The problem is in Strauss, Partial Diferential Equations 2nd edition, 12.1 Ex.5.

I want to verify by direct substitution that Heaviside DISTRIBUTION $H(x-ct)$ is a weak solution of wave equation $u_{tt}=c^2u_{xx}$.

I found almost identical question at weak solution of wave equation here but I think my problem is slightly different from this, since I considered that $(x,t)\in\mathbb R\times(\mathbb R^+\cup\{0\})$.

$$\int_{\infty}^{\infty} \int_0^{\infty} H(x-ct)(\phi_{tt}-c^2\phi_{xx})dtdx=0$$

should be true for all $\phi\in\mathcal D(\mathbb R\times(\mathbb R^+))$.

Using $\phi$ is a $\mathcal C^\infty$ function with compact support, I found it can be reduced to

$$ \begin{align} & \int_{0}^{\infty}\int_{0}^{x/c} \phi_{tt}dtdx - c^2\int_{0}^{\infty}\int_{ct}^{\infty} \phi_{xx}dxdt \\ = & \int_{0}^{\infty} (\phi_t (x, \dfrac{x}{c}) - \phi_t (x, 0))dx + c\int_{0}^{\infty}\phi_x(t,\dfrac{t}{c})dt \\ = & \int_{0}^{\infty} c\dfrac{d\phi}{ds}(s,\dfrac{s}{c})ds-\int_{0}^{\infty} \phi_t (x, 0)dx \\ = & -c\phi(0,0) - \int_{0}^{\infty} \phi_t (x, 0)dx. \end{align}$$

Since the second term cannot be integrated explicitly, I considered a closed curve on $\mathbb R\times(\mathbb R^+\cup\{0\})$ which connects $(0,0),(X,0),(X,T),(0,T)$ by line segments, then send $X$ and $T$ to $+\infty$. Since $\phi$ and its derivatives vanish over some radius R, the second term would be $-\phi(0,0)$, not $-c\phi(0,0)$.

Maybe I has made a mistake on integration, but can't find that. Could anyone give me some help?

  • $\begingroup$ If $\phi \in \mathcal D(\mathbb R_x \times \mathbb R^+_t)$, does this mean that the support of $\phi$ in time is bounded away from 0? $\endgroup$ – Calvin Khor Mar 15 at 18:56
  • $\begingroup$ @CalvinKhor I meant that $\phi$ is a test function in distribution space en.wikipedia.org/wiki/Distribution_(mathematics). So $\phi$ and its derivatives will be identically zero outside some bounded interval. $\endgroup$ – Jinmu You Mar 16 at 2:23
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    $\begingroup$ I know what a distribution and what a test function is, I'm asking you to check that you mean $t=0$ is included in the support of $\phi$. In this case, i think the definition of a weak solution should also have some terms with $t=0$ from the integration by parts $\endgroup$ – Calvin Khor Mar 16 at 8:31
  • $\begingroup$ @CalvinKhor Sorry for misunderstanding. I think I cannot be sure that $t=0$ is included in the support of $\phi$. Maybe $\phi(0,0)$ can get some nonzero value. My expectation was that the last integral term would be $-c\phi(0,0)$ and make the integration zero, regardless of the $\phi$. But I can't justify that. $\endgroup$ – Jinmu You Mar 16 at 14:26

If you want your test functions to have support containing $t=0$, your definition of a weak solution should be adjusted since if $u\in C^2$,

$$ \int_{\mathbb R}\int_0^\infty u_{tt}(\tau) \phi(\tau) d\tau dx = \int_{\mathbb R} \left( -u_t(0)\phi(0) +u(0)\phi_t(0) +\int_0^\infty u(\tau)\phi_{tt}(\tau) d\tau \right)dx$$ So for test functions $\mathfrak D:=\mathcal D(\mathbb R \times [0,\infty))$, we should say that $u\in\mathfrak D'$ is a weak solution of the IVP for $u_0,u_1\in\mathcal D(\mathbb R)'$, $$ u_{tt} = c^2 u_{xx}\\ u(0)=u_0\\ u_t(0)=u_1$$ if $$ -\int_\mathbb R u_1(x)\phi(x,0)dx + \int_\mathbb R u_0(x)\phi_t(x,0)dx +\int_{\mathbb R} \int_0^\infty u(x,t) (\phi_{tt}(x,t)-c^2\phi_{xx}(x,t)) dxdt = 0$$ When $u(x,t)=H(x-ct)$, then $u(x,0)= H(x),\ u_t(x,0) = -c\delta_0(x),$ so we should use $$u_0(x) = H(x),\\ u_1(x) = -c\delta_0(x). $$ This causes the two extra terms you found to exactly cancel.

For the definition of a weak solution you suggested, you should use the fact that functions in $\mathcal D(\mathbb R \times (0,\infty))$ vanish at $t=0$.

  • $\begingroup$ Thanks for perfect explanations. Aside from this problem, does the conclusion $ \int_{0}^{\infty} \phi_t(x,0)dx=-\phi(0,0) $ when $\phi\in\mathcal D$ makes sense? I got the idea from contour integral in complex analysis, but I feel unsure of using this. $\endgroup$ – Jinmu You Mar 17 at 12:53
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    $\begingroup$ @JinmuYou if I understand correctly, the Cauchy's theorem that is required to say integrals on loops are 0 is unavailable in this setting. $\endgroup$ – Calvin Khor Mar 17 at 13:16

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