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The Research of Solution of Levinson — Smith Equation
ISSN 1812-3368. Вестник МГТУ им. Н.Э. Баумана. Сер. Естественные науки. 2017. № 1
21
◄
Without loss of generality, we can assume that
*
= .
x
It follows from Lemma 2 and the condition (6) that, for
,
there exist numbers
0
0
t
and
> 0
such that
f
on
once
0
.
t t
Since
2
0
=
t
fy dt
0
( ) =
t
h dt
0
*
0
| = ( ) ,
t
h h t
h
the integral
2
0
t
fy dt
is convergent. On the other hand,
,
f
2
2
0
fy y
once
0
t t
and, thus, the integral
2
0
t
y dt
is convergent. Hence, the
integrals
2
0
t
y dt
and
2
0
0
(
( )) =
2
t
t
y
h G x dt
dt
are also convergent.
On
,
we have
*
*
( )
h h G G x
implying
*
*
( )
0.
h G x h G
Since the
integral
0
(
( ))
t
h G x dt
is convergent,
*
*
= .
h G
Therefore, by Proposition 1,
dist( ( ), )
0.
t
t
As said above, on
,
we have
*
( ),
h G G x
*
( )
( ) 0,
h G x G G x
*
( )
( ) 0,
h G x G G x
2
*
=
( )
( )
0,
2
y
h G x y G G x y
and, since the
integral
2
0
t
y dt
is convergent, so is the integral
*
0
( )
.
t
G G x y dt
Also, recall that
the function
( ) = ( )
y t x t
is of constant sign and the function
( )
x t
is strictly
monotonous. It follows from above that the integral
*
0
( ) =
t
G G x ydt
*
*
( )0
( )
x
x t
G G x dx
is convergent and, thus, so is the integral
*
*
0
( ) .
x
G G x dx
This completely proves Theorem 1.
►
Corollaries and examples. Corollary 4.
Suppose that the condition (6) holds and
the integrals
*
0
( )
G G x dx
and
0
*
( )
G G x dx
are divergent. Then any trajectory
either enters the set
and then never leaves it or tends to some equilibrium point
.
p
Corollary 5.
Suppose that the condition (6) holds, the function g has period
> 0,
T
0,
g
and
0
( ) = 0.
T
g t dt
Then any trajectory either enters the set
and then never
leaves it or tends to some equilibrium point
.
p
◄
The function
( )
G x
is upper bounded since, for all
,
x