Is $H(x − ct)$ a weak solution of the wave equation? This problem is from the book by Strauss (1), §12.1, Exercise 5 p. 337.


  
*Verify, directly from the definition of a distribution, that the discontinuous function $u(x, t) = H(x − ct)$ is a weak solution of the wave
  equation.
  

A function is called a weak solution of the wave equation $u_{tt}=c^2 u_{xx}$ if $\iint _{\mathbb R^2}u(x,t) (\phi_{tt}-c^2 \phi_{xx})dxdt=0$ for every $C^\infty$ function with compact support $\phi$.
I tried that $$
\begin{aligned}
\iint _{\mathbb R^2}u(x,t) (\phi_{tt}-c^2 \phi_{xx})dxdt &=\iint_{x>ct} (\phi_{tt}-c^2\phi_{xx})dxdt \\
&= \int_{-\infty}^{\infty}\int_{-\infty}^{x/c}\phi_{tt}dtdx - c^2 \int _{-\infty}^{\infty}\int_{ct}^{\infty}\phi_{xx}dxdt \\
&= \int_{-\infty}^{\infty}\phi_t(x,x/c)dx+c^2 \int_{-\infty}^{\infty}\phi_x(ct,t)dt
\end{aligned}
$$
but I don't know how to proceed.
Does anyone have an idea?

(1) W.A. Strauss, Partial Differential Equations: An Introduction, 2nd ed., John Wiley & Sons, 2008.
 A: Of course, one can go back to the general definition of weak solutions, as proposed in OP (see related post for complements). I will propose a solution which may be more in the spirit of the book (Example 10 p. 336). The derivatives of the discontinuous function $u(x,t) = H(x-ct)$ viewed as a distribution are
\begin{aligned}
u_x &= \delta(x-ct) , & u_t &= -c \delta(x-ct) ,\\
u_{xx} &= \delta'(x-ct) , & u_{tt} &= c^2 \delta'(x-ct) .
\end{aligned}
Hence, $u_{tt} = c^2 u_{xx}$ and $u$ is called a “weak” solution of the wave
equation.
A: $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}H(x-ct) (\phi_{tt}-c^2 \phi_{xx})dxdt = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}H(x-ct)\phi_{tt}dtdx-c^2\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}H(x-ct)\phi_{xx}dxdt$
Now evaluate the inner integral of the first integral.
$\int_{-\infty}^{\infty}H(x-ct)\phi_{tt}dt=H(x-ct)\phi_t|_{-\infty}^\infty +c\int_{-\infty}^{\infty}\delta(x-ct)\phi_{t}dt=c\int_{-\infty}^{\infty}\delta(x-ct)\phi_{t}dt=c[\delta(x-ct)\phi|_{-\infty}^{\infty}+c\int_{-\infty}^{\infty}\delta'(x-ct)\phi dt]=c^2\int_{-\infty}^{\infty}\delta'(x-ct)\phi dt$
Evaluating the inner integral of the second integral,
$\int_{-\infty}^{\infty}H(x-ct)\phi_{xx}dx=H(x-ct)\phi_x|_{-\infty}^\infty -\int_{-\infty}^{\infty}\delta(x-ct)\phi_{x}dx=-\int_{-\infty}^{\infty}\delta(x-ct)\phi_{t}dt=-\delta(x-ct)\phi|_{-\infty}^{\infty}+\int_{-\infty}^{\infty}\delta'(x-ct)\phi dt=\int_{-\infty}^{\infty}\delta'(x-ct)\phi dt$
Therefore, $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}H(x-ct)\phi_{tt}dtdx-c^2\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}H(x-ct)\phi_{xx}dxdt=c^2\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\delta'(x-ct)\phi dtdx-c^2\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\delta'(x-ct)\phi dxdt=0$
Am I right?
