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Klein-Gordon PDE

  1. Apr 22, 2010 #1
    Hi.
    I'm following the solution of a Klein-Gordon PDE in a textbook. The equation is
    [tex]\begin{align}
    k_{xx}(x,y) - k_{yy}(x,y) &= \lambda k(x,y) \\
    k(x,0) &= 0 \\
    k(x,x) &= - \frac{\lambda}{2} x
    \end{align}
    [/tex]
    The book uses a change of variables
    [tex]$\xi = x+y$, $\eta = x-y$[/tex]
    to write
    [tex]\begin{align}
    k(x,y) &= G(\xi,\eta)\\
    k_{xx} &= G_{\xi \xi} + 2G_{\xi \eta} + G_{\eta \eta}\\
    k_{yy} &= G_{\xi \xi} - 2G_{\xi \eta} + G__{\eta \eta}
    \end{align}[/tex]
    and then they write the original PDE as
    [tex]\begin{align}
    G_{\xi \eta}(\xi,\eta) &= \frac{\lambda}{4} G(\xi,\eta),\\
    G(\xi,\xi) &= 0,\\
    G(\xi,0) &= - \frac{\lambda}{4} \xi
    \end{align}
    [/tex]
    I'm fine with the first line in the new PDE, but the other two, the boundary conditions, i don't get how they arrive at.

    Can somebody help me understand? I'll be much appreciative :-)

    J.
     
    Last edited: Apr 22, 2010
  2. jcsd
  3. Apr 23, 2010 #2
    Basically just check the definition!

    k(x,y)=G(xi,eta)

    In the first boundary condition, k(x,0)=0.
    now what does y=0 mean? xi=x+y and eta=x-y, hence y=0 means: xi=x and eta=x, hence xi=eta, hence k(x,0)=G(xi,xi)=0

    similarly for the second one..
     
  4. Apr 23, 2010 #3
    I see, thanks. Would it then be the same, for the case k(x,0)=0, where y=0 and xi=eta, to write G(eta,eta)=0?
     
  5. Apr 23, 2010 #4
    Absolutely! :)
     
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