Simple PDE using Chain Rule (Galilean transformation)

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I am trying to assist another student with a question but have just come to a roadblock, so to speak.

The problem is to use the 2D wave equation:

$$u_{tt} = c^2u_{xx}$$

and the transforms $p = x-ct, q = x+ct$

and that $u(x,t) = \hat{u}(p,q)$

To show that $\hat{u}_{pq} = 0$

Some hints would be helpful. Thank you.

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First, this is the 1D wave equation.

Next, using the change of variables $p=x-ct, q=x+ct$, we see that \begin{align} x= \frac{1}{2}(q+p), \ \ t=\frac{1}{2c}(q-p) \end{align} which means \begin{align} \frac{\partial }{\partial p} =& \frac{\partial t}{\partial p}\frac{\partial}{\partial t}+\frac{\partial x}{\partial p}\frac{\partial}{\partial x} = -\frac{1}{2c}\frac{\partial}{\partial t}+\frac{1}{2}\frac{\partial}{\partial x}\\ \frac{\partial }{\partial q} =& \frac{\partial t}{\partial q}\frac{\partial}{\partial t}+\frac{\partial x}{\partial q}\frac{\partial}{\partial x} = \frac{1}{2c}\frac{\partial}{\partial t}+\frac{1}{2}\frac{\partial}{\partial x}. \end{align} Thus, it follows \begin{align} \frac{\partial^2}{\partial p \partial q}\hat u = \frac{1}{4}\left( -\frac{1}{c}\frac{\partial}{\partial t}+\frac{\partial}{\partial x}\right) \left( \frac{1}{c}\frac{\partial}{\partial t}+\frac{\partial}{\partial x}\right)u = -\frac{1}{4c^2}(u_{tt}-c^2u_{xx}) =0. \end{align}