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Math Help - 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

    \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

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

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

    \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

    <br />
\beta = -3/2 \; ?<br />

    <br />
u_x=w_x+\beta w<br />

    <br />
u_{xx}=w_{xx}+2 \beta w_x +\beta^2w<br />

    <br />
u_{yy}=w_{yy}<br />
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  3. #3
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    Quote Originally Posted by zzzoak View Post
    If

    <br />
\beta = -3/2 \; ?<br />
    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

    <br />
u_x=w_x+\beta w<br />

    <br />
u_{xx}=w_{xx}+2 \beta w_x +\beta^2w<br />

    <br />
u_{yy}=w_{y}<br />
    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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    <br />
u=w(x,y) \; exp(\beta x)<br />

    <br />
\frac{d}{dx}u=(\frac{d}{dx}w)exp(\beta x)+w(\frac{d}{dx}exp(\beta x))<br />
    Last edited by zzzoak; January 12th 2011 at 04:07 PM.
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