Lemma-1 (Gronuolla).
Faraz qilaylik,
]
,
[
0
x
x
kesmada
)
(
),
(
x
v
x
u
funksiyalar uzluksiz va manfiy bo‘lmasin. Agar ular
uchun, ushbu
0
,
)
(
)
(
)
(
0
A
dt
t
v
t
u
A
x
u
x
x
(1.13.13)
baho o‘rinli bo‘lsa, u holda
x
x
dt
t
v
A
x
u
0
)
(
exp
)
(
(1.13.14)
tengsizlik bajariladi.
Isbot.
Aytaylik,
0
0,
A
x
x
bo‘lsin. U holda (1.13.13)
tengsizlikda modul ishorasini tashlab va uni
)
(
x
v
ga ko‘paytirsak,
)
(
)
(
)
(
)
(
)
(
0
x
v
dt
t
v
t
u
A
x
v
x
u
x
x
(1.13.15)
hosil bo‘ladi.
Oxirgi (1.13.15) tengsizlikni ushbu
)
(
)
(
)
(
)
(
0
x
v
x
u
dt
t
v
t
u
A
dx
d
x
x
munosabatdan foydalanib
dx
x
v
dt
t
v
t
u
A
dt
t
v
t
u
A
d
x
x
x
x
)
(
)
(
)
(
)
(
)
(
0
0
ko‘rinishda yozish mumkin. Bu tengsizlikning ikkala tomonini
integrallab
x
x
x
x
dt
t
v
A
dt
t
v
t
u
A
0
0
)
(
ln
)
(
)
(
ln
munosabatni hosil qilamiz. Bundan
x
x
x
x
x
x
dt
t
v
A
dt
t
v
A
dt
t
v
t
u
A
0
0
0
)
(
exp
)
(
exp
)
(
)
(
kelib chiqadi.
Lemma shartidagi (1.13.13) tengsizlikka asosan
x
x
x
x
x
x
dt
t
v
A
dt
t
v
t
u
A
dt
t
v
t
u
A
x
u
0
0
0
)
(
exp
)
(
)
(
)
(
)
(
)
(
baho
hosil bo‘ladi. Bu baho
0
,
0
x
x
A
larda ham o‘rinli. Chunki
0
x
x
larda (1.13.13) tengsizlikni quyidagi
0
0
)
(
)
(
)
(
)
(
)
(
x
x
x
x
dt
t
v
t
u
A
dt
t
v
t
u
A
x
u
ko‘rinishda yozish mumkin. Bundan ham
x
x
x
x
dt
t
v
A
dt
t
v
A
x
u
0
0
)
(
exp
)
(
exp
)
(
kelib chiqadi.
Agar
0
A
bo‘lsa, u holda
0
)
(
x
u
bo‘ladi.
Haqiqatan ham
0
,
)
(
)
(
)
(
0
x
x
dt
t
v
t
u
x
u
bo‘lsa, (1.13.14) dan
x
x
dt
t
v
x
u
0
)
(
exp
)
(
bahoga ega bo‘lamiz. Bundan
0
da
0
)
(
x
u
kelib chiqadi, bu
esa
0
)
(
x
u
shartga zid. Shuning uchun
0
)
(
x
u
. ■
Yagonaligi.
Aytaylik,
)
(
1
x
y
,
)
(
2
x
y
funksiyalar (1.13.1)
differensial tenglamani va (1.13.2) boshlang‘ich shartni
qanoatlantirsin. Bundan tashqari ularning grafiklari
P
to‘g‘ri
to‘rtburchakda joylashsin, ammo
),
(
)
(
2
1
x
y
x
y
]
,
[
0
0
h
x
h
x
x
bo‘lsin.
U holda ushbu
,
)
(
)),
(
,
(
)
(
0
0
1
1
1
y
x
y
x
y
x
f
dx
x
dy
0
0
2
2
2
)
(
)),
(
,
(
)
(
y
x
y
x
y
x
f
dx
x
dy
tengliklardan, avvalo
,
0
)
(
)
(
0
2
0
1
x
y
x
y
so‘ngra
))
(
,
(
))
(
,
(
))
(
)
(
(
2
1
2
1
x
y
x
f
x
y
x
f
dx
x
y
x
y
d
munosabatni olamiz. Bu tenglikning ikki tomonini integrallab
x
x
dt
t
y
t
f
t
y
t
f
x
y
x
y
0
))]
(
,
(
))
(
,
(
[
)
(
)
(
2
1
2
1
ifodani olamiz. Lipshits shartidan foydalanib, oxirgi munosabatni
baholaymiz:
,
)
(
)
(
))
(
,
(
))
(
,
(
)
(
)
(
0
0
2
1
2
1
2
1
x
x
x
x
dt
t
y
t
y
N
dt
t
y
t
f
t
y
t
f
x
y
x
y
ya’ni
x
x
dt
t
y
t
y
N
x
y
x
y
0
)
(
)
(
)
(
)
(
2
1
2
1
bahoni olamiz. Ushbu
0
,
0
)
(
,
0
)
(
)
(
)
(
2
1
A
N
x
v
x
y
x
y
x
u
belgilashlarni
olib,
Gronuolla
tengsizligidan
foydalansak,
)
(
)
(
,
0
)
(
2
1
x
y
x
y
x
u
ekanligiga ishonch hosil qilamiz. Teorema to‘la
isbot bo‘ldi. ■
Ko‘pchilik hollarda (1.13.1)-(1.13.2) Koshi masalasining
)
(
x
y
yechimi
bilan
(1.13.6)
tengliklar
orqali
aniqlangan
( ),
n
y x
n
yaqinlashish orasidagi farqni hisoblashga to‘g‘ri keladi.
Buning uchun ushbu
)
(
)
(
x
y
x
y
n
ayirmani baholashga to‘g‘ri
keladi. Avvalo biz
)
(
x
y
n
funksiyani quyidagi
n
k
k
k
n
x
y
x
y
y
x
y
1
1
0
)]
(
)
(
[
)
(
ko‘rinishda yozib olamiz. So‘ngra bu tenglikning ikki tomonida
n
da limitga o‘tib
1
1
0
)]
(
)
(
[
)
(
k
k
k
x
y
x
y
y
x
y
munosabatni hosil qilamiz.
Bundan va (1.13.10) tengsizlikdan foydalanib, quyidagi ayirmani
baholaymiz:
1
1
0
)
(
)]
(
)
(
[
)
(
)
(
k
n
k
k
n
x
y
x
y
x
y
y
x
y
x
y
1
1
1
1
)
(
)
(
)]
(
)
(
[
n
k
k
k
n
k
k
k
x
y
x
y
x
y
x
y
,
)!
1
(
)
(
!
)
(
)!
1
(
)
(
!
0
1
1
j
n
hN
j
n
n
k
k
k
n
Nh
Mhe
j
Nh
n
Nh
Mh
k
h
MN
ya’ni
.
)!
1
(
)
(
)
(
)
(
n
Nh
Mhe
x
y
x
y
n
hN
n
(1.13.16)
Bu yerda
,
,
min
,
,
)
,
(
0
)
,
(
max
M
b
a
h
h
x
x
y
x
f
M
P
y
x
N
Lipshits o‘zgarmasi.
Mustaqil yechish uchun mashqlar: [5], §7, №221-223,
225, 226, 228.
E’tibor uchun
tashakkur !
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