helmholtz zero in R^3

Question

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 $k = 0$.

Answer 1

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).