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I read this:

For $v$, $w$ in $L^2(0,T;H^1(S))$ (with weak derivatives in $H^{-1}(S)$ for each time), the product $(v(t), w(t))_{L^2(S)}$ is absolutely continuous wrt. $t \in [0,T]$ and $$\frac{d}{dt}\int_S v(t)w(t) = \langle v', w \rangle_{H^{-1}(S), H^1(S)} + \langle v, w' \rangle_{H^{1}(S), H^{-1}(S)}$$ holds a.e. in $(0,T)$. As a consequence, the formula of partial integration holds $$\int_0^T \langle v', w \rangle_{H^{-1}, H^1} = (v(T), w(T))_{L^2(S)} - (v(0), w(0))_{L^2(S)} - \int_0^T \langle v, w' \rangle_{H^{1}, H^{-1}}$$

I am wondering what role absolute continuity plays here. I know if a real-valued (normal in the undergrad sense) function is abs. cont. then it satisfies an equation similar to the one above but this one involves dual space pairing so I can't see how it is analgous. I would appreciate someone explaining me this. Thanks.

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up vote 1 down vote accepted

Here's a somewhat rough sketch of what's going on. Absolute continuity plays the same role as usual: since $(v(t),w(t))_{L^2}$ is an absolutely continuous real valued function of $t$, $\frac{d}{dt}[(v,w)_{L^2}]$ exists for almost every $t\in (0,T)$ by the Lebesgue differentiation theorem. Now to compute this derivative, we use Reynold's transport theorem (assuming $S$ does not change with respect to time):

$$ \frac{d}{dt} \int_S v(t)w(t) dV = \int_S\frac{d}{dt}( v(t)w(t))=\int_S(v^\prime w+vw^\prime)=\langle v^\prime,w\rangle+\langle v,w^\prime\rangle$$

Since we're working with weak derivatives, we interpret $\int_Sv^\prime w$ as the dual pairing $\langle v^\prime,w\rangle_{H^{-1},H^1}$ and similarly we interpret $\int_S vw^\prime$ as $\langle v,w^\prime\rangle_{H^1,H^{-1}}$.

Hope this helps.

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Thank you! Are you sure it's the Lebesgue diff. theorem? That has something to do with taking a limit over a region.. – soup Nov 15 '12 at 21:18
Sure no prob! It might have another name, but they're essentially equivalent - Lebesgue diff thm covers all types of 'differentiating the integral' theorems. – icurays1 Nov 15 '12 at 21:51

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