On quasi-periodic solutions to a higher-order Emden-Fowler type differential equation

Abstract

The paper is devoted to the existence of oscillatory and non-oscillatory quasi-periodic, in some sense, solutions to a higher-order Emden-Fowler type differential equation.

1 Introduction

The paper is devoted to the existence of oscillatory and non-oscillatory quasi-periodic, in some sense, solutions to the higher-order Emden-Fowler type differential equation

$$y^{(n)}+p_0{|y|}^{k}sgny=0,n>2,k\in \mathbb{R},k>1,p_0\ne 0.$$

(1)

The fact of the existence of such solutions answers the two questions posed by IT Kiguradze:

Question 1

Can we describe more precisely qualitative properties of oscillatory solutions to (1)?

Question 2

Do all blow-up solutions to this equation (and similarly all Kneser solutions) have the power asymptotic behavior?

A lot of results on the asymptotic behavior of solutions to (1) are described in detail in [1]. In particular (see Ch. IV, §15), the existence of oscillatory solutions to a generalization of this equation was proved (see also [2] Ch. I, §6.1). In [3] a result was formulated on non-extensibility of oscillatory solutions to (1) with odd n and $p_0>0$. In the cases $n=3$ and $n=4$ the asymptotic behavior of all oscillatory solutions is described in [4]–[6]. Some results on the existence of blow-up solutions are in [1] (Ch. IV, §16), [2] (Ch. I, §5), [7], [8]. Some results on the existence of some special solutions to this equation are in [2], [4], [5], [7], [9]–[13].

2 On existence of quasi-periodic oscillatory solutions

In this section some results will be obtained on the existence of special oscillatory solutions. The main results of this section were formulated in [14].

Theorem 1

For any integer $n>2$ and real $k>1$ there exists a periodic oscillatory function h such that for any $p_0>0$ and $x^{\ast }\in \mathbb{R}$ the function

$$y(x)=p_0^{\frac{1}{k-1}}{(x^{\ast }-x)}^{-\alpha }h(log(x^{\ast }-x)),-\mathrm{\infty }<x<x^{\ast },$$

(2)

with$\alpha =\frac{n}{k-1}$is a solution to (1). (See Figure1.)

Figure 1

A quasi-periodic solution for the equation ${\mathit{y}}^{\mathbf{‴}}\mathbf{+}{\mathit{y}}^{\mathbf{3}}\mathbf{=}\mathbf{0}$ .

Proof

For any $q=(q_0,\dots ,q_{n-1})\in {\mathbb{R}}^n$ let $y_q(x)$ be the maximally extended solution to the equation

$$y^{(n)}(x)+{|y(x)|}^k=0$$

(3)

satisfying the initial conditions $y^{(j)}(0)=q_j$ with $j=0,\dots ,n-1$.

For $0\le j<n$ put $B_j=\frac{nk}{n+j(k-1)}>1$ and ${\beta }_j=\frac{1}{B_j}$.

Consider the function $N:{\mathbb{R}}^n\to \mathbb{R}$ defined by the formula

$$N(q_0,\dots ,q_{n-1})=\sum _{j=0}^{n-1}|q_j{|}^{B_j}$$

(4)

and the mapping $\tilde{N}:{\mathbb{R}}^n\setminus \{0\}\to {\mathbb{R}}^n\setminus \{0\}$ defined by the formula

$$\tilde{N}{(q)}_j=N{(q)}^{-{\beta }_j}{q}_j,j=0,\dots ,n-1,$$

and satisfying the equality $N(\tilde{N}(q))=1$ for all $q\in {\mathbb{R}}^n\setminus \{0\}$.

Next, consider the subset $Q\subset {\mathbb{R}}^n$ consisting of all $q=(q_0,\dots ,q_{n-1})\in {\mathbb{R}}^n$ satisfying the following conditions:

• (1) $q_0=0$,

• (2) $q_j\ge 0$ for all $j\in \{1,\dots ,n-1\}$,

• (3) $N(q)=1$.

The restriction of the projection $(q_0,\dots ,q_{n-1})\mapsto (q_1,\dots ,q_{n-2})$ to the set Q is a homeomorphism of Q onto the convex compact subset

$$\{(q_1,\dots ,q_{n-2}):\sum _{j=1}^{n-2}|q_j{|}^{B_j}\le 1\text{ and }{q}_j\ge 0,j=1,\dots ,n-2\}\subset {\mathbb{R}}^{n-2}.$$

Lemma 1

For any$q\in Q$there exists$a_q>0$such that$y_q(a_q)=0$and$y_q^{(j)}(a_q)<0$for all$j\in \{1,\dots ,n-1\}$.

Proof

Put $J=max\{j\in \mathbb{Z}:0\le j<n,q_j>0\}$. This J exists and is positive due to the definition of Q. On some interval $(0;\epsilon )$ all derivatives $y_q^{(j)}(x)$ with $0\le j\le J$ are positive. Those with $J<j\le n$, due to (3), are negative on the same interval.

While keeping this sign combination, the function $y_q$ and its derivatives are bounded, which provides extensibility of $y_q(x)$ as the solution to (3) outside the interval $(0;\epsilon )$.

On the other hand, this sign combination cannot take place up to +∞. Indeed, in that case $y_q(x)$ would increase providing $y_q^{(n)}(x)<-y_q{(\epsilon )}^k<0$ for all $x>\epsilon$, which is impossible for any positive function on the unbounded interval $(0;+\mathrm{\infty })$.

So, $y_q(x)$ must change the sign combination of its derivatives. The only possible combination to be the next one corresponds to the positive derivatives $y_q^{(j)}(x)$ with $0\le j\le J-1$ and the negative ones with $J\le j\le n$.

The same arguments show that the new sign combination must also change and finally, after J changes, we arrive at the case with $y_q(x)>0$ and $y_q^{(j)}(x)<0$ with $1\le j\le n$. Now, contrary to the previous cases, the function $y_q(x)$ does not increase, but its first derivative is negative and decreases (recall that $n>2$). Hence this sign combination also cannot take place on an unbounded interval and therefore it must change to the case with all negative $y_q^{(j)}(x)$, $0\le j\le n$. By the way, the function $y_q(x)$ must vanish at some point $a_q>0$, which completes the proof of Lemma 1. □

Note that $a_q$ is not only the first positive zero of $y_q(x)$, but the only positive one. Indeed, all $y_q^{(j)}(x)$ with $0<j<n$ are negative at $a_q$, whence, according to (3), all $y_q^{(j)}(x)$ with $0\le j<n$ decrease and are negative for all $x>a_q$ in the domain of $y_q(x)$.

To continue the proof of Theorem 1, consider the function $\xi :q\mapsto a_q$ taking each $q\in Q$ to the first positive zero of the function $y_q$. To prove its continuity, we apply the implicit function theorem. The function $\xi (q)$ can be considered as a local solution $X(q)$ to the equation $S_0(q,X)=0$, where

$$S:(q,x)\mapsto (S_0(q,x),S_1(q,x),\dots ,S_{n-1}(q,x))=(y_q(x),y_q^{\prime }(x),\dots ,y_q^{(n-1)}(x))$$

is the $C^1$ ‘solution’ mapping defined on a domain including ${\mathbb{R}}^n\times \{0\}$. The necessary for the implicit function theorem condition $\frac{\partial S_0}{\partial X}(q_0,\dots ,q_{n-1},a_q)\ne 0$ is satisfied since the left-hand side of the last inequality is equal to $y_q^{\prime }(a_q)<0$. Besides, any function $X(q)$ implicitly defined near a point $(q_0,a_{q_0})$ must be positive in some its neighborhood. Hence locally $X(q)$ must be equal to $\xi (q)$, but neither to a non-positive zero of $y_q(x)$ nor to a non-first positive one, which does not exist. Hence the function $\xi (q)$ is continuous as well as $X(q)$.

Now we can consider the mapping $\tilde{S}:q\mapsto \tilde{N}(-S(q,\xi (q)))$, which maps Q into itself. Since $\tilde{S}$ is continuous and Q is homeomorphic to a convex compact subset of ${\mathbb{R}}^{n-2}$, the Brouwer fixed-point theorem can be applied. Thus, there exists $\hat{q}\in Q$ such that $\tilde{S}(\hat{q})=\hat{q}$.

According to the definitions of the functions $\tilde{N}$, S, and ξ, this yields the result that there exists a non-negative solution $\hat{y}(x)=y_{\hat{q}}(x)$ to (3) defined on a segment $[0;a_1]$ with $a_1=a_{\hat{q}}$, positive on the open interval $(0;a_1)$, and such that

$${\lambda }^{-{\beta }_j}{\hat{y}}^{(j)}(a_1)=-{\hat{y}}^{(j)}(0),j=0,\dots ,n-1,$$

(5)

with

$$\lambda =N\left(S(\hat{q},\xi (\hat{q}))\right)=\sum _{j=0}^{n-1}|{\hat{y}}^{(j)}(a_1){|}^{B_j}>0.$$

(6)

Since $\hat{y}(x)$ is non-negative, it is also a solution to the equation

$$y^{(n)}(x)+{|y(x)|}^{k}sgny(x)=0.$$

(7)

Note that for any solution $y_1(x)$ to (7) the function $y_2(x)=-b^{\alpha }{y}_1(bx+c)$ with arbitrary constants $b>0$ and c is also a solution to (7). Indeed, we have $\alpha +n=k\alpha$ and $y_2^{(j)}(x)=-b^{\alpha +j}{y}_1^{(j)}(bx+c)$ for all $j=0,\dots ,n$, whence

$$\begin{array}{r}{y}_2^{(n)}(x)+{|y_2(x)|}^{k}sgn{y}_2(x)\\ =-b^{\alpha +n}{y}_1^{(n)}(bx+c)-b^{k\alpha }{|y_1(bx+c)|}^{k}sgn{y}_1(bx+c)\\ =-b^{k\alpha }(y_1^{(n)}(bx+c)+{|y_1(bx+c)|}^{k}sgn{y}_1(bx+c))=0.\end{array}$$

So, the function $z(x)=-b^{\alpha }\hat{y}(bx-a_{1}b)$ is a solution to (7) and is defined on the segment $[a_1;a_2]$ with $a_2=a_1+\frac{a_1}{b}$.

Put $b={\lambda }^{\frac{k-1}{nk}}$ with λ defined by (6). Then

$$b^{\alpha +j}={\lambda }^{\frac{k-1}{nk}\cdot (\frac{n}{k-1}+j)}={\lambda }^{\frac{n+(k-1)j}{nk}}={\lambda }^{{\beta }_j},$$

whence, taking into account (5), we obtain $z^{(j)}(a_1)=-{\lambda }^{{\beta }_j}{\hat{y}}^{(j)}(0)={\hat{y}}^{(j)}(a_1)$. Thus, $z(x)$ can be used to extend the solution $\hat{y}(x)$ on $[0;a_2]$. Since $z(x)$ satisfies the conditions similar to (5), namely,

$${\lambda }^{-{\beta }_j}{z}^{(j)}(a_2)=-{\lambda }^{-{\beta }_j}{b}^{\alpha +j}{\hat{y}}^{(j)}(a_1)=-z^{(j)}(a_1),$$

the procedure of extension can be repeated on $[0;a_3]$, $[0;a_4]$, and so on with $a_{s+1}=a_s+\frac{a_s-a_{s-1}}{b}$. In the same way the solution $\hat{y}(x)$ can be extended to the left. Its restrictions to the neighboring segments satisfy the following equality:

$$\hat{y}(x)=-b^{\alpha }\hat{y}(b(x-a_s)+a_{s-1}),$$

(8)

where $x\in [a_s;a_{s+1}]$ and hence $b(x-a_s)+a_{s-1}\in [a_{s-1};a_s]$.

Now we will investigate whether b is greater or less than 1.

Let $a_{j,s}$ be the zero of the derivative ${\hat{y}}^{(j)}(x)$ belonging to the interval $(a_{s-1};a_s)$. Note that according to the above consideration on changing the sign combinations, we have

$$a_{j+1,s}<a_{j,s}<\cdots <a_{0,s}=a_s<a_{n-1,s+1}<a_{n-2,s+1}<\cdots .$$

Lemma 2

In the above notation the solution$y(x)=\hat{y}(x)$satisfies the following inequalities:

$$|y(a_{1,s})|<|y(a_{n-1,s+1})|,$$

(9)

$$|y(a_{j+1,s})|<|y(a_{j,s})|,0<j<n-1.$$

(10)

Proof

Indeed,

$$\begin{array}{r}\frac{1}{k+1}\left(\right|y(a_{n-1,s+1}){|}^{k+1}-|y(a_{1,s}){|}^{k+1})\\ ={\int }_{a_{1,s}}^{a_{n-1,s+1}}{y}^{\prime }(x)|y(x){|}^{k}sgny(x)dx=-{\int }_{a_{1,s}}^{a_{n-1,s+1}}{y}^{\prime }(x)y^{(n)}(x)dx\\ =-y^{\prime }(x)y^{(n-1)}(x){|}_{a_{1,s}}^{a_{n-1,s+1}}+{\int }_{a_{1,s}}^{a_{n-1,s+1}}{y}^{″}(x)y^{(n-1)}(x)dx>0\end{array}$$

since $y^{\prime }(a_{1,s})=y^{(n-1)}(a_{n-1,s+1})=0$ and $y^{″}(x)y^{(n-1)}(x)>0$ on the interval $(a_{1,s};a_{n-1,s+1})$, where only $y(x)$ itself changes its sign, while all other $y^{(j)}(x)$ with $0<j<n$ keep the same one. Recall that $n>2$, which makes $y^{″}(x)$ to be one of these others. Inequality (9) is proved.

Inequality (10) follows from $y(x)y^{\prime }(x)>0$ on the interval $(a_{j+1,s},a_{j,s})$, where the derivatives $y^{(j)}(x)$ and $y^{(j+1)}(x)$ with $0<j<n-1$ keep different signs, while all lower-order derivatives keep the same sign as $y^{(j)}(x)$. □

From the lemma proved it follows that $|\hat{y}(a_{1,s})|<|\hat{y}(a_{1,s+1})|=b^{\alpha }|\hat{y}(a_{1,s})|$, whence it follows that $b>1$ and $a_s-a_{s-1}=b(a_{s+1}-a_s)>a_{s+1}-a_s$.

Now we see that

$$\sum _{s=-\mathrm{\infty }}^0(a_{s+1}-a_s)=a_1\sum _{s=0}^{\mathrm{\infty }}{b}^s=\mathrm{\infty }\text{and}\sum _{s=0}^{\mathrm{\infty }}(a_{s+1}-a_s)=a_1\sum _{s=0}^{\mathrm{\infty }}{b}^{-s}=a^{\ast }<\mathrm{\infty }.$$

So, the solution $\hat{y}(x)$ is extended on the half-bounded interval $(-\mathrm{\infty };a^{\ast })$ and cannot be extended outside it since

$$\underset{x\to a^{\ast }}{lim\hspace{0.17em}sup}|\hat{y}(x)|=\underset{s\to +\mathrm{\infty }}{lim}|\hat{y}(a_{1,s})|=|\hat{y}(a_{1,0})|\underset{s\to +\mathrm{\infty }}{lim}{b}^{s\alpha }=+\mathrm{\infty }.$$

Now consider the function

$$h(t)=e^{t\alpha }\hat{y}(a^{\ast }-e^t),$$

(11)

which is periodic. Indeed, if $a_{\ast }-e^t\in [a_s;a_{s+1}]$ for some $s\in \mathbb{Z}$, then

$$h(t+logb)=e^{t\alpha }{b}^{\alpha }\hat{y}(a^{\ast }-b{e}^t)$$

and, according to (8),

$$h(t)=e^{\alpha t}\hat{y}(a^{\ast }-e^t)=-e^{\alpha t}{b}^{\alpha }\hat{y}(b{a}^{\ast }-b{e}^t-b{a}_s+a_{s-1}).$$

The expression in the last parentheses is equal to

$$b(a^{\ast }-a_s)-b{e}^t+a_{s-1}=b\cdot \frac{a_{s+1}-a_s}{1-b^{-1}}-b{e}^t+a_{s-1}=\frac{a_s-a_{s-1}}{1-b^{-1}}+a_{s-1}-b{e}^t=a^{\ast }-b{e}^t.$$

So, $h(t+logb)=-h(t)$ for all $t\in \mathbb{R}$ and hence the function $h(t)$ is periodic with period $2logb$.

Now, according to (11), we can express the solution $\hat{y}(x)$ to (7) just as $\hat{y}(x)={(a^{\ast }-x)}^{-\alpha }h(log(a^{\ast }-x))$. Multiplying it by $p_0^{\frac{1}{k-1}}$ we obtain a solution to (3) having the form needed. It still will be a solution to (3) after replacing $a^{\ast }$ by arbitrary $x^{\ast }\in \mathbb{R}$.  □

The substitution $x\mapsto -x$ produces the following.

Corollary 1

For any integer $n>2$ and real $k>1$ there exists a periodic oscillatory function h such that for any $p_0\in \mathbb{R}$ satisfying ${(-1)}^n{p}_0>0$ and any $x^{\ast }\in \mathbb{R}$ the function

$$y(x)=|p_0{|}^{\frac{1}{k-1}}{(x-x^{\ast })}^{-\alpha }h(log(x-x^{\ast })),x^{\ast }<x<\mathrm{\infty },$$

is a solution to (1).

Note that the following theorem was earlier proved in [4], [5].

Theorem 2

For$n=3$, there exists a constant$B\in (0,1)$such that any oscillatory solution$y(x)$to (1) with$p_0<0$satisfies the conditions

$$\begin{array}{r}(1)\frac{x_{i+1}-x_i}{x_i-x_{i-1}}=B^{-1},i=2,3,\dots ,\\ (2)\frac{y(x_{i+1}^{\prime })}{y(x_i^{\prime })}=-B^{\alpha },i=1,2,3,\dots ,\\ (3)\frac{y^{\prime }(x_{i+1})}{y^{\prime }(x_i)}=-B^{\alpha +1},i=1,2,3,\dots ,\\ (4)\left|y\left(x_i^{\prime }\right)\right|=M{(x_i^{\prime }-x_{\ast })}^{-\alpha },i=1,2,3,\dots ,\end{array}$$

for some$M>0$and$x_{\ast }$, where$x_1<x_2<\cdots <x_i<\cdots$and$x_1^{\prime }<x_2^{\prime }<\cdots <x_i^{\prime }<\cdots$are sequences satisfying$y(x_j)=0$, $y^{\prime }(x_j^{\prime })=0$, $y(x)\ne 0$if$x\in (x_i,x_{i+1})$, $y^{\prime }(x)\ne 0$if$x\in (x_i^{\prime },x_{i+1}^{\prime })$.

With the help of this theorem, another one can be proved, namely the following.

Theorem 3

For$n=3$and any real$k>1$there exists a periodic oscillatory function h such that the functions$y(x)=p_0^{\frac{1}{k-1}}{|x-x_{\ast }|}^{-\alpha }h(log|x-x_{\ast }|)$with$\alpha =\frac{n}{k-1}$and arbitrary$x_{\ast }$are solutions, respectively, to (1) with$p_0<0$if defined on$(-\mathrm{\infty };x_{\ast })$and to (1) with$p_0>0$if defined on$(x_{\ast };+\mathrm{\infty })$.

3 On existence of positive solutions with non-power asymptotic behavior

For (1) with $p_0=-1$ it was proved [11] that for any N and $K>1$ there exist an integer $n>N$ and $k\in \mathbf{R}$ such that $1<k<K$ and (1) has a solution of the form

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

(12)

where $\alpha =\frac{n}{k-1}$ and h is a positive periodic non-constant function on R.

A similar result was also proved [11] about Kneser solutions, i.e. those satisfying $y(x)\to 0$ as $x\to \mathrm{\infty }$ and ${(-1)}^j{y}^{(j)}(x)>0$ for $0\le j<n$. Namely, if $p_0={(-1)}^{n-1}$, then for any N and $K>1$ there exist an integer $n>N$ and $k\in \mathbf{R}$ such that $1<k<K$ and (1) has a solution of the form

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

where 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 positive solutions.

Theorem 4

[13]

If$12\le n\le 14$, then there exists$k>1$such that (1) with$p_0=-1$has a solution$y(x)$such that

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

where$\alpha =\frac{n}{k-1}$and$h_j$are periodic positive non-constant functions on R.

Remark

Computer calculations give approximate values of α. They are, with the 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 $\alpha \approx 2.37$, $k\approx 6.9$.

Corollary 2

If$12\le n\le 14$, then there exists$k>1$such that (1) with$p_0={(-1)}^{n-1}$has a Kneser solution$y(x)$satisfying

$$y^{(j)}(x)={(x-x_0)}^{-\alpha -j}{h}_j(log(x-x_0)),j=0,1,\dots ,n-1,$$

with periodic positive non-constant functions$h_j$on R.

4 Conclusions, concluding remarks, and open problems

• 1. So, we give the negative answer to Question 1 and prove the existence of oscillatory solutions with special qualitative properties for Question 2.

• 2. It would be interesting to know if positive solutions like (12) exist for $n\ge 15$ and for $5\le n\le 11$.

• 3. If a positive solution like (12) exists for some $k_0>1$, does it follow, for the same n, that such solutions exist for all $k>k_0$?