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Thread: u_x-3u_y=0

  1. #1
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    u_x-3u_y=0

    $\displaystyle u_x-3u_y=0$

    $\displaystyle u(x,x)=x^2$

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

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

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

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

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

    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 $\displaystyle B = \frac{1}{3}$ and $\displaystyle 0$. - it looks nice that way.

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

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

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

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

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

    $\displaystyle x=3x+y$

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

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

    How do I incorporate $\displaystyle g(4x)=x^2$ into the general solution $\displaystyle u(x,y)=g(3x+y)\mbox{?}$
    Last edited by dwsmith; Jan 11th 2011 at 10:16 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: $\displaystyle g(x) = \left(\frac{x}{4}\right)^2$ - from the equation you have there.

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

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