I can't find any way to get to that expression. Honestly I don't how to start this. Any recommendations?
Consider the One Dimension Wave Equation $$ \frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}} $$ where $u:=u(x, t) .$ If now we make a change of variable $\xi=x+c t, \eta=x-c t,$ show that the wave equation can be written as $$\frac{\partial^{2} u}{\partial \eta \partial \xi}=0$$


If you want to approach this from the other direction and take the brute force approach, note that by the chain rule,

  • $\frac{\partial u}{\partial t} = \frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial t}+\frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial t}$,
  • $\frac{\partial u}{\partial x} = \frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial x}+\frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial x}$.

Since $\xi=x+c t, \eta=x-c t$,

$ \frac{\partial \eta}{\partial t}=-c, \frac{\partial \eta}{\partial x}=1,\\ \frac{\partial \xi}{\partial t}=c, \frac{\partial \xi}{\partial x}=1. $

Replacing these in the first two equations:

  • $\frac{\partial u}{\partial t} = -c\frac{\partial u}{\partial \eta}+c\frac{\partial u}{\partial \xi}$,
  • $\frac{\partial u}{\partial x} = \frac{\partial u}{\partial \eta}+\frac{\partial u}{\partial \xi}$.

Now, take the derivatives again,

  • $\frac{\partial^2 u}{\partial t^2} = -c\frac{\partial^2 u}{\partial \eta \partial t}+c\frac{\partial^2 u}{\partial \xi \partial t}$,
  • $\frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 u}{\partial \eta \partial x}+\frac{\partial^2 u}{\partial \xi \partial x},$

and use the chain rule again:

  • $\frac{\partial^2 u}{\partial t^2} = c^2\frac{\partial^2 u}{\partial^2 \eta}-c^2\frac{\partial^2 u}{\partial \eta \partial \xi}-c^2\frac{\partial^2 u}{\partial \xi \partial \eta}+c^2\frac{\partial^2 u}{\partial^2 \xi }$,
  • $\frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 u}{\partial^2 \eta} +\frac{\partial^2 u}{\partial \eta \partial \xi} +\frac{\partial^2 u}{\partial \xi \partial \eta} +\frac{\partial^2 u}{\partial^2 \xi}$.

Finally, replace these in $$\frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}}$$ to get $$ c^2\frac{\partial^2 u}{\partial^2 \eta}-c^2\frac{\partial^2 u}{\partial \eta \partial \xi}-c^2\frac{\partial^2 u}{\partial \xi \partial \eta}+c^2\frac{\partial^2 u}{\partial^2 \xi } = \frac{\partial^2 u}{\partial x^2} = c^2\frac{\partial^2 u}{\partial^2 \eta} +c^2\frac{\partial^2 u}{\partial \eta \partial \xi} +c^2\frac{\partial^2 u}{\partial \xi \partial \eta} +c^2\frac{\partial^2 u}{\partial^2 \xi},$$ which simplifies to $$\frac{\partial^2 u}{\partial \xi \partial \eta}=0$$ after cancellations.

  • $\begingroup$ The proof from the other side is far shorter. $\endgroup$
    – ConvexHull
    Nov 11 '20 at 15:09
  • $\begingroup$ True, though you would have to start from this side if you didn't know where you would end up. (I realize the question gives it to you in this case.) $\endgroup$ Nov 11 '20 at 19:59
  • $\begingroup$ True! Otherwise you are not able to proof anything. $\endgroup$
    – ConvexHull
    Nov 11 '20 at 20:53

The solution is quite simple: Try to substitude $\xi$ and $\eta$ in

$\frac{\partial^2 u}{\partial \eta \partial \xi}=0$,

with the given relations and try to recover your first equation. The rest is up to you. It is done in one second.



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