By eigenvalue criteria, Webb 1 obtained the existence of multiple positive solutions of a Hammerstein integral equation of the form

u ( t ) = ∫ 0 1 k ( t , s ) g ( s ) f ( s , u ( s ) ) d s ,

where k can have discontinuities and g ∈ L ¹. Then, some articles have studied different BVPs by this way (see 2 3 4 5 ). Webb 4 introduced an unified method to study existence of at least one nonzero solution for higher order boundary value problems

u ( n ) ( t ) + g ( t ) f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( n ) ( 0 ) = 0 , 0 ≤ k ≤ n - 2 , u ( 1 ) = ∫ 0 1 u ( s ) d A ( s ) .

In 2010, Hao 5 considered the existence of positive solutions of the nth-order BVP

u ( n ) ( t ) + λ a ( t ) f ( t , u ( t ) ) = 0 , t ∈ ( 0 , 1 ) , u ( 0 ) = u ′ ( 0 ) = ⋯ = u ( n - 2 ) ( 0 ) = 0 , u ( 1 ) = ∫ 0 1 u ( s ) d A ( s ) .

Guo 6 studied the existence of positive solutions for the there-point boundary problem with the first-order derivative.

x ″ + f ( t , x , x ′ ) = 0 , 0 < t < 1 , x ( 0 ) = 0 , x ( 1 ) = α x ( η ) ,

where f is a nonnegative continuous function. In 2011, Zhao 7 studied third-order differential equations:

x ‴ + f ( t , x ( t ) ) = θ , t ∈ [ 0 , 1 ] ,

subject to integral boundary condition of the form

x ( 0 ) = θ , x ″ ( 0 ) = θ , x ( 1 ) = ∫ 0 1 g ( t ) x ( t ) d t ,

where f ∈ C([0, 1] × P, P).

In this article, we study the existence of positive solutions for the following boundary value problem

u ‴ ( t ) + f ( t , u ( t ) , u ″ ( t ) ) = 0 , t ∈ ( 0 , 1 ) , u ( 0 ) = 0 , u ″ ( 0 ) = 0 , u ( 1 ) = ∫ 0 1 g ( t ) u ( t ) d t .

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 second-order 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 φ r , r ′ ( t ) ∈ L ∞ [ 0 , 1 ] , such that 0 ≤ f ( t , u , v ) ≤ φ r , r ′ ( t ) , where (u, v) ∈ [0, r] × [-r', 0], a.e. t ∈ [0, 1];

(H ₂) g ∈ L[0, 1] is nonnegative, b ∈ [0, 1), where b = ∫ 0 1 s g ( s ) d s .

Lemma 2.1 7 . Let y ∈ L ¹[0, 1] and y ≥ 0, the problem

u ‴ ( t ) + y ( t ) = 0 , t ∈ ( 0 , 1 ) , u ( 0 ) = 0 , u ″ ( 0 ) = 0 , u ( 1 ) = ∫ 0 1 g ( t ) u ( t ) d t

has a unique solution

u ( t ) = ∫ 0 1 H ( t , s ) y ( t ) d s ,

where H ( t , s ) = G ( t , s ) + t 1 - b ∫ 0 1 G ( τ , s ) y ( τ ) d τ , b = ∫ 0 1 s g ( s ) d s ,

G ( t , s ) = 1 2 t ( 1 - s ) 2 - 1 2 ( t - s ) 2 , 0 ≤ s ≤ t ≤ 1 , 1 2 t ( 1 - s ) 2 , 0 ≤ t ≤ s ≤ 1 .

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

u ″ ( t ) - u ″ ( 0 ) = - ∫ 0 1 y ( s ) d s , u ″ ( t ) = - ∫ 0 1 y ( s ) d s ≤ 0 .

Let X = C ²[0, 1] with u = max 0 ≤ t ≤ 1 u ( t ) + max 0 ≤ t ≤ 1 u ″ ( t ) . Obviously, (X, ||·||) is a Banach space. Define the cone P ⊂ X by

P = u ∈ X u ( t ) ≥ 0 , u ″ ( t ) ≤ 0 , P r = u ∈ P u < r , r > 0 .

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

u ″ = u - - u + ,

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

u ′ ( t ) = ∫ 0 t u - ( s ) d s - ∫ 0 t u + ( s ) d s + u ′ ( 0 ) , u ( t ) = ∫ 0 t d s ∫ 0 s u - ( τ ) d τ - ∫ 0 t d s ∫ 0 s u + ( τ ) d τ + u ′ ( 0 ) t + u ( 0 ) .

If u(0) ≥ 0, u'(0)t ≥ 0, let

u 1 = - ∫ 0 t d s ∫ 0 s u + ( τ ) d τ + u ′ ( 0 ) t + u ( 0 ) , u 2 = - ∫ 0 t d s ∫ 0 s u - ( τ ) d τ .

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

u 1 = - ∫ 0 t d s ∫ 0 s u + ( τ ) d τ , u 2 = - ∫ 0 t d s ∫ 0 s u - ( τ ) d τ - u ′ ( 0 ) t - u ( 0 ) .

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

u 1 = - ∫ 0 t d s ∫ 0 s u + ( τ ) d τ + u ( 0 ) , u 2 = - ∫ 0 t d s ∫ 0 s u - ( τ ) d τ - u ′ ( 0 ) t .

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

u 1 = - ∫ 0 t d s ∫ 0 s u + ( τ ) d τ + u ′ ( 0 ) t , u 2 = - ∫ 0 t d s ∫ 0 s u - ( τ ) d τ - u ( 0 ) .

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 (Krein-Rutman) 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 A : Ω ( P ) ¯ → P 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 fixed-point index i(A, Ω(P), P) = 0.

(2) If 0 ∈ Ω(P), Au ≠ λu, for any u ∈ ∂Ω(P), λ ≥ 1, then the fixed-point index i(A, Ω(P), P) = 1.

Define the operator A: X → X, L: X → X, by

A u ( t ) = ∫ 0 1 H ( t , s ) f ( s , u ( s ) , u ″ ( s ) ) d s ,

L u ( t ) = ∫ 0 1 H ( t , s ) ( u ( s ) - u ″ ( s ) ) d s ,

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 δ ∈ 0 , 1 2 , for all t ∈ [δ, 1 - δ],v, s ∈ [0, 1], we have

G ( t , s ) ≥ ρ G ( v , s ) , H ( t , s ) ≥ ρ H ( v , s ) ,

where ρ = 4δ ²(1 - δ).

Note: r(L) is the spectral radius of L. h = min t ∈ [ δ , 1 - δ ] ∫ δ 1 - δ H ( t , s ) d s , where δ ∈ 0 , 1 2 . 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

L u ( t ) ≥ ∫ δ 1 - δ H ( t , s ) d s ≥ h > ( 0 ) . L 2 u ( t ) ≥ ∫ δ 1 - δ H ( t , s ) L u ( s ) d s ≥ h ∫ δ 1 - δ H ( t , s ) d s ≥ h 2 > 0 .

Repeating the process gives

L k u ( t ) ≥ h k .

So, we get L k ≥ h k , r ( L ) = lim k → ∞ L k 1 k ≥ h > 0 . The proof is completed.

By Lemma 2.4, then there is φ ₁ ∈ P\{0} such that Lφ ₁ = r(L)φ ₁.

In the following, we use the notation:

  f ¯ ( u , v ) = sup t ∈ [ 0 , 1 ] \ E f ( t , u , v ) ,   f ( u , v ) = inf t ∈ [ 0 , 1 ] \ E f ( t , u , v ) , f ∞ = max { lim u → ∞ sup { sup v ∈ R − f ¯ ( u , v ) u − v } ,   lim v → − ∞ sup { sup u ∈ R + f ¯ ( u , v ) u − v } } , f 0 d = max { lim u → 0 + inf { inf v ∈ [ − d , 0 ] f ( u , v ) u − v } ,   lim v → 0 − inf { inf u ∈ [ 0 , d ] f ( u , v ) u − v } } ,

where E is a fixed subset of [0, 1] of measure zero, d > 0.

Lemma 3.1. Suppose

0 ≤ f ∞ < μ ,

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

f ( t , u , v ) ≤ ( μ - ε ) ( u - v ) ,

for all u > r ₁ or v < -r ₁ and a.e. t ∈ [0, 1].

By (H ₁), there exists φ r 1 ∈ L ∞ [ 0 , 1 ] such that

0 ≤ f ( t , u , v ) ≤ φ r 1 ( t ) ,

for all (u, v) ∈ [0, r ₁] × [-r ₁, 0] and a.e. t ∈ [0, 1]. Hence, we have

f ( t , u , v ) ≤ ( μ - ε ) ( u - v ) + φ r 1 ( t ) ,

for all u ∈ R⁺, v ∈ R ^- and a.e. t ∈ [0, 1].

Since 1 μ is the spectrum radius of L. It follows from 1 μ - ε I - L - 1 = ∑ n = 0 ∞ ( μ - ε ) n + 1 L n , (I/(μ - ε) - L)^-l exists, let

C = ∫ 0 1 H ( t , s ) φ r 1 ( s ) d s , R 0 = 1 μ - ε I - L - 1 C μ - ε 3 2 - 1 2 t 2 .

For r > R ₀, by Lemma 2.5 we will prove

A u ≠ λ u ,

for each u ∈ ∂P_r and λ ≥ 1.

In fact, if not, there exist u ₀ ∈ ∂P_r and λ₀ ≥ 1 such that Au ₀ = λ₀ u ₀.

Together with (3.2) implies

u 0 ( t ) ≤ A u 0 ( t ) ≤ ∫ 0 1 H ( t , s ) [ ( μ − ε ) u 0 ( s ) + φ r 1 ( t ) ] d s           ≤ ∫ 0 1 H ( t , s ) ( μ − ε ) [ u 0 ( s ) − v 0 ( s ) ) + φ r 1 ( s ) ] d s .

So

u 0 ( t ) ≤ ( μ - ε ) L u 0 ( t ) + C , u 0 ″ ( t ) ≥ λ 0 u 0 ″ ( t ) = ( A u 0 ( t ) ) ″ ≥ ( μ - ε ) ( L u 0 ( t ) ) ″ - C .

Then

1 μ - ε I - L u 0 ( t ) ≤ C μ - ε 3 2 - 1 2 t 2 , 1 μ - ε I - L u 0 ( t ) ″ ≥ C μ - ε 3 2 - 1 2 t 2 ″ .

So

C μ - ε 3 2 - 1 2 t 2 - 1 μ - ε I - L u 0 ( t ) ∈ P .

Then

u 0 ( t ) ≤ ( I μ − ε − L ) − 1 C μ − ε ( 3 2 − 1 2 t 2 ) ,     u 0 ' ' ( t ) ≥ [ ( I μ − ε − L ) − 1 C μ − ε ( 3 2 − 1 2 t 2 ) ] ' ' ,                           ‖ u 0 ( t ) ‖ ≤ R 0 < r .

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

μ < f 0 d ≤ ∞ .

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 f 0 d ≥ μ + ε , there exist d ≥ ρ ₀ > 0 such that

f ( t , u , v ) ≥ ( μ + ε ) ( u - v ) ,

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 1μ . In fact, if not, there exist u ₀ ∈ ∂P_ρ , λ₀ > 0 such that u ₀ = Au ₀ + λ₀ φ ₁. This implies

u 0 ≥ λ 0 φ 1 a n d u 0 ″ ≤ λ 0 φ 1 ″ .

Let: λ * = sup λ | u 0 ≥ λ φ 1 , u 0 " ≤ λ φ 1 " .

So 0 < λ₀ < λ* < ∞ and u 0 ≥ λ * φ 1 , u 0 ″ ≤ λ * φ 1 ″ . Then, u ₀ - λ*φ ₁ ∈ P.

For L(P) ⊂ P, we get

μ L u 0 ≥ λ * μ L φ 1 = λ * φ 1 , μ ( L u 0 ) ″ ≤ λ * μ ( L φ 1 ) ″ = λ * φ 1 ″ .

By (3.4), we get

A u 0 = ∫ 0 1 H ( t ,   s ) f ( s ,   u 0 ( s ) ,   u 0 ' ' ( s ) ) d s ≥ ( μ + ε ) L u 0 .           ( A u 0 ) ″ ≤ ( μ + ε ) ( L u 0 ) ″ .

So, we know

u 0 = A u 0 + λ 0 φ 1 ≥ ( μ + ε ) L u 0 + λ 0 φ 1 ≥ ( λ * + λ 0 ) φ 1 . ( u 0 ) ″ = ( A u 0 ) ″ + λ 0 φ 1 ″ ≤ ( μ + ε ) ( L u 0 ) ″ + λ 0 φ 1 ″ ≤ ( λ * + λ 0 ) φ 1 ″ .

which contradicts the definition of λ*.

Lemma 3.3. Suppose there is ρ ₁ > 0 such that

f ( t , u , v ) ≤ d 1 ρ 1 ,

for u ∈ [0, ρ ₁] and v ∈ [-ρ ₁, 0] a.e. t ∈ [0, 1], where d 1 = 1 ∫ 0 1 H ( t , s ) d s , if Au ≠ u for u ∈ ∂ P ρ 1 , then i ( A , P ρ 1 , P ) = 1 .

Proof. Suppose u∈∂Pρ1 , by Lemma 2.2, we get

A u = max 0 ≤ t ≤ 1 A u ( t ) - min 0 ≤ t ≤ 1 ( A u ( t ) ) ″ = max 0 ≤ t ≤ 1 ∫ 0 1 H ( t , s ) f ( t , u ( t ) , u ″ ( t ) ) d s + max 0 ≤ t ≤ 1 ∫ 0 1 H ( t , s ) f ( t , u ( t ) , u ″ ( t ) ) d s ″ ≤ d 1 ρ 1 max 0 ≤ t ≤ 1 ∫ 0 1 H ( t , s ) d s + max 0 ≤ t ≤ 1 ∫ 0 1 H ( t , s ) d s ″ ≤ ρ 1 .

That is Au ≠ λu for each u∈∂Pρ1 , λ > 1. If Au ≠ u for u∈∂Pρ1 , by Lemma 2.5, then i(A,Pρ1,P)=1 .

Lemma 3.4. Suppose there is ρ ₂ > 0 such that

f ( t , u , v ) ≥ d 2 ρ 2 ,

for u ∈ [0, _ρ2] and v ∈ [- _ρ2, 0] a.e. t ∈ [0, 1], where d 2 = 1 min t ∈ [ δ , 1 - δ ] ∫ 0 1 H ( t , s ) d s - max t ∈ [ δ , 1 - δ ] ∫ 0 1 H ( t , s ) d s ″ . . If Au ≠ u for u ∈ ∂ P ρ 2 , then i ( A , P ρ 2 , P ) = 0 .

Proof. For u∈∂Pρ2 , t ∈ [δ, 1 - δ], by Lemma 2.2, we get

A u + A u ″ = ∫ 0 1 H ( t , s ) f ( t , u ( t ) , u ″ ( t ) ) d s + ∫ 0 1 H ( t , s ) f ( t , u ( t ) , u ″ ( t ) ) d s ″ ≥ d 2 ρ 2 ∫ 0 1 H ( t , s ) d s + ∫ 0 1 H ( t , s ) d s ″ ≥ ρ 2 .

This implies that u ≠ Au + λφ for each u∈∂Pρ2 , λ > 0, where φ ∈ P\{0} is the eigenfunction of L corresponding to r(L). Suppose u ≠ Au for u∈∂Pρ2 , by Lemma 2.5, then i(A,Pρ2,P)=0 .

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 μ < f 0 b ≤ ∞ , then there exists 0 < ρ < min{r, d} such that either there exists u ∈ ∂P_ρthat 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.

The authors declare that they have no competing interests.

The authors declare that the study was realized in collaboration with same responsibility. All authors read and approved the final manuscript.