Proving a Solution of \[u_{tt} - c^2 u_{xx} = 0\]

[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 don't 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?
Thanks in advance.

 

[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.

 

[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...