0
$\begingroup$

So my lecturer proves the mean value property which poses that under a few conditions on $u$ and

$\overline{B_r(x_0)} \subset \Omega \subset \mathbb{R}^n$:

$$u(x_0) = \dfrac{1}{\text{Vol}(\partial B_r(x_0))}\oint_{\partial B_r(x_0)} u(y)d\sigma(y)$$

In the first part of the proof he writes a coordinate transformation: $y \mapsto x_0 + z \mapsto x_0 + r\zeta$ (because he wants to write the integral in question in a unit disk) which corresponds to an integral transformation:

$$\oint_{\partial B_r(x_0)}u(y)d\sigma(y) = \oint_{\partial B_r(0)}u(x_0 + z)d\sigma(z) = r^{n-1}\oint_{\partial B_1(0)}u(x_0 + r\zeta)d\sigma(\zeta)$$

The first question I have is where does this $r^{n-1}$ come from exactly and how?

I am struggling to understand the concept of 'Volume' in terms of these boundary surfaces which may be related to this; for instance, I don't know why my lecturer writes $\text{Vol}(\partial B_r(x_0)) = \sigma_{n-1}r^{n-1}$ later on in the proof either - where does the $n-1$ power come from? I see that the boundary exists in $n-1$ dimensions, but don't get what this means exactly, if that makes sense?

This also brings me onto his latter, much shorter, proof of the 'Bulk Mean Value Property', namely that now having proved the mean value property, he can say: $$u(x_0) = \dfrac{1}{\text{Vol}(B_r(x_0))}\int_{B_r(x_0)}u(x)dx$$ Here, he starts by saying: $$\int_{B_r(x_0)}u(x)dx = \int_0^r ds\oint_{\partial B_s(x_0)}u(y)d\sigma(y)$$ How are they equal though? I don't understand that at all.

Thanks for any help anyone can provide! I'm really lost.

$\endgroup$
1
  • $\begingroup$ Regarding the factor $r^{n-1}$: we have $z = r\zeta,$ where $z$ ranges over the sphere $\partial B_r(0)$ and $\zeta$ ranges over the sphere $\partial B_1(0).$ Since $\partial B_r(0)$ is obtained from $\partial B_1(0)$ through multiplication with $r$, and since those spheres have dimension $n-1$, the surface area resp. surface element on $\partial B_r(0)$ is $r^{n-1}$ times the surface area resp. surface element on $\partial B_1(0)$. Hence the factor $r^{n-1}$ when changing the domain of integration. $\endgroup$ – jflipp May 20 '15 at 10:24
0
$\begingroup$

The factor $r^{n-1}$ comes from the Jacobian of the transformation: see Wikipedia.

As to your second question, you need to apply Fubini's theorem to interpret the n-tuple integral $\int_{B_r(x_0)}$ as an iterated integral.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.