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\begin{document}

NOTE ANSWER IS NOT COMPLETE!!!!!

{\bf Question}

Laplace's equation can be written in polar coordinates as

$$\nabla ^2 \phi=\ds\frac{\pl^2\phi}{\pl
r^2}+\ds\frac{1}{r}\ds\frac{\pl \phi}{\pl
r}+\ds\frac{1}{r^2}\ds\frac{\pl^2\phi}{\pl \theta^2}=0$$

Suppose $\nabla ^2 \phi=0$ inside the unit semi-circle $0<r\leq
1,\ 0 \leq \theta \leq \pi$. Deduce that a possible solution can
take the form $\phi=A+B\theta$ where $A,\ B$ are constants.
Confirm that this is indeed the unique solution when the following
boundary conditions are specified and find $A$ and $B$.

$$\ds\frac{\pl \phi}{\pl r}=0\ \rm{when}\ r=1;\ \ \phi=+1\
\rm{when}\ \theta=0;\ \ \phi=-1\ \rm{when}\ \theta=\pi$$

Show that $\phi=Re(A-iB\log z),\ z=x+iy=r\exp(i\theta)$. Now use
the method of conformal transformations to solve the following
boundary-value problems, where $\nabla ^2 \phi=0$ in the unit
semi-circle $0<r\leq 1,\ 0 \leq \theta \leq \pi$.

\begin{description}
\item[(a)]
If the boundary conditions are now:

$$\phi=-1\ \rm{when}\ r=1, 0 \leq \theta < \ds\frac{\pi}{2};\ \
\phi=+1\ \rm{when}\ r=1,\ \ds\frac{\pi}{2} \leq \theta <\pi$$

$$\ds\frac{\pl \phi}{\pl y}=0\ \rm{when}\ \theta=0\ \rm{or}\
\theta=\pi,\ 0 \leq r \leq 1$$

In particular, show that on $\theta=\pi,\
\phi=1-\ds\frac{2}{\pi}\arctan\left(\ds\frac{1-r^2}{2r}\right)$.

(Hint: consider the transformation $w=\ds\frac{(z-i)}{(iz-1)}$.)

\item[(b)]
If the boundary conditions are now

$\begin{array} {l} \ds\frac{\pl \phi}{\pl r}=0\ \rm{when}\ r=1,\ 0
\leq \theta <\pi;\\ \phi=-1\ \rm{when}\ \theta=\pi, 0 \leq r \leq
1;\\ \phi=-1\ \rm{when}\ \theta=0,\ 0 \leq r < \ds\frac{1}{2};\\
\phi=+1\ \rm{when}\ \theta=0,\ \ds\frac{1}{2} \leq r \leq 1
\end{array}$

\newpage

In particular, show that on $r=1$, for a suitably chosen range of
$\arctan$,

$$\phi=1-\ds\frac{2}{\pi}\arctan\left(\ds\frac{2\sin\theta}{5\cos\theta-4}\right)$$

(Hint: consider the transformation $w=\ds\frac{(2z-1)}{(2-z)}$.)
\end{description}

\medskip

{\bf Answer}

PICTURE \vspace{2in}

Check: $\phi(r,\theta)=A+B\theta \Rightarrow \phi_r=\phi_{rr}=0,\
\phi_{\theta}=B,\ \phi_{\theta\theta}=0$

Therefore $\nabla^2\phi(x,y)=\phi_{rr}+\ds\frac{1}{r}\phi_r
+\ds\frac{1}{r^2}\phi_{\theta\theta}=0+\ds\frac{0}{r}+\ds\frac{0}{r^2}=0,\
r \ne 0 \surd\surd$

Now satisfy boundary conditions

$\phi_r=0$ at $r=1 \Rightarrow \ds\frac{\pl _r}{\pl
r}(a+B\theta)=0$

(which it does everywhere so boundary condition is satisfied)

$\begin{array} {rcl} \phi_r=+1\ \rm{when}\  \theta=0 & \Rightarrow
& B\times 0 + A\\ & \Rightarrow & A=1 \end{array}$

$\begin{array} {rcl} \phi=-1\ \rm{when}\ \theta=\pi & \Rightarrow
& B\pi+A=-1\\ & \Rightarrow & B=-\ds\frac{2}{\pi} \end{array}$

so \un{$\phi(r,\theta)=1-\ds\frac{2}{\pi}\theta$ satisfies
$\nabla^2\phi=0$ in $P$}

Consider $z=x+iy$

$\Phi(z)=A-iB\log z,\ \ A,\ B \in {\bf{R}}$

$Re[\Phi(z)]=Re(A+iB\log |z|+B\theta)=A+B\theta=\phi(r,\theta)$ as
required.

\begin{description}
\item[(a)]
PICTURE \vspace{2in}

consider $w=\left(\ds\frac{z-i}{iz-1}\right)$:

$y=0\ 0 \leq x \leq 1$:

\begin{eqnarray*} w & = & \ds\frac{x-i}{ix-1}\\ & = &
\ds\frac{(x-i)}{x^2+1}(-1-ix)\\ & = &
\ds\frac{-x-ix^2+i-x}{x^2+1}\\ & = & \ds\frac{-2x-i(x^2-1)}{x^2+1}
\end{eqnarray*}

\begin{eqnarray*} (iz-1) & = & \left(\ds\frac{z-i}{w}\right)\\
\left(i-\ds\frac{1}{w}\right)^2 & = & 1-\ds\frac{i}{w}\\ z & = &
\ds\frac{w-i}{(iw-1)} \end{eqnarray*}

$\begin{array}{rcl} |iw-1| & = & |w-i|\\ |w+i| & = & |w-i|
\end{array}$

$z=Re^{i\theta}$

$w=\left(\ds\frac{Re^{i\theta}-i}{iRe^{i\theta}-1}\right):
w=\left(\ds\frac{R-i}{iR-1}\right)=\ds\frac{R-i}{R+i}\ds\frac{1}{i}$
where $\theta=0$

$\ds\frac{1}{i}\ds\frac{(z-i)}{(z+i)}=
\ds\frac{1}{i}\ds\frac{(z-i)^2}{|z|^2+1}=
\ds\frac{1}{i}\ds\frac{1}{i}\ds\frac{z^2-2iz-1}{(|z|^2+1)}$

$\ds\frac{(Re^{i\theta}-i)}{i(Re^{i\theta}+i)}=\ds\frac{1}{i}
\ds\frac{(Re^{2i\theta}-2iRe^{i\theta}-1)}{(R^2+1)}$

\begin{eqnarray*} w(0) & = & +i\\ w(i) & = & 0\\ w(1) & = &
\ds\frac{1-i}{-(1+i)}=-1\\ w(-1) & = & \ds\frac{-(1+i)}{-(1+i)}=+1
\end{eqnarray*}

\begin{eqnarray*} w & = & \ds\frac{(x-i)}{-1+ix}\\ & = &
\ds\frac{(x-i)(-1-ix)}{1+x^2}\\ & = & -x-ix^2+i-x\\ & = &
\ds\frac{-2x+i(1-x^2)}{1+x^2} \end{eqnarray*}
\end{description}
\end{document}
