• Feb 9th 2010, 11:09 AM
yungman
This is not a school problem. I just learning PDE on my own. The problem is:

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

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

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

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

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

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

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

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

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

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

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

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

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 $\alpha$ and $\beta$ are independent variable, $\frac{\partial G(\beta)}{\partial \alpha}$ not necessary equal 0.

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

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

Please tell me if I am correct.
Thanks a million.
Alan
• Feb 9th 2010, 09:00 PM
yungman