Abstract
By using the fixedpoint index theory in a cone and defining a linear operator, we obtain the existence of at least one positive solution for the thirdorder boundary value problem with integral boundary conditions
where f : [0, 1] × R^{+ }× R^{ }→ R^{+ }is a nonnegative function. The associated Green's function for the above problem is also used, and a new reproducing cone also used.
Keywords:
fixedpoint index theory; Green's function; positive solution; boundary value problem1 Introduction
By eigenvalue criteria, Webb [1] obtained the existence of multiple positive solutions of a Hammerstein integral equation of the form
where k can have discontinuities and g ∈ L^{1}. Then, some articles have studied different BVPs by this way (see [25]). Webb [4] introduced an unified method to study existence of at least one nonzero solution for higher order boundary value problems
In 2010, Hao [5] considered the existence of positive solutions of the nthorder BVP
Guo [6] studied the existence of positive solutions for the therepoint boundary problem with the firstorder derivative.
where f is a nonnegative continuous function. In 2011, Zhao [7] studied thirdorder differential equations:
subject to integral boundary condition of the form
where f ∈ C([0, 1] × P, P).
In this article, we study the existence of positive solutions for the following boundary value problem
The results are proved by applying the fixed point index theory in a cone and spectral radius of a linear operator. Unlike reference [7], the nonlinear part f involves the secondorder derivative and just satisfies Caratheodory conditions.
The following conditions are satisfied throughout this article:
(H_{1}) f : [0, 1] × R^{+ }× R^{ }→ R^{+ }satisfies Caratheodory conditions, that is, f(·,u, v) is measurable for each fixed u ∈ R^{+}, v ∈ R^{}, and f(t, ·,·) is continuous for a.e. t ∈ [0, 1]. For any r, r' > 0, there exists , such that , where (u, v) ∈ [0, r] × [r', 0], a.e. t ∈ [0, 1];
2 Preliminaries
Lemma 2.1 [7]. Let y ∈ L^{1}[0, 1] and y ≥ 0, the problem
has a unique solution
Lemma 2.2. Let y ∈ L^{1}[0, 1] and y ≥ 0, the unique solution of the boundary value problem (2.1) satisfies the following conditions: u(t) ≥ 0, u"(t) ≤ 0, for t ∈ [0, 1].
Proof. By Lemma 2.1, u(t) ≥ 0. By differential equations u'"(t) + y(t) = 0, t ∈ (0, 1), we get
Let X = C^{2}[0, 1] with . Obviously, (X, ·) is a Banach space. Define the cone P ⊂ X by
Obviously P is a cone in X, and P_{r}is a bounded open subset in P.
Definition 2.1 [1]. Let P be a cone in a Banach space X. If for any x ∈ X and x^{+}, x^{ }∈ P, writing x = x^{+ }+ x^{ }shows that P is a reproducing cone.
Lemma 2.3. P is a reproducing cone in X.
Proof. Suppose u ∈ X, so u" ∈ C[0, 1] and
where u^{ }= min{u"(t), 0}, u^{+ }= min{u"(t), 0}. Obviously u^{+},u^{ }∈ C[0, 1] and u^{+ }≤ 0,u^{ }≤ 0. For (2.2), we get
If u(0) ≥ 0, u'(0)t ≥ 0, let
So u_{1 }≥ 0, u_{2 }≥ 0, then u_{1}, u_{2 }∈ P and u = u_{1 } u_{2}.
If u(0) ≤ 0, u'(0)t ≤ 0, let
So u_{1 }≥ 0, u_{2 }≥ 0, then u_{1}, u_{2 }∈ P and u = u_{1 } u_{2}.
If u(0) ≥ 0, u'(0)t ≤ 0, let
So u_{1 }≥ 0, u_{2 }≥ 0, then u_{1}, u_{2 }∈ P and u = u_{1 } u_{2}.
If u(0) ≤ 0, u'(0)t ≥ 0, let
So u_{1 }≥ 0, u_{2 }≥ 0, then u_{1}, u_{2 }∈ P and u = u_{1 } u_{2}.
Then P is a reproducing cone in X.
Lemma 2.4 (KreinRutman) [8]. Let K be a reproducing cone in a real Banach space X and let L : K → K be a compact linear operator with L(K) ⊂ K. r(L) is the spectral radius of L. If r(L) > 0, then there is φ_{1 }∈ K\{0} such that Lφ_{1 }= r(L)φ_{1}.
Lemma 2.5 [9]. Let X be a Banach space, P be a cone in X and Ω(P) be a bounded open subset in P. Suppose that is a completely continuous operator. Then the following results hold
(1) If there exists u_{0 }∈ P\{0} such that u ≠ Au + λu_{0}, for any u ∈ ∂Ω(P), λ ≥ 0, then the fixedpoint index i(A, Ω(P), P) = 0.
(2) If 0 ∈ Ω(P), Au ≠ λu, for any u ∈ ∂Ω(P), λ ≥ 1, then the fixedpoint index i(A, Ω(P), P) = 1.
Define the operator A: X → X, L: X → X, by
So A : P → P is completely continuous operator; L : P → P is a compact linear operator.
Lemma 2.6 [7]. Assume that (H_{2}) holds, then choose , for all t ∈ [δ, 1  δ],v, s ∈ [0, 1], we have
where ρ = 4δ^{2}(1  δ).
Note: r(L) is the spectral radius of L. , where . By Lemma 2.6, obviously h > 0.
Lemma 2.7. Suppose conditions (H_{1}), (H_{2}) hold, then r(L) > 0.
Proof. Take u(t) ≡ 1, then u"(t) = 0, for any t ∈ [δ, 1  δ] we get
Repeating the process gives
So, we get . The proof is completed.
By Lemma 2.4, then there is φ_{1 }∈ P\{0} such that Lφ_{1 }= r(L)φ_{1}.
3 Main results
In the following, we use the notation:
where E is a fixed subset of [0, 1] of measure zero, d > 0.
Lemma 3.1. Suppose
where μ = 1/r(L), then there exists R_{0 }> 0 such that i(A, P_{r}, P) = 1 for each r > R_{0}.
Proof. Let ε > 0 satisfy f^{∞ }≤ μ  ε, then there exist r_{1 }> 0 such that
for all u > r_{1 }or v < r_{1 }and a.e. t ∈ [0, 1].
By (H_{1}), there exists such that
for all (u, v) ∈ [0, r_{1}] × [r_{1}, 0] and a.e. t ∈ [0, 1]. Hence, we have
for all u ∈ R^{+}, v ∈ R^{ }and a.e. t ∈ [0, 1].
Since is the spectrum radius of L. It follows from , (I/(μ  ε)  L)^{l }exists, let
For r > R_{0}, by Lemma 2.5 we will prove
for each u ∈ ∂P_{r}and λ ≥ 1.
In fact, if not, there exist u_{0 }∈ ∂P_{r}and λ_{0 }≥ 1 such that Au_{0 }= λ_{0}u_{0}.
Together with (3.2) implies
So
Then
So
Then
This is a contradiction. By Lemma 2.5 (2), we get that i(A, P_{r}, P) = 1 for each r > R_{0}. The proof is completed.
Lemma 3.2. Suppose there exists d > 0 such that
Then there exists ρ_{0 }> 0 and d ≥ ρ_{0 }such that for each ρ∈ (0, ρ_{0}], if u ≠ Au for u ∈ ∂Pρ, then i(A, P_{ρ}, P) = 0.
Proof. Let ε > 0 satisfy , there exist d ≥ ρ_{0 }> 0 such that
for u ∈ [0, ρ_{0}],v ∈ [ρ_{0},0] and a.e. t ∈ [0, 1].
Let ρ ∈ (0,ρ_{0}], by Lemma 2.5 (1), we prove that: u ≠ Au + λφ_{1 }for all u ∈ ∂Pρ, λ > 0, where φ_{1 }∈ P\{0} is the eigenfunction of L corresponding to the eigenvalue . In fact, if not, there exist u_{0 }∈ ∂P_{ρ}, λ_{0 }> 0 such that u_{0 }= Au_{0 }+ λ_{0}φ_{1}. This implies
So 0 < λ_{0 }< λ* < ∞ and . Then, u_{0 } λ*φ_{1 }∈ P.
For L(P) ⊂ P, we get
By (3.4), we get
So, we know
which contradicts the definition of λ*.
Lemma 3.3. Suppose there is ρ_{1 }> 0 such that
for u ∈ [0, ρ_{1}] and v ∈ [ρ_{1}, 0] a.e. t ∈ [0, 1], where , if Au ≠ u for , then .
Proof. Suppose , by Lemma 2.2, we get
That is Au ≠ λu for each , λ > 1. If Au ≠ u for , by Lemma 2.5, then .
Lemma 3.4. Suppose there is ρ_{2 }> 0 such that
for u ∈ [0, _{ρ2}] and v ∈ [_{ρ2}, 0] a.e. t ∈ [0, 1], where . If Au ≠ u for , then .
Proof. For , t ∈ [δ, 1  δ], by Lemma 2.2, we get
This implies that u ≠ Au + λφ for each , λ > 0, where φ ∈ P\{0} is the eigenfunction of L corresponding to r(L). Suppose u ≠ Au for , by Lemma 2.5, then .
Theorem 3.1. The boundary value problem (1.1) has at least one positive solution if one of the following conditions holds.
(C1) There exists d > 0 such that (3.3) and (3.1) hold.
(C2) There exists d > 0, ρ_{1 }> 0 such that (3.3) and (3.5) hold.
(C3) There exists ρ_{2 }> 0 such that (3.6) and (3.1) hold.
(C4) There exists ρ_{1}, ρ_{2 }> 0 with 0 < ρ_{2 }< ρ_{1}d_{1}/d_{2 }such that (3.5) and (3.6) hold.
Proof. When condition (C1) holds, by Lemma 3.1 and 0 ≤ f^{∞ }< μ, we get that there exists r > 0 such that i(A, P_{r}, P) = 1. It follows from Lemma 3.2 and , then there exists 0 < ρ < min{r, d} such that either there exists u ∈ ∂P_{ρ}that i(A, P_{ρ}, P) = 0 or u = Au. So BVP (1.1) has at least one positive solution u ∈ P with ρ ≤ u < r.
When one of other conditions holds, the results can be proved similarly.
Competing interests
The authors declare that they have no competing interests.
Authors' contributions
The authors declare that the study was realized in collaboration with same responsibility. All authors read and approved the final manuscript.
Acknowledgements
The project is supported by the Natural Science Foundation of China (10971045) and the Natural Science Foundation of Hebei Province (A2009000664, A2011208012). The research item financed by the talent training project funds of Hebei Province. The authors would like to thank the referee for helpful comments and suggestions.
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