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Thread: DE Change of Variables

  1. #1
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    DE Change of Variables

    u_x-u_y+u=\exp{(x+2y)}

    u(x,0)=0

    The solutions is u(x,y)=-y\exp{(x+2y)}

    \displaystyle\omega_{\xi}(\cos{\alpha}-\sin{\alpha})-\omega_{\eta}(\sin{\alpha}+\cos{\alpha})+\omega=ex  p{[\xi(\cos{\alpha}+2\sin{\alpha})+\eta(2\cos{\alpha}-\sin{\alpha})]

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

    \displaystyle -\sqrt{2}\omega_{\xi}+\omega=\exp{\left(\frac{\sqrt  {2}}{2}\xi-\frac{3\sqrt{2}}{2}\eta}\right)\Rightarrow \omega_{\xi}-\frac{1}{\sqrt{2}}\omega=\frac{\exp{\left(\frac{\s  qrt{2}}{2}\xi-\frac{3\sqrt{2}}{2}\eta}\right)}{-\sqrt{2}}

    x=A\xi+B\eta \ \ \mbox{and} \ \ y=C\xi+D\eta

    \omega_{\xi}=Au_x+Cu_y=u_x-u_y\Rightarrow A=1 \ \ B=-1

    x=\xi+B\eta \ \ \mbox{and} \ \ y=-\xi+D\eta

    \omega_{\xi}+\omega=\mbox{?}

    Should this be set equal to \displaystyle\exp{(x+2y)} \ \mbox{or} \ \exp{\left(\frac{\sqrt{2}}{2}\xi-\frac{3\sqrt{2}}{2}\eta}\right)
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  2. #2
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    \omega_{\xi}+\omega=\exp{(x+2y)}

    \displaystyle\int\left(\frac{\partial}{\partial\xi  }\left[\exp{(\xi)}\omega\right]\right)d\xi=\int\exp{(\xi+x+2y)}d\xi

    \exp{(\xi)}\omega=\exp{(\xi+x+2y)}+g(\eta)\Rightar  row\omega(\xi,\eta)=\exp{(x+2y)}+\exp{(-\xi)}g(\eta)

    x=\xi+B\eta \ \ \mbox{and} \ \ 0=-\xi+D\eta

    Let D = 0 and B = 1 since y = 0, both D and B are arbitrary, and AD - BC\neq 0.


    x=\xi+\eta \ \ \mbox{and} \ \ y=0=-\xi

    \eta=x-\xi=x+y

    \omega(\xi,\eta)=u(x,y)=\exp{(x+2y)}+\exp{(y)}g(x+  y)

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

    u(x,y)=\exp{(x+2y)}-\exp{(y)}\exp{(x+y)}=0

    This is obviously wrong but I don't know what to do differently.
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  3. #3
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    You got two things going on that seems to be confusing you - (1) a rotation of coordinates and (2) a general change of coordinates. Let me see if I can help.

    Your problem is

    u_x - u_y + u = e^{x+2y}

    u(x,0)=0

    You have introduced a linear change of variables (this can be more general but it's not necessary here)

    x = A \xi + B \eta,\;\;\;y = C \xi + D \eta.

    What you want is to transform your PDE to essentially an ODE by choosing the constant A,B,C and D with the requirement that AD - BC \ne 0 (why?).

    As you've said

    u_{\xi} = Au_x + Cu_y

    and picked A = 1 and C = -1 (that's good!) so

    u_{\xi} = u_x - u_y.

    You still have the flexibility to choose B and D anything. You further chose B =1 and D = 0 so

    x =  \xi + \eta,\;\;\;y = -\xi .

    This is also good b/c the boundary

    y = 0\;\;\;\Rightarrow\;\;\; \xi = 0.

    Now you must switch everthing over in terms of your new variables so since

    u_{\xi} = u_x - u_y

    then your new problem is

    u_{\xi} + u = e^{\eta - \xi} with u = 0 when \xi = 0.

    Now follow what you did in post #2.
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  4. #4
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    Quote Originally Posted by Danny View Post

    What you want is to transform your PDE to essentially an ODE by choosing the constant A,B,C and D with the requirement that AD - BC \ne 0 (why?).



    u_{\xi} + u = e^{\eta - \xi} with u = 0 when \xi = 0.
    The book said that \displaystyle\begin{vmatrix}A&B\\C&D\end{vmatrix}\  ne 0\Rightarrow AD-BC\ne 0.

    I don't understand how you obtain e^{\eta-\xi}
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  5. #5
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    The right hand side is

    e^{x+2y}

    From your change of variables

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

    so

    x+2y = \xi + \eta - 2 \xi = \eta - \xi.
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