12 / 23
since
y
0
(
x
)
y
IV
(
x
) =
−
p
0
|
y
|
k
−
1
y
(
x
)
y
0
(
x
)
<
0
on
x
000
j
, x
0
j
and
y
0
(
x
0
j
) =
=
y
000
(
x
000
j
) = 0
.
The rest ones follow from
y
00
(
x
)
y
000
(
x
)
>
0
on
x
0
j
, x
000
j
+1
.
Finally, for the first of (12) we have
y
000
(
x
j
)
2
−
y
000
(
x
00
j
+1
)
2
=
−
2
x
00
j
+1
Z
x
j
y
000
(
x
)
y
IV
(
x
)
dx
=
= 2
p
0
x
00
j
+1
Z
x
j
y
000
(
x
)
y
(
x
)
|
y
(
x
)
|
k
−
1
dx
=2
p
0
y
00
(
x
)
y
(
x
)
|
y
(
x
)
|
k
−
1
x
00
j
+1
x
j
−
−
2
kp
0
Z
x
00
j
+1
x
j
y
00
(
x
)
y
0
(
x
)
|
y
(
x
)
|
k
−
1
dx <
0
,
since
y
0
(
x
)
y
00
(
x
)
>
0
on
x
j
, x
00
j
+1
and
y
(
x
j
) =
y
00
(
x
00
j
+1
) = 0
,
whereas the
rest inequalities follow from
y
000
(
x
)
y
IV
(
x
)
>
0
on
x
00
j
+1
, x
j
+1
.
I
So, the absolute values of the local extrema of any typical solution to
equation (1) form a strictly increasing sequence. The same holds for its
first, second, and third derivatives.
Hereafter we need some extra notations. Put
Ω
1
+
=Traj
1
(Ω
+
∩
S
3
,
1)
⊂
S
3
.
This is a compact subset of the interior of
Ω
+
containing ultimate parts
of all trajectories generated by maximally extended typical solutions to
equation (1) with
p
0
= 1
.
As for solutions generating the curves in
R
4
completely lying in
Ω
+
,
the trajectories related completely lie in
Ω
1
+
.
Besides, we define the compact sets
K
i
=
a
∈
Ω
1
+
:
a
i
= 0
and the
functions
ξ
j
:
R
4
\{
0
} →
R
,
j
= 0
,
1
,
2
,
3
,
taking each
a
∈
R
4
\{
0
}
to the
minimal positive zero of the derivative
y
(
j
)
(
x
)
of the solution to the initial
data problem
y
IV
(
x
) +
y
(
x
)
|
y
(
x
)
|
k
−
1
= 0;
y
(
j
)
(0) =
a
j
,
j
= 0
,
1
,
2
,
3
.
(13)
Further, to each solution
y
(
x
)
to equation (1) we associate the function
F
y
(
x
) =
3
X
j
=0
ρy
(
j
)
(
x
)
1
j
(
k
−
1)+4
with
ρ
=
p
1
k
−
1
0
.
The notation
F
y
does not
use
p
0
,
since non-trivial functions cannot be solutions to equation (1) with
different
p
0
.
Lemma 7.
The restrictions
ξ
i
|
K
j
, i, j
= 0
,
1
,
2
,
3
,
are continuous.
J
First we prove continuity of
ξ
i
at
a
∈
Ω
+
with
a
i
>
0
.
Suppose
ξ
i
(
a
) =
x
i
and
ε >
0
.
We can assume that
ε
is sufficiently small to be less than
x
i
and to
provide, for the solution
y
(
x
)
to (13), the inequalities
y
(
i
)
(
x
−
ε
)
>
0
on
[0
, x
i
−
ε
]
and
y
(
i
)
(
x
i
+
ε
)
<
0
.
In this case the point
a
has a neighborhood
14
ISSN 1812-3368. Вестник МГТУ им. Н.Э. Баумана. Сер. “Естественные науки”. 2015. № 2