About the Structure of Attractors for a Nonlocal Chafee-Infante Problem

In this paper, we study the structure of the global attractor for the multivalued semiflow generated by a nonlocal reaction-diffusion equation in which we cannot guarantee the uniqueness of the Cauchy problem. First, we analyse the existence and properties of stationary points, showing that the problem undergoes the same cascade of bifurcations as in the Chafee-Infante equation. Second, we study the stability of the fixed points and establish that the semiflow is a dynamic gradient. We prove that the attractor consists of the stationary points and their heteroclinic connections and analyse some of the possible connections.


Introduction
Ordinary and partial differential equations play a key role in modelling for all sciences: engineering, physics, chemistry, biology, medicine, economics and many others. The right understanding of the behaviour of solutions (in particular, well-posedness versus blowup) means not only to predict the future of trajectories but also to establish strategies for control (i.e., optimisation). Concerning PDE and economics, it is interesting to cite the nice survey [1] and the references therein on many different problems dealing with effects such as aggregation and repulsion, optimal control, mean-field games and so on as applications.
Parabolic PDE models reflect the diffusion phenomena due to local touching of molecules and dissipation of energy, and when different internal and external factors come into play, they link naturally to some reaction-diffusion models, such as the growth versus capacity of the environment in biology or the endogenous growth versus the neoclassical theories in economics. In particular, capital accumulation distribution in space and time following spatial extensions of the continuous Ramsey model [2] by Brito [3][4][5] and others later use the semilinear parabolic PDE This spatiality introduces important issues about the steady states distribution and the dynamic evolution, convergence, local interaction among local agents and so on.
Not for the sake of generality but for real modelling purposes, in the last two decades the increment of nonlocal PDE models that attempt to capture in a more accurate way the real spreading of the problem (density of population, capital accumulation, consumption or prices and innovation indexes and so on) has been very important. Firstly we might (here the functional Φ Ω may represent a general nonlocal functional acting over the whole domain Ω, for instance, u(t) 2 or Ω g(y)u(t, y)dy), equilibria are difficult to analyse.
Oppositely to ordinary differential equations, the analysis of the existence of stationary states for the above problem is much more involved. Additionally, comparing the reactiondiffusion equations with local diffusion, another difficulty is that in general a Lyapunov functional is not known to exist in most cases. The dynamical analysis of problem (1) and in particular the existence of global attractors has been established till now in several papers (cf. [17][18][19][20][21]). Other differential operators such as the p-Laplacian coupled with nonlocal viscosity has also been considered (cf. [21][22][23]). However, in general little is known about the internal structure of the attractor, which is very important as it gives us a deep insight into the long-term dynamics of the problem. When we manage to obtain a Lyapunov functional some insights can be obtained.
If we consider the non-local equation (2) with Dirichlet boundary conditions, then it is possible to define a suitable Lypaunov functional. In [18] it is shown that regular and strong solutions generate (possibly) multivalued semiflows having a global attractor which is described by the unstable set of the stationary points. Although this is already a good piece of information, our goal is to describe the structure of the attractor as accurately as possible. For this aim we need to study the particular situation where the domain is one-dimensional and the function f is of the type of the standard Chafee-Infante problem, for which the dynamics inside the attractor has been completely understood [24]. The first step when studying the structure of the attractor consists of analysing the stationary points. In the case where the function f is odd and Equation (2) generates a continuous semigroup, the existence of fixed points of the type given in the Chafee-Infante problem was established in [25]. Moreover, if a is non-decreasing, then they coincide with the ones in the Chafee-Infante problem, and moreover, in [26] the stability and hyperbolicity of the fixed points was studied. In this paper we extend these results for a more general function f (not necessarily odd and for which we do not known whether the Cauchy problem has a unique solution or not), showing that Equation (2) undergoes the same cascade of bifurcations as the Chafee-Infante equation. Moreover, when we allow the function a to decrease, though the problem possesses at least the same fixed point as in the Chafee-Infante problem, we show that more equilibria can appear. For a non-decreasing function a and an odd function f we prove also that even when uniqueness fails, the stability of the fixed points is the same as for the corresponding ones in the Chafee-Infante problem. Finally, we are able to prove that in this last case we have a dynamically gradient semiflow with respect to the disjoint family of isolated weakly invariant sets generated by the equilibria, which is ordered by the number of zeros of the fixed points. More precisely, the attractor consists of the set of equilibria and their heteroclinic connections and a connection from a fixed point to another is allowed only if the number of zeros of the first one is greater.
In Section 3 we study the existence of strong solutions of the Cauchy problem in the space H 1 0 . In Section 4 we prove that strong solutions generate a multivalued semiflow in H 1 0 having a global attractor which is equal to the unstable set of the stationary points. In Section 5 we study the existence and properties of equilibria. In Section 6 we analyse the stability of the fixed points and establish that the semiflow is dynamically gradient.

Setting of the Problem
Let us consider the following problem: where Ω = (0, 1) and λ > 0. Throughout the paper we will use the following conditions (but not all of them at the same time): (A1) f ∈ C(R). Growth and dissipation conditions: for p ≥ 2, C i > 0, i = 1, .., 4, we have lim sup The function a ∈ C(R + ) satisfies: The function a ∈ C(R + ) satisfies: The function a ∈ C(R + ) is non-decreasing. (A9) h ∈ L 2 loc 0, +∞; L 2 (Ω) . (A10) h does not depend on time and h ∈ L 2 (Ω).
We define the function F (u) = u 0 f (s)ds. We observe that from (4) we have |F (s)| ≤ C(1 + |s| p ) ∀s ∈ R, (7) whereas (5) implies Additionally, from condition (6) it follows that for all ε > 0, there exists a constant M > 0 such that In addition, it follows that where C ε > 0. These two inequaities are also true under condition (5).
The main aim of this paper consists of describing in as much detail as possible the internal structure of the global attractor in a similar way as for the classical Chafee-Infante equation.
Some of these conditions will be used all the time, whereas other ones will be used only in certain results. In particular, the function h will be considered as a time-dependent function satisfying (A9) only for establishing the existence of solution for problem (3). However, since we will study the asymptotic behaviour of solutions in the autonomous situation, for the second part concerning the existence and properties of global attractors, the function h will be time-independent, so assumption (A10) will be used instead. Finally, in order to study the structure of the global attractors in terms of the stationary points and their possible heteroclinic connections, we will assume that h ≡ 0.
Throughout the paper, · X will denote the norm in the Banach space X.

Existence of Solutions
In this section we will establish the existence of strong solutions for problem (3) with an initial condition in the phase space H 1 0 (Ω). Although we will follow along the same lines as a similar result given in [18], we would like to point out that in the present case, as we are working in a one-dimensional problem, the assumptions for the function f are much weaker. In particular, we do not need to impose a growth assumption of any kind. Definition 1. For u 0 ∈ L 2 (Ω), a weak solution to (3) is an element u ∈ L ∞ (0, T; L 2 (Ω)) ∩ L 2 (0, T; H 1 0 (Ω)), for any T > 0, such that where the equation is understood in the sense of distributions.

Theorem 1.
Assume conditions (A1), (A6) and (A9). Assume also the existence of constants β, γ > 0 such that Then, for any u 0 ∈ H 1 0 (Ω), problem (3) has at least one strong solution. (13) is weaker than the dissipative property (9) as the constant ε is arbitrarily small. Due to the fact that we are working in a one-dimensional domain, no growth condition of the type given in (A5) is necessary in order to prove existence of solutions. Additionally, (13) implies that F(u) ≤ γ + βu 2 (14) for some constants γ, β > 0.

Remark 3. Assumption
Proof. Consider a fixed value T > 0. In order to use the Faedo-Galerkin method, let {w j } j≥1 be the sequence of eigenfunctions of −∆ in H 1 0 (Ω) with homogeneous Dirichlet boundary conditions, which forms a special basis of L 2 (Ω). Since Ω is a bounded regular domain, it is known that {w j } ⊂ H 1 0 (Ω) and that ∪ n∈N V n is dense in the spaces L 2 (Ω) and H 1 0 (Ω), where V n = span[w 1 , . . . , w n ]. As usual, P n will be the orthogonal projection in L 2 (Ω), that is, and λ j will be the eigenvalues associated with the eigenfunctions w j . For each integer n ≥ 1, we consider the Galerkin approximations u n (t) = n ∑ j=1 γ nj (t)w j , which are given by the following nonlinear ODE system: We observe that P n u 0 → u 0 in H 1 0 (Ω). This Cauchy problem possesses a solution on some interval [0, t n ) and by the estimates in the space L 2 (Ω) of the sequence {u n } given below for any T > 0, such a solution can be extended to the whole interval [0, T].
Now, we multiply Equation (3) by du n dt to obtain we have Integrating the previous expression between 0 and t we get By (A6), (14) and (17) it follows that Since dim(Ω) = 1, Thus, as f maps bounded sets of R into bounded ones, F (u n (0)) is bounded in L ∞ (Ω) as well. Therefore, we deduce that {u n } is bounded in L ∞ (0, T; H 1 0 (Ω)) and du n dt is bounded in L 2 (0, T; L 2 (Ω)).
By using again the embedding H 1 0 (Ω) ⊂ L ∞ (Ω) we obtain that u n is bounded in the space L ∞ (0, T; L ∞ (Ω)). Thus, Additionally, we deduce that u n (t) 2 is uniformly bounded in [0, T], and then by the continuity of the function a(·) we get that the sequence a u n (t) 2 Finally, multiplying (15) by λ j γ ni (t) and summing from i = 1 to n we obtain By (23) and applying the Young inequality, we get Integrating the previous expression between 0 and t, it follows that Taking into account (23), the last inequality implies that so {−∆u n } and {a( u n 2 As a consequence, there exist u ∈ L ∞ (0, T; H 1 0 (Ω)) and a subsequence u n (relabelled the same) such that u n * u in L ∞ (0, T; H 1 0 (Ω)), u n u in L 2 (0, T; D(A)), where ( * ) stands for the weak (weak star) convergence. By (22) and (24) the Aubin-Lions compactness lemma gives that u n → u in L 2 (0, T; H 1 0 (Ω)), so u n (t) → u(t) in H 1 0 (Ω) a.e. on (0, T). Consequently, there exists a subsequence u n , relabelled the same, such that u n (t, x) → u(t, x) a.e. in Ω × (0, T).
As a consequence, by the continuity of a we get that a( u n (t) 2 ) a.e. on (0, T).
Since the sequence is uniformly bounded, by Lebesgue's theorem this convergence takes place in L 2 (0, T), so b = a( u 2 Therefore, we can pass to the limit to conclude that u is a strong solution. It remains to show that u(0) = u 0 which makes sense since u ∈ C([0, T]; H 1 0 (Ω)) (see Remark 4). Indeed, let be φ ∈ C 1 ([0, T]; H 1 0 (Ω)) with φ(T) = 0, φ(0) = 0. We multiply the equation in (3) and (15) by φ and integrate by parts in the t variable to obtain that In view of the previous convergences, we can pass to the limit in (27). Taking into account (26) and bearing in mind u n (0) = P n u 0 → u 0 , since φ(0) ∈ H 1 0 (Ω) is arbitrary, we infer that u(0) = u 0 .

The Existence and Structure of Attractors
In this section, we will prove the existence of a global attractor for the semiflow generated by strong solutions in the autonomous case. Thus, the function h will be an independent of time function satisfying (A10) instead of (A9). Additionally, we will establish that the attractor is equal to the unstable set of the stationary points (see the definition in (45)).
Throughout this section, for a metric space X with metric d we will denote by dist X (C, D) the Hausdorff semidistance from C to D, that is, Let us consider the phase space X = H 1 0 (Ω) and the sets Denote by P(X) the class of nonempty subsets of X. We define the (possibly multivalued) map G : R + × X → P(X) by In order to study the map G let us consider the following axiomatic properties of the set R: (K2) φ τ (·) := φ(· + τ) ∈ R for every τ ≥ 0 and φ ∈ R (translation property). (K3) Let φ 1 , φ 2 ∈ R be such that φ 2 (0) = φ 1 (s) for some s > 0. Then, the function φ defined by belongs to R (concatenation property). (K4) For every sequence {φ n } ⊂ R satisfying φ n (0) → x 0 in X, there is a subsequence {φ n k } and φ ∈ R such that φ n k (t) → φ(t) for every t ≥ 0.
Moreover, (K3) implies that the m-semiflow is strict, that is, G(t + s, x) = G(t, G(s, x)) for all t, s ≥ 0 and x ∈ X.
We will show first that the m-semiflow G possesses a bounded absorbing set in the space L 2 (Ω) and that property (K4) is satisfied. Lemma 1. Assume conditions (A1), (A6), (A10) and (13). Given {u n } ⊂ R, u n (0) → u 0 weakly in H 1 0 (Ω), there exists a subsequence of {u n } (relabelled the same) and u ∈ K(u 0 ) such that and u n ∈ C([0, T]; H 1 0 (Ω)). Additionally, as f (u n ) ∈ L ∞ (0, T; L ∞ (Ω)), by regularization one can show that (F(u n (t)), 1) is an absolutely continuous function on [0, T] and By a similar argument as in Theorem 1, there is a subsequence of u n such that Therefore, arguing as in the proof of Theorem 1, there exist u ∈ K(u 0 ) and a subsequence u n , relabelled the same, such that We also need to prove that To that end, we multiply (3) by u n t , and using (A10), (29) and (31) we have Thus, we obtain Since this inequality is also true for u(·), the functions Q n (t) = A( u n (t) 2 ) − 2Ct are continuous and non-increasing in [0, T]. Moreover, from (32) we deduce that Q n (t) → Q(t) for a.e. t ∈ (0, T).
Take 0 < t ≤ T and 0 < t j < t such that t j → t and Q n (t j ) → Q(t j ) for all j. Then For any δ > 0 there exist j(δ) and N(j(δ)) such that , which follows by contradiction using the continuity of the function A(s). As u n (t) → u(t) weakly in H 1 0 (Ω) implies that lim inf u n (t) 2 Repeating the above argument, we infer that u n (t n ) → u 0 strongly in H 1 0 (Ω).

Corollary 1.
Assume the conditions of Lemma 1. Then the set R satisfies condition (K4).
The map t → G(t, x) is said to be upper semicontinuous if for every x ∈ X and for an arbitrary neighbourhood O(G(t, x)) in X there is δ > 0 such that as soon as d(y, x) < δ, we have G(t, y) ⊂ O. Proposition 1. Assume the conditions of Lemma 1. The multivalued semiflow G is upper semicontinuous for all t ≥ 0. Additionally, it has compact values.
Proof. By contradiction let us assume that there exist t ≥ 0, u 0 ∈ H 1 0 (Ω), a neighbourhood O(G(t, u 0 )) and sequences {y n }, {u n 0 } such that y n ∈ G(t, u n 0 ), u n 0 converges strongly to u 0 in H 1 0 (Ω) and y n / ∈ O(G(t, u n )) for all n ∈ N. Thus, there exists u n ∈ K(u n 0 ) such that y n = u n (t). From Lemma 1 there exists a subsequence of y n which converges to some y ∈ G(t, u 0 ). This contradicts y n / ∈ O(G(t, u 0 )) for any n ∈ N.
In order to prove the existence of an absorbing set in the space L 2 (Ω) we need to use the stronger condition (A5) instead of (13).

Proposition 2.
Assume that conditions (A1), (A5), (A6) and (A10) hold. Then the m-semiflow G has a bounded absorbing set in L 2 (Ω); that is, there exists a constant K > 0 such that for any Proof. By multiplying Equation (3) by u and using (A6) and (9), we get By using the Poincaré inequality it follows that We take a small enough ε > 0 so that δ > 0. Then Gronwall's lemma gives Hence, taking On the other hand, using again the Poincaré inequality from (35) we get and integrating from t to t + 1 we obtain Therefore, applying (33) and (34) follows.
Further, in order to obtain an absorbing set in H 1 0 (Ω) we need to assume additionally that either the function a(·) is bounded above or that it is non-decreasing. Proposition 3. Assume the conditions in Proposition 2 and that either (A7) or (A8) holds true. Then there exists an absorbing set B 1 for G, which is compact in H 1 0 (Ω).
Proof. In view of Proposition 2 we have an absorbing set B 0 in L 2 (Ω). Let K > 0 be such that y ≤ K for all y ∈ B 0 . Through multiplying (3) by u and using (9) and (36) we get Thus, integrating between t and t + r, 0 < r ≤ 1, we deduce by using (36) again that Additionally, if p > 2 in (A5), we multiply again by (3) by u and use (5) and (A6) to obtain Integrating over (t, t + r) we have If we assume (A7), by (37) and (A6) we have that t+r t A( u(s) 2 If we assume (A8), by (37) we obtain t+r t A( u(s) 2 On the other hand, by (7) we get By using (29) and (30) we can argue as in Theorem 1 to obtain Since (38)-(41) imply that we can apply the uniform Gronwall lemma to get so by condition (A6), (10) and (36) it follows that for all t ≥ 0. In particular, for any strong solution u(·) with initial condition u(0).
Then there exists M > 0 such that the closed ball B M in H 1 0 (Ω) centred at 0 with radius M is absorbing for G.
By Lemma 1 the set for all t ≥ 0, and strictly invariant (or, simply, invariant) if it is both negatively and positively invariant.
We recall that a set A ⊂ X is called a global attractor for the m-semiflow G if it is negatively invariant and attracts all bounded subsets; i.e., dist X (G(t, B), A) → 0 as t → +∞. When A is compact, it is the minimal closed attracting set ([32] Remark 5). We recall some concepts which are necessary to study the structure of the global attractor.
We denote the set of all fixed points by R R .
An element z ∈ X is a fixed point of G if z ∈ G(t, z) for every t ≥ 0.
Several properties concerning fixed points, complete trajectories and global attractors are summarised in the following results [33]. The standard well-known result in the single-valued case for describing the attractor as the union of bounded complete trajectories reads in the multivalued case as follows.
Theorem 3. Suppose that (K1) and (K2) are satisfied and that either (K3) or (K4) holds true. The semiflow G is assumed to have a compact global attractor A. Then where K stands for the set of all bounded complete trajectories in R.
In view of Theorem 3, as R satisfies (K3) and (K4) (by Corollary 1), the global attractor is characterised in terms of bounded complete trajectories, so (42) follows.
The set B is said to be weakly invariant if for any x ∈ B there exists a complete trajectory γ of R contained in B such that γ(0) = x. Characterisation (42) implies that the attractor A is weakly invariant.
The set of fixed points R R is characterised as follows.
Finally, we shall obtain the characterisation of the global attractor in terms of the unstable and stable sets of the stationary points.

Theorem 4. Assume the conditions of Proposition 3. Then it holds that
and F denotes the set of all complete trajectories of R (see Definition 3).

Remark 4.
In (45) it is equivalent to use K instead of F because all the solutions are bounded forward in time.
Proof. We consider the function E : A → R Note that E(y) is continuous in H 1 0 (Ω). Indeed, the maps y → 1 2 A( y 2 H 1 0 ) and y → Ω h(x)y(x)dx are obviously continuous in H 1 0 (Ω). On the other hand, by the embedding H 1 0 (Ω) ⊂ L ∞ (Ω) and using Lebesgue's theorem, the continuity of y → Ω F(y(x))dx follows.
By using (29)-(30) and multiplying Equation (3) by du dt for any u ∈ R, we can obtain the following energy equality: Hence, E(u(t)) is non-increasing, and by (A6), (10) and the boundedness of A, it is bounded from below. Thus E(u(t)) → l, and t → +∞, for some l ∈ R. Let z ∈ A and u ∈ K be such that u(0) = z. By contradiction, suppose the existence of ε > 0 and u(t n ), where t n → +∞, for which dist H 1 0 (u(t n ), R) > ε. Since A is compact in H 1 0 (Ω), we can take a converging subsequence (relabelled the same) such that u(t n ) → y in H 1 0 (Ω), where t n → ∞. By the continuity of the function E, it follows that E(y) = l. We will obtain a contradiction by proving that y ∈ R. Define v n (·) = u(· + t n ). By Lemma 1, there exist v ∈ R and a subsequence satisfying v(0) = y and v n (t) → v(t) in H 1 0 (Ω) for t ≥ 0. Thus, from E(v n (t)) → E(v(t)) we infer that E(v(t)) = l. Additionally, v(·) satisfies the energy equality, so that Therefore, dv dt (s) = 0 for a.a. s, and then by Lemma 3 we have y ∈ R R = R. As a consequence, A ⊂ M + (R). The converse inclusion follows from (42). As before, take arbitrary z ∈ A and u ∈ K satisfying u(0) = z. Since by the embedding H 1 0 (Ω) ⊂ C([0, 1]) the energy function is bounded from above in A, E(u(t)) → l, as t → −∞, for some l ∈ R. Suppose that there are ε > 0 and u(t n ), where t n → +∞, such that dist H 1 0 (u(−t n ), R) > ε. Up to a subsequence we have that u(−t n ) → y in H 1 0 (Ω), E(y) = l. Moreover, for v n (·) = u(· − t n ) there are v ∈ R and a subsequence such that v(0) = y and v n (t) → v(t) in H 1 0 (Ω) for t ≥ 0. Therefore, E(v n (t)) → E(v(t)) gives E(v(t)) = l and then by the above arguments we get a contradiction because y ∈ R. Hence, A ⊂ M − (R) and we deduce the converse inclusion from (42).
Finally, we are able to obtain that the global attractor is compact in the space C 1 ([0, 1]). This property will be important in order to study a more precise structure of the global attractor in terms of the stationary points and their heteroclinic connections.
We define the function is under the conditions of Proposition 3 (see [18] for more details) a strong solution to the problem Let V 2r = D(A r ), r ≥ 0. We will prove first that the attractor is compact in any space V 2r with 0 ≤ r < 1. For this aim we will need the concept of mild solution. We consider the auxiliary problem where g ∈ L 2 loc 0, +∞; L 2 (Ω) . The function u ∈ C([0, +∞), L 2 (Ω)) is called a mild solution to problem (48) if In the same way as in Lemma 2 in [34] we obtain that a strong solution to problem (47) is a mild solution to problem (48) with g(t) = ( f (w(t)) + h)/a( w(t) 2 ).

Lemma 4.
Assume the conditions of Proposition 3. Then the global attractor A is compact in V 2r for every 0 ≤ r < 1.
Proof. Let z ∈ A be arbitrary. Since A is invariant, there exist u 0 ∈ A and u ∈ R such that z = u(1) and u(t) ∈ A for all t ≥ 0. Since w(t) = u α −1 (t) is a mild solution of (48) with ), the variation of constants formula (49) gives As A is bounded in H 1 0 (Ω) (and then in L ∞ (Ω)), condition (A6) and the continuity of f imply that where C > 0 does not depend on z. The standard estimate e −At L(L 2 (Ω),D(A r )) ≤ M r t −r e −at , M r , a > 0 ( [27] Theorem 37.5), implies that From the compact embedding V α ⊂ V β , for α > β, and the fact that A is closed in any V 2r we obtain the result. Proof. We obtain by Lemma 37.8 in [27] the continuous embedding Hence, the statement follows from Lemma 4.

Fixed Points
In this section we are interested in studying the fixed points of problem (3) when h ≡ 0, that is, the solutions of the boundary-value problem For this aim we will use the properties of the fixed points of the standard Chafee-Infante equation. In order to do that, for any d ≥ 0 we will study the )-(Aollowing boundaryvalue problem. as it is obvious that u(·) is solution to problem (50) if and only if u(·) is a solution to problem (51) with d = u 2
We need to study the dependence of the norm of these fixed points on the parameter λ. First, we will show that the H 1 -norm of the fixed points of problem (52) is strictly increasing with respect to the parameter λ.

Lemma 5. Assume conditions (A1)-(A5). Let v
Proof. We consider the equivalent norm in H 1 0 (Ω) given by v L 2 . The fixed points are the solutions of the initial value problem such that u(1) = 0. The solutions of (53) satisfy the relation (u (x)) 2 2 . By Theorem 7 in [35] we have that u λ is associated with a unique value E = E + k ( λ) > 0. Moreover, E + k ( λ) is a solution of one of the following equations: where either k = 2m − 1 or k = 2m and being U + (E) (U − (E)) the positive (negative) inverse of F at E. It is obvious that for E fixed the functions τ λ + (E), τ λ − (E) are strictly decreasing with respect to λ. Then from (55) we deduce that the root E + k ( λ) is strictly increasing with respect to λ. Thus, If λ 1 < λ 2 , we have We will prove now that u λ L 2 is strictly increasing in λ. The function u λ has k + 1 simple zeros in [0, 1] and u λ is positive in the first subinterval. Let T + (E + k (λ)) be the x-time necessary to go from the initial condition u λ (0) = 0 to the point where u λ (T + (E + k (λ))) = 0. Then the length of the first subinterval is 2T + (E + k (λ)) [35]. By (54), By the change of variable v = u λ (x) we obtain .
Since λ → U + (E + k ( λ)) is strictly increasing and by using (58), we conclude that the function g( λ) is strictly increasing. Hence, by putting x 1 ( λ) = 2T + (E + k ( λ)) we obtain that the norm of u λ in the first subinterval, u , is strictly increasing. By arguing in the same way as for the other subintervals, we obtain that λ → u λ L 2 is strictly increasing. Let us prove the same result but with respect to the norm u λ L p with p ≥ 1.

Lemma 6. Assume conditions (A1)-(A5) and let f be odd. Let v
Then v 1 L p < v 2 L p for any p ≥ 1.

Proof. As in the previous lemma, denote u
. The function u λ has k + 1 zeros in [0, 1] at the points 0 < x 1 < x 2 < ... < x k−1 < 1. When f is odd, by symmetry, the length of all subintervals has to be the same, so x j = j k regardless the value of λ. We shall prove that in the first subinterval we have that u λ 1 (x) < u λ 2 (x), for all , so (58) yields , if x ∈ (0, 1 2k ].
Thus, u λ 1 (x) < u λ 2 (x), for all x ∈ (0, 1 2k ]. By symmetry we obtain that the inequality is true in 0, 1 k . Repeating the same argument in the other subintervals we get that This implies that u λ 1 L p < u λ 2 L p for any p ≥ 1.

Remark 5. The statements in Lemmas 5 and 6 are also true for v
so the H 1 0 and L p norms of v − k, λ and v + k, λ are the same.

Nonlocal Fixed Points
Although in this paper we are mainly interested in problem (3), we will study the existence of stationary points for an elliptic problem with a more general nonlocal term than in (50). Namely, let us consider the following problem: where l(u) = u r Then for any d < d k there exists the fixed point u d k of (51), where u d k is either equal to u + k or u − k .
It is obvious that any solution of (59) is a solution of (51) with d = l(u). Therefore, all the solutions to problem (59) have to be solutions u d k to problem (51) for a suitable d.

Theorem 5.
Assume conditions (A1)-(A6) and, additionally, that Then: • For any 1 ≤ j ≤ k there exists d * j < d k such that u d * j j is a fixed point of problem (59).

•
If N ≥ k is the first integer such that λ ≤ inf s≥0 {a(s)π 2 (N + 1) 2 }, there are no fixed points for j > N.
Consider first the case where d k is finite. We need to obtain the existence of d * Additionally, we know that l u d k k = 0. Through multiplying (51) by u d k and using (9), (A6) and the Poincaré inequality we obtain so, by using the embedding H 1 0 (Ω) ⊂ L ∞ (Ω), l u d k is bounded in d.  1]) it follows that v cannot be a point with less than k + 1 simple zeros in [0, 1] and then λ/a(d k ) = k 2 π 2 implies that v ≡ 0. As the limit is the same for every converging Second, let d k = +∞. Then the existence of d * k < +∞ follows by the same argument as before.
The last two statements are a consequence of the first two statements and of the fact that the points of intersection of the functions g(d) = l u d k and y(d) = d has to be unique, because if a is non-decreasing, then g(d) is non-increasing by Lemmas 5 and 6.
In view of this theorem, we have exactly the same equilibria and bifurcations as in the classical Chafee-Infante equation (see [24,35]) when the function a(d) is non-decreasing, because in this case in view of the monotone dependence between the functions a(d) and g(d), there is only one intersection point of the function g(d) with the bisector, as it is shown in Figure 1. This follows from the fact that g(d) − d is strictly decreasing, but there may be weaker conditions on a(·) that would lead g(d) − d to be strictly decreasing.
When the function a(·) is not assumed to be monotone, an interesting situation appears. More precisely, it is possible to have more than two equilibria with the same number of zeros. If l(u) = u 2 H 1 0 , for the equilibria with k + 1 zeros in [0, 1] this happens when the equation has more than one solution. For instance, if a(0) = a(d) for some 0 <d < g(0), then g(0) = g(d). Assuming that there are 0 < d 1 , obtaining six fixed points with k + 1 zeros in [0,1]. This situation is shown in Figure 2, where d * 1 , d * 2 and d * 3 are solutions of (61), that is, there are three intersection points with the bisector. We notice that when a(d) > λ/(π 2 k 2 ), the function g(d) is not defined since the condition for such equilibria to exist is not satisfied, but we can make this function continuous by putting g(d) = 0 whenever a(d) ≥ λ/(π 2 k 2 ). This procedure establishes that, having fixed a natural number k, for any j ∈ N we may construct a(·) in such a way that we have 2(2j + 1) equilibria with k + 1 zeros in [0, 1].
At least there is always one intersection point with the bisector, but the function g(d) could be even tangent to the bisector at some point or not cut it again.

Lap Number and Some Forbidden Connections
With Theorem 5 at hand we can improve the description of the global attractor given in Theorem 4.
Under conditions (A1)-(A6), (A8) and h ≡ 0, if a(0)π 2 n 2 < λ ≤ a(0)π 2 (n + 1) 2 (62) then problem (3) possesses exactly 2n + 1 fixed points Let φ be a bounded complete trajectory. We know by Theorem 4 that As the number of fixed points is finite, we will prove that in fact the solution has to converge to one fixed point forwards and backwards. We recall the omega and alpha limit sets of φ, given by Thus, we have established the following result.
where n is given in (62) and , ... In other words, the global attractor A consists of the set of stationary points R (which has 2n + 1 elements) and the bounded complete trajectories that connect them (the heteroclinic connections).

Remark 6.
As the Lyapunov function (46) is strictly decreasing along a trajectory φ which is not a fixed point, then there cannot exist homoclinic connections for any fixed point. This implies in particular that if n = 0, then A = {0}. Remark 7. If we use condition (A7) instead of (A8), then we cannot guarantee that the number of fixed points is finite. However, if we suppose that this is the case, then the result remains valid. In this situation, there could be more than two fixed points with the same number of zeros.
, where E is the Lyapunov function (46). Since this function is strictly decreasing along a trajectory φ which is not a fixed point, there cannot exist a heteroclinic connection between these two points.

Remark 8.
In the case where condition (A7) is assumed, there could be more than two equilibria with k + 1 zeros in [0, 1]. In this case there could exist connections between fixed points with different values of the constant d.
By using the concept of lap number of the solutions we can discard some more heteroclinic connections.
We consider the function w(t) = u(α −1 (t)), which is a strong solution to problem (47). For any strong solution u(·) conditions (A1), (A3), (A6) and u ∈ C([0, +∞), H 1 0 (Ω)) imply that the function is continuous and w(·) is a solution of the linear equation Thus, by Theorem A3 in the Appendix A (see also Theorem C in [37]) the number of zeros of w(t) in [0, 1] is a nonincreasing function of t. Since α −1 (t) is an increasing function of time, the result is also true for the solution u(·). Making use of this property we will prove the following result.
Proof. By contradiction assume that such complete trajectory exists. Denote by l(z) the number of zeros of z in [0, 1]. By using the compactness of the attractor in C 1 ([0, 1]) (see Corollary 2) we obtain that Then, as the zeros are simple, we can choose t 1 > 0 large enough such that l(φ(−t 1 )) = l u ± k,d * k = k + 1. Put u(t) = φ(t − t 1 ), which is a strong solution of (3). Now we choose t 2 > 0 such that l(u(t 2 )) = l u ± n,d * n = n + 1. Then l(u(0)) = k + 1 and l(u(t 2 )) = n + 1 > k + 1. This contradicts the fact that the number of zeros of u(t) is non-increasing.

Morse Decomposition
In this section we study in more detail the structure of the global attactor in the case where the function f is odd. More precisely, we obtain a dynamically gradient m-semiflow G, which is equivalent to saying that there is a Morse decomposition of the attractor [38], and we study the stability of the fixed points.

Approximations
We consider now the situation when conditions (A1)-(A6), h = 0 and either (A7) or (A8) are satisfied, and moreover, the function f is odd. uniformly bounded in any compact set of R. Then for u ∈ [−1, 1] the Hölder inequality and R ρ ε n (s)ds = 1 give If f satisfies (5), then where we have used |u| p ≤ 2 p−1 |s p | + |u − s| p and the Young inequality.
Our next aim is to focus on the convergence of solutions of the approximations.
Proof. By using (29) and (30) we can repeat the same lines of the proof of Theorem 1 and obtain the existence of a function u(·) and a subsequence of u ε n such that )∆u in L 2 (0, T; L 2 (Ω)).
Additionally, in the same way we prove that u(·) is a strong solution to problem (3) such that u(0) = u 0 .
The uniform estimate in the space H 1 0 (Ω) implies also that if t n → t 0 , then u ε n (t n ) u(t 0 ) in H 1 0 (Ω). We need to prove that this convergence is in fact strong, proving then the convergence in C([0, T], H 1 0 (Ω)) for any T > 0. In the same way as in the proof of Lemma 1 we deduce that for some C > 0 the functions Q n (t) = A( u ε n (t) 2 ) − 2Ct are continuous and non-increasing in [0, T]. Moreover, Q n (t) → Q(t) for a.e. t ∈ (0, T). Let first t 0 >0. Then we take 0 < t j < t 0 such that t j → t 0 and Q n (t j ) → Q(t j ) for all j. Then For any δ > 0 there exist j(δ) and N(j(δ)) such that Q n (t n ) − Q(t 0 ) ≤ δ if n ≥ N, so lim sup Q n (t n ) ≤ Q(t 0 ). Hence, a contradiction argument using the continuity of A(s) shows that lim sup u ε n (t n ) 2 For the case when t 0 = 0 we use the same argument as in Lemma 1.
We denote by A ε n the global attractor for the semiflow G ε n corresponding to problem (64).

Lemma 10.
Assume the condition of Theorem 7. Then ∪ n∈N A ε n is bounded in H 1 0 (Ω). Hence, the set ∪ n∈N A ε n is compact in L 2 (Ω).
Proof. By Lemma 9 inequality (9) is satisfied for any n with constants which are independent of ε n , so inequality (36) holds true with constants independent of ε n . Thus, there a exists a common absorbing ball B 0 in L 2 (Ω) (with radius K > 0) for problems (64). Further, by repeating the same steps as in Proposition 3 we obtain a common absorbing ball in H 1 0 (Ω) (with radius K > 0), as by Lemma 9 the constants which are involved are independent of ε n . Thus, y H 1 0 ≤ K for any y ∈ ∪ n∈N A ε n . Lemma 11. Assume the condition of Theorem 7. Then ∪ n∈N A ε n is bounded in V 2r for any 0 ≤ r < 1. Hence, ∪ n∈N A ε n is compact in V 2r and C 1 ([0, 1]).
Proof. By using Lemma 10 we obtain the boundedness of ∪ n∈N A ε n in V 2r by repeating the same lines in Lemma 4. The rest of the proof follows from the compact embedding V α ⊂ V β , α > β, and the continuous embedding Corollary 3. Assume the condition of Theorem 7. Then any sequence ξ n ∈ A ε n with ε n → 0 is relatively compact in C 1 ([0, 1]).

Lemma 12.
Assume the condition of Theorem 7. Then up to a subsequence any bounded complete trajectory u ε n of (64) converges to a bounded complete trajectory u of (3) in C([−T, T], H 1 0 (Ω)) for any T > 0. On top of that, if y n ∈ A ε n , then passing to a subsequence y n → y ∈ A in Proof. Let us fix T > 0. By Corollary 3 u ε n (−T) → y in H 1 0 (Ω) up to a subsequence. Theorem 7 implies that u ε n converges in C([−T, T], H 1 0 (Ω)) to some solution u of (3). If we choose successive subsequences for −2T, −3T . . . and apply the standard diagonal procedure, we obtain that a subsequence u ε n converges to a complete trajectory u of (3) in C([−T, T], H 1 0 (Ω)) for any T > 0. Finally, from Lemma 10 this trajectory is bounded. If y n ∈ A ε n , by Corollary 3 we can extract a subsequence converging to some y. If we take a sequence of bounded complete trajectories φ n (·) of (64) such that φ n (0) = y n , then by the previous result it converges in C([−T, T], H 1 0 (Ω)) to some bounded complete trajectory φ(·) of (3), so y ∈ A.
Finally, if (67) was not true, there would exist δ > 0 and a sequence y n ∈ A ε n such that dist H 1 0 (y, A) > δ. However, passing to a subsequence y n → y ∈ A, which is a contradiction.
Lemma 13. Assume the conditions of Theorem 7. Let τ d n ,ε n ± be the functions (56)-(57) for problem (51) but replacing f by f ε n and d by d n . Let d n , E n → 0 as n → ∞. Then Proof. Let us consider f d n ,ε n (u) = λ f εn (u) a(d n ) . In view of property (B4) and (66), since f ε n (0) = f (0) = 1 and f ε n (0) = f (0) = 0, given γ ∈ (0, 1) there exists δ > 0 (independent of ε n ) such that The sequence F ε n (·) converges uniformly to F (·) in compact sets. Moreover, as U + (E) is continuous and using ( [39] p. 60), given δ > 0, there exists η > 0 such that U ε n + (E) ≤ δ for any 0 < E ≤ η. Now, if we integrate the first inequality in (68) between 0 and u we obtain By using the change of variable E n y 2 = F ε n (u), we have Dividing the previous expression by λ a(d n ) f d n ,ε n (u) and using (68) we obtain Now if we multiply by 2 √ E n (1 − y 2 ) − 1 2 and integrate from 0 to 1, we get Then the theorem follows as a(d n ) → a(0) when n → ∞. The proof for τ ε n − is analogous.
Under the conditions of Theorem 7, if (A8) is satisfied and (0), so E n → 0. We will show that this is not possible. We know by Lemma 13 that is a fixed point with d = d ε n m one of the following conditions has to be satisfied (see (55)): Since E n → 0 and λ > k 2 π 2 a(0) ≥ m 2 π 2 a(0), there exists ε n 0 such that Hence, neither of (70)-(72) is possible.
and v is a solution of (50), so v is a fixed point of (3). We need to prove that

Instability
We will prove that the fixed points 0 and u ± k,d * k , k ≥ 2, are unstable under some additional assumptions on the functions f and a. For that aim we need to use the approximative problems (64). Theorem 8. Assume that the conditions (A1)-(A8), h = 0 and (69) with k ≥ 1 are satisfied; and let the function f (·) be odd and a(·) be globally Lipschitz continuous. Then the equilibria v 0 = 0 and u ± j,d * j , 2 ≤ j ≤ k (if k ≥ 2), are unstable.

Remark 9.
The condition that a(·) is globally Lipschitz continuous could be dropped, as we can replace a(·) in (64) by a sequence a ε n (·) of globally Lipschitz continuous functions.
Proof. Problem (64) generates a single-valued semigroup {T ε n (t); t ≥ 0} with a finite number of fixed points [26]. We know by Theorems 3.5 and 3.6 in [26] that for any v + j,d εn j with j ≥ 2 and v 0 there exists a bounded complete trajectory u ε n such that On the other hand, by Lemma 15 where u ± j,d * j is a fixed point of problem (3) with j + 1 zeros in [0, 1]. We prove the result for and v 0 the proof is the same.
By Lemma 12 we obtain that up to a subsequence u ε n converges to a bounded complete trajectory u of problem (3) In the second case, if v −1 = u + j,d * j , the proof would be finished, so let assume the opposite.
Assume first that either u(·) is not a fixed point or it is a fixed point but We consider r 0 > 0 such that the neighbourhood O 2r 0 (v −1 ) does not contain any other fixed point of problem (3). For any r ≤ r 0 we can choose t r → −∞ and n r such that u ε n (t r ) ∈ O r (v −1 ) for all n ≥ n r . On the other hand, since u ε n (t) Let first t t − t r → +∞. We define the sequence u ε nr 1 (t) = u ε nr (t + t r ), which passing to a subsequence converges to a bounded complete trajectory where v −2 is a fixed point different from v −1 . Second, let |t t − t r | ≤ C. Then put u ε nr 1 (t) = u ε nr (t + t r ). Passing to a subsequence we have that u ε nr Additionally, u ε nr 1 (·) converges to a bounded complete trajectory u 1 (·) of problem (3) such that u 1 (0) = v −1 . Let is not a fixed point. Then ψ 1 is a bounded complete trajectory of problem (3) such that , the proof is finished.
, we continue constructing by the same procedure a chain of connections in which the new fixed point is always different from the previous ones, because the existence of the Lyapunov function (46) avoids the existence of a cyclic chain of connections. Since the number of fixed points is finite, at some moment we obtain a bounded complete Defining the neighbourhood O 2r 0 (v −1 ) as before, for any r ≤ r 0 we can choose n r such that u ε n (0) ∈ O r (v −1 ) for all n ≥ n r . Additionally, since u ε n (t) → z n 0 , as t → +∞, where z n is a fixed point of (64), there exists t r > 0 such that The sequence {t r } cannot be bounded. Indeed, if t r → t 0 , then u ε nr (t r ) → u(t 0 ) = v −1 , which is a contradiction with u ε nr (t 0 ) − v −1 H 1 0 = r 0 . Then t r → +∞. We define the functions u ε nr 1 (t) = u ε nr (t + t r ), which satisfy that u ε nr 1 (t) ∈ O r 0 (v −1 ) for all t ∈ [−t r , 0). Passing to a subsequence it converges to a bounded complete trajectory φ(·) such that φ(t) ∈ O r 0 (v −1 ) for all t ≤ 0. This trajectory is not a fixed point as Further, we will prove that there is also a connection from 0 to the point u ± k,d * Proof. We start with the case where k = 1. We have three fixed points: 0, u + 1,d * whereas Theorem 4 and Remark 6 imply that it has to converge forward to a fixed point different from 0, that is, to either u + 1,d * Further we consider the problem where f k (u) = √ k f u/ √ k satisfies (A1)-(A5). In this problem, condition (69) implies that there are again three fixed points: 0, u + dµ < +∞.
We are ready to prove the lap-number property, saying that the number of zeros is a non-increasing function of time.
Theorem A3. Let r(t, x) be a continuous function and u ∈ C([t 0 , t 1 ], H 1 0 (Ω)) ∩ L 2 t 0 , t 1 ; H 2 (Ω) be such that du dt ∈ L 2 t 0 , t 1 ; L 2 (Ω) and satisfies the equation Then the number of components of is a non-increasing function of t.
Proof. We follow similar lines as in ([24] Theorem 6). Denote Q(t) = {x ∈ (0, 1) : u(t, x) = 0}. We need to show that there is an injective map from the components of Q(t 1 ) to the components of Q(t 0 ) if t 1 > t 0 . If we denote by C a component of Q(t 1 ) and by S C the component of [t 0 , t 1 ] × (0, 1) ∩ {u(t, x) = 0)} which contains C, then in order to obtain the injective map it is necessary to prove two facts: 1.
If C 1 , C 2 are two components of Q(t 1 ), then S C 1 ∩ S C 2 = ∅.
Let us prove the first statement by contradiction, so assume that S C ∩ Q(t 0 ) = ∅. We can assume without loss of generality that r(t, x) < 0, because this property is satisfied for the function W(t, x) = u(t, x)e −λt with λ > 0 large enough and the components of these two functions coincide. Consider, for example, that u(t, x) > 0 in S C . Let M = max (t,x)∈S C u(t, x). By hypothesis and the Dirichlet boundary conditions this maximum has to be attained at a point (t , x ) such that t 0 < t ≤ t 1 , 0 < x < 1. Additionally, there has to be an ε > 0 such that if (t, x) ∈ S C and t 0 < t ≤ t 0 + ε, then u(t, x) < M, as otherwise there would be a sequence (t n , x n ) ∈ S C , t n > t 0 , such that t n → t 0 and u(t n , x n ) = M. By the continuity of u this would imply that u(t 0 , x 0 ) = M for some (t 0 , x 0 ) ∈ S C , which is a contradiction. Then we can choose t as the first time when the maximum is attained, so u(t, x) < M for all (t, x) ∈ S C , t 0 < t < t . By the continuity of u there exists a rectangle R = [t − δ, t ] × [x − γ, x + γ] such that R belongs to S C . In order to apply Theorem A1 we put O = R and Q γ,δ = {(t, x) : t ∈ (t − δ, t ), x − x < γ}.
We have that sup (t,x)∈Q νγ,σ 1 u(t, x) = M, for some 0 < ν < 1 and any 0 < σ 1 < δ. Since u satisfies (A1), we conclude from Theorem A1 that u(t, x) = M for all (t, x) ∈ Q ρ,σ , which is a contradiction. For the second statement suppose the existence of two disjoints components C 1 , C 2 of Q(t 1 ) such that S C 1 ∩ S C 2 = ∅, which implies in fact that S C 1 = S C 2 . In this case we can assume that r(t, x) > 0, being this justified by the function W(t, x) = u(t, x)e λt with λ > 0 large enough. Let, for example, u(t, x) > 0 in S C 1 and assume that the interval C 1 has lesser values than the interval C 2 . Additionally, it is clear that between C 1 and C 2 there must exist a point (t 1 , x 0 ) such that u(t 1 , x 0 ) = 0. On the other hand, the set S C 1 ∩ (t 0 , t 1 ) × [0, 1] is path connected. Thus, there exists a simple path ξ such that one end point is in {t 1 } × C 1 and the other one is in {t 1 } × C 2 . Let us consider the set L of all points which are above the curve ξ and such that the function u vanishes at them. This set is non-empty because (t 1 , x 0 ) ∈ L. Since L is compact, the function g : L → [t 0 , t 1 ] given by g(t, x) = t attains it minimum at a certain point (t , x ) ∈ L such that t 0 < t . Then there exists a set R = [t − δ, t ) × [x − γ, x + γ] which belongs to S C 1 . Let O = R and We have that inf (t,x)∈Q νγ,σ 1 u(t, x) = 0, for some 0 < ν < 1 and any 0 < σ 1 < δ. Since u satisfies (A2), we conclude from Theorem A2 that u(t, x) = 0 for all (t, x) ∈ Q ρ,σ , which is a contradiction.