Question 12.18: Prove that the function E : R × R → R given by E(x,t) = 1/2 ......
Prove that the function E : R × R → R given by
E(x, t)=\frac{1}{2} H(t-|x|)= \begin{cases}\frac{1}{2} & \text { if } t>|x| \\ 0 & \text { if } t<|x|\end{cases} (12.36)
is the fundamental solution of the one-dimensional wave equation. In other words, it holds
\frac{\partial^2 E}{\partial t^2}-\frac{\partial^2 E}{\partial x^2}=\delta
As before, H is the Heaviside’s function, i.e., the characteristic function of the interval (0, +∞).
Let us note first that E from (12.36) is a locally integrable function on the set 0 = {(x, t)| x ∈ R, t > 0}, and thus defines a regular distribution which we simply denote also by E. For any \varphi \in \mathcal {D}( \mathbf{R}) it holds
Putting y = -x in the second integral and taking y instead of x in the other three integrals of the last right-hand side, we get
\left\langle\frac{\partial^2 E}{\partial t^2}-\frac{\partial^2 E}{\partial x^2}, \varphi\right\rangle=-\frac{1}{2} \int_0^{+\infty} \frac{d}{d y}(\varphi(y, y)) d y-\frac{1}{2} \int_0^{+\infty} \frac{d}{d y}(\varphi(-y, y)) d y ( 12.37)
From (12.37) it finally follows
\left\langle\frac{\partial^2 E}{\partial t^2}-\frac{\partial^2 E}{\partial x^2}, \varphi\right\rangle=\frac{1}{2} \varphi(0,0)+\frac{1}{2} \varphi(0,0)=\langle\delta, \varphi\rangle.