CC BY
0Central Asian Journal of
Education and Innovation
HILFER HOSILASI VA UNI HISOBLASH USULLARI.
Latipova Shahnoza Salim qizi
Osiyo Xalqaro Universiteti "Umumtexnik fanlar" kafedrasi o'qituvchisi
slatipova543@gmail.com https://doi.org/10.5281/zenodo.10652624
ARTICLE INFO
ABSTRACT
Qabul qilindi: 05-February 2024 yil Ma'qullandi: 10- February 2023 yil Nashr qilindi: 13- February 2023 yil
KEY WORDS
Uzluksizlik, Caputo hosilasi, Hilfer hosilasi, Reiman - Liuvell integrali
Hilfer ma'nosida kasr tartibli hosila qatnashgan tenglamalar o'rganilib, qaralayotgan tenglamaning elliptik qismi o'z-o'ziga qo'shma, musbat aniqlangan ixtiyoriy operatordan iborat. O'rganilayotgan masalani yechish uchun o'zgaruvchilarni ajratish usulidan foydalaniladi..
Hilfer hosilasi
Hilfer ma'nosidagi kasr tartibli hosila Reiman - Liuvell ma'nosidagi kasr tartibli hosila va Caputo ma'nosidagi kasr tartibli hosilalarning umumiy holi hisoblanadi. Shuning uchun yuqorida Reiman - Liuvell ma'nosidagi kasr tartibli hosila va Caputo ma'nosidagi kasr tartibli hosilalar haqida ma'lumotlar keltirdim. Endi Hilfer ma'nosidagi kasr tartibli hosila va tenglamalar haqida ma'lumotlar keltiraman. Ta'rif. Ushbu
dn
Da,ß f (t) = Iß(n-œ) ^—I (1-ß)(1-a) f (t)
dtn
ko'rinishdagi tenglikka Hilfer ma'nosidagi kasr tartibli hosila deyiladi. Bu yerda n - 1 < Œ < n, 0 <ß< 1 oraliqlardagi sonlar.
Agar ß = 0 va n - 1 < Œ < n bo'lsa (1) tenglik quyidagi ko'rinishga keladi:
(1)
(2)
jn
DŒf (t ) = d-I(n -Œ)f (t ) dt
ya'ni, bu biz bilgan Reiman - Liuvell ma'nosidagi kasr tasrtibli hosila hisoblanadi. ß = 1 va n - 1 <Œ< n bo'lganda
esa
CDŒf (t ) = I(n-Œ) d-f (t )
dt
(3)
Caputo ma'nosidagi kasr tartibli hosilaga kelar ekan. Demak, yuqorida aytganimizdek Hilfer ma'nosidagi kasr tartibli hosila Reiman - Liuvell va Caputo ma'nosidagi kasr tartibli hosilalarning umumiy holi ekan.
Hilfer hosilasi qatnashan differensial tenglamalar.
Hilfer ma'nosidagi kasr tartibli hosila qatnashgan differensial tenglamalar juda ham ko'p bo'lib ulardan ba'zilarini ko'rsatamiz.
Lemma 1. Agar JLl > n — 1 bo'lsa u holda quyidagi tenglik o'rinli bo'ladi:
(4)
L — T(l + 1)
T(l +1 — a)
Isbot. L > n — 1 bo'lsin. (2.1) tenglikdan m vaqtda farq qilsin
dn
-I ötL =
T(l + 1)
(L + ô)...(l + ö — n)t
L+ö—n
(5)
dtn T(l +1 + ö)
Endi Gamma funksiyaning F(z + 1) = zF(z) xossasidan foydalanamiz.
T(jU + S +1) = (ju + S)*(ju + S-l)...(jU + S-n + l)r(// + S - n +1).
Endi yuqoridagi tenglikni (2.3.5) ga qo'llaymiz u holda biz quyidagiga ega bo'lamiz
dn ^ ^ ^
ÖL _ T (l + 1) L+Ö—n dtn TL + ö — n +1)
Boshqa tomondan
(6)
jß(n—a)t v+5—n_ T(l + ö + 1— n) L+ö—n+ß( n—a)
T(l + ö — n +1 + ß(n — a))
(7)
va
ö + ß(n — a) = (1 — ß)(n — a) + ß(n — a) = n — a,
Bundan biz quyidagiga erishamiz
jg,ß L — T (L + 1) jß—a T(l +1 — a)
tasdiq isbotlandi.
Ta'rif. (Kasr tartibli hosilaning limit nuqtasi haqida). —to < a < uchun kasr tartibli hosilaning darajalari 0 < a < 1 va 0 < ß < 1 bo'lganda f (x) a nuqtadagi limiti deb quyidagi tenglikka aytiladi
Daßf (a) = ± lim Daaßf (x).
a, ß
(8)
x ^a ±
Tasdiq 1. / : [tf, —>• M bo'lsa a < A < oo oraliqda bo'lsin. Agar 0 <ß< 1 — a bo'lsa
La+
( D^ff )(x) = ( Daa,f )(x)
A . 1 — a — ß
qachonki 0 < ß < A <-, va
1 — a
(9)
I
P
a+
( D"U f )( x) — ( D";(1-a-p)/(1-a) f ) ( x) + ■ ( a+ J 1 a+ j ) r(1 -a-P)(x - a)a+p
0 1 -a-P
Barcha jU < A —-lar uchun o'rinli bo'ladi.
(10)
1 -a
Isbot.
( di\a ) ( x)
f ( a )
T(1 -a)(x - a)a
+
( Ia+D ) (x) (11)
g (x) ni vaqtinchalik quyidagicha belgilaymiz g (x) — (Ia+ )( a)f ) (x) va integral
a jP _ ja+P
operatorning quyidagi xossasidan foydalanamiz I Ip — I A(1-a) m(1-A)(1-a)
( D"Af )( x) — ( IaA<1-a) DIa —(
u(1-a) j(A-j)(1-a)
) Dg ) ( x)
_ I J(1-a)
La+
( diA-M-") g )( x) -
( x - a )
(A-j)(1-a)-1
V
r((A-j)(1 -a))
g(a)
J
A(1-a)-1
— ( D"+Uf )( x) - -— g (a). (12)
r(A(1 - a))
Kasr tartibli integraldan foydalanamiz
1^+ (x - ^ — 1>1. (x - a)"+P (13)
r(a + P +1)
1 -a-p 1 -a-p
A <- uchun (9) tenglik o'rinli chunki g (a) — 0 bo'ladi, va A —-
1 -a 1 -a
bo'lganda esa (10) tenglik o'rinli.
Misol 1. D",P - Hilfer ma'nosida kasr tartibli hosilali, Iq-P - R-L kasr tartibli integrali. Bu yerda 0 <a < 1 va 0 < P< 1 oraliqlarda bo'lsin. f (t ) E C (( 0, œ) ;H ) va (E H.
D",PT ( t ) + AT (t ) — f,
lim I0+PT(t) — p. t ^+0
t > 0,
(14)
bu yerda p, f eH. Koshi masalasi berilgan bo'lsin. Tenglamaning yechimi quyidagi ko'rinishda bo'ladi:
t t \ T (t) = ptß—lEaß (-Ata) + {( t -rf-1 Ea.al—A( t — zf ) f (r)dr . (15)
0
Bizda manba funsiyasi f (t) vaqtga bog'liq bo'lmasin. U holda (2.3.14) masala quyidagi masalaga aylanadi:
'd^T ( t ) +AT (t ) = f,
t > 0,
(16)
lim /¿+ßT(t) = q>.
j
(16) u holda masalani yechishda bizga Mittag - Liffler funksiyasing integrali yordam berdi. Ya'ni quyidagicha:
t , V
J( t - r)a-1 Ea,a U( t -r)a\ dr = taEaa+i (-Xta). (17)
0
(15) yechimdan foydalansak
t 1 \ T(t) = <ptß-1Eatß(-Äta) + Kt - r)a-1 Eaa ("A(t - r)a) f (r)dr
0
t l \ = <ptß-XEaß(-lta) + f J(t-r)a-1 Eaa (-A(t-r)a)dr =
0
= cptß-1Ea,ß(-Äta) + ftaEa a+1 (-¿ta ).
T(t) = <ptß~XEaß(-lta ) + ftaEa a+i (-Zta ).
Demak, (16) masalaning yechimi (18) ko'rinishida bo'lar ekan.
(18)
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