J. Math. Anal. Appl. 32 26) 578 59 www.elsevier.com/locate/jmaa Positive solutions of three-point boundary value problems for systems of nonlinear second order ordinary differential equations Youming Zhou, Yan Xu Department of Basic Science, Jiangsu Teachers University of Technology, Changzhou, Jinagsu 235, PR China Received 8 May 25 Available online 24 August 25 Submitted by S. Heikkilä Abstract In this paper, we study the three-point boundary value problems for systems of nonlinear second order ordinary differential equations of the form u = ft,v), t, ), v = gt,u), t, ), u) = v) =, αu)= u), αv) = v). Under some conditions, we show the existence and multiplicity of positive solutions of the above problem by applying the fixed point index theory in cones. 25 Published by Elsevier Inc. Keywords: Systems of ordinary differential equations; Three-point boundary value problem; The fixed point index; Positive solutions Supported by the Qing-Lan Project of Jiangsu Province. * Corresponding author. E-mail address: ymzhou@jstu.edu.cn Y. Zhou). 22-247X/$ see front matter 25 Published by Elsevier Inc. doi:.6/j.jmaa.25.7.4
Y. Zhou, Y. Xu / J. Math. Anal. Appl. 32 26) 578 59 579. Introduction Recently an increasing interest has been observed in investigating the existence of positive solutions of boundary value problems for systems of partial differential equations [ 3]. The study of positive radial solutions for elliptic problems in annular regions can usually be transformed into that of positive solutions of two-point boundary value problems for ordinary differential equations. The multi-point boundary value problems for ordinary differential equations arise in a variety of different areas of applied mathematics and physics. For example, the vibrations of a guy wire of a uniform cross-section and composed of N parts of different densities can be set up as a multi-point boundary value problem [4]; also many problems in the theory of elastic stability can be handled as multi-point problems [5]. The study of multi-point boundary value problems for linear second order ordinary differential equations was initiated by Il in and Moiseev [6]. Since then, nonlinear multi-point boundary value problems have been studied by several authors using the Leray Schauder continuation theorem, nonlinear alternatives of Leray Schauder, coincidence degree theory, and fixed point theorem in cones. We refer the readers to [7 5] for some recent results of nonlinear multi-point boundary value problems. Motivated by the work mentioned above, in this paper we consider the existence and multiplicity of positive solutions to the system of nonlinear second order ordinary differential equations { u = ft,v), t, ), v.) = gt,u), t, ), subject to the three-point boundary condition u) = v) =, αu)= u), αv) = v),.2) where f C[, ] [, ), [, )), g C[, ] [, ), [, )), ft,), gt,), <<, α>, and α <. By a positive solution of.),.2) we understand a pair of functions u, v) C 2 [, ], [, )) C 2 [, ], [, )) which satisfies.),.2) and ut) >, vt) >, t, ). Our purpose here is to give some existence and multiplicity results of positive solutions to.),.2), assuming that A ) There exists a positive constant p, ] such that i) lim inf u ft,u) u p >, ii) lim inf u uniformly on [, ]. A 2 ) There exists a positive constant q, ) such that i) lim sup u + uniformly on [, ]. gt,u) = u/p ft,u) gt,u) u q <, ii) lim sup u + u /q =
58 Y. Zhou, Y. Xu / J. Math. Anal. Appl. 32 26) 578 59 A 3 ) There exists a positive constant r, ) such that lim sup u uniformly on [, ]. A 4 ) Conditions lim inf u + ft,u) gt,u) u r <, lim sup u u /r = ft,u) u >, lim inf u + gt,u) u = hold uniformly on [, ]. A 5 ) f t, u), gt, u) are all nondecreasing with respect to u and there exists a constant N>such that ) f t,m gs,n)ds < N, t [, ], m where m = max{,α} 4 α). The following are the main results of this paper. Theorem. Assume A ) and A 2 ) hold. Then the problem.),.2) has at least one positive solution u, v) C 2 [, ], [, )) C 2 [, ], [, )) with ut) >, vt) >, t, ). Theorem 2. Assume A 3 ) and A 4 ) hold. Then the problem.),.2) has at least one positive solution u, v) C 2 [, ], [, )) C 2 [, ], [, )) with ut) >, vt) >, t, ). Theorem 3. Assume A ), A 4 ) and A 5 ) hold. Then the problem.),.2) has at least two positive solutions u,v ), u 2,v 2 ) C 2 [, ], [, )) C 2 [, ], [, )) with u i t) >, v i t) > i =, 2), t, ). In what follows, we assume that E be a real Banach space, P E a cone and the partial ordering defined by P. θ denotes the zero element in E.Forρ>, let B ρ ={u E u <ρ}. The proofs of the above theorems are based upon the application of the following fixed point index theorems. Lemma. [6] Let A : B ρ P P be a completely continuous operator which has no fixed point on B ρ P.If Au u, u B ρ P. Then ia,b ρ P,P)=. Lemma 2. [6] Let A : B ρ P P be a completely continuous operator. If there exists u P \{θ} such that u Au λu, λ, u B ρ P. Then ia,b ρ P,P)=.
Y. Zhou, Y. Xu / J. Math. Anal. Appl. 32 26) 578 59 58 Lemma 3. Let A : B ρ P P be a completely continuous operator which has no fixed point on B ρ P. If there exists a linear operator L : P P and u P \{θ} such that i) u Lu, ii) Lu Au, u B ρ P. Then ia,b ρ P,P)=. Proof. By Lemma 2, we need only prove that u Au λu, u B ρ P, λ..3) In fact, if not, then there exist x B ρ P and λ such that x Ax = λ u. Hence x = Ax + λ u λ u. Since A has no fixed point on B ρ P, λ >. Let λ = sup{λ> x λu }, then x λ u and λ λ >. On the other hand, from the conditions i) and ii) we have x = Ax + λ u Lx + λ u λ Lu + λ u λ ) + λ u, which contradicts the definition of λ. Thus.3) holds. 2. The preliminary lemmas Lemma 4. [] Let <<, <α</, then for y C[, ], the problem { u + yt) =, t, ), u) =, αu)= u) has a unique solution ut) = 2.) ys)ds, 2.2) where : [, ] [, ] [, ) is defined by t s) α αt s) α t s) for s t and s, t s) α = αt s) α for t s, t s) α for t s and s, t s) for s t. t s) α Lemma 5. [] Let <<, <α</, if y C[, ], yt), t [, ], then the unique solution u of the problem 2.) satisfies ut), t [, ] and min ut) γ u, t [,] where { γ = min α, } α ) α,. 2.3)
582 Y. Zhou, Y. Xu / J. Math. Anal. Appl. 32 26) 578 59 Lemma 6. [3] i) a s s), t, s) [, ] [, ], ii) a 2 s s), t, s) [,] [, ], where a = max{,α} α, a 2 = min{,} min{,α}. α Now we consider the problem.),.2). Obviously, u, v) C 2 [, ] C 2 [, ] is a solution of the problem.),.2) iff u, v) C[, ] C[, ] is a solution of the following system of nonlinear integral equations: { ut) = fs,vs))ds, vt) = gs,us))ds. 2.4) Moreover, the system 2.4) can be written as the nonlinear integral equation ut) = f s, ks,τ)g τ,uτ) ) ) dτ ds. 2.5) Let E = C[, ] be endowed with the norm u =max t [,] ut), let P ={u E ut), t [, ]}, then E, ) is a real Banach space, P is a cone in E. Define Au)t) = Bu)t) = f s, ks,τ)g τ,uτ) ) ) dτ ds, us)ds. 2.6) It is easy to see from Lemmas 4 and 5 that A and B are completely continuous from P to P. Thus, the existence and multiplicity of positive solutions of the problem.),.2) are equivalent to the existence and multiplicity of fixed points of the operator A. Lemma 7. Let <<, <α</, then there exists x,π) such that sin x α sin x =. Proof. Let ϕt) = sin t α sin t, then we have ϕπ) = αsin π <, ϕ) =, and ϕ ) = α >, hence there exists δ,π) such that ϕδ) >. It follows from the intermediate value theorem that there exists x δ, π),π)such that ϕx ) =. This completes the proof.
Y. Zhou, Y. Xu / J. Math. Anal. Appl. 32 26) 578 59 583 By Lemma 7, we know that the three-point eigenvalue problem { u = λu, 2.7) u) =, αu)= u) has an eigenvalue λ = x 2, where x,π)satisfies sin x α sin x =, and sin x t is a positive eigenfunction corresponding to λ = x 2, i.e., sin x t = x 2 sin x sds= x 2 Bsin x t), t [, ]. 2.8) 3. Proof of the theorems Proof of Theorem. From i) of A ), we know that there exist constants C >, C 2 > such that ft,u) C u p C 2, t, u) [, ] [, ). Hence, for u P, by using the Jensen inequality we have Au)t) = C f s, ks,τ)g τ,uτ) ) ) dτ ds [C ks,τ)g τ,uτ) ) ) p ] dτ C 2 ds [ ] ) p )) p ks,τ) g τ,uτ) dτ ds C 3, t [, ], 3.) where C 3 > is a constant. Take u t) = sin x t, where x is as in Lemma 7, then u P.Let { P = u P } min ut) γ u, t [,] where γ is defined by 2.3), then, from Lemma 5 and the definition of the operator B, we have BP) P.Let M ={u P u = Au + λu,λ }. In the following we show that M is a bounded subset of E. Ifu M, then there exists λ such that ut) = Au)t) + λ sin x t. From 2.8) we have ut) = Au)t) + λ sin x t = BFut)+ λx 2 Bsin x t) = B Fut)+ λx 2 sin x t ) P,
584 Y. Zhou, Y. Xu / J. Math. Anal. Appl. 32 26) 578 59 where F : P P is defined by F u)t) = f t, g s,us) ) ) ds, hence, M P and from the definition of P we have u γ min ut), u M. 3.2) t [,] From ii) of A ), there exists a constant C 4 > such that ) p 2a gt,u) C a p u C 4, t, u) [, ] [, ), 3.3) 2 γm2 )2 where m = min t [,] ds = 2 α) min{,α} >,a and a 2 are as in Lemma 6. For u M and t [,], it follows from 3.), Lemma 6 and 3.3) that ut) = Au)t) + λ sin x t Au)t) C C C a p 2 C a p 2 C a p 2 a C a p 2 a 2 γm 2 [ [ ] ) p )) p ks,τ) g τ,uτ) dτ ds C 3 ] ) p )) p ks,τ) g τ,uτ) dτ ds C 3 [ [ [ [ ] ) p )) p τ τ) g τ,uτ) dτ ds C 3 τ τ) g τ,uτ) )) p dτ ] ds C 3 ks,τ) g τ,uτ) )) p dτ ] ds C 3 ks,τ) 2a C a p 2 γm2 uτ) C 4 ) ] dτ ds C 3 Bus) ) ds C 5, 3.4) where C 5 > is a constant. Since Bu P,wehave
Y. Zhou, Y. Xu / J. Math. Anal. Appl. 32 26) 578 59 585 min Bu)s) γ Bu γ s [,] γ kt,τ)uτ)dτ kt,τ)uτ)dτ γ ) kt,τ)dτ min uτ) τ [,] γm min uτ), t [,]. 3.5) τ [,] From 3.4) and 3.5), we have ) ut) 2 γm 2 ds min Bu)s) C 5 2 min uτ) C 5, s [,] τ [,] t [,], hence, min t [,] ut) C 5, u M. It follows from 3.2) that M is a bounded subset of E, and there exists a sufficiently large G> such that ut) Au)t) + λ sin x t, u B G P, λ. From Lemma 2, we have ia,b G P,P)=. 3.6) On the other hand, from i) of A 2 ),wehave { ft,u) } b := sup t, u) [, ], ] u q <. Let { ε = min, m 2bm q+ ) /q }, where m = max t [,] From ii) of A 2 ) we know that there exists δ, ) such that gt,u) ε u /q, t, u) [, ] [,δ ]. Hence, we have g s,us) ) ds u B δ P, t [, ], ds >. ε us) ) /q ds ε m u /q,
586 Y. Zhou, Y. Xu / J. Math. Anal. Appl. 32 26) 578 59 Au)t) = b f s, bε q mq u ks,τ)g τ,uτ) ) ) dτ ds ks,τ)g τ,uτ) ) dτ) q ds ds bε q mq+ u 2 u, u B δ P, t [, ]. This implies that Au 2 u, u B δ P.FromLemmawehave ia,b δ P,P)=. 3.7) Combining 3.6) with 3.7), we have i A, ) ) B G \ B δ P,P = ia,bg P,P) ia,b δ P,P)=. Then A has at least one fixed point u B G \ B δ ) P.Let v t) = g s,u s) ) ds, then u,v ) P P is a solution of the problem.),.2). Now we show that u t) >, v t) >, t, ). In fact, because of u t) = ft,v t)), we know that the graph of u t) is concave down on [, ]. This, together with the fact that u ) min u t) γ u >, t [,] implies that u t) > for all t, ). From the facts that u t) = f s,v s) ) ds and u >, we have that v > otherwise, from v t) it follows fs,v s)) = fs,), and this implies u t), which contradicts u > ). Hence, it is easy to see that v t) >, t, ). This completes the proof of Theorem. Proof of Theorem 2. From A 3 ), we know that there exist constants ε 2 >, C 6 > and C 7 > such that ft,u) ε 2 u r + C 6, t, u) [, ] [, ), ) /r gt,u) 2ε 2 m r+ u /r + C 7, t, u) [, ] [, ),
Y. Zhou, Y. Xu / J. Math. Anal. Appl. 32 26) 578 59 587 where m is as in the proof of Theorem. Hence, for u P,wehave Au)t) [ε 2 ks,τ)g τ,uτ) ) ) r ] dτ + C 6 ds ε 2 ε 2 [ [ ) uτ) /r ) r ks,τ) 2ε 2 m r+ + C 7 dτ] ds + m C 6 ) u /r ) r ks,τ) 2ε 2 m r+ + C 7 dτ] ds + m C 6 [ ) u /r ] r r ε 2 2ε 2 m r+ + C 7 ks,τ)dτ) ds + m C 6 [ ) u /r ] r ε 2 m r+ 2ε 2 m r+ + C 7 + m C 6, t [, ]. 3.8) Through elementary calculation, we have ε 2 m r+ [ u ) /r ] r + C7 + m C 2ε 2 m r+ 6 lim = u u 2, so, there exists a sufficiently large R> such that [ u ε 2 m r+ ) /r ] r + C 7 + m C 6 < 3 u, u P with u R. 4 2ε 2 m r+ Hence, from 3.8) we have Au < u, u B R P,andfromLemmawehave ia,b R P,P)=. 3.9) On the other hand, from A 4 ) we know that there exist constants β>, ξ>such that ft,u) βu, t, u) [, ] [,ξ], 3.) gt,u) x4 u, t, u) [, ] [,ξ], 3.) β where x,π) and x 2 is an eigenvalue of the problem 2.7). From the fact gt,) and the continuity of gt,u), we know that there exists a sufficiently small δ 2,ξ)such that Hence, gt,u) ξ m, t, u) [, ] [,δ 2 ]. ks,τ)g τ,uτ) ) dτ ξ, u B δ2 P, s [, ]. 3.2)
588 Y. Zhou, Y. Xu / J. Math. Anal. Appl. 32 26) 578 59 From 3.), 3.) and 3.2), we have Au)t) = f s, ks,τ)g τ,uτ) ) ) dτ ds β x 4 ks,τ)g τ,uτ) ) ) dτ ds ) ks,τ)uτ)dτ ds = Lu)t), u B δ2 P, t [, ], 3.3) where the linear operator L : P P is defined by Lu)t) = x 4 ks,τ)uτ)dτ ds. From 2.8) we know that u t) = sin x t satisfies Lu )t) = x 4 = x 2 ks,τ)sin x τdτds sin x sds= sin x t = u t), t [, ]. 3.4) We may suppose that A has no fixed point on B δ2 P otherwise, the proof is finished). From 3.3), 3.4) and Lemma 3, we have ia,b δ2 P,P)=. 3.5) Hence, from 3.9) and 3.5) we have i A,B R \ B δ2 ) P,P ) = ia,b R P,P) ia,b δ2 P,P)=. Then A has at least one fixed point in B R \ B δ2 ) P, thus, the problem 2.), 2.2) has at least one positive solution u, v) P P with ut) >, vt) >, t, ). This completes the proof of Theorem 2. Proof of Theorem 3. From i) of Lemma 6, we have a s s) a 4 = m, t, s) [, ] [, ]. Hence, from A 5 ) we have
Au)t) = < N m Y. Zhou, Y. Xu / J. Math. Anal. Appl. 32 26) 578 59 589 f s, f s,m ks,τ)g τ,uτ) ) ) dτ ds ) gτ,n)dτ ds ds N, u B N P, t [, ], so, Au < u for all u B N P. It follows from Lemma that ia,b N P,P)=. 3.6) On the other hand, from A ) and A 4 ), we know that there exists a sufficiently large G>N and a sufficiently small δ 2 with <δ 2 <N such that 3.6) and 3.5) hold, respectively. Hence, from 3.6), 3.5) and 3.6), we have i A,B G \ B N ) P,P ) = ia,b G P,P) ia,b N P,P)=, i A,B N \ B δ2 ) P,P ) = ia,b N P,P) ia,b δ2 P,P)=. Then A has at least one fixed point u in B G \ B N ) P and has one fixed point u 2 in B N \ B δ2 ) P, respectively. Therefore, the problem.),.2) has two distinct positive solutions u,v ), u 2,v 2 ) P P with u i t) >,v i t) > i =, 2) for all t, ). The proof of Theorem 3 is completed. Finally, we give some examples to illustrate our results. Example. Consider the problem.),.2). Let ft,v)= v 2, gt,u) = u 3, p = /2, q = /2, then A ) and A 2 ) are satisfied. Hence, by Theorem, the problem.),.2) has at least one positive solution. In this example, f and g are all superlinear. Example 2. Consider the problem.),.2). Let ft,v)= v /2, gt,u) = u 3, p = /2, q = /2, then A ) and A 2 ) are satisfied. Hence, by Theorem, the problem.),.2) has at least one positive solution, where f is sublinear and g is superlinear. Example 3. Consider the problem.),.2). Let ft,v)= v /2, gt,u) = u /2, r = /2, then A 3 ) and A 4 ) are satisfied. Hence, by Theorem 2, the problem.),.2) has at least one positive solution. In this example, f and g are all sublinear. Example 4. Consider the following problem: u = v 2 + v /2, v = u 3 + u /2, u) = v) =, /2)u/2) = u), /2)v/2) = v). In this example, 3.7)
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