In this article, we study the existence and multiplicity of positive solution for the following nonlinear sixth-order boundary value problem (BVP for short) with three variable parameters

- u ( 6 ) - C ( t ) u ( 4 ) + B ( t ) u ″ - A ( t ) u = f ( t , u ) , t ∈ ( 0 , 1 ) , u ( 0 ) = u ( 1 ) = u ″ ( 0 ) = u ″ ( 1 ) = u ( 4 ) ( 0 ) = u ( 4 ) ( 1 ) = 0 ,

where A(t), B(t), C(t) ∈ C[0,1], f(t, u) : [0,1] × [0, ∞) → [0. ∞) is continuous.

In recent years, BVPs for sixth-order ordinary differential equations have been studied extensively, see 1 2 3 4 5 6 7 and the references therein. For example, Tersian and Chaparova 1 have studied the existence of positive solutions for the following systems (1.2):

u ( 6 ) + A u ( 4 ) + B u ″ + C u - f ( t , u ) = 0 . 0 < x < L , u ( 0 ) = u ( L ) = u ″ ( 0 ) = u ″ ( L ) = u ( 4 ) ( 0 ) = u ( 4 ) ( L ) = 0 ,

where A, B, and C are some given real constants and f(x, u) is a continuous function on R ², is motivated by the study for stationary solutions of the sixth-order parabolic differential equations

∂ u ∂ t = ∂ 6 u ∂ x 6 + A ∂ 4 u ∂ x 4 + B ∂ 2 u ∂ x 2 + f ( x , u ) .

This equation arose in the formation of the spatial periodic patterns in bistable systems and is also a model for describing the behaviour of phase fronts in materials that are undergoing a transition between the liquid and solid state. When f(x, u) = u - u ³, it was studied by Gardner and Jones 2 as well as by Caginalp and Fife 3 . In 1 , existence of nontrivial solutions for (1.2) is proved using a minimization theorem and a multiplicity result using Clarks theorem when C = 1 and f(x, u) = u ³. The authors have studied also the homoclinic solutions for (1.2) when C = -1 and f(x, u) = -a(x)u|u|^σ , where a(x) is a positive periodic function and σ is a positive constant by the mountain-pass theorem of Brezis-Nirenberg and concentration-compactness arguments. In 4 , by variational tools, including two Brezis-Nirenbergs linking theorems, Gyulov et al. have studied the existence and multiplicity of nontrivial solutions of BVP (1.2).

Recently, in 5 , the existence and multiplicity of positive solutions of sixth-order BVP with three parameters

- u ( 6 ) - γ u ( 4 ) + β u ″ - α u = f ( t , u ) , t ∈ [ 0 , 1 ] , u ( i ) ( 0 ) = u ( i ) ( 1 ) = 0 , i = 0 , 1 , 2 , 3 , 4 , 5

has been studied under the hypothesis of

(A ₁) f : [0,1] × [0, ∞) → [0. ∞) is continuous.

(A ₂) α, β, γ ∈ R and under the condition of satisfying

α π 6 + β π 4 + γ π 2 < 1 , 3 π 4 - 2 γ π 2 - β > 0 , γ < 3 π 2 , 18 α β γ - β 2 γ 2 + 4 α γ 3 + 27 α 2 - 4 β 3 ≤ 0 ,

the existence and multiplicity for positive solution of BVP (1.3) are established by using fixed point index theory. In this article, we consider more general BVP (1.1), based upon spectral theory of operators and fixed point theorem in cone, we will establish the existence and multiplicity positive solution of BVP (1.1) and extend the result of 5 under appropriate conditions. Our ideas mainly come from 5 8 9 10 .

We list the following conditions for convenience:

(H ₁) f : [0,1] × [0, +∞) → [0. +∞) is continuous.

(H ₂) A(t), B(t), C(t) ∈ C[0,1], α = min_0≤t≤1 A(t), β = min_0≤t≤1 B(t), γ = min_0≤t≤1 C(t), and satisfies

α π 6 + β π 4 + γ π 2 < 1 , 3 π 4 - 2 γ π 2 - β > 0 , γ < 3 π 2 , 18 α β γ - β 2 γ 2 + 4 α γ 3 + 27 α 2 - 4 β 3 ≤ 0 .

Let Y = C[0,1], Y ₊ = {u ∈ Y : u(t) ≥ 0, t ∈ [0,1]}. It is well known that Y is a Banach space equipped with the norm ||u||₀ = sup_0≤t≤1 |u(t)|, u ∈ Y. Set X = { u ∈ C ⁴[0,1] : u(0) = u(1) = u''(0) = u''(1) = 0}, then X also is a Banach space equipped with the norm ||u|| _X = max {||u(t)||₀, ||u"(t)||₀, ||u ⁽⁴⁾(t)||₀}. If u ∈ C ⁴[0,1] ∩ C ⁶(0,1) fulfils BVP (1.1), then we call u is a solution of BVP (1.1). If u is a solution of BVP (1.1), and u(t) > 0, t ∈ (0, 1), then we say u is a positive solution of BVP (1.1).

In this section, we will make some preliminaries which are needed to show our main results.

Lemma 2.1. Let u ∈ X, then ||u||₀ ≤ ||u"||₀ ≤ ||u ⁽⁴⁾||₀ ≤ ||u|| _X .

Proof. The proof is similar to the Lemma 1 in 8 , so we omit it. □

Lemma 2.2. 5 Let λ ₁, λ ₂, and λ ₃ be the roots of the polynomial P (λ) = λ³ + γλ ² - βλ + α. Suppose that condition (H ₂) holds, then λ ₁, λ ₂, and λ ₃ are real and greater than -π ².

Note : Based on Lemma 2.2, it is easy to learn that when the three parameters satisfy the condition of (H ₂), they satisfy the condition of non-resonance.

Let G_i (t, s)(i = 1, 2, 3) be the Green's function of the linear BVP

-u"(t) + λ_iu(t) = 0, u(0) = u(1) = 0,

Lemma 2.3. 10 G_i (t, s)(i = 1, 2, 3) has the following properties

(c1) G_i (t, s) > 0, ∀t, s ∈ (0, 1).

(c2) G_i (t, s) <C_iG_i (s, s), ∀t, s ∈ [0,1], in which C_i > 0 is constant.

(c3) G_i (t, s) ≥ δ_iG_i (t, t)G_i (s, s), ∀t, s ∈ [0,1], in which δ_i > 0 is constant.

We set

M i = max 0 ≤ t ≤ 1 G i ( s , s ) , m i = min 1 4 ≤ t ≤ 3 4 G i ( s , s ) , i = 1 , 2 , 3 .

C i j = ∫ 0 1 G i ( δ , δ ) G j ( δ , δ ) d δ , c i j = ∫ 1 4 3 4 G i ( δ , δ ) G j ( δ , δ ) d δ , i , j = 1 , 2 , 3 .

D i = max 0 ≤ t ≤ 1 ∫ 0 1 G i ( t , s ) d s , d i = max 1 4 ≤ t ≤ 3 4 ∫ 1 4 3 4 G i ( t , s ) d s i = 1 , 2 , 3 ,

then starting from Lemma 2.3 we know M_i , m_i , C_ij > 0.

For any h ∈ Y, take into consideration of linear BVP:

- u ( 6 ) - γ u ( 4 ) + β u ″ - α u = h ( t ) , t ∈ [ 0 , 1 ] , u ( 0 ) = u ( 1 ) = u ″ ( 0 ) = u ″ ( 1 ) = u ( 4 ) ( 0 ) = u ( 4 ) ( 1 ) = 0 ,

where α, β, γ satisfy assumption (H ₂). Since

- u ( 6 ) - γ u ( 4 ) + β u ″ - α u = - d 2 d t 2 + λ 1 - d 2 d t 2 + λ 2 - d 2 d t 2 + λ 3 u ,

then for any h ∈ Y, the LBVP(2.4) has a unique solution u, which we denoted by Ah = u. The operator A can be expressed by

u ( t ) = ( A h ) ( t ) : = ∫ 0 1 ∫ 0 1 ∫ 0 1 G 1 ( t , δ ) G 2 ( δ , τ ) G 3 ( τ , s ) h ( t ) d s d τ d δ .

Lemma 2.4. The linear operator A : Y → X is completely continuous and ||A|| ≤ ϖ, where ϖ = |λ ₂ +λ ₃ |(C ₁ C ₂ C ₃ M ₁ M ₂ M ₃|λ ₃|+C ₁ C ₂ M ₁ M ₂)+| λ ₂ λ ₃|(C ₁ C ₂ C ₃ M ₁ M ₂ M ₃+C ₁ M ₁).

Proof. It is easy to show that the operator A : Y → X is linear operator. ∀h ∈ Y, u = Ah ∈ X, u(0) = u(1) = u"(0) = u"(1) = u ⁽⁴⁾(0) = u ⁽⁴⁾(1) = 0. Let v = - d 2 d t 2 + λ 2 - d 2 d t 2 + λ 3 u , that is

v = - d 2 d t 2 + λ 2 - d 2 d t 2 + λ 3 u = u ( 4 ) - ( λ 2 + λ 3 ) u ″ + λ 2 λ 3 u ,

by (2.5) and (2.7), we have

- v ″ + λ 1 v = h ( t ) , t ∈ ( 0 , 1 ) , v ( 0 ) = v ( 1 ) = 0 ,

and v ( t ) = ∫ 0 1 G 1 ( t , s ) h ( s ) d s , t ∈ [ 0 , 1 ] , so

u ( 4 ) - ( λ 2 + λ 3 ) u ″ + λ 2 λ 3 u = ∫ 0 1 G 1 ( t , s ) h ( s ) d s , t ∈ [ 0 , 1 ] .

By (2.6), for any t ∈ [0,1], we have

| u ( t ) | ≤ ∫ 0 1 ∫ 0 1 ∫ 0 1 G 1 ( t , δ ) G 2 ( δ , τ ) G 3 ( τ , s ) | h ( t ) | d s d τ d δ ≤ C 1 C 2 C 3 M 1 M 2 M 3 | | h | | 0 .

Again, let ω = -u" + λ ₃ u, then ω(0) = ω(1) = ω"(0) = ω"(1) = 0, by (2,5), we have

ω ( 4 ) - ( λ 1 + λ 2 ) ω ″ + λ 1 λ 2 ω = h ( t ) , t ∈ ( 0 , 1 ) , ω ( 0 ) = ω ( 1 ) = ω ″ ( 0 ) = ω ″ ( 1 ) = 0 .

Then ω ( t ) = ∫ 0 1 ∫ 0 1 G 1 ( t , τ ) G 2 ( τ , s ) h ( s ) d s d τ ,   t ∈ [ 0 , 1 ] , that is

- u ″ + λ 3 u = ∫ 0 1 ∫ 0 1 G 1 ( t , τ ) G 2 ( τ , s ) h ( s ) d s d τ , t ∈ [ 0 , 1 ] .

So

| u ″ | ≤ C 1 C 2 M 1 M 2 ( 1 + | λ 3 | C 3 M 3 ) | | h | | 0 , t ∈ [ 0 , 1 ] .

Based on (2.8), (2.9), and (2.12), we have

| u ( 4 ) ( t ) | ≤ | λ 2 + λ 3 | | u ″ ( t ) | + | λ 2 λ 3 | | u ( t ) | + ∫ 0 1 G 1 ( t , s ) | h ( s ) | d s ≤ | λ 2 + λ 3 | ( C 1 C 2 C 3 M 1 M 2 M 3 | λ 3 | + C 1 C 2 M 1 M 2 ) | | h | | 0 + | λ 2 λ 3 | ( C 1 C 2 C 3 M 1 M 2 M 3 | | h | | 0 + C 1 M 1 ) | | h | | 0 ≤ ϖ | | h | | 0 , t ∈ [ 0 , 1 ] ,

where

ϖ = | λ 2 + λ 3 | ( C 1 C 2 C 3 M 1 M 2 M 3 | λ 3 | + C 1 C 2 M 1 M 2 ) + | λ 2 λ 3 | ( C 1 C 2 C 3 M 1 M 2 M 3 + C 1 M 1 ) .

So, ||u ⁽⁴⁾(t)|| ≤ ϖ||h||₀, by Lemma2.1, ||u|| _X ≤ ϖ||h||₀, then

| | A h | | X ≤ ϖ | | h | | 0 ,

so A is continuous, and ||A|| ≤ ϖ.

Next, we will show that A is compact with respect to the norm ||·|| _X on X.

Suppose {h_n }(n = 1, 2, . . .) an arbitrary bounded sequence in Y, then there exists K ₀ >0 such that ||h_n ||₀ ≤ K ₀, n = 1, 2, . . . . Let u_n = Ah_n , 1, 2, ...By (2.8), ∀t ₁, t ₂ ∈ [0, 1], t ₁ < t ₂, we have

u n ( 4 ) ( t 2 ) - u n ( 4 ) ( t 1 ) ≤ | λ 2 + λ 3 | | u ″ n ( t 2 ) - u ″ n ( t 1 ) | + | λ 2 λ 3 | | u n ( t 2 ) - u n ( t 1 ) | + ∫ 0 1 | G 1 ( t 2 , s ) - G 1 ( t 1 , s ) | | h n ( s ) | d s . ≤ | λ 2 + λ 3 | | λ 3 | | u n ( t 2 ) - u n ( t 1 ) | + ∫ 0 1 ∫ 0 1 | G 1 ( t 2 , τ ) - G 1 ( t 1 , τ ) | | G 2 ( τ , s ) | | h n ( s ) | d s d τ + | λ 2 λ 3 | | u n ( t 2 ) - u n ( t 1 ) | + ∫ 0 1 | G 1 ( t 2 , s ) - G 1 ( t 1 , s ) | | h n ( s ) | d s . ≤ ( λ 3 2 + 2 | λ 2 λ 3 | ) ∫ 0 1 ∫ 0 1 ∫ 0 1 | G 1 ( t 2 , δ ) - G 1 ( t 1 , δ ) | | G 2 ( δ , τ ) | | G 3 ( τ , s ) | | h n ( s ) | d s d τ d δ . + | λ 2 + λ 3 | ∫ 0 1 ∫ 0 1 | G 1 ( t 2 , τ ) - G 1 ( t 1 , τ ) | | G 2 ( τ , s ) | | h n ( s ) | d s d τ + ∫ 0 1 | G 1 ( t 2 , s ) - G 1 ( t 1 , s ) | | h n ( s ) | d s . ≤ ( λ 3 2 + 2 | λ 2 λ 3 | ) ∫ 0 1 ∫ 0 1 ∫ 0 1 | G 1 ( t 2 , δ ) - G 1 ( t 1 , δ ) | G 2 ( δ , τ ) G 3 ( τ , s ) d s d τ d δ . + | λ 2 + λ 3 | ∫ 0 1 ∫ 0 1 | G 1 ( t 2 , τ ) - G 1 ( t 1 , τ ) G 2 ( τ , s ) d s d τ + ∫ 0 1 | G 1 ( t 2 , s ) - G 1 ( t 1 , s ) | d s K 0 .

Because G_i (t, s)(i = 1, 2, 3) is uniform continuity on [0,1] × [0,1], based on the above demonstration, it is easy to proof that u n ( 4 ) n = 1 ∞ is equicontinuous on [0,1]. From (2.15), we know ||u||₀, ||u"||₀, ||u ⁽⁴⁾||₀ ≤ ||u|| _X ≤ ϖ||h_n ||₀ ≤ ϖK ₀, so u n ( t ) , { u n ″ ( t ) } and u n ( 4 ) ( t ) are relatively compact in R. Based on Lemma 1.2.7 in 11 , we know { u n } n = 1 ∞ is the relatively compact in X, so A is compact operator. □

The main tools of this article are the following well-known fixed point index theorems.

Let E be a Banach Space and K ⊂ E be a closed convex cone in E. Assume that Ω is a bounded open subset of E with boundary ∂Ω, and K ∩ Ω ≠ ∅. Let A : K ∩ Ω ̄ → K be a completely continuous mapping. If Au ≠u for every u ∈ K ∩ ∂Ω, then the fixed point index i(A, K ∩ Ω, K) is well defined. We have that if i(A, K ∩ Ω, K) ≠0, then A has a fixed point in K ∩ Ω.

Let K_r = {u ∈ K |||u|| <r} and ∂K_r = {u ∈ K |||u|| <r} for every r >0.

Lemma 2.5. 12 Let A : K → K be a completely continuous mapping. If μAu ≠u for every u ∈ ∂K_r and 0 < μ ≤ 1, then i(A, K_r , K) = 1.

Lemma 2.6. 12 Let A : K → K be a completely continuous mapping. Suppose that the following two conditions are satisfied:

(i) inf u ∈ ∂ K r | | A u | | > 0 ,

(ii) μAu ≠u for every u ∈ ∂K_r and μ ≥ 1,

then i(A, K_r , K) = 0.

Lemma 2.7. 12 Let X be a Banach space, and let K ⊆ X be a cone in X. For p >0, define K p = { u ∈ K | | | u | | < p } . Assume that A : K_p → K is a completely continuous mapping such that Au ≠u for every u ∈ ∂K_p = {u ∈ K|||u|| = p}.

(i) If ||u|| ≤ ||Au||, for every u ∈ ∂K_p , then i(A, K_p , K) = 0.

(ii) If ||u|| ≥ ||Au||, for every u ∈ ∂K_p , then i(A, K_p , K) = 1.

We bring in following notations in this section:

f 0 = lim u → 0 + inf min 0 ≤ t ≤ 1 ( f ( t , u ) / u ) , f ̄ ∞ = lim u → + ∞ sup max 0 ≤ t ≤ 1 ( f ( t , u ) / u ) , f ̄ 0 = lim u → 0 + sup max 0 ≤ t ≤ 1 ( f ( t , u ) / u ) , f ∞ = lim u → + ∞ inf min 0 ≤ t ≤ 1 ( f ( t , u ) / u ) . a ( t ) = A ( t ) - α , b ( t ) = B ( t ) - β , c ( t ) = C ( t ) - γ , Γ = π 6 - γ π 4 - β π 2 - α , K = max 0 ≤ t ≤ 1 [ a ( t ) + b ( t ) + c ( t ) ] ,

Suppose that:

(H ₃) L = ϖK <1, where ϖ is defined as in (2.14).

Theorem 3.1. Assume that (H ₁)-(H ₃) hold, and b ( t ) ≥ ( λ 2 + λ 3 ) c ( t ) , λ 3 b ( t ) - a ( t ) ≤ λ 3 2 c ( t ) , then in each of the following cases:

(i) f 0 > Γ , f ̄ ∞ < ( 1 - L ) Γ , (ii) f ̄ 0 < ( 1 - L ) Γ , f ∞ > Γ , the BVP (1.1) has at least one positive solution.

Proof. ∀h ∈ Y, consider the LBVP

- u ( 6 ) - C ( t ) u ( 4 ) + B ( t ) u ″ - A ( t ) u = h ( t ) , u ( 0 ) = u ( 1 ) = u ″ ( 0 ) = u ″ ( 1 ) = u ( 4 ) ( 0 ) = u ( 4 ) ( 1 ) = 0 , 0 < t < 1

It is easy to prove (3.1) is equivalent to the following BVP

- u ( 6 ) - γ u ( 4 ) + β u ″ - α = G u + h ( t ) , u ( 0 ) = u ( 1 ) = u ″ ( 0 ) = u ″ ( 1 ) = u ( 4 ) ( 0 ) = u ( 4 ) ( 1 ) = 0 , 0 < t < 1

where Gv := (C(t) - γ)v ⁽⁴⁾ - (B(t)- β)v" + (A(t) - α)v, ∀v ∈ X. Obviously, the operator G : X → Y is linear, and ∀v ∈ X, t ∈ [0,1], we have |Gv(t)| ≤ K ||v|| _X . Hence ||Gv||₀ ≤ K ||v|| _X , and so ||G|| ≤ K. On the other hand, u ∈ C ⁴[0,1]⋂C ⁶(0,1), t ∈ [0,1] is a solution of (3.2) iff u ∈ X satisfies u = A(Gu + h), i.e.,

u ∈ X , ( I - A G ) u = A h .

Owing to G : X → Y and A : Y → X, the operator I - AG maps X into Y. From A ≤ ϖ (by Lemma 2.4) together with ||G|| ≤ K and condition (H ₃), applying operator spectral theorem, we have that the operator (I - AG)^-1 exists and is bounded. Let H = (I - AG)^-1 A, then (3.3) is equivalent to u = Hh. By the Neumann expansion formula, H can be expressed by

H = ( I + A G + ⋯ + ( A G ) n + ⋯ ) A = A + ( A G ) A + ⋯ + ( A G ) n A + ⋯ .  

The complete continuity of A with the continuity of (I - AG)^-1 yields that the operator H : Y → X is completely continuous. If we restrict H : Y ₊ → Y, ∀h ∈ Y ₊ and mark u = Ah, then u ∈ X ∩ Y ₊. Based on equation (2.8), (2.11) and Lemma 2.4, we have

u ″ = λ 3 u - ∫ 0 1 ∫ 0 1 G 1 ( t , τ ) G 2 ( τ , s ) h ( s ) d s d τ ≤ λ 3 u , t ∈ [ 0 , 1 ] , u ( 4 ) = ( λ 2 + λ 3 ) u ″ - λ 2 λ 3 u + ∫ 0 1 G 1 ( t , s ) h ( s ) d s ≥ ( λ 2 + λ 3 ) u ″ - λ 2 λ 3 u , t ∈ [ 0 , 1 ] ,

by b(t) ≥ (λ ₂ + λ ₃)c(t) and λ 3 b ( t ) - a ( t ) ≤ λ 3 2 c ( t ) , we have

( G u ) ( t ) = c ( t ) u ( 4 ) − b ( t ) u ' ' + a ( t ) u ≥ [ ( λ 2 + λ 3 ) c ( t ) − b ( t ) ] u ' ' − [ λ 2 λ 3 c ( t ) − a ( t ) ] u ≥ λ 3 [ ( λ 2 + λ 3 ) c ( t ) − b ( t ) ] u − [ λ 2 λ 3 c ( t ) − a ( t ) ] u ≥ [ λ 3 2 c ( t ) − λ 3 b ( t ) + a ( t ) ] u ≥ 0 ,   t ∈ [ 0 , 1 ] .

Hence

∀ h ∈ Y + , ( G A h ) ( t ) ≥ 0 , ∀ t ∈ 0 ,  1 ,

and so (AG)(Ah)(t) = A(GAh)(t) ≥ 0, ∀t ∈ [0,1]. Suppose that ∀h ∈ Y ₊, (AG) ^k (Ah)(t) ≥ 0, ∀t ∈ [0,1]. For any h ∈ Y ₊, let h ₁ = GAh, by (3.5) we have h ₁ ∈ Y ₊, and so

( A G ) k + 1 ( A h ) ( t ) = ( A G ) k ( A G A h ) ( t ) = ( A G ) k ( A h 1 ) ( t ) ≥ 0 , ∀ t ∈ [ 0 , 1 ] .

Thus by induction it follows that ∀n ≥ 1, ∀h ∈ Y ₊, (AG) ⁿ (Ah)(t) ≥ 0, ∀t ∈ [0,1]. By (3.4), we have

∀ h ∈ Y + , ( H h ) ( t ) = ( A h ) ( t ) + ( A G ) ( A h ) ( t ) + ⋯ + ( A G ) n ( A h ) ( t ) + ⋯ ≥ ( A h ) ( t ) , ∀ t ∈ [ 0 , 1 ] .

So H : Y ₊ → Y ₊ ∩ X.

On the other hand, we have

∀ h ∈ Y + , ( H h ) ( t ) ≤ ( A h ) ( t ) + | | ( A G ) | | ( A h ) ( t ) + ⋯ + | | ( A G ) n | | ( A h ) ( t ) + ⋯ ≤ ( 1 + L + ⋯ + L n + ⋯ ) ( A h ) ( t ) ≤ 1 1 - L ( A h ) ( t ) , ∀ t ∈ [ 0 , 1 ] .

So the following inequalities hold

| | H h | | 0 ≤ 1 1 - L | | A h | | 0 , ∀ t ∈ [ 0 , 1 ] .

For any u ∈ Y ₊, define Fu = f(t, u). Based on condition (H ₁), it is easy to show F : Y ₊ → Y ₊ is continuous. By (3.1)-(3.3), It is easy to see that u ∈ C ⁴[0,1] ∩ C ⁶(0, 1) is a positive solution of BVP (1.1) iff u ∈ Y ₊ is a nonzero solution of an operator equation as follows

u = H F u .

Let Q = HF. Obviously, Q : Y ₊ → Y ₊ is completely continuous. We next show that the operator Q has at least one nonzero fixed point in Y ₊.

Let

P = { u ∈ Y + | u ( t ) ≥ σ | | u | | 0 , ∀ t ∈ [ 0 , 1 ] } .

In which

σ = δ 1 δ 2 δ 3 C 12 C 23 C 1 C 2 C 3 M 1 M 2 ( 1 - L ) G 1 ( t , t ) .

Here M ₁ and M ₂ can be defined as that in (2.1), C ₁₂ and C ₂₃ can be defined as that in (2.2), C_i ,δ_i (i = 1, 2, 3) can be defined as that in Lemma 2.3. It is easy to prove that P is a cone in Y. We will prove QP ⊂ P next.

For any u ∈ P, let h = Fu, then h ∈ Y ₊. By (3.6) and Lemma 2.3, we have

( Q u ) ( t ) = ( H F u ) ( t ) ≥ ( A F u ) ( t ) , ∀ t ∈ [ 0 , 1 ] } .

By Lemma 2.3, for all u ∈ P, we have

( A F u ) ( t ) = ∫ 0 1 ∫ 0 1 ∫ 0 1 G 1 ( t , δ ) G 2 ( δ , τ ) G 3 ( τ , s ) ( F u ) ( s ) d s d τ d δ ≤ C 1 C 2 C 3 M 1 M 2 ∫ 0 1 G 3 ( s , s ) ( F u ) ( s ) d s .

And accordingly we have | | A F u | | 0 ≤ C 1 C 2 C 3 M 1 M 2 ∫ 0 1 G 3 ( s , s ) ( F u ) ( s ) d s , that is

∫ 0 1 G 3 ( s , s ) ( F u ) ( s ) d s ≥ | | A F u | | 0 C 1 C 2 C 3 M 1 M 2 .

By using (c3) in Lemma 2.3, (3.8) and (3.11), we have

( A F u ) ( t ) ≥ δ 1 δ 2 δ 3 C 12 C 23 G 1 ( t , t ) ∫ 0 1 G 3 ( s , s ) ( F u ) ( s ) d s ≥ δ 1 δ 2 δ 3 C 12 C 23 G 1 ( t , t ) C 1 C 2 C 3 M 1 M 2 | | A F u | | 0 ≥ δ 1 δ 2 δ 3 C 12 C 23 G 1 ( t , t ) C 1 C 2 C 3 M 1 M 2 ( 1 - L ) | | H F u | | 0 ≥ δ 1 δ 2 δ 3 C 12 C 23 G 1 ( t , t ) C 1 C 2 C 3 M 1 M 2 ( 1 - L ) | | Q u | | 0 .

So ( Q u ) ( t ) ≥ δ 1 δ 2 δ 3 C 12 C 23 G 1 ( t , t ) C 1 C 2 C 3 M 1 M 2 ( 1 - L ) | | Q u | | 0 = σ | | Q u | | 0 . Thus QP ⊂ P.

Let

ρ = δ 1 δ 2 δ 3 C 12 C 23 m 1 ( 1 - L ) C 1 C 2 C 3 M 1 M 2 ,

in which m ₁ can be defined as that in (2.1). It's easy to prove

∀ u ∈ P ⇒ u ( t ) ≥ ρ | | u | | 0 , ∀ t ∈ 1 4 , 3 4 .

Case (i), since f 0 > Γ , there exist ε >0 and r ₀ >0 such that f(t, x) ≥ (Γ + ε)x, 0 ≤ t ≤ 1, 0 < × ≤ r ₀. Let r ∈ (0, r ₀) and Ω _r = {u ∈ P | ||u||₀ ≤ r}, then for every u ∈ ∂Ω _r , we have ||u||₀ = r, 0 < u(t) ≤ r, t ∈ (0, 1), and so f(t, u(t)) ≥ (Γ + ε)u(t), t ∈ (0,1). By (3.13), it follows that

f ( t , u ( t ) ) > ( Γ + ε ) u ( t ) ≥ ( Γ + ε ) ρ r , ∀ t ∈ 1 4 , 3 4 .

From (3.6) and (3.14), we have

| | Q u | | 0 ≥ Q u 1 2 = ( H F u ) 1 2 ≥ ( A F u ) 1 2 = ∫ 0 1 ∫ 0 1 ∫ 0 1 G 1 1 2 , δ G 2 ( δ , τ ) G 3 ( τ , s ) f ( s , u ( s ) ) d s d τ d δ ≥ ∫ 1 4 3 4 ∫ 1 4 3 4 ∫ 1 4 3 4 G 1 1 2 , δ G 2 ( δ , τ ) G 3 ( τ , s ) ( Γ + ε ) ρ r d s d τ d δ ≥ δ 1 δ 2 δ 3 m 1 C 12 C 23 ( Γ + ε ) ρ r ∫ 1 4 3 4 G 3 ( s , s ) d s . ≥ 1 2 δ 1 δ 2 δ 3 m 1 m 3 C 12 C 23 ( Γ + ε ) ρ r > 0 .