Об асимптотической классификации решений не линейных уравнений третьего и четвертого порядков со степенной нелинейностью

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to (1) and generates a trajectory completely lying in Ω+. So, we have to
prove existence of a trajectory of the first type.
Assume the converse. Then any trajectory passing through a point
s ∈ Ω− ∩ S3 must reach the boundary ∂Ω − ∩ S3. Thus we obtain the
mapping Ω− ∩ S3 → ∂Ω− ∩ S3.
To prove its continuity we represent it as s∈ Ω−∩S3 7→ Trajp0 ( s, ξ( s)) ∈
∈ ∂Ω− ∩ S3.
Here Trajp0 (s, t) is the point in S3 reached at the time t by the trajectory
of the dynamical system on the sphere that passed s at the time 0 . The
mapping Trajp0 : S3 × R → S3 is continuous according to the general
properties of differential equations.
The function ξ : Ω− ∩ S3 → R gives the time t at which the trajectory
passing through the given point of Ω− at t0 = 0 reaches ∂Ω−. Now we
prove continuity of ξ.
Suppose ξ(s 1) = t1 and ε > 0. Then, since Trajp0
(s 1, t1 + ε ) is inside
Ω+, there exists a neighborhood U + of s1 such that for any s ∈ U+ the
point Trajp0
(s, t1 + ε) is also inside Ω+. So, we have ξ( s ) < t 1 + ε for all
s ∈ U+.
Similarly, since Trajp0
(s1 , t1 − ε ) is inside Ω−, there exists a neighbor-
hood U− of s 1 such that for any s ∈ U− the point Trajp0 (s, t1 − ε) is also
inside Ω−, whence ξ(s ) > t1 − ε.
So, for all s ∈ U− ∩U+ we have |ξ(s) − t1| < ε. Thus ξ ( s) is continuous
on Ω− ∩ S3 and we have the continuous mapping Ω− ∩ S3 → ∂Ω− ∩ S3
whose restriction to ∂Ω− ∩ S 3 is the identity map. In other words, we
have the composition ∂Ω− ∩ S3 → Ω − ∩ S 3 → ∂Ω− ∩ S3, which is
the identity map, inducing the identity map on the homology groups:
H2(∂ Ω− ∩ S3) → H2(Ω − ∩ S3) → H 2(∂ Ω− ∩ S3).
Since Ω− ∩ S3 and ∂ Ω− ∩ S3 are homeomorphic to the solid torus
and the torus surface respectively, the above composition can be written
as Z → 0 → Z, which cannot be the identity mapping. This contradiction
proves the lemma. I
Lemma 5. Suppose y (x) is a non-trivial solution to equation (1)
maximally extended to the right. Then neither y (x) nor any of its derivatives
y0(x), y 00(x), y000 (x) can have constant sign near the right boundary of their
domain.
J We prove it for y (x). For the derivatives the proof is just similar.
Suppose y(x) is defined on an interval (x−, x +) , bounded or not, and is
positive in a neighborhood of x+. Then y000 (x), due to (1), is monotonically
decreasing to a finite or infinite limit as x → x+. Then y000 (x) ultimately
has a constant sign. In the same way, y00 (x), y 0( x), and y (x) itself are all
ultimately monotone and have finite or infinite limits as x → x+.
ISSN 1812-3368. Вестник МГТУ им. Н.Э. Баумана. Сер. “Естественные науки”. 2015. № 2 11