Consider the equation which can be seen as a backwards heat equation with diffusivity depending on time and space:
$$ \frac{\partial^2}{\partial t^2}\Phi(x,t)=-\frac{x}{t}\frac{\partial}{\partial x}\Phi(x,t) $$
$x \in (0,1),$ $t\in(0,\infty)$ and $\Phi(x,0)=\rho(x)$
where we define a new distribution as $\rho(x):= \lim_{t\to 0} \exp \frac{t}{\log x}.$
Note that:
$$ \int_0^1 \rho(x)~dx=1.$$
I discovered this equation when I was trying to model the diffusive properties of a string stretching over time $t.$ $\Phi(x,t)=\exp \frac{t}{\log x}$ is a solution to the equation. $\Phi(x,t)$ is also the solution to the Killing field $\vec X=\langle x\log x,-y\log y \rangle$ for $g=\frac{dxdy}{xy}$ and represents infinitesimal isometries of the Lorentzian manifold $(M,g)$ (which is isometric to Minkowski space). But change the metric to $ds^2=dx^2+dy^2$ and you get this strange "reversed" heat equation, for which $\Phi$ is also a solution.
Using the ansatz $\Phi(x,t)=X(x)Y(t)$ the PDE can be simplified to ODE's and solved.
Is my solution the fundamental solution? Can you derive my solution from the ansatz I gave?
I believe that as written, the problem is underconstrained. Perform a change of variables $(X,t)=(-\log x,t)$ and it can be shown that the function $\Psi(X,t)=\Phi(e^{-X},t)$ satisfies a diffusion equation in the 1st quadrant:
$$t\frac{\partial^2\Psi}{\partial t^2}=\frac{\partial\Psi}{\partial X}~,~~X,t\geq 0$$
Since the problem is defined in $\Omega=[0,\infty)\times[0,\infty)$, then suitable boundary conditions have to be defined over the entire boundary and not just the line $(X,0)$. We will see in the following that an arbitrary function can be added to the solution when the BC on $(0,t)$ has not been specified.
We will proceed assuming the boundary conditions wlog $\Psi(X,0)=\rho(X), \Psi(0,t)=q(t)$. Performing a Laplace transform in the $t$ coordinate yields the first order PDE
$$\frac{\partial \hat\Psi(X,s)}{\partial X}+\frac{\partial(s^2 \hat\Psi(X,s))}{\partial s}=\rho(X)$$
where $\hat{\Psi}(x,s)=\int_0^\infty \Psi(x,t)e^{-st}$. This has the general solution
$$s^2 \hat\Psi(x,s)=F(x+1/s)+\int_0^x\frac{\rho(w)}{(x+1/s-w)^2}dw $$
for some arbitrary function $F$. This function cannot be completely fixed by the $\rho$ boundary condition, so we will use the $q$ one instead to find that
$$F(1/s)=s^2\hat\Psi(0,s)=s^2\hat q(s)\iff F(t)=\frac{\hat q(1/t)}{t^2}$$
which specifies the unique solution to be
$$\hat\Phi(x,s)=\frac{\hat q\left(\frac{s}{1-s\log(x)}\right)}{(1-s\log x)^2}+\int_0^{-\log x}\frac{\rho(w)}{(1-s\log{x}-sw)^2}dw $$
This general solution recovers the specific solution found above when $\rho\equiv 1, q\equiv 0$. The solution would be entirely different and less singular if a different boundary condition on $(x,t=1)$ (in the old coordinate system) had been chosen!