# 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