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Consider the weak formulation: find $u \in L^2(0,T;V)$ with $u' \in L^2(0,T;V')$ such that for all $v \in L^2(0,T;V)$, $$\langle u'(t), v(t) \rangle_{V',V} + \langle Au(t), v(t) \rangle_{V',V} = \langle f(t), v(t) \rangle_{V',V}$$ holds for almost every $t \in [0,T].$ Here $A:V \to V'$ is some operator (eg. Laplacian).

Presumably the null set on which this equation doesn't hold is the same for every $v.$ What would happen in the null set is different for every $v$? How would the theory change?

I have never seen null sets being discussed, except for one line in Zeidler where he says the null sets are the same. A similar question was asked here: https://math.stackexchange.com/questions/418693/need-explanation-of-passage-about-lebesgue-bochner-space

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  • $\begingroup$ I don't understand the notation. Is $V$ a vector space, perhaps finite dimensional? Is $V'$ the dual space? Is $b$ a constant coefficient bilinear form? $\endgroup$
    – Ben McKay
    Commented Sep 1, 2013 at 9:11
  • $\begingroup$ @BenMcKay Yes, $V$ is some Banach space (eg. $H^1(\Omega)$), $V'$ is its dual. The specific equation is not that important though. $\endgroup$
    – maximumtag
    Commented Sep 1, 2013 at 9:58
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    $\begingroup$ The integrals in your bilinear forms are time integrals, thus the weak formulation may hold for all $v$, but cannot hold for almost every $t$. You would have to look at Dirac-sequences of test functions to obtain a result of the type "for almost every t". But this almost certainly will require more regularity of $u$. $\endgroup$
    – Guido Kanschat
    Commented Sep 1, 2013 at 9:59
  • $\begingroup$ @GuidoKanschat I'm sorry, please see my edited post. My notation was misleading. $\endgroup$
    – maximumtag
    Commented Sep 1, 2013 at 10:04
  • $\begingroup$ Then your test function should be $v\in V$. $\endgroup$
    – Guido Kanschat
    Commented Sep 1, 2013 at 10:49

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The meaning of your question is not completely clear to me. Anyway, the following remark might be useful: given a weak solution $u(t)$, the set of test functions $v(t)$ satisfying the identity against $u$ is a closed subset of $L^2(0,T;V)$, thus it is sufficient to use test functions belonging to $C([0,T];V)$ to check for a weak solution (at least when $V$ is Hilbert). This shows that the times where the identity does not hold depend only on the (representative chosen for the) function $u$ itself and not on the choice of the test function.

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  • $\begingroup$ Do you mean the displayed equation in the OP by "the identity" that $v$ satisfies? If so then the set of test functions is the whole of $L^2(0,T;V)$ by definition, right? Presumably the sufficiency of testing with $C(0,T;V)$ comes from some density result. Also couldn't it be the case that even though testing with $v \in C(0,T;V)$ is well-defined in that $v(t)$ is well-defined for all $t$, the equality may still not hold on a null set in $[0,T]?$ Sorry for so many questions! $\endgroup$
    – aere
    Commented Sep 29, 2013 at 16:16
  • $\begingroup$ I think you are right with your questions, in the sense that my remark does not clarify these points. It only suggests that one can drastically restrict the test functions used and focus on the singularities of $u$ itself. E.g., if $V$ is separable one might use as test functions step functions taking values in a dense countable set $\endgroup$
    – Piero D'Ancona
    Commented Oct 2, 2013 at 11:00

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