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Math Help - Linear Equations in two independent variables

  1. #1
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    Linear Equations in two independent variables

    2.2.2 3u_x+4u_y-2u=1\Rightarrow \omega_{\xi}+k\omega=\varphi(\xi,\eta)

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

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

    u_y=\omega_{\xi}\sin{\alpha}+\omega_{\eta}\cos{\al  pha}

    3(\omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha})+4(\omega_{\xi}\sin{\alp  ha}+\omega_{\eta}\cos{\alpha})-2w=1

    \omega_{\xi}(3\cos{\alpha}+4\sin{\alpha})+\omega_{  \eta}(4\cos{\alpha}-3\sin{\alpha})-2\omega=1

    \displaystyle 4\cos{\alpha}-3\sin{\alpha}=0\Rightarrow \tan{\alpha}=\frac{4}{3}

    We have a 3,4,5 right triangle.

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

    Substitution:

    \displaystyle \omega_{\xi}\left(3\frac{3}{5}+4\frac{4}{5}\right)  +\omega_{\eta}(0)-2\omega=1

    \displaystyle \omega_{\xi}\left(\frac{9+16}{5}\right)-2\omega=1\Rightarrow\omega_{\xi}-\frac{2}{5}\omega=1

    I understand everything posted above; however, I don't understanding anything below. Everything here is understood now.

    Let

    2.2.8 x=A\xi+B\eta \ \mbox{and} \ y=C\xi+D\eta

    A,B,C,D are constants to be determined, and set u(x,y)=\omega(\xi,\eta) 2.2.9

    Then, from \displaystyle\frac{\partial\omega}{\partial\xi}=u_  x\frac{\partial x}{\partial\xi}+u_y\frac{\partial y}{\partial\xi} and 2.2.8, 2.2.9, we see that with the choice A = 3, C = 4, Equation 2.2.2 becomes \omega_{\xi}-2\omega=1
    Last edited by dwsmith; January 8th 2011 at 05:29 PM. Reason: This post is all understood.
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  2. #2
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    I have a follow-up question now though.

    The choice for B and D is arbitrary, except that AD - BC\neq 0.

    Looking ahead to the effort to satisfy the auxiliary condition u(x,0)=u_0(x), we shall choose B and D so that the line \xi=0 is the line on which the auxiliary data is prescribed, namely y = 0. This requires D = 0, and, since B is arbitrary we make the convenient choice B = 1.

    Can someone explain this and why?
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