Bulletin of Mathematical Analysis and Applications
ISSN: 1821-1291, URL: http://www.bmathaa.org
Volume 6 Issue 3(2014), Pages 60-78.
POSITIVE SOLUTION OF SYSTEM OF THIRD-ORDER
BOUNDARY VALUE PROBLEM WITH THREE-POINT AND
INTEGRAL BOUNDARY CONDITIONS
(COMMUNICATED BY DOUGLAS R. ANDERSON)
ROCHDI JEBARI
Abstract. This paper is concerned with the nonlinear third-order boundary
value problem
0
0
u000
i (t) + fi (t, u1 (t), ..., un (t), u1 (t), ..., un (t)) = 0, 0 < t < 1
with three-point and integral conditions

Z 1


ui (0) =
h1,i (s, u1 (s), ..., un (s))ds



0
0
ui (0) = 0
Z 1




 αi u0i (1) = βi ui (ηi ) +
h2,i (s, u1 (s), ..., un (s))ds
0
where for i ∈ {1, ..., n}, αi > 0, βi > 0, 1 > ηi ≥ 0, 2αi > βi ηi2 , fi :
[0, 1] × Rn × Rn → R, for all k ∈ {1, 2}, hk,i : [0, 1] × Rn → R are continuous
functions. By using Guo-Krasnosel’skii fixed point theorem in cone, we discuss
the existence of positive solution of this problem. We also prove nonexistence
of positive solution and we give some examples to illustrate our results.
1. Introduction
Third-order ordinary differential equations arise in a variety of different areas
of applied mathematics and physics, in the deflection of a curved beam having
a constant or varying cross section, a three-layer beam, electromagnetic waves or
gravity driven flows and so on [1]. The main propose of the present paper is to
investigate sufficient conditions for the existence of positive solution of following
problem:
0
0
u000
i (t) + fi (t, u1 (t), ..., un (t), u1 (t), ..., un (t)) = 0, 0 < t < 1
(1.1)
2000 Mathematics Subject Classification. 34-XX,44-XX, 34Bxx, 34Lxx, 34B15, 34B18, 34B27.
Key words and phrases. Third-order boundary value problem, three-point and integral boundary conditions, Guo-Krasnosel’skii fixed point in cone, positive solution.
c
2014
Universiteti i Prishtinës, Prishtinë, Kosovë.
Submitted May 18, 2014. Published August 11, 2014.
60
SYSTEM OF THIRD-ORDER BOUNDARY VALUE PROBLEM
with three-point and integral conditions

Z 1


h1,i (s, u1 (s), ..., un (s))ds
u (0) =


 i
0
0
ui (0) = 0
Z 1




0
 αi ui (1) = βi ui (ηi ) +
h2,i (s, u1 (s), ..., un (s))ds
61
(1.2)
0
where for i ∈ {1, ..., n}, for k ∈ {1, 2}, αi > 0, βi > 0, 1 > ηi ≥ 0, 2αi > βi ηi2 ,
fi : [0, 1]×Rn ×Rn → R, hk,i : [0, 1]×Rn → R are continuous functions. The integral
conditions has physical significance such as total mass, moment, etc. Sometimes it
is better to impose integral conditions to get a more accurate measure than a local
condition see [9].
Various types of boundary value problems were studied by many authors using fixed point theorems on cones, fixed point index theory, upper and lower
solutions method, differential inequality, topological transversality and LeggettWilliams fixed point theorem [2-8,12]. In [2], Yao and Feng used the upper and
lower solutions method to prove some existence results for the following third-order
two-point boundary value problem:
u000 (t) + f (t, u(t)) = 0, 0 < t < 1
(1.3)
0
(1.4)
0
u(0) = u (0) = u (1) = 0.
In [3], Sanyang Liu and Yuqiang Feng used the upper and lower solutions method
and a new maximum principle to prove some existence results for the more general
third-order two-point boundary value problem:
u000 (t) + f (t, u(t), u0 (t)) = 0, 0 < t < 1
0
0
u(0) = u (0) = u (1) = 0.
(1.5)
(1.6)
In [8] Guo, Sun and Zhao, considered the third-order three-point boundary value
problem
u000 (t) + a(t)f (u(t)) = 0, 0 < t < 1
0
0
0
u(0) = u (0) = 0, u (1) = αu (η)
(1.7)
(1.8)
1
η,
f : [0, 1] × R → R is given function. The existence
where 0 < η < 1 and 1 < α <
of at least one positive solution for (1.7)-(1.8) was proved when f was superlinear or
sublinear. By the use of some fixed point theorem in cones, A. Guezane-Lakoud et
al. [12], investigated the existence of positive solutions for the following third-order
eigenvalue problem
u000 (t) + f (t, u(t), u0 (t)) = 0, 0 < t < 1
0
0
u(0) = u (0) = 0, αu (1) = βu(η)
R∗+ , β
(1.9)
(1.10)
R∗+ ,
where α ∈
∈
0 < η < 1, by Guo-Krasnosel’skii fixed point theorem in
cone the author studied the existence of least positive solution of (1.9)-(1.10) when
2α > βη 2 and f (t, u(t), u0 (t)) = a(t)g(u(t), u0 (t)). In [7], Sun et al considered the
following third-order boundary value problem:
u000 + a(t)f (t, u(t)) = 0, 0 < t < 1
0
0
0
u(0) = u (0) = 0, u (1) − αu (η) = λ
(1.11)
(1.12)
by employing the Guo-Krasnosel’skii fixed point theorem and Schauder’s fixed point
theorem, the author studied the existence and nonexistence of positive solutions
62
ROCHDI JEBARI
to the third-order three-point nonhomogeneous boundary condition (1.11)-(1.12).
For more knowledge about the boundary value problem, we refer the reader to
(see [13-21]). Our aim is to use the Guo-Krasnosel’skii fixed point theorem to
prove the existence of least one positive solutions of our problem. For this, we
formulate the boundary value problem as the fixed point problem. The particularity
of our equation (1.1)-(1.2) is that the boundary condition involving three-point
boundary condition and integral condition which leads to extra difficulties. To the
best of our knowledge, no one has studied the existence and nonexistence of positive
solution for nonlinear differential equation (1.1) jointly with conditions (1.2). Our
work is new and more general than [2, 3, 12], for example our problem reduces
to the problem (1.9)-(1.10) in the case n = 1, h1,1 ≡ h2,1 ≡ 0, α1 = α, β1 = β,
f1 (t, u1 (t), u01 (t)) = f (t, u(t), u0 (t)).
This paper is organized as follows. In Section 2, we present some preliminaries
that will be used to prove our results. In Section 3, the study the positivity of
solution is based on a Guo-Krasnosel’skii fixed point theorem. In Section 4, the
nonexistence of positive solution is studied. Finally, we shall give two examples to
illustrate our main results.
2. Preliminaries and lemmas
The cartesian power of C 1 ([0, 1]; R) can be defined as:
E
= C 1 ([0, 1]; R)n
= C 1 ([0, 1]; R) × ... × C 1 ([0, 1]; R)
{z
}
|
n times
equipped with the norm kukE =
n
X
kui k where kui k = max(kui k∞ , ku0i k∞ ), kui k∞ =
i=1
max |ui (t)| and u = (u1 , .., un ) ∈ E. The space E is a Banach space. We denote
t∈[0,1]
by for x ∈ Rn , kxk1 =
n
X
|xi |.
i=1
Definition 1. We give the following definitions:
(1) The function u = (u1 , .., un ) is called nonnegative solution of the system
(1.1)-(1.2) if and only if, u satisfies (1.1)-(1.2) and for all i ∈ {1, ..., n},
ui (t) ≥ 0 for t ∈ [0, 1].
(2) The function u = (u1 , .., un ) is called positive solution of the system (1.1)(1.2) if and only if, u satisfies (1.1)-(1.2) and for all i ∈ {1, ..., n}, ui (t) > 0
for t ∈]0, 1[.
Lemma 2.1. Let i ∈ {1, ..., n}, k ∈ {1, 2}, gk,i ∈ C([0, 1]; R),hi ∈ C([0, 1]; R), then
the problem
 000
ui (t) + hi (t) = 0, 0 < t < 1, i ∈ {1, ..., n}


Z 1




g1,i (s)ds
 ui (0) =
0
(2.1)
0
ui (0) = 0


Z

1



 αi u0i (1) = βi ui (ηi ) +
g2,i (s)ds
0
SYSTEM OF THIRD-ORDER BOUNDARY VALUE PROBLEM
63
has an unique solution
Z
u(t) =
1
H(t, s)h(s)ds + ϕ(t)
0
where u(t) = (u1 (t), ..., un (t)), ϕ(t) = (ϕ1 (t), ..., ϕn (t)),
1
Z
Z
H(t, s)h(s)ds = (
0
1
1
Z
H1 (t, s)h1 (s)ds, ...,
0
Hn (t, s)hn (s)ds)
0
with
Hi (t, s) = G(t, s) + t2 Ki G(ηi , s)
G(t, s) =
1
2
Z
1
(1 − s)t2
(−s + 2t − t2 )s
Z
ϕi (t) = [Ki
g1,i (s)ds + Ci
0
(2.2)
if 0 ≤ t ≤ s
if s ≤ t ≤ 1
1
2
Z
g2,i (s)ds]t +
0
(2.3)
1
g1,i (s)ds
(2.4)
0
Ki =
βi
2αi − βi ηi2
(2.5)
Ci =
1
.
2αi − βi ηi2
(2.6)
Proof. Let i ∈ {1, ..., n}, integrating the equation (2.1), it yields
ui (t) = −
1
2
Z
0
t
1
(t − s)2 hi (s)ds + C1,i t2 + C2,i t + C3,i .
2
Z
1
From the boundary condition ui (0) =
g1,i (s)ds, u0i (0) = 0 we deduce that C3,i =
0
Z 1
Z 1
g1,i (s)ds and C2,i = 0. From the condition αi u0i (1) = βi ui (ηi ) +
g2,i (s)ds,
0
0
we have
C1,i
=
+
Z 1
2αi
(1 − s)hi (s)ds
2αi − βi ηi2 0
0
Z 1
Z 1
2βi
2
g1,i (s)ds +
g2,i (s)ds.
2αi − βi ηi2 0
2αi − βi ηi2 0
−βi
2αi − βi ηi2
Z
ηi
(ηi − s)2 hi (s)ds +
64
ROCHDI JEBARI
Therefore
Z ηi
t2
−βi
(ηi − s)2 hi (s)ds
ui (t) =
(t − s) hi (s)ds + [
2
2
2α
−
β
η
i
i i 0
0
Z 1
Z 1
2αi
2βi
+
(1 − s)hi (s)ds +
g1,i (s)ds
2αi − βi ηi2 0
2αi − βi ηi2 0
Z 1
Z 1
2
g
(s)ds]
+
g1,i (s)ds
+
2,i
2αi − βi ηi2 0
0
Z
Z
1 t
t2 1
βi t 2
= −
(t − s)2 hi (s)ds +
(1 − s)hi (s)ds +
2 0
2 0
2αi − βi ηi2
Z
Z
η2 1
1 ηi
× [ i
(1 − s)h(s)ds −
(ηi − s)2 hi (s)ds]
2 0
2 0
Z 1
Z 1
Z 1
2βi
2
t2
+ [
g1,i (s)ds +
g2,i (s)ds] +
g1,i (s)ds
2αi − βi ηi2 0
2αi − βi ηi2 0
2
0
Z 1
Z 1
t2 βi
G(ηi , s)hi (s)ds + Pi (t)
=
G(t, s)hi (s)ds +
2αi − βi ηi2 0
0
Z 1
Z 1
=
G(t, s)hi (s)ds + Ki t2
G(ηi , s)hi (s)ds + ϕi (t)
1
−
2
Z
t
2
0
0
1
Z
=
Hi (t, s)hi (s)ds + ϕi (t)
0
where G(t, s), Hi (t, s), Pi (t), Ki and Ci are given by (2.2), (2.3), (2.4), (2.5) and
(2.6) which achieve the proof of Lemma 1.
Lemma 2.2. Let i ∈ {1, ..., n}, k ∈ {1, 2}, hk,i ∈ C([0, 1] × Rn ; R), fi ∈ C([0, 1] ×
Rn ×Rn , R). Then u is a solution of (1.1)-(1.2) if and only if for all, T (u)(t) = u(t).
Where T (u) = (T1 (u), ..., Tn (u)) and for all t ∈ [0, 1], for all i ∈ {1, ..., n}
Z 1
Ti (u)(t) = Pi (t) +
Hi (t, s)fi (s, u(s), u0 (s))ds
0
with Hi (t, s) is given by (2.2)-(2.4) and
Z 1
Z 1
Z
2
Pi (t) = [Ki
h1,i (s, u(s))ds + Ci
h2,i (s, u(s))ds]t +
0
0
1
h1,i (s, u(s))ds.
0
3. Existence of positive solution
In this section, we will give some preliminary considerations and some lemmas
which are essential to establish sufficient conditions for the existence of least one
positive solution for our problem. We make the following additional assumption.
(H1) For all i ∈ {1, ..., n}, k ∈ {1, 2}, fi ∈ C([0, 1] × Rn × Rn , R+ ), hk,i ∈
C([0, 1] × Rn , R+ ).
Lemma 3.1. For all t ∈ [0, 1], s ∈ [0, 1], we have
(1) 0 ≤ G(t, s) ≤ ϕ(s)
≤ 2ϕ(s)
(2) 0 ≤ ∂G(t,s)
∂t
where ϕ(s) =
(1−s)s
.
2
SYSTEM OF THIRD-ORDER BOUNDARY VALUE PROBLEM
65
Proof. It is easy to see that, if t ≤ s, G(t, s) = 21 (1 − s)t2 ≥ 0
.
and G(t, s) = 12 (1 − s)t2 ≤ (1−s)s
2
If s ≤ t, G(t, s) = 12 (2t − t2 − s)s = 21 [(1 − s) − (1 − t)2 ]s ≥ 0
and G(t, s) ≤ 21 (1 − s)(1 − 1 + s) = 12 (1 − s)s
then the proof of (1) is complete.
∂G(t, s)
=
∂t
(1 − s)t
(1 − t)s
if 0 ≤ t ≤ s
if s ≤ t ≤ 1
= (1 − s)t ≤ (1 − s)s.
If t ≤ s, ∂G(t,s)
∂t
If s ≤ t, then −t ≤ −s this implies that ∂G(t,s)
= (1 − t)s ≤ (1 − s)s.
∂t
We deduce that proof of (2) is complete.
Lemma 3.2. Let a ∈]0, 1[ and b ∈]0, 1[, then for all (t, s) ∈ [a, b] × [a, b]
(1) G(t, s) ≥ ω1 (s)
where ω1 (s) = 12 (1 − s)a2
(2) ∂G(t,s)
≥ ω2 (s)
∂t
where ω2 (s) = 12 (1 − b)a.
Proof. Let a ∈]0, 1[ and b ∈]0, 1[, then for all (t, s) ∈ [a, b] × [a, b].
If t ≤ s,
G(t, s)
=
=
=
≥
≥
1
(−s + 2t − t2 )s
2
1
((1 − s) − (1 − t)2 )s
2
1
((1 − s) − (1 − s)2 )s
2
1
(1 − s)s2
2
1
(1 − s)a2 .
2
If t ≥ s,
G(t, s)
1
(1 − s)t2
2
1
(1 − s)a2 .
2
=
≥
Then proof of (1) is complete.
If t ≤ s,
∂G(t, s)
∂t
=
(1 − s)t
≥
(1 − s)a
≥
(1 − b)a.
(3.1)
66
ROCHDI JEBARI
If t ≥ s,
∂G(t, s)
∂t
=
(1 − t)s
≥
(1 − b)s
≥
(1 − b)a.
Then proof of (2) is complete, and the proof of Lemma 3.2 is complete.
Lemma 3.3. Suppose that (H 1) holds and let a ∈]0, 1[, b ∈]0, 1[, then the solution
u = (u1 , ..., un ) of the problem (1.1)-(1.2) is nonnegative and satisfies
min
t∈[a,b]
n
X
(ui (t) + u0i (t)) ≥ γ(a, b)kuk
i=1
where
max
γ(a, b) =
Z
max
i∈{1,...,n}
n
b
(ω1 (s) + ω2 (s))fi (s, u(s), u0 (s))ds
max(A, B)
1
Z
(1 +
γi (a, b)
a
γi (a, b) =
A =
i∈{1,...,n}
ϕ(s)fi (s, u(s), u0 (s))ds +
Ki )
0
max
i∈{1,...,n}
Pi (1) + 1
and
1
Z
B
=
2(1 +
max
ϕ(s)fi (s, u(s), u0 (s))ds +
Ki )
i∈{1,...,n}
0
max
i∈{1,...,n}
Pi0 (1) + 1.
Proof. Suppose that u = (u1 , ..., un ) is a solution of (1.1)-(1.2), then from lemma
2.2, (H1) and G(t, s) ≥ 0 for all (s, t) ∈ [0, 1] × [0, 1], it is obvious that for all
i ∈ {1, ..., n}, for all t ∈ [0, 1], ui (t) ≥ 0. For all i ∈ {1, ..., n}, for all t ∈ [0, 1], we
have
Z 1
|ui (t)| ≤
Hi (t, s)fi (s, u(s), u0 (s))ds + Pi (t)
0
Z
1
≤ (1 + Ki )
ϕ(s)fi (s, u(s), u0 (s))ds + Pi (1)
0
Z
≤
(1 +
max
i∈{1,...,n}
1
ϕ(s)fi (s, u(s), u0 (s))ds +
Ki )
0
max
i∈{1,...,n}
Pi (1).
This implies that, for all i ∈ {1, ..., n}
Z
kui k∞ ≤ (1 +
max
i∈{1,...,n}
Ki )
0
1
ϕ(s)fi (s, u(s), u0 (s))ds +
max
i∈{1,...,n}
Pi (1).
SYSTEM OF THIRD-ORDER BOUNDARY VALUE PROBLEM
67
For all i ∈ {1, ..., n}, for all t ∈ [0, 1], we have
|u0i (t)|
Z
≤
≤
1
∂Hi (t, s)
fi (s, u(s), u0 (s))ds + Pi0 (1)
∂t
0
Z 1
2(1 + Ki )
ϕ(s)fi (s, u(s), u0 (s))ds + Pi0 (1)
0
Z
≤
2(1 +
max
i∈{1,...,n}
1
ϕ(s)fi (s, u(s), u0 (s))ds +
Ki )
0
max
i∈{1,...,n}
Pi0 (1).
This implies that, for all i ∈ {1, ..., n}
ku0i k∞ ≤ 2(1 +
1
Z
max
i∈{1,...,n}
ϕ(s)fi (s, u(s), u0 (s))ds +
Ki )
max
i∈{1,...,n}
0
Pi0 (1).
We denote by
1
Z
A =
(1 +
max
i∈{1,...,n}
ϕ(s)fi (s, u(s), u0 (s))ds +
Ki )
max
i∈{1,...,n}
0
Pi (1) + 1
and
Z
B
=
2(1 +
max
i∈{1,...,n}
1
ϕ(s)fi (s, u(s), u0 (s))ds +
Ki )
0
max
i∈{1,...,n}
Then for all i ∈ {1, ..., n},
kui k = max(kui k∞ , ku0i k∞ ) ≤ max(A, B).
Using lemma 3.2-(1) we have, for all i ∈ {1, ..., n}, for all t ∈ [a, b]
b
Z
ω1 (s)fi (s, u(s), u0 (s))ds + Pi (t)
ui (t) ≥
a
b
Z
ω1 (s)fi (s, u(s), u0 (s))ds
≥
a
Z
≥
ω1 (s)fi (s, u(s), u0 (s))ds
a
max(A, B)
Z
≥
b
× max(A, B)
b
ω1 (s)fi (s, u(s), u0 (s))ds
a
max(A, B)
kui k.
Then for all i ∈ {1, ..., n}
Z
min ui (t) ≥
t∈[a,b]
b
ω1 (s)fi (s, u(s), u0 (s))ds
a
max(A, B)
kui k.
Pi0 (1) + 1.
68
ROCHDI JEBARI
Similarly, using lemma 3.2-(2) for all i ∈ {1, ..., n}, for all t ∈ [a, b]
Z b
u0i (t) ≥
ω2 (s)fi (s, u(s), u0 (s))ds + P 0 (t)
a
b
Z
ω2 (s)fi (s, u(s), u0 (s))ds
≥
a
Z
b
ω2 (s)fi (s, u(s), u0 (s))ds
a
≥
× max(A, B)
max(A, B)
Z
≥
b
ω2 (s)fi (s, u(s), u0 (s))ds
a
kui k.
max(A, B)
Then for all i ∈ {1, ..., n},
b
Z
min
t∈[a,b]
u0i (t)
≥
ω2 (s)fi (s, u(s), u0 (s))ds
a
kui k.
max(A, B)
Then for all i ∈ {1, ..., n},
b
Z
min (ui (t) +
t∈[a,b]
u0i (t))
≥
=
(ω1 (s) + ω2 (s))fi (s, u(s), u0 (s))ds
a
γi (a, b)kui k.
We deduce that
n
X
min
(ui (t) + u0i (t)) ≥
t∈[a,b]
kui k
max(A, B)
i=1
n
X
γi (a, b)kui k
i=1
≥
max [γj (a, b) × kuj k]
j∈{1,...,n}
≥
max
j∈{1,...,n}
γj (a, b) ×
max
j∈{1,...,n}
kuj k.
(3.2)
We deduce that for all j ∈ {1, ..., n},
min
t∈[a,b]
n
X
(ui (t) + u0i (t)) ≥
i=1
max
i∈{1,...,n}
γi (a, b) × kuj k.
Then
min
t∈[a,b]
n
X
max
(ui (t) +
u0i (t))
≥
i∈{1,...,n}
n
i=1
=
γi (a, b) X
n
kuj k
j=1
γ(a, b)kukE .
The proof is complete.
Definition 2. We denote by E + the following set
E + = {u = (u1 , ..., un ) ∈ E, ui (t) ≥ 0, t ∈ [0, 1], i ∈ {1, ..., n}}.
Definition 3. Let E be a Banach space, Con nonempty closed convex subset Con ⊂
E, is called a cone if it satisfies the following two conditions:
SYSTEM OF THIRD-ORDER BOUNDARY VALUE PROBLEM
69
(1) x ∈ Con, λ ≥ 0 implies λx ∈ Con
(2) x ∈ Con, −x ∈ Con implies x = 0.
Remark. For all a, b ∈]0, 1[, the set defined by
Con(a, b) = {u ∈ E + , min
t∈[a,b]
n
X
(ui (t) + u0i (t)) ≥ γ(a, b)kukE }
i=1
is a cone subset of E.
Theorem 3.4. (Guo-Krasnosel’skii fixed point theorem) [10]
Let E be a Banach space, and let Con ⊂ E be a cone. Assume Ω1 and Ω2 be two
bounded open subsets in E, with 0 ∈ Ω1 , Ω1 ⊂ Ω2 and let A : Con∩(Ω2 \Ω1 ) → Con
by a completely continuous operator such that:
(1) kA(u)k ≤ kuk, u ∈ Con ∩ ∂Ω1 and kA(u)k ≥ kuk, u ∈ Con ∩ ∂Ω2 or
(2) kA(u)k ≥ kuk, u ∈ Con ∩ ∂Ω1 and kA(u)k ≤ kuk, u ∈ Con ∩ ∂Ω2 .
Then A has a fixed point in Con ∩ (Ω2 \Ω1 ).
Now, we give the following assumptions
(H2) For all i ∈ {1, ..., n}, k ∈ {1, 2}, fi ∈ C([0, 1] × Rn+ × Rn , R+ ), hk,i ∈
C([0, 1] × Rn+ , R+ ).
(H3) There exists i0 ∈ {1, ..., n} and there exists t0 ∈]0, 1[ such that fi0 (t0 , x, y) >
0, for all (x, y) ∈ Rn+ × Rn .
Theorem 3.5. Suppose that (H2) and (H3) hold. Then the problem (1.1)-(1.2)
has at least one nonnegative solution in the case:
For all i ∈ {1, ..., n},
fi (t, u, v)
= +∞
kuk1 +kvk1 →0 t∈[0,1] kuk1 + kvk1
lim
min
and
fi (t, u, v)
=0
lim
max
kuk1 + kvk1 → +∞ t∈[0,1] kuk1 + kvk1
where u = (u1 , ..., un ) ∈ Rn and v = (v1 , ..., vn ) ∈ Rn .
Proof. Step 1: From (H2) and (H3), there exists [α, β] ⊂]0, 1[, such that t0 ∈ [α, β]
and for all t ∈ [α, β], (x, y) ∈ Rn+ × Rn we have fi0 (t, x, y) > 0. Then for all u ∈ E + ,
Z β
Z β
ω1 (s)fi0 (s, u(s), u0 (s))ds > 0 and
ω2 (s)fi0 (s, u(s), u0 (s))ds > 0.
α
α
This implies that
max
γi0 (α, β) > 0 and γ(α, β) =
i∈{1,...,n}
γi (α, β)
n
From Remark, Con(α, β) = {u ∈ E + , min
t∈[α,β]
n
X
≥
γi0 (α, β)
> 0.
n
(ui (t) + u0i (t)) ≥ γ(α, β)kukE } is
i=1
a cone subset of E. By using Arzela-Ascoli theorem [11], T : Con(α, β) → E is a
completely continuous mapping.
70
ROCHDI JEBARI
We will show that T (Con(α, β)) ⊂ Con(α, β). In fact, from (H2) and G(t, s) ≥ 0
for all [0, 1]×[0, 1], it is obvious that for all i ∈ {1, ..., n}, for all t ∈ [0, 1], Ti (u)(t) ≥
0. Using lemma 3.1-(1), for all i ∈ {1, ..., n}, for all t ∈ [0, 1], we have
1
Z
|Ti (u)(t)|
Hi (t, s)fi (s, u(s), u0 (s))ds + Pi (t)
≤
0
1
Z
≤
ϕ(s)fi (s, u(s), u0 (s))ds + Pi (1)
(1 + Ki )
0
1
Z
≤
(1 +
max
ϕ(s)fi (s, u(s), u0 (s))ds +
Ki )
i∈{1,...,n}
0
max
i∈{1,...,n}
Pi (1).
This implies that for all i ∈ {1, ..., n}
1
Z
kTi (u)k∞ ≤ (1 +
max
i∈{1,...,n}
ϕ(s)fi (s, u(s), u0 (s))ds +
Ki )
0
max
i∈{1,...,n}
Pi (1).
Using lemma 3.1-(2), for all i ∈ {1, ..., n}, for all t ∈ [0, 1], we have
Z
|Ti (u)0 (t)|
≤
≤
1
∂Hi (t, s)
fi (s, ui (s), u0i (s))ds + Pi0 (1)
∂t
0
Z 1
2(1 + Ki )
ϕ(s)fi (s, ui (s), u0i (s))ds + Pi0 (1)
0
Z
≤
2(1 +
max
i∈{1,...,n}
Ki )
1
ϕ(s)fi (s, u(s), u0 (s))ds +
0
max
i∈{1,...,n}
Pi0 (1).
This implies that for all i ∈ {1, ..., n}
Z
0
kTi (u) k∞ ≤ 2(1 +
max
i∈{1,...,n}
Ki )
1
ϕ(s)fi (s, u(s), u0 (s))ds +
0
max
i∈{1,...,n}
Pi0 (1).
We denote by
1
Z
A =
(1 +
max
i∈{1,...,n}
ϕ(s)fi (s, u(s), u0 (s))ds +
Ki )
0
max
i∈{1,...,n}
Pi (1) + 1
and
Z
B
=
2(1 +
max
i∈{1,...,n}
Ki )
0
1
ϕ(s)fi (s, u(s), u0 (s))ds +
max
i∈{1,...,n}
Then for all i ∈ {1, ..., n},
kTi (u)k = max(kTi (u)k∞ , kTi (u)0 k∞ ) ≤ max(A, B).
Pi0 (1) + 1.
SYSTEM OF THIRD-ORDER BOUNDARY VALUE PROBLEM
71
Using lemma 3.2-(1), we have for all i ∈ {1, ..., n}, for all t ∈ [α, β]
Z β
ω1 (s)fi (s, u(s), u0 (s))ds + Pi (t)
Ti (u)(t) ≥
α
β
Z
ω1 (s)fi (s, u(s), u0 (s))ds
≥
α
β
Z
ω1 (s)fi (s, u(s), u0 (s))ds
α
≥
× max(A, B)
max(A, B)
β
Z
ω1 (s)fi (s, u(s), u0 (s))ds
α
≥
max(A, B)
kTi (u)k.
Then for all i ∈ {1, ..., n},
β
Z
ω1 (s)fi (s, u(s), u0 (s))ds
α
min Ti (u)(t) ≥
kTi (u)k.
max(A, B)
t∈[α,β]
Similarly by use lemma 3.2-(2), we have for all i ∈ {1, ..., n}, for all t ∈ [α, β]
Z β
Ti (u)0 (t) ≥
ω2 (s)fi (s, u(s), u0 (s))ds + P 0 (t)
α
β
Z
ω2 (s)fi (s, u(s), u0 (s))ds
≥
α
Z
β
ω2 (s)fi (s, u(s), u0 (s))ds
α
≥
max(A, B)
Z
β
ω2 (s)fi (s, u(s), u0 (s))ds
α
≥
× max(A, B)
max(A, B)
kTi (u)k.
Then for all i ∈ {1, ..., n},
Z
0
β
ω2 (s)fi (s, u(s), u0 (s))ds
α
min Ti (u) (t) ≥
max(A, B)
t∈[α,β]
kTi (u)k.
We deduce that
Z
0
min (Ti (u)(t) + Ti (u) (t)) ≥
t∈[α,β]
=
β
(ω1 (s) + ω2 (s))fi (s, u(s), u0 (s))ds
α
max(A, B)
γi (α, β)kTi (u)k
where
Z
γi (α, β) =
β
(ω1 (s) + ω2 (s))fi (s, u(s), u0 (s))ds
α
max(A, B)
kT (ui )k
72
ROCHDI JEBARI
we deduce that
n
X
min
(Ti (u)(t) + Ti (u)0 (t)) ≥
t∈[α,β]
n
X
i=1
γi (α, β)kTi (u)k
i=1
≥
max [γj (α, β)kTj (u)k]
j∈{1,...,n}
≥
max
j∈{1,...,n}
This implies that for all j ∈ {1, ..., n},
n
X
min
(Ti (u)(t) + Ti (u)0 (t)) ≥
t∈[α,β]
i=1
γj (α, β) ×
max
i∈{1,...,n}
max
j∈{1,...,n}
kTj (u)k.
γi (a, b)kTj (u)k.
Then
min
t∈[α,β]
n
X
max
(Ti (u)(t) + Ti (u)0 (t))
i∈{1,...,n}
≥
γi (α, β) X
n
n
i=1
kTj (u)k
j=1
= γ(α, β)kT (u)kE .
Step 2: Let i ∈ {1, ..., n} we have
fi (t, u, v)
= +∞. Then for
kuk1 +kvk1 →0 t∈[0,1] kuk1 + kvk1
lim
min
all M > 0, ∃Ri,1 > 0 such that
min fi (t, u, v) ≥ M (kuk1 + kvk1 ), for kuk1 + kvk1 ≤ Ri,1 .
t∈[0,1]
This implies that, for all M > 0 ∃Ri,1 > 0 such that for all t ∈ [0, 1],
fi (t, u, v) ≥ min fi (t, u, v) ≥ M (kuk1 + kvk1 ) for kuk1 + kvk1 ≤ Ri,1 ,
t∈[0,1]
we choose
1
M=
Z
β
n γ(α, β) min(
.
β
Z
ω1 (s)ds,
ω2 (s)ds)
α
α
Let Ω1 be the bounded open set in E defined by Ω1 = {u ∈ E, kukE <
min
i∈{1,...,n}
Then for all u ∈ Con(α, β) ∩ ∂Ω1 , it yields, for all s ∈ [α, β],
n
X
(ui (s) + u0i (s)) ≥ γ(α, β) min Ri,1 .
i∈{1,...,n}
i=1
Then for all i ∈ {1, ..., n}, s ∈ [α, β],
fi (s, u(s), u0 (s)) ≥ M (ku(s)k1 + ku0 (s)k1 ) ≥ M γ(α, β)
This implies that, for all i ∈ {1, ..., n}, t ∈ [0, 1]
Z β
|Ti (u)(t)| ≥ M
ω1 (s)ds γ(α, β)
min
i∈{1,...,n}
α
and for all i ∈ {1, ..., n}
min
kTi (u)k ≥ kTi (u)k∞ ≥
i∈{1,...,n}
Ri,1
n
We deduce that
kT (u)k ≥
min
i∈{1,...,n}
Ri,1 = kukE .
.
min
i∈{1,...,n}
Ri,1
Ri,1 .
Ri,1 }.
SYSTEM OF THIRD-ORDER BOUNDARY VALUE PROBLEM
Step 3: Now, let i ∈ {1, ..., n} we have
73
fi (t, u, v)
= 0. Then
kuk1 +kvk1 →+∞ t∈[0,1] kuk1 + kvk1
lim
max
for all ε > 0, ∃R0fi > 0 such that
max fi (t, u, v) ≤ ε(kuk1 + kvk1 ) for kuk1 + kvk1 ≥ R0fi .
t∈[0,1]
This implies that, for all ε > 0, ∃R0fi > 0 such that for all t ∈ [0, 1],
fi (t, u, v) ≤ max fi (t, u, v) ≤ ε(kuk1 + kvk1 ) for kuk1 + kvk1 ≥ R0fi .
t∈[0,1]
From (H2), there exist three constants M1,i , M2,i ≥ 0 such that, for k ∈ {1, 2},
|hk,i (t, u(t))| ≤ Mk,i , for each t ∈ [0, 1].
We choose
1
ε=
Z
4n(1 +
max
i∈{1,...,n}
.
1
Ki )
ϕ(s)ds
0
Let
R2
=
max{2
min
i∈{1,...,n}
,
2n
Ri,1 ,
max
i∈{1,...,n}
R0fi , 2n
max [Ki M1,i + Ci M2,i + M1,i + 1]
i∈{1,...,n}
max [2(Ki M1,i + Ci M2,i ) + 1]}
i∈{1,...,n}
and Ω2 be the bounded open set in E defined by Ω2 = {u ∈ E, kukE < R2 }. Then
for all u ∈ Con(α, β) ∩ ∂Ω2 , we have for all i ∈ {1, ..., n} for all s ∈ [0, 1]
fi (s, u(s), u0 (s)) ≤ max fi (s, u(s), u0 (s)) ≤ ε(ku(s)k1 + ku0 (s)k1 ) ≤ εR2 .
s∈[0,1]
Using lemma 3.1-(1) we have, for all t ∈ [0, 1]
1
Z
|Ti (u)(t)|
Hi (t, s)fi (s, u(s), u0 (s))ds + Pi (t) + 1
≤
0
Z
≤ ε R2 (1 +
+
max
i∈{1,...,n}
Ki )
≤
ϕ(s)ds
0
max [Ki M1,i + Ci M2,i + M1,i + 1]
i∈{1,...,n}
≤
1
R2
R2
+
4n
2n
R2
.
n
Then, for all i ∈ {1, ..., n},
kTi (u)k∞ ≤
R2
.
n
74
ROCHDI JEBARI
Using lemma 3.1-(2) we have, for all t ∈ [0, 1]
Z 1
∂Hi (t, s)
|Ti (u)0 (t)| ≤
fi (s, u(s), u0 (s))ds + Pi0 (t) + 1
∂t
0
Z 1
ϕ(s)fi (s, u(s), u0 (s))ds
≤ ε R2 (1 + 2 max Ki )
i∈{1,...,n}
+
0
max [2(Ki M1,i + Ci M2,i ) + 1]
i∈{1,...,n}
R2
R2
R2
+
=
.
2n
2n
n
Then for all i ∈ {1, ..., n},
≤
kTi (u)0 k∞ ≤
Therefore
kTi (u)k ≤
R2
.
n
R2
.
n
We deduce that
kT (u)kE ≤ R2 = kukE .
Step 4: Let u ∈ Ω1 then kuk ≤ min Ri,1 < 2 min
i∈{1,...,n}
i∈{1,...,n}
Ri,1 ≤ R2 . This implies
that kuk < R2 , then u ∈ Ω2 . We deduce that Ω1 ⊂ Ω2 . By theorem 3.4, T has at
least one fixed point in K ∩ (Ω2 \Ω1 ). Then (1.1)-(1.2) has a least one nonnegative
solution u.
Theorem 3.6. Under assumptions of theorem 3.5 and adding the following condition:
For all i ∈ {1, . . . , n}, there exist t0,i ∈]0, 1[ such that fi (t0,i , x, y) > 0
Rn+ ,
(3.3)
n
for all x ∈
for all y ∈ R .
Then the problem (1.1)-(1.2) has at least one positive solution.
Proof. Consider the nonnegative solution u for problem (1.1)-(1.2) whose existence
is guaranteed by theorem 3.5. Notice that ui satisfied for all i ∈ {1, ..., n} ui (t) =
Z 1
Hi (t, s)fi (s, u(s), u0 (s))ds + Pi (t). From (H2) and condition (3.3), there exist
0
[αi , βi ] ⊂]0, 1[, such that, for all t ∈ [αi , βi ] and x ∈ Rn+ , y ∈ Rn , fi (t, x, y) > 0.
Then for all i ∈ {1, . . . , n} and for all t ∈]0, 1[,
Z 1
Z βi
ui (t) =
Hi (t, s)fi (s, u(s), u0 (s))ds + Pi (t) ≥
Hi (t, s)fi (s, u(s), u0 (s))ds >
0
αi
0. The proof is complete.
4. Nonexistence results
In this section, we give some sufficient conditions for Zthe nonexistence of pos1
itive solutions. We define the constants Bi = (1 + Ki )
ϕ(s)ds and C(a, b) =
0
Z b
ω1 (s)ds.
a
Theorem 4.1. Suppose that (H2) holds. There exists i0 ∈ {1, ..., n} such that for
all t ∈]0, 1[, for all xi0 ∈]0, +∞[, for all i ∈ {1, ..., n}\{i0 }, xi ∈ R
SYSTEM OF THIRD-ORDER BOUNDARY VALUE PROBLEM
75
x
(1) (Ki0 + 1)h1,i0 (t, x1 , ..., xi0 , ..., xn ) < 3i0
x
(2) Ci0 h2,i0 (t, x1 , ..., xi0 , ..., xn ) < 3i0
x
(3) Bi0 fi0 (t, x1 , ..., xi0 , ..., xn , y1 , ..., yn ) < 3i0 .
Then the problem (1.1)-(1.2) has no positive solutions.
Proof. Assume, to the contrary, that u(t) is a positive solution of (1.1)-(1.2). We denote by u(s) = (u1 (s), ..., ui0 (s), ..., un (s)) and u0 (s) = (u01 (s), ..., u0i0 (s), ..., u0n (s)).
Then for all s ∈]0, 1[ we have
ui0 (s) −1
B i0 .
3
Then for all t ∈ [0, 1] and for all s ∈]0, 1[ we have
fi0 (t, u(s), u0 (s)) <
ui0 (s)Bi−1
0
.
3
Multiplying this by Hi0 (t, s) and integrating on [0, 1] we deduce that
Z 1
Z 1
Bi−1
Hi0 (t, s)ui0 (s)ds
0
0
Hi0 (t, s)fi0 (t, u(s), u0 (s))ds <
3
0
Z 1
ui0 (s)ds
0
.
≤
3
Since, for all t ∈ [0, 1]
Z 1
Z 1
Pi0 (t) ≤ Ci0
h2,i0 (s, u(s))ds + (Ki0 + 1)
h1,i0 (s, u(s))ds
Hi0 (t, s)fi0 (t, u(s), u0 (s)) < Hi0 (t, s)
0
Z
0
1
Z
ui0 (s)ds
ui0 (s)ds
≤
0
+
3
1
0
=
3
2
3
Z
1
ui0 (s)ds.
0
Then for all t ∈ [0, 1]
Z
ui0 (t) <
2
3
Z
ui0 (s)ds +
0
1
ui0 (s)ds
1
0
3
Z
=
1
ui0 (s)ds.
0
Z
By mean value theorem there exists s0 ∈]0, 1[ such that
which is a contradiction. The proof is complete.
1
ui0 (s)ds = ui0 (s0 )
0
Theorem 4.2. Suppose that (H2) holds. There exists i0 ∈ {1, ..., n}, there exists
a, b ∈]a, b[ such that for all t ∈ [0, 1], for all xi0 ∈]0, +∞[, for all i ∈ {1, ..., n}\{i0 },
xi ∈ R
C(a, b)fi0 (t, x1 , ..., xi0 , ..., xn , y1 , ..., yn ) > xi0 .
(4.1)
Then the problem (1.1)-(1.2) has no positive solutions.
Proof. Assume, to the contrary, that u(t) is a positive solution of (1.1)-(1.2). We denote by u(s) = (u1 (s), ..., ui0 (s), ..., un (s)) and u0 (s) = (u01 (s), ..., u0i0 (s), ..., u0n (s)).
Then for all s ∈]0, 1[ we have
fi0 (s, u1 (s), ..., ui0 (s), ..., un (s), u01 (s), ..., u0n (s)) > ui0 (s)C(a, b)−1 .
76
ROCHDI JEBARI
Then for all t ∈ [0, 1] for all s ∈]0, 1[ we have
Hi0 (t, s)fi0 (t, u(s), u0 (s)) > Hi0 (t, s)ui0 (s)C(a, b)−1 .
Multiplying this by Hi0 (t, s) and integrating on [0, 1] we deduce that
Z 1
Z 1
Hi0 (t, s)ui0 (s)ds
Hi0 (t, s)fi0 (t, u(s), u0 (s))ds > C(a, b)−1
0
0
Z
1
≥
ui0 (s)ds.
0
Then for all t ∈ [0, 1]
Z
ui0 (t)
1
Hi0 (t, s)fi0 (t, u(s), u0 (s))ds + Pi0 (t)
=
0
Z
1
Hi0 (t, s)fi0 (t, u(s), u0 (s))ds
≥
0
Z
>
1
ui0 (s)ds.
0
Z
By mean value theorem there exists s0 ∈]0, 1[ such that
which is a contradiction. The proof is complete.
1
ui0 (s)ds = ui0 (s0 )
0
Example 1. Consider the following system of boundary value problem.

3
√ t0 +1 2 + e−u2 (t) = 0

u000
1 (t) +

(u

1 (t)) +1

p
0

000
−t


|u1 (t)| + |u02 (t)| + e−u2 (t) = 0
 u2 (t) + e
u1 (0) = u2 (0) = 1


u01 (0) = u02 (0) = 0




u0 (1) = 2u1 ( 16 ) + 3

 10
u2 (1) = 2u2 ( 18 ) + 1.
3
We have f1 (t, x1 , x2 , y1 , y2 ) = √t
+1
y12 +1
+ e−x2
f1 (t, x1 , x2 , x1 , x2 )
= e−x2
(|x1 | + |x2 | + |y2 | + |y2 |)
f1 (t, x1 , x2 , y1 , y2 )
then
lim
min
= +∞
(|x1 |+|x2 |+|y2 |+|y2 |)→0 t∈[0,1] (|x1 | + |x2 | + |y2 | + |y2 |)
f1 (t, x1 , x2 , y1 , y2 )
2
max
=p 2
+ e−x2 then
t∈[0,1] (|x1 | + |x2 | + |y2 | + |y2 |)
y1 + 1
f1 (t, x1 , x2 , y1 , y2 )
lim
max
= 0.
(|x1 |+|x2 |+|y2 |+|y2 |)→+∞ t∈[0,1] (|x1 | + |x2 | + |y2 | + |y2 |)
p
f2 (t, x1 , x2 , y1 , y2 ) = e−t |x1 | + |x2 | + e−y2
p
f2 (t, x1 , x2 , y1 , y2 )
= e−1 |x1 | + |y2 | + e−y2
min
t∈[0,1] (|x1 | + |x2 | + |y2 | + |y2 |)
f2 (t, x1 , x2 , y1 , y2 )
then
lim
min
= +∞
(|x1 |+|x2 |+|y2 |+|y2 |)→0 t∈[0,1] (|x1 | + |x2 | + |y2 | + |y2 |)
p
f2 (t, x1 , x2 , y1 , y2 )
= |x1 | + |y2 | + e−y2
max
t∈[0,1] (|x1 | + |x2 | + |y2 | + |y2 |)
min
t∈[0,1]
(4.2)
SYSTEM OF THIRD-ORDER BOUNDARY VALUE PROBLEM
77
f2 (t, x1 , x2 , y1 , y2 )
= 0.
(|x1 |+|x2 |+|y2 |+|y2 |)→+∞ t∈[0,1] (|x1 | + |x2 | + |y2 | + |y2 |)
By using Theorem 3.6 the problem (4.2) has no positive solution.
Example 2. Consider the following system of boundary value problem.
then
lim
max
 000
u1 (t) 2
) + |u2 (t))| + (u01 (t))2 = 0
u1 (t) + (1 + 1024

5 e


2u1 (t)

u000
) + |u02 (t)|3 = 0

2 (t) + e

Z 1


|u2 (s)|ds


u
(0)
=
u
(0)
=

2

 1
0 |u1 (s)| + 1
u01 (0) = u02 (0) = 0
Z 1


 0
1


)
+
|u2 (s)|ds
u
(1)
=
2u
(
1 6

 1

0

Z
1

 0

 u2 (1) = 2u2 ( 1 ) +
(u1 (s) + 1)2 ds
9
(4.3)
0
1
1
2
x1 2
we denote by f1 (t, x1 , x2 , y1 , y2 ) = (1 + 1024
5 e ) + |x2 | + y1 , a = 4 , b = 2 , B =
Z 1
Z b
1
5
ϕ(s)ds =
and C(a, b) =
ω1 (s)ds =
.
12
1024
0
a
5((1+ 1024 ex1 )2 +|x |+y 2
2
1
5
C(a, b)[f1 (t, x1 , x2 , y1 , y2 )] =
> x1
1024
By using Theorem 4.2 the problem (4.3) has no positive solution.
Acknowledgments. The authors would like to thank the anonymous referee for
his/her comments that helped us improve this article.
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Tunis-El Manar University, Faculty of sciences of Tunis,, Campus Universitaire 2092
- El Manar, TUNISIA.,
E-mail address: rjebari@yahoo.fr