Derivative with respect to $x + t$ I am reading through Princeton's lectures in analysis and I am on the 10th page of the first book on Fourier series. 
In analyzing the wave equation, they state that $\xi = x + t $ and $\eta = x -t$ and $v(\xi,\eta) = u(x,t)$ and then $$ \frac{\partial^2 v}{\partial \xi \partial \eta} = 0.$$
The partials are confusing me, but I need to know why that equates to zero. It might also be useful to note that $$ \frac{ \partial^2 u}{\partial x^2} = \frac{ \partial^2 u}{\partial t^2}. $$
Even a hint in the right direction would suffice. My differential calculus is rusty.
 A: This is just an application of the chain rule.
$$v_\xi = v_xx_\xi + v_t t_\xi = u_x + u_t \\ v_{\xi\eta} = (u_x + u_t)_\eta = u_{xx} x_{\eta} + u_{xt}{t_\eta}+u_{tx}x_\eta + u_{tt}t_\eta = (x_\eta+t_\eta)(u_{xx}+u_{xt})$$
But $x_\eta+t_\eta = (1)+(-1) = 0$.  Therefore $v_{\xi\eta}=0$.
A: This is one big application of the chain rule!
Note:
$x= \xi - t \Rightarrow \frac {\partial x}{\partial \xi} = 1$,
$x = \eta +t \Rightarrow \frac {\partial x}{\partial \eta} = 1$
$t=\xi -x \Rightarrow \frac {\partial t}{\partial \xi} = 1$,
$t=x-\eta \Rightarrow \frac {\partial t}{\partial \eta} = -1$
Now,
$\frac {\partial v}{\partial \eta} = \frac {\partial u}{\partial \eta} = \frac {\partial u}{\partial t}\frac {\partial t}{\partial \eta} + \frac {\partial u}{\partial x}\frac {\partial x}{\partial \eta}  = -\frac {\partial u}{\partial t} +  \frac {\partial u}{\partial x}$ 
$\frac {\partial^2 v}{{\partial \xi}{\partial \eta}} = -\frac {\partial^2 u}{{\partial \xi}{\partial t}} + \frac {\partial^2 u}{{\partial \xi}{\partial x}} = -\frac {\partial^2 u}{{\partial^2 t}}\frac {\partial t}{\partial \xi} -\frac {\partial^2 u}{{\partial x}{\partial t}}\frac {\partial x}{\partial \xi} + \frac {\partial^2 u}{{\partial^2 x}}\frac {\partial x}{\partial \xi} + \frac {\partial^2 u}{{\partial t}{\partial x}}\frac {\partial t}{\partial \xi} $
$= -\frac {\partial^2 u}{{\partial^2 t}} + \frac {\partial^2 u}{{\partial^2 x}} + \frac {\partial^2 u}{{\partial t}{\partial x}} - \frac {\partial^2 u}{{\partial x}{\partial t}} = 0 + \frac {\partial^2 u}{{\partial x}{\partial t}} - \frac {\partial^2 u}{{\partial x}{\partial t}} = 0 $
Since u satisfies the wave equation, $\frac {\partial^2 u}{{\partial^2 t}} = \frac {\partial^2 u}{{\partial^2 x}} \Rightarrow -\frac {\partial^2 u}{{\partial^2 t}} + \frac {\partial^2 u}{{\partial^2 x}} = 0$
and because u is continuous,$\frac {\partial^2 u}{{\partial x}{\partial t}} = \frac {\partial^2 u}{{\partial x}{\partial t}} \Rightarrow \frac {\partial^2 u}{{\partial x}{\partial t}} - \frac {\partial^2 u}{{\partial x}{\partial t}}= 0$   
A: Chain Rule:
$$
\begin{align}
\frac{\partial}{\partial x}
&=\frac{\partial\xi}{\partial x}\frac{\partial }{\partial\xi}+\frac{\partial\eta}{\partial x}\frac{\partial }{\partial\eta}\tag{1}\\
\frac{\partial}{\partial t}
&=\frac{\partial\xi}{\partial t}\frac{\partial }{\partial\xi}+\frac{\partial\eta}{\partial t}\frac{\partial }{\partial\eta}\tag{2}
\end{align}
$$
Composing $(1)$ with $(1)$:
$$
\begin{align}
\frac{\partial^2}{\partial x^2}
&=\frac{\partial^2\xi}{\partial x^2}\frac{\partial }{\partial\xi}+\frac{\partial^2\eta}{\partial x^2}\frac{\partial }{\partial\eta}\\
&+\left(\frac{\partial\xi}{\partial x}\right)^2\frac{\partial^2 }{\partial\xi^2}+2\frac{\partial\xi}{\partial x}\frac{\partial\eta}{\partial x}\frac{\partial^2}{\partial\xi\partial\eta}+\left(\frac{\partial\eta}{\partial x}\right)^2\frac{\partial^2 }{\partial\eta^2}\tag{3}
\end{align}
$$
Composing $(2)$ with $(2)$:
$$
\begin{align}
\frac{\partial^2}{\partial t^2}
&=\frac{\partial^2\xi}{\partial t^2}\frac{\partial }{\partial\xi}+\frac{\partial^2\eta}{\partial t^2}\frac{\partial }{\partial\eta}\\
&+\left(\frac{\partial\xi}{\partial t}\right)^2\frac{\partial^2 }{\partial\xi^2}+2\frac{\partial\xi}{\partial t}\frac{\partial\eta}{\partial t}\frac{\partial^2}{\partial\xi\partial\eta}+\left(\frac{\partial\eta}{\partial t}\right)^2\frac{\partial^2 }{\partial\eta^2}\tag{4}
\end{align}
$$
since
$$
\frac{\partial(\xi,\eta)}{\partial(x,t)}=\begin{bmatrix}1&1\\1&-1\end{bmatrix}\tag{5}
$$
is constant, higher derivatives must be $0$. Applying $(5)$ to $(3)$ and $(4)$ gives
$$
\begin{align}
\frac{\partial^2}{\partial x^2}
&=\frac{\partial^2 }{\partial\xi^2}+2\frac{\partial^2}{\partial\xi\partial\eta}+\frac{\partial^2 }{\partial\eta^2}\tag{6}
\end{align}
$$
and
$$
\begin{align}
\frac{\partial^2}{\partial t^2}
&=\frac{\partial^2 }{\partial\xi^2}-2\frac{\partial^2}{\partial\xi\partial\eta}+\frac{\partial^2 }{\partial\eta^2}\tag{7}
\end{align}
$$
Therefore, the Laplacian becomes
$$
\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial t^2}
=2\left(\frac{\partial^2}{\partial\xi^2}+\frac{\partial^2}{\partial\eta^2}\right)\tag{8}
$$
and the Wave Operator becomes
$$
\frac{\partial^2}{\partial x^2}-\frac{\partial^2}{\partial t^2}
=4\frac{\partial^2}{\partial\xi\partial\eta}\tag{9}
$$
Equation $(9)$ gives the Wave Operator in the rotated coordinates.
