2
$\begingroup$

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 $u_0$.

$\endgroup$
9
  • $\begingroup$ 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 Jun 28 '10 at 12:14
  • 1
    $\begingroup$ @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 Jun 28 '10 at 12:39
  • $\begingroup$ The solution must be harmonic in $B-K$ for all $s$. $\endgroup$ – Kaminoite Jun 28 '10 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 Jun 28 '10 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 Jun 28 '10 at 13:12
1
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ 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 Jun 28 '10 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 Jun 28 '10 at 19:51
  • 1
    $\begingroup$ 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 Jun 28 '10 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 Jun 28 '10 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 Jun 28 '10 at 20:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.