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

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.

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 $\beta^{k}_{0}$.