**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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