TURLI CHEGARALANISHLI HOL UCHUN П STRATEGIYANING QURILISHI HAQIDA Текст научной статьи по специальности «Математика»

Oriental Renaissance: Innovative, (E)ISSN: 2181-1784

educational, natural and social sciences 4(9), Oct., 2024

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TURLI CHEGARALANISHLI HOL UCHUN П STRATEGIYANING

QURILISHI HAQIDA

Abduraximova Zulayxo Ikromjon qizi

Turan International University o'qituvchisi [email protected]

ANNOTATSIYA

Ushbu maqolada П strategiyaning geometrik chegaralanish, integro -geometrik chegaralanishli holatlar uchun qurilishi yoritilgan. П strategiyaning qurilishi turli chegaralanishli hol uchun quvish - qochish maslasining yechimlariga uzviy bog'liqligi isbotlangan.

Kalit so'zlar: П strategiya, geometrik chegaralanish, integro - geometrik chegaralanish.

О ПОСТРОЕНИИ П - СТАТЕГИИИ ДЛЯ РАЗЛИЧНЫХ ГРАНИЧНЫХ СЛУЧАЕВ

АННОТАЦИЯ

В статье описано построение П стратегии для геометрически ограниченных, интегро-геометрически ограниченных случаев. Доказано, что построение П-стратегии неразрывно связано с решением задачи погони-убегания для различных предельных случаев.

Ключевые слава: П стратегии, геометрическое ограничение, интегро -геометрическое ограничение

ON THE CONSTRUCTION OF П STRATEGY FOR DIFFERENT

BOUNDARY CASES

ABSTRACT

This article describes the construction of strategy P for geometric bounded, integro-geometrically bounded cases. It has been proved that the construction of P strategy is inextricably linked to the solutions of the chase-escape problem for different limiting cases.

Keywords: П strategy, geometric limitation, integro - geometric limitation.

Differensial o'yinlar nazariyasida chegaralanishlar uchun П strategiya muhim ahamiyatga ega. Har bir chegaralanish uchun alohida П strategiyalar mavjud. Quyida ularni har birini ko'rib chiqamiz.

Fazoda P va E obyekt harakatlanyapti

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E

= {

x U x(0) = x0 y = V y(o) = y0

1. Geometrik chegaralanish

(1) (2)

a) \U\ < a ; \V\ < ß ; \U\ = Ju2 + - + U2 ; \V\ = Jv2 +

V2

bu yerda

(3)

(4)

n strategiya : U(v) = v-ÀG(v)^0 , ^ .„ ^ .

lxo ïdI

fr > ß da tutish masalasi yechiladi ; a < ß da qochish masalasi yechiladi. 2. Integro - geometrik chegaralanishli xol :

b) J>i00l2ds < pi0, £\U2(s)\2ds < p20

P: x = U(t) , x(0) = x0 x(t) = x0 + J0tU(s)ds E : y = V(t) , y(0) = y0 ^ y(t) = y0 + J*'V(s)ds

Tutish masalasi : Vt* : x (t* ) = y (t* ) Qochish masalasi : x (ty (t) , t > 0

z

z + Tu = Tv , Tu = Tv - z , u = v - —

J\U\2 ds = p, T\U\2 = A\U\2 =|

(5)

(6)

(7)

(8)

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U 2 2 v

Я P _ v Z

z

- 2Я( vj) + Ä7

Я2 - 2Ä(-P + {vj) + v2 = 0

'2 z

Quyidagicha belgilash kiritib olamiz : j, = Я ^U = v Я |U| = Я z

- 2À( vj) + h2

(9)

P

P

Я.2 =P + ( ^ +

2 z

"2 z

Ы) )

2 VI2

Я va Я2 ham musbat . Savol tuo'iladi qaysi Я ni olamiz ?

¿2 >0agar , tt-t bo'lsa yechim mavjud bo'lishi uchun ildiz osti

(10)

aniqlangan bo'lishi kerak ya'ni

D =

(P, z, v):(P+ (v,j)2 - |v|2 > 0 2 z

Integro - geometrik holda quvlovchini parallel yaqinlashtirish orqali (G)x, û)2 ) oraliqda tutish mumkin bo'lar ekan . Biz bitta quvlovchi bitta qochuvchi bo'lgan holda ko'ramiz ya'ni Я+ ni olamiz . Shuning uchun (10) ga yechim sifatida Я ni tanlaymiz . Natijada hal qiluvchi funksiya deb,

Я(р, z, v) =

P+M+J (2P+< vj )2 -

v

z\ V 2\z\

funksiyani olamiz . n o'lchovli fazoda 2n +1 ta yechimi bor. (11)- funksiyani

P

(11)

aniqlanish sohasini tahlil qilamiz

D =

(P, z, v):(p+< vj )2 -

v

2

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(P+ (v,)) + |v|)(-f+ (v))- |v|) > 0

P

2 z

2 z

demak

P

2 z

+ (v,)-|v| > 0

(12)

bo'lishi kerak . (12) tengsizlikning geometrik mohiyati ( v ga nisbatan ): 2- P , a2 + v) +... + v) -Jv,2 +... + v2„ > 0

a

2 z

figura hosil bo'ladi

Soddalashtirish uchun tekislikda ko'raylik . a + v

(1;0)

+ v)1 + v2^2 -\lv12 + v22 > 0

) = XA, X0 = (0;0), y = (1;

v

));) (-1;0 ) ,) = -1,) = 0

X0 y0|

a2 -v1 -J v2 + v22 > 0 , a2 > v +y[v[+vî > 0 , a = 1 desak

1 +Vvf + v22 < 1, Vvf + v22 < 1 - v1

1 - V2

1 - 2v + vf > vf + v2,1 - 2v > v22, v < 2

2

D

D sohamiz parabolaning ichi ekan . Uch o'lchovl

a

2

1 v

fazoda ko'rsak , paraboloid

bo'lib qolar ekan . Agar a * 1 bo'lsa uchi — da bo'lar ekan tekislikda .

) (-1,0,0) ; a2 - v-yfvf + v2 + v2 > 0, a2 - v

2 ' v3

0, a2 - v ^

v2 + v2 + v32

a4 - 2a2v + v2 > vf + v22 + v32, a4 - 2a2v > v22 + v32,v1 < At - v2 v , v < —■ v2 v3

2a2 2a

2 ' V1

2 2a

2

v

2

v

2

1

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±

ß- ^ = p ^P- 4ß| Z. Demak , xulosa (11) Â(p, z, v) funksiya 2 4 z

p - 4ß|z| bo'lganda V\v\ - ß uchun aniqlangan .

(11) ko'rinishni yuqoridagi U = v |U| = ¿¿p tenglikka olib kelib qo'yamiz

va quyidagi strategiyani hosil qilamiz :

U (p, z, v) = v - X(p, z, v)£

\U (p, z, v)2 = A(p, z, v)

p \z

(13)

(1) va (2) tenglamadan quyidagi tenglamani hosil qilamiz

z = C/-v z(0) = z0

buyerda z = x-y,x = U;z0 = x0-y0,y = v;x-y = U-v (14) ko'rinishdagi U ni o'rniga (13) ni 1-qatorini qo'yamiz natijada :

z = —A(p, z,v)^ z( 0) = z0

t

p(t) = Po -\ \U(s)|2 ds

0

(16) tenglamaga |U| ni o'rniga (13) ni 2-qatorini qo'yamiz

t ( \ P(t) = p0 - {¿(p(s),z(s),v(s)) pp(S) ds tenglikni hosil qilamiz . Endi har ikki

tomondan hosila olaylik

(14)

(15)

(16)

p(t) = -A,(p,z,v) P(0) = Po

(15) va (17) lardan quyidagi natijaga ega bo'lamiz

P

z

(17)

<

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i = -A{p{t\z{t\v{ t))

p = -X{p{t\z{t\v{ t)) z (0) = z 0, p(0) = po

z (t )

|z (t )| p(t ) |z (t )|

(1S)

(1S) dan (p,z) ga nisbatan differensial tenglamalar sistemasi hosil bo'lishi kelib chiqdi.

(1S) sistema nochiziqli differensial sistema . Agar z Ф 0 bo'lsa bu sistemaning o'ng tomonidagi funksiya p > 4ß|z| shartda uzluksiz ( (p,z) ga nisbatan) . Funksiya

t ga nisbatan esa o'lchovli funksiya ( ya'ni Karatedore shartlari bajarilyapti ) deyiladi. Karatedore sharti bo'lishi uchun , Karatedore tenglamasining o'ng tomoni z bo'yicha Lipshist shartini qanoatlantirishi kerak ya'ni

f(zi, t) - f ( z2, t)| < L|z - z2| (19)

(18) tenglamaning o'ng tomonidagi funksiyalar (p, z) bo'yicha z ^ 0 holda

Karatedore shartlari o'rinli . Shuning uchun (18) sistemani yagona yechimi mavjud bo'ladi . (18) tenglamadan quyidagicha almashtirish hosil qilamiz

<

z2 =-Ä{p{t\z{t\v{ t))

zi( t )

|z (t )|

Z2(t)

|z ( t )|

z

^ i z

1

z (t )| 1

zn =-Ä(p(0,z(0,v( t)) endi integrallasak ,

z n (t )

|z ( t )|

( t )l

z 1

-s- = -A(/>(0,*(0,v( t))

z

n

|z ( t )|

ln z.

f 1

* 1 -jÄ(p(t),z(t),v(t))—ds -

-\A(p(t), z(t), v(t))T— ds; zt (t) =z (0)e 0 1 ( * , i = 1, n

о |z(t)|

•л

si

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-\Up(t ), z (t ),v (t))^ids

e o Z = h(t) deb olsak ,

z (t ) = Zo h(t ) (20)

hosil bo'ladi . Xuddi shu kabi ,

fp> (t ) = ^o h(t )

[ Z (t ) = Zo h(t ) (21)

bu yerda h(t) = h(z, p, v, t) . (21) ni (11) ga olib borib qo'yaylik

^(P' Z'V) =

^(P^v,^)2-|v|2 =-Po^ Jv,-^)+ (( <v4)2-|v|2 =^(Po,Zo,v)

2|Z 2|z0|h \ '|z0|h/ V 2|z

Demak , boshlano'ich holatdagi (p0, z0 ) ni bilsak bo'ldi ekan .

Ta'rif : Integro -geometrik differensial o'yinda n strategiya deb , quyidagi funksiyaga aytamiz

U (v) = v -AjG (v)4o (22)

bu yerda (v) = ^+ (v, ft ^ + (^ + (v, ft ^)2 -

v

Mavzu bo'yicha tarixiy ma'lumot : Strategiyalar quvish masalasi qadimdan olimlarni qiziqtirgan . Eramizdan 2000 yil avval Xitoy qo'lyozmalarida "Burgut va O'lja masalasi " o'rganilgan bunda burgut o'z o'ljasini izma -iz quvish strategiyasi orqali harakatlanib ushlash masalasi ko'rilgan . Lekin matematika rivoji u davrda bu masalani yechishga yetarli bo'lmagan . 1732- yil fransus gidrogrifi va matematigi Bagauer " Tulki va Quyon " masalasini yechadi . Bunda quyin ma'lim to'o'ri ch iziq bo'yicha qochib borganda tulki uni izma - iz quvish natijasida qaysi trayektoriya orqali harakat qilishi mumkinligi m a asaasi yechildi . Bu masala amerikalik olim Lokning " Zambaraklarni boshqarish " kitobida bafurcha tahlil qilinib yechimlari keltirilgan .

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