# Klein-Gordon PDE

1. Apr 22, 2010

### yonatan

Hi.
I'm following the solution of a Klein-Gordon PDE in a textbook. The equation is
\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}
The book uses a change of variables
$$\xi = x+y, \eta = x-y$$
to write
\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}
and then they write the original PDE as
\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}
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. Apr 23, 2010

### Thaakisfox

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..

3. Apr 23, 2010

### yonatan

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?

4. Apr 23, 2010

### Thaakisfox

Absolutely! :)