Laplace equation on the disk with Robin boundary condition

Question

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.

Answer 1

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 $$ f=\lambda T f. $$

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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 $u\in W^{1-1/p,p}(\partial D)$.

Answer 2

This is a partial answer. Consider a (real ) function $u$ satifying the problem above and note that $z^k, \bar{z}^k$ satisfy a similar one but with $b=k$. Then using Green's identity on $D$ one obtains $$\int_{\partial D} (b-k)u z^k=\int_{\partial D} (b-k)u \bar{z}^k=0$$ that is $\hat{(bu)}(k)=|k| \hat{u}(k)$. In particular $bu$ has zero mean and $\|bu\|_2=\|u'\|_2<\infty$, where the $L^2$ norms are taken on $\partial D$. This would be the end of the story if we had $k$ instead of $|k|$ but now the Hilbert transform appears and I treat only the case where $b=\chi_I$. I may assume that $I=(-a,a)$ with $a<\pi$ and observe that $g=u\chi_I$ has zero mean. Then we may bound the $L^2$ norm of $g$ with that of $g'$: the simplest inequality is $\|g\|^2_{2,I} \le 2a^2 \|g'\|^2_{2,I} $ which gives $\|u'\|_2^2=\|g\|_{2,I}^2 \le 2a^2 \|g'\|_{2,I}^2 \le 2a^2\|u'\|_2^2$ and $u=0$ if $2a^2 <1$. This is not very satisfactory but can be improved to $a<\pi$ if we can prove that $u(-a)=u(a)$. In this case, in fact, expanding $g$ in Fourier series in $(-a,a)$, the constant $2a^2$ can be replaced by $a^2/\pi^2$ and one concludes as before. The equality $u(a)=u(-a)$ depends on the regularity of $u$. Assuming $u \in H^2(D)$, then $\nabla u \in H^1(D)$ and $ \chi_i u=\partial u/\partial n \in H^{1/2}(\partial D)$. However, since $u' \in L^2(\partial D)$ the values $u(\pm a)$ exists finite. If one of these values is non-zero, then the Fourier coefficients of $\chi_I u$ would behave like $1/n$ for large $n$ and $\chi_i u$ would not be in $H^{1/2}(\partial D)$. This gives $u(\pm a)=0$ and then $u=0$ by the above argument. The assumption $u \in H^2(D)$ seems reasonable in a strong formulation of the problem but too strong in the weak one. Maybe one sees how to weaken it, if what written above is correct (I admit that some details are not written).