- Email: [email protected]
Unbounded connected component of the positive solution set of nonlinear operator equation and its applications
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.1 (1-16)
J. Math. Anal. Appl. ••• (••••) •••–•••
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Unbounded connected component of the positive solution set of nonlinear operator equation and its applications ✩ Xian Xu, Xunxia Zhu ∗ Department of Mathematics, Jiangsu Normal University, Xuzhou, Jiangsu, 221116, PR China
a r t i c l e
i n f o
Article history: Received 7 April 2012 Available online xxxx Submitted by D. O’Regan Keywords: Bifurcation theories Positive solution Connected component
a b s t r a c t In this paper, by using global bifurcation theories we obtain some results for structure of positive solution set of some nonlinear equations with parameters. As a result, we obtain some existence results for positive solutions of the nonlinear operator equation. The main result can be applied to various of differential boundary value problems to obtain the existence results for positive solutions without the assumption that the nonlinearities are positone. © 2014 Elsevier Inc. All rights reserved.
1. Introduction Recently, many problems arising in biological mathematics have been studied extensively. See [11–14,17,18] and the references therein. From an application viewpoint only positive solution is of interest for most problems arising in biological mathematics. As is well known, to show the existence of positive solutions that the nonlinearity has the cone mapping property always be assumed. However, usually the nonlinearities of many problems arising in biological mathematics don’t have this property. As an example, let us consider a population model of the form: ∂N = cΔN + (B − SN )N − H; Ω × (0, ∞) ∂t N (x, 0) = A; Ω N (x, t) = 0; ∂Ω × [0, ∞), where Ω is a bounded domain in R3 , c, B, S are positive constants, H(x) denotes the quantity harvested per unit time, which is independent of time, and A denotes the initial population, N (x, t) denotes the population of a species which is harvested at a constant rate. To find steady-state solutions of the model, we need consider the existence of positive solutions of the problem ✩ This paper is supported by NSFC 10971179, Natural Science Foundation of Jiangsu Education Committee (09KJB110008), Qing Lan Project and Innovation Project of Jiangsu Province postgraduate training project (CXLX12_0979). * Corresponding author.
http://dx.doi.org/10.1016/j.jmaa.2014.05.080 0022-247X/© 2014 Elsevier Inc. All rights reserved.
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.2 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
2
cΔN + (B − SN )N − H = 0; Ω N = 0; ∂Ω. Here, the nonlinearity f (N ) = (B − SN )N − H may be negative as N ≥ 0. Generally speaking, it is difficult to show the existence of positive solutions when the nonlinearities don’t have the cone mapping property. However, there are still many advances on these problems. Especially, people have extensively studied the existence of positive solutions of the problems with nonlinearities which are bounded from below. We call it a semi-positone problem when we seek the existence of positive solutions of a problem which has a nonlinearity that is bounded from below. In addition to biological mathematics, semi-positone problems arise in many different areas of applied mathematics and physics, such as the buckling of mechanical systems, the design of suspension bridges, chemical reactions and management of natural resources. During the last twenty years finding positive solutions to semi-positone problems has been actively pursued and significant progress has taken place; see [1–3,5–8,15,20] and the references therein. For example, the authors of [3,4,6,9] studied some semi-positone problems with parameters and super-linear nonlinearities by the topological degree method. Naturally it would be interesting to consider semi-positone problems in abstract Banach spaces. However, to our best knowledge, there were only a few papers considering semi-positone problems in abstract Banach spaces; see [16,22,23]. The main purpose of this paper is to study an abstract semi-positone problem with parameters in a Banach space. By using bifurcation theory, we show the existence of unbounded connected component of positive solution set of the abstract semi-positone problem. As some applications, by using the main results of the paper we obtain some main results concerning the existence of the positive solutions of some semi-positone problems considered in literature. Therefore, the main results of this paper are generalizations of some existence results for positive solutions of some previous papers (see Remarks 3.1, 3.3–3.5). 2. Main results Let E be a real Banach space which is ordered by a cone P , that is, x ≤ y if and only if y − x ∈ P . We write x < y if x ≤ y and x = y. Let θ denote the zero element of the real Banach space E. The cone P is said to be normal, if there exists N > 0 such that x ≤ N y for every θ ≤ x ≤ y. The Banach space E is said to be a lattice in the partial ordering ≤, if for arbitrary x, y ∈ E, sup{x, y} and inf{x, y} exist. For x ∈ E, let x+ = sup{x, θ}, x− = sup{−x, θ}, x+ and x− are called the positive part and the negative part of x, respectively. For arbitrary λ ∈ R+ , let Qλ = x ∈ E x ≥ βx − g(λ) e , where β > 0, e ∈ P \ {θ}, g(λ) is increasing with respect to λ. Here the requirement that g(λ) is increasing with respect to λ is not essential as we can replace it with a weaker condition that g(λ) is bounded on λ ∈ [0, s] for each s ≥ 0. For convenience we make the following assumptions: (H1 ) A(·, ·) : R+ × P → E is a completely continuous operator and A(λ, ·) : R+ × P → Qλ for each λ ∈ R+ ; (H2 ) For arbitrary λ , λ ∈ R+ \{0} with λ < λ , lim
x∈Qλ ,x→∞
uniformly with λ ∈ [λ , λ ].
A(λ, x) =∞ x
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.3 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
3
Consider the nonlinear operator equation x = A(λ, x),
(λ, x) ∈ R+ × P \ {θ}.
(2.1)
Let L = (λ, x) ∈ R+ × P x = A(λ, x), x = θ . Now we give the main results of this paper. Theorem 2.1. Let P be a normal cone, and E a lattice in the partial ordering ≤ satisfying the following property (H): |xn | → |x|
whenever xn → x,
∀{xn } ⊂ E, x ∈ E.
Assume that (H1 ) and (H2 ) hold. Then L ∩ ([0, 1] × P ) possesses an unbounded connected component C ∗ such that Projλ C ∗ ⊃ (0, λ∗ ] for some λ∗ > 0 and lim
λ→0+ ,(λ,x)∈C ∗
x = +∞,
where Projλ C ∗ denotes the projection of C ∗ onto λ-axis. Corollary 2.1. Assume that all conditions of Theorem 2.1 hold. Then (2.1) has at least one positive solution for each 0 < λ ≤ λ∗ . Now we establish some preliminary lemmas and definitions. From [19, Theorem 18.1] we have the following Lemma 2.1. Lemma 2.1. Let E1 and E2 be two real Banach spaces, D ⊂ E1 be a closed set and let A be a completely over all of E1 as a completely continuous continuous operator from D to E. Then A admits an extension A operator with values in the closed convex hull CoA(D). From [10, Lemma 29.1] we have the following Lemma 2.2. Lemma 2.2. Let X be a compact metric space. Assume that A and B are two disjoint closed subsets of X. Then either there exists a connected component of X meeting both A and B or X = ΩA ∪ΩB , where ΩA , ΩB are disjoint compact subsets of X containing A and B, respectively. Let U be an open and bounded subset of the metric space [a, b] × E. We set U (λ) = {x ∈ E : (λ, x) ∈ U }, whose boundary is denoted by ∂U (λ). Consider a map h(λ, x) = x − k(λ, x), such that k(λ, ·) is completely continuous and θ ∈ / h(∂U ). Such a map h will also be called an admissible homotopy on U . If h is an admissible homotopy, for every λ ∈ [a, b] and every x ∈ ∂U (λ), one has that hλ (x) := h(λ, x) = θ and it makes sense to evaluate deg(hλ , U (λ), θ). Lemma 2.3. If h is an admissible homotopy on U ⊂ [a, b] × E, the deg(hλ , U (λ), θ) is constant for all λ ∈ [a, b].
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.4 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
4
From [21] we have the following Definition 2.1. Definition 2.1. Let A(·, ·) : R+ × P → E be a nonlinear operator and E a lattice in the partial ordering ≤. Let A∗ (λ, x) = A(λ, x+ ) for each λ ∈ R+ and x = x+ − x− ∈ E. Then we call A∗ an extension of A with respect to the lattice structure. Lemma 2.4. Let E be a lattice satisfying the property (H), P a normal cone and A : R+ × P → E a completely continuous operator. Then A∗ : R+ × E → E is a completely continuous operator. Proof. Obviously, A∗ is well defined and bounded. Let (λn , xn ) → (λ0 , x0 )(n → ∞). Then we have + x+ n → x0 ,
− x− n → x0 (n → ∞),
and thus ∗ A∗ (λn , xn ) = A λn , x+ → A λ0 , x+ 0 = A (λ0 , x0 )(n → ∞). That implies that A∗ : E → E is continuous. In addition, it is easy to show that A is a compact operator, so A∗ is a completely continuous operator. The proof is complete. 2 Lemma 2.5. Let h ∈ E \ {θ} such that h ≥ βhe. Assume that (H1 )–(H2 ) hold, and P is a normal cone. Then for arbitrary λ ∈ (0, 1], there exists Rλ > 0 such that for each λ ∈ [λ, 1], R ≥ Rλ and μ ≥ 0, x = A∗ λ , x + μh,
∀x ∈ ∂B θ, R ,
(2.2)
where B(θ, R ) = {x : x < R } and A∗ (λ , x) is an extension of A(λ , x) with respect to the lattice structure. Proof. As the cone P is normal, we may assume without loss of generality that y ≥ x for y ≥ x ≥ θ. From (H2 ), for M2 > 1 with M2 βe > 2, there exists R λ > 0 such that A λ , x ≥ M2 x,
x ∈ Q1 ,
x ≥ Rλ ,
λ ∈ [λ, 1].
(2.3)
Let
g(1) , g(1)e + 1. Rλ = max R λ , β Assume by contradiction that x0 = A∗ (λ0 , x0 ) + μ0 h for some R ≥ Rλ , x0 ∈ ∂B(θ, R ), λ0 ∈ [λ, 1] and μ0 ≥ 0. Let y0 = A∗ (λ0 , x0 ). Then, we have y0 = A∗ (λ0 , x0 ) = A λ0 , x+ 0 ∈ Qλ0 ⊂ Q1 , and so
(2.4)
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.5 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
y0 ≥ βy0 − g(1) e.
5
(2.5)
From (2.4) and (2.5), we have x0 = y0 + μ0 h ≥ βy0 − g(1) e + μ0 βhe ≥ βy0 + μ0 h − g(1) e = βx0 − g(1) e, and so x0 ∈ Q1 . Then we have x0 > θ because βx0 > g(1). By (2.3) we have x0 = A∗ (λ0 , x0 ) + μ0 h ≥ β A∗ (λ0 , x0 ) − g(1) e + μ0 h ≥ β A∗ (λ0 , x0 ) − g(1) e − g(1) e = β A λ0 , x+ 0 = β A(λ0 , x0 ) − g(1) e ≥ βM2 x0 − g(1) e > θ, and so x0 + g(1)e ≥ M2 βx0 e ≥ 2x0 . Thus, g(1)e ≥ x0 ≥ Rλ , which is a contradiction. The proof is complete. 2 Lemma 2.6. Let T : R+ × E → E is a completely continuous operator, and L∗ = (λ, x) ∈ R+ × E : x = T (λ, x), x = θ . Assume that there exist λ∗ ∈ (0, ∞) and a bounded subset U of {λ∗ } × E such that Fix T λ∗ , · ⊂ U ⊂ λ∗ × E, deg I − T λ∗ , · , U, θ = 0, L∗ ∩ R+ × {θ} ∪ {0} × E = ∅.
(2.6) (2.7) (2.8)
Let Fix T (λ∗ , ·) = {Dα |α ∈ Γ }, where Dα is a connected component of Fix T (λ∗ , ·). Denote by Eα− the connected component of L∗ ∩ ([0, λ∗ ] × E) containing Dα , and Eα+ the connected component of L∗ ∩([λ∗ , +∞) × E) containing Dα , respectively. Then there exist α0 , α1 ∈ Γ such that Eα−0 and Eα+1 are unbounded. Proof. Obviously, Γ = ∅. Assume by contradiction that Eα− is bounded for each α ∈ Γ . From (2.8) we may (1) take a bounded open neighborhood Uα in [0, λ∗ ] × E for each Eα− such that Cl [0,λ∗ ]×E Uα(1) ∩ 0, λ∗ × {θ} = ∅, Uα(1) ∩ {0} × E = ∅,
(2.9) (2.10)
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.6 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
6
where Cl [0,λ∗ ]×E Uα denotes the closure of Uα in the metric space [0, λ∗ ]×E. Obviously, Cl [0,λ∗ ]×E Uα ∩L∗ (1) (1) is a compact subset. Let ∂[0,λ∗ ]×E Uα denote the boundary of Uα in the metric space [0, λ∗ ] × E. Assume that (1)
(1)
(1)
∂[0,λ∗ ]×E Uα(1) ∩ L∗ = ∅. From the maximal connectedness of Eα− , there are no connected subset of Cl [0,λ∗ ]×E Uα ∩ L∗ meeting (1) both ∂[0,λ∗ ]×E Uα ∩ L∗ and Eα− . From Lemma 2.2, there exist compact disjoint subsets ΩA and ΩB of (1) Cl [0,λ∗ ]×E Uα ∩ L∗ such that (1)
Cl [0,λ∗ ]×E Uα(1) ∩ L∗ = ΩA ∪ ΩB , Eα− ⊂ ΩA ,
∂[0,λ∗ ]×E Uα(1) ∩ L∗ ⊂ ΩB , (2)
and ΩA ∩ ΩB = ∅. Let d = d(ΩA , ΩB ) > 0 and Uα [0, λ∗ ] × E. Set
∗
U (Dα ) =
be the
d 3 -neighborhood
of ΩA in the metric space
∅; Uα ∩ Uα , if ∂[0,λ∗ ]×E Uα ∩ L∗ = (1) (1) ∗ Uα , if ∂[0,λ∗ ]×E Uα ∩ L = ∅. (1)
(2)
(1)
(2.11)
Then U ∗ (Dα ) is a bounded neighborhood of Eα− in the metric space [0, λ∗ ] × E such that ∂[0,λ∗ ]×E U ∗ (Dα ) ∩ L∗ = ∅.
(2.12)
Obviously, the collection of the subsets
U ∗ (Dα ) ∩ λ∗ × E : α ∈ Γ
is an open cover of L∗ ∩ ({λ∗ } × E), then there exist finite subsets, say U ∗ (Dα1 ), U ∗ (Dα2 ), ..., U ∗ (Dαm ) such that L∗ ∩
m λ∗ × E ⊂ U ∗ (Dαj ) ∩ λ∗ × E . j=1
Let U ∗∗ =
m j=1
U ∗ (Dαj ). Then U ∗∗ is a bounded open subset of [0, λ∗ ] × E, and deg I − T λ∗ , · , U ∗∗ λ∗ , θ = deg I − T λ∗ , · , U, θ = 0.
From (2.9) and (2.12), we have ∂[0,λ∗ ]×E U ∗∗ ∩ S = ∅, where S = (λ, x) ∈ R+ × E : x = T (λ, x) . So by Lemma 2.3 we have deg I − T λ∗ , · , U ∗∗ λ∗ , θ = deg I − T (0, ·), U ∗∗ (0), θ . From (2.10), we have U ∗∗ (0) = ∅, and so
(2.13)
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.7 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
7
deg I − T (0, ·), U ∗∗ (0), θ = 0, which contradicts to (2.7) and (2.13). So, there must exist a α0 ∈ Γ such that Eα−0 is unbounded. Similarly, we can show that there must exist a α1 ∈ Γ such that Eα+1 is unbounded. The proof is complete. 2 Proof of Theorem 2.1. Let
T (λ, x) =
A∗ (λ, x), ([0, 1] × E) ∩ M [2, +∞), θ, ([0, 1] × B(θ, 1)) ∪ ({0} × E),
(2.14)
where B(θ, 1) = {x ∈ E : x ≤ 1} and M [2, +∞) = (λ, x) ∈ R+ × E : 2 ≤ x < ∞ . By Lemma 2.1, T (λ, x) admits an extension over all of [0, 1] × E, denote it also by T (·, ·) for simplicity. Obviously, T (·, ·) is a completely continuous operator. Note that T (λ, x) = A∗ (λ, x) for all λ ∈ [0, 1] and x ∈ E with x ≥ 2. From Lemma 2.5, there exists R1 > 2 large enough such that Fix T (1, ·) ⊂ B(θ, R1 ) and deg I − T (1, ·), B(θ, R1 ), θ = 0. Obviously, deg I − T (1, ·), B(θ, 1), θ = 1. Therefore, deg I − T (1, ·), B(θ, R1 )\B(θ, 1), θ = deg I − T (1, ·), B(θ, R1 ), θ − deg I − T (1, ·), B(θ, 1), θ = 0 − 1 = −1.
(2.15)
So, Fix T (1, ·) = ∅ and Fix T (1, ·) ⊂ U := B(θ, R1 )\B(θ, 1). From (2.14) we see that (2.8) holds. Now from Lemma 2.6 there exists an unbounded connected component C in the metric space L∗ ∩ ([0, 1] × E) containing some connected component of L∗ ∩ ({1} × E). Let r0 = 2 + β −1 g(1) and D1 = {1} × E ∩ M [r0 , +∞), D2 = [0, 1] × E ∩ M (r0 ), X ∗ = [0, 1] × E ∩ M [r0 , +∞), where M (r0 ) = (λ, x) ∈ R+ × E : x = r0 . Obviously, C ∩ (D1 ∪ D2 ) = ∅. For each p ∈ C ∩ (D1 ∪ D2 ), denote by E(p) the connected component of the metric space C ∩X ∗ which passes the point p. Now we shall prove that, there must exists a p0 ∈ C ∩(D1 ∪D2 ) such that E(p0 ) is an unbounded connected component of the metric space C ∩X ∗ . On the contrary, assume that E(p) is bounded for each p ∈ C ∩ (D1 ∪ D2 ). Then, for each p ∈ C ∩ (D1 ∪ D2 ), in the same way as
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.8 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
8
in the construction of U ∗ (Dα ) in (2.11) we can show that there exists a neighborhood U ∗ (p) of E(p) in X ∗ such that ∂X ∗ U ∗ (p) ∩ C = ∅.
(2.16)
Obviously, the set of {U ∗ (p) ∩ (D1 ∪ D2 )|p ∈ C ∩ (D1 ∪ D2 )} is an open cover of the set C ∩ (D1 ∪ D2 ) and C ∩ (D1 ∪ D2 ) is a compact set. Thus, there exist finite subsets of {U ∗ (p) ∩ (D1 ∪ D2 )|p ∈ C ∩ (D1 ∪ D2 )}, say, U ∗ (p1 ) ∩ (D1 ∪ D2 ), U ∗ (p2 ) ∩ (D1 ∪ D2 ), ..., U ∗ (pn ) ∩ (D1 ∪ D2 ), which is also an open cover of C ∩ (D1 ∪ D2 ), that is n
U ∗ (pi ) ∩ (D1 ∪ D2 ) ⊃ C ∩ (D1 ∪ D2 ).
(2.17)
i=1
Let U0 =
n i=1
U ∗ (pi ), then U0 is a bounded open subset of X ∗ . Since ∂X ∗ U0 ⊂
n
∂X ∗ U (pi ),
i=1
then by (2.16) we have ∂X ∗ U0 ∩ C = ∅.
(2.18)
Let W1 =
[0, 1] × E ∩ M [0, r0 ) ∪ U0 ,
and W2 = ([0, 1] × E)\Cl [0,1]×E W1 , where M [0, r0 ) = {x ∈ E : 0 ≤ x < r0 }. It is easy to see that ∂[0,1]×E W1 ⊂ ∂X ∗ U0 ∪
M (r0 ) ∩ [0, 1] × E \ U0 ∩ (D1 ∪ D2 ) .
From (2.17) and (2.18) we see that ∂[0,1]×E W1 ∩ C = ∅. Obviously, W1 ∩ C = ∅ and W1 ∩ W2 = ∅. Note the unboundedness of C, then W2 ∩ C = ∅. Now we have C = (W1 ∩ C) ∪ (W2 ∩ C), which is a contradiction of the connectedness of C. Therefore, there must exist p0 ∈ C ∩ (D1 ∪ D2 ) such that E(p0 ) is an unbounded connected component of C ∩ X ∗ . Since x = A∗ (λ, x) ∈ Qλ for each (λ, x) ∈ E(p0 ), we have x ≥ βx − g(λ) e ≥ βr0 − g(1) e > θ. So, x = A(λ, x) for each (λ, x) ∈ E(p0 ). This implies E(p0 ) ⊂ L is an unbounded subset. Let C ∗ be the connected component of L containing E(p0 ). Obviously, there exists λ∗ > 0 such that Proj Cλ∗ ⊃ (0, λ∗ ]. As C ∗ is unbounded and by Lemma 2.5, we easily see that lim
λ→0+ ,(λ,x)∈C ∗
The proof is complete. 2
x = +∞.
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.9 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
9
3. Applications of main results As some applications of Theorem 2.1, let us first consider the boundary value problem
(rN −1 φ(u (r))) + λrN −1 f (u(r)) = 0, u (0) = u(1) = 0,
0 < r < 1;
(3.1)
where λ ∈ R+ is a parameter, N is a natural number, f is a continuous function on R1 . We make the following assumptions concerning (3.1). (A1 ) φ is an odd, increasing homeomorphism on R1 , and for each ζ > 0, there exists a positive number (ζ) > 0 such that (ζ) → ∞ as ζ → ∞, and (ζ)u ≤ φ−1 (ζu),
u ∈ R+ .
(3.2)
(A2 ) f : R+ → R1 is continuous and lim
u→+∞
f (u) = +∞. φ(u)
Let E = C[0, 1] be the well known real Banach space of all continuous functions on [0, 1] with the maximum norm · , and P = {u ∈ C[0, 1] : u(r) ≥ 0, r ∈ [0, 1]}. Then P is a normal cone of E and E is a lattice in the partial ordering ≤ induced by cone P satisfying the property (H). From (A2 ), one easily find m > 0 such that f (u) ≥ −m for u ∈ R+ . Let
u ≥ 0, u < 0.
f (u), f (0),
f ∗ (u) = Then, f ∗ (u) ≥ −m for u ∈ R1 . Define operator A : R+ × E → E by
1 A(λ, u)(r) =
−1
φ
τ
sN −1
r
s
λ
N −1 ∗
f (u)dτ ds.
0
Let L2 = (λ, u) ∈ R+ × E : u is a positive solution of (3.1), u = θ . Theorem 3.1. Assume that (A1 )–(A2 ) hold and φ−1 is concave on R1 . Then L2 possesses an unbounded connected component C ∗ such that Projλ C ∗ ⊃ (0, λ∗ ] for some λ∗ > 0, and lim
λ→0+ ,(λ,x)∈C ∗
x = +∞.
Proof. For each (λ, u) ∈ R+ × P , let 1 y(r) = A(λ, u)(r) =
−1
φ r
λ
s τ
sN −1 0
N −1 ∗
f (u)dτ ds.
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.10 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
10
Clearly,
1
−1
y(r) =
φ
s
λ
τ
sN −1
r
N −1
s
λm f(u)dτ − N −1 s
τ
0
N −1
dτ ds,
0
where f(u) = f ∗ (u) + m ≥ 0. As φ−1 is concave on R+ , there exists β > 0 such that (here we may take β = 1, see Lemma 2.2 in [6]) φ−1 (x − y) ≥ βφ−1 (x) − 2φ−1 (y),
x, y ≥ 0.
(3.3)
And so, 1 y(r) ≥ β
s
λ
φ−1
sN −1
r
τ N −1 f(u)dτ ds − 2
1
φ−1
λm sN −1
r
0
s
τ N −1 dτ ds.
(3.4)
0
Clearly, for 0 ≤ r ≤ 1, 1
−1
φ
s
λm sN −1
τ
r
N −1
dτ ds ≤ φ
−1
λm (1 − r). N
0
And so, −1
y(r) ≥ β u(r) − 2φ
λm −1 λm (1 − r) ≥ −2φ (1 − r), N N
(3.5)
1 s where u (r) = r φ−1 ( sNλ−1 0 τ N −1 f(u)dτ )ds ≥ 0 for each r ∈ [0, 1]. Clearly, u (r) ≥ y(r) for each r ∈ [0, 1], u = u (0) and
(rN −1 φ( u )) + λrN −1 f(u(r)) = 0, u (1) = 0, u (0) = 0.
r ∈ (0, 1);
We will prove that u (r) ≥ ω(r) for each r ∈ [0, 1], where ω satisfies
−(rN −1 φ(ω )) = 0, r ∈ (0, 1); ω(1) = 0, ω(0) = u.
Suppose that u (r0 ) < ω(r0 ) for some r0 ∈ (0, 1), then u − ω has a negative absolute minimum at τ ∈ (0, 1). Since u (1) − ω(1) = u (0) − ω(0) = 0, there exist τ0 , τ1 ∈ [0, 1] such that u (τ0 ) − ω(τ0 ) = u (τ1 ) − ω(τ1 ) = 0, and u (r) − ω(r) < 0,
r ∈ (τ0 , τ1 ).
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.11 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
11
Then
N −1 rN −1 φ u (r) − r φ ω (r) = −λrN −1 f(u) ≤ 0,
r ∈ (τ0 , τ1 ).
Let ν(t) = u (t) − ω(t) < 0, r ∈ (τ0 , τ1 ), then τ1
N −1 rN −1 φ u ν(t)dt ≥ 0. (r) − r φ ω (r)
τ0
On the other hand, using the inequality
φ(b) − φ(a) (b − a) ≥ 0,
a, b ∈ R,
and the fact that there exists τ ∗ ∈ [τ0 , τ1 ] such that u (τ ∗ ) = ω(τ ∗ ), we have τ1
N −1 (r) − r φ ω (r) rN −1 φ u ν(t)dt
τ0
=−
τ1
N −1 (r) − r φ ω (r) rN −1 φ u u − ω dt < 0,
τ0
which is a contradiction. So, u (r) ≥ ω(r) for r ∈ [0, 1]. Clearly, ω(r) = u(1 − r). As u (r) ≥ y(r) for each r ∈ [0, 1], we have from (3.5) that u (r) + 2φ
−1
λm −1 λm (1 − r) ≥ y(r) + 2φ (1 − r) ≥ 0 N N
for r ∈ [0, 1]. So, −1 λm −1 λm u (r) + 2φ ≥ y(r) − 2φ . N N It follows that −1 λm u (r) ≥ y(r) − 4φ . N From (3.4) and (3.6) we have
λm (1 − r) y(r) ≥ β u(r) − 2φ N λm ≥ β u − 2φ−1 (1 − r) N −1 λm ≥ β y(r) − (4β + 2)φ (1 − r). N −1
So, A(λ, u)(r) ≥ β A(λ, u) − g(λ) e(r),
(3.6)
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.12 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
12
+ where g(λ) := (4β + 2)φ−1 ( λm N ). Let Qλ = {u ∈ E : u ≥ (βu − g(λ))e(r)} for λ ∈ R and e(r) = (1 − r) for each r ∈ [0, 1]. This implies that (H1 ) holds. From (A2 ), for arbitrary m0 > 0, there exists m1 > 0 such that f ∗ (u) ≥ m0 φ(u) for u ≥ m1 . Let λ , λ > 0 with λ < λ . Then, for each λ ∈ [λ , λ ] and u ∈ Qλ with u ≥ max{β −1 (2m1 + g(λ )), 2β −1 g(λ )}, we have
1 u(r) ≥ βu − g(λ) (1 − r) ≥ βu − g λ ≥ m1 2 for each r ∈ [0, 12 ]. From (3.2) and using the inequality (3.3) we have 1 s 1 λ −1 N −1 ∗ = φ A(λ, u) τ f (u)dτ ds 2 sN −1 0
1 2
1 ≥β
−1
φ
s
λ
≥β
−1
φ
≥β
1
N −1
−1
1 2
−1 mλ f (u) + m dτ ds − 2φ N ∗
m0 λ sN −1
2 τ
N −1
φ(u)dτ ds − 2φ−1
mλ N
0
−1
φ
≥β
1
1
m0 λ φ( 12 (βu − g(λ ))) sN −1
2 τ
N −1
dτ ds − 2φ
−1
mλ N
0
1 2
1
0
0
1 2
≥β
τ
sN −1
φ 1
1 s mλ −1 N −1 τ dτ ds f (u) + m dτ ds − 2 φ sN −1 ∗
1 2
2
λ
1 2
1
0
1 2
1
τ
sN −1
N −1
φ−1
m0 λ φ( 12 (βu − g(λ ))) −1 mλ ds − 2φ 2N · N N
β mλ m0 λ βu − g λ N − 2φ−1 2 2 ·N N 2 β mλ m0 λ u N − 2φ−1 . ≥ 2 2 ·N N
≥
Thus A(λ, u)( 12 ) β2 2 −1 mλ A(λ, u) m0 λ ≥ ≥ N − φ . u u 2 2 ·N u N Note that 2 −1 mλ φ →0 u N as u → ∞. So, A(λ, u) = +∞, u u∈Qλ ,u→∞ lim
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.13 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
13
uniformly with λ ∈ [λ , λ ]. This implies that (H2 ) holds. Now by applying Theorem 2.1 one can easily prove that Theorem 3.1 holds. The proof is complete. 2 Corollary 3.1. Let (A1 )–(A2 ) hold. Then there exists a positive number λ > 0 such that for 0 < λ ≤ λ, (3.1) has a positive solution uλ such that lim uλ = ∞.
λ→0+
Remark 3.1. D.D. Hai and K. Schmitt in [15] have obtained a very similar result to Corollary 3.1. See Theorem 2.1 in [15]. To show Theorem 2.1 in [15], besides condition (A2 ), the authors employed the following two assumptions: (A∗1 ) φ is an odd, increasing homeomorphism on R1 , and for each ζ > 0, there exists a positive number Mζ > 0 such that φ(ζu) ≤ Mζ φ(u),
u ∈ R1 .
(3.7)
(A∗3 ) There exists θ ∈ (0, 1) such that
F (θx) Φ(x) > max N lim sup − 1, 0 , N lim inf x→∞ x→∞ xf (x) xφ(x) x x where F (x) = 0 f (s)ds and Φ(x) = 0 φ(s)ds. Here we have removed the assumption (A∗3 ). Compare (A1 ) with (A∗1 ), here we replace (3.7) with (3.2). We should point out, a condition similar to (3.2) has been employed in [16] (see assumption (A.2) in [16]). As for the case φ(x) = xp−1 for some p ≥ 1, the assumption (A∗1 ) is needn’t. In fact we have the following result. Corollary 3.2. Let φ(x) = xp−1 for some p ≥ 1. Assume that (A2 ) holds. Then there exists a positive number λ > 0 such that for 0 < λ ≤ λ, (3.1) has a positive solution uλ such that lim uλ = ∞.
λ→0+
Next we consider Sturm–Lioville boundary value problem
(p(t)u ) + λf (t, u(t)) = 0,
r < t < R;
au(r) − bu (r) = cu(R) + du (R) = 0
(3.8)
where λ ∈ R+ ,f (t, u) is a continuous function on [r, R] × R1 , and satisfies (A3 ) There exists M0 > 0 such that f (t, u) > −M0 , for all (t, u) ∈ [r, R] × R+ ; (A4 ) p(t) ∈ C 1 [r, R], p(t) > 0 for t ∈ [r, R]; (A5 ) There exists [α, β] ⊂ (r, R) such that lim
u→+∞
f (t, u) = +∞ u
uniformly with t ∈ [α, β].
Let E = C[r, R] be the well known real Banach space of all continuous functions on [r, R] with the maximum norm · , and P = {u ∈ C[0, 1] : u(r) ≥ 0, r ∈ [r, R]}. Then, P is a normal cone of E.
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.14 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
14
Also E is a lattice in the partial ordering ≤ induced by the cone P which satisfies the property (H). Let Q = {u ∈ E : u ≥ ue}, where e(t) = min
b+a b+
t
r R a r
R
p−1 c + d t p−1 . , R p−1 c + d r p−1
Let ⎧ ⎨ α−1 (b + a s p−1 )(d + c R p−1 ), r t G(t, s) = ⎩ α−1 (b + a t p−1 )(d + c R p−1 ), r s
s ≤ t, s > t,
R R R where α = ad + bc + ac r p−1 and r p−1 = r p−1 (s)ds. Let us define the operators K : E → E, F : P → E and A : R+ × P → E by R (Ku)(t) =
G(t, s)u(s)ds,
x ∈ E, t ∈ [r, R],
r
(F u)(t) = f t, u(t) ,
u ∈ P, t ∈ [r, R],
and A(λ, u) = λKF u. Let ω0 = M0 for all t ∈ [r, R] and R ω1 (t) = M0
G(t, s)ds,
t ∈ [r, R].
r
Remark 3.2. It is easy to know that if u > θ, and u = A(λ, u), then u is a positive solution of (3.8). From the proof in [2] we have the following Lemmas 3.1 and 3.2. Lemma 3.1. K : P → Q is a linear completely operator. Lemma 3.2. There exists σ > 0 such that ω1 ≤ σe. Let L3 = (λ, u) ∈ R+ × P : u is a positive solution of (3.8), u = θ . Theorem 3.2. Assume that (A3 )–(A5 ) hold. Then L3 possesses an unbounded connected component C ∗ such that Projλ C ∗ ⊃ (0, λ∗ ] for some λ∗ > 0, and lim
λ→0+ ,(λ,x)∈C ∗
x = +∞.
Proof. For each (λ, u) ∈ R+ × P , by Lemmas 3.1 and 3.2 we have A(λ, u)(t) = λKF u(t) R = r
G(t, s) λf s, u(s) + λM0 ds − λM0
R G(t, s)ds r
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.15 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
R =
15
G(t, s) λf s, u(s) + λM0 ds − λω1 (t)
r
R ≥ G(t, s) λf s, u(s) + λM0 ds e − λω1 (t) r
≥ A(λ, u)e − λω1 e − λσe = A(λ, u)e − g(λ)e, where g(λ) = λω1 + λσ. Thus (H1 ) holds. By condition (A5 ) one can easily prove (H2 ) using the same method proving Theorem 3.1. A very similar detailed proof can be found in Theorem 3.3 of [23]. Now by applying Theorem 2.1 one can easily prove that Theorem 3.2 holds. The proof is complete. 2 Corollary 3.3. Assume that all conditions of Theorem 3.2 hold. Then there exists λ∗ > 0 such that (3.8) has at least one positive solution for 0 < λ ≤ λ∗ . Remark 3.3. Corollary 3.3 is the main result of [2]. V. Anuradha, D.D. Hai and R. Shivaji using the degree method obtained this result. Here we obtain this result by using a different method—bifurcation theories. Moreover, Theorem 3.2 gave some informations concerning the structure of positive solution set of (3.8). Remark 3.4. D.D. Hai, K. Schmitt and R. Shivaji in [16] considered the following boundary value problem:
(p(t)φ(u )) + λp(t)f (t, u) = 0, u(a) = u(b) = 0,
a < t < b;
where f is φ-superlinear or φ-sub-linear at ∞ and f (t, 0) may be negative and p is a continuous function. By using degree method, some existence results for positive solutions be obtained. Obviously, the main results in [16] concerning the existence results for positive solutions in the case where f is φ-superlinear can also be derived from our Theorem 2.1. Remark 3.5. Very recently, by using bifurcation method we studied the structure of positive solution set of some semi-positone problems in a Banach space. We should point out that, our main results here also are an improvement of the main results in [23]. The main results of [23] is applicable only when A(λ, x) has the form of A(λ, x) = λKF x, where K is a linear completely continuous operator and F is a nonlinear continuous operator. Here we removed this requirement. Therefore, the main results would have more applications than that of the results in [23]. References [1] A. Ambrosetti, D. Arcoya, B. Buffoni, Positive solutions for some semipositone problems via bifurcation theory, Differential Integral Equations 7 (3) (1994) 655–663. [2] V. Anuradha, D.D. Hai, R. Shivaji, Existence results for superlinear semipositone BVP’S, Proc. Amer. Math. Soc. 124 (1996) 757–763. [3] A. Castro, R. Shivaji, Nonnegative solutions for a class of nonpositone problems, Proc. Roy. Soc. Edinburgh Sect. A 108 (1988) 291–302. [4] A. Castro, R. Shivaji, Semipositone problems, in: J.A. Goldstein, G. Goldstein (Eds.), Semigroups of Linear and Nonlinear Operators and Applications, Kluwer Academic Publishers, New York, 1993, pp. 109–119 (invited review paper). [5] A. Castro, R. Shivaji, Positive solutions for a concave semipositone Dirichlet problem, Nonlinear Anal. 31 (1/2) (1998) 91–98. [6] A. Castro, S. Gadam, R. Shivaji, Branches of radial solutions for semipositone problems, J. Differential Equations 120 (1) (1995) 30–45.
JID:YJMAA 16
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.16 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
[7] A. Castro, M. Hassanpour, R. Shivaji, Uniqueness of nonnegative solutions for a semipositone problem with concave nonlinearity, Comm. Partial Differential Equations 20 (11–12) (1995) 1927–1936. [8] A. Castro, S. Gadam, R. Shivaji, Positive solution curves of semipositone problems with concave nonlinearities, Proc. Roy. Soc. Edinburgh Sect. A 127 (1997) 921–934. [9] M. Chhetri, P. Girg, Existence and nonexistence of positive solutions for a class of superlinear semipositone systems, Nonlinear Anal. 71 (2009) 4984–4996. [10] K. Deimling, Nonlinear Functional Analysis, Springer-Verlag, Berlin, Heidelberg, New York, 1985. [11] Y. Du, Effects of a degeneracy in the competition model. I. Classical and generalized steady-state solutions, J. Differential Equations 181 (1) (2002) 92–132. [12] Y. Du, Realization of prescribed patterns in the competition model, J. Differential Equations 193 (1) (2003) 147–179. [13] Y. Du, X. Liang, A diffusive competition model with a protection zone, J. Differential Equations 244 (2008) 61–86. [14] Y. Du, J. Shi, A diffusive predator–prey model with a protection zone, J. Differential Equations 229 (2006) 63–91. [15] D.D. Hai, K. Schmitt, On radial solutions of quasilinear boundary value problems, Progr. Nonlinear Differential Equations Appl. 35 (1999) 349–361. [16] D.D. Hai, K. Schmitt, R. Shivaji, Positive solutions of quasilinear boundary value problems, J. Math. Anal. Appl. 217 (1998) 672–686. [17] V. Hutson, Y. Lou, K. Mischaikow, P. Polacik, Competing species near a degenerate limit, SIAM J. Math. Anal. 35 (2003) 453–491. [18] V. Hutson, Y. Lou, K. Mischaikow, Convergence in competition models with small diffusion coefficients, J. Differential Equations 211 (1) (2005) 135–161. [19] M.A. Krasnosel’skii, P.P. Zabreiko, Geometrical Methods of Nonlinear Analysis, Springer-Verlag, Berlin, Heidelberg, New York, Tokyo, 1984. [20] Ruyun Ma, Yulian An, Global structure of positive solutions for superlinear second order m-point boundary value problems, Topol. Methods Nonlinear Anal. 34 (2009) 279–290. [21] Jingxian Sun, Xian Xu, Positive fixed points of semi-positone nonlinear operator and its applications, Acta Math. Sin. 55 (2012) 55–64 (in Chinese). [22] Xian Xu, D. O’Regan, Existence of positive solutions for operator equations and applications to semipositone problems, Positivity 10 (2) (2006) 315–328. [23] Xian Xu, Jingxian Sun, Unbounded connected component of the positive solutions set of some semi-position problems, Topol. Methods Nonlinear Anal. 37 (2011) 283–302.
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.1 (1-16)
J. Math. Anal. Appl. ••• (••••) •••–•••
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Unbounded connected component of the positive solution set of nonlinear operator equation and its applications ✩ Xian Xu, Xunxia Zhu ∗ Department of Mathematics, Jiangsu Normal University, Xuzhou, Jiangsu, 221116, PR China
a r t i c l e
i n f o
Article history: Received 7 April 2012 Available online xxxx Submitted by D. O’Regan Keywords: Bifurcation theories Positive solution Connected component
a b s t r a c t In this paper, by using global bifurcation theories we obtain some results for structure of positive solution set of some nonlinear equations with parameters. As a result, we obtain some existence results for positive solutions of the nonlinear operator equation. The main result can be applied to various of differential boundary value problems to obtain the existence results for positive solutions without the assumption that the nonlinearities are positone. © 2014 Elsevier Inc. All rights reserved.
1. Introduction Recently, many problems arising in biological mathematics have been studied extensively. See [11–14,17,18] and the references therein. From an application viewpoint only positive solution is of interest for most problems arising in biological mathematics. As is well known, to show the existence of positive solutions that the nonlinearity has the cone mapping property always be assumed. However, usually the nonlinearities of many problems arising in biological mathematics don’t have this property. As an example, let us consider a population model of the form: ∂N = cΔN + (B − SN )N − H; Ω × (0, ∞) ∂t N (x, 0) = A; Ω N (x, t) = 0; ∂Ω × [0, ∞), where Ω is a bounded domain in R3 , c, B, S are positive constants, H(x) denotes the quantity harvested per unit time, which is independent of time, and A denotes the initial population, N (x, t) denotes the population of a species which is harvested at a constant rate. To find steady-state solutions of the model, we need consider the existence of positive solutions of the problem ✩ This paper is supported by NSFC 10971179, Natural Science Foundation of Jiangsu Education Committee (09KJB110008), Qing Lan Project and Innovation Project of Jiangsu Province postgraduate training project (CXLX12_0979). * Corresponding author.
http://dx.doi.org/10.1016/j.jmaa.2014.05.080 0022-247X/© 2014 Elsevier Inc. All rights reserved.
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.2 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
2
cΔN + (B − SN )N − H = 0; Ω N = 0; ∂Ω. Here, the nonlinearity f (N ) = (B − SN )N − H may be negative as N ≥ 0. Generally speaking, it is difficult to show the existence of positive solutions when the nonlinearities don’t have the cone mapping property. However, there are still many advances on these problems. Especially, people have extensively studied the existence of positive solutions of the problems with nonlinearities which are bounded from below. We call it a semi-positone problem when we seek the existence of positive solutions of a problem which has a nonlinearity that is bounded from below. In addition to biological mathematics, semi-positone problems arise in many different areas of applied mathematics and physics, such as the buckling of mechanical systems, the design of suspension bridges, chemical reactions and management of natural resources. During the last twenty years finding positive solutions to semi-positone problems has been actively pursued and significant progress has taken place; see [1–3,5–8,15,20] and the references therein. For example, the authors of [3,4,6,9] studied some semi-positone problems with parameters and super-linear nonlinearities by the topological degree method. Naturally it would be interesting to consider semi-positone problems in abstract Banach spaces. However, to our best knowledge, there were only a few papers considering semi-positone problems in abstract Banach spaces; see [16,22,23]. The main purpose of this paper is to study an abstract semi-positone problem with parameters in a Banach space. By using bifurcation theory, we show the existence of unbounded connected component of positive solution set of the abstract semi-positone problem. As some applications, by using the main results of the paper we obtain some main results concerning the existence of the positive solutions of some semi-positone problems considered in literature. Therefore, the main results of this paper are generalizations of some existence results for positive solutions of some previous papers (see Remarks 3.1, 3.3–3.5). 2. Main results Let E be a real Banach space which is ordered by a cone P , that is, x ≤ y if and only if y − x ∈ P . We write x < y if x ≤ y and x = y. Let θ denote the zero element of the real Banach space E. The cone P is said to be normal, if there exists N > 0 such that x ≤ N y for every θ ≤ x ≤ y. The Banach space E is said to be a lattice in the partial ordering ≤, if for arbitrary x, y ∈ E, sup{x, y} and inf{x, y} exist. For x ∈ E, let x+ = sup{x, θ}, x− = sup{−x, θ}, x+ and x− are called the positive part and the negative part of x, respectively. For arbitrary λ ∈ R+ , let Qλ = x ∈ E x ≥ βx − g(λ) e , where β > 0, e ∈ P \ {θ}, g(λ) is increasing with respect to λ. Here the requirement that g(λ) is increasing with respect to λ is not essential as we can replace it with a weaker condition that g(λ) is bounded on λ ∈ [0, s] for each s ≥ 0. For convenience we make the following assumptions: (H1 ) A(·, ·) : R+ × P → E is a completely continuous operator and A(λ, ·) : R+ × P → Qλ for each λ ∈ R+ ; (H2 ) For arbitrary λ , λ ∈ R+ \{0} with λ < λ , lim
x∈Qλ ,x→∞
uniformly with λ ∈ [λ , λ ].
A(λ, x) =∞ x
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.3 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
3
Consider the nonlinear operator equation x = A(λ, x),
(λ, x) ∈ R+ × P \ {θ}.
(2.1)
Let L = (λ, x) ∈ R+ × P x = A(λ, x), x = θ . Now we give the main results of this paper. Theorem 2.1. Let P be a normal cone, and E a lattice in the partial ordering ≤ satisfying the following property (H): |xn | → |x|
whenever xn → x,
∀{xn } ⊂ E, x ∈ E.
Assume that (H1 ) and (H2 ) hold. Then L ∩ ([0, 1] × P ) possesses an unbounded connected component C ∗ such that Projλ C ∗ ⊃ (0, λ∗ ] for some λ∗ > 0 and lim
λ→0+ ,(λ,x)∈C ∗
x = +∞,
where Projλ C ∗ denotes the projection of C ∗ onto λ-axis. Corollary 2.1. Assume that all conditions of Theorem 2.1 hold. Then (2.1) has at least one positive solution for each 0 < λ ≤ λ∗ . Now we establish some preliminary lemmas and definitions. From [19, Theorem 18.1] we have the following Lemma 2.1. Lemma 2.1. Let E1 and E2 be two real Banach spaces, D ⊂ E1 be a closed set and let A be a completely over all of E1 as a completely continuous continuous operator from D to E. Then A admits an extension A operator with values in the closed convex hull CoA(D). From [10, Lemma 29.1] we have the following Lemma 2.2. Lemma 2.2. Let X be a compact metric space. Assume that A and B are two disjoint closed subsets of X. Then either there exists a connected component of X meeting both A and B or X = ΩA ∪ΩB , where ΩA , ΩB are disjoint compact subsets of X containing A and B, respectively. Let U be an open and bounded subset of the metric space [a, b] × E. We set U (λ) = {x ∈ E : (λ, x) ∈ U }, whose boundary is denoted by ∂U (λ). Consider a map h(λ, x) = x − k(λ, x), such that k(λ, ·) is completely continuous and θ ∈ / h(∂U ). Such a map h will also be called an admissible homotopy on U . If h is an admissible homotopy, for every λ ∈ [a, b] and every x ∈ ∂U (λ), one has that hλ (x) := h(λ, x) = θ and it makes sense to evaluate deg(hλ , U (λ), θ). Lemma 2.3. If h is an admissible homotopy on U ⊂ [a, b] × E, the deg(hλ , U (λ), θ) is constant for all λ ∈ [a, b].
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.4 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
4
From [21] we have the following Definition 2.1. Definition 2.1. Let A(·, ·) : R+ × P → E be a nonlinear operator and E a lattice in the partial ordering ≤. Let A∗ (λ, x) = A(λ, x+ ) for each λ ∈ R+ and x = x+ − x− ∈ E. Then we call A∗ an extension of A with respect to the lattice structure. Lemma 2.4. Let E be a lattice satisfying the property (H), P a normal cone and A : R+ × P → E a completely continuous operator. Then A∗ : R+ × E → E is a completely continuous operator. Proof. Obviously, A∗ is well defined and bounded. Let (λn , xn ) → (λ0 , x0 )(n → ∞). Then we have + x+ n → x0 ,
− x− n → x0 (n → ∞),
and thus ∗ A∗ (λn , xn ) = A λn , x+ → A λ0 , x+ 0 = A (λ0 , x0 )(n → ∞). That implies that A∗ : E → E is continuous. In addition, it is easy to show that A is a compact operator, so A∗ is a completely continuous operator. The proof is complete. 2 Lemma 2.5. Let h ∈ E \ {θ} such that h ≥ βhe. Assume that (H1 )–(H2 ) hold, and P is a normal cone. Then for arbitrary λ ∈ (0, 1], there exists Rλ > 0 such that for each λ ∈ [λ, 1], R ≥ Rλ and μ ≥ 0, x = A∗ λ , x + μh,
∀x ∈ ∂B θ, R ,
(2.2)
where B(θ, R ) = {x : x < R } and A∗ (λ , x) is an extension of A(λ , x) with respect to the lattice structure. Proof. As the cone P is normal, we may assume without loss of generality that y ≥ x for y ≥ x ≥ θ. From (H2 ), for M2 > 1 with M2 βe > 2, there exists R λ > 0 such that A λ , x ≥ M2 x,
x ∈ Q1 ,
x ≥ Rλ ,
λ ∈ [λ, 1].
(2.3)
Let
g(1) , g(1)e + 1. Rλ = max R λ , β Assume by contradiction that x0 = A∗ (λ0 , x0 ) + μ0 h for some R ≥ Rλ , x0 ∈ ∂B(θ, R ), λ0 ∈ [λ, 1] and μ0 ≥ 0. Let y0 = A∗ (λ0 , x0 ). Then, we have y0 = A∗ (λ0 , x0 ) = A λ0 , x+ 0 ∈ Qλ0 ⊂ Q1 , and so
(2.4)
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.5 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
y0 ≥ βy0 − g(1) e.
5
(2.5)
From (2.4) and (2.5), we have x0 = y0 + μ0 h ≥ βy0 − g(1) e + μ0 βhe ≥ βy0 + μ0 h − g(1) e = βx0 − g(1) e, and so x0 ∈ Q1 . Then we have x0 > θ because βx0 > g(1). By (2.3) we have x0 = A∗ (λ0 , x0 ) + μ0 h ≥ β A∗ (λ0 , x0 ) − g(1) e + μ0 h ≥ β A∗ (λ0 , x0 ) − g(1) e − g(1) e = β A λ0 , x+ 0 = β A(λ0 , x0 ) − g(1) e ≥ βM2 x0 − g(1) e > θ, and so x0 + g(1)e ≥ M2 βx0 e ≥ 2x0 . Thus, g(1)e ≥ x0 ≥ Rλ , which is a contradiction. The proof is complete. 2 Lemma 2.6. Let T : R+ × E → E is a completely continuous operator, and L∗ = (λ, x) ∈ R+ × E : x = T (λ, x), x = θ . Assume that there exist λ∗ ∈ (0, ∞) and a bounded subset U of {λ∗ } × E such that Fix T λ∗ , · ⊂ U ⊂ λ∗ × E, deg I − T λ∗ , · , U, θ = 0, L∗ ∩ R+ × {θ} ∪ {0} × E = ∅.
(2.6) (2.7) (2.8)
Let Fix T (λ∗ , ·) = {Dα |α ∈ Γ }, where Dα is a connected component of Fix T (λ∗ , ·). Denote by Eα− the connected component of L∗ ∩ ([0, λ∗ ] × E) containing Dα , and Eα+ the connected component of L∗ ∩([λ∗ , +∞) × E) containing Dα , respectively. Then there exist α0 , α1 ∈ Γ such that Eα−0 and Eα+1 are unbounded. Proof. Obviously, Γ = ∅. Assume by contradiction that Eα− is bounded for each α ∈ Γ . From (2.8) we may (1) take a bounded open neighborhood Uα in [0, λ∗ ] × E for each Eα− such that Cl [0,λ∗ ]×E Uα(1) ∩ 0, λ∗ × {θ} = ∅, Uα(1) ∩ {0} × E = ∅,
(2.9) (2.10)
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.6 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
6
where Cl [0,λ∗ ]×E Uα denotes the closure of Uα in the metric space [0, λ∗ ]×E. Obviously, Cl [0,λ∗ ]×E Uα ∩L∗ (1) (1) is a compact subset. Let ∂[0,λ∗ ]×E Uα denote the boundary of Uα in the metric space [0, λ∗ ] × E. Assume that (1)
(1)
(1)
∂[0,λ∗ ]×E Uα(1) ∩ L∗ = ∅. From the maximal connectedness of Eα− , there are no connected subset of Cl [0,λ∗ ]×E Uα ∩ L∗ meeting (1) both ∂[0,λ∗ ]×E Uα ∩ L∗ and Eα− . From Lemma 2.2, there exist compact disjoint subsets ΩA and ΩB of (1) Cl [0,λ∗ ]×E Uα ∩ L∗ such that (1)
Cl [0,λ∗ ]×E Uα(1) ∩ L∗ = ΩA ∪ ΩB , Eα− ⊂ ΩA ,
∂[0,λ∗ ]×E Uα(1) ∩ L∗ ⊂ ΩB , (2)
and ΩA ∩ ΩB = ∅. Let d = d(ΩA , ΩB ) > 0 and Uα [0, λ∗ ] × E. Set
∗
U (Dα ) =
be the
d 3 -neighborhood
of ΩA in the metric space
∅; Uα ∩ Uα , if ∂[0,λ∗ ]×E Uα ∩ L∗ = (1) (1) ∗ Uα , if ∂[0,λ∗ ]×E Uα ∩ L = ∅. (1)
(2)
(1)
(2.11)
Then U ∗ (Dα ) is a bounded neighborhood of Eα− in the metric space [0, λ∗ ] × E such that ∂[0,λ∗ ]×E U ∗ (Dα ) ∩ L∗ = ∅.
(2.12)
Obviously, the collection of the subsets
U ∗ (Dα ) ∩ λ∗ × E : α ∈ Γ
is an open cover of L∗ ∩ ({λ∗ } × E), then there exist finite subsets, say U ∗ (Dα1 ), U ∗ (Dα2 ), ..., U ∗ (Dαm ) such that L∗ ∩
m λ∗ × E ⊂ U ∗ (Dαj ) ∩ λ∗ × E . j=1
Let U ∗∗ =
m j=1
U ∗ (Dαj ). Then U ∗∗ is a bounded open subset of [0, λ∗ ] × E, and deg I − T λ∗ , · , U ∗∗ λ∗ , θ = deg I − T λ∗ , · , U, θ = 0.
From (2.9) and (2.12), we have ∂[0,λ∗ ]×E U ∗∗ ∩ S = ∅, where S = (λ, x) ∈ R+ × E : x = T (λ, x) . So by Lemma 2.3 we have deg I − T λ∗ , · , U ∗∗ λ∗ , θ = deg I − T (0, ·), U ∗∗ (0), θ . From (2.10), we have U ∗∗ (0) = ∅, and so
(2.13)
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.7 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
7
deg I − T (0, ·), U ∗∗ (0), θ = 0, which contradicts to (2.7) and (2.13). So, there must exist a α0 ∈ Γ such that Eα−0 is unbounded. Similarly, we can show that there must exist a α1 ∈ Γ such that Eα+1 is unbounded. The proof is complete. 2 Proof of Theorem 2.1. Let
T (λ, x) =
A∗ (λ, x), ([0, 1] × E) ∩ M [2, +∞), θ, ([0, 1] × B(θ, 1)) ∪ ({0} × E),
(2.14)
where B(θ, 1) = {x ∈ E : x ≤ 1} and M [2, +∞) = (λ, x) ∈ R+ × E : 2 ≤ x < ∞ . By Lemma 2.1, T (λ, x) admits an extension over all of [0, 1] × E, denote it also by T (·, ·) for simplicity. Obviously, T (·, ·) is a completely continuous operator. Note that T (λ, x) = A∗ (λ, x) for all λ ∈ [0, 1] and x ∈ E with x ≥ 2. From Lemma 2.5, there exists R1 > 2 large enough such that Fix T (1, ·) ⊂ B(θ, R1 ) and deg I − T (1, ·), B(θ, R1 ), θ = 0. Obviously, deg I − T (1, ·), B(θ, 1), θ = 1. Therefore, deg I − T (1, ·), B(θ, R1 )\B(θ, 1), θ = deg I − T (1, ·), B(θ, R1 ), θ − deg I − T (1, ·), B(θ, 1), θ = 0 − 1 = −1.
(2.15)
So, Fix T (1, ·) = ∅ and Fix T (1, ·) ⊂ U := B(θ, R1 )\B(θ, 1). From (2.14) we see that (2.8) holds. Now from Lemma 2.6 there exists an unbounded connected component C in the metric space L∗ ∩ ([0, 1] × E) containing some connected component of L∗ ∩ ({1} × E). Let r0 = 2 + β −1 g(1) and D1 = {1} × E ∩ M [r0 , +∞), D2 = [0, 1] × E ∩ M (r0 ), X ∗ = [0, 1] × E ∩ M [r0 , +∞), where M (r0 ) = (λ, x) ∈ R+ × E : x = r0 . Obviously, C ∩ (D1 ∪ D2 ) = ∅. For each p ∈ C ∩ (D1 ∪ D2 ), denote by E(p) the connected component of the metric space C ∩X ∗ which passes the point p. Now we shall prove that, there must exists a p0 ∈ C ∩(D1 ∪D2 ) such that E(p0 ) is an unbounded connected component of the metric space C ∩X ∗ . On the contrary, assume that E(p) is bounded for each p ∈ C ∩ (D1 ∪ D2 ). Then, for each p ∈ C ∩ (D1 ∪ D2 ), in the same way as
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.8 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
8
in the construction of U ∗ (Dα ) in (2.11) we can show that there exists a neighborhood U ∗ (p) of E(p) in X ∗ such that ∂X ∗ U ∗ (p) ∩ C = ∅.
(2.16)
Obviously, the set of {U ∗ (p) ∩ (D1 ∪ D2 )|p ∈ C ∩ (D1 ∪ D2 )} is an open cover of the set C ∩ (D1 ∪ D2 ) and C ∩ (D1 ∪ D2 ) is a compact set. Thus, there exist finite subsets of {U ∗ (p) ∩ (D1 ∪ D2 )|p ∈ C ∩ (D1 ∪ D2 )}, say, U ∗ (p1 ) ∩ (D1 ∪ D2 ), U ∗ (p2 ) ∩ (D1 ∪ D2 ), ..., U ∗ (pn ) ∩ (D1 ∪ D2 ), which is also an open cover of C ∩ (D1 ∪ D2 ), that is n
U ∗ (pi ) ∩ (D1 ∪ D2 ) ⊃ C ∩ (D1 ∪ D2 ).
(2.17)
i=1
Let U0 =
n i=1
U ∗ (pi ), then U0 is a bounded open subset of X ∗ . Since ∂X ∗ U0 ⊂
n
∂X ∗ U (pi ),
i=1
then by (2.16) we have ∂X ∗ U0 ∩ C = ∅.
(2.18)
Let W1 =
[0, 1] × E ∩ M [0, r0 ) ∪ U0 ,
and W2 = ([0, 1] × E)\Cl [0,1]×E W1 , where M [0, r0 ) = {x ∈ E : 0 ≤ x < r0 }. It is easy to see that ∂[0,1]×E W1 ⊂ ∂X ∗ U0 ∪
M (r0 ) ∩ [0, 1] × E \ U0 ∩ (D1 ∪ D2 ) .
From (2.17) and (2.18) we see that ∂[0,1]×E W1 ∩ C = ∅. Obviously, W1 ∩ C = ∅ and W1 ∩ W2 = ∅. Note the unboundedness of C, then W2 ∩ C = ∅. Now we have C = (W1 ∩ C) ∪ (W2 ∩ C), which is a contradiction of the connectedness of C. Therefore, there must exist p0 ∈ C ∩ (D1 ∪ D2 ) such that E(p0 ) is an unbounded connected component of C ∩ X ∗ . Since x = A∗ (λ, x) ∈ Qλ for each (λ, x) ∈ E(p0 ), we have x ≥ βx − g(λ) e ≥ βr0 − g(1) e > θ. So, x = A(λ, x) for each (λ, x) ∈ E(p0 ). This implies E(p0 ) ⊂ L is an unbounded subset. Let C ∗ be the connected component of L containing E(p0 ). Obviously, there exists λ∗ > 0 such that Proj Cλ∗ ⊃ (0, λ∗ ]. As C ∗ is unbounded and by Lemma 2.5, we easily see that lim
λ→0+ ,(λ,x)∈C ∗
The proof is complete. 2
x = +∞.
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.9 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
9
3. Applications of main results As some applications of Theorem 2.1, let us first consider the boundary value problem
(rN −1 φ(u (r))) + λrN −1 f (u(r)) = 0, u (0) = u(1) = 0,
0 < r < 1;
(3.1)
where λ ∈ R+ is a parameter, N is a natural number, f is a continuous function on R1 . We make the following assumptions concerning (3.1). (A1 ) φ is an odd, increasing homeomorphism on R1 , and for each ζ > 0, there exists a positive number (ζ) > 0 such that (ζ) → ∞ as ζ → ∞, and (ζ)u ≤ φ−1 (ζu),
u ∈ R+ .
(3.2)
(A2 ) f : R+ → R1 is continuous and lim
u→+∞
f (u) = +∞. φ(u)
Let E = C[0, 1] be the well known real Banach space of all continuous functions on [0, 1] with the maximum norm · , and P = {u ∈ C[0, 1] : u(r) ≥ 0, r ∈ [0, 1]}. Then P is a normal cone of E and E is a lattice in the partial ordering ≤ induced by cone P satisfying the property (H). From (A2 ), one easily find m > 0 such that f (u) ≥ −m for u ∈ R+ . Let
u ≥ 0, u < 0.
f (u), f (0),
f ∗ (u) = Then, f ∗ (u) ≥ −m for u ∈ R1 . Define operator A : R+ × E → E by
1 A(λ, u)(r) =
−1
φ
τ
sN −1
r
s
λ
N −1 ∗
f (u)dτ ds.
0
Let L2 = (λ, u) ∈ R+ × E : u is a positive solution of (3.1), u = θ . Theorem 3.1. Assume that (A1 )–(A2 ) hold and φ−1 is concave on R1 . Then L2 possesses an unbounded connected component C ∗ such that Projλ C ∗ ⊃ (0, λ∗ ] for some λ∗ > 0, and lim
λ→0+ ,(λ,x)∈C ∗
x = +∞.
Proof. For each (λ, u) ∈ R+ × P , let 1 y(r) = A(λ, u)(r) =
−1
φ r
λ
s τ
sN −1 0
N −1 ∗
f (u)dτ ds.
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.10 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
10
Clearly,
1
−1
y(r) =
φ
s
λ
τ
sN −1
r
N −1
s
λm f(u)dτ − N −1 s
τ
0
N −1
dτ ds,
0
where f(u) = f ∗ (u) + m ≥ 0. As φ−1 is concave on R+ , there exists β > 0 such that (here we may take β = 1, see Lemma 2.2 in [6]) φ−1 (x − y) ≥ βφ−1 (x) − 2φ−1 (y),
x, y ≥ 0.
(3.3)
And so, 1 y(r) ≥ β
s
λ
φ−1
sN −1
r
τ N −1 f(u)dτ ds − 2
1
φ−1
λm sN −1
r
0
s
τ N −1 dτ ds.
(3.4)
0
Clearly, for 0 ≤ r ≤ 1, 1
−1
φ
s
λm sN −1
τ
r
N −1
dτ ds ≤ φ
−1
λm (1 − r). N
0
And so, −1
y(r) ≥ β u(r) − 2φ
λm −1 λm (1 − r) ≥ −2φ (1 − r), N N
(3.5)
1 s where u (r) = r φ−1 ( sNλ−1 0 τ N −1 f(u)dτ )ds ≥ 0 for each r ∈ [0, 1]. Clearly, u (r) ≥ y(r) for each r ∈ [0, 1], u = u (0) and
(rN −1 φ( u )) + λrN −1 f(u(r)) = 0, u (1) = 0, u (0) = 0.
r ∈ (0, 1);
We will prove that u (r) ≥ ω(r) for each r ∈ [0, 1], where ω satisfies
−(rN −1 φ(ω )) = 0, r ∈ (0, 1); ω(1) = 0, ω(0) = u.
Suppose that u (r0 ) < ω(r0 ) for some r0 ∈ (0, 1), then u − ω has a negative absolute minimum at τ ∈ (0, 1). Since u (1) − ω(1) = u (0) − ω(0) = 0, there exist τ0 , τ1 ∈ [0, 1] such that u (τ0 ) − ω(τ0 ) = u (τ1 ) − ω(τ1 ) = 0, and u (r) − ω(r) < 0,
r ∈ (τ0 , τ1 ).
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.11 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
11
Then
N −1 rN −1 φ u (r) − r φ ω (r) = −λrN −1 f(u) ≤ 0,
r ∈ (τ0 , τ1 ).
Let ν(t) = u (t) − ω(t) < 0, r ∈ (τ0 , τ1 ), then τ1
N −1 rN −1 φ u ν(t)dt ≥ 0. (r) − r φ ω (r)
τ0
On the other hand, using the inequality
φ(b) − φ(a) (b − a) ≥ 0,
a, b ∈ R,
and the fact that there exists τ ∗ ∈ [τ0 , τ1 ] such that u (τ ∗ ) = ω(τ ∗ ), we have τ1
N −1 (r) − r φ ω (r) rN −1 φ u ν(t)dt
τ0
=−
τ1
N −1 (r) − r φ ω (r) rN −1 φ u u − ω dt < 0,
τ0
which is a contradiction. So, u (r) ≥ ω(r) for r ∈ [0, 1]. Clearly, ω(r) = u(1 − r). As u (r) ≥ y(r) for each r ∈ [0, 1], we have from (3.5) that u (r) + 2φ
−1
λm −1 λm (1 − r) ≥ y(r) + 2φ (1 − r) ≥ 0 N N
for r ∈ [0, 1]. So, −1 λm −1 λm u (r) + 2φ ≥ y(r) − 2φ . N N It follows that −1 λm u (r) ≥ y(r) − 4φ . N From (3.4) and (3.6) we have
λm (1 − r) y(r) ≥ β u(r) − 2φ N λm ≥ β u − 2φ−1 (1 − r) N −1 λm ≥ β y(r) − (4β + 2)φ (1 − r). N −1
So, A(λ, u)(r) ≥ β A(λ, u) − g(λ) e(r),
(3.6)
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.12 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
12
+ where g(λ) := (4β + 2)φ−1 ( λm N ). Let Qλ = {u ∈ E : u ≥ (βu − g(λ))e(r)} for λ ∈ R and e(r) = (1 − r) for each r ∈ [0, 1]. This implies that (H1 ) holds. From (A2 ), for arbitrary m0 > 0, there exists m1 > 0 such that f ∗ (u) ≥ m0 φ(u) for u ≥ m1 . Let λ , λ > 0 with λ < λ . Then, for each λ ∈ [λ , λ ] and u ∈ Qλ with u ≥ max{β −1 (2m1 + g(λ )), 2β −1 g(λ )}, we have
1 u(r) ≥ βu − g(λ) (1 − r) ≥ βu − g λ ≥ m1 2 for each r ∈ [0, 12 ]. From (3.2) and using the inequality (3.3) we have 1 s 1 λ −1 N −1 ∗ = φ A(λ, u) τ f (u)dτ ds 2 sN −1 0
1 2
1 ≥β
−1
φ
s
λ
≥β
−1
φ
≥β
1
N −1
−1
1 2
−1 mλ f (u) + m dτ ds − 2φ N ∗
m0 λ sN −1
2 τ
N −1
φ(u)dτ ds − 2φ−1
mλ N
0
−1
φ
≥β
1
1
m0 λ φ( 12 (βu − g(λ ))) sN −1
2 τ
N −1
dτ ds − 2φ
−1
mλ N
0
1 2
1
0
0
1 2
≥β
τ
sN −1
φ 1
1 s mλ −1 N −1 τ dτ ds f (u) + m dτ ds − 2 φ sN −1 ∗
1 2
2
λ
1 2
1
0
1 2
1
τ
sN −1
N −1
φ−1
m0 λ φ( 12 (βu − g(λ ))) −1 mλ ds − 2φ 2N · N N
β mλ m0 λ βu − g λ N − 2φ−1 2 2 ·N N 2 β mλ m0 λ u N − 2φ−1 . ≥ 2 2 ·N N
≥
Thus A(λ, u)( 12 ) β2 2 −1 mλ A(λ, u) m0 λ ≥ ≥ N − φ . u u 2 2 ·N u N Note that 2 −1 mλ φ →0 u N as u → ∞. So, A(λ, u) = +∞, u u∈Qλ ,u→∞ lim
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.13 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
13
uniformly with λ ∈ [λ , λ ]. This implies that (H2 ) holds. Now by applying Theorem 2.1 one can easily prove that Theorem 3.1 holds. The proof is complete. 2 Corollary 3.1. Let (A1 )–(A2 ) hold. Then there exists a positive number λ > 0 such that for 0 < λ ≤ λ, (3.1) has a positive solution uλ such that lim uλ = ∞.
λ→0+
Remark 3.1. D.D. Hai and K. Schmitt in [15] have obtained a very similar result to Corollary 3.1. See Theorem 2.1 in [15]. To show Theorem 2.1 in [15], besides condition (A2 ), the authors employed the following two assumptions: (A∗1 ) φ is an odd, increasing homeomorphism on R1 , and for each ζ > 0, there exists a positive number Mζ > 0 such that φ(ζu) ≤ Mζ φ(u),
u ∈ R1 .
(3.7)
(A∗3 ) There exists θ ∈ (0, 1) such that
F (θx) Φ(x) > max N lim sup − 1, 0 , N lim inf x→∞ x→∞ xf (x) xφ(x) x x where F (x) = 0 f (s)ds and Φ(x) = 0 φ(s)ds. Here we have removed the assumption (A∗3 ). Compare (A1 ) with (A∗1 ), here we replace (3.7) with (3.2). We should point out, a condition similar to (3.2) has been employed in [16] (see assumption (A.2) in [16]). As for the case φ(x) = xp−1 for some p ≥ 1, the assumption (A∗1 ) is needn’t. In fact we have the following result. Corollary 3.2. Let φ(x) = xp−1 for some p ≥ 1. Assume that (A2 ) holds. Then there exists a positive number λ > 0 such that for 0 < λ ≤ λ, (3.1) has a positive solution uλ such that lim uλ = ∞.
λ→0+
Next we consider Sturm–Lioville boundary value problem
(p(t)u ) + λf (t, u(t)) = 0,
r < t < R;
au(r) − bu (r) = cu(R) + du (R) = 0
(3.8)
where λ ∈ R+ ,f (t, u) is a continuous function on [r, R] × R1 , and satisfies (A3 ) There exists M0 > 0 such that f (t, u) > −M0 , for all (t, u) ∈ [r, R] × R+ ; (A4 ) p(t) ∈ C 1 [r, R], p(t) > 0 for t ∈ [r, R]; (A5 ) There exists [α, β] ⊂ (r, R) such that lim
u→+∞
f (t, u) = +∞ u
uniformly with t ∈ [α, β].
Let E = C[r, R] be the well known real Banach space of all continuous functions on [r, R] with the maximum norm · , and P = {u ∈ C[0, 1] : u(r) ≥ 0, r ∈ [r, R]}. Then, P is a normal cone of E.
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.14 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
14
Also E is a lattice in the partial ordering ≤ induced by the cone P which satisfies the property (H). Let Q = {u ∈ E : u ≥ ue}, where e(t) = min
b+a b+
t
r R a r
R
p−1 c + d t p−1 . , R p−1 c + d r p−1
Let ⎧ ⎨ α−1 (b + a s p−1 )(d + c R p−1 ), r t G(t, s) = ⎩ α−1 (b + a t p−1 )(d + c R p−1 ), r s
s ≤ t, s > t,
R R R where α = ad + bc + ac r p−1 and r p−1 = r p−1 (s)ds. Let us define the operators K : E → E, F : P → E and A : R+ × P → E by R (Ku)(t) =
G(t, s)u(s)ds,
x ∈ E, t ∈ [r, R],
r
(F u)(t) = f t, u(t) ,
u ∈ P, t ∈ [r, R],
and A(λ, u) = λKF u. Let ω0 = M0 for all t ∈ [r, R] and R ω1 (t) = M0
G(t, s)ds,
t ∈ [r, R].
r
Remark 3.2. It is easy to know that if u > θ, and u = A(λ, u), then u is a positive solution of (3.8). From the proof in [2] we have the following Lemmas 3.1 and 3.2. Lemma 3.1. K : P → Q is a linear completely operator. Lemma 3.2. There exists σ > 0 such that ω1 ≤ σe. Let L3 = (λ, u) ∈ R+ × P : u is a positive solution of (3.8), u = θ . Theorem 3.2. Assume that (A3 )–(A5 ) hold. Then L3 possesses an unbounded connected component C ∗ such that Projλ C ∗ ⊃ (0, λ∗ ] for some λ∗ > 0, and lim
λ→0+ ,(λ,x)∈C ∗
x = +∞.
Proof. For each (λ, u) ∈ R+ × P , by Lemmas 3.1 and 3.2 we have A(λ, u)(t) = λKF u(t) R = r
G(t, s) λf s, u(s) + λM0 ds − λM0
R G(t, s)ds r
JID:YJMAA
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.15 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
R =
15
G(t, s) λf s, u(s) + λM0 ds − λω1 (t)
r
R ≥ G(t, s) λf s, u(s) + λM0 ds e − λω1 (t) r
≥ A(λ, u)e − λω1 e − λσe = A(λ, u)e − g(λ)e, where g(λ) = λω1 + λσ. Thus (H1 ) holds. By condition (A5 ) one can easily prove (H2 ) using the same method proving Theorem 3.1. A very similar detailed proof can be found in Theorem 3.3 of [23]. Now by applying Theorem 2.1 one can easily prove that Theorem 3.2 holds. The proof is complete. 2 Corollary 3.3. Assume that all conditions of Theorem 3.2 hold. Then there exists λ∗ > 0 such that (3.8) has at least one positive solution for 0 < λ ≤ λ∗ . Remark 3.3. Corollary 3.3 is the main result of [2]. V. Anuradha, D.D. Hai and R. Shivaji using the degree method obtained this result. Here we obtain this result by using a different method—bifurcation theories. Moreover, Theorem 3.2 gave some informations concerning the structure of positive solution set of (3.8). Remark 3.4. D.D. Hai, K. Schmitt and R. Shivaji in [16] considered the following boundary value problem:
(p(t)φ(u )) + λp(t)f (t, u) = 0, u(a) = u(b) = 0,
a < t < b;
where f is φ-superlinear or φ-sub-linear at ∞ and f (t, 0) may be negative and p is a continuous function. By using degree method, some existence results for positive solutions be obtained. Obviously, the main results in [16] concerning the existence results for positive solutions in the case where f is φ-superlinear can also be derived from our Theorem 2.1. Remark 3.5. Very recently, by using bifurcation method we studied the structure of positive solution set of some semi-positone problems in a Banach space. We should point out that, our main results here also are an improvement of the main results in [23]. The main results of [23] is applicable only when A(λ, x) has the form of A(λ, x) = λKF x, where K is a linear completely continuous operator and F is a nonlinear continuous operator. Here we removed this requirement. Therefore, the main results would have more applications than that of the results in [23]. References [1] A. Ambrosetti, D. Arcoya, B. Buffoni, Positive solutions for some semipositone problems via bifurcation theory, Differential Integral Equations 7 (3) (1994) 655–663. [2] V. Anuradha, D.D. Hai, R. Shivaji, Existence results for superlinear semipositone BVP’S, Proc. Amer. Math. Soc. 124 (1996) 757–763. [3] A. Castro, R. Shivaji, Nonnegative solutions for a class of nonpositone problems, Proc. Roy. Soc. Edinburgh Sect. A 108 (1988) 291–302. [4] A. Castro, R. Shivaji, Semipositone problems, in: J.A. Goldstein, G. Goldstein (Eds.), Semigroups of Linear and Nonlinear Operators and Applications, Kluwer Academic Publishers, New York, 1993, pp. 109–119 (invited review paper). [5] A. Castro, R. Shivaji, Positive solutions for a concave semipositone Dirichlet problem, Nonlinear Anal. 31 (1/2) (1998) 91–98. [6] A. Castro, S. Gadam, R. Shivaji, Branches of radial solutions for semipositone problems, J. Differential Equations 120 (1) (1995) 30–45.
JID:YJMAA 16
AID:18582 /FLA
Doctopic: Applied Mathematics
[m3L; v 1.134; Prn:3/06/2014; 9:28] P.16 (1-16)
X. Xu, X. Zhu / J. Math. Anal. Appl. ••• (••••) •••–•••
[7] A. Castro, M. Hassanpour, R. Shivaji, Uniqueness of nonnegative solutions for a semipositone problem with concave nonlinearity, Comm. Partial Differential Equations 20 (11–12) (1995) 1927–1936. [8] A. Castro, S. Gadam, R. Shivaji, Positive solution curves of semipositone problems with concave nonlinearities, Proc. Roy. Soc. Edinburgh Sect. A 127 (1997) 921–934. [9] M. Chhetri, P. Girg, Existence and nonexistence of positive solutions for a class of superlinear semipositone systems, Nonlinear Anal. 71 (2009) 4984–4996. [10] K. Deimling, Nonlinear Functional Analysis, Springer-Verlag, Berlin, Heidelberg, New York, 1985. [11] Y. Du, Effects of a degeneracy in the competition model. I. Classical and generalized steady-state solutions, J. Differential Equations 181 (1) (2002) 92–132. [12] Y. Du, Realization of prescribed patterns in the competition model, J. Differential Equations 193 (1) (2003) 147–179. [13] Y. Du, X. Liang, A diffusive competition model with a protection zone, J. Differential Equations 244 (2008) 61–86. [14] Y. Du, J. Shi, A diffusive predator–prey model with a protection zone, J. Differential Equations 229 (2006) 63–91. [15] D.D. Hai, K. Schmitt, On radial solutions of quasilinear boundary value problems, Progr. Nonlinear Differential Equations Appl. 35 (1999) 349–361. [16] D.D. Hai, K. Schmitt, R. Shivaji, Positive solutions of quasilinear boundary value problems, J. Math. Anal. Appl. 217 (1998) 672–686. [17] V. Hutson, Y. Lou, K. Mischaikow, P. Polacik, Competing species near a degenerate limit, SIAM J. Math. Anal. 35 (2003) 453–491. [18] V. Hutson, Y. Lou, K. Mischaikow, Convergence in competition models with small diffusion coefficients, J. Differential Equations 211 (1) (2005) 135–161. [19] M.A. Krasnosel’skii, P.P. Zabreiko, Geometrical Methods of Nonlinear Analysis, Springer-Verlag, Berlin, Heidelberg, New York, Tokyo, 1984. [20] Ruyun Ma, Yulian An, Global structure of positive solutions for superlinear second order m-point boundary value problems, Topol. Methods Nonlinear Anal. 34 (2009) 279–290. [21] Jingxian Sun, Xian Xu, Positive fixed points of semi-positone nonlinear operator and its applications, Acta Math. Sin. 55 (2012) 55–64 (in Chinese). [22] Xian Xu, D. O’Regan, Existence of positive solutions for operator equations and applications to semipositone problems, Positivity 10 (2) (2006) 315–328. [23] Xian Xu, Jingxian Sun, Unbounded connected component of the positive solutions set of some semi-position problems, Topol. Methods Nonlinear Anal. 37 (2011) 283–302.