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On quasiperiodic solutions to a higherorder EmdenFowler type differential equation
Boundary Value Problems volume 2014, Article number: 174 (2014)
Abstract
The paper is devoted to the existence of oscillatory and nonoscillatory quasiperiodic, in some sense, solutions to a higherorder EmdenFowler type differential equation.
1 Introduction
The paper is devoted to the existence of oscillatory and nonoscillatory quasiperiodic, in some sense, solutions to the higherorder EmdenFowler type differential equation
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 blowup 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 nonextensibility 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 blowup 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 quasiperiodic 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
with$\alpha =\frac{n}{k1}$is a solution to (1). (See Figure1.)
Proof
For any $q=({q}_{0},\dots ,{q}_{n1})\in {\mathbb{R}}^{n}$ let ${y}_{q}(x)$ be the maximally extended solution to the equation
satisfying the initial conditions ${y}^{(j)}(0)={q}_{j}$ with $j=0,\dots ,n1$.
For $0\le j<n$ put ${B}_{j}=\frac{nk}{n+j(k1)}>1$ and ${\beta}_{j}=\frac{1}{{B}_{j}}$.
Consider the function $N:{\mathbb{R}}^{n}\to \mathbb{R}$ defined by the formula
and the mapping $\tilde{N}:{\mathbb{R}}^{n}\setminus \{0\}\to {\mathbb{R}}^{n}\setminus \{0\}$ defined by the formula
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 $q=({q}_{0},\dots ,{q}_{n1})\in {\mathbb{R}}^{n}$ satisfying the following conditions:

(1)
${q}_{0}=0$,

(2)
${q}_{j}\ge 0$ for all $j\in \{1,\dots ,n1\}$,

(3)
$N(q)=1$.
The restriction of the projection $({q}_{0},\dots ,{q}_{n1})\mapsto ({q}_{1},\dots ,{q}_{n2})$ to the set Q is a homeomorphism of Q onto the convex compact subset
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$j\in \{1,\dots ,n1\}$.
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 $0\le j\le J1$ 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
is the ${C}^{1}$ ‘solution’ mapping defined on a domain including ${\mathbb{R}}^{n}\times \{0\}$. The necessary for the implicit function theorem condition $\frac{\partial {S}_{0}}{\partial X}({q}_{0},\dots ,{q}_{n1},{a}_{q})\ne 0$ is satisfied since the lefthand 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 nonpositive zero of ${y}_{q}(x)$ nor to a nonfirst 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 ${\mathbb{R}}^{n2}$, the Brouwer fixedpoint theorem can be applied. Thus, there exists $\stackrel{\u02c6}{q}\in Q$ such that $\tilde{S}(\stackrel{\u02c6}{q})=\stackrel{\u02c6}{q}$.
According to the definitions of the functions $\tilde{N}$, S, and ξ, this yields the result that there exists a nonnegative solution $\stackrel{\u02c6}{y}(x)={y}_{\stackrel{\u02c6}{q}}(x)$ to (3) defined on a segment $[0;{a}_{1}]$ with ${a}_{1}={a}_{\stackrel{\u02c6}{q}}$, positive on the open interval $(0;{a}_{1})$, and such that
with
Since $\stackrel{\u02c6}{y}(x)$ is nonnegative, it is also a solution to the equation
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
So, the function $z(x)={b}^{\alpha}\stackrel{\u02c6}{y}(bx{a}_{1}b)$ 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 $b={\lambda}^{\frac{k1}{nk}}$ with λ defined by (6). Then
whence, taking into account (5), we obtain ${z}^{(j)}({a}_{1})={\lambda}^{{\beta}_{j}}{\stackrel{\u02c6}{y}}^{(j)}(0)={\stackrel{\u02c6}{y}}^{(j)}({a}_{1})$. Thus, $z(x)$ can be used to extend the solution $\stackrel{\u02c6}{y}(x)$ on $[0;{a}_{2}]$. Since $z(x)$ satisfies the conditions similar to (5), namely,
the procedure of extension can be repeated on $[0;{a}_{3}]$, $[0;{a}_{4}]$, and so on with ${a}_{s+1}={a}_{s}+\frac{{a}_{s}{a}_{s1}}{b}$. In the same way the solution $\stackrel{\u02c6}{y}(x)$ can be extended to the left. Its restrictions to the neighboring segments satisfy the following equality:
where $x\in [{a}_{s};{a}_{s+1}]$ and hence $b(x{a}_{s})+{a}_{s1}\in [{a}_{s1};{a}_{s}]$.
Now we will investigate whether b is greater or less than 1.
Let ${a}_{j,s}$ be the zero of the derivative ${\stackrel{\u02c6}{y}}^{(j)}(x)$ belonging to the interval $({a}_{s1};{a}_{s})$. Note that according to the above consideration on changing the sign combinations, we have
Lemma 2
In the above notation the solution$y(x)=\stackrel{\u02c6}{y}(x)$satisfies the following inequalities:
Proof
Indeed,
since ${y}^{\prime}({a}_{1,s})={y}^{(n1)}({a}_{n1,s+1})=0$ and ${y}^{\u2033}(x){y}^{(n1)}(x)>0$ on the interval $({a}_{1,s};{a}_{n1,s+1})$, 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 ${y}^{\u2033}(x)$ 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 $0<j<n1$ keep different signs, while all lowerorder derivatives keep the same sign as ${y}^{(j)}(x)$. □
From the lemma proved it follows that $\stackrel{\u02c6}{y}({a}_{1,s})<\stackrel{\u02c6}{y}({a}_{1,s+1})={b}^{\alpha}\stackrel{\u02c6}{y}({a}_{1,s})$, whence it follows that $b>1$ and ${a}_{s}{a}_{s1}=b({a}_{s+1}{a}_{s})>{a}_{s+1}{a}_{s}$.
Now we see that
So, the solution $\stackrel{\u02c6}{y}(x)$ is extended on the halfbounded interval $(\mathrm{\infty};{a}^{\ast})$ and cannot be extended outside it since
Now consider the function
which is periodic. Indeed, if ${a}_{\ast}{e}^{t}\in [{a}_{s};{a}_{s+1}]$ for some $s\in \mathbb{Z}$, then
and, according to (8),
The expression in the last parentheses is equal to
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 $\stackrel{\u02c6}{y}(x)$ to (7) just as $\stackrel{\u02c6}{y}(x)={({a}^{\ast}x)}^{\alpha}h(log({a}^{\ast}x))$. Multiplying it by ${p}_{0}^{\frac{1}{k1}}$ 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
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
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$y(x)={p}_{0}^{\frac{1}{k1}}{x{x}_{\ast}}^{\alpha}h(logx{x}_{\ast})$with$\alpha =\frac{n}{k1}$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 nonpower 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
where $\alpha =\frac{n}{k1}$ and h is a positive periodic nonconstant 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 ${p}_{0}={(1)}^{n1}$, 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
where h is a positive periodic nonconstant 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
where$\alpha =\frac{n}{k1}$and${h}_{j}$are periodic positive nonconstant 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${p}_{0}={(1)}^{n1}$has a Kneser solution$y(x)$satisfying
with periodic positive nonconstant 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 $k>{k}_{0}$?
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The research was supported by RFBR (grant 110100989).
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Keywords
 EmdenFowler type equation
 quasiperiodic solutions
 oscillatory and nonoscillatory solutions