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Math Help - u_x-3u_y=0

  1. #1
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    u_x-3u_y=0

    u_x-3u_y=0

    u(x,x)=x^2

    \omega_{\xi}(\cos{\alpha}-3\sin{\alpha})-\omega_{\eta}(\sin{\alpha}+3\cos{\alpha})=0

    \displaystyle\cos{\alpha}=\frac{1}{\sqrt{10}} \ \ \ \sin{\alpha}=\frac{-3}{\sqrt{10}}

    \displaystyle\sqrt{10}\omega_{\xi}=0\Rightarrow\om  ega_{\xi}=0

    x=\xi+B\eta \ \ \ y=-3\xi+D\eta

    \displaystyle\int\omega_{\xi}d\xi=\int 0d\xi\Rightarrow\omega(\xi,\eta)=g(\eta)

    I am lost with initial condition and what to set D and B equal too.
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  2. #2
    Super Member PaulRS's Avatar
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    You may choose any numbers you like for B and D as long as the change of variables is invertible. - because these do not affect your original equation.

    Say you could choose B = \frac{1}{3} and 0. - it looks nice that way.

    Once you have undone your change of variables, set x = y to use your initial condition.

    Why do you use the \cos and \sin for coefficients?
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  3. #3
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    Quote Originally Posted by PaulRS View Post

    Why do you use the \cos and \sin for coefficients?
    That is how my book has taught me. Here is what is in the book.

    \xi=x\cos{\alpha}+y\sin{\alpha}, \ \ \ x=\xi\cos{\alpha}-\eta\sin{\alpha}, \ \ \ \eta=-x\sin{\alpha}+y\cos{\alpha}, \ \ \ y=\xi\sin{\alpha}+\eta\cos{\alpha}

    u(x,y)=u(\xi\cos{\alpha}-\eta\sin{\alpha}, \ \xi\sin{\alpha}+\eta\cos{\alpha})=\omega(\xi,\eta)
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  4. #4
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    x=\xi+\frac{1}{3}\eta\Rightarrow\eta=3x-3\xi \ \ \ y=-3\xi=x

    x=3x+y

    \omega(\xi,\eta)=u(x,y)=g(3x+y)

    u(x,x)=g(3x+x)=g(4x)=x^2

    How do I incorporate g(4x)=x^2 into the general solution u(x,y)=g(3x+y)\mbox{?}
    Last edited by dwsmith; January 11th 2011 at 11:16 AM. Reason: forgot to with into
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  5. #5
    Super Member PaulRS's Avatar
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    Well, note that then: g(x) = \left(\frac{x}{4}\right)^2 - from the equation you have there.

    Now to get u(x,y), plug 3x+y into g.
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  6. #6
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    Quote Originally Posted by PaulRS View Post
    Well, note that then: g(x) = \left(\frac{x}{4}\right)^2 - from the equation you have there.

    Now to get u(x,y), plug 3x+y into g.
    Wouldn't it be \displaystyle\left(\frac{x}{2}\right)^2\mbox{?}
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  7. #7
    Super Member PaulRS's Avatar
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    We want g(u) , but to calculate this we let u = 4\cdot x and then g(u) = g(4\cdot x) = x^2 = ...\text{in terms of } u
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