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
$$\begin{align} k_{xx}(x,y) - k_{yy}(x,y) &= \lambda k(x,y) \\ k(x,0) &= 0 \\ k(x,x) &= - \frac{\lambda}{2} x \end{align}$$
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.

 

[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! :)

 

[blue_raver22]

Hello,

I can help you understand the derivation of the boundary conditions in the new PDE using the change of variables. First, let's take a look at the original PDE and its boundary conditions:
\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}

Using the change of variables $\xi = x+y$ and $\eta = x-y$, we can rewrite the original PDE as:
\begin{equation}
k_{\xi \xi}(\xi,\eta) - k_{\eta \eta}(\xi,\eta) = \lambda k(\xi,\eta)
\end{equation}

Now, let's look at the boundary conditions. The first one, $k(x,0) = 0$, can be rewritten as $k(\xi,\eta) = 0$ when $\eta = 0$ (since $\xi = x+y$ and $\eta = x-y$, $x = \frac{\xi+\eta}{2}$). This means that at the line $\eta = 0$, the function $k$ is equal to 0 for all values of $\xi$.

For the second boundary condition, $k(x,x) = - \frac{\lambda}{2} x$, we can use the same substitution to rewrite it as $k(\xi,\eta) = -\frac{\lambda}{2} \xi$ when $\xi = \eta$ (since $x = \frac{\xi+\eta}{2}$ and $y = \frac{\xi-\eta}{2}$). This means that at the line $\xi = \eta$, the function $k$ is equal to $-\frac{\lambda}{2} \xi$ for all values of $\xi$.

Substituting these boundary conditions into the rewritten PDE, we get:
\begin{align}
k_{\xi \xi}(\xi,\eta) - k_{\eta \eta}(\xi,\eta) &= \lambda k(\xi,\eta) \\
k(\xi,0) &= 0 \\
k(\xi,\xi) &= - \frac{\lambda}{2} \xi
\end{align}

I hope this helps you understand the