I've been reviewing the Wave equation, and there are a few things that I don't understand in Kevorkian's book. It says the fundamental solution to the wave equation is defined by the solution to:
$$\begin{align} u_{tt} - u_{xx} &= \delta(x)\delta(t),\quad -\infty < x < \infty, t \geq 0 \\ u(x,0^{-}) &= 0\\ u_t(x,0^{-}) &= 0\\ \end{align}$$
He continues to say this is equivalent to the homogeneous system:
$$\begin{align} u_{tt} - u_{xx} &= 0, \quad -\infty < x < \infty, t > 0 \\ u(x,0^{+}) &= 0\\ u_t(x,0^{+}) &= \delta(x) \end{align}$$
He says that you can get from one system to the other by simply integrating with respect to t from $-\epsilon$ to $\epsilon$ and letting $\epsilon \to 0$. What's going on here?
I was able to find a partial answer (see below), but it's not quite clear how to go from the second system to the first system.
While typing up my question, I was able to work out some of the details with Kevorkian's assertion. Kevorkian's claim that these two systems of equations are equivalent is meant to be interpreted in the strongest sense, that the solutions $u$ of both equation are the same.
There are a few things worth noting. The $0^{-}$ in the initial conditions in the first equation suggest that $u$ is actually defined on some open domain containing $-\infty < x < \infty, t \geq 0$. That is, $u$ is actually defined for $t > \epsilon$ for some small $\epsilon$ for instance.
In the second system, $u$ is only define for $t > 0$. This second system makes more sense in terms of $u$ actually being a function, while the first system is more appealing to demonstrate that $u$ is actually a fundamental solution. This primary distinction is what I originally overlooked.
Why are they equavalent?
(I $\to$ II) To get from the first one to the second, integrate $u_{tt} - u_{xx} = \delta(x)\delta(t)$ from $t = -\epsilon$ to $t = \epsilon$ to get:
$$ u_t(x,t) \big|_{t = -\epsilon}^{t = \epsilon} - \int_{-\epsilon}^{\epsilon} u_{xx}\,dt = \delta(x)$$
Since $u_t(x,0^{-}) = 0$, and since $\int_{-\epsilon}^{\epsilon} u_{xx}\,dt \to 0$ we get:
$$u_t(x,0^{+}) = \delta(x).$$
Integrating twice, we have (a little care is need to make this rigorous, I took the $-\epsilon$ from the first equation, and set that equal to 0 first, then integrated with respect to $\epsilon$):
$$u(x,0+) = 0.$$
(II $\to$ I) Now, vice versa. To get from the second equation to the first equation. The boundary conditions of the second allow us to extend $u$ continuously to $t \geq 0$, and we can simply just define $u \equiv 0$ for $t > \epsilon$ for some small $\epsilon$. This makes the boundary conditions trivially satisfied for the first system. Does $u$, in fact, solve the PDE though?