Let $L=\Delta + c$ in 3 dimensions, where $c$ is a positive constant.


I met this modified mean value property of a solution $u$ of $Lu=0$ as 
$$u(\xi)=\frac{\sqrt{c}\rho}{sin(\sqrt{c}\rho)}\frac{1}{4\pi \rho^2}\int\_{\partial B(\xi,\rho)} u(x)d\sigma(x)$$
where $sin(\sqrt{c}\rho) \ne 0$, and $\sigma$ denotes the surface measure.


I'm trying to prove it. The following is what I did.


First I figured out that $$K(x,\xi)=-\frac{cos(\sqrt{c}r)}{4\pi r}, r=|x-\xi|$$ is a fundamental solution of $L$ with pole $\xi$. Then I tried to apply the second Green identity (which is also true for $L$ for sure)
$$\int\_{\Omega}(LG) u-\int\_{\Omega}G (Lu)=\int\_{\partial \Omega} \frac{\partial G}{\partial \nu} u- G \frac{\partial u}{\partial \nu}$$ For fixed $\rho$, let $G=K+\frac{cos(\sqrt{c}\rho)}{4\pi \rho}$ and $\Omega=B\_{\rho}(\xi)$, so that $G$ vanishes on the boundary of $\Omega$, and $LG=\delta\_{\xi}+c\frac{cos(\sqrt{c}\rho)}{4\pi \rho}$, where $\delta\_{\xi}$ denotes the Dirac mass at $\xi$. Then the identity above becomes $$u(\xi)+c\frac{cos(\sqrt{c}\rho)}{4\pi \rho}\int\_{B\_{\rho}(\xi)}u(x)dx=\frac{\sqrt{c}\rho sin(\sqrt{c}\rho)+cos(\sqrt{c}\rho)}{4 \pi \rho^2} \int\_{\partial B(\xi,\rho)} u(x)d\sigma(x)$$


Then no matter how hard I tried, I just could not get the desired form. 


I also tried to take the derivative with respect to $\rho$ of the desired form, and I got a complicated form which doesn't seem to be zero. Then I took the second derivative, and things got worse...


Can anyone either give the right reference about the solution of this operator $L$, or help me figure out how to obtain the desired form? Thanks!



Modified mean value property
Ah, it is one of these cases when you better know your formula in all dimensions. 
 It is 
 $$
 \int\_{|x-y|=\rho} u(x) d\sigma(x) = (2\pi)^{n/2} \cdot \frac{J\_{n/2-1}(\sqrt{c}\rho)}{(\sqrt{c}\rho)^{n/2-1}} \cdot u(y)
 $$  

 One of the ways to prove it:
 $$
 \Delta u + c u = 0,
 $$
 $$
 \int\_{|x-y|\leq R} \Delta u(x) + c u(x) dx = 0.
 $$
 Apply Green's formula to the first integrand, 
 $$
 \int\_{|x-y|= R} \frac{\partial u}{\partial r}(x) d\sigma(x) + \int\_{|x-y|\leq R} c u(x) dx = 0,
 $$
 (here $\frac{\partial u}{\partial r}(x) = \nabla u(x)\cdot x/|x|$).
 $$
 R^{n-1}\int\_{|x|=1} \frac{\partial u}{\partial r}(y-Rx) d\sigma + c \int\_{|x-y|\leq R} u(x) dx = 0.
 $$  

 Differentiate w.r. to $R$ 
 $$
 R^{n-1}\int\_{|x|=1} \frac{\partial^2 u}{\partial r^2}(y-Rx) d\sigma +(n-1)R^{n-2} \int\_{|x|=1} \frac{\partial u}{\partial r}(y-Rx) d\sigma + c R^{n-1} \int\_{|x|=1} u(y-Rx) d\sigma = 0,
 $$
 $$
 \int\_{|x|=1} \frac{\partial^2 u}{\partial r^2}(y-Rx) d\sigma +\frac{n-1}{R}\int\_{|x|=1} \frac{\partial u}{\partial r}(y-Rx) d\sigma + c \int\_{|x|=1} u(y-Rx) d\sigma = 0.
 $$
 $$
 \frac{d^2}{dR^2}\int\_{|x|=1} u (y-Rx) d\sigma +\frac{n-1}{R} \frac{d}{dR}\int\_{|x|=1} u(y-Rx) d\sigma + c \int\_{|x|=1} u(y-Rx) d\sigma = 0.
 $$
 So $v(R) = \int\_{|x|=1} u (y-Rx) d\sigma $ is a solution of 
 $$
 v''(R) + \frac{n-1}{R} v'(R) + c v(R)=0.
 $$
 This equation is one of the forms of Bessel's equation 
 $$
 x^2 y''+ xy' + (x^2-\alpha^2) y = 0. 
 $$
 In particular, it is not difficult to show that 
 $$
 v(R) = C\_1 \frac{J\_{n/2-1}(R\sqrt{c})}{R^{n/2-1}},
 $$
 here $J\_\nu$ is the Bessel function of the first kind.
 To get the right constant just send $R\to 0$: on the one hand 
 $$
 \lim\_{R\to 0} v(R) = \omega\_{n-1} \cdot u(y),
 $$ 
 on the other 
 $$
 \lim\_{R\to 0} v(R) = C\_1\lim\_{R\to 0} \frac{J\_{n/2-1}(R\sqrt{c})}{R^{n/2-1}} = C\_1 \frac{(\sqrt{c}/2)^{n/2-1}}{2\pi^{n/2}} \omega\_{n-1}. 
 $$
 So $C\_1 = (2\pi)^{n/2} c^{1/2-n/4} u(y)$.


