Change of variables

1. Dec 7, 2005

stunner5000pt

define $$L = a \frac{\partial^2u}{\partial t^2} + B \frac{\partial^2 u}{\partial x \partial t} + C \frac{\partial^2u}{\partial x^2} = 0$$

show that if L is hyperbolic then and A is not zero the transofmartion to moving coordinates
$$x' = x - \frac{B}{2A} t$$
$$t' = t$$
tkaes L into a multiple of the wave operator
now how would igo about changing the variables in L to x' and t'?

i mean i could certainly find out
$$\frac{\partial x}{\partial u}$$ amd $$\frac{\partial t'}{\partial u}$$ and use this identitity that
$$\frac{\partial u}{\partial x} = \frac{1}{\frac{\partial x}{\partial u}}$$

but im not sure how to proceed from there

2. Dec 7, 2005

Fermat

I'm afraid I'm not sure what is meant by "multiple of the wave operator", but when doing these things I use the chain rule procedure like this,

$$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial x'}\cdot \frac{\partial x'}{\partial x} + \frac{\partial u}{\partial t'}\cdot \frac{\partial t'}{\partial x}$$
$$\frac{\partial u}{\partial t} = \frac{\partial u}{\partial x'}\cdot \frac{\partial x'}{\partial t} + \frac{\partial u}{\partial t'}\cdot \frac{\partial t'}{\partial t}$$

You can get $$\frac{\partial x'}{\partial x},\frac{\partial x'}{\partial t},\frac{\partial t'}{\partial x},\frac{\partial t'}{\partial t}$$ from your transformation expressions.

Then you can continue to work out the 2nd partial derivatives and substitute into the original expression for L.

I got,

$$L = \left(C - \frac{B^2}{4A}\right)\cdot \frac{\partial ^2 u}{\partial x'^2} + A\cdot \frac{\partial ^2 u}{\partial t'^2} = 0$$
(not checked)

Last edited: Dec 7, 2005
3. Dec 7, 2005

saltydog

Really I think that x prime, y prime makes it more confussing. Better to let:

$$w=x-\frac{B}{2A}t$$

$$z=t$$

Easier to follow this way and less likely to make a mistake by forgetting a prime or putting one where one doesn't belong.

4. Dec 7, 2005

Fermat

Agreed.

Normally, I would use,

$$\zeta = \zeta (x,t)$$
$$\phi = \phi(x,t)$$

5. Dec 8, 2005

saltydog

Alright, now I understand what's goin' on and I wish for the record to modify my statements above and perhaps help Stunner as well: The use of the notation:

$$x^{'}=x-\frac{B}{2A}t$$

$$t^{'}=t$$

maintains the standard notation for the wave equation which I'll write in operator notation:

$$\left(\frac{\partial^2}{\partial t^2}-v^2\frac{\partial^2}{\partial x^2}\right)u=0$$

Fermat stated above which I verified:

$$\left(C-\frac{B^2}{4A}\right)\frac{\partial^2 u}{\partial x'^2}+A\frac{\partial^2 u}{\partial t'^2}=0$$

or:

$$A\frac{\partial^2 u}{\partial t'^2}+\left(\frac{4AC-B^2}{4A}\right)\frac{\partial^2 u}{\partial x'^2}=0$$

$$A\left[\frac{\partial^2 u}{\partial t'^2}+\left(\frac{4AC-B^2}{4A^2}\right)\frac{\partial^2 u}{\partial x'^2}\right]=0$$

$$A\left[\frac{\partial^2 u}{\partial t'^2}-\left(\frac{B^2-4AC}{4A^2}\right)\frac{\partial^2 u}{\partial x'^2}\right]=0$$

$$A\left[\frac{\partial^2}{\partial t'^2}-\left(\frac{\sqrt{B^2-4AC}}{2A}\right)^2\frac{\partial^2}{\partial x'^2}\right]u=0$$

Thus being the wave operator multiplied by A. Note also how this form shows why the requirement that the mixed-partial operator must be hyperbolic, that is:

$$B^2-4AC>0$$

Last edited: Dec 8, 2005