Expressing the wave equation $\partial_x^2 u - \partial_y^2 u = 0$ in different forms. Show that the wave equation $\partial_x^2 u - \partial_y^2 u = 0$ can be written


*

*in factored form: $(\partial_x - \partial_y)(\partial_x + \partial_y) = 0$

*as a first-order system $u_x = v_y, u_y = v_x$

*as $u_{st} = 0$ under the change of variables $s = x + y$ and $t = x - y$. Use the form (c) to determine the general solution.


Attempt:


*

*$\partial_x^2 u - \partial_y^2 u = \partial_x u_x - \partial_y u_y + \partial_x \dot \partial_y - \partial_y \partial_x = (\partial_x - \partial_y)(\partial_x + \partial_y)$. Can I assume smoothness to change the order of differentiation?

*I am not certain how to use the provided hint: $u_x = v_y, u_y = v_x$.

*Same as number 2, I am completely lost.

 A: For the second part, they have used a confusing notation, because the $u$ in the desired form is not the same as the $u$ in the original equation.  I will show how to get the coupled equations 
$w_x = v_y, w_y = v_x$.
Let $v(x,y) = \frac{\partial u(x,y)}{\partial x}$ and let $w(x,y) = \frac{\partial u(x,y)}{\partial y}$. Then 
$$w_x = \frac{\partial^2 u(x,y)}{\partial x\partial y}\\
v_y = \frac{\partial^2 u(x,y)}{\partial y\partial x} $$
and making a smoothness assumption, you can change the order of partial differentiation, so these two expressions are equal: $w_x = v_y$.  And
$$w_y = \frac{\partial^2 u(x,y)}{\partial y^2}\\
v_x = \frac{\partial^2 u(x,y)}{\partial x^2}  \\
v_x - w_y = \frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial y^2} = 0
$$
For the third part, $x=\frac{s+t}{2}$ and $y=\frac{s-t}{2}$.
Then
$$
\frac{\partial u}{\partial x} = \frac12 \left( \frac{\partial u}{\partial s} + \frac{\partial u}{\partial t } \right) \\
\frac{\partial u}{\partial y} = \frac12 \left( \frac{\partial u}{\partial s} - \frac{\partial u}{\partial t } \right) 
$$
whence
$$
\frac{\partial^2 u}{\partial x^2} = \frac14 \left( \frac{\partial^2 u}{\partial s^2} + 2\frac{\partial^ u}{\partial s \partial t }  + \frac{\partial^2 u}{\partial t^2} \right)\\
\frac{\partial^2 u}{\partial y^2} = \frac14 \left( \frac{\partial^2 u}{\partial s^2} - 2\frac{\partial^ u}{\partial t \partial s }  + \frac{\partial^2 u}{\partial t^2} \right)\\
u_{xx}-u_{yy} = \frac14 \left(4\frac{\partial^ u}{\partial s \partial t } \right) = u_{st}
$$
The solution to $u_{st}=0$ is $u(s,t)=f(s)+g(t)$, for arbitrary differentiable functions $f$ and $g$.  This translates to a solution to the original equation
$$ u(x,y) = f(x+y) + g(x-y)$$
Try it out on some particular cases, like $u(x,y) = (x+y)^3 - 2(x-y)^2$ and you will see ilt alwyas satisfies the original equation.  
