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Thread: u_x-u_y+u=1

  1. #1
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    u_x-u_y+u=1

    $\displaystyle u_x-u_y+u=1$

    Auxiliary condition $\displaystyle u(x,0)=\sin{x}$

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

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

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

    $\displaystyle \omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha}-(\omega_{\xi}\sin{\alpha}+\omega_{\eta}\cos{\alpha })+\omega=1$

    $\displaystyle \omega_{\xi}(\cos{\alpha}-\sin{\alpha})-\omega_{\eta}(\sin{\alpha}+\cos{\alpha})+\omega=1$

    $\displaystyle \sin{\alpha}+\cos{\alpha}=0$

    $\displaystyle \displaystyle\tan{\alpha}=-1\Rightarrow\alpha=\frac{3\pi}{4}+\pi k \ k\in\mathbb{Z}$

    $\displaystyle \displaystyle\cos{\alpha}=\frac{-\sqrt{2}}{2} \ \mbox{and} \ \sin{\alpha}=\frac{\sqrt{2}}{2}$

    $\displaystyle \displaystyle -\sqrt{2}\omega_{\xi}+\omega=1\Rightarrow\omega_{\x i}-\frac{1}{\sqrt{2}}\omega=-\frac{1}{\sqrt{2}}$

    $\displaystyle x=A\xi+B\eta$

    $\displaystyle y=C\xi+D\eta$

    $\displaystyle \omega_{\xi}=Au_x+Cu_y=u_x-u_y$

    $\displaystyle A=1 \ \mbox{and} \ B=-1$

    $\displaystyle x=\xi+B\eta$

    $\displaystyle y=-\xi+D\eta$

    $\displaystyle u_x-u_y+u=1\Rightarrow\omega_{\xi}+\omega=1$

    $\displaystyle P(\xi)=1 \ \ Q(\xi)=1$

    $\displaystyle \exp{\left(\int P(\xi)d\xi\right)}=\exp{\left(\int d\xi\right)}=\exp{\xi}$

    $\displaystyle \displaystyle\int\frac{\partial}{\partial\xi}\left[\exp{\left(\int P(\xi)d\xi\right)}\omega\right]d\xi=\int\exp{\left(\int P(\xi)d\xi\right)}Q(\xi)d\xi$

    $\displaystyle \displaystyle\int\frac{\partial}{\partial\xi}\left[\exp{(\xi)}\omega\right]d\xi=\int\exp{(\xi)}d\xi\Rightarrow\exp{(\xi)}\ome ga=\exp{(\xi)}+g(\eta)$

    $\displaystyle \omega(\xi,\eta)=1+\exp{(-\xi)}g(\eta)$

    How do I solve for the auxiliary equation?
    Last edited by dwsmith; Jan 8th 2011 at 08:38 PM. Reason: + to -
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  2. #2
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    I think you made a typo, should be

    $\displaystyle \omega_{\xi}(\cos{\alpha}-\sin{\alpha})-\omega_{\eta}(\sin{\alpha}+\cos{\alpha})+\omega=1$

    but it seems to correct itself later.

    Quote Originally Posted by dwsmith View Post
    Auxiliary condition $\displaystyle u(x,0)=\sin{x}$

    $\displaystyle \omega(\xi,\eta)=1+\exp{(-\xi)}g(\eta)$

    How do I solve for the auxiliary equation?
    You can use the inverse rotation to get your solution in terms of x and y, and then solve using the auxiliary condition.
    Or you can transform the auxiliary condition into your new coordinate system.
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  3. #3
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    Can you show me how that is done? Yes, I made a typo.
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  4. #4
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    $\displaystyle y=0$ is the line $\displaystyle -\xi \tan\alpha = \eta$.

    Taking the partial w.r.t x at y=0 for the constraint gives:
    $\displaystyle \cos(x) = u_x = \omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha}$
    when $\displaystyle -\xi \tan\alpha = \eta$

    Take the partials. Substitute to get rid of $\displaystyle \xi$, and solve for $\displaystyle g(\eta)$.
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  5. #5
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    Quote Originally Posted by snowtea View Post
    $\displaystyle y=0$ is the line $\displaystyle \xi \tan\alpha = \eta$.
    Where did you get this equation?
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  6. #6
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    $\displaystyle 0 = y = \xi\sin{\alpha}+\eta\cos{\alpha}$

    implies $\displaystyle -\xi\sin{\alpha} = \eta\cos{\alpha}$
    $\displaystyle -\xi\tan{\alpha} = \eta$

    actually I made a mistake

    Let me fix that.
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  7. #7
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    Quote Originally Posted by snowtea View Post
    $\displaystyle \cos(x) = u_x = \omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha}$
    Can you explain this part, too?
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  8. #8
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    Sorry about the poor explanations

    $\displaystyle u(x,0)=\sin(x)$ so $\displaystyle u_x(x,0)=\cos(x)$
    This is the same thing as the pair of constraints:
    $\displaystyle
    y=0
    $
    $\displaystyle
    u_x = \cos(x)
    $

    Transforming each equation gives:
    $\displaystyle
    -\xi \tan\alpha = \eta
    $
    $\displaystyle
    \omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha} = \cos(\xi\cos{\alpha}-\eta\sin{\alpha})
    $

    Last part inside the cosine is because:
    $\displaystyle x = \xi\cos{\alpha}-\eta\sin{\alpha}$
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  9. #9
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    Quote Originally Posted by snowtea View Post
    Sorry about the poor explanations

    $\displaystyle u(x,0)=\sin(x)$ so $\displaystyle u_x(x,0)=\cos(x)$
    This is the same thing as the pair of constraints:
    $\displaystyle
    y=0
    $
    $\displaystyle
    u_x = \cos(x)
    $

    Transforming each equation gives:
    $\displaystyle
    -\xi \tan\alpha = \eta
    $
    $\displaystyle
    \omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha} = \cos(\xi\cos{\alpha}-\eta\sin{\alpha})
    $

    Last part inside the cosine is because:
    $\displaystyle x = \xi\cos{\alpha}-\eta\sin{\alpha}$
    I am still not sure how to solve for $\displaystyle \eta$ though.
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  10. #10
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    Quote Originally Posted by dwsmith View Post
    I am still not sure how to solve for $\displaystyle \eta$ though.
    You are trying to solve for $\displaystyle g(\eta)$.

    If you just simplify with the constraints (using what you found for $\displaystyle \omega$), you should get something of the form:
    $\displaystyle Ag'(\eta) + Bg(\eta) = e^{C\eta}cos(D\eta)$ where $\displaystyle A,B,C,D$ are known constants (expressed in terms of $\displaystyle \alpha$).
    Remeber $\displaystyle \alpha$ is just a constant, and so is $\displaystyle sin(\alpha)$ and $\displaystyle cos(\alpha)$.
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  11. #11
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    Quote Originally Posted by snowtea View Post
    $\displaystyle Ag'(\eta) + Bg(\eta) = e^{C\eta}cos(D\eta)$
    I don't know how you obtained this equation though.

    I already found A and C which are 1 and -1, respectively.
    Last edited by dwsmith; Jan 9th 2011 at 12:09 PM.
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  12. #12
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    Snowtea,

    unfortunately, I have no idea how to solve the problem using your method but I was able to obtain the solution.

    $\displaystyle \displaystyle
    \begin{vmatrix}A&B\\C&D\end{vmatrix}\neq 0$

    B and D are arbitrary and since $\displaystyle y=0\Rightarrow y=0=-\xi+D\eta$, let D = 0.

    $\displaystyle y=0=-\xi$

    Now, let B = 1.

    $\displaystyle x=\xi+\eta\Rightarrow\eta=x-\xi=x+y$

    $\displaystyle \omega(\xi,\eta)=1+\exp{(-\xi)}g(\eta)\Rightarrow u(x,y)=1+\exp{(y)}g(x+y)$

    $\displaystyle u(x,0)=1+g(x)=\sin{x}\Rightarrow g(x)=\sin{(x)}-1$

    $\displaystyle u(x,y)=1+\exp{(y)}[\sin{(x+y)}-1]$
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  13. #13
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    Quote Originally Posted by dwsmith View Post
    I don't know how you obtained this equation though.

    I already found A and C which are 1 and -1, respectively.
    A,B,C,D were actually new constants (just to indicate the form of the DE for g(eta)). They are different from the ones you used above. I forgot that you used them.
    Sorry for the confusion.

    Glad you got the right solution though.
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