Electronic Journal of Differential Equations, Vol. 2008(2008), No. 62, pp. 1–11.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
ON BOUNDARY-VALUE PROBLEMS FOR HIGHER-ORDER
DIFFERENTIAL INCLUSIONS
MYELKEBIR AITALIOUBRAHIM, SAID SAJID
Abstract. We show the existence of solutions to boundary-value problems for
higher-order differential inclusion x(n) (t) ∈ F (t, x(t)), where F (., .) is a closed
multifunction, measurable in t and Lipschitz continuous in x. We use the fixed
point theorem introduced by Covitz and Nadler for contraction multivalued
maps.
1. Introduction
The aim of this paper is to establish the existence of solutions of the following
higher-order boundary-value problems:
• For n ≥ 2
x(n) (t) ∈ F (t, x(t))
(i)
a.e. on [0, 1];
0 ≤ i ≤ n − 2;
x (0) = 0,
(1.1)
x(η) = x(1).
• For n ≥ 2
x(n) (t) ∈ F (t, x(t))
0
x(0) = x (η);
a.e. on [0, 1];
x(1) = x(τ ).
(1.2)
• For n ≥ 4
x(n) (t) ∈ F (t, x(t))
(i)
(i+1)
x (0) = x
(η),
0
x(0) = x (η);
a.e. on [0, 1];
2 ≤ i ≤ n − 2;
(1.3)
x(1) = x(τ ).
• For n ≥ 2
x(n) (t) ∈ F (t, x(t))
a.e. on [0, 1];
x(i) (0) = x(i+1) (η),
0 ≤ i ≤ n − 2.
(1.4)
where F : [0, 1] × R → 2R is a closed multivalued map, measurable with respect to
the first argument and Lipschitz with respect to the second argument, and (η, τ ) ∈
]0, 1[2 .
2000 Mathematics Subject Classification. 34A60, 34B10, 34B15.
Key words and phrases. Boundary value problems; contraction; measurability; multifunction.
c
2008
Texas State University - San Marcos.
Submitted March 14, 2007. Published April 22, 2008.
1
2
M. AITALIOUBRAHIM, S. SAJID
EJDE-2008/62
Three and four-point boundary-value problems for second-order differential inclusions was initiated by Benchohra and Ntouyas, see [4, 5, 6]. The authors investigate the existence of solutions on compact intervals for the problems (1.1) and
(1.2) in the particular case n = 2. In order to obtain solutions of (1.1) and (1.2)
when F is not necessarily convex values, Benchohra and Ntouyas (see [6]) reduce
the existence of solutions to the search for fixed points of a suitable multivalued
map on the Banach space C([0, 1], R). Indeed, they used the fixed point theorem
for contraction multivalued maps, due to Covitz and Nadler [3].
In this paper, we give an extension of the Benchohra and Ntouyas’s result [6]
to the n−order non-convex boundary-value problems and we prove the existence of
solutions for (1.3) and (1.4). We shall adopt the technique used by Benchohra and
Ntouyas in the previous paper.
2. Preliminaries and statement of the main results
Let (E, d) be a complete metric space. We denote by C([0, 1], E) the Banach
space
of continuous
functions
from
[0,
1]
to
E
with
the
norm
kx(.)k
:=
sup
kx(t)k;
t∈
∞
[0, 1] , where k · k is the norm of E. For x ∈ E and for nonempty sets A, B of
E we denote d(x, A) = inf{d(x, y); y ∈ A}, e(A, B) := sup{d(x, B); x ∈ A} and
H(A, B) := max{e(A, B), e(B, A)}. A multifunction is said to be measurable if its
graph is measurable. For more detail on measurability theory, we refer the reader
to the book of Castaing and Valadier [2].
Definition 2.1. Let T : E → 2E be a multifunction with closed values.
(1) T is k-Lipschitz if and only if
H T (x), T (y) ≤ kd(x, y),
for each x, y ∈ E.
(2) T is a contraction if and only if it is k-Lipschitz with k < 1.
(3) T has a fixed point if there exists x ∈ E such that x ∈ T (x).
Let us recall the following results that will be used in the sequel.
Lemma 2.2. [3] If T : E → 2E is a contraction with nonempty closed values, then
it has a fixed point.
Lemma 2.3. [7] Assume that F : [a, b] × R → 2R is a multifunction with nonempty
closed values satisfying:
• For every x ∈ R, F (., x) is measurable on [a, b];
• For every t ∈ [a, b], F (t, .) is (Hausdorff ) continuous on R.
Then for any measurable function x(.) : [a, b] → R, the multifunction F (., x(.)) is
measurable on [a, b].
Definition 2.4. A function x(.) : [0, 1] → R is said to be a solution of (1.1)
(resp. (1.2), (1.3), (1.4)) if x(.) is (n−1)-times differentiable, x(n−1) (.) is absolutely
continuous and x(.) satisfies the conditions of (1.1) (resp. (1.2), (1.3), (1.4)).
Let η ∈ R and n ∈ N \ {0, 1}. Define a sequence of functions (ϕp (.))2≤p≤n by:
For all t ∈ [0, 1]
ϕ2 (t) = 1;
ϕ3 (t) = t + ϕ2 (η);
EJDE-2008/62
BOUNDARY VALUE PROBLEMS
3
p−1
ϕp (t) =
X
tp−2
tp−k
+
+ ϕp−1 (η).
ϕk−1 (η)
(p − 2)!
(p − k)!
k=3
We remark that
(k)
(a) For t ∈ [0, 1] and k ∈ {0, . . . , n − 2}, ϕn (t) = ϕn−k (t);
(b) For k ∈ {0, . . . , n − 3}, ϕn−k (0) = ϕn−k−1 (η);
(k)
(c) For k ∈ {0, . . . , n − 2} the function ϕn (.) is increasing.
Assumptions. We will use the following hypotheses:
(H1) F : [0, 1] × R → 2R is a multivalued map with nonempty closed values
satisfying
(i) For each x ∈ R, t 7→ F (t, x) is measurable;
(ii) There exists a function m(.) ∈ L1 ([0, 1], R+ ) such that for all t ∈ [0, 1]
and for all x1 , x2 ∈ R,
H F (t, x1 ), F (t, x2 ) ≤ m(t)|x1 − x2 |.
(H2) For η ∈]0, 1[,
L(η) + L(1) 1
< 1
L(1) +
(n − 1)!
1 − η n−1
Rt
where L(t) = 0 m(s)ds for all t ∈ [0, 1];
(H3) For (η, τ ) ∈]0, 1[2 ,
n−2
(3 − τ )L(1) + 2L(τ ) X L(η) (k)
+
(3 − τ )ϕ(k)
n (1) + 2ϕn (τ ) < 1;
(1 − τ )(n − 1)!
(1 − τ )k!
k=0
(H4) For η ∈]0, 1[,
n−2
(k)
X ϕn (1)
L(1)
+ L(η)
< 1.
(n − 1)!
k!
k=0
Main results. We shall prove the following results.
Theorem 2.5. If assumptions (H1) and (H2) are satisfied, then problem (1.1) has
at least one solution on [0, 1].
Theorem 2.6. If assumptions (H1) and (H3) are satisfied, then problems (1.2)
and (1.3) have at least one solution on [0, 1].
Theorem 2.7. If assumptions (H1) and (H4) are satisfied, then problem (1.4) has
at least one solution on [0, 1].
3. Proof of the main results
Proof of Theorem 2.5. For y(.) ∈ C([0, 1], R), set
SF,y(.) := g ∈ L1 ([0, 1], R) : g(t) ∈ F (t, y(t)) for a.e. t ∈ [0, 1] .
By Lemma 2.3, for y(.) ∈ C([0, 1], R), F (., y(.)) is closed and measurable, then it
has a selection. Thus SF,y(.) is nonempty. Let us transform the problem into a fixed
point problem. Consider the multivalued map T : C([0, 1], R) → 2C([0,1],R) defined
4
M. AITALIOUBRAHIM, S. SAJID
EJDE-2008/62
as follows: for y(.) ∈ L1 ([0, 1], R), T (y(.)) is the set of all z(.) ∈ C([0, 1], R), such
that
Z t
Z η
(t − s)n−1
(η − s)n−1
tn−1
z(t) =
g(s)ds +
g(s)ds
n−1
(n − 1)!
1−η
(n − 1)!
0
0
Z 1
(1 − s)n−1
tn−1
−
g(s)ds,
n−1
1−η
(n − 1)!
0
where g ∈ SF,y(.) .
We shall show that T satisfies the assumptions of Lemma 2.2. The proof will be
given in two steps:
Step 1: T has non-empty closed values. Indeed, let (yp (.))p≥0 ∈ T (y(.)) converges
to ȳ(.) in C([0, 1], R). Then ȳ(.) ∈ C([0, 1], R) and for each t ∈ [0, 1],
Z η
Z t
tn−1
(η − s)n−1
(t − s)n−1
F (s, y(s))ds +
F (s, y(s))ds
yp (t) ∈
(n − 1)!
1 − η n−1 0 (n − 1)!
0
Z 1
(1 − s)n−1
tn−1
F (s, y(s))ds.
−
1 − η n−1 0 (n − 1)!
Since the sets
Z t
Z η
(t − s)n−1
tn−1
(η − s)n−1
F (s, y(s))ds ,
F (s, y(s))ds ,
(n − 1)!
1 − η n−1 0 (n − 1)!
0
Z 1
tn−1
(1 − s)n−1
F (s, y(s))ds
n−1
1−η
(n − 1)!
0
are closed for all t ∈ [0, 1], we have
Z t
Z η
(t − s)n−1
tn−1
(η − s)n−1
ȳ(t) ∈
F (s, y(s))ds +
F (s, y(s))ds
(n − 1)!
1 − η n−1 0 (n − 1)!
0
Z
1
tn−1
(1 − s)n−1
−
F (s, y(s))ds.
1 − η n−1 0 (n − 1)!
Then ȳ(.) ∈ T (y(.)). So T (y(.)) is closed for each y(.) ∈ C([0, 1], R).
Step 2: T is a contraction. Indeed, let y1 (.), y2 (.) ∈ C([0, 1], R) and z1 (.) ∈
T (y1 (.)). Then
Z t
Z η
(t − s)n−1
tn−1
(η − s)n−1
g1 (s)ds +
g1 (s)ds
z1 (t) =
n−1
(n − 1)!
1−η
(n − 1)!
0
0
Z 1
tn−1
(1 − s)n−1
−
g1 (s)ds,
1 − η n−1 0 (n − 1)!
where g1 ∈ SF,y1 (.) . Consider the multivalued map U : [0, 1] → 2R , defined by
U (t) = x ∈ R : |g1 (t) − x| ≤ m(t)|y1 (t) − y2 (t)| .
For each t ∈ [0, 1], U (t) is nonempty. Indeed, let t ∈ [0, 1], from (H1) we have
H F (t, y1 (t)), F (t, y2 (t)) ≤ m(t)|y1 (t) − y2 (t)|.
Hence, there exists x ∈ F (t, y2 (t)), such that
|g1 (t) − x| ≤ m(t)|y1 (t) − y2 (t)|.
By [2, Proposition III.4], the multifunction
V : t → U (t) ∩ F (t, y2 (t))
(3.1)
EJDE-2008/62
BOUNDARY VALUE PROBLEMS
5
is measurable. Then there exists a measurable selection of V denoted g2 such that
g2 (t) ∈ F (t, y2 (t))
and |g1 (t) − g2 (t)| ≤ m(t)|y1 (t) − y2 (t)|,
for each t ∈ [0, 1].
Now, for t ∈ [0, 1] set
Z t
Z η
(t − s)n−1
(η − s)n−1
tn−1
z2 (t) =
g2 (s)ds +
g2 (s)ds
n−1
(n − 1)!
1−η
(n − 1)!
0
0
Z 1
tn−1
(1 − s)n−1
−
g2 (s)ds.
1 − η n−1 0 (n − 1)!
Then
t
(t − s)n−1
|g1 (t) − g2 (s)|ds
(n − 1)!
0
Z η
(η − s)n−1
tn−1
|g1 (s) − g2 (s)|ds
+
n−1
1−η
(n − 1)!
0
Z 1
tn−1
(1 − s)n−1
+
|g1 (s) − g2 (s)|ds
1 − η n−1 0 (n − 1)!
Z t
(t − s)n−1
≤
m(s)|y1 (s) − y2 (s)|ds
(n − 1)!
0
Z η
tn−1
(η − s)n−1
+
m(s)|y1 (s) − y2 (s)|ds
1 − η n−1 0 (n − 1)!
Z 1
(1 − s)n−1
tn−1
m(s)|y1 (s) − y2 (s)|ds
+
n−1
1−η
(n − 1)!
0
Z 1
1
≤
ky1 (.) − y2 (.)k∞
m(s)ds
(n − 1)!
0
Z η
1
ky
(.)
−
y
(.)k
m(s)ds
+
1
2
∞
(1 − η n−1 )(n − 1)!
0
Z 1
1
+
ky
(.)
−
y
(.)k
m(s)ds
1
2
∞
(1 − η n−1 )(n − 1)!
0
1
L(η) + L(1) ≤
ky1 (.) − y2 (.)k∞ .
L(1) +
(n − 1)!
1 − η n−1
Z
|z1 (t) − z2 (t)| ≤
So, we conclude that
kz1 (.) − z2 (.)k∞ ≤
1
L(η) + L(1) L(1) +
ky1 (.) − y2 (.)k∞ .
(n − 1)!
1 − η n−1
By the analogous relation, obtained by interchanging the roles of y1 (.) and y2 (.), it
follows that
1
L(η) + L(1) H T (y1 (.)), T (y2 (.)) ≤
L(1) +
ky1 (.) − y2 (.)k∞ .
(n − 1)!
1 − η n−1
Consequently, T is a contraction. Hence, by Lemma 2.2, T has a fixed point y(.).
Proposition 3.1. y(.) is a solution of (1.1).
Proof. We have
Z
y(t) =
0
t
(t − s)n−1
tn−1
g(s)ds +
(n − 1)!
1 − η n−1
Z
0
η
(η − s)n−1
g(s)ds
(n − 1)!
6
M. AITALIOUBRAHIM, S. SAJID
−
tn−1
1 − η n−1
Z
0
1
EJDE-2008/62
(1 − s)n−1
g(s)ds,
(n − 1)!
where g ∈ SF,y(.) . Then
Z η
Z η
(η − s)n−1
η n−1
(η − s)n−1
g(s)ds +
g(s)ds
y(η) =
(n − 1)!
1 − η n−1 0 (n − 1)!
0
Z 1
η n−1
(1 − s)n−1
−
g(s)ds
n−1
1−η
(n − 1)!
0
Z η
Z 1
(η − s)n−1
(1 − s)n−1
1
η n−1
=
g(s)ds
−
g(s)ds
n−1
n−1
1−η
(n − 1)!
1−η
(n − 1)!
0
0
and
Z η
1
(η − s)n−1
(1 − s)n−1
g(s)ds +
g(s)ds
(n − 1)!
1 − η n−1 0 (n − 1)!
0
Z 1
1
(1 − s)n−1
−
g(s)ds
n−1
1−η
(n − 1)!
0
Z η
Z 1
1
(η − s)n−1
η n−1
(1 − s)n−1
=
g(s)ds
−
g(s)ds,
1 − η n−1 0 (n − 1)!
1 − η n−1 0 (n − 1)!
Z
1
y(1) =
hence y(1) = y(η). On the other hand, for 0 ≤ i ≤ n − 2, we have
Z t
Z
(t − s)n−i−1
(n − 1) . . . (n − i)tn−i−1 η (η − s)n−1
y (i) (t) =
g(s)ds +
g(s)ds
1 − η n−1
(n − 1)!
0 (n − i − 1)!
0
Z
(n − 1) . . . (n − i)tn−i−1 1 (1 − s)n−1
g(s)ds,
−
1 − η n−1
(n − 1)!
0
hence y (i) (0) = 0. Finally, it is clear that y (n) (t) = g(t), so y (n) (t) ∈ F (t, y(t)). Proof of Theorem 2.6. We transform the problem into a fixed point problem.
For t ∈ [0, 1], set
Z t
Z η
n−2
X
(t − s)n−1
(η − s)k
ψng (t) =
g(s)ds +
ϕ(k)
(t)
g(s)ds,
n
(n − 1)!
k!
0
0
k=0
where g ∈ SF,y(.) . Consider the multivalued map, T : C([0, 1], R) → 2C([0,1],R)
defined as follows: for y(.) ∈ C([0, 1], R),
1+t g
T (y(.)) := z(.) ∈ C([0, 1], R) : z(t) = ψng (t) +
ψn (τ ) − ψng (1) .
1−τ
We shall show that T satisfies the assumptions of Lemma 2.2. The proof will be
given in two steps:
Step 1: T has non-empty closed values. Indeed, let (yp (.))p≥0 ∈ T (y(.)) converges
to ȳ(.) in C([0, 1], R). Then ȳ(.) ∈ C([0, 1], R) and for each t ∈ [0, 1],
Z t
Z η
n−2
X
(t − s)n−1
(η − s)k
(k)
yp (t) ∈
F (s, y(s))ds +
ϕn (t)
F (s, y(s))ds
(n − 1)!
k!
0
0
k=0
Z
1 + t h τ (τ − s)n−1
+
F (s, y(s))ds
1 − τ 0 (n − 1)!
EJDE-2008/62
+
BOUNDARY VALUE PROBLEMS
n−2
X
ϕ(k)
n (τ )
Z
0
k=0
Z
−
0
1
η
7
(η − s)k
F (s, y(s))ds
k!
n−2
X
(1 − s)n−1
F (s, y(s))ds −
ϕ(k)
n (1)
(n − 1)!
Z
k=0
0
η
i
(η − s)k
F (s, y(s))ds .
k!
Since the set
Z
0
t
(t − s)k
F (s, y(s))ds
k!
is closed for all t ∈ [0, 1] and 0 ≤ k ≤ n − 1, we have
Z t
Z η
n−2
X
(t − s)n−1
(η − s)k
(k)
ȳ(t) ∈
F (s, y(s))ds +
F (s, y(s))ds
ϕn (t)
(n − 1)!
k!
0
0
k=0
Z
1 + t h τ (τ − s)n−1
F (s, y(s))ds
+
1 − τ 0 (n − 1)!
Z η
n−2
X
(η − s)k
+
ϕ(k)
(τ
)
F (s, y(s))ds
n
k!
0
k=0
Z 1
Z η
n−2
i
X
(1 − s)n−1
(η − s)k
−
F (s, y(s))ds −
ϕ(k)
(1)
F
(s,
y(s))ds
.
n
(n − 1)!
k!
0
0
k=0
Then ȳ(.) ∈ T (y(.)). So T (y(.)) is closed for each y(.) ∈ C([0, 1], R).
Step 2: T is a contraction. Indeed, let y1 (.), y2 (.) ∈ C([0, 1], R) and z1 (.) ∈
T (y1 (.)). Then
1 + t g1
z1 (t) = ψng1 (t) +
ψn (τ ) − ψng1 (1) ,
1−τ
where g1 ∈ SF,y1 (.) . By (3.1), there exists g2 such that
g2 (t) ∈ F (t, y2 (t))
and |g1 (t) − g2 (t)| ≤ m(t)|y1 (t) − y2 (t)|,
for each t ∈ [0, 1].
Now, set for all t ∈ [0, 1],
z2 (t) = ψng2 (t) +
On the other hand, we have
Z
g2
g1
|ψn (t) − ψn (t)| ≤
t
1 + t g2
ψn (τ ) − ψng2 (1) .
1−τ
(t − s)n−1
|g1 (s) − g2 (s)|ds
(n − 1)!
0
Z η
n−2
X
(η − s)k
(k)
+
ϕn (t)
|g1 (s) − g2 (s)|ds
k!
0
k=0
Z t
1
≤
m(s)|y1 (s) − y2 (s)|ds
(n − 1)! 0
Z
n−2
X
1 η
(k)
+
ϕn (t)
m(s)|y1 (s) − y2 (s)|ds
k! 0
k=0
Z t
1
≤
ky1 (.) − y2 (.)k∞
m(s)ds
(n − 1)!
0
8
M. AITALIOUBRAHIM, S. SAJID
+
n−2
X
ϕ(k)
n (t)
k=0
≤
1
ky1 (.) − y2 (.)k∞
k!
EJDE-2008/62
Z
η
m(s)ds
0
n−2
L(1)
X ϕ(k)
n (1)
ky1 (.) − y2 (.)k∞ .
+ L(η)
(n − 1)!
k!
k=0
Then, by (c)
1 + t h g2
|ψn (τ ) − ψng1 (τ )| + |ψng2 (1) − ψng1 (1)|
1−τ
n−2
L(1)
X ϕ(k)
n (1)
≤
ky1 (.) − y2 (.)k∞
+ L(η)
(n − 1)!
k!
|z2 (t) − z1 (t)| ≤ |ψng2 (t) − ψng1 (t)| +
k=0
n−2
X ϕ(k)
2 h L(τ )
n (τ )
+
+ L(η)
ky1 (.) − y2 (.)k∞
1 − τ (n − 1)!
k!
k=0
≤
n−2
i
L(1)
X ϕ(k)
n (1)
+ L(η)
ky1 (.) − y2 (.)k∞
+
(n − 1)!
k!
k=0
h (3 − τ )L(1) + 2L(τ )
(1 − τ )(n − 1)!
+
n−2
X
k=0
i
L(η) (k)
(3 − τ )ϕ(k)
n (1) + 2ϕn (τ ) ky1 (.) − y2 (.)k∞ .
(1 − τ )k!
By the analogous relation, obtained by interchanging the roles of y1 (.) and y2 (.), it
follows that
X L(η) h (3 − τ )L(1) + 2L(τ ) n−2
+
(3 − τ )ϕ(k)
H T (y1 (.)), T (y2 (.)) ≤
n (1)
(1 − τ )(n − 1)!
(1 − τ )k!
k=0
i
+ 2ϕ(k)
(τ
)
ky1 (.) − y2 (.)k∞ .
n
Consequently, T is a contraction. Thus, by Lemma 2.2, T has a fixed point y(.).
Proposition 3.2. y(.) is a solution of (1.2) and (1.3).
Proof. We have
y(t) = ψng (t) +
1+t g
ψ (τ ) − ψng (1) ,
1−τ n
where g ∈ SF,y(.) . Then
y(1) = ψng (1) +
−1 − τ g
2
2
ψ g (τ ) − ψng (1) =
ψ (1) +
ψ g (τ )
1−τ n
1−τ n
1−τ n
y(τ ) = ψng (τ ) +
−1 − τ g
1+τ g
2
ψn (τ ) − ψng (1) =
ψn (1) +
ψ g (τ ),
1−τ
1−τ
1−τ n
and
hence y(1) = y(τ ). On the other hand, for 0 ≤ i ≤ n − 2 and t ∈ [0, 1], we have
Z t
Z η
n−2
X
(t − s)n−i−1
(η − s)k
g (i)
(k+i)
[ψn ] (t) =
g(s)ds +
ϕn
(t)
g(s)ds
k!
0 (n − i − 1)!
0
k=0
EJDE-2008/62
BOUNDARY VALUE PROBLEMS
t
Z
=
0
t
Z
=
0
9
Z η
n+i−2
X
(t − s)n−i−1
(η − s)l−i
g(s)ds +
g(s)ds
ϕ(l)
(t)
n
(n − i − 1)!
(l − i)!
0
l=i
Z η
n−2
X
(t − s)n−i−1
(η − s)l−i
(l)
g(s)ds +
g(s)ds.
ϕn (t)
(n − i − 1)!
(l − i)!
0
l=i
Then, by (a) and (b)
Z η
Z η
n−3
X
(η − s)n−i−2
(η − s)l−i
[ψng ](i) (0) =
g(s)ds +
g(s)ds
ϕ(l)
(0)
n
(n − i − 2)!
(l − i)!
0
0
l=i
Z η
Z η
n−3
X
(η − s)n−i−2
(η − s)l−i
g(s)ds +
g(s)ds
=
ϕn−l−1 (η)
(n − i − 2)!
(l − i)!
0
0
l=i
and by (a)
[ψng ](i+1) (η) =
Z
η
0
Z
=
0
η
n−3
X
(η − s)n−i−2
ϕ(l+1)
(η)
g(s)ds +
n
(n − i − 2)!
n−i−2
(η − s)
g(s)ds +
(n − i − 2)!
Z
η
l=i
0
n−3
X
Z
ϕn−l−1 (η)
0
l=i
(η − s)l−i
g(s)ds
(l − i)!
η
(η − s)l−i
g(s)ds,
(l − i)!
consequently
[ψng ](i+1) (η) = [ψng ](i) (0),
0
(i)
(3.2)
(i+1)
which implies that y(0) = y (η) and y (0) = y
(η) for 2 ≤ i ≤ n − 2 whenever
if n ≥ 4. Finally, it is clear that y (n) (t) = g(t), hence y (n) (t) ∈ F (t, y(t)).
Proof of Theorem 2.7. Consider the multivalued map T : C([0, 1], R) → 2C([0,1],R)
defined as follows: for y(.) ∈ C([0, 1], R),
T (y(.)) := z(.) ∈ C([0, 1], R) : z(t) = ψng (t) .
We shall show that T satisfies the assumptions of Lemma 2.2. The proof will be
given in two steps:
Step 1: T has non-empty closed values. Indeed, let (yp (.))p≥0 ∈ T (y(.)) converges
to ȳ(.) in C([0, 1], R). Then ȳ(.) ∈ C([0, 1], R) and for each t ∈ [0, 1],
Z η
Z t
n−2
X
(t − s)n−1
(η − s)k
F (s, y(s))ds +
ϕ(k)
(t)
F (s, y(s))ds.
yp (t) ∈
n
(n − 1)!
k!
0
0
k=0
Since the set
t
(t − s)k
F (s, y(s))ds
k!
0
is closed for all t ∈ [0, 1] and 0 ≤ k ≤ n − 1, we have
Z t
Z η
n−2
X
(t − s)n−1
(η − s)k
ȳ(t) ∈
F (s, y(s))ds +
ϕ(k)
(t)
F (s, y(s))ds.
n
(n − 1)!
k!
0
0
Z
k=0
Then ȳ(.) ∈ T (y(.)). So T (y(.)) is closed for each y(.) ∈ C([0, 1], R).
Step 2: T is a contraction. Indeed, let y1 (.), y2 (.) ∈ C([0, 1], R) and z1 (.) ∈
T (y1 (.)). Then
z1 (t) = ψng1 (t),
10
M. AITALIOUBRAHIM, S. SAJID
EJDE-2008/62
where g1 ∈ SF,y1 (.) . By (3.1), there exists g2 such that
g2 (t) ∈ F (t, y2 (t))
and |g1 (t) − g2 (t)| ≤ m(t)|y1 (t) − y2 (t)|,
for each t ∈ [0, 1].
ψng2 (t).
Now, for t ∈ [0, 1], we set z2 (t) =
On the other hand, we have
Z t
(t − s)n−1
g2
g1
|ψn (t) − ψn (t)| ≤
|g1 (s) − g2 (s)|ds
(n − 1)!
0
Z η
n−2
X
(η − s)k
(k)
|g1 (s) − g2 (s)|ds
+
ϕn (t)
k!
0
k=0
Z t
1
m(s)|y1 (s) − y2 (s)|ds
≤
(n − 1)! 0
Z
n−2
X
1 η
+
ϕ(k)
m(s)|y1 (s) − y2 (s)|ds
(t)
n
k! 0
k=0
Z t
1
ky1 (.) − y2 (.)k∞
≤
m(s)ds
(n − 1)!
0
Z η
n−2
X
1
+
ϕ(k)
(t)
ky
(.)
−
y
(.)k
m(s)ds
1
2
∞
n
k!
0
k=0
n−2
L(t)
X ϕ(k)
n (t)
+ L(η)
ky1 (.) − y2 (.)k∞ .
≤
(n − 1)!
k!
k=0
Then, by (c)
|z2 (t) − z1 (t)| ≤
n−2
L(1)
X ϕ(k)
n (1)
+ L(η)
ky1 (.) − y2 (.)k∞ .
(n − 1)!
k!
k=0
By the analogous relation, obtained by interchanging the roles of y1 (.) and y2 (.), it
follows that
n−2
X ϕ(k)
L(1)
n (1)
H T (y1 (.)), T (y2 (.)) ≤
+ L(η)
ky1 (.) − y2 (.)k∞ .
(n − 1)!
k!
k=0
Consequently, T is a contraction. Hence, by Lemma 2.2, T has a fixed point y(.). Proposition 3.3. y(.) is a solution of (1.4).
Proof. By (3.2), we have y (i) (0) = y (i+1) (η), for 0 ≤ i ≤ n − 2. Since y (n) (t) = g(t),
we have y (n) (t) ∈ F (t, y(t)).
References
[1] A. Boucherif and S. M. Bouguima; Nonlinear second order ordinary differential equations with
nonlocal boundary conditions, Commu. Appl. Nonl. Anal., 5(2), (1998), 73-85.
[2] C. Castaing and M. Valadier; Convex Analysis and Measurable Multifunctions, Lecture Notes
in Mathematics 580, Springer-Verlag, Berlin-Heidelberg-New York, (1977).
[3] H. Covitz and S. B. Jr. Nadler; Multivalued contraction mappings in generalized metric spaces,
Israel J. Math., 8, (1970), 5-11.
[4] M. Benchohra and S. K. Ntouyas; A note on a three-point boundary-value problem for second
order differential inclusions, Mathematical Notes, Miskolc. 2, (2001), 39-47.
[5] M. Benchohra and S. K. Ntouyas; Multi-point boundary-value problems for second order differential inclusions, Math. Vesnik, to appear.
EJDE-2008/62
BOUNDARY VALUE PROBLEMS
11
[6] M. Benchohra and S. K. Ntouyas; On three and for point boundary-value problems for second
order differential inclusions, Mathematical Notes, Miskolc. 2, (2001), 93-101.
[7] Q. Zhu; On the solution set differential inclusions in Banach spaces, J. Diff. eqs., No. 41,
(2001), 1-8.
[8] S. A. Marano; A remark on a second order three-point boundary-value problem, J. Math. Anal.
Appl., 183, (1994), 518-522.
Myelkebir Aitalioubrahim
University Hassan II-Mohammedia, U.F.R Mathematics and applications, F.S.T, BP 146,
Mohammedia, Morocco
E-mail address: aitalifr@yahoo.fr
Said Sajid
University Hassan II-Mohammedia, U.F.R Mathematics and applications, F.S.T, BP 146,
Mohammedia, Morocco
E-mail address: saidsajid@hotmail.com