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Math Help - Second Order

  1. #1
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    Second Order

    3u_{xx}+4u_{yy}-u=0

    a=3, \ \ b=0, \ \ c=4

    ac-b^2=3*4-0=12>0

    \mbox{Elliptic} \ \ \omega_{\xi\xi}+\omega_{\eta\eta}+\gamma\omega=0

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

    u_{xx}=(\omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha})^2=\omega_{\xi\xi}\cos^2  {\alpha}-2\omega_{\xi\eta}\sin{\alpha}\cos{\alpha}+\omega_{  \eta\eta}\sin^2{\alpha}

    u_{yy}=(\omega_{\xi}\sin{\alpha}+\omega_{\eta}\cos  {\alpha})^2=\omega_{\xi\xi}\sin^2{\alpha}+2\omega_  {\xi\eta}\sin{\alpha}\cos{\alpha}+\omega_{\eta\eta  }\cos^2{\alpha}

    \omega_{\xi\xi}(3\cos^2{\alpha}+4\sin^2{\alpha})+\  omega_{\eta\eta}(3\sin^2{\alpha}+4\cos{\alpha})+2\  sin{\alpha}\cos{\alpha} \ \omega_{\xi\eta}+\omega=0

    Since I am trying to eliminate \omega_{\xi\eta}, does it matter what alpha I pick?

    2\sin{\alpha}\cos{\alpha}=0\Rightarrow \alpha=\frac{\pi}{2}+\pi k \ \mbox{or} \ \pi k \ \ \ k\in\mathbb{Z}
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  2. #2
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    \cos{\alpha}=1 \ \ \ \sin{\alpha}=0

    \displaystyle 3\omega_{\xi\xi}+4\omega_{\eta\eta}-\omega=0\Rightarrow\omega_{\xi\xi}+\frac{4}{3}\ome  ga_{\eta\eta}-\frac{1}{3}\omega=0

    \displaystyle u_{xx}+\frac{4}{3}u_{yy}-\frac{1}{3}u=0

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

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

    2\beta=0\Rightarrow \beta=0

    \omega_{xx}+\omega_{yy}-\omega=0

    x=\mu\xi \ \ \ y=\nu\eta

    \displaystyle\frac{1}{\mu^2}\omega_{\xi\xi}+\frac{  1}{\nu^2}\omega_{\eta\eta}-\omega=0

    \displaystyle\frac{1}{\mu^2}=\frac{1}{\nu^2}=-1\Rightarrow\mu=\pm i \ \ \ \nu=\pm i

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

    However, the solution is u_{xx}+u_{yy}-u=0

    Why is it changed to u from omega?
    Last edited by dwsmith; January 13th 2011 at 04:49 PM. Reason: Forgot ^2
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