We know by mean value property that harmonic functions satisfies the equalities \begin{equation} u(x) = \dfrac{1}{|B_r|}\int_{B_r}f dx = \dfrac{1}{|\partial B_r|}\int_{ \partial B_r}udS. \end{equation} I'm wondering if we do not have harmonic functions, instead we have $\Delta u =f$. Is there some equality like above? For example adding some term depending on $f$ or its intergral on the right hand side?

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    $\begingroup$ There is indeed such an equality. This is usually called $Green's\;representation\;formula$. $\endgroup$ – Etienne Jun 13 '13 at 18:56

Consider the problem in $\mathbb{R}^n$. If $$ u(x) = \frac{1}{|\partial B(x,r)|}\int_{ \partial B(x,r)}u(y) \,dS(y) ,\tag{1} $$ then $u$ has to be harmonic. Please refer to the proof in Evans' book about mean value property. For if define $$ \phi(r) = \frac{1}{|\partial B(x,r)|}\int_{ \partial B(x,r)}u(y)\,dS(y) = \frac{1}{|\partial B(0,1)|}\int_{ \partial B(0,1)}u(x+rz)\,dS(z), $$ and $$ \phi'(r) = \frac{1}{|\partial B(0,1)|}\int_{ \partial B(0,1)}\nabla u(x+rz)z\,dS(z) \\ = \frac{1}{|\partial B(x,r)|}\int_{ \partial B(x,r)} \nabla u(y) \cdot n\,dS(y)\\ =\frac{1}{|\partial B(x,r)|} \int_{ B(x,r)} \Delta u(y) \,dy = \frac{1}{|\partial B(x,r)|} \int_{ B(x,r)} f(y) \,dy.$$

If $f\neq 0$, $\phi'(r)\neq 0$, and $\phi(r)$'s value depends on $r$. If (1) holds, then $u(x)$'s value is not unique, contradiction. If $u$ is not harmonic we have to find something else.

If $f$ is radially symmetric with respect to the point $p$, then you have $u$ is radially symmetric with respect to point $p$. We can set $$u(x) = v(|x-p|) = v(r), \quad \text{ and }\quad f(x) = f(|x-p|) = h(r),$$ then $$ \Delta u = v''(r) + \frac{n-1}{r}v'(r) = h(r), $$ Assuming $u(x)$ is smooth everywhere, then $$ v(r) = v(0) + \int^r_0\tau^{1-n}\int^{\tau}_0 \xi^{n-1}h(\xi) d\xi, $$ and translates to $$ u(x) = u(p) + \int^{|x-p|}_0\frac{1}{|\partial B(p,\tau)|} \left(\int^{\tau}_0 \Big(\int_{\partial B(p,r)} \Delta u(y)dy\Big)dr \right) d\tau, $$ which is a formula similar to mean value formula for we are integrating on $(n-1)$-spheres layer by layer around a point $p$, also with an unspecified boundary value.

Lastly, you can represent the solution using Green's formula if boundary value is given in a domain, please refer to Evans' PDE book Chapter 2 Theorem 12.


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