Universal International Scientific Journal
77
Universaljurnal.uz
Kurbanov Akmal Abdumutolib o’g’li
O’zbekiston Respublikasi Fanlar akademiyasi
V. I. Romanovskiy nomidagi Matematika instituti
Uzbekistan
RIKATTI TENGLAMASI YECHIMINING HOSILALARI UCHUN REKURRENT-
DIFFERENSIAL TENGLAMALAR SISTEMASI
Abstract:
This is
𝑦'' + 𝑥𝑦 = 0
equivalent to the Liouville
𝑧'(𝑥) = 𝑥 + 𝑧
2
(𝑥)
equation. The system of
recurrent-differential equations for the derivatives of the solution of the Riccati equation is studied.
Key words:
recurrent-differential equation, polynomial triangle, infinite system.
Annotatsiya:
Ushbu
𝑦'' + 𝑥𝑦 = 0
Liuvill tenglamasiga
𝑧'(𝑥) = 𝑥 + 𝑧
2
(𝑥)
tengkuchli. Rikkati
tenglamasi yechimining hosilalari uchun rekurrent-differensial tenglamalar sistemasi o‘rganiladi.
Kalit so’zlar:
rekurrent-differensial tenglama, polinomial uchburchak, cheksiz sistema.
Аннотация:
Это эквивалентно
𝑦'' + 𝑥𝑦 = 0
уравнению Лиувилля. Исследована
𝑧'(𝑥) = 𝑥 + 𝑧
2
(𝑥)
система рекуррентно-дифференциальных уравнений для производных решения уравнения Риккати.
Ключевые слова:
рекуррентно-дифференциальное уравнение, полиномиальный треугольник,
бесконечная система.
Language:
Uzbek
Universal Xalqaro Ilmiy Jurnal
Universal International Scientific Journal
Year: 2024 Issue: 1 Volume: 6
International indexes
Universal International Scientific Journal
Universaljurnal.uz
7
8
Citation:
Kurbanov , A. (2024). SYSTEM OF RECURRENT-DIFFERENTIAL EQUATIONS FOR THE
DERIVATIVES OF THE SOLUTION OF THE RICATTI EQUATION. Universal International Scientific Journal,
https://universaljurnal.uz/index.php/jurnal/article/view/1113
Doi:
https://doi.org/10.5281/zenodo.13929631
Parametrga
bog‘liq
chiziqli
tenglamalarni kichik parametr usuli bilan
𝑦'
𝑛
(𝑥) = 𝑎
𝑛
(𝑥)𝑦
𝑛
(𝑥) + 𝑏
𝑛
(𝑥)𝑦
𝑛−1
(𝑥) + 𝑐
𝑛
(𝑥), 𝑛 =
0,1,2, . ..
(1)
ko’rinishidagi
rekurrent-differensial
tenglamalar cheksiz sistemasiga keltirish
mumkinligi yaxshi ma’lum, bunda
𝑦
0
(𝑥)
berilgan deb qaraladi. Bu sistema aslida sof
rekurrent tabiatli bo‘lib, no’malumlar
birin-ketin integrallash yo‘li hisoblanadi.
Amaliyotda (1) sistemadan farqli
𝑃
𝑛+1,𝑘
(𝑥) = 𝑎
𝑛,𝑘
(𝑥)𝑃
𝑛,𝑘
(𝑥) + 𝑏
𝑛,𝑘
(𝑥)(𝑃
𝑛,𝑘−1
(𝑥))
′
+ 𝑐
𝑛,𝑘
(𝑥)𝑃
𝑛,𝑘−2
(𝑥)(2)
ko‘rinishdagi tenglamalar sistemasiga
keltiriladigan masalalar uchraydi. Xususan,
𝑎
𝑛,𝑘
(𝑥) = 𝑛 − 𝑘, 𝑏
𝑛,𝑘
(𝑥) = 1, 𝑐
𝑛,𝑘
(𝑥) = (𝑛 − 𝑘 +
2)𝑥
bo’lgan hol Liuvill tenglamasi uchun
kombinatorik
masalaga
tengkuchli.
Boshlang‘ich shart
𝑃
𝑛,0
= 𝑛!
bo’lgan holni
qaraymiz (bu yerda
𝑘 < 𝑛
). Shunda
𝑃
𝑛,𝑘
(𝑥)
ko‘phadlardan iborat bo‘lib, u binomial
koeffitsientlar kabi uchburchak hosil
qiladi.
(2) munosabatdan bevosita quyidagi
xossalar kelib chiqadi:
1) Agar
𝑘 −
juft
bo‘lsa
𝑃
𝑛,𝑘
(𝑥)
bu
(
𝑘
2
− 1)
-darajali, agar toq
bo’lsa
([
𝑘
2
] + 1)
-darajali ko‘phad;
2)
𝑑
𝑛
𝑘
=
𝑑𝑒𝑔 𝑃
𝑛
𝑘
(𝑥)
deyilsa,
𝑑
𝑛+1
𝑘+2
= 𝑑
𝑛
𝑘
+ 1
,
𝑛 =
2,3, . . . , 𝑛 − 1; 𝑘 = 1,2, . . . , 𝑛 − 1.
Endi
𝑦'' + 𝑥𝑦 = 0
tenglamani qaraymiz.
Uning yechimi elementar funksiyalar
orqali, ular ustida algebraik amallar va
integrallash amali bilan ifodalanmaydi
(J.Liuvil teoremasi) [1]. U
𝑧'(𝑥) = 𝑥 + 𝑧
2
(𝑥)
(3)
Rikkati tenglamasiga teng kuchli.
Teorema.
(3) tenglama uchun
𝑧(0) =
𝑧
0
Koshi masalasining yechimi
𝑧(𝑥)
𝑧
(𝑛)
(𝑥) = 𝑛! 𝑧
𝑛+1
+
1
3
(𝑛 + 1)! 𝑥𝑧
𝑛−1
+
1
12
(𝑛
+ 1)! 𝑧
𝑛−2
+
Universal International Scientific Journal
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9
+ ∑[(𝑛 − 𝑘 + 1)𝑥𝑃
𝑛−1,𝑘−2
+ (𝑃
𝑛−1,𝑘−1
)
′
+ (𝑛 − 1
𝑛
𝑘=3
− 𝑘)𝑃
𝑛−1,𝑘
]𝑧
𝑛−𝑘
xossaga ega.
Isbot.
(3) tenglamadan
𝑧
″
= 2𝑧
3
+
2𝑥𝑧 + 1
,
𝑧
‴
= 6𝑧
4
+ 8𝑞𝑧
2
+ 2𝑧 + 2𝑥
2
bo‘lishi ravshan[2].
So‘ng
𝑧
(𝑛)
= 𝑛! 𝑧
𝑛+1
+ 𝑃
𝑛,1
𝑧
𝑛−1
+ 𝑃
𝑛,2
𝑧
𝑛−2
+
𝑃
𝑛,3
𝑧
𝑛−3
+. . . +𝑃
𝑛,𝑛−2
𝑧
2
+ 𝑃
𝑛,𝑛−1
𝑧
1
+ 𝑃
𝑛,𝑛
(4)
o‘rinli bo‘lsin deb faraz qilaylik. U holda
𝑧
(𝑛+1)
= (𝑛 + 1)! 𝑧
𝑛+2
+ [(𝑛 − 1)𝑃
𝑛,1
+ (𝑛
+ 1)! 𝑥]𝑧
𝑛
+ [(𝑛 − 2)𝑃
𝑛,2
+ (𝑃
𝑛,2
)']𝑧
𝑛−1
+
+[(𝑛 − 1)𝑥𝑃
𝑛,1
+ (𝑛 − 3)𝑃
𝑛,3
+ (𝑃
𝑛,2
)']𝑧
𝑛−2
+ [(𝑛
− 2)𝑥𝑃
𝑛,2
+ (𝑛 − 4)𝑃
𝑛,4
+ (𝑃
𝑛,3
)']𝑧
𝑛−3
+ (5)
+[(𝑛 − 3)𝑥𝑃
𝑛,3
+ (𝑛 − 5)𝑃
𝑛,5
+ (𝑃
𝑛,4
)']𝑧
𝑛−4
+ [(𝑛
− 4)𝑥𝑃
𝑛,4
+ (𝑛 − 6)𝑃
𝑛,6
+ (𝑃
𝑛,5
)']𝑧
𝑛−5
+
+[(𝑛 − 𝑘 + 2)𝑥𝑃
𝑛,𝑘−2
+ (𝑛 − 𝑘)𝑃
𝑛,𝑘
+ (𝑃
𝑛,𝑘−1
)']𝑧
𝑛−𝑘+1
+ [(𝑛 − 𝑘
+ 1)𝑥𝑃
𝑛,𝑘−1
+ (𝑛 − 𝑘 − 1)𝑃
𝑛,𝑘+1
+
+(𝑃
𝑛,𝑘
)']𝑧
𝑛−𝑘
+. . . +[2𝑥𝑃
𝑛,𝑛−2
+ (𝑃
𝑛,𝑛−1
)']𝑧
+ 𝑥𝑃
𝑛,𝑛−1
+ (𝑃
𝑛,𝑛
)'.
Bu ifodani (4) formulada
𝑛
o‘rniga
𝑛 + 1
qo‘yilgani bilan taqqoslab, quyidagi
tengliklarga ega bo‘lamiz:
𝑃
𝑛+1,0
= (𝑛 + 1)!,
1,1
,1
(
1)!
(
1)
,
n
n
P
n
x
n
P
+
=
+
+
−
𝑃
𝑛+1,2
= (𝑛 −
2)𝑃
𝑛,2
+ (𝑃
𝑛,1
)
′
,
va
𝑘 = 3, . . . , 𝑛
,
𝑃
𝑛+1,𝑘
=
(𝑛 − 𝑘 + 2)𝑥𝑃
𝑛,𝑘−2
+ (𝑃
𝑛,𝑘−1
)
′
+ (𝑛 − 𝑘)𝑃
𝑛,𝑘
.
Xususan,
𝑃
𝑛+1,𝑛+1
= 𝑥𝑃
𝑛,𝑛−1
+ (𝑃
𝑛,𝑛
)
′
.
Boshlang‘ich shart
𝑃
𝑛,0
= 𝑛!
bo‘lgan
holdan keyingi
𝑃
𝑛,1
ko‘phadni topamiz.
Yuqorida keltirilgan
𝑃
𝑛,1
= 𝑛! 𝑥 + (𝑛 − 2)𝑃
𝑛−1,1
(6)
tenglamaning bir jinsli qismini quyidagicha
yechamiz:
𝑃
𝑛,1
= (𝑛 − 2)𝑃
𝑛−1,1
tenglamada
𝑃
𝑛,1
= 𝑦
𝑛
kabi belgilash kiritsak,
𝑦
𝑛
= (𝑛 −
2)𝑦
𝑛−1
ga ega bo’lamiz. Uning yechimi
𝑦
𝑛
=
(𝑛 − 2)! 𝐶
ko‘rinishda bo‘ladi
.
O‘zgarmasni
variatsiyalash yo‘li orqali
𝑦
𝑛
= (𝑛 − 2)! 𝐶
𝑛
(7)
ifodani (6) tenglamaga qo‘yamiz va
𝐶
𝑛
ga
nisbatan
𝐶
𝑛
= 𝐶
𝑛−1
+ (𝑛 − 1)𝑛𝑥
rekurrent formulaga ega bo‘lamiz. Keyingi
qadamda
𝑛
ga
1,2, . . . , 𝑛
qiymatlarni berib
tenglamalar sistemasini hosil qilamiz va
uni yechamiz:
𝐶
𝑛
= (1 ⋅ 2 + 2 ⋅ 3+. . . +(𝑛 − 1) ⋅
𝑛)𝑥 =
1
3
𝑛(𝑛 − 1)(𝑛 + 1)𝑥.
𝐶
𝑛
uchun topilgan
ifodani (7) ga qo‘yamiz va quyidagiga ega
bo‘lamiz :
𝑃
𝑛,1
= 𝑦
𝑛
= (𝑛 − 2)! 𝐶
𝑛
=
1
3
(𝑛 − 2)! 𝑛(𝑛 − 1)(𝑛 +
1)𝑥 =
1
3
(𝑛 + 1)! 𝑥
.
Yuqoridagi qadamlarni davom ettirib,
Universal International Scientific Journal
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80
,2
1
(
1)!,
12
n
P
n
=
+
𝑃
𝑛,3
=
1
90
(𝑛 + 1)! (5𝑛 − 8)𝑥
2
,
,4
1
(
1)!(5
12) ,
180
n
P
n
n
x
=
+
−
𝑃
𝑛,5
=
1
5670
(𝑛 +
1)! (35𝑛
2
− 203𝑛 + 264)𝑥
3
+
1
10080
(𝑛 +
1)! (35𝑛 − 106), . ..
larni hosil qilamiz.
Ushbu
𝑃
𝑛,𝑘
= (𝑛 − 𝑘 + 1)𝑥𝑃
𝑛−1,𝑘−2
+
(𝑃
𝑛−1,𝑘−1
)
′
+ (𝑛 − 𝑘 − 1)𝑃
𝑛−1,𝑘
lar
𝑛 > 𝑘 ≥ 3
uchun o‘rinli. Topilgan ifodalarni (4) ga
qo‘yib quyidagiga ega bo‘lamiz:
𝑧
(𝑛)
= 𝑛! 𝑧
𝑛+1
+
1
3
(𝑛 + 1)! 𝑥𝑧
𝑛−1
+
1
12
(𝑛
+ 1)! 𝑧
𝑛−2
+ 𝑃
𝑛,3
𝑧
𝑛−3
+. . . +𝑃
𝑛,𝑛−2
𝑧
2
+ 𝑃
𝑛,𝑛−1
𝑧
1
+ 𝑃
𝑛,𝑛
=
= 𝑛! 𝑧
𝑛+1
+
1
3
(𝑛 + 1)! 𝑥𝑧
𝑛−1
+
1
12
(𝑛 + 1)! 𝑧
𝑛−2
+ ∑ 𝑃
𝑛,𝑘
𝑧
𝑛−𝑘
=
𝑛
𝑘=3
𝑛! 𝑧
𝑛+1
+
1
3
(𝑛
+ 1)! 𝑥𝑧
𝑛−1
+
+
1
12
(𝑛 + 1)! 𝑧
𝑛−2
+ ∑[(𝑛 − 𝑘 + 1)𝑥𝑃
𝑛−1,𝑘−2
𝑛
𝑘=3
+ (𝑃
𝑛−1,𝑘−1
)
′
+ (𝑛 − 1
− 𝑘)𝑃
𝑛−1,𝑘
]𝑧
𝑛−𝑘
.
Xulosa.
Ushbu maqolada Rikkati tenglamasi
yechimining hosilalari uchun rekurrent-
differensial
tenglamalar
sistemasi
o’rganildi.
Bunda
𝑧'(𝑥) = 𝑥 + 𝑧
2
(𝑥)
ko’rinishidagi tenglama uchun
𝑧(0) = 𝑧
0
Koshi masalasining yechimi
𝑧(𝑥)
𝑧
(𝑛)
(𝑥) = 𝑛! 𝑧
𝑛+1
+
1
3
(𝑛 + 1)! 𝑥𝑧
𝑛−1
+
1
12
(𝑛
+ 1)! 𝑧
𝑛−2
+
+ ∑[(𝑛 − 𝑘 + 1)𝑥𝑃
𝑛−1,𝑘−2
+ (𝑃
𝑛−1,𝑘−1
)
′
+ (𝑛 − 1
𝑛
𝑘=3
− 𝑘)𝑃
𝑛−1,𝑘
]𝑧
𝑛−𝑘
xossaga ekanligi isbotlandi.
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P. T. Leung and G. J. Ni, 2020, " A New Look at the Quantum Liouville Theorem " J. Found.
Appl. Phys. 7: 25-31
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А.О.Гельфонд. Исчиcление конечных разност. Государственное издптелбство
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