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Perturbative behaviour of solutions of the solutions of the Dirichlet problem for the Laplacian:

Lets consider $ B = B(0, 1) \in \mathbb{R}^2$ be the unit circle with center at $0\in\mathbb{R}^2$. Let $u_0$ be an harmonic function on $B$ also harmonic at the boundary, that is, $u_0$ is harmonic in the ball $B(0, 1+\varepsilon)$ for $\varepsilon > 0$ small. Then, if we denote by $f = {u_0}_{|\partial B}$ we have that $u_0$ satisies (trivially) the Dirichlet problem

$$ \begin{array} {rcl} \Delta u_0(x) & = & 0 \newline {u_0}_{|\partial B}(x) &= &f(x) \end{array} $$

Now, let $K\subset B$ be a compact set and $g:K\rightarrow \mathbb{R}$ be a smooth function (real analytic, for instance), and consider the one parameter family of Dirichlet problems

$$ \begin{array} {rcl} \Delta u_s(x) & = & 0 \newline {u_s}_{|\partial B}(x) &= &f(x)\newline {u_s}_{|K}(x) &= & {u_0}_{|K}(x)+sg(x)\newline \end{array} $$

It is clear that for $s=0$ the solution of this problem is the same as the original problem stated above, so we consider this as a perturbative problem.

MY QUESTION IS:

How does $u_s$ behaves near the compact set $K$? It is known that $u_s$ is continuous in all the unit ball (also in $K$) but it is hoped that is not differentiable near $K$. It is possible to show that, generically, there exists an $\alpha\in\mathbb{R}$ such that it is satisfied

$$ \lim_{x\longrightarrow z}\frac{|u_s(x)-u_s(z)|}{||x-z||^{\alpha}} = C(s, z) \neq 0, $$ where $C(s, z)$ is a constant, depending on $s$ and $z\in K$?

Note that for $s=0$, the above limit exists when $\alpha = 1$ and $C(0)$ is the Lipschitz constant of of $u_0$.

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  • $\begingroup$ I don't understand the question. The condition ${u_s}_{|K}(x) = {u_0}_{|K}(x)+sg(x)$ is redundant since a harmonic function on $B$ is uniquely determined by its trace on $\partial B$. $\endgroup$
    – Andrey Rekalo
    Commented Jun 28, 2010 at 12:14
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    $\begingroup$ @Kaminoite: for the perturbed problem, do you actually want $\triangle u_s = 0$ only on $B\setminus K$? If $u_s$ is not differentiable near $\partial K$ (as indicated by the bit after "MY QUESTION IS"), it can hardly be a harmonic function in $B$. If this is the case, aren't you just looking at the Dirichlet problem on $B\setminus K$ with $u | \partial B = 0$ and $u | \partial K = s g$? Then you are just comparing arbitrary extensions of $g$ into $K$ against harmonic extensions of $g$ into $B\setminus K$... $\endgroup$
    – Willie Wong
    Commented Jun 28, 2010 at 12:39
  • $\begingroup$ The solution must be harmonic in $B-K$ for all $s$. $\endgroup$
    – Kaminoite
    Commented Jun 28, 2010 at 12:55
  • $\begingroup$ @Andrey Rekalo: The condition ${u_s}_{|K}(x) = {u_0}_{|K}(x)+sg(x)$ is not redundant. In fact, you can think that $K$ is part of the boundary for a new $\bar{\Omega} = \Omega-K$. $\endgroup$
    – Kaminoite
    Commented Jun 28, 2010 at 12:57
  • $\begingroup$ @Kaminoite: Thank you for the comment. I thought $u_s$ was supposed to be harmonic everywhere in $B$. $\endgroup$
    – Andrey Rekalo
    Commented Jun 28, 2010 at 13:12

1 Answer 1

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No, it will not be differentiable in the whole ball. To see this, let $u$ be the zero function and $g$ be nearly anything nonnegative and not identically zero in $K$. For example $g=1$. Then recall Hopf's lemma.

This will also work to show that differentiability fails at any point on the boundary of $K$, at which $g$ achieves its maximum (on the whole of $K$).

However, it will be $C^\alpha$ in the ball, which is the last question you stated. This follows from the smoothness of $g$ and Holder estimates for $u$. For this you also need something about $K$ itself being smooth of course-- all hope is lost if the boundary of $K$ is irregular.

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    $\begingroup$ Adding... it helps to think about explicit examples. Let $K$ be the smaller ball $B(0,\epsilon)$ and $g=1$. Then $u_s$ is going to be the fundamental solution of Laplace's equation, with its singularity chopped off (and you need to subtract a little constant to make it zero on the boundary, and also multiply by a little constant so that it is equal to $s$ on the boundary of $B(0,\epsilon)$. $\endgroup$
    – Scott Armstrong
    Commented Jun 28, 2010 at 19:17
  • $\begingroup$ @Scott Armstrong: I have two questions: 1.- Does the value $\alpha$ depends on $K$ and $g$? 2.- What happens if $K$ is a Cantor set? and $g$ is a smooth function on the ball $B$ restricted to $K$? Can we recover the $C^\alpha$ smoothness? $\endgroup$
    – Kaminoite
    Commented Jun 28, 2010 at 19:51
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    $\begingroup$ As long as $K$ is smooth, I do not believe the value of $\alpha$ depends on $K$, and the same goes for $g$. But to be sure you should look it up in Gilbarg and Trudinger. However, if $K$ is irregular then you lose regularity. I think you need $K$ to be $C^{1,\beta}$ for some $\beta > 0$, as simply $C^1$ does not work. There is some paper of Safonov on the arxiv from 2008 that I was reading on how Hopf's lemma needs $C^{1,\beta}$ domains and not simply $C^1$ domains, and this is related. As for $g$, it doesn't matter what it is, provided it is itself $C^{1,\alpha}$ on $K$. $\endgroup$
    – Scott Armstrong
    Commented Jun 28, 2010 at 19:58
  • $\begingroup$ So, if $K$ is smooth, you state that $\alpha=1/2$. This is because there are examples in the unit ball with $K$ an interval on the real axis where $u$ is $C^{1/2}$ on the adherence of the ball. $\endgroup$
    – Kaminoite
    Commented Jun 28, 2010 at 20:08
  • $\begingroup$ Well, if $K$ is an interval on the real axis, then it isn't smooth. If $K$ is completely smooth and so is $g$, then the optimal value of $\alpha$ is $1$. That is, $u$ is $C^{0,1}$ (also known as Lipchitz) which implies it is differentiable almost everywhere, but not everywhere. And it will generally fail to be differentiable on the boundary of $K$. You seem to be interested in more irregular $K$. I am sure there are some things known about that, and the number of papers could fill volumes-- but I can't help you, because I don't really know anything about that. $\endgroup$
    – Scott Armstrong
    Commented Jun 28, 2010 at 20:16

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