Surface integral computation

Okay, so this is a step in the proof to Liouville's theorem, and it comes from Fritz John's *Partial Differential Equations* 4th ed (p109, sec. 4.3). The problem is, it requires computation of a surface integral (I think), and I don't know how to do that.

Here's the setup:

Let $\displaystyle u$ be harmonic in $\displaystyle \mathbb{R}^n$ with $\displaystyle \xi$ in the ball of radius $\displaystyle a>0$ about the origin. Let $\displaystyle \omega_n=2\pi^{n/2}/\Gamma(n/2)$ be the surface area of the unit sphere $\displaystyle S^{n-1}\subset\mathbb{R}^n$ (i.e. $\displaystyle \omega_2=2\pi$, $\displaystyle \omega_3=4\pi$, and so on.)

We are also given the following identity:

$\displaystyle u_{\xi_i}(0)=\frac{n}{\omega_n a^{n+1}}\int_{|x|=a}x_i u(x)\;dS_x$

Frankly, I'm not sure what $\displaystyle dS_x$ is supposed to denote. I assume it's the "surface integral" with respect to x, but then I'm not sure what the definition of a surface integral is, much less how to evaluate them routinely.

Anyway, we have to use the above information to prove the following identity:

$\displaystyle |u_{\xi_i}(0)|\leq\frac{n}{a}\max_{|x|=a}|u(x)|$

Any ideas on how to do this? Any help would be much appreciated!

Re: Surface integral computation

Quote:

I'm not sure what the definition of a surface integral is, much less how to evaluate them routinely.

Maybe you should revisit your notes, it is quite helpful to be able to deal with them.

Now for the estimate on the integral. Let $\displaystyle \omega_n a^{n-1}=\int_{|x|=a}dS_x$ be the area of the $\displaystyle a-$radius n-sphere.

Use the Cauchy-Schwartz inequality and basic properties of the integral to obtain

$\displaystyle |u_{\xi_i}(0)|\leq\frac{n}{\omega_n a^{n+1}}\left(\int_{|x|=a}\sum|x_i|^2dS_x\right)^{ 1/2}\left(\int_{|x|=a}|u|^2dS_x\right)^{1/2}$

or

$\displaystyle |u_{\xi_i}(0)|\leq\frac{n}{\omega_n a^{n+1}}\left(a\left[\omega_n a^{n-1}\right]^{1/2}\right)\left((\sup|u|)\left[\omega_n a^{n-1}\right]^{1/2}\right)$

from where the result follows.