Boundary Value Problems

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Asymptotic problems for fourth-order nonlinear differential equations

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  1. Jean Mawhin's Achievements in Nonlinear Analysis

Abstract

We study vanishing at infinity solutions of a fourth-order nonlinear differential equation. We state sufficient and/or necessary conditions for the existence of the positive solution on the half-line [ 0 , ) Open image in new window which is vanishing at infinity and sufficient conditions ensuring that all eventually positive solutions are vanishing at infinity. We also discuss an oscillation problem.

Keywords

Quasilinear Equation Homoclinics Solution Nonoscillatory Solution Oscillation Problem Asymptotic Boundary Condition 

Dedication

Dedicated to Jean Mawhin on occasion of his seventieth birthday.

1 Introduction

In this paper we study the fourth-order nonlinear differential equation
x ( 4 ) ( t ) + q ( t ) x ( t ) + r ( t ) | x ( t ) | λ sgn x ( t ) = 0 ( t R + ) , Open image in new window
(1)

where λ 1 Open image in new window, q C 3 ( R + ) Open image in new window, q ( t ) > 0 Open image in new window for large t, r C ( R + ) Open image in new window such that r ( t ) 0 Open image in new window for large t and R + = [ 0 , ) Open image in new window.

Jointly with (1), we consider a more general equation
where f C 3 ( R ) Open image in new window satisfies f ( u ) u > 0 Open image in new window for u 0 Open image in new window, and the associated linear second-order equation
h ( t ) + q ( t ) h ( t ) = 0 . Open image in new window
(2)

By a solution of (1) we mean a function x C 4 [ T x , ) Open image in new window, T x 0 Open image in new window, which satisfies (1) on [ T x , ) Open image in new window. A solution is said to be nonoscillatory if x ( t ) 0 Open image in new window for large t; otherwise, it is said to be oscillatory. Observe that if λ 1 Open image in new window, according to [[1], Theorem 11.5], all nontrivial solutions of (1) satisfy sup { | x ( t ) | : t T } > 0 Open image in new window for T T x Open image in new window, on the contrary to the case λ < 1 Open image in new window, when nontrivial solutions satisfying x ( t ) 0 Open image in new window for large t may exist.

Fourth-order differential equations have been investigated in detail during the last years. The periodic boundary value problem for the superlinear equation x ( 4 ) = g ( x ) + e ( t ) Open image in new window has been studied in [2]. In [3], the fourth-order linear eigenvalue problem, together with the nonlinear boundary value problem x ( 4 ) f ( t , x ) = 0 Open image in new window, has been investigated. Oscillatory properties of solutions for self-adjoint linear differential equations can be found in [4]. Equation (1) with q ( t ) 0 Open image in new window can be viewed as a prototype of even-order two-term differential equations, which are the main object of monographs [1, 5, 6].

Equation (1′) with q ( t ) 1 Open image in new window for t R + Open image in new window is a special case of higher-order differential equations investigated in [7]. Equation (1′) with q near to a nonzero constant as t Open image in new window has been considered in [8] as a perturbation of the linear equation y ( 4 ) ( t ) + q ( t ) y ( t ) = 0 Open image in new window, and the existence of oscillatory solutions of (1′) has been proved. In [9], necessary and sufficient conditions for the existence of asymptotically linear solutions of (1′) have been given.

In the recent paper [10], the equation
x ( 4 ) ( t ) + k x ( t ) + f ( x ( t ) ) = 0 ( t R ) , Open image in new window

where k R Open image in new window, f ( u ) u > 0 Open image in new window for u 0 Open image in new window, and f Lip loc ( R ) Open image in new window has been investigated and applications to the biharmonic PDE’s can be found there. In particular, the so called homoclinics solutions, which are defined as nontrivial solutions x such that lim t ± x ( t ) = 0 Open image in new window, are studied.

The goal of this paper is to investigate asymptotic problems associated with (1) and the asymptotic boundary condition
x ( t ) > 0 for large  t , lim t x ( t ) = 0 . Open image in new window
(3)

A solution x of (1) satisfying (3) is said to be vanishing at infinity.

We start with the Kneser problem for (1). The Kneser problem is a problem concerning the existence of solutions of (1) subject to the boundary conditions on the half-line [ 0 , ) Open image in new window
x ( 0 ) = c > 0 , ( 1 ) i x ( i ) ( t ) > 0 for  t 0 , i = 0 , 1 , 2 . Open image in new window
(4)

We establish necessary and/or sufficient conditions for the solvability of the boundary value problem (1), (3), (4). In the light of these results, as the second problem, we study when all eventually positive solutions x of (1) are vanishing at infinity assuming that λ > 1 Open image in new window and (2) is oscillatory. As a consequence, we give a bound for the set of all nonoscillatory solutions. Finally, we discuss when problem (1), (3) is not solvable and solutions to (1) are oscillatory.

A systematic analysis of solutions of (1) satisfying (3) is made according to whether (2) is nonoscillatory or oscillatory. If (2) is nonoscillatory, then the following approach will be used. Equation (1) can be rewritten as the two-term equation
( h 2 ( t ) ( x ( t ) h ( t ) ) ) + h ( t ) r ( t ) | x ( t ) | λ sgn x ( t ) = 0 , Open image in new window
(5)
where h is a positive solution of (2). According to [11], a solution h of (2) is said to be a principal solution if h 2 ( t ) d t = Open image in new window, and such a solution is determined uniquely up to a multiple constant. Since q ( t ) > 0 Open image in new window, every eventually positive solution of (2) is nondecreasing for large t. Hence there exists a principal solution h of (2) such that h ( t ) > 0 Open image in new window for t a 0 Open image in new window and
a 1 h 2 ( t ) d t = , a h ( t ) d t = . Open image in new window
(6)

Therefore, we can use the known results [12, 13] stated for systems of differential equations or in [14] for fourth order differential equations.

If (2) is oscillatory, then our approach is based on the choice of a suitable transformation. The main idea is based on a transformation of (1) to the fourth-order quasilinear equation and the use of the estimates for positive solutions of such an equation on a compact interval stated in [15]. This, together with an energy function associated with (1), enables us to state an oscillation theorem. In the final section, some extensions of our results to (1′) are given.

2 The Kneser problem

In this section we present necessary and/or sufficient conditions for solvability of boundary value problem (1), (3), (4).

2.1 Case r ( t ) < 0 Open image in new window

Proposition 1 Let λ 1 Open image in new window, (2) be disconjugate on [ 0 , ) Open image in new window, and r ( t ) < 0 Open image in new window for t R + Open image in new window. Then boundary value problem (1), (4) is solvable for any c > 0 Open image in new window.

To prove this theorem, we use Chanturia’s result [[12], Theorem 1] for the system of differential equations
d y d t = f ( t , y ) , Open image in new window
(7)

where we restrict to the case that f = ( f 1 , f 2 , f 3 , f 4 ) : R + × R 4 R 4 Open image in new window are continuous functions, n = 4 Open image in new window, m = 1 Open image in new window, l = 2 Open image in new window, a = 1 Open image in new window and r = c Open image in new window. Then this result reads as follows.

Theorem A ([12])

Let there exist c > 0 Open image in new window such that
f i ( t , y 1 , y 2 , y 3 , y 4 ) 0 ( i = 1 , , 4 ) Open image in new window
for t R + Open image in new window, y 1 [ 0 , c ) Open image in new window, y i R + Open image in new window ( i = 2 , 3 , 4 Open image in new window). Suppose
f i ( t , y 1 , y 2 , y 3 , y 4 ) φ i ( t , y i + 1 ) ( i = 1 , 2 , 3 ) i = 2 4 f i ( t , y 1 , y 2 , y 3 , y 4 ) ψ ( t ) ω ( i = 2 4 y i ) Open image in new window
for t [ 0 , 1 ] Open image in new window, y 1 [ 0 , c ] Open image in new window, y i R + Open image in new window ( i = 2 , 3 , 4 Open image in new window), where functions φ i : [ 0 , 1 ] × R + R + Open image in new window ( i = 1 , 2 , 3 Open image in new window) are continuous and nondecreasing in the second argument such that
lim inf x 0 1 φ 1 ( t , x ) d t > c , lim x 0 1 φ i ( t , x ) d t = ( i = 2 , 3 ) , Open image in new window
ψ : [ 0 , 1 ] R + Open image in new window is a continuous function and ω : R + ( 0 , ) Open image in new window is a continuous nondecreasing function such that
0 d u ω ( u ) = . Open image in new window
Then, for any x 0 [ 0 , c ] Open image in new window, system (7) has a solution satisfying
y 1 ( 0 ) = x 0 , y i ( t ) 0 , y i ( t ) 0 for t R + ( i = 1 , 2 , 3 , 4 ) . Open image in new window
(8)
Proof of Proposition 1 Assume r ( t ) < 0 Open image in new window for t R + Open image in new window. Since (2) is disconjugate, it has a positive solution h on R + Open image in new window, and (1) can be written as (5) where h ( t ) r ( t ) < 0 Open image in new window on R + Open image in new window. Let x be a solution of (5) and denote
y 1 ( t ) = x ( t ) , y 2 ( t ) = y 1 ( t ) , y 3 ( t ) = 1 h ( t ) y 2 ( t ) , y 4 ( t ) = h 2 ( t ) y 3 ( t ) . Open image in new window
(9)
Then (5) is equivalent to the system
{ y 1 ( t ) = y 2 ( t ) , y 2 ( t ) = h ( t ) y 3 ( t ) , y 3 ( t ) = 1 h 2 ( t ) y 4 ( t ) , y 4 ( t ) = h ( t ) | r ( t ) | | y 1 ( t ) | λ ( t ) sgn y 1 ( t ) ( t R + ) . Open image in new window
(10)
Let c > 0 Open image in new window be from (4). We apply Theorem A choosing
φ 1 ( t , x ) = x , φ i ( t , x ) = k i x , i = 2 , 3 , ψ ( t ) = k 4 , ω ( u ) = c λ + u , Open image in new window
where k 2 = min t [ 0 , 1 ] h ( t ) Open image in new window, k 3 = min t [ 0 , 1 ] 1 / h 2 ( t ) Open image in new window and k 4 = max t [ 0 , 1 ] [ h ( t ) + 1 / h 2 ( t ) + h ( t ) r ( t ) ] Open image in new window. By this result, system (10) has a solution such that
y 1 ( 0 ) = c > 0 , y i ( t ) 0 , y i ( t ) 0 for  t 0 , i = 1 , 2 , 3 , 4 . Open image in new window

Since λ 1 Open image in new window, system (10) has no solutions such that y i 0 Open image in new window for some i = 1 , 2 , 3 , 4 Open image in new window and large t; see [16] or [[17], Lemma 2, Theorem 2]. Thus, for any c > 0 Open image in new window, equation (1) has a solution x such that x ( 0 ) = c Open image in new window, x ( t ) = y 1 ( t ) > 0 Open image in new window, x ( t ) = y 2 ( t ) < 0 Open image in new window and x ( t ) = h ( t ) y 3 ( t ) > 0 Open image in new window for t 0 Open image in new window. □

Now we state conditions for the existence of a solution for problem (1), (3), (4).

Theorem 1 Let λ 1 Open image in new window, r ( t ) < 0 Open image in new window and
q ( t ) ( t + 1 ) 2 1 4 Open image in new window
(11)
on R + Open image in new window. If
0 t 2 | r ( t ) | d t = , Open image in new window
(12)

then problem (1), (3), (4) is solvable for any c > 0 Open image in new window.

In addition, if
0 t q ( t ) d t < , Open image in new window
(13)
then the condition
0 t 3 | r ( t ) | d t = Open image in new window
(14)

is necessary and sufficient for the solvability of problem (1), (3), (4).

For the proof, the following lemma will be needed.

Lemma 1 Consider system (10) on [ a , ) Open image in new window ( a 0 Open image in new window), where h ( t ) > 0 Open image in new window for t a Open image in new window and h is a principal solution of (2). Let y = ( y 1 , y 2 , y 3 , y 4 ) Open image in new window be a solution of (10) such that y i ( t ) 0 Open image in new window and y i ( t ) 0 Open image in new window for i = 1 , 2 , 3 , 4 Open image in new window, and t a Open image in new window. Then lim t y i ( t ) = 0 Open image in new window for i = 2 , 3 , 4 Open image in new window, and if
a h ( t ) | r ( t ) | a t 1 h 2 ( s ) a s h ( τ ) τ d τ d s d t = , Open image in new window
(15)

then lim t y 1 ( t ) = 0 Open image in new window, too. Vice versa, if λ 1 Open image in new window and lim t y 1 ( t ) = 0 Open image in new window, then (15) holds.

Proof In view of the monotonicity of y i Open image in new window, there exist lim t y i ( t ) = y i ( ) 0 Open image in new window, i = 1 , 2 , 3 , 4 Open image in new window. Since h is the principal solution, (6) holds, and integrating the first three equations in (10) from a to t, we get y i ( ) = 0 Open image in new window for i = 2 , 3 , 4 Open image in new window. Now integrating (10) from t to ∞, we have
y 2 ( t ) = t h ( s ) y 3 ( s ) d s , y 3 ( t ) = t 1 h 2 ( t ) y 4 ( s ) d s , y 4 ( t ) = t h ( s ) | r ( s ) | y 1 λ ( s ) d s . Open image in new window
Let (15) hold and assume, by contradiction, that y 1 ( ) > 0 Open image in new window. Then
y 1 ( a ) y 1 ( t ) y 1 λ ( ) a t s h ( u ) u 1 h 2 ( v ) v h ( τ ) | r ( τ ) | d τ d v d u d s . Open image in new window
(16)

Letting t Open image in new window and using the change of the order of integration, we get a contradiction with the boundedness of y 1 Open image in new window. This proves that y 1 ( ) = 0 Open image in new window.

Let the integral in (15) be convergent and assume, by contradiction, that y 1 ( ) = 0 Open image in new window. Then we have
y 1 ( t ) = t s h ( u ) u 1 h 2 ( v ) v h ( τ ) | r ( τ ) | y 1 λ ( τ ) d τ d v d u d s , Open image in new window
so
y 1 ( t ) y 1 λ ( t ) t s h ( u ) u 1 h 2 ( v ) v h ( τ ) | r ( τ ) | d τ d v d u d s . Open image in new window

Since λ 1 Open image in new window, then using the change of the order of integration, we get a contradiction for large t. This proves that y 1 ( ) > 0 Open image in new window. □

Proof of Theorem 1 In view of (11), (2) is disconjugate on R + Open image in new window. By Proposition 1, equation (1) has a solution x satisfying (4). Therefore, system (10) has a solution such that y i ( t ) > 0 Open image in new window and y i ( t ) < 0 Open image in new window for t R + Open image in new window. Choose h in (10) as a principal solution of (2). The Euler equation
h ˜ + 1 4 ( t + 1 ) 2 h ˜ = 0 ( t R + ) Open image in new window
(17)
is the majorant of (2) on R + Open image in new window and has the principal solution h ˜ ( t ) = t + 1 Open image in new window. By the comparison theorem, for the minimal solution of the Riccati equation related to (2) and (17), we have
0 < h ( t ) h ( t ) h ˜ ( t ) h ˜ ( t ) Open image in new window

for large t; see, e.g., [11]. Thus there exists > 0 Open image in new window such that h ( t ) t + 1 Open image in new window for t 0 Open image in new window. Assume (12). Then (15) holds, and by Lemma 1 a solution x satisfies (3).

Assume (13). Then the principal solution h of (2) satisfies h ( t ) Open image in new window for large t (see, e.g., [11]). Hence, condition (15) reads as (14), and by Lemma 1 this condition is equivalent to the property (3). □

As a consequence of Lemma 1, we get the following result.

Corollary 1 Let (2) be disconjugate on [ 0 , ) Open image in new window, and r ( t ) < 0 Open image in new window for t R + Open image in new window. Then any solution x of (1) satisfying
x ( 0 ) = c > 0 , x ( t ) > 0 , t 0 , lim t x ( t ) = 0 Open image in new window
(18)

is a solution of the Kneser problem, i.e., ( 1 ) i x ( i ) ( t ) > 0 Open image in new window for t 0 Open image in new window and i = 1 , 2 Open image in new window.

Proof Let h be a positive solution on R + Open image in new window satisfying (6), and let x be a solution of (1) satisfying (18). Then y = ( y 1 , y 2 , y 3 , y 4 ) Open image in new window, where y 1 Open image in new window are defined by (9), is a solution of system (10). Since y 1 ( t ) > 0 Open image in new window for t 0 Open image in new window and (6) holds, we have by the Kiguradze lemma (see, e.g., [1]) that either y i ( t ) > 0 Open image in new window or y i ( t ) < 0 Open image in new window for i = 1 , 2 , 3 Open image in new window and large t, say for t a 0 Open image in new window. Since x is positive and tends to zero, we have y 1 0 Open image in new window for t a Open image in new window, so also y i 0 Open image in new window ( i = 2 , 3 Open image in new window) for t a Open image in new window. By Lemma 1, we get y i ( ) = 0 Open image in new window ( i = 2 , 3 Open image in new window) for t a Open image in new window. Since x ( t ) > 0 Open image in new window for t 0 Open image in new window, we have y 4 ( t ) < 0 Open image in new window for t 0 Open image in new window and y 4 Open image in new window is positive and decreasing on R + Open image in new window. Hence, proceeding by the same argument, y i Open image in new window ( i = 2 , 3 Open image in new window) is positive and decreasing on R + Open image in new window. Now the conclusion follows from (9). □

2.2 Case r ( t ) > 0 Open image in new window

First we show that the sign condition posed on r is necessary for the solvability of problem (1), (4).

A function g, defined in a neighborhood of infinity, is said to change sign if there exists a sequence { t k } Open image in new window such that g ( t k ) g ( t k + 1 ) < 0 Open image in new window.

Theorem 2 Let r ( t ) > 0 Open image in new window for large t. Then problem (1), (4) has no solution and the following hold:
  1. (a)

    If (2) is nonoscillatory, then every nonoscillatory solution x of (1) satisfies x ( t ) x ( t ) > 0 Open image in new window and x Open image in new window is of one sign for large t.

     
  2. (b)

    If (2) is oscillatory, then every nonoscillatory solution x of (1) satisfies either x ( t ) x ( t ) 0 Open image in new window, or x ( t ) Open image in new window changes sign. In addition, if a solution x satisfies (3), then x Open image in new window changes sign.

     
Proof Claim (a). Let x be a positive solution of (1) on [ a , ) Open image in new window, or, equivalently, of (5) on [ a , ) Open image in new window, where h satisfies (6). Denote
y 1 ( t ) = x ( t ) , y 2 ( t ) = y 1 ( t ) , y 3 ( t ) = 1 h ( t ) y 2 ( t ) , y 4 ( t ) = h 2 ( t ) y 3 ( t ) . Open image in new window
(19)
Then (5) is equivalent to the system
{ y 1 ( t ) = y 2 ( t ) , y 2 ( t ) = h ( t ) y 3 ( t ) , y 3 ( t ) = 1 h 2 ( t ) y 4 ( t ) , y 4 ( t ) = h ( t ) r ( t ) y 1 λ ( t ) sgn y 1 ( t ) ( t a ) . Open image in new window
(20)
We have y 1 ( t ) > 0 Open image in new window for t a Open image in new window. Assume by contradiction that y 2 ( t ) < 0 Open image in new window for t a Open image in new window. Let y 3 ( t ) > 0 Open image in new window and y 4 ( t ) < 0 Open image in new window. Since y 4 Open image in new window is nonincreasing, y 4 ( t ) y 4 ( a ) < 0 Open image in new window and
y 3 ( t ) y 3 ( a ) y 4 ( a ) a t 1 / h 2 ( s ) d s . Open image in new window

Letting t Open image in new window, we get a contradiction with the positiveness of y 3 Open image in new window. The remaining case y 3 ( t ) < 0 Open image in new window can be eliminated in a similar way using (6). Observe that system (20) is a special case of the Emden-Fowler system investigated in [13], and the proof follows also from [[13], Lemma 2.1].

Claim (b). Without loss of generality, suppose that r ( t ) > 0 Open image in new window for t T Open image in new window and there exists a solution x of (1) such that x ( t ) > 0 Open image in new window and x ( t ) 0 Open image in new window on [ T , ) Open image in new window, T 0 Open image in new window. Borůvka [18] proved that if (2) is oscillatory, then there exists a function α C 3 [ T , ) Open image in new window, called a phase function, such that α ( t ) > 0 Open image in new window and
3 ( α ( t ) ) 2 4 ( α ( t ) ) 4 α ( t ) 2 ( α ( t ) ) 3 + q ( t ) α 2 ( t ) = 1 . Open image in new window
(21)
Using this result, we can consider the change of variables
s = α ( t ) , x ( t ) = 1 α ( t ) X ( s ) ( ˙ = d d s ) Open image in new window
(22)
for t [ T , ) Open image in new window, s [ T , ) Open image in new window, T = α ( T ) Open image in new window. Thus, t = α 1 ( s ) Open image in new window and
x ( t ) = 1 2 ( α ( t ) ) 3 2 α ( t ) X ( s ) + ( α ( t ) ) 1 / 2 X ˙ ( s ) , x ( 4 ) ( t ) = 1 2 ( α ( t ) ) 3 2 α ( t ) X ( s ) + 3 4 ( α ( t ) ) 5 2 ( α ( t ) ) 2 X ( s ) + ( α ( t ) ) 3 2 X ¨ ( s ) . Open image in new window
Substituting into (1), we obtain the second-order equation
X ¨ ( s ) + X ( s ) ( 3 4 ( α ( t ) ) 4 ( α ( t ) ) 2 1 2 ( α ( t ) ) 3 α ( t ) + q ( t ) ( α ( t ) ) 2 ) + ( α ( t ) ) 3 2 r ( t ) x λ ( t ) = 0 . Open image in new window
From here and (21), we obtain
X ¨ ( s ) + X ( s ) = ( α ( t ) ) 3 2 r ( t ) x λ ( t ) < 0 . Open image in new window
(23)
Since x ( t ) 0 Open image in new window, (22) yields X ( s ) 0 Open image in new window and so X ¨ ( s ) < 0 Open image in new window, that is, X ˙ Open image in new window is decreasing. If there exists s 1 T Open image in new window such that X ˙ ( s 1 ) < 0 Open image in new window, X becomes eventually negative, which is a contradiction. Then X ˙ ( s ) 0 Open image in new window and X ( s ) Open image in new window is nondecreasing. Let T 1 T Open image in new window be such that X ( s ) > 0 Open image in new window on [ T 1 , ) Open image in new window. Thus, using (23) we obtain
X ˙ ( s ) X ˙ ( T 1 ) = T 1 s X ¨ ( u ) d u T 1 s X ( u ) d u X ( T 1 ) ( s T 1 ) . Open image in new window

Hence, lim s X ˙ ( s ) = Open image in new window, which contradicts the nonnegativity of X ˙ ( s ) Open image in new window. Finally, the case X ( s ) 0 Open image in new window on [ T , ) Open image in new window cannot occur, because if x ( t ) 0 Open image in new window on [ T , ) Open image in new window, then from (1) and r ( t ) > 0 Open image in new window, we have x 0 Open image in new window on [ T , ) Open image in new window, which is a contradiction.

Finally, let x be a positive solution of (1) satisfying (3). Then x Open image in new window is either oscillatory or x ( t ) < 0 Open image in new window for large t. Assume x ( t ) < 0 Open image in new window on some J = [ T , ) Open image in new window, then x Open image in new window is decreasing and either x ( t ) 0 Open image in new window or x ( t ) < 0 Open image in new window for large t. If x ( t ) 0 Open image in new window for large t, then we get a contradiction with (3). If x ( t ) < 0 Open image in new window, then x ( t ) x ( T 1 ) < 0 Open image in new window for t T 1 T Open image in new window and x becomes negative for large t. Hence x Open image in new window must be oscillatory. □

For λ > 1 Open image in new window, the analogous result to Theorem 1 is the following oscillation result.

Proposition 2 Let λ > 1 Open image in new window, r ( t ) > 0 Open image in new window for large t. Assume either (11) for large t, (12), or (13), (14). Then all the solutions of (1) are oscillatory.

Proof Let x be a solution of (1) and h be the principal solution of (2). Then y = ( y 1 , y 2 , y 3 , y 4 ) Open image in new window, where y i Open image in new window are given by (19), is a solution of system (20). Proceeding by the similar way as in the proof of Theorem 1, we have that (15) holds. Using the change of the order of integration in (15), we can check that conditions of Theorem 4.3 in [13] applied to system (20) are verified. Hence by this result all the solutions of (20) are oscillatory, which gives the conclusion. □

The following result follows from [[7], Theorem 1.5] and completes Proposition 2 in the case when (2) is oscillatory.

Proposition 3 Let λ > 1 Open image in new window, q ( t ) 1 Open image in new window and r ( t ) > 0 Open image in new window for t R + Open image in new window. Then the condition t r ( t ) d t = Open image in new window is necessary and sufficient for every solution of (1) to be oscillatory.

In the light of these results, in the sequel, we study asymptotic and oscillation problems to (1) when (2) is oscillatory.

3 Vanishing at infinity solutions

In this section we study when all nonoscillatory solutions of (1) are vanishing at infinity.

Theorem 3 Let λ > 1 Open image in new window and (2) be oscillatory. Assume that q ( t ) K t 2 Open image in new window for large t and some K > 0 Open image in new window, the functions
q q 3 / 2 , q q 2 , q q 5 / 2 are bounded on R + , Open image in new window
(24)
and
| r ( t ) | q 2 ( t ) as t . Open image in new window
(25)

Then any eventually positive solution of (1) is vanishing at infinity.

The proof of Theorem 3 is based on the following auxiliary results.

Consider the fourth-order quasi-linear differential equation
y ( 4 ) ( s ) + i = 1 3 Q i ( s ) y ( i ) ( s ) + R ( s ) | y | λ ( s ) sgn y ( s ) = 0 , Open image in new window
(26)

where Q i Open image in new window and R are continuous functions on [ 0 , ) Open image in new window. In [[15], Theorem 2.4], the following uniform estimate for positive solutions of (26) with a common domain was proved.

Proposition 4 ([[15], Theorem 3.4, Corollary 3.6])

Assume λ > 1 Open image in new window. Let y be a positive solution of (26) defined on [ 0 , b ] Open image in new window and
| R ( s ) | r , | Q i ( s ) | Q 4 i , i = 1 , 2 , 3 , Open image in new window
(27)
on [ 0 , b ] Open image in new window for some constants r > 0 Open image in new window and Q > 0 Open image in new window. Then there exists a positive constant M = M ( λ ) Open image in new window such that
y ( s ) M r 1 λ 1 δ 4 λ 1 ( s ) for s [ 0 , b ] , Open image in new window
(28)
where
δ ( s ) = min { s , α , b s } , α = 2 19 Q 1 . Open image in new window

Remark 1 In [[15], Theorem 3.4] the constant M is explicitly calculated.

Lemma 2 Let λ > 1 Open image in new window. Assume that (27) holds on [ 0 , ) Open image in new window. Then any positive solution of (26) defined on [ 0 , ) Open image in new window satisfies
y ( s ) M r 1 λ 1 α 4 λ 1 for s [ α , ) , Open image in new window
(29)

where α and M are constants from Proposition  4.

Proof Let b α Open image in new window. By Proposition 4, applied on [ 0 , b ] Open image in new window, we have δ = δ ( s ) α Open image in new window for s [ α , b α ] Open image in new window and
y ( s ) M r 1 λ 1 α 4 λ 1 for  s [ α , b α ] . Open image in new window

Letting b Open image in new window, we get (29). □

The next lemma describes the transformation between solutions of (1) and a certain quasi-linear equation.

Lemma 3 Let q ( t ) > 0 Open image in new window on [ a , ) Open image in new window be such that
a q ( t ) d t = Open image in new window
and consider the transformation
s = a t q ( t ) d t , x ( t ) = y ( s ) , ˙ = d d s . Open image in new window
Then x = x ( t ) Open image in new window is a solution of equation (1) on [ a , ) Open image in new window if and only if y = y ( s ) Open image in new window is a solution of the equation
d 4 y d s 4 + 3 q ( t ) q 3 / 2 ( t ) d 3 y d s 3 + [ 2 q ( t ) q 2 ( t ) ( q ( t ) ) 2 4 q 3 ( t ) + 1 ] y ¨ + [ q ( t ) 2 q 5 / 2 ( t ) 3 4 q ( t ) q ( t ) q 7 / 2 ( t ) + 3 8 ( q ( t ) ) 3 q 9 / 2 ( t ) + q ( t ) 2 q 3 / 2 ( t ) ] y ˙ + r ( t ) q 2 ( t ) | y | λ sgn y = 0 on [ 0 , ) , Open image in new window
(30)

where t = t ( s ) Open image in new window is the inverse function to s = s ( t ) Open image in new window.

Proof We have
x ( t ) = y ˙ ( s ) q ( t ) , x ( t ) = y ¨ ( s ) q ( t ) + y ˙ ( s ) q ( t ) 2 q ( t ) , x ( t ) = d 3 y ( s ) d s 3 q 3 / 2 ( t ) + 3 2 y ¨ ( s ) q ( t ) + y ˙ ( s ) [ q ( t ) 2 q ( t ) ( q ( t ) ) 2 4 q 3 / 2 ( t ) ] , x ( 4 ) ( t ) = d 4 y ( s ) d s 4 q 2 ( t ) + 3 d 3 y d s 3 q ( t ) q ( t ) + y ¨ ( s ) [ 2 q ( t ) ( q ( t ) ) 2 4 q ( t ) ] x ( 4 ) ( t ) = + y ˙ ( s ) [ q ( t ) 2 q 1 / 2 ( t ) 3 4 q ( t ) q ( t ) q 3 / 2 ( t ) + 3 8 ( q ( t ) ) 3 q 5 / 2 ( t ) ] . Open image in new window

Substituting into (1), we get the conclusion. □

Proof of Theorem 3 Let x be a positive solution of (1) on I = [ a , ) Open image in new window ( a > 0 Open image in new window). Suppose, for simplicity, that q ( t ) K t 2 Open image in new window for t a Open image in new window. Let
| q ( t ) | q 3 2 ( t ) C 1 , | q ( t ) | q 2 ( t ) C 2 , | q ( t ) | q 5 2 ( t ) C 3 Open image in new window
(31)
on [ 0 , ) Open image in new window for some positive constants C 1 Open image in new window, C 2 Open image in new window, C 3 Open image in new window and
Q = max { 3 C 1 , ( 2 C 2 + 1 4 C 1 2 + 1 ) 1 / 2 , ( C 3 2 + 3 4 C 1 C 2 + 3 8 C 1 3 + C 1 2 ) 1 / 3 } . Open image in new window
(32)
Denote
α = 2 19 Q 1 , a ¯ = exp ( α / K ) . Open image in new window
(33)
Define the function t = t ( t ) Open image in new window such that
t t q ( s ) d s = α Open image in new window
(34)
for t a ¯ Open image in new window. Then, according to q ( t ) K t 2 Open image in new window, we have
t t exp ( α / K ) 1 . Open image in new window
(35)
Let r / q 2 Open image in new window be nondecreasing on [ a ˜ , ) Open image in new window and put T max { a , a ¯ , a ˜ } Open image in new window. Choose t 0 [ T , ) Open image in new window arbitrarily fixed. Since q ( t ) K t 2 Open image in new window, we can consider the transformation from Lemma 3 with a = t ( t 0 ) Open image in new window, i.e.,
s = t ( t 0 ) t q ( τ ) d τ , x ( t ) = y ( s ) , ˙ = d d s . Open image in new window
(36)
Then equation (1) is transformed into equation (30) which is a quasilinear equation of the form (26), where
| Q 1 ( s ) | Q 3 , | Q 2 ( s ) | Q 2 , | Q 3 ( s ) | Q , R ( s ) = r ( t ( s ) ) q 2 ( t ( s ) ) Open image in new window
and Q is defined by (31) and (32). Choose t 0 T Open image in new window arbitrarily. We apply Lemma 2 to equation (30) with
r = min s 0 | r ( t ( s ) ) | q 2 ( t ( s ) ) = | r ( t ( t 0 ) ) | q 2 ( t ( t 0 ) ) . Open image in new window
Hence estimate (29) with s = α Open image in new window reads as
x ( t 0 ) = y ( α ) C ( q 2 ( t ( t 0 ) ) | r ( t ( t 0 ) ) | ) 1 λ 1 , Open image in new window
(37)

where C = M α 4 λ 1 Open image in new window. Letting t 0 Open image in new window, we have by (35) that t ( t 0 ) Open image in new window and the conclusion follows from (25) and (37). □

From the proof of Theorem 3, we get the estimate for the set of all nonoscillatory solutions of (1) which will be used in the next section.

Corollary 2 Let λ > 1 Open image in new window, lim t q ( t ) = Open image in new window, (24) and (25) hold. Then, for any ε > 0 Open image in new window, there exists a positive constant C = C ( λ , q ( t ) , ε ) Open image in new window and T 0 Open image in new window such that every nonoscillatory solution x of (1) satisfies
| x ( t ) | C ( q 2 ( t ε ) | r ( t ε ) | ) 1 λ 1 for t T . Open image in new window
(38)
Proof Let ε > 0 Open image in new window be fixed and let T 0 Open image in new window be such that
q ( t ) α 2 ε 2 for  t T , Open image in new window
where α is given by (33). Let t T Open image in new window be fixed. Using estimate (37) with t 0 = t T Open image in new window, we have
x ( t ) C ( q 2 ( t ( t ) ) | r ( t ( t ) ) | ) 1 λ 1 ( t T ) , Open image in new window
(39)
where t Open image in new window is given by (34), i.e.,
α = t ( t ) t q ( τ ) d τ α ε ( t t ( t ) ) . Open image in new window

Therefore t ( t ) t ε Open image in new window and estimate (38) follows from (25) and (39). □

Example 1 Consider the equation
x ( 4 ) ( t ) + ( t + 1 ) x ( t ) ( 24 ( t + 1 ) 4 + 2 ( t + 1 ) 7 ) x 9 ( t ) = 0 ( t R + ) . Open image in new window
(40)

Then r / q 2 Open image in new window and by Theorem 3 all eventually positive solutions are vanishing at infinity. One can check that x = 1 t + 1 Open image in new window is such a solution of (40).

Open problem It is an open problem to find conditions for the solvability of boundary value problem (1), (3), (4) in case r ( t ) < 0 Open image in new window and (2) is oscillatory.

In view of Theorem 2, Corollary 2 and Proposition 1, it is a question whether (1) can have vanishing at infinity solutions in case r ( t ) > 0 Open image in new window and (2) is oscillatory.

In the next section, we show that under certain additional assumptions the answer is negative.

4 Oscillation

Here we consider (1) in case r ( t ) > 0 Open image in new window for large t. When (2) is nonoscillatory, we have established the oscillation criterion in Proposition 2. When (2) is oscillatory, the following oscillation theorem holds.

Theorem 4 Let λ > 1 Open image in new window, r ( t ) > 0 Open image in new window and assumptions (24), (25) hold. Assume
lim t q ( t ) = , q ( t ) 0 , q ( t ) 0 for large t Open image in new window
(41)
and
lim sup t q 1 + σ ( t ) ( q 2 ( t ε ) r ( t ε ) ) 1 λ 1 < Open image in new window
(42)

for some ε > 0 Open image in new window and σ > 0 Open image in new window. Then problem (1), (3) is not solvable and all the solutions of (1) are oscillatory.

Proof Suppose that (25) holds on [ a , ) Open image in new window. First, observe that the assumption (42) implies that
0 q ( t ) ( q 2 ( t ϵ ) r ( t ϵ ) ) 1 λ 1 d t < . Open image in new window
(43)
Indeed, putting H ( t ) = ( q 2 ( t ) / r ( t ) ) 1 λ 1 Open image in new window, we have
a t q ( s ) H ( s ) d s < H ( a ) q 1 + σ ( t ) a t q ( s ) q 1 + σ ( s ) d t , Open image in new window
and thus, in view of (42), we get (43). Consider a solution x of (1) such that x ( t ) > 0 Open image in new window for t t 0 0 Open image in new window. According to Corollary 2, there exists t ¯ 0 Open image in new window such that
0 < x ( t ) C ( q 2 ( t ϵ ) r ( t ϵ ) ) 1 λ 1 for  t t ¯ 0 t 0 , Open image in new window
(44)
and in view of (25) we get lim t x ( t ) = 0 Open image in new window. Consider the function
F ( t ) = x ( t ) q ( t ) x ( t ) + q ( t ) x ( t ) . Open image in new window
Then
F ( t ) = r ( t ) x λ ( t ) + q ( t ) x ( t ) , Open image in new window
and in view of (41) the function F is increasing for large t. Hence, there exists t 2 t 1 Open image in new window such that either
F ( t ) < 0 for  t t 2 , Open image in new window
(45)
or
F ( t ) F ( t 2 ) > 0 for  t t 2 . Open image in new window
(46)

According to Theorem 2(b), x Open image in new window oscillates. Define by { τ k } k = 1 Open image in new window an increasing sequence of zeros of x Open image in new window tending to ∞ with τ 1 t 2 Open image in new window.

Define
Z ( t ) = x ( t ) q ( t ) x ( t ) 2 t q ( s ) x ( s ) d s . Open image in new window
(47)
In view of (44) and (43) the function Z is well defined and
Z ( t ) = x ( t ) q ( t ) x ( t ) + q ( t ) x ( t ) = F ( t ) Open image in new window
(48)
on [ t 0 , ) Open image in new window. Moreover, we have from (42), (44) and (47)
lim k Z ( τ k ) = 0 . Open image in new window
(49)
If (45) holds, then Z ( t ) < 0 Open image in new window and because
Z ( τ 1 ) = q ( τ 1 ) x ( τ 1 ) 2 τ 1 q ( s ) x ( s ) d s < 0 , Open image in new window

we get Z ( t ) Z ( τ 1 ) < 0 Open image in new window for t τ 1 Open image in new window. This is a contradiction with (49), so (45) is impossible.

If (46) holds, then Z ( t ) > 0 Open image in new window and lim t Z ( t ) = Open image in new window. This is again a contradiction with (49), so also this case is impossible. □

Example 2 Consider the equation
x ( 4 ) ( t ) + c t 2 x ( t ) + σ | x | 3 ( t ) sgn x ( t ) = 0 ( t R + ) , Open image in new window

where σ = ± 1 Open image in new window. If σ = 1 Open image in new window and c ( 0 , 1 / 4 ) Open image in new window, then by Theorem 1 this equation has a solution satisfying (3) and (4). If σ = ± 1 Open image in new window and c > 1 / 4 Open image in new window, then by Theorem 3 any nonoscillatory solution (if any) satisfies (3).

5 Extensions

As it was mentioned in [10], a certain nonlinear PDE leads to the fourth-order equation with the exponential nonlinearity. In the sequel, we show that the results of this paper can be extended to the nonlinear equation
where q, r are as for (1) and f ( u ) u > 0 Open image in new window for u 0 Open image in new window such that
f ( u ) k u λ for  u R + Open image in new window
(50)

for some λ 1 Open image in new window and k > 0 Open image in new window. The prototype of such an extension is the function f ( u ) = e u 1 Open image in new window for u 0 Open image in new window.

Theorems 1-4 read for (1′) as follows.

Theorem 1′ Let λ 1 Open image in new window, r ( t ) < 0 Open image in new window and (11) hold for t R + Open image in new window. Assume that either (i) (12), or (ii) (13) and (14) hold. Then problem (1′), (3), (4) has a solution for any c > 0 Open image in new window.

Proof of Theorem 1′ It is analogous to the proofs of Proposition 1 and Theorem 1 replacing the nonlinearity | y 1 ( t ) | λ ( t ) sgn y 1 ( t ) Open image in new window in system (10) by f ( y 1 ( t ) ) Open image in new window. Lemma 1 remains to hold as a sufficient condition for (3). □

Theorem 2′ Theorem  2 remains to hold for (1′) without assuming (50).

Proof of Theorem 2′ In the proof of claim (a) of Theorem 2, we consider system (20) where the nonlinearity | y 1 ( t ) | λ ( t ) sgn y 1 ( t ) Open image in new window is replaced by f ( y 1 ( t ) ) Open image in new window. The proof of claim (b) of Theorem 2 is the same for the nonlinearity f. □

Theorem 3′ Theorem  3 remains to hold for (1′).

Proof of Theorem 3′ Let x be a positive solution of (1′) on [ a , ) Open image in new window. Then v = x Open image in new window is a solution of the equation
v ( 4 ) ( t ) + q ( t ) v ( 2 ) ( t ) + R ( t ) v λ ( t ) = 0 , Open image in new window
(51)
where
R ( t ) = r ( t ) f ( x ( t ) ) x λ ( t ) k r ( t ) for  t a . Open image in new window
(52)

Now we apply Theorem 3 to (51). □

Theorem 4′ Let the assumptions of Theorem  4 hold. Then (1′) has no eventually positive solutions.

Proof of Theorem 4′ It is similar to the one of Theorem 4. In view of (52), the estimate (38) holds and the energy function F is the same. □

Notes

Acknowledgements

Supported by the grant GAP 201/11/0768 of the Czech Grant Agency.

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© Bartušek and Došlá; licensee Springer. 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Authors and Affiliations

  1. 1.Faculty of ScienceMasaryk University BrnoBrnoThe Czech Republic

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