# Waves Equation

1. Apr 8, 2005

### Palindrom

Hi everyone.

I tried a bit, but got stuck.

Let $$$u\left( {x,t} \right)$$$ be a solution of $$$u_{tt} - c^2 u_{xx} = 0$$$, and suppose $$$u\left( {x,t} \right)$$$ is constant along the line $$$x = 2 + ct$$$. Then $$$u\left( {x,t} \right)$$$ must keep:$$$u_t + cu_x = 0$$$
I can prove it for any point $$$\left( {x,t} \right)$$$ which is right to the line $$$x = 2 - ct$$$. I dont see any way to prove it for the points left to that line.
Is there a simpler way, or more general one that doesn't make that last line special?

Last edited: Apr 8, 2005
2. Apr 8, 2005

### dextercioby

U know that

$$u\left(2+ct,t\right)=C$$

Take the PD wrt "t"

$$\frac{\partial u}{\partial (2+ct)}\frac{d(2+ct)}{dt}+\frac{\partial u}{\partial t} =0$$

Equivalently,using that $2+ct=x$

$$\frac{\partial u}{\partial x} c+\frac{\partial u}{\partial t} = 0$$

Q.e.d.

Daniel.

3. Apr 8, 2005

### Palindrom

Well it's the first thing I did, but then it only proves it along the line $$$x = 2 + ct$$$.
$$$\begin{array}{l} \frac{d}{{dt}}\left( {u\left( {2 + ct,t} \right)} \right) = 0 \\ \frac{{\partial \left( {u\left( {x\left( t \right),t} \right)} \right)}}{{\partial x}}\frac{{\partial x\left( t \right)}}{{\partial t}} + u_t \left( {2 + ct,t} \right) = 0 \\ u_t \left( {2 + ct,t} \right) + cu_x \left( {2 + ct,t} \right) = 0 \\ \end{array}$$$
I need to prove it for all $$$\left( {x,t} \right) \in \Re ^2$$$

It's driving me a little crazy... :surprised