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Thread: au_x+bu_y+cu=0

  1. #1
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    au_x+bu_y+cu=0

    $\displaystyle au_x+bu_y+cu=0$

    $\displaystyle a,b,c\in\mathbb{C} \ \ \mbox{and} \ \ b\ne 0$

    a,b,c aren't necessarily Complex numbers.

    $\displaystyle u(x,0)=u_0(x)$

    $\displaystyle u_0(x)$ is differentiable.

    $\displaystyle \omega_{\xi}(a\cos{\alpha}+b\sin{\alpha})+\omega_{ \eta}(b\cos{\alpha}-a\sin{\alpha})+c\omega=0$

    $\displaystyle \displaystyle\tan{\alpha}=\frac{b}{a}$

    Right Triangle:

    Legs a and b

    Hypotenuse $\displaystyle d=\sqrt{a^2+b^2}$

    $\displaystyle \displaystyle\cos{\alpha}=\frac{a}{d} \ \ \sin{\alpha}=\frac{b}{d}$

    $\displaystyle \displaystyle\frac{a^2+b^2}{d}\omega_{\xi}+c\omega = 0$

    $\displaystyle \displaystyle\omega_{\xi}+\frac{c}{d}\omega=0$

    $\displaystyle x=A\xi+B\eta \ \ \ y=C\xi+D\eta$

    $\displaystyle au_x+bu_y=\omega_{\xi}=Au_x+Cu_y\Rightarrow \ \ A=a \ \ C=b$

    Let $\displaystyle D=0 \ \ B=1$

    $\displaystyle \displaystyle x=a\xi+\eta \ \ y=b\xi\Rightarrow \eta=x-\frac{ay}{b} \ \ \xi=\frac{y}{b}$

    $\displaystyle \omega_{\xi}+c\omega=0$

    $\displaystyle \exp{(c\xi)}\omega=g(\eta)\Rightarrow \omega(\xi,\eta)=\exp{(-c\xi)}g(\eta)$

    $\displaystyle \displaystyle\omega(\xi,\eta)=u(x,y)=\exp{\left(\f rac{-cy}{b}\right)}g\left(x-\frac{ay}{b}\right)$

    $\displaystyle \displaystyle u(x,0)=g(x)=u_0(x)$

    $\displaystyle \displaystyle u(x,y)=\exp{\left(\frac{-cy}{b}\right)}u_0\left(x-\frac{ay}{b}\right)$

    Is this correct?
    Last edited by dwsmith; Jan 10th 2011 at 02:04 PM. Reason: changed an x to a c
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  2. #2
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    An easy way of determining whether your answer is correct is

    (1) Does it satisfy the IC?

    (2) Does it satisfy the PDE when you substitute the solution in?
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  3. #3
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    I wasn't sure how to differentiate $\displaystyle \displaystyle u(x,y)=\exp{\left(\frac{-cy}{b}\right)}u_0\left(x-\frac{ay}{b}\right)$ due to the u_0 piece.
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  4. #4
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    $\displaystyle u_0$ is an arbitrary function of its argument so

    $\displaystyle \dfrac{\partial}{\partial x} u_0\left(x - \dfrac{a}{b} y\right) = u_0'\left(x - \dfrac{a}{b} y\right)$

    $\displaystyle \dfrac{\partial}{\partial y} u_0\left(x - \dfrac{a}{b} y\right) = u_0'\left(x - \dfrac{a}{b} y\right)\left(-\dfrac{a}{b}\right)$.
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