I'm looking at the linear PDE in 3+1 dimensions, $$ \left[ -(\partial_t - \xi)^2 - \partial_k \partial_k \right] \phi(t,x) = 4\pi^2 \delta(t)\delta(x)\label{1} \tag{1} $$ Where $\xi$ is generally a complex parameter and $k=1,2,3$. So I'm after a particular Green's function $\phi(t,x) = \phi(t,x,\xi)$ and would like to study the $\xi$-dependence. The solution is required to be $SO(3)$ invariant and should fall off in all directions at infinity.
Now for purely imaginary parameter $\xi = i \alpha$ where $\alpha$ is real it's easy to see that a solution is $$ \phi(t,x) = \frac{e^{it\alpha}}{t^2 + r^2}\label{2}\tag{2}$$ with $r = \sqrt{x_1^2 + x_2^2 + x_3^2}$. So far so good :)

Now if $\xi$ is real, or has a non-zero real part, we would have $$ \phi(t,x) = \frac{e^{\xi t}}{t^2 + r^2}\label{3}\tag{3} $$ as the analytic continuation of \eqref{2} but this blows up for $t\to\pm\infty$, depending on the sign of $\xi$. So this analytically continued solution is not okay.

The solution which does satisfy the boundary conditions can still be found and is $$\phi(t,x) = \frac{\cos(\xi r) + \frac{t}{r} \sin(\xi r)}{t^2 + r^2}\label{4}\tag{4}$$ which does fall off in all directions at infinity. Notice that if $\xi$ is purely imaginary the analytic continuation of the solution \eqref{4} will behave badly and will blow up for large $r$.

I was wondering if it's possible to construct one of the solutions if the other is known. In other words if I know the Green's function on the imaginary line, can I somehow guess or construct the Green's function everywhere on the complex plane? Or if I know it everywhere except the imaginary axis, can I find the form of the Green's function along it?

Or one just have to consider both problems totally independently and work hard twice? :)

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    $\begingroup$ the typical practice here is one question per post; for a second question you might want to start a new post. $\endgroup$ Dec 4, 2022 at 20:54

1 Answer 1


Q: Do I have to consider both problems (real $\xi$ or imaginary $\xi$) totally independently and work hard twice?.

A: A single calculation suffices, you could just do the inverse Fourier transform of $[(\omega+i\xi)^2+k^2]^{-1}$ for complex $\xi=i\alpha+\beta$ to arrive at the solution that works for both real and imaginary $\xi$, $$\phi(t,x) = e^{i\alpha t}\frac{r\cos\beta r + t\sin\beta r}{r(t^2 + r^2)}.$$ This is not an analytic function of $\xi$, so you cannot reconstruct it by analytic continuation from the real or imaginary axis.


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