I asked this question on Mathematics Stackexchange, but got no answer.

In Pavel's book: Nonlinear Evolution Operators and Semigroups - Applications to Partial Differential Equations, we have the following definitions and result:

Consider the following semilinear equation:

$$\tag{5.59}\begin{cases}u_t(t)=Au(t)+F(u(t)), ~0 \leq t \leq T,\\u(0)=u_0 \in H. \end{cases} $$ where $A:D(A)\subset H \rightarrow H$ be the infinitesimal generator of $C_0$-semigroup $S(t)$ with $\|S(t)\|\leq 1$.

Definition 5.1. (1)A function $u:[0,T]\rightarrow H$ is said to be a classical solution of $(5.59)$ if $u(0)=u_0$ and $$\tag{5.68}u \in C([0,T];D(A))\cap C^1([0,T];H)$$ and $(5.59)$ is satisfied. (2) The function $u$ is a classical solution on $[0,\infty[$ of $(5.59)$ if $(5.68)$ holds for every $T>0$ and $(5.59)$ is satisfied for every $t>0$.

By a mild solution of $(5.59)$ we mean a continuous function $u:[0,T] \rightarrow H$ satisfying $$\tag{5.69} u(t)=S(t)u_0+\int_{0}^{t} S(t-s)F(u(s))\, ds, ~0 \leq t \leq T.$$

Theorem 5.8 If $F:H\rightarrow H$ is locally Lipschitz then for every $u \in H$, there is a unique mild solution $u:[0,T_{\max}[\rightarrow H$ of $(5.59)$ with either $T_{\max}=+\infty$, or $T_{\max}<\infty$ and $\lim_{t \uparrow T_{\max}}\|u(t)\|=+\infty$. If $u_0 \in D(A)$, the mild solution $u$ is classical. (2) If $F$ is Lispschitz continuous, then $T_{max}=+\infty.$

My question:

I would like to know if Theorem 5.8 is valid for $ A $ being just the infinitesimal generator of a $C_0$-semigroup, that is, without the hypothesis that $\|S(t)\|\leq 1$.

I imagine that the answer is positive in the case that $S(t)$ is uniformly bounded, that is, there is $M>0$ such that $\|S(t)\|\leq M$ for $t \geq 0$.

Assuming I have proven that $A-\lambda I$ is the infinitesimal generator of a $C_0$-semigroup of contractions, for some $\lambda>0$. Then, $A$ is the infinitesimal generator of a $C_0$-semigroup $T(t)$ satisfyng $\|T(t)\|\leq e^{\lambda t}$, for all $t \geq 0$. I don't know if the result is valid in this case.

Can anyone help me? Please.


3 Answers 3


Yes the theorem is true in the general case. The semigroup is assumed to be contractive for the sake of simplicity. Of course the assumption is not restrictive since there is always an equivalent norm which makes the semigroup contractive (this is the same idea in the proof of Hille-Yosida theorem to pass from the uniformly bounded case to the contractive case). You can find the general statement (for Lipschitz continuous functions) of the theorem in Theorem 1.4 pp 185. (see also Theorem 1.2) of the book

A. Pazy, Semigroups of linear operators and applications to partial differential equations. Springer 1983.

Edit: As for the second part on the improved regularity, the result is true for locally Lipschitz functions, when $X$ is reflexive (in your case it is a Hilbert space). See Proposition 4.3.9 pp 60 (and Remark 4.3.10) in the book

T. Cazenave and A. Haraux, An Introduction to Semilinear Evolution Equations, Oxford 1998.

PS: One should note the definition of Lipschitz continuous function in pp 55.

  • $\begingroup$ By using Theorem 1.4 I agree with the first part of Theorem 5.8, that is, If $F:H\rightarrow H$ is locally Lipschitz then for every $u \in H$, there is a unique mild solution $u:[0,T_{\max}[\rightarrow H$ of $(5.59)$ with either $T_{\max}=+\infty$, or $T_{\max}<\infty$ and $\lim_{t \uparrow T_{\max}}\|u(t)\|=+\infty$. I am still in doubt with respect to the second and third part of the theorem, I have not found the respective results in the Pazy's book. $\endgroup$
    – Math
    Mar 29, 2020 at 11:55
  • $\begingroup$ @VictorHugo for the second part see Theorem 1.7 (and theorems before) and for the third part see Theorem 1.2 (for global solution) as I said in my answer. $\endgroup$
    – S. Maths
    Mar 29, 2020 at 12:35
  • 1
    $\begingroup$ In Theorem $1.7$ you have the hypothesis that $f$ is Lipschitz continuous from $Y=D(A)$ into $Y$ with the graph norm (which is a strong condition.). After Theorem $1.7$ there is a remark saying that: If in the previous theorem we assume only that $f: [t_ 0, T] \times Y \rightarrow Y$ is locally Lipschitz continuous in $Y$ uniformly in $[t_0, T]$ we obtain, using Theorem 1.4, that for every $u_0 \in D(A) $ the initial value problem possesses a classical solution on a maximal interval $[t_0, t_\max[$. It would be interesting if this comment was valid with $f$ only locally Lipschitz in $X$. $\endgroup$
    – Math
    Mar 29, 2020 at 12:46
  • $\begingroup$ @VictorHugo see my edit $\endgroup$
    – S. Maths
    Mar 29, 2020 at 15:29
  • $\begingroup$ Thank you! @S.Maths $\endgroup$
    – Math
    Mar 29, 2020 at 15:31

Your claim is true. Observe that there is no loss of generality to assume $\|T(t)\|\leq M$ for all $t\geq 0$ since you can add a constant to $A$, which you then absorb into $F$. Now the standard fixed point argument works like charm and $[0,T]\ni t\mapsto u(t)$ is seen to be Lipschitz for every $T>0$.


This is a comment to complement @S.Maths response.

Suppose we show that $$\|u(t)-u(t')\|_X\leq C|t-t'| \hbox{ for all } t,t' \in [0,T]$$ and some $C>0$.

Since $F$ is locally Lipschitz, there is $L_M>0$ such that $$\|F(v)-F(w)\|\leq L_M\|v-w\|$$ for all $v,w \in X$ such that $\|v\| \leq M$ and $\|w\| \leq M$.

Taking into account the continuity of the norm, the function $\|u(\cdot)\|:[0,T] \rightarrow \mathbb{R}$ is continuous on the compact set $[0,T]$.

Therefore, there is $M>0$ such that $\|u(t)\| \leq M$ for all $t \in [0,T]$. Thus, $$\|F(u(t))-F(u(t'))\|\leq L_M\|u(t)-u(t')\|\leq CL_M|t-t'|$$ for all $t,t' \in [0,T]$, which proves that $t \in [0,T] \mapsto F(u(t))$ is Lipschitz continuous, no matter if $F$ is locally Lipschitz or globally Lipschitz.

  • $\begingroup$ What do you mean by "no matter if $F$ is locally Lipschitz or locally Lipschitz." $\endgroup$
    – S. Maths
    Mar 29, 2020 at 23:00
  • $\begingroup$ Sorry, it should be "no matter if $F$ is locally Lipschitz or globally Lipschitz." $\endgroup$
    – Math
    Mar 29, 2020 at 23:26
  • $\begingroup$ I wanted to say that the same proof applies in the case where$F$ is globally lipschitz. $\endgroup$
    – Math
    Mar 29, 2020 at 23:30
  • $\begingroup$ Yes, this is obvious, because in locally compact metric spaces "locally Lipschitz" is equivalent to "globally Lipschitz on every compact subset". In your case the space is $[0,T]$ and the function is $F(u)$. $\endgroup$
    – S. Maths
    Mar 30, 2020 at 0:47

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