topsquark said:
I'm not prepared to discuss this fully, but this is not a simple matter of a change of variables. The function u(x, y) has to solve the Partial Differential Equation:
u_{tt} - c^2u_{xx} = 0
This is the 1D wave equation with wave propagation speed c. The change of variable of the characteristic lines leads to the replacement
x + ct = \xi and x - ct = \eta
which changes the PDE to
u_{\xi \eta} = 0
More explanation than this should probably be given by someone else until I bone up on the solution.
-Dan
I already solved that. I wondering how I get back to $u(x,t)$.
Here is where I am at
$$
\frac{\partial^2 u}{\partial\xi\partial\eta} = 0.
$$
Let's take the standard form of the wave equation $u_{tt} = c^2u_{xx}$.
Then by the chain rule
$$
\frac{\partial}{\partial x} = \frac{\partial\xi}{\partial x}\frac{\partial}{\partial\xi} + \frac{\partial\eta}{\partial x}\frac{\partial}{\partial\eta} = \frac{\partial}{\partial\xi} + \frac{\partial}{\partial\eta}
$$
and
$$
\frac{\partial}{\partial t} = \frac{\partial\xi}{\partial t}\frac{\partial}{\partial\xi} + \frac{\partial\eta}{\partial t}\frac{\partial}{\partial\eta} = c\frac{\partial}{\partial\xi} - c\frac{\partial}{\partial\eta}.
$$
Therefore, the second derivatives are
\begin{alignat*}{3}
\frac{\partial^2 u}{\partial x^2} & = & \frac{\partial^2 u}{\partial\xi^2} + 2\frac{\partial^2 u}{\partial\xi\partial\eta} + \frac{\partial^2 u}{\partial\eta^2}\\
\frac{\partial^2 u}{\partial t^2} & = & c^2\frac{\partial^2 u}{\partial\xi^2} - 2c^2\frac{\partial^2 u}{\partial\xi\partial\eta} + c^2\frac{\partial^2 u}{\partial\eta^2}
\end{alignat*}
Making the appropriate substitution, we have
\begin{alignat*}{3}
c^2\frac{\partial^2 u}{\partial\xi^2} - 2c^2\frac{\partial^2 u}{\partial\xi\partial\eta} + c^2\frac{\partial^2 u}{\partial\eta^2} & = & c^2\frac{\partial^2 u}{\partial\xi^2} + 2c^2\frac{\partial^2 u}{\partial\xi\partial\eta} + c^2\frac{\partial^2 u}{\partial\eta^2}\\
4c^2\frac{\partial^2 u}{\partial\xi\partial\eta} & = & 0\\
\frac{\partial^2 u}{\partial\xi\partial\eta} & = & 0
\end{alignat*}Show that by integrating twice, once with respect to $\eta$ and once with respect to $\xi$, that the general form for $u$ is
\begin{alignat*}{3}
u(\xi,\eta) & = & F(\xi) + G(\eta)\\
u(x,t) & = & F(x + ct) + G(x - ct)
\end{alignat*}
where the forms $F$ and $G$ are arbitrary. This result lies at the heart of the method of characteristics for linear wave motion.
\begin{alignat*}{3}
\iint u_{\xi\eta} d\eta d\xi & = & \iint 0 d\eta d\xi\\
\int u_{\xi} & = & \int f(\xi)\\
u(\xi,\eta) & = & F(\xi) + G(\eta)
\end{alignat*}