Euler-Poisson-Darboux equation - PDE Evans

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 ?

• Looks potentially like an application of the Leibniz rule Commented Apr 17, 2018 at 11:33

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