Distributional solution of this equation

Question

I am having trouble finding the distributional solutions $u$ of: $x^2u = \delta$. Could somebody help? Thanks in advance

Answer 1

Intuitively you a solution to $x^2 u = \delta_0$ is a distribution of the form : "some distribution that is killed by the multiplication by $x^2$" + "a particular solution to $x^2u = \delta_0"$.

That is to say that a general solution is "sln to $x^2u = 0$" + "sln to $x^2u = \delta_0$".

Solution to the homogeneous system $x^2u = 0$

What does a solution to $x^2 u = 0$ look like ? Well the support of $u$ must be $\{0\}$ since it was killed by the function $x^2$ which is non-zero everywhere except in 0. It is known that only distributions with a point support are (finite) linear combinations of the delta-function and its derivatives: $f(x)=\sum_{k=0}^m c_k \delta^{(k)}(x)$.

It is easy to check that $$\langle x^2 c_1\delta_0, \varphi \rangle = c_1\langle \delta_0, x^2 \varphi\rangle = 0$$ and that $$\langle x^2 c_2\delta_0', \varphi \rangle = -c_2\langle \delta_0, 2x \varphi + x^2 \varphi\rangle = 0.$$

But $$\langle x^2 \delta_0'', \varphi \rangle = \langle \delta_0, 2\varphi + 4x \varphi' + x^2 \varphi''\rangle = 2\varphi(0)$$ which is not zero in general.

Therefore a solution to the homogenous system is of the form $c_1\delta_0 + c_2\delta_0'$. That is to say $$\langle c_1\delta_0 + c_2\delta_0', \varphi\rangle$$ for every constant $c_1, c_2$ and every test function $\varphi$.

Particular solution to $x^2u = \delta_0$

From the computation of $\langle x^2\delta_0'', \varphi\rangle$ we can easily see that $$ \left\langle x^2\left(\frac{1}{2}\delta_0''\right), \varphi\right\rangle = \varphi(0)$$ for every test function $\varphi$. Therefore $\frac{1}{2}\delta_0''$ is a solution to $x^2 u = \delta_0$.

General solution

$u = c_1\delta_0 + c_2 \delta_0' + \frac{1}{2}\delta_0''$