Positive Solutions of Three-Point Nonlinear Second Order Boundary Value Problem

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1 Positive Solutions of Three-Point Nonlinear Second Order Boundary Value Problem YOUSSEF N. RAFFOUL Department of Mathematics University of Dayton, Dayton, OH Abstract In this paper we apply a cone theoretic fixed point theorem and obtain conditions for the existence of positive solutions to the three-point nonlinear second order boundary value problem u (t) + a(t)f(u(t)) =, t (, ) where < < and < α <. u() =, αu() = u(), AMS Subject Classifications: 34B2. Keywords: Cone theory; Three-point; Nonlinear second order boundary value problem; Positive solutions. Introduction In this paper, we are concerned with determining values for so that the three-point nonlinear second order boundary value problem where < <, (A) the function f : [, ) [, ) is continuous, u (t) + a(t)f(u(t)) =, t (, ) (.) u() =, αu() = u(), (.2) (A2) a : [, ] [, ) is continuous and does not vanish identically on any subinterval, (L) lim x x =, (L2) lim x x =, (L3) lim x x =, EJQTDE, 22 No. 5, p.

2 (L4) (L5) lim (L6) lim x x =, x and x = l with < l <, lim x x = L with < L < has positive solutions. In the case =, Ruyun Ma [] showed the existence of positive solutions of (.)-(.2) when f is superlinear (l = and L = ), or f is sublinear (l = and L = ). In this research it is not required that f be either sublinear or superlinear. As in [8] and [], the arguments that we present here in obtaining the existence of a positive solution of (.)-(.2), rely on the fact that solutions are concave downward. In arriving at our results, we make use of Krasnosel skii fixed point theorem []. The existence of positive periodic solutions of nonlinear functional differential equations have been studied extensively in recent years. For some appropriate references we refer the reader to [], [2], [3], [4], [5], [6], [8], [9], [2], [3], [4], [5], [6] and the references therein. In section 2, we state some known results and Krasnosel skii fixed point theorem []. In section 3, we construct the cone of interest and present a lemma, four theorems and a corollary. In each of the theorems and the corollary, an open interval of eigenvalues is determined, which in return, imply the existence of a positive solution of (.)-(.2) by appealing to Krasnosel skii fixed point theorem. We say that u(t) is a solution of (.)-(.2) if u(t) C[, ] and u(t) satisfies (.)-(.2). 2 Preliminaries Theorem 2. (Krasnosel skii) Let B be a Banach space, and let P be a cone in B. Suppose Ω and Ω 2 are bounded open subsets of B such that Ω Ω Ω 2 and suppose that T : P (Ω 2 \Ω ) P is a completely continuous operator such that (i) T u u, u P Ω, and T u u, u P Ω 2 ; or (ii) T u u, u P Ω, and T u u, u P Ω 2. Then T has a fixed point in P (Ω 2 \Ω ). In arriving at our results, we need to state four preliminary Lemmas. Consider the boundary value problem u (t) + y(t) =, t (, ), (I) u() =, αu() = u(), (II) EJQTDE, 22 No. 5, p. 2

3 Lemma 2.2 Let α. Then, for y C[, ], the boundary value problem (I) (II) has the unique solution [ u(t) = t + t (t s)y(s)ds αt ( s)y(s)ds ] ( s)y(s)ds. (2.) The proof of (2.) follows along the lines of the proof that is given in [7] in the case =, and hence we omit it. The proofs of the next three lemmas can be found in []. Lemma 2.3 Let < α < and assume (A) and (A2) hold. Then, the unique solution of (I) (II) is non-negative for all t (, ). Lemma 2.4 Let α > and assume (A) and (A2) hold. Then, (I) (II) has no positive solution. Lemma 2.5 Let < α < (I) (II) satisfies where γ = min{α, α( ) α, }. and assume (A) and (A2) hold. Then, the unique solution of inf u(t) t [,] γ u, The proofs of Lemmas 2.3, 2.4 and 2.5 depend on the fact that under conditions (A) and (A2) the solution u(t) concave downward for t (, ). 3 Main Results Assuming (A) and (A2), it follows from Lemmas 2.3 and 2.4, that (.)-(.2) has a non-negative solution if and only if α <. Therefore, throughout this paper we assume that α <. Let B = C[, ], with y = sup y(t). t [,] Define a cone, P, by P = {y C[, ] : y(t), t (, ) and min y(t) γ y }. t [,] Define an integral operator T : P B [ T u(t) = t + t (t s)a(s)f(u(s))ds αt ( s)a(s)f(u(s))ds ] ( s)a(s)f(u(s))ds. (3.) EJQTDE, 22 No. 5, p. 3

4 By Lemma 2.2, (.)-(.2) has a solution u = u(t) if and only if u solves the operator defined by (3.). Note that, for < α < /, the first two terms on the right of (3.) are less than or equal to zero. We seek a fixed point of T in the cone P. For the sake of simplicity, we let and A = B = ( s)a(s)ds, (3.2) ( s)a(s)ds. (3.3) Lemma 3. Assume that (A) and (A2) hold. If T is given by (3.), then T : P P and is completely continuous. Proof: Let φ, ψ C[, ]. In view of A, given an ɛ > there exists a δ > such that for φ ψ < δ we have ɛ sup f(φ) f(ψ) < t [,] A[2 + α( )]. Using (3.) we have for t (, ), (T φ)(t) (T ψ)(t) + ( s)a(s) f(φ(s)) f(φ(s)) ds α ( s)a(s) f(φ(s)) f(φ(s)) ds + ( s)a(s) f(φ(s)) f(φ(s)) ds [()A + αa + A] f(φ(s)) f(φ(s)) A[2 + α( )] sup f(φ) f(ψ) < ɛ. t [,] Thus, T is continuous. Notice from Lemma 2.3 that, for u P, T u(t) on [, ]. Also, by Lemma 2.5, T P P. Thus, we have shown that T : P P. Next, we show that f maps bonded sets into bounded sets. Let D be a positive constant and define the set K = {x C[, ] : x D}. Since A holds, for any x, y K, there exists a δ > such that if x y < δ, implies f(y) <. We choose a positive integer N so that δ > D N. For x(t) C[, ], define x j(t) = jx(t) N, for j =,, 2,..., N. For x K, x j x j = sup jx(t) (j )x(t) N N t [,] x N D N < δ. EJQTDE, 22 No. 5, p. 4

5 Thus, f(x j ) f(x j ) <. As a consequence, we have which implies that f() = N j= ( ) f(x j ) f(x j ), N f(x j ) f(x j ) + f() j= < N + f(). Thus, f maps bounded sets into bounded sets. It follows from the above inequality and (3.), that Next, for t (, ), we have Hence, [ (T x) (t) = t (T x)(t) t + ( s)a(s) f(x(s)) ds ( s)a(s)(n + f() ) A(N + f() ). a(s)f(u(s))ds α ] ( s)a(s)f(u(s))ds. ( s)a(s)f(u(s))ds (T x) (t) ( s)a(s) f(x(s)) ds A(N + f() ). Thus, the set {(T x) : x P, x D} is a family of uniformly bounded and equicontinuous functions on the set t [, ]. By Ascoli- Arzela Theorem, the map T is completely continuous. This completes the proof. Theorem 3.2 Assume that (A), (A2), (L5) and (L6) hold. Then, for each satisfying γbl < < Al (3.4) (.)-(.2) has at least one positive solution. EJQTDE, 22 No. 5, p. 5

6 Proof: We construct the sets Ω and Ω 2 in order to apply Theorem 2.. Let be given as in (3.4), and choose ɛ > such that γb(l ɛ) A(l + ɛ). By condition (L5), there exists H > such that f(y) (l + ɛ)y, for < y H. So, choosing u P with u = H, we have t (T u)(t) Consequently, T u u. So, if we set then t ( s)a(s)f(u(s))ds ( s)a(s)(l + ɛ)u(s)ds ( s)a(s)(l + ɛ) u ds = ( s)a(s)(l + ɛ)h ds A(l + ɛ) u u. Ω = {y P : y < H }, T u u, for u P Ω. (3.5) Next we construct the set Ω 2. Considering (L6) there exists H 2 such that f(y) (L ɛ)y, for all y H 2. Let H 2 = max{2h, H2 γ } and set If u P with u = H 2, then Thus, by a similar argument as in [], we have (T u)() Ω 2 = {y P : y < H 2 }. min t [,] y(t) γ y H 2. = γ Bγ(L ɛ) u u. ( s)a(s)f(u(s))ds ( s)a(s)(l ɛ)u(s)ds ( s)a(s)(l ɛ)γ u ds ( s)a(s)(l ɛ)h 2 ds EJQTDE, 22 No. 5, p. 6

7 Thus, T u u. Hence T u u, for u P Ω 2. (3.6) Applying (i) of Theorem 2. to (3.5) and (3.6) yields that T has a fixed point u P (Ω 2 \Ω ). The proof is complete. Theorem 3.3 Assume that (A), (A2), (L5) and (L6) hold. Then, for each satisfying γbl < < AL (3.7) (.)-(.2) has at least one positive solution. Proof: We construct the sets Ω and Ω 2 in order to apply Theorem 2.. Let be given as in (3.7), and choose ɛ > such that γb(l ɛ) A(L + ɛ). By condition (L5), there exists H > such that f(y) (l ɛ)y, for < y H. So, choosing u P with u = H, we have Thus, T u u. So, if we let then (T u)() = γ Bγ(l ɛ) u u. ( s)a(s)f(u(s))ds ( s)a(s)(l ɛ)u(s)ds ( s)a(s)(l ɛ)γ u ds ( s)a(s)(l ɛ)h ds Ω = {y P : y < H }, T u u, for u P Ω. (3.8) Next we construct the set Ω 2. Considering (L6) there exists H 2 such that f(y) (L + ɛ)y, for all y H 2. We consider two cases; f is bounded and f is unbounded. The case where f is bounded is straight forward. If f(y) is bounded by Q >, set H 2 = max{2h, QA}. EJQTDE, 22 No. 5, p. 7

8 Then if u P and u = H 2, we have t (T u)(t) Q = AQ H 2 = u. Consequently, T u u. So, if we set then ( s)a(s)f(u(s))ds ( s)a(s)ds Ω 2 = {y P : y < H 2 }, T u u, for u P Ω 2. (3.9) When f is unbounded, we let H 2 > max{2h, H 2 } be such that f(y) f(h 2 ), for < y H 2. For u P with u = H 2, t (T u)(t) ( s)a(s)f(u(s))ds ( s)a(s)f(h 2 )ds ( s)a(s)(l + ɛ)h 2 ds = ( s)a(s)(l + ɛ) u ds = A(L + ɛ) u u. Consequently, T u u. So, if we set then Ω 2 = {y P : y < H 2 }, T u u, for u P Ω 2. (3.) Applying (ii) of Theorem 2. to (3.8) and (3.9) yields that T has a fixed point u P (Ω 2 \Ω ). Also, applying (ii) of Theorem 2. to (3.8) and (3.) yields that T has a fixed point u P (Ω 2 \Ω ). The proof is complete. Theorem 3.4 Assume that (A), (A2), (L) and (L6) hold. Then, for each satisfying < < AL (3.) EJQTDE, 22 No. 5, p. 8

9 (.)-(.2) has at least one positive solution. Proof: Apply (L) and choose H > such that if < y < H, then f(y) y γb. Define Ω = {y P : y < H }. If y P Ω, then (T u)() u. ( s)a(s)f(u(s))ds ( s)a(s) u(s) γb ds ( s)a(s) γ u γb ds In particular, T u u, for all u P Ω. In order to construct Ω 2, we let be given as in (3.), and choose ɛ > such that A(L + ɛ). The construction of Ω 2 follows along the lines of the construction of Ω 2 in Theorem 3.3, and hence we omit it. Thus, by (ii) of Theorem 2., (.)-(.2) has at least one positive solution. Theorem 3.5 Assume that (A), (A2), (L2) and (L5) hold. Then, for each satisfying < < Al (3.2) (.)-(.2) has at least one positive solution. Proof: Assume (L5) holds. Then, we may take the set Ω to be the one obtained for Theorem 3.. That is, Ω = {y P : y < H }. Hence, we have T u u, for u P Ω. Next, we assume (L2). Choose H 2 > such that f(y) y γb, for y H 2. Let H 2 = max{2h, H2 γ } and set Ω 2 = {y P : y < H 2 }. EJQTDE, 22 No. 5, p. 9

10 If u P with u = H 2, Consequently, (T u)() u. ( s)a(s)f(u(s))ds ( s)a(s) u(s) γb ds ( s)a(s) γ u γb ds T u u, for u P Ω 2. Applying (i) of Theorem 2. yields that T has a fixed point u P (Ω 2 \Ω ). We state the next results as corollary, because by now, its proof can be easily obtained from the proofs of the previous results. Corollary 3.6 Assume that (A) and (A2) hold. Also, if either (L3) and (L6) hold, or, (L4) and (L5) hold, then (.)-(.2) has at least one positive solution if satisfies either /(γbl) <, or, /(γbl) <. References [] A. Datta and J. Henderson, Differences and smoothness of solutions for functional difference equations, Proceedings Difference Equations (995), [2] S. Cheng and G. Zhang, Existence of positive periodic solutions for non-autonomous functional differential equations, Electronic Journal of Differential Equations 59 (2), -8. [3] J. Henderson and A. Peterson, Properties of delay variation in solutions of delay difference equations, Journal of Differential Equations (995), [4] R.P. Agarwal and P.J.Y. Wong, On the existence of positive solutions of higher order difference equations, Topological Methods in Nonlinear Analysis (997) 2, [5] P.W. Eloe, Y. Raffoul, D. Reid and K. Yin, Positive solutions of nonlinear Functional Difference Equations, Computers and Mathematics With applications 42 (2), [6] C.P. Gupta and S. I. Trofimchuk, A sharper condition for the stability of a three-point second order boundary value problem, Journal of Mathematical Analysis and Applications, (997). [7] C.P. Gupta, Solvability of a three-point boundary value problem for a second order ordinary differential equation, Journal of Mathematical Analysis and Applications, (997). EJQTDE, 22 No. 5, p.

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