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Prove that if $L(u)=0$ and $L(v)=0$ then $$ L(u_tv_t+c^2u_xv_x)=0. $$ Where $L$ is defined by $$ L(u)=u_{tt}-c^2u_{xx}=0. $$

I don't know how I would arrive to the conclusion. I have gotten close such that since $v=u=0$ then we can do $$ L(u)*L(v) = u_tv_t+c^2u_xv_x - cv_tu_x - cu_tv_x $$

Thank you

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    Thanks for fixing that @chinny84. should have known it was latex coding. – 2017-02-08
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    So if $u,v$ satisfy the wave equation, applying $L$ to $u_{tt}-c^2 u_{xx}$ means you need to evaluate \begin{align} L(u_{t}v_{t}-c^{2} u_{x}v_{x}) &= (u_{t}v_{t}-c^{2} u_{x}v_{x})_{tt}-c^{2}(u_{t}v_{t}-c^{2} u_{x}v_{x})_{xx} \\ \end{align} – 2017-02-08
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    How would i evaluate that? – 2017-02-08
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    After a few attempts using the product rule, you'll find that you can use the fact that both $L(u),L(v)$ vanish and things will begin to cancel. It may be long winded but it's a good exercise in the product rule, if nothing else! – 2017-02-08
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    For example $(u_t v_t)_{tt} = (u_{tt}v_{t}+u_{t}v_{tt})_{t}= \ldots$ and soon the wave equations on both $u,v$ are present – 2017-02-08
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    ahh okay i understand. I will try that and see if i come up with that solution – 2017-02-08
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    @Tilly - Everyone gets one edit from me! :) – 2017-02-08
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    @Tilly How did you do Tilly, all good? – 2017-02-09

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