Electronic Journal of Differential Equations, Vol. 2008(2008), No. 144, pp. 1–16.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
MONOTONE SOLUTIONS FOR A NONCONVEX FUNCTIONAL
DIFFERENTIAL INCLUSIONS OF SECOND ORDER
AHMED G. IBRAHIM, FERYAL A. AL-ADSANI
Abstract. We give sufficient conditions for the existence of a monotone solution for second-order functional differential inclusions. No convexity condition,
on the values of the multifunction defining the inclusion, is involved in this
construction.
1. Introduction
Let K be a closed subset of Rn , Ω an open subset of Rn , and P a lower semicontinuous set-valued map (multifunction) from K to the family of all non-empty
subsets of K, with closed graph satisfying the following two conditions:
(i) for all x ∈ K, x ∈ P (x)
(ii) for all x, y ∈ K, y ∈ P (x) ⇒ P (y) ⊆ P (x).
Under these conditions, a preorder (reflexive and transitive relation) on K is defined
as
x y ⇔ y ∈ P (x) .
Let σ > 0 and C([−σ, 0], Rn ) be the space of continuous functions from [−σ, 0]
to Rn with the uniform norm kxkσ = sup{kx(t)k : t ∈ [−σ, 0]}. For each t ∈
[0, T ]; T > 0, we define the operator τ (t) from C([−σ, T ], Rn ) to C([−σ, 0], Rn ) as
(τ (t)x)(s) = x(t + s),
for all s ∈ [−σ, 0].
Here, τ (t)x represents the history of the state from the time t − σ to the present
time t.
Let K0 = {ϕ ∈ C([−σ, T ], Rn ) : ϕ(0) ∈ K} and F be a set-valued map (multifunction) defined from K0 × Ω to the family of non-empty compact subsets (not
necessarily convex) in Rn and (ϕ0 , y0 ) be a given element in K0 × Ω. We consider
2000 Mathematics Subject Classification. 34A60, 49K24.
Key words and phrases. Functional differential inclusion; monotone solution;
second-order contingent cone.
c
2008
Texas State University - San Marcos.
Submitted June 10, 2008. Published October 24, 2008.
1
2
A. G. IBRAHIM, F. A. AL-ADSANI
EJDE-2008/144
the second-order functional differential inclusion
x00 (t) ∈ F (τ (t)x, x0 (t)),
x(t) = ϕ0 (t),
a.e. on [0, T ]
∀t ∈ [−σ, 0]
0
x (0) = y0
x(t) ∈ P (x(t)) ⊂ K,
x(s) x(t)
(1.1)
∀t ∈ [0, T ]
whenever 0 ≤ s ≤ t ≤ T
In the present work, we prove under reasonable conditions that there are a
positive real number T and a continuous function x : [−σ, T ] → Rn such that
(1) the function x is absolutely continuous on [0, T ] with absolutely continuous
derivative
(2) τ (t)x ∈ K0 , for all t ∈ [0, T ]
(3) x0 (t) ∈ Ω, a.e. on [0, T ]
(4) the functions x, x0 , x00 satisfy (1.1).
Ibrahim and Alkulaibi [13] proved the existence of a monotone solution for (1.1)
without delay. They consider the problem
x00 (t) ∈ F (x(t), x0 (t)),
a.e. on [0, T ]
0
x(0) = x0 ,
x (0) = y0
x(t) ∈ K,
∀t ∈ [0, T ]
x(s) x(t),
whenever 0 ≤ s ≤ t ≤ T .
Further, Lupulescu [17] proved the existence of a local solution, not necessarily
monotone, for (1.1) in the particular case P (x) = K, for all x ∈ K. Thus, the
result, we are going to prove, generalizes the results of Ibrahim and Alkulaibi [13]
and Lupulescu [17].
We mention, among others the works, [9, 11, 12, 13, 18] for the proof of existence of monotone solutions for differential inclusions or functional differential
inclusions and the works [3, 4, 7, 8, 10, 14, 15, 16, 17, 19] for solutions not necessarily monotone. Note that the case where the solutions are not necessarily monotone
has been widely investigated compared with that of monotone solutions which has
been rarely investigated.
The present paper is organized as follows: In section 2, some definitions and
facts to be used later are introduced. In section 3, the main result is proved.
2. Preliminaries
Let Rn be the n-dimensional Euclidean space with norm k · k and scalar product
h·, ·i. For x ∈ Rn and r > 0 let B(x, r) = {y ∈ Rn : ky − xk < r} denote the open
ball centered at x of radius r, and B(x, r) its closure.
For ϕ ∈ C([−σ, 0], Rn ) let Bσ (ϕ, r) = {ψ ∈ C([−σ, 0], Rn ) : kψ − ϕkσ < r} and
B σ (ϕ, r) = {ψ ∈ C([−σ, 0], Rn ) : kψ − ϕkσ ≤ r}.
We also, denote by d(x, A) = inf{kx − yk : y ∈ A} the distance from x ∈ Rn to
a closed subset A ⊆ Rn .
A function V : Rn → R ∪ {∞} is said to be proper if its effective domain
D(V ) = {x ∈ Rn : V (x) < ∞} is non-empty.
EJDE-2008/144
MONOTONE SOLUTIONS
3
The subdifferential of a proper convex lower semicontinuous function V : Rn → R
at a point x ∈ Rn is defined (in the sense of convex analysis) by
∂V (x) = ξ ∈ Rn : V (y) − V (x) ≥ hξ, y − xi, ∀y ∈ Rn
The second-order contingent cone of a non-empty closed subset C ⊂ Rn at a point
(x, y) ∈ C × Rn is defined by
2
d(x + ty + t2 z, C)
= z ∈ R : lim inf
=0 .
t2
t→0+
For the properties of the second-order contingent cone see for example [2, 3, 7, 14].
n
A multifunction F : K0 × Ω → 2R is said to be upper semicontinuous at a point
(ϕ, y) ∈ K0 × Ω if for every ε > 0 there exists δ > 0, such that
TC2 (x, y)
n
F (ψ, z) ⊂ F (ϕ, y) + B(0, ε),
for all (ψ, z) ∈ Bσ (ϕ, δ) × B(y, δ). For more information about the continuity
properties for multifunctions we refer the reader to [1, 2, 6].
3. Main Result
Lemma 3.1. Let K be a non-empty closed subset of Rn , Ω a non-empty open
subset of Rn , P a set-valued map from K to the family of non-empty closed subsets
of K and K0 = {ϕ ∈ C([−σ, 0], Rn ), ϕ(0) ∈ K}. Let F be an upper semicontinuous
set-valued map from K0 × Ω to the family of non-empty compact subsets of Rn .
Assume also the following conditions:
(H1) For all x ∈ K, x ∈ P (x)
(H2) There exists a proper convex lower semicontinuous function V : Rn → R
such that F (ϕ, y) ⊆ ∂V (y), for every (ϕ, y) ∈ K0 × Ω
(H3) For (ϕ, y) ∈ K0 × Ω, F (ϕ, y) ⊆ TP2 (ϕ(0)) (ϕ(0), y), where TP2 (ϕ(0)) (ϕ(0), y) is
the second order contingent cone of the closed subset P (ϕ(0)) at the point
(ϕ(0), y).
Let (ϕ0 , y0 ) be a fixed element in K0 × Ω. Then there are two positive numbers
r and T such that for each positive integer m there are:
(1) A positive integer νm .
(2) A set of points
m
m
m
Pm = {tm
0 = 0 < t1 < · · · < tνm −1 ≤ T < tνm }
(3) Three sets of elements in Rn :
Xm = {xm
p : p = 0, 1, . . . , νm − 1},
Ym = {ypm : p = 0, 1, . . . , νm − 1},
Zm = {zpm : p = 0, 1, . . . , νm − 1},
m
with xm
0 = ϕ0 (0) and y0 = y0
(4) A continuous function xm : [−σ, T ] → Rn with xm (t) = ϕ0 (t), for all
t ∈ [−σ, 0], such that for each p = 0, 1, . . . , νm − 1, the following properties
are satisfied:
1
m
m
(i) hm
p+1 = tp+1 − tp < m
1
m
m
m
m
m
m
(ii) zp = up + wp where um
p ∈ F (τ (tp )xm , yp ) and wp ∈ m B(0, 1)
1
m m
m 2 m
m m
(iii) xm (t) = xm
p + (t − tp )yp + 2 (t − tp ) zp , for all t ∈ [tp , tp+1 ]
1
m
m
m
m
m
2 m
m
(iv) xp+1 = xp + hp+1 yp + 2 (hp+1 ) zp = xm (tp+1 )
4
A. G. IBRAHIM, F. A. AL-ADSANI
EJDE-2008/144
m
(v) xm
p+1 ∈ P (xp ) ∩ B(ϕ0 (0), r) ⊆ K and
m
m
yp+1
= ypm + hm
p+1 zp ∈ B(y0 , r) ⊆ Ω
m
(vi) xm (t) ∈ B(ϕ0 (0), r), for all t ∈ [tm
p , tp+1 ].
m
(vii) τ (tp+1 )xm ∈ Bσ (ϕ0 , r) ∩ K0 .
Proof. We follow the techniques developed in [18]. From [6, Prop. I.26], for each
y ∈ Rn , the subset ∂V (y) is closed, convex and bounded. Moreover, by [1, Thm.
0.7.2] the multifunction y → ∂V (y) is upper semicontinuous. So, by [1, Prop. 1.1.3]
there are two positive real numbers r and M such that
sup{kzk : z ∈ ∂V (y)} ≤ M ,
for all y ∈ B(y0 , r). Using condition (H2), we get
sup{kzk : z ∈ F (ψ, y)} < M ,
(3.1)
for all (ψ, y) ∈ (K0 ∩ Bσ (ϕ0 , r)) × B(y0 , r). Since Ω is open we can choose r such
that B(y0 , r) ⊆ Ω. It is obvious that the closedness of K implies the closedness of
K0 in C([−σ, 0], Rn ). From the continuity of ϕ0 on [−σ, 0], there is µ > 0 such that
for all t, s ∈ [−σ, 0] we have
r
|t − s| < µ =⇒ kϕ0 (t) − ϕ0 (s)k < .
(3.2)
4
Put
r
n
o
r
r
r
T = min µ,
,
,
(3.3)
4(M + 1) 8(ky0 k + 1)
4(M + 1)
Thus the numbers r and T are well defined. Now let m be a fixed positive integer.
m
m
We put tm
0 = 0, x0 = ϕ0 (0) and y0 = y0 . The sets Pm , Xm , Ym and Zm will be
m
m
m
m
defined by induction. We first define xm
1 , t1 , y1 , z0 and xm on [0, t1 ] such that
the properties (i)–(vii) are satisfied for p = 0.
Using condition (H3), there is um
0 ∈ F (ϕ0 , y0 ) such that
lim inf
h↓0
h2
1
d(ϕ0 (0) + h y0m + um
, P (ϕ0 (0))) = 0.
2
h
2 0
1
m
So, a positive number hm
1 is found such that h1 ≤ min{ m , T } and
2
2
(hm
(hm
1 )
1 )
um
.
0 , P (ϕ0 (0))) ≤
2
4m
Since P (ϕ0 (0)) is closed, there is xm
1 ∈ P (ϕ0 (0)) with
m
d(ϕ0 (0) + hm
1 y0 +
m
kϕ0 (0) + hm
1 y0 +
2
2
(hm
(hm
1 )
1 )
m
um
.
0 − x1 k ≤
2
4m
Consequently there is w0m ∈ Rn such that kw0m k ≤
m m
xm
1 = ϕ0 (0) + h1 y0 +
1
2m
and
2
2
(hm
(hm
1 )
1 )
um
w0m .
0 +
2
2
1
m
2m B(0, 1) and x1 =
m
m
tm
1 = t0 + h1 and for
m
m
Now we define z0m = um
0 + w0 , therefore z0 ∈ F (ϕ0 , y0 ) +
2
(hm
1 )
m
m
m
m
m m
ϕ0 (0) + hm
1 y0 +
2 z0 . We put y1 = y0 + h1 z0 and
m m
t ∈ [t0 , t1 ] we define
m
xm (t) = ϕ0 (0) + (t − tm
0 )y0 +
2
(t−tm
0 )
z0m .
2
EJDE-2008/144
MONOTONE SOLUTIONS
5
Thus, the properties (i)–(iv) are clearly satisfied for p = 0.
Since τ (tm
0 )xm = ϕ0 , using relation (3.1), we obtain
sup{kvk : v ∈ F (τ (tm
0 )xm , y0 )} ≤ M.
1
Therefore, kz0m k ≤ M + 2m
< M +1. We get from the definition of y1m , ky1m −y0m k ≤
m
m
m
m
hm
1 kz0 k ≤ T (M + 1) ≤ r. Thus y1 ∈ B(y0 , r). Since, x1 ∈ P (x0 ) ⊆ K then to
prove property (v) for p = 0, it is sufficient to show that
kxm
1 − ϕ0 (0)k < r.
We get using (3.1), (3.3)
m
m
kxm
1 − ϕ0 (0)k = h1 ky0 k +
≤ T ky0m k +
2
(hm
1 )
kz0m k
2
T2
(M + 1)
2
r
r
ky0 k +
(M + 1)
8(ky0 k + 1)
8(M + 1)
r
r
< + <r
8 8
and hence (v) is satisfied for p = 0. To prove (vi) for p = 0, we note that, for
m
t ∈ [tm
0 , t1 ],
≤
m
kxm (t) − ϕ0 (0)k ≤ (t − tm
0 )ky0 k +
m
≤ hm
1 ky0 k +
2
(t − tm
0 )
kz0m k
2
2
(hm
1 )
kz0m k
2
T2
(M + 1)
2
r
1
r
≤
ky0 k +
(M + 1)
8(ky0 k + 1)
2 4(M + 1)
r
r
< + < r,
8 8
≤ T ky0 k +
which proves (vi) for p = 0.
m
To prove property (vii) for p = 0, we note that if −σ ≤ s ≤ −tm
1 , then t1 + s ≤ 0
and by (3.2), (3.3), we get
r
kτ (tm
sup kxm (tm
sup kϕ0 (tm
1 )xm −ϕkσ =
1 +s)−ϕ0 (s)k =
1 +s)−ϕ0 (s)k < .
4
−σ≤s≤0
−σ≤s≤0
m
m
while if −tm
1 ≤ s ≤ 0, then 0 ≤ t1 + s ≤ t1 and hence by (3.3) we get
m
kxm (tm
1 + s) − ϕ0 (s)k ≤ kxm (t1 + s) − ϕ0 (0)k + kϕ0 (0) − ϕ0 (s)k
2
(hm
r
1 + s)
kz0m k +
2
4
T2 m
r
m
≤ T ky0 k +
kz k +
2 0
4
r
r
r
≤
ky0 k +
(M + 1) +
8(ky0 k + 1)
8(M + 1)
4
r
r
r
r
< + + = ,
8 8 4
2
∈ Bσ (ϕ0 , r) and hence (vii) is proved.
m
≤ (hm
1 + s)ky0 k +
which shows that τ (tm
1 )xm
6
A. G. IBRAHIM, F. A. AL-ADSANI
EJDE-2008/144
m
m
m
Now we suppose that tm
p+1 , xp+1 , yp+1 , zp are well defined for p = 0, 1, . . . , (q−1)
m
and xm is defined on the interval [−σ, tq ] such that all the properties (i)–(vii) are
satisfied for p = 0, 1, . . . , (q − 1).
m
m
m
m m
We define tm
q+1 , xq+1 , yq+1 , zq and xm on [tq , tq+1 ] such that the properties
1
[ for which
(i)–(vii) are satisfied for p = q. We denote by Hqm the set of all h ∈]0, m
the following conditions are satisfied:
(a) h < T − tm
q .
m
m
(b) there exists um
q ∈ F (τ (tq )xm , yq ) such that
h2 m
h2
uq , p(xm
.
q )) ≤
2
4m
From the fact that (v) and (vii) are true for p = q − 1, we get yqm ∈ Ω and
τ (tm
q )xm ∈ K0 . Moreover, since (iv) is true for p = q − 1, then
m
d(xm
q + hyq +
m
m
τ (tm
q )xm (0) = xm (tq ) = xq .
So, the condition (H3) gives:
m
2
m
F (τ (tm
(xm
q )xm , yq ) ⊆ TP (xm
q , yq ).
q )
m
m
Therefore there is um
q ∈ F (τ (tq )xm , yq ) such that
lim inf
h↓0
1
h2 m
m
d(xm
u , P (xm
q + h yq +
q )) = 0,
2
h
2 q
1
which shows that there is a positive number h such that h < min{ m
, T − tm
q } and
h2 m
h2
uq , P (xm
.
q )) ≤
2
4m
is bounded by the number T , there is a number dm
q
m
d(xm
q + hyq +
Hence h ∈ Hqm . Since Hqm
m
m
such that dm
q = sup{α : α ∈ Hq }. Since Hq ∩ [
Hqm ∩
dm
[ 2q , dm
q ]
dm
q
m
2 , dq ]
6= φ, an element hm
q+1 ∈
is found such that
2
2
(hm
(hm
q+1 )
q+1 )
m
um
,
P
(x
))
≤
.
q
q
2
4m
m
m
From the closedness of P (xm
q ), there is xq+1 ∈ P (xq ) ⊆ K with
m
m
d(xm
q + hq+1 yq +
m
m
kxm
q + hq+1 yq +
2
2
(hm
(hm
q+1 )
q+1 )
m
um
−
x
k
≤
.
q
q+1
2
4m
Consequently, there is wqm ∈ Rn with kwqm k ≤
1
2m
<
1
m
such that
2
2
(hm
(hm
q+1 )
q+1 )
um
+
wqm
q
2
2
2
(hm
q+1 )
m
m
m
(um
= xm
q + wq ).
q + hq+1 yq +
2
m
= um
q + wq . So that
m
m
m
xm
q+1 = xq + hq+1 yq +
We define zqm
1
B(0, 1),
m
m
(hq+1 )2 m
m
m
= xm
+
h
y
+
zq .
q
q+1 q
2
m
zqm ∈ F (τ (tm
q )xm , yq ) +
xm
q+1
EJDE-2008/144
MONOTONE SOLUTIONS
7
m
m
m
m
m
m m
We put yq+1
= yqm + hm
q+1 zq and tq+1 = tq + hq+1 and for t ∈ [tq , tq+1 ], we define
m m
xm (t) = xm
q + (t − tq )yq +
2
(t − tm
q )
zqm .
2
Obviously the relations (i)–(iv) are satisfied for p = q.
Now we prove that (v) is true for p = q. Since (v) and (vii) are true for p = q − 1,
m
m
then τ (tm
q )xm ∈ Bσ (ϕ0 , r) and yq ∈ B(y0 , r), and hence by (3.1) we get kzq k ≤
M + 1.
m
m
m
Let us prove that kyq+1
− y0 k < r. We note that yq+1
= yqm + hm
q+1 zq =
m
m m
m
m
yq−1 + hq zq−1 + hq+1 zq . By iterating we get
m
yq+1
=
y0m
+
q
X
m
hm
s+1 zs .
(3.4)
s=0
Thus,
m
kyq+1
− y0m k ≤
q
X
m
hm
s+1 kzs k ≤ (M + 1)
s=0
To prove that
relation
xm
q+1
xm
p+1 = ϕ0 (0) +
q
X
s=0
hm
s+1 ≤ (M + 1)T <
r
< r.
4
∈ Bσ (ϕ0 (0), r) we first use the induction technique to prove the
p
X
p
p−1
p
1X m 2 m X X m m m
hm
y
+
(h
)
z
+
hi+1 hj+1 zi ,
j+1 0
j
2 j=0 j+1
j=0
i=0 j=i+1
(3.5)
for p = 1, . . . , q. For p = 1 we note that
1 m 2 m
m
m m
xm
2 = x1 + h2 y1 + (h2 ) z1
2
1 m 2 m
m m
m m
= xm
1 + h2 (y0 + h1 z0 ) + (h2 ) z1
2
1 m 2 m
1 m 2 m
m m
m m
m m
= (xm
0 + h1 y0 + (h1 ) z0 ) + h2 (y0 + h1 z0 ) + (h2 ) z1
2
2
1 m 2 m
m
m m
m 2 m
m m m
= xm
0 + (h1 + h2 )y0 + ((h1 ) z0 + (h2 ) z1 ) + h1 h2 z0
2
1
1
1
X
1X m 2 m X m m m
m
m
(h ) zj +
h1 hj+1 z0 .
= ϕ0 (0) + (
hj+1 )y0 +
2 j=0 j+1
j=1
j=0
Then relation (3.5) is true for p = 1. Suppose that (3.5) is true for p = q − 1. This
gives us
xm
q = ϕ0 (0) +
q−1
X
j=0
hm
j+1 )y0 +
q−1
q−2 q−1
1X m 2 m X X m m m
(hj+1 ) zj +
hi+1 hj+1 zi .
2 j=0
i=0 j=i+1
So, according to the definition of xm
q+1 we have
1 m 2 m
m
m
m
xm
q+1 = xq + hq+1 yq + (hq+1 ) zq
2
q−1
q−1
q−2 q−1
X
1X m 2 m X X m m m
m
= ϕ0 (0) + (
hm
)y
+
(h
)
z
+
hi+1 hj+1 zi
j+1 0
j
2 j=0 j+1
j=0
i=0 j=i+1
8
A. G. IBRAHIM, F. A. AL-ADSANI
+ hm
q+1 (y0 +
EJDE-2008/144
q−1
X
1 m
m
m
hm
s+1 zs ) + (hq+1 )zq
2
s=0
q
q−2 q−1
q
X
1X m 2 m X X m m m
m
(hj+1 ) zj +
hi+1 hj+1 zi
= ϕ0 (0) + (
hm
j+1 )y0 +
2 j=0
i=0 j=i+1
j=0
m m
m m
m m
+ hm
q+1 (h1 z0 + h2 z1 + · · · + hq zq−1 )
= ϕ0 (0) +
q
X
m
hm
j+1 )y0 +
j=0
q
q−1
q
1X m 2 m X X m m m
(hj+1 ) zj +
hi+1 hj+1 zi .
2 j=0
i=0 j=i+1
This implies that the relation (3.5) is true for p = q. Now, from the fact that
kzpm k ≤ M + 1, for all p = 0, 1, . . . , q we get
kxm
q+1 − ϕ0 (0)k
q
q
q−1 X
q
X
X
1X m 2
m
≤ ky0 k(
hm
)
+
(h
)
(M
+
1)
+
(M
+
1)
hm
j+1
j+1
i+1 hj+1
2
j=0
i=0 j=i+1
j=0
q
q
h
X
X
1
m
m
≤ ky0 kT + (M + 1)T 2 + (M + 1) hm
h
+
h
hm
1
j+1
2
j+1 + . . .
2
j=1
j=2
+
hm
q
q
X
hm
j+1
i
j=q
1
≤ ky0 kT + (M + 1)T 2 + (M + 1)T 2
2
3
= ky0 kT + (M + 1)T 2
2
3r
r
r
= .
< +
8
8
2
Thus (v) is true for p = q.
m
Let us prove (vi) for p = q, namely kxm (t) − ϕ0 (0)k < r, for all t ∈ [tm
q , tq+1 ].
m m
Let t ∈ [tq , tq+1 ]. We have
1
m m
m 2 m
xm (t) = xm
q + (t − tq )yq + (t − tq ) zq
2
q−1
q−1
q−2 q−1
X
1X m 2 m X X m m m
(h
)
z
+
hi+1 hj+1 zi
= ϕ0 (0) + (
hm
)y
+
j+1 0
j
2 j=0 j+1
j=0
i=0 j=i+1
+ (t − tm
q )(y0 +
q−1
X
1
m 2 m
m
hm
j+1 zj ) + (t − tq ) zq .
2
j=0
Thus,
kxm (t) − ϕ0 (0)k
q−1
q−1
X
1X m 2 m
≤ ky0 k(
hm
)
+
(h ) kzj k
j+1
2 j=0 j+1
j=0
EJDE-2008/144
+
MONOTONE SOLUTIONS
q−2 X
q−1
X
m
m
m
hm
i+1 hj+1 kzi k + hq+1 (y0 +
i=0 j=i+1
9
q−1
X
1 m 2 m
m
hm
j+1 kzj k) + (hq+1 ) kzq k
2
j=0
q
q−1 X
q
q
X
X
1X m 2 m
m
m
(h
)
kz
k
+
hm
≤ ky0 k(
hm
)
+
j+1
j
i+1 hj+1 kzi k
j+1
2
j=0
i=0 j=i+1
j=0
1
≤ ky0 kT + T 2 (M + 1) + T 2 (M + 1)
2
3(M + 1) 2
T
≤ ky0 kT +
2
r
r
r
< + = .
8 8
4
We prove (vii) for p = q.
kτ (tm
q+1 )xm − ϕ0 kσ
sup kτ (tm
q+1 )xm (s) − ϕ0 (s)k.
=
−σ≤s≤0
sup kxm (tm
q+1 + s) − ϕ0 (s)k.
=
−σ≤s≤0
≤
sup
−σ≤s≤−tm
q+1
≤
sup
−σ≤s≤−tm
q+1
+
kτ (tm
q+1 )xm (s) − ϕ0 (s)k +
kϕ0 (tm
q+1 + s) − ϕ0 (s)k +
sup
−tm
q+1 ≤s≤0
≤
sup
−tm
q+1 ≤s≤0
sup
−tm
q+1 ≤s≤0
kτ (tm
q+1 )xm (s) − ϕ0 (s)k.
kxm (tm
q+1 + s) − ϕ0 (s)k
kϕ0 (0) − ϕ0 (s)k.
r
r
r
+ + < r.
4 4 4
m
It remains to show that there is a positive number νm such that tm
νm −1 ≤ T < tνm .
Therefore, we have to prove that the iterative process is finite. For this purpose
suppose that the iterative process is not finite. So, for each non negative integer
m
m
m
number p, there are tm
p ∈ [0, T [, xp , yp , zp such that the relations (i)–(vii) are
m
satisfied. Since the sequence {tp }p≥1 is bounded and increasing, there is tm
α ∈
m
m
m
]0, T ] such that limp→∞ tm
p = tα . Let us show that {xp }p≥1 , {yp }p≥1 are Cauchy
sequences. Let p and q be two positive integers such that p > q. From the relation
(3.5) we have
m
kxm
p − xq k
p−1
p−1
p−2 p−1
X
1X m 2 m X X m m m
= k(
hm
)y
+
(h
)
z
+
hi+1 hj+1 zi
0
j+1
j
2 j=0 j+1
j=0
i=0 j=i+1
q−1
q−1
q−2 q−1
X
1X m 2 m X X m m m
−(
hm
)y
−
(h
)
z
−
hi+1 hj+1 zi k.
j+1 0
j
2 j=0 j+1
j=0
i=0 j=i+1
p−1
p−1
q−1 p−1
X
1 X m 2 m XX m m m
= k(
hm
)y
+
(h
)
z
+
hi+1 hj+1 zi k.
j+1 0
2 j=q j+1 j
j=q
i=0 j=q
≤
p−1
X
j=q
hm
j+1 )ky0 k +
p−1
p−2 X
p−1
X
1X m 2 m
m
m
(hj+1 ) kzj k +
hm
i+1 hj+1 kzi k.
2 j=q
i=q−1 j=i+1
10
A. G. IBRAHIM, F. A. AL-ADSANI
EJDE-2008/144
p−1
X
1
m
m
m 2
m
≤ ky0 k(tm
−
t
)
+
(M
+
1)(t
−
t
)
+
(M
+
1)
hm
p
q
p
q
q hj+1
2
j=q
+ (M + 1)
p−1
X
m
hm
q+1 hj+1 + · · · + (M + 1)
j=q+1
p−1
X
m
hm
p−2 hj+1 .
j=p−2
1
m
m 2
m m
m
m
= ky0 k(tm
p − tq ) + (M + 1)(tp − tq ) + (M + 1)hq (tp − tq )
2
m
m
m
m
m
+ (M + 1)hm
q+1 (tp − tq ) + · · · + (M + 1)hp−2 (tp − tq ).
1
m
m 2
m
= ky0 k(tm
p − tq ) + (M + 1)(tp − tq )
2
m
m
m
m
+ (M + 1)(tm
p − tq )(hq + hq+1 + · · · + hp−2 ).
1
m
m
m 2
m
m 2
= ky0 k(tm
p − tq ) + (M + 1)(tp − tq ) + (M + 1)(tp − tq ) .
2
m
Since the sequence {tm
p }p≥1 is convergent, the sequence {xp }p≥1 is Cauchy. Then
m
n
m
m
there is xα ∈ R such that limp→∞ xp = xα . Also,
kypm − yqm k = k
p−1
X
m
m
m
hm
s+1 zs k ≤ (M + 1)(tp − tq ).
s=q
is a Cauchy sequence in Rn . Hence there is yαm ∈ Rn
Thus the
such that
From property (v) we note that
sequence {ypm }p≥1
yαm = limp→∞ ypm .
m
xm
p ∈ P (xp ) ∩ B(ϕ0 (0), r) ⊆ K,
(3.6)
and
ypm ∈ B(y0 , r) ⊂ Ω.
m
Thus xm
α ∈ K and yα ∈ B(y0 , r) ⊂ Ω.
m
m
m
Now we put xm (tα ) = xm
α . To show that xm is continuous at tα let {sp : p ≥ 1}
m
m
m
m
m
m
be a sequence in [0, tα [ such that limp→∞ sp = tα and tp ≤ sp ≤ tp+1 for every
p ≥ 1. We have
m
m
m
m
m
kxm (sm
p ) − xm (tα )k ≤ kxm (sp ) − xm (tp )k + kxm (tp ) − xα k
1 m
m 2
m
m
m
m
≤ (sm
p − tp )kyp k + (sp − tp ) (M + 1) + kxp − xα k.
2
By taking the limit as p → ∞, we obtain
m
lim kxm (sm
p ) − xm (tα )k = 0
p→∞
m
which prove that xm is continuous at tm
α . Hence xm is continuous on [−σ, tα ].
Consequently,
m
lim τ (tm
p )xm = τ (tα )xm .
p→∞
Note that from (vii),
is closed, we obtain
τ (tm
p )xm
∈ K0 ∩ Bσ (ϕ0 , r). Since the subset K0 ∩ Bσ (ϕ0 , r),
τ (tm
α )xm ∈ K0 ∩ Bσ (ϕ0 , r) .
Furthermore, by (ii) and the relation (3.1), the sequences {zp }p≥1 and {up }p≥1
are bounded in Rn . So, there are two convergent subsequences, denoted again
EJDE-2008/144
MONOTONE SOLUTIONS
11
n
by, {zp }p≥1 , {up }p≥1 . Thus there are two elements zαm , um
such that
α of R
m
m
m
m
limp→∞ zp = zα , limp→∞ up = uα .
Now since F is upper semicontinuous on K0 × Ω with compact values and since
m
m
m
m
m
um
p ∈ F (τ (tp )xm , yp ), for all p ≥ 1, it follows that uα ∈ F (τ (tα )xm , yα ). Applying condition (H3),
m
lim d(xm (tm
α ) + hyα +
h→0+
h2 m
u , P (xm (tm
α ))) = 0.
2 α
Hence, there is h ∈]0, T − tm
α [ such that
h2 m
h2
uα , P (xm
.
(3.7)
α )) ≤
2
16m
We prove that h belongs to Hpm for every p sufficient large. Since {tm
p }p is an
m
m
m
increasing sequence to tm
and
since
lim
x
=
x
,
lim
y
=
yαm and
p→∞ p
p→∞ p
α
α
m
m
limp→∞ up = uα . Then we can find a natural number p1 such that for every
m
m
m
p > p1 we have tm
p < tα < tp + h < tα + h,
m
d(xm
α + hyα +
h2
,
(3.8)
24m
h
kypm − yαm k ≤
,
(3.9)
24m
1
m
kum
.
(3.10)
p − uα k ≤
12m
From the lower semicontinuity of P at xm
p , there is a natural number p2 such
m
kxm
p − xα k ≤
m
that P (xm
α ) ⊆ P (xp ) +
h2
16m B(0, 1),
for all p > p2 . This gives that if z ∈ Rn , then
h2
, ∀p > p2 .
16m
Now let p > max(p1 , p2 ). By (3.7)–(3.11), we have
m
d(z, P (xm
p )) ≤ d(z, P (xα )) +
(3.11)
h2 m
u , P (xm
p ))
2 p
h2
h2 m m
m
up , xα + hyαm + um
)
≤ d(xm
p + hyp +
2
2 α
h2 m
h2
m
+ d xm
uα , P (xm
α + hyα +
α) +
2
16m
2
h
h2
h2
m
m
m
m
≤ kxm
kum
+
p − xα k + hkyp − yα k +
p − uα k +
2
8m 16m
h2
h2
h2
h2
h2
≤
+
+
+
+
24m 24m 24m 16m 16m
h2
h2
h2
=
+
=
.
8m 8m
4m
Thus h ∈ Hpm , for all p ≥ max(p1 , p2 ). From the choice of hm
p we have
m
d(xm
p + hyp +
1
m
sup Hpm ≤ hm
p ≤ sup Hp .
2
h
h
m
Hence, hm
p ≥ 2 > 4 for all p ≥ max(p1 , p2 ). This means that limp→∞ hp =
m
m
limp→∞ (tp+1 − tp ) can not equal to zero, which contradicts with the fact that the
sequence {tm
p }p≥1 is convergent. So, the process must be finite.
12
A. G. IBRAHIM, F. A. AL-ADSANI
EJDE-2008/144
Theorem 3.2. In addition to the assumptions of lemma 3.1 we suppose that the
graph of P is closed and the following condition is satisfied.
(H4) for all x ∈ K and all y ∈ P (x) we have P (y) ⊆ P (x).
Then for all (ϕ0 , y0 ) ∈ K0 × Ω there exist T > 0 and an absolutely continuous
function x : [0, T ] → K, with absolutely continuous derivative such that
x00 (t) ∈ F (τ (t)x, x0 (t))
x(t) = ϕ0 (t),
a.e. on [0, T ]
∀t ∈ [−σ, 0]
0
x (0) = y0 .
Moreover, x is monotone with respect to P in the sense that for all t ∈ [0, T ] and
all s ∈ [t, T ] we have x(s) ∈ P (x(t)); i.e., 0 ≤ t ≤ s ≤ T ⇒ x(t) x(s).
Proof. According to the definition of xm , for all m ≥ 1, all p = 0, 1, 2, . . . , νm − 1
m
and all t ∈ [tm
p , tp+1 ] we have
m
x0m (t) = ypm + (t − tm
p )zp ,
m
x00m (t) = zpm ∈ F (τ (tm
p )xm , yp ) +
1
B(0, 1).
m
Then from (ii) and (v) of lemma 3.1 we get
m
kx0m (t)k ≤ kypm k + hm
p+1 kzp k ≤ ky0 k + r + T (M + 1)
r
≤ ky0 k + r + , ∀t ∈ [0, T ]
4
(3.12)
and
1
≤ M + 1, ∀t ∈ [0, T ].
(3.13)
m
Then the sequences (xm ) and (x0m ) are equicontinuous in C([0, T ], Rn ). Applying
Ascoli-Arzela theorem, there is a subsequence of (xm ),denoted again by (xm ), and
an absolutely continuous function x : [0, T ] → Rn with absolutely continuous derivative x0 such that (xm ) converges uniformly to x on [0, T ] and (x0m ) converges
uniformly to x0 on [0, T ] and (x00m ) converges weakly to x00 in L2 ([0, T ], Rn ). Furthermore, since all the functions xm equal ϕ0 on [−σ, 0], we can say that xm converges
uniformly to x on [−σ, T ] where x = ϕ0 on [−σ, 0].
m
Now, for each t ∈ [0, T ] and each m ≥ 1, let δm (t) = tm
p , θm (t) = tp+1 , if
m m
m m
t ∈]tp , tp+1 ] and δm (0) = θm (0) = 0. For t ∈]tp , tp+1 ] we get
kx00m (t)k ≤ M +
1
B(0, 1).
m
1
= F (τ (δm (t))xm , x0m (tm
B(0, 1).
p )) +
m
Thus for all m ≥ 1 and a.e. on [0, T ],
m
x00m (t) = zpm ∈ F (τ (tm
p )xm , yp ) +
x00m (t) ∈ F (τ (δm (t))xm , x0m (δm (t))) +
1
B(0, 1)
m
(3.14)
Also, for all m ≥ 1 and all t ∈ [0, T ],
τ (θm (t))xm ∈ Bσ (ϕ0 , r) ∩ K0
(3.15)
xm (t) ∈ B(ϕ0 (0), r)
(3.16)
xm (θm (t)) ∈ P (xm (δm (t))) ⊆ K
(3.17)
EJDE-2008/144
MONOTONE SOLUTIONS
13
Claim: For each t ∈ [0, T ], limm→∞ τ (θm (t))xm = τ (t)x in C([−σ, 0], Rn ). Let
t ∈ [0, T ]. then
kτ (θm (t))xm − τ (t)xkσ
≤ kτ (θm (t))xm − τ (t)xm kσ + kτ (t)xm − τ (t)xkσ
≤
sup kxm (θm (t) + s) − xm (t + s)k + kτ (t)xm − τ (t)xkσ
−σ≤s≤0
≤
kxm (s2 ) − xm (s1 )k + kτ (t)xm − τ (t)xkσ
sup
1
−σ≤s1 ≤s2 ≤T ,|s2 −s1 |≤ m
≤
kxm (s2 ) − xm (s1 )k
sup
1
−σ≤s1 ≤s2 ≤0, |s2 −s1 |≤ m
+
kxm (s2 ) − xm (s1 )k
sup
1
−σ≤s1 ≤0≤s2 ≤T, |s2 −s1 |≤ m
+
kxm (s2 ) − xm (s1 )k + kτ (t)xm − τ (t)xkσ
sup
1
0≤s1 ≤s2 ≤T, |s2 −s1 |≤ m
≤
kϕ0 (s2 ) − ϕ0 (s1 )k
sup
1
−σ≤s1 ≤s2 ≤0 ,|s2 −s1 |≤ m
+
kxm (0) − xm (s1 )k +
sup
1
−σ≤s1 ≤0, |s1 |≤ m
+
sup
kxm (s2 ) − xm (0)k
1
0≤s2 ≤T, |s2 |≤ m
kxm (s2 ) − xm (s1 )k + kτ (t)xm − τ (t)xkσ
sup
1
0≤s1 ≤s2 ≤T, |s2 −s1 |≤ m
≤2
kϕ0 (s2 ) − ϕ0 (s1 )k
sup
1
−σ≤s1 ≤s2 ≤0, |s2 −s1 |≤ m
+2
kxm (s2 ) − xm (s1 )k + kτ (t)xm − τ (t)xkσ .
sup
1
0≤s1 ≤s2 ≤T, |s2 −s1 |≤ m
Using the continuity of ϕ0 , the fact that (x0m ) is uniformly bounded, the uniform
convergence of (xm ) towards x and the preceding estimate, we get
lim kτ (θm (t))xm − τ (t)xkσ = 0.
m→∞
Similarly, for each t ∈ [0, T ], limm→∞ τ (δm (t))xm = τ (t)x in C([−σ, 0], Rn ). Also,
since limm→∞ δm (t) = t and (x00m ) is uniformly bounded, then
lim x0m (δm (t)) = x0 (t) ∀t ∈ [0, T ].
m→∞
(3.18)
Thus by the upper semicontinuity of F , and by (3.12), we obtain
x00 (t) ∈ CoF (τ (t)x, x0 (t)) ⊆ ∂V (x0 (t))a.e. on [0, T ].
(3.19)
Our aim now is proving the relation
x00 (t) ∈ F (τ (t)x, x0 (t))
a.e. on [0, T ].
(3.20)
Since F is upper semicontinuous with closed values, then by [1, prop. 1.1.2], the
graph of F is closed in [0, T ] × Rn × Rn . So, if we prove that the sequence (x00m ) has
a subsequence converges strongly point wise to x00 then the relation (3.14) assures
that the relation (3.20) is true.
In order to show that (x00m ) has a subsequence converges strongly point wise to
00
x , we note that the condition (H2) and property (ii) of Lemma 3.1 give
m
m
0
m
zpm − wpm ∈ F (τ (tm
p )xm , yp ) ⊆ ∂V (yp ) = ∂V (xm (tp )),
for p = 0, 1, 2, . . . , νm − 2.
(3.21)
14
A. G. IBRAHIM, F. A. AL-ADSANI
EJDE-2008/144
From the definition of the subdifferential of V , for p = 0, 1, 2, . . . , νm − 2, we
have
0
m
m
m
0
m
0
m
V (x0m (tm
p+1 )) − V (xm (tp )) ≥ hzp − wp , xm (tp+1 ) − xm (tp )i
Z tm
p+1
= hzpm − wpm ,
x00m (s) dsi
tm
p
=
=
hzpm , zpm (tm
p+1
−
m 2
hm
p+1 kzp k
hwpm ,
tm
p+1
Z
=
tm
p
−
tm
p )i
kx00m (s)k2 ds
hwpm ,
−
Z
tm
p+1
tm
p
−
Z
tm
p+1
tm
p
x00m (s) dsi
x00m (s)dsi
hwpm ,
Z
tm
p+1
x00m (s)dsi
tm
p
(3.22)
Analogously,
V
(x0m (T ))
−V
(x0m (tm
νm −1 ))
≥
hzνmm −1
Z
−
wνmm −1 ,
Z
T
tm
νm −1
x00m (s) dsi
T
=
tm
νm −1
kx00m (s)k2
ds −
hwνmm −1 ,
T
Z
x00m (s) dsi
tνm −1
(3.23)
By adding the νm − 1 inequalities from (3.22) and the inequality (3.23), we get
V (x0m (T )) − V (x0m (0))
0
m
0
m
= V (x0m (T )) − V (x0m (tm
νm −1 )) + V (xm (tνm −1 )) − V (xm (tνm −2 )) + . . .
0
+ V (x0m (tm
1 )) − V (xm (0))
Z T
Z
νX
m −2
≥
kx00m (s)k2 ds −
hwpm ,
0
p=0
tm
p+1
tm
p
x00m (s)dsi − hwνmm −1 ,
Z
T
tm
ν
x00m (s) dsi
m −1
(3.24)
Now,
νX
m −2
|hwpm ,
p=0
≤
νX
m −2
Z
tm
p+1
tm
p
x00m (s) dsi| + |hwνmm −1 ,
Z
T
tm
νm −1
x00m (s) dsi|
m
m
m
kwpm k(M + 1)(tm
p+1 − tp ) + kwνm −1 k(M + 1)(T − tνm −1 )
p=0
T (M + 1)
m
Hence, by passing to the limit as m → ∞ in (3.24) we obtain
Z T
V (x0m (T )) − V (y0 ) ≥ lim sup
kx00m (s)k2 ds.
≤
m→∞
0
On the other hand from relation (3.19) and [5, Lemma 3.3], we obtain
d
V (x0 (t)) = kx00 (t)k2 ,
dt
a.e. on [0, T ] .
(3.25)
EJDE-2008/144
MONOTONE SOLUTIONS
Thus, V (x0 (T )) − V (x0 (0)) =
15
RT
kx00 (s)k2 ds, which yields directly that
Z T
V (x0 (T )) − V (y0 ) =
kx00 (s)k2 ds
0
(3.26)
0
Therefore, by (3.25) and (3.26), we get
Z
Z T
kx00 (s)k2 ds ≥ lim sup
T
kx00m (s)k2 ds
(3.27)
Since (x00m ) converges weakly to x00 in L2 ([0, T ], Rn ). hence
Z T
Z T
00
2
kx00m (s)k2 ds
kx (s)k ds ≤ lim inf
(3.28)
m→∞
0
0
m→∞
0
0
By (3.27) and (3.28), we obtain
Z T
Z
lim
kx00m (s)k2 ds =
m→∞
0
T
kx00 (s)k2 ds,
0
this means that the sequences (x00m ) converges strongly to x00 in L2 ([0, T ], Rn ). Consequently there is a subsequence of (x00m ), denoted again by (x00m ), converges point
wise to x00 . From the facts that the graph of F is closed, τ (t)xm converges uniformly to τ (t)x, (x0m ) converges uniformly to x0 and (x00m ) converges point wise to
x00 the relation(18) is proved.
It remains to prove the following two properties:
(1) (x(t), x0 (t)) ∈ K × Ω, for all t ∈ [0, T ].
(2) x(s) ∈ P (x(t)) for all t, s ∈ [0, T ] and t ≤ s.
To prove the first property we note that the property (iii) of l Lemma 3.1 implies that xm (δm (t)) ∈ B(ϕ0 (0), r) ∩ K and x0m (δm (t)) ∈ B(y0 , r) ∩ Ω. Since
limm→∞ xm (δm (t)) = x(t) and limm→∞ x0m (δm (t)) = x0 (t) then x(t) ∈ B(ϕ0 (0), r)∩
K and x0 (t) ∈ B(y0 , r) ∩ Ω.
To prove the second property, let t, s ∈ [0, T ] be such that t ≤ s. Then for m
m
large enough, we can find p, q ∈ {0, 1, 2, . . . , νm − 2} such that p > q, t ∈ [tm
q , tq+1 ]
m m
and s ∈ [tp , tp+1 ]. Assume that j = p − q. Using property (v) of Lemma 3.1 and
condition (H4) we get
m
m
m
P (xm (tm
p )) ⊆ P (xm (tp−1 )) ⊆ P (xm (tp−2 )) ⊆ · · · ⊆ P (xm (tq )).
This implies P (xm (δm (s))) ⊆ P (xm (δm (t))). Since xm (δm (s)) ∈ P (xm (δm (s))), it
follows that P (xm (δm (t))) and hence the second property is proved.
AS an example, let K = R and P (x) = [x, ∞). Then x y if and only if
y ∈ P (x); i.e., if and only if x ≤ y. Then the solution obtained above is monotone
in the usual sense.
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A. G. IBRAHIM, F. A. AL-ADSANI
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Ahmed G. Ibrahim
Department of Mathematics, Faculty of Science, Cairo University, Egypt
E-mail address: agamal2000@yahoo.com
Feryal A. Al-Adsani
Department of Mathematics, Faculty of Educations for Women, King Faisal University,
Sudi Arabia
E-mail address: faladsani@hotmail.com