Find a solution that satisfies Laplace's equation in polar coordinates

Question

How may I find a solution that solves Laplace's equation in polar coordinates, subject to the boundary conditions? In particular, I need to find one solution that satisfies $$\Delta u = 0,$$ subject to $u(2,\theta)=0$ and $u(4,\theta)=\sin \theta$. Recall that, in polar coordinates, the Laplacian of $u$ is $$\Delta u = u_{rr}+\frac 1r u_{r}+\frac 1{r^2}u_{\theta \theta}.$$

Answer 1

The general form of a harmonic function in an annulus is $$ u(r,\theta) = (A_0 +B_0\log r)+ \sum_{n\ne 0} r^n (A_n \cos |n|\theta+B_n\sin |n|\theta)\tag{1} $$ The stated boundary conditions are:

$$ 0 = (A_0 +B_0\log 2)+ \sum_{n\ne 0} 2^n (A_n \cos |n|\theta+B_n\sin |n|\theta) \tag{2}$$ and $$ \sin\theta = (A_0 +B_0\log 4)+ \sum_{n\ne 0} 4^n (A_n \cos |n|\theta+B_n\sin |n|\theta) \tag{3}$$ To match the coefficient of $\sin \theta$ in $(3)$, you need $4^1B_1+4^{-1}B_{-1}=1$. To match its (zero) coefficient in $(2)$, you need $2^1B_1+2^{-1}B_{-1}=0$. Solve this linear system to find $B_1$ and $B_{-1}$. The rest of coefficients are set to zero since no other harmonics appear on the left side of $(2) $ or $(3)$.

The final answer is: $$ u(r,\theta) = (r B_1+r^{-1}B_{-1})\sin \theta $$ where $B_1$ and $B_{-1}$ are as above.

Answer 2

According to this user, the general form of a harmonic function in an annulus is $$u(r,\theta)=a_0\ln r+b_0 +\sum_{n=1}^\infty (a_n r^n+ b_n r^{-n} \cos(n\theta)+\sum_{n=1}^\infty (c_n r^n + d_n r^{-n}) \sin(n\theta).$$

At $r=2$, we integrate with respect to $\theta$ over $[0,2\pi]$ to find $$b_0=-a_0 \ln 2. \tag{A}$$

At $r=4$, we integrate with respect to $\theta$ over $[0,2\pi]$ to find $$a_0=0 \implies b_0=0. \tag{B}$$

At $r=2$, we multiply by $\cos (m\theta)$ and integrate with respect to $\theta$ over $[0,2\pi]$ to find $$b_m=-4^m a_m. \tag{C}$$

At $r=2$, we multiply by $\sin (m\theta)$ and integrate with respect to $\theta$ over $[0,2\pi]$ to find $$d_m=-4^m c_m. \tag{D}$$

At $r=4$, we multiply by $\cos (m\theta)$ and integrate with respect to $\theta$ over $[0,2\pi]$ to find $$b_m=-16^m a_m. \tag{E}$$

At $r=4$, we multiply by $\sin (m\theta)$ and integrate with respect to $\theta$ over $[0,2\pi]$ to find $$d_m=\begin{cases}-4^m c_m & \text{if }m \not=1 \\ 4-16c_m & \text{if }m=1. \end{cases} \tag{F}$$

From equations $\text{(C)}$, $\text{(D)}$, $\text{(E)}$, and $\text{(F)}$, we find that $$a_m=b_m=0 \, \forall m, c_m=d_m=0 \, \forall m \not=1, c_1=\frac 13, d_1=-\frac 43.$$

We reindex to $n$ all constants that are based on $m$. Substituting these into our general form, our solution is therefore $$u(r,\theta)=\left(\frac r3 - \frac 4{3r}\right) \sin(\theta).$$