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Change of variables

  1. Dec 7, 2005 #1
    define [tex] 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 [/tex]

    show that if L is hyperbolic then and A is not zero the transofmartion to moving coordinates
    [tex] x' = x - \frac{B}{2A} t [/tex]
    [tex] t' = t [/tex]
    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
    [tex] \frac{\partial x}{\partial u} [/tex] amd [tex] \frac{\partial t'}{\partial u} [/tex] and use this identitity that
    [tex] \frac{\partial u}{\partial x} = \frac{1}{\frac{\partial x}{\partial u}} [/tex]

    but im not sure how to proceed from there
    please help
     
  2. jcsd
  3. Dec 7, 2005 #2

    Fermat

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    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,

    [tex]\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}[/tex]
    [tex]\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}[/tex]

    You can get [tex]\frac{\partial x'}{\partial x},\frac{\partial x'}{\partial t},\frac{\partial t'}{\partial x},\frac{\partial t'}{\partial t}[/tex] 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,

    [tex]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[/tex]
    (not checked)
     
    Last edited: Dec 7, 2005
  4. Dec 7, 2005 #3

    saltydog

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

    [tex]w=x-\frac{B}{2A}t[/tex]

    [tex]z=t[/tex]

    Easier to follow this way and less likely to make a mistake by forgetting a prime or putting one where one doesn't belong.
     
  5. Dec 7, 2005 #4

    Fermat

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    Agreed.

    Normally, I would use,

    [tex]\zeta = \zeta (x,t)[/tex]
    [tex]\phi = \phi(x,t)[/tex]
     
  6. Dec 8, 2005 #5

    saltydog

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    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:

    [tex]x^{'}=x-\frac{B}{2A}t[/tex]

    [tex]t^{'}=t[/tex]

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

    [tex]\left(\frac{\partial^2}{\partial t^2}-v^2\frac{\partial^2}{\partial x^2}\right)u=0[/tex]

    Fermat stated above which I verified:

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

    or:

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

    [tex]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[/tex]

    [tex]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[/tex]

    [tex]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[/tex]

    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:

    [tex]B^2-4AC>0[/tex]
     
    Last edited: Dec 8, 2005
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