Minimal periods for semilinear parabolic equations

Abstract

We show that, if $-A$ generates a bounded holomorphic semigroup in a Banach space X, $\alpha \in [0,1)$, and $f:D(A)\rightarrow X$ satisfies $\Vert f(x)-f(y)\Vert \le L\Vert A^\alpha (x-y)\Vert$, then a non-constant T-periodic solution of the equation ${\dot{u}}+Au=f(u)$ satisfies $LT^{1-\alpha }\ge K_\alpha$ where $K_\alpha >0$ is a constant depending on $\alpha$ and the semigroup. This extends results by Robinson and Vidal-Lopez, which have been shown for self-adjoint operators $A\ge 0$ in a Hilbert space. For the latter case, we obtain - with a conceptually new proof - the optimal constant $K_\alpha$, which only depends on $\alpha$, and we also include the case $\alpha =1$. In Hilbert spaces H and for $\alpha =0$, we present a similar result with optimal constant where Au in the equation is replaced by a possibly unbounded gradient term $\nabla _H{\mathscr {E}}(u)$. This is inspired by applications with bounded gradient terms in a paper by Mawhin and Walter.

1 Introduction and main results

In this paper, we study periodic solutions of equations

$$\begin{aligned} {\dot{u}}(t)+Au(t)=f(u(t)),\quad t\in \mathbb {R}, \end{aligned}$$

(1.1)

where $-A$ is the generator of a bounded analytic semigroup in a complex Banach space $(X,\Vert \cdot \Vert )$, $X\ne \{0\}$, with domain D(A) and $f: D(A)\rightarrow X$ is a function which is Lipschitz continuous in the sense that

$$\begin{aligned} \Vert f(x)-f(y)\Vert \le L\Vert A^\alpha (x-y)\Vert ,\qquad x,y\in D(A), \end{aligned}$$

(1.2)

for a fixed $\alpha \in [0,1)$ and some constant $L\ge 0$. Here $A^\alpha$ denotes the fractional power of A of order $\alpha$. We shall relate the minimal period of a non-constant T-periodic solution u of (1.1) to the Lipschitz constant L in (1.2).

The ODE case $A=0$ and $\alpha =0$, i.e.,

$$\begin{aligned} {\dot{u}}(t) = f(u(t)),\quad t\in \mathbb {R},\quad \hbox {where}\ \Vert f(x)-f(y)\Vert \le L\Vert x-y\Vert ,\quad x,y\in X,\nonumber \\ \end{aligned}$$

(1.3)

has been addressed in several papers: If u is a T-periodic solution to (1.3) and $LT<6$, then u has to be constant [2]. The constant 6 is known to be optimal in general Banach spaces [10]. In a Hilbert space, a T-periodic solution u is constant if $LT<2\pi$ and $2\pi$ is optimal [2, 8, 17]. These results rely on the estimates in Lemma 3.1 below, in particular, the Hilbert space result uses Wirtinger’s inequality in Lemma 3.1(a). $L^p$-versions of Wirtinger’s inequality with optimal constants have been established in [4]. They have been used in [11] to improve the constant 6 in case $X=L^p(\Omega )$ for p in a certain symmetric interval around 2, which is strictly contained in $(\frac{4}{3},4)$. For further details on the $L^p$-case, we refer to Remark 4.1(c) below. In strictly convex Banach spaces X, a T-periodic solution u to (1.3) is constant if $LT\le 6$; see [11].

For the special case that $X=H$ is a Hilbert space and A is self-adjoint with $A\ge 0$, the problem has been studied in [13] under the additional restrictions $\alpha \in [0,\frac{1}{2})$ and A invertible with $A^{-1}$ compact. These additional restrictions have been removed in [14]. It is shown in [13, 14] that there exists a constant $K_\alpha >0$ only depending on $\alpha$ such that $LT^{1-\alpha }<K_\alpha$ implies that a T-periodic solution u to (1.1) is constant. The proofs given there rely in an essential way on properties of spectral projections for A provided by the spectral theorem and study the mild formulation (4.1) of the abstract Cauchy problem corresponding to (1.1), also known as Duhamel’s principle or variation-of-constants formula. In a remark [14, p. 4286], conditions are given that allow to extend this method of proof to Banach spaces. These conditions involve existence and certain estimates for spectral projections of the operator A and seem rather restrictive. Hence, the extension to the situation “when A is a sectorial operator, as treated by Henry [7]” ([13, p. 402]) is still missing.

The new contributions to the problem in the present paper are the following.

• We modify the argument in [14] in such a way that it works in arbitrary Banach spaces X under the sole assumption that $-A$ generates a bounded analytic semigroup. In particular, no assumptions on spectral projections are needed; see Theorem 1.1. We thus provide the extension conjectured on [13, p. 402].

• In case $X=H$ is a Hilbert space and A is self-adjoint with $A\ge 0$, we present a new argument, which yields the optimal constants for the result in [14]. Our proof is based on refined energy type estimates, inspired by the applications in [9], and we can also include the limit case $\alpha =1$. See Theorem 1.2 and Sect. 2.

• For $X=H$ a real Hilbert space and $\alpha =0$, we replace the term Au(t) in (1.1) by a possibly unbounded nonlinear gradient term $\nabla _H{\mathscr {E}}(u(t))$. This is inspired by finite-dimensional applications in [9]. We get the same bound as for the case $\alpha =0$ in Theorem 1.2, namely $LT<2\pi$; see Theorem 1.4.

We recall that, if $-A$ is the generator of a bounded analytic semigroup, one can define fractional powers $A^\gamma$ of A for any $\gamma \ge 0$ and has

$$\begin{aligned} c_\gamma :=\sup _{t>0}\Vert t^\gamma A^\gamma e^{-tA}\Vert < \infty ,\qquad \gamma \ge 0; \end{aligned}$$

(1.4)

see, e.g., [6, 12]. We denote by [D(A)] the space D(A) equipped with the graph norm. Our main results read as follows.

Theorem 1.1

Let $-A$ be the generator of a bounded analytic semigroup in a Banach space X. Let $\alpha \in [0,1)$ and suppose that $f:D(A)\rightarrow X$ satisfies (1.2) for some $L\ge 0$. If $T\in (0,\infty )$ and $u\in C^1(\mathbb {R},X)\cap C(\mathbb {R},[D(A)])$ is a T-periodic solution of (1.1) with

$$\begin{aligned} LT^{1-\alpha } < \left( 1-\frac{c_1}{6k}\right) \left( \frac{c_\alpha }{6k^\alpha }+\frac{c_\alpha k^{1-\alpha }}{1-\alpha }\right) ^{-1} \end{aligned}$$

(1.5)

for some $k\in \mathbb {N}$ with $k>\frac{c_1}{6}$, then u is constant. In a Hilbert space, the conclusion holds if

$$\begin{aligned} LT^{1-\alpha } < \left( 1-\frac{c_1}{2\pi k}\right) \left( \frac{c_\alpha }{2\pi k^\alpha }+\frac{c_\alpha k^{1-\alpha }}{1-\alpha }\right) ^{-1} \end{aligned}$$

(1.6)

for some $k\in \mathbb {N}$ with $k>\frac{c_1}{2\pi }$. Here we understand that the right hand side in (1.5) or (1.6) is $=\infty$ if $c_\alpha =0$.

The proof of Theorem 1.1 is inspired in principle by the approach in [14]. In case $X=H$ is a Hilbert space and A is self-adjoint in H with $A\ge 0$, then $c_0=1$ and, for $A\ne 0$, the spectral theorem (see, e.g., [15]) allows us to calculate

$$\begin{aligned} c_\gamma =\sup _{t\ge 0,\lambda \in \sigma (A)} (t\lambda )^\gamma e^{-t\lambda } = \sup _{s\ge 0}\left( s^\gamma e^{-s} \right) =\gamma ^\gamma e^{-\gamma },\quad \gamma >0, \end{aligned}$$

where $\sigma (A)$ denotes the spectrum of A, which by $A\ge 0$ satisfies $\sigma (A)\subseteq [0,\infty )$. Hence $c_1=e^{-1}<1$, we can take $k=1$, and the condition (1.6) reads

$$\begin{aligned} L T^{1-\alpha } < \left( 1-\frac{1}{2\pi e}\right) \left( \alpha ^\alpha e^{-\alpha }\left( \frac{1}{2\pi }+\frac{1}{1-\alpha }\right) \right) ^{-1}. \end{aligned}$$

(1.7)

But one can do better.

Theorem 1.2

Let $X=H$ be a Hilbert space and A be self-adjoint in H with $A\ge 0$. Let $\alpha \in [0,1]$ and suppose that $f:D(A)\rightarrow X$ satisfies (1.2) for some $L\ge 0$. If $T\in (0,\infty )$ and $u\in C^1(\mathbb {R},X)\cap C(\mathbb {R},[D(A)])$ is a T-periodic solution of (1.1) with

$$\begin{aligned} LT^{1-\alpha } < \frac{(2\pi )^{1-\alpha }}{\sqrt{\alpha ^\alpha (1-\alpha )^{1-\alpha }}}, \end{aligned}$$

(1.8)

then u is constant. The bound (1.8) is sharp.

Remark 1.3

In the situation of Theorem 1.2, one can compare the constants in (1.7) and in (1.8) with the constant in [14, p. 4286] where (recalling $\gamma$ from p. 4285) the corresponding condition reads

$$\begin{aligned} LT^{1-\alpha }< \left( 2^{1-2\alpha }+\frac{\alpha ^\alpha e^{-\alpha }}{(1-\alpha )(1-e^{-1/2})}\right) ^{-1}; \end{aligned}$$

(1.9)

see Fig. 1.

Fig. 1

Different bounds for $LT^{1-\alpha }$ for a self-adjoint operator $A\ge 0$ in a Hilbert space H as a function of $\alpha \in [0,1]$. The upper curve is the bound in (1.8), the intermediate curve is the bound in (1.7), and the lower curve is the bound in (1.9) from [14]

We can replace Au(t) in (1.1) by a gradient term $\nabla _H{\mathscr {E}}(u(t))$. Here, H is a real Hilbert space, V is a Banach space that is densely and continuously embedded into H, and ${\mathscr {E}}:V\rightarrow \mathbb {R}$ is continuously differentiable. For the precise definition of the H-gradient $\nabla _H{\mathscr {E}}$ and some remarks on existence, we refer to Sect. 6. We look at periodic solutions of the equation

$$\begin{aligned} {\dot{u}}(t)+\nabla _H{\mathscr {E}}(u(t))=f(u(t)),\quad t\in \mathbb {R}, \end{aligned}$$

(1.10)

where $f:H\rightarrow H$ is Lipschitz continuous, i.e., there exists a constant $L\ge 0$ such that

$$\begin{aligned} \Vert f(x)-f(y)\Vert _{H}\le L\Vert x-y\Vert _{H},\qquad x,y\in H. \end{aligned}$$

(1.11)

Then we have the following result.

Theorem 1.4

In the situation described above, let $T\in (0,\infty )$ and let $u\in C^1(\mathbb {R},V)$ be a T-periodic solution of (1.10). If $LT<2\pi$, then u is constant. The bound $2\pi$ is sharp.

Remark 1.5

(a) The regularity assumptions on u in our results are made for simplicity. We do not say much on existence in this paper. From the proofs, one can see that the natural assumption in Theorem 1.1 is $u\in W^{1,1}_\textrm{loc}(\mathbb {R};X)\cap L^1_\textrm{loc}(\mathbb {R};[D(A)])$ for the statement in Banach spaces. For the Hilbert space statement in Theorem 1.1, the natural assumption here is $u\in W^{1,2}_\textrm{loc}(\mathbb {R};X)\cap L^2_\textrm{loc}(\mathbb {R};[D(A)])$, and in Theorem 1.2, it is $u\in W^{1,2}_\textrm{loc}(\mathbb {R};H)\cap L^2_\textrm{loc}(\mathbb {R};[D(A)])$. In Theorem 1.4, we can relax the condition to $u\in W^{1,2}_\textrm{loc}(\mathbb {R};V)$, provided the derivative ${\mathscr {E}}':V\rightarrow V'$ maps bounded sets into bounded sets; we refer to Remark 6.1.

(b) A special case of the situation in Theorem 1.4 is given by ${\mathscr {E}}(v)=\Vert A^{1/2}v\Vert _{H}^2$ where A is a self-adjoint operator in H with $A\ge 0$ and $V=D(A^{1/2})$. However, in this case neither the regularity assumptions on u in Theorem 1.2 and Theorem 1.4 nor their respective relaxations in part (a) are comparable.

We remark that in [13, 14], applications are given to the two-dimensional Navier–Stokes equation with periodic boundary conditions. In this context, we also refer to the recent existence results on time-periodic solutions in [1, 5].

The paper is organized as follows. In Sect. 2, we show optimality of the bounds in Theorems 1.2 and 1.4. In Sect. 3, we collect the basic inequalities we shall use. Then we present the proofs of Theorem 1.1, Theorem 1.2, and Theorem 1.4 in Sects. 4, 5, and 6, respectively.

2 Optimality in Theorem 1.2 and Theorem 1.4

We start with examples for the cases $\alpha =0$ and $\alpha =1$ in Theorem 1.2.

Example 2.1

(Case $\alpha =0$) Let $A=0$ in $X=H=\mathbb {C}$, $f(x)=2\pi i x$. Then f is Lipschitz continuous with $L=2\pi$ and the ordinary differential equation ${\dot{u}}(t)=2\pi i u(t)$ has the 1-periodic solution $u(t)=e^{2\pi i t}$, $t\in \mathbb {R}$. Hence, for $\alpha =0$, the condition $LT<2\pi$ in Theorem 1.2 is optimal.

Example 2.2

For $\alpha =1$, the condition $L<1$ in Theorem 1.2 is optimal: Let $X=H=\mathbb {C}$ and $Ax=ax$ where $a>0$. Let $f(x)=(a+ib)x$ where $b=\varepsilon a$ and $\varepsilon >0$. Then the ordinary differential equation

$$\begin{aligned} {\dot{u}}(t)+au(t)=f(u(t))=(a+ib)u(t)\ \Longleftrightarrow \ {\dot{u}}(t)=ib u(t) \end{aligned}$$

has the solution $u(t)=e^{ibt}$, which is T-periodic for $T=\frac{2\pi }{b}=\frac{2\pi }{\varepsilon a}$, and f satisfies (1.2) for $\alpha =1$ with constant $L=\frac{\sqrt{a^2+b^2}}{a}=\sqrt{1+\varepsilon ^2}$ which tends to 1 as $\varepsilon \rightarrow 0$. Observe that, by adjusting a for fixed $\varepsilon >0$, we can arrange for any period $T>0$. It seems unclear what happens for $L=1$.

As a preparation for our examples on optimality in Theorem 1.2 for $\alpha \in (0,1)$, we note the following consequence for linear f.

Corollary 2.3

Let $X=H$ be a Hilbert space, A be self-adjoint with $A\ge 0$, U be unitary in H, and $L>0$. If $\alpha \in (0,1)$ and $T>0$ are such that (1.8) holds, then purely imaginary eigenvalues of $A+LUA^\alpha$ belong to $i(-\frac{2\pi }{T},\frac{2\pi }{T})$.

Proof

The linear operator $B:D(A)\rightarrow X$ given by $B=-LUA^\alpha$ satisfies (1.2) in place of f. Let $\lambda \in \mathbb {R}$ be such that $i\lambda$ is an eigenvalue of $A-B$ with eigenvector x. We may assume that $\lambda \ne 0$. Then $u(t)=e^{-i\lambda t}x$ is a non-constant $\frac{2\pi }{|\lambda |}$-periodic solution of ${\dot{u}}(t)+Au(t)=Bu(t)$. By Theorem 1.2 and (1.8), we infer that $\frac{2\pi }{|\lambda |}>T$, i.e., $|\lambda |<\frac{2\pi }{T}$. $\square$

Example 2.4

Let $\alpha \in (0,1)$, $X=H=\mathbb {C}$, $Ax=ax$ where $a>0$, $L=\frac{a^{1-\alpha }}{\sqrt{\alpha }}$, $f(x)=-Le^{i\varphi }a^\alpha x$, and $\lambda =a\sqrt{\frac{1-\alpha }{\alpha }}$. Then, f satisfies (1.2) and we have

$$\begin{aligned} |a-i\lambda |^2= a^2+\lambda ^2=a^2\left( 1+\frac{1-\alpha }{\alpha }\right) =\frac{a^2}{\alpha }=(La^\alpha )^2, \end{aligned}$$

and we find $\varphi \in \mathbb {R}$ such that $a-i\lambda =-Le^{i\varphi }a^\alpha$ which means that $a+Le^{i\varphi }a^\alpha =i\lambda$. But then $u(t)=e^{-i\lambda t}$ defines a $\frac{2\pi }{\lambda }$-periodic solution of (1.1), and for $T=\frac{2\pi }{\lambda }=\frac{2\pi }{a}\sqrt{\frac{\alpha }{1-\alpha }}$, we have

$$\begin{aligned} (LT^{1-\alpha })^2=\frac{a^{2(1-\alpha )}}{\alpha }\frac{(2\pi )^{2(1-\alpha )}}{a^{2(1-\alpha )}}\left( \frac{\alpha }{1-\alpha }\right) ^{1-\alpha } =\frac{(2\pi )^{2(1-\alpha )}}{\alpha ^\alpha (1-\alpha )^{1-\alpha }}. \end{aligned}$$

(2.1)

Thus the bound in (1.8) is sharp.

The example can be modified to work in the real Hilbert space $\mathbb {R}^2$. We simply use the representation of complex numbers $x+iy$ as matrices $\left( {x\atop y} {-y\atop x}\right)$. To be more precise, let $I\in \mathbb {R}^{2\times 2}$ denote the identity matrix and set $A:=aI$ where $a>0$. Let L, $\lambda$ be as before and $f(x)=-L U_\varphi (a^\alpha x)$, $x\in \mathbb {R}^2$, where $U_\varphi =\left( {\cos \varphi \atop \sin \varphi }{-\sin \varphi \atop \cos \varphi }\right)$. Then f satisfies (1.2) and we have

$$\begin{aligned} A+LU_\varphi (a^\alpha I)=a(I+\frac{1}{\sqrt{\alpha }}U_\varphi ), \quad \sigma (A+LU_\varphi (a^\alpha I)) =\left\{ a\left( 1+\frac{1}{\sqrt{\alpha }} e^{\pm i\varphi }\right) \right\} . \end{aligned}$$

As before, we find $\varphi$ such that $a(1+\frac{1}{\sqrt{\alpha }} e^{i\varphi })=ia\sqrt{\frac{1}{\alpha }-1}=:i\lambda$. Finally, $u(t)=\left( {\cos (\lambda t) \atop -\sin (\lambda t)}\right)$ is a $\frac{2\pi }{\lambda }$-periodic solution of (1.1), and (2.1) holds as before.

Finally, we present an example for sharpness of the bound in Theorem 1.4.

Example 2.5

We rephrase Example 2.1 in $H=\mathbb {R}^2$, a real Hilbert space. Let $V=\mathbb {R}^2$, ${\mathscr {E}}(x)=0$, $f(x)=\left( {0\atop 2\pi }{-2\pi \atop 0}\right)$. Then f satisfies (1.2) for $\alpha =0$ and $L=2\pi$. We have $\nabla _H{\mathscr {E}}(x)=0$ for all $x\in \mathbb {R}^2$. Hence $u(t)=\left( {\cos (2\pi t)\atop -\sin (2\pi t)}\right)$ defines a 1-periodic solution of (1.10).

Concerning the context of Remark 6.1, observe that ${\mathscr {E}}'$ maps bounded sets of $V=\mathbb {R}^2$ into bounded sets of $V'=\mathbb {R}^2$.

3 Basic inequalities

The following lemma is a basic tool in the proofs.

Lemma 3.1

Let $v:\mathbb {R}\rightarrow X$ be continuously differentiable and T-periodic.

• (a) If $X=H$ is a Hilbert space and $\int _0^T v(r)\,dr=0$, then

  $$\begin{aligned} \left( \int \limits _0^T \Vert v(r)\Vert ^2\,dr\right) ^{1/2} \le \frac{T}{2\pi } \left( \int \limits _0^T \Vert {\dot{v}}(r)\Vert ^2\,dr\right) ^{1/2}. \end{aligned}$$

• (b) In the general Banach space case, we have

  $$\begin{aligned} \int \limits _0^T\int \limits _0^T\Vert v(t)-v(s)\Vert \,ds\,dt \le \frac{T}{6} \int \limits _0^T\int \limits _0^T\Vert {\dot{v}}(t)-{\dot{v}}(s)\Vert \,ds\,dt. \end{aligned}$$

We include a proof for (a), which is called Wirtinger’s inequality, and refer to [2] or [10] for the proof of (b).

Proof

(a): By scaling, it is sufficient to study the case $T=1$. We expand v in a Fourier series

$$\begin{aligned} v(t)=\sum _{j\in \mathbb {Z}\setminus \{0\}} c_j e^{2\pi i j t},\qquad {\dot{v}}(t)=\sum _{j\in \mathbb {Z}\setminus \{0\}}2\pi ij c_j e^{2\pi i j t}, \end{aligned}$$

and use Plancherel in H, e.g., $\int _0^T \Vert v(r)\Vert ^2\,dr=\sum _{j\ne 0}\Vert c_j\Vert ^2$. $\square$

Remark 3.2

If, in the situation of Lemma 3.1, $X=H$ is a Hilbert space, we also have

$$\begin{aligned} \left( \int \limits _0^T\int \limits _0^T\Vert v(t)-v(s)\Vert ^2\,ds\,dt\right) ^{1/2} \le \frac{T}{2\pi } \left( \int \limits _0^T\int \limits _0^T\Vert {\dot{v}}(t)-{\dot{v}}(s)\Vert ^2\,ds\,dt\right) ^{1/2}. \end{aligned}$$

This follows easily from Lemma 3.1(a).

4 The general Banach space case

We recall the well-known fact that a solution $u\in C^1(\mathbb {R},X)\cap C(\mathbb {R},[D(A)])$ of the abstract Cauchy problem

$$\begin{aligned} {\dot{u}}(t)+Au(t)=f(u(t)),\quad t\in \mathbb {R}, \end{aligned}$$

satisfies, by Duhamel’s formula, the equation

$$\begin{aligned} u(t) = e^{-tA}u(0)+\int \limits _0^t e^{-(t-s)A} f(u(s))\,ds, \quad t\ge 0. \end{aligned}$$

(4.1)

Proof of Theorem 1.1

Suppose that $u\in C^1(\mathbb {R},X)\cap C(\mathbb {R},[D(A)])$ is a solution of (1.1) that is T-periodic. We write, for $s,t\in [0,T]$,

$$\begin{aligned} A^\alpha (u(t)-u(s))&= A^\alpha e^{-kTA} (u(t)-u(s)) +A^\alpha (I-e^{-kTA}) (u(t)-u(s))\\&=:v_1(t,s)+v_2(t,s), \end{aligned}$$

where $k\in \mathbb {N}$ is such that $\frac{c_1}{6 k}<1$. Observe that the operator $A^\alpha e^{-kTA}$ is bounded. We shall give an estimate on

$$\begin{aligned} d_T(u):=\int \limits _0^T \int \limits _0^T \Vert A^\alpha (u(t)-u(s))\Vert \,ds\,dt \end{aligned}$$

(4.2)

from which our result will follow. Clearly,

$$\begin{aligned} d_T(u)\le \int \limits _0^T \int \limits _0^T \Vert v_1(t,s)\Vert \,ds\,dt+\int \limits _0^T\int \limits _0^T\Vert v_2(t,s)\Vert \,ds\,dt =: I_1+I_2, \end{aligned}$$

and we start by applying Lemma 3.1(b) to $v(t)=A^\alpha e^{-kTA}u(t)$. By (1.1), we then obtain

$$\begin{aligned} \frac{6}{T} I_1&\le \int \limits _0^T \int \limits _0^T \Vert A^\alpha e^{-kTA}({\dot{u}}(t)-{\dot{u}}(s))\Vert \,ds\,dt \\&\le \int \limits _0^T \int \limits _0^T \Vert A^\alpha e^{-kTA}A(u(t)-u(s))\Vert \,ds\,dt \\&\ \ + \int \limits _0^T \int \limits _0^T \Vert A^\alpha e^{-kTA}(f(u(t))-f(u(s)))\Vert \,ds\,dt \\&\le \frac{c_1}{kT}\int \limits _0^T\Vert A^\alpha (u(t)-u(s))\Vert \,ds\,dt \\&\ \ + \frac{c_\alpha }{k^\alpha T^\alpha } \int \limits _0^T\int \limits _0^T\Vert f(u(t))-f(u(s))\Vert \,ds\,dt \\&\le \left( \frac{c_1}{kT} + \frac{c_\alpha L}{k^\alpha T^\alpha }\right) \int \limits _0^T \int \limits _0^T\Vert A^\alpha (u(t)-u(s))\Vert \,ds\,dt. \end{aligned}$$

Hence, we have shown

$$\begin{aligned} I_1&\le \left( \frac{c_1}{6k} + \frac{c_\alpha }{6k^\alpha }\, LT^{1-\alpha }\right) \,d_T(u). \end{aligned}$$

In order to get an estimate on $I_2$, we use (4.1) and the T-periodicity of u to write

$$\begin{aligned} u(t)=u(t+kT)=e^{-kTA}u(t)+\int \limits _0^{kT} e^{-(kT-r)A}f(u(r+t))\,dr. \end{aligned}$$

Hence, we have

$$\begin{aligned} (I-e^{-kTA})(u(t)-u(s))=\int \limits _0^{kT} e^{-(kT-r)A}\left( f(u(r+t))-f(u(r+s))\right) \,dr. \end{aligned}$$

For $I_2$, we thus obtain, by Minkowski’s inequality,

$$\begin{aligned} I_2&=\int \limits _0^T \int \limits _0^T \Vert A^\alpha (I-e^{-kTA})(u(t)-u(s))\Vert \,ds\,dt\\&\le \int \limits _0^T \int \limits _0^T \left\| \int \limits _0^{kT}A^\alpha e^{-(kT-r)A}(f(u(r+t))-f(u(r+s)))\,dr\right\| \,ds\,dt\\&\le \int \limits _0^{kT} \int \limits _0^T \int \limits _0^T \left\| A^\alpha e^{-(kT-r)A}(f(u(r+t))-f(u(r+s)))\right\| \,ds\,dt\,dr\\&\le c_\alpha \int \limits _0^{kT} (kT-r)^{-\alpha } \int \limits _0^T \int \limits _0^T \Vert f(u(r+t))-f(u(r+s))\Vert \,ds\,dt\,dr\\&\le c_\alpha L \int \limits _0^{kT} r^{-\alpha }\,dr \int \limits _0^T \int \limits _0^T \Vert A^\alpha (u(t)-u(s))\Vert \,ds\,dt\\&= \frac{c_\alpha k^{1-\alpha }}{1-\alpha } LT^{1-\alpha } d_T(u). \end{aligned}$$

Hence, we have proved

$$\begin{aligned} d_T(u)\le \left( \frac{c_1}{6k}+\left( \frac{c_\alpha }{6k^\alpha }+\frac{c_\alpha k^{1-\alpha }}{1-\alpha }\right) L T^{1-\alpha } \right) d_T(u), \end{aligned}$$

and conclude $d_T(u)=0$ if

$$\begin{aligned} \left( \frac{c_\alpha }{6k^\alpha }+\frac{c_\alpha k^{1-\alpha }}{1-\alpha }\right) L T^{1-\alpha } < 1-\frac{c_1}{6k}. \end{aligned}$$

(4.3)

We see that (4.3) implies that $A^\alpha u$ is constant. Then $Au=A^{1-\alpha }A^\alpha u$ is constant and, by (1.2), f(u) is constant. Hence (1.1) implies that also ${\dot{u}}$ is constant. But since u is periodic, it has to be constant, too.

In case $X=H$ is a Hilbert space, we can run a nearly identical argument, letting

$$\begin{aligned} d_T(u)=\left( \int \limits _0^T \int \limits _0^T\Vert A^\alpha (u(t)-u(s))\Vert ^2\,ds\,dt\right) ^{1/2} \end{aligned}$$

and using Remark 3.2 to obtain that u is constant if

$$\begin{aligned} \left( \frac{c_\alpha }{2\pi k^\alpha }+\frac{c_\alpha k^{1-\alpha }}{1-\alpha }\right) L T^{1-\alpha } < 1-\frac{c_1}{2\pi k}. \end{aligned}$$

(4.4)

Of course, we have to take $k\in \mathbb {N}$ with $\frac{c_1}{2\pi k}<1$ here. $\square$

Remark 4.1

If $X=L^p(\Omega )$ for a $\sigma$-finite measure space $(\Omega ,\mu )$ and $p\in (1,\infty )$, it is tempting to use

$$\begin{aligned} d_T(u)&=\left( \int \limits _0^T \int \limits _0^T \Vert A^\alpha (u(t)-u(s))\Vert _{L^p(\Omega )}^p\,ds\,dt\right) ^{1/p} \\&=\left\| \left( \int \limits _0^T \int \limits _0^T \big |A^\alpha (u(t)-u(s))(\cdot )\big |^p\,ds\,dt\right) ^{1/p}\right\| _{L^p(\Omega )} \end{aligned}$$

and an analogue of Lemma 3.1(a) or (b) in $L^p$ for scalar-valued T-periodic functions in order to improve the constant $\frac{T}{6}$.

The best constant of the $L^p$-analogue of Lemma 3.1(a) is known (see [4]). This has been used in [11] to give estimates on minimal periods, but only leads to an improvement over the constant $\frac{T}{6}$ for p in an interval $I\ni 2$, which is strictly contained in $(\frac{4}{3},4)$ and is symmetric in the sense that $p\in I\ \Leftrightarrow \ \frac{p}{p-1}\in I$. For example, the best constant in the $L^1$-analogue of Lemma 3.1(a) is $\frac{T}{4}$.

To the best of our knowledge, the best constant in the $L^p$-analogue of Lemma 3.1(b) is not known apart from the cases $p=1,2,\infty$. Here, it is tempting to resort to interpolation. However, for such an inequality, one has to interpolate closed subspaces of $L^p$, which is possible in this case by a retraction-coretraction argument (see [16, 1.2.4]). The corresponding operators will bring in other constants which do not seem to lead to improvements over the $L^p$-result in [11].

5 Self-adjoint operators in a Hilbert space

Proof of Theorem 1.2

We take a T-periodic solution $u\in C^1(\mathbb {R},H)\cap C(\mathbb {R},[D(A)])$ of (1.1) and put

$$\begin{aligned} v(t):=u(t)-u(t-\tau ),\quad t\in \mathbb {R}, \end{aligned}$$

where $\tau \in (0,T)$ is arbitrary. Then $r\mapsto g(r):=\left\langle Av(r),v(r)\right\rangle$ is T-periodic and differentiable with

$$\begin{aligned} \frac{d}{dr}\left\langle Av(r),v(r)\right\rangle =2\,\textrm{Re}\,\left\langle Av(r),{\dot{v}}(r)\right\rangle ,\quad r\in \mathbb {R}. \end{aligned}$$

For a proof, observe that, by the self-adjointness of A,

$$\begin{aligned} \frac{g(r+h)-g(r)}{h}=&\left\langle Av(r+h),\frac{1}{h}(v(r+h)-v(r))\right\rangle \\&+\left\langle \frac{1}{h}(v(r+h)-v(r)),Av(r)\right\rangle , \end{aligned}$$

and take the limit as $h\rightarrow 0$.

Hence, we have $\int _0^T\textrm{Re}\left\langle Av(r),{\dot{v}}(r)\right\rangle \,dr=0$, and we obtain

$$\begin{aligned} \int \limits _0^T\Vert {\dot{v}}(r)\Vert ^2+\Vert Av(r)\Vert ^2\,dr&=\int \limits _0^T \Vert {\dot{v}}(r)+Av(r)\Vert ^2\,dr \\&= \int \limits _0^T \Vert f(u(r))-f(u(r-\tau ))\Vert ^2\,dr. \end{aligned}$$

Thus, by (1.2), we have

$$\begin{aligned} \int \limits _0^T\Vert {\dot{v}}(r)\Vert ^2+\Vert Av(r)\Vert ^2\,dr \le L^2 \int \limits _0^T \Vert A^\alpha v(r)\Vert ^2\,dr. \end{aligned}$$

(5.1)

We shall exploit (5.1) for the different cases of $\alpha$.

If $\alpha =0$, we obtain, by Lemma 3.1(a),

$$\begin{aligned} d_T(v):=\int \limits _0^T\Vert {\dot{v}}(r)\Vert ^2+\Vert Av(r)\Vert ^2\,dr\le \left( \frac{LT}{2\pi }\right) ^2 \int \limits _0^T \Vert {\dot{v}}(r)\Vert ^2\,dr. \end{aligned}$$

Hence, if $LT< 2\pi$, then ${\dot{v}}$ vanishes and, since $\tau$ was arbitrary, ${\dot{u}}$ is constant. Since u is periodic, ${\dot{u}}$ has to vanish and u is constant.

If $\alpha =1$ and $L<1$, we see from (5.1) that Av vanishes, and then also ${\dot{v}}$ vanishes. Again, u has to be constant. In other words, if $\alpha =1$ and $L<1$, any periodic solution has to be constant.

In case $\alpha \in (0,1)$, we use the following two lemmata.

Lemma 5.1

Let $v\in C^1(\mathbb {R},H)\cap C(\mathbb {R},[D(A)])$ be T-periodic with $\int _0^T v(r)\,dr=0$ and $\alpha \in (0,1)$. Then

$$\begin{aligned} \Vert A^\alpha v\Vert _{L^2((0,T);H)} \le \left( \frac{T}{2\pi }\right) ^{1-\alpha } \Vert {\dot{v}}\Vert _{L^2((0,T);H)}^{1-\alpha } \Vert Av\Vert _{L^2((0,T);H)}^{\alpha }. \end{aligned}$$

Proof

We have the moment inequality $\Vert A^\alpha x\Vert \le \Vert x\Vert ^{1-\alpha }\Vert Ax\Vert ^\alpha$ for $x\in D(A)$ (using the spectral theorem (see, e.g., [15]) write $\Vert A^\beta x\Vert ^2=\int _0^\infty \lambda ^{2\beta }\,d\mu _x$ for $\beta \in \{0,\alpha ,1\}$, where $\mu _x$ is the spectral measure for x and use Hölder’s inequality with exponent $\frac{1}{\alpha }$ and dual exponent $(\frac{1}{\alpha })'=\frac{1}{1-\alpha }$). Thus we have

$$\begin{aligned} \int \limits _0^T\Vert A^\alpha v(r)\Vert ^2\,dr\le \int \limits _0^T \Vert v(r)\Vert ^{2(1-\alpha )}\Vert Av(r)\Vert ^{2\alpha }\,dr. \end{aligned}$$

We use Hölder again with exponent $\frac{1}{\alpha }$ and dual exponent $(\frac{1}{\alpha })'=\frac{1}{1-\alpha }$ and obtain

$$\begin{aligned} \int \limits _0^T\Vert A^\alpha v(r)\Vert ^2\,dr\le \left( \int \limits _0^T \Vert v(r)\Vert ^2\,dr\right) ^{1-\alpha } \left( \int \limits _0^T\Vert Av(r)\Vert ^{2}\,dr\right) ^{\alpha }. \end{aligned}$$

Finally, we use Lemma 3.1(a). $\square$

Lemma 5.2

For all $a,b\ge 0$ and $\alpha \in [0,1]$, we have

$$\begin{aligned} a^\alpha b^{1-\alpha }\le \alpha ^\alpha (1-\alpha )^{1-\alpha }(a+b). \end{aligned}$$

Proof

The assertion is clear for $\alpha \in \{0,1\}$, so let $\alpha \in (0,1)$. Letting $x=\frac{a}{\alpha }$, $y=\frac{b}{1-\alpha }$, the assertion is equivalent to $x^\alpha y^{1-\alpha }\le \alpha x+(1-\alpha )y$, which again is clear if $0\in \{x,y\}$. For $x,y\ne 0$, it is equivalent to

$$\begin{aligned} \alpha \ln x+(1-\alpha )\ln y\le \ln (\alpha x+(1-\alpha )y), \end{aligned}$$

and this holds since $\ln$ is concave. $\square$

We continue the proof of Theorem 1.2. For $\alpha \in (0,1)$, we have, combining (5.1), Lemmas 5.1, and 5.2,

$$\begin{aligned} d_T(v) \le L^2 \left( \frac{T}{2\pi }\right) ^{2(1-\alpha )} \alpha ^\alpha (1-\alpha )^{1-\alpha }\, d_T(v). \end{aligned}$$

Hence, if

$$\begin{aligned} LT^{1-\alpha } < \frac{(2\pi )^{1-\alpha }}{\sqrt{\alpha ^\alpha (1-\alpha )^{1-\alpha }}}, \end{aligned}$$

then ${\dot{v}}$ and Av vanish, and we conclude that u is constant as before. $\square$

6 Gradient systems

In this section, we replace Au(t) in (1.1) by a gradient term. For the setting, we follow [3]. So let H be a real Hilbert space and V be a Banach space with a dense and continuous embedding $V\hookrightarrow H$. Let ${\mathscr {E}}:V\rightarrow \mathbb {R}$ be differentiable with continuous derivative ${\mathscr {E}}':V\rightarrow V'$, where $V'$ denotes the dual space of V. Identifying $h\in H$ with the linear functional $v\mapsto \left\langle h,v\right\rangle _H$, we can consider H as a subspace of $V'$. Then the gradient $\nabla _H{\mathscr {E}}$ of ${\mathscr {E}}$ with respect to H is defined by

$$\begin{aligned} D(\nabla _H{\mathscr {E}})&=\{u\in V:\exists h\in H\,\forall v\in V: {\mathscr {E}}'(u)v=\left\langle h,v\right\rangle _H\},\\ \nabla _H{\mathscr {E}}(u)&=h\quad \hbox {for}\, u\in D(\nabla _H{\mathscr {E}}). \end{aligned}$$

We recall the usual solution concept from [3, Sect. 6] for the equation

$$\begin{aligned} {\dot{u}}(t)+\nabla _H{\mathscr {E}}(u(t))=g(t),\quad t\in \mathbb {R}, \end{aligned}$$

(6.1)

where $g\in L^2_{\textrm{loc}}(\mathbb {R};H)$: u is a solution of (6.1) if $u\in W^{1,2}_\textrm{loc}(\mathbb {R};H)\cap L^\infty _{\textrm{loc}}(\mathbb {R};V)$ and

$$\begin{aligned} \left\langle {\dot{u}}(t),v\right\rangle _H+{\mathscr {E}}'(u)v=\left\langle f(t),v\right\rangle _H\quad \hbox {for all }\,v\in V\, \text { and almost every }t\in \mathbb {R}. \end{aligned}$$

Observe that this implies $u(t)\in D(\nabla _H{\mathscr {E}})$ for almost every $t\in \mathbb {R}$. One has existence and uniqueness of solutions to the corresponding initial value problem on finite time intervals if the following conditions (i)–(iii) on ${\mathscr {E}}$ hold (see [3, Theorem 6.1]):

• (i) ${\mathscr {E}}:V\rightarrow \mathbb {R}$ is convex,

• (ii) ${\mathscr {E}}:V\rightarrow \mathbb {R}$ is coercive, i.e., $\{u\in V:{\mathscr {E}}(u)\le c\}$ is bounded in V for every $c\in \mathbb {R}$,

• (iii) ${\mathscr {E}}':V\rightarrow V'$ maps bounded sets into bounded sets.

Here, we do not go into details as in Theorem 1.4 we are interested in periodic solutions to (1.10) that are more regular, namely solutions $u\in C^1(\mathbb {R},V)$.

Proof of Theorem 1.4

Let $u\in C^1(\mathbb {R},V)$ be a solution to (1.10) that is T-periodic. Put $v(t):=u(t)-u_0$, where $u_0=\frac{1}{T}\int _0^T u(r)\,dr$. Then $C^1(\mathbb {R}, V)$ is T-periodic as well and $\int _0^T v(r)\,dr=0$. Since

$$\begin{aligned} \frac{d}{dr}{\mathscr {E}}(u(r))={\mathscr {E}}'(u(r)){\dot{u}}(r)=\nabla _H{\mathscr {E}}(u(r)){\dot{u}}(r) \end{aligned}$$

and u is T-periodic, we have

$$\begin{aligned} \int \limits _0^T \left\langle \nabla _H{\mathscr {E}}(u(r))-f(u_0),{\dot{u}}(r)\right\rangle _H\,dr=0. \end{aligned}$$

Hence we obtain

$$\begin{aligned}&\int \limits _0^T \Vert {\dot{v}}(r)\Vert _{H}^2+\Vert \nabla _H{\mathscr {E}}(u(r))-f(u_0)\Vert _{H}^2\,dr\\&=\int \limits _0^T\Vert {\dot{u}}(r)+\nabla _H{\mathscr {E}}(u(r))-f(u_0)\Vert _{H}^2\,dr \\&=\int \limits _0^T\Vert f(v(r)+u_0)-f(u_0)\Vert _{H}^2\,dr\\&\le L^2 \int \limits _0^T \Vert v(r)\Vert _{H}^2\,dr \le \left( \frac{LT}{2\pi }\right) ^2 \int \limits _0^T \Vert {\dot{v}}(r)\Vert _{H}^2\,dr. \end{aligned}$$

If $LT<2\pi$, then ${\dot{v}}$ vanishes, $u=u_0$ is constant, and $\nabla _H{\mathscr {E}}(u_0)=f(u_0)$. $\square$

Remark 6.1

We can relax the regularity of u in Theorem 1.4 to $u\in W^{1,2}_{\textrm{loc}}(\mathbb {R};V)$ provided the property (iii) above holds. Under this assumption, we can prove Lemma 6.2 below. With Lemma 6.2 at hand, we can run the same argument as before for $u\in W^{1,2}_{\textrm{loc}}(\mathbb {R};V)$.

Lemma 6.2

Let $I\subseteq \mathbb {R}$ be an interval and $u\in W^{1,2}(I;V)$ be such that $u(r)\in D(\nabla _H{\mathscr {E}})$ for almost every $r\in I$. Then, for all $s,t\in I$ with $s<t$, we have

$$\begin{aligned} \int \limits _s^t \nabla _H{\mathscr {E}}(u(r)){\dot{u}}(r)\,dr={\mathscr {E}}(u(t))-{\mathscr {E}}(u(s)). \end{aligned}$$

Proof

It clearly suffices to show that $r\mapsto {\mathscr {E}}'(u(r)){\dot{u}}(r)$ is the weak derivative of $r\mapsto {\mathscr {E}}(u(r))$. This is clear for $u\in C^1(I,V)$. Using mollifiers and passing to a subsequence if necessary, we approximate u by a sequence $(u_n)$ in $C^1(I, V)$ such that $u_n \rightarrow u$ in $W^{1,2}(I;V)$ and such that we have pointwise almost everywhere ${\dot{u}}_n\rightarrow {\dot{u}}$ in V. By the inclusion $W^{1,2}(I;V)\hookrightarrow C_b(I,V)$, we have $u_n\rightarrow u$ uniformly on I. Then the sequence $({\mathscr {E}}'(u_n))$ is bounded in $V'$ by (iii) and converges pointwise in $V'$ to ${\mathscr {E}}'(u)$. Since $({\dot{u}}_n)$ converges to ${\dot{u}}$ in $L^2(I;V)$, we have ${\mathscr {E}}'(u_n){\dot{u}}_n\rightarrow {\mathscr {E}}'(u){\dot{u}}=\nabla _H{\mathscr {E}}(u){\dot{u}}$ in $L^2(I;\mathbb {R})$. Since also ${\mathscr {E}}(u_n)\rightarrow {\mathscr {E}}(u)$ pointwise, the assertion follows. $\square$