Convert a radially symmetric PDE into ODE

Suppose we know $u(x)$ is a radially symmetric function in $B_1 =\{ x\in \mathbb{R}^N: |x|< 1 \}$ and $u(x)=0$ for $|x|=1$.

Suppose $-\operatorname{div}(v(|x|)\nabla u(x))=0$ in $B_1 .$

My professor wrote that, by using polar coordinates, we can convert the PDE above into an ODE as follows: $$(-vr^{N-1}u^\prime)^\prime=0 \mbox{ on } [0,1).$$

I cannot see why this is true and how to implement the polar coordinates. I didn't learn the multivariable calculus very well... Any help will be appreciated. :)

My attempt: I have done some calculations though I feel like the result may not be useful.

$-\operatorname{div}(v(|x|)\nabla u(x))=-(\nabla v \cdot \nabla u + v\Delta u)$

Solution 1:

One way to prove this by directly computing it. Let $u(x) = w(\vert x \vert)$. Then $$D_i u = w'(\vert x \vert ) \frac{x_i}{ \vert x \vert}.$$ Moreover, if $ f(r) = v(r) w'(r)/r$ then $$D_i \bigg ( v (\vert x \vert ) w'(\vert x \vert ) \frac{x_i}{ \vert x \vert}\bigg ) = x_i D_if(\vert x \vert ) + f(\vert x \vert) = f'(\vert x \vert ) \frac{x_i^2}{\vert x \vert}+f(\vert x \vert)$$

Hence, \begin{align*} \text{div}(v(\vert x \vert ) \nabla u(x) ) &= \text{div} \bigg(v(\vert x \vert )w'(\vert x \vert ) \frac{x_i}{ \vert x \vert} \bigg ) \\ &= \sum_{i=1}^n D_i \bigg ( v (\vert x \vert ) w'(\vert x \vert ) \frac{x_i}{ \vert x \vert}\bigg ) \\ &= \vert x \vert f'(\vert x \vert ) + n f(\vert x \vert ) \end{align*} Then using an integrating factor $$rf'(r)+nf(r)=r \bigg( f'(r)+ \frac{n}{r}f(r) \bigg ) = r^{1-n} \big (r^n f(r)\big)' .$$ Thus, writing $r=\vert x \vert$ gives $$ \text{div}(v(\vert x \vert ) \nabla u(x) ) = r^{1-n} (r^n f(r))' = r^{1-n} (vr^{n-1}w')'. $$

Note that your professor is using a common (slight) abuse of notation namely they write $u(x)=u(\vert x\vert)$. Replacing $w$ with $u$ gives the desired result.