A variation on the Cauchy-Euler equation

Question

In my research, I have managed to formulate the following PDE: for a function $V(p, w)$, I have $$ A\, p^2\, \partial^2_p V + (B\,w + C\,p)\ \partial_p V - E\, V = 0\ , $$ for $A, B, C, D, E$ positive constants.

This is almost, save for the $B\,w$ term, a Cauchy-Euler equation. And it is also almost a Kummer equation, save for the extra power of $p$ on the first term. Indeed, trying to solve this with Mathematica produces a Frankenstein's monster combination of these solutions which seems like overkill.

If $B=0$, the Cauchy equation admits a solution of the type $V(p) = p^x$; this is the one that is relevant to me. Bringing $w$ into the picture, I had hoped for a sort of "correction" to this solution. It would be elegant for my problem if $V(p, w) = V(p/w)$, and the equation above (and Mathematica) suggest that something like this may be afoot.

Does anyone see an elegant solution? Many thanks!

Answer 1

$$A\, p^2\, \frac{\partial^2 V}{\partial p^2} + (B\,w + C\,p)\frac{ \partial V}{\partial p} - E\, V = 0 $$ There is no $\frac{ \partial }{\partial w}$ in the equation. Thus this is not a PDE, this is an ODE (In fact a "degenerated" PDE ). Then $w$ is a parameter. The derivatives are wrt the variable $p$ only. So, we can write : $$A\, p^2\, \frac{d^2V}{dp^2} + (\beta + C\,p)\frac{dV}{dp} - E\, V = 0 \quad\text{with }\beta=B\,w$$ $A,\beta,C,E$ can be functions of any parameters that you want, but NOT function of $p$.

Solving the ODE is tiresome. The arduous method is :

First, change of variable $x=\frac{\beta}{A\:p}$ which leads to a second order linear ODE where the unknown is $V(x)$ instead of $V(t)$.

Second, change of function on the form $V(x)=x^\gamma y(x)$ and determine $\gamma$ so that the PDE is reduced to the standard form of confluent hypergeometric Kummer equation of first kind : $$xy''(x)+(c-x)y'(x)-ay(x)= 0$$ in which the $a$ and $c$ are obtained as functions of $A,\beta,C,E$.

From the solutions $y(x)$ obtained (where two new arbitrary constant appear) come back to $V(x)$ then to $V(p)$. Finally, replace $\beta$ by $(B\,w)$. for the solution $V(p,w)$.