# Solving wave equation in $\mathbb{R}^3$ .

The wave equation in $\mathbb{R}^3$ , ie $u_{tt}-\Delta u=0$ for $x\in \mathbb{R}^3, t>0$

$u=g, u_t=h$ for $x\in \mathbb{R}^3, t=0$

Define an average: $U(x,r,t)= \frac {1}{|\partial B|} \int_{\partial B(x,r) }u(y,t)dS_y$ and similarly $G(x,r,t) =\frac {1}{|\partial B|} \int_{\partial B(x,r) }g(y)dS_y$

and

$H(x,r,t) =\frac {1}{|\partial B|} \int_{\partial B(x,r) }h(y)dS_y$

We fix $x \in R^n , n\ge2$ and suppose $u \in C^m(R^m\times \mathbb{R}_+$ for $m\ge2$

Claim : $U$ solves the following initial value problem . $U_{tt}-U_{rr}-\frac {(n-1)}{r} U_r=0, r>0, t>0$

$U=G, U_t=H , r>0, t=0$

Proof : $U_r =\frac {\partial}{\partial r}\frac {1}{|\partial B|} \int_{\partial B(x,r) }u(y,t)dS_y$. First derivative was easy for me to find but now My aim is to find $U_{rr}$ . There is another way of proving the claim without computation of second derivative . But i am looking forward to know how to differentiate it twice and many more times .

Any help will be appreciated.

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Since you didn't like my answer to your previous question, I'll give you the details. Let $B=B(0,1)$, and $y=x+r\,\xi$ with $\xi=1$. Then $$U(x,r,t)=\frac{1}{|\partial B(x,r)|} \int_{\partial B(x,r)}u(y,t)\,dS_y=\frac{1}{|\partial B|} \int_{\partial B}u(x+r\,\xi,t)\,dS_\xi.$$ Derivating with respect to $r$ and using the divergence theorem \begin{align*} \frac{\partial U}{\partial r}&=\frac{1}{|\partial B|} \int_{\partial B}\xi\cdot\nabla_xu(x+r\,\xi,t)\,dS_\xi\\ &=\frac{r}{|B|} \int_{B}\Delta_xu(x+r\,\xi,t)\,d\xi\\ &=\frac{r^{1-n}}{|B|}\Delta_x \int_{B(x,r)}u(y,t)\,dy\\ &=r^{1-n}\Delta_x \int_0^r\rho^{n-1}U(x,\rho,t)\,d\rho. \end{align*} From here it is easy to get $\partial^2 U/\partial r^2$, or better yet, $$\frac{\partial}{\partial r}\Bigl(r^{n-1}\frac{\partial U}{\partial r}\Bigr)=r^{n-1}\Delta_xU(x,r,t).$$
Here basically we differentiate in a unit radius Ball in order to make sure that $u$ doesn't blowup right ? – Theorem Jul 10 '12 at 19:24