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I have looked through some MO and ME posts, and the common opinion is that weak and mild solutions are equivalent for "many" cases of linear parabolic equation.

However, detailed proofs can only be found for $L^2\bigl([0,T],L^2(\Omega)\bigr)$ where $\Omega \subset \mathbb{R}^3$ is a bounded region with smooth boundaries.

I would like to know if such equivalnce holds for $L^q\bigl([0,T],L^p(\Omega)\bigr)$ with $1 <q <\infty$ and $1 < p \leq 6/5$ and $T \in (0,\infty)$.

More precisely, let us consider the following abstract Cauchy problem: \begin{equation} \partial_t U - \Delta U = f \text{ for } t>0 \text{ and } U(\cdot,0)=u_0 \end{equation} where $f \in L^q\bigl([0,T],L^p(\Omega)\bigr)$ with $1 <q <\infty$ and $1 < p \leq 6/5$ while $u_0 \in W^{2, p }(\Omega)$.

Then, it is known, cf. https://arxiv.org/pdf/2110.10442.pdf, that there exists a unique "mild" solution $U(t) \in W^{1,q}\bigl([0,T],L^p(\Omega)\bigr)$ satisfying the "maximal regularity estimate" \begin{equation} \lVert \partial_t U \rVert_{ L^q\bigl([0,T],L^p(\Omega)\bigr)}+\lVert \Delta U \rVert_{ L^q\bigl([0,T],L^p(\Omega)\bigr)} \leq C \Bigl( \lVert u_0 \rVert_{W^{2, p }(\Omega)} + \lVert f \rVert_{ L^q\bigl([0,T],L^p(\Omega)\bigr)}\Bigr) \end{equation} for some universal constant $C>0$.

Now, I define a "weak" solution of the above Cauchy problem as $G \in L^q\bigl([0,T],H^1_0(\Omega) \bigr)$ with $\partial_t G \in L^q\bigl([0,T],H^{-1}(\Omega) \bigr)$ such that \begin{equation} \bigl\langle [\partial_t G](t_0), v \bigr\rangle + \int_{\Omega} \nabla G(x,t_0), \nabla v(x)dx=\int_{\Omega} f(x,t_0) \cdot v(x) dx \end{equation} for each $v \in H^1_0(\Omega)$ and a.e. $t_0 \in [0,T]$, while $G(\cdot,0)=u_0$. Here, $\langle, \rangle$ is the bilinear pairing.

Note that since $1 < p \leq 6/5$, the integral $\int_{\Omega} f(x,t_0) \cdot v(x) dx$ makes sense for any $v \in H^1_0(\Omega)$.

I repeat my question : Are the above $U$ and $G$ equivalent notions? In other words, is $G$ the "unique mild" solution of the Cauchy problem and satisfies the above maximal regularity estimate?

I tried to prove it directly myself but cannot proceed easily.. Could anyone please help me?

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    $\begingroup$ I think the basic problem with this is that there's no reason for $G$ to exist (as a function in the spaces you mention): ignoring the $t$ variable entirely, you have $f \in L^p$ for some small $p$, so you expect that $\nabla u \in L^{\frac{n}{n - 1} + \delta}$ for a small $\delta$ at best, so like $3/2 + \delta < 2$ in 3D. If you do have a $G$, you can show that $U$ satisfies the distributional form with a more restrictive set of test functions, so $G - U$ is a distributional solution to the heat equation. You would then argue these are smooth ... $\endgroup$
    – user378654
    Commented Aug 30, 2023 at 22:49
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    $\begingroup$ for example maybe by explicitly verifying the heat equation mean value property. This then implies that $U - G = 0$, as smooth solutions are unique (I'm ignoring technicalities with the boundary). There are famous examples where if you have nonsmooth coefficients, this program actually fails, as in there are distributional solutions to the homogeneous equation (with gradient not in $L^2$) which are not as regular as optimal for that equation (i.e. not bounded, say). So in that sense, your notion of solution $U$ is "stronger." $\endgroup$
    – user378654
    Commented Aug 30, 2023 at 22:54
  • $\begingroup$ @user378654 Thank you so much for your comment. I think your concern regarding $\nabla u$ would be resolved if we just fix $p=6/5$. $\endgroup$
    – Isaac
    Commented Aug 31, 2023 at 4:19

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