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Math Help - Hyperbolic second order

  1. #1
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    Hyperbolic second order

    u_{xx}-3u_{yy}+2u_x-u_y+u=0

    u(x,y)=u(\xi\cos{\alpha}-\eta\sin{\alpha},\xi\sin{\alpha}+\eta\cos{\alpha})  =\omega(\xi,\eta)

    \displaystyle\omega_{\xi\xi}(\cos^2{\alpha}-3\sin^2{\alpha})+\omega_{\eta\eta}(\sin^2{\alpha}-3\cos^s{\alpha})-8\omega_{\xi\eta}\sin{\alpha}\cos{\alpha}+\omega_{  \xi}(2\cos{\alpha}-\sin{\alpha})-\omega_{\eta}(2\sin{\alpha}+\cos{\alpha})+\omega=0

    \cos{\alpha}=1 \ \ \ \sin{\alpha}=0

    \omega_{\xi\xi}-3\omega_{\eta\eta}+2\omega_{\xi}-\omega_{\eta}+\omega=0

    u_{xx}-3u_{yy}+2u_x-u_y+u=0

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

    \exp{(\beta x)}[\omega_{xx}-3\omega_{yy}+\omega_x(2\beta-3)-\omega_y+\omega(\beta^2+\beta+1)]=0

    \displaystyla\beta=\frac{3}{2}

    \displaystyle\omega_{xx}-3\omega_{yy}-\omega_y+\frac{19}{4}\omega=0

    I wasn't able to eliminate omega y. If I do u=\exp{(\beta y)}\omega(x,y), I won't be able to eliminate omega x.

    Should I continue or how should I approach this problem?
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  2. #2
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    For equation

    <br />
u_{xx}-3u_{yy}+2u_x-u_y+u=0<br />

    you may try this

    <br />
u=\exp{(\alpha x)} \exp{(\beta y)} \: \omega(x,y).<br />
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  3. #3
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    Quote Originally Posted by zzzoak View Post
    For equation

    <br />
u_{xx}-3u_{yy}+2u_x-u_y+u=0<br />

    you may try this

    <br />
u=\exp{(\alpha x)} \exp{(\beta y)} \: \omega(x,y).<br />
    So then I am setting w_x and w_y = 0 to obtain the respected alpha and beta?
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  4. #4
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    \exp{(\alpha x+\beta y)}[\omega_{xx}-3\omega_{yy}+\omega_{x}(2\alpha+2)-\omega_y(6\beta+1)+\omega(\alpha^2-3\beta^2+2\alpha-\beta+1)]=0

    \displaystyle\alpha=-1 \ \ \ \beta=-\frac{1}{6}

    \displaystyle\omega_{xx}-3\omega_{yy}+\frac{1}{12}\omega=0

    \displaystyle\frac{1}{\mu^2}\omega_{\xi\xi}-\frac{3}{\nu^2}\omega_{\eta\eta}+\frac{1}{12}\omeg  a=0

    \displaystyle\frac{1}{\mu^2}=\frac{3}{\nu^2}=\frac  {1}{12} \ \ \ \ \mu=\pm\sqrt{12} \ \ \ \ \nu=\pm\sqrt{36}

    \displaystyle u_{xx}-u_{yy}+u=0
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  5. #5
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    Seems good to me.
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