Is $H(x − ct)$ a weak solution of the wave equation?

Question

This problem is from the book by Strauss (1), §12.1, Exercise 5 p. 337.

5. 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?

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(1) W.A. Strauss, Partial Differential Equations: An Introduction, 2nd ed., John Wiley & Sons, 2008.

Answer 1

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}H(x-ct) (\phi_{tt}-c^2 \phi_{xx})dxdt = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}H(x-ct)\phi_{tt}dtdx-c^2\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}H(x-ct)\phi_{xx}dxdt$

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?