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Math Help - 2nd Order PDE problem

  1. #1
    Senior Member bugatti79's Avatar
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    2nd Order PDE problem

    Hi all,
    I am in the process of working out a solution to a 2nd order PDE.

    However I am stuck on calculating u_x_y given that

    \xi=y-sin(x)-x and \eta=y-sin(x)+x and \omega(\xi,\eta)=u(x,y)

    I can calculate uxx which is

    u_x_x=\frac{\partial^{2}\xi}{\partial x^{2}}\frac{\partial u}{\partial \xi}}+\frac{\partial \xi}{\partial x}(\frac{\partial^{2} u}{\partial \xi^{2}}\frac{\partial \xi}{\partial x}+\frac{\partial^{2} u}{\partial \eta \partial \xi}\frac{\partial \eta}{\partial x}) +\frac{\partial^{2}\eta}{\partial x^{2}}\frac{\partial u}{\partial \eta}}+\frac{\partial \eta}{\partial x}(\frac{\partial^{2} u}{\partial \eta^{2}}\frac{\partial \eta}{\partial x}+\frac{\partial^{2} u}{\partial \xi \partial \eta}\frac{\partial \xi}{\partial x})

    and uy,uxx and uyy etc however I get stuck on uxy. I just cant get the answer in the book which leads me to belive my uxy is wrong. Can anyone give me a start on this uxy derivation like above for uxx?
    Thanks
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  2. #2
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    I get

    u_{xy} = \xi_{y}\xi_y u_{\xi \xi}  + (\xi_x \eta_y + \xi_y \eta_x)u_{\xi \eta} + \eta_x \eta_y u_{\eta \eta} + \xi_{xy} u_\xi + \eta_{xy} u_\eta.
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    I get

    u_{xy} = \xi_{y}\xi_y u_{\xi \xi}  + (\xi_x \eta_y + \xi_y \eta_x)u_{\xi \eta} + \eta_x \eta_y u_{\eta \eta} + \xi_{xy} u_\xi + \eta_{xy} u_\eta.
    hmmmm, can you explain how you got this? Is it a chain rule within a product rule? I started my calculation to be
    \frac{\partial}{\partial x}[u_y]=\frac{\partial}{\partial x} [\omega_\xi \cdot \xi_y+ \omega_\eta \cdot \eta_y] then my next step goes to pieces..

    The final answer in the book is given as uxy =
    \omega_\xi_\xi(-cos(x)-1)+\omega_\xi_\eta(-cos(x)+1)+\omega_\eta_\xi(-cos(x)-1)+\omega_\eta_\eta(-cos(x)+1)

    it doesnt show the general form of uxy before the above line. The calculation of this seems to be different to that of uxx, uyy etc.
    Any help is appreciated
    Thanks
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  4. #4
    Senior Member yeKciM's Avatar
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    you don't need to check "general form" for u_{xy} if you can find u_x to check in the book and if your u_x is correct where you have problem with  (u_x)'_y derivate that function on "y" ?
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  5. #5
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    Quote Originally Posted by bugatti79 View Post
    hmmmm, can you explain how you got this? Is it a chain rule within a product rule? I started my calculation to be
    \frac{\partial}{\partial x}[u_y]=\frac{\partial}{\partial x} [\omega_\xi \cdot \xi_y+ \omega_\eta \cdot \eta_y] then my next step goes to pieces..

    Thanks
    I can. First you'll need the first derivative transforms

    \dfrac{\partial}{\partial x} = \xi_x \dfrac{\partial}{\partial \xi} + \eta_x \dfrac{\partial}{\partial \eta}

    \dfrac{\partial}{\partial y} = \xi_y \dfrac{\partial}{\partial \xi} + \eta_y \dfrac{\partial}{\partial \eta}.

    Now expand what you have (although I prefer to the keep the u's)

    \frac{\partial}{\partial x}[u_y]=\frac{\partial}{\partial x} [u_\xi \cdot \xi_y+ u_\eta \cdot \eta_y]

    =\frac{\partial}{\partial x}\left(u_\xi  \right)\cdot \xi_y + u_\xi \cdot \xi_{xy} + \frac{\partial}{\partial x}\left(u_\eta  \right) \cdot \eta_y+ u_\eta \cdot \eta_{xy}.

    Now bring in the x transform

    \frac{\partial}{\partial x}\left( u_\xi  \right) =  \xi_x \dfrac{\partial}{\partial \xi} \left(u_\xi\right)  + \eta_x \dfrac{\partial}{\partial \eta} \left(u_\xi\right)

    \frac{\partial}{\partial x}\left(u_\eta  \right) =  \xi_x \dfrac{\partial}{\partial \xi} \left(u_\eta\right)  + \eta_x \dfrac{\partial}{\partial \eta} \left(u_\eta\right).

    Substituting and expanding will give you the desired result.
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  6. #6
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    I can. First you'll need the first derivative transforms

    \dfrac{\partial}{\partial x} = \xi_x \dfrac{\partial}{\partial \xi} + \eta_x \dfrac{\partial}{\partial \eta}

    \dfrac{\partial}{\partial y} = \xi_y \dfrac{\partial}{\partial \xi} + \eta_y \dfrac{\partial}{\partial \eta}.

    Now expand what you have (although I prefer to the keep the u's)

    \frac{\partial}{\partial x}[u_y]=\frac{\partial}{\partial x} [u_\xi \cdot \xi_y+ u_\eta \cdot \eta_y]

    =\frac{\partial}{\partial x}\left(u_\xi  \right)\cdot \xi_y + u_\xi \cdot \xi_{xy} + \frac{\partial}{\partial x}\left(u_\eta  \right) \cdot \eta_y+ u_\eta \cdot \eta_{xy}.

    Now bring in the x transform

    \frac{\partial}{\partial x}\left( u_\xi  \right) =  \xi_x \dfrac{\partial}{\partial \xi} \left(u_\xi\right)  + \eta_x \dfrac{\partial}{\partial \eta} \left(u_\xi\right)

    \frac{\partial}{\partial x}\left(u_\eta  \right) =  \xi_x \dfrac{\partial}{\partial \xi} \left(u_\eta\right)  + \eta_x \dfrac{\partial}{\partial \eta} \left(u_\eta\right).

    Substituting and expanding will give you the desired result.
    So i get

    u_x_y=\omega_\xi \cdot \xi_x_y+\xi_y \cdot [\xi_x \cdot \omega_\xi_\xi+\eta_x \cdot \omega_\eta_\xi] +\omega_\eta \cdot \eta_x_y+\eta_y \cdot[\xi_x \cdot \omega_\xi_\eta+\eta_x \cdot \omega_\eta_\eta]

    I can finally get the answer in the book. However, I just have 2 small queries:

    1) How did you anticipate \dfrac{\partial}{\partial x} = \xi_x \dfrac{\partial}{\partial \xi} + \eta_x \dfrac{\partial}{\partial \eta}

    Is it a setup for the chain rule?

    2) As part of the above calculations we had \frac{\partial}{\partial x}[\frac{\partial u}{\partial \xi} \cdot \frac{\partial \xi}{\partial y}] and \omega(\xi,\eta)=u(x(\xi,\eta), y(\xi, \eta)))

    I am not 100% clear why this is differentiated as a product rule because I understand the product rule to be of the form y=uv where u and v are both functions of x.
    Yet in the above equation I dont see how the product should be used because the denominator \partial y is not a function of x but only \xi and \eta.....and even the denominator \partial\xi is not a function of x for the matter I dont think.....?
    Im learning slowly but surely!
    bugatti
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