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// Наука и школа. 2013, № 4, с.81- 83.
2.
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560 s.
4.
Э.Б.Винберг.Алгебра многочленов. М.,“ Просвещение”,1980 – 175 c
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Əkbərov M.S. Cəbr və ədədlər nəzəriyyəsi. Bakı, “Nurlar”,
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Okunyov L.Y. Ali cəbr, Bakı, Azərbaycan Dövlət nəşriyyatı, 1955-468 s
7.
Винберг Э.Б. Курс алгебры. - М., МЦНМО, 2011 – 592 c.
8.
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М. : Физматгиз, 1962 – 200 с.
ABSTRACT
R.A.Hasanov
The misunderstanding of the training of the algebraic definitions
and about their foundation
The article deals with the misunderstanding of the training of the algebraic definitions and
about their foundation. The typical examples are shown also there/
РЕЗЮМЕ
Р.А.Гасанов
О недоразумениях возникновении их при обучение
алгебраических понятий
В работе исследуется причины возникновении недостатков при обучения
алгебраических понятиях в высших школах с педогогическим профилем. Указывается
типичные примеры уяснящие сущность статьи.
NDU-nun Elmi Şurasının 30 may 2015-ci il tarixli qərarı ilə çapa tövsiyə
olunmuşdur (protokol № 10)
Məqaləni çapa təqdim etdi: Riyaziyyat üzrə fəlsəfə doktoru, dosent
T.Nəcəfov
NAXÇIVAN DÖVLƏT UNİVERSİT ET İ. ELMİ ƏSƏRLƏR, 2015, № 9 (65)
-
31 -
NAKHCHIVAN ST AT E UNIVERSIT Y
.
SC IENTIFIC WO RKS, 2015, № 9 (65)
НАХЧЫВАНСКИЙ ГОСУДАРСТ ВЕННЫЙ УНИВЕРСИТ ЕТ . НАУЧНЫЕ ТРУДЫ, 2015, № 9 (65)
ELSHAD AGAYEV
e-mail:
agayev.elshad@gmail.com
SAHIB ALIYEV
Nakhchivan State University
SEFA ALIYEV
Nakhchivan University
UOT: 517
ON NONLINEAR ELLIPTIC SECOND ORDER
EQUATION`S SOLUTION BEHAVIOUR IN UNBOUNDED DOMAIN
In this parer the behavior in infinity of the positive solution u(x) of nonlinear elliptic equation of the
second order in a narrow area theth parameter , turning into zero on the baundary of the area is
considered.
The incuasing speed of the solution is determined depending on the equation and parameters of the
area.
Let
n
R
G
be an unfounded domain and there are such
4
1
0
,
0
R
R
that for arbitrary
G
x
0
,
\
0
G
B
G
B
C
x
R
x
R
S
Here
x
R
B is an open sphere with the cente
G
x
in
n
R . We denote S-capacity of
G
B
x
R
\
as
)
(E
C
S
. Let us call the domain G having the afove conditions as “ narrow” domain.
Assume that in G the positive solution of equation
n
j
i
x
x
j
i
u
u
x
u
u
u
x
a
Pu
j
i
1
,
0
)
,
,
(
)
,
,
(
(1)
is defined.
Here
ji
ij
j
i
n
j
i
ij
a
a
x
x
u
u
x
a
L
,
)
,
,
(
2
,
is a continuous elliptic operator: that is there is such
0
that in all G
2
1
1
,
2
)
,
,
(
j
i
n
j
i
ij
p
x
a
is true for arbitrary
n
n
R
P
R
R
,
,
.And function
satisfies conditions
1
1
1
,
)
2
,
1
min(
1
)
,
,
(
,
sgn
sgn
1
2
1
1
S
S
u
C
u
C
u
u
x
u
(2)
Defined as
n
j
i
j
i
ij
n
i
ii
G
x
x
a
x
a
1
,
1
1
,
)
(
)
(
sup
is called the constant of ellipticity of operator L and assume that s is positive and satisfies
inequality
2
s
when talking about the solution of (1) we shall understand its classic solution .
-
32 -
In order to investigate the solution of equation (1) in G satisfying the condition (2) we shall
give the following form of “ Principle of maxsimum “ and “ Lemma about incuasing “.
19
15
.
.
1
c
Principle of maximum. Let u(x) be a positive solution of (1) defined in an open domain
and continuous in
. Function
satisfies
u
sgn
sgn
.Then
u
u
max
sup
is true.
Proof: Assume countrary. Let us suppose that
0
0
,
)
(
max
x
x
u
u
.Then as
0
x
is a
maximum point.
n
j
i
x
x
x
x
ij
u
U
x
a
j
i
1
,
0
0
)
(
,
0
)
(
0
From
u
u
u
x
sgn
)
,
,
(
sgn
and
0
)
(
x
u
we get
0
)
,
,
(
0
x
x
u
u
x
.That is whu (1) is
not true.This contrary fact shows that our contrary assumption is not true.
Lemma about Increasing. Let
4
1
0
,
0
4
R
B
D
R
is an open set and
G
B
H
x
R
\
.
Let us take
0
4R
B
D
and assume that u(x) is a positive solution of (1) defined in domain D
and continuous in
D
and satisfying the condition
0
Г
u
in boundary Г . Let function φ satisfy
condition (2).
Then the equality
0
)
(
max
)
(
1
)
(
max
R
S
S
D
B
D
x
u
R
H
C
x
u
is true. Here
0
is a constant depending on S.
Proff: Here we will give main points off the proff.
Consider function
S
x
x
0
1
. Here
0
x
is a fixed point. Then according to
Lemma 2.1 (se[1] page 21) if
2
S
.Then
2
0
1
0
1
S
S
x
x
C
x
x
L
Here C
1
is some constant and
1
0
0
1
S
S
x
x
S
x
x
u
. Traking into consideration the condition
1
2
1
1
)
,
,
(
u
C
u
C
u
u
x
we will get
)
1
)(
1
(
0
2
)
1
(
0
1
1
0
2
1
0
1
0
0
1
1
1
1
1
,
1
,
S
S
S
S
S
x
x
C
x
x
C
x
x
C
x
x
C
x
x
x
x
x
substitute
r
x
x
0
.
If
2
)
1
(
S
S
and
2
)
1
)(
1
(
S
S
are true, then
S
r
1
will be subelliptik in
0
\ x
D
for arfitrary
n
R
x
0
as a function depending on x.
In this case let us find the conditions laying on
and
-
33 -
S
S
S
S
S
2
2
2
1
1
1
1
2
1
S
S
S
S
S
Thus all the conditions of E.M.Landis`s Lemma about increasing is true.
Theorem: Let G be a narrow domain a positive solution of (1) continuous in G and
equal to zero on the boundary of this domain is defned. Further, let function
)
,
,
(
u
u
x
satisfy (2)
.
Then there is such a constant
0
C
that
1
)
(
0
)
(
C
r
r
M
C
is true. Here
)
(
max
)
(
x
u
r
M
r
x
, and C depends on
e
,
0
and the dimension n of the space
(beginning from some
)
(r
M
A
r
is possible)
Proff: If we apply the lemma about increasing for
x
R
B and
x
R
B
4
we shall get
x
R
S
x
R
S
x
R
B
G
x
u
R
G
B
C
B
G
x
u
)
(
max
\
(
1
)
(
max
4
Here
0
does not depend on S.
As G is a narrow domain
0
x
R
S
B
C
is satisfied.
Denote a as
a
B
G
x
u
x
R
)
(
max
.
Then we will get
a
B
G
x
u
S
x
R
4
1
1
)
(
max
0
S
0
a
B
G
x
u
x
R
0
`
1
)
(
max
here
S
4
`
.
Assume that according to principle of maximum function u(x) gets its highest valume in
compact
x
R
B
G
4
in some point x
1
on this sphere. If we apply Lemma about increasing again
we will get
a
B
G
x
u
x
R
2
0
16
`
1
)
(
max
1
If we apply Lemma about increasing and principle of maximum k times we willl get
a
B
G
x
u
k
x
R
i
k
0
4
`
1
)
(
max
Denote
x
r
as
r
and take
R
r
k
4
.
Then we will get
a
r
M
k
0
`
1
)
(
from
R
r
k
4
we can define K as follows:
R
r
k
4
log
-
34 -
It is clear that
1
`
1
0
If we take into account this condition, we can write inequalty (3) as follows:
)
(
1
1
log
)
(
)
(
0
4
0
0
4
4
)
4
(
)
(
C
R
r
C
k
C
k
r
C
a
a
R
M
r
M
Thus we get
1
)
(
0
)
(
C
r
r
M
C
.
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