2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
1
$\begingroup$

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.

Regards

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.