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

**huberscher** · February 19th 2013, 06:19 AM

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

**huberscher** · February 20th 2013, 03:45 AM

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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