A class of second order BVPs on infinite intervals

In this work, we are concerned with a boundary value problem associated with a generalized Fisher-like equation. This equation involves an eigenvalue and a parameter which may be viewed as a wave speed. According to the behavior of the nonlinear source term, existence results of bounded solutions, positive solutions, classical as well as weak solutions are provided. We mainly use fixed point arguments.


INTRODUCTION
The aim of this paper is to prove existence theorems for the boundary value problem −u + cu + λu = h(x, u), −∞ < x < +∞.lim |x|→+∞ u(x) = 0. (1.1) The parameter c > 0 is a real positive constant while h: R × R → R is a continuous function satisfying lim |x|→+∞ h(x, 0) = 0; the parameter λ > 0 may be seen as an eigenvalue of the problem.In the linear case h(x, u) = f (x)u, the problem arises in the study of a reaction-diffusion system involved in disease propagation throughout a given population [2]; the sublinear case h(x, u) ≤ f (x)u was also studied in [2] where existence and non existence results are given.For other recent developments in solvability to boundary value problems on unbounded domains see [1,7] and references therein.
In this work, we investigate the nonlinear case; the study of problem (1.1) depends on the growth type of the source term h with respect to the second argument.In Section 2, we prove existence of bounded classical solutions in case the nonlinear right-hand term h obeys a generalized polynomial growth condition.Section 3 is devoted to proving existence of positive solutions under integral restrictions on the nonlinear function h.A general existence principle is given in Section 4. In Section 5, we show existence of positive solutions on the half-line under polynomial-like growth condition on the function h.Finally, existence of weak solutions is discussed in Section 6.
Our arguments will be based on fixed point theory.So, let us recall for the sake of completeness, respectively Schauder's and Schauder-Tichonoff's fixed point theorems [11]: Theorem 1.Let E be a Banach space and K ⊂ E a bounded, closed and convex subset of E. Let F : K −→ K be a completely continuous operator.Then F has a fixed point in K.
Theorem 2. Let K be a closed, convex subset of a locally convex, Hausdorff space E. Assume that T : K −→ K is continuous, and T (K) is relatively compact in E. Then T has at least one fixed point in K.
In sequel, C k (I, R) (k ∈ N) will refer to the space of k th continuously differentiable functions defined on an interval I of the real line.C 0 (R, R) stands for the space of continuous functions defined on the real line and vanishing at infinity; throughout this article, we will shorten the notation of this space to E 0 .Endowed with the sup-norm u = sup x∈R |u(x)| , it is a Banach space.Recall that L p (R) is the Banach space of p th power integrable functions on R. Hereafter, R + * refers to the set of positive real numbers and the notation : = means throughout to be defined equal to.

A GENERALIZED POLYNOMIAL GROWTH CONDITION
The main existence result of this section is Theorem 2.1 The Green function being defined by (2.3), assume the following assumptions hold true: x 2 +1 : = q(x) and |g(y)| ≤ |y| + 1: = Ψ(|y|).The real numbers r 1 , r 2 being defined in (2.4), we have 0 ≤ G(x, y) ≤ 1 r 1 −r 2 ; then the constant α introduced in Assumptions 2.1 satisfies the estimate 0 We infer the existence of some positive number M 0 large enough such that π and so Assumptions 2.1 are satisfied.For instance, the following problem has at least one nontrivial solution: Remark 2.1 (a) Assumptions (2.1) encompass the case where the nonlinear function h satisfies the polynomial growth condition (2.2) 0 , then we can only take ρ < 1 in Assumption (2.2).This particular case was studied in [9]; Theorem 2.1 then improves a similar result obtained in [9].
(c) Since we work on the whole real line, Theorem 2.1, as well as the other existence theorems in this paper, provide solutions which are not in general known to be nontrivial.To ensure existence of nontrivial solutions, one must add assumptions on the nonlinear function h such as h(x, 0) ≡ 0 further to lim (d) If we consider instead the autonomous case h(x, u) = g(u) with g(0) = 0, then the trivial solution u ≡ 0 is the unique solution.Let us prove this in two steps: • For any solution u to Problem (1.1), note that lim |x|→+∞ u (x) = 0. We check this when x → +∞.Indeed, let Then by a classical fluctuation lemma [6], there exist two sequences (x n ) n∈N and (y n ) n∈N converging to positive infinity such that = lim with G(u): = u 0 g(s) ds.Then, multiplying the equation in Problem (1.1) by u and making an integration by part, we find that (d) As for the separated-variable case h(x, u) = f (x)g(u), we must impose g(0) = 0 both with f (±∞) = 0 otherwise we could also obtain u ≡ 0 as a solution.
Under Hypothesis (2.1), we will make use of Schauder's fixed point theorem to prove existence of a solution in a closed ball B(0, R) with some radius R > 0.
Proof of Theorem 2.1 It is clear that Problem (1.1) is equivalent to the integral equation: and characteristic roots Define the mapping T : In view of Schauder's fixed point theorem, we look for fixed points for the operator T in the Banach space E 0 .The proof is split into four steps.
• Claim 1: The mapping T is well defined; indeed, for any u ∈ E 0 , we get, by Assumptions (2.1), the following estimates: The convergence of the integral defining T u(x) is then established.In addition for any y ∈ R, G(±∞, y) = 0, and then, taking the limit in T u(x), we get, by l'Hospital Theorem, T u(±∞) = 0. Therefore, the mapping T : E 0 → E 0 is well defined.
Claim 2: The operator T is continuous.Let be a sequence (u n ) n ∈ E 0 converge uniformly to u 0 on all compact subinterval of R. For some fixed a > 0, we will prove the uniform convergence of (T u n ) n to some limit T u 0 on the interval [−a, a].Let ε > 0 and choose some b > a large enough.By the uniform convergence of the sequence (by Cauchy Convergence Criterion and lim |y|→+∞ h(y, u 0 (y)) = 0.) This proves the uniform convergence of the sequence (T u n ) n to the limit T u 0 on the interval [−a, a].
Claim 3: For any M > 0, the set {T u, u ≤ M } is relatively compact in E 0 .By Ascoli-Arzela Theorem, it is sufficient to prove that all the functions of this set are equicontinuous on every subinterval [−a, a] and that there exists a function γ ∈ E 0 such that for any a] ; we have successively the estimates: By continuity of the Green function G, the latter term tends to 0, when x 2 tends x 1 ; whence comes the equicontinuity of the functions {T (u); u ≤ M }.Now, we check analogously the second statement: By l'Hopital Theorem, we have that γ ∈ E 0 .

EJQTDE, 2006 No. 4, p. 5
Claim 4: There exists some R > 0 such that T maps the closed ball B(0, R) into itself.From assumption (2.1), we know that there is some positive number M 0 such that αΨ(M 0 ) so that it is enough to take R = M 0 .The proof of Theorem 2.1 then follows from Schauder's fixed point theorem.

EXISTENCE OF POSITIVE SOLUTIONS
Making use of Schauder-Tichonov's theorem, we prove here existence of a positive solution under an integral condition on the nonlinear term: Theorem 3.1 Problem (1.1) has a positive solution provided the following mean growth assumption on the nonlinear function h is fulfilled The function h is positive and satisfies h(x, u) ≤ H(x, |u|) where H : R × R + → R + is continuous, nondecreasing with respect to the second argument and verifies has at least one positive nontrivial solution.Indeed, the function H(x, y) = y n x 2 +y 2 + 1 x 2 +1 satisfies lim |x|→0 H(x, 0) = 0, H(x, 0) ≡ 0 and is nondecreasing in the second argument y for any integer n > 2.Moreover, ) so that Assumption (3.1) may be satisfied.For instance, if we take n = 3, then there exists To prove Theorem 3.1, we proceed as in Theorem 2.1, and reformulate Problem (1.1) as a fixed point problem for the mapping T defined in 2.5.Here, we appeal to Schauder-Tichonov's fixed point theorem.Let K be the closed convex subset of E 0 defined by: Using Assumption (3.1) and the fact that the mapping H is nondecreasing in the second argument, we find that T maps K into itself.Indeed, taking into account the bound 0 < G(x, y) ≤ 1 r 1 −r 2 , we derive the straightforward estimates: y)H(y, c * ) dy and G(±∞, y) = 0, ∀ y ∈ R, we have that T u(±∞) = 0 and so T (E 0 ) ⊂ E 0 .In addition, the mapping T is continuous as can easily be seen.It remains to check that T (K) is relatively compact.By Ascoli-Arzela Theorem, it is sufficient to prove that all the functions of this set are equicontinuous on every subinterval [−a, a] and that there exists a function γ ∈ E 0 such that for any By continuity of the function G, we deduce from Lebesgue dominated convergence theorem that the last right-hand term tends to 0 when x 2 tends to x 1 .Whence comes the compactness of T (K) by Ascoli-Arzela Lemma and then the claim of Theorem 3.1 follows.Now, we check analogously the second statement: with γ ∈ E 0 for G(±∞, y) = 0, ∀ y ∈ R.

A FURTHER TYPE OF GROWTH
In this section, we prove existence of bounded, solutions to Problem (1.1) under new growth conditions on the nonlinearity h; by the way we show that polynomial-like growth condition may be relaxed.The proof of our existence result relies on the following fixed point theorem by Furi and Pera [3].This theorem was also used in [1] to deal with a BVP on an infinite interval.
Theorem 3. Let E be a Fréchet space, Q a closed convex subset of E, 0 ∈ Q and let T : Q → E be a continuous compact mapping.Assume further that, for any sequence (u j , µ j ) j≥1 from ∂Q × [0, 1] that converges to (u, µ) with u = µT u, 0 ≤ µ < 1, one has µ j T u j ∈ Q for all j large enough.EJQTDE, 2006 No. 4, p. 7 Then, T has a fixed point in Q.
Our aim is now to prove Theorem 4.1 First, assume the following sign assumption is fulfilled: Then Problem (1.1) has a bounded solution provided either one of the following growth assumptions is satisfied: (H2) 1 There exists a function H : R × R + → R + continuous and nondecreasing with respect to the second argument such that: with ψ : R + → R + continuous and nondecreasing, q ∈ E 0 positive, continuous and α: x 2 +1 dx = 2π < ∞, Assumptions (H1) and (H2) 1 are satisfied with M 0 = 1.Since h(x, 0) ≡ 0, Problem (1.1) has a nontrivial solution for such a nonlinear right-hand term.We may notice that the function h can be written as h(x, y) = θ(y)−y x 2 +1 where θ(y) = H(y − 1) − H(−y − 1), the function H being the Heaviside function.• Claim 1. T is continuous.Let (u n ) n∈N be a sequence in Q such that u n → u in Q; we show that T u n → T u in Q, as n goes to infinity.We have that for any x ∈ R, |h(x, u n (x))| ≤ H(x, r 0 ), |h(x, u(x))| ≤ H(x, r 0 ) and that lim n→∞ h(x, u n (x)) = h(x, u(x)).By the Dominated Convergence Lebesgue's Theorem, we deduce that lim In addition, for x 1 , x 2 ∈ [−x m , x m ], the following estimates hold true Furthermore, the convergence is uniform since T u n (x) → T u(x) on [−x m , x m ] as n → +∞, and the claim follows.
• Claim 2. Using Ascoli-Arzela Theorem, we are going to prove that with x 1 < x 2 ; we have: which also tends to 0 as x 2 tends to x 1 , for any u ∈ Q.
• Claim 3. Now, we check the last assumption in Furi-Pera's Theorem.Consider some sequence (u j , µ j ) j≥0 ∈ ∂Q × [0, 1] such that, when j → ∞, µ j → µ and u j → u with u = µT u and µ ∈ [0, 1].We must show that Let us consider the case x ∈ [−x * , x * ].Since µ j → µ and T (Q) is bounded in E, the sequence µ j T u j converges uniformly to µT u on [−x * , x * ]; thus there exists some We have also by Remak 4.1(a) that |µT u(x)| ≤ M 0 .Therefore, for j large enough, it holds that We then conclude the estimate The claim of Theorem 4.1 then follows from Theorem 3.

Setting of the problem
In this section, we consider the problem posed on the positive half-line: (5.2) Equivalently, the unknown v satisfies the integral equation: 2 s h(s, e Notice that K is different from the one in (2.3) and that the unknown u is now solution of the integral equation: The following lemma provides estimates of the Green function K and will play an important role in the sequel; we omit the proof: (5.4)Here m: = min e −kδ , e kγ − e −kγ .

EJQTDE, 2006 No. 4, p. 10
Under suitable assumptions on the nonlinear function h, we shall prove the existence of a positive solution to Problem (5.1).The proof relies on Krasnosels'kii fixed point theorem in cones ( [5], [8]) and Zima's compactness criterion [12]; but first of all, let us recall some

Preliminaries
Definition 5.1 A nonempty subset C of a Banach space X is called a cone if C is convex, closed, and satisfies (i) αx ∈ C for all x ∈ C and any real positive number α, (ii) x, −x ∈ C imply x = 0. Definition 5.2 A set of functions u ∈ Ω ⊂ X are said to be almost equicontinuous on I if they are equicontinuous on each interval [0, T ] , 0 < T < +∞.
Next we state Krasnosels'kii Fixed Point Theorem in cones.

Proof of Theorem 5.1:
We follow the method used in [2,14].Let θ ∈ R be as in Hypothesis (5.6)(a) First step.In the following, we study the properties of this operator: 2 )s a(s) ds • Claim 2: Next, we prove that F is completely continuous: • Claim 3: Let Ω 1 = {u ∈ X, u θ < r}, Ω 2 = {u ∈ X, u θ < R}, the constants 0 < r < R being real positive numbers to be selected later on.Consider some u ∈ C ∩ Ω 2 ; then F u is uniformly bounded.Indeed, as in claim 1, we have that F u θ ≤ 1 2k (M 1 + M 2 R p ), for any u ∈ C ∩ Ω 2 .
• Claim 4: The functions {F u} for u ∈ CK ∩ Ω 2 are almost equicontinuous on I; indeed In the following, we derive lengthy estimates of each of the summands in the right-hand side.We have successively: s (e ks − e −ks )a(s) ds s (e ks − e −ks )b(s) ds.
The right-hand term in the last inequality tends to 0 when x 2 −→ x 1 , for any u ∈ C ∩ Ω2 .In addition, we have the bounds: x 1 e − c 2 s (e ks − e −ks )a(s) ds Again, all of the terms in the right side tend to 0 when x 2 −→ x 1 , for all u ∈ C ∩ Ω2 .
At last, we have: And all of the terms in the right side tend to 0 when x 2 −→ x 1 , for all u ∈ C ∩ Ω2 .
According to Lemma 5.2, we conclude that the operator F is completely continuous on C ∩ Ω 2 .
(b) Second step.Now, we check the first alternative in Theorem 4.
Thanks to Theorem 4, the operator F has a fixed point in C ∩ (Ω 2 \ Ω 1 ) and so Problem (5.1) admits a positive solution u in the cone C.

WEAK SOLUTIONS
In contrast with the previous results, we look here for solutions in the Lebesgue Space L p (R).We first need a compactness criterion in L p (R) due to Fréchet-Kolmogorov: EJQTDE, 2006 No. 4, p. 15

(3. 1 )
Remark 3.1 It is easy to check that in the separated-variable case, Assumption (3.1) leads to Assumption (2.1).Example 3.1 The problem

Remark 4 .
1 (a) The sign condition (H1) implies that any solution u of Problem (1.1) satisfies |u(x)| ≤ M 0 , ∀ x ∈ R. Indeed, on the contrary, assume max x∈R |u(x)| leading to a contradiction.(b) Assumption (H2) 1 is weaker than the one in Theorem 3.1.(c) Assumption (H2) 2 is weaker than the one in Theorem 2.1.(d) Then, when Assumption (H1) is satisfied, Theorem 4.1 improves both of Theorems 2.1 and 3.1.Proof of Theorem 4.1.For the sake of clarity, we only do the proof in case Assumptions (H1) and (H2) are simultaneously satisfied.The other cases can be treated similarly.In the Fréchet space E: = C(R, R), set r 0 : = M 0 +1, consider the closed, convex set Q = {u ∈ E : sup x∈R |u(x)| ≤ r 0 }, and define the mapping T : Q → E as in (2.5).In three steps, we carry over the EJQTDE, 2006 No. 4, p. 8 proof.

c 2 s
v(s)) ds with new Green function K(x, s) = 1 2k e −ks (e kx − e −kx ) x ≤ s e −kx (e ks − e −ks ) x ≥ s.

11 Theorem 5 . 1
has at least one fixed point in C ∩ (Ω 2 − Ω 1 ).Now, let p: I −→]0, +∞[ be a continuous function.Denote by X the Banach space consisting of all weighted functions u continuous on I and satisfying sup x∈I {|u(x)|p(x)} < ∞, equipped with the norm u = sup x∈I {|u(x)|p(x)}.We have Lemma 5.2 ([13]) If the functions u ∈ Ω are almost equicontinuous on I and uniformly bounded in the sense of the normu q = sup x∈I {|u(x)|q(x)}where the function q is positive, continuous on I and satisfieslim x→+∞ p(x) q(x) = 0, then Ω is relatively compact in X.Having disposed of these auxiliary results, we are ready to prove EJQTDE, 2006 No. 4, p. Suppose that: where a, b : I −→ R + are continuous positive functions vanishing at positive infinity and k+ c 2 )s a(s) ds < ∞, M 2 : = +∞ 0 e (pθ−k− c 2 )s b(s) ds < ∞.
) the Green function K being defined in (5.3) and m: = min e −kδ , e kγ − e −kγ .Then Problem (5.1) has at least one positive solution u ∈ C(I; R + ).