On the Wikipedia page here , it states that the Green's function for 3D relativistic heat conduction (with $c=1$)

$$[\partial_t^2 + 2\gamma\partial_t -\Delta_{3D}] u(t,x) = \delta(t,x) = \delta(t)\delta(x)$$

is given by $$u(t,x) = \frac{e^{-\gamma t}}{20\pi}\bigg( \big[8 - 3e^{-\gamma t} + 2\gamma t + 4\gamma^2t^2\big]\frac{\delta(t-|x|)}{|x|^2} + \gamma^2\Theta(t - |x|)\big[ \frac{I_1\big(\gamma\sqrt{t^2 - |x|^2}\big)}{\sqrt{t^2 - |x|^2}} + t\frac{I_2\big(\gamma\sqrt{t^2 - |x|^2}\big)}{t^2 - |x|^2}\big] \bigg)$$

Where $I_1$ and $I_2$ are modified Bessel functions of the 1st and 2nd kind. Furthermore, $\Theta$ is the Heaviside step function . I am looking for a derivation of this or a reference to one.

Soft Attempt: Taking the Fourier transform with respect to $x$ we obtain $$[\partial_t^2 + 2\gamma\partial_t + (4\pi^2|\xi|^2)] \hat{u}(t,\xi) = \delta(t)$$

On the same Wikipedia page, this is a 1D damped harmonic oscillator and they claim the Green's function is given by $$\hat{u}(t,\xi) = \Theta(t)e^{-\gamma t} \frac{\sin\big(t\sqrt{4\pi^2|\xi|^2 - \gamma^2}\big)}{\sqrt{4\pi^2|\xi|^2 - \gamma^2}}$$ So I should expect that

$$u(t,x) = \int e^{2\pi i \xi\cdot x}\Theta(t) e^{-\gamma t}\frac{\sin\big(t\sqrt{4\pi^2|\xi|^2 - \gamma^2}\big)}{\sqrt{4\pi^2|\xi|^2 - \gamma^2}}d\xi$$

But I am not sure how we would get the formula above or if this is even the correct approach.

  • $\begingroup$ Certainly, going into Fourier space makes sense here - you essentially get yourself a spectral representation of the Green's function. I'd expect the transformation back to go via the residue theorem. Doesn't that let you evaluate the integral in your final expression? $\endgroup$ Oct 8 '19 at 17:56
  • $\begingroup$ crossposted: math.stackexchange.com/q/3385735/87355 ; please don't crosspost to different sites within a short period, in particular not without disclosing the crossposting, since that may well lead to wasteful duplication of efforts. $\endgroup$ Oct 8 '19 at 20:11
  • $\begingroup$ @MichaelEngelhardt I have tried playing around with the integral, but I still can't seem to get a tractable solution, especially one looking like the one given. $\endgroup$
    – Dayton
    Oct 8 '19 at 20:39
  • $\begingroup$ @Carlo Beenakker okay, I have deleted the other post. I did not know that was not acceptable, as these are separate sites. $\endgroup$
    – Dayton
    Oct 8 '19 at 20:39

The "relativistic" heat equation is more generally known as the Telegrapher's equation, $$\frac{\partial f}{\partial t}+\tau\frac{\partial^2 f}{\partial t^2}=\kappa\nabla^2 f.$$ The Green's function is calculated in Application of the three-dimensional telegraph equation to cosmic-ray transport (2016), see equation 10. In equation 11 it is checked that the $\tau\rightarrow 0$ limit gives the familiar Gaussian Green's function of the 3D diffusion equation.

I note that the result in that 2016 paper is different from the formula in the OP, taken from Wikipedia.

  • 2
    $\begingroup$ Nice find - additionally, in the appendix, they explicitly go through the derivation through Fourier space. $\endgroup$ Oct 10 '19 at 19:41
  • 2
    $\begingroup$ Something else that worried me about the Wikipedia solution is that it doesn't reduce to the well-known Green's function of the D'Alembertian (also given there) in the limit $\gamma \rightarrow 0$. Since that limit could be subtle, I didn't draw a firm conclusion, but seeing the different behavior shown in this paper, which does seem more consistent with the D'Alembertian limit, makes me more suspicious. $\endgroup$ Oct 10 '19 at 20:03

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