The following two functions are the fundamental solutions of the wave equation and its Laplace transform (the modified Helmholtz equation) in two dimensions, respectively, when the speed of wave propagation is $1$:

$$ g(x,t) = \frac{H(t - |x|)}{2\pi \sqrt{t^2 - |x|^2}}, $$ $$ G(x,s) = \frac{i}{4}H_0^{(1)}(i s |x|). $$ Here $H$ is the Heavside function and $H_0^{(1)}$ is the Hankel function of the first kind of order $0$.

Question 1: Is it possible to take the Laplace transform of the function $g(x,t)$ in Mathematica and show that it equals $G(x,s)$?

Now the reason I asked Question 1 is because it is the basis for the following question.

The fundamental solution of the damped wave equation is (page 4 here): $$ g_\text{damped}(x,t) = \frac{1}{2\pi}\cosh\bigg(\frac{\alpha}{2}\sqrt{t^2-|x|^2}\bigg)\frac{H(t - |x|)}{2\pi \sqrt{t^2 - |x|^2}} e^{\frac{-\alpha}{2}t}, $$ where $\alpha$ is the damping parameter. When $\alpha = 0$ this reduces to the fundamental solution of the undamped equation $g(x,t)$ above.

Now I 'think' the Laplace transform of this fundamental solution should look something like $$ G_\text{damped}(x,s) = \frac{i}{4} H_0^{(1)}(i s (1-\beta i) |x|), $$ where $\beta$ is some damping parameter analogous to $\alpha$ in the time domain equation. I would like to verify if I am correct about this.

Question 2: Is it possible to take the Laplace transform of the function $g_\text{damped}(x,t)$ in Mathematica so that I can check if the form of the resulting expression looks like $G_\text{damped}(x,s)$?

  • 1
    $\begingroup$ Try:{LaplaceTransform[HeavisideTheta[t - Abs[x]]/( 2 Pi*Sqrt[t^2 - Abs[x]^2]), t, s, Assumptions -> {x > 0}], Simplify[I/4*HankelH1[0, I*s Abs[x]] // FunctionExpand, Assumptions -> {x > 0, s > 0}]} $\endgroup$ Jul 11, 2020 at 11:57
  • $\begingroup$ Thanks, I see that it confirms Question 1. When I try the same approach for Question 2 however Mathematica seems to stall and just say 'Running...' $\endgroup$ Jul 11, 2020 at 12:11
  • 1
    $\begingroup$ lpt = LaplaceTransform[ With[{α = 0}, 1/(2 π) Cosh[α/2 Sqrt[t^2 - Abs[x]^2]] HeavisideTheta[ t - Abs[x]]/(2 π Sqrt[t^2 - Abs[x]^2]) Exp[-α t/2] ], t, s, Assumptions -> {x > 0}] gives BesselK[0, s x]/(4 Pi^2) $\endgroup$
    – flinty
    Jul 11, 2020 at 12:41
  • $\begingroup$ I am looking for the case when $\alpha$ is not equal to zero so that I can see where the $\alpha$ ends up after the Laplace transform is performed. $\endgroup$ Jul 11, 2020 at 12:47


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