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Suppose we are interested in the existence of a $u \in L^2(0,T;V)$ with $u' \in L^2(0,T;V^*)$ such that $$(u(T),\varphi(t))_H -\int_0^T \langle \varphi'(t), u(t) \rangle_{V^*,V} + \int_0^T a(u(t),\varphi(t)) = (u_0, \varphi(0))_H$$ holds for all $\varphi \in L^2(0,T;V)$ with $\varphi \in L^2(0,T;V^*)$.

What happens if instead look for solutions with requiring $\varphi' \in L^2(0,T;H)$, i.e., the weak formulation becomes: find $u \in L^2(0,T;V)$ with $u' \in L^2(0,T;V^*)$ such that $$(u(T),\varphi(t))_H -\int_0^T (\varphi'(t), u(t))_H + \int_0^T a(u(t),\varphi(t)) = (u_0, \varphi(0))_H$$ holds for all $\varphi \in L^2(0,T;V)$ with $\varphi' \in L^2(0,T;H)$.

The reason I want to do this because I consider an approximate problem $$(u_n(T),\varphi(t))_H -\int_0^T \langle \varphi'(t), A_nu_n(t) \rangle_{V^*,V} + \int_0^T a(u_n(t),\varphi(t)) = (u_0, \varphi(0))_H$$ where $u_n$ satisfies some estimates which allow us to pass to the limit in every term except the second one, where the only estimate I have is $Au_n \rightharpoonup u$ in $L^2(0,T;H)$. So this is why I want to change the test space, so that this estimate will suffice (the duality product turns into an inner product). Is there something bad about doing this?

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    $\begingroup$ It is quite common to have different definitions of weak solutions to the same equation. In particular, the space of test functions quite often vary in different papers for the same equation. Often it turns out that the definitions are equivalent and the authors use the definition which suits them best. But of course some definitions are more standard than others. To be safe, you should prove the equivalence of these different definitions (if true). $\endgroup$
    – Juhana Siljander
    Commented Jun 2, 2014 at 16:32

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