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\begin{document}


\begin{center}
Laplace Equation-Polar Coordinates
\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 r^{2}}+\frac{1}{r}\frac{\partial u}{%
\partial r}+\frac{1}{r^{2}}\frac{\partial ^{2}u}{\partial \theta ^{2}} \\
& =\frac{1}{r}\frac{\partial }{\partial r}(r\frac{\partial u}{\partial r})+%
\frac{1}{r^{2}}\frac{\partial ^{2}u}{\partial \theta ^{2}}=0\qquad
\end{align*}

\noindent Consider a solution only of $r$. So%
\begin{align*}
\frac{1}{r}\frac{\partial }{\partial r}(r\frac{\partial u}{\partial r}) &=
0\qquad \frac{\partial }{\partial r}(r\frac{\partial u}{\partial r})=0 \\
r\frac{\partial u}{\partial r} &= C_{1} \\
u(r) &= C_{1}\log (r)+C_{2}
\end{align*}

\bigskip

\begin{center}
Spherical Coordinates
\end{center}

\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} \\
&= \frac{\partial^2 u}{\partial r^2}+\frac{2}{r}\frac{\partial u}{\partial r}
+\frac{1}{r^2} \left[ \frac{\partial^2 u}{\partial \theta^2} +\cot (\theta )%
\frac{\partial u}{\partial \theta }+\frac{1}{\sin^2 (\theta)} \frac{%
\partial^2 u}{\partial \varphi^2 }\right] =0 \\
&= \frac{1}{r^2} \frac{\partial }{\partial r} (r^2 \frac{\partial u}{%
\partial r}) +\frac{1}{r^2} \left[ \frac{\partial^2 u}{\partial \theta^2}
+\cot (\theta )\frac{\partial u}{\partial \theta }+ \frac{1}{\sin^2(\theta)} 
\frac{\partial^2 u}{\partial \varphi^2}\right] =0
\end{align*}

\bigskip \noindent Consider a solution only of $r$. So%
\begin{align*}
\frac{1}{r^2} \frac{\partial }{\partial r}(r^2 \frac{\partial u}{\partial r}%
) &= 0\qquad \frac{\partial }{\partial r}(r^{2}\frac{\partial u}{\partial r}%
)=0 \\
r^{2}\frac{\partial u}{\partial r} &= C_{1} \\
u(r) &= \frac{C_{1}}{r}+C_{2}
\end{align*}

\newpage

\begin{center}
Poisson Formula
\end{center}

\bigskip

\noindent Consider the Laplace equation inside a circle of radius $a$ \ and
on the boundary $u(a,\theta )=h(\theta )$.

\noindent Using separation of variables%
\begin{equation*}
u(r,\theta )=R(r)\Theta (\theta )
\end{equation*}

\noindent Substituting into the Laplace equation%
\begin{align*}
R^{\prime \prime }\Theta +\frac{1}{r}R^{\prime }\Theta +\frac{1}{r^{2}}%
R\Theta ^{\prime \prime }& =0 \\
r^{2}\frac{R^{\prime \prime }+\frac{1}{r}R^{\prime }}{R}& =-\frac{\Theta
^{\prime \prime }}{\Theta }=\lambda _{n}^{2}
\end{align*}

\noindent So%
\begin{align*}
\Theta ^{\prime \prime}+\lambda _{n}^{2}\Theta &=0 \\
\Theta (\theta ) &=A\cos (\lambda _{n}\theta )+B\sin (\lambda _{n}\theta )
\end{align*}

\noindent In order for the functions to be periodic with period $2\pi $ we
require that $\lambda _{n}=n$. So%
\begin{equation*}
\Theta (\theta )=A\cos (n\theta )+B\sin (n\theta )
\end{equation*}

\noindent Then%
\begin{equation*}
r^{2}R^{\prime \prime}+rR^{\prime}-nR=0
\end{equation*}

\noindent guess%
\begin{align*}
R(r) &= r^{\alpha } \\
\left( \alpha (\alpha -1)+\alpha -n^{2}\right) r^{\alpha } &= 0 \\
\alpha &= \pm n
\end{align*}

\noindent However $r^{-n}$ is not bounded at $r=0$ and so we are left with%
\begin{equation*}
u(r,\theta )=\frac{A_0}{2}+\tsum\limits_{n=1}^{\infty } r^n (A_n \cos
(n\theta )+B_n \sin (n\theta ))
\end{equation*}

\noindent Using the boundary condition at $r=a$ we get

\begin{align*}
h(\theta )& =\frac{A_{0}}{2}+\tsum\limits_{n=1}^{\infty }a^{n}(A_{n}\cos
(n\theta )+B_{n}\sin (n\theta )) \\
A_{n}& =\frac{1}{\pi a^{n}}\tint\limits_{0}^{2\pi }h(\varphi )\cos (n\varphi
)d\varphi  \\
B_{n}& =\frac{1}{\pi a^{n}}\tint\limits_{0}^{2\pi }h(\varphi )\sin (n\varphi
)d\varphi 
\end{align*}

\noindent Substituting the formula for $A$ and $B$ we get%
\begin{align*}
u(r,\theta )& =\frac{1}{2\pi }\tint\limits_{0}^{2\pi }h(\varphi )d\varphi +%
\frac{1}{\pi }\tsum\limits_{n=1}^{\infty }\left( \frac{r}{a}\right) ^{n}%
\left[ \tint\limits_{0}^{2\pi }h(\varphi )\cos (n\varphi )d\varphi \cos
(n\theta )+\tint\limits_{0}^{2\pi }h(\varphi )\sin (n\varphi )d\varphi \sin
(n\theta )\right]  \\
& =\frac{1}{2\pi }\tint\limits_{0}^{2\pi }h(\varphi )d\varphi +\frac{1}{\pi }%
\tsum\limits_{n=1}^{\infty }\left( \frac{r}{a}\right)
^{n}\tint\limits_{0}^{2\pi }h(\varphi )\left[ \cos (n\varphi )\cos (n\theta
)+\sin (n\varphi )\sin (n\theta )\right] d\varphi  \\
& =\frac{1}{2\pi }\tint\limits_{0}^{2\pi }h(\varphi )d\varphi +\frac{1}{\pi }%
\tsum\limits_{n=1}^{\infty }\left( \frac{r}{a}\right)
^{n}\tint\limits_{0}^{2\pi }h(\varphi )\cos (n(\varphi -\theta ))d\varphi  \\
& =\frac{1}{2\pi }\tint\limits_{0}^{2\pi }h(\varphi )\left[
1+2\tsum\limits_{n=1}^{\infty }\left( \frac{r}{a}\right) ^{n}\cos (n(\varphi
-\theta ))\right] d\varphi 
\end{align*}

\noindent but%
\begin{align*}
1+2\tsum\limits_{n=1}^{\infty }\left( \frac{r}{a}\right) ^{n}\cos (n\psi )&
=1+\tsum\limits_{n=1}^{\infty }\left( \frac{r}{a}\right) ^{n}\left(
e^{in\psi }+e^{-in\psi }\right)  \\
& =1+\tsum\limits_{n=1}^{\infty }\left( \frac{r}{a}\right) ^{n}e^{in\psi
}+\tsum\limits_{n=1}^{\infty }\left( \frac{r}{a}\right) ^{n}e^{-in\psi } \\
& =1+\tsum\limits_{n=1}^{\infty }\left( \frac{r}{a}e^{i\psi }\right)
^{n}+\tsum\limits_{n=1}^{\infty }\left( \frac{r}{a}e^{-i\psi }\right) ^{n}
\end{align*}

\noindent This is a geometric series. So%
\begin{align*}
u(r,\theta ) &= \frac{1}{2\pi }\tint\limits_{0}^{2\pi }h(\varphi ) \left[ 1+%
\frac{re^{i\left( \varphi -\theta \right) }}{a-re^{i\left( \varphi -
\theta\right) }}+ \frac{re^{-i\left( \varphi -\theta \right) }} {%
a-re^{-i\left(\varphi -\theta \right) }}\right] d\varphi \\
&=\frac{a^{2}-r^{2}}{2\pi }\tint\limits_{0}^{2\pi }\frac{h(\varphi )} {%
a^{2}-2ar\cos \left( \varphi -\theta \right) +r^{2}}d\varphi \\
&=\frac{a^{2}-|x|^{2}}{2\pi a}\toint \frac{u(x^{\prime})}{|x-x^{\prime}|^{2}}%
ds^{\prime}\qquad ds^{\prime}=ad\varphi
\end{align*}

\newpage

\begin{center}
Mean Value
\end{center}

\bigskip

\noindent We choose $r=0$ in the Poisson formula. We then get%
\begin{equation*}
u(0,\theta )=\frac{1}{2\pi }\tint\limits_{0}^{2\pi }h(\varphi )d\varphi =%
\text{average over circumference}
\end{equation*}

\bigskip

\noindent This is another proof of the maximum principle!

\begin{center}
\bigskip 

Solution in annulus
\end{center}

\bigskip

\noindent We repeat the above derivation but now in the annulus .%
\begin{align*}
\Delta u &= 0\qquad a\leq r\leq b \\
u(a,\theta ) &= g(\theta ) \\
u(b,\theta ) &= h(\theta )
\end{align*}

\noindent As before we do separation of variables and get%
\begin{align*}
u(r,\theta ) &= \frac{1}{2}(C_{0}+D_{0})\log (r) \\
& +\tsum\limits_{n=1}^{\infty }\left( C_{n}r^{n}+\frac{D_{n}}{r^{n}}\right)
(A_{n}\cos (n\theta )+B_{n}\sin (n\theta ))
\end{align*}

\noindent Imposing boundary conditions we get%
\begin{align*}
g(\theta ) &= u(a,\theta )=\frac{1}{2}(C_{0}+D_{0})\log (a) \\
& +\tsum\limits_{n=1}^{\infty }\left( C_{n}a^{n}+\frac{D_{n}}{a^{n}}\right)
(A_{n}\cos (n\theta )+B_{n}\sin (n\theta ))
\end{align*}

\noindent So%
\begin{align*}
(C_{0}+D_{0})\log (a) &= \frac{1}{\pi }\tint\limits_{-\pi }^{\pi
}g(\theta)d\theta \\
\left( C_{n}a^{n}+\frac{D_{n}}{a^{n}}\right) A_{n} &= \frac{1}{\pi }%
\tint\limits_{-\pi }^{\pi }g(\theta )\cos (n\theta )d\theta \\
\left( C_{n}a^{n}+\frac{D_{n}}{a^{n}}\right) B_{n} &= \frac{1}{\pi }%
\tint\limits_{-\pi }^{\pi }g(\theta )\sin (n\theta )d\theta
\end{align*}

\noindent Similarly%
\begin{eqnarray*}
(C_{0}+D_{0})\log (b) &=&\frac{1}{\pi }\tint\limits_{-\pi }^{\pi }h(\theta
)d\theta \\
\left( C_{n}b^{n}+\frac{D_{n}}{b^{n}}\right) A_{n} &=&\frac{1}{\pi }%
\tint\limits_{-\pi }^{\pi }h(\theta )\cos (n\theta )d\theta \\
\left( C_{n}b^{n}+\frac{D_{n}}{b^{n}}\right) B_{n} &=&\frac{1}{\pi }%
\tint\limits_{-\pi }^{\pi }h(\theta )\sin (n\theta )d\theta
\end{eqnarray*}

\noindent Using the first equation from each set this yields%
\begin{align*}
C_{0} &= \frac{1}{\pi }\tint\limits_{-\pi }^{\pi }g(\theta )d\theta
-D_{0}\log (a) \\
D_{0} &= \frac{1}{\pi \log (\frac{a}{b})}\tint\limits_{-\pi }^{\pi }\left[
g(\theta )-h(\theta )\right] d\theta
\end{align*}

\noindent Similarly%
\begin{align*}
\frac{A_{n}}{B_{n}}& =\frac{\tint\limits_{-\pi }^{\pi }g(\theta )\cos
(n\theta )d\theta }{\tint\limits_{-\pi }^{\pi }g(\theta )\sin (n\theta
)d\theta } \\
\frac{C_{n}a^{n}+\frac{D_{n}}{a^{n}}}{C_{n}b^{n}+\frac{D_{n}}{b^{n}}}& =%
\frac{\tint\limits_{-\pi }^{\pi }g(\theta )\cos (n\theta )d\theta }{%
\tint\limits_{-\pi }^{\pi }h(\theta )\cos (n\theta )d\theta }
\end{align*}

\end{document}
