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

In summary, the problem can be transformed by making the substitution $\rho = \frac{r}{R}$, resulting in the function $u(r,\theta)$ becoming a function $U(\rho,\theta)$ for $0\leq\rho\leq 1$. This leads to the solution of Laplace's equation on a disk, given by $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)\
  • #1
Dustinsfl
2,281
5
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 $0\leq\rho\leq 1$.
 
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  • #2
dwsmith said:
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.
 

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

What is the Laplace general solution for the equation (x-x_0)^2 + (y-y_0)^2\leq R^2?

The Laplace general solution for this equation is a circle with center (x_0, y_0) and radius R. This means that all points (x, y) within the circle satisfy the equation.

How is the Laplace general solution derived?

The Laplace general solution is derived using the Laplace equation, which is a second-order partial differential equation that describes the behavior of a scalar field in space. By solving the Laplace equation for the given boundary conditions (x=x_0, y=y_0 at R), we can obtain the general solution in the form of a circle.

What is the significance of the Laplace general solution?

The Laplace general solution is significant because it provides a solution to the Laplace equation for a given set of boundary conditions. It is a fundamental solution that can be used to solve a wide range of physical problems, such as heat transfer, electrostatics, and fluid flow.

Can the Laplace general solution be applied to other shapes besides a circle?

Yes, the Laplace general solution can be applied to other shapes besides a circle, such as a rectangle or an ellipse. However, the equation will be different for each shape and will depend on the specific boundary conditions.

Are there any limitations to using the Laplace general solution?

Yes, there are limitations to using the Laplace general solution. It is only applicable to linear, homogeneous boundary value problems, which means that the boundary conditions must be constant and the coefficients in the Laplace equation must be constant. It may also not be accurate for more complex physical systems that do not follow the assumptions of the Laplace equation.

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