Open Access

On power and non-power asymptotic behavior of positive solutions to Emden-Fowler type higher-order equations

Advances in Difference Equations20132013:220

https://doi.org/10.1186/1687-1847-2013-220

Received: 1 March 2013

Accepted: 21 June 2013

Published: 23 July 2013

Abstract

For the equation

y ( n ) = y k , k > 1 , n = 12 , 13 , 14 ,

the existence of positive solutions with non-power asymptotic behavior is proved, namely

y = ( x x ) α h ( log ( x x ) ) , α = n k 1 , x < x ,

where x is an arbitrary point, h is a positive periodic non-constant function on R.

To prove this result, the Hopf bifurcation theorem is used.

Keywords

asymptotic behaviorEmden-Fowler higher-order equations

Introduction

For the equation
y ( n ) = p ( x , y , y , , y ( n 1 ) ) | y | k sgn y , n 2 , k > 1 ,
(1)
Kiguradze posed the problem on the asymptotic behavior of its positive solutions such that
lim x x 0 y ( x ) = .
(2)
He found an asymptotic formula for these solutions to (1) with n = 2 (see [1]) and supposed all such solutions to have power asymptotic behavior for other n, too. The problem was solved for n = 3 and n = 4 [2]. For these n, it was proved that all such solutions behave as
y ( x ) = C ( x x ) α ( 1 + o ( 1 ) ) , x x 0 ,
(3)
with
α = n k 1 , C = ( α ( α + 1 ) ( α + n 1 ) p 0 ) 1 k 1 ,
(4)

p 0 = const > 0 - is a limit of p ( x , y 0 , , y n 1 ) as x x 0 , y 0 , , y n 1 .

So, the hypothesis of Kiguradze was confirmed in this case.

The existence of solutions satisfying (3) was proved for arbitrary n 2 . For 2 n 11 , an ( n 1 ) -parametric family of such solutions to equation (1) was proved to exist (see [2], [3], Ch.I(5.1)).

For the equation
y ( n ) = y k , k > 1 ,
(5)
a negative answer to the conjecture of Kiguradze for large n was obtained. It was proved [4] that for any N and K > 1 , there exist an integer n > N and k R , 1 < k < K , such that equation (5) has a solution
y = ( x x ) α h ( log ( x x ) ) ,
(6)

where α is defined by (4), h is a positive periodic non-constant function on R.

Still, it was not clear how large n should be for the existence of that type of solutions.

Preliminary results

Suppose the following conditions hold:
  1. (A)
    The continuous positive function p ( x , y 0 , , y n 1 ) has a limit p 0 = const > 0 as x x 0 , y 0 , , y n 1 , and for some γ > 0 , it holds
    p ( x , y 0 , , y n 1 ) p 0 = O ( ( x x ) γ + j = 0 n 1 y j γ ) .
    (7)
     
  2. (B)
    For some K 1 > 0 and μ > 0 in a neighborhood of x for sufficiently large y 0 , , y n 1 , z 0 , , z n 1 , it holds
    | p ( x , y 0 , , y n 1 ) p ( x , z 0 , , z n 1 ) | K 1 max j | y j μ z j μ | .
    (8)
     
Then equation (1) can be transformed (see [2] or [3], Ch.I(5.1)) by using the substitution
x x = e t , y = ( C + v ) e α t ,
(9)
where C and α are defined by (4). The derivatives y ( j ) , j = 0 , 1 , , n 1 , become
e ( α + j ) t L j ( v , v , , v ( j ) ) ,
where v ( j ) = d j v d t j and L j is a linear function with
L j ( 0 , 0 , , 0 ) = C α ( α + 1 ) ( α + j 1 ) 0

and the coefficient of v ( j ) equal to 1.

Thus (1) is transformed into
e ( α + n ) t L n ( v , v , , v ( n ) ) = ( C + v ) k e α k t p ˜ ( t , v , v , , v ( n 1 ) ) ,
(10)

where the function p ˜ ( t , v 0 , , v n 1 ) is obtained from p ( x , y 0 , , y n 1 ) with x , y 0 , , y n 1 properly expressed in terms of t , v 0 , , v n 1 . This function tends to p 0 as t , v 0 , , v ( n 1 ) 0 .

Due to condition (8) for the function p ( x , y 0 , , y n 1 ) , we obtain the following inequalities for sufficiently large t and sufficiently small v 0 , , v n 1 , w 0 , , w n 1 :
| p ˜ ( t , v 0 , , v n 1 ) p ˜ ( t , w 0 , , w n 1 ) | K 1 max j e μ ( α + j ) t | L j μ ( v 0 , , v n 1 ) L j μ ( w 0 , , w n 1 ) | .
Since L j ( 0 , 0 , , 0 ) 0 , the function L j μ is a C one in a neighborhood of 0 and
| p ˜ ( t , v 0 , , v n 1 ) p ˜ ( t , w 0 , , w n 1 ) | K 2 e μ α t max j | v j w j |

for some K 2 > 0 .

Solving (10) for v ( n ) and using formulae (4), we obtain the equation
v ( n ) = ( C + v ) k p ˜ ( t , v , v , , v ( n 1 ) ) p 0 C k j = 0 n 1 a j v ( j ) ,
(11)
where a j are the coefficients of the linear function L n . Equation (11) can be written as
v ( n ) = k C k 1 p 0 v j = 0 n 1 a j v ( j ) + f ( v ) + g ( t , v , v , , v ( n 1 ) ) ,
(12)
where
f ( v ) = p o ( ( C + v ) k C k k C k 1 v ) = O ( v 2 ) as  v 0 , f ( v ) = O ( v ) as  v 0 , g ( t , v 0 , , v n 1 ) = ( C + v 0 ) k ( p ˜ ( t , v 0 , , v n 1 ) p 0 ) = O ( e γ t + j = 0 n 1 e γ ( α + j ) t ) = O ( exp ( γ min ( α , 1 ) t ) ) as  t , v 0 0 , , v n 1 0 .
Besides, for sufficiently large t and sufficiently small v 0 , , v n 1 , w 0 , , w n 1 , it holds
| g ( t , v 0 , , v n 1 ) g ( t , w 0 , , w n 1 ) | | ( C + v 0 ) k ( C + w 0 ) k | | p ˜ ( t , v 0 , , v n 1 ) p 0 | + ( C + w 0 ) k | p ˜ ( t , v 0 , , v n 1 ) p ˜ ( t , w 0 , , w n 1 ) | K 3 max j | w j v j | e min ( γ , μ ) min ( α , 1 ) t .
Suppose that V is the vector with coordinates V j = v ( j ) , j = 0 , , n 1 . Then equation (12) can be written as
d V d t = A V + F ( V ) + G ( t , V ) ,
(13)
where A is a constant n × n matrix
A = ( 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 a ˜ 0 a 1 a 2 a 3 a n 1 ) ,
with
a ˜ 0 = a 0 k c k 1 p 0 = a 0 k α ( α + 1 ) ( α + n 1 ) = a 0 ( α + 1 ) ( α + n 1 ) ( α + n )
and eigenvalues satisfying the equation
0 = det ( A λ E ) = ( 1 ) n + 1 ( a ˜ 0 a 1 λ a n 1 λ n 1 λ n ) = ( 1 ) n + 1 ( ( α + 1 ) ( α + 2 ) ( α + n ) ( λ + α ) ( λ + α + n 1 ) ) ,
which is equivalent to
j = 0 n 1 ( λ + α + j ) = j = 0 n 1 ( 1 + α + j ) .
(14)
The mappings F : R n R n and G : R × R n R n satisfy the following estimates as t :
{ F ( V ) = O ( V 2 ) , F V ( V ) = O ( V ) , G ( t , V ) = O ( e 2 β t ) , G ( t , V ) G ( t , W ) K V W e 2 β t
(15)

with some constants β > 0 , K > 0 .

Lemma 1 [3]

Suppose that (15) holds and A is an arbitrary constant n × n matrix. Then there exists a solution V ( t ) to equation (13) tending to zero as t .

Lemma 2 [3]

Let the conditions of Lemma  1 hold. If equation (14) has m roots with negative real part, then there exists an m-parametric family of solutions V ( t ) to equation (13) tending to zero as t .

If equation (13) has a solution V ( t ) tending to 0 as t and V 0 ( t ) is its first coordinate, then the function
y ( x ) = ( V 0 ( log ( x x ) ) + C ) ( x x ) α

with C and α defined by (4) is a solution to (1) such that (2) and (3) hold.

Theorem 1 [2, 3]

Suppose that conditions (A) and (B) are satisfied. Then for such x there exists a solution to (1) with power asymptotic behavior (3).

Investigating the signs of the real parts of the roots of equation (14), by the Routh-Hurwitz criterion, we can prove the following theorem.

Theorem 2 [2, 3]

Suppose that 3 n 11 and conditions (A) and (B) are satisfied. Then there exists an ( n 1 ) -parametric family of solutions to equation (1) with power asymptotic behavior (3).

Theorem 3 [2, 3, 5]

Suppose that n = 3 or n = 4 in equation (1), the continuous positive function p ( x , y 0 , , y n 1 ) is Lipschitz continuous in y 0 , , y n 1 and has a limit p 0 > 0 as x x 0 , y 0 , , y n 1 . Then any positive solution to this equation with a vertical asymptote x = x has asymptotic behavior (3).

To prove the main results of this article, we use the Hopf bifurcation theorem [6].

Theorem (Hopf)

Consider the α-parameterized dynamical system x ˙ = L α x + Q α ( x ) in a neighborhood of 0 R n with linear operators L α and smooth enough functions Q α ( x ) = O ( | x | 2 ) as x 0 . Let λ α and λ ¯ α be simple complex conjugated eigenvalues of the operators  L α . Suppose that Re λ α ˜ = Re λ ¯ α ˜ = 0 for some α ˜ and the operator L α ˜ has no other eigenvalues with zero real part.

If Re d λ α d α ( α ˜ ) 0 , then there exist continuous mappings ε α ( ε ) R , ε T ( ε ) R , and ε b ( ε ) R n defined in a neighborhood of 0 and such that α ( 0 ) = α ˜ , T ( 0 ) = 2 π / Im λ α ˜ , b ( 0 ) = 0 , b ( ε ) 0 for ε 0 , and the solutions to the problems
x ˙ = L α ( ε ) x + Q α ( ε ) ( x ) , x ( 0 ) = b ( ε )

are T ( ε ) -periodic and non-constant.

Main results

In this section, the result about the existence of solutions with non-power asymptotic behavior is proved for equation (5) with n = 12 , 13 , 14 .

Theorem 4 For n = 12 , 13 , 14 , there exists k > 1 such that equation (5) has a solution y ( x ) with
y ( j ) ( x ) = ( x x ) α j h j ( log ( x x ) ) , j = 0 , 1 , , n 1 ,

where α is defined by (4) and h j are periodic positive non-constant functions on R.

Proof To apply the Hopf bifurcation theorem, we investigate equation (13) with G ( t , V ) 0 corresponding to the case of the constant function p and the roots of the algebraic equation (14). F is a vector function with all zero components F ( V ) = ( 0 , , 0 , F n 1 ( V ) ) , V = ( V 0 , , V n 1 ) , and
F n 1 ( V ) = ( ( C + V 0 ) k C k k C k 1 V 0 ) = O ( V 0 2 ) , V 0 0 , d d V F n 1 ( V ) = O ( | V 0 | ) , V 0 0 .

If equation (14) has a pair of pure imaginary roots, we have to check other conditions of this theorem and then apply it.

Proposition 1 For any integer n > 11 , there exist α > 0 and q > 0 such that
j = 0 n 1 ( q i + α + j ) = j = 0 n 1 ( 1 + α + j )
(16)

with i 2 = 1 .

Remark 1 In the particular case n = 12 , this result was obtained by Vyun [7].

Proof Consider the positive functions ρ n ( α ) and σ n ( α ) defined for all α > 0 via the equations
j = 0 n 1 ( ρ n ( α ) 2 + ( α + j ) 2 ) = j = 0 n 1 ( 1 + α + j ) 2
(17)
and
j = 0 n 1 arg ( σ n ( α ) i + α + j ) = 2 π
(18)

supposing arg z [ 0 , 2 π ) for all z C { 0 } .

First, we prove the functions to be well defined for all α > 0 .

The product j = 0 n 1 ( q 2 + ( α + j ) 2 ) is continuous and strictly increasing as a function of q > 0 .

It tends to j = 0 n 1 ( α + j ) 2 < j = 0 n 1 ( 1 + α + j ) 2 as q 0 and to +∞ as q + . Hence, for any α > 0 , there exists a unique q > 0 such that j = 0 n 1 ( q 2 + ( α + j ) 2 ) = j = 0 n 1 ( 1 + α + j ) 2 .

In the same way, for any α > 0 , the sum j = 0 n 1 arg ( q i + α + j ) is a continuous function of q > 0 strictly increasing from 0 to π n 2 > 2 π . So, there exists a unique q > 0 such that the sum is equal to 2π.

Since both the product and the sum considered are C 1 -functions with positive partial derivative in q > 0 , the implicit function theorem provides both ρ n ( α ) and σ n ( α ) to be C 1 -functions, too.

Now it is sufficient to prove the existence of α > 0 such that ρ n ( α ) and σ n ( α ) are equal to the same value q, which makes the two sides of (16) be equal.

Compare the functions ρ n ( α ) and σ n ( α ) near the boundaries of their common domain.

Equation (17) defining the function ρ n ( α ) may be written as
j = 0 n 1 ( 1 + 2 j α + j 2 α 2 + ( ρ n ( α ) α ) 2 ) = j = 0 n 1 ( 1 + j + 1 α ) 2 .

This shows that ρ n ( α ) α 0 as α + .

Equation (18) defining the function σ n ( α ) may be written as
j = 0 n 1 arctan σ n ( α ) α 1 + j α = 2 π .

This shows that σ n ( α ) α tan 2 π n > 0 as α + . Thus, ρ n ( α ) < σ n ( α ) for sufficiently large α.

Now, to prove Proposition 1, it is sufficient to show that ρ n ( α ) > σ n ( α ) for sufficiently small α. To compare the functions ρ n ( α ) and σ n ( α ) for small α > 0 , we need some lemmas.

Lemma 3 For all α > 0 , it holds ρ n ( α ) 2 < 2 ( α + n ) 1 .

Proof Suppose that ρ n ( α ) 2 2 ( α + n ) 1 for some α > 0 . Then
j = 0 n 1 ( ρ n ( α ) 2 + ( α + j ) 2 ) j = 0 n 1 ( 2 ( α + n ) 1 + ( α + j ) 2 ) > j = 0 n 1 ( 2 ( α + j + 1 ) 1 + ( α + j ) 2 ) = j = 0 n 1 ( 1 + ( α + j ) ) 2 .

This contradiction with the definition of ρ n ( α ) completes the proof of Lemma 3. □

Lemma 4 For all α > 0 , it holds ρ n + 1 ( α ) > ρ n ( α ) .

Proof According to the definition of ρ n ( α ) by (17) and Lemma 3, we have
j = 0 n ( ρ n ( α ) 2 + ( α + j ) 2 ) = j = 0 n 1 ( 1 + α + j ) 2 ( ρ n ( α ) 2 + ( α + n ) 2 ) < j = 0 n 1 ( 1 + α + j ) 2 ( 2 ( α + n ) 1 + ( α + n ) 2 ) < j = 0 n ( 1 + α + j ) 2 .

In order to make the first and the last products be equal, we have to replace ρ n ( α ) in the first one by a greater value. This means that ρ n + 1 ( α ) > ρ n ( α ) and Lemma 4 is proved. □

Lemma 5 For all α > 0 , it holds σ n + 1 ( α ) < σ n ( α ) .

Proof According to the definition of σ n ( α ) by (18), we have
j = 0 n arg ( σ n ( α ) i + α + j ) = 2 π + arg ( σ n ( α ) i + α + n ) > 2 π .

In order to make the sum equal 2π, we have to replace σ n ( α ) by a smaller value. So, σ n + 1 ( α ) < σ n ( α ) and Lemma 5 is proved. □

Due to Lemmas 3, 4, 5 proved, it is sufficient now for the proof of Proposition 1 to show that ρ 12 ( α ) > σ 12 ( α ) for sufficiently small α > 0 .

Lemma 6 It holds ρ 12 ( α ) > 2 for all sufficiently small α > 0 .

Proof Straightforward exact calculations show that
lim α 0 j = 0 11 ( 2 2 + ( α + j ) 2 ) = j = 0 11 ( 4 + j 2 ) = 192 , 175 , 659 , 520 , 000 , 000 < 2 10 17
and
lim α 0 j = 0 11 ( 1 + α + j ) 2 = ( 12 ! ) 2 = 229 , 442 , 532 , 802 , 560 , 000 > 2 10 17 .
So, for sufficiently small α > 0 , we have
j = 0 11 ( 2 2 + ( α + j ) 2 ) < 2 10 17 < j = 0 11 ( 1 + α + j ) 2 .

Hence, for these α, in order to avoid contradiction with the definition of ρ 12 ( α ) , the inequality ρ 12 ( α ) 2 > 2 2 is necessary. Lemma 6 is proved. □

Lemma 7 It holds σ 12 ( α ) < 2 for sufficiently small α > 0 .

Proof Consider the limit
lim α 0 j = 0 11 arg ( 2 i + α + j ) = arg 2 i + arctan 2 + arctan 1 + arctan 2 3 + arctan 1 2 + j = 5 11 arctan 2 j = 5 π 4 + arctan 2 3 + j = 5 11 arctan 2 j = 5 π 4 + arctan 2 3 + 2 5 1 2 3 2 5 + arctan 2 6 + 2 7 1 2 6 2 7 + arctan 2 8 + 2 9 1 2 8 2 9 + arctan 2 10 + 2 11 1 2 10 2 11 = 5 π 4 + arctan 16 11 + arctan 13 19 + arctan 1 2 + arctan 21 53 = 5 π 4 + arctan 16 11 + 13 19 1 16 11 13 19 + arctan 1 2 + 21 53 1 1 2 21 53 = 5 π 4 + arctan 447 + arctan 19 17 .
Note that
tan ( arctan 447 + arctan 19 17 ) = 447 + 19 17 1 447 19 17 = 3 , 809 4 , 238 .

Hence, arctan 447 + arctan 19 17 > 3 π 4 and j = 0 11 arg ( 2 i + α + j ) > 2 π for sufficiently small α > 0 . Thus, for these α, we have σ 12 ( α ) < 2 , which completes the proof of Lemma 7. □

Now Proposition 1 is also proved.  □

Proposition 2 For any α > 0 and any integer n > 1 , all roots λ C to equation (14) are simple.

Proof Since we consider a polynomial equation of degree n, it is sufficient to prove the existence of n different roots to (14). We will show that for any integer m such that n < m n , there exists μ m C satisfying
j = 0 n 1 | μ m + j | = j = 0 n 1 ( 1 + α + j )
(19)
and
j = 0 n 1 arg ( μ m + j ) = m π
(20)

with argz denoting the principal value of the argument lying in the open-closed interval ( π , π ] . Surely, all these 2n complex numbers μ m are different. Those with even m generate, via the relation λ m + α = μ m , just n different roots λ m to (14).

We begin to accomplish this plan by noting that the set of μ satisfying equation (20) with m = 0 is the real semi-axis ( 0 , + ) containing a single point satisfying (19), namely μ 0 = 1 + α .

Similarly, the set of μ satisfying equation (20) with m = n is the real unbounded interval ( , 1 n ) containing a single point satisfying (19), namely μ n = α n .

Now consider the cases 0 < m < n and the upper complex half-plane. For any ω > 0 , the smooth function
ϕ ω ( r ) = j = 0 n 1 arg ( r + ω i + j ) = j = 0 n 1 arccot r + j ω

monotonically decreases from to 0 as r increases from −∞ to +∞. So, for any ω > 0 and b ( 0 , n π ) , there exists a unique value r such that ϕ ω ( r ) = b . Due to the inequality d ϕ ω d r ( r ) < 0 , the implicit function theorem provides the existence of the smooth functions r m ( ω ) satisfying ϕ ω ( r m ( ω ) ) = m π .

Note that if r m , then r + j < 0 for all j < m and r + m 0 . Hence,
lim _ ω + 0 j = 0 n 1 arccot r + j ω lim _ ω + 0 j = 0 m 1 arccot r + j ω + lim _ ω + 0 arccot 0 ω = m π + π 2 > m π

and such r cannot be the value of r m ( ω ) for sufficiently small ω > 0 .

Similarly, if r 1 m , then r + j > 0 for all j > m 1 and r + m 1 0 . Hence,
lim ω + 0 j = 0 n 1 arccot r + j ω lim ω + 0 j = 0 m 2 arccot r + j ω + π 2 + lim ω + 0 j = m n 1 arccot r + j ω ( m 1 ) π + π 2 + 0 < m π

and such r cannot be the value of r m ( ω ) for sufficiently small ω > 0 .

So, if ω > 0 is sufficiently small, then r m ( ω ) satisfies the inequality m < r m ( ω ) < 1 m and thereby is negative.

Consider the product j = 0 n 1 | r m ( ω ) + ω i + j | with 0 < m < n and investigate its behavior for small ω > 0 .

If j m , then for sufficiently small ω > 0 , we have | r m ( ω ) + j | = r m ( ω ) + j < j and
j = m n 1 | r m ( ω ) + j | j = m n 1 j < j = m n 1 ( 1 + j ) .
(21)
If j m 1 , then for sufficiently small ω > 0 , we have | r m ( ω ) + j | = r m ( ω ) j < m j = 1 + ( m 1 j )
j = 0 m 1 | r m ( ω ) + j | j = 0 m 1 | 1 + ( m 1 j ) | = J = 0 m 1 ( 1 + J ) , J = m 1 j .
(22)
Combining (21) and (22), we obtain, for sufficiently small ω > 0 ,
j = 0 n 1 | r m ( ω ) + j | < j = 0 n 1 ( 1 + j ) ,
and
j = 0 n 1 | r m ( ω ) + ω i + j | < j = 0 n 1 ( 1 + α + j ) .
As for large ω, the left-hand side of the above inequality evidently tends to +∞ as ω + and hence is greater than its right-hand side for sufficiently large ω. By continuity there exists ω m > 0 such that
j = 0 n 1 | r m ( ω m ) + ω m i + j | = j = 0 n 1 ( 1 + α + j ) .

Thus, we can take μ m = r m ( ω m ) + ω m i C to satisfy (19) and (20) for 0 < m < n . For n < m < 0 , we can take the conjugates μ m = μ m ¯ . Thus, the existence of all μ m needed is proved. This completes the proof of Proposition 2. □

Lemma 8 If 12 n 14 , α > 0 , and q > 0 satisfy the polynomial equation
j = 0 n 1 ( ( α + j ) 2 + q 2 ) = j = 0 n 1 ( α + j + 1 ) 2 ,

then 2 α + 4 < q 2 < 3 α + 5 .

Proof It can be proved in the same way for all n mentioned. We show this for n = 12 .

First, compute the right-hand side of the equation:
j = 0 11 ( α + j + 1 ) 2 = α 24 + 156 α 23 + 11 , 518 α 22 + 535 , 392 α 21 + 17 , 581 , 135 α 20 + 433 , 823 , 676 α 19 + 8 , 353 , 410 , 208 α 18 + 128 , 665 , 048 , 512 α 17 + 1 , 612 , 229 , 817 , 055 α 16 + 16 , 625 , 859 , 652 , 116 α 15 + 142 , 196 , 061 , 481 , 318 α 14 + 1 , 013 , 438 , 536 , 648 , 512 α 13 + 6 , 032 , 418 , 472 , 347 , 265 α 12 + 29 , 989 , 851 , 619 , 249 , 236 α 11 + 124 , 253 , 074 , 219 , 885 , 468 α 10 + 427 , 135 , 043 , 298 , 835 , 872 α 9 + 1 , 209 , 806 , 045 , 835 , 003 , 760 α 8 + 2 , 795 , 060 , 589 , 044 , 133 , 696 α 7 + 5 , 194 , 030 , 186 , 679 , 450 , 688 α 6 + 7 , 613 , 724 , 634 , 416 , 755 , 712 α 5 + 8 , 564 , 233 , 279 , 835 , 510 , 784 α 4 + 7 , 096 , 936 , 674 , 284 , 421 , 120 α 3 + 4 , 059 , 952 , 667 , 309 , 260 , 800 α 2 + 1 , 424 , 017 , 035 , 657 , 216 , 000 α + 229 , 442 , 532 , 802 , 560 , 000 .
Now, estimate the left-hand side supposing q 2 3 α + 5 > 0 :
j = 0 11 ( ( α + j ) 2 + q 2 ) j = 0 11 ( ( α + j ) 2 + 3 α + 5 ) α 24 + 168 α 23 + 13 , 216 α 22 + 647 , 658 α 21 + 22 , 191 , 136 α 20 + 565 , 650 , 624 α 19 + 11 , 143 , 609 , 279 α 18 + 174 , 022 , 752 , 156 α 17 + 2 , 192 , 303 , 359 , 180 α 16 + 22 , 557 , 120 , 652 , 044 α 15 + 191 , 221 , 185 , 335 , 728 α 14 + 1 , 343 , 463 , 278 , 373 , 840 α 13 + 7 , 851 , 135 , 965 , 424 , 751 α 12 + 38 , 226 , 775 , 470 , 470 , 448 α 11 + 155 , 030 , 143 , 411 , 290 , 136 α 10 + 522 , 520 , 458 , 095 , 057 , 994 α 9 + 1 , 457 , 064 , 439 , 886 , 002 , 624 α 8 + 3 , 337 , 255 , 633 , 900 , 992 , 816 α 7 + 6 , 209 , 925 , 089 , 367 , 687 , 345 α 6 + 9 , 237 , 499 , 888 , 429 , 090 , 764 α 5 + 10 , 723 , 421 , 856 , 201 , 549 , 372 α 4 + 9 , 360 , 016 , 963 , 404 , 522 , 912 α 3 + 5 , 777 , 193 , 048 , 791 , 013 , 360 α 2 + 2 , 247 , 088 , 906 , 508 , 241 , 600 α + 413 , 920 , 896 , 501 , 672 , 000 .
The difference of this polynomial and the previous one is equal to
j = 0 11 ( ( α + j ) 2 + 3 α + 5 ) j = 0 11 ( α + j + 1 ) 2 = 12 α 23 + 1 , 698 α 22 + 112 , 266 α 21 + 4 , 610 , 001 α 20 + 131 , 826 , 948 α 19 + 2 , 790 , 199 , 071 α 18 + 45 , 357 , 703 , 644 α 17 + 580 , 073 , 542 , 125 α 16 + 5 , 931 , 260 , 999 , 928 α 15 + 49 , 025 , 123 , 854 , 410 α 14 + 330 , 024 , 741 , 725 , 328 α 13 + 1 , 818 , 717 , 493 , 077 , 486 α 12 + 8 , 236 , 923 , 851 , 221 , 212 α 11 + 30 , 777 , 069 , 191 , 404 , 668 α 10 + 95 , 385 , 414 , 796 , 222 , 122 α 9 + 247 , 258 , 394 , 050 , 998 , 864 α 8 + 542 , 195 , 044 , 856 , 859 , 120 α 7 + 1 , 015 , 894 , 902 , 688 , 236 , 657 α 6 + 1 , 623 , 775 , 254 , 012 , 335 , 052 α 5 + 2 , 159 , 188 , 576 , 366 , 038 , 588 α 4 + 2 , 263 , 080 , 289 , 120 , 101 , 792 α 3 + 1 , 717 , 240 , 381 , 481 , 752 , 560 α 2 + 823 , 071 , 870 , 851 , 025 , 600 α + 184 , 478 , 363 , 699 , 112 , 000 ,

which is positive for any α 0 . This shows that the polynomial equation cannot be satisfied by α > 0 and q > 0 with q 2 3 α + 5 .

In the same way, compute
j = 0 11 ( α + j + 1 ) 2 j = 0 11 ( ( α + j ) 2 + 2 α + 4 ) = 96 α 22 + 13 , 156 α 21 + 844 , 624 α 20 + 33 , 778 , 316 α 19 + 943 , 838 , 852 α 18 + 19 , 590 , 096 , 240 α 17 + 313 , 464 , 915 , 984 α 16 + 3 , 960 , 996 , 926 , 744 α 15 + 40 , 162 , 617 , 066 , 616 α 14 + 330 , 203 , 929 , 721 , 796 α 13 + 2 , 215 , 299 , 128 , 334 , 800 α 12 + 12 , 163 , 303 , 361 , 220 , 828 α 11 + 54 , 651 , 209 , 110 , 677 , 476 α 10 + 200 , 323 , 721 , 839 , 107 , 240 α 9 + 595 , 229 , 721 , 350 , 941 , 648 α 8 + 1 , 419 , 051 , 246 , 703 , 474 , 880 α 7 + 2 , 673 , 079 , 829 , 956 , 829 , 568 α 6 + 3 , 889 , 993 , 689 , 940 , 050 , 432 α 5 + 4 , 228 , 750 , 706 , 659 , 177 , 984 α 4 + 3 , 257 , 831 , 645 , 648 , 401 , 920 α 3 + 1 , 625 , 109 , 784 , 526 , 284 , 800 α 2 + 437 , 271 , 322 , 981 , 376 , 000 α + 37 , 266 , 873 , 282 , 560 , 000 .

Hence, j = 0 11 ( α + j + 1 ) 2 > j = 0 11 ( ( α + j ) 2 + q 2 ) if 2 α + 4 q 2 .

This contradiction yields 2 α + 4 < q 2 < 3 α + 5 . So, Lemma 8 is proved. □

The condition Re d λ α d α ( α ˜ ) 0 needed for the Hopf theorem, expressed explicitly by means of the implicit function theorem, looks like
[ j = 0 n 1 α + j q 2 + ( α + j ) 2 ] 2 + [ j = 0 n 1 q q 2 + ( α + j ) 2 ] 2 j = 0 n 1 α + j q 2 + ( α + j ) 2 j = 0 n 1 1 1 + α + j .
Lemma 9 If 12 n 14 , α > 0 and 0 < q 2 < 3 α + 5 , then
[ j = 0 n 1 α + j q 2 + ( α + j ) 2 ] 2 + [ j = 0 n 1 q q 2 + ( α + j ) 2 ] 2 > j = 0 n 1 α + j q 2 + ( α + j ) 2 j = 0 n 1 1 1 + α + j .
(23)

Proof Hereafter all sums and products with no limits indicated are over j = 0 , 1 , , n 1 .

Multiplying inequality (23) by U = ( 1 + α + j ) and then twice by V = [ q 2 + ( α + j ) 2 ] , we obtain the following equivalent inequality provided α > 0 :
U [ ( ( α + j ) V j ) 2 + q 2 ( V j ) 2 ] > V ( α + j ) V j U j
(24)

with the polynomials U j = U 1 + α + j and V j = V q 2 + ( α + j ) 2 .

Put q 2 = 3 α + 5 1 + w , w > 0 . Substituting this into inequality (24) and multiplying the result by ( 1 + w ) 2 n 1 , we obtain another equivalent one:
U [ ( 1 + w ) ( ( α + j ) P j ) 2 + ( 3 α + 5 ) ( P j ) 2 ] > P ( α + j ) P j U j
(25)

with P = [ 3 α + 5 + ( 1 + w ) ( α + j ) 2 ] and P j = P 3 α + 5 + ( 1 + w ) ( α + j ) 2 .

Both sides of inequality (25) are polynomials of α and w with non-negative integer coefficients. So, they can be computed exactly, with no rounding. This rather cumbersome computation gives the following result for the difference of the left- and right-hand sides of (25) expressed as
U [ ( 1 + w ) ( ( α + j ) P j ) 2 + ( 3 α + 5 ) ( P j ) 2 ] P ( α + j ) P j U j = j = 0 5 n 2 Δ j α j
(26)

with polynomials Δ j R [ w ] . Straightforward though very cumbersome calculations show that Δ 5 n 2 = 0 , and all other Δ j in (26) are polynomials with positive coefficients.

This completes the proof of Lemma 9. □

To apply the Hopf bifurcation theorem, we need to check that equation (14) cannot have more than a single pair of imaginary conjugated roots. It can be easily obtained by considering equation (16).

Now, the Hopf bifurcation theorem and the lemmas proved provide, for n = 12 , 13 , 14 , the existence of a family α ε > 0 such that equation (14) with α = α 0 has imaginary roots λ = ± q i and for sufficiently small ε, system (13) with α = α ε has a periodic solution V ε ( t ) with period T ε T = 2 π q as ε 0 . In particular, the coordinate V ε , 0 ( t ) = v ( t ) of the vector V ε ( t ) is also a periodic function with the same period. Then, taking into account (9), we obtain
y ( x ) = ( C + v ( ln ( x x ) ) ) ( x x ) α .
Put h ( s ) = C + v ( s ) , which is a non-constant continuous periodic and positive for sufficiently small ε function and obtain the required equality
y ( x ) = ( x x ) α h ( ln ( x x ) ) .

In the similar way, we obtain the related expressions for y ( j ) ( x ) , j = 0 , 1 , , n 1 .

Theorem 4 is proved. □

Conclusions, concluding remarks and open problems

  1. 1.

    Computer calculations give approximate values of α providing equation (14) to have a pure imaginary root λ. They are, with corresponding values of k, as follows:

     

if n = 12 , then α 0.56 , k 22.4 ;

if n = 13 , then α 1.44 , k 10.0 ;

if n = 14 , then α 2.37 , k 6.9 .
  1. 2.

    Note that equation (14) has no pure imaginary roots if n 11 . So, the Hopf bifurcation theorem cannot be applied, but it does not follow that Theorem 4 cannot be proved for some n < 12 .

     
  2. 3.

    Equation (5) with n = 3 has solutions of type (6) with oscillatory h (see [3, 5]).

     
  3. 4.

    If n 15 , then the inequality needed for the Hopf bifurcation theorem Re d λ α d α ( α ˜ ) 0 cannot be proved in the same way because the estimate q 2 < 3 α + 5 does not hold.

     

Declarations

Acknowledgements

The research was supported by RFBR (grant 11-01-00989).

Authors’ Affiliations

(1)
Department of Mechanics and Mathematics, Lomonosov Moscow State University, Moscow, Russia
(2)
Department of Higher Mathematics, Moscow State University of Economics, Statistics and Informatics, Moscow, Russia

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