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Math Help - u_x-u_y+u=1

  1. #1
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    u_x-u_y+u=1

    u_x-u_y+u=1

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

    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}

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

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

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

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

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

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

    x=A\xi+B\eta

    y=C\xi+D\eta

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

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

    x=\xi+B\eta

    y=-\xi+D\eta

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

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

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

    \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\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)

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

    How do I solve for the auxiliary equation?
    Last edited by dwsmith; January 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

    \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 u(x,0)=\sin{x}

    \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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    y=0 is the line -\xi \tan\alpha = \eta.

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

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

    implies -\xi\sin{\alpha} = \eta\cos{\alpha}
    -\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
    \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

    u(x,0)=\sin(x) so u_x(x,0)=\cos(x)
    This is the same thing as the pair of constraints:
    <br />
y=0<br />
    <br />
u_x = \cos(x)<br />

    Transforming each equation gives:
    <br />
-\xi \tan\alpha = \eta<br />
    <br />
\omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha} = \cos(\xi\cos{\alpha}-\eta\sin{\alpha})<br />

    Last part inside the cosine is because:
    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

    u(x,0)=\sin(x) so u_x(x,0)=\cos(x)
    This is the same thing as the pair of constraints:
    <br />
y=0<br />
    <br />
u_x = \cos(x)<br />

    Transforming each equation gives:
    <br />
-\xi \tan\alpha = \eta<br />
    <br />
\omega_{\xi}\cos{\alpha}-\omega_{\eta}\sin{\alpha} = \cos(\xi\cos{\alpha}-\eta\sin{\alpha})<br />

    Last part inside the cosine is because:
    x = \xi\cos{\alpha}-\eta\sin{\alpha}
    I am still not sure how to solve for \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 \eta though.
    You are trying to solve for g(\eta).

    If you just simplify with the constraints (using what you found for \omega), you should get something of the form:
    Ag'(\eta) + Bg(\eta) = e^{C\eta}cos(D\eta) where A,B,C,D are known constants (expressed in terms of \alpha).
    Remeber \alpha is just a constant, and so is sin(\alpha) and cos(\alpha).
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  11. #11
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    Quote Originally Posted by snowtea View Post
    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; January 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<br />
\begin{vmatrix}A&B\\C&D\end{vmatrix}\neq 0

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

    y=0=-\xi

    Now, let B = 1.

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

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

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

    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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