Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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.

share|improve this answer
    
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.