1

I am stuck at proving the mean value property of harmonic functions for an open domain $X \subset \mathbb{R^n}$: If $u \in C^2(X)$ is harmonic in $X$ and $\overline{B_r(y)} \subset X$, then:

$$u(y)=\frac{1}{\overline{B_1(0)}r^{n-1}}\int_{\partial B_r(y)} u \, ds$$

where $ds$ is the surface element.

Here's the proof I am going through:

$$0= \int_{B_r(y)} \Delta u\, dx=\int_{B_r(y)}\nabla \cdot \nabla u \,dx=\int_{\partial B_r(y)} \nabla u \cdot n \,ds=\int_{\partial B_r(y)} \nabla u\cdot \frac{x-y}{r} \,ds $$

In my notes, this is followed by the substitution $w=\frac{x-y}{r}$, apparently giving us: $$ =r^{n-1}\int_{|w|=1} \nabla u(y+rw)\cdot w dw=\cdots $$

Can anyone please tell me how did we get that expression? How did our surface integral changed to a standard integral? What exactly is the Jacobian here?

Thank you!

miyagi_do
  • 1,663
  • 2
    It seems that it didn't change to a standard integral. It's still a surface integral, only that it's over the unit sphere. – Michał Miśkiewicz Aug 24 '20 at 19:23
  • 1
    @MichałMiśkiewicz Thank you for the insight. Can you please tell me how did the surface element $ds$ changed to $dw$, as $w \in \mathbb{R^n}$? How come the jacobian is $r^{n-1}$? – miyagi_do Aug 25 '20 at 07:27
  • 2
    I think there is some abuse of notation. Note that in your first line of integrals you integrate over $x$ in the sense of $ds = ds(x)$, whereas after the substitution $dw$ should probably be interpreted as $ds(w)$. Which is still a surface integral as pointed out by Michał Miśkiewicz. – SC2020 Aug 25 '20 at 08:18
  • 1
    @SC2020 How do we justify the factor $r^{n-1}$ then? – miyagi_do Aug 26 '20 at 05:55
  • 2
    This is still the Jacobian. See for example here. – SC2020 Aug 26 '20 at 08:14

0 Answers0