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My research has brought me to the following linear parabolic second order PDE:

$$ \frac{\partial^2}{\partial x^2}\Psi(t,x)=c(t,x)\frac{\partial}{\partial t}\Psi(t,x) $$

for $c(t,x)=-\frac{t}{x}$ and $x\in (0,\infty)$ and time $t\in (0, 1).$

Using the ansatz $\Psi(t,x)=X(x)Y(t)$ this can be simplified to ODE's and solved.

However, an important solution class is realized in the following way:

Let $(M,g)$ be the Minkowski plane with null coordinates. Take the isometry $f:M \to \zeta $ with $f(x,y)=(e^x,e^y).$ Take the natural Cauchy foliation of $\zeta$ given by $\log x \log y = s.$ Essentially when you furnish $\zeta$ with $g$ and the induced measure by means of the volume form, a natural solution to the above equation becomes clear:

$$ \Psi(t,x)=\exp \frac{x}{\log t} $$

Just like the free Schrodinger equation can be seen as a heat equation with an imaginary constant, there is an analogous PDE here too:

$$ \frac{\partial^2}{\partial x^2}\Psi(t,x)=\frac{t}{ix} \frac{\partial}{\partial t}\Psi(t,x) $$

We get a similar looking solution:

$$ \Psi(t,x)=\exp \frac{ix}{\log t} $$

Are there any papers in the literature that deal with these specific PDE's where I can read more about them?

If there are no papers in the literature, as I suspect, what is the interpretation of these equations and my solution, and how might they be physically relevant?

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Sorry to bring bad news. Usually, physicists are interested in the Cauchy problem: given initial data, determine the evolution of the state.

Alas, it is well known that PDE's whose order in the time variable (here 2) exceeds the order in the space variable (here 0) are not globally well-posed. Well, you can invoque Cauchy-Kovalevska's theorem in the analytic class, but it is useless for practical applications.

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  • $\begingroup$ Ah I'm sorry, it looks like I mixed up the spatial and temporal variables. I'll edit the question so that it's well posed $\endgroup$ Commented Apr 15, 2023 at 20:47
  • $\begingroup$ @53Demonslayer --- in your edit you only changed the symbol $x\leftrightarrow t$, didn't you? I don't think that resolves the issue. $\endgroup$ Commented Apr 15, 2023 at 21:16
  • $\begingroup$ @CarloBeenakker the order of the time variable no longer exceeds the order of the space variable. $\endgroup$ Commented Apr 15, 2023 at 21:31

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