Derivative with respect to $x + t$

Question

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.

Answer 1

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 $v_{\xi\eta}=0$.

Answer 2

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,$\frac {\partial^2 u}{{\partial x}{\partial t}} = \frac {\partial^2 u}{{\partial x}{\partial t}} \Rightarrow \frac {\partial^2 u}{{\partial x}{\partial t}} - \frac {\partial^2 u}{{\partial x}{\partial t}}= 0$

Answer 3

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.