Boundedness and Asymptotic Behaviour of Solutions of a Second Order Nonlinear Di ff erential Equation

This paper considers asymptotic behaviour of solutions of a nonlinear differential equation of second order where the coefficient of nonlinearity is a bounded function for arbitrarily large values of x in R. Here we obtained sufficient conditions for boundedness, convergence of solutions to zero as x→ ∞, and unboundedness of solutions.


Introduction
Mathematical modelling of most natural, scientific and industrial phenomena is based on nonlinear differential equations which are not easily solvable.Therefore, the determination of behaviour of solutions has attracted the attention of researchers over years (Ademole & Arawomo, 2011).
The monographs by many authors contain good results, including existence Theorems, asymptotic behaviour of solutions stability theory and oscillation theory.Waltman (1964) proved that if a(t) is continuous and ∞ 0 t 2n+1 |a(t)| dt then the equation x + a(t)x 2n+1 = 0, where n is a nonnegative integer, has a solution x(t) satisfying lim t→∞ x(t) t = a 0.Moreover, the solutions of that equation have asymptotic form x(t) = at + b + O(1), a, b constants Kiguradze (1965) had obtained sufficient conditions for existence solutions that tend to ±∞ as t → ∞ of a nonlinear differential equation of Emden-Fowler type u − a(t) |u| n sgnu = 0, where a(t) > 0 and n > 1.
The study (Tong, 1982) showed that the nonlinear differential equation u + f (t, u)sgnu = 0 has solutions which are asymptotic to a + bt where a, b are constants and b 0.
Sufficient criteria of integral type were obtained by Mingareilli and Sadarangani in (2007) for non-oscillation and asymptotic behaviour of solutions of nonlinear differential equation They proved the existence of a positive asymptotically linear solution y(x) Lipovan in her study (2003) has obtained for a class of second order nonlinear differential equations, sufficient conditions that are presented to ensure that some, respectively all solutions are asymptotic to lines.More explicitly, she has given sufficient conditions of integral type guaranteeing that for all a, b ∈ R, there is a solution to the equation u Briefly the asymptotic behaviour of solutions of nonlinear differential equation of second order has been widely studied by Wong (1975), Trench (1976), Kusano, Naito and Usam (1986), Constantin (2000), Hovhannisyan (2004), Saker (2006) and many others.
Thus our study is a continuation of preceding contributions to qualitative theory of differential equations.The aim of this paper is to get sufficient conditions for boundedness, asymptotic convergence to zero as x → ∞ and unboundedness solutions of the equation (1).In this paper we will impose integral conditions on the nonlinear coefficient function h(x).Here we do not impose on h(x) any additional conditions like monotonicity or definiteness of sign.
Motivation for this study comes from the work of Bucur (2006) where a boundedness criterion was obtained for the linear equation y + y = α(x)y, where α(x) → 0, as x → ∞ It should be note that we apply a similar approach to the one used by Bucur, to establish the boundedness of solutions of nonlinear equation.
In this paper, we consider a nonlinear differential equation of second order where It should be note here that we may assume that z > 0, because if z < 0 we set z = −u, u > 0. So we will consider in future the equation
Example 1 Consider the equation Substituting z(x) , z (x), z (x) in the equation Note that the condition 1 Theorem 1 Suppose that h(x)is a continuously differentiable function, bounded for all sufficiently large x ∈ R, and that the integral Proof.Multiplying both sides of the equation ( 2) by y and integrate the result with respect to x from some positive Integrating by parts the integral on the right-hand side we get where A x 0 ≥ 0 is an expression dependent only on x 0 .
Let M = max x 0 ≤t≤x |y(t)|, and without loss of generality we may assume that M ≥ a 0 > 0, otherwise the theorem is proved.Since h(x) is bounded we get

Now we give an example illustrating the theorem.
Example 2 Consider the differential equation x We will show that all its solutions are bounded for x → ∞.
x 2 dx converges to1.Applying the same approach used in theorem to this equation, we get, Which completes the proof.
Theorem 2 Assume that h(x) satisfies the conditions of theorem1 and

then for any solution y(x) of equation (3) holds the asymptotic formula y(x) =
where A, ω 0 are some constants.
Proof.We treat the right-hand side of the equation (3) as a known function.Then using the method of variations of parameters we get the solution of the equation ( 3) By Theorem 1 the function y(x) is bounded, hence we get Therefore formula ( 5) is done.
Corollary 1 Suppose that h(x) is a continuously differentiable function and bounded for all sufficiently large x ∈ R.

If the integrals
p(x)dx converge then any solution of the the equation ( 3) is bounded as x → ∞.
Proof.From Theorem 1 it follows that the solution y(x) of the equation ( 3) is bounded and by virtue of the substitution in Lemma1 it follows that all solutions of the the equation ( 2) are also bounded as x → ∞.
Corollary 2 Suppose that h(x) is a continuously differentiable function and bounded for all sufficiently large x ∈ R. Proof.From Theorem 1 it follows that the solution y(x) of the equation ( 3) is bounded and by virtue of the substitution in Lemma1 it follows that all solutions of the the equation (2) converge to zero as x → ∞.

If the integral
Example 3 Consider the equation x in the equation ( 6) we obtain Since the conditions of Theorem 1 holds, the any solution y(x)of the equation ( 7) is bounded for sufficiently large x.Vol. 4, No. 3;2012 Now since the integral ∞ 1 p(x)dx = ∞, by Lemma 1 it follows that all solutions of the equation ( 6) converge to zero as x → ∞.

Unboundedness of Solutions
Theorem 3 Suppose that h(x) is a continuously differentiable function, bounded for all sufficiently large x ∈ R, and that the integral Proof.Since the function h(x) is bounded then there exists a positive constant L such that |h(x)| ≤ L for x sufficiently large.Assume that the solution y(x) of the equation ( 3) is bounded for sufficiently large x.It follows that there exists a positive constant M such that max Multiplying both sides of the equation ( 3) by y and integrating the result with respect to x from some positive x 0 to x Using the Mean Value Theorem for the integral on the right hand side we obtain in the last inequality we get that M = ∞.This contradiction proves that the solution y(x) is not bounded as x → ∞.
Corollary 3 Suppose that h(x) is a continuously differentiable function and bounded for all sufficiently large x ∈ R. Remark If in the equation (2) the function P(x) is negative then we set P(x) = −P 0 (x), where P 0 (x) > 0, and then all results obtained above will hold.
Lemma 2 If we let z(x) = y(x) x λ , λ 0, then the equation ( 8) is reduced to the equation (3).Proof.Substituting the expressions z(x), z (x), z (x) in the equation and dividing both sides of equation by x −λ we get x 2 y + x 2 y = h(x)x 2+λα x −λα y α Implies that y + y = h(x)y α .(9) Theorem 4 If the conditions of Theorem 1 hold then any solution of the equation ( 9) is bounded as x → ∞.Moreover If the conditions of Theorem 1 hold then any solution of the equation ( 8) converges to zero as x → ∞.
Proof.It follows from Lamma 2 and Theorem 1.
dx = ∞, then any solution of the equation (2) converges to zero as x → ∞.
)| = M for all sufficiently large x.
dx ≺ ∞ , then any solution of the equation (2) is unbounded as x → ∞.Proof.From Theorem 3 it follows that the solution y(x) of the equation (3) is unbounded.Since ∞ x 0 p(x)dx < ∞ then by virtue of the substitution in Lemma1 it follows that all solutions of the equation (2) are unbounded as x → ∞.