Change of Variables in L[u]: Hyperbolic Transformation to Wave Operator

[stunner5000pt]

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

show that if L is hyperbolic then and A is not zero the transofmartion to moving coordinates
x&#039; = x - \frac{B}{2A} t
t&#039; = 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&#039;}{\partial u} and use this identity that
\frac{\partial u}{\partial x} = \frac{1}{\frac{\partial x}{\partial u}}

but I am not sure how to proceed from there
please help

 

[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&#039;}\cdot \frac{\partial x&#039;}{\partial x} + \frac{\partial u}{\partial t&#039;}\cdot \frac{\partial t&#039;}{\partial x}
\frac{\partial u}{\partial t} = \frac{\partial u}{\partial x&#039;}\cdot \frac{\partial x&#039;}{\partial t} + \frac{\partial u}{\partial t&#039;}\cdot \frac{\partial t&#039;}{\partial t}

You can get \frac{\partial x&#039;}{\partial x},\frac{\partial x&#039;}{\partial t},\frac{\partial t&#039;}{\partial x},\frac{\partial t&#039;}{\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<u> = \left(C - \frac{B^2}{4A}\right)\cdot \frac{\partial ^2 u}{\partial x&#039;^2} + A\cdot \frac{\partial ^2 u}{\partial t&#039;^2} = 0</u>
(not checked)

 

[saltydog]

define L<u> = a \frac{\partial^2u}{\partial t^2} + B \frac{\partial^2 u}{\partial x \partial t} + C \frac{\partial^2u}{\partial x^2} = 0 </u>
show that if L is hyperbolic then and A is not zero the transofmartion to moving coordinates
x&#039; = x - \frac{B}{2A} t
t&#039; = t

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.

 

[Fermat]

Really I think that x prime, y prime makes it more confussing...

Agreed.

Normally, I would use,

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

 

[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^{&#039;}=x-\frac{B}{2A}t

t^{&#039;}=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&#039;^2}+A\frac{\partial^2 u}{\partial t&#039;^2}=0

or:

A\frac{\partial^2 u}{\partial t&#039;^2}+\left(\frac{4AC-B^2}{4A}\right)\frac{\partial^2 u}{\partial x&#039;^2}=0

A\left[\frac{\partial^2 u}{\partial t&#039;^2}+\left(\frac{4AC-B^2}{4A^2}\right)\frac{\partial^2 u}{\partial x&#039;^2}\right]=0

A\left[\frac{\partial^2 u}{\partial t&#039;^2}-\left(\frac{B^2-4AC}{4A^2}\right)\frac{\partial^2 u}{\partial x&#039;^2}\right]=0

A\left[\frac{\partial^2}{\partial t&#039;^2}-\left(\frac{\sqrt{B^2-4AC}}{2A}\right)^2\frac{\partial^2}{\partial x&#039;^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&gt;0