Chiziqli boshqaruv sistemasining erishish toβplami va uning tayanch funksiyasi.
(1) chiziqli sistemaning π’(β) β π(π) joyiz boshqaruvlarga mos keluvchi va π₯(π‘0) = π₯0
boshlangβich shartni qanoatlantiruvchi barcha trayektoriyalarining ong chetlari boβlgan
π₯(π‘1, π₯0, π’(β)) nuqtalardan tuzilgan
π(π‘1, π₯0) = {π β π
π: π = π₯(π‘1, π₯0, π’(β)), π’(β) β π(π)} (5) toβplamni qaraymiz.
taβrif. π(π‘1, π₯0) toβplamga qaralayotgan (1) boshqaruv sistemasining π₯0 boshlangβich holatdan π‘1 vaqt momentida erishiladigan terminal holatlar toβplami, yoki qisqacha erishish toβplami deb aytiladi.
Taβrifdan ravshanki, π(π‘1, π₯0) = {π β π
π: π = π₯(π‘1), π₯(β) β π»(π₯0, π) }.
Chiziqli differensial tenglamalar nazariyasidan maβlumki, (1) tenglamaning π₯(π‘0) = π₯0
shartni qanoatlantiruvchi π₯ = π₯(π‘, π₯0, π’(β)),π‘ β π = [π‘0, π‘1] absolyut usluksiz yechimini π₯Μ =
β«
β«
π΄(π‘)π₯ bir jinsli sistemaning fundamental yechimlar matritsasi yordamida quyidagi Koshi formulasi
π₯(π‘, π₯0, π’(β)) = πΉ(π‘, π‘0
)π₯0 +
π‘ πΉ(π‘, π)π΅(π)π’(π)ππ +
π‘0
π‘ πΉ(π‘, π)π(π)ππ
π‘0
(6)
bilan ifodalash mumkin.
Har bir π’ (β ) β π(π) boshqaruv uchun π (π‘ ) = πΉ (π‘ 1, π‘ )π΅ (π‘ )π’(π‘) funksiyaning π = [π‘ 0, π‘ 1] oraliqda Lebeg integrali mavjud, yaβni
{π β π
π: π =
π‘1 πΉ(π‘
, π‘)π΅(π‘)π’(π‘)ππ‘ , π’(β) β π(π) } (7)
β«π‘0 1
toβplam boβsh emas. (7) toβplamga πΊ(π‘) = πΉ(π‘1, π‘)π΅(π‘)π, π‘ β π = [π‘0, π‘1] koβrinishdagi koβp qiymatli akslantirishning integrali [10] deb ataladi va
π‘1 πΉ(π‘
, π‘)π΅(π‘)πππ‘
(8)
β« π‘0 1
kabi belgilanadi. Koβp qiymatli akslantirishlar nazariyasidan maβlumki, πΊ (π‘ ) =
πΉ (π‘ 1, π‘ )π΅ (π‘ )π, π‘ β π akslantirish integraliu π
π fazoning qavariq va kompakt toβplami boβladi. Bizga bundan keyin zarur boβladigan tayanch funksiya tuchunchasini keltiramiz.
π
π fazoning π kompakt toβplami tayanch funksiyasi deb quyidagi
π (π, π ) = max(π€, π) , π β π
π (9)
π€βπ
formula boβyicha aniqlangan funksiyaga aytiladi [5]. Tayanch funksiyalar qavariq tahlilda muhim ahamiyatga ega. Ularning baβzi zarur xossalarini keltiramiz.
(9) dan π (π, π ) funksiyaning π β π
π boβyicha qavariqligi, yaβni β π 1 β π
π, π 2 β
π
π, πΌ 1 β₯ 0, πΌ 2 β₯ 0, πΌ 1 + πΌ 2 = 1 uchun
π (π, πΌ 1π 1 + πΌ 2π 2) β€ πΌ 1π (π, π 1) + πΌ 2π (π, π 2)
tengsizlikning bajarilishi kelib chiqadi. π (π, π ) tayanch funksiya birinchi argument boβyicha ham muhim xossaga ega: ixtiyoriy π 1, π 2 β kompakt toβplamlar, πΌ 1 β₯ 0, πΌ 2 β₯ 0 sonlar va π β π
π uchun
π (πΌ 1π 1 + πΌ 2π 2, π ) = πΌ 1π (π 1, π ) + πΌ 2π (π 2, π ) (10)
tenglik bajariladi. π kompakt toβplam va uning πππ qavariq qobigining tayanch funksiyalari teng:
π (π, π ) = π (πππ, π ). (11)
(8) integralning tayanch funsiyasi uchun
π‘1 π‘1
π ( πΉ(π‘ , π‘)π΅(π‘)πππ‘ , π) = π(πΉ(π‘
, π‘)π΅(π‘)π, π)ππ‘
(12)
β«π‘0 1 β«π‘0 1
tenglik bajariladi.
π(π‘1, π₯0) erishish toβplami qavariq va kompakt toβplam boβlib, u uchun quyidagi
π(π‘ , π₯0) = πΉ(π‘ , π‘
)π₯0 +
π‘1 πΉ(π‘
, π‘)π΅(π‘)πππ‘ +
π‘1 πΉ(π‘
, π‘)π(π‘)π
(13)
1 1 0
formula oβrinli.
β«π‘0 1
β«π‘0 1
Haqiqatan ham, π(π‘1, π₯0) erishish toβplamining qavariq va kompakt toβplam ekanligi haqidagi tasdiq (13) formulani hisobga olgan holda, (8) integralninq qavariq va kompaktligidan hamda qavariq va kompakt toβplamlarning algebraik yigβindisi yana shunday xususiyatli toβplam boβlishidan kelib chiqadi. Bu tasdiq toβgβriligi uchun π toβplamning qavariq boβlishi shart enas. Chunki (8) integralning qavariqlik xossasi π toβplam qavariq boβlmagan holda ham saqlanib qolaveradi. Demak, faqat (13) formula toβgβriligini koβrsatish qoldi. Buning uchun esa (6) Koshi formulasidan foydalanamiz. Shunga koβra
π₯(π‘ , π₯0, π’(β)) = πΉ(π‘ , π‘
)π₯0 + β«π‘1 πΉ(π‘
, π‘)π΅(π‘)π’(π‘)ππ‘ +
π‘1 πΉ(π‘
, π‘)π(π‘)π
(14)
1 1 0
π‘0 1
β«π‘0 1
Endi π (π‘ 1, π₯ 0) erishish toβplamining (5) taβrifini va πΊ (π‘ ) = πΉ (π‘ 1, π‘ )π΅ (π‘ )π, π‘ β π = [π‘ 0, π‘ 1] koβp qiymatli akslantirishning integrali tushunchasidan foydalanib [10], (14) tenglikdan
formulaga ega boβlamiz.
π (π‘ 1, π₯ 0) erishish toβplamining tayanch funksiyasi uchun
π(π(π‘ , π₯0), π)= (πΉ(π‘ , π‘
)π₯0, π) +
π‘1 π(πΉ(π‘
, π‘)π΅(π‘)π, π)ππ‘ +
1 1 0
+ π‘1(πΉ(π‘ , π‘)π(π‘), π)ππ‘
β«π‘0 1
(15)
β« π‘0
1
formula oβrinli.
Haqiqatan ham, (13) formuladan va tayanch funksiyaning (10) xossasidan foydalanib,
π(π(π‘ , π₯0), π)= (πΉ(π‘
, π‘ )π₯0, π) + π (
π‘1 πΉ(π‘
, π‘)π΅(π‘)πππ‘ , π) +
1 1
+ π‘1(πΉ(π‘
0
, π‘)π(π‘), π)ππ‘
β«π‘0 1
β« π‘0 1
tenglikni hosil qilamiz. Endi (8) integralning tayanch funksiyasu uchun (12) tenglikdan foydalansak, (15) formulaga ega boβlamiz.
TADQIQOT NATIJALARI
Terminal funksionalning xossalari. (1) chiziqli boshqaruv sistemasining trayektoriyalarida aniqlangan (2) koβrinishdagi, yaβni
π (π₯(π‘ 1, π₯ 0, π’ (β ))) = βπ max(π§ π, π₯(π‘ 1, π₯ 0, π’ (β ))) , π’(β) β π(π)
π=1 π§πβππ
terminal funksionalning baβzi xossalarini keltiramiz. Bu funksionalni π π, π = 1,2, β¦ , π
toβplamlar tayanch funksiyalari orqali
π=1
π (π₯(π‘1, π₯0, π’(β))) = βπ
π(ππ, π₯(π‘1, π₯0, π’(β))) , π’(β) β π(π)
π=1
koβrinishda, yoki π = βπ π π deb olganda esa,
π (π₯(π‘ 1, π₯ 0, π’ (β ))) = π(π, π₯(π‘ 1, π₯ 0, π’ (β ))) , π’(β) β π(π)
kabi yozish ham mumkin.Tayanch funksiyalarning (11) xossasiga koβra
π (π, π₯(π‘ 1, π₯ 0, π’ (β ))) = π (πππ, π₯(π‘ 1, π₯ 0, π’ (β ))).
Shunday qilib,
π (π₯(π‘ 1, π₯ 0, π’ (β ))) = π(πππ, π₯(π‘ 1, π₯ 0, π’ (β ))) , π’(β) β π(π) . (16)
formuladan foydalanib, ixtiyoriy zβ π uchun
(π§, π₯(π‘ , π₯0, π’(β))) = (πΉ(π‘ , π‘
)π₯0, π§) +
π‘1(πΉ(π‘
, π‘)π΅(π‘)π’(π‘), π§)ππ‘+
+ β« (πΉ(π‘1, π‘)π(π‘), π§)ππ‘
π‘0
tenglikni yozamiz. Bundan va (16) dan quyidagi tasdiqga ega boβlamiz.
lemma. (2) funksional quyidagi
π (π₯(π‘1, π₯0, π’(β))) = max (π§, π₯(π‘1, π₯0, π’(β))) = max [(πΉ(π‘1, π‘0)π₯0, π§) +
π§βπππ π§βπππ
+ π‘1(πΉ(π‘ , π‘)π΅(π‘)π’(π‘), π§)ππ‘ + π‘1(πΉ(π‘ , π‘)π(π‘), π§)ππ‘] (17)
β«π‘0 1 β«π‘0 1
Formula bilan ifodalanadi.
lemma. (1) sistema trayektoriyalarida aniqlangan
π½(π’(β)) = π (π₯(π‘1, π₯0, π’(β))) = βπ max(π§π, π₯(π‘1, π₯0, π’(β))), π’(β) β π(π)
π=1 π§πβππ
funksional π(π) toβplamda qavariq boβladi va ixtiyoriy π’1(β) β π(π) , π’2(β) β π(π)
uchun quyidagi tengsizlik
|π½(π’ (β)) β π½(π’ (β))| β€ π β«π‘1βπ’ (π‘) β π’
(π‘)βππ‘, (18)
1 2 π‘0 1 2
bajariladi. Bu yerda
π = max βπΉ (π‘ 1, π‘ )π΅(π‘) β βπ
βπ β, βπ β = max βπ§ β .
π‘βπ
π=1
zβZ
Isboti. Quyidagi funsionalni qaraymiz:
π(π’(β), π§) = (πΉ(π‘ , π‘ )π₯0, π§) + π‘1(πΉ(π‘
, π‘)π΅(π‘)π’(π‘), π§)ππ‘+
1 0 β«π‘0 1
+ π‘1(πΉ(π‘
, π‘)π(π‘), π§)ππ‘.
(19)
β« π‘0 1
(17) formulani hisobga olib va (19) funksionaldan foydalanib, π½(π’ (β )) =
π (π₯(π‘ 1, π₯ 0, π’ (β ))) funksionalni
π½(π’ (β )) = max π (π’ (β ), π§ ) (20)
π§βπππ
kabi yozish mumkin. π (π’ (β ), π§ ) funksional har bir argument boβyicha chiziqlidir. Shuni hisobga olib, ixtiyoriy π’ 1(β) β π(π), π’ 2(β) β π(π), πΌ 1 β₯ 0, πΌ 2 β₯ 0, πΌ 1 + πΌ 2 = 1, uchun quyidagi
π½ (πΌ 1π’ 1(β ) + πΌ 2π’ 2(β) ) = max π (πΌ 1π’ 1(β ) + πΌ 2π’ 2(β), π§ ) =
π§βπππ
= max [ πΌ 1π (π’ 1(β ), π§ ) + πΌ 2π (π’ 2(β), π§ )] β€
π§βπππ
β€ πΌ 1max π (π’ 1(β ), π§ )+ πΌ 2max π (π’ 2(β), π§ )=πΌ 1 π½ (π’ 1(β )) + πΌ 2π½(π’ 2(β) )
π§βπππ π§βπππ
tengsizlikni olamiz, yaβni π½(π’ (β )) funksional π(π) da qavariqdir.
Endi (18) tengsizlikning toβgβriligini koβrsatamiz. Ixtiyoriy π’ 1(β) β π(π), π’ 2(β) β π(π)
olib, (19) funksional uchun quyidagi tengsizlikni yozamiz:
|π (π’ 1(β ), π§ ) β π (π’ 2(β), π§ )|=
=|β«π‘1(πΉ(π‘
, π‘)π΅(π‘)
(π‘), π§)ππ‘ β
π‘1(πΉ(π‘ , π‘)π΅(π‘)π’
(π‘), π§)ππ‘| β€
π‘0 1 1
β«π‘0 1
2
π‘1
β€ maxβπΉ(π‘1, π‘)π΅(π‘)β max βπ§β β« βπ’1(π‘) β π’2(π‘)β dt
π‘βπ
zβcoZ
π‘0
Bu yerdan (20) tenglikni hisobga olib,
|π½(π’1(β)) β π½(π’2(β))| = | max π(π’1(β), π§) β max π(π’2(β), π§)| β€
π§βπππ π§βπππ
β€ max |π(π’ (β), π§) βπ(π’ (β), π§)| β€ π π‘1βπ’ (π‘) β π’ (π‘)β dt,
π§βπππ 1 2
β«π‘0 1 2
yaβni (18) tengsizlikni hosil qilamiz.
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