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Sulaymonov M.M. o'qituvchi Qo'qon DPI Ummatova M.A. katta o'qituvchi Qo'qon DPI
KASR TARTIBLI DIFFERENSIAL OPERATOR ISHTIROK ETGAN INTEGRO-DIFFERENSIAL TENGLAMALAR UCHUN INTEGRAL
SHARTLI MASALALAR
Annotatsiya. Ushbu maqolada kasr tartibli differensial operator qatnashgan integro-differensial tenglama uchun ikkinchi tur integral shartli masala o 'rganilgan.
Kalit so^zlar: tenglama, chegaraviy shartlar, chegaralangan va bo'lakli uzluksiz bo 'lgan ma 'lum funksiyalar, uzluksiz ma'lum funksiyalar, ikkinchi tur Fredgolm integral tenglamasi.
Sulaymanov M.M.
teacher Kokand SPI Ummatova M.A. senior teacher Kokand SPI
FRACTIONAL DIFFERENTIAL OPERATOR PARTICIPATION FOR INTEGRO-DIFFERENTIAL EQUATIONS INTEGRAL CONDITIONAL PROBLEMS
Abstract. In this article, the integral conditional problem of the second type for the integro-differential equation involving a fractional differential operator is studied.
Key words: equation, boundary conditions, certain bounded and piecewise continuous functions, certain continuous functions, Fredholm integral equation of the second kind.
1-masala.
y" (x) + pi (x) y' (x) + pi (x) y (x) + p3 (x) D> (x) y (x) = f (x),
x e (a, b ) (1)
tenglamaning [ a, b] segmentda aniqlangan, uzluksiz va
p
y(a) = k1, y'(b) + hy(b) = hJy(t)dt + k2 (2)
a
shartlarni qanoatlantiruvchi yechimi topilsin, bu yerda k, k2, h,a, ß-berilgan sonlar bo'lib, a <a< ß<b.
(2) dan ko'rinib turibdiki h = 0 da 1-masaladan y (a) = ki y' (b) = k2 (3)
chegaraviy shartlarni qanoatlantiruvchi masala kelib chiqadi. Agar 0< h < 1 vaa<a<ß<b bo'lsa, (2) shartlarning ikkinchisini undagi
integralga o'rta qiymat haqidagi teoremani tatbiq qilib, y \b) + hy(%) = k2
ko'rinishda yozib olish mumkin bo'ladi, bu yerda % - [a, b] segmentdagi qandaydir tayinlangan son. Demak, bu holda, 1 -masala y(a) = ki, y'(b) + hy(%) = k2 (4)
shartlarni qanoatlantiruvchi yechimi topilganidek o'rganiladi. 0 < q < 1 va [a,ß] = [a,b] bo'lgan holda ham 1-masala 4-shartga keltirib, o'rganiladi.
Yuqoridagilarni e'tiborga olgan holda 1 -masalani h = 1, a = a, ß = b bo'lgan holda, ya'ni (2) shartlar
b
y(a) = k y'(b) + y(b) = Jy (t)dt + k2 (5)
a
ko'rinishga ega bo'lgan holda o'rganamiz.
Bu masalaning yechimi mavjud va yagonaligini ko'rsatish uchun xuddi 3-shartdagi kabi, (1) tenglamani [a, x] oraliqda ikki marta integrallab,
P2 (t) - p (t) ■+ [ P3 ( x ) (x - t y" -
x
y'(x) + P1 (x) y (x ) + J
Jp3 (z)(z -1)-a dz }y (t)dt = ]f (t)dt + y'(a) + k,px (a),
t ^ a
(6)
va
y i x ) + J{ pi(t ) + [ p1 (t ) - p"(t ) ] ( x -1 ) +
it)
x
' r(l - a)^ -1 " t p3 if) - p -3 if )i x - f ) <1 y i t ) dt
x
= Ji x -1 ) f i t ) dt + y ' i a )i x - a ) + kxpx i a )i x - a ) + kx
tengliklarga ega bo'lamiz. (6) va (7) tengliklarni quyidagicha yozib olamiz;
x
y i x ) = -J{ pi (t ) + [ p2 (t ) - p"(t )] ( x -1 ) +
it)
x
' r(l-«)'Jif-trtbp"x-f)df]yit)d
x
+ Ji x -1 ) f i t ) dt + y ' a )i x - a ) + kxpx i a )i x - a ) + kx,
+
x
y ' i x ) = - pi i x ) y i x )-J
p2 it )- p"it ) +
ю
it)
г(1 - " )
p3 ix)ix - t)-
x
-Jp"iz)iz -1)- " dz
x
y i t ) dt + J f i t ) dt + y'i a ) + kxpx i a )
Bulardan quyidagiga ega bo'lamiz:
x
y i x ) = J K2 i x, t ) y i t ) dt + f2 i x ) + y'i a )i x - a ), (S)
a
x
y ' i x ) = - pi i x ) y i x ) - J Kl i x, t ) y i t ) dt + fi i x ) + y ' i a ), (9)
a
bu yerda
xx
f i x ) = J f i t ) dt + kp i a ), f2 i x ) = Ji x -1 ) f i t ) dt + kp i a )i x - a ) + kx,
a a
Ki i x, t ) = pi(t ) + [ p2(t ) - p"(t )]( x -1 ) +
J if-1 "[ p" if)-p '" if)i x-f) df] ■
K i x, t ) = - pi i t ) + [ p, i t )- pi i t )]i x -1 )-
-f^b Jif-1 "[ P" if)- P '"Ш x -f)] •
a
a
a
a
a
a
Kj (x, t), K2 (x, t)-{(x, t) : a < x < to'rtburchakda chegaralangan
va bo'lakli uzluksiz bo'lgan ma'lum funksiyalar, f (x), f2 (x) esa [a,b] da
uzluksiz ma'lum funksiyalar.
(S) dan quyidagini topamiz:
b Ъ Ъ Ъ 1
J y (t) dt = J J K2 (x, t)y(t)dt +J f2 (t)dt + - y'(a)(b - a)..
a a a a 2
Buni va
(S) va (9) tengliklarda x = b deb,
b
y ' ( b ) = - pi ( b ) y ( b )-J K, (b, t ) y ( t ) dt + f (b ) + y '( a ), (10)
a
b
y ( b ) = J K2 ( b, t ) y (t ) dt + f. (b ) + y '( a )(b - a ), (11)
a
tengliklarni topamiz. (10) va (11) ni (5) shartga qo'yib, y '(a)•{(Ъ - a)[-Pi (Ъ) +1] +1} =
b b = -[-Pi ( Ъ ) + l}J K 2 ( Ъ, t ) y ( t ) dt + J K, ( b, t ) y (t ) dt -
b
-f. ( b )[-Pi ( b ) +1]- f, ( b ) + J y (t ) dt + к.
tenglikka ega bo'lamiz.
Agar (b - a)[-p (b) +1] +1Ф 0 bo'lsa y'(a) oxirgi tenglikdan bir qiymatli topiladi
y '(a ) =
Ъ Ъ
-[-p, ( b ) + l]-J K 2 ( b, t ) y ( t ) dt + J K, ( b, t ) y (t ) dt -
f2( b )[-p,( b ) +f ( b ) + fy (t ) dt + к2 b - a }[_ pl ( b ) +1] + !
Uni (S) ga qo'yib,
x b b
y ( x ) = J K 2 ( x, t ) y ( t ) dt + f, ( x ) + ]-[-P; ( b ) + l]-J K 2 (b, t ) y (t ) dt + J K- (b, t ) y (t ) dt -
a
b
- f2 ( b )[-P; (b ) +1] - f; (b ) + J y (t ) dt + к.
x - a
(b - a)[-p- (b) +1] +1 (12) y (x) ga nisbatan ikkinchi tur Fredgolm integral tenglamasiga ega bo'lamiz.
a
4
a
a
Agar berilganlarga qo'yilgan f (x) e C[a,b], p(x) e C1[a,b], p2(x),p (x)e C2[a,b] shartlarda |K2(x,t)|< 1 bo'lsa, (12) tenglama va,
demak, 1-masala yagona yechimga ega bo'ladi.
Foydalanilgan adabiyotlar:
1. O'rinov A.Q. Oddiy differensial tenglamalar uchun chegaraviy masalalar. -Toshkent: MUMTOZ SO'Z, 2014, 115 bet.
2. Salohiddinov M.S. Integral tenglamalar. -Toshkent: Yangiyo'l poligraf servis, 2007, 256 bet.
3. Ummatova, Mahbuba Axmedovna, and Olimaxon Oxunjonovna Rahmonova. "ELEMENTAR MATEMATIKADA ANTISIMMETRIK KO 'PHADLAR." INTERNATIONAL CONFERENCE DEDICATED TO THE ROLE AND IMPORTANCE OF INNOVATIVE EDUCATION IN THE 21ST CENTURY. Vol. 1. No. 10. 2022.
4. Axmedovna, Ummatova Mahbuba, and Ilhomjonova Shahnozaxon Ilhomjonovna. "TALIMDA BIOLOGIYA VA MATEMATIKA FANLARINING OZARO ALOQASI HAQIDA." BARQARORLIK VA YETAKCHI TADQIQOTLAR ONLAYN ILMIY JURNALI 2.12 (2022): 816817.
5. Ummatova, Mahbuba Axmedovna, and Abdurahim Tursunboyevich Mamatqulov. "AL-XORAZMIY ASARLARINING AMALIY AHAMIYATI HAQIDA." INTERNATIONAL CONFERENCE DEDICATED TO THE ROLE AND IMPORTANCE OF INNOVATIVE EDUCATION IN THE 21ST CENTURY. Vol. 1. No. 10. 2022.
