Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose $$u'(t) + A(t)u = f(t)$$ holds in the sense of $L^2(0,T;V')$.

Is the problem

Find $u \in L^2(0,T;V)$ with $u' \in L^2(0,T;V')$ such that for all $v \in L^2(0,T;V)$, $$\langle u'(t), v(t) \rangle + a(u(t),v(t)) = \langle f(t), v(t) \rangle$$ for a.e. $t \in [0,T]$

equivalent to

Find $u \in L^2(0,T;V)$ with $u' \in L^2(0,T;V')$ such that for all $v \in V$, $$\langle u'(t), v \rangle + a(u(t),v) = \langle f(t), v \rangle$$ for a.e. $t \in [0,T]$

The second one is more common in the literature, but the first one seems a lot more natural to me. Any comments?

share|improve this question
    
Try approximating $v(t)$ from the first definition by a piecewice constant function. –  timur Jun 12 '13 at 19:48
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.