This is not a school problem. I just learning PDE on my own. The problem is:

Transform $\displaystyle \frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}$ in to $\displaystyle \frac{\partial^2 u}{\partial \alpha \partial \beta}=0$

where $\displaystyle u(x,t)=F(x+ct) + G(x-ct)$

and $\displaystyle \alpha = x+ct,\beta = x-ct$

$\displaystyle \alpha = x+ct,\beta = x-ct \Rightarrow x=\frac{\alpha + \beta}{2}, t=\frac{\alpha - \beta}{2c}$

$\displaystyle \alpha = x+ct,\beta = x-ct \Rightarrow \frac{\partial \alpha}{\partial x}= 1, \frac{\partial \alpha}{\partial t}= c,$ and also $\displaystyle \frac{\partial \beta}{\partial x}= 1, \frac{\partial \beta}{\partial t}= -c$

$\displaystyle \frac{\partial x}{\partial \alpha} = \frac{\partial (\alpha + \beta)}{2} = \frac{1}{2} [\frac{\partial \alpha}{\partial \alpha} + \frac{\partial \beta}{\partial \alpha} = \frac{1}{2}[ 1 + 0] = \frac{1}{2} $

$\displaystyle \Rightarrow \frac{\partial x}{\partial \alpha} = \frac{1}{2}$

$\displaystyle \frac{\partial u}{\partial \alpha}= \frac{\partial u}{\partial x} \frac{\partial x}{\partial \alpha} + \frac{\partial u}{\partial t} \frac{\partial t}{\partial \alpha} = \frac{\partial F(\alpha)}{\partial \alpha}+ \frac{\partial G(\beta)}{\partial \alpha}$

I gave this a lot of thoughts. I still cannot accept $\displaystyle \frac{\partial G(\beta)}{\partial \alpha}=0 $

Please bear with me. Let's take a look at this example:

Let $\displaystyle G(\beta)=-\beta = \beta - 2\beta = x-ct-2x+2ct=(x+ct)-2x=\alpha -2x$

$\displaystyle \Rightarrow\frac{\partial G(\beta)}{\partial \alpha}= \frac{\partial \alpha}{\partial \alpha} + \frac{\partial 2x}{\partial \alpha}= 1+1=2 $

Yes I know I am playing around and the argument is very thin, but never the less, it is valid. My whole point is just because $\displaystyle \alpha$ and $\displaystyle \beta$ are independent variable, $\displaystyle \frac{\partial G(\beta)}{\partial \alpha}$ not necessary equal 0.

$\displaystyle \frac{\partial \beta}{\partial \alpha}=0 $ do not imply at all $\displaystyle \frac{\partial G(\beta)}{\partial \alpha}=0 $

I don't think the question is good to say $\displaystyle \frac{\partial^2 u}{\partial \alpha \partial \beta}=0$ I have been struggling on this very point for two days!!!!

Please tell me if I am correct.

Thanks a million.

Alan