# About definition of weak derivative in abstract PDE problems

I'm confused about weak derivative definition.

$u \in L^2(0,T;V)$ has weak derivative $u'\in L^2(0,T;V')$ iff

$$\int_0^T u(t)\varphi'(t) = -\int_0^T u'(t)\varphi(t)$$

holds for all $\varphi \in C_0^\infty(0,T).$

The LHS is an object in $V$. What is the RHS? Should the integrand be $\langle u'(t), \varphi(t) \rangle_{V',V}$ (edit: I guess not)? If so then it is a real number. If not, it is in $V'$.

Help would be appreciated

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This question belongs in math.stackexchange.com and not here. –  Deane Yang Apr 27 '13 at 15:49
@DeaneYang I read math.stackexchange.com/questions/250690/… and I have read conflicting statements on MSE so it seemed better to ask it here :| –  martin_e Apr 27 '13 at 15:53
I've posted an answer on MSE. –  Deane Yang Apr 27 '13 at 18:35
Hi @AaronHoffman my test functions are real-valued. I disagree that it will be $V$-valued (unless it lies in that stronger space). Please see Definition 5.7 here, books.google.co.uk/…, which i believe answered my doubts. I would say that both integrals will be $V*$ valued. –  martin_e Apr 27 '13 at 19:11
@DeaneYang thank you for your answer. But why restrict the derivative $u'(t)$ to lie in such a strong space? $H^{-1}$ is better and seems more widely used. As I understand it now (after reading the book I linked to above), $u'(t)\varphi(t)$ makes sense because you're just multiplying an element of $V^*$ by some real-valued function, so no problems. It's not the dual pairing in this case. –  martin_e Apr 27 '13 at 19:14