Positive Solutions of Operator Equations and Nonlinear Beam Equations with a Perturbed Loading Force

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1 Positive Solutions of Operator Equations Nonlinear Beam Equations with a Perturbed Loading Force WEN-XIA WANG Taiyuan Normal University Department of Mathematics Taiyuan 312 P. R. China wwxgg@126.com XI-LAN LIU Shanxi Datong University Department of mathematics Computational Science Datong 37 P. R. China doclanliu22@yahoo.com.cn Abstract: In this paper we are concerned with the existence uniqueness of positive solutions for an operator equation x = Ax + λbx on an order Banach space, where A B are nonlinear operators λ is a parameter. By properties of cones we obtain that there exists a λ > such that the operator equation has a unique positive solution which is increasing in λ for λ [, λ ], further, we give an estimate for λ. In addition, we discuss the existence uniqueness of positive solutions for an elastic beam equation with three parameters one perturbed loading force. Key Words: Nonlinear operator equation; positive solution; elastic beam equation; perturbed loading force 1 Introduction Preliminaries It is well known that nonlinear operator equations defined on a cone in Banach spaces play an important role in theory of nonlinear differential integral equations has been extensively studied over the past several decades (see [1]-[14]. In this paper, we consider the operator equation on a Banach space E x = λax + Bx, (1 where A is an increasing convexity operator, B is a increasing concavity operator λ is a parameter. Many nonlinear problems with a parameter, such as initial value problems, boundary value problems, impulsive problems, can be transformed into Eq.(1, which shows the importance to study the operator equation (1 both in theory applications. There are many recent discussions to positive solutions of operator equations. For example [1], [2], [3]-[9] [1]-[12] investigated operator equations Ax = λx (λ >, Ax = x Ax + Bx = x, respectively, which are special forms of the Eq.(1. To our knowledge, little has been done on the Eq.(1 in literature, especially on the solution s dependence on the parameter λ, thus it is worthwhile doing this work. By properties of cones, we study the existence uniqueness of the positive solutions for the operator equation (1. Moreover we find the value λ such that the operator equation has a unique positive solution for λ [, λ ], on the other h, we discuss the elastic beam equation with three parameters a perturbed loading force, obtain the concrete interval I such that the problem has a unique positive solution for the parameter λ I. It may be the first time that the simply supported beam equation with three parameters one perturbed loading force is studied. Let E be a real Banach space which is partially ordered by a cone P of E, i.e., x y if y x P. By θ we denote the zero element of E. Recall that a nonempty closed convex set P E is a cone if it satisfies x P, r rx P ; x P, x P x = θ. Recall that a cone P is said to be normal if there exists a positive number N, called the normal constant of P, such that θ x y implies x N y. For a given e > θ, that is, e θ e θ, let P e = {x E there exist τ 1 (x > τ 2 (x > such that τ 1 (xe x τ 2 (xe}. (2 Then it is easy to see that (a P e P ; (b for any given x, y P e, there exist < τ 1 1 τ 2 < such that τ 1 y x τ 2 y. Let D E P be a cone of E. An operator T : D E is said to be increasing if for x 1, x 2 D, with x 1 x 2 we have T x 1 T x 2. An element x D is called a fixed point of T if T x = x. E-ISSN: Issue 3, Volume 11, March 212

2 All the concepts discussed above can be found in [13]. Lemma 1. Suppose that P is a normal cone of E T : P P be an increasing operator. Assume that (L1 there exist y, z P e with y z such that y T y, T z z ; (L2 for any t (, 1, there exists η(t > such that T (tx t(1 + η(tt x, x [y, z ]. Then the following statements hold (a T has a unique fixed point x [y, z ]; (b T has not any fixed point in P e \ [y, z ]; (c for any u P e, the sequence {u n, n 1} generated by u n = T u n 1 has limit x, i.e., lim u n x =. Proof. Set y n = T y n 1 z n = T z n 1 for n = 1, 2,. The condition (L1 the fact that T is increasing yield to Let y y 1 y n z n z 1 z. (3 µ n = sup{τ > y n τz n }, n = 1, 2,. (4 In view of the property (b of P e we get < µ n 1, y n µ n z n, n = 1, 2,. (5 From (3 (5 we infer that < µ µ 1 µ n 1, which means that lim µ n = µ 1. We assert that µ = 1. If it is not true, i.e., < µ n µ < 1 for n 1, then by (L2 (3 we deduce that y n+1 = T y n T (µ n z n T ( µ n µ µz n µ ( n µ T (µz n µ n 1 + η(µ z n+1. By (4, we have µ n+1 µ n (1 + η(µ, n =, 1, 2,, µ n+1 µ (1 + η(µ n+1, n =, 1, 2,. This gives rise to the contradiction 1 > µ +. Note that P is normal. By (3 (5 we have z n y n N(1 µ n z as n, which implies that both {y n } {z n } are Cauchy sequences, there exist y, z P such that y n y, z n z y n y z z n. Thus y = z [y, z ]. Since T is increasing, then y n y n+1 T y = T z z n+1 z n. Therefore, we have y T y = T z z, which implies that y is a fixed point of T in [y, z ]. Now, let y is a fixed point of T in [y, z ] µ = sup {τ > y τy }. Then < µ 1 y µy. If µ 1, we have y = T y T ( µy µ(1 + η( µy, which implies that µ µ(1 + η( µ. This is a contradiction. Hence, µ = 1. This means that y y. Similar argument show that y y. Consequently, we have y = y. Next to prove (b. Assume that y is a fixed point of T in P e \ [y, z ]. Let µ = sup {τ > τy y 1τ } y. (6 Then < µ 1. We assert that µ = 1. If < µ < 1, by (L2 we have y = T y T (µy µ(1 + η(µy y T ( 1 µ y 1 µ(1 + η(µ y. Thus, from (6 we have µ µ(1 + η(µ, which is a contradiction. Thus, (6 implies that y = y, which means a contradiction [y, z ] y [y, z ]. This end the proof of the conclusion (b. Note that y [x, y ] is unique fixed point of T in P e T (ty t(1 + η(tt y, t (, 1. the conclusion (c can be proved by similar way to the proof of Theorem 3.4 of [13], here is omitted. The proof is complete. Lemma 2. ([4, 7] Let T : P e P e be an increasing operator. Suppose that (L3 there exists α (, 1 such that T (tx t α T x, x P e, t (, 1. Then T has a unique fixed point x in P e. Moreover, for any u P e, letting u n = T u n 1, n = 1, 2,, one has lim u n x =. E-ISSN: Issue 3, Volume 11, March 212

3 2 Positive solutions for operator e- quation Throughout this section, we assume that E is a real Banach space, P is a normal cone in E with the normal constant N P e is defined by (2, e > θ. In this section, we investigate the existence uniqueness of positive solutions of the operator equation (1, where A is a convexity operator B is a constant operator or an α concave operator. Firstly, we discuss the case of Eq.(1 with B x (x E which can be widely applied to various problems for differential equations. We have the following result. Theorem 3. Let x P e. Suppose that the operator A : P P is increasing satisfies conditions: (H1 Ae > θ there exists l > such that Ae le; (H2 there exists a real number > 1 such that A(tx = t Ax, t (, 1, x P e. Then the following statements are true: (a there exists λ > such that the equation x = x + λax has a unique solution x λ [x, 1 x ] in P e for λ [, λ ]. Moreover, for any u P e, set C λ = x + λa u n = C λ u n 1, n = 1, 2,, then lim u n x λ = ; (b x x λ 1 x for λ [, λ ]; (c x λ is increasing in λ for λ [, λ ]; (d lim x λ x = ; λ x λ x N 1 x, λ [, λ ]. Proof. We prove all statements by five steps. Step 1. Define a mapping ρ : P e [, + by ρ(x = inf{τ > Ax τx }, x P e. (7 By the property (b of P e we get < ρ(x < +. In addition, for any x 1, x 2 P e, x 1 x 2, we have which implies that Ax 1 Ax 2 ρ(x 2 x, ρ(x 1 ρ(x 2, i.e., ρ(x is increasing in x P e. Let C λ = x + λa. It is obvious from (H1 that C λ (P e P e for λ. Set Γ = {λ there exists S > 1 such that C λ (Sx Sx 1 1 λs ρ(x } (8 λ = sup Γ. (9 Take S = Then λ(s = 1 > 1, set λ(s = S 1 S ρ(x. (1 S 1 = ( 1 1 S ρ(x ρ(x S 1 S ρ(x = λ(s >, S > 1. (11 Moreover, for any λ [, λ(s ] we have ( C λ (S x 1 + λ(s S ρ(x x S x, (12 λs ρ(x λ(s S ρ(x = S 1 = 1 1. Therefore, [, λ(s ] Γ. Step 2. Now, we show that Suppose to the contrary that λ = λ(s. (13 λ > λ(s. (14 By the definition of λ, there exists a increasing sequence {λ n } n=1 Γ such that lim λ n = λ. That is, there exists a nonincreasing sequence {S n } n=1 with S n > 1 such that This means that C λn (S n x S n x. S n x x + λ n S nax. (15 Set lim S n = S, then (15 implies that So, S x x + λ S Ax. Ax S 1 λ S x S > 1. Thus, from (7 we have ρ(x S 1 λ (S. E-ISSN: Issue 3, Volume 11, March 212

4 Combining (11 (14 gives λ S 1 S ρ(x λ(s < λ, which is a contradiction. Hence, Γ = [, λ(s ] = [, λ ]. Step 3. Conclusion (a holds. Let z = S x, that is, z = 1 x. Note that (12 x C λ x we obtain that C λ satisfies the condition (L1 in Lemma 1 for λ [, λ ]. Note that (8 implies that 1 t t(1 t 1 > 1, t (, t t(1 t 1 > λs ρ(x, λ [, λ ], t (, 1, For any t (, 1, let where ( 1 t η(t = t λ(1 t 1 S ρ(x q(λ, q(λ = sup{τ > x τc λ z }, λ [, λ ]. Then, η(t >. Hence, for any λ [, λ ] t (, 1, from (H2 we get that C λ (tx = x + λt Ax tc λ x + (1 tx λt(1 t 1 Az ( 1 t = tc λ x + t λ(1 t 1 S t ρ(x t(1 + η(tc λ x, x [x, z ]. Thus, C λ satisfy the condition (L2 in Lemma 1 for λ [, λ ]. Consequently, the conclusion (a follows from Lemma 1. Step 4. Conclusions (b (c hold. From the above proof of the conclusion (a, it is easy to see that the conclusion (b holds. Next we prove (c. Let λ 1, λ 2 [, λ ] with λ 1 λ 2. Noting that C λ1 x λ2 = x + λ 1 Ax λ2 x + λ 2 Ax λ2 = x λ2, we have x C λ1 x C λ1 x λ2 x λ2. Similar to the above proof, we know that C λ1 have a unique fixed point x [x, x λ2 ] in P e, which implies that x λ1 = x x λ2. Step 5. Finally, we prove (d. x For any λ [, λ ], in virtue of the conclusion (a, there exists a unique x λ [x, z ] in P e such that x λ = x + λax λ. Thus, from (7 we have θ x λ x = λax λ λaz λρ(z x, which implies that lim λ x λ x =. By the conclusion (b, we have θ x λ x 1 1 x. This means that x λ x 1 x. The proof is complete. Remark 4. From (1 (13, we can give the expression of λ in Theorem 3 that is, λ = ( 1 1 ρ(x. Corollary 5. Let operator A : P P is increasing satisfies (H1 (H2. Suppose that h P e, < M ( 1 ( κ 1 1, where κ = inf{τ > Ah τh}. Then the operator equation x = Mh + Ax has a unique solution x P e. Moreover, for any u P e, set C = Mh + A u n = Cu n 1, n = 1, 2,, then lim u n x =. Proof. In the proof of Theorem 3, let x = Mh. So, ρ(x = M 1 κ. Just note that λ = λ(s 1 when < M ( 1 ( κ 1 1. Taking x = Mh λ = 1 in Theorem 3 finishes the proof. If B α concavity operator, we can obtain: Theorem 6. Let A, B : P P be increasing operators. Suppose that the operator A satisfies (H1 (H2, the operator B satisfies the following conditions: (H3 B(P e P e ; (H4 there exists a real number α (, 1 such that B(tx = t α Bx, t (, 1, x P e. Then (a there exists λ > such that the operator equation x = λax + Bx has a unique fixed point x λ in P e for λ [, λ ]. Moreover, for any u P e, set C λ = λa + B u n = C λ u n 1, n = 1, 2,, then lim u n x λ = ; (b there exist x, z P e with x z such that x λ [x, z ]; (c x λ is increasing in λ for λ [, λ ]. Proof. By virtue of Lemma 2, B has a unique fixed point x in P e. For any x P e λ from (H1 (H3 we have That is λl(xe + Bx C λ x = λax + Bx Bx. N C λ (P e P e, λ. (16 E-ISSN: Issue 3, Volume 11, March 212

5 Since x is the unique fixed point of the operator B in P e, then x C λ x, λ. (17 The proof of Part (a. Set { R > 1, s.t C λ (Rx Rx Ω = λ 1 α 1 λr ρ(x where ρ(x is defined by (7. Set }. (18 λ = sup Ω. (19 Now, we show that λ > Ω = [, λ ]. Let y = Rx for R > 1, then we have x y. Set λ 1 (R = R Rα R ρ(x. (2 Then, for any λ λ 1 (R, we have Set Then Let C λ y = λr Ax + R α Bx λ 1 (RR ρ(x x + R α x = Rx = y. λ 2 (R = 1 α 1 1 R ρ(x. 1 α 1 = λ 2(RR ρ(x λr ρ(x, λ [, λ 2 (R]. λ(r = min{λ 1 (R, λ 2 (R}. (21 (22 Taking F (R = R R α 1 α 1 for any R > 1, it is easy to check that lim F (R < R 1 + ( F ( α α >, which implies that there exists R (1, ( α α such that F (R =. Note that F (R is increasing, we obtain that λ(r = min{λ 1 (R, λ 2 (R} λ 1 (R, 1 < R < R, = λ 1 (R = λ 2 (R, R = R, λ 2 (S, R > R. (23 Since λ 1 (R is increasing in intervals (1, R λ 2 (R are decreasing in (R, +, then λ(r = λ(r >. Thus, from (21 (22 we have max R>1 C λ (R x R x, 1 α 1 λr ρ(x (24 for any λ [, λ(r ]. Therefore, [, λ(r ] Ω. Now, we show that λ = λ(r. (25 Suppose to the contrary that λ > λ(r. By (19, there exists a increasing sequence {λ n } n=1 Ω with λ n λ(r such that lim λ n = λ. This means that there exists a nonincreasing sequence {R n } n=1 (1, R ] such that C λn (R n x R n x. Moreover, we have R n x λ n A(R n x + B(R n x = λ n R nax + R α nx. Set lim R n = R, then (26 implies that So, R x λ R Ax + R α x. Ax R R α λ R 1 < R < R. This means that ρ(x R R α λ R Combining (23 (25 gives x λ R R α R ρ(x = λ 1(R = λ(r λ(r < λ,. (26 which is a contradiction. Hence, Ω = [, λ ]. Now, let z = R x, then, for any fixed λ Ω, by (17 (24, we know that x C λ x C λ (z z 1 α 1 λr ρ(x. Noting that we have t α t 1 > 1 α, t (, 1, 1 t α 1 1 λ(1 t 1 R ρ(x >, t (, 1. Thus, from (H2 (H4 we get that C λ (tx = tc λ x λt(1 t 1 Ax + t(t α 1 1Bx tc λ x λt(1 t 1 Az + t(t α 1 1Bx ( tc λ x + t t α 1 1 (1 t 1 λr ρ(x t(1 + η(tc λ x, t (, 1, x [x, z ], x E-ISSN: Issue 3, Volume 11, March 212

6 where η(t = ( t α 1 1 λ(1 t 1 R ρ(x q(λ, q(λ = sup{τ > x τc λ z }, λ [, λ ]. The application of Lemma 1 concludes the proof of part (a. From above proof, it is easy to see x x λ z, where x is the unique fixed point of B in P e z = R x, R is the unique solution of F (R = R in (1,. This ends the proof of part (b. Next we prove part (C. Let λ 1, λ 2 [, λ ] with λ 1 λ 2. Noting that C λ1 x λ2 = λ 1 Ax λ2 + Bx λ2 λ 2 Ax λ2 + Bx λ2 = x λ2, we have x C λ1 x R α 1 α 1 C λ1 x λ2 x λ2. From the above proof, we know that C λ1 have a unique fixed point x [x, x λ2 ] in P e, which implies that x λ1 = x x λ2. The proof is complete. Remark 7. From (23 (25, we can obtain the expression of λ in Theorem 6, that is, λ = λ 2 (R = 1 α 1 1 R ρ(x, where R is the unique solution of F (R = R R α 1 α 1 in (1, x is the unique fixed point of B in P e. Theorem 8. Let A, B : P P be increasing operators. Suppose that the operator A satisfies (H1 (H2, the operator B satisfies (H3 (H5 there exists a real number α (, 1 such that B(tx t α Ax, t (, 1, x P e. Then (a there exists an interval I with [, λ(r ] I [, + such that x = λax + Bx has a unique solution x λ in P e for λ I. Moreover, for any u P e, set C λ = λa + B u n = C λ u n 1, n = 1, 2,, then lim u n x λ = ; (b x λ is increasing in λ for λ I; (c there exist x, z P e with x z such that x λ [x, z ] for λ [, λ(r ]. Where λ(r is defined (23 R is the unique solution of F (R = R R α 1 α 1 in (1,. Proof. By virtue of Lemma 2, B has a unique fixed point x in P e. It is easy to see that x C λ x, C λ (P e P e, λ. Similar to the proof of Theorem 6 we obtain [, λ(r ] Ω, where Ω is defined (18. Moreover, x = λax + Bx has a unique solution x λ in P e for λ [, λ(r ]. On the other h, for any λ Ω it is evident that [, λ] Ω x = λax + Bx has a unique solution x λ in P e for λ [, λ]. Thus, the conclusion (a can be proved. The proof of the conclusions (b (c is the same as the proof of Theorem 6. The proof is complete. 3 Positive solutions for beam equation In this section, we apply the results of Section 2 to study the existence uniqueness of positive solutions for the following perturbed elastic beam equations with three parameters u (4 (t + ηu (t ζu(t = λf(t, u(t + φ(t, < t < 1, u( = u(1 = u ( = u (1 =, u (4 (t + ηu (t ζu(t = λf(t, u(t + g(u(t, < t < 1, u( = u(1 = u ( = u (1 =, (27 (28 where ζ, η λ are parameters. It is well-known that the deformation of the equilibrium state an elastic beam, its two ends of which are simply supported, can be described by a boundary value problem for a fourth-order ordinary differential equation [15]. The existence multiplicity of positive solutions for the elastic beam equations without perturbations have been studied extensively, see for example [16]-[27] references therein. However, there are few papers concerned with the uniqueness of positive solutions for the problem (27 the problem (28with three parameters one perturbed loading force in literatures. In this section, we consider the problems for (27 (28, give an example to illustrate the result. In what follows, set E = C[, 1], the Banach s- pace of continuous functions on [, 1] with the norm x = max t [,1] x(t. P = {x C[, 1] x(t, t [, 1]}. It is clear that P is a normal cone of which the normality constant is 1. The following hypotheses are needed in this section. (H6 f C[[, 1] [,, [, + ] is increasing in u [, + for fixed t [, 1] f(t, 1 for t [, 1]. (H7 φ : [, 1] [, + is an integrable function, m = inf φ(t >, M = sup φ(t < +. t [,1] t [,1] E-ISSN: Issue 3, Volume 11, March 212

7 (H8 there exists a constant > 1 such that f(t, ru = r f(t, u, t [, 1], r (, 1, u [, +. (H9 g C[[,, (, + ] is increasing. (H1 there exists a constant α (, 1 such that g(ru r α g(u, r (, 1, u [, +. (H11 ζ, η R η < 2π 2, ζ η2 4, ζ/π4 + η/π 2 < 1. Let γ 1 γ 2 be the roots of the polynomial γ 2 + ηγ ζ, i.e., γ 1, γ 2 = 1 { η ± } η ζ. In view of (H11 it is easy to see that γ 1 γ 2 > π 2. Let G i (t, s (i = 1, 2 be the Green s functions corresponding to the boundary value problems u (t + γ i u(t =, u( = u(1 =. (29 Moreover, G i (t, s = for γ i > ; for γ i = ; G i (t, s = G i (t, s = { sinh νi t sinh ν i (1 s ν i sinh ν i, t s 1 sinh ν i s sinh ν i (1 t ν i sinh ν i, s t 1 { t(1 s, t s 1 s(1 t, s t 1 { sin νi t sin ν i (1 s ν i sin ν i, t s 1 sin ν i s sin ν i (1 t ν i sin ν i, s t 1 for π 2 < γ i <, where ν i = γ i, inf <t,s<1 G i (t, s G i (t, tg i (s, s = δ i >, (3 where δ i = ν i sinh ν i if γ i > ; δ i = 1 if γ i = ; δ i = ν i sin ν i if π 2 < γ i <. For σ 1, σ 2 (, 1 with σ 1 σ 2, let ϵ i = min G i (s, s. σ 1 s σ 2 Then ϵ i > (i = 1, 2. For more information on the Green s function of (29, we refer to [19, 22]. Let e(t = G 1 (t, τg 2 (τ, sdsdτ, t [, 1], then (H11 implies that e P. Moreover, e(t σ 2 σ 1 σ2 σ 1 G 1 (t, τg 2 (τ, sdsdτ (σ 2 σ 1 δ 1 δ 2 ϵ 2 1 ϵ2 2 >, t (, 1. (31 Hence, e > θ. Define P e as (2. Theorem 9. Assume that (H6-(H8 (H11 hold. Then (a there exists λ > such that (27 has a u- nique positive solution u λ (t P e for λ [, λ ]. Moreover, for any u P e, set u n (t = λ G 1(t, τg 2 (τ, sf(s, u n 1 (sdsdτ +x (t, t [, 1], n = 1, 2,, then lim u n u λ = ; (b x (t u λ (t 1 x (t(t [, 1] for λ [, λ ]; (c u λ is increasing in λ for λ [, λ ]; (d lim λ u λ x = ; where x (t = u λ x 1 1 x, λ [, λ ], G 1 (t, τg 2 (τ, sφ(sdsdτ. (32 Proof. It is easy to see that the problem (27 has an integral formulation given by u(t = x (t + λ G 1(t, τg 2 (τ, s f(s, u(sdsdτ, t [, 1]. where x (t is defined by (32. Define operator A : P E by Au(t = G 1 (t, τg 2 (τ, sf(s, u(sdsdτ for t [, 1]. It is easy to prove that u λ is a solution of the problem (27 if only if u λ is a fixed point of the operator x + λa. In virtue of (H6 (H7, we know that x P A : P P is an increasing operator. Further, from (H7 we have x (t m x (t M G 1(t, τg 2 (τ, sdsdτ = me(t, G 1(t, τg 2 (τ, sdsdτ = Me(t. That is, x P e. From (H6, there exist σ 1, σ 2 (, 1 with σ 1 < σ 2 such that inf f(t, 1 >. t [σ 1,σ 2 ] E-ISSN: Issue 3, Volume 11, March 212

8 Moreover, from (H8 (31, we obtain Ae(t Ae(t σ2 σ2 σ 1 σ 1 G 1 (t, τg 2 (τ, sf(s, e(sdsdτ ( (σ 2 σ 1 δ 1 δ 2 ϵ 2 1ϵ 2 2 inf t [σ 1,σ 2 ] σ2 d G 1 (t, τg 2 (τ, sdsdτ σ 1 c >, t (, 1, ( e f(t, 1 G 1 (t, τg 2 (τ, sf(s, e dsdτ sup f(t, 1 e(t. t [,1] That is, A satisfies (H1. For any r (, 1 u P e, by (H8 we obtain A(ru(t = = r G 1 (t, τg 2 (τ, sf(s, ru(sdsdτ = r Au(t, u P e. G 1 (t, τg 2 (τ, sf(s, u(sdsdτ That is, A satisfies (H2. Thus, the results of Theorem 9 follows from Theorem 3. The proof is complete. Remark 1. There exist many functions which satisfy (H6 (H8. For example, f(t, x = ψ(tx, where > 1, ψ C[, 1] ψ(t ψ(t for t [, 1]. We give a simple example to illustrate Theorem 9 give an estimate for parameter λ. Consider equation (27 with f(t, x = x 2, φ(t = 1 η = ζ =. Then, G 1 (t, s = G 2 (t, s { t(1 s, t s 1, = s(1 t, s t 1, e(t = x (t = = t(1 t3. 24 It is not hard to verify that G 1 (t, τg 2 (τ, sdsdτ x = max t [,1] x (t = It is easy to check that (H6-(H8 (H11 hold, where = 2. Hence, Theorem 9 implies that there exists λ > such that (27 has a unique positive solution u λ (t P e for λ [, λ ]. Furthermore, such a solution u λ (t satisfies the following properties: (a for any u (t P e, set u n (t = λ G 1 (t, τg 2 (τ, su 2 n 1(sdsdτ + t(1 t3, n = 1, 2,, 24 then lim u n(λ u λ = ; (b t(1 t3 24 u λ (t t(1 t3 12 (t [, 1] for λ [, λ ]; (c u λ (t is increasing in λ for λ [, λ ]. (d lim u λ x = ; λ u λ x , λ [, λ ]. Now, we give an estimate for λ. Since x (t = e(t x = Therefore, Ax (t = x 2 x (t = G 1 (t, τg 2 (τ, sx 2 (sdsdτ x (t. 1 This means that ρ(x Hence, from Remark 4 we obtain that λ Theorem 11. Assume that (H6,(H8-(H11 hold. Then (a there exists an interval I with [, λ(r ] I [, such that (28 has a unique positive solution u λ (t P e for λ I. Moreover, for any u P e, set 1 u n (t = λ G 1(t, τg 2 (τ, sf(s, u n 1 (sdsdτ + G 1(t, τg 2 (τ, sg(u n 1 (sdsdτ, t [, 1], n = 1, 2,, then lim u n u λ = ; (b u λ is increasing in λ for λ I; (c there exist x, z P e with x z such that u λ [x, z ] for λ [, λ(r ]. Where λ(r is defined by Theorem 8. Proof. It is easy to see that the problem (28 has an integral formulation given by u(t = λ + G 1(t, τg 2 (τ, sf(s, u(sdsdτ G 1(t, τg 2 (τ, sg(u(sdsdτ. E-ISSN: Issue 3, Volume 11, March 212

9 Define operator A, B : P E by Au(t = Bu(t = G 1 (t, τg 2 (τ, sf(s, u(sdsdτ, G 1 (t, τg 2 (τ, sg(u(sdsdτ for t [, 1]. It is easy to prove that u λ is a solution of the problem (28 if only if u λ is a fixed point of the operator λa + B. In virtue of (H6,(H8 (H9, we know that A, B : P P is an increasing operator. By the proof of Theorem 9 we obtain that A satisfies (H1 (H2. From (H9 we have Be(t g( On the other h, Be(t G 1 (t, τg 2 (τ, sg(e(sdsdτ > g(e(t, t (, 1. g( e e(t, G 1 (t, τg 2 (τ, sdsdτ G 1 (t, τg 2 (τ, sg( e dsdτ That is, B satisfies (H3. For any r (, 1 u P e, by (H1 we have B(ru(t = r α G 1 (t, τg 2 (τ, sg(ru(sdsdτ = r α Bu(t, u P e. G 1 (t, τg 2 (τ, sg(u(sdsdτ That is, B satisfies (H5. Thus, Theorem 11 follows from Theorem 8. The proof is complete. Remark 12. We can give a simple example to illustrate Theorem 11. Consider equation (28 with f(t, x = x 2, g(x = 1 + x 1 2 η = ζ =. From Remark 1, f(t, x satisfies (H6 (H8 with = 2. G 1 (t, s, G 2 (t, s e(t are the same as Remake 1. It is easy to check that g(x satisfies (H9 (H1 with α = 1 2. Moreover, integral equation u(t = G 1 (t, τg 2 (τ, s(1 + u 1 2 (sdsdτ. has a unique solution x (t in P e. Note that 1 α 1 = 1 2 R = is the unique solution of algebraic equation R R 1 2 in (1, +, Then = 1 2 λ(r = 1 α 1 1 R ρ(x = 2( ρ(x Hence, Theorem 11 implies that there exists interval [, 2(7 4 3 ρ(x ] I [, such that (28 has a u- nique positive solution u λ (t P e for λ I. Furthermore, such a solution u λ (t satisfies the following properties: (a for any u (t P e λ I, set u n (t = λ G 1(t, τg 2 (τ, su 2 n 1 (sdsdτ + G 1(t, τg 2 (τ, s(1 + u 1 2 n 1 (sdsdτ, then lim u n(λ u λ = ; (b u λ (t is increasing in λ for λ I; (c x (t u λ (t x (t(t [, 1] for λ [, 2(7 4 3 ρ(x ]. Remark 13. The problem discussed by [19], [22]- [25] is the special case of the problem (28 where g(u. Acknowledgements: The authors are grateful to the referees for their comments suggestions. This research was partially supported by the NNSF of China (19612, the Key Project of Chinese Ministry of Education (28154, the Chunhui Project of Chinese Ministry of Education (Z , the Science Foundation of Shanxi Province of China ( the University Natural Science Research Develop Foundation of Shanxi Province of China( References: [1] H. Amann, Fixed point equations nonlinear eigenvalue problems in ordered Banach spaces, SIAM Review, Vol.18, 1976, pp [2] A. J. B. Botter, Applications of Hilbert s projective metric to certain classes of nonhomogenous operators, Quart. J. Math. Oxford Ser., Vol.28, 1977, (2: pp [3] K. Li, J. Liang, T. J. Xiao, A fixed point theorem for convex decreasing operators, Nonlinear Anal., Vol.63, 25, pp. e29 e216. [4] W. X. Wang, Z. D. Liang, Fixed point theorems of a class of nonlinear operators applications, Acta Mathematica Sinica, Vol.48A, 25, (4:pp E-ISSN: Issue 3, Volume 11, March 212

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