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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∈R,k>1, p 0 ≠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 ∗ ∈R the function

y(x)= p 0 1 k − 1 ( x ∗ − x ) − α h ( log ( x ∗ − x ) ) ,−∞<x< x ∗ ,
(2)

withα= n k − 1 is a solution to (1). (See Figure1.)

Figure 1
figure 1

A quasi-periodic solution for the equation y ‴ + y 3 =0 .

Proof

For any q=( q 0 ,…, q n − 1 )∈ 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,…,n−1.

For 0≤j<n put B j = n k n + j ( k − 1 ) >1 and β j = 1 B j .

Consider the function N: R n →R defined by the formula

N( q 0 ,…, q n − 1 )= ∑ j = 0 n − 1 | q j | B j
(4)

and the mapping N ˜ : R n ∖{0}→ R n ∖{0} defined by the formula

N ˜ ( q ) j =N ( q ) − β j q j ,j=0,…,n−1,

and satisfying the equality N( N ˜ (q))=1 for all q∈ R n ∖{0}.

Next, consider the subset Q⊂ R n consisting of all q=( q 0 ,…, q n − 1 )∈ R n satisfying the following conditions:

  1. (1)

    q 0 =0,

  2. (2)

    q j ≥0 for all j∈{1,…,n−1},

  3. (3)

    N(q)=1.

The restriction of the projection ( q 0 ,…, q n − 1 )↦( q 1 ,…, q n − 2 ) to the set Q is a homeomorphism of Q onto the convex compact subset

{ ( q 1 , … , q n − 2 ) : ∑ j = 1 n − 2 | q j | B j ≤ 1  and  q j ≥ 0 , j = 1 , … , n − 2 } ⊂ R n − 2 .

Lemma 1

For anyq∈Qthere exists a q >0such that y q ( a q )=0and y q ( j ) ( a q )<0for allj∈{1,…,n−1}.

Proof

Put J=max{j∈Z:0≤j<n, q j >0}. This J exists and is positive due to the definition of Q. On some interval (0;ε) all derivatives y q ( j ) (x) with 0≤j≤J are positive. Those with J<j≤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;ε).

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 ( ε ) k <0 for all x>ε, which is impossible for any positive function on the unbounded interval (0;+∞).

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≤j≤J−1 and the negative ones with J≤j≤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≤j≤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≤j≤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≤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 ξ:q↦ a q taking each q∈Q to the first positive zero of the function y q . To prove its continuity, we apply the implicit function theorem. The function ξ(q) can be considered as a local solution X(q) to the equation S 0 (q,X)=0, where

S:(q,x)↦ ( S 0 ( q , x ) , S 1 ( q , x ) , … , S n − 1 ( q , x ) ) = ( y q ( x ) , y q ′ ( x ) , … , y q ( n − 1 ) ( x ) )

is the C 1 ‘solution’ mapping defined on a domain including R n ×{0}. The necessary for the implicit function theorem condition ∂ S 0 ∂ X ( q 0 ,…, q n − 1 , a q )≠0 is satisfied since the left-hand side of the last inequality is equal to y q ′ ( 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 ξ(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 ξ(q) is continuous as well as X(q).

Now we can consider the mapping S ˜ :q↦ N ˜ (−S(q,ξ(q))), which maps Q into itself. Since S ˜ is continuous and Q is homeomorphic to a convex compact subset of R n − 2 , the Brouwer fixed-point theorem can be applied. Thus, there exists q ˆ ∈Q such that S ˜ ( q ˆ )= q ˆ .

According to the definitions of the functions N ˜ , S, and ξ, this yields the result that there exists a non-negative solution y ˆ (x)= y q ˆ (x) to (3) defined on a segment [0; a 1 ] with a 1 = a q ˆ , positive on the open interval (0; a 1 ), and such that

λ − β j y ˆ ( j ) ( a 1 )=− y ˆ ( j ) (0),j=0,…,n−1,
(5)

with

λ=N ( S ( q ˆ , ξ ( q ˆ ) ) ) = ∑ j = 0 n − 1 | y ˆ ( j ) ( a 1 ) | B j >0.
(6)

Since y ˆ (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 α y 1 (bx+c) with arbitrary constants b>0 and c is also a solution to (7). Indeed, we have α+n=kα and y 2 ( j ) (x)=− b α + j y 1 ( j ) (bx+c) for all j=0,…,n, whence

y 2 ( n ) ( x ) + | y 2 ( x ) | k sgn y 2 ( x ) = − b α + n y 1 ( n ) ( b x + c ) − b k α | y 1 ( b x + c ) | k sgn y 1 ( b x + c ) = − b k α ( y 1 ( n ) ( b x + c ) + | y 1 ( b x + c ) | k sgn y 1 ( b x + c ) ) = 0 .

So, the function z(x)=− b α y ˆ (bx− a 1 b) is a solution to (7) and is defined on the segment [ a 1 ; a 2 ] with a 2 = a 1 + a 1 b .

Put b= λ k − 1 n k with λ defined by (6). Then

b α + j = λ k − 1 n k ⋅ ( n k − 1 + j ) = λ n + ( k − 1 ) j n k = λ β j ,

whence, taking into account (5), we obtain z ( j ) ( a 1 )=− λ β j y ˆ ( j ) (0)= y ˆ ( j ) ( a 1 ). Thus, z(x) can be used to extend the solution y ˆ (x) on [0; a 2 ]. Since z(x) satisfies the conditions similar to (5), namely,

λ − β j z ( j ) ( a 2 )=− λ − β j b α + j y ˆ ( 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 + a s − a s − 1 b . In the same way the solution y ˆ (x) can be extended to the left. Its restrictions to the neighboring segments satisfy the following equality:

y ˆ (x)=− b α y ˆ ( b ( x − a s ) + a s − 1 ) ,
(8)

where x∈[ a s ; a s + 1 ] and hence b(x− a s )+ a s − 1 ∈[ 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 y ˆ ( 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 <⋯< a 0 , s = a s < a n − 1 , s + 1 < a n − 2 , s + 1 <⋯.

Lemma 2

In the above notation the solutiony(x)= y ˆ (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,

1 k + 1 ( | y ( a n − 1 , s + 1 ) | k + 1 − | y ( a 1 , s ) | k + 1 ) = ∫ a 1 , s a n − 1 , s + 1 y ′ ( x ) | y ( x ) | k sgn y ( x ) d x = − ∫ a 1 , s a n − 1 , s + 1 y ′ ( x ) y ( n ) ( x ) d x = − y ′ ( x ) y ( n − 1 ) ( x ) | a 1 , s a n − 1 , s + 1 + ∫ a 1 , s a n − 1 , s + 1 y ″ ( x ) y ( n − 1 ) ( x ) d x > 0

since y ′ ( 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 ′ (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 | y ˆ ( a 1 , s )|<| y ˆ ( a 1 , s + 1 )|= b α | y ˆ ( 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

∑ s = − ∞ 0 ( a s + 1 − a s )= a 1 ∑ s = 0 ∞ b s =∞and ∑ s = 0 ∞ ( a s + 1 − a s )= a 1 ∑ s = 0 ∞ b − s = a ∗ <∞.

So, the solution y ˆ (x) is extended on the half-bounded interval (−∞; a ∗ ) and cannot be extended outside it since

lim sup x → a ∗ | y ˆ (x)|= lim s → + ∞ | y ˆ ( a 1 , s )|=| y ˆ ( a 1 , 0 )| lim s → + ∞ b s α =+∞.

Now consider the function

h(t)= e t α y ˆ ( a ∗ − e t ) ,
(11)

which is periodic. Indeed, if a ∗ − e t ∈[ a s ; a s + 1 ] for some s∈Z, then

h(t+logb)= e t α b α y ˆ ( a ∗ − b e t )

and, according to (8),

h(t)= e α t y ˆ ( a ∗ − e t ) =− e α t b α y ˆ ( b a ∗ − b e t − b a s + a s − 1 ) .

The expression in the last parentheses is equal to

b ( a ∗ − a s ) −b e t + a s − 1 =b⋅ a s + 1 − a s 1 − b − 1 −b e t + a s − 1 = a s − a s − 1 1 − b − 1 + a s − 1 −b e t = a ∗ −b e t .

So, h(t+logb)=−h(t) for all t∈R and hence the function h(t) is periodic with period 2logb.

Now, according to (11), we can express the solution y ˆ (x) to (7) just as y ˆ (x)= ( a ∗ − x ) − α h(log( a ∗ −x)). Multiplying it by p 0 1 k − 1 we obtain a solution to (3) having the form needed. It still will be a solution to (3) after replacing a ∗ by arbitrary x ∗ ∈R.  □

The substitution x↦−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 ∈R satisfying ( − 1 ) n p 0 >0 and any x ∗ ∈R the function

y(x)=| p 0 | 1 k − 1 ( x − x ∗ ) − α h ( log ( x − x ∗ ) ) , x ∗ <x<∞,

is a solution to (1).

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

Theorem 2

Forn=3, there exists a constantB∈(0,1)such that any oscillatory solutiony(x)to (1) with p 0 <0satisfies the conditions

( 1 ) x i + 1 − x i x i − x i − 1 = B − 1 , i = 2 , 3 , … , ( 2 ) y ( x i + 1 ′ ) y ( x i ′ ) = − B α , i = 1 , 2 , 3 , … , ( 3 ) y ′ ( x i + 1 ) y ′ ( x i ) = − B α + 1 , i = 1 , 2 , 3 , … , ( 4 ) | y ( x i ′ ) | = M ( x i ′ − x ∗ ) − α , i = 1 , 2 , 3 , … ,

for someM>0and x ∗ , where x 1 < x 2 <⋯< x i <⋯and x 1 ′ < x 2 ′ <⋯< x i ′ <⋯are sequences satisfyingy( x j )=0, y ′ ( x j ′ )=0, y(x)≠0ifx∈( x i , x i + 1 ), y ′ (x)≠0ifx∈( x i ′ , x i + 1 ′ ).

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

Theorem 3

Forn=3and any realk>1there exists a periodic oscillatory function h such that the functionsy(x)= p 0 1 k − 1 | x − x ∗ | − α h(log|x− x ∗ |)withα= n k − 1 and arbitrary x ∗ are solutions, respectively, to (1) with p 0 <0if defined on(−∞; x ∗ )and to (1) with p 0 >0if defined on( x ∗ ;+∞).

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∈R such that 1<k<K and (1) has a solution of the form

y= ( x ∗ − x ) − α h ( log ( x ∗ − x ) ) ,
(12)

where α= 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)→0 as x→∞ and ( − 1 ) j y ( j ) (x)>0 for 0≤j<n. Namely, if p 0 = ( − 1 ) n − 1 , then for any N and K>1 there exist an integer n>N and k∈R such that 1<k<K and (1) has a solution of the form

y(x)= ( x − x ∗ ) − α h ( log ( x − x ∗ ) ) ,

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]

If12≤n≤14, then there existsk>1such that (1) with p 0 =−1has a solutiony(x)such that

y ( j ) (x)= ( x ∗ − x ) − α − j h j ( log ( x ∗ − x ) ) ,j=0,1,…,n−1,

whereα= 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 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.

Corollary 2

If12≤n≤14, then there existsk>1such that (1) with p 0 = ( − 1 ) n − 1 has a Kneser solutiony(x)satisfying

y ( j ) (x)= ( x − x 0 ) − α − j h j ( log ( x − x 0 ) ) ,j=0,1,…,n−1,

with periodic positive non-constant functions h j on R.

4 Conclusions, concluding remarks, and open problems

  1. 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. 2.

    It would be interesting to know if positive solutions like (12) exist for n≥15 and for 5≤n≤11.

  3. 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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Acknowledgements

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

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Astashova, I. On quasi-periodic solutions to a higher-order Emden-Fowler type differential equation. Bound Value Probl 2014, 174 (2014). https://doi.org/10.1186/s13661-014-0174-7

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