Consider the following two dimensional Laplace equation on the unit disk $D$ with homogeneous Robin boundary condition:
$$\Delta u = 0, ~~\frac{\partial u}{\partial n} = b(x) u(x)~~ \forall x \in \partial D.$$ 
Here $n$ denotes the outer normal direction to the bounary $\partial D$. Assume that $b(x)$ is **piecewisely constant**. Does this equation have nontrivial solution (a solution that is not a constant) for general $b(x)$ that is piecewisely constant?  

I know if $b(x) \equiv k$ for some positive integer $k$, then the above equation has nontrivial solutions. Also, if $b(x) < 0$ and satisfies some integrability conditions, the equation only has zero solutions. But what can we say for generic piecewisely constant function $b(x)$? Is it possible for this equation to have nontrivial solutions? For example, what happens when $b(x) = \chi\_{I}$? That is, $b(x) = 1$ when $x \in I$, $b(x) = 0$ when $x \notin I$ and $I$ is a subinterval on the circle.



Laplace equation on the disk with Robin boundary condition
The normal trick is to set it up as an eigenvalue problem, namely to look instead at
$$\Delta u = 0, ~~\frac{\partial u}{\partial n} = \lambda b(x) u(x)~~ \forall x \in \partial D. (1)$$
You first establish that this corresponds to an eigenvalue problem on $L^2(\partial \Omega)$ for a compact operator, thanks to the compact embedding of $H^{1/2}(\partial D)$ into $L^2((\partial D)$.


Then the usual machinery rolls out. There is a discrete spectrum, and there are eigenvalues. And if it so happens that $\lambda=1$ is one of those, then your problem has non trivial solutions.


The advantage of this point of view is that the problem is well posed, there is a clear functional analytic setup, and you don't need much on $b$, some integrability (if it is bounded as you liked to consider, it is fine).


It is done in the following way. Given $f \in H^{-1/2}(\partial D)$ such that $\int\_{\partial\Omega} f d\sigma= 0$ consider the so called Neumann-to-Dirichlet map
\begin{eqnarray\*}
NtD : H^{-1/2}(\partial D)/\mathbb R &\to& H^{-1/2}(\partial D)/\mathbb R \\
f &\to& u|\_{\partial D} : \begin{cases} -\Delta u =0& \text{ in }D \\ \partial\_n u =f &\text{ on }\partial D \end{cases}
\end{eqnarray\*}
the quotient over $\mathbb R$ means that we impose $\int\_{\partial\Omega} f d\sigma= \int\_{\partial\Omega} u d\sigma= 0.$


The operator $NtD$ is compact : we are going to use this operator on a restricted domain, where the compactness is very easy to see.


Restrict this operator to $L^2(\partial D)/ \mathbb R$, and you can write it explicitely in terms of Fourier coefficients.
Write
$$
f=\sum\_{n=1}^\infty a\_n \cos n \theta + b\_n \sin n \theta, \sum |a\_n|^2 + |b\_n|^2 <\infty
$$
then
$$
NtD(f)= \sum\_{n=1}^\infty \frac{a\_n}{n} \cos n \theta + \frac{b\_n}{n} \sin n \theta.
$$
It is easy to see that it is a compact operator from $L^{2}(\partial D)/\mathbb R$ to $L^{2}(\partial D)/\mathbb R$ (the $n^{-1}$ helps) .


Supppose $\| b \|\_{L^\infty(\partial D)} <\infty$. Then
\begin{eqnarray\*}
B : L^{2}(\partial D)/\mathbb R &\to& L^{2}(\partial D)/\mathbb R \\
f &\to& b f - \frac{1}{\left|\partial D\right|} \int\_{\partial D}bf \end{eqnarray\*}
is a continuous, bounded operator from $ L^{2}(\partial D)/\mathbb R$ into itself.


So the map
\begin{eqnarray\*}
T : L^{2}(\partial D)/\mathbb R &\to& L^{2}(\partial D)/\mathbb R \\
f &\to& B\left(NtD \left(f\right)\right) 
\end{eqnarray\*}
is a compact, linear operator. Its spectrum is discrete, it has a maximal eigenvalue etc. and (1) is
$$
f=\lambda T f.
$$




---


Regarding regularity of the solutions. Naturally, they are analytic inside $D(0,\rho)$ for any $\rho<1$, this is just interior regularity of harmonic functions. At the boundary, the regularity is dictated by $b$. Indeed, suppose $b$ is piecewise constant. We have $Du= \partial\_r u e\_r + \frac{1}{r} \partial\_\theta u e\_\theta= \lambda b u e\_r + \frac{1}{r} \partial\_\theta u e\_\theta $. If $u$ is regular, this formula can be "extended" a little inside. Then you see immediately that you cannot differentiate $u$ a second time: all terms are differentiable, except one, $b$. However, nothing forbid $Du$ to be very integrable, so it is: $u\in W^{1,p}(D)$ for every $p<\infty$ (and maybe $\infty$ as well). So $u\in W^{1-1/p,p}(\partial D)$.


