How do I prove the "mean value property" for Helmholtz equation?

Question

Suppose $u: \mathbb{R}^2 \to \mathbb{R}$ is a $C^2$ function that satisfies the Helmholtz equation $-\Delta u = \lambda u$ for $\lambda \in \mathbb{R}$.

I am trying to prove something that looks like a modified mean value property, namely that

$$\frac{1}{2\pi r}\int_{\partial B(z_0,r)} u \,\mathrm{d}\Gamma = \begin{cases} u(z_0) \cdot J_0(\sqrt{\lambda} r) &\quad \text{if } \lambda > 0\\ u(z_0) &\quad\text{if } \lambda = 0\\ u(z_0) \cdot I_0(\sqrt{-\lambda} r) &\quad\text{if }\lambda < 0\end{cases}$$

I used polar coordinates and tried to approach the equation with separation of variables, saying that $u(r, \theta) = R(r) \Theta(\theta)$. I got $\Theta(\theta) = C_1 e^{i m \theta} + C_2 e^{- i m \theta}$, from the equation $$ \partial_{\theta}^2 \Theta = - C \Theta $$

For $R(r)$, I got the (modified / standard, depending on $\lambda$) Bessel equation. However, I don't know how to proceed. If $\Theta(\theta)$ was a harmonic function, I could simply integrate and put the Bessel part before the integral and then apply the mean value theorem. However, I am stuck, since it is not a harmonic function.

Ultimately, I have three questions;

1. What is the justification for using separation of variables? I have used it, but I don't know if I can, since the statements is phrased for every solution of the Helmholtz equation; what if there are solutions that don't have separated variables?
2. When solving the Bessel equation, we get multiple solutions $J_m$ and when solving the modified Bessel equations, we get multiple solutions $I_m$. Why are there only $J_0$ and $I_0$ in the final result?
3. As mentioned above, how do I proceed to actually show this property?

Answer 1

I'll provide a completely different approach. Let $\lambda >0 $. Define the elliptic operator $L:= \Delta + \lambda I$.

1. $ K(x,z_0) = - \dfrac{cos(\sqrt{\lambda}r)}{4\pi r}$ for $r=|x-z_0| $ is a fundamental solution of $Lu = 0$. (check the details)

2. Same goes for $S(x,z_0) = c \dfrac{sin(\sqrt{\lambda}r)}{r}$ for some $c$.

You can gain $K$ and $S$ by finding all the solutions of $Lu=0$ with spherical symmetry. Now by using $G(x,z_0)= K(x,z_0) + S(x,z_0)$ as the Green function for the problem $Lu=0$ in $B(z_0,r)$ we have:

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$$\Delta G(r)+ \lambda G(r) = \delta_{z_0}(r), \text{in} \; \;B(z_0,r) \\ G= 0, \; \text{in} \; \; \partial B(z_0,r)$$

where $\delta_{z_0}(r)$ is the dirac function centered at $z_0$.Then use the Green Identity in $B(z_0,r)$

$$ \int _{B(z_0,r)}u\Delta G - G\Delta u dx = \int_{\partial B(z_0,r)} u \dfrac{\partial G}{\partial n} - G \dfrac{\partial u}{\partial n}d\Gamma $$

Now due to the fact that $\Delta u = -\lambda u$ we have

$$\int _{B(z_0,r)}u\Delta G - G\Delta u dx = \int _{B(z_0,r)}u\Delta G + G\lambda u dx = \int _{B(z_0,r)}u(\Delta G + \lambda G )dx = \int _{B(z_0,r)}u \delta_{z_0} dx = u(z_0)$$

Now since $G= 0, \; \text{in} \; \; \partial B(z_0,r)$, we obtain

$$ u(z_0) = \int_{\partial B(z_0,r)}u \dfrac{\partial G}{\partial n} d \Gamma $$

Now we can just calculate the last integral, and by using $G= 0, \; \text{in} \; \; \partial B(z_0,r)$, we can also compute the constant $c$ and at last we obtain:

$$ u(z_0) = \dfrac {\sqrt{\lambda}r}{2 r sin(\sqrt{\lambda}r)} \dfrac{1}{2 \pi r} \int_{\partial B(z_0,r)}u(x)d\Gamma$$

where the last equation is undoubtedly what a "mean value property" of the Helmholtz equation should look like. If $\lambda <0$ we obtain an analogous result, particularly:

$$u(z_0)=\dfrac{\sqrt{-\lambda}r}{2rsinh(\sqrt{-\lambda}r)}\dfrac{1}{2 \pi r} \int_{\partial B(z_0,r)}u(x)d\Gamma$$.