### Author Topic: HA7 problem 3  (Read 2796 times)

#### Victor Ivrii

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##### HA7 problem 3
« on: March 11, 2015, 08:24:19 AM »
a.  Solve
\begin{align*}
& \Delta :=u_{xx}+u_{yy}=0&& \text{in } r<a\\[3pt]
& u|_{r=a}=f(\theta).
\end{align*}
where we use polar coordinates $(r,\theta)$ and  f(\theta)=\left\{\begin{aligned} &\sin(\theta) &&0<\theta<\pi\\ &0 &&\pi<\theta<2\pi. \end{aligned}\right.

b.  Solve
\begin{align*}
& \Delta :=u_{xx}+u_{yy}=0&& \text{in }  r>a\\[3pt]
& u|_{r=a}=f(\theta),\\[3pt]
& \max |u| <\infty.
\end{align*}
where we use polar coordinates  $(r,\theta)$ and  f(\theta)=\left\{\begin{aligned} &\sin(\theta) &&0<\theta<\pi\\ &0 &&\pi<\theta<2\pi. \end{aligned}\right.

#### Biao Zhang

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##### Re: HA7 problem 3
« Reply #1 on: March 12, 2015, 09:22:17 PM »
(a)
Take  $r \ge 0$. Let $a >0$.  $(x,y) \mapsto (r,\theta)$ with Dirichlet BC: $\{u = h(\theta) : r = a\}$, $u(r,\theta)$ has a solution of the form:

$$u(r,\theta) = \frac{1}{2} A_0 + \sum_{n=1}^{\infty} r^n (A_n \cos(n \theta) + B_n \sin(n \theta))$$

$$A_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}h(\phi)\cos(n\phi)d\phi$$

$$B_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}h(\phi)\sin(n\phi)d\phi$$

Substituting in $f(\theta)$ and splitting the integral between $(0,\pi)$ and $(\pi,2\pi)$:

$$\implies A_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}f(\phi)\cos(n\phi)d\phi = \frac{1}{\pi a^n}( \int_{0}^{\pi}\sin(\phi)\cos(n\phi)d\phi + \int_{\pi}^{2\pi}0\cos(n\phi)d\phi)$$

$$= \frac{1}{\pi a^n}( \frac{-(-1)^n-1}{n^2-1}) \text{, as } n \in \mathbb{N}$$

$$A_0=\frac{-(-1)^0-1}{\pi a^0(0-1)}=\frac{2}{\pi}$$

$$\implies B_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}f(\phi)\sin(n\phi)d\phi = \frac{1}{\pi a^n}( \int_{0}^{\pi}\sin(\phi)\sin(n\phi)d\phi + \int_{\pi}^{2\pi}0\sin(n\phi)d\phi)$$

$$= \frac{1}{\pi a^n}( 0) = 0 ; \text{, as } n \in \mathbb{N}$$

$$\implies u(r,\theta) =\frac{1}{\pi} + \sum_{n=1}^{\infty} (\frac{r}{a})^n \frac{-1-(-1)^n}{(n^2-1) \pi } \cos(n \theta) \phantom{\ }$$
For $n=2m-1$,$m=1,2,3,...$ as 'odd' coefficients are 0.

(b)

Let $a >0$. BC $h(\theta) = f(\theta)$.  $r\mapsto \frac{1}{r}$ : $\{ z : z \in \mathbb{R}_\infty , z \ge a \} \mapsto \{ \frac{1}{z} : \frac{1}{z} \in \mathbb{R}_{\infty} , \frac{1}{z} \le \frac{1}{a}\}$. Then $\frac{1}{r} \rightarrow 0$ as $r \rightarrow \infty$.

BC: $\{u = h(\theta) : r = a\}$, $u(r,\theta)$ has a solution of the form:

$$u(r,\theta) = \frac{1}{2} A_0 + \sum_{n=1}^{\infty} r^n (A_n \cos(n \theta) + B_n \sin(n \theta))$$

$$A_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}h(\phi)\cos(n\phi)d\phi$$

$$B_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}h(\phi)\sin(n\phi)d\phi$$

Substituting in $f(\theta)$, mapping $\{r \mapsto \frac{1}{r}, a \mapsto \frac{1}{a}\}$, and splitting the integral between $(0,\pi)$ and $(\pi,2\pi)$, gives us the solution on the exterior:

$$\implies u(r,\theta) = \frac{1}{2} A_0 + \sum_{n=1}^{\infty} r^{-n} (A_n \cos(n \theta) + B_n \sin(n \theta))$$

$$\implies A_n = \frac{1}{\pi a^{-n}} \int_{0}^{2\pi}f(\phi)\cos(n\phi)d\phi = \frac{a^{n}}{\pi }( \int_{0}^{\pi}\sin(\phi)\cos(n\phi)d\phi + \int_{\pi}^{2\pi}0\cos(n\phi)d\phi)$$

$$= \frac{a^{n}}{\pi}( \frac{-1-(-1)^n}{(n^2-1) \pi}) \text{, as } n \in \mathbb{N}$$

so we have::

$$A_0=\frac{2}{\pi}$$

$$\implies B_n = \frac{1}{\pi a^{-n}} \int_{0}^{2\pi}f(\phi)\sin(n\phi)d\phi = \frac{a^{n}}{\pi }( \int_{0}^{\pi}\sin(\phi)\sin(n\phi)d\phi + \int_{\pi}^{2\pi}0\sin(n\phi)d\phi)$$

$$= \frac{a^{n}}{\pi }(0) =0 \text{, as } n \in \mathbb{N}$$

$$\implies u(r,\theta) = \frac{1}{\pi}+\sum_{n=1}^{\infty} (\frac{a}{r})^n \frac{-1-(-1^n)}{(n^2-1) \pi } \cos(n \theta) \phantom{\ }$$