Intuitively, why is the solution to Dirichlet Problem on open disk a convolution?

Question

We know that, given a continuous function $f$ on unit circle $\Bbb T$, we can find the solution to Dirichlet problem on the unit disk by $$ U(re^{2\pi it}) = P_r \star f $$ where $P_r$ is the Poisson kernel and $0<r<1$.

Is there an intuitive reason why this formula is a convolution for each fixed $r$? I understand that the Poisson kernels form a set of test functions that approach the $\delta$ - function as $r \to 1^-$; hence near the boundary $U$ approximates $f$ very well.

Is there an intuitive reason why inside the disk these convolutions string together to give us a harmonic function?

Answer 1

Depends on your intuition...

Say $D[f]$ is the solution to the Dirichlet problem with boundary data $f$.

Fix $\theta$ and define the rotation $\rho$ by $$\rho(z)=e^{i\theta}z.$$Now if $f\in C(\Bbb T)$ then $D[f]\circ\rho$ is a harmonic function with boundary values $f\circ\rho$, so uniqueness says $$D[f]\circ\rho=D[f\circ\rho].$$

That is, $D$ commutes with rotations. Now it seems "intuitive" to me that a rotation-invariant operator should be a convolution; my intuition has been warped by various examples of such.

Ah, come to think of it the version we want here is trivial:

Lemma. If $T:C(\Bbb T)\to C(\Bbb T)$ is linear, bounded, and commutes with rotations then there exists a complex measure $\mu$ with $Tf=f*\mu$.

Proof: If $\lambda f=Tf(e^{i0})$ then $\lambda$ is a bounded linear functional on $C(\Bbb T)$, so there exists $\mu$ with $$Tf(1)=\int_{[0,2\pi]}f(e^{-it})\,d\mu(t).$$Now use the rotation-invariance of $T$.

Answer 2

An answer with a more physics-based intuition.

The solution(s) of a differential equation locally (point $\mathbf{r}$) depend on the data and geometry away from the point considered (point $\mathbf{r}'$). The decay of that influence is embedded in the shape of the corresponding kernel. The screened Poisson equation is a good example of this, one sees how increasing $\lambda$ makes the decay of the kernel steeper, and correspondingly, the magnitude of the solution decreases at a given distance from a point source.

This should give you an intuition on the usage of a convolution product, which makes the distance between $\mathbf{r}'$ and $\mathbf{r}$ appear: $(\mathcal{K}*f)(\mathbf{r}) = \int_{\Omega} \mathcal{K}(\mathbf{r}-\mathbf{r}')f(\mathbf{r}')\mathrm{d}\mathbf{r}'$

On the other hand, as David C. Ullrich shows, the solution operator should commute with rotations, so it is the distance only that should appear, not the direction. This is true provided that $\mathcal{K}(\mathbf{r})$ depends only on $|\mathbf{r}|$, which is the case with the Poisson kernel and other kernels of isotropic PDEs.