Fundamental solutions and initial conditions (for d'Alembert operator)

Question

I would like to understand the plane wave solution for the (3+1-dimensional) d'Alembert operator $$ \square = \nabla^2 - \frac 1{c^2}\frac{\partial^2}{\partial t^2}\tag{1} $$ in terms of its fundamental solutions. I know that the “function” (distribution) $$ G(\vec x, \, t) = \frac{c}{4\pi r}\delta(ct - r),\tag{2} $$ where $r = \lvert\vec x\rvert$, is a fundamental solution, that is, it satisfies $$ \square G(\vec x, \, t) = -\delta^3(\vec x)\delta(t)\tag{3}. $$ This fundamental solution can be seen as a mathematical statement of Huygens principle, since it states that the response to an instantaneous impulse at $t = 0$ and $\vec x = 0$ is a spherical propagation of that impulse with an attenuation given by the $1/r$ which ensures conservation of energy. So I wanted to see this principle in action in the simple example of a plane wave, but I'm having a hard time.

A plane wave $f(\vec x, \, t)$ propagating satisfies $\square f = 0$, but the fundamental solution method only allows me to find particular solutions of an inhomogeneous equation $\square f = g$, by convolution with the fundamental solution, i.e. $f = G \circledast g$. I figured that I could choose some singular $g$, i.e. $g \propto \delta(t)$, in order to have $g = 0$ most of the time. I thought that $g$ would then have the meaning of an “initial condition” or “initial impulse”. In particular, for a plane wave traveling in the $z$ direction, $g$ should also have translational symmetry with respect to $x$ and $y$, i.e. $$g(\vec x, \, t) = g(0, \, 0, \, z, \, t) \equiv g(z, \, t) = h(z)\delta(t).\tag{4}$$

But this doesn't seem to work since I'm not getting solutions of the form $g(ct - z)$.

Question: is this method correct in principle? i.e. is choosing a source proportional to $\delta(t)$ akin to choosing initial conditions? If not, is there a different way of doing that?

Answer 1

The solution $u(\vec{r},t)$ at a future spacetime point $(\vec{r},t)$ to the 2nd-order homogeneous wave equation depends on Cauchy data (initial displacement & initial velocity) at the intersection between the past light-cone and the Cauchy surface $\{t=0\}$

$$\begin{align} u(\vec{r},t) ~=~& \int_{\mathbb{R}^3} \!d^3r^{\prime}~\partial_tG_{\rm ret}(\vec{r}\!-\!\vec{r}^{\prime},t) u(\vec{r}^{\prime},0)\cr ~+~&\int_{\mathbb{R}^3} \!d^3r^{\prime}~G_{\rm ret}(\vec{r}\!-\!\vec{r}^{\prime},t) \partial_tu(\vec{r}^{\prime},0) \cr ~=~&\int_{\mathbb{R}^3} \!d^3r^{\prime}\frac{\delta^{\prime}(t-|\vec{r}-\vec{r}^{\prime}|)}{4\pi|\vec{r}-\vec{r}^{\prime}|}u(\vec{r}^{\prime},0)\cr ~+~&\int_{\mathbb{R}^3} \!d^3r^{\prime}\frac{\delta(t-|\vec{r}-\vec{r}^{\prime}|)}{4\pi|\vec{r}-\vec{r}^{\prime}|} \partial_tu(\vec{r}^{\prime},0), \cr &\quad \text{for}\quad t~>~0, \tag{A} \end{align}$$ cf. Huygens' principle and Ref. 1. Here$^1$ $$ G_{\rm ret}(\vec{r},t)~=~\frac{\theta(t)\delta(t^2-r^2)}{2\pi }~=~\frac{\delta(t-r)}{4\pi r}\tag{B}$$ is the retarded Greens function,

$$ \Box G_{\rm ret}(\vec{r},t)~=~-\delta^3(\vec{r})\delta(t).\tag{C} $$

References:

1. P.M. Morse & H. Feshbach, Methods of Theoretical Physics, Vol. 1, 1953; section 7.3.

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$^1$ In this answer we use the Minkowski signature $(-,+,+,+)$ and works in units where the wave speed is $c=1$.

Answer 2

Question: is this method correct in principle? i.e. is choosing a source proportional to δ(t) akin to choosing initial conditions? If not, is there a different way of doing that?

D'Alembert did more than deriving the wave equation, $\square$. He is credited with also supplying its solution: https://en.wikipedia.org/wiki/D%27Alembert%27s_formula . Referring to this web page, the first two terms are due to the initial displacement at $t=0$. The third term (the integral) is due to the initial velocity at $t=0$. So these are the initial conditions and their effects on the resulting wave. My https://www.nature.com/articles/s41598-021-99049-7 should help. Also I derive the third term geometrically in the notes in the supplementary information. (There is some confusion as to whether this third term came from Euler) Note that the third term is very important as it is the reason that the backward wave from the second term is canceled when the source speed is equal to the wave propagation speed.