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Let

  • $H$ be a separable $\mathbb R$-Hilbert space
  • $S:[0,\infty)\to H$ be an uniformly continuous$^1$ semigroup
  • $-A$ be the infinitesimal generator of $S$
  • $f:H\to H$ be Lipschitz continuous with sublinear growth
  • $t>0$
  • $u\in C^0([0,T],H)$

I've read (in An Introduction to Computational Stochastic PDEs on page 111) that $$\frac{{\rm d}}{{\rm d}t}\int_0^tS(t-s)f(u(s))\:{\rm d}s=-\int_0^tAS(t-s)f(u(s))\:{\rm d}s+f(u(t))\;.\tag 1$$

How can we prove $(1)$?

I've tried to write $$\frac{{\rm d}}{{\rm d}t}\int_0^tS(t-s)f(u(s))\:{\rm d}s=\lim_{h\to0+}\left(\int_0^t\frac{S(h)-\operatorname{id}_H}hS(t-s)f(u(s))\:{\rm d}s+S(h)\frac1h\int_t^{t+h}S(t-s)f(u(s))\:{\rm d}s\right)\tag 2$$ and it's clear that the first term on the right-hand side of $(2)$ converges to $-\int_0^tAS(t-s)f(u(s))\:{\rm d}s$. However, the second term is not even well-defined, cause $S$ is evaluated at negative points. So, what's the correct approach?

$^1$ i.e. $S\in C^0([0,\infty),\mathfrak L(H))$.

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    By the way, can you recommend the book you cited as useful for learning how to use stochastic differential equations? (Not necessarily PDE). – 2017-01-01
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    @s.harp I can recommend this book if you're interested in the *numerical* simulation of S(P)DEs. If you're interested in an elegant and general theory of stochastic integration and SDEs, then I can recommend *Foundations of Modern Probability* by *Olav Kallenberg*. However, for the beginning, I would recommend *PDE and Martingale Methods in Option Pricing* by *Andrea Pascucci*. Don't let you confuse by the title. It's a really good book for an introduction into stochastic integration (in $\mathbb R^d)$ and SDEs. The corresponding chapters are (almost) pure mathematics. – 2017-01-01

2 Answers 2

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Define $G:\mathbb R\to \mathfrak{L}(H)$ by $$G(t)=\left\{\begin{align} S(t),&\quad t\geq 0\\ [S(-t)]^{-1},&\quad t<0 \end{align}\right.$$ As $S$ is uniformly continuous, $S(t)$ is invertible for all $t\geq 0$ and thus $G$ is well defined. In addition, $G$ is a uniformly continuous group.

Using your idea we obtain:

$$\begin{align} \frac{{\rm d}}{{\rm d}t}\int_0^tS(t-s)f(u(s))\:{\rm d}s=\frac{{\rm d}}{{\rm d}t}\int_0^tG(t-s)f(u(s))\:{\rm d}s\\ =\lim_{h\to0+}\left(\int_0^t\frac{G(h)-\operatorname{id}_H}hG(t-s)f(u(s))\:{\rm d}s+G(h)\frac1h\int_t^{t+h}G(t-s)f(u(s))\:{\rm d}s\right)\\ =\lim_{h\to0+}\left(\int_0^t\frac{S(h)-\operatorname{id}_H}hS(t-s)f(u(s))\:{\rm d}s+S(h)\frac1h\int_t^{t+h}G(t-s)f(u(s))\:{\rm d}s\right) \end{align}$$

Now the second term is well-defined and converges to $$S(0)G(t-s)f(u(s))\big|_{s=t}=f(u(t)).$$

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    The second term in the limit on the right-hand side of $(2)$ should be $$\frac1h\underbrace{\int_t^{t+h}S(t+h-s)f(u(s))\:{\rm d}s}_{=:\:I_2(h)}\;.$$ Since $S$ is uniformly continuous, $$\left\|S(r)-\text{id}_H\right\|_{\mathfrak L(H)}\xrightarrow{r\to0+}0$$ and hence $$\left\|S(t+h-s)f(u(s))-f(u(t))\right\|_H\le\left\|S(t+h-s)-\text{id}_H\right\|_{\mathfrak L(H)}\left\|f(u(s))\right\|+\left\|f(u(s))-f(u(t))\right\|_H\xrightarrow{h\to0+}0$$ for all $h\ge 0$ and $s\in[t,t+h]$. – 2017-01-01
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    So, wen can conclude that $$\left\|\frac1h I_2(h)-f(u(t))\right\|_H\le\frac1h\int_t^{t+h}\left\|S(t+h-s)f(u(s))-f(u(t))\right\|_H{\rm d}s\xrightarrow{h\to0+}0\;.$$ The limit from the other side can be calculated in the same way. So, this approach should be correct to. Do you agree? – 2017-01-01
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    @0xbadf00d I did not understand why $$\left\|S(t+h-s)-\text{id}_H\right\|_{\mathfrak L(H)}\left\|f(u(s))\right\|+\left\|f(u(s))-f(u(t))\right\|_H\xrightarrow{h\to0+}0.$$ – 2017-01-01
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    If $h\to0+$, then $[t,t+h]\ni s\to t+$ and hence $t+h-s\to0+$. Now use the uniform continuity of $S$ and the continuity of $f\circ u$. – 2017-01-02
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    @0xbadf00d Your approach looks right, I don't see nothing wrong. – 2017-01-02
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    In the book I've linked to in the question ([page 111](http://tinyurl.com/hr7futa)), the authors claim that $\frac{{\rm d}u}{{\rm d}t}\in\mathcal D(A^{-1/2})$. Isn't that a trivial statement, since $H\subseteq\mathcal D(A^\beta)$ for all $\beta<0$? Note that $A$ and $S(t)=e^{-tA}$ are defined as in [my other question](http://math.stackexchange.com/questions/2077755/regularity-of-the-mild-solution-of-a-semilinear-pde). – 2017-01-02
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    Asked for that in a [separate question](http://math.stackexchange.com/questions/2080594/regularity-of-the-derivative-of-the-mild-solution-to-a-semilinear-pde). – 2017-01-02
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    @0xbadf00d Maybe there is something wrong, or maybe I'm missing something. Wouldn't we need uniform continuity for $f\circ u$ to pass the limit in the integral? – 2017-01-02
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    $f\circ u$ *is* uniformly continuous on $[0,T]$, since $f\circ u$ is continuous on $[0,T]$ and $[0,T]$ is compact. – 2017-01-02
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    @0xbadf00d You didn't said explicitly that the convergence $\left\|S(t+h-s)f(u(s))-f(u(t))\right\|_H\xrightarrow{h\to0+}0$ is uniform, but it is important to the conclusion $$\frac1h\int_t^{t+h}\left\|S(t+h-s)f(u(s))-f(u(t))\right\|_H{\rm d}s\xrightarrow{h\to0+}0\;.$$ – 2017-01-02
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    Yes, you're right. But let me ask again, do you know a solution to [my other problem](http://math.stackexchange.com/questions/2080594/regularity-of-the-derivative-of-the-mild-solution-to-a-semilinear-pde)? In light of your help so far, I guess that you know it. There must be something trivial, that I don't see. – 2017-01-02
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    @0xbadf00d In view of Lemma 1.89(iv), which you proved in a prior post, the said statement really seems trivial. If you are missing something, I'm missing it too. – 2017-01-02
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    Just to be sure: So, you agree to me that $\frac{{\rm d}u}{{\rm d}t}\in\mathcal D(A^{\beta})$ for all $\beta\le0$ and, most importantly, $u\in C^1((0,T),H)$, right? – 2017-01-03
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    It's still unclear to me why the authors claim that $u(t)\not\in\mathcal D(A)$. I've asked for that in [another question](http://mathoverflow.net/questions/258687/regularity-of-the-mild-solution-of-a-semilinear-evolution-equation). – 2017-01-03
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    Maybe I've found what I (and in that case the book too) did wrong: Let $$g(s):=S(t-s)f(u(s))\;\;\;\text{for }s\in[0,t]\;.$$ Then it's clear that $$g(s)\in\mathcal D(A)$$ and hence $$\frac{S(h)-\text{id}_H}hg(s)\xrightarrow{h\to0+}-Ag(s)\tag3$$ for all $s\in[0,t)$. Due to their representation of $u'$, I guess that the authors (and so did I) conclude from $(3)$ that $$\int_0^t\frac{S(h)-\text{id}_H}hg(s)\:{\rm d}s\xrightarrow{h\to0+}-\int_0^tAg(s)\:{\rm d}s\;.\tag4$$ But why can we do it? – 2017-01-03
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    If I'm not terribly wrong, the convergence in $(3)$ is **not** uniform. And I don't think that the left-hand side of $(3)$ is bounded by a function integrable over $[0,t]$. Moreover, I'm only aware of the estimate $$\left\|Ag(s)\right\|\le\frac C{t-s}\;\;\;\text{for all }s\in[0,t)\tag5$$ for some $C\ge 0$. And since $$\int_0^t\frac1{(t-s)^\alpha}\:{\rm d}s$$ won't exist, unless $\alpha<0$, $(5)$ is not strong enough to conclude the existence of the right-hand side of $(4)$. – 2017-01-03
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I will write $f$ instead of $f\circ u$. You have not said what $u$ is, but I will assume it to be continuous.

You have made a mistake, the differential quotient is: $$\frac1h\left(\int_0^{t+h}S(t+h-s)f(s)\,ds - \int_0^t S(t-s)f(s)\,ds\right) \\= \int_0^t\frac{S(t+h-s)-S(t-s)}hf(s)\,ds+\frac1h\int_t^{t+h}S(t+h-s)f(s)\,ds$$ Although this is only true if $h>0$. If $h<0$ one must reduce to $$\int_0^{t-|h|}\frac{S(t-|h|-s)-S(t-s)}{-|h|}f(s)\,ds+\frac1{-|h|}\int_{t-|h|}^tS(t-s)f(s)\,ds$$ instead because otherwise the terms are not well defined. I'll only consider the quotient for $h>0$, the other case should be analog.

Now the interior of the first sum is, if $h>0$: $\frac{S(h)-1}hS(t-s)f(s)$ and the $\frac{S(h)-1}h$ term can be pulled out of the integral as you correctly noted.

For the other term: $$\left\|\frac1h\int_t^{t+h}S(t+h-s)f(s)-f(t)\,ds\right\|≤\frac hh\sup_{s\in[t,t+h]}\|S(t+h-s)f(s)-f(s)+f(s)-f(t)\|\\ ≤\sup_{s\in[t,t+h]}\|S(t+h-s)-1\|\cdot\sup_{s\in[t,t+h]}\|f(s)\|+\sup_{s\in[t,t+h]}\|f(s)-f(t)\|$$ From continuity of the functions this goes to $0$ as $h\to0$.

Put everything together and you get the derivative stated.

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    There is no mistake in my differential quotient. By the semigroup property, $$\frac{S(t+h-s)-S(t-s)}hf(u(s))=\frac{S(h)-\operatorname{id}_H}hS(t-s)f(u(s))$$ and $$\int_t^{t+h}S(t+h-s)f(u(s))\:{\rm d}s=S(h)\int_t^{t+h}S(t-s)f(u(s))\:{\rm d}s\;.$$ And note that I'm taking the limit from the right, $h\to0+$. – 2017-01-01
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    I did not see the $S(h)$ that had been pulled out. But you cannot do it, because (when $h>0$) it is the same as writing $S(0)=S(1)S(-1)$, where the right-hand side does not exist. – 2017-01-01
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    However, as I said before, the last integral is not well-defined. The correct approach is to write $$\frac{{\rm d}}{{\rm d}t}\int_0^tS(t-s)f(u(s))\:{\rm d}s=\lim_{h\to 0+}=\lim_{h\to0+}\left(\int_0^t\frac{S(h)-\operatorname{id}_H}hS(t-s)f(u(s))\:{\rm d}s+\frac1h\underbrace{\int_t^{t+h}S(t+h-s)f(u(s))\:{\rm d}s}_{=:I_2(h)}\right)$$ and use $$I_2(h)=\int_0^hS(h-r)f(u(t+r))\:{\rm d}r\;.$$ – 2017-01-01
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    $u\in C^0([0,T],H)$. I've added that to the question. – 2017-01-01
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    Strictly speaking the derivative is not only the right-sided limit of the quotient, but the last formula is well defined and the same that I have^^. – 2017-01-01
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    s.harp: with respect to the "forbidden step", @0xbadf00d is implicitly using the extension of $S$ (see my answer). – 2017-01-01
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    ... which is wrong if it is not explicitly stated. – 2017-01-01