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Let the convection equation $\frac{\delta u}{\delta t} + c\frac{\delta u}{\delta x} = 0$. I want to show that $$\frac{\delta^2 u}{\delta t^2}-c\frac{\delta^2 u}{\delta x^2}=0$$

I think it would be possible to differentiate the original convection equation with respect to t and x. Then the following two equations can be obtained. $$\frac{\delta^2 u}{\delta t^2} + c\frac{\delta^2 u}{\delta t \delta x}=0$$ $$\frac{\delta^2 u}{\delta x \delta t} + c\frac{\delta^2 u}{\delta x^2}=0$$ If the two mixed derivatives can be interchanged, I can conclude $\frac{\delta^2 u}{\delta t^2}-c\frac{\delta^2 u}{\delta x^2}=0$ by subtracting above two equations, but I don't know if it is possible to interchange it. Is it possible?

alryosha
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    This is true if your solution is regular enough. And given that only a shift takes place you need regular initial datum – GReyes Dec 12 '19 at 02:36
  • Could you elaborate more what's the meaning of "regular enough"? – alryosha Dec 12 '19 at 02:40
  • For example if $u$ is not even twice differentiable, then there is no way to conclude the equation you want. – Arctic Char Dec 12 '19 at 02:54
  • You need $u$ to be twice differentiable. What you are saying is essentially this: let $u=f(x-ct)$ be a solution to the transport equation with twice differentiable $f$. Then it is a solution to the wave equation (classical solutions to the wave equation have the form $u=f(x-ct)+g(x+ct)$ with twice differentiable $f$ and $g$). Here you do not have a backwards moving wave, $g=0$ and the one moving forward is smooth enough. – GReyes Dec 12 '19 at 06:50

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