It is well known that solutions to the Laplace equation in a region $\Omega\subseteq\mathbb{R}^n$, $\nabla^2u=0$ satisfy the mean value property, namely for all $x\in\Omega$, and for all $r>0$,

$$ u(x) = \frac{1}{\left|B(x,r)\right|}\int\limits_{B(x,r)}u\ dV = \frac{1}{\left|\partial B(x,r)\right|}\int\limits_{\partial B(x,r)}u\ dS $$

where $B(x,r)$ is a ball of radius $r$ around $x$ (we assume $B(x,r)\subseteq \Omega$).

I was wondering whether there is a generalized property for solutions to the Poisson equation, namely suppose $\nabla^2u=f$ in some region $\Omega\subseteq\mathbb{R}^n$.

Can we find some function $w$ in $\Omega$ such that for all $x\in\Omega$ and for all $r>0$, the following is true?

$$ u(x) = \frac{\int\limits_{B(x,r)}uw\ dV}{\int\limits_{B(x,r)}w\ dV} = \frac{\int\limits_{\partial B(x,r)}uw\ dS}{\int\limits_{\partial B(x,r)}w\ dS} $$

The only thing I know is that we are constrained to the fact that if $\nabla^2u \geq 0$, then

$$ u(x) \leq \frac{1}{\left|B(x,r)\right|}\int\limits_{B(x,r)}u\ dV = \frac{1}{\left|\partial B(x,r)\right|}\int\limits_{\partial B(x,r)}u\ dS $$ and the other way around if $\nabla^2u\leq 0$.


1 Answer 1


Note that in your wish equality would imply that if $u$ is constant on a sphere, it is also constant on the whole ball. This is certainly not true.

I think what you're looking for is the Feynman-Kac formula, which says $$ u(x)=\frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)}u \;dS+\int_{B(x,r)}f\;d\mu, $$ where the measure $\mu$ can be expressed by a Brownian motion $W$ and its exit time $\tau$ from the ball of radius $r$: $$ \int_{B(x,r)}f\;d\mu=\mathbb{E}\int_0^{\tau}f(x+W_t)\;dt. $$

  • $\begingroup$ Hi, can you please elaborate? I am not sure I understand how this is related to random processes. $\endgroup$
    – Joshhh
    Jan 10, 2022 at 7:52
  • $\begingroup$ Strictly speaking for answering your question random processes are not needed, you can just take my statement as the existence of a measure $\mu$ (depending on $x$ and $r$) such that this property holds. The relation to random processes is nontrivial, and very much depends on your background in stochastic analysis, so I would just suggest to look up Feynman-Kac and go from there. $\endgroup$
    – m7e
    Jan 11, 2022 at 21:27

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