<s>[INST] Laplace's equation: separation of variables Question:
Let $(r,\theta)$ denote plane polar coordinates. Show that there are countably infinitely many $k \in \Bbb R$ for which
$$\nabla^2 u=0 \qquad 1≤r≤2 \\ ku + \frac{\partial u}{\partial r}=0 \qquad r=1,2$$
has a non-trivial solution of the form $u(r,\theta) = f(r)g(\theta)$.

Attempt:
Writing out the Laplacian in plane polars and plugging in $u(r,\theta) = f(r)g(\theta)$, we get
$$\nabla^2 u = \frac{\partial^2 u}{\partial r^2} + \frac 1r \frac{\partial u}{\partial r}+ \frac{1}{r^2} \frac{\partial u^2}{\partial \theta^2} = g\frac{d^2f}{dr^2} + \frac gr \frac{df}{dr} + \frac{f}{r^2}\frac{d^2g}{d\theta^2}$$
Thus
\begin{align}
\nabla^2 u = 0 & \implies g\frac{d^2f}{dr^2} + \frac gr \frac{df}{dr} + \frac{f}{r^2}\frac{d^2g}{d\theta^2} = 0 \\
& \implies \frac{r^2}{f}\frac{d^2f}{dr^2} + \frac rf \frac{df}{dr} = - \frac 1g \frac{d^2g}{d\theta^2} = \text{constant} : = \lambda
\end{align}
First assuming that $\lambda \neq 0$, I solved the $r$-equation:
$$\frac{r^2}{f}\frac{d^2f}{dr^2} + \frac rf \frac{df}{dr} = \lambda \implies r^2\frac{d^2f}{dr^2} + r\frac{df}{dr}-\lambda f = 0 \implies f(r) = Ar^{\sqrt \lambda} + Br^{-\sqrt \lambda}$$
The boundary condition given is
$$ku + \frac{\partial u}{\partial r} = 0 \implies kfg + g\frac{df}{dr}=0 \implies g(\theta) \big(k f(r) + f'(r)\big)=0 \qquad r=1,2 \; \; , \; \; \theta \in \Bbb R$$
$g(\theta) \equiv 0$ gives only non-trivial solutions, thus we must have $kf(1)+f'(1) = kf(2)f'(2)=0$. That is:
\begin{align*}
& k f(1)+f'(1)=0 \implies k (A+B) + \sqrt\lambda(A-B)=0 \\
& k f(2)+f'(2)=0 \implies k \big(A2^{\sqrt\lambda} + B2^{-\sqrt{\lambda}} \big) + \frac {\sqrt\lambda}{2} \big(A2^{\sqrt\lambda} - B2^{-\sqrt{\lambda}} \big)=0
\end{align*}
In matrix form, this is
$$\begin{pmatrix} k + \sqrt \lambda & k - \sqrt \lambda \\ \Big(k + \frac {\sqrt\lambda}{2}\Big) 2^{\sqrt \lambda} & \Big(k - \frac {\sqrt\lambda}{2}\Big) 2^{-\sqrt \lambda} \end{pmatrix} \begin{pmatrix} A \\ B \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$
and we get a non-trivial solution iff the determinant of the matrix is zero.
However, it seems to me that for every $k \in \Bbb R$ there is a corresponding $\lambda$ that gives a non-trivial solution. Have I done something wrong? Or am I misunderstanding something? Any help would be much appreciated.
 [/INST] From the periodicity condition, we get a general solution
$$ u_n(r,\theta) = f(r)g(\theta) = (Ar^n + Br^{-n})(C\cos n\theta + D\sin n \theta) $$
where $n=1,2,3,\dots$
Since the boundary conditions are radially symmetric, it follows that $ kf(r) +f'(r) =0 $ on $r=1,2$. Plugging this in
\begin{align}
k(A+B) + n(A-B) = 0 \\
k(A2^n + B2^{-n}) +n(A2^{n-1} -B2^{-n-1})=0
\end{align}
Converting into a matrix equation
$$ \begin{pmatrix} k+n & k-n \\ 2^nk + 2^{n-1}n & 2^{-n}k - 2^{-n-1}n \end{pmatrix}\begin{pmatrix} A \\ B \end{pmatrix} = 0 $$
There's a non-trivial solution for $(A, B)$ if the determinant of the coefficient matrix is zero
$$ (k+n)(2k-n) - 2^{2n}(k-n)(2k+n) = 0 $$
In quadratic form this is
$$ 2(2^{2n}-1)k^2 - n(2^{2n}+1)k - (2^{2n}-1)n^2 = 0 $$
where $\Delta = b^2-4ac = n^2(2^{2n}+1)^2 + 8n^2(2^{2n}-1)^2 > 0$
So there exists two values of $k$ for each corresponding positive integer $n$, thus "countably infinite"
</s>