MHB Laplace general soln (x-x_0)^2 + (y-y_0)^2\leq R^2

[Dustinsfl]

I don't see how this $\rho = \frac{r}{R}$ helps.Let $(x_0,y_0)$ be a point in the plane, and suppose $u(x,y)$ is continuous for $(x - x_0)^2 + (y - y_0)^2 \leq R^2$, and $u_{xx} + u_{yy} = 0$ for $(x - x_0)^2 + (y - y_0)^2 < R^2$.
Show that, for $0\leq r < R$,
$$
u(x_0 + r\cos\theta, y_0 + r\sin\theta) = \sum_{n = -\infty}^{\infty}a_n\left(\frac{r}{R}\right)^{|n|}e^{in\theta}
$$
with
$$
a_n = \frac{1}{2\pi}\int_{-\pi}^{\pi}u(x_0 + R\cos\theta, y_0 + R\sin\theta)e^{-in\theta}d\theta.
$$
In particular,
$$
u(x_0,y_0) = \frac{1}{2\pi}\int_{-\pi}^{\pi}u(x_0 + R\cos\theta, y_0 + R\sin\theta)d\theta.
$$Transform the problem by making the substitution $\rho = \frac{r}{R}$ for $0\leq r\leq R$.
The function $u(r,\theta)$ that was defined for $0\leq r\leq R$ then becomes a function $U(\rho,\theta)$ for $0\leq\rho\leq 1$.

 

[Sudharaka]

I don't see how this $\rho = \frac{r}{R}$ helps.Let $(x_0,y_0)$ be a point in the plane, and suppose $u(x,y)$ is continuous for $(x - x_0)^2 + (y - y_0)^2 \leq R^2$, and $u_{xx} + u_{yy} = 0$ for $(x - x_0)^2 + (y - y_0)^2 < R^2$.
Show that, for $0\leq r < R$,
$$
u(x_0 + r\cos\theta, y_0 + r\sin\theta) = \sum_{n = -\infty}^{\infty}a_n\left(\frac{r}{R}\right)^{|n|}e^{in\theta}
$$
with
$$
a_n = \frac{1}{2\pi}\int_{-\pi}^{\pi}u(x_0 + R\cos\theta, y_0 + R\sin\theta)e^{-in\theta}d\theta.
$$
In particular,
$$
u(x_0,y_0) = \frac{1}{2\pi}\int_{-\pi}^{\pi}u(x_0 + R\cos\theta, y_0 + R\sin\theta)d\theta.
$$Transform the problem by making the substitution $\rho = \frac{r}{R}$ for $0\leq r\leq R$.
The function $u(r,\theta)$ that was defined for $0\leq r\leq R$ then becomes a function $U(\rho,\theta)$ for $0\leq\rho\leq 1$.

Hi dwsmith, :)

So you have,

\[u(x_0 + r\cos\theta, y_0 + r\sin\theta) = \sum_{n = -\infty}^{\infty}a_n\left(\frac{r}{R}\right)^{|n|}e^{in\theta}\]

where,

\[a_n = \frac{1}{2\pi}\int_{-\pi}^{\pi}u(x_0 + R\cos\theta, y_0 + R\sin\theta)e^{-in\theta}d\theta\]

I will take, \(u(r,\theta)=u(x_0 + r\cos\theta, y_0 + r\sin\theta)\) and \(f(\theta)=u(x_0 + R\cos\theta, y_0 + R\sin\theta)\).

Then we can write,

\begin{eqnarray}

u(r,\theta)&=&\sum_{n = -\infty}^{\infty}a_n\left(\frac{r}{R}\right)^{|n|}e^{in\theta}\\

&=&a_{0}+\sum_{n = 1}^{\infty}\left[a_{-n}\left(\frac{r}{R}\right)^{n}e^{-in\theta}+a_n\left(\frac{r}{R}\right)^{n}e^{in \theta}\right]\\

&=&a_{0}+\frac{1}{2\pi}\sum_{n = 1}^{\infty}\left(\frac{r}{R}\right)^{n}\left[e^{-in\theta}\int_{-\pi}^{\pi}f(\theta)e^{in\theta}d\theta+e^{in\theta}\int_{-\pi}^{\pi}f(\theta)e^{-in\theta}d\theta\right]\\

\end{eqnarray}

Simplify this using the Euler's formula and you'll get,

\begin{eqnarray}

u(r,\theta)&=&a_{0}+\frac{1}{\pi}\sum_{n = 1}^{\infty}\left(\frac{r}{R}\right)^{n}\left[\cos(n\theta)\int_{-\pi}^{\pi}f(\theta)\cos(n\theta)d\theta+\sin(n \theta)\int_{-\pi}^{\pi}f(\theta)\sin(n \theta)d\theta\right]\\

&=&a_{0}+\frac{1}{\pi}\sum_{n = 1}^{\infty}\left(\frac{r}{R}\right)^{n}\left[\cos(n\theta)\int_{-\pi}^{\pi}f(\phi)\cos(n\phi)d\phi+\sin(n\theta) \int_{-\pi}^{\pi}f(\phi)\sin(n\phi)d\phi\right]\\

&=&a_{0}+\frac{1}{\pi}\sum_{n = 1}^{\infty}\left(\frac{r}{R}\right)^{n}\int_{-\pi}^{\pi}f(\phi)\cos n(\theta-\phi)d\phi

\end{eqnarray}

Since \(\displaystyle a_{0}=\frac{1}{2\pi}\int_{\pi}^{\pi}f(\phi)d\phi\) we get,

\[u(r,\theta)=\frac{1}{2\pi}\int_{\pi}^{\pi}f(\phi)d\phi+\frac{1}{\pi}\sum_{n = 1}^{\infty}\left(\frac{r}{R}\right)^{n}\int_{-\pi}^{\pi}f(\phi)\cos n(\theta-\phi)d\phi\]

Read pages 1 to 7 in >>this lecture note<< (Or you can refer Example 3 >>here<<). It describes how to obtain the solution of the Laplace's equation on a disk. As you can see the final result is exactly the one that we have obtained above. So what you are given to show is another form of the solution to the Laplace's equation on a disk.

Kind Regards,
Sudharaka.