I have come across some difficulties in this question.
Given the Laplacian equation in $\mathbb{R}^n$: $\Delta u = 0$,set $$\bar{u}(r) = \int_{|x| = r}u(x)d\sigma$$ with $x = \left(r, \sigma\right)$ is the spherical coordinate of $x$. How can I deduce the equation for $\bar{u}$ from the Laplace equation?
I think I can't compute $u$ according to $\bar{u}$. Is there any way to work around this?
Thank you very much in advance for your help.
It's not entirely clear what you're looking for, but if you are allowed to use the mean value property of harmonic functions, that is, $$u(0)= \frac1{n \omega_n r^{n-1}}\int_{\partial B_r} u(x) \, d \mathcal H^{n-1}_x$$ where $\omega_n$ is the volume of the unit ball in $\mathbb R^n$ then $$\bar u (r) = n \omega_n u(0) r^{n-1} $$ for all $r>0$.
If you a trying to prove the mean-value formula then you've written the formula for $\bar u$ wrong: it should be $$\bar u (r) =\frac1{n \omega_n r^{n-1}}\int_{\partial B_r} u(x) \, d \mathcal H^{n-1}_x. $$ Then you want to change coordinates to obtain $$\bar u (r) = \frac1{n \omega_n r^{n-1}}\int_{\partial B_r} u(x) \, d \mathcal H^{n-1}_x = \frac1{n \omega_n } \int_{\partial B_1} u(ry) \, d \mathcal H^{n-1}_y $$ then differentiate which gives \begin{align*} \bar u '(r) &= \frac1{n \omega_n } \int_{\partial B_1} \nabla u(ry)\cdot y \, d \mathcal H^{n-1}_y\\ &=\frac1{n \omega_n r^{n-1}} \int_{\partial B_r} \nabla u(x)\cdot \frac x r \, d \mathcal H^{n-1}_x. \end{align*} Since the outward point unit normal $\nu$ is precisely $x/r$, it follows that \begin{align*} \bar u '(r) &=\frac1{n \omega_n r^{n-1}} \int_{\partial B_r} \nabla u(x)\cdot \nu(x) \, d \mathcal H^{n-1}_x \\ &= \frac1{n \omega_n r^{n-1}} \int_{ B_r} \Delta u(x)\, d x\\ &=0 \end{align*} using Green's formula. Hence, $\bar u$ is constant, so $$\bar u(r) = \lim_{r\to 0^+}\bar u(r) = u(0). $$