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Thread: Hyperbolic DE

  1. #1
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    Hyperbolic DE

    I don't understand the second step in the change of variables for this hyperbolic DE.

    After a change of variables and some algebra, we have

    $\displaystyle \mbox{2.3.11} \ \ u_{xx}-4u_{yy}+3u_{x}+u=0$

    I am then told that the second step is a change of dependent variable

    $\displaystyle \mbox{2.3.12} \ \ u=\exp{(\beta x)}\omega$,

    where beta is chosen so that in the transformed equation the coefficient of $\displaystyle \omega_{x}$ vanishes. Differentiating 2.3.12 and substituting in 2.3.11, we obtain for $\displaystyle \omega$ the equation

    $\displaystyle \omega_{xx}-4\omega_{yy}+(2\beta+3)\omega_{x}+(\beta^2+3\beta+ 1)\omega=0$

    I can't obtain this equation (directly above) when I follow the directions.
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  2. #2
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    If

    $\displaystyle
    \beta = -3/2 \; ?
    $

    $\displaystyle
    u_x=w_x+\beta w
    $

    $\displaystyle
    u_{xx}=w_{xx}+2 \beta w_x +\beta^2w
    $

    $\displaystyle
    u_{yy}=w_{yy}
    $
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  3. #3
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    Quote Originally Posted by zzzoak View Post
    If

    $\displaystyle
    \beta = -3/2 \; ?
    $
    The book has that but I can't get the omega equation which is the first priority.
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  4. #4
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    Quote Originally Posted by zzzoak View Post

    $\displaystyle
    u_x=w_x+\beta w
    $

    $\displaystyle
    u_{xx}=w_{xx}+2 \beta w_x +\beta^2w
    $

    $\displaystyle
    u_{yy}=w_{y}
    $
    How did you get that?
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  5. #5
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    Sorry, I write all derivatives without exp term.
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  6. #6
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    $\displaystyle
    u=w(x,y) \; exp(\beta x)
    $

    $\displaystyle
    \frac{d}{dx}u=(\frac{d}{dx}w)exp(\beta x)+w(\frac{d}{dx}exp(\beta x))
    $
    Last edited by zzzoak; Jan 12th 2011 at 03:07 PM.
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