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Reading Non-linear Elliptic and Parabolic Equations Involving Measure Data by Boccardo$\&$Gallouet , I had trouble understanding the following:

Why is $\psi(u_n)\chi_{(0,t)}$ admissible as a test function in $(62)$?

The reason I'm asking is because characteristic\indicator functions have no smooth derivatives and plus I don't understand in which function space of test functions the authors define the weak formulation of $(59)$.

Below you can have a quick look at the set up of the problem and the point in which I'm stuck. Throughout this paper $\Omega$ is a bounded open set of $\mathbb R^N$ with $N \ge 2$ and $p$ is such that $p \in (2-1/(N+1),\infty)$.

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Any help is much appreciated. Thanks in advance!

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The reason I'm asking is because characteristic\indicator functions have no smooth derivatives and plus I don't understand in which function space of test functions the authors define the weak formulation of (59).

I think we can agree that whatever the authors mean by "$\psi(u_n(t)\chi_{(0,t)}$", they must realise that it isn't a 'test function' in the traditional (smooth, compactly supported) sense, given its insufficient smoothness in both the $x$ and $t$ variables. However, it is possible that 'test function' means something more general here. The sequences $(u_n)$ and $(f_n)$ being considered here are test functions in the traditional sense, so we can still plug things into the natural bilinear pairing between $\mathscr{D}(Q) = C_0^\infty(Q)$ and its dual $\mathscr{D}'(Q)$ (the space of distributions on $Q$) and draw conclusions from this.

The fact that $P(u_n) = f_n$ in $\mathscr{D}(Q)$ implies (tautologically) that

$$ \langle \Psi, P(u_n)\rangle = \langle \Psi,f_n\rangle $$

for all distributions $\Psi\in \mathscr{D}'(Q)$. We can therefore learn things about $u_n$ and $f_n$ by making clever choices of 'test distribution' $\Psi$. When $\Psi$ is integration against a function, said function may arguably be termed a 'test function', which I think may be what is being proposed here.

This sort of terminology occurs in other settings where one has a non-degenerate bilinear pairing; 'test function' often just means 'thing we're free to choose which will fit into the other slot of the pairing from what we already have'.

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  • $\begingroup$ First of all, thank you very much for your time and your reply. Now, if I understood correctly, in the case of this paper $\psi(u_n(t))\chi_{(0,t)}$ is a distribution and since $(u_n)$ and $(f_n)$ are considered as test functions in the traditional sense, the duality pairing is satisfied. Thus $\psi(u_n(t))\chi_{(0,t)}$ is basically the function we use to integrate against $P(u_n),f_n$ and for that reason the wide term "test function" is adopted. Am I right? $\endgroup$
    – kaithkolesidou
    Commented May 9, 2019 at 8:20
  • $\begingroup$ "Thus $\psi(u_n(t))\chi_{(0,t)}$is basically the function we use to integrate against $P(u_n), f_n$ and for that reason the wide term "test function" is adopted." - that is my understanding yes. $\endgroup$
    – DCM
    Commented May 9, 2019 at 17:44

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