How to change variables in the function $\phi(r)=\int_{\partial B(x,r)}u(y)\;dS(y)$ to take out $r$ of the domain of integration?

Question

Let $\alpha(n)$ be the volume of the unit ball in $\mathbb{R}^n$, $U\subset\mathbb{R}^n$ an open set, $u\in C^2(U)$ a harmonic function, $x\in U$ and, for $r$ small enough, $$\phi(r)=\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)}u(y)\;dS(y).$$

In order to calculate $\phi'(r)$ we need to change variables. According to the PDE Evans book (p. 26), if we take $y=x+rz$ then we get

$$\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)}u(y)\;dS(y)=\frac{1}{n\alpha(n)}\int_{\partial B(0,1)}u(x+rz)\;dS(z).$$

Therefore, the determinant $|J|$ of the Jacobian matrix $J$ is $|J|=r^{n-1}$.

My question is: since $u$ is a function of $n$ variables, shouldn't $J$ be a $n\text{-by-}n$ matrix? Thus, shouldn't we get $|J|=r^n$? I think it's a very elementary question, but I hope you help me.

Thanks.

Answer 1

Note that $n\alpha(n)=:\omega_{n-1}$ is the $(n-1)$-dimensional surface area of the unit sphere $S^{n-1}=\partial B(0,1)\subset{\mathbb R}^n$. Therefore $\omega_n r^{n-1}$ is nothing else but the surface area of the sphere $\partial B(x,r)$ with center $x$ and radius $r$. It follows that $\phi(r)$ denotes the average value of $u$ over this sphere (which is equal to $u(x)$, since $u$ was assumed harmonic).

In the second version of this integral we integrate over the unit sphere $S^{n-1}$, which is used as a parameter domain for the representation $$\psi:\quad S^{n-1}\to\partial B(x,r),\qquad z\mapsto x+r\>z\ .$$ The representation $\psi$ concerns $(n-1)$-dimensional manifolds, and it is obvious that it multiplies $(n-1)$-dimensional areas on $S^{n-1}$ with $r^{n-1}$. It follows that $$\int\nolimits_{\partial B(x,r)} u\ {\rm d}S= \int\nolimits_{S^{n-1}} u\bigl(\psi(z)\bigr)\ r^{n-1}\ {\rm d}S(z)\ .$$

Answer 2

The change of variables formula as you've used it applies only when you're doing an integral of full dimension - for surface integrals you've used the wrong determinant. If you were doing a volume integral, then you would indeed have a factor of $r^n$ when you did this change of variables. However, since you are doing a surface integral, this is not quite correct. You're scaling length by $r$, so the area element of the hypersphere will scale as $r^{n-1}$.

If you wanted to see this by computing a Jacobian determinant then you'd choose ($n-1$-dimensional!) coordinate systems on the spheres themselves, resulting in a transformation $\partial B(0,1) \to \partial B(x,r)$ whose Jacobian matrix at each point is just $r$ times the $n-1\times n-1$-dimensional identity matrix.