Consider the wave equation \begin{align} (\partial_t^2-\nabla^2)\phi(t,x)=0, \end{align} with Cauchy boundary conditions $\phi(0,x)=f(x)$ and $\dot{\phi}(0,x)=g(x)$. Suppose we perturb the system with an external source $J=J(t,x)$. The solution to this is given by \begin{align} \phi(t,x)=\phi_0+\int_{-\infty}^\infty dt'dx'\,G(t,t';x,x')J(t',x'), \end{align} where $\phi_0$ is the homogeneous solution without the external source that fulfills the Cauchy boundary conditions, and $G(t,t',x,x')$ is the Greens function, compatible with the differential equation \begin{align} (\partial_t^2-\nabla^2)G(t,t';x,x')=\delta(t-t')\delta(x-x'). \end{align} A common approach to solve this is to Fourier expand the Greens function, i.e., to consider \begin{align} G(t,t';x,x')=\int_{-\infty}^\infty \frac{d\omega d^3k}{(2\pi)^4}\tilde{G}(\omega,\vec{k})e^{i\left[\vec{k}\cdot(\vec{x}-\vec{x}\,')-\omega(t-t')\right]}, \end{align} and replace it in the Greens equation where we can invert the differential operator in momentum and frequency space. Regarding the wave operator, we can find \begin{align} \tilde{G}(\omega,\vec{k})=\frac{1}{-\omega^2+k^2}, \end{align} where $k=\left|\vec{k}\right|$, and then if we are able to solve the integral \begin{align} G(t,t';x,x')=\int_{-\infty}^\infty \frac{d\omega d^3k}{(2\pi)^4}\frac{e^{i\left[\vec{k}\cdot(\vec{x}-\vec{x}\,')-\omega(t-t')\right]}}{k^2-\omega^2}, \end{align} we are "done"—we obtained the Greens function, and then if we integrate it with the external source we can obtain the total solution of the system, in principle. However, regarding this method, how does the Greens function, or the total solution "feel" the Cauchy boundary conditions in this example? Moreover, where can I find literature about this discussion?

  • $\begingroup$ You have the contour integral, but to solve this integral, you will have to specify the contour as well. In what way do you think that the choice of contour will affect your answer? $\endgroup$
    – KP99
    Nov 27, 2022 at 17:53
  • $\begingroup$ With the retarded prescription the result will be proportional to $\Theta(t-t')$, with the advanced to $\Theta(t'-t)$ while the Feynman's one will be ordered in time. I know that, however, I don't see how those quantities are involved in the discussion of my question. $\endgroup$ Nov 27, 2022 at 19:11

1 Answer 1


Your problem is not specific to d'Alembert's PDE, but can be traced back to ODE's, even the simple harmonic oscillator. Say you want to solve: $$ \frac{d^2G}{dt^2}+G = \delta $$ Going to Fourier transform: $$ \tilde G(\omega)=\int\frac{d\omega}{2\pi}e^{i\omega t}G(t) \\ G(t)=\int dte^{-i\omega t}\tilde G(t) \\ (-\omega^2+1)\tilde G = 1 $$ You want to divide by $1-\omega^2$. However, when $\omega=\pm1$ this is zero, so the "value" there is undefined. Using the theory of distributions, you can prove that $\hat G$ is necessarily: $$ \tilde G = PV\frac{1}{1-\omega^2}+C_+\delta(\omega-1)+C_-\delta(\omega+1) $$ with $PV$ denoting the Cauchy principal value and $C_\pm$ being constants. The converse is easy to check, using the fact that: $$ f(t)\delta(t) = f(0)\delta(t) $$ Going back to time: $$ G = \frac{\sin |t|}{2} +C_+e^{-it}+C_-e^{it} $$ with $H$ you see that $C_\pm$ is specified by the initial conditions $G(t_0),\frac{dG}{dt}(t_0)$.

Back to your case, the similar situation happens when you divide by $k^2-\omega^2$. From: $$ (-\omega^2+k^2)\tilde G = 1 $$ you deduce: $$ \tilde G = PV\frac{1}{-\omega^2+k^2}+C_+(k)\delta(\omega-|k|)+C_-(k)\delta(\omega+|k|) $$ with now $C_\pm$ are functions defined on $R^3$. They can be determined by the Fourier transforms of $G,\partial_tG$ at $t_0$ which will give the unique solution to the Cauchy problem.

In classical physics, you are often interested in the causal solutions, i.e. $G$ is supported on positive times (in field theory, you are rather more interested in the Feynman propagator which you obtain by Wick rotation or time ordering). It turns out, using Jordan's lemma and the residue theorem, that it amounts to taking the limit of $\epsilon\to 0^+$ of: $$ \tilde G = \frac{1}{k^2-(\omega-i\epsilon)^2} $$ and more generally making the substitution $\omega\to \omega-i\epsilon$.

Hope this helps.

Answer to comment

a) Actually the prescription retarded/advanced/Feynman will determine the $C_\pm$. Once again, there is an ambiguity in defining $\frac{1}{k^2-\omega^2}$ which these prescriptions help resolve. If you want to interpret it as the Cauchy $PV$, then you need to add the appropriate $C_\pm$.

However, in this case, it is easier to think directly in terms of contour integration since you are not interested in solving the Cauchy problem. Indeed, writing explicitly what the prescriptions mean for the $C_\pm$ is needlessly complicated.

b) It depends. The problem arises when you divide by zero, in which case you need additional information to uniquely define your Gren's function (boundary condition, causality ...). However, this is not always necessary. Take for example the operator $-\Delta+m^2$ with $m$ real in arbitrary dimensions. Then its Green's function is (in Fourier): $$ \tilde G = \frac{1}{k^2+m^2} $$ which makes sense no matter what since the denominator never vanishes. Adding boundary conditions would overdetermine it.

An important caveat is that when you are taking Fourier transforms, you are restricting your attention to tempered solutions. In general, you could have more solutions thus requiring further restrictions. Take for example a modification of the previous simple case: $$ -\frac{d^2G}{dt^2}+G = \delta $$ By Fourier transform, you obtain the unique tempered Green's function

$$ \tilde G = \frac{1}{1+\omega^2} \\ G = \frac{e^{-|t|}}{2} $$

however, if you are looking for distribution solutions in general, you could get for example: $$ G = \frac{e^{-|t|}}{2}+C_+e^t+C_-e^{-t} $$ with $C_\pm$ arbitrary constants and can be resolved by Cauchy boundary conditions. Note that when the coefficients are not zero, $G$ is not tempered, so the usual Fourier transform is ill defined.

In physics, you usually sweep under the rug such technical assumption (temperedness ...). To really guarantee unicity you'll need to do some rigorous math. Check out any course on distributions for more information.

  • $\begingroup$ Thank you for your answer. I know what the retarded, advanced, and Feynman's prescriptions are. However, a) If I compute one of them, these constants $C_\pm$ will not appear? I have computed them, and it seems that they do not. And b) so it seems that when I compute Green's functions by means of Fourier transform, it is necessary to specify boundary conditions for it apart? (I know that they can be read from the boundary conditions for the dynamical function (see en.wikipedia.org/wiki/Green's_function#Theorem)). I am confused. $\endgroup$ Nov 27, 2022 at 19:18

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