Page 254 - Mathematical Techniques of Fractional Order Systems

Controllability of Fractional Chapter | 8 243

Since y n -y ; it follows from Lemma 8.3 that, for some y AN F;x ; then
q2p q2p q2p q2p
Ψ ðtÞ2E q2p ðAt Þx 0 1At E q2p;q2p11 ðAt Þx 0 2tE q2p;2 ðAt Þx 0
0
ð t ð s
q21 q2p
2 ðt2sÞ E q2p;q ðAðt2sÞ Þ Bu x; ðsÞ1v ðsÞ1 Gðθ;x ðθÞÞdwðθÞ ds AΛðN F;x Þ:
0 0
Clearly, for each tAJ; one can set
q2p q2p q2p q2p
: Ψ n ðtÞ 2 E q2p ðAt Þx 0 1 At E q2p;q2p11 ðAt Þx 0 2 tE q2p;2 ðAt Þx 0
0
ð t ð s
q21 q2p
2 ðt2sÞ E q2p;q ðAðt2sÞ Þ Bu x;n ðsÞ 1 v n ðsÞ 1 Gðθ; x n ðθÞÞdwðθÞ ds
0
0
2 Ψ ðtÞ 2 E q2p ðAt q2p Þx 0 1 At q2p E q2p;q2p11 ðAt q2p Þx 0 2 tE q2p;2 ðAt q2p Þx 0 0
t s
ð ð
q21 q2p 2
2 ðt2sÞ E q2p;q ðAðt2sÞ Þ Bu x; ðsÞ1v ðsÞ1 Gðθ; x ðθÞÞdwðθÞ ds : -0

B
0 0
as n-N: From Lemma 8.2, it can be concluded that Φ is u.s.c. As a conse-
quence of Lemma 8.4, it is clear that Φ has a fixed point which is the solu-
tion of the system (8.1). &
Theorem 8.3: (Nonconvex Case) Assume that conditions (H4) (H5) are sat-
isfied, then the system (8.1) has at least one solution on J; provided that
2q
T 2 2 2
4mðtÞC 2 1 1 n :B: :B : l , 1: ð8:5Þ
4
q 2
Proof: Under the assumption (H5) it is easy to see that for each xAB; the set
S F;x is nonempty. Therefore, F has a nonempty measurable selection (by
Theorem 8.1). First show that Φ defined in Theorem 8.2 satisfies the assump-
tion of Lemma 8.5. The proof will be given in two steps.
Step 1. ΦðxÞAP cl ðBÞ for each xAB:
Indeed, let ðΨ n Þn $ 0AΦðxÞ such that Ψ n -Ψ: Then, ΨAB and there exists
v n AN F;x such that, for each tAJ; Ψ n ðtÞ is defined in (8.4). Using (H5), one
can have
2
jv n ðtÞj # mðtÞ 1 mðtÞ:x: ; nAℕ
for a.e. tAJ:
The Lebesgue dominated convergence theorem implies that
:v n 2 v: 2 -0as n-N:
L