Solving Klein-Gordon PDE w/ Change of Variables

[yonatan]

Hi.
I'm following the solution of a Klein-Gordon PDE in a textbook. The equation is
$$\begin{align}<br /> k_{xx}(x,y) - k_{yy}(x,y) &= \lambda k(x,y) \\<br /> k(x,0) &= 0 \\<br /> k(x,x) &= - \frac{\lambda}{2} x<br /> \end{align}$$
The book uses a change of variables
$$\xi = x+y, \eta = x-y$$
to write
$$\begin{align}<br /> k(x,y) &= G(\xi,\eta)\\<br /> k_{xx} &= G_{\xi \xi} + 2G_{\xi \eta} + G_{\eta \eta}\\<br /> k_{yy} &= G_{\xi \xi} - 2G_{\xi \eta} + G__{\eta \eta}<br /> \end{align}$$
and then they write the original PDE as
$$\begin{align}<br /> G_{\xi \eta}(\xi,\eta) &= \frac{\lambda}{4} G(\xi,\eta),\\<br /> G(\xi,\xi) &= 0,\\<br /> G(\xi,0) &= - \frac{\lambda}{4} \xi<br /> \end{align}$$
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.

 

[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..

 

[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?

 

[Thaakisfox]

Absolutely! :)