suppose $n \geq 2,\;m \geq 2$ and $u \in C^{m}(\mathbb{R}^n \times[0,+\infty))$ solves the following IVP :
\begin{cases}
u_{tt} – \Delta u = 0 \; & \text{in} \; \mathbb{R}^n \times(0,+\infty) \\
u = g, \; u_t = h \; & \text{on} \; \mathbb{R}^n \times\{t = 0\}
\end{cases}
let $x \in \mathbb{R}^n, \; t > 0, \; r > 0 $. Define
\begin{align}
U(x;r,t) & = \frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)}u(y,t)dS(y) \\
G(x;r) & = \frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)}g(y)dS(y) \\
H(x;r) & = \frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)}h(y)dS(y) \\
\end{align}
$$\alpha(n) = \frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+ 1)} = \text{volume of the unit ball in } \mathbb{R}^n $$
(Euler-Poisson-Darboux equation) fix $x$ in $\mathbb{R}^n$ and let $u$ satisfies the IVP above. then :
\begin{cases}
U_{tt} – U_{rr} – \frac{n-1}{r}U_r= 0 \; & \text{in} \; \mathbb{R}_{+}\times(0,+\infty) \\
U = G, \; U_t = H \; & \text{on} \; \mathbb{R}_{+}\times\{t = 0\}
\end{cases}
the author states that :
\begin{align}
U_r(x;,r,t) & = \frac{1}{n\alpha(n)r^{n-1}}\int_{ B(x,r)}\Delta u(y,t)dy \\
U_{rr}(x;,r,t) & = \frac{1}{n\alpha(n)r^{n-1}}\int_{ \partial B(x,r)}\Delta u(y,t)dS(y) +(\frac1n – 1)(\frac{1}{\alpha(n)r^n})\int_{ B(x,r)}\Delta u(y,t)dy \\
\end{align}
I could derive $U_r$ on my own but I'm having trouble trying to compute $U_{rr}$
if I differentiate $U_r$ with respect to $r$ I get :
\begin{align}
U_{rr}(x;,r,t) & = \frac{1}{n\alpha(n)r^n}\int_{ B(x,r)}\Delta u(y,t)dy + \frac{r}{n} [\frac{1}{\alpha(n)r^n}\int_{ B(x,r)}\Delta u(y,t)dy]'_{r} \\
\end{align}
this means we must have :
\begin{align}
\frac{r}{n}[\frac{1}{\alpha(n)r^n}\int_{ B(x,r)}\Delta u(y,t)dy]'_{r} & = \frac{1}{n\alpha(n)r^{n-1}}\int_{ \partial B(x,r)}\Delta u(y,t)dS(y) -\frac{1}{\alpha(n)r^n}\int_{ B(x,r)}\Delta u(y,t)dy \\
\end{align}
how do you prove this ?
Best Answer
I had the same problem studying from Evans!
The trick is to differentiate $$(U_r r^{n-1})=\frac{1}{n\alpha(n)}\int_{ B(x,r)}\Delta u(y,t)dy.$$ The left hand side gives $$U_{rr}r^{n-1}+(n-1)U_r r^{n-2}=U_{rr}r^{n-1}+(n-1)\frac{1}{rn\alpha(n)}\int_{ B(x,r)}\Delta u(y,t)dy.$$ The right hand side gives $$\frac{1}{n\alpha(n)}\int_{ \partial B(x,r)}\Delta u(y,t)dS(y).$$ And you get the result