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

$$y={(x^{\ast }-x)}^{-\alpha }h(log(x^{\ast }-x)),\alpha =\frac{n}{k-1},x<x^{\ast },$$

where $x^{\ast }$ 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

$$y^{(n)}=p(x,y,y^{\prime },\dots ,y^{(n-1)}){|y|}^{k}sgny,n\ge 2,k>1,$$

(1)

Kiguradze posed the problem on the asymptotic behavior of its positive solutions such that

$$\underset{x\to x^{\ast }-0}{lim}y(x)=\mathrm{\infty }.$$

(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

$$y(x)=C{(x^{\ast }-x)}^{-\alpha }(1+o(1)),x\to x^{\ast }-0,$$

(3)

with

$$\alpha =\frac{n}{k-1},C={\left(\frac{\alpha (\alpha +1)\cdots (\alpha +n-1)}{p_0}\right)}^{\frac{1}{k-1}},$$

(4)

$p_0=\mathrm{const}>0$ - is a limit of $p(x,y_0,\dots ,y_{n-1})$ as $x\to x^{\ast }-0$, $y_0\to \mathrm{\infty },\dots ,y_{n-1}\to \mathrm{\infty }$.

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

The existence of solutions satisfying (3) was proved for arbitrary $n\ge 2$. For $2\le n\le 11$, 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 $k\in \mathbf{R}$, $1<k<K$, such that equation (5) has a solution

$$y={(x^{\ast }-x)}^{-\alpha }h(log(x^{\ast }-x)),$$

(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:

• (A) The continuous positive function $p(x,y_0,\dots ,y_{n-1})$ has a limit $p_0=\mathrm{const}>0$ as $x\to x^{\ast }-0$, $y_0\to \mathrm{\infty },\dots ,y_{n-1}\to \mathrm{\infty }$, and for some $\gamma >0$, it holds

  $$p(x,y_0,\dots ,y_{n-1})-p_0=O({(x^{\ast }-x)}^{\gamma }+\sum _{j=0}^{n-1}{y}_j^{-\gamma }).$$

  (7)

• (B) For some $K_1>0$ and $\mu >0$ in a neighborhood of $x^{\ast }$ for sufficiently large $y_0,\dots ,y_{n-1}$, $z_0,\dots ,z_{n-1}$, it holds

  $$|p(x,y_0,\dots ,y_{n-1})-p(x,z_0,\dots ,z_{n-1})|\le K_1\underset{j}{max}|y_j^{-\mu }-z_j^{-\mu }|.$$

  (8)

Then equation (1) can be transformed (see [2] or [3], Ch.I(5.1)) by using the substitution

$$x^{\ast }-x=e^{-t},y=(C+v)e^{\alpha t},$$

(9)

where C and α are defined by (4). The derivatives $y^{(j)}$, $j=0,1,\dots ,n-1$, become

$$e^{(\alpha +j)t}\cdot L_j(v,v^{\prime },\dots ,v^{(j)}),$$

where $v^{(j)}=\frac{d^{j}v}{d{t}^j}$ and $L_j$ is a linear function with

$$L_j(0,0,\dots ,0)=C\alpha (\alpha +1)\cdots (\alpha +j-1)\ne 0$$

and the coefficient of $v^{(j)}$ equal to 1.

Thus (1) is transformed into

$$e^{(\alpha +n)t}\cdot L_n(v,v^{\prime },\dots ,v^{(n)})={(C+v)}^k{e}^{\alpha kt}\tilde{p}(t,v,v^{\prime },\dots ,v^{(n-1)}),$$

(10)

where the function $\tilde{p}(t,v_0,\dots ,v_{n-1})$ is obtained from $p(x,y_0,\dots ,y_{n-1})$ with $x,y_0,\dots ,y_{n-1}$ properly expressed in terms of $t,v_0,\dots ,v_{n-1}$. This function tends to $p_0$ as $t\to \mathrm{\infty }$, $v\to 0,\dots ,v^{(n-1)}\to 0$.

Due to condition (8) for the function $p(x,y_0,\dots ,y_{n-1})$, we obtain the following inequalities for sufficiently large t and sufficiently small $v_0,\dots ,v_{n-1}$, $w_0,\dots ,w_{n-1}$:

$$\begin{array}{c}|\tilde{p}(t,v_0,\dots ,v_{n-1})-\tilde{p}(t,w_0,\dots ,w_{n-1})|\hfill \\ \le K_1\underset{j}{max}{e}^{-\mu (\alpha +j)t}|L_j^{-\mu }(v_0,\dots ,v_{n-1})-L_j^{-\mu }(w_0,\dots ,w_{n-1})|.\hfill \end{array}$$

Since $L_j(0,0,\dots ,0)\ne 0$, the function $L_j^{-\mu }$ is a $C^{\mathrm{\infty }}$ one in a neighborhood of 0 and

$$|\tilde{p}(t,v_0,\dots ,v_{n-1})-\tilde{p}(t,w_0,\dots ,w_{n-1})|\le K_2{e}^{-\mu \alpha t}\underset{j}{max}|v_j-w_j|$$

for some $K_2>0$.

Solving (10) for $v^{(n)}$ and using formulae (4), we obtain the equation

$$v^{(n)}={(C+v)}^k\tilde{p}(t,v,v^{\prime },\dots ,v^{(n-1)})-p_0{C}^k-\sum _{j=0}^{n-1}{a}_j{v}^{(j)},$$

(11)

where $a_j$ are the coefficients of the linear function $L_n$. Equation (11) can be written as

$$v^{(n)}=k{C}^{k-1}{p}_{0}v-\sum _{j=0}^{n-1}{a}_j{v}^{(j)}+f(v)+g(t,v,v^{\prime },\dots ,v^{(n-1)}),$$

(12)

where

$$\begin{array}{c}f(v)=p_o({(C+v)}^k-C^k-k{C}^{k-1}v)=O\left(v^2\right)\text{as }v\to 0,\hfill \\ f^{\prime }(v)=O(v)\text{as }v\to 0,\hfill \\ \begin{array}{rl}g(t,v_0,\dots ,v_{n-1})=& {(C+v_0)}^k(\tilde{p}(t,v_0,\dots ,v_{n-1})-p_0)\\ =& O(e^{-\gamma t}+\sum _{j=0}^{n-1}{e}^{-\gamma (\alpha +j)t})=O(exp(-\gamma min(\alpha ,1)t))\\ \text{as }t\to \mathrm{\infty },v_0\to 0,\dots ,v_{n-1}\to 0.\end{array}\hfill \end{array}$$

Besides, for sufficiently large t and sufficiently small $v_0,\dots ,v_{n-1}$, $w_0,\dots ,w_{n-1}$, it holds

$$\begin{array}{r}|g(t,v_0,\dots ,v_{n-1})-g(t,w_0,\dots ,w_{n-1})|\\ \le |{(C+v_0)}^k-{(C+w_0)}^k|\cdot |\tilde{p}(t,v_0,\dots ,v_{n-1})-p_0|\\ +{(C+w_0)}^k|\tilde{p}(t,v_0,\dots ,v_{n-1})-\tilde{p}(t,w_0,\dots ,w_{n-1})|\\ \le K_3\underset{j}{max}|w_j-v_j|e^{-min(\gamma ,\mu )\cdot min(\alpha ,1)t}.\end{array}$$

Suppose that V is the vector with coordinates $V_j=v^{(j)}$, $j=0,\dots ,n-1$. Then equation (12) can be written as

$$\frac{dV}{dt}=AV+F(V)+G(t,V),$$

(13)

where A is a constant $n\times n$ matrix

$$A=\left(\begin{array}{cccccc}0& 1& 0& 0& \cdots & 0\\ 0& 0& 1& 0& \cdots & 0\\ 0& 0& 0& 1& \cdots & 0\\ \cdot & \cdot & \cdot & \cdot & \cdots & \cdot \\ 0& 0& 0& 0& \cdots & 1\\ -{\tilde{a}}_0& -a_1& -a_2& -a_3& \cdots & -a_{n-1}\end{array}\right),$$

with

$$\begin{array}{rcl}-{\tilde{a}}_0& =& a_0-k{c}^{k-1}{p}_0=a_0-k\alpha (\alpha +1)\cdots (\alpha +n-1)\\ =& a_0-(\alpha +1)\cdots (\alpha +n-1)(\alpha +n)\end{array}$$

and eigenvalues satisfying the equation

$$\begin{array}{rcl}0& =& det(A-\lambda E)={(-1)}^{n+1}(-{\tilde{a}}_0-a_1\lambda -\cdots -a_{n-1}{\lambda }^{n-1}-{\lambda }^n)\\ =& {(-1)}^{n+1}((\alpha +1)(\alpha +2)\cdots (\alpha +n)-(\lambda +\alpha )\cdots (\lambda +\alpha +n-1)),\end{array}$$

which is equivalent to

$$\prod _{j=0}^{n-1}(\lambda +\alpha +j)=\prod _{j=0}^{n-1}(1+\alpha +j).$$

(14)

The mappings $F:{\mathbf{R}}^n\to {\mathbf{R}}^n$ and $G:\mathbf{R}\times {\mathbf{R}}^n\to {\mathbf{R}}^n$ satisfy the following estimates as $t\to \mathrm{\infty }$:

$$\{\begin{array}{c}\parallel F(V)\parallel =O({\parallel V\parallel }^2),\hfill \\ \parallel F_V^{\mathrm{\prime }}(V)\parallel =O(\parallel V\parallel ),\hfill \\ \parallel G(t,V)\parallel =O(e^{-2\beta t}),\hfill \\ \parallel G(t,V)-G(t,W)\parallel \le K\parallel V-W\parallel e^{-2\beta t}\hfill \end{array}$$

(15)

with some constants $\beta >0$, $K>0$.

Lemma 1 [3]

Suppose that (15) holds and A is an arbitrary constant $n\times n$ matrix. Then there exists a solution $V(t)$ to equation (13) tending to zero as $t\to \mathrm{\infty }$.

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 $t\to \mathrm{\infty }$.

If equation (13) has a solution $V(t)$ tending to 0 as $t\to \mathrm{\infty }$ and $V_0(t)$ is its first coordinate, then the function

$$y(x)=(V_0(-log(x^{\ast }-x))+C)\cdot {(x^{\ast }-x)}^{-\alpha }$$

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 $x^{\ast }$ 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 $3\le n\le 11$ 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 $p(x,y_0,\dots ,y_{n-1})$ is Lipschitz continuous in $y_0,\dots ,y_{n-1}$ and has a limit $p_0>0$ as $x\to x^{\ast }-0$, $y_0\to \mathrm{\infty },\dots ,y_{n-1}\to \mathrm{\infty }$. Then any positive solution to this equation with a vertical asymptote $x=x^{\ast }$ 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 $\dot{x}=L_{\alpha }x+Q_{\alpha }(x)$ in a neighborhood of $0\in {\mathbf{R}}^n$ with linear operators $L_{\alpha }$ and smooth enough functions $Q_{\alpha }(x)=O({|x|}^2)$ as $x\to 0$. Let ${\lambda }_{\alpha }$ and ${\overline{\lambda }}_{\alpha }$ be simple complex conjugated eigenvalues of the operators $L_{\alpha }$. Suppose that $Re{\lambda }_{\tilde{\alpha }}=Re{\overline{\lambda }}_{\tilde{\alpha }}=0$ for some $\tilde{\alpha }$ and the operator $L_{\tilde{\alpha }}$ has no other eigenvalues with zero real part.

If $Re\frac{d{\lambda }_{\alpha }}{d\alpha }(\tilde{\alpha })\ne 0$, then there exist continuous mappings $\epsilon \mapsto \alpha (\epsilon )\in \mathbf{R}$, $\epsilon \mapsto T(\epsilon )\in \mathbf{R}$, and $\epsilon \mapsto b(\epsilon )\in {\mathbf{R}}^n$ defined in a neighborhood of 0 and such that $\alpha (0)=\tilde{\alpha }$, $T(0)=2\pi /Im{\lambda }_{\tilde{\alpha }}$, $b(0)=0$, $b(\epsilon )\ne 0$ for $\epsilon \ne 0$, and the solutions to the problems

$$\dot{x}=L_{\alpha (\epsilon )}x+Q_{\alpha (\epsilon )}(x),x(0)=b(\epsilon )$$

are $T(\epsilon )$-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

$$\begin{array}{c}{y}^{(j)}(x)={(x^{\ast }-x)}^{-\alpha -j}{h}_j(log(x^{\ast }-x)),\hfill \\ j=0,1,\dots ,n-1,\hfill \end{array}$$

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 $G(t,V)\equiv 0$ 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 $F(V)=(0,\dots ,0,F_{n-1}(V))$, $V=(V_0,\dots ,V_{n-1})$, and

$$\begin{array}{c}{F}_{n-1}(V)=({(C+V_0)}^k-C^k-k{C}^{k-1}{V}_0)=O\left({V_0}^2\right),V_0\to 0,\hfill \\ \frac{d}{dV}{F}_{n-1}(V)=O(|V_0|),V_0\to 0.\hfill \end{array}$$

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 $\alpha >0$ and $q>0$ such that

$$\prod _{j=0}^{n-1}(qi+\alpha +j)=\prod _{j=0}^{n-1}(1+\alpha +j)$$

(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 ${\rho }_n(\alpha )$ and ${\sigma }_n(\alpha )$ defined for all $\alpha >0$ via the equations

$$\prod _{j=0}^{n-1}({\rho }_n{(\alpha )}^2+{(\alpha +j)}^2)=\prod _{j=0}^{n-1}{(1+\alpha +j)}^2$$

(17)

and

$$\sum _{j=0}^{n-1}arg({\sigma }_n(\alpha )i+\alpha +j)=2\pi$$

(18)

supposing $argz\in [0,2\pi )$ for all $z\in \mathbb{C}\setminus \{0\}$.

First, we prove the functions to be well defined for all $\alpha >0$.

The product ${\prod }_{j=0}^{n-1}(q^2+{(\alpha +j)}^2)$ is continuous and strictly increasing as a function of $q>0$.

It tends to ${\prod }_{j=0}^{n-1}{(\alpha +j)}^2<{\prod }_{j=0}^{n-1}{(1+\alpha +j)}^2$ as $q\to 0$ and to +∞ as $q\to +\mathrm{\infty }$. Hence, for any $\alpha >0$, there exists a unique $q>0$ such that ${\prod }_{j=0}^{n-1}(q^2+{(\alpha +j)}^2)={\prod }_{j=0}^{n-1}{(1+\alpha +j)}^2$.

In the same way, for any $\alpha >0$, the sum ${\sum }_{j=0}^{n-1}arg(qi+\alpha +j)$ is a continuous function of $q>0$ strictly increasing from 0 to $\frac{\pi n}{2}>2\pi$. 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 ${\rho }_n(\alpha )$ and ${\sigma }_n(\alpha )$ to be $C^1$-functions, too.

Now it is sufficient to prove the existence of $\alpha >0$ such that ${\rho }_n(\alpha )$ and ${\sigma }_n(\alpha )$ are equal to the same value q, which makes the two sides of (16) be equal.

Compare the functions ${\rho }_n(\alpha )$ and ${\sigma }_n(\alpha )$ near the boundaries of their common domain.

Equation (17) defining the function ${\rho }_n(\alpha )$ may be written as

$$\prod _{j=0}^{n-1}(1+\frac{2j}{\alpha }+\frac{j^2}{{\alpha }^2}+{\left(\frac{{\rho }_n(\alpha )}{\alpha }\right)}^2)=\prod _{j=0}^{n-1}{(1+\frac{j+1}{\alpha })}^2.$$

This shows that $\frac{{\rho }_n(\alpha )}{\alpha }\to 0$ as $\alpha \to +\mathrm{\infty }$.

Equation (18) defining the function ${\sigma }_n(\alpha )$ may be written as

$$\sum _{j=0}^{n-1}arctan\frac{\frac{{\sigma }_n(\alpha )}{\alpha }}{1+\frac{j}{\alpha }}=2\pi .$$

This shows that $\frac{{\sigma }_n(\alpha )}{\alpha }\to tan\frac{2\pi }{n}>0$ as $\alpha \to +\mathrm{\infty }$. Thus, ${\rho }_n(\alpha )<{\sigma }_n(\alpha )$ for sufficiently large α.

Now, to prove Proposition 1, it is sufficient to show that ${\rho }_n(\alpha )>{\sigma }_n(\alpha )$ for sufficiently small α. To compare the functions ${\rho }_n(\alpha )$ and ${\sigma }_n(\alpha )$ for small $\alpha >0$, we need some lemmas.

Lemma 3 For all $\alpha >0$, it holds ${\rho }_n{(\alpha )}^2<2(\alpha +n)-1$.

Proof Suppose that ${\rho }_n{(\alpha )}^2\ge 2(\alpha +n)-1$ for some $\alpha >0$. Then

$$\begin{array}{rcl}\prod _{j=0}^{n-1}({\rho }_n{(\alpha )}^2+{(\alpha +j)}^2)& \ge & \prod _{j=0}^{n-1}(2(\alpha +n)-1+{(\alpha +j)}^2)\\ >& \prod _{j=0}^{n-1}(2(\alpha +j+1)-1+{(\alpha +j)}^2)=\prod _{j=0}^{n-1}{(1+(\alpha +j))}^2.\end{array}$$

This contradiction with the definition of ${\rho }_n(\alpha )$ completes the proof of Lemma 3. □

Lemma 4 For all $\alpha >0$, it holds ${\rho }_{n+1}(\alpha )>{\rho }_n(\alpha )$.

Proof According to the definition of ${\rho }_n(\alpha )$ by (17) and Lemma 3, we have

$$\begin{array}{rcl}\prod _{j=0}^n({\rho }_n{(\alpha )}^2+{(\alpha +j)}^2)& =& \prod _{j=0}^{n-1}{(1+\alpha +j)}^2\cdot ({\rho }_n{(\alpha )}^2+{(\alpha +n)}^2)\\ <& \prod _{j=0}^{n-1}{(1+\alpha +j)}^2\cdot (2(\alpha +n)-1+{(\alpha +n)}^2)<\prod _{j=0}^n{(1+\alpha +j)}^2.\end{array}$$

In order to make the first and the last products be equal, we have to replace ${\rho }_n(\alpha )$ in the first one by a greater value. This means that ${\rho }_{n+1}(\alpha )>{\rho }_n(\alpha )$ and Lemma 4 is proved. □

Lemma 5 For all $\alpha >0$, it holds ${\sigma }_{n+1}(\alpha )<{\sigma }_n(\alpha )$.

Proof According to the definition of ${\sigma }_n(\alpha )$ by (18), we have

$$\sum _{j=0}^{n}arg({\sigma }_n(\alpha )i+\alpha +j)=2\pi +arg({\sigma }_n(\alpha )i+\alpha +n)>2\pi .$$

In order to make the sum equal 2π, we have to replace ${\sigma }_n(\alpha )$ by a smaller value. So, ${\sigma }_{n+1}(\alpha )<{\sigma }_n(\alpha )$ 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 ${\rho }_{12}(\alpha )>{\sigma }_{12}(\alpha )$ for sufficiently small $\alpha >0$.

Lemma 6 It holds ${\rho }_{12}(\alpha )>2$ for all sufficiently small $\alpha >0$.

Proof Straightforward exact calculations show that

$$\underset{\alpha \to 0}{lim}\prod _{j=0}^{11}(2^2+{(\alpha +j)}^2)=\prod _{j=0}^{11}(4+j^2)=192\text{,}175\text{,}659\text{,}520\text{,}000\text{,}000<2\cdot {10}^{17}$$

and

$$\underset{\alpha \to 0}{lim}\prod _{j=0}^{11}{(1+\alpha +j)}^2={(12!)}^2=229\text{,}442\text{,}532\text{,}802\text{,}560\text{,}000>2\cdot {10}^{17}.$$

So, for sufficiently small $\alpha >0$, we have

$$\prod _{j=0}^{11}(2^2+{(\alpha +j)}^2)<2\cdot {10}^{17}<\prod _{j=0}^{11}{(1+\alpha +j)}^2.$$

Hence, for these α, in order to avoid contradiction with the definition of ${\rho }_{12}(\alpha )$, the inequality ${\rho }_{12}{(\alpha )}^2>2^2$ is necessary. Lemma 6 is proved. □

Lemma 7 It holds ${\sigma }_{12}(\alpha )<2$ for sufficiently small $\alpha >0$.

Proof Consider the limit

$$\begin{array}{c}\underset{\alpha \to 0}{lim}\sum _{j=0}^{11}arg(2i+\alpha +j)\hfill \\ =arg2i+arctan2+arctan1+arctan\frac{2}{3}+arctan\frac{1}{2}+\sum _{j=5}^{11}arctan\frac{2}{j}\hfill \\ =\frac{5\pi }{4}+arctan\frac{2}{3}+\sum _{j=5}^{11}arctan\frac{2}{j}\hfill \\ =\frac{5\pi }{4}+arctan\frac{\frac{2}{3}+\frac{2}{5}}{1-\frac{2}{3}\cdot \frac{2}{5}}+arctan\frac{\frac{2}{6}+\frac{2}{7}}{1-\frac{2}{6}\cdot \frac{2}{7}}+arctan\frac{\frac{2}{8}+\frac{2}{9}}{1-\frac{2}{8}\cdot \frac{2}{9}}+arctan\frac{\frac{2}{10}+\frac{2}{11}}{1-\frac{2}{10}\cdot \frac{2}{11}}\hfill \\ =\frac{5\pi }{4}+arctan\frac{16}{11}+arctan\frac{13}{19}+arctan\frac{1}{2}+arctan\frac{21}{53}\hfill \\ =\frac{5\pi }{4}+arctan\frac{\frac{16}{11}+\frac{13}{19}}{1-\frac{16}{11}\cdot \frac{13}{19}}+arctan\frac{\frac{1}{2}+\frac{21}{53}}{1-\frac{1}{2}\cdot \frac{21}{53}}=\frac{5\pi }{4}+arctan447+arctan\frac{19}{17}.\hfill \end{array}$$

Note that

$$tan(arctan447+arctan\frac{19}{17})=\frac{447+\frac{19}{17}}{1-447\cdot \frac{19}{17}}=-\frac{3\text{,}809}{4\text{,}238}.$$

Hence, $arctan447+arctan\frac{19}{17}>\frac{3\pi }{4}$ and ${\sum }_{j=0}^{11}arg(2i+\alpha +j)>2\pi$ for sufficiently small $\alpha >0$. Thus, for these α, we have ${\sigma }_{12}(\alpha )<2$, which completes the proof of Lemma 7. □

Now Proposition 1 is also proved.  □

Proposition 2 For any $\alpha >0$ and any integer $n>1$, all roots $\lambda \in \mathbb{C}$ 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 $-n<m\le n$, there exists ${\mu }_m\in \mathbb{C}$ satisfying

$$\prod _{j=0}^{n-1}|{\mu }_m+j|=\prod _{j=0}^{n-1}(1+\alpha +j)$$

(19)

and

$$\sum _{j=0}^{n-1}arg({\mu }_m+j)=m\pi$$

(20)

with argz denoting the principal value of the argument lying in the open-closed interval $(-\pi ,\pi ]$. Surely, all these 2n complex numbers ${\mu }_m$ are different. Those with even m generate, via the relation ${\lambda }_m+\alpha ={\mu }_m$, just n different roots ${\lambda }_m$ 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 $(0,+\mathrm{\infty })$ containing a single point satisfying (19), namely ${\mu }_0=1+\alpha$.

Similarly, the set of μ satisfying equation (20) with $m=n$ is the real unbounded interval $(-\mathrm{\infty },1-n)$ containing a single point satisfying (19), namely ${\mu }_n=\alpha -n$.

Now consider the cases $0<m<n$ and the upper complex half-plane. For any $\omega >0$, the smooth function

$${\varphi }_{\omega }(r)=\sum _{j=0}^{n-1}arg(r+\omega i+j)=\sum _{j=0}^{n-1}arccot\frac{r+j}{\omega }$$

monotonically decreases from nπ to 0 as r increases from −∞ to +∞. So, for any $\omega >0$ and $b\in (0,n\pi )$, there exists a unique value r such that ${\varphi }_{\omega }(r)=b$. Due to the inequality $\frac{d{\varphi }_{\omega }}{dr}(r)<0$, the implicit function theorem provides the existence of the smooth functions $r_m(\omega )$ satisfying ${\varphi }_{\omega }(r_m(\omega ))=m\pi$.

Note that if $r\le -m$, then $r+j<0$ for all $j<m$ and $r+m\le 0$. Hence,

$$\underset{\omega \to +0}{\underset{\_}{lim}}\sum _{j=0}^{n-1}arccot\frac{r+j}{\omega }\ge \underset{\omega \to +0}{\underset{\_}{lim}}\sum _{j=0}^{m-1}arccot\frac{r+j}{\omega }+\underset{\omega \to +0}{\underset{\_}{lim}}arccot\frac{0}{\omega }=m\pi +\frac{\pi }{2}>m\pi$$

and such r cannot be the value of $r_m(\omega )$ for sufficiently small $\omega >0$.

Similarly, if $r\ge 1-m$, then $r+j>0$ for all $j>m-1$ and $r+m-1\ge 0$. Hence,

$$\begin{array}{rcl}\underset{\omega \to +0}{\overline{lim}}\sum _{j=0}^{n-1}arccot\frac{r+j}{\omega }& \le & \underset{\omega \to +0}{\overline{lim}}\sum _{j=0}^{m-2}arccot\frac{r+j}{\omega }+\frac{\pi }{2}+\underset{\omega \to +0}{\overline{lim}}\sum _{j=m}^{n-1}arccot\frac{r+j}{\omega }\\ \le & (m-1)\pi +\frac{\pi }{2}+0<m\pi \end{array}$$

and such r cannot be the value of $r_m(\omega )$ for sufficiently small $\omega >0$.

So, if $\omega >0$ is sufficiently small, then $r_m(\omega )$ satisfies the inequality $-m<r_m(\omega )<1-m$ and thereby is negative.

Consider the product ${\prod }_{j=0}^{n-1}|r_m(\omega )+\omega i+j|$ with $0<m<n$ and investigate its behavior for small $\omega >0$.

If $j\ge m$, then for sufficiently small $\omega >0$, we have $|r_m(\omega )+j|=r_m(\omega )+j<j$ and

$$\prod _{j=m}^{n-1}|r_m(\omega )+j|\le \prod _{j=m}^{n-1}j<\prod _{j=m}^{n-1}(1+j).$$

(21)

If $j\le m-1$, then for sufficiently small $\omega >0$, we have $|r_m(\omega )+j|=-r_m(\omega )-j<m-j=1+(m-1-j)$

$$\prod _{j=0}^{m-1}|r_m(\omega )+j|\le \prod _{j=0}^{m-1}|1+(m-1-j)|=\prod _{J=0}^{m-1}(1+J),J=m-1-j.$$

(22)

Combining (21) and (22), we obtain, for sufficiently small $\omega >0$,

$$\prod _{j=0}^{n-1}|r_m(\omega )+j|<\prod _{j=0}^{n-1}(1+j),$$

and

$$\prod _{j=0}^{n-1}|r_m(\omega )+\omega i+j|<\prod _{j=0}^{n-1}(1+\alpha +j).$$

As for large ω, the left-hand side of the above inequality evidently tends to +∞ as $\omega \to +\mathrm{\infty }$ and hence is greater than its right-hand side for sufficiently large ω. By continuity there exists ${\omega }_m>0$ such that

$$\prod _{j=0}^{n-1}|r_m({\omega }_m)+{\omega }_{m}i+j|=\prod _{j=0}^{n-1}(1+\alpha +j).$$

Thus, we can take ${\mu }_m=r_m({\omega }_m)+{\omega }_{m}i\in \mathbb{C}$ to satisfy (19) and (20) for $0<m<n$. For $-n<m<0$, we can take the conjugates ${\mu }_m=\overline{{\mu }_{-m}}$. Thus, the existence of all ${\mu }_m$ needed is proved. This completes the proof of Proposition 2. □

Lemma 8 If $12\le n\le 14$, $\alpha >0$, and $q>0$ satisfy the polynomial equation

$$\prod _{j=0}^{n-1}({(\alpha +j)}^2+q^2)=\prod _{j=0}^{n-1}{(\alpha +j+1)}^2,$$

then $2\alpha +4<q^2<3\alpha +5$.

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:

$$\begin{array}{c}\prod _{j=0}^{11}{(\alpha +j+1)}^2\hfill \\ ={\alpha }^{24}+156{\alpha }^{23}+11\text{,}518{\alpha }^{22}+535\text{,}392{\alpha }^{21}+17\text{,}581\text{,}135{\alpha }^{20}\hfill \\ +433\text{,}823\text{,}676{\alpha }^{19}+8\text{,}353\text{,}410\text{,}208{\alpha }^{18}+128\text{,}665\text{,}048\text{,}512{\alpha }^{17}\hfill \\ +1\text{,}612\text{,}229\text{,}817\text{,}055{\alpha }^{16}+16\text{,}625\text{,}859\text{,}652\text{,}116{\alpha }^{15}+142\text{,}196\text{,}061\text{,}481\text{,}318{\alpha }^{14}\hfill \\ +1\text{,}013\text{,}438\text{,}536\text{,}648\text{,}512{\alpha }^{13}+6\text{,}032\text{,}418\text{,}472\text{,}347\text{,}265{\alpha }^{12}\hfill \\ +29\text{,}989\text{,}851\text{,}619\text{,}249\text{,}236{\alpha }^{11}\hfill \\ +124\text{,}253\text{,}074\text{,}219\text{,}885\text{,}468{\alpha }^{10}+427\text{,}135\text{,}043\text{,}298\text{,}835\text{,}872{\alpha }^9\hfill \\ +1\text{,}209\text{,}806\text{,}045\text{,}835\text{,}003\text{,}760{\alpha }^8+2\text{,}795\text{,}060\text{,}589\text{,}044\text{,}133\text{,}696{\alpha }^7\hfill \\ +5\text{,}194\text{,}030\text{,}186\text{,}679\text{,}450\text{,}688{\alpha }^6+7\text{,}613\text{,}724\text{,}634\text{,}416\text{,}755\text{,}712{\alpha }^5\hfill \\ +8\text{,}564\text{,}233\text{,}279\text{,}835\text{,}510\text{,}784{\alpha }^4+7\text{,}096\text{,}936\text{,}674\text{,}284\text{,}421\text{,}120{\alpha }^3\hfill \\ +4\text{,}059\text{,}952\text{,}667\text{,}309\text{,}260\text{,}800{\alpha }^2+1\text{,}424\text{,}017\text{,}035\text{,}657\text{,}216\text{,}000\alpha \hfill \\ +229\text{,}442\text{,}532\text{,}802\text{,}560\text{,}000.\hfill \end{array}$$

Now, estimate the left-hand side supposing $q^2\ge 3\alpha +5>0$:

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

The difference of this polynomial and the previous one is equal to

$$\begin{array}{c}\prod _{j=0}^{11}({(\alpha +j)}^2+3\alpha +5)-\prod _{j=0}^{11}{(\alpha +j+1)}^2\hfill \\ =12{\alpha }^{23}+1\text{,}698{\alpha }^{22}+112\text{,}266{\alpha }^{21}+4\text{,}610\text{,}001{\alpha }^{20}+131\text{,}826\text{,}948{\alpha }^{19}\hfill \\ +2\text{,}790\text{,}199\text{,}071{\alpha }^{18}+45\text{,}357\text{,}703\text{,}644{\alpha }^{17}+580\text{,}073\text{,}542\text{,}125{\alpha }^{16}\hfill \\ +5\text{,}931\text{,}260\text{,}999\text{,}928{\alpha }^{15}+49\text{,}025\text{,}123\text{,}854\text{,}410{\alpha }^{14}+330\text{,}024\text{,}741\text{,}725\text{,}328{\alpha }^{13}\hfill \\ +1\text{,}818\text{,}717\text{,}493\text{,}077\text{,}486{\alpha }^{12}+8\text{,}236\text{,}923\text{,}851\text{,}221\text{,}212{\alpha }^{11}\hfill \\ +30\text{,}777\text{,}069\text{,}191\text{,}404\text{,}668{\alpha }^{10}+95\text{,}385\text{,}414\text{,}796\text{,}222\text{,}122{\alpha }^9\hfill \\ +247\text{,}258\text{,}394\text{,}050\text{,}998\text{,}864{\alpha }^8+542\text{,}195\text{,}044\text{,}856\text{,}859\text{,}120{\alpha }^7\hfill \\ +1\text{,}015\text{,}894\text{,}902\text{,}688\text{,}236\text{,}657{\alpha }^6+1\text{,}623\text{,}775\text{,}254\text{,}012\text{,}335\text{,}052{\alpha }^5\hfill \\ +2\text{,}159\text{,}188\text{,}576\text{,}366\text{,}038\text{,}588{\alpha }^4+2\text{,}263\text{,}080\text{,}289\text{,}120\text{,}101\text{,}792{\alpha }^3\hfill \\ +1\text{,}717\text{,}240\text{,}381\text{,}481\text{,}752\text{,}560{\alpha }^2+823\text{,}071\text{,}870\text{,}851\text{,}025\text{,}600\alpha \hfill \\ +184\text{,}478\text{,}363\text{,}699\text{,}112\text{,}000,\hfill \end{array}$$

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

In the same way, compute

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

Hence, ${\prod }_{j=0}^{11}{(\alpha +j+1)}^2>{\prod }_{j=0}^{11}({(\alpha +j)}^2+q^2)$ if $2\alpha +4\ge q^2$.

This contradiction yields $2\alpha +4<q^2<3\alpha +5$. So, Lemma 8 is proved. □

The condition $Re\frac{d{\lambda }_{\alpha }}{d\alpha }(\tilde{\alpha })\ne 0$ needed for the Hopf theorem, expressed explicitly by means of the implicit function theorem, looks like

$$\begin{array}{c}{\left[\sum _{j=0}^{n-1}\frac{\alpha +j}{q^2+{(\alpha +j)}^2}\right]}^2+{\left[\sum _{j=0}^{n-1}\frac{q}{q^2+{(\alpha +j)}^2}\right]}^2\hfill \\ \ne \sum _{j=0}^{n-1}\frac{\alpha +j}{q^2+{(\alpha +j)}^2}\sum _{j=0}^{n-1}\frac{1}{1+\alpha +j}.\hfill \end{array}$$

Lemma 9 If $12\le n\le 14$, $\alpha >0$ and $0<q^2<3\alpha +5$, then

$$\begin{array}{r}{\left[\sum _{j=0}^{n-1}\frac{\alpha +j}{q^2+{(\alpha +j)}^2}\right]}^2+{\left[\sum _{j=0}^{n-1}\frac{q}{q^2+{(\alpha +j)}^2}\right]}^2\\ >\sum _{j=0}^{n-1}\frac{\alpha +j}{q^2+{(\alpha +j)}^2}\sum _{j=0}^{n-1}\frac{1}{1+\alpha +j}.\end{array}$$

(23)

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

Multiplying inequality (23) by $U_{\ast }=\prod (1+\alpha +j)$ and then twice by $V_{\ast }=\prod [q^2+{(\alpha +j)}^2]$, we obtain the following equivalent inequality provided $\alpha >0$:

$$U_{\ast }[{(\sum (\alpha +j)V_j)}^2+q^2{(\sum V_j)}^2]>V_{\ast }\sum (\alpha +j)V_j\sum U_j$$

(24)

with the polynomials $U_j=\frac{U_{\ast }}{1+\alpha +j}$ and $V_j=\frac{V_{\ast }}{q^2+{(\alpha +j)}^2}$.

Put $q^2=\frac{3\alpha +5}{1+w}$, $w>0$. Substituting this into inequality (24) and multiplying the result by ${(1+w)}^{2n-1}$, we obtain another equivalent one:

$$\begin{array}{r}{U}_{\ast }[(1+w){(\sum (\alpha +j)P_j)}^2+(3\alpha +5){(\sum P_j)}^2]\\ >P_{\ast }\cdot \sum (\alpha +j)P_j\cdot \sum U_j\end{array}$$

(25)

with $P_{\ast }=\prod [3\alpha +5+(1+w){(\alpha +j)}^2]$ and $P_j=\frac{P_{\ast }}{3\alpha +5+(1+w){(\alpha +j)}^2}$.

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

$$\begin{array}{r}{U}_{\ast }[(1+w){(\sum (\alpha +j)P_j)}^2+(3\alpha +5){(\sum P_j)}^2]\\ -P_{\ast }\sum (\alpha +j)P_j\sum U_j=\sum _{j=0}^{5n-2}{\mathrm{\Delta }}_j{\alpha }^j\end{array}$$

(26)

with polynomials ${\mathrm{\Delta }}_j\in \mathbb{R}[w]$. Straightforward though very cumbersome calculations show that ${\mathrm{\Delta }}_{5n-2}=0$, and all other ${\mathrm{\Delta }}_j$ 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 ${\alpha }_{\epsilon }>0$ such that equation (14) with $\alpha ={\alpha }_0$ has imaginary roots $\lambda =\pm qi$ and for sufficiently small ε, system (13) with $\alpha ={\alpha }_{\epsilon }$ has a periodic solution $V_{\epsilon }(t)$ with period $T_{\epsilon }\to T=\frac{2\pi }{q}$ as $\epsilon \to 0$. In particular, the coordinate $V_{\epsilon ,0}(t)=v(t)$ of the vector $V_{\epsilon }(t)$ is also a periodic function with the same period. Then, taking into account (9), we obtain

$$y(x)=(C+v(-ln(x^{\ast }-x))){(x^{\ast }-x)}^{-\alpha }.$$

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

$$y(x)={(x^{\ast }-x)}^{-\alpha }h(ln(x^{\ast }-x)).$$

In the similar way, we obtain the related expressions for $y^{(j)}(x)$, $j=0,1,\dots ,n-1$.

Theorem 4 is proved. □

Conclusions, concluding remarks and open problems

• 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 $\alpha \approx 0.56$, $k\approx 22.4$;

if $n=13$, then $\alpha \approx 1.44$, $k\approx 10.0$;

if $n=14$, then $\alpha \approx 2.37$, $k\approx 6.9$.

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

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

• 4. If $n\ge 15$, then the inequality needed for the Hopf bifurcation theorem $Re\frac{d{\lambda }_{\alpha }}{d\alpha }(\tilde{\alpha })\ne 0$ cannot be proved in the same way because the estimate $q^2<3\alpha +5$ does not hold.