I need some help with a problem (a homework/programming exercise) regarding the adjoint operator of the Frechet derivative of an operator.

I have the forward operator $F(a) = L_a ^{-1}f$ where $L_au = -\text{div}(a \; \text{grad} \; u)$, $u$ is known and $a \in L^{\infty}(\Omega)$, $\text{essinf}(a) > 0$. $L_a$ is bounded and linear and one can show that the operator $F(a)$ has the Frechet derivative $$F[a]'h = -L_a^{-1} \text{div}( h \; \text{grad} \; u)$$

I now need to find the adjoint of this operator. Using that $L_a$ is self-adjoint and $\text{div}^* = -\text{grad}$ I arrive at the following form for the adjoint: $$<h \cdot \text{grad} \; u, \; \text{grad}(L_a^{-1} v)>$$

My problem how to go on and get the final expression for the adjoint. I have been able to get the adjoint for the one dimensional case by discretizing the problem first and using the properties of the matrices, but somehow it feels like there should be a better way than relying on properties of the discretization.

Can someone help me out with this?

Edit: I have tried to use $(F[a]')^* = -\text{div} (u \; \text{grad}(L_a^{-1}v))$ but it doesn't seem to work, i.e. it does not pass the one point test $<F[a]'h, v> = <h, (F[a]')^*>$.

Edit: SOLVED! :D During my first tries to solve this I had totally forgotten that the multiplication operator $M_{\text{grad}(u)}$ used above is self-adjoint. This means that the adjoint operator of $F[a]'h$ is (with $\cdot$ as pointwise multiplication) $$(F'[a])^*v = \text{grad}(u) \cdot \text{grad}(L_a^{-1}v)$$

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Please post the answer to the question not as an edit inside the question but as an answer, and accept the answer; this will mark the question as resolved. (It also obviates the need to mark it thus by changing the title.) – joriki Dec 2 '12 at 17:54