A Fourth-Order Boundary Value Problem with One-Sided Nagumo Condition

Boundary Value Problems, Mar 2011

The aim of this paper is to study a fourth-order separated boundary value problem with the right-hand side function satisfying one-sided Nagumo-type condition. By making a series of a priori estimates and applying lower and upper functions techniques and Leray-Schauder degree theory, the authors obtain the existence and location result of solutions to the problem.

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A Fourth-Order Boundary Value Problem with One-Sided Nagumo Condition

Boundary Value Problems Hindawi Publishing Corporation A Fourth-Order Boundary Value Problem with One-Sided Nagumo Condition Wenjing Song 0 1 Wenjie Gao 1 I. T. Kiguradze 0 Institute of Applied Mathematics, Jilin University of Finance and Economics , Changchun 130017 , China 1 Institute of Mathematics, Jilin University , Changchun 130012 , China Correspondence should be addressed to Wenjing Song, The aim of this paper is to study a fourth-order separated boundary value problem with the righthand side function satisfying one-sided Nagumo-type condition. By making a series of a priori estimates and applying lower and upper functions techniques and Leray-Schauder degree theory, the authors obtain the existence and location result of solutions to the problem. 1. Introduction nonlinear equation f t, u t , u t , u t , u t , with f : 0, 1 × 4 → being a continuous function. This equation can be used to model the deformations of an elastic beam, and the type of boundary conditions considered depends on how the beam is supported at the two endpoints 1, 2 . We consider the separated boundary conditions au 0 − bu with a, b, c, d ∈ 0, ∞ , A, B ∈ . For the fourth-order differential equation f t, u t , u t , u t , u t , 0 < t < 1, define a pair of lower and upper functions of problems 1.1 and 1.2 if the following conditions are satisfied: iii α 0 − β 0 ≤ min{β 0 − β 1 , α 1 − α 0 , 0}. Remark 2.2. By integration, from iii and 2.1 , we obtain dα 1 ≤ B, β 0 ≥ 0, β 1 ≥ the authors in 3 obtained the existence of solutions with the assumption that f satisfies the two-sided Nagumo-type conditions. For more related works, interested readers may refer to 1–14 . The one-sided Nagumo-type condition brings some difficulties in studying this kind of problem, as it can be seen in 15–18 . Motivated by the above works, we consider the existence of solutions when f satisfies one-sided Nagumo-type conditions. This is a generalization of the above cases. We apply lower and upper functions technique and topological degree method to prove the existence of solutions by making a priori estimates for the third derivative of all solutions of problems 1.1 and 1.2 . The estimates are essential for proving the existence of solutions. The outline of this paper is as follows. In Section 2, we give the definition of lower and upper functions to problems 1.1 and 1.2 and obtain some a priori estimates. Section 3 will be devoted to the study of the existence of solutions. In Section 4, we give an example to illustrate the conclusions. 2. Definitions and A Priori Estimates Upper and lower functions will be an important tool to obtain a priori bounds on u, u , and u . For this problem we define them as follows. α t ≤ β t , ∀t ∈ 0, 1 , α t ≤ β t , α t ≤ β t , ∀t ∈ 0, 1 , that is, lower and upper functions, and their first derivatives are also well ordered. and consider the set Lemma 2.4. Let Γi t , γi t ∈ C 0, 1 , f t, x0, x1, x2, x3 ≤ hE |x3| , ∀ t, x0, x1, x2, x3 ∈ E, ∞. ∀t ∈ 0, 1 , i Definition 2.3. Given a set E ⊂ 0, 1 × 4 , a continuous f : E → is said to satisfy the one-sided Nagumo-type condition in E if there exists a real continuous function hE : 0 → k, ∞ , for some k > 0, such that ds > max Γ2 t − min γ2 t . t∈ 0,1 t∈ 0,1 t, x0, x1, x2, x3 ∈ 0, 1 × 4 : γi t ≤ xi ≤ Γi t , i γi t ≤ u i t ≤ Γi t , 0, 1, 2 and every t ∈ 0, 1 , one has u ∞ < R. Proof. Let u be a solution of problems 1.1 and 1.2 such that 2.7 and 2.8 hold. Define η : max Γ2 1 − γ2 0 , Γ2 0 − γ2 1 . Assume that ρ ≥ η, and suppose, for contradiction, that |u t | > ρ for every t ∈ 0, 1 . If u t > ρ for every t ∈ 0, 1 , then we obtain the following contradiction: Γ2 1 − γ2 0 ≥ u 1 − u 0 If u t < −ρ for every t ∈ 0, 1 , a similar contradiction can be derived. So there is a t ∈ 0, 1 such that |u t | ≤ ρ. By 2.4 we can take R1 > ρ such that −u t1 −u 4 t dt f t, u t , u t , u t , u t dt −u t −u t t0 hE −u t t1 hE −u t −u t dt u t1 − u t0 ≤ tm∈0a,1x Γ2 t − tm∈0i,n1 γ2 t < ρ hE s If |u t | ≤ ρ for every t ∈ 0, 1 , then we have trivially |u t | < R1. If not, then we can take t1 ∈ 0, 1 such that u t1 < −ρ or t1 ∈ 0, 1 such that u t1 > ρ. Suppose that the first case holds. By 2.7 we can consider t1 < t0 ≤ 1 such that −ρ, u t < −ρ, ∀t ∈ t1, t0 . Applying a convenient change of variable, we have, by 2.3 and 2.11 , In a similar way, we may show that max{R1, R2}, we have u ∞ < R. Remark 2.5. Observe that the estimation R depends only on the functions hE, γ2, Γ2, and ρ and it does not depend on the boundary conditions. 3. Existence and Location Result In the presence of an ordered pair of lower and upper functions, the existence and location results for problems 1.1 and 1.2 can be obtained. u t < R1, ∀t ∈ 0, 1 . η hE s ds > tm∈0a,1x Γ2 t − tm∈0i,n1 γ2 t . u t < R2, ∀t ∈ 0, 1 . Theorem 3.1. Suppose that there exist lower and upper functions α t and β t of problems 1.1 and 1.2 , respectively. Let f : 0, 1 × 4 → be a continuous function satisfying the one-sided Nagumo-type conditions 2.3 and 2.4 in E∗ t, x0, x1, x2, x3 ∈ 0, 1 × ≤ x0, x1 ≤ β t , β t , for t ∈ 0, 1 . Proof. Define the auxiliary functions α i t ≤ xi ≤ β i t , i u t − λδ2 t, u t , with the boundary conditions Step 1. Every solution u t of problems 3.6 and 3.7 satisfies −r1 < α t ≤ β t < r1, f t, α t , α t , α t , 0 − r1 − α t < 0, ∀t ∈ 0, 1 0 ≥ u 4 t0 ≥ λf t0, β t0 , β t0 , β t0 , 0 0 ≥ u 4 t0 u t0 ≥ r1 > 0. and u 0 u 0 ≤ 0. If λ 0, then u 0 0 and so u 4 0 ≤ 0. Therefore, the above computations with t0 replaced by 0 yield a contradiction. For λ ∈ 0, 1 , by 3.11 , we get the following contradiction: 0 ≥ u 0 u s ds < r1|t − ξ| ≤ r1, u s ds < r1t ≤ r1. u t < R, ∀t ∈ 0, 1 , Step 2. There is an R > 0 such that for every solution u t of problems 3.6 and 3.7 with R independent of λ ∈ 0, 1 . Consider the set and for λ ∈ 0, 1 the function Fλ : Er1 → In the following we will prove that the function Fλ satisfies the one-sided Nagumo-type conditions 2.3 and 2.4 in Er1 independently of λ ∈ 0, 1 . Indeed, as f verifies 2.3 in E∗, then ≤ hE∗ |x3| So, defining hEr1 t hE∗ |x3| 2r1 in 0 , we see that Fλ verifies 2.3 with E and hE replaced by Er1 and hEr1 , respectively. The condition 2.4 is also verified since ∞. ρ : max ar1 |A| , |B| cr1 , b d every solution u of 3.6 and 3.7 satisfies γi t : −r1, Γi t : r1, for i 0, 1, 2. B − cu 1 . The hypotheses of Lemma 2.4 are satisfied with E replaced by Er1 . So there exists an R > 0, depending on r1 and hEr1 , such that |u t | < R for every t ∈ 0, 1 . As r1 and hEr1 do not depend on λ, we see that R is maybe independent of λ. Define the operators and for λ ∈ 0, 1 , Nλ : C3 0, 1 Observe that L has a compact inverse. Therefore, we can consider the completely continuous operator For R given by Step 2, take the set ∞ < R . By Steps 1 and 2, degree d I −Tλ, Ω, 0 is well defined for every λ ∈ 0, 1 and by the invariance with respect to a homotopy The equation x T0 x is equivalent to the problem and has only the trivial solution. Then, by the degree theory, So the equation T1 x x has at least one solution, and therefore the equivalent problem au 0 − bu 0 Step 4. The function u1 t is a solution of the problems 1.1 and 1.2 . The proof will be finished if the above function u1 t satisfies the inequalities α t ≤ u1 t ≤ β t . tm∈0a,1x u1 t − β t : u1 t2 − β t2 > 0. ≥ f t2, β t2 , β t2 , β t2 , β t2 ≥ β 4 t2 . If t2 0, then we have By Definition 2.1 this yields a contradiction tm∈0a,1x u1 t − β t : u1 0 − β 0 > 0, u1 0 − β 0 ≤ 0. 1 1 b au1 0 − A > b aβ 0 − A ≥ β 0 . Then t2 / 0 and, by similar arguments, we prove that t2 / 1. Thus, u1 t ≤ β t , ∀t ∈ 0, 1 . α t ≤ u1 t ≤ β t . On the other hand, by 1.2 , 0 u1 1 − u1 0 Applying the same technique, we have and then by Definition 2.1 iii , 3.44 and 3.46 , we obtain 1 t β s ds dt ≤ − 0 0 u1 s ds dt u1 0 , 1 t α s ds dt ≥ − 0 0 u1 s ds dt u1 0 , α 0 ≤ u1 0 ≤ β 0 . β t − u1 t ≥ β 0 − u1 0 ≥ 0, β t − u1 t ≥ β 0 − u1 0 4. An Example The following example shows the applicability of Theorem 3.1 when f satisfies only the onesided Nagumo-type condition. with A, B ∈ . The nonlinear function are, respectively, lower and upper functions of 4.1 and 4.2 . Moreover, define t, x0, x1, x2, x3 ∈ 0, 1 × Then f satisfies condition 3.2 and the one-sided Nagumo-type condition with hE |x3| in E. −t2 − t ≤ u t ≤ t2 t, −2t − 1 ≤ u t ≤ 2t 1, −2 ≤ u t ≤ 2. Example 4.1. 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Wenjing Song, Wenjie Gao. A Fourth-Order Boundary Value Problem with One-Sided Nagumo Condition, Boundary Value Problems, 2011, 569191,