[I asked a similar question, Linear PDE, analytic continuation, Green's function and boundary conditions, and was told that a follow-up question should be a separate post.]

I'm interested in a heat conductivity type of equation in 4D, $$\frac{\partial u}{\partial t} = \left(\frac{\partial}{\partial x_4} - \xi\right)^2 u + \frac{\partial^2 u}{\partial x_k^2}$$ where $k=1,2,3$ and $u=u(t,x) = u(t,x_1,x_2,x_3,x_4)$ and the initial condition is, $$u(t=0,x) = 4\pi^2\delta(x)$$ where generally $\xi$ is a complex parameter. Clearly, if $\xi = i \alpha$ with $\alpha$ real we have, $$u(t,x) = e^{i\alpha x_4} \frac{e^{-\frac{x^2}{4t}}}{4t^2}$$ However, if $\xi$ is real, the above form is not okay, because upon integration over $t$ we do not get the Green's function referenced in Linear PDE, analytic continuation, Green's function and boundary conditions but rather something which does not fall off for $x_4 \to \pm \infty$. What would be the solution for $u$ for real $\xi$ which does lead to the correct Green's function?

The relationship between $u$ and the Green's function $\phi$ is simply, as usual, $\phi(x) = \int_0^\infty dt u(t,x)$. With $\xi=i\alpha$ and $\alpha$ real this gives the correct Green's function, but not with a real $\xi$ it appears. And the general theory of heat kernels and Green's functions would guarantee, I think, that $\int_0^\infty dt u(t,x) = \phi(x)$ should be the correct Green's function. Formally, $u(t,x) = 4\pi^2 e^{-t {\cal D}} \delta(x)$ satisfies $\frac{\partial u}{dt} = - {\cal D} u$ and the boundary condition too, $u(0,x) = 4\pi^2\delta(x)$, and $\int_0^\infty dt u = 4\pi^2 {\cal D}^{-1}$ which means that $\int_0^\infty dt u = \phi$, i.e. the Green's function.

So it appears we can't obtain the correctly decaying Green's function from $u$ for real $\xi$?