11 / 23
Fig. 2. Zeroes of the derivatives of a typical solution
y
0
(
x
00
j
)
<
|
y
0
(
x
j
)
|
< y
0
(
x
000
j
+1
)
< y
0
(
x
00
j
+1
) ;
(10)
y
00
(
x
000
j
)
< y
00
(
x
0
j
)
<
|
y
00
(
x
j
)
|
< y
00
(
x
000
j
+1
) ;
(11)
|
y
000
(
x
j
)
|
< y
000
(
x
00
j
+1
)
< y
000
(
x
0
j
+1
)
<
|
y
000
(
x
j
+1
)
|
.
(12)
J
Indeed,
p
0
k
+ 1
y
(
x
0
j
)
k
+1
−
y
(
x
000
j
+1
)
k
+1
=
−
p
0
x
000
j
+1
Z
x
0
j
y
0
(
x
)
|
y
(
x
)
|
k
−
1
y
(
x
)
dx
=
=
x
000
j
+1
Z
x
0
j
y
0
(
x
)
y
IV
(
x
)
dx
=
y
0
(
x
)
y
000
(
x
)
x
000
j
+1
x
0
j
−
x
000
j
+1
Z
x
0
j
y
00
(
x
)
y
000
(
x
)
dx <
0
,
since
y
00
(
x
)
y
000
(
x
)
>
0
for all
x
∈
x
0
j
, x
000
j
+1
and
y
0
(
x
0
j
) =
y
000
(
x
000
j
+1
) = 0
.
This gives the first of inequalities (9), whereas the rest inequalities follow
from
y
(
x
)
y
0
(
x
)
>
0
on the interval
x
000
j
+1
, x
0
j
+1
.
Similarly, for the first of inequalities (10) we have
y
0
(
x
00
j
)
2
−
y
0
(
x
j
)
2
=
=
−
2
x
j
Z
x
00
j
y
0
(
x
)
y
00
(
x
)
dx
=
−
2
y
(
x
)
y
00
(
x
)
x
j
x
00
j
+ 2
x
j
Z
x
00
j
y
(
x
)
y
000
(
x
)
dx <
0
, since
y
(
x
j
) =
y
00
(
x
00
j
) = 0
and
y
(
x
)
y
000
(
x
)
<
0
on
x
00
j
, x
j
.
The rest ones follow
from the inequality
y
0
(
x
)
y
00
(
x
)
>
0
on
x
j
, x
00
j
+1
.
In the same way, for the first of (11) we have
y
00
(
x
000
j
)
2
−
y
00
(
x
0
j
)
2
=
−
2
x
0
j
Z
x
000
j
y
00
(
x
)
y
000
(
x
)
dx
=
=
−
2
y
0
(
x
)
y
000
(
x
)
x
0
j
x
000
j
+ 2
x
0
j
Z
x
000
j
y
0
(
x
)
y
IV
(
x
)
dx <
0
,
ISSN 1812-3368. Вестник МГТУ им. Н.Э. Баумана. Сер. “Естественные науки”. 2015. № 2
13