Multiple positive solutions for a nonlinear three-point integral boundary-value problem

Int. J. Open Problems Compt. Math., Vol. 8, No. 1, March 215 ISSN 1998-6262; Copyright c ICSRS Publication, 215 www.i-csrs.org Multiple positive solutions for a nonlinear three-point integral boundary-value problem Faouzi Haddouchi 1,2, Slimane Benaicha 1 1 Department of Mathematics, University of Oran, Es-senia, 31 Oran, Algeria e-mail: fhaddouchi@gmail.com 2 Faculty of Physics, University of Sciences and Technology of Oran, El Mnaouar, BP 155, 31 Oran, Algeria e-mail: slimanebenaicha@yahoo.fr Abstract We investigate the existence of positive solutions to the nonlinear second-order three-point integral boundary value problem. u (t) + f(t, u(t)) =, < t < T, u() = βu(), u(t ) = α u(s)ds, 2T α2 α 2 2+2T where < < T, < α < 2T, < β < are given 2 constants. We establish the existence of at least three positive solutions by using the Leggett-Williams fixed-point theorem. Keywords: Positive solutions, Three-point boundary value problems, multiple solutions, Fixed points, Cone. 1 Introduction The study of the existence of solutions of multipoint boundary value problems for linear second-order ordinary differential equations was initiated by II in and Moiseev [16]. Then Gupta [6] studied three-point boundary value problems for nonlinear second-order ordinary differential equations. Since then, nonlinear

3 F. Haddouchi, S. Benaicha second-order three-point boundary value problems have also been studied by several authors. We refer the reader to [5, 7, 8, 9, 1, 11, 12, 13, 18, 19, 2, 21, 22, 23, 25, 26, 27, 28, 3, 31, 32, 33, 34, 35, 36, 37, 39, 4, 41] and the references therein. This paper is a continuation of our study in [15] and is concerned with the existence and multiplicity of positive solutions of the problem u (t) + f(t, u(t)) =, t (, T ), (1) with the three-point integral boundary condition u() = βu(), u(t ) = α u(s)ds, (2) Throughout this paper, we assume the following hypotheses: (H1) f C([, T ] [, ), [, )) and f(t,.) does not vanish identically on any subset of [, T ] with positive measure. (H2) (, T ), < α < 2T 2 and < β < 2T α2 α 2 2+2T. In this paper, by using the Leggett-Williams fixed-point theorem [17], we will show the existence of at least three positive solutions for a three-point integral boundary value problem. Some papers in this area include [24, 29, 38, 1, 2, 3, 14, 4]. 2 Background and definitions Definition 2.1 Let E be a real Banach space. A nonempty closed convex set P E is called a cone if it satisfies the following two conditions: (i) x P, λ implies λx P ; (ii) x P, x P implies x =. Every cone P E induces an ordering in E given by x y if and only if y x E. Definition 2.2 An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets. Definition 2.3 A map ψ is said to be a nonnegative continuous concave functional on a cone P of a real Banach space E if ψ : P [, ) is continuous and ψ(tx + (1 t)y) tψ(x) + (1 t)ψ(y)

Positive solutions 31 for all x, y P and t [, 1]. Similarly we say the map ϕ is a nonnegative continuous convex functional on a cone P of a real Banach space E if ϕ : P [, ) is continuous and for all x, y P and t [, 1]. ϕ(tx + (1 t)y) tϕ(x) + (1 t)ϕ(y) Definition 2.4 Let ψ be a nonnegative continuous concave functional on the cone P. Define the convex sets P c and P (ψ, a, b) by P c = {x P : x < c}, for c > P (ψ, a, b) = {x P : a ψ(x), x b}, for < a < b. Next we state the Leggett-Williams fixed-point theorem. Theorem 2.5 ([17]) Let A : P c P c be a completely continuous operator and let ψ be a nonnegative continuous concave functional on P such that ψ(x) x for all x P c. Suppose that there exist < a < b < d c such that the following conditions hold, (C1) {x P (ψ, b, d) : ψ(x) > b} and ψ(ax) > b for all x P (ψ, b, d); (C2) Ax < a for x a; (C3) ψ(ax) > b for x P (ψ, b, c) with Ax > d. Then A has at least three fixed points x 1, x 2 and x 3 in P c satisfying x 1 < a, ψ(x 2 ) > b, a < x 3 with ψ(x 3 ) < b. 3 Some preliminary results In order to prove our main result, we need some preliminary results. Let us consider the following boundary value problem u (t) + y(t) =, t (, T ), (3) u() = βu(), u(t ) = α u(s)ds (4) For problem (3), (4), we have the following conclusions which are derived from [15].

32 F. Haddouchi, S. Benaicha 2T α2 Lemma 3.1 (See [15]) Let β. Then for y C([, T ], R), the α 2 2+2T problem (3)-(4) has the unique solution u(t) = β(2t α 2 ) 2β(1 α)t (α 2 2T ) β(2 α 2 2T ) αβ α(β 1)t + (α 2 2T ) β(2 α 2 2T ) 2(β 1)t 2β + (α 2 2T ) β(2 α 2 2T ) t (t s)y(s)ds. ( s)y(s)ds T ( s) 2 y(s)ds (T s)y(s)ds 2T α2 Lemma 3.2 (See [15]) Let < α < 2T, β <. If y 2 α 2 2+2T C([, T ], [, )), then the unique solution u of problem (3), (4) satisfies u(t) for t [, T ]. 2T α2 Lemma 3.3 (See [15]) Let < α < 2T, β <. If y 2 α 2 2+2T C([, T ], [, )), then the unique solution u of the problem (3), (4) satisfies min u(t) γ u, u = max u(t), (5) t [,T ] t [,T ] where { γ := min T α(β + 1)2,, 2T } α(β + 1)(T ) (, 1). (6) 2T α(β + 1) 2 4 Existence of triple solutions In this section, we discuss the multiplicity of positive solutions for the general boundary-value problem (1), (2) In the following, we denote := (2T α 2 ) β(α 2 2 + 2T ), (7) ( T 2 (2T (β + 1) + β(α + 2) + αβt 2 ) ) 1, m := (8) 2 { (T ) 2 δ := min, α2 (1 + β)(t ) 2 }. (9) 2 Using Theorem 2.5, we established the following existence theorem for the boundary-value problem (1), (2).

Positive solutions 33 Theorem 4.1 Assume (H1) and (H2) hold. Suppose there exists constants < a < b < b/γ c such that (D1) f(t, u) < ma for t [, T ], u [, a]; (D2) f(t, u) b δ for t [, T ], u [b, b γ ]; (D3) f(t, u) mc for t [, T ], u [, c], where γ, m, δ are as defined in (6), (8) and (9), respectively. Then the boundaryvalue problem (1)-(2) has at least three positive solutions u 1, u 2 and u 3 satisfying u 1 < a, min u 2(t) > b, a < u 3 with min u 3(t) < b. t [,T ] t [,T ] Proof. Let E = C([, T ], R) be endowed with the maximum norm, u = max t [,T ] u(t), define the cone P C([, T ], R) by P = {u C([, T ], R) : u concave down and u(t) on [, T ]}. (1) Let ψ : P [, ) be defined by ψ(u) = min u(t), u P. (11) t [,T ] then ψ is a nonnegative continuous concave functional and ψ(u) u, u P. Define the operator A : P C([, T ], R) by Au(t) = β(2t α2 ) 2β(1 α)t ( s)f(s, u(s))ds αβ α(β 1)t ( s) 2 f(s, u(s))ds 2(β 1)t 2β T (T s)f(s, u(s))ds t (t s)f(s, u(s))ds. Then the fixed points of A just are the solutions of the boundary-value problem (1)-(2) from Lemma 3.1. Since (Au) (t) = f(t, u(t)) for t (, T ), together with (H1) and Lemma 3.2, we see that Au(t), t [, T ] and (Au) (t), t (, T ). Thus A : P P. Moreover, A is completely continuous.

34 F. Haddouchi, S. Benaicha We now show that all the conditions of Theorem 2.5 are satisfied. From (11), we know that ψ(u) u, for all u P. Now if u P c, then u c, together with (D3), we find t [, T ], Au(t) 2β(1 α)t β(2t α2 ) ( s)f(s, u(s))ds α(β 1)t αβ + ( s) 2 f(s, u(s))ds 2β 2(β 1)t T + (T s)f(s, u(s))ds 2βT + αβ2 ( s)f(s, u(s))ds + αβt ( s) 2 f(s, u(s))ds 2β + 2T T + (T s)f(s, u(s))ds 2T (β + 1) + β(α + 2) T (T s)f(s, u(s))ds + αβt ( s) 2 f(s, u(s))ds 2T (β + 1) + β(α + 2) T (T s)f(s, u(s))ds + αβt T T (T s)f(s, u(s))ds = 2(β + 1) + T 1 β(α + 2) + αβt T T (T s)f(s, u(s))ds mc 2(β + 1) + T 1 β(α + 2) + αβt T T (T s)ds = mc T 2 (2T (β + 1) + β(α + 2) + αβt 2 ) 2 = c Thus, A : P c P c. By (D1) and the argument above, we can get that A : P a P a. So, Au < a for u a, the condition (C2) of Theorem 2.5 holds. Consider the condition (C1) of Theorem 2.5 now. Since ψ(b/γ) = b/γ > b, let d = b/γ, then {u P (ψ, b, d) : ψ(u) > b} =. For u P (ψ, b, d), we have b u(t) b/γ, t [, T ]. Combining with (D2), we get f(t, u) b δ, t [, T ]. Since u P (ψ, b, d), then there are two cases, (i) ψ(au)(t) = Au(T ) and (ii) ψ(au)(t) = Au(). In case (i), we have

Positive solutions 35 ψ(au)(t) = Au(T ) = β(2t α2 ) 2β(1 α)t ( s)f(s, u(s))ds αβ α(β 1)T ( s) 2 f(s, u(s))ds 2(β 1)T 2β T (T s)f(s, u(s))ds T (T s)f(s, u(s))ds αβ( 2T ) = ( s)f(s, u(s))ds α(β 1)T αβ + ( s) 2 f(s, u(s))ds + α2 (β + 1) T (T s)f(s, u(s))ds = α2 (β + 1) T (T s)f(s, u(s))ds α2 T (β + 1) α(β + 2T ) + sf(s, u(s))ds α(β 1)T αβ + s 2 f(s, u(s))ds = α2 T (β + 1) α2 (β + 1) T T f(s, u(s))ds α2 (β + 1) sf(s, u(s))ds + α(β + 2T ) sf(s, u(s))ds f(s, u(s))ds sf(s, u(s))ds α(β 1)T αβ + s 2 f(s, u(s))ds = α2 (β + 1) T α(2t ) (T s)f(s, u(s))ds + sf(s, u(s))ds αβ(t ) αt + s 2 f(s, u(s))ds > α2 (β + 1) T (T s)f(s, u(s))ds + αt sf(s, u(s))ds αt = α2 (β + 1) s 2 f(s, u(s))ds T (T s)f(s, u(s))ds + αt s( s)f(s, u(s))ds

36 F. Haddouchi, S. Benaicha > α2 (β + 1) b δ α 2 (β + 1) T T = b α 2 (β + 1)(T ) 2 δ 2 b. (T s)f(s, u(s))ds (T s)ds In case (ii), we have ψ(au)(t) = Au() = β(α2 2 + 2T ) + 2 = 2 α T T ( s)f(s, u(s))ds (T s)f(s, u(s))ds α 2T α2 (T s)f(s, u(s))ds T ( s) 2 f(s, u(s))ds ( s) 2 f(s, u(s))ds ( s)f(s, u(s))ds ( s)f(s, u(s))ds = 2 (T s)f(s, u(s))ds (2T α2 ) f(s, u(s))ds 2T α2 + sf(s, u(s))ds α3 f(s, u(s))ds + 2α2 sf(s, u(s))ds α s 2 f(s, u(s))ds T = 2 (T s)f(s, u(s))ds 2T f(s, u(s))ds 2T + α2 + sf(s, u(s))ds α s 2 f(s, u(s))ds = 2T T f(s, u(s))ds 2 sf(s, u(s))ds 2 T sf(s, u(s))ds = 2 2T + α2 + α T sf(s, u(s))ds α s 2 f(s, u(s))ds 2(T ) + α2 sf(s, u(s))ds (T s)f(s, u(s))ds + s 2 f(s, u(s))ds

Positive solutions 37 > 2 = 2 > 2 T α b δ (T s)f(s, u(s))ds + α2 T T 2 T = b (T ) 2 δ b. s 2 f(s, u(s))ds (T s)f(s, u(s))ds + α (T s)f(s, u(s))ds (T s)ds sf(s, u(s))ds s( s)f(s, u(s))ds So, ψ(au) > b ; u P (ψ, b, b/γ). For the condition (C3) of the Theorem 2.5, we can verify it easily under our assumptions using Lemma 3.3. Here ψ(au) = min t [,T ] Au(t) γ Au > γ b γ = b as long as u P (ψ, b, c) with Au > b/γ. Since all conditions of Theorem 2.5 are satisfied. Then problem (1)-(2) has at least three positive solutions u 1, u 2, u 3 with u 1 < a, ψ(u 2 ) > b, a < u 3 with ψ(u 3 ) < b. 5 Some examples In this section, in order to illustrate our result, we consider some examples. Example 5.1 Consider the boundary value problem u (t) + 4u2 u 2 + 1 u() = 1 2 u(1 3 ), u(1) = 3 1 3 Set β = 1/2, α = 3, = 1/3, T = 1, and =, < t < 1, (12) u(s)ds. (13)

38 F. Haddouchi, S. Benaicha f(t, u) = f(u) = 4u2 u 2 + 1, u. It is clear that f(.) is continuous and increasing on [, ). We can also show that < α = 3 < 18 = 2T 2, < β = 1 2 < 1 = 2T α2 α 2 2 + 2T. Now we check that (D1), (D2) and (D3) of Theorem 4.1 are satisfied. By (6), (8), (9), we get γ = 1/4, m = 1/3, δ = 2/15. Let c = 124, we have f(u) 4 < mc = 124 3 41, 33, u [, c], from lim u f(u) = 4, so that (D3) is met. Note that f(2) = 32, when we set b = 2, f(u) b δ = 15, u [b, 4b], holds. It means that (D2)is satisfied. To verify (D1), as f( 1 ) = 4, we 12 1441 take a = 1, then 12 f(u) < ma = 1, u [, a], 36 and (D1) holds. Summing up, there exists constants a = 1/12, b = 2, c = 124 satisfying < a < b < b γ c, such that (D1), (D2) and (D3) of Theorem 4.1 hold. So the boundary-value problem (12)-(13) has at least three positive solutions u 1, u 2 and u 3 satisfying u 1 < 1 12, min t [,T ] u 2(t) > 2, 1 12 < u 3 with min t [,T ] u 3(t) < 2. Example 5.2 Consider the boundary value problem u (t) + f(t, u) =, < t < 1, (14) u() = u( 1 2 ), u(1) = 1 2 Set β = 1, α = 1, = 1/2, T = 1, f(t, u) = e t h(u) where u(s)ds. (15)

Positive solutions 39 2 u u 1 25 2173 u 2167 1 u 4 75 75 h(u) = 87 4 u 544 u 544 u 546 87 544 39(3u+189) (16) u 546. u+27 By (6), (8), (9) and after a simple calculation, we get γ = 1/4, m = 4/25, δ = 1/8. We choose a = 1/4, b = 4, and c = 544; consequently, f(t, u) = e t 2 25 u 2 25 u < 4 25 1 = ma, t 1, u 1/4, 4 f(t, u) = e t 87 87 e > 32 = b, 1/2 t 1, 4 u 16, δ f(t, u) = e t h(u) 87 < 4 544 = mc, t 1, u 544. 25 That is to say, all the conditions of Theorem 4.1 are satisfied. Then problem (14), (15) has at least three positive solutions u 1 u 2, and u 3 satisfying u 1 < 1 4, ψ(u 2) > 4, u 3 > 1 4 with ψ(u 3) < 4. 6 Open Problem Is it possible to generalize the above results for multipoint integral boundary value problems? In this work, we have assumed that f is continuous function, it will be interesting to consider the same problem but with singularities. References [1] D. R. Anderson; Multiple Positive Solutions for a Three-Point Boundary Value Problem, Math. Comput. Modelling, Vol. 27, No. 6, (1998), pp. 49-57. [2] D. R. Anderson, R. Avery and A. Peterson; Three positive solutions to a discrete focal boundary value problem, In Positive Solutions of Nonlinear Problems, (Edited by R. Agarwal), J. Comput. Appl. Math., (1998). [3] R. I. Avery, D. R. Anderson; Existence of three positive solutions to a second-order boundary value problem on a measure chain, J. Comput. Appl. Math., 141, (22), pp. 65-73.

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