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This problem is from Strauss's "Partial Differential Equations", 2/E, Chapter 12.1 Exercise 5.

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 $\phi$. Show that the function $u(x,t)=H(x-ct) $ is a weak solution of the wave equation.

I tried that $\iint _{\mathbb R^2}u(x,t) (\phi_{tt}-c^2 \phi_{xx})dxdt=\iint_{x>ct} (\phi_{tt}-c^2\phi_{xx})dxdt = \int_{-\infty}^{\infty}\int_{-\infty}^{x/c}\phi_{tt}dtdx - c^2 \int _{-\infty}^{\infty}\int_{ct}^{\infty}\phi_{xx}dxdt= \int_{-\infty}^{\infty}\phi_t(x,x/c)dx+c^2 \int_{-\infty}^{\infty}\phi_x(ct,t)dt$

but I don't know how to proceed.

Does anyone have an idea?

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  • $\begingroup$ Why is the inner integral over $\mathbb R^2$ and not $\mathbb R$? $\endgroup$ – rubik Apr 6 '17 at 19:49
  • $\begingroup$ @rubik That's not the inner integral, but just the double integral for all plane. $\endgroup$ – bellcircle Apr 6 '17 at 19:49
  • $\begingroup$ I see. Isn't it better to use $\iint$ then? I find the current form a bit confusing, even if the would-be-outer doesn't have bounds. $\endgroup$ – rubik Apr 6 '17 at 19:50
  • $\begingroup$ Okay, I edited it. Thanks. $\endgroup$ – bellcircle Apr 6 '17 at 19:52
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    $\begingroup$ Potentially, using integration by parts or a Green's Identity could simplify this, keeping in mind that $H^\prime(x) = \delta(x)$. $\endgroup$ – AlexanderJ93 Apr 6 '17 at 19:55
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$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}H(x-ct) (\phi_{tt}-c^2 \phi_{xx})dxdt = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}H(x-ct)\phi_{tt}dtdx-c^2\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}H(x-ct)\phi_{xx}dxdt$

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?

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  • $\begingroup$ Why is $\phi_x$ compactly supported? $\endgroup$ – Jason Moore Jul 17 '17 at 23:06

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