$\frac{\partial^2 u}{\partial t^2}-c^2\frac{\partial^2u}{\partial x^2}=0\text{ implies } \frac{\partial^2 u}{\partial z \partial y}=0$

Question

I have the following question:

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Show that if $z=x+ct$ and $y=x-ct$ then:

$$\frac{\partial^2 u}{\partial t^2}-c^2\frac{\partial^2u}{\partial x^2}=0\text{ implies } \frac{\partial^2 u}{\partial z \partial y}=0$$

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If I had $u$ it would follow trivially I imagine, but I don't have $u$. It would seem that $u$ is linear in $t,x$ and since $t,x$ are linear in $y,z$ it seems logically consistent, but I am not sure what to do. Perhaps this is related to harmonic functions.

What should I first do to get this started?

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Perhaps I notice that $c=\frac{z-x}{t} \text{ and } c=\frac{x-y}{t}$

and thus $\frac{z-x}{t} =\frac{x-y}{t}\implies x-y=z-x\implies 2x=y+z$

Answer 1

Compute the $(x,t)$ derivatives in terms of the new variables $(z,y)$ by means of your given transformation. Then, the chain rule reads:

\begin{align} \frac{\partial u}{\partial t} = & \frac{\partial u}{\partial z } \frac{\partial z }{\partial t } + \frac{\partial u}{\partial y } \frac{\partial y }{\partial t } = c \, (u_z - u_y) \\ \frac{\partial u}{\partial x} = & \frac{\partial u}{\partial z } \frac{\partial z }{\partial x } + \frac{\partial u}{\partial y } \frac{\partial y }{\partial x } = u_z + u_y \\ \frac{\partial^2 u}{\partial t^2} = & \frac{\partial}{\partial t}\left( c \, (u_z - u_y) \right) = c^2 (u_{zz} -2 u_{zy} + u_{yy}) \end{align}

Can you derive the second $x$-derivative and substitute back in the original PDE?

Hope this helps. Cheers!

Answer 2

Write the variables $x$ and $t$ in terms of $z$ and $y$. Next use the chain rule by writing

$$u(x,t) = u(\phi(z,y), \psi(z,y)).$$

Next compute

$$\frac{\partial}{\partial y} u(\phi(z,y), \psi(z,y))$$

and

$$\frac{\partial}{\partial z} \left( \frac{\partial}{\partial y} u(\phi(z,y), \psi(z,y)) \right).$$

You will use the equivalent form of the wave equation given by

$$\frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} =0.$$