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I already asked the question on MSE, and have tried to figure it out myself.

But the problem seems trickier than expected, so I guess MO is a better place to ask..

For the sake of completeness, I repeat the question.

For each $n \in \mathbb{N}$, let us consider the Cauchy problem \begin{equation} \partial_t u_n -\Delta u_n = f_n \text{ on } [0,T] \times \mathbb{T}^m \text{ with } u_n(t=0,x)=u^{0}_n(x) \end{equation} where $f_n(t,x) \in L^p_t\bigl( [0,T], L^q_x(\mathbb{T}^m) \bigr)$ for some fixed $p,q \in (1,\infty)$ and $u^{0}_n(x) \in W^{2,q}(\mathbb{T}^m)$. $T \in (0,\infty)$ is assumed to be fixed and $\mathbb{T}^m=(\mathbb{R}/\mathbb{Z})^m$ is the $m$-torus.

Then, it is well-known that there exists a unique weak solution, denoted as $u_n(t,x)$ for the above Cauchy problem.

Now, suppose that $u^{0}_n \to u^0$ in $W^{2,q}(\mathbb{T}^4)$ and $f_n \to f$ in $L^p_t\bigl( [0,T], L^q_x(\mathbb{T}^m) \bigr)$. Denote by $u(t,x)$ the unique weak solution of the Cauchy problem \begin{equation} \partial_t u -\Delta u = f \text{ on } [0,T] \times \mathbb{T}^m \text{ with } u(t=0,x)=u^0(x) \end{equation}

My question is that, do we have \begin{equation} u_n \to u \text{ in } W^{1,p}_t\bigl([0,T], L^q_x(\mathbb{T}^m) \bigr) \text{ as } n \to \infty? \end{equation}

I strongly suspect so, but cannot find a way to prove using maximal regularity or well-posedness. Could anyone please help me?

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    $\begingroup$ I am confused, when you say, "it is well-known that there exists a unique weak solution [...] for the above Cauchy problem", how can you have that not but not a stability estimate w.r.t. to the data (and a suitable function space for the solution)? $\endgroup$
    – Hannes
    Commented Dec 7, 2023 at 7:45
  • $\begingroup$ @Hannes OK, I took the definition of a weak solution as well as equivalence to a mild solution for granted....I will add the details soon. $\endgroup$
    – Isaac
    Commented Dec 7, 2023 at 8:34

1 Answer 1

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Edited to make it correct for future reference.

You might want to look into this expository paper on maximal regularity for linear parabolic equations: https://people.math.ethz.ch/~salamon/PREPRINTS/parabolic.pdf

Then, if you had an estimate of the form $$\|u_n\|_{W^{1,p}_t(L^q_x)} \le C\left(\|u_{n,0}\|_{W^{2,q}} + \|f_n\|_{L^p(L^q)}\right),$$

for the same price, using that the equation is linear, you could use that $$\|u_n - u_m\|_{W^{1,p}_t(L^q_x)} \le C\left(\|u_{n,0} - u_{m,0}\|_{W^{2,q}} + \|f_n - f_m\|_{L^p(L^q)}\right),$$ and thus you have a Cauchy sequence.

Note that $W^{2,q}$ is enough initial regularity to carry over, the optimal space is a Besov (check it in the link), but you can have the same kind of estimate with a more restrictive class.

I don't think you are going to have a stability/convergence result without having some existence + quantitative estimate.

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    $\begingroup$ I set the initial data to be quite regular on purpose, so that maximal regularity can be carried out with $\lVert u^0_n \rVert_{W^{2,q}}$. Then, it seems quite clear that your argument works with $\lVert u^0_n - u^0 \rVert_{W^{2,q}}$! Thank you for your insights! $\endgroup$
    – Isaac
    Commented Dec 7, 2023 at 19:41

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