Page 226 - Mathematical Techniques of Fractional Order Systems

Controllability of Single-valued Chapter | 7 215

0 0
Set C 5 fzAC b ; z 0 5 0AC h g. For any zAC ,
b b
1 1

2 2 2 2
1 sup EjzðsÞj 5 sup EjzðsÞj
OzO b 5 Oz 0 O C h
sA½0;bŠ sA½0;bŠ
0
thus ðC ; OUO b Þ is a Banach space. For each positive number q, set
b
0
2
B q 5 fyAC ; OyO # qg, then for each q, B q is clearly a bounded closed con-
b b
0
vex set in C . For zAB q , from Lemma 7.4, one can get
b
Oz t 1 y t O 2 # 2ðOz t O 1 Oy t O Þ
2
2
C h C h C h

2
2
2
2
2
# 4 l sup EjzðsÞj 1 Oz 0 O 1 l sup EjyðsÞj 1 Oy 0 O 2
sA½0;tŠ C h sA½0;tŠ C h
2
2
# 4l q 1 4OφO 5 q :

C h
0
Let the operator Π on C be defined by
b
8
0; tAð2N;0Š
>
>
1
> ð t ð t
>
φð0Þ1 BðsÞuðsÞds1 fðs;z s 1y s Þds
> α21
>
> ðt2sÞ
>
> ΓðαÞ 0 0
>
>
>
> "
> ð t ð s ð t
>
> 1
> α21 0
> 1 ðt2sÞ σðτ;z τ 1y τ ÞdWðτÞ ds1 S ðt2sÞ φð0Þ
<
ðΠzÞðtÞ5 ΓðαÞ 0 2N 0
> ð s ð s
> 1
> α21
> 1 BðτÞuðτÞdτ1 fðτ;z τ 1y τ Þdτ
> ðs2τÞ
>
ΓðαÞ
>
> 0 0
>
>
> #
>
> ð s ð τ
> 1
> α21
> 1 ðs2τÞ σðη;z η 1y η ÞdWðηÞ dτ ds; tAJ:
>
>
: ΓðαÞ 0 2N
It is obvious that the operator Φ has a fixed point if and only if Π
has a fixed point. In order to prove that Π is continuous, decompose
Π as Π 5 Π 1 1 Π 2 , where the operators Π 1 and Π 2 are defined respectively
by
t ð t
ð
1
ðΠ 1 zÞðtÞ 5 φð0Þ 1 ðt2sÞ α21 BðsÞuðsÞds 1 fðs; z s 1 y s Þds
ΓðαÞ 0 0
1 ð t α21 ð s
1 ðt2sÞ σðτ; z τ 1 y τ ÞdWðτÞ ds
ΓðαÞ 0 2N
Ð t
ðΠ 2 zÞðtÞ 5 S ðt 2 sÞðΠ 1 zÞðsÞds:
0
0
One needs to prove that Π 1 and Π 2 are completely continuous. It follows
from the hypothesis ðH 14 Þ that S ðtÞ is compact for all t . 0 (see Herna ´ndez
0
et al., 2010).