Journal of Inequalities in Pure and Applied Mathematics

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1 Journal of Inequalities in Pure and Applied Mathematics THE METHOD OF LOWER AND UPPER SOLUTIONS FOR SOME FOURTH-ORDER EQUATIONS ZHANBING BAI, WEIGAO GE AND YIFU WANG Department of Applied Mathematics, Beijing Institute of Technology, Beijing , People s Republic of China. baizhanbing@263.net volume 5, issue 1, article 13, Received 17 September, 2003; accepted 23 January, Communicated by: A.M. Fink Abstract Home Page c 2000 Victoria University ISSN (electronic):

2 Abstract In this paper, by combining a new maximum principle of fourth-order equations with the theory of eigenline problems, we develop a monotone method in the presence of lower and upper solutions for some fourth-order ordinary differential equation boundary value problem. Our results indicate there is a relation between the existence of solutions of nonlinear fourth-order equation and the first eigenline of linear fourth-order equation Mathematics Subject Classification: 34B15, 34B10. Key words: Maximum principle; Lower and upper solutions; Fourth-order equation. This work is sponsored by the National Nature Science Foundation of China ( ) and the Doctoral Program Foundation of Education Ministry of China ( ). The authors thank the referees for their careful reading of the manuscript and useful suggestions. 1 Introduction Maximum Principle The Monotone Method References Page 2 of 17

3 1. Introduction This paper consider solutions of the fourth-order boundary value problem (1.1) u (4) (x) = f(x, u(x), u (x)), 0 < x < 1, (1.2) u(0) = u(1) = u (0) = u (1) = 0, where f : [0, 1] R R R is continuous. Many authors [1] [8], [10], [11], [13] [17] have studied this problem. In [1, 4, 6, 8, 10, 16], Aftabizadeh et al. showed the existence of positive solution to (1.1) (1.2) under some growth conditions of f and a non-resonance condition involving a two-parameter linear eigenvalue problem. These results are based upon the Leray Schauder continuation method and topological degree. In [2, 5, 7, 11, 15], Agarwal et al. considered an equation of the form u (4) (x) = f(x, u(x)), with diverse kind of boundary conditions by using the lower and upper solution method. Recently, Bai [3] and Ma et al. [14] developed the monotone method for the problem (1.1) (1.2) under some monotone conditions of f. More recently, with using Krasnosel skii fixed point theorem, Li [13] showed the existence results of positive solutions for the following problem u (4) + βu αu = f(t, u), 0 < t < 1, Page 3 of 17

4 u(0) = u(1) = u (0) = u (1) = 0, where f : [0, 1] R + R + is continuous, α, β R and β < 2π 2, α β 2 /4, α/π 4 + β/π 2 < 1. In this paper, by the use of a new maximum principle of fourth-order equation and the theory of the eigenline problem, we intend to further relax the monotone condition of f and get the iteration solution. Our results indicate there exists some relation between the existence of positive solutions of nonlinear fourth-order equation and the first eigenline of linear fourth-order equation. Page 4 of 17

5 2. Maximum Principle In this section, we prove a maximum principle for the operator L : F C[0, 1] defined by Lu = u (4) au + bu. Here a, b R satisfy (2.1) u F and a π + b 2 π + 1 > 0, 4 a2 4b 0, a > 2π 2 ; F = {u C 4 [0, 1] u(0) = 0, u(1) = 0, u (0) 0, u (1) 0}. Lemma 2.1. [12] Let f(x) be continuous for a x b and let c < λ 1 = π 2 /(b a) 2. Let u satisfies u (x) + cu(x) = f(x), u(a) = u(b) = 0. for x (a, b), Assume that u(x 1 ) = u(x 2 ) = 0 where a x 1 < x 2 b and u(x) 0 for x 1 x x 2. If either f(x) 0 for all x [x 1, x 2 ] or f(x) 0 for all x [x 1, x 2 ] and f(x) is not identically zero on [x 1, x 2 ], then u(x)f(x) 0 for all x [x 1, x 2 ]. Lemma 2.2. If u(x) satisfies u + cu(x) 0, for x (a, b) u(a) 0, u(b) 0, where c < λ 1 = π 2 /(b a) 2. Then u(x) 0, in [a, b]. Page 5 of 17

6 Proof. It follows by Lemma 2.1. Lemma 2.3. If u F satisfies Lu 0, then u 0 in [0, 1]. Proof. Set Ax = x. As a, b R satisfy (2.1), we have that Lu = u (4) au + bu = (A r 2 )(A r 1 )u 0, where r 1,2 = (a ± a 2 4b)/2 π 2. In fact, r 1 = (a + a 2 4b)/2 r 2 = (a a 2 4b)/2. By a/π 2 + b/π > 0, we have aπ 2 + b + π 4 > 0, thus a 2 + 4aπ 2 + 4π 4 > a 2 4b, because a 2 4b 0, so (a + 2π 2 ) 2 > ( a 2 4b) 2. Combining this together with a > 2π 2, we can conclude a + 2π 2 > a 2 4b. Then, r 1 r 2 = (a a 2 4b)/2 > π 2. Let y = (A r 1 )u = u r 1 u, then i.e., On the other hand, u F yields that (A r 2 )y 0, y r 2 y 0. (2.2) y(0) = u (0) r 1 u(0) 0, y(1) = u (1) r 1 u(1) 0. Page 6 of 17

7 Therefore, by the use of Lemma 2.2, there exists i.e., y(x) 0, x [0, 1], u (x) r 1 u(x) = y(x) 0. This together with Lemma 2.2 and the fact that u(0) = 0, that u(x) 0 in [0, 1]. Remark 2.1. Observe that a, b R satisfies (2.1) if and only if (2.3) b 0, or a π + b 2 π + 1 > 0, a > 4 2π2 ; (2.4) b > 0, a > 0, a 2 4b 0; u(1) = 0 implies or (2.5) b > 0, 0 > a > 2π 2, a π + b 2 π + 1 > 0, 4 a2 4b 0. From (2.3) and (2.4), we can easily conclude r 1 = a + a 2 4b 0. 2 Therefore, (2.2) can be obtained under u(0) 0, u(1) 0, u (0) 0, u (1) 0, and F can be defined as F = {u C 4 [0, 1] u(0) 0, u(1) 0, u (0) 0, u (1) 0}, we refer the reader to [3, 13]. Page 7 of 17

8 Lemma 2.4. [7] Given (a, b) R 2, the following problem (2.6) (2.7) u (4) au + bu = 0, u(0) = u(1) = u (0) = u (1) = 0, has a non-trivial solution if and only if for some k N. a (kπ) + b 2 (kπ) + 1 = 0, 4 Page 8 of 17

9 3. The Monotone Method In this section, we develop the monotone method for the fourth order two-point boundary value problem (1.1) (1.2). For given a, b R satisfying a/π 2 +b/π 4 +1 > 0, a 2 4b 0, a > 2π 2 and f : [0, 1] R R R, let (3.1) f 1 (x, u, v) = f(x, u, v) + bu av. Then (1.1) is equal to (3.2) Lu = f 1 (x, u, u ). Definition 3.1. Letting α C 4 [0, 1], we say that α is an upper solution for the problem (1.1) (1.2) if α satisfies α (4) (x) f(x, α(x), α (x)), for x (0, 1), α(0) = 0, α(1) = 0, α (0) 0, α (1) 0. Definition 3.2. Letting β C 4 [0, 1], we say β is a lower solution for the problem (1.1) (1.2) if β satisfies β (4) (x) f(x, β(x), β (x)), for x (0, 1), β(0) = 0, β(1) = 0, β (0) 0, β (1) 0. Page 9 of 17

10 Remark 3.1. If a, b satisfy (2.3) or (2.4), the boundary values can be replaced by α(0) 0, α(1) 0; β(0) 0, β(1) 0. It is clear that if α, β are upper and lower solutions of the problem (1.1) (1.2) respectively, α, β are upper and lower solutions of the problem (3.2) (1.2) respectively, too. Theorem 3.1. If there exist α and β, upper and lower solutions, respectively, for the problem (1.1) (1.2) which satisfy (3.3) β α and β + r(α β) α, and if f : [0, 1] R R R is continuous and satisfies (3.4) f(x, u 2, v) f(x, u 1, v) b(u 2 u 1 ), for β(x) u 1 u 2 α(x), v R, and x [0, 1]; (3.5) f(x, u, v 2 ) f(x, u, v 1 ) a(v 2 v 1 ), for v 2 + r(α β) v 1, α r(α β) v 1, v 2 β + r(α β), u R, and x [0, 1], where a, b R satisfy a/π 2 + b/π > 0, a 2 4b 0, a > 2π 2 and r = (a a 2 4b)/2, then there exist two monotone sequences {α n } and {β n }, non-increasing and non-decreasing, respectively, with α 0 = α and β 0 = β, which converge uniformly to the extremal solutions in [β, α] of the problem (1.1) (1.2). Page 10 of 17

11 Proof. Consider the problem (3.6) u (4) (x) au (x) + bu(x) = f 1 (x, η(x), η (x)), for x (0, 1), (3.7) u(0) = u(1) = u (0) = u (1) = 0, with η C 2 [0, 1]. Since a/π 2 + b/π > 0, with the use of Lemma 2.4 and Fredholm Alternative [9], the problem (3.6) (3.7) has a unique solution u. Define T : C 2 [0, 1] C 4 [0, 1] by (3.8) T η = u. Now, we divide the proof into three steps. Step 1. We show (3.9) T C C. Here, C = {η C 2 [0, 1] β η α, α r(α β) η β + r(α β)} is a nonempty bounded closed subset in C 2 [0, 1]. In fact, for ζ C, set ω = T ζ. By the definition of α, β and C, combining (3.1), (3.4), and (3.5), we have that (3.10) (α ω) (4) (x) a(α ω) (x) + b(α ω)(x) f 1 (x, α(x), α (x)) f 1 (x, ζ(x), ζ (x)) = f(x, α(x), α (x)) f(x, ζ(x), ζ (x)) 0, a(α ζ) (x) + b(α ζ)(x) Page 11 of 17

12 (3.11) (α ω)(0) = 0, (α ω)(1) = 0, (3.12) (α ω) (0) 0, (α ω) (1) 0. With the use of Lemma 2.3, we obtain that α ω. Analogously, there holds ω β. By the proof of Lemma 2.3, combining (3.10), (3.11), and (3.12), we have that (α ω) (x) r(α ω)(x) 0, x (0, 1), hence, ω (x) + r(α β)(x) ω (x) + r(α ω)(x) α (x), for x (0, 1), i.e., Analogously, Thus, (3.9) holds. ω (x) α (x) r(α β)(x), for x (0, 1). ω (x) β (x) + r(α β)(x), for x (0, 1). Step 2. Let u 1 = T η 1, u 2 = T η 2, where η 1, η 2 C satisfy η 1 η 2 and η r(α β) η 2. We show (3.13) u 1 u 2, u 1 + r(α β) u Page 12 of 17

13 In fact, by (3.4), (3.5), and the definition of u 1, u 2, L(u 2 u 1 )(x) = f 1 (x, η 2 (x), η 2(x)) f 1 (x, η 1 (x), η 1(x)) 0, (u 2 u 1 )(0) = (u 2 u 1 )(1) = 0, (u 2 u 1 ) (0) = (u 2 u 1 ) (1) = 0. With the use of Lemma 2.3, we get that u 1 u 2. Similar to Step 1, we can easily prove u 1 + r(α β) u 2. Thus, (3.13) holds. Step 3. The sequences {α n } and {β n } are obtained by recurrence: α 0 = α, β 0 = β, α n = T α n 1, β n = T β n 1, n = 1, 2,.... From the results of Step 1 and Step 2, we have that (3.14) β = β 0 β 1 β n α n α 1 α 0 = α, (3.15) β = β 0, α = α 0, α r(α β) α n, β n β +r(α β). Moreover, from the definition of T (see (3.8)), we get i.e., α n (4) (x) aα n(x) + bα n (x) = f 1 (x, α n 1 (x), α n 1(x)), α n (4) (x) = f 1 (x, α n 1 (x), α n 1(x)) + aα n(x) bα n (x) (3.16) f 1 (x, α n 1 (x), α n 1(x)) + a[β + r(α β)](x) bβ(x), Page 13 of 17

14 (3.17) α n (0) = α n (1) = α n(0) = α n(1) = 0. Analogously, β n (4) (x) = f 1 (x, β n 1 (x), β n 1(x)) + aβ n(x) bβ n (x) (3.18) f 1 (x, β n 1 (x), β n 1(x)) + a[β + r(α β)](x) bβ(x), (3.19) β n (0) = β n (1) = β n(0) = β n(1) = 0. From (3.14), (3.15), (3.16), and the continuity of f 1, we have that there exists M α,β > 0 depending only on α and β (but not on n or x) such that (3.20) α (4) n (x) M α,β, for all x [0, 1]. Using the boundary condition (3.17), we get that for each n N, there exists ξ n (0, 1) such that (3.21) α n (ξ n ) = 0. This together with (3.20) yields (3.22) α n (x) = α n (ξ n ) + x ξ n α (4) n (s)ds M α,β. Page 14 of 17

15 By combining (3.15) and (3.17), we can similarly get that there is C α,β > 0 depending only on α and β (but not on n or x) such that (3.23) α n(x) C α,β, for all x [0, 1], (3.24) α n(x) C α,β, for all x [0, 1]. Thus, from (3.14), (3.22), (3.23), and (3.24), we know that {α n } is bounded in C 3 [0, 1]. Similarly, {β n } is bounded in C 3 [0, 1]. Now, by using the fact that {α n } and {β n } are bounded in C 3 [0, 1], we can conclude that {α n }, {β n } converge uniformly to the extremal solutions in [0, 1] of the problem (3.2) (1.2). Therefore, {α n }, {β n } converge uniformly to the extremal solutions in [0, 1] of the problem (1.1) (1.2), too. Example 3.1. Consider the boundary value problem (3.25) u (4) (x) = 5u (x) (u(x) + 1) 2 + sin 2 πx + 1, (3.26) u(0) = u(1) = u (0) = u (1) = 0. It is clear that the results of [3, 7, 13, 14] can t apply to the example. On the other hand, it is easy to check that α = sin πx, β = 0 are upper and lower solutions of (3.25) (3.26), respectively. Letting a = 5, b = 4, then all assumptions of Theorem 3.1 are fulfilled. Hence the problem (3.25) (3.26) has at least one solution u, which satisfies 0 u sin πx. Page 15 of 17

16 References [1] A.R. AFTABIZADEH, Existence and uniqueness theorems for fourthorder boundary value problems, J. Math. Anal. Appl., 116 (1986), [2] R.P. AGARWAL, On fourth-order boundary value problems arising in beam analysis, Differential Integral Equations, 2 (1989), [3] Z.B. BAI, The Method of lower and upper solutions for a bending of an elastic beam equation, J. Math. Anal. Appl., 248 (2000), [4] Z.B. BAI AND H.Y. WANG, On the positive solutions of some nonlinear fourth-order beam equations, J. Math. Anal. Appl., 270 (2002), [5] A. CABADA, The method of lower and upper solutions for second, third, fourth and higher order boundary value problems, J. Math. Anal. Appl., 185 (1994), [6] C. DE COSTER, C. FABRY AND F. MUNYAMARERE, Nonresonance conditions for fourth-order nonlinear boundary value problems, Internat. J. Math. Sci., 17 (1994), [7] C. DE COSTER AND L. SANCHEZ, Upper and lower solutions, Ambrosetti-Prodi problem and positive solutions for fourth-order O. D. E., Riv. Mat. Pura Appl., 14 (1994), [8] M.A. DEL PINO AND R.F. MANASEVICH, Existence for a fourth-order boundary value problem under a two parameter nonresonance condition, Proc. Amer. Math. Soc., 112 (1991), Page 16 of 17

17 [9] D. GILBARG AND N.S. TRUDINGER, Elliptic Partial Differential Equations of Second-Order, Springer Verlag, New York, [10] C.P. GUPTA, Existence and uniqueness theorem for a bending of an elastic beam equation, Appl. Anal., 26 (1988), [11] P. KORMAN, A maximum principle for fourth-order ordinary differential equations, Appl. Anal., 33 (1989), [12] A.C. LAZER AND P.J. MCKENNA, Global bifurcation and a theorem of Tarantello, J. Math. Anal. Appl., 181 (1994), [13] Y.X. LI, Positive solutions of fourth-order boundary value problems with two parameters, J. Math. Anal. Appl., 281 (2003), [14] R.Y. MA, J.H. ZHANG AND S.M. FU, The method of lower and upper solutions for fourth-order two-point boundary value problems, J. Math. Anal. Appl., 215 (1997), [15] J. SCHRODER, Fourth-order two-point boundary value problems; estimates by two side bounds, Nonl. Anal., 8 (1984), [16] R.A. USMANI, A uniqueness theorem for a boundary value problem, Proc. Amer. Math. Soc., 77 (1979), [17] Q.L. YAO AND Z.B. BAI, Existence of positive solutions of boundary value problems for u (4) (t) λh(t)f(u(t)) = 0, Chinese. Ann. Math. Ser. A, 20 (1999), [In Chinese] Page 17 of 17

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