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?

  • $\begingroup$ Behind a convolution, there is often a formulation in terms of averaging as explianed in this question and its answers. $\endgroup$
    – Jean Marie
    Oct 5, 2021 at 9:08
  • $\begingroup$ @JeanMarie I have seen the post before but I can't seem to relate it to U being harmonic inside the disk $\endgroup$ Oct 5, 2021 at 9:16

2 Answers 2


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 $T$.

  • $\begingroup$ That's a wonderful answer! Thank you. (On a side note, I have seen and often benefited from your answers on other questions on this site, I'm quite happy that you answered the very first question I ever asked here!) $\endgroup$ Oct 5, 2021 at 14:53
  • 2
    $\begingroup$ Don't miss the spot in Rudin Real and Complex Analysis where he answers the same question from the point of view of Fourier series... $\endgroup$ Oct 6, 2021 at 23:58

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.

  • $\begingroup$ There is something called nonlocal differential equation and the harmonic equation is not one of them. $\endgroup$ Oct 5, 2021 at 9:37
  • $\begingroup$ @ArcticChar be that as it may, the heuristic seems correct to me? $\endgroup$ Oct 5, 2021 at 10:38
  • $\begingroup$ @CalvinKhor I really don't know. But this does not explain why it is related by a convoluted operator (I did not downvote BTW). $\endgroup$ Oct 5, 2021 at 11:31
  • $\begingroup$ @ArcticChar: there is a set of DEs called nonlocal equations, but this doesn't mean that other problems are local. I'll reword when I have time. I believe it does give an intuition of the relation with convolution, as written - again, if this is not clear enough I'll reword. $\endgroup$
    – Joce
    Oct 5, 2021 at 11:34
  • $\begingroup$ @ArcticChar sure. Since the OP is asking for intuition, I would suppose it is permissible to avoid theorems. In the hopes of adding clarity - my sense is that Joce argues (heuristically) for the existence of the $\mu$ in David's answer (some sort of integral kernel) $\endgroup$ Oct 5, 2021 at 11:55

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