Mean value property for solution of Helmholtz equation I am trying to prove that the solution of $3$d Helmholtz equation $\nabla_{x}^2u+cu=0$  on $ B(s,l)$ satisfies the the following property
$$u(s)=\frac{\sqrt{c}l}{sin{\sqrt{c}l}}\frac{1}{4πl^2}\int_{\partial_{B}}u(x)dS$$
My idea is to use the Fundamental solution  $G(r)=\frac{e^{ir}}{4πr}$ ,  $r=|x-s| $ and Green's second identity for the functions $u,G$ on the ball,relying on $\nabla_{x}^2G+cG=\delta(x-s)$ and the equation to get to 
$$\int{u\nabla^2G-G\nabla^2u}=u(s)=\int{u\frac{\partial{G}}{\partial{n}}}-G\frac{\partial{u}}{\partial{n}} dS$$ and bound the last term on the surface integral but i don't quite get $\frac{\sqrt{c}l}{sin{\sqrt{c}l}}$ factor.
I might be completely wrong here but any help would be appreciated.
 A: I think you've got the right idea, but need a different Green's function. Let $\nabla^2 u + m^2u=0$, and $\nabla^2 G+m^2G=\delta(x-s)$, and consider $B(s,R)$. We have, using these equations and Green's second identity,
$$ 0= \int_{B(s,R) \setminus \{s\}} (u\nabla^2 G-G\nabla^2 u) \, dV = \int_{\partial B(s,R)} \left(u\frac{\partial G}{\partial r}-G \frac{\partial u}{\partial r} \right) dS - u(s), $$
provided that $G$ satisfies
$$ \lim_{\varepsilon \to 0} \int_{\partial B(s,\varepsilon)} \frac{\partial G}{\partial r} \, dS = 1, \qquad \lim_{\varepsilon \to 0} \int_{\partial B(s,\varepsilon)} G \, dS = 0. $$
We ask that we have spherical symmetry $G(r,\theta,\varphi)=G(r)$ to simplify matters, so in 3D the previous conditions translate to
$$ 4\pi r^2 G'(r) \to 1, \qquad 4\pi r^2 G(r) \to 0 $$
as $r \to 0$. Now, we want to choose $G$ so that $G(R)=0$, so that the second term in the integral vanishes. Putting all the conditions on $G$ together we have
$$ \frac{1}{r^2}(r^2G')' + m^2 G=0, \quad G(r) \sim \frac{1}{4\pi r} \text{ as } r \to 0, \quad G(R)=0. $$
Solving these equations in the usual way, we obtain a different Green's function from normal,
$$ G(r) = \frac{m}{\sin{mR}} \frac{\sin{m(r-R)}}{4\pi r} $$
Now we get $G'(R) = \frac{mR}{\sin{mR}} \frac{1}{4\pi R^2}$, and the result follows since the surface integral has constant $R$, so we can factor out the $G(R)$.
Essentially, the key lies in the freedom we have to add $(\sin{mr})/r$ terms to the Green's function without affecting the local boundary condition at $r=0$; this is the Helmholtz version of being able to add harmonic functions (and in particular, constants) to the Laplace Green's function.
