Laplace's equation in Polar coordinate, an example?

Question

Consider Laplace's equation in polar coordinates $$ \frac {1}{r} \frac {\partial} {\partial r} (r \frac {\partial U} {\partial r}) + \frac {1} {r^2} \frac {\partial^2 U} {\partial \theta^2} = 0$$ with $U(r,\theta)$ as the solution, subject to the boundary conditions:

$$U(a,\theta)=\begin{cases} 2\theta && 0 < \theta <\pi\\ 0 &&\pi< \theta < 2\pi\\ \end{cases}$$

How can we calculate $\,U(0, \theta)$ ?

Answer 1

If we use separation of variables, we can write: $u(r,\theta) = P(r) Q(\theta)$. The boundary conditions for $u$ yield $P(a) = U(a,\theta)$ and, because of periodicity, $Q(0) = Q(2\pi)$ and $Q'(0) = Q'(2\pi)$. Plugging the new definition for $u$ into the PDE results in:

$$r^2 \left( \frac{1}{r P} \frac{\mathrm{d}}{\mathrm{d}r } \, r \frac{\mathrm{d}P}{\mathrm{d}r} \right) = -\frac{Q''}{Q} = \lambda \in \mathbb{R}, $$ where the primes denote differentiation with respect the independent variable. The problem for $Q(\theta)$ has as solution:

\begin{align} \lambda = 0: \quad & Q = A \theta + B, \\ \lambda > 0: \quad & Q = A \cos{k \theta} + B \sin{k \theta}, \\ \lambda < 0: \quad & Q = A \cosh{\sqrt{|\lambda|} \theta} + B \sinh{\sqrt{|\lambda|} \theta}, \end{align} $k^2 = \lambda$, where the periodicity condition remains to be satisfied. For $\lambda = 0$ we find $A = 0$, while for $\lambda < 0$, $A = B = 0$. For the case $\lambda > 0$, the only nontrivial solution is that corresponding to $k = n = 1,2, \ldots$ (i.e., $k$ is an integer) and therefore:

$$ Q_n(\theta) = A_n \cos{n \theta} + B_n \sin{n \theta}$$

The problem for $P(r)$ reduces to solving an Euler equation, which upon trying solutions of the type $P(r) = r^s$ yields $s^2-n^2= 0$ and therefore $s = \pm n$. The bounded solution (that for which $|P(0)|<\infty$) is therefore given by:

$$ P_n(r) = C_n r^n $$

When you impose the boundary condition at $r=a$ you can find the constant of integration $C_n$.

I hope you find this useful.

If somebody finds any typo or any other type of fail in my discussing of the problem, please let me know. Cheers!

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Note that we can let $n = 0$ as a solution for the case $\lambda = 0$ since $Q_0(\theta) = A_0$ (which plays the role of the constant $B$)

Answer 2

The simplest approach to evaluating the value at $r=0$ (and due to the way that Laplace's equation works, the value is independent of $\theta$) is to start by simplifying down the equation system. First, we integrate along $\theta$. This gives $$ \frac1r\frac{d}{dr}\left(r \frac{d}{dr}\int Ud\theta\right)+\frac1{r^2}\left[\frac{\partial U}{\partial\theta}\right]_b^{b+2\pi} =0 $$ Again, through the properties of Laplace's equation, we have that $\frac{\partial U}{\partial\theta}|_{\theta=b}=\frac{\partial U}{\partial\theta}|_{\theta=b+2\pi}$ except at the surface, where this relationship may not hold depending on the boundary conditions and the choice of $b$. If we choose $b=0$ or $b=\pi$, then the gradient is different on the two sides, and this could cause a problem (it actually doesn't, but it's better to be rigorous). Choosing any other value within the domain for $b$, we can see that the second term above is zero, and thus $$ \frac1r\frac{d}{dr}\left(r\frac{d}{dr}\int U d\theta\right) = 0 $$ This can be solved easily to obtain $$ \int U d\theta = A + \frac{B}r $$ for some integration constants $A$ and $B$. By the properties of Laplace's equation again, we know that the solution will not go infinite, and thus $B=0$. This gives us that $\int Ud\theta$ must remain constant for all $r$.

In particular, we have $$ \int_0^{2\pi} U(a,\theta)d\theta = \pi^2 $$ and thus $$ \int_0^{2\pi} U(0,\theta)d\theta = \pi^2 $$ but $U(0,\theta)=U(0,0)$ is a constant, and thus we have $$ U(0,\theta) = \frac\pi2 $$

Answer 3

There is a much simpler approach for a question like this, since the function $U$ is harmonic it satisfies the mean value property (my assumption by the way is that $0\le r\le a$), and so $$U(0,\theta)=U(0,0)=\frac{1}{|B_a|}\int_{B_a(0)}U(r,\theta)\,dr\,d\theta=\frac{1}{|B_1(0)|a}\int_{\partial B_a(0)}U(a,\theta)\,d\theta$$ $$=\frac{1}{a\pi}\int_0^{2\pi}U(a,\theta)\,d\theta=\frac{1}{a\pi}\int_0^{\pi}2\theta\,d\theta=\frac{\pi^2}{a\pi}=\frac{\pi}{a}.$$ Note that $U(0,\theta)=U(0,0)$, because rotating a radius of zero does nothing.