I am trying to solve the following problem:
Let $\Omega$ denote the region $\{ (r, \theta) : r>1, 0 < \theta < \pi \}$. Find the bounded solution to the problem:
$\begin{align}\nabla^{2} u(r, \theta) = 0 && \text{in}\,\Omega \\ u(r, \pi) = u(r, 0) = 0, && r>1 \\ u(1,\theta) = 1 && \text{for}\, 0 < \theta < \pi \end{align}$
So, because we're looking at this problem on a semicircle of unbounded radius, I need to make sure to take steps to ensure that any solutions I get are bounded.
The problem is that the solution I am getting doesn't match the one in the back of the book, and I don't know where my mistake is.
Here's what I've done:
First, I rewrote the Laplace equation, $\nabla^{2} u(r, \theta) = 0$, as $r^{-1}[ru_{r}(r, \theta)]_{r} + r^{-2}u_{\theta\theta} (r, \theta) = 0$, and then finally as $\displaystyle u_{rr}(r, \theta) + \frac{1}{r}u_{r}(r, \theta) + \frac{1}{r^{2}}u_{\theta \theta}(r, \theta) = 0$.
Next, I assumed a separable solution of the form $u(r, \theta) = R(r)\Theta(\theta)$, and taking the appropriate partial derivatives and substituting them into the Laplace equation, and I obtained $\displaystyle R^{\prime\prime}(r)\Theta (\theta) + \frac{1}{r}R^{\prime}(r)\Theta^{\prime}(\theta) + \frac{1}{r^{2}}R(r)\Theta^{\prime\prime}(\theta) = 0$, which separates out as $\displaystyle \frac{r^{2}R^{\prime\prime}(r)}{R(r)} + \frac{rR^{\prime}(r)}{R(r)} = -\frac{\Theta^{\prime\prime}(\theta)}{\Theta(\theta)}$.
Then, in order for the two sides to remain equal as $r$ and $\theta$ range through their possible values, both sides must equal the same separation constant, call it $\mu$.
So, with that I get the two ODE problems: $\mathbf{r^{2}R^{\prime\prime}(r)+rR^{\prime}(r) = \mu R(r)}$, $\mathbf{R(r)}$ bounded for $\mathbf{r \to \infty}$, and $\mathbf{\Theta^{\prime\prime}(\theta) = -\mu \Theta(\theta)}$, $\mathbf{\Theta(0) = \Theta(\pi) = 0}$.
Then, I went through the 3 possible cases with the eigenvalues:
- Case I: $\mu = 0$. Then $r^{2}R^{\prime\prime}(r) + rR^{\prime}(r)=0$. Rewriting it as $\displaystyle \frac{r(rR^{\prime}(r))^{\prime}}{R(r)}=0$, and integrating twice, I get that $R(r)= c_{1}\ln(r) + c_{2}$. Since we want all of our solutions to be bounded, I must choose $c_{1} = 0$, because otherwise, as $r \to \infty$, $\ln(r)$ will become unbounded. So, $R(r) = c_{2}$, a constant.
Then, I solve $\Theta^{\prime\prime}(\theta) = 0$ by integrating twice, and I obtain $\Theta(\theta) = c_{3}\theta + c_{4}$. Appplying the boundary condition $\Theta(0) = 0$, we get that $\Theta(0) = c_{3}(0)+c_{4} = 0$ implies that $c_{4} = 0$. Applying the boundary condition $\Theta(\pi) = 0$, we get that $\Theta(\pi) = c_{3}\pi + 0 = 0$, and so $c_{3} = 0$ as well. So, the only solution to $\Theta(\theta) = 0$, the trivial solution.
Combining these, we see that when $\mu = 0$, $u(r,\theta) = R_{0}(r)\Theta_{0}(\theta) = c_{2}\cdot 0 = 0$, the trivial solution, so $\mu = 0$ is not an eigenvalue.
- Case II: $\mu < 0$. Let $\mu = -\beta^{2}$, $\beta$ real and nonzero. Then, $\Theta^{\prime\prime}(\theta) = -(-\beta^{2})\Theta(\theta) \, \implies \, \Theta^{\prime\prime}(\theta)-\beta^{2}\Theta(\theta) = 0$. The characteristic equation is $m^{2}-\beta^{2} = 0$, which has the two distinct real roots, $m = \pm \beta$. So, the general solution takes the form $\Theta(\theta) = c_{5}\sinh(\beta \theta) + c_{6}\cosh (\beta \theta)$. Plugging in the boundary condition $\Theta(0) = 0$, we have that $\Theta(0) = c_{5}\sinh (0) + c_{6}\cosh(0) = c_{6}\cosh(0) = 0\, \implies \, c_{6} = 0$. Plugging in the boundary condition $\Theta(\pi) = c_{5}\sinh(\beta \pi) + c_{6}\cosh(\beta \pi) = c_{5}\sinh(\beta \pi) + 0 = c_{5}\sinh(\beta \pi) = 0$. So, either $c_{5} = 0$ or $\sinh(\beta \pi) = 0$, but $\beta \neq 0$, so we must also have $c_{5} = 0$. Therefore, when $\mu < 0$, the only solution for $\Theta(\theta)$ is the trivial solution, $\Theta(\theta) = 0$.
Then, I solve $r^{2}R^{\prime\prime}(r) + rR^{\prime}(r) = -\beta^{2}R(r) \, \implies \, r^{2}R^{\prime\prime}(r) + rR^{\prime}(r) + \beta^{2}R(r) = 0$. Letting $R(r) = r^{\alpha}$, taking first and second derivatives of this, and plugging them back into the equation, I obtain $\alpha(\alpha-1)r^{\alpha}+\alpha r^{\alpha}+\beta^{2}r^{\alpha}=0$, which simplifies to $\alpha^{2} + \beta^{2} = 0$. We have complex conjugate roots, $\alpha = \pm \beta i$, so the general solution for $R(r)$ is of the form $R(r) = c_{7}r^{0}\cos(\beta \ln r) + c_{8} r^{0} \sin(\beta \ln r) = c_{7}\cos (\beta \ln r) + c_{8} \sin (\beta \ln r)$. In order for my solution $R(r)$ to be bounded as $r \to \infty$, I will need $c_{7} = c_{8} = 0$. So, when $\mu < 0$, I get the trivial solution for $R(r)$ as well.
Putting these together, I have that for $\mu < 0$, the solution $u(r, \theta) = \Theta(\theta)R(r) = 0 \cdot 0 = 0$, the trivial solution, so there are no negative eigenvalues.
- Case III: $\mu > 0$. Let $\mu = \beta^{2}$, $\beta$ real and nonzero. Then, solving for $\Theta(\theta)$, we obtain that $\Theta^{\prime\prime}(\theta) = -\beta^{2}\Theta(\theta) \, \implies \, \Theta^{\prime\prime}(\theta) + \beta^{2} \Theta(\theta) = 0$. The characteristic equation is $m^{2}+\beta^{2} = 0$, which has complex conjugate roots $m = \pm \beta i$. Applying the boundary condition $\Theta(0) = 0$, we obtain $\Theta(0) = c_{9} \cos(0) + c_{10} \sin (0) = c_{9} + 0 = c_{9} = 0$. And applying the boundary condition $\Theta(\pi) = 0$, we obtain $\Theta(\pi) = c_{9} \cos(\beta pi)+ c_{10}\sin(\beta \pi) = 0 + c_{10}\sin(\beta \pi) = c_{10}\sin(\beta \pi) = 0$. So, either $c_{10}=0$ or $\sin(\beta \pi) = 0$. Supposing that $c_{10} \neq 0$, we must have that $\sin(\beta \pi) = 0$, so $\beta \pi = n \pi$, which implies that $\beta_{n} = n$, $n = 1, 2, 3, \cdots$. So, $\mu_{n} = \beta_{n}^{2} = n^{2}$ are the eigenvalues.
Since $c_{9}$ vanished, our solution for $\displaystyle \Theta(\theta) = c_{10_{n}}\sin(n \theta)$.
Then, we look at the $R$ equation in this case: $r^{2}R^{\prime\prime}(r) + rR^{\prime}(r) = \beta^{2}R(r) \, \implies \, r^{2}R^{\prime\prime}(r)+rR^{\prime}(r) - \beta^{2}R(r) = 0$. Letting $R(r) = r^{\alpha}$, taking first and second derivatives, and plugging them back into this equation, we obtain $r^{2}(\alpha(\alpha-1)r^{\alpha - 2} + r \alpha r^{\alpha - 1} - \beta^{2}r^{\alpha} = 0$, which reduces to $\alpha^{2} - \beta^{2} = 0$, where we have the distinct real roots, $\alpha = \pm \beta$.
Therefore, the solution takes on the form $R(r) = c_{11}r^{\beta} + c_{12}r^{-\beta}$. In order for our solution to be bounded as $r \to \infty$, we must have $c_{11} = 0$, so $R(r) = c_{12}r^{-\beta}$, and since earlier, we found $\beta_{n} = n$, this becomes $\displaystyle R(r) = c_{12_{n}}r^{-n}$.
Putting our solutions together, we have that $\displaystyle u(r,\theta) = R_{n}\Theta_{n} = c_{12_{n}}r^{-n}\cdot c_{10_{n}}\sin(n \theta)$. Letting $A = c_{12}\cdot c_{10}$, we have that $\displaystyle u(r, \theta) = A_{n}r^{-n}\sin(n \theta)$, $n = 1,2,3,\cdots$
Then, by the superposition principle, we obtain the general solution by summing over all the eigenfunctions $\displaystyle u(r,\theta) = \sum_{n=1}^{\infty} A_{n}r^{-n}\sin (n\theta )$.
Finally, we apply the inhomogeneous boundary condition: $\displaystyle u(1,0) = \sum_{n=1}^{\infty} A_{n} \sin(n \theta) = 1$, where $\displaystyle A_{n} = \frac{2}{r^{n}\pi}\int_{0}^{\pi}\sin(n \theta)d\theta = \left[-\frac{2}{nr^{n}\pi}\cos(n \theta) \right]_{0}^{\pi} = -\frac{2}{n r^{n}\pi}\cos(n\pi)- \left(-\frac{2}{n r^{n} \pi}\cos(0) \right)= \frac{-2}{nr^{n}\pi}(-1) + \frac{2}{nr^{n}\pi}= \frac{4}{nr^{n}\pi}$.
So, my over-all solution is $\mathbf{\displaystyle u(r,\theta) = \frac{4}{\pi} \sum_{n=1}^{\infty} n^{-1} r^{-2n} \sin (n \theta)}$.
The only problem with this is that the book's solution is given as $\mathbf{\displaystyle \frac{4}{\pi}\sum_{n=1}^{\infty} (2n-1)^{-1}r^{-(2n-1)}\sin(2n-1)\theta}$
Which one of us is wrong? And if I'm the one who's wrong, can you please tell me what my mistake(s) was(were) and how to fix it(them)?
Thank you!