Induction proof for identity used to solve Euler-Poisson-Darboux PDE in Evans

Question

In Partial Differential Equations by Evans there's this one identity that is used to solve the Euler-Poisson-Darboux equation for odd $n \geq 3$:

$$ \left( \frac{1}{r} \frac{d}{dr} \right)^{k-1} \left( r^{2k-1} \phi(r)\right)=\sum_{j=0}^{k-1} \beta^k_j r^{j+1} \frac{d^j \phi(r)}{dr^j}$$

where $\phi: \mathbb{R} \rightarrow \mathbb{R}$ is $C^{k+1}$, the $\beta_j^k$ are independent of $\phi$ and $$\beta_0^k=1 \cdot3 \cdot 5 \cdots (2k-1) \,.$$

The proof (by induction) is omitted and my classmates and I are stumped. Has anyone else done this proof before, or seen it done? I've looked everywhere but the proof is always omitted in lecture notes and textbooks.

Answer 1

Every time you apply $$\left(\frac{1}{r}\frac{d}{dr}\right)$$ to a function of the form $$r^{a}\phi(r)$$ you either apply it to $\phi(r)$ and reduce the degree of $r^{a}$ by 1, or you apply it to $r^{a}$ and reduce the degree of $r^{a}$ by $2$ while keeping $\phi(r)$. Since we are differentiating $k-1$ times while the highest power of $r$ is $2k-1$, you are going to get a constant term. And so is every other term if you count the degree carefully. The coefficients cannot be depended on $\phi(r)$.

Now let $\phi(r)=1$ and calculating the constant term in $$\left(\frac{1}{r}\frac{d}{dr}\right)^{k-1}r^{2k-1}$$ by a similar argument as above gives you the value of $\beta^{k}_{0}$.