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Math Help - Show that a transformation can be hapen of a system of 2nd order

  1. #1
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    Show that a transformation of a system of 2nd order is possible

    Hi there

    This is the exercise I'm trying to solve:
    $$
    Show that the general ansatz

    -\sum_{1\le k,l \le n} A_{k,l}(x) \partial_k \partial_l u(x)+\sum_{1 \le k \le n}b_i(x) \partial u(x)+c(x)*u(x) (1)

    can be transformed to the following

    -div(A(x) \nabla u(x)) + \langle b(x),\nabla u(x) \rangle  + c(x)u(x)=g(x) (2)

    when u has continuous derivatives of second order
    $$

    by div(u) the divergence of u is meant and when I calculate (2) and just write it down I get all the same as in (1) excepted the part with the matrix (notice: A does depend on x... is a matrix of functions):   -div(A \nabla u)=-\sum_{1\le k,l \le n} A_{k,l}(x) \partial_k \partial_l u(x)-\sum_{1\le k,l \le n} \partial_k A_{k,l}(x) \partial_l u(x)


    Well so do I just have to set g(x)=\sum_{1\le k,l \le n} \partial_k A_{k,l}(x) \partial_l u(x) and the proof is complete? Not really, yeah? I did not substitute or transform anything???

    What do I have to do here?

    Regards
    Huberscher
    Last edited by huberscher; February 19th 2013 at 12:42 PM.
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  2. #2
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    Re: Show that a transformation can be hapen of a system of 2nd order

    I tried the transformation

    \tilde{x} =Ax where A is the matrix

     \begin{bmatrix}1 &-1 & 0           \\0 &           1 & -1 \\1           & 0 & -1\end{bmatrix}

    but that one isnt invertible. Cant anybody give me a hint how to show this?

    Regards
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