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 \[u_{tt} - c^2 u_{xx} = 0<br /> \], and suppose \[u\left( {x,t} \right)\] is constant along the line \[<br /> x = 2 + ct<br /> \]<br />. Then \[u\left( {x,t} \right)\] must keep:\[<br /> u_t + cu_x = 0<br /> \]<br />
I can prove it for any point \[<br /> \left( {x,t} \right)<br /> \]<br /> which is right to the line \[<br /> x = 2 - ct<br /> \]<br />. 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 \[<br /> x = 2 + ct<br /> \]<br />.
\[<br /> \begin{array}{l}<br /> \frac{d}{{dt}}\left( {u\left( {2 + ct,t} \right)} \right) = 0 \\ <br /> \frac{{\partial \left( {u\left( {x\left( t \right),t} \right)} \right)}}{{\partial x}}\frac{{\partial x\left( t \right)}}{{\partial t}} + u_t \left( {2 + ct,t} \right) = 0 \\ <br /> u_t \left( {2 + ct,t} \right) + cu_x \left( {2 + ct,t} \right) = 0 \\ <br /> \end{array}<br /> \]<br />
I need to prove it for all \[<br /> \left( {x,t} \right) \in \Re ^2 <br /> \]<br />

It's driving me a little crazy...