a friend and i are working on a research problem and we just hit a wall:


if $u$ is a solution of the 3-dimensional helmholtz operator $\Delta + k^2$ in the ball $B\_r(x)$ of radius $r > 0$, then does $u$ satisfy anything like the mean value for harmonic functions? basically, we are wondering what an analogous "mean value property" would be for a nonzero $k$ term. we can't find anything obvious like a harmonic function, where $k = 0$.



helmholtz zero in R^3
If $u$ is a solution to the equation $\triangle u +k^2 u=0$ in a 3D domain $\Omega$, then 
for any $x\in\Omega$ and any $r>0$ such that $\{y\in\mathbb R^3:\ |x-y|\leq r \}\subset\Omega$, we have
$$u(x)=\frac {p(r)}{4\pi r^2}\int\_{|x-y|=r} u(y)dS\_y,\qquad\qquad\qquad(1)$$
where
$$p(r)=\frac{rk}{\sin rk}.$$


Formula (1) is an analogue of the mean value theorem for harmonic functions (in the case of spherical means).


**Edit added:** relation (1) is valid for all $r\_1\leq r$. If we multiply it by $4\pi r^2/p(r)$ and integrate between $0$ and $r$ we will obtain that
$$u(x)=\frac{k^3}{4\pi(\sin rk-rk\cos rk)}\int\_{|x-y|\leq r} u(y)dy.$$
The latter formula generalizes the property that the value of a harmonic function at $x\in\Omega$ is equal to function's average value over a ball with the center at $x$.


A short derivation of formula (1) can be found in chapter IV of *Methods of Mathematical Physics* (Vol. 2)
by Courant and Hilbert (or see Harald's comment below).


