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I have seen Wick rotation used to describe the relationship between the heat and Schrodinger equations. That is, if $u(t,x)$ solves the heat equation then $v(t,x):=u(it,x)$ solves the Schrodinger equation.

I also noticed the same relationship between the wave and Laplace equation. If you take the solution to either one, send $t\mapsto it$ you get the solution to the other.

Has this been studied? Is there a good reference on Wick rotation for wave/Laplace equations?

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  • $\begingroup$ Note that the wave equation does not have a maximum principle under general initial conditions. $\endgroup$
    – Hollis Williams
    Commented Nov 26, 2023 at 17:32

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The transformation to imaginary time is used to relate the Green's function of the Laplacian to the Green's function of the d'Alembertian (therefore relating Laplace equation and heat equation). See for example these MSE posts -- 1 and 2 -- with references to the literature.

The transformation of an elliptic to a hyperbolic PDE breaks many methods of solution, variational methods based on a maximum principle do not survive the Wick rotation. This is likely why you will not find much use of that transformation.

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  • $\begingroup$ Do you know how long Wick rotation survives? Because you can rotate poles. $\endgroup$
    – user479223
    Commented Dec 17, 2023 at 17:23
  • $\begingroup$ Because there's a pole. For example $1/(1+(x+t)^2)$ solves the wave equation for all time but $1/(1+(x+it)^2)$ only solves the Laplace equation until $t=1$ at $x=0$. $\endgroup$
    – user479223
    Commented Dec 17, 2023 at 18:05
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A new question has appeared, so I'll start a new answer: How can we ensure that the Wick rotated solution does not blow up?

Wick rotation $t\mapsto i\tau$ of a solution $f$ is an analytic continuation which requires the absence of poles in either quadrants I and III, or in quadrants II and IV. In that case Cauchy's theorem ensures that $I_1=i\int_{-\infty}^\infty f(i\tau)\,d\tau$ remains finite if $I_2=\int_{-\infty}^\infty f(t)\,dt$ is finite. More precisely, $I_1=I_2$ if there are no poles in quadrants I and III, while $I_1=-I_2$ if there are no poles in quadrants II and IV.

If these two cases do not apply, the life time of the solution depends on the location of the poles. I don't think you can make a generic statement.

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  • $\begingroup$ This doesn't quite seem right. $\int_{-\infty}^\infty \frac{1}{1+t^2} dt<\infty$ but $i\int_{-\infty}^\infty \frac{1}{1-\tau^2} d\tau$ blows up. $\endgroup$
    – user479223
    Commented Dec 17, 2023 at 18:58
  • $\begingroup$ this integrand has a pole on the imaginary axis, so it's ambiguous in which quadrant it lies. $\endgroup$
    – Carlo Beenakker
    Commented Dec 17, 2023 at 19:01

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