Directory UMM :Journals:Journal_of_mathematics:EJQTDE:
Electronic Journal of Qualitative Theory of Differential Equations
2010, No. 74, 1-15; http://www.math.u-szeged.hu/ejqtde/
EXISTENCE OF POSITIVE SOLUTIONS FOR A CLASS OF
HIGHER-ORDER m-POINT BOUNDARY VALUE
PROBLEMS
Rodica Luca
Department of Mathematics, Gh. Asachi Technical University
11 Blvd. Carol I, Iasi 700506, Romania
E-mail: rluca@math.tuiasi.ro
Abstract. We investigate the existence of positive solutions with respect to a cone
for a higher-order nonlinear differential system, subject to some boundary conditions
which involve m points.
Keywords: Higher-order differential system, boundary conditions, positive solutions, fixed point theorem.
AMS Subject Classification: 34B10, 34B18.
1
Introduction
We consider the higher-order nonlinear differential system
(
u(n) (t) + λb(t)f (v(t)) = 0, t ∈ (0, T ),
(S)
v (n) (t) + µc(t)g(u(t)) = 0, t ∈ (0, T ), n ≥ 2,
with the m-point boundary conditions
m−2
X
′
(n−2)
u(0)
=
u
(0)
=
·
·
·
=
u
(0)
=
0,
u(T
)
=
ai u(ξi),
i=1
(BC)
m−2
X
′
(n−2)
v(0)
=
v
(0)
=
·
·
·
=
v
(0)
=
0,
v(T
)
=
ai v(ξi ), m ≥ 3,
i=1
where 0 < ξ1 < · · · < ξm−2 < T , ai > 0, i = 1, m − 2.
In this paper we shall investigate the existence of positive solutions with respect
to a cone of (S), (BC), where λ, µ > 0. The existence of positive solutions for (S)
m−2
X
′
ai u(ξi) + b0 ,
with n = 2 and the boundary conditions βu(0) − γu (0) = 0, u(T ) =
i=1
EJQTDE, 2010 No. 74, p. 1
βv(0) − γv ′ (0) = 0, v(T ) =
m−2
X
ai v(ξi ) + b0 has been investigated in [19] for b0 = 0 and in
i=1
[18] for b0 > 0 and λ = µ = 1. The corresponding discrete case, namely the system with
second-order differences
(
∆2 un−1 + λbn f (vn ) = 0, n = 1, N − 1
∆2 vn−1 + µcn g(un ) = 0, n = 1, N − 1,
with the m+1-point boundary conditions βu0 −γ∆u0 = 0, uN =
m−2
X
ai uξi , βv0 −γ∆v0 = 0,
i=1
vN =
m−2
X
ai vξi , m ≥ 3 has been studied in [17]. We also mention the paper [15] where the
i=1
authors investigated the existence of positive solutions to the n-th order m-point boundary
value problem u(n) (t) + f (t, u, u′) = 0, t ∈ (0, 1), u(0) = u′ (0) = · · · = u(n−2) (0) =
m−2
X
0, u(1) =
ki u(ξi).
i=1
Due to applications in different areas of applied mathematics and physics, the ex-
istence of positive solutions of multi-point boundary value problems for second-order or
higher-order differential or difference equations has been the subject of investigations by
many authors (see [1]–[14], [16], [20]–[25]).
In Section 2, we shall present several auxiliary results which investigate a boundary
value problem for a n-th order equation (the problem (1), (2) below), some of them from
the paper [15]. In Section 3, we shall give sufficient conditions on λ and µ such that
positive solutions with respect to a cone for our problem (S), (BC) exist. In Section
4, we shall present an example that illustrates the obtained results. Our main results
(Theorem 2 and Theorem 3) are based on the Guo-Krasnoselskii fixed point theorem,
presented below.
Theorem 1. Let X be a Banach space and let C ⊂ X be a cone in X. Assume Ω1 and
Ω2 are bounded open subsets of X with 0 ∈ Ω1 ⊂ Ω1 ⊂ Ω2 and let A : C ∩ (Ω2 \ Ω1 ) → C
be a completely continuous operator such that, either
i) kAuk ≤ kuk, u ∈ C ∩ ∂Ω1 , and kAuk ≥ kuk, u ∈ C ∩ ∂Ω2 , or
ii) kAuk ≥ kuk, u ∈ C ∩ ∂Ω1 , and kAuk ≤ kuk, u ∈ C ∩ ∂Ω2 .
Then A has a fixed point in C ∩ (Ω2 \ Ω1 ).
EJQTDE, 2010 No. 74, p. 2
2
Auxiliary results
In this section, we shall study the n-th order differential equation with the boundary
conditions
u(n) (t) + y(t) = 0, 0 < t < T
′
u(0) = u (0) = · · · = u
(n−2)
(1)
(0) = 0, u(T ) =
m−2
X
ai u(ξi).
(2)
i=1
We denote by d = T
n−1
−
m−2
X
ai ξin−1.
i=1
Lemma 1. If d 6= 0, 0 < ξ1 < · · · < ξm−2 < T and y ∈ C([0, T ]), then the solution
of (1), (2) is given by
tn−1
u(t) =
d(n − 1)!
Z
T
(T − s)
n−1
0
1
−
(n − 1)!
Z
t
Z ξi
m−2
tn−1 X
ai
y(s) ds −
(ξi − s)n−1 y(s) ds
d(n − 1)! i=1
0
(t − s)
n−1
(3)
y(s) ds, 0 ≤ t ≤ T.
0
Proof. By (1) and the first relations from (2) we deduce
Z t
1
Ctn−1
u(t) = −
(t − s)n−1 y(s) ds +
.
(n − 1)! 0
(n − 1)!
From the above relation and the condition u(T ) =
m−2
X
(4)
ai u(ξi) we obtain
i=1
1
−
(n − 1)!
Cξin−1
+
(n − 1)!
or
C
and so
T
n−1
Z
−
T
(T − s)
n−1
0
m−2
X
i=1
ai ξin−1
!
Z ξi
m−2
X
1
CT n−1
y(s) ds +
=
ai −
(ξi − s)n−1 y(s) ds
(n − 1)!
(n
−
1)!
0
i=1
=
Z
0
T
(T − s)
n−1
y(s) ds −
m−2
X
i=1
ai
Z
ξi
(ξi − s)n−1 y(s) ds,
0
Z
Z ξi
m−2
1 T
1X
n−1
ai
C=
(T − s) y(s) ds −
(ξi − s)n−1 y(s) ds.
d 0
d i=1
0
Therefore from (4) and the above expression for C we obtain the relation (3).
✷
Lemma 2. If d 6= 0, 0 < ξ1 < · · · < ξm−2 < T then the Green’s function for the
EJQTDE, 2010 No. 74, p. 3
boundary value problem (1), (2) is given by
#
"
m−2
n−1
X
1
t
(T − s)n−1 −
ai (ξi − s)n−1 −
(t − s)n−1 ,
d(n − 1)!
(n − 1)!
i=j+1
if ξj ≤ s < ξj+1 , s ≤ t,
#
"
m−2
n−1
X
t
(T − s)n−1 −
ai (ξi − s)n−1 ,
d(n − 1)!
G(t, s) =
i=j+1
if ξj ≤ s < ξj+1 , s ≥ t, j = 0, m − 3,
n−1
1
t
(T − s)n−1 −
(t − s)n−1 , if ξm−2 ≤ s ≤ T, s ≤ t,
d(n
−
1)!
(n
−
1)!
tn−1
(T − s)n−1 , if ξm−2 ≤ s ≤ T, s ≥ t, (ξ0 = 0).
d(n − 1)!
Proof. Using the relation (3) we obtain
Z ξi
m−2 Z
m−2
tn−1 X ξj+1
tn−1 X
n−1
u(t) =
ai
(T − s) y(s) ds −
(ξi − s)n−1 y(s) ds
d(n − 1)! j=0 ξj
d(n − 1)! i=1
0
Z t
1
(t − s)n−1 y(s) ds
−
(n − 1)! 0
"Z m−2
m−2 Z
ξ1 X
n−1
t
tn−1 X ξj+1
n−1
ai (ξi − s)n−1 y(s) ds
(T − s) y(s) ds −
=
d(n − 1)! j=0 ξj
d(n − 1)! 0 i=1
#
Z
Z ξ2 m−2
ξm−2
X
am−2 (ξm−2 − s)n−1 y(s) ds
ai (ξi − s)n−1 y(s) ds + · · · +
+
ξm−3
ξ1 i=2
Z t
1
−
(t − s)n−1 y(s) ds
(n − 1)! 0
m−2 Z
m−3 Z
m−2
tn−1 X ξj+1 X
tn−1 X ξj+1
n−1
(T − s) y(s) ds −
ai (ξi − s)n−1 y(s) ds
=
d(n − 1)! j=0 ξj
d(n − 1)! j=0 ξj i=j+1
Z t
1
−
(t − s)n−1 y(s) ds,
(n − 1)! 0
where we denoted ξ0 = 0 and ξm−1 = T.
Therefore, we obtain
"Z
#
Z ξj+1 m−2
m−3
ξj+1
X tn−1
X
u(t) =
(T − s)n−1 y(s) ds −
ai (ξi − s)n−1 y(s) ds
d(n
−
1)!
ξj
ξj
j=0
Z T
Z i=j+1
t
n−1
1
t
n−1
(T − s) y(s) ds −
(t − s)n−1 y(s) ds.
+
d(n − 1)! ξm−2
(n − 1)! 0
By (5) we have u(t) =
of this lemma.
Z
0
(5)
T
G(t, s)y(s) ds, where G is of the form given in the statement
✷
EJQTDE, 2010 No. 74, p. 4
Lemma 3. If ai > 0 for all i = 1, m − 2, 0 < ξ1 < · · · < ξm−2 < T , d > 0 and
y ∈ C([0, T ]), y(t) ≥ 0 for all t ∈ [0, T ], then the unique solution u of problem (1), (2)
satisfies u(t) ≥ 0 for all t ∈ [0, T ].
Proof. We first show that u(T ) ≥ 0. Indeed we have
Z T
Z ξi
m−2
T n−1 X
T n−1
n−1
ai
(T − s) y(s) ds −
(ξi − s)n−1 y(s) ds
u(T ) =
d(n − 1)! 0
d(n − 1)! i=1
0
Z T
1
(T − s)n−1 y(s) ds
−
(n − 1)! 0
Z
Z ξi
m−2
T n−1 − d T
T n−1 X
n−1
=
ai
(T − s) y(s) ds −
(ξi − s)n−1 y(s) ds
d(n − 1)! 0
d(n − 1)! i=1
0
m−2
X
ai ξin−1 Z
Z ξi
m−2
T
T n−1 X
i=1
n−1
ai
=
(T − s) y(s) ds −
(ξi − s)n−1 y(s) ds
d(n − 1)! 0
d(n − 1)! i=1
0
"m−2
Z
Z
ξi
T
X
1
n−1
n−1
n−1
ai ξi
(T − s) y(s) ds +
(T − s) y(s) ds
=
d(n − 1)! i=1
0
ξi
#
m−2
X Z ξi
ai
−T n−1
(ξi − s)n−1 y(s) ds
i=1 ( 0
Z ξi m−2
X
1
=
ai ξin−1 (T − s)n−1 − T n−1 (ξi − s)n−1 y(s) ds
d(n − 1)!
0
i=1
)
Z
m−2
T
X
+
ai ξin−1
(T − s)n−1 y(s) ds ≥ 0,
i=1
ξi
because for s ∈ [0, ξi ] we have ξin−1 (T − s)n−1 − T n−1(ξi − s)n−1 = (ξi T − ξi s)n−1 − (ξiT −
T s)n−1 > 0.
Using a result from [6] (see also Theorem 1.1 from [15]), we deduce that u(t) ≥ 0 for
all t ∈ [0, T ].
✷
Lemma 4. ([15]) If d > 0, ai > 0 for all i = 1, m − 2, 0 < ξ1 < · · · < ξm−2 < T ,
then G(t, s) ≥ 0 for all t, s ∈ [0, T ].
Remark Z
1. Under the assumptions of Lemma 3, by using Lemma 4 and the expresT
sion of u(t) =
G(t, s)y(s) ds, we can also deduce that u(t) ≥ 0 for all t ∈ [0, T ].
0
Lemma 5. If ai > 0 for all i = 1, m − 2, 0 < ξ1 < · · · < ξm−2 < T , d > 0,
y ∈ C([0, T ]), y(t) ≥ 0 for all t ∈ [0, T ], then the solution of problem (1), (2) satisfies
Z T
T n−1
(T − s)n−1 y(s) ds, ∀t ∈ [0, T ],
d(n − 1)! 0
Z T
n−1
ξ
j
u(ξj ) ≥
(T − s)n−1 y(s) ds, ∀ j = 1, m − 2.
d(n − 1)! ξm−2
u(t) ≤
(6)
EJQTDE, 2010 No. 74, p. 5
Proof. By (3) we have
Z T
Z T
T n−1
tn−1
n−1
(T − s) y(s) ds ≤
(T − s)n−1 y(s) ds,
u(t) ≤
d(n − 1)! 0
d(n − 1)! 0
for all t ∈ [0, T ].
Then
ξjn−1
u(ξj ) =
G(ξj , s)y(s) ds ≥
G(ξj , s)y(s) ds =
d(n − 1)!
0
ξm−2
for all j = 1, m − 2.
Z
T
Z
T
Z
T
(T −s)n−1 y(s) ds,
ξm−2
✷
From the proof of Lemma 2.2 in [15] we obtain the following result.
Lemma 6. We assume that 0 < ξ1 < · · · < ξm−2 < T , ai > 0 for all i = 1, m − 2,
d > 0 and y ∈ C([0, T ]), y(t) ≥ 0 for all t ∈ [0, T ]. Then the solution of problem (1), (2)
satisfies
inf
t∈[ξm−2 ,T ]
u(t) ≥ γkuk, where
m−2
n−1
X
am−2 (T − ξm−2 ) am−2 ξm−2
min
,
if
,
ai < 1,
T − am−2 ξm−2
T n−1
i=1
γ=
n−1 n−1
m−2
X
ξm−2
a1 ξ1
min
,
,
if
ai ≥ 1
T n−1 T n−1
i=1
and kuk = sup |u(t)|.
t∈[0,T ]
Remark 2. From the above expression for γ, we see that γ < 1.
3
The existence of positive solutions
In this section we shall give sufficient conditions on λ and µ such that positive solutions
with respect to a cone for problem (S), (BC) exist.
We present the assumptions that we shall use in the sequel.
(H1) 0 < ξ1 < · · · < ξm−2 < T , ai > 0, i = 1, m − 2, d = T
n−1
−
m−2
X
ai ξin−1 > 0.
i=1
(H2) The functions b, c : [0, T ] → [0, ∞) are continuous and each does not vanish
identically on any subinterval of [0, T ].
(H3) The functions f, g : [0, ∞) → [0, ∞) are continuous and the limits
f0 = lim+
x→0
g(x)
f (x)
g(x)
f (x)
, g0 = lim+
, f∞ = lim
, g∞ = lim
x→∞ x
x→∞ x
x→0
x
x
exist and are positive numbers.
Using the Green’s function given in Lemma 2, a pair (u(t), v(t)), t ∈ [0, T ] is a
EJQTDE, 2010 No. 74, p. 6
solution of the eigenvalue problem (S), (BC) if and only if
Z T
Z T
G(t, s)b(s)f µ
G(s, τ )c(τ )g(u(τ )) dτ ds, 0 ≤ t ≤ T,
u(t) = λ
0
0
Z T
v(t) = µ
G(t, s)c(s)g(u(s)) ds, 0 ≤ t ≤ T.
0
We consider the Banach space X = C([0, T ]) with supremum norm k · k and define
the cone C ⊂ X by
C = {u ∈ X, u(t) ≥ 0, ∀ t ∈ [0, T ] and
inf
t∈[ξm−2 ,T ]
u(t) ≥ γkuk},
where γ is defined in Lemma 6.
For our first
the positive numbers L1 and L2 by
(result we define
−1
n−1 Z T
γ 2 ξm−2
n−1
(T − s) b(s)f∞ ds
,
L1 = max
d(n − 1)! ξm−2
2 n−1 Z T
−1 )
γ ξm−2
(T − s)n−1 c(s)g∞ ds
,
d(n − 1)! ξm−2
(
−1
Z T
T n−1
n−1
(T − s) b(s)f0 ds
,
L2 = min
d(n − 1)! 0
−1 )
Z T
T n−1
(T − s)n−1 c(s)g0 ds
.
d(n − 1)! 0
Theorem 2. Assume that (H1)–(H3) hold and L1 < L2 . Then for each λ and µ
satisfying λ, µ ∈ (L1 , L2 ), there exist a positive solution with respect to a cone, (u(t), v(t)),
t ∈ [0, T ], of problem (S), (BC).
Proof. Let λ, µ ∈ (L1 , L2 ) and we choose a positive number ε such that ε < f∞ ,
ε < g∞ , (
2 n−1 Z T
−1
γ ξm−2
n−1
max
(T − s) b(s)(f∞ − ε) ds
,
d(n − 1)! ξm−2
−1 )
2 n−1 Z T
γ ξm−2
(T − s)n−1 c(s)(g∞ − ε) ds
≤ min(λ, µ)
d(n − 1)! ξm−2
and
(
−1
Z T
T n−1
n−1
(T − s) b(s)(f0 + ε) ds
max(λ, µ) ≤ min
,
d(n − 1)! 0
−1 )
Z T
T n−1
(T − s)n−1 c(s)(g0 + ε) ds
.
d(n − 1)! 0
We now define
: ZC → X, by
Z the operator A
T
A(u)(t) = λ
T
G(t, s)b(s)f
0
µ
G(s, τ )c(τ )g(u(τ )) dτ
ds, 0 ≤ t ≤ T, u ∈ C.
0
By Lemma 6, we have A(C) ⊂ C. By using the Arzela-Ascoli theorem we deduce
that the operator A is completely continuous (compact and continuous). By definitions
of f0 and g0 there exists K1 > 0 such that
EJQTDE, 2010 No. 74, p. 7
f (x) ≤ (f0 + ε)x and g(x) ≤ (g0 + ε)x, 0 < x ≤ K1 .
Using (H3) we have f (0) = g(0) = 0 and the above inequalities are also valid for x = 0.
Z T
Let u ∈ C with kuk = K1 . Because v(t) = µ
G(t, s)c(s)g(u(s)) ds, t ∈ [0, T ]
0
satisfies the problem (1), (2) with y(t) = µc(t)g(u(t)), t ∈ [0, T ], then by (6) and the
above property of g, we deduce for t ∈ [0, T ]
Z T
Z T
µT n−1
(T − s)n−1 c(s)g(u(s)) ds
v(t) = µ
G(t, s)c(s)g(u(s)) ds ≤
d(n
−
1)!
0
Z 0T
µT n−1
n−1
(T − s) c(s)(g0 + ε)u(s) ds
≤
d(n − 1)! 0
Z
T
µT n−1
≤
(T − s)n−1 c(s)(g0 + ε)kuk ds ≤ kuk = K1 .
d(n − 1)! 0
By using once again Lemma 5 (relations (6)) and the properties of the function f ,
we have
T
Z T
A(u)(t) = λ
G(t, s)b(s)f µ
G(s, τ )c(τ )g(u(τ )) dτ ds
0
0
Z T
Z T
λT n−1
n−1
≤
(T − s) b(s)f µ
G(s, τ )c(τ )g(u(τ )) dτ ds
d(n − 1)! 0
0 Z
Z
T
T
λT n−1
n−1
(T − s) b(s)(f0 + ε) µ
G(s, τ )c(τ )g(u(τ )) dτ ds
≤
d(n − 1)! 0
0
Z T
λT n−1
≤
(T − s)n−1 b(s)(f0 + ε)K1 ds ≤ K1 = kuk, 0 ≤ t ≤ T.
d(n − 1)! 0
Then kA(u)k ≤ kuk, for all u ∈ C with kuk = K1 . If we denote by Ω1 = {u ∈
Z
C, kuk < K1 }, then we obtain kA(u)k ≤ kuk for all u ∈ C ∩ ∂Ω1 .
¯ 2 > 0 such that
Next, by the definitions of f∞ and g∞ , there exists K
¯ 2.
f (x) ≥ (f∞ − ε)x and g(x) ≥ (g∞ − ε)x, x ≥ K
¯ 2 /γ . For u ∈ C with kuk = K2 , we obtain by using
We consider now K2 = max 2K1 , K
Lemma 6, that
u(t) ≥
inf
s∈[ξm−2 ,T ]
¯ 2 , ∀ t ∈ [ξm−2 , T ].
u(s) ≥ γkuk = γK2 ≥ K
Then, by using (6), Lemma 6, and the above relations, we obtain for t ≥ ξm−2
Z T
v(t) = µ
G(t, s)c(s)g(u(s)) ds ≥ γkvk ≥ γv(ξm−2 )
0
Z T
= γµ
G(ξm−2 , s)c(s)g(u(s)) ds
0
n−1 Z T
γµξm−2
≥
(T − s)n−1 c(s)g(u(s)) ds
d(n − 1)! ξm−2
n−1 Z T
γµξm−2
≥
(T − s)n−1 c(s)(g∞ − ε)u(s) ds
d(n − 1)! ξm−2
n−1 Z T
γ 2 µξm−2
(T − s)n−1 c(s)(g∞ − ε)kuk ds ≥ kuk = K2
≥
d(n − 1)! ξm−2
and
EJQTDE, 2010 No. 74, p. 8
A(u)(ξm−2)
Z T
Z T
n−1
ξm−2
n−1
(T − s) b(s)f µ
≥λ
G(s, τ )c(τ )g(u(τ )) dτ ds
0
ξm−2 d(n − 1)!
Z T
Z T
n−1
λξm−2
n−1
(T − s) b(s)(f∞ − ε) µ
G(s, τ )c(τ )g(u(τ )) dτ ds
≥
d(n − 1)! ξm−2
0
Z T
n−1
λξm−2
(T − s)n−1 b(s)(f∞ − ε)K2 ds
≥
d(n − 1)! ξm−2
n−1 Z T
γ 2 λξm−2
(T − s)n−1 b(s)(f∞ − ε)K2 ds ≥ K2 = kuk.
≥
d(n − 1)! ξm−2
Therefore kA(u)k ≥ A(u)(ξm−2 ) ≥ kuk, for all u ∈ C with kuk = K2 . We denote by
Ω2 = {u ∈ C, kuk < K2 }. Then kA(u)k ≥ kuk, for all u ∈ C ∩ ∂Ω2 .
¯2 \
We now apply Theorem 1 i) and we deduce that A has a fixed point u ∈ C ∩ (Ω
Z T
Ω1 ). This element together with v(t) = µ
G(t, s)c(s)g(u(s)) ds, t ∈ [0, T ] represent a
0
positive solution of (S), (BC) with respect to cone C, for the given λ and µ.
✷
Remark 3. The condition L1 < L2 from Theorem 2 is equivalent to
Z T
−1
Z T
d(n − 1)!
n−1
n−1
min
(T − s) b(s)f∞ ds,
(T − s) c(s)g∞ ds
n−1
γ 2 ξm−2
ξm−2
ξm−2
Z T
−1
Z T
d(n − 1)!
n−1
n−1
max
<
(T − s) b(s)f0 ds,
(T − s) c(s)g0 ds
T n−1
0
0
or
Z T
Z T
n−1
n−1
max
(T − s) b(s)f0 ds,
(T − s) c(s)g0 ds
n−1
γ 2 ξm−2
0
0
.
<
Z T
Z T
T n−1
n−1
n−1
(T − s) c(s)g∞ ds
(T − s) b(s)f∞ ds,
min
ξm−2
ξm−2
In what follows we shall present another existence result for (S), (BC). Let us
consider positive
numbers
(
−1
Z T
n−1
γξm−2
n−1
(T − s) b(s)f0 ds
,
L3 = max
d(n − 1)! ξm−2
−1 )
Z T
n−1
γξm−2
n−1
(T − s) c(s)g0 ds
,
d(n − 1)! ξm−2
(
−1
Z T
T n−1
n−1
(T − s) b(s)f∞ ds
,
L4 = min
d(n − 1)! 0
−1 )
Z T
T n−1
(T − s)n−1 c(s)g∞ ds
.
d(n − 1)! 0
Theorem 3. Assume the assumptions (H1)–(H3) hold and L3 < L4 . Then for each
λ and µ satisfying λ, µ ∈ (L3 , L4 ), there exists a positive solution with respect to a cone,
(u(t), v(t)), t ∈ [0, T ], of (S), (BC).
Proof. Let λ and µ with λ, µ ∈ (L3 , L4 ). We select a positive number ε such that
EJQTDE, 2010 No. 74, p. 9
ε < f0 , ε (
< g0 and
−1
Z T
n−1
γξm−2
n−1
max
(T − s) b(s)(f0 − ε) ds
,
d(n − 1)! ξm−2
−1 )
Z T
n−1
γξm−2
(T − s)n−1 c(s)(g0 − ε) ds
≤ min(λ, µ)
d(n − 1)! ξm−2
and
(
−1
Z T
T n−1
n−1
max(λ, µ) ≤ min
(T − s) b(s)(f∞ + ε) ds
,
d(n − 1)! 0
−1 )
Z T
T n−1
(T − s)n−1 c(s)(g∞ + ε) ds
.
d(n − 1)! 0
We also consider the operator A defined in the proof of Theorem 2. From the defini¯ 3 > 0 such that
tions of f0 and g0 , we deduce that there exists K
¯ 3.
f (x) ≥ (f0 − ε)x and g(x) ≥ (g0 − ε)x, 0 < x ≤ K
Using the properties of f and g the above inequalities are also valid for x = 0.
In addition, because g is a continuous function with g0 > 0, then g(0) = 0 and there
¯ 3 ) such that
exists K3 ∈ (0, K
¯3
K
g(x) ≤
, 0 < x ≤ K3 .
Z
T
µT n−1
n−1
(T − s) c(s) ds
d(n − 1)! 0
For u ∈ C with kuk = K3 , by (6) and the above inequality, we deduce that for all
t ∈ [0, T ]
T
Z T
µT n−1
v(t) = µ
G(t, s)c(s)g(u(s)) ds ≤
(T − s)n−1 c(s)g(u(s)) ds
d(n
−
1)!
0
Z 0T
¯3
K
µT n−1
n−1
¯ 3.
(T − s) c(s)
≤
ds = K
Z T
d(n − 1)! 0
µT n−1
(T − τ )n−1 c(τ ) dτ
d(n − 1)! 0
By using (6), Lemma 6 and the properties of f,g we then obtain
Z T
Z T
n−1
ξm−2
n−1
(T − s) b(s)f µ
G(s, τ )c(τ )g(u(τ )) dτ ds
A(u)(ξm−2) ≥ λ
0
ξm−2 d(n − 1)!
Z T
Z T
n−1
λξm−2
n−1
(T − s) b(s)(f0 − ε) µ
G(s, τ )c(τ )g(u(τ )) dτ ds
≥
d(n − 1)! ξm−2
0
Z T
n−1
λξm−2
(T − s)n−1 b(s)(f0 − ε)γkvk ds
≥
d(n − 1)! ξm−2
n−1 Z T
λγξm−2
≥
(T − s)n−1 b(s)(f0 − ε)v(ξm−2) ds
d(n − 1)! ξm−2
n−1 Z T
λγξm−2
n−1
(T − s) b(s)(f0 − ε) ds
≥
d(n − 1)! ξm−2
Z T
n−1
µξm−2
n−1
(T − s) c(s)g(u(s)) ds
×
d(n − 1)! ξm−2
Z
EJQTDE, 2010 No. 74, p. 10
n−1 Z T
λγξm−2
n−1
(T − s) b(s)(f0 − ε) ds
≥
d(n − 1)! ξm−2
Z T
n−1
µξm−2
n−1
(T − s) c(s)(g0 − ε)u(s) ds
×
d(n − 1)! ξm−2
n−1 Z T
λγξm−2
n−1
(T − s) b(s)(f0 − ε) ds
≥
d(n − 1)! ξm−2
n−1 Z T
µγξm−2
n−1
×
(T − s) c(s)(g0 − ε) ds kuk ≥ kuk.
d(n − 1)! ξm−2
Hence, kA(u)k ≥ A(u)(ξm−2) ≥ kuk, for u ∈ C with kuk = K3 . We denote by
Ω3 = {u ∈ C, kuk < K3 }, and then we have kA(u)k ≥ kuk for all u ∈ C ∩ ∂Ω3 .
We now consider the functions f ∗ , g ∗ : [0, ∞) → [0, ∞) defined by f ∗ (x) = sup f (y),
0≤y≤x
f ∗ (x)
∗
∗
∗
= f∞ ,
g (x) = sup g(y). By (H2) we obtain for f and g the relations lim
x→∞
x
0≤y≤x
g ∗ (x)
lim
= g∞ .
x→∞
x
¯ 4 > 0 such
We also have f (x) ≤ f ∗ (x), g(x) ≤ g ∗(x), for all x ≥ 0. Then there exists K
that
¯ 4.
f ∗ (x) ≤ (f∞ + ε)x, g ∗(x) ≤ (g∞ + ε)x, for all x ≥ K
¯ 4 }. Then for u with kuk = K4 we obtain
Let K4 > max{2K3 , K
Z T
Z T
n−1
T
n−1
A(u)(t) ≤
(T − s) b(s)λf µ
G(s, τ )c(τ )g(u(τ )) dτ ds
d(n − 1)! 0
0
Z
Z
T
T
λT n−1
n−1
∗
(T − s) b(s)f µ
G(s, τ )c(τ )g(u(τ )) dτ ds
≤
d(n − 1)! 0
0
Z T
Z T
λT n−1
µT n−1
n−1
∗
n−1
≤
(T − s) b(s)f
(T − τ ) c(τ )g(u(τ )) dτ ds
d(n − 1)! 0
d(n
−
1)!
0
Z T
Z T
µT n−1
λT n−1
n−1
∗
n−1
∗
(T − s) b(s)f
(T − τ ) c(τ )g (u(τ )) dτ ds
≤
d(n − 1)! 0
d(n − 1)! 0
Z
Z
T
T
λT n−1
µT n−1
n−1
∗
n−1
∗
≤
(T − s) b(s)f
(T − τ ) c(τ )g (K4 ) dτ ds
d(n − 1)! 0
d(n
−
1)!
0
Z T
Z T
µT n−1
λT n−1
n−1
∗
n−1
(T − s) b(s)f
(T − τ ) c(τ )(g∞ + ε)K4 dτ ds
≤
d(n − 1)! 0
d(n − 1)! 0
Z T
λT n−1
≤
(T − s)n−1 b(s)f ∗ (K4 ) ds
d(n − 1)! 0
Z T
λT n−1
≤
(T − s)n−1 b(s)(f∞ + ε)K4 ds ≤ K4 = kuk.
d(n − 1)! 0
So kA(u)k ≤ kuk, for all u ∈ C with kuk = K4 . If we denote by Ω4 = {u ∈ C, kuk <
K4 }, then we obtain kA(u)k ≤ kuk, for all u ∈ C ∩ ∂Ω4 .
¯ 4 \Ω3 ), which together
By Theorem 1 ii) we deduce that A has a fixed point u ∈ C ∩(Ω
Z T
with v(t) = µ
G(t, s)c(s)g(u(s)) ds, t ∈ [0, T ] give us a positive solution of (S), (BC)
0
with respect to cone C, for the chosen values λ and µ.
✷
EJQTDE, 2010 No. 74, p. 11
Remark 4. The condition L3 < L4 is equivalent to
Z T
−1
Z T
d(n − 1)!
n−1
n−1
min
(T − s) b(s)f0 ds,
(T − s) c(s)g0 ds
n−1
γξm−2
ξm−2
ξm−2
Z T
−1
Z T
d(n − 1)!
n−1
n−1
<
max
(T − s) b(s)f∞ ds,
(T − s) c(s)g∞ ds
T n−1
0
0
or
Z T
Z T
n−1
n−1
max
(T − s) b(s)f∞ ds,
(T − s) c(s)g∞ ds
n−1
γξm−2
0
0
.
Z T
< n−1
Z T
T
n−1
n−1
min
(T − s) b(s)f0 ds,
(T − s) c(s)g0 ds
ξm−2
4
ξm−2
An example
As in Example
( 4.1 in [12], let us consider the functions
f (x) = p2 | sin x| + p1 xe−1/x , x ∈ [0, ∞),
g(x) = q2 | sin x| + q1 xe−1/x , x ∈ [0, ∞),
with p1 , p2 , q1 , q2 > 0.
g(x)
f (x)
g(x)
f (x)
= p2 , lim
= p1 , lim+
= q2 , lim
= q1 .
We have lim+
x→∞ x
x→∞ x
x→0
x→0
x
x
Let T = 1, n = 3, m = 4, b(t) = b0 t, c(t) = c0 t, t ∈ [0, 1], with b0 , c0 > 0 and
ξ1 = 31 , ξ2 = 32 , a1 = 1, a2 = 21 .
We consider the third-order differential system
(
u′′′ (t) + λb0 t p2 | sin v(t)| + p1 v(t)e−1/v(t) = 0, t ∈ (0, 1)
(S0 )
v ′′′ (t) + µc0 t q2 | sin u(t)| + q1 |u(t)|e−1/u(t) = 0, t ∈ (0, 1),
with the boundary conditions
(
u(0) = u′ (0) = 0, u(1) = u( 31 ) + 12 u( 23 )
(BC0 )
v(0) = v ′ (0) = 0, v(1) = v( 31 ) + 12 v( 32 ).
2
X
2
X
2
3
We also have d = 1 −
= > 0,
ai = > 1 and γ = min{a1 ξ12 , ξ22} = 91 .
3
2
i=1
i=1
The condition
form given
Z L1 < L2 or the equivalent
in Remark 3 is
Z
ai ξi2
1
or
max
Z
min
1
2
(1 − s)2 c0 sq2 ds
(1 − s) b0 sp2 ds,
0
1
(1 − s)2 b0 sp1 ds,
2/3
Z 01
2/3
(1 − s)2 c0 sq1 ds
<
4
729
max{b0 p2 , c0 q2 }
4
<
.
min{b0 p1 , c0 q1 }
6561
EJQTDE, 2010 No. 74, p. 12
Therefore if the above condition is verified, then by Theorem 2 we deduce that for
all numbers λ, µ ∈ (L1 , L2 ) the problem (S0 ), (BC0 ) has positive solutions.
Acknowledgment. The author would like to thank the referee for the careful reading of the manuscript and for his comments which led to this revised version.
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(Received September 5, 2010)
EJQTDE, 2010 No. 74, p. 15
2010, No. 74, 1-15; http://www.math.u-szeged.hu/ejqtde/
EXISTENCE OF POSITIVE SOLUTIONS FOR A CLASS OF
HIGHER-ORDER m-POINT BOUNDARY VALUE
PROBLEMS
Rodica Luca
Department of Mathematics, Gh. Asachi Technical University
11 Blvd. Carol I, Iasi 700506, Romania
E-mail: rluca@math.tuiasi.ro
Abstract. We investigate the existence of positive solutions with respect to a cone
for a higher-order nonlinear differential system, subject to some boundary conditions
which involve m points.
Keywords: Higher-order differential system, boundary conditions, positive solutions, fixed point theorem.
AMS Subject Classification: 34B10, 34B18.
1
Introduction
We consider the higher-order nonlinear differential system
(
u(n) (t) + λb(t)f (v(t)) = 0, t ∈ (0, T ),
(S)
v (n) (t) + µc(t)g(u(t)) = 0, t ∈ (0, T ), n ≥ 2,
with the m-point boundary conditions
m−2
X
′
(n−2)
u(0)
=
u
(0)
=
·
·
·
=
u
(0)
=
0,
u(T
)
=
ai u(ξi),
i=1
(BC)
m−2
X
′
(n−2)
v(0)
=
v
(0)
=
·
·
·
=
v
(0)
=
0,
v(T
)
=
ai v(ξi ), m ≥ 3,
i=1
where 0 < ξ1 < · · · < ξm−2 < T , ai > 0, i = 1, m − 2.
In this paper we shall investigate the existence of positive solutions with respect
to a cone of (S), (BC), where λ, µ > 0. The existence of positive solutions for (S)
m−2
X
′
ai u(ξi) + b0 ,
with n = 2 and the boundary conditions βu(0) − γu (0) = 0, u(T ) =
i=1
EJQTDE, 2010 No. 74, p. 1
βv(0) − γv ′ (0) = 0, v(T ) =
m−2
X
ai v(ξi ) + b0 has been investigated in [19] for b0 = 0 and in
i=1
[18] for b0 > 0 and λ = µ = 1. The corresponding discrete case, namely the system with
second-order differences
(
∆2 un−1 + λbn f (vn ) = 0, n = 1, N − 1
∆2 vn−1 + µcn g(un ) = 0, n = 1, N − 1,
with the m+1-point boundary conditions βu0 −γ∆u0 = 0, uN =
m−2
X
ai uξi , βv0 −γ∆v0 = 0,
i=1
vN =
m−2
X
ai vξi , m ≥ 3 has been studied in [17]. We also mention the paper [15] where the
i=1
authors investigated the existence of positive solutions to the n-th order m-point boundary
value problem u(n) (t) + f (t, u, u′) = 0, t ∈ (0, 1), u(0) = u′ (0) = · · · = u(n−2) (0) =
m−2
X
0, u(1) =
ki u(ξi).
i=1
Due to applications in different areas of applied mathematics and physics, the ex-
istence of positive solutions of multi-point boundary value problems for second-order or
higher-order differential or difference equations has been the subject of investigations by
many authors (see [1]–[14], [16], [20]–[25]).
In Section 2, we shall present several auxiliary results which investigate a boundary
value problem for a n-th order equation (the problem (1), (2) below), some of them from
the paper [15]. In Section 3, we shall give sufficient conditions on λ and µ such that
positive solutions with respect to a cone for our problem (S), (BC) exist. In Section
4, we shall present an example that illustrates the obtained results. Our main results
(Theorem 2 and Theorem 3) are based on the Guo-Krasnoselskii fixed point theorem,
presented below.
Theorem 1. Let X be a Banach space and let C ⊂ X be a cone in X. Assume Ω1 and
Ω2 are bounded open subsets of X with 0 ∈ Ω1 ⊂ Ω1 ⊂ Ω2 and let A : C ∩ (Ω2 \ Ω1 ) → C
be a completely continuous operator such that, either
i) kAuk ≤ kuk, u ∈ C ∩ ∂Ω1 , and kAuk ≥ kuk, u ∈ C ∩ ∂Ω2 , or
ii) kAuk ≥ kuk, u ∈ C ∩ ∂Ω1 , and kAuk ≤ kuk, u ∈ C ∩ ∂Ω2 .
Then A has a fixed point in C ∩ (Ω2 \ Ω1 ).
EJQTDE, 2010 No. 74, p. 2
2
Auxiliary results
In this section, we shall study the n-th order differential equation with the boundary
conditions
u(n) (t) + y(t) = 0, 0 < t < T
′
u(0) = u (0) = · · · = u
(n−2)
(1)
(0) = 0, u(T ) =
m−2
X
ai u(ξi).
(2)
i=1
We denote by d = T
n−1
−
m−2
X
ai ξin−1.
i=1
Lemma 1. If d 6= 0, 0 < ξ1 < · · · < ξm−2 < T and y ∈ C([0, T ]), then the solution
of (1), (2) is given by
tn−1
u(t) =
d(n − 1)!
Z
T
(T − s)
n−1
0
1
−
(n − 1)!
Z
t
Z ξi
m−2
tn−1 X
ai
y(s) ds −
(ξi − s)n−1 y(s) ds
d(n − 1)! i=1
0
(t − s)
n−1
(3)
y(s) ds, 0 ≤ t ≤ T.
0
Proof. By (1) and the first relations from (2) we deduce
Z t
1
Ctn−1
u(t) = −
(t − s)n−1 y(s) ds +
.
(n − 1)! 0
(n − 1)!
From the above relation and the condition u(T ) =
m−2
X
(4)
ai u(ξi) we obtain
i=1
1
−
(n − 1)!
Cξin−1
+
(n − 1)!
or
C
and so
T
n−1
Z
−
T
(T − s)
n−1
0
m−2
X
i=1
ai ξin−1
!
Z ξi
m−2
X
1
CT n−1
y(s) ds +
=
ai −
(ξi − s)n−1 y(s) ds
(n − 1)!
(n
−
1)!
0
i=1
=
Z
0
T
(T − s)
n−1
y(s) ds −
m−2
X
i=1
ai
Z
ξi
(ξi − s)n−1 y(s) ds,
0
Z
Z ξi
m−2
1 T
1X
n−1
ai
C=
(T − s) y(s) ds −
(ξi − s)n−1 y(s) ds.
d 0
d i=1
0
Therefore from (4) and the above expression for C we obtain the relation (3).
✷
Lemma 2. If d 6= 0, 0 < ξ1 < · · · < ξm−2 < T then the Green’s function for the
EJQTDE, 2010 No. 74, p. 3
boundary value problem (1), (2) is given by
#
"
m−2
n−1
X
1
t
(T − s)n−1 −
ai (ξi − s)n−1 −
(t − s)n−1 ,
d(n − 1)!
(n − 1)!
i=j+1
if ξj ≤ s < ξj+1 , s ≤ t,
#
"
m−2
n−1
X
t
(T − s)n−1 −
ai (ξi − s)n−1 ,
d(n − 1)!
G(t, s) =
i=j+1
if ξj ≤ s < ξj+1 , s ≥ t, j = 0, m − 3,
n−1
1
t
(T − s)n−1 −
(t − s)n−1 , if ξm−2 ≤ s ≤ T, s ≤ t,
d(n
−
1)!
(n
−
1)!
tn−1
(T − s)n−1 , if ξm−2 ≤ s ≤ T, s ≥ t, (ξ0 = 0).
d(n − 1)!
Proof. Using the relation (3) we obtain
Z ξi
m−2 Z
m−2
tn−1 X ξj+1
tn−1 X
n−1
u(t) =
ai
(T − s) y(s) ds −
(ξi − s)n−1 y(s) ds
d(n − 1)! j=0 ξj
d(n − 1)! i=1
0
Z t
1
(t − s)n−1 y(s) ds
−
(n − 1)! 0
"Z m−2
m−2 Z
ξ1 X
n−1
t
tn−1 X ξj+1
n−1
ai (ξi − s)n−1 y(s) ds
(T − s) y(s) ds −
=
d(n − 1)! j=0 ξj
d(n − 1)! 0 i=1
#
Z
Z ξ2 m−2
ξm−2
X
am−2 (ξm−2 − s)n−1 y(s) ds
ai (ξi − s)n−1 y(s) ds + · · · +
+
ξm−3
ξ1 i=2
Z t
1
−
(t − s)n−1 y(s) ds
(n − 1)! 0
m−2 Z
m−3 Z
m−2
tn−1 X ξj+1 X
tn−1 X ξj+1
n−1
(T − s) y(s) ds −
ai (ξi − s)n−1 y(s) ds
=
d(n − 1)! j=0 ξj
d(n − 1)! j=0 ξj i=j+1
Z t
1
−
(t − s)n−1 y(s) ds,
(n − 1)! 0
where we denoted ξ0 = 0 and ξm−1 = T.
Therefore, we obtain
"Z
#
Z ξj+1 m−2
m−3
ξj+1
X tn−1
X
u(t) =
(T − s)n−1 y(s) ds −
ai (ξi − s)n−1 y(s) ds
d(n
−
1)!
ξj
ξj
j=0
Z T
Z i=j+1
t
n−1
1
t
n−1
(T − s) y(s) ds −
(t − s)n−1 y(s) ds.
+
d(n − 1)! ξm−2
(n − 1)! 0
By (5) we have u(t) =
of this lemma.
Z
0
(5)
T
G(t, s)y(s) ds, where G is of the form given in the statement
✷
EJQTDE, 2010 No. 74, p. 4
Lemma 3. If ai > 0 for all i = 1, m − 2, 0 < ξ1 < · · · < ξm−2 < T , d > 0 and
y ∈ C([0, T ]), y(t) ≥ 0 for all t ∈ [0, T ], then the unique solution u of problem (1), (2)
satisfies u(t) ≥ 0 for all t ∈ [0, T ].
Proof. We first show that u(T ) ≥ 0. Indeed we have
Z T
Z ξi
m−2
T n−1 X
T n−1
n−1
ai
(T − s) y(s) ds −
(ξi − s)n−1 y(s) ds
u(T ) =
d(n − 1)! 0
d(n − 1)! i=1
0
Z T
1
(T − s)n−1 y(s) ds
−
(n − 1)! 0
Z
Z ξi
m−2
T n−1 − d T
T n−1 X
n−1
=
ai
(T − s) y(s) ds −
(ξi − s)n−1 y(s) ds
d(n − 1)! 0
d(n − 1)! i=1
0
m−2
X
ai ξin−1 Z
Z ξi
m−2
T
T n−1 X
i=1
n−1
ai
=
(T − s) y(s) ds −
(ξi − s)n−1 y(s) ds
d(n − 1)! 0
d(n − 1)! i=1
0
"m−2
Z
Z
ξi
T
X
1
n−1
n−1
n−1
ai ξi
(T − s) y(s) ds +
(T − s) y(s) ds
=
d(n − 1)! i=1
0
ξi
#
m−2
X Z ξi
ai
−T n−1
(ξi − s)n−1 y(s) ds
i=1 ( 0
Z ξi m−2
X
1
=
ai ξin−1 (T − s)n−1 − T n−1 (ξi − s)n−1 y(s) ds
d(n − 1)!
0
i=1
)
Z
m−2
T
X
+
ai ξin−1
(T − s)n−1 y(s) ds ≥ 0,
i=1
ξi
because for s ∈ [0, ξi ] we have ξin−1 (T − s)n−1 − T n−1(ξi − s)n−1 = (ξi T − ξi s)n−1 − (ξiT −
T s)n−1 > 0.
Using a result from [6] (see also Theorem 1.1 from [15]), we deduce that u(t) ≥ 0 for
all t ∈ [0, T ].
✷
Lemma 4. ([15]) If d > 0, ai > 0 for all i = 1, m − 2, 0 < ξ1 < · · · < ξm−2 < T ,
then G(t, s) ≥ 0 for all t, s ∈ [0, T ].
Remark Z
1. Under the assumptions of Lemma 3, by using Lemma 4 and the expresT
sion of u(t) =
G(t, s)y(s) ds, we can also deduce that u(t) ≥ 0 for all t ∈ [0, T ].
0
Lemma 5. If ai > 0 for all i = 1, m − 2, 0 < ξ1 < · · · < ξm−2 < T , d > 0,
y ∈ C([0, T ]), y(t) ≥ 0 for all t ∈ [0, T ], then the solution of problem (1), (2) satisfies
Z T
T n−1
(T − s)n−1 y(s) ds, ∀t ∈ [0, T ],
d(n − 1)! 0
Z T
n−1
ξ
j
u(ξj ) ≥
(T − s)n−1 y(s) ds, ∀ j = 1, m − 2.
d(n − 1)! ξm−2
u(t) ≤
(6)
EJQTDE, 2010 No. 74, p. 5
Proof. By (3) we have
Z T
Z T
T n−1
tn−1
n−1
(T − s) y(s) ds ≤
(T − s)n−1 y(s) ds,
u(t) ≤
d(n − 1)! 0
d(n − 1)! 0
for all t ∈ [0, T ].
Then
ξjn−1
u(ξj ) =
G(ξj , s)y(s) ds ≥
G(ξj , s)y(s) ds =
d(n − 1)!
0
ξm−2
for all j = 1, m − 2.
Z
T
Z
T
Z
T
(T −s)n−1 y(s) ds,
ξm−2
✷
From the proof of Lemma 2.2 in [15] we obtain the following result.
Lemma 6. We assume that 0 < ξ1 < · · · < ξm−2 < T , ai > 0 for all i = 1, m − 2,
d > 0 and y ∈ C([0, T ]), y(t) ≥ 0 for all t ∈ [0, T ]. Then the solution of problem (1), (2)
satisfies
inf
t∈[ξm−2 ,T ]
u(t) ≥ γkuk, where
m−2
n−1
X
am−2 (T − ξm−2 ) am−2 ξm−2
min
,
if
,
ai < 1,
T − am−2 ξm−2
T n−1
i=1
γ=
n−1 n−1
m−2
X
ξm−2
a1 ξ1
min
,
,
if
ai ≥ 1
T n−1 T n−1
i=1
and kuk = sup |u(t)|.
t∈[0,T ]
Remark 2. From the above expression for γ, we see that γ < 1.
3
The existence of positive solutions
In this section we shall give sufficient conditions on λ and µ such that positive solutions
with respect to a cone for problem (S), (BC) exist.
We present the assumptions that we shall use in the sequel.
(H1) 0 < ξ1 < · · · < ξm−2 < T , ai > 0, i = 1, m − 2, d = T
n−1
−
m−2
X
ai ξin−1 > 0.
i=1
(H2) The functions b, c : [0, T ] → [0, ∞) are continuous and each does not vanish
identically on any subinterval of [0, T ].
(H3) The functions f, g : [0, ∞) → [0, ∞) are continuous and the limits
f0 = lim+
x→0
g(x)
f (x)
g(x)
f (x)
, g0 = lim+
, f∞ = lim
, g∞ = lim
x→∞ x
x→∞ x
x→0
x
x
exist and are positive numbers.
Using the Green’s function given in Lemma 2, a pair (u(t), v(t)), t ∈ [0, T ] is a
EJQTDE, 2010 No. 74, p. 6
solution of the eigenvalue problem (S), (BC) if and only if
Z T
Z T
G(t, s)b(s)f µ
G(s, τ )c(τ )g(u(τ )) dτ ds, 0 ≤ t ≤ T,
u(t) = λ
0
0
Z T
v(t) = µ
G(t, s)c(s)g(u(s)) ds, 0 ≤ t ≤ T.
0
We consider the Banach space X = C([0, T ]) with supremum norm k · k and define
the cone C ⊂ X by
C = {u ∈ X, u(t) ≥ 0, ∀ t ∈ [0, T ] and
inf
t∈[ξm−2 ,T ]
u(t) ≥ γkuk},
where γ is defined in Lemma 6.
For our first
the positive numbers L1 and L2 by
(result we define
−1
n−1 Z T
γ 2 ξm−2
n−1
(T − s) b(s)f∞ ds
,
L1 = max
d(n − 1)! ξm−2
2 n−1 Z T
−1 )
γ ξm−2
(T − s)n−1 c(s)g∞ ds
,
d(n − 1)! ξm−2
(
−1
Z T
T n−1
n−1
(T − s) b(s)f0 ds
,
L2 = min
d(n − 1)! 0
−1 )
Z T
T n−1
(T − s)n−1 c(s)g0 ds
.
d(n − 1)! 0
Theorem 2. Assume that (H1)–(H3) hold and L1 < L2 . Then for each λ and µ
satisfying λ, µ ∈ (L1 , L2 ), there exist a positive solution with respect to a cone, (u(t), v(t)),
t ∈ [0, T ], of problem (S), (BC).
Proof. Let λ, µ ∈ (L1 , L2 ) and we choose a positive number ε such that ε < f∞ ,
ε < g∞ , (
2 n−1 Z T
−1
γ ξm−2
n−1
max
(T − s) b(s)(f∞ − ε) ds
,
d(n − 1)! ξm−2
−1 )
2 n−1 Z T
γ ξm−2
(T − s)n−1 c(s)(g∞ − ε) ds
≤ min(λ, µ)
d(n − 1)! ξm−2
and
(
−1
Z T
T n−1
n−1
(T − s) b(s)(f0 + ε) ds
max(λ, µ) ≤ min
,
d(n − 1)! 0
−1 )
Z T
T n−1
(T − s)n−1 c(s)(g0 + ε) ds
.
d(n − 1)! 0
We now define
: ZC → X, by
Z the operator A
T
A(u)(t) = λ
T
G(t, s)b(s)f
0
µ
G(s, τ )c(τ )g(u(τ )) dτ
ds, 0 ≤ t ≤ T, u ∈ C.
0
By Lemma 6, we have A(C) ⊂ C. By using the Arzela-Ascoli theorem we deduce
that the operator A is completely continuous (compact and continuous). By definitions
of f0 and g0 there exists K1 > 0 such that
EJQTDE, 2010 No. 74, p. 7
f (x) ≤ (f0 + ε)x and g(x) ≤ (g0 + ε)x, 0 < x ≤ K1 .
Using (H3) we have f (0) = g(0) = 0 and the above inequalities are also valid for x = 0.
Z T
Let u ∈ C with kuk = K1 . Because v(t) = µ
G(t, s)c(s)g(u(s)) ds, t ∈ [0, T ]
0
satisfies the problem (1), (2) with y(t) = µc(t)g(u(t)), t ∈ [0, T ], then by (6) and the
above property of g, we deduce for t ∈ [0, T ]
Z T
Z T
µT n−1
(T − s)n−1 c(s)g(u(s)) ds
v(t) = µ
G(t, s)c(s)g(u(s)) ds ≤
d(n
−
1)!
0
Z 0T
µT n−1
n−1
(T − s) c(s)(g0 + ε)u(s) ds
≤
d(n − 1)! 0
Z
T
µT n−1
≤
(T − s)n−1 c(s)(g0 + ε)kuk ds ≤ kuk = K1 .
d(n − 1)! 0
By using once again Lemma 5 (relations (6)) and the properties of the function f ,
we have
T
Z T
A(u)(t) = λ
G(t, s)b(s)f µ
G(s, τ )c(τ )g(u(τ )) dτ ds
0
0
Z T
Z T
λT n−1
n−1
≤
(T − s) b(s)f µ
G(s, τ )c(τ )g(u(τ )) dτ ds
d(n − 1)! 0
0 Z
Z
T
T
λT n−1
n−1
(T − s) b(s)(f0 + ε) µ
G(s, τ )c(τ )g(u(τ )) dτ ds
≤
d(n − 1)! 0
0
Z T
λT n−1
≤
(T − s)n−1 b(s)(f0 + ε)K1 ds ≤ K1 = kuk, 0 ≤ t ≤ T.
d(n − 1)! 0
Then kA(u)k ≤ kuk, for all u ∈ C with kuk = K1 . If we denote by Ω1 = {u ∈
Z
C, kuk < K1 }, then we obtain kA(u)k ≤ kuk for all u ∈ C ∩ ∂Ω1 .
¯ 2 > 0 such that
Next, by the definitions of f∞ and g∞ , there exists K
¯ 2.
f (x) ≥ (f∞ − ε)x and g(x) ≥ (g∞ − ε)x, x ≥ K
¯ 2 /γ . For u ∈ C with kuk = K2 , we obtain by using
We consider now K2 = max 2K1 , K
Lemma 6, that
u(t) ≥
inf
s∈[ξm−2 ,T ]
¯ 2 , ∀ t ∈ [ξm−2 , T ].
u(s) ≥ γkuk = γK2 ≥ K
Then, by using (6), Lemma 6, and the above relations, we obtain for t ≥ ξm−2
Z T
v(t) = µ
G(t, s)c(s)g(u(s)) ds ≥ γkvk ≥ γv(ξm−2 )
0
Z T
= γµ
G(ξm−2 , s)c(s)g(u(s)) ds
0
n−1 Z T
γµξm−2
≥
(T − s)n−1 c(s)g(u(s)) ds
d(n − 1)! ξm−2
n−1 Z T
γµξm−2
≥
(T − s)n−1 c(s)(g∞ − ε)u(s) ds
d(n − 1)! ξm−2
n−1 Z T
γ 2 µξm−2
(T − s)n−1 c(s)(g∞ − ε)kuk ds ≥ kuk = K2
≥
d(n − 1)! ξm−2
and
EJQTDE, 2010 No. 74, p. 8
A(u)(ξm−2)
Z T
Z T
n−1
ξm−2
n−1
(T − s) b(s)f µ
≥λ
G(s, τ )c(τ )g(u(τ )) dτ ds
0
ξm−2 d(n − 1)!
Z T
Z T
n−1
λξm−2
n−1
(T − s) b(s)(f∞ − ε) µ
G(s, τ )c(τ )g(u(τ )) dτ ds
≥
d(n − 1)! ξm−2
0
Z T
n−1
λξm−2
(T − s)n−1 b(s)(f∞ − ε)K2 ds
≥
d(n − 1)! ξm−2
n−1 Z T
γ 2 λξm−2
(T − s)n−1 b(s)(f∞ − ε)K2 ds ≥ K2 = kuk.
≥
d(n − 1)! ξm−2
Therefore kA(u)k ≥ A(u)(ξm−2 ) ≥ kuk, for all u ∈ C with kuk = K2 . We denote by
Ω2 = {u ∈ C, kuk < K2 }. Then kA(u)k ≥ kuk, for all u ∈ C ∩ ∂Ω2 .
¯2 \
We now apply Theorem 1 i) and we deduce that A has a fixed point u ∈ C ∩ (Ω
Z T
Ω1 ). This element together with v(t) = µ
G(t, s)c(s)g(u(s)) ds, t ∈ [0, T ] represent a
0
positive solution of (S), (BC) with respect to cone C, for the given λ and µ.
✷
Remark 3. The condition L1 < L2 from Theorem 2 is equivalent to
Z T
−1
Z T
d(n − 1)!
n−1
n−1
min
(T − s) b(s)f∞ ds,
(T − s) c(s)g∞ ds
n−1
γ 2 ξm−2
ξm−2
ξm−2
Z T
−1
Z T
d(n − 1)!
n−1
n−1
max
<
(T − s) b(s)f0 ds,
(T − s) c(s)g0 ds
T n−1
0
0
or
Z T
Z T
n−1
n−1
max
(T − s) b(s)f0 ds,
(T − s) c(s)g0 ds
n−1
γ 2 ξm−2
0
0
.
<
Z T
Z T
T n−1
n−1
n−1
(T − s) c(s)g∞ ds
(T − s) b(s)f∞ ds,
min
ξm−2
ξm−2
In what follows we shall present another existence result for (S), (BC). Let us
consider positive
numbers
(
−1
Z T
n−1
γξm−2
n−1
(T − s) b(s)f0 ds
,
L3 = max
d(n − 1)! ξm−2
−1 )
Z T
n−1
γξm−2
n−1
(T − s) c(s)g0 ds
,
d(n − 1)! ξm−2
(
−1
Z T
T n−1
n−1
(T − s) b(s)f∞ ds
,
L4 = min
d(n − 1)! 0
−1 )
Z T
T n−1
(T − s)n−1 c(s)g∞ ds
.
d(n − 1)! 0
Theorem 3. Assume the assumptions (H1)–(H3) hold and L3 < L4 . Then for each
λ and µ satisfying λ, µ ∈ (L3 , L4 ), there exists a positive solution with respect to a cone,
(u(t), v(t)), t ∈ [0, T ], of (S), (BC).
Proof. Let λ and µ with λ, µ ∈ (L3 , L4 ). We select a positive number ε such that
EJQTDE, 2010 No. 74, p. 9
ε < f0 , ε (
< g0 and
−1
Z T
n−1
γξm−2
n−1
max
(T − s) b(s)(f0 − ε) ds
,
d(n − 1)! ξm−2
−1 )
Z T
n−1
γξm−2
(T − s)n−1 c(s)(g0 − ε) ds
≤ min(λ, µ)
d(n − 1)! ξm−2
and
(
−1
Z T
T n−1
n−1
max(λ, µ) ≤ min
(T − s) b(s)(f∞ + ε) ds
,
d(n − 1)! 0
−1 )
Z T
T n−1
(T − s)n−1 c(s)(g∞ + ε) ds
.
d(n − 1)! 0
We also consider the operator A defined in the proof of Theorem 2. From the defini¯ 3 > 0 such that
tions of f0 and g0 , we deduce that there exists K
¯ 3.
f (x) ≥ (f0 − ε)x and g(x) ≥ (g0 − ε)x, 0 < x ≤ K
Using the properties of f and g the above inequalities are also valid for x = 0.
In addition, because g is a continuous function with g0 > 0, then g(0) = 0 and there
¯ 3 ) such that
exists K3 ∈ (0, K
¯3
K
g(x) ≤
, 0 < x ≤ K3 .
Z
T
µT n−1
n−1
(T − s) c(s) ds
d(n − 1)! 0
For u ∈ C with kuk = K3 , by (6) and the above inequality, we deduce that for all
t ∈ [0, T ]
T
Z T
µT n−1
v(t) = µ
G(t, s)c(s)g(u(s)) ds ≤
(T − s)n−1 c(s)g(u(s)) ds
d(n
−
1)!
0
Z 0T
¯3
K
µT n−1
n−1
¯ 3.
(T − s) c(s)
≤
ds = K
Z T
d(n − 1)! 0
µT n−1
(T − τ )n−1 c(τ ) dτ
d(n − 1)! 0
By using (6), Lemma 6 and the properties of f,g we then obtain
Z T
Z T
n−1
ξm−2
n−1
(T − s) b(s)f µ
G(s, τ )c(τ )g(u(τ )) dτ ds
A(u)(ξm−2) ≥ λ
0
ξm−2 d(n − 1)!
Z T
Z T
n−1
λξm−2
n−1
(T − s) b(s)(f0 − ε) µ
G(s, τ )c(τ )g(u(τ )) dτ ds
≥
d(n − 1)! ξm−2
0
Z T
n−1
λξm−2
(T − s)n−1 b(s)(f0 − ε)γkvk ds
≥
d(n − 1)! ξm−2
n−1 Z T
λγξm−2
≥
(T − s)n−1 b(s)(f0 − ε)v(ξm−2) ds
d(n − 1)! ξm−2
n−1 Z T
λγξm−2
n−1
(T − s) b(s)(f0 − ε) ds
≥
d(n − 1)! ξm−2
Z T
n−1
µξm−2
n−1
(T − s) c(s)g(u(s)) ds
×
d(n − 1)! ξm−2
Z
EJQTDE, 2010 No. 74, p. 10
n−1 Z T
λγξm−2
n−1
(T − s) b(s)(f0 − ε) ds
≥
d(n − 1)! ξm−2
Z T
n−1
µξm−2
n−1
(T − s) c(s)(g0 − ε)u(s) ds
×
d(n − 1)! ξm−2
n−1 Z T
λγξm−2
n−1
(T − s) b(s)(f0 − ε) ds
≥
d(n − 1)! ξm−2
n−1 Z T
µγξm−2
n−1
×
(T − s) c(s)(g0 − ε) ds kuk ≥ kuk.
d(n − 1)! ξm−2
Hence, kA(u)k ≥ A(u)(ξm−2) ≥ kuk, for u ∈ C with kuk = K3 . We denote by
Ω3 = {u ∈ C, kuk < K3 }, and then we have kA(u)k ≥ kuk for all u ∈ C ∩ ∂Ω3 .
We now consider the functions f ∗ , g ∗ : [0, ∞) → [0, ∞) defined by f ∗ (x) = sup f (y),
0≤y≤x
f ∗ (x)
∗
∗
∗
= f∞ ,
g (x) = sup g(y). By (H2) we obtain for f and g the relations lim
x→∞
x
0≤y≤x
g ∗ (x)
lim
= g∞ .
x→∞
x
¯ 4 > 0 such
We also have f (x) ≤ f ∗ (x), g(x) ≤ g ∗(x), for all x ≥ 0. Then there exists K
that
¯ 4.
f ∗ (x) ≤ (f∞ + ε)x, g ∗(x) ≤ (g∞ + ε)x, for all x ≥ K
¯ 4 }. Then for u with kuk = K4 we obtain
Let K4 > max{2K3 , K
Z T
Z T
n−1
T
n−1
A(u)(t) ≤
(T − s) b(s)λf µ
G(s, τ )c(τ )g(u(τ )) dτ ds
d(n − 1)! 0
0
Z
Z
T
T
λT n−1
n−1
∗
(T − s) b(s)f µ
G(s, τ )c(τ )g(u(τ )) dτ ds
≤
d(n − 1)! 0
0
Z T
Z T
λT n−1
µT n−1
n−1
∗
n−1
≤
(T − s) b(s)f
(T − τ ) c(τ )g(u(τ )) dτ ds
d(n − 1)! 0
d(n
−
1)!
0
Z T
Z T
µT n−1
λT n−1
n−1
∗
n−1
∗
(T − s) b(s)f
(T − τ ) c(τ )g (u(τ )) dτ ds
≤
d(n − 1)! 0
d(n − 1)! 0
Z
Z
T
T
λT n−1
µT n−1
n−1
∗
n−1
∗
≤
(T − s) b(s)f
(T − τ ) c(τ )g (K4 ) dτ ds
d(n − 1)! 0
d(n
−
1)!
0
Z T
Z T
µT n−1
λT n−1
n−1
∗
n−1
(T − s) b(s)f
(T − τ ) c(τ )(g∞ + ε)K4 dτ ds
≤
d(n − 1)! 0
d(n − 1)! 0
Z T
λT n−1
≤
(T − s)n−1 b(s)f ∗ (K4 ) ds
d(n − 1)! 0
Z T
λT n−1
≤
(T − s)n−1 b(s)(f∞ + ε)K4 ds ≤ K4 = kuk.
d(n − 1)! 0
So kA(u)k ≤ kuk, for all u ∈ C with kuk = K4 . If we denote by Ω4 = {u ∈ C, kuk <
K4 }, then we obtain kA(u)k ≤ kuk, for all u ∈ C ∩ ∂Ω4 .
¯ 4 \Ω3 ), which together
By Theorem 1 ii) we deduce that A has a fixed point u ∈ C ∩(Ω
Z T
with v(t) = µ
G(t, s)c(s)g(u(s)) ds, t ∈ [0, T ] give us a positive solution of (S), (BC)
0
with respect to cone C, for the chosen values λ and µ.
✷
EJQTDE, 2010 No. 74, p. 11
Remark 4. The condition L3 < L4 is equivalent to
Z T
−1
Z T
d(n − 1)!
n−1
n−1
min
(T − s) b(s)f0 ds,
(T − s) c(s)g0 ds
n−1
γξm−2
ξm−2
ξm−2
Z T
−1
Z T
d(n − 1)!
n−1
n−1
<
max
(T − s) b(s)f∞ ds,
(T − s) c(s)g∞ ds
T n−1
0
0
or
Z T
Z T
n−1
n−1
max
(T − s) b(s)f∞ ds,
(T − s) c(s)g∞ ds
n−1
γξm−2
0
0
.
Z T
< n−1
Z T
T
n−1
n−1
min
(T − s) b(s)f0 ds,
(T − s) c(s)g0 ds
ξm−2
4
ξm−2
An example
As in Example
( 4.1 in [12], let us consider the functions
f (x) = p2 | sin x| + p1 xe−1/x , x ∈ [0, ∞),
g(x) = q2 | sin x| + q1 xe−1/x , x ∈ [0, ∞),
with p1 , p2 , q1 , q2 > 0.
g(x)
f (x)
g(x)
f (x)
= p2 , lim
= p1 , lim+
= q2 , lim
= q1 .
We have lim+
x→∞ x
x→∞ x
x→0
x→0
x
x
Let T = 1, n = 3, m = 4, b(t) = b0 t, c(t) = c0 t, t ∈ [0, 1], with b0 , c0 > 0 and
ξ1 = 31 , ξ2 = 32 , a1 = 1, a2 = 21 .
We consider the third-order differential system
(
u′′′ (t) + λb0 t p2 | sin v(t)| + p1 v(t)e−1/v(t) = 0, t ∈ (0, 1)
(S0 )
v ′′′ (t) + µc0 t q2 | sin u(t)| + q1 |u(t)|e−1/u(t) = 0, t ∈ (0, 1),
with the boundary conditions
(
u(0) = u′ (0) = 0, u(1) = u( 31 ) + 12 u( 23 )
(BC0 )
v(0) = v ′ (0) = 0, v(1) = v( 31 ) + 12 v( 32 ).
2
X
2
X
2
3
We also have d = 1 −
= > 0,
ai = > 1 and γ = min{a1 ξ12 , ξ22} = 91 .
3
2
i=1
i=1
The condition
form given
Z L1 < L2 or the equivalent
in Remark 3 is
Z
ai ξi2
1
or
max
Z
min
1
2
(1 − s)2 c0 sq2 ds
(1 − s) b0 sp2 ds,
0
1
(1 − s)2 b0 sp1 ds,
2/3
Z 01
2/3
(1 − s)2 c0 sq1 ds
<
4
729
max{b0 p2 , c0 q2 }
4
<
.
min{b0 p1 , c0 q1 }
6561
EJQTDE, 2010 No. 74, p. 12
Therefore if the above condition is verified, then by Theorem 2 we deduce that for
all numbers λ, µ ∈ (L1 , L2 ) the problem (S0 ), (BC0 ) has positive solutions.
Acknowledgment. The author would like to thank the referee for the careful reading of the manuscript and for his comments which led to this revised version.
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(Received September 5, 2010)
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