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Math Help - change of variable

  1. #1
    Junior Member
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    change of variable

    I'm learning about the second order PDEs, and I've come across the transformation of variables I don't understand.

    We have u=u(x, y) and a, b, c functions of class C^2, which are not at the same time all equal to zero.
    The equation in question is:
    (*) a \frac{\partial^2 u}{\partial x^2} + 2b \frac{\partial^2 u}{\partial x \partial y} +c \frac{\partial ^2 u}{\partial y^2} + f(x, u, \nabla u)=0
    We introduce the change of variables:
    \xi=x+y, \eta=x-y

    How does the equation (*) now looks like?
    I've tried googling change of variables, but I'm still lost.
    Please, help.

    Thank you.
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  2. #2
    MHF Contributor

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    By the chain rule, \frac{\partial f}{\partial x}= \frac{\partial f}{\partial\xi}\frac{\partial xi}{\partial x}+ \frac{\partial f}{\partial\eta}\frac{\partial\eta}{\partial x}
    Here, it is easy to see that \frac{\partial\xi}{\partial x}= 1 and \frac{\partial\eta}{\partial x}= 1 so \frac{\partial f}{\partial x}= \frac{\partial f}{\partial\xi}+ \frac{\partial f}{\partial\eta}.

    Similarly, \frac{\partial f}{\partial y}= \frac{\partial f}{\partial\xi}- \frac{\partial f}{\partial\eta}

    You find the second derivatives by using those in succession:
    \frac{\partial^2 f}{\partial x^2}= \frac{\partial \left(\frac{\partial f}{\partial\xi}+ \frac{\partial f}{\partial\eta}\right)}{\partial\xi}+ \frac{\partial\left(\frac{\partial f}{\partial\xi}+ \frac{\partial f}{\partial\eta}\right)}{\partial\eta}.

    \frac{\partial^2 f}{\partial y^2}= \frac{\partial\left(\frac{\partial f}{\partial\xi}- \frac{\partial f}{\partial\eta}\right)}{\partial\xi}- \frac{\partial \left(\frac{\partial f}{\partial\xi}- \frac{\partial f}{\partial\eta}\right)}{\partial\eta}

    and finally
    \frac{\partial^2 f}{\partial x\partial y}= \frac{\partial \left(\frac{\partial f}{\partial\xi}- \frac{\partial f}{\partial\eta}\right)}{\partial\xi}+ \frac{\partial\left(\frac{\partial f}{\partial\xi}- \frac{\partial f}{\partial\eta}\right)}{\partial\eta}

    Of course, since \xi= x+ y and \eta= x- y, adding those equations, 2x= \xi+ \eta and so x= \frac{\xi- \eta}{2}. Use that on the right hand side of the equation.
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