existence and multiplicity for positive solutions of a system of higher-order multi-point boundary...

Nonlinear Differ. Equ. Appl.c© 2012 Springer Basel AGDOI 10.1007/s00030-012-0195-9

Nonlinear Differential Equationsand Applications NoDEA

Existence and multiplicity for positivesolutions of a system of higher-ordermulti-point boundary value problems

Johnny Henderson and Rodica Luca

Abstract. We investigate the existence and multiplicity of positive solu-tions of multi-point boundary value problems for systems of nonlinearhigher-order ordinary differential equations.

Mathematics Subject Classification. 34B10, 34B18.

Keywords. higher-order differential system, multi-point boundaryconditions, positive solutions.

1. Introduction

We consider the system of nonlinear higher-order ordinary differential equa-tions {

u(n)(t) + f(t, v(t)) = 0, t ∈ (0, T ), n ∈ N, n ≥ 2,v(m)(t) + g(t, u(t)) = 0, t ∈ (0, T ), m ∈ N, m ≥ 2,

(S)

with the multi-point boundary conditions⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

u(0) = u′(0) = · · · = u(n−2)(0) = 0, u(T ) =

p−2∑i=1

aiu(ξi), p ∈ N, p ≥ 3,

v(0) = v′(0) = · · · = v(m−2)(0) = 0, v(T ) =

q−2∑i=1

biv(ηi), q ∈ N, q ≥ 3.

(BC)

Under sufficient assumptions on f and g, we prove the existence and mul-tiplicity of positive solutions of the above problem, by applying the fixed pointindex theory. By a positive solution of (S) − (BC), we understand a pair offunctions (u, v) ∈ Cn([0, T ])×Cm([0, T ]) satisfying (S) and (BC) with u(t) ≥0, v(t) ≥ 0 for all t ∈ [0, T ], and supt∈[0,T ] u(t) > 0, supt∈[0,T ] v(t) > 0. Thisproblem is a generalization of the problem studied in [5], where n = m = 2. In[16], the authors investigated the existence and multiplicity of positive solu-tions for system (S) with n = m = 2 and T = 1 with the boundary conditions

J. Henderson and R. Luca NoDEA

u(0) = 0, u(1) = αu(η), v(0) = 0, v(1) = αv(η), η ∈ (0, 1), 0 < αη < 1. We alsomention the paper [14] where the authors used the fixed point index theory toprove the existence of positive solutions for the system (S) with f(t, v(t)) andg(t, u(t)) replaced by c(t)f(u(t), v(t)) and d(t)g(u(t)), v(t)), respectively, and(BC), where 1

2 ≤ ξ1 < ξ2 < · · · < ξp−2 < 1, 12 ≤ η1 < η2 < · · · < ηq−2 < 1,

(T = 1). Some multi-point boundary value problems for systems of ordinarydifferential equations which involve positive eigenvalues were studied in recentyears by Henderson, Luca, Ntouyas and Purnaras, by using the Guo–Krasno-sel’skii fixed point theorem. In [3], the authors give sufficient conditions forλ, μ, f and g such that the system,{

u(n)(t) + λc(t)f(u(t), v(t)) = 0, t ∈ (0, T ), n ∈ N, n ≥ 2,v(m)(t) + μd(t)g(u(t), v(t)) = 0, t ∈ (0, T ), m ∈ N, m ≥ 2,

(S1)

with the boundary conditions (BC), has positive solutions (u(t) ≥ 0, v(t) ≥ 0for all t ∈ [0, T ] and ‖u‖+ ‖v‖ > 0). The system (S1) with n = m = 2 and themulti-point boundary conditions⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

αu(0) − βu′(0) = 0, u(T ) =p−2∑i=1

aiu(ξi), p ∈ N, p ≥ 3,

γv(0) − δv′(0) = 0, v(T ) =q−2∑i=1

biv(ηi), q ∈ N, q ≥ 3,

(BC1)

has been investigated in [4]. Some particular cases of the above problems havebeen studied in [6–9,12,13,15].

In recent years, the multi-point boundary value problems for second-order or higher-order differential or difference equations/systems have beeninvestigated by many authors, by using different methods such as fixed pointtheorems in cones, the Leray–Schauder continuation theorem and its nonlinearalternatives and the coincidence degree theory.

In Sect. 2, we shall present some auxiliary results which investigate twoboundary value problems for higher-order equations (the problems (1)–(2) and(10)–(11) below). In Sect. 3, inspired by the paper [16], we shall prove someexistence and multiplicity results for positive solutions with respect to a conefor our problem (S) − (BC), which are based on three fixed point index theo-rems. Finally, in Sect. 4, we shall present some examples which illustrate ourmain results.

2. Auxiliary results

2.1. Fixed point index theorems

First, we shall recall some theorems concerning the index fixed point theory.Let E be a real Banach space, P ⊂ E a cone, “≤” the partial ordering definedby P and θ the zero element in E. For > 0, let B� = {u ∈ E, ‖u‖ < }. Theproofs of our results are based on the following fixed point index theorems.

Existence and multiplicity for positive solutions

Theorem 2.1. [1] Let A : B� ∩ P → P be a completely continuous operatorwhich has no fixed point on ∂B� ∩ P . If ‖Au‖ ≤ ‖u‖ for all u ∈ ∂B� ∩ P , theni(A,B� ∩ P, P ) = 1.

Theorem 2.2. [1] Let A : B� ∩ P → P be a completely continuous operator. Ifthere exists u0 ∈ P \{θ} such that u−Au = λu0, for all λ ≥ 0 and u ∈ ∂B�∩P ,then i(A,B� ∩ P, P ) = 0.

Theorem 2.3. [16] Let A : B� ∩ P → P be a completely continuous operatorwhich has no fixed point on ∂B�∩P . If there exists a linear operator L : P → Pand u0 ∈ P \ {θ} such that

(i) u0 ≤ Lu0, (ii) Lu ≤ Au, ∀u ∈ ∂B� ∩ P,

then i(A,B� ∩ P, P ) = 0.

2.2. Higher-order differential equations with multi-point boundary conditions

In this subsection, we shall present some auxiliary results (for the proofs see,for example, [11,12]) related to the nth-order differential equation with p-pointboundary conditions

u(n)(t) + y(t) = 0, t ∈ (0, T ), (1)

u(0) = u′(0) = · · · = u(n−2)(0) = 0, u(T ) =p−2∑i=1

aiu(ξi). (2)

Lemma 2.1. If d = Tn−1 − ∑p−2i=1 aiξ

n−1i = 0, 0 < ξ1 < · · · < ξp−2 < T and

y ∈ C([0, T ]), then the solution of (1)–(2) is given by

u(t) =tn−1

d(n − 1)!

∫ T

0

(T − s)n−1y(s) ds − tn−1

d(n − 1)!

p−2∑i=1

ai

∫ ξi

0

(ξi − s)n−1y(s) ds

− 1(n − 1)!

∫ t

0

(t − s)n−1y(s) ds, 0 ≤ t ≤ T.

Lemma 2.2. Under the assumptions of Lemma 2.1, the Green’s function forthe boundary value problem (1)–(2) is given by

G1(t, s) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

tn−1

d(n−1)!

⎡⎣(T − s)n−1 −

p−2∑i=j+1

ai(ξi − s)n−1

⎤⎦ − 1

(n−1)! (t − s)n−1,

if ξj ≤ s < ξj+1, s ≤ t,

tn−1

d(n−1)!

⎡⎣(T − s)n−1 −

p−2∑i=j+1

ai(ξi − s)n−1

⎤⎦ ,

if ξj ≤ s < ξj+1, s ≥ t, j = 0, . . . p − 3,tn−1

d(n−1)! (T − s)n−1− 1(n−1)! (t − s)n−1, if ξp−2 ≤ s ≤ T, s ≤ t,

tn−1

d(n−1)! (T − s)n−1, if ξp−2 ≤ s ≤ T, s ≥ t, (ξ0 = 0),

for all (t, s) ∈ [0, T ] × [0, T ].


Using the Heaviside function on R,H(x) = 1 for x ≥ 0, and H(x) = 0for x < 0, the above Green’s function can be written in a compact form

G1(t, s) =tn−1

d(n − 1)!

[(T − s)n−1 −

p−2∑i=1

ai(ξi − s)n−1H(ξi − s)

]

− 1(n − 1)!

(t − s)n−1H(t − s), (t, s) ∈ [0, T ] × [0, T ].

By using the above Green’s function the solution of problem (1)–(2) isexpressed as u(t) =

∫ T

0G1(t, s)y(s) ds.

Lemma 2.3. Under the assumptions of Lemma 2.1, the Green’s function forthe boundary value problem (1)–(2) can be expressed as

G1(t, s) = g1(t, s) +tn−1

d

p−2∑i=1

aig1(ξi, s), (3)

where

g1(t, s) =1

(n − 1)!T n−1

{tn−1(T − s)n−1 − T n−1(t − s)n−1, 0 ≤ s ≤ t ≤ T,tn−1(T − s)n−1, 0 ≤ t ≤ s ≤ T.

(4)

Proof. If s ≤ t and ξj ≤ s < ξj+1 for j ∈ {0, 1, . . . , p − 3}, we have

G1(t, s) =1

d(n − 1)!

⎡⎣tn−1(T − s)n−1 − tn−1

p−2∑i=j+1

ai(ξi − s)n−1

−(t − s)n−1T n−1 + (t − s)n−1p−2∑i=1

aiξn−1i

]

=1

d(n − 1)!

[tn−1(T − s)n−1 − T n−1(t − s)n−1 + (t − s)n−1

p−2∑i=1

aiξn−1i

− tn−1(T − s)n−1

T n−1

p−2∑i=1

aiξn−1i +

tn−1(T − s)n−1

T n−1

j∑i=1

aiξn−1i

+tn−1(T − s)n−1

T n−1

p−2∑i=j+1

aiξn−1i − tn−1

p−2∑i=j+1

ai(ξi − s)n−1

⎤⎦

=1

d(n−1)!T n−1

{(T n−1−

p−2∑i=1

aiξn−1i

)[tn−1(T −s)n−1−T n−1(t−s)n−1]

+tn−1(T − s)n−1j∑

i=1

aiξn−1i + tn−1

p−2∑i=j+1

ai

[ξn−1

i (T − s)n−1

−T n−1(ξi − s)n−1]}

=1

(n − 1)!T n−1

[tn−1(T − s)n−1 − T n−1(t − s)n−1]


+tn−1

d(n − 1)!T n−1

j∑i=1

aiξn−1i (T − s)n−1

+tn−1

d(n − 1)!T n−1

p−2∑i=j+1

ai

[ξn−1

i (T − s)n−1 − T n−1(ξi − s)n−1]

= g1(t, s) +tn−1

d

p−2∑i=1

aig1(ξi, s).

We use the convention that∑0

i=1 αi = 0.If s ≥ t, ξj ≤ s < ξj+1 for j ∈ {0, 1, . . . , p − 3}, we obtain

G1(t, s) =1

d(n − 1)!

⎡⎣tn−1(T − s)n−1 − tn−1

p−2∑i=j+1

ai(ξi − s)n−1

⎤⎦

=1

d(n − 1)!

[tn−1(T − s)n−1 − tn−1(T − s)n−1

Tn−1

p−2∑i=1

aiξn−1i

+tn−1(T − s)n−1

Tn−1

j∑i=1

aiξn−1i +

tn−1(T − s)n−1

Tn−1

×p−2∑

i=j+1

aiξn−1i − tn−1

p−2∑i=j+1

ai(ξi − s)n−1

⎤⎦ =

tn−1(T − s)n−1

d(n − 1)!Tn−1

×(

Tn−1 −p−2∑i=1

aiξn−1i

)+

tn−1

d(n − 1)!Tn−1

j∑i=1


+tn−1

d(n − 1)!Tn−1

p−2∑i=j+1

ai

[ξn−1i (T − s)n−1 − Tn−1(ξi − s)n−1

]

=1

(n − 1)!Tn−1tn−1(T − s)n−1 +

tn−1

d(n − 1)!

j∑i=1


+tn−1

d(n − 1)!Tn−1

p−2∑i=j+1

ai

[ξn−1i (T − s)n−1 − Tn−1(ξi − s)n−1

]

= g1(t, s) +tn−1

d

p−2∑i=1

aig1(ξi, s).

If s ≤ t and ξp−2 ≤ s ≤ T , we have

G1(t, s) =tn−1

d(n − 1)!(T − s)n−1 − 1

(n − 1)!(t − s)n−1

=tn−1

d(n − 1)!(T − s)n−1 − 1

d(n − 1)!Tn−1tn−1(T − s)n−1

p−2∑i=1

aiξn−1i


− 1(n − 1)!

(t − s)n−1 +tn−1

d(n − 1)!Tn−1

p−2∑i=1


=

(Tn−1 −

p−2∑i=1

aiξn−1i

)1

d(n − 1)!Tn−1tn−1(T − s)n−1

− 1(n − 1)!

(t − s)n−1 +tn−1

d(n − 1)!Tn−1

p−2∑i=1


=1

(n − 1)!Tn−1

[tn−1(T − s)n−1 − Tn−1(t − s)n−1

]

+tn−1

d(n − 1)!Tn−1

p−2∑i=1


= g1(t, s) +tn−1

d

p−2∑i=1

aig1(ξi, s).

If s ≥ t and ξp−2 ≤ s ≤ T , we obtain

G1(t, s) =tn−1(T − s)n−1

d(n − 1)!=

tn−1(T − s)n−1

d(n − 1)!− tn−1(T − s)n−1

d(n − 1)!T n−1

p−2∑i=1

aiξn−1i

+tn−1(T − s)n−1

d(n − 1)!T n−1

p−2∑i=1

aiξn−1i

=tn−1(T − s)n−1

d(n − 1)!T n−1

(T n−1 −

p−2∑i=1

aiξn−1i

)

+tn−1

d(n − 1)!T n−1

p−2∑i=1

aiξn−1i (T − s)n−1 =

1

(n − 1)!T n−1tn−1(T − s)n−1

+tn−1

d(n − 1)!T n−1

p−2∑i=1


= g1(t, s) +tn−1

d

p−2∑i=1

aig1(ξi, s).

Therefore, in every case from above, we obtain the relation (3). �

If ai = 0 for i = 1, . . . , p − 2, then G1(t, s) = g1(t, s) is the Green’sfunction for the problem

w(n)(t) + y0(t) = 0, t ∈ (0, T ), (5)

w(0) = w′(0) = · · · = w(n−2)(0) = 0, w(T ) = 0. (6)


The solution of the above problem is

w(t) =tn−1

(n − 1)!Tn−1

∫ T

0

(T − s)n−1y0(s) ds

− 1(n − 1)!

∫ t

0

(t − s)n−1y0(s) ds, t ∈ [0, T ]

⇔ w(t) =∫ T

0

g1(t, s)y0(s) ds, t ∈ [0, T ],

where g1 is given in (4).Therefore, the function g1 satisfies the conditions g1(0, s) = 0, ∂g1

∂t (0, s) =

0, . . . ∂n−2g1∂tn−2 (0, s) = 0 for all s ∈ [0, T ]. Moreover, we have g1(t, 0) = 0 and

g1(t, T ) = 0 for all t ∈ [0, T ]. We also observe that g1 is a continuous func-tion on [0, T ] × [0, T ] and g1(t, s) ≥ 0 for all (t, s) ∈ [0, T ] × [0, T ]. Indeed, if0 ≤ t ≤ s ≤ T , it is obvious that g1(t, s) ≥ 0. For 0 ≤ s ≤ t ≤ T , we have

g1(t, s) =1

(n − 1)!Tn−1[(tT − ts)n−1 − (tT − sT )n−1] ≥ 0,

because tT − ts ≥ tT − sT ⇔ t ≤ T .By using the above properties of g1 and similar arguments as those used

in the proof of Lemma 3.3 from [10], we obtain the following result.

Lemma 2.4. The Green’s function g1 of problem (5)–(6) satisfies the inequali-ties

(a) g1(t, s) ≤ g1(θ1(s), s), for all (t, s) ∈ [0, T ] × [0, T ],(b) For any c ∈ (0, T

2 ),

mint∈[c,T−c]

g1(t, s) ≥ cn−1

Tn−1g1(θ1(s), s), for all s ∈ [0, T ]

where θ1(s) = s if n = 2 and θ1(s) =

⎧⎨⎩

s

1−(1− sT )

n−1n−2

, s ∈ (0, T ],

T (n−2)n−1 , s = 0,

if n ≥ 3.

In the case n ≥ 3, we choose the values of θ1 in s = 0 and s = T suchthat θ1 be a continuous function on [0, T ].

Lemma 2.5. Assume that d > 0, 0 < ξ1 < · · · < ξp−2 < T, ai ≥ 0 for alli = 1, . . . , p − 2. Then the Green’s function G1 of the problem (1)–(2) is con-tinuous on [0, T ]× [0, T ] and satisfies G1(t, s) ≥ 0 for all (t, s) ∈ [0, T ]× [0, T ].Moreover, if y ∈ C([0, T ]) satisfies y(t) ≥ 0 for all t ∈ [0, T ], then the uniquesolution u of problem (1)–(2) satisfies u(t) ≥ 0 for all t ∈ [0, T ].

The proof of Lemma 2.5 is immediate by using the nonnegativity propertyof g1 and the expressions for G1 and u.

Lemma 2.6. Assume that d > 0, 0 < ξ1 < · · · < ξp−2 < T, ai ≥ 0 for alli = 1, . . . , p − 2. Then the Green’s function G1 of problem (1)–(2) satisfies the


inequalities

(a) G1(t, s) ≤ I1(s), ∀(t, s) ∈ [0, T ] × [0, T ],

where I1(s) = g1(θ1(s), s) + T n−1

d

p−2∑i=1

aig1(ξi, s), ∀ s ∈ [0, T ];(7)

(b) For every c ∈ (0, T2 ),

mint∈[c,T−c]

G1(t, s) ≥ cn−1

Tn−1I1(s), ∀ s ∈ [0, T ]. (8)

Proof. From (3) and Lemma 2.4 we obtain the inequality (a), and for anyc ∈ (0, T

2 ), we have

G1(t, s) ≥ cn−1

Tn−1g1(θ1(s), s) +

cn−1

d

p−2∑i=1

aig1(ξi, s)

=cn−1

Tn−1

[g1(θ1(s), s) +

Tn−1

d

p−2∑i=1

aig1(ξi, s)

]

=cn−1

Tn−1I1(s), ∀ s ∈ [0, T ], t ∈ [c, T − c],

and then mint∈[c,T−c] G1(t, s) ≥ cn−1

T n−1 I1(s) for all s ∈ [0, T ]. �

Lemma 2.7. Assume that d > 0, 0 < ξ1 < · · · < ξp−2 < T, ai ≥ 0 for alli = 1, . . . , p− 2, c ∈ (0, T

2 ) and y ∈ C([0, T ]) satisfies y(t) ≥ 0 for all t ∈ [0, T ].Then the solution u(t), t ∈ [0, T ] of problem (1)–(2) satisfies the inequality

mint∈[c,T−c]

u(t) ≥ cn−1

Tn−1max

t′∈[0,T ]u(t′). (9)

Proof. By using the inequalities (7) and (8), we obtain

u(t) =∫ T

0

G1(t, s)y(s) ds ≥ cn−1

Tn−1

∫ T

0

I1(s)y(s) ds

≥ cn−1

Tn−1

∫ T

0

G1(t′, s)y(s) ds =cn−1

Tn−1u(t′), ∀ t ∈ [c, T − c], t′ ∈ [0, T ].

Therefore, we conclude

mint∈[c,T−c]

u(t) ≥ cn−1

Tn−1max

t′∈[0,T ]u(t′),

that is the inequality (9). �

We can also formulate similar results as Lemmas 2.1–2.7 above for theboundary value problem

v(m)(t) + h(t) = 0, t ∈ (0, T ), (10)

v(0) = v′(0) = · · · = v(m−2)(0) = 0, v(T ) =q−2∑i=1

biv(ηi), (11)


where 0 < η1 < · · · < ηq−2 < T, bi ≥ 0 for i = 1, . . . , q − 2 and h ∈ C([0, T ]).If e = Tm−1 − ∑q−2

i=1 biηm−1i = 0, we denote by G2 the Green’s function asso-

ciated to problem (10)–(11) and defined in a similar manner as G1. We alsodenote by g2, θ2 and I2 the corresponding functions for (10)–(11) defined in asimilar manner as g1, θ1 and I1, respectively.

3. Main results

In this section, we shall investigate the existence and multiplicity of positivesolutions for our problem (S) − (BC), under various assumptions on f and g.

We present the assumptions that we shall use in the sequel(H1) 0 < ξ1 < · · · < ξp−2 < T, ai ≥ 0, i = 1, . . . , p − 2, d = Tn−1 −∑p−2

i=1 aiξn−1i > 0, 0 < η1 < · · · < ηq−2 < T, bi ≥ 0, i = 1, . . . , q − 2, e =

Tm−1 − ∑q−2i=1 biη

m−1i > 0.

(H2) The functions f, g ∈ C([0, T ] × [0,∞), [0,∞)) and f(t, 0) = 0, g(t, 0) = 0for all t ∈ [0, T ].(H3) There exists a positive constant p1 ∈ (0, 1] such that

(i) f i∞ =lim inf

u→∞ inft∈[0,T ]

f(t, u)up1

∈ (0,∞]; (ii) gi∞ = lim inf


g(t, u)u1/p1

=∞.

(H4) There exists a positive constant q1 ∈ (0,∞) such that

(i) fs0 = lim sup

u→0+sup

t∈[0,T ]

f(t, u)uq1

∈ [0,∞); (ii) gs0 = lim sup

u→0+sup

t∈[0,T ]

g(t, u)u1/q1

= 0.

(H5) There exists a positive constant r ∈ (0,∞) such that

(i) fs∞ = lim sup

u→∞sup

t∈[0,T ]

f(t, u)ur

∈ [0,∞); gs∞ = lim sup

u→∞sup

t∈[0,T ]

g(t, u)u1/r

= 0.

(H6) The following conditions are satisfied

f i0 = lim inf

u→0+inf

t∈[0,T ]

f(t, u)u

∈ (0,∞]; gi0 = lim inf

u→0+inf

t∈[0,T ]

g(t, u)u

= ∞.

(H7) For each t ∈ [0, T ], f(t, u) and g(t, u) are nondecreasing with respect tou, and there exists a constant N > 0 such that

f

(t,m0

∫ T

0

g(s,N) ds

)<

N

m0, ∀ t ∈ [0, T ],

where m0 = max{K1T,K2},K1 = maxs∈[0,T ] I1(s),K2 = maxs∈[0,T ] I2(s) andI1, I2 are defined in Sect. 2.

The pair of functions (u, v) ∈ Cn([0, T ])×Cm([0, T ]) is a solution for ourproblem (S) − (BC) if and only if (u, v) ∈ C([0, T ]) × C([0, T ]) is a solutionfor the nonlinear integral system⎧⎪⎪⎨

⎪⎪⎩u(t) =

∫ T

0

G1(t, s)f(s, v(s)) ds, t ∈ [0, T ],

v(t) =∫ T

0

G2(t, s)g(s, u(s)) ds, t ∈ [0, T ].(12)


Besides, the system (12) can be written as the nonlinear integral system⎧⎪⎪⎪⎨⎪⎪⎪⎩

u(t) =∫ T

0

G1(t, s)f

(s,

∫ T

0

G2(s, τ)g(τ, u(τ)) dτ

)ds, t ∈ [0, T ],

v(t) =∫ T

0

G2(t, s)g(s, u(s)) ds, t ∈ [0, T ].

We consider the Banach space X = C([0, T ]) with supremum norm ‖ · ‖and define the cone P ⊂ X by P = {u ∈ X,u(t) ≥ 0,∀ t ∈ [0, T ]}.

We also define the operators A : P → X by

(Au)(t) =∫ T

0

G1(t, s)f

(s,

∫ T

0


)ds, t ∈ [0, T ],

and B : P → X,C : P → X by

(Bu)(t) =∫ T

0

G1(t, s)u(s) ds, (Cu)(t) =∫ T

0

G2(t, s)u(s) ds, t ∈ [0, T ].

Under the assumptions (H1) and (H2), using also Lemma 2.5, it is easyto see that A,B and C are completely continuous from P to P . Thus theexistence and multiplicity of positive solutions of the system (S) − (BC) areequivalent to the existence and multiplicity of fixed points of the operator A.

Theorem 3.1. Assume that (H1)–(H4) hold. Then the problem (S)−(BC) hasat least one positive solution (u(t), v(t)), t ∈ [0, T ].

Proof. From assumption (i) of (H3), we conclude that there exist C1, C2 > 0such that

f(t, u) ≥ C1up1 − C2, ∀ (t, u) ∈ [0, T ] × [0,∞). (13)

Then for u ∈ P , by using (13), the reverse form of Holder’s inequalityand Lemma 2.6, we have for p1 ∈ (0, 1)

(Au)(t) =∫ T

0

G1(t, s)f

(s,

∫ T

0


)ds

≥∫ T

0

G1(t, s)

[C1

(∫ T

0


)p1

− C2

]ds

≥∫ T

0

G1(t, s)

⎡⎣C1

∫ T

0

(G2(s, τ)g(τ, u(τ)))p1 dτ

(∫ T

0

dτ

)p1/q0⎤⎦ ds

−C2

∫ T

0

I1(s) ds = C1Tp1/q0

∫ T

0

G1(t, s)

×(∫ T

0

(G2(s, τ))p1(g(τ, u(τ)))p1 dτ

)ds − C3, ∀ t ∈ [0, T ],

where q0 = p1/(p1 − 1) and C3 = C2

∫ T

0I1(s) ds.


Therefore, for u ∈ P and p1 ∈ (0, 1], we have

(Au)(t) ≥ C1

∫ T

0

G1(t, s)

(∫ T

0

(G2(s, τ))p1(g(τ, u(τ)))p1 dτ

)ds − C3,

∀ t ∈ [0, T ], (14)

where C1 = C1Tp1/q0 for p1 ∈ (0, 1) and C1 = C1 for p1 = 1.

For c ∈ (0, T2 ), we define the cone

P0 ={

u ∈ P ; inft∈[c,T−c]

u(t) ≥ γ‖u‖}

,

where γ = min{cn−1/Tn−1, cm−1/Tm−1

}.

From our assumptions and Lemma 2.7, it can be shown that for anyy ∈ P the functions u(t) = (By)(t) and v(t) = (Cy)(t) satisfy the inequalities

inft∈[c,T−c]

u(t) ≥ cn−1

Tn−1‖u‖ ≥ γ‖u‖, inf

t∈[c,T−c]v(t) ≥ cm−1

Tm−1‖v‖ ≥ γ‖v‖.

So, u = By ∈ P0, v = Cy ∈ P0. Therefore, we deduce that B(P )⊂P0, C(P )⊂P0.

Now we consider the function u0(t), t ∈ [0, T ], the solution of problem (1)–(2) with y = y0, where y0(t) = 1 for all t ∈ [0, T ]. Then u0(t) =

∫ T

0G1(t, s) ds =

(By0)(t), t ∈ [0, T ]. Obviously, we have u0(t) ≥ 0 for all t ∈ [0, T ]. We also con-sider the set

M = {u ∈ P ; there exists λ ≥ 0 such that u = Au + λu0}.

We will show that M ⊂ P0 and M is a bounded subset of X. If u ∈ M , thenthere exists λ ≥ 0 such that u(t) = (Au)(t)+λu0(t), t ∈ [0, T ]. From definitionof u0, we have

u(t)=(Au)(t) + λ(By0)(t)=B(Fu(t)) + λ(By0)(t)=B(Fu(t)+λy0(t)) ∈ P0,

where F : P → P is defined by (Fu)(t) = f(t,∫ T

0G2(t, s)g(s, u(s)) ds

).

Therefore, M ⊂ P0, and from the definition of P0, we have

‖u‖ ≤ 1γ

inft∈[c,T−c]

u(t), ∀u ∈ M. (15)

From (ii) of assumption (H3), we conclude that for ε0 =(

2T p1(m−1)+n−1

C1m1m2cp1(m−1)+n−1

)1/p1

>0 there exists C4 >0 such that

(g(t, u))p1 ≥ εp10 u − C4, ∀ (t, u) ∈ [0, T ] × [0,∞), (16)

where m1 =∫ T−c

cI1(τ) dτ > 0,m2 =

∫ T−c

c(I2(τ))p1 dτ > 0.


For u ∈ M and t ∈ [c, T − c], by using Lemma 2.6 and the relations (14),(16), it follows that

u(t) = (Au)(t) + λu0(t) ≥ (Au)(t)

≥ C1

∫ T

0

G1(t, s)

[∫ T

0

(G2(s, τ))p1(g(τ, u(τ)))p1 dτ

]ds − C3

≥ C1

∫ T−c

c

G1(t, s)

[∫ T−c

c

(G2(s, τ))p1(g(τ, u(τ)))p1 dτ

]ds − C3

≥ C1cp1(m−1)+n−1

T p1(m−1)+n−1

∫ T−c

c

I1(s)

(∫ T−c

c

(I2(τ))p1(g(τ, u(τ)))p1 dτ

)ds − C3

≥ C1cp1(m−1)+n−1

T p1(m−1)+n−1

(∫ T−c

c

I1(s) ds

)(∫ T−c

c

(I2(τ))p1 (εp10 u(τ) − C4) dτ

)− C3

=C1ε

p10 cp1(m−1)+n−1

T p1(m−1)+n−1

(∫ T−c

c

I1(s) ds

)(∫ T−c

c

(I2(τ))p1u(τ) dτ

)− C5

≥ C1εp10 cp1(m−1)+n−1

T p1(m−1)+n−1

(∫ T−c

c

I1(s) ds

)(∫ T−c

c

(I2(τ))p1 dτ

)inf

τ∈[c,T−c]u(τ) − C5

= 2 infτ∈[c,T−c]

u(τ) − C5,

where C5 = C3 + C4C1m1m2cp1(m−1)+n−1

T p1(m−1)+n−1 > 0.Hence, inft∈[c,T−c] u(t) ≥ 2 inft∈[c,T−c] u(t) − C5, and so

inft∈[c,T−c]

u(t) ≤ C5, ∀u ∈ M. (17)

Now from relations (15) and (17), it can be shown that ‖u‖ ≤1γ inft∈[c,T−c] u(t) ≤ C5

γ , for all u ∈ M , that is, M is a bounded subset of X.Besides, there exists a sufficiently large L > 0 such that

u = Au + λu0, ∀u ∈ ∂BL ∩ P, ∀λ ≥ 0.

From Theorem 2.2, we deduce that

i(A,BL ∩ P, P ) = 0. (18)

Next, from (i) of assumption (H4), we conclude that there exists M0 > 0such that

f(t, u) ≤ M0uq, ∀ (t, u) ∈ [0, T ] × [0, 1]. (19)

From (ii) of assumption (H4) and (H2), it can be shown that for ε1 =

min{

1M2

,(

12M0M1M

q12

)1/q1}

> 0, there exists δ1 ∈ (0, 1) such that g(t, u) ≤ε1u

1/q1 for all (t, u) ∈ [0, T ] × [0, δ1], where M1 =∫ T

0I1(s) ds > 0,M2 =∫ T

0I2(s) ds > 0. Hence, we obtain

∫ T

0

G2(t, s)g(s, u(s)) ds ≤ ε1

∫ T

0

I2(s)(u(s))1/q1 ds

≤ ε1M2‖u‖1/q1 ≤1, ∀u∈Bδ1 ∩P, ∀ t∈ [0, T ]. (20)


Therefore, by (19) and (20) we deduce

(Au)(t) =∫ T

0

G1(t, s)f

(s,

∫ T

0


)ds

≤ M0

∫ T

0

G1(t, s)

(∫ T

0


)q1

ds

≤ M0εq11 Mq1

2 ‖u‖∫ T

0

I1(s) ds = M0εq11 M1M

q12 ‖u‖ ≤ 1

2‖u‖,

∀u ∈ Bδ1 ∩ P, t ∈ [0, T ].

This implies that ‖Au‖ ≤ 12‖u‖, ∀u ∈ ∂Bδ1 ∩ P . From Theorem 2.1,

we conclude

i(A,Bδ1 ∩ P, P ) = 1. (21)

Combining (18) and (21), we obtain

i(A, (BL \ Bδ1) ∩ P, P ) = i(A,BL ∩ P, P ) − i(A,Bδ1 ∩ P, P ) = −1.

We conclude that A has at least one fixed point u1 ∈ (BL \ Bδ1) ∩ P , that isδ1 < ‖u1‖ < L.

Let v1(t) =∫ T

0G2(t, s)g(s, u1(s)) ds. Then (u1, v1) ∈ P × P is a solution

of (S) − (BC). In addition ‖v1‖ > 0. Indeed, if we suppose that v1(t) = 0,for all t ∈ [0, T ], then by using (H2) we have f(s, v1(s)) = f(s, 0) = 0, for alls ∈ [0, T ]. This implies u1(t) =

∫ T

0G1(t, s)f(s, v1(s)) ds = 0, for all t ∈ [0, T ],

which contradicts ‖u1‖ > 0. By using Theorem 1.1 from [11] (see [2]), weobtain u1(t) > 0 and v1(t) > 0 for all t ∈ (0, T − c]. The proof of Theorem 3.1is completed. �

Theorem 3.2. Assume that (H1), (H2), (H5) and (H6) hold. Then the problem(S) − (BC) has at least one positive solution (u(t), v(t)), t ∈ [0, T ].

Proof. From the assumption (H5)(i), we deduce that there exist C6, C7 > 0such that

f(t, u) ≤ C6ur + C7, ∀ (t, u) ∈ [0, T ] × [0,∞). (22)

From (H5)(ii), we conclude that for ε2 =(

12C6M1Mr

2

)1/r

there existsC8 > 0 such that

g(t, u) ≤ ε2u1/r + C8, ∀ (t, u) ∈ [0, T ] × [0,∞). (23)


Hence, for u ∈ P , by using (22) and (23), we obtain

(Au)(t) =∫ T

0

G1(t, s)f

(s,

∫ T

0


)ds

≤∫ T

0

G1(t, s)

[C6

(∫ T

0


)r

+ C7

]ds

≤ C6

∫ T

0

G1(t, s)

[∫ T

0

G2(s, τ)(ε2(u(τ))1/r + C8

)dτ

]r

ds + M1C7

≤ C6

∫ T

0

G1(t, s)

[∫ T

0

G2(s, τ)(ε2‖u‖1/r + C8

)dτ

]r

ds + M1C7

= C6

(ε2‖u‖1/r + C8

)r(∫ T

0

I1(s) ds

)(∫ T

0

I2(τ) dτ

)r

+ M1C7,

∀ t ∈ [0, T ].

Therefore, we have

(Au)(t) ≤ C6M1Mr2

(ε2‖u‖1/r + C8

)r

+ M1C7, ∀ t ∈ [0, T ]. (24)

After some computations, it can be shown that

lim‖u‖→∞

[C6M1M

r2

(ε2‖u‖1/r + C8

)r

+ M1C7

] /‖u‖ =12,

and so, there exists a sufficiently large R > 0 such that

C6M1Mr2

(ε2‖u‖1/r + C8

)r

+ M1C7 ≤ 34‖u‖, ∀u ∈ P, ‖u‖ ≥ R. (25)

Hence, from (24) and (25), we obtain ‖Au‖ ≤ 34‖u‖ < ‖u‖, for all u ∈

∂BR ∩ P , and from Theorem 2.1, we have

i(A,BR ∩ P, P ) = 1. (26)

On the other hand, from (H6)(i) we deduce that there exist positive con-stants C9 > 0 and u1 > 0 such that f(t, u) ≥ C9u,∀ (t, u) ∈ [0, T ] × [0, u1].From (H6)(ii), for ε = C0/C9 > 0 with C0 = T n+m−2

cn+m−2m1m3> 0 and m3 =∫ T−c

cI2(τ) dτ > 0, we conclude that there exists ˜u1 > 0 such that g(t, u) ≥ C0

C9u

for all (t, u) ∈ [0, T ]×[0, ˜u1]. We consider u1 = min{u1, ˜u1} and then we obtain

f(t, u) ≥ C9u, g(t, u) ≥ C0

C9u, ∀ (t, u) ∈ [0, T ] × [0, u1]. (27)

Because g(t, 0) = 0 for all t ∈ [0, T ], and g is continuous, it can be shownthat there exists a sufficiently small δ2 ∈ (0, u1) such that g(t, u) ≤ u1

M2for all

(t, u) ∈ [0, T ] × [0, δ2]. Hence,∫ T

0

G2(s, τ)g(τ, u(τ)) dτ ≤∫ T

0

I2(τ)g(τ, u(τ)) dτ ≤ u1, ∀u ∈ Bδ2 ∩ P,

s ∈ [0, T ]. (28)


From (27), (28) and Lemma 2.6, we deduce that for any u ∈ Bδ2 ∩ P, wehave

(Au)(t) =∫ T

0

G1(t, s)f

(s,

∫ T

0


)ds

≥ C9

∫ T

0

G1(t, s)

(∫ T

0


)ds

≥ C0

∫ T

0

G1(t, s)

(∫ T

0

G2(s, τ)u(τ) dτ

)ds

≥ C0

∫ T−c

c

G1(t, s)

(∫ T

0

G2(s, τ)u(τ) dτ

)ds

≥ C0cm−1

Tm−1

∫ T−c

c

G1(t, s)

(∫ T

0

I2(τ)u(τ) dτ

)ds=(Lu)(t), ∀ t∈ [0, T ],

where the linear operator L : P → P is defined by

(Lu)(t) =C0c

m−1

Tm−1

(∫ T

0

I2(τ)u(τ) dτ

)(∫ T−c

c

G1(t, s) ds

), t ∈ [0, T ].

Hence, we obtain

Au ≥ Lu, ∀u ∈ ∂Bδ2 ∩ P. (29)

For w0(t) =∫ T−c

cG1(t, s) ds, t ∈ [0, T ], we have w0 ∈ P \ {θ0} and

(Lw0)(t) =C0c

m−1

T m−1

[∫ T

0

I2(τ)

(∫ T−c

c

G1(τ, s) ds

)dτ

](∫ T−c

c

G1(t, s) ds

)

≥ C0cn+m−2

T n+m−2

(∫ T−c

c

I2(τ) dτ

)(∫ T−c

c

I1(τ) dτ

)(∫ T−c

c

G1(t, s) ds

)

=C0c

n+m−2m1m3

T n+m−2

∫ T−c

c

G1(t, s) ds =

∫ T−c

c

G1(t, s) ds = w0(t), t ∈ [0, T ].

Therefore,

Lw0 ≥ w0. (30)

We may suppose that A has no fixed point on ∂Bδ2 ∩ P (otherwise theproof is finished). From (29), (30) and Theorem 2.3 (with u0 = w0), we con-clude that

i(A,Bδ2 ∩ P, P ) = 0. (31)

Therefore, from (26) and (31), we have

i(A, (BR \ Bδ2) ∩ P, P ) = i(A,BR ∩ P, P ) − i(A,Bδ2 ∩ P, P ) = 1.

Then A has at least one fixed point in (BR \ Bδ2) ∩ P . Thus, the problem(S) − (BC) has at least one positive solution (u, v) ∈ P × P with u(t) ≥0, v(t) ≥ 0 for all t ∈ [0, T ] and ‖u‖ > 0, ‖v‖ > 0. This completes the proof ofTheorem 3.2. �


Theorem 3.3. Assume that (H1)–(H3), (H6) and (H7) hold. Then the problem(S) − (BC) has at least two positive solutions (u1(t), v1(t)), (u2(t), v2(t)), t ∈[0, T ].

Proof. From Sect. 2, we have 0 ≤ G1(t, s) ≤ I1(s) ≤ K1 and G2(t, s) ≤ I2(s) ≤K2, for all (t, s) ∈ [0, T ] × [0, T ]. By using (H7), for any u ∈ ∂BN ∩ P , weobtain

(Au)(t) =∫ T

0

G1(t, s)f

(s,

∫ T

0


)ds

≤∫ T

0

G1(t, s)f

(s,K2

∫ T

0

g(τ, u(τ)) dτ

)ds

≤∫ T

0

G1(t, s)f

(s,m0

∫ T

0

g(τ,N) dτ

)ds

<N

m0

∫ T

0

G1(t, s) ds ≤ N

m0K1T ≤ N, ∀ t ∈ [0, T ].

So, ‖Au‖ < ‖u‖ for all u ∈ ∂BN ∩ P .By Theorem 2.1, we conclude that

i(A,BN ∩ P, P ) = 1. (32)

On the other hand, from (H3), (H6) and the proofs of Theorems 3.1 and3.2, we know that there exists a sufficiently large L > N and a sufficientlysmall δ2 with 0 < δ2 < N such that

i(A,BL ∩ P, P ) = 0, i(A,Bδ2 ∩ P, P ) = 0. (33)

From the relations (32) and (33), we obtain

i(A, (BL \ BN ) ∩ P, P ) = i(A,BL ∩ P, P ) − i(A,BN ∩ P, P ) = −1,i(A, (BN \ Bδ2) ∩ P, P ) = i(A,BN ∩ P, P ) − i(A,Bδ2 ∩ P, P ) = 1.

Then A has at least one fixed point u1 in (BL \BN )∩P and has one fixedpoint u2 in (BN \Bδ2)∩P , respectively. Therefore, the problem (S)−(BC) hastwo distinct positive solutions (u1, v1), (u2, v2) ∈ P×P with ui(t) ≥ 0, vi(t) ≥ 0for all t ∈ [0, T ] and ‖ui‖ > 0, ‖vi‖ > 0, i = 1, 2. The proof of Theorem 3.3 iscompleted. �

4. Examples

In this section, we shall present some examples which illustrate our results.

Example 1. Let f(t, u) = (atα + b)u1/2(2+sin u) and g(t, u) = (ctβ +d)u3(3+cos u) with a, c ≥ 0, b, d > 0, α, β > 0 and p1 = 1/2, q1 = 1/2. Then the


assumptions (H3) and (H4) are satisfied; indeed, we have

f i∞ = lim inf


(atα + b)u1/2(2 + sin u)u1/2

= b,

gi∞ = lim inf


(ctβ + d)u3(3 + cos u)u2

= ∞,

fs0 = lim sup

u→0+sup

t∈[0,T ]

(atα + b)u1/2(2 + sinu)u1/2

= 2(aTα + b),

gs0 = lim sup

u→0+sup

t∈[0,T ]

(ctβ + d)u3(3 + cos u)u2

= 0.

Under the assumption (H1), by Theorem 3.1, we deduce that the problem(S) − (BC) has at least one positive solution.

Example 2. Let f(t, u) = u1/2, g(t, u) = u1/3 and r = 1/2. Then the assump-tions (H5) and (H6) are satisfied; indeed, we have

fs∞ = lim

u→∞u1/2

u1/2= 1, gs

∞ = limu→∞

u1/3

u2= 0, f i

0 = limu→0+

u1/2

u= ∞,

gi0 = lim

u→0+

u1/3

u= ∞.

Under the assumption (H1), by Theorem 3.2, we conclude that the problem(S) − (BC) has at least one positive solution.

Example 3. We consider the following problem

{u(3)(t) + a(vα + vβ) = 0, t ∈ (0, 1),v(4)(t) + b(uγ + uδ) = 0, t ∈ (0, 1),

(S0)

with the multi-point boundary conditions

{u(0) = u′(0) = 0, u(1) = 2u(1

4 ) + u( 12 ) + 1

2u( 34 ),

v(0) = v′(0) = v′′(0) = 0, v(1) = v(13 ) + 3v( 2

3 ), (BC0)

where α > 1, β < 1, γ > 2, δ < 1, a, b > 0. Here T = 1, n = 3,m = 4, p = 5, q =4, ξ1 = 1

4 , ξ2 = 12 , ξ3 = 3

4 , a1 = 2, a2 = 1, a3 = 12 , η1 = 1

3 , η2 = 23 , b1 = 1, b2 = 3.

Then, we have d = 1−∑3i=1 aiξ

2i = 11

32 > 0, e = 1−∑3i=1 biη

3i = 2

27 > 0. Here,

θ1(s) =1

2 − s, s ∈ [0, 1]; θ2(s) =

{ s1−(1−s)3/2 , s ∈ (0, 1],23 , s = 0.


For the functions I1 and I2, we obtain

I1(s) = g1(θ1(s), s) +1d

[2g1

(14, s

)+ g1

(12, s

)+

12g1

(34, s

)]

=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

2s − 5s2 + 4s3 − s4

2(2 − s)2+ 1

22 (−91s2 + 46s), 0 ≤ s < 14 ,

2s − 5s2 + 4s3 − s4

2(2 − s)2+

122

(−27s2 + 14s + 4), 14 ≤ s < 1

2 ,

2s − 5s2 + 4s3 − s4

2(2 − s)2+ 1

22 (5s2 − 18s + 12), 12 ≤ s < 3

4 ,

2s − 5s2 + 4s3 − s4

2(2 − s)2+ 21

22 (1 − 2s + s2), 34 ≤ s ≤ 1,

I2(s) = g2(θ2(s), s) +1e

[g2

(13, s

)+ 3g2

(23, s

)]

=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

0, s = 0,s3(1 − s)3 − s3(1 − s)9/2

6[1 − (1 − s)3/2]3+ 1

12 (83s3 − 114s2 + 42s), 0 < s < 13 ,

s3(1 − s)3 − s3(1 − s)9/2

6[1 − (1 − s)3/2]3+ 1

12 (56s3 − 87s2 + 33s + 1), 13 ≤ s < 2

3 ,

s3(1 − s)3 − s3(1 − s)9/2

6[1 − (1 − s)3/2]3+ 25

12 (1 − 3s + 3s2 − s3), 23 ≤ s ≤ 1.

We have K1 = maxs∈[0,1] I1(s) ≈ 0.306,K2 = maxs∈[0,1] I2(s) ≈ 0.398. Thenm0 = max{K1T,K2} = K2. The functions f(t, u) and g(t, u) are nondecreas-ing with respect to u, for any t ∈ [0, 1], and for p1 = 1/2 the assumptions (H3)and (H6) are satisfied; indeed we obtain

f i∞ = limu→∞

a(uα+uβ)u1/2 = ∞, gi

∞ = limu→∞b(uγ+uδ)

u2 = ∞,

f i0 = limu→0+

a(uα+uβ)u = ∞, gi

0 = limu→0+b(uγ+uδ)

u = ∞.

We take N = 1 and then∫ 1

0g(s, 1) ds = 2b and f(t, 2bm0) = a[(2bm0)α +

(2bm0)β ]. If a[(2bm0)α + (2bm0)β ] < 1m0

⇔ a[mα+1

0 (2b)α + mβ+10 (2b)β

]< 1,

then the assumption (H7) is satisfied. For example, if α = 2, β = 1/2, b = 1/2and a < 1

m30+m

3/20

(e.g. a ≤ 3), then the above inequality is satisfied. By

Theorem 3.3, we deduce that the problem (S0)−(BC0) has at least two positivesolutions.

Acknowledgments

The authors thank the referee for his/her valuable comments and suggestions.The work of R. Luca was supported by the CNCS Grant PN-II-ID-PCE-2011-3-0557, Romania.


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J. HendersonDepartment of MathematicsBaylor UniversityWacoTX 76798-7328USAe-mail: Johnny [email protected]

R. LucaDepartment of MathematicsGh. Asachi Technical University700506 IasiRomaniae-mail: [email protected];

[email protected]

Received: 25 April 2012.

Accepted: 4 August 2012.

existence and multiplicity for positive solutions of a system of higher-order multi-point boundary...

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