+y^2}+iy\ds\frac{(x^2+y^2-1)}{x^2+y^2}\right|\\
&amp; = &amp; \ds\frac{1}{2}
\sqrt{x^2\ds\frac{(x^2+y^2+1)^2}{(x^2+y^2)^2}+y^2\ds\frac{(x^2+y^2-1)^2}{(x^2+y^2)^2}}\\
&amp; = &amp; \ds\frac{1}{2(x^2+y^2)} \times
\sqrt{x^2(x^2+y^2+1)^2+y^2(x^2+y^2-1)^2} \end{eqnarray*}

so

$\phi(x,y)$

\un{$=-\ds\frac{V}{\log 2}\log\left[\ds\frac{1}{2(x^2+y^2)} \times
\sqrt{x^2\ds\frac{(x^2+y^2+1)^2}{(x^2+y^2)^2}+y^2\ds\frac{(x^2+y^2-1)^2}{(x^2+y^2)^2}}\right]$}

UGH!!
\end{document}
</pre></body></html>