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Math Help - Second Order Elliptic

  1. #1
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    Second Order Elliptic

    u_{xx}+u_{yy}+3u_x-4u_y+25u=0

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

    u_{yy}=\omega_{\xi\xi}\sin^{2}{\alpha}+\omega_{\xi  \eta}2\cos{\alpha}\sin{\alpha}+\omega_{\eta\eta}\c  os^{2}{\alpha}

    u_x=\omega_{\xi}\cos{\alpha}-\omega{\eta}\sin{\alpha}

    u_y=\omega_{\xi}\sin{\alpha}+\omega{\eta}\cos{\alp  ha}

    \displaystyle\omega_{\xi\xi}(\cos^2{\alpha}+\sin^2  {\alpha})+\omega_{\eta\eta}(\cos^2{\alpha}+\sin^2{  \alpha})+\omega_{\eta\xi}(2\sin{\alpha}\cos{\alpha  }-2\sin{\alpha}\cos{\alpha})+\omega_{\xi}(3\cos{\alp  ha}-4\sin{\alpha})+\omega_{\eta}(-4\cos{\alpha}-3\sin{\alpha})+25\omega=0

    \omega_{\xi\xi}+\omega_{\eta\eta}+\omega_{\xi}(3\c  os{\alpha}-4\sin{\alpha})+\omega_{\eta}(-4\cos{\alpha}-3\sin{\alpha})+25\omega=0

    Since \omega_{\xi\eta} is already gone, should I be working to eliminate \omega_{\xi} \ \mbox{and} \ \omega_{\eta}\mbox{?}
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  2. #2
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    Yes, that's the idea but let me say that it's not always possible :-)
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  3. #3
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    So continuing....

    \displaystyle\tan{\alpha}=\frac{3}{4} \ \ \cos{\alpha}=\frac{4}{5} \ \ \sin{\alpha}=\frac{3}{5}

    \omega_{\xi\xi}+\omega_{\eta\eta}-5\omega_{\eta}+25\omega=0

    u_{xx}+u_{yy}-5u_y+25u=0

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

    u_{xx}=\exp{(\beta x)}[\beta^2\omega+2\beta\omega_x+\omega_{xx}], \ \ u_{yy}=\exp{(\beta x)\omega_{yy}, \ \ u_y=\exp{(\beta x)}\omega_y

    \exp{(\beta x)}[\omega_{xx}+\omega_{yy}+2\beta\omega_x-5\omega_y+(\beta^2+25)\omega]=0

    \beta=0

    Doing this didn't eliminate \omega_y

    \omega_{xx}+\omega_{yy}-5\omega_y+25\omega=0

    The answer is u_{xx}+u_{yy}+u=0

    How do I solve this equation correctly?
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  4. #4
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    Try  u = e^{\beta y} \omega(x,y)
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  5. #5
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    Quote Originally Posted by Danny View Post
    Try  u = e^{\beta y} \omega(x,y)
    \exp{(\beta y)}[\omega_{xx}+\omega_{yy}+\omega_y(2\beta-5)+(\beta^2+\beta+25)\omega=0

    \displaystyle\beta =\frac{5}{2}

    \displaystyle\omega_{xx}+\omega_{yy}+\frac{135}{4}  \omega=0

    \displaystyle\frac{1}{\mu^2}\omega_{xx}+\frac{1}{\  nu^2}\omega_{yy}+\frac{135}{4}\omega=0

    \displaystyle\frac{1}{\mu^2}=\frac{1}{\nu^2}=\frac  {135}{4}\Rightarrow \mu=\nu=\pm\frac{2}{3\sqrt{15}}

    How from this am I able to say u_{xx}+u_{yy}+u=0\mbox{?}

    I know I am supposed to conclude that once I solve for mu and nu, but I have no clue why.
    Last edited by dwsmith; January 15th 2011 at 01:39 PM. Reason: Fixed an error
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  6. #6
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    Let's start over. First let

    u = e^{a x + b y} v(x,y)

    and choose a and b so the first order terms vanish. Then scale x and y so that you can cancel the coefficient on the v term.
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  7. #7
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    Quote Originally Posted by Danny View Post
    Let's start over. First let

    u = e^{a x + b y} v(x,y)

    and choose a and b so the first order terms vanish. Then scale x and y so that you can cancel the coefficient on the v term.
    Why?

    You last suggestion worked fine.
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  8. #8
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    Why. Because (1) You have three different transformations when you can use two and (2), you seem a little fixated on using a rotation of coordinates. I personally won't rely on this so much on this (it really only works for constant coefficient PDEs.)
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  9. #9
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    Quote Originally Posted by Danny View Post
    Why. Because (1) You have three different transformations when you can use two and (2), you seem a little fixated on using a rotation of coordinates. I personally won't rely on this so much on this (it really only works for constant coefficient PDEs.)
    I have to rely on what the book presents, because I am not going to be able to guess another form. Also, I haven't made it to variable coefficients yet.
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