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[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 $\xi$?

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  • $\begingroup$ @Carlo Beenakker, I added some reasoning why I think it should be possible to reproduce the correct Green's function for real $\xi$. It's possible that I'm misunderstanding something of course. $\endgroup$
    – Fetchinson0234
    Dec 6, 2022 at 11:30
  • $\begingroup$ the issue is not whether $\phi=\int_0^\infty u\,dt$ satisfies the equation for the Green's function, which it does, but whether the function $\phi$ you obtain in this way decays when $x_4\rightarrow\pm\infty$; the integrand does decay, but there is no guarantee this decay is conserved upon integration. $\endgroup$
    – Carlo Beenakker
    Dec 6, 2022 at 12:23
  • $\begingroup$ That may very well be, but then the statement is that some Green's functions (non-decaying ones) can be obtained from the heat kernel but some other Green's functions (decaying ones) can not? It may be, but I was not aware of such a subtlety. And what singles out the non-decaying Green's functions over the decaying Green's functions? So this entire state of affairs is a bit mysterious to me, but I might be overlooking something obvious. $\endgroup$
    – Fetchinson0234
    Dec 6, 2022 at 12:55
  • $\begingroup$ Also, in the earlier question about Green's functions, there is an analytic Green's function, $\phi(x) = e^{\xi x_4} / x^2$. This is analytic in $\xi$ but does not decay correctly for real $\xi$. For real $\xi$ there is another one with the correct decay. Similarly, I'd think there is the analytic heat conduction solution, $u(t,x) = e^{\xi t} e^{-x^2/4/t} / 4 / t^2$ but for real $\xi$ there should be another one, which leads to the aforementioned correct Green's function for real $\xi$. $\endgroup$
    – Fetchinson0234
    Dec 6, 2022 at 13:04

1 Answer 1

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Unlike in your earlier question, the function $u(t,x)$ is analytic in $\xi$, so there are no complications arising from a nonzero real part of $\xi$.

You can simply invert the Fourier transform of $\exp\bigl[-t\bigl((\omega+i\xi)^2+k^2\bigr)\bigr]$ for any complex $\xi$, to arrive at $$u(t,x_4,r) = e^{\xi x_4} e^{-(x_4^2+r^2)/4t}\frac{1}{4t^2}.$$ This decays at large $x_4$ or large $r$ for any $t>0$, irrespective of whether $\xi$ is real or imaginary.

The integral $\int_0^\infty u(t,x_4,r)\,dt$ does not decay as a function of $x_4$ when ${\rm Re}\,\xi\neq 0$, but there is no reason it should.

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    $\begingroup$ I am bit at a loss to understand the down vote: the OP asks for the solution of the 4+1 dimensional heat equation with a complex parameter $\xi$, for a given initial condition and subject to the requirement that the solution decays at infinity in each coordinate; the solution given in the answer satisfies those requirements; I presume the solution is unique, so what other answer could be possible? Am I missing something? $\endgroup$
    – Carlo Beenakker
    Dec 6, 2022 at 15:55
  • $\begingroup$ Respectfully, the down vote came from me, because the crux of the problem is not addressed I believe. The issue is this: on general grounds one can obtain the Green's function from $u$. For purely imaginary $\xi$ we know what the appropriate (decaying) Green's function is and we can construct $u$. For real $\xi$ we also know what the appropriate (decaying) Green's function is but we cannot construct $u$. The decay property requirements are for the Green's function, not $u$. $\endgroup$
    – Fetchinson0234
    Dec 6, 2022 at 16:01
  • $\begingroup$ I presume you don't question that $u$ in the answer solves the heat equation with all boundary conditions; so if there is a second solution the heat equation would not have a unique solution? (not impossible, but I always thought the heat equation uniquely determined the flow) $\endgroup$
    – Carlo Beenakker
    Dec 6, 2022 at 16:05
  • $\begingroup$ You are absolutely right, I don't question that $u$ as given is always a solution for any $\xi$. Similarly, $\phi = e^{\xi x_4} / x^2$ is always a Green's function for any $\xi$. But the decay properties are set up for $\phi$, not $u$. So the puzzle I'd like to understand is this: why is it that apparently we have no problem with constructing a $u$-solution for imaginary $\xi$ such that the corresponding Green's function is decaying, but we cannot construct a $u$-solution for real $\xi$ such that the corresponding Green's function is decaying. Is this really the case? Kind of strange if true. $\endgroup$
    – Fetchinson0234
    Dec 6, 2022 at 16:12

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