Generalized singular differential equation

Question

I'm interested to understand what is the general solution for the differential equation $$x^{n}\frac{d^{m}y}{dx^{m}}\,=\,0$$ where $n,m\in\mathbb{N}$.

If $m=1$, then the solution can be found in Kanwal's generalized function book and is given by: $$y\,=\,c_{1}\,+\,c_{2}\Theta(x)\,+\,c_{3}\delta(x)\,+\,c_{4}\delta^{\prime}(x)\,+\,\ldots\,+\,c_{n+1}\delta^{(n-2)}(x)$$ where $\delta^{(n-2)}(x)$ denotes a Dirac delta function differentiated $n-2$ times. How can I generalize this result?

Answer 1

I can't even guarantee that these operations are correct since we are working with distributions but here's an attempt to play with the problem.

If we know how to solve

$$ x^n \frac{dy}{dx} = 0 $$

Then we can differentiate this to find

$$ nx^{n-1} \left( \frac{dy}{dx} \right)^2 + x^n \frac{d^2y}{dx^2} = 0 $$

Now we can multiply by $x$ to yield:

$$ nx^{n} \left( \frac{dy}{dx} \right)^2 + x^{n+1} \frac{d^2y}{dx^2} = 0 $$

So by substitution then it must be that $x^{n+1} \frac{d^2 y}{dx^2}$ = 0

So we have that the solutions to:

$$ x^n \frac{dy}{dx} = 0 $$

Also solve:

$$ x^{n+1} \frac{d^2 y}{dx^2}$$

By this token I would be inclined to try if the solutions to

$$ x^{n} \frac{d^m y}{dx^m} $$

Are solved by the solutions to

$$ x^{n-m+1} \frac{dy}{dx} = 0 $$

Which you say already has a solution.

Answer 2

This is a charming question. First, we can ask what distributions $u$ satisfy $x^m\cdot u=0$. This would mean that $u(x^m\cdot f)=0$ for all test functions $f$. By various means (depending on your context) this requires that $u$ is a linear combination of $\delta, \delta',\ldots,\delta^{(m-1)}$.

Then the more complicated questions asks what distribution $u$ has $n$th derivative equal to one of these derivatives of $\delta$. Locally away from $0$, this requires that $u$ is a polynomial, of degree less than $n$, so that its derivative will locally be $0$ away from $0$. (This locality property is something that should be verified...)

If the order of derivative is larger than the power of $x$, no derivatives of $\delta$ can enter...

Etc.

EDIT: in particular, the $(d/dx)^n$ will also annihilate polynomials of degrees $<n$. Functions which are polynomials in $x>0$ and also in $x<0$, but not quite matching up at $0$, can have derivatives that are various linear combinations of derivatives of $\delta$. The Heaviside step function is already an example...