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

Boundary Value Problems20142014:174

https://doi.org/10.1186/s13661-014-0174-7

Received: 5 February 2014

Accepted: 30 June 2014

Published: 25 September 2014

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.

Keywords

Emden-Fowler type equationquasi-periodic solutionsoscillatory and non-oscillatory solutions

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 sgn y = 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)
with α = n k 1 is a solution to (1). (See Figure1.)
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 any q Q there exists a q > 0 such that y q ( a q ) = 0 and y q ( j ) ( a q ) < 0 for all j { 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 sgn y ( x ) = 0 .
(7)
Note that for any solution y 1 ( x ) to (7) the function y 2 ( x ) = b α y 1 ( b x + c ) with arbitrary constants b > 0 and c is also a solution to (7). Indeed, we have α + n = k α and y 2 ( j ) ( x ) = b α + j y 1 ( j ) ( b x + 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 ( b x 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 solution y ( 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 + log b ) = 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 + log b ) = h ( t ) for all t R and hence the function h ( t ) is periodic with period 2 log b .

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

For n = 3 , there exists a constant B ( 0 , 1 ) such that any oscillatory solution y ( x ) to (1) with p 0 < 0 satisfies 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 some M > 0 and x , where x 1 < x 2 < < x i < and x 1 < x 2 < < x i < are sequences satisfying y ( x j ) = 0 , y ( x j ) = 0 , y ( x ) 0 if x ( x i , x i + 1 ) , y ( x ) 0 if x ( x i , x i + 1 ) .

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 1 k 1 | x x | α h ( log | x x | ) with α = n k 1 and arbitrary x are solutions, respectively, to (1) with p 0 < 0 if defined on ( ; x ) and to (1) with p 0 > 0 if 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]

If 12 n 14 , then there exists k > 1 such that (1) with p 0 = 1 has a solution y ( x ) such that
y ( j ) ( x ) = ( x x ) α j h j ( log ( x x ) ) , j = 0 , 1 , , n 1 ,

where α = 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

If 12 n 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 ) 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 ?

     

Declarations

Acknowledgements

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

Authors’ Affiliations

(1)
Department of Mechanics and Mathematic, Lomonosov Moscow State University, Moscow, Russia
(2)
Department of Higher Mathematics, Moscow State University of Economics, Statistics and Informatics, Moscow, Russia

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© Astashova; licensee Springer. 2014

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