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Let $u \in L^2(0,T;V)$ with $u_t \in L^2(0,T;V^*)$ be a solution of $$\langle u_t(t), v \rangle + a(t;u(t), v) = \langle f(t), v \rangle$$ where $f \in L^2(0,T;V^*)$ and we have the usual assumptions on $V$ and $a$ (i.e. $V \subset H \subset V^*$ forms a Gelfand triple and $a(t;.,.)$ is a bilinear form associated to some elliptic coercive linear operator).

Now this equality holds almost everywhere for all $v$, i.e., for all $v \in V$ and all $t \in [0,T]/Z$ where $Z$ is a set of measure zero which does not depend on $v$ (see Zeidler).

What happens if the null set were to depend on $v$? Does anything go wrong? I read the thread Null sets in PDE but it didn't answer this question.

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  • $\begingroup$ I'm confused; can you clarify how this question is different from Null sets in PDE? Also, can you elaborate on "the usual assumptions"? $\endgroup$ Commented Oct 2, 2014 at 2:00
  • $\begingroup$ @NateEldredge I edited to include the assumptions. The thread is similar (the variational formulation is slightly different) and it seems to me (if I understood) that there is not a satisfactory answer to that question. I think it important to get a definitive understanding of this because it is overlooked often. $\endgroup$
    – MMML
    Commented Oct 2, 2014 at 7:04

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