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I took a basic multivariable calculus class as an undergrad where I saw Green's theorem, Stokes theorem, and Divergence theorem without proof. I know how to use and make sense of them in dimension $n \leq 3$.

I have recently started learning PDEs from Evans and take the 'Guass-Green' theorem for granted since I haven't learnt about differential forms/manifolds yet.

$$\int_{U}{\frac{\partial u}{\partial x_i}}dx=\int_{\partial U}u\nu^{i}dS\;\;\;\;(i=1,\ldots,n),$$

I am trying to understand the proof of MVP for a harmonic function $u:U \to \mathbb{R}$, where $U \subseteq \mathbb{R}^n$ and $u \in C^2(\overline{U})$ which says $$u(x)=\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{\partial B(x,r)} u(y) d\sigma $$ for all balls $\overline{B(x,r)} \subseteq U$ where $d\sigma$ is the surface element.

The proof begins by fixing $x$ and defining

$$ \phi(r):= \def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{\partial B(x,r)} u(y) d\sigma$$ and showing $\phi'(r)=0$ to get

$$\phi(r)=\phi(0)=u(x)=\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{\partial B(x,r)} u(y) d\sigma$$.

My question is what is the expression for $d \sigma$? How does it depend on $y$ in the above integrals?

I know when using hyperspeherical coordinates, we get $$dy=dy_1dy_2 \ldots dy_n=r^{n-1}\sin^{n-2}(\theta_1)\sin^{n-3}(\theta_2) \ldots \sin^{n-2}(\theta_{n-2})dr d\theta_1d\theta_2 \ldots d\theta_{n-1}=r^{n-1}dr d\sigma$$

How can I write $d \sigma$ explicitly in terms of $y$ and make sure I am using change of variables correctly?

approximation
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  • You never need to know an explicit (local coordinates) formula for the surface area element of a sphere. What you need is the generalization of the polar coordinates $dA = r,dr,d\theta$ to $dV = r^{n-1}dr,d\sigma$ in $\Bbb R^n$. – Ted Shifrin Oct 04 '22 at 17:45
  • @TedShifrin How do you use a change of variables then? One writes $$ \phi(r)= \def\avint{\mathop{,\rlap{-}!!\int}\nolimits} \avint_{\partial B(x,r)} u(y) d\sigma (y)$$ and uses $z=\frac{y-x}{r}$. How does this change of variables affect $d\sigma (y)$? If I knew $\sigma(y)$ I could calculate the jacobian $J$ of this transformation and write $d\sigma(y)=d\sigma(z)|J|$ – approximation Oct 04 '22 at 18:09
  • In my formula, $d\sigma$ was the surface area element of the unit sphere.Your change of variables maps the sphere of radius $r$ centered at $x$ to the unit sphere centered at $0$. So the integral becomes $\int_{\partial B(0,1)} u(x+rz) r^{n-1} d\sigma(z)$. – Ted Shifrin Oct 04 '22 at 18:15
  • @TedShifrin Yes, I understand how the domain changes. But how do I show that $d\sigma(y)=r^{n-1}d\sigma(z)$ – approximation Oct 04 '22 at 18:19
  • Because when you scale a sphere by a factor of $r$, each of its inherent $n-1$ dimensions is scaled by $r$. Without a rigorous discussion of $d\sigma$ in the first place, I'm not inclined to go any further. – Ted Shifrin Oct 04 '22 at 18:21
  • @TedShifrin Could you please point me to a reference that discusses $d\sigma$ rigorously or give some insights yourself? – approximation Oct 04 '22 at 19:53
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    The approach I would take is based on differential forms. $d\sigma$ on a hypersurface in $\Bbb R^n$ is an $(n-1)$-form given by contracting the volume $n$-form $dx_1\wedge\dots\wedge dx_n$ with the unit outward normal $\nu$ to the hypersurface. For the sphere(s) you can write this down explicitly since $\nu(x)=x/r$. So you get $d\sigma = \sum (-1)^{i-1} \frac{x_i}r dx_1\wedge\dots\widehat{dx_i}\dots\wedge dx_n$. If you map $x\rightsquigarrow rx$, it's easy to pull back this form and see what happens. Change of variables = pullback. If you're unfamiliar with forms, look at my YouTube lectures. – Ted Shifrin Oct 04 '22 at 21:12
  • For the relationship between $d\sigma_r$ and $d\sigma_1$ (thought of as measures, since you already got a response for dealing with them as differential forms), it boils down to the definition of integrating on submanifolds (i.e parametrize appropriately) and the change of variables theorem (in this case, in $\Bbb{R}^{n-1}$). See e.g About Lebesgue area measure $d\sigma(\omega)$ in spatial integration / spherical coordinates for more details. – peek-a-boo Oct 06 '22 at 14:21

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