I'm learning for an exam and I'm surprised by the following statement that is given without proof or example:

Let $\Omega\subset\mathbb{R}^n$ be open, bounded and connected and let $f\in C^0(\overline{\Omega})$ (i.e. $f$ is continuous up to the boundary of $\Omega$). Then, in general, there is no $u\in C^2(\Omega)\cap C^0(\overline \Omega)$ satisfying $-\Delta u = f$ in $\Omega$ and $u=0$ on $\partial\Omega$.

I thought that there would always be such a solution under the given prerequisites and that one could even write it down in a closed formula using Green's function...

So what am I missing? Are there sets $\Omega$ as above such that no Green's function exists for them? And can anyone give me an example where no $u$ as above exists?

Thank you very much for your help!

  • 2
    $\begingroup$ No, the Green's function for the domain may be fine. It's that if the source term isn't at least Dini continuous, it's possible to construct a function whose Newtonian potential is not $C^2$. That's what Problem 4.9 in Gilbarg-Trudinger (referenced below in the answer by Tomas and written out by Willie) refers to. $\endgroup$ – Ray Yang Feb 14 '13 at 16:12

The problem is not the domain, the problem is that you are asking only $f\in C^0(\overline{\Omega})$. Take a look in Problem 4.9 of Gilbard-Trudinger.

Update 1: I found this example in the book of Qung Han, Fang Hua Lin - Elliptic pArtial Differential Equations (page 65):

Let $R<1$ and $B_R(0)=B_R$ the ball in $\mathbb{R}^N$ with center in origin. Let $x=(x_1,...,x_N)$ and define $$f(x)=\frac{x_2^2-x_1^2}{2|x|^2}\Bigg[\frac{N+2}{(-\log{|x|}^{1/2})}+\frac{1}{2(-\log{|x|})^{3/2}}\Bigg]$$



You can verify that $f\in C(\overline{B}_R)$, $u\in C(\overline{B}_R)\cap C^\infty (\overline{B}_R\setminus\{0\})$. Also,

$$ \left\{ \begin{array}{rl} \Delta u=f &\mbox{ in $B_R$} \\ u=\phi &\mbox{ in $\partial B_R$ } \end{array} \right. $$

Nonetheless, $\lim_{|x|\to 0} D_{11}u(x)=0$, which implies that $u\notin C^2(B_r)$.

  • 4
    $\begingroup$ ... a sketch of its solution is given in my answer here. $\endgroup$ – Willie Wong Feb 14 '13 at 16:10
  • $\begingroup$ Thank you for pointing me to the problem. But isn't there a simpler counterexample? :) $\endgroup$ – Sh4pe Feb 14 '13 at 16:31
  • 2
    $\begingroup$ Maybe there is, but I dont know, sorry. $\endgroup$ – Tomás Feb 14 '13 at 16:46

You can write the solution as the convolution of Green's function with $f$. It will have the correct Laplacian (in some sense), but it will not necessarily have all second-order partial derivatives as continuous functions. The underlying reason is that inverting the Laplacian and then taking two derivatives amounts to applying a couple of Riesz transforms. These are singular integral operators of Calderón-Zygmund type, so they map $L^p$ into itself for $1<p<\infty$, and also map $C^{k,\alpha}$ into itself as long as $\alpha$ is not an integer.

At the integer exponents there is trouble, which already manifests itself in Calculus 2: the formula $\int x^p\,dx = \frac{x^{p+1}}{p+1}$ fails when $p=-1$. The power $x^{-1}$ is unique in that its integral is spread out evenly over all scales. This causes trouble in integral estimates: despite having expected control on each individual scale, we end up with a divergent integral anyway.

Other instances of this: Hilbert transform, harmonic conjugates. E.g., the conjugate of a harmonic function that is continuous on closed disk is not necessarily continuous. But the Hölder (and Dini) continuity is preserved under conjugation.


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