c
Applied Mathematics E-Notes, 7(2007), 9-15 Available free at mirror sites of http://www.math.nthu.edu.tw/∼amen/
ISSN 1607-2510
On A Second-Order Differential Inclusion With
Constraints∗
Aurelian Cernea†
Received 27 January 2006
Abstract
We prove the existence of viable solutions to the Cauchy problem x00 ∈
F (x, x0 ) + f (t, x, x0 ), x(0) = x0 , x0 (0) = y0 , x(t) ∈ K, where K ⊂ Rn is a closed
set, F is a set-valued map contained in the Fréchet subdifferential of a φ- convex
function of order two and f is a Carathéodory map.
1
Introduction
In this note we consider the second order differential inclusions of the form
x00 ∈ F (x, x0) + f(t, x, x0),
x(0) = x0, x0(0) = y0 ,
(1)
where F (., .) : D ⊂ Rn × Rn → P(Rn ) is a given set-valued map, f(., ., .) : D1 ⊂
R × Rn × Rn → P(Rn ) is a given function and x0, y0 ∈ Rn .
Existence of solutions of problem (1.1) that satisfy a constraint of the form x(t) ∈ K,
∀t, well known as viable solutions, has been studied by many authors, mainly in the
case when the multifunction is convex valued and f ≡ 0 ([2], [6], [8], [10] etc.).
Recently in [1], the situation when the multifunction is not convex valued is considered. More exactly, in [1] it is proved the existence of viable solutions of the problem
(1) when F (., .) is an upper semicontinuous, compact valued multifunction contained
in the subdifferential of a proper convex function. The result in [1] extends the result
in [9] obtained for problems without constraints (i.e., K = Rn ).
The aim of this note is to prove existence of viable solutions of the problem (1) in
the case when the set-valued map F (., .) is upper semicontinuous compact valued and
contained in the Fréchet subdifferential of a φ- convex function of order two.
On one hand, since the class of proper convex functions is strictly contained into
the class of φ- convex functions of order two, our result generalizes the result in [1]. On
the other hand, our result may be considered as an extension of our previous viability
result for second-order nonconvex differential inclusions in [5] obtained for a problem
without perturbations (i.e., f ≡ 0).
∗ Mathematics
Subject Classifications: 34A60
† Faculty of Mathematics and Informatics, University of Bucharest, Academiei 14,
Romania
9
010014 Bucharest,
10
Differential Inclusion with Constraints
The proof of our result follows the general ideas in [1] and [5]. We note that in
the proof we pointed out only the differences that appeared with respect to the other
approaches.
The paper is organized as follows: in Section 2 we recall some preliminary facts
that we need in the sequel and in Section 3 we prove our main result.
2
Preliminaries
We denote by P(Rn ) the set of all subsets of Rn and by R+ the set of all positive
real numbers. For > 0 we put B(x, ) = {y ∈ Rn; ||y − x|| < } and B(x, ) =
{y ∈ Rn ; ||y − x|| ≤ }. With B we denote the unit ball in Rn. By cl(A) we denote
the closure of the set A ⊂ Rn, by co(A) we denote the convex hull of A and we put
||A|| = sup{||a||; a ∈ A}.
Let Ω ⊂ Rn be an open set and let V : Ω → R ∪ {+∞} be a function with domain
D(V ) = {x ∈ Rn; V (x) < +∞}.
DEFINITION 2.1. The multifunction ∂F V : Ω → P(Rn), defined as:
∂F V (x) = {α ∈ Rn, lim inf
y→x
V (y) − V (x) − hα, y − xi
≥ 0} if V (x) < +∞
||y − x||
and ∂F V (x) = ∅ if V (x) = +∞ is called the Fréchet subdifferential of V .
According to [4] the values of ∂F V are closed and convex.
DEFINITION 2.2. Let V : Ω → R ∪ {+∞} be a lower semicontinuous function. We
say that V is a φ-convex of order 2 if there exists a continuous map φV : (D(V ))2 ×R2 →
R+ such that for every x, y ∈ D(∂F V ) and every α ∈ ∂F V (x) we have
V (y) ≥ V (x) + hα, x − yi − φV (x, y, V (x), V (y))(1 + ||α||2)||x − y||2.
(2)
In [4], [7] there are several examples and properties of such maps. For example,
according to [4], if M ⊂ R2 is a closed and bounded domain, whose boundary is a C 2
regular Jordan curve, the indicator function of M
0,
if x ∈ M
V (x) = IM (x) =
+∞, otherwise
is φ- convex of order 2.
In what follows we assume the next assumptions.
HYPOTHESIS 2.3. i) Ω = K × 0, where K ⊂ Rn is a closed set and O ⊂ Rn is a
nonempty open set.
ii) F (., .) : Ω → P(Rn ) is upper semicontinuous (i.e., ∀z ∈ Ω, ∀ > 0 there exists
δ > 0 such that ||z − z 0 || < δ implies F (z 0 ) ⊂ F (z) + B) with compact values.
iii) f(., ., .) : R ×Ω → Rn is a Carathèodory function, i.e., ∀(x, y) ∈ Ω, t → f(t, x, y)
is measurable, for all t ∈ R f(t, .) is continuous and there exists m(.) ∈ L2 (R, R+) such
that ||f(t, x, y)|| ≤ m(t) ∀(t, x, y) ∈ R × Ω.
A. Cernea
11
iv) For all (t, x, v) ∈ R × Ω, there exists w ∈ F (x, v) such that
lim inf
h→0
Z
1
h2
d(x
+
hv
+
w+
h2
2
t+h
f(s, x, v)ds, K) = 0.
t
v) There exists a proper lower semicontinuous φ- convex function of order two
V : Rn → R ∪ {+∞} such that
F (x, y) ⊂ ∂F V (y),
3
∀(x, y) ∈ Ω.
The Mmain Result
In order to prove our result we need the following lemmas.
LEMMA 3.1 ([1]). Assume that Hypotheses 2.3 i)-iv) are satisfied. Consider
(x0, y0 ) ∈ Ω, r > 0 such that B(x0 , r) ⊂ O, M := sup{||F (t, x)||; (t, x) ∈ Ω0 :=
RT
[K × B(y0 , r)] ∩ B((x0, y0 ), r)}, T1 > 0 such that 0 1 (m(s) + M + 1) < r3 , T2 =
min{ 3(Mr+1) , 3(||y2r
} and T ∈ (0, min{T1 , T2}). Then for every > 0 there exists
0 ||+r)
η ∈ (0, ) and p ≥ 1 such that for all i = 1, ..., p − 1 there exists (hi, (xi , yi), wi) ∈
[η, ] × Ω0 × Rn with the following properties
xi = xi−1 + hi−1yi−1 +
h2i−1
wi−1 +
2
Z
yi = yi−1 + hi−1wi−1,
and
hi−2 +hi−1
f(s, xi−1, yi−1 )ds ∈ K,
hi−2
wi ∈ F (xi, yi ) +
p−1
(xi , yi ) ∈ Ω0,
X
hi < T ≤
i=0
p
X
B,
T
hi .
i=0
Pp−1 h2
Pp−1
Moreover, for > 0 sufficiently small we have i=0 2i ≤ i=0 hi < T .
For k ≥ 1 and q = 1, ..., p denote by hkq the real number associated to = k1 and
(t, x, y) = (hkq−1 , xq , yq ) given by Lemma 3.1. Define t0k = 0, tpk = T , tqp = hk0 + ... + hkq−1
q
n
and consider the sequence xk (.) : [tq−1
k , tk ] → R , k ≥ 1 defined by
xk (0) = x0 ,
1
q−1 2
xk (t) = xq−1 + (t − tq−1
k )yq−1 + (t − tk ) wq−1 +
2
Z
t
tq−1
k
(t − s)f(s, xq−1 , yq−1)ds.
LEMMA 3.2 ([1]). Assume that Hypotheses 2.3 i)-iv) are satisfied and consider
xk (.) the sequence constructed above. Then there exists a subsequence, still denoted
by xk (.) and an absolutely continuous function x(.) : [0, T ] → Rn such that
i) xk (.) converges uniformly to x(.),
ii) x0k (.) converges uniformly to x0(.),
iii) x00k (.) converges weakly in L2 ([0, T ], Rn) to x00(.),
12
Differential Inclusion with Constraints
iv) The sequence
RT
0
P
R tqk
p
q=1 tq−1
k
0 q−1
< x00k (s), f(s, xk (tq−1
),
x
(t
))
>
ds
converges to
k k
k
k
< x00(s), f(s, x(s), x0 (s)) > ds,
v) For every t ∈ (0, T ) there exists q ∈ {1, ..., p} such that
0 q−1
lim d((xk (t), x0k (t), x00k (t) − f(t, xk (tq−1
k ), xk (tk ))), graph(F )) = 0,
k→∞
vi) x(.) is a solution of the convexified problem
x00 ∈ coF (x, x0) + f(t, x, x0),
x(0) = x0, x0(0) = v0.
We are now able to prove our result.
THEOREM 3.3. Assume that Hypothesis 2.3 is satisfied. Then, for every (x0, y0 ) ∈
Ω there exist T > 0 and x(.) : [0, T ] → Rn a solution of problem (1) that satisfies
x(t) ∈ K ∀t ∈ [0, T ].
PROOF. Let (x0 , y0) ∈ Ω and consider r > 0, T > 0 as in Lemma 3.1 and xk (.) :
[0, T ] → Rn , x(.) : [0, T ] → Rn as in Lemma 3.2. Let φV the continuous function
appearing in Definition 2.2. Since V (.) is continuous on D(V ) (e.g. [7]), by possibly
decreasing r one can assume that for all y ∈ B(y0 , r) ∩ D(V )
|V (y) − V (y0 )| ≤ 1.
Set
S := sup{φv (y1 , y2 , z1, z2); yi ∈ B(y0 , r), zi ∈ [V (y0 ) − 1, V (y0 ) + 1], i = 1, 2}.
From the statement vi) in Lemma 3.2 and Hypothesis 2.3 v) it follows that for almost
all t ∈ [0, T ],
x00(t) − f(t, x(t), x0(t)) ∈ ∂F V (x0 (t)).
(3)
Since the mapping x(.) is absolutely continuous, from (3) and Theorem 2.2 in [4]
we deduce that there exists T3 > 0 such that the mapping t → V (x0 (t)) is absolutely
continuous on [0, min{T, T3 }] and
(V (x0(t)))0 = hx00(t), x00(t) − f(t, x(t), x0 (t))i
a.e. [0, min{T, T3 }].
(4)
Without loss of generality we may assume that T = min{T, T3 }. From (4) we have
V (x0(T )) − V (y0 ) =
Z
T
||x00(s)||2 ds −
0
Z
T
hx00(s), f(s, x(s), x0 (s)i ds.
0
q
On the other hand, for q = 1, ..., p and t ∈ [tq−1
k , tk )
q−1
0 q−1
0 q−1
x00k (t) − f(t, xk (tq−1
k ), xk (tk )) ∈ F (xk (tk ), xk (tk )) +
1
B
kT
and therefore
0 q−1
0 q−1
x00k (t) − f(t, xk (tq−1
k ), xk (tk )) ∈ ∂F V (xk (tk )) +
1
B.
kT
(5)
A. Cernea
13
We deduce the existence of bqk ∈ B such that
0 q−1
x00k (t) − f(t, xk (tq−1
k ), xk (tk )) −
bqk
∈ ∂F V (x0k (tq−1
k )).
kT
Taking into account Definition 2.2 we obtain
*
V
(x0k (tqk ))
−V
(x0k (tq−1
k ))
≥
x00k (t)
−
0 q−1
f(t, xk (tq−1
k ), xk (tk ))
bq
− k,
kT
Z
tqk
tq−1
k
+
x00k (s)ds
0 q
0 q−1
−φV x0k (tqk ), x0k (tq−1
),
V
(x
(t
)),
V
(x
(t
))
k k
k k
k
!
q 00
2
b
q−1
q−1
0
k
× 1+
xk (t) − f(t, xk (tk ), xk (tk )) − kT 2
× x0k (tqk ) − x0k (tq−1
k ) .
q
Using the fact that x00k (.) is constant on [tq−1
k , tk ] one may write
Z tqk
Z tqk bqk
00
00
00
V (x0k (tqk )) − V (x0k (tq−1
x
))
≥
hx
(s),
x
(s)i
ds
−
(s),
ds
k
k
k
k
kT
tq−1
tq−1
k
k
Z tqk D
E
0 q−1
−
x00k (s), f(s, xk (tq−1
k ), xk (tk )) ds
tq−1
k
0 q
0 q−1
−φV x0k (tqk ), x0k (tq−1
),
V
(x
(t
)),
V
(x
(t
))
k k
k k
k
!
q 00
2
b
q−1
q−1
0
k
× 1+
xk (t) − f(t, xk (tk ), xk (tk )) − kT 2
× x0k (tqk ) − x0k (tq−1
k ) .
By adding on q the last inequalities we get
Z T
V (x0k (T )) − V (y0 ) ≥
||x00k (s)||2 ds + a(k) + b(k)
0
−
p Z
X
q=1
where
b(k) =
tqk
tq−1
k
D
E
0 q−1
x00k (s), f(s, xk (tq−1
),
x
(t
))
ds,
k k
k
(6)
Z tq
p
X
k
1
a(k) = −
hx00k (s), bqk i ds,
q−1
kT
tk
q=1
−
p
X
q=1
0 q
0 q−1
φV x0k (tqk ), x0k (tq−1
k ), V (xk (tk ))), V (xk (tk )
!
q 2
00
2 b
0 q
q−1
q−1
0
0 q−1 k
× 1+
x
(t)
−
f(t,
x
(t
),
x
(t
))
−
(t
)
−
x
(t
)
x
.
k k
k k
k k
k k
k
kT 14
Differential Inclusion with Constraints
On the other hand, one has
Z tk
p
1 X q
||bk||.
||x00k (s)||ds
kT q=1
tq−1
k
Z T
Z T
1
1
1
00
||xk (s)||ds ≤
[M + + m(s)]ds
kT 0
kT 0
T
q
|a(k)| ≤
≤
and
|b(k)| ≤
p
X
S(1 + M 2)||
Z
tq−1
k
q=1
≤
≤
tqk
x00k (s)ds||2
Z p
Z
p
X
1 tk
1 T 00
||x00k (s)||2 ds ≤ S(1 + M 2)
||xk (s)||2ds
p−1
k
k
t
0
k
q=1
Z T
1
1
[M + + m(s)]2 ds.
S(1 + M 2)
k
T
0
S(1 + M 2)
We infer that
lim a(k) = lim b(k) = 0.
k→∞
k→∞
Hence using also statement iv) in Lemma 3.2 and the continuity of the function V (.)
by passing to the limit as k → ∞ in (6) we obtain
V (x0 (T )) − V (y0 ) ≥ lim sup
Z
k→∞
T
||x00k (s)||2ds −
Z
0
T
00
x (s), f(s, x( s), x0(s)) ds.
(7)
0
Using (4) we infer that
lim sup
Z
k→∞
T
||x0k(t)||2dt
≤
Z
0
T
||x00(t)||2dt
0
and, since x00k (.) converges weakly in L2 ([0, T ], Rn) to x00(.), by the lower semicontinuity
of the norm in L2 ([0, T ], Rn) (e.g. Prop. III.30 in [3]) we obtain that
lim
k→∞
Z
T
||x00k (t)||2dt =
0
Z
T
||x00(t)||2dt
0
i.e., x00k (.) converges strongly in L2 ([0, T ], Rn). Hence, there exists a subsequence (still
denoted) x00k (.) that converges pointwise to x00(.). From the statement v) in Lemma 3.2
it follows that
d((x(t), x0(t), x00(t) − f(t, x(t), x0(t))), graph(F )) = 0
a.e. [0, T ].
and since by Hypothesis 2.4 graph(F ) is closed we obtain
x00(t) ∈ F (x(t), x0(t)) + f(t, x(t), x0 (t)) a.e. [0, T ].
A. Cernea
15
In order to prove the viability constraint satisfied by x(.) fix t ∈ [0, T ]. There
exists a sequence (tqk )k such that t = limk→∞ tqk . But limk→∞ ||x(t) − xk (tqk )|| = 0 and
xk (tqk ) ∈ K. So the fact that K is closed gives x(t) ∈ K and the proof is complete.
REMARK 3.4. If V (.) : Rn → R is a proper lower semicontinuous convex function
then (e.g. [7]) ∂F V (x) = ∂V (x), where ∂V (.) is the subdifferential in the sense of
convex analysis of V (.), and Theorem 3.3 yields the result in [1]. At the same time if
in Theorem 3.3 f ≡ 0 then Theorem 3.3 yields the result in [5].
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