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Uzoqbayev A.H. Assistent.Jizzax Politexnika institute
INTEGRAL BELGI OSTIDA ARALASH MAKSIMUMLI DIFFERENSIAL TENGLAMALAR UCHUN SHARTLI BOSHLANG'ICH MASALA
Annotatsiya: Ushbu maqolada quyidagi chiziqli bo'lmagan maksimumli differensial tenglama uchun boshlang'ich masala ,Chiziqli bo'lmagan filtrlash masalasini sonli yechib cho 'kmaning oshib borishi ham monoton ortib borishi ko'rsatildi., integral belgi ostida aralash maksimumli differensial tenglamalar uchun ulash shartli boshlang'ich masala, o'rganilgan.
Kalit so'zlar: Boshlang'ich shart, vector funksiyasi, cheklangn yopiq to 'plam, differensial tenglama, yevklid normasi, matematik induksiya metodi.
Uzokbayev A.H. Assistant. Jizzakh Polytechnic Institute
AN INITIAL PROBLEM WITH A CONNECTION CONDITION FOR MIXED MAXIMAL DIFFERENTIAL EQUATIONS UNDER THE
INTEGRAL SIGN
Abstract: In this article, the initial problem for the following nonlinear maximum differential equation is 3. By numerically solving the nonlinear filtering problem, it was shown that the increase in precipitation also increases monotonically. initial issue, studied.
Keywords: Initial condition, vector function, finite closed set, differential equation, Euclidean norm, mathematical induction method.
x'(t) = F(t, x(t), max {x(r) | r e[ f | g]}), t e [0; T]
Bizga quyidagicha boshlang'ich shart berilgan
x(0) = p0<(x>
Bu yerda x e X ^ Rn noma'lum vector funksiyasi, X -cheklangn yopiq to'plam
0 < f = f ^t, J K(t, x(s))ds j < T 0 < g = g ^t, J Q(t, s, x(s))ds
t
x(1) (t) = A + J F(s, x(1) (s), max {x(1) (r) | r e [ f(1), g(1) ]})ds, t e T(1), 0
t
x(2) (t) = B + JF(s, x(2) (s), max {x(2) (r) | r e [g(2), f(2) ]})ds, t e T(2),
< T
(1) Boshlang'ich shartdan foydalanib, (3) quydagicha ko'rinishga keltiramiz
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x(1)(t) = I (x(1); t) = % +
t
-J F(s, x(1) (s), max {x(1) (r) | r e [f(1), g(1) ]})ds, t e T(1),
t
+JF (s,x'''(s),max{x(1)(r) | r e I
0
(4) differensial tenglamada B koeffesentni topish uchun qo'shimcha shart kiritib olamiz
x(2) (+1) = ax(1) (-1), a - const.
(3) Va (6) asosida (4) ni quyidagicha yozamiz
*
x(2) (t) = J (x(2); x(1); t) = ax(r> (t) + t
+J F (s, xw(s),max {x(2)(r)| r e|
0
Keling isbotlaylik.
Lemma . Quyidagilar o'rinli bo'lsin
1. 0 < f(t, y) < g(t, y) < T, t e T(1), y = 0
2. F(t, x, z) e C(T(1) x X x X) nBnd(Mx) nLip(L1|x z)
t
-JF(s, x(2)(s),max {x(2)(r) | r e [g(2), f(2) ]})ds, t e T(2),
-(i).
3. f(t, y) e C (T(1) x Rn) nLip(L2y);
4. g(t, y) e C(T(1) x Rn) n Lip(L3|y);
5. K(t, s, x) e C(T(1) x T(1) x X ) n Lip(px (t, s));
6. Q(t, s, x) e C(T(1) x T(1) x X) n Lip(P2|x(t, s)); Bu yerda || yevkelid normasi Rn sohada.
T(1) segmentda (5) differensial tenglamaning yagona yechimi mavjud.
Isbot. T(1) segmentda (5) differensial tenglama uchun integral jarayon
quyidagicha tuziladi.
x0(1)(t) = x0, xk) = I(xk+1(1);t), k e N0
Bu yerda N = u{0}, N -natural sonlar to'plami.
x-,
(1)
(t), k e N ketma-ketlik X ^ Rn sohaga tegishli.
Biz quyidagicha belgilashlarni olamiz
PiU1(t) = \\xt0)(t)-x,.-1(j)(t)|, fkU) = f ft, jK(t,s,xku)(s))ds ,
V 0 J
f t \ qk(j)(t) = qI t, JQ(t,s,xk(j)(s))ds I, k = 1,2,.....
pir(j) (t) = ||max {x(j) (r) | r e fj) | q(j) ]} - max {x') (r) | r e [f-/j) | qjj) ]} Pi[i]( j) (t) = 11max {x(j) (r) | r e fj) | q(j) ]} - max {x^ j) (r) | r e fj) | q(j) ]}|
(8) shartdan foydalanib, (14) ni k = 1 uchun quyidagicha olamiz
t
P2(1)(t) < L J[p1(1)(s) + Pr(1)(s)>, t e T(1) 0
Chunki
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p(1)(t) < max {p(1)(t) \ t e T(1)} < Mt
U holda (15) quyidagicha ko'rinishga ega bo'ladi
Pz(1)(t) < 2MlLl -, t e T(1)
Xuddi shunday (15) ni (14) da к = 2 dan kelib chiqib quyidagicha olamiz
t
P3(1)(t) < L j[pz(1)(s) + Pzr(1)(s)]ds, t e T(1)
g
(17) ning o'ng tomonidagi ifoda uchun ikkinchi shart asosida quyidagicha baho o'rinli
P2r(1)(t) <
P
(1)
2r [2]
(t ) + Pzr [l](1)(t) _
t eT
(l)
(17) ni hisobga olgan holda , (18) ning o'ng tomonini birinchi shart asosida quyidagicha baholaylik
Pzr[2](1)(t) < 2MlLl -, t e T(1)
(9)-(13) va (17) hisobga olgan holda, (18) ning o'ng tomonini ikkinchi shart asosida quyidagicha baholaylik
(1)- ?l(1)
P r[2](1)(t ) < Ml [| fz(1) - fl(1) I +
t
< Ml j [L, IIP (t, s)|| + L31P2 (t, s)||]P2(1) (s)ds
G
< Mßß max {p(1)(t)\ t e T(1) }<
<
12
< 2M2 L-, t e T(1) 1 1 2
(18) ga (19) va (20) almashtirishlarni olib, quyidagini hosil qilaylik
12
P2r(1)(t) < 2MlLl(1 +Mlß)-, t e T(1)
U holda (17) quyidagicha ko'rinishga keladi
13
P3r(1)(t) < 2MlLl(2 + Mlß)-, t e T(1)
Xuddi shunday (17)-(21) к = 3 uchun olamiz .
P4
(1)
I
(t ) < Li j
2 max P (1)(s)
G < s < t < t ! +
+M1ß max ^p3 r (1)(s)
G < s < t < t !
ds <
< 2M1L31(2 + Mß)2 —, t e T(1)
Ushbu jarayonni davom ettirib, V, e uchun to'liq matematik induksiya
metodidan foydalanib, quyidagini olamiz
^+1
pk+l(1)(t ) < IM1L 1(2 + Mlß)
к-1
(к + l)!
t eT(1)
2
2
G
4
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(22) da {xk(1)(t)} ketma-ketlik t da bir xilda yaqinlashishini ko'ramiz. Bundan
kelib chiqadiki, {xk(1)(tx(1)(t) uchun k ^«bo'lsa, x(1)(t) (5) tenglamaning
yechimi, demak, (5) sistemaning yechimi mavjudligini isbotladik. Endi biz bu yechimning yagona ekanligini isbotlaylik. Xuddi shunday (5) differensial tenglama boshlang'ich shartga ko'ra T(1) segmentda boshqa y(t) e X1 yechimga ega bo'lsin. y(t) va ketma-ketlik x(1)(t), k e N0 larni farqini ko'raylik, ular uchun quyidagicha baho o'rinli
tk+1
||y(t) - xk(1)(t)|| < 2MlLkl(2 + M^)k-1 ^^, t e T(1)
(23) da quyidagicha ko'rinishga keladi ||y(t) - xk(1)(t)||
k da teng ravishda t e T(1). T(1) segmentda (5) differensial tenglama yagona yechimga ega bo'ladi.
ADABIYOTLAR;
1. Uzoqboyev, Azizbek, Sarvar Abdullayev, and Nematillo Abriyev. "ROBOTOTEXNIK MEXANIZMLARNING MAXSUSLIKLARINI IZLASHDA MATRITSAVIY USULNING QO'LLANISHI." Eurasian Journal of Mathematical Theory and Computer Sciences 3.1 (2023): 92-100.
2. Uzoqbayev, Azizbek, Abbos Samandarov, and Kamoliddin Ne'matov. "ROBOTOTEXNIK MEXANIZMLARNING MAXSUSlKLARINI TOPISH ALGORITMI." Eurasian Journal of Academic Research 3.1 Part 6 (2023): 150153..
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