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

## 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.

## Introduction

For the equation

$y ( n ) =p ( x , y , y ′ , … , y ( n − 1 ) ) | y | k sgny,n≥2,k>1,$
(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 ) and supposed all such solutions to have power asymptotic behavior for other n, too. The problem was solved for $n=3$ and $n=4$ . 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 , , 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  that for any N and $K>1$, there exist an integer $n>N$ and $k∈R$, $1, 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  or , 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

$L j (0,0,…,0)=Cα(α+1)⋯(α+j−1)≠0$

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

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

$d V d t =AV+F(V)+G(t,V),$
(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 

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 

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 .

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 $α(0)= α ˜$, $T(0)=2π/Im λ α ˜$, $b(0)=0$, $b(ε)≠0$ 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

$∏ j = 0 n − 1 (qi+α+j)= ∏ j = 0 n − 1 (1+α+j)$
(16)

with $i 2 =−1$.

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

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

$∑ j = 0 n − 1 arg ( σ n ( α ) i + α + j ) =2π$
(18)

supposing $argz∈[0,2π)$ 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 $∑ j = 0 n − 1 arg(qi+α+j)$ is a continuous function of $q>0$ strictly increasing from 0 to $π n 2 >2π$. 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

$∑ j = 0 n − 1 arctan σ n ( α ) α 1 + j α =2π.$

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

$∑ j = 0 n arg ( σ n ( α ) i + α + j ) =2π+arg ( σ n ( α ) i + α + n ) >2π.$

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, $arctan447+arctan 19 17 > 3 π 4$ and $∑ j = 0 11 arg(2i+α+j)>2π$ 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, there exists $μ m ∈C$ satisfying

$∏ j = 0 n − 1 | μ m +j|= ∏ j = 0 n − 1 (1+α+j)$
(19)

and

$∑ j = 0 n − 1 arg( μ m +j)=mπ$
(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 and the upper complex half-plane. For any $ω>0$, the smooth function

$ϕ ω (r)= ∑ j = 0 n − 1 arg(r+ωi+j)= ∑ j = 0 n − 1 arccot r + j ω$

monotonically decreases from to 0 as r increases from −∞ to +∞. So, for any $ω>0$ and $b∈(0,nπ)$, 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 $ϕ ω ( r m (ω))=mπ$.

Note that if $r≤−m$, then $r+j<0$ for all $j and $r+m≤0$. Hence,

$lim _ ω → + 0 ∑ j = 0 n − 1 arccot r + j ω ≥ lim _ ω → + 0 ∑ j = 0 m − 1 arccot r + j ω + lim _ ω → + 0 arccot 0 ω =mπ+ π 2 >mπ$

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 $∏ j = 0 n − 1 | r m (ω)+ωi+j|$ with $0 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 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

$∏ j = 0 m − 1 | r m (ω)+j|≤ ∏ j = 0 m − 1 |1+(m−1−j)|= ∏ J = 0 m − 1 (1+J),J=m−1−j.$
(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

$∏ j = 0 n − 1 | r m (ω)+ωi+j|< ∏ j = 0 n − 1 (1+α+j).$

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. For $−n, 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 $2α+4< q 2 <3α+5$.

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 $q 2 ≥3α+5>0$:

$∏ j = 0 11 ( ( α + j ) 2 + q 2 ) ≥ ∏ j = 0 11 ( ( α + j ) 2 + 3 α + 5 ) ≥ α 24 + 168 α 23 + 13 , 216 α 22 + 647 , 658 α 21 + 22 , 191 , 136 α 20 + 565 , 650 , 624 α 19 + 11 , 143 , 609 , 279 α 18 + 174 , 022 , 752 , 156 α 17 + 2 , 192 , 303 , 359 , 180 α 16 + 22 , 557 , 120 , 652 , 044 α 15 + 191 , 221 , 185 , 335 , 728 α 14 + 1 , 343 , 463 , 278 , 373 , 840 α 13 + 7 , 851 , 135 , 965 , 424 , 751 α 12 + 38 , 226 , 775 , 470 , 470 , 448 α 11 + 155 , 030 , 143 , 411 , 290 , 136 α 10 + 522 , 520 , 458 , 095 , 057 , 994 α 9 + 1 , 457 , 064 , 439 , 886 , 002 , 624 α 8 + 3 , 337 , 255 , 633 , 900 , 992 , 816 α 7 + 6 , 209 , 925 , 089 , 367 , 687 , 345 α 6 + 9 , 237 , 499 , 888 , 429 , 090 , 764 α 5 + 10 , 723 , 421 , 856 , 201 , 549 , 372 α 4 + 9 , 360 , 016 , 963 , 404 , 522 , 912 α 3 + 5 , 777 , 193 , 048 , 791 , 013 , 360 α 2 + 2 , 247 , 088 , 906 , 508 , 241 , 600 α + 413 , 920 , 896 , 501 , 672 , 000 .$

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 $q 2 ≥3α+5$.

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 $2α+4≥ q 2$.

This contradiction yields $2α+4< q 2 <3α+5$. 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 $0< q 2 <3α+5$, 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 $P ∗ =∏[3α+5+(1+w) ( α + j ) 2 ]$ 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 $λ=±qi$ 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 $q 2 <3α+5$ does not hold.

## References

1. 1.

Kiguradze IT, Chanturia TA: Asymptotic Properties of Solutions of Nonautonomous Ordinary Differential Equations. Kluwer Academic, Dordrecht; 1993.

2. 2.

Astashova IV: Asymptotic behavior of solutions of certain nonlinear differential equations. 1(3). In Reports of Extended Session of a Seminar of the I. N. Vekua Institute of Applied Mathematics. Tbilis. Gos. Univ., Tbilisi; 1985:9-11. (Russian)

3. 3.

Astashova IV: Qualitative properties of solutions to quasilinear ordinary differential equations. In Qualitative Properties of Solutions to Differential Equations and Related Topics of Spectral Analysis. Edited by: Astashova IV. UNITY-DANA, Moscow; 2012:22-290. (Russian)

4. 4.

Kozlov VA: On Kneser solutions of higher order nonlinear ordinary differential equations. Ark. Mat. 1999, 37(2):305-322. 10.1007/BF02412217

5. 5.

Astashova IV: Application of dynamical systems to the study of asymptotic properties of solutions to nonlinear higher-order differential equations. J. Math. Sci. 2005, 126(5):1361-1391.

6. 6.

Marsden JE, McCracken M: The Hopf Bifurcation and Its Applications. Springer, Berlin; 1976. XIII

7. 7.

Astashova IV, Vyun SA: On positive solutions with non-power asymptotic behavior to Emden-Fowler type twelfth order differential equation. Differ. Equ. 2012, 48(11):1568-1569. (Russian)

## Acknowledgements

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

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Correspondence to Irina Astashova.

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Astashova, I. On power and non-power asymptotic behavior of positive solutions to Emden-Fowler type higher-order equations. Adv Differ Equ 2013, 220 (2013). https://doi.org/10.1186/1687-1847-2013-220 