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Waves Equation

  1. Apr 8, 2005 #1
    Hi everyone.

    I tried a bit, but got stuck.

    Let [tex]\[u\left( {x,t} \right)\][/tex] be a solution of [tex]\[u_{tt} - c^2 u_{xx} = 0
    \][/tex], and suppose [tex]\[u\left( {x,t} \right)\][/tex] is constant along the line [tex]\[
    x = 2 + ct
    \]
    [/tex]. Then [tex]\[u\left( {x,t} \right)\][/tex] must keep:[tex]\[
    u_t + cu_x = 0
    \]
    [/tex]
    I can prove it for any point [tex]\[
    \left( {x,t} \right)
    \]
    [/tex] which is right to the line [tex]\[
    x = 2 - ct
    \]
    [/tex]. 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?
    Thanks in advance.
     
    Last edited: Apr 8, 2005
  2. jcsd
  3. Apr 8, 2005 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    U know that

    [tex] u\left(2+ct,t\right)=C [/tex]

    Take the PD wrt "t"

    [tex]\frac{\partial u}{\partial (2+ct)}\frac{d(2+ct)}{dt}+\frac{\partial u}{\partial t} =0 [/tex]

    Equivalently,using that [itex]2+ct=x [/itex]

    [tex] \frac{\partial u}{\partial x} c+\frac{\partial u}{\partial t} = 0 [/tex]

    Q.e.d.

    Daniel.
     
  4. Apr 8, 2005 #3
    Well it's the first thing I did, but then it only proves it along the line [tex]\[
    x = 2 + ct
    \]
    [/tex].
    [tex]\[
    \begin{array}{l}
    \frac{d}{{dt}}\left( {u\left( {2 + ct,t} \right)} \right) = 0 \\
    \frac{{\partial \left( {u\left( {x\left( t \right),t} \right)} \right)}}{{\partial x}}\frac{{\partial x\left( t \right)}}{{\partial t}} + u_t \left( {2 + ct,t} \right) = 0 \\
    u_t \left( {2 + ct,t} \right) + cu_x \left( {2 + ct,t} \right) = 0 \\
    \end{array}
    \]
    [/tex]
    I need to prove it for all [tex]\[
    \left( {x,t} \right) \in \Re ^2
    \]
    [/tex]

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