Hi there

This is the exercise I'm trying to solve:

$$

Show that the general ansatz

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

can be transformed to the following

$\displaystyle -div(A(x) \nabla u(x)) + \langle b(x),\nabla u(x) \rangle + c(x)u(x)=g(x)$ (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): $\displaystyle -div(A \nabla u)=-\sum_{1\le k,l \le n} A_{k,l}(x) \partial_k \partial_l u(x)-\sum_{1\le k,l \le n} \partial_k A_{k,l}(x) \partial_l u(x)$

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

What do I have to do here?

Regards

Huberscher