7
$\begingroup$

Consider the PDE $-\Delta u = f$ on a bounded domain $\Omega \subset \mathbb{R}^n$, where $f \in C^\infty(\bar{\Omega})$. I wish to consider both the boundary conditions $u = 0$ and $\frac{\partial u}{\partial n} = 0$. My question is, are there reasonably well-known "compatibility" conditions under which such equations admit a solution?

$\endgroup$
5
$\begingroup$

Here is one study Overdetermined elliptic problems in physics.
It is proven in $\mathbb{R}^2$ that the Poisson equation $\Delta u=-{\rm constant}$ in $\Omega$, with boundary conditions $u=0$, $\partial u/\partial n={\rm constant}$ on $\delta\Omega$, only has a solution for a circular domain. [That solution is $u(x,y)\propto R^2-x^2-y^2$.]

$\endgroup$
2
$\begingroup$

We claim the following: a solution $u$ exists if and only if $f$ is orthogonal to the Poisson kernel with pole at every $x \in \partial \Omega$.


Suppose that $u$ is a solution. Since $u = 0$ on the boundary, we have $$u(x) = \int_\Omega G_\Omega(x, y) f(y) dy,$$ where $G_\Omega(x, y)$ is the Green function. Assuming $\Omega$ is sufficiently regular, one can differentiate under the integral sign and write $$0 = \partial_n u(x) = \int_\Omega \partial_n G_\Omega(x, y) f(y) dy,$$ where $\partial_n$ denotes the derivative in $x$ in the direction normal to the boundary; here of course $x \in \partial \Omega$. The normal derivative of the Green function defines a Poisson kernel. Thus, $f$ is orthogonal to the Poisson kernel with pole at every $x \in \partial \Omega$.

Conversely, if $f$ is orthogonal to the Poisson kernel with pole at every $x \in \partial \Omega$, then $u(x) = \int_\Omega G_\Omega(x, y) f(y) dy$ has all desired properties. Therefore, these two conditions are equivalent, as desired.


Alternatlively, one could argue as follows: if $u$ is a solution and $h$ is a harmonic function in $\Omega$ which is $C^1$ in $\overline{\Omega}$, then $$\int_\Omega h(x) f(x) dx = \int_\Omega h(x) \Delta u(x) = \int_\Omega \Delta h(x) u(x) dx = 0$$ by Green's first identity and the boundary conditions imposed on $u$. Therefore, $f$ is necessarily orthogonal to all harmonic functions.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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