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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?

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Try approximating $v(t)$ from the first definition by a piecewice constant function. –  timur Jun 12 '13 at 19:48
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