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Let

  • $T>0$
  • $H$ be a separable $\mathbb R$-Hilbert space
  • $u_0\in H$
  • $(e_n)_{n\in\mathbb N}\subseteq\mathcal D(A)$ be an orthonormal basis of $H$ with $$Ae_n=\lambda_ne_n\;\;\;\text{for all }n\in\mathbb N$$ for some $(\lambda_n)_{n\in\mathbb N}\subseteq(0,\infty)$ with $$\lambda_{n+1}\ge\lambda_n\;\;\;\text{for all }n\in\mathbb N$$
  • $S(t):=e^{-tA}$ with $$e^{-tA}x:=\sum_{n\in\mathbb N}e^{-t\lambda_n}\langle x,e_n\rangle_He_n\;\;\;\text{for }x\in H$$ for $t\ge 0$
  • $f:H\to H$ with $$\left\|f(x)\right\|_H\le L(1+\left\|x\right\|_H)\;\;\;\text{for all }x\in H\tag1$$ and $$\left\|f(x)-f(y)\right\|_H\le L\left\|x-y\right\|_H\;\;\;\text{for all }x,y\in H\tag2$$

Note that $S$ is a $C^0$-semigroup on $H$ and $-A$ is the infinitesimal generator of $S$.

It's well known that there is a unique $u\in C^0([0,T],H)$ with $$u(t)=S(t)u_0+\underbrace{\int_0^tS(t-s)f(u(s))\:{\rm d}s}_{=:\:v(t)}\;\;\;\text{for all }t\in[0,T]\tag 3\;.$$

Let $t\in (0,T)$ and $h\in[0,T-t)$. Then, $$\frac{v(t+h)-v(t)}h=\frac{S(h)-\operatorname{id}_H}hv(t)+\underbrace{\frac1h\int_t^{t+h}S(t+h-s)f(u(s))\:{\rm d}s}_{=:\:I_h(t)}\tag 4$$ with $$I_{\tilde h}(t)\xrightarrow{\tilde h\to0}f(u(t))\tag 5\;.$$

Question: What can we say about the regularity of $v$? Is $v$ (continuously) differentiable? And if so, in which sense is it a solution of $$\frac{{\rm d}v}{{\rm d}t}(\tilde t)=-Av(\tilde t)+f(u(\tilde t))\tag 6$$ for all $\tilde t\in[0,T]$?

The question arose as I read page 111 of An Introduction to Computational Stochastic PDEs. The authors claim that $u$ is not "smooth enough to interpret $(6)$ directly, as we don't know whether $u(t)\in\mathcal D(A)$".

Now, $$\frac{S(h)-\operatorname{id}_H}hv(t)=\int_0^t\frac{S(h)-\operatorname{id}_H}hS(t-s)f(u(s))\:{\rm d}s\tag 7$$ and (since $S(t-s)f(u(s))\in\mathcal D(A)$ for all $s\in[0,t)$) $$\frac{S(\tilde h)-\operatorname{id}_H}{\tilde h}S(t-s)f(u(s))\xrightarrow{\tilde h\to0}-AS(t-s)f(u(s))\;\;\;\text{for all }s\in[0,t)\tag 8\;.$$ Since the convergence in $(8)$ is uniformly with respect to $s$ and $t$, the right-hand side of $(7)$ is convergent as $h\to0+$. Thus, the left-hand side of $(7)$ is convergent as $h\to0+$ and hence $v(t)\in\mathcal D(A)$ by the very definition of $\mathcal D(A)$.

So, we should be able to conclude $u(t)\in\mathcal D(A)$. Did I made any mistake or are the authors wrong?

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  • $\begingroup$ Your (7) looks suspicious to me. $\endgroup$
    – Fan Zheng
    Jan 3 '17 at 19:09
  • $\begingroup$ @FanZheng $S(h)-\text{id}_H$ is a bounded linear operator on $H$ and it's easy to see that $$[0,t]\ni s\mapsto S(t-s)f(u(s))\tag 9$$ is integrable over $[0,t]$. It's well-known that $(7)$ holds in this situation. $\endgroup$
    – 0xbadf00d
    Jan 3 '17 at 19:24
  • $\begingroup$ Maybe there is an other problem: Denote the function $(9)$ by $g$. We have the estimate $$\left\|Ag(s)\right\|\le\frac C{t-s}\;\;\;\text{for all }s\in[0,t)\tag{10}$$ for some $C\ge 0$. However, $$\int_0^t\frac1{(t-s)^\alpha}\:{\rm d}s$$ won't exist, unless $\alpha<1$. So, if there is no stronger estimate than $(10)$, $$\int_0^tAg(s)\:{\rm d}s$$ might be undefined. However, in that case the book would be heavily wrong and that's why I guess that I'm missing something. $\endgroup$
    – 0xbadf00d
    Jan 3 '17 at 19:51
  • $\begingroup$ I guess you are missing some terms as when f=0 the RHS is 0 but the RHS is not. $\endgroup$
    – Fan Zheng
    Jan 3 '17 at 19:54
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    $\begingroup$ @0xbadf00d: No, this semigroup is NOT uniformly continuous. Uniform continuity fails at t=0! $\endgroup$ Jan 4 '17 at 11:35
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As I don't know exactly what you would like to ask on $A$, the answer I give can only be very vague and not much better than the comments given by Hannes. It would be good if you could specify your question (e.g. do you mean regularity in space or in time?) and give a link to the version of the book you are referring to?

In my version of the book you refer to, Lemma 3.35 on page 111 tells you that $u(t) \in \mathcal{D}(A^{1/2})$. You can see why by just going through the proof of Lemma 3.22. on page 102, where the regularity of solutions is derived via the properties of the semigroup.

As already indicated by Hannes, without further assumption we cannot prove that $u(t) \in \mathcal{D}(A)$, as this would require the finiteness of $\| A u(t) \|$ instead of $\| A^{1/2} u(t) \| < \infty$, as shown in Lemma 3.35. The condition $\| A u(t) \| < \infty$ is true e.g. in the case of a fast enough growth of the eigenvalues of A (which depends also on the dimension of the open set $D$).

Taking for example the $A= \Delta$ with Dirichlet boundary conditions, the domain of $A$ is $\mathcal{D}(A) = H^2 \cap H^1_0$, as pointed out also in Example 3.24.

Now I get even more vague in w.r.t. answering your original question: How regular a solution to a semilinear (S)PDE is depends on the operator $A$ and how much $A$ "smoothes" the initial data over time. Also, as you pointed out already, the type of solution that one is looking for makes a difference in the results one can get. So there is no quick and general answer to this question. If you are looking for regularity results on second order operators, best is probably to check Gilbarg-Trudinger. Showalter has results in case that $A$ is a monotone operator.

The problem is also that the chapter of this book has just merely informative character, it just states a couple of background results from the deterministic setting needed in later chapters (when we do have a noise term). So it does not aim in being concise. The setup given is also aligned to the setup one classically uses in SPDE theory, where you will see that the condition $\| A u(t) \| < \infty$ appears in the definition of so-called "strong solutions" to SPDEs.

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  • $\begingroup$ There are two questions: (a) I claim that $u(t)\in\mathcal D(A^α)$ for all $α<1$ by the same argument as in the book. Can we show more? $\endgroup$
    – 0xbadf00d
    Jan 6 '17 at 22:30
  • $\begingroup$ (b) What can we say about $\frac{{\rm d}u}{{\rm d}t}$? The book states $\frac{{\rm d}u}{{\rm d}t}∈\mathcal D(A^{-1/2})$ which doesn't make much sense, since $\mathcal D(A^β)=H$ for all $β≤0$; if I'm not terribly wrong. Moreover, the book states that $\frac{{\rm d}u}{{\rm d}t}$ satisfies $(16)$ (from the comments below the question), which should imply $\frac{{\rm d}u}{{\rm d}t}\in H$. And I'm curious whether $(19)$ can be concluded from $(18)$, since I see no other way to obtain $(16)$. $\endgroup$
    – 0xbadf00d
    Jan 6 '17 at 22:30
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This problem has been addressed by many authors. Here I present a general result (Theorem 4.1) from the "Abstract parabolic equations and their applications" by A. Yagi. Let us consider the following semilinear equation in $X$: \begin{equation} \left\{\begin{array}{l} u'(t)+Au(t)=F(u)+g(t), t>0\\ u(0)=u_0, \end{array}\right. \end{equation} where $F:D(A^\eta)\to X$, ($0<\eta< 1$) satisfies $$ \|F(u)-F(v)\|\leq \varphi(\|A^\beta u\|+\|A^\beta v\|)\|A^\eta(u-v)\|+(\|A^\eta u\|+\|A^\eta u\|)(\|A^\beta (u-v)\|)\quad \forall~u,v\in D(A^\eta). $$ with some $0<\beta \leq \eta<1$. Then we have the following result: Given any $g\in \mathcal{F}^{\beta,\sigma}((0,T];D(A^\beta))$, where $0<\sigma<1-\eta$, and any $u_0\in D(A^\beta)$. The equation above possesses a unique local solution $U$ in the function space: $$ u\in C((0,T_{g,u_0}];D(A))\cap C([0,T_{g,u_0}];D(A^\beta))\cap C^{1}((0,T_{g,u_0}];X), $$ and $$ \frac{d u}{dt},~Au\in \mathcal{F}^{\beta,\sigma}((0,T];X), $$ where $T_{g,u_0}$ depends only on the norms of $g$ and $A^\beta u_0$. You can apply this result directly. Below is the definition of the function space used here. (It is midnight in Hong Kong, and I really need to sleep. This is just a draft. Sorry to use the picture instead of typing myself. If I have time, then I may update more).

enter image description here

enter image description here

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  • $\begingroup$ The problem is that your theorem requires $u_0\in\mathcal D(A^\beta)$ for some $\beta\in(0,1)$; but we only know that $u_0\in H$. Moreover, you say that I can imply this result directly, but it only gives me a local solution, while I'm searching for a global one. $\endgroup$
    – 0xbadf00d
    Jan 8 '17 at 12:23

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