São Paulo J. Math. Sci. (2015) 9:181–194 DOI 10.1007/s40863-015-0018-0
Asymptotic behavior for a nonlocal model of neural fields Severino H. da Silva1 · Antônio L. Pereira2
Published online: 8 September 2015 © Instituto de Matemática e Estatística da Universidade de São Paulo 2015
Abstract In this paper we consider the non local evolution problem ∂t u = −u + K ( f ◦ u) in , u = 0 in R N \, where is a smooth bounded domain in R N , f : R → R and K is an integral operator with a symmetric kernel. We prove existence and some regularity properties of the global attractor. We also characterize the global attractor, using the properties of a Lyapunov functional for this model. Keywords
Global attractors · Non local equations · Lyapunov functional
Mathematics Subject Classification
35B40 · 35B41 · 37B55
Severino H. da Silva: Research partially supported by CAPES/CNPq. Antônio L. Pereira: Research partially supported by CNPq.
B
Antônio L. Pereira
[email protected] Severino H. da Silva
[email protected]
1
Unidade Acadêmica de Matemática, Universidade Federal de Campina Grande, Campina Grande, PB 58051-900, Brazil
2
Instituto de Matemática e Estatística, Universidade de São Paulo, São Paulo, SP 13565-905, Brazil
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1 Introduction We consider here the non local evolution problem ⎧ ⎪ ⎨∂t u(t, x) = −u(t, x) + K ( f ◦ u)(t, x) for t ≥ 0 and x ∈ , u(0, x) = u 0 (x) in , ⎪ ⎩ u(t, x) = 0 for t ≥ 0 and x ∈ R N \,
(1.1)
where is a bounded smooth domain in R N (N ≥ 1), f : R → R is a (sufficiently smooth) function and K is an integral operator with symmetric nonnegative kernel K v(x) :=
RN
J (x, y)v(y)dy.
(1.2)
We will suppose, without loss of generality, that R N J (x, y) d y = R N J (x, y) d x = 1. The understanding of the dynamics of non local evolution equations like in (1.1) has attracted the attention of many researchers in recent years; see for instance [1–18]. However, the approach taken here seems to be new for this model and was motivated by similar approaches in [19–21]. The basic idea is to find an abstract way to impose Dirichlet boundary conditions in non local evolutions equations. The problem (1.1) is also a way to describe the neural model with short-range excitation and long-range inhibition. This description illustrates, for example, the ability of neural field models to generate intricate spreading labyrinthine patterns (see [4]). The paper is organized as follows. In Sect. 2, assuming a growth condition in the nonlinearity f , we prove that (1.1) generates a C 1 flow in a space X which is isometric to L p (). In Sect. 3, we prove existence of a global attractor and establish some regularity properties for it. Finally, in Sect. 4, we exhibit a continuous Lyapunov functional for the flow generated by (1.1), and we use it to prove that the flow is gradient in the sense of [22].
2 Well posedness In this section we show, assuming a growth condition in the nonlinearity f , that the problem (1.1) is well posed in a suitable phase space. Consider, for any 1 ≤ p ≤ ∞, the subspace X of L p (R N ) given by
X = u ∈ L p (R N ) : u(x) = 0, if x ∈ R N \ with the induced norm. The space X is canonically isomorphic to L p () and we usually identify the two spaces, without further comment. We also use the same notation for a function in R N and its restriction to for simplicity, wherever we believe the intention is clear from the context.
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In order to obtain well posedness of (1.1), we consider the Cauchy problem
∂t u = −u + F(u), u(t0 ) = u 0 ,
(2.1)
where the map F : X → X is defined by F(u)(x) =
K ( f ◦ u)(x), 0,
x ∈ , x ∈ R N \.
(2.2)
We prove below (Proposition 2.4) that F is well defined under appropriate growth conditions on f . The map given by (1.2) is well defined as a bounded linear operator in various function spaces, depending on the properties assumed for J . We collect here some estimates for this map which will be used in the sequel. Lemma 2.1 Let K be the map defined by (1.2) and J r := supx∈ J (x, ·) L r () , 1 ≤ r ≤ ∞. If u ∈ L p (), 1 ≤ p ≤ ∞, then K u ∈ L ∞ (), |K u(x)| ≤ J q u L p () , for all x ∈ ,
(2.3)
where 1 ≤ q ≤ ∞ is the conjugate exponent of p, and K u L p () ≤ J 1 u L p () ≤ u L p () .
(2.4)
Moreover, if u ∈ L 1 (), then K u ∈ L p (), 1 ≤ p ≤ ∞, and K u L p () ≤ J p u L 1 () .
(2.5)
Proof Estimate (2.3) follows easily from Hölder inequality. Estimate (2.4) follows from the generalized Young’s inequality (see [23] ). The proof of (2.5) is similar to (2.4), but we include it here for the sake of completeness. Suppose 1 < p < ∞ and let q be its the conjugate exponent. Then, by Hölder inequality |K u(x)| ≤ ≤
1 p
|J (x, y)u(y) u(y) | dy 1 |J (x, y)| |u(y)| d y p
1 q
1 q
≤ u L 1 ()
1 |u(y)| d y
|J (x, y)| |u(y)| d y p
p
q
1
p
.
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Raising both sides to the p − th power and integrating, we obtain
|J (x, y)| p |u(y)| d x d y p |u(y)| |J (x, y)| p d x d y ≤ u Lq 1 () p
|K u(x)| p d x ≤ u Lq 1 ()
p q
p
≤ u L 1 () u L 1 () J p p+q
p
≤ u L 1q() J p . The inequality (2.5) then follows by taking p − th roots. The case p = 1 is similar but easier, and the case p = ∞ is trivial.
Definition 2.2 If E is a normed space, we say that a function F : E → E is locally Lipschitz continuous (or simply locally Lipschitz) if, for any x0 ∈ E, there exists a constant C and a ball B = {x ∈ E : x − x0 < b} such that, if x and y belong to B then F(x) − F(y) ≤ Cx − y; we say that F is Lipschitz continuous on bounded sets if the ball B in the previous definition can be chosen as any bounded ball in E. Remark 2.3 The two definitions in 2.2 are equivalent if the normed space E is locally compact. Proposition 2.4 Suppose, in addition to the hypotheses of Lemma 2.1, that the function f satisfies the growth condition | f (x)| ≤ C1 (1 + |x| p ), for any x ∈ R, with 1 ≤ p < ∞. Then the function F given by (2.2) is well defined in L p (). If f is locally bounded, F is well defined in L ∞ (), Additionaly, if | f (x) − f (y)| ≤ C2 (1 + |x| p−1 + |y| p−1 )|x − y|, for any (x, y) ∈ R × R, the function F is bounded Lipschitz continuous in L p (), 1 ≤ p < ∞. If p = ∞, this is true if f is locally Lipschitz. Proof Suppose 1 ≤ p < ∞. Then, if u ∈ L p () f (u) L 1 () ≤
p
C1 (1 + |u| p ) d x ≤ C1 (|| + u L p () ).
(2.6)
From estimates (2.5) and (2.6), it follows that p
F(u) L p () = K f (u) L p () ≤ J p f (u) L 1 () ≤ C1 J p (|| + u L p () ) showing that F is well defined. The proof for p = ∞ is straightforward.
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Suppose now that | f (x) − f (y)| ≤ C2 (1 + |x| p−1 + |y| p−1 )|x − y|, for some 1 < p < ∞ Then, if u, v ∈ L p () f (u) − f (v) L 1 () ≤
≤ C2
C2 (1 + |u| p−1 + |v| p−1 )|u − v| d x 1
(1 + |u|
+ |v|
) dx
q
1
p
|u − v| d x
≤ C2 1 L q () + u p−1 L q () + v p−1 L q () u − v L p () p p 1 ≤ C2 || q + u Lq p () + v Lq p () u − v L p () , p−1
p−1 q
p
where q is the conjugate exponent of p. Using (2.5) once again and the hypothesis on f , it follows that F(u) − F(v) L p () = K ( f (u) − f (v)) L p () ≤ J p f (u) − f (v) L 1 () p p 1 ≤ C2 J p || q + u Lq p () + v Lq p () u − v L p () showing that F is Lipschitz in bounded sets of L p () as claimed. If p = 1, the proof is similar, but simpler. Suppose finally that u L ∞ () ≤ R, v L ∞ () ≤ R and let M be the Lipschitz constant of f in the interval [−R, R] ⊂ R. Then | f (u(x)) − f (v(x))| ≤ M|u(x) − v(x)|, for any x ∈ , so f (u) − f (v) L ∞ () ≤ Mu − v L ∞ () . Thus, by (2.4) F(u) − F(v) L ∞ () = K ( f (u) − f (v)) L ∞ () ≤ MJ 1 u − v L ∞ () .
From the result above, it follows from well known results that the problem (2.1) has a local solution for any initial condition in X . For the global existence, we need the following result ([24]—Theorem 5.6.1) Theorem 2.5 Let X be a Banach space, and suppose that g : [t0 , ∞) × X → X is continuous and g(t, u) ≤ h(t, u); for all (t, u) ∈ [t0 , ∞) × X , where h : [t0 , ∞) × R+ → R+ is continuous and h(t, r ) is non decreasing in r ≥ 0, for each t ∈ [t0 , ∞). Then, if the maximal solution r (t, t0 , r0 ) of the scalar initial value problem r = h(t, r ), r (t0 ) = r0 , exists throughout [t0 , ∞), the maximal interval of existence of any solution u(t, t0 , u 0 ) of the initial value problem
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du = g(t, u), t > t0 , u(t0 ) = u 0 , dt with u 0 ≤ r0 , also contains [t0 , ∞). Corollary 2.6 Suppose, in addition to the hypotheses of Proposition 2.4, that f satisfies: lim sup |x|→∞
| f (x)| < k1 , for some constant k1 ∈ R. |x|
(2.7)
Then the problem (2.1) has a unique globally defined solution for any initial condition in X , which is given, for t ≥ t0 , by the “variation of constants formula” t ⎧ ⎨e−(t−t0 ) u (x) + e−(t−s) K ( f ◦ (u(s, ·))(x)) ds, 0 u(t, x) = t0 ⎩ 0,
x ∈ ,
(2.8) x ∈ R N \.
Proof From Proposition 2.4, it follows that the right-hand-side of 2.1 is Lipschitz continuous in bounded sets of X and, therefore, the Cauchy problem 2.1 is well posed in X , with a unique local solution u(t, x), given by (2.8) (see [25]). From condition (2.7), it follows that there is a constant k2 , such that | f (x)| ≤ k2 + k1 |x|, for any x ∈ R.
(2.9)
If 1 ≤ p < ∞, we obtain from (2.4) and (2.9) that K ( f ◦ u) L p () ≤ f ◦ u L p () ≤ k2 ||1/ p + k1 u L p () . For p = ∞, we obtain by the same arguments (or by making p → ∞), that K ( f ◦ u) L ∞ () ≤ k2 + k1 u L ∞ () . Defining h : [t0 , ∞)×R+ → R+ , by h(t, r ) = k2 ||1/ p +(k1 +1)r , it follows that Problem (2.1) satisfies the hypothesis of Theorem 2.5 and the global existence follows immediately. The variation of constants formula can be verified by direct derivation.
The result below can be found in [26]. Proposition 2.7 Let Y and Z be normed linear spaces, F : Y → Z a map and suppose that the Gateaux derivative of F, D F : Y → L(Y, Z ) exists and is continuous at y ∈ Y . Then the Fréchet derivative F of F exists and is continuous at y.
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Proposition 2.8 Suppose, in addition to the hypotheses of Corollary 2.6 that the function f is continuously differentiable in R and f satisfies the growth condition | f (x)| ≤ C1 (1 + |x| p−1 ), for any x ∈ R, if 1 ≤ p < ∞. Then F is continuously Fréchet differentiable on X with derivative given by D F(u)v(x) :=
K ( f (u)v)(x), 0,
x ∈ , x ∈ R N \.
Proof From a simple computation, using the fact f is continuously differentiable on R, it follows that the Gateaux’s derivative of F is given by D F(u)v(x) :=
K ( f (u)v)(x), 0,
x ∈ , x ∈ R N \.
The operator D F(u) is clearly a linear operator in X . Suppose 1 ≤ p < ∞ and q is the conjugate exponent of p. Then, if u ∈ L p ()
f (u) L q () ≤
1 [C1 (1 + |u|
)] d x
p−1 q
1 q
q
1
≤ C1 || + C1 |u| d x p 1 q q = C1 || + u L p () 1 p−1 = C1 || q + u L p () . p
q
(2.10)
From Hölder inequality and (2.10), it follows that 1
f (u) · v L 1 () ≤ C1 (|| q + u L p () )v L p () . p−1
From estimate (2.5), it follows that 1
D F(u) · v L p () = K ( f (u)v) L p () = J p f (u)v L 1 () ≤ C1 J p (|| q p−1
+ u L p () )v L p () showing that D F(u) is a bounded operator. In the case p = ∞, we have that | f (u)| is bounded by a constant C2 , for each u ∈ L ∞ (). Therefore f (u)v L ∞ () ≤ C2 v L ∞ () and thus, from (2.4), we obtain D F(u) · v L ∞ () = K ( f (u)v) L ∞ () ≤ J 1 f (u)v L ∞ () ≤ J 1 C2 v L ∞ () . showing the boundeness of D F(u) also in this case.
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Suppose now that u 1 and u 2 and v belong to L p (), 1 ≤ p < ∞. From (2.5) and Hölder inequality, it follows that (D F(u 1 ) − D F(u 2 ))v L p () = K [( f (u 1 ) − f (u 2 ))v] L p ()
≤ J p ( f (u 1 ) − f (u 2 ))v L 1 () ≤ J p f (u 1 ) − f (u 2 ) L q () v L p () .
Thus to prove continuity of the derivative, we only have to show that f (u 1 ) − f (u 2 ) L q () → 0 as u 1 −u 2 L p () → 0. Now, from the growth condition we obtain | f (u 1 )(x) − f (u 2 )(x)|q ≤ [C1 (2 + |u 1 (x)| p−1 + |u 2 (x)| p−1 )]q and a computation similar to (2.10) above shows that the right-hand-side is integrable. The result then follows from Lebesgue’s convergence theorem. In the case p = ∞, we obtain from (2.4) (D F(u 1 ) − D F(u 2 ))v L ∞ () = K [( f (u 1 ) − f (u 2 ))v] L ∞ ()
≤ J 1 f (u 1 ) − f (u 2 ) L ∞ () v L ∞ ()
and the continuity of D F follows from the continuity of f . Therefore, it follows from Proposition 2.7 above that F is Fréchet differentiable with continuous derivative in X .
Remark 2.9 Since the right-hand side of (2.1) is a C 1 function, the semigroup generated by (2.1) in X is C 1 with respect to initial conditions, (see [27]). From the results above, we have that, for each t0 ≥ 0 and u 0 ∈ X , the unique solution of (2.1) with initial condition u 0 exists for all t ≥ t0 and this solution (t, x) → u(t, x) = u(t, x, t0 , u 0 ) [(defined by (2.8)] gives rise to a family of nonlinear C 1 semigroup on X given by S(t)u 0 (x) := u(t, x), t ≥ 0.
3 Existence and regularity of global attractor We prove the existence of a global attractor A in X for the semigroup {S(t); t ≥ 0} 1
when 1 ≤ p ≤ ∞. In the estimates below, the constant || p should be taken as 1 when p = ∞. Lemma 3.1 Suppose that the hypotheses of Proposition 2.8 hold with the constant k1 in (2.7) satisfying k1 < 1 Then the ball of L p (), 1 ≤ p ≤ ∞, centered at the origin 1 p
|| with radius (1 + δ) k21−k , where k1 and k2 are the constants appearing in (2.9) and 1 δ is any positive number, absorbs bounded subsets of X under the semigroup {S(t)} 1
generated by (2.1) (with || p replaced by 1 if p = ∞). Proof If u(t, x) is a solution of (2.1) with initial condition u 0 then, for 1 ≤ p < ∞ d dt
123
|u(t, x)| d x = p
p|u(t, x)| p−1 sgn(u(t, x))u t (t, x) d x
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= −p
+p
|u(t, x)| p (t, x)d x |u(t, x)| p−1 sgn(u(t, x))K ( f ◦ u)(x)d x.
Using Hölder’s inequality, estimate (2.4) and condition (2.9), we obtain |u(t, x)| p−1 sgn(u(t, x))K ( f ◦ u)(x)d x
≤
1
(|u(t, x)| p−1 )q
≤
1
|K ( f ◦ u)(t, x)| p
p
q
J 1 ( f ◦ u)(t, ·) L p () 1 p−1 k1 u(t, ·) L p () + k2 || p , ≤ u(t, ·) L p ()
|u(t, x)| p
1
q
where q is the conjugate exponent of p. Hence d p p p u(t, ·) L p () ≤ − pu(t, ·) L p () + pk1 u(t, ·) L p () dt 1
p−1
+ pk2 || p u(t, ·) L p () =
p pu(t, ·) L p ()
1 k2 || p −1 + k1 + . u(t, ·) L p () 1 p
, we have Let ε = 1 − k1 > 0. Then, while u(t, ·) L p () ≥ (1 + δ) k2 || ε ε d p p u(t, ·) L p () ≤ pu(t, ·) L p () (−ε + ) dt 1+δ δ p = −p εu(t, ·) L p () . 1+δ 1 p
|| Therefore, while u(t, ·) L p () ≥ (1 + δ) k21−k , we have 1 εδp
u(t, ·) L p () ≤ e− (1+δ) t u 0 L p () p
δp
p
= e− (1+δ) (1−k1 )t u 0 L p () . p
(3.1)
From this, the result follows easily for 1 ≤ p < ∞. Since the estimates are uniform in p, it also follows for p = ∞, by taking the limit with p → ∞.
Theorem 3.2 In addition to the conditions of Lemma 3.1, suppose that Jx p = supx∈ ∂x J (x, ·) L q () < ∞. Then there exists a global attractor A for the
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semigroup {S(t), t ≥ 0} generated by (2.1) in X = L p () and A is contained in the 1
k2 || p 1−k1
ball of radius
in L p (), for any t ∈ R and 1 ≤ p ≤ ∞.
Proof From (2.8), it follows that S(t)u = T (t)u + U (t)u, t ≥ 0, x ∈ where T (t)u(x) := e−t u(x) and
t
U (t)u(x) :=
e−(t−s) K ( f ◦ u(s, ·)(x)) ds.
0
Suppose u ∈ B, where B is a bounded subset of X . We may suppose that B is contained in the ball centered at the origin of radius r > 0. Then T (t)u L p () ≤ r e−t , t ≥ 0. From (3.1), we have that u(t, ·) L p () ≤ M, for t ≥ 0, where M = 1 p
2 || max{r, 2k1−k }. Hence, using (2.6), we obtain 1
p
f (u) L 1 () ≤ C1 (|| + u L p () ) ≤ C1 (|| + M p ). From estimate (2.5) (applied to Jx in the place of J ), it follows that ∂x K ( f ◦ u) L p () ≤ Jx p f (u) L 1 () ≤ C1 Jx p (|| + M p ). Thus, we obtain ∂x U (t)u
L p ()
t
≤ 0
e−(t−s) ∂x K ( f (u(s, ·))) L p () d s
≤ C1 Jx p (|| + M p ).
(3.2)
Therefore, for t ≥ 0 and any u ∈ B, the value of ∂x U (t)u L p () is bounded by a constant (independent of t and u ∈ B). It follows that U (t)u belongs to a (fixed) ball of W 1, p () for all u ∈ B. From Sobolev’s Embedding Theorem, it follows that U (t) is a compact operator, for any 0 ≤ t. Therefore it follows from Lemma 3.1 and Theorem 1.1, Chapter 1, of [28], that the global attractor A exists set of any bounded subset of X containing and is the ω-limit 2 || Bδ , where Bδ = B 0, (1+δ)k 1−k1
123
1 p
, for any δ > 0. From this, since Bδ absorbs
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bounded subsets of X , it also follows that A is contained in the ball centered at the 1
origin of radius
k2 || p 1−k1
in L p (R N ), for any t ∈ R and 1 ≤ p ≤ ∞.
Theorem 3.3 Assume the same conditions as in Theorem 3.2. Then there exists a bounded set of W 1, p (), 1 ≤ p ≤ ∞ containing A. Proof From Theorem 3.2, we obtain that A is contained in the ball centered at the 1 p
|| origin and radius k21−k in L p (). Now, if u(t, x) is a solution of (2.1) such that 1 u(t, x) ∈ A for all t ∈ R, then
u(t, x) =
t
−∞
e−(t−s) K (( f ◦ u)(s, ·))(x) ds,
where the equality above is in the sense of L p (R N ). Proceeding as in the proof of the Theorem 3.2 [see estimate (3.2) above], we obtain ∂x u(t, ·) L p () ≤ ≤
t
−∞ t −∞
e−(t−s) ∂x K ( f ◦ u)(s, ·) L p () d s Jx p f (u(s, ·)) d s
≤ C1 Jx p (|| + M p ), 1 p
2 || where now M = 2k1−k . 1 It follows that A = S(t)A is in a bounded set of W 1, p (), as claimed.
4 Existence of a Lyapunov functional We define an “energy” functional F : L p () → R, 1 ≤ p ≤ ∞ motivated by similar functionals appearing in [2,8,9,11,29,30] S(x) 1 −1 J (x, y)S(y)dy + f (r )dr d x, − S(x) F(u) = 2 0
(4.1)
where S(x) = f (u(x)). a Remark 4.1 Assuming that | f (x)| ≤ a, for all x ∈ , and | 0 f −1 (r )dr | < L < ∞, for some constants a, L > 0, it is easy to see that the functional given in (4.1) is defined in the whole space L p () and it is lower bounded. Theorem 4.2 In addition a to the conditions of Theorem 3.2, assume that | f (x)| ≤ a, for all x ∈ , and | 0 f −1 (r )dr | < L < ∞. Then the functional given in (4.1) is continuous in the topology of L p ().
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Proof Let (u n ) be a sequence converging to u in the norm of L p (). We can extract a subsequence (u n k ), such that, u n k (x) −→ u(x) a.e. in . Since f is continuous, we have Sn k (x) = f (u n k (x)) −→ f (u(x)) = S(x) a.e. Thus lim
Sn k (x)
k→∞ 0
f
−1
S(x)
(r )dr =
f −1 (r )dr.
0
Sn (x) −1 a Since | 0 k f (r )dr | ≤ 0 | f −1 (r )|dr , it follows from Lebesgue’s dominated convergence theorem that lim
Sn k (x)
k→∞ 0
f
−1
(r )dr d x =
0
S(x)
f −1 (r )dr d x.
It follows also from Lebesgue’s dominated convergence theorem that lim J (x, y)Sn k (y)dy = J (x, y)S(y)dy k→∞
and 1 1 − Sn k (x) J (x, y)Sn k (y)dy d x = − S(x) lim J (x, y)S(y)dy d x. k→∞ 2 2 Thus F(u n k ) converges to F(u), as k → ∞. Therefore (F(u n )) is a sequence such that every subsequence has a subsequence that converges to F(u). Hence F(u n ) → F(u), as n → ∞.
Theorem 4.3 Beyond of the hypotheses from Theorem 4.2 suppose that f has positive derivative. Let u(t, ·) be a solutions of (2.1) Then F(u(t, ·)) is differentiable with respect to t and dF = − [−u(t, x) + K ( f ◦ u)(t, x)]2 f (u(t, x))d x ≤ 0. dt S(s,x) −1 Proof Let ϕ(s, x) = − 21 S(s, x) J (x, y)S(s, y)dy + 0 f (r )dr , where S(s, x) = f (u(s, x)). From hypotheses under f follows that ∂ϕ(s,·) ∂s L 1 () < ∞, for all s ∈ R+ . Hence, deriving under the integration sign, obtain
1 ∂ S(t, x) J (x, y)S(t, y)dy 2 ∂t 1 ∂ S(t, y) ∂ S(t, x) − S(t, x) dy + f −1 (S(t, x)) ]d x J (x, y) 2 ∂t ∂t 1 ∂ S(t, x) =− d yd x J (x, y)S(t, y) 2 ∂t 1 ∂ S(t, y) ∂ S(t, x) − d yd x + d x. J (x, y)S(t, x) u(t, x) 2 ∂t ∂t
d F(u(·, t)) = dt
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[−
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Since 1 ∂ S(t, x) ∂ S(t, y) 1 J (x, y)S(t, y) J (x, y)S(t, x) d yd x = d yd x, 2 ∂t 2 ∂t it follows that
= = = =
∂ S(t, x) ∂ S(t, x) J (x, y)S(t, y) u(t, x) d yd x + dx ∂t ∂t ∂ S(t, x) dx − [−u(t, x) + J (x, y)S(t, y)dy] ∂t ∂ f (u(t, x)) − [−u(t, x) + K ( f ◦ u)(t, x)] dx ∂t ∂u(t, x) − [−u(t, x) + K ( f ◦ u)(t, x)] f (u(t, x)) dx ∂t − [−u(t, x) + K ( f ◦ u)(t, x)]2 f (u(t, x))d x.
d F(u(·, t)) = − dt
Since by hypothesis f > 0, the result follows.
Remark 4.4 From Theorem 4.2 it follows that, if F(T (t)u) = F(u) for t ∈ R, then u is an equilibrium point for T (t). Proposition 4.5 Assume the same hypotheses of Theorem 4.2 Then the flow generated by equation (2.1)is gradient. Proof The precompacity of the orbits follows from Theorem 3.2. From Remark 4.1, Theorem 4.2, Theorem 4.2 and Remark 4.4 it follows that the functional given in (4.1) is a continuous Lyapunov functional.
Remark 4.6 As a consequence of the Proposition 4.5, we have that the global attractor given in the Theorem 3.2 coincides with the unstable set of the equilibria (see [22]— Theorem 3.8.5), that is A = W u (E), where E = {u ∈ L p () : u(x) = K ( f ◦ u)(x)}. Acknowledgments The authors would like to thank the anonymous referee for his careful reading of the manuscript and his helpful suggestions.
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