ON THE STABILITY OF THE SOLUTIONS OF CERTAIN FIFTH ORDER NON-AUTONOMOUS DIFFERENTIAL EQUATIONS

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1 ARCHIVUM MATHEMATICUM (BRNO) Tomus 41 (25), ON THE STABILITY OF THE SOLUTIONS OF CERTAIN FIFTH ORDER NON-AUTONOMOUS DIFFERENTIAL EQUATIONS A. I. SADEK Our aim in this paper is to present sufficient conditions under which all solutions of (1.1) tend to zero as t. 1. Introduction The equation studied here is of the form (1.1) x (5) + f(t, ẋ, ẍ,... x )x (4) + φ(t, ẍ,... x ) + ψ(t, ẍ) + g(t, ẋ) + e(t)h(x) =, where f, φ, ψ, g, e and h are continuous functions which depend only on the displayed arguments, φ(t,, ) = ψ(t, ) = g(t, ) = h() =. The dots indicate differentiation with respect to t and all solutions considered are assumed real. Chukwu [3] discussed the stability of the solutions of the differential equation x (5) + ax (4) + f 2 (... x ) + cẍ + f 4 (ẋ) + f 5 (x) =. In [1], sufficient conditions for the uniform global asymptotic stability of the zero solution of the differential equation x (5) + f 1 (... x )x (4) + f 2 (... x ) + f 3 (ẍ) + f 4 (ẋ) + f 5 (x) = were investigated. Tiryaki & Tunc [6] and Tunc [7] studied the stability of the solutions of the differential equations x (5) + φ(x, ẋ, ẍ,... x, x (4) )x (4) + b... x + h(ẋ, ẍ) + g(x, ẋ) + f(x) =, x (5) + φ(x, ẋ, ẍ,... x, x (4) )x (4) + ψ(ẍ,... x ) + h(ẍ) + g(ẋ) + f(x) =. We shall present here sufficient conditions, which ensure that all solutions of (1.1) tend to zero as t. Many results have been obtained on asymptotic properties of non-autonomous equations of third order in Swich [5], Hara [4] and Yamamoto [8]. 2 Mathematics Subject Classification: 34C, 34D. Key words and phrases: asymptotic stability, Lyapunov function, nonautonomous differential equations of fifth order. Received April 23, 23.

2 94 A. I. SADEK 2. Assumptions and theorems We shall state the assumptions on the functions f, φ, ψ, g, e and h appeared in the equation (1.1). Assumptions: (1) h(x) is a continuously differentiable function in R 1, and e(t) is a continuously differentiable function in R + = [, ). (2) The function g(t, y) is continuous in R + R 1, and for the function g(t, y) there exist non-negative functions d(t), g (y) and g 1 (y) which satisfy the inequalities d(t)g (y) g(t, y) d(t)g 1 (y) for all (t, y) R + R 1. The function d(t) is continuously differentiable in R +. Let g(y) 1 2 g (y) + g 1 (y), g(y) and g (y) are continuous in R 1. (3) The function ψ(t, z) is continuous in R + R 1. For the function ψ(t, z) there exist non-negative functions c(t), ψ (z) and ψ 1 (z) which satisfy the inequalities c(t)ψ (z) ψ(t, z) c(t)ψ 1 (z) for all (t, z) R + R 1. The function c(t) is continuously differentiable in R +. Let ψ(z) 1 2 ψ (z) + ψ 1 (z), ψ(z) is continuous in R 1. (4) The function φ(t, z, w) is continuous in R + R 2. For the function φ(t, z, w) there exist non-negative functions b(t), φ (z, w) and φ 1 (z, w) which satisfy the inequalities b(t)φ (z, w) φ(t, z, w) b(t)φ 1 (z, w) for all (t, z, w) R + R 2. The function b(t) is continuously differentiable in R +. Let φ(z, w) 1 2 φ (z, w) + φ 1 (z, w), φ(z, w) and φ(z, w)/ z are continuous in R 2. (5) The function f(t, y, z, w) is continuous in R + R 3, and for the function f(t, y, z, w) there exist functions a(t), f (y, z, w) and f 1 (y, z, w) which satisfy the inequality a(t)f (y, z, w) f(t, y, z, w) a(t)f 1 (y, z, w) for all (t, y, z, w) R + R 3. Further the function a(t) is continuously differentiable in R +, and let f(y, z, w) 1 2 f (y, z, w) + f 1 (y, z, w), f(y, z, w) is continuous in R 3.

3 NON AUTONONOUS DIFFERENTIAL EQUATIONS OF FIFTH ORDER 95 Theorem 1. Further to the basic assumptions (1) (5), suppose the following (ɛ, ɛ 1,..., ɛ 5 are small positive constants): (i) A a(t) a 1, B b(t) b 1, C c(t) c 1, D d(t) d 1, E e(t) e 1, for t R +. (ii),..., α 5 are some constants satisfying (2.1) (2.2) >, α 2 α 3 >, ( α 2 α 3 )α 3 ( ) >, δ : = (α 4 α 3 α 2 α 5 )( α 2 α 3 ) ( ) 2 >, α 5 > ; 1 : = (α 4α 3 α 2 α 5 )( α 2 α 3 ) d(t) g (y) α 5 > 2ɛα 2, for all y and all t R + ; (2.3) 2 : = α 4α 3 α 2 α 5 (α 4 α 5 )γd(t) α 4 ( α 2 α 3 ) ɛ >, for all y and all t R +, where g(y)/y, y (2.4) γ := g (), y =. (iii) ɛ f(y, z, w) ɛ 1 for all z and w. (iv) φ(, ) =, φ(z, w)/w α 2 ɛ 2 (w ), φ(z, z w). (v) ψ() =, ψ(z)/z α3 ɛ 3 (z ). (vi) g() =, g(y)/y Eα4 d (y ), α 4 g (y) ɛ 4 for all y and g (y) g(y)/y α 5 δ /Dα 2 4 (α 2 α 3 ) (y ). (vii) h() =, h(x) sgn x > (x ), H(x) x h(ξ)dξ as x and α 5 h (x) ɛ 5 for all x. (viii) β (t)dt <, e (t) as t, where β (t) : = b +(t) + c +(t) + d (t) + e (t), b +(t) : = maxb (t), and c +(t) := maxc (t),. (ix) A(f 1 f ) + B(φ 1 φ ) + C(ψ 1 ψ ) + D(g 1 g ) (y 2 + z 2 + w 2 + u 2 ) 1/2, where is a non-negative constant.

4 96 A. I. SADEK Then every solution of (1.1) satisfies x(t), ẋ(t), ẍ(t),... x (t), x (4) (t) as t. Next, considering the equation (2.5) x (5) + a(t)f(ẋ, ẍ,... x )x (4) + b(t)φ(ẋ, ẍ) + c(t)ψ(ẍ) + d(t)g(ẋ) + e(t)h(x) =, we can take the function g(y) in place of g (y) and g 1 (y); the function φ(y, z) in place of φ (y, z) and φ 1 (y, z); the function ψ(z) in place of ψ (z) and ψ 1 (z), and the function f(y, z, w) in place of f (y, z, w) and f 1 (y, z, w) in the Assumptions (2) (5). Thus in this case the functions g(y), φ(y, z), ψ(z), f(y, z, w) coincide with g(x, y), φ(y, z), ψ(z), f(y, z, w) respectively. Thus from Theorem 1, we have Theorem 2. Suppose that the functions a(t), b(t), c(t), d(t) and e(t) are continuously differentiable in R +, and the functions h(x), g(x, y), φ(y, z), ψ(z), f(y, z, w), g (y), h (x), z φ(y, z) and that these functions satisfy the following conditions: (i) A a(t) a 1, B b(t) b 1, C c(t) c 1, D d(t) d 1, E e(t) e 1 for t R +. (ii),..., α 5 are some constants satisfying >, α 2 α 3 >, ( α 2 α 3 )α 3 ( ) >, δ := (α 4 α 3 α 2 α 5 )( α 2 α 3 ) ( ) 2 >, α 5 > ; 1 : = (α 4α 3 α 2 α 5 )( α 2 α 3 ) d(t) g (y) α 5 > 2ɛα 2, for all y and all t R + ; 2 : = α 4α 3 α 2 α 5 (α 4 α 5 )γd(t) α 4 ( α 2 α 3 ) for all y and all t R +, where g(y)/y, y γ : = g (), y =. (iii) ɛ f(y, z, w) ɛ 1, for all z and w. ɛ >, (iv) φ(, ) =, φ(z, w)/w α 2 ɛ 2 (w ), z φ(z, w). (v) ψ() =, ψ(z)/z α 3 ɛ 3 (z ). (vi) g() =, g(y)/y Eα4 d (y ), α 4 g (y) ɛ 4 for all y and g (y) g(y)/y α 5 δ /Dα 2 4 (α 2 α 3 ) (y ).

5 NON AUTONONOUS DIFFERENTIAL EQUATIONS OF FIFTH ORDER 97 (vii) h() =, h(x) sgn x > (x ), H(x) x h(ξ) dξ as x and α 5 h (x) ɛ 5 for all x. (viii) β (t) dt <, e (t) as t, where β (t) : = b + (t) + c + (t) + d (t) + e (t), b + (t) : = maxb (t), and c + (t) := maxc (t),. Then every solution of (2.5) satisfies x(t), ẋ(t), ẍ(t),... x (t), x (4) (t) as t. 3. The Lyapunov function V (t, x, y, z, w, u) We consider, in place of (1.1), the equivalent system (3.1) ẋ = y, ẏ = z, ż = w, ẇ = u, u = f(t, y, z, w)u φ(t, z, w) ψ(t, z) g(t, y) e(t)h(x). The proof of the theorem is based on some fundamental properties of a continuously differentiable function V = V (t, x, y, z, w, u) defined by 2V = u uw + 2α 4( α 2 α 3 ) w uz + 2δyu + 2b(t) + α 2 1 α 4( α 2 α 3 ) w α 3 + α 4 ( α 2 α 3 ) + 2 δwy + 2d(t)w g(y) + 2e(t)wh(x) + 2 c(t) z φ(z, ω) dω ψ(ζ) dζ δ wz α2 α 4 ( α 2 α 3 ) + α 4 δ z 2 + 2δα 2 yz + 2 d(t)z g(y) 2α 5 yz + 2 e(t)zh(x) + 2α 4( α 2 α 3 ) y d(t) g(η) dη + (δα 3 α 5 )y 2 (3.2) + 2α 4( α 2 α 3 ) x e(t)yh(x) + 2δe(t) h(ξ) dξ, where (3.3) δ := α 5 ( α 2 α 3 )/( ) + ɛ. The properties of the function V = V (t, x, y, z, w, u) are summarized in Lemma 1 and Lemma 2.

6 98 A. I. SADEK Lemma 1. Subject to the hypotheses (i) (vii) of the theorem, there are positive constants D 7 and D 8 such that (3.4) D 7 H(x) + y 2 + z 2 + w 2 + u 2 V D 8 H(x) + y 2 + z 2 + w 2 + u 2. Proof. We observe that 2V in (3.2) can be rearranged as (3.5) where 2V = u + w + α 4( α 2 α 3 ) + α 4( ) ( α 2 α 3 )γd(t) 2 z + δy α1 α 2 α 3 e(t)h(x) + α 2 α 3 γd(t)y + γd(t)z α α 4 δ ( γd(t)w + α 4 ( ) 2 z + α ) 2 5 y + 2 (w + z) 2 α 4 ( α4 α 3 α 2 α ) ɛ yz + S i, x i=1 S 1 : = 2δe(t) h(ξ) dξ α 4( α 2 α 3 ) ( )γd(t) e2 (t)h 2 (x), S 2 : = α 4( α 2 α 3 )d(t) y 2 g(η) dη y g(y) α δα 3 α 5 δ α 4 ( ) 2 δ2 y 2, S 3 : = ɛ w 2 + 2b(t) S 4 : = 2 c(t) z w φ(z, ω) dω α 2 w 2, ψ(ζ) dζ α 3 z 2. It can be seen from the estimates arising in the course of the proof of [2; Lemma 1] that x (3.6) 2α 5 h(ξ) dξ h 2 (x), Since y g(y) x S 1 2ɛe h(ξ) dξ. y g(η) dη + y η g (η) dη,

7 NON AUTONONOUS DIFFERENTIAL EQUATIONS OF FIFTH ORDER 99 we have S 2 = α 4( α 2 α 3 )d(t) y 2 g(η) dη y g(y) [ α 5 δ + α 4 ( ) ɛ ɛ + 2α 5( α 2 α 3 ) ] α 3 y 2 y [ 2α 5 δ = α 4 ( ) α 4( α 2 α 3 )d(t) g (η) g(η) η 2ɛ ɛ + 2α 5( α 2 α 3 ) ] α 3 η dη α 4 α 5 y [ α 5 δ α 4 ( ) 2ɛ ɛ + 2α 5( α 2 α 3 ) ] α 3 η dη, by (vi) and (i) α 5 δ 4α 4 ( ) y2, provided that (3.7) α 5 δ 4α 4 ( ) ɛ ɛ + 2α 5( α 2 α 3 ) α 3, which we now assume. From (i), (iv) and (v) we find S 3 = ɛ w 2 + 2b(t) ɛ w S 4 = 2 c(t) w z w φ(z, ω)dω α 2 w 2 φ(z, ω) α 2 ω dω ɛ w 2, ω z ψ(ζ) ψ(ζ)dζ α 3 z 2 2 α 3 ζ dζ. ζ On gathering all of these estimates into (3.5) we deduce 2V u + w + α 4( α 2 α 3 ) x + 2 (w + z) 2 + 2ɛe ( α4 α 3 α 2 α ) 5 + 2ɛ yz, 2 α 4 δ ( z + δy + ( ) 2 z + α ) 2 5 y α 4 h(ξ) dξ + α 5 δ 4α 4 ( ) y2 + ɛ w 2 by (ii) and (vi). It is clear that there exist sufficiently small positive constants D 1,..., D 5 such that ( 2V D 1 H(x) + 2D 2 y 2 + 2D 3 z 2 + D 4 w 2 + D 5 u 2 α4 α 3 α 2 α ) ɛ yz.

8 1 A. I. SADEK Let ( S 5 := D 2 y 2 α4 α 3 α 2 α ) 5 + 2ɛ yz + D 3 z 2. By using the inequality yz 1 2 (y2 + z 2 ), we obtain ( S 5 D 2 y 2 + D 3 z 2 α4 α 3 α 2 α ) 5 ɛ (y 2 + z 2 ) D 6 (y 2 + z 2 ), for some D 6 >, D 6 = 1 2 mind 2, D 3, if (3.8) ɛ ( )/ ( 2(α 4 α 3 α 2 α 5 ) ) mind 2, D 3, which we also assume. Then 2V D 1 H(x) + (D 2 + D 6 )y 2 + (D 3 + D 6 )z 2 + D 4 w 2 + D 5 u 2. Consequently there exists a positive constant D 7 such that V D 7 H(x) + y 2 + z 2 + w 2 + u 2, provided ɛ is so small that (3.7) and (3.8) hold. From (i), (iv), (v),(vi) and (3.6) we can verify that there exists a positive constant D 8 satisfying V D 8 H(x) + y 2 + z 2 + w 2 + u 2. Thus (3.4) follows. Lemma 2. Assume that all conditions of the theorem hold. positive constants D i (i = 11, 12) such that Then there exist (3.9) V D 12 (y 2 + z 2 + w 2 + u 2 ) + D 11 β V.

9 NON AUTONONOUS DIFFERENTIAL EQUATIONS OF FIFTH ORDER 11 Proof. From (3.2) and (3.1) it follows that (for y, z, w ) d dt V u 2 a(t) f(y, z, w) (3.1) w 2[ b(t) φ(z, w) w z 2[ α 4 ( α 2 α 3 )c(t) y 2 δd(t) g(y) y w + wb(t) α 3 + α 4 ( α 2 α 3 ) ψ(z) z ] δ ] δα 2 + d(t) g (y) α 5 α 4( α 2 α 3 ) e(t)h (x) z φ(z, ω) dω wua(t) f(y, z, w) uzc(t) α 4( α 2 α 3 ) uza(t) f(y, z, w) α 4( α 2 α 3 ) wzb(t) φ(z, w) α 2 w ψ(z) α 3 z wzd(t)α 4 g (y) δyua(t) f(y, z, w) ywe(t)α 5 h (x) φ(z, w) ψ(z) δywb(t) α 2 yze(t)α 5 h (x) δyzc(t) α 3 w z + α 2 1uw + α 4 ( α 2 α 3 ) uz + δyu 1 a(t) α2 α 4 ( α 2 α 3 ) + wz + δα 2 yw 1 b(t) + (α 3 uz + δα 3 yz)1 c(t) α 4 wz1 d(t) (α 5 yw + α 5 yz)1 e(t) a(t)(f 1 f ) + b(t)(φ 1 φ ) + c(t)(ψ 1 ψ ) + d(t)(g 1 g ) u + w + α 4( α 2 α 3 ) z + δy + V t. By (i) and (iii), a(t) f(y, z, w) ɛ. From (i), (iv) and (3.3) we have (for w ) b(t) φ(z, w) w α 3 + α 4 ( α 2 α 3 ) δ φ(z, w) w α 2 + α 2 α 3 + δ α 4 ( α 2 α 3 ) ɛ. By using (i), (v), (3.3) and (2.2) we obtain (for z ) α 4 ( α 2 α 3 )c(t) ψ(z) z δα 2 + d(t) g (y) α 5 (α 4α 3 α 2 α 5 )( α 2 α 3 ) d(t) g (y) α 5 ɛα 2 ɛα 2.

10 12 A. I. SADEK From (i), (vi) and (vii) we find (for y ) δd(t) g(y) α 4( α 2 α 3 ) e(t)h (x) y ɛα 4 E + α 4E( α 2 α 3 ) α 5 h (x) ɛα 4 E. Therefore, the first four terms involving u 2, w 2, z 2 and y 2 in (3.1) are majorizable by (ɛ u 2 + ɛw 2 + ɛα 2 z 2 + ɛα 4 Ey 2 ). Let R(t, x, y, z, w, u) denote the sum of the remaining terms in (3.1). By using hypotheses (i), (iii) (vii) and the inequalities uw 1 2 (u2 + w 2 ), uz 1 2 (u2 + z 2 ), uy 1 2 (u2 + y 2 ), wz 1 2 (w2 + z 2 ), wy 1 2 (w2 + y 2 ), yz 1 2 (y2 + z 2 ) ; it follows that R(t, x, y, z, w, u) D 9 (ɛ 1 + ɛ 2 + ɛ 3 + ɛ 4 + ɛ 5 )(y 2 + z 2 + w 2 + u 2 ) a(t)(f 1 f )+b(t)(φ 1 φ )+c(t)(ψ 1 ψ )+d(t)(g 1 g ) u + w + α 4( α 2 α 3 ) z + δy + V t, for some D 9 >. Thus, after substituting in (3.1), one obtains V (ɛ u 2 +ɛw 2 +ɛα 2 z 2 +ɛα 4 Ey 2 )+D 9 (ɛ 1 + ɛ 2 +ɛ 3 +ɛ 4 +ɛ 5 )(y 2 +z 2 +w 2 +u 2 ) a(t)(f 1 f ) + b(t)(φ 1 φ ) + c(t)(ψ 1 ψ ) + d(t)(g 1 g ) u + w + α 4( α 2 α 3 ) V z + δy + t 1 2 minɛ, ɛ, ɛα 2, ɛα 4 E(y 2 + z 2 + w 2 + u 2 ) a(t)(f 1 f ) + b(t)(φ 1 φ ) + c(t)(ψ 1 ψ ) + d(t)(g 1 g ) (3.11) u + w + α 4( α 2 α 3 ) V z + δy + t, provided that (3.12) D 9 (ɛ 1 + ɛ 2 + ɛ 3 + ɛ 4 + ɛ 5 ) 1 2 minɛ, ɛ, ɛα 2, ɛα 4 E.

11 NON AUTONONOUS DIFFERENTIAL EQUATIONS OF FIFTH ORDER 13 Now we assume that D 9 and ɛ 1,..., ɛ 5 are so small that (3.12) holds. The case y, z, w = is trivially dealt with. From (3.2) we find V t = b (t) w + d (t) + e (t) φ(z, ω) dω + c (t) z ψ(ζ) dζ w g(y) + z g(y) + α 4( α 2 α 3 ) wh(x) + zh(x) + α 4( α 2 α 3 ) y g(η) dη yh(x) + 2δ x h(ξ) dξ. From (iv), (v), (vi), (3.6) and (3.4) we can find a positive constant D 1 which satisfies V D 1 b t +(t) + c +(t) + d (t) + e (t) H(x) + y 2 + z 2 + w 2 (3.13) D 11 β V, where D 11 = D1 D 7. Let D 12 = 1 4 minɛ, ɛ, ɛα 2, ɛα 4 E, and D 13 = max 1,, α 4( α 2 α 3 ), δ then from (3.11), (3.13) and (ix) we obtain the estimate V 2D 12 (y 2 + z 2 + w 2 + u 2 ) + 2D 13 (y 2 + z 2 + w 2 + u 2 ) + D 11 β V. Let be fixed, in what follows, to satisfy = D12 2D 13. With this limitation on we find (3.14) V D 12 (y 2 + z 2 + w 2 + u 2 ) + D 11 β V. Now (3.9) is verified and the lemma is proved. 4. Completion of the proof of Theorem 1 Define the function V (t, x, y, z, w, u) as follows (4.1) V (t, x, y, z, w, u) = e t D11β(τ)dτ V (t, x, y, z, w, u). Then one can verify that there exist two functions U 1 and U 2 satisfying (4.2) U 1 ( x ) V (t, x, y, z, w, u) U 2 ( x ), for all x = (x, y, z, w, u) R 5 and t R + ; where U 1 is a continuous increasing positive definite function, U 1 (r) as r and U 2 is a continuous increasing function. Along any solution (x,y,z,w,u) of (3.1) we have V = e t D11β(τ)dτ V β(t)v D 12 e t D11β(τ)dτ (y 2 + z 2 + w 2 + u 2 ). Thus we can find a positive constant D 14 such that (4.3) V D14 (y 2 + z 2 + w 2 + u 2 ). From the inequalities (4.2) and (4.3), we obtain the uniform boundedness of all solutions (x, y, z, w, u) of (3.1) [9; Theorem 1.2].

12 14 A. I. SADEK Auxiliary lemma Consider a system of differential equations (4.4) x = F (t, x), where F (t, x) is continuous on R + R n, F (t, ) =. The following lemma is well-known [9]. Lemma 3. Suppose that there exists a non-negative continuously differentiabla scalar function V (t, x) on R + R n such that V(4.4) U( x ), where U( x ) is positive definite with respect to a closed set Ω of R n. Moreover, suppose that F (t, x) of system (4.4) is bounded for all t when x belongs to an arbitrary compact set in R n and that F (t, x) satisfies the following two conditions with respect to Ω: (1) F (t, x) tends to a function H( x) for x Ω as t, and on any compact set in Ω this convergence is uniform. (2) corresponding to each ɛ > and each ȳ Ω, there exist a δ, δ = δ(ɛ, ȳ) and T, T = T (ɛ, ȳ) such that if t T and x ȳ < δ, we have F (t, x) F (t, ȳ) < ɛ. Then every bounded solution of (4.4) approaches the largest semi-invariant set of the system x = H( x) contained in Ω as t. From the system (3.1) we set y z F (t, x) = w. u f(t, y, z, w)u φ(t, z, w) ψ(t,z) g(t, y) e(t)h(x) It is clear that F satisfies the conditions of Lemma 3. Let U( x ) = D 14 (y 2 + z 2 + w 2 + u 2 ), then (4.5) V (t, x, y, z, w, u) U( x ) and U( x ) is positive definite with respect to the closed set Ω := (x, y, z, w, u) x R, y =, z =, w =, u =. It follows that in Ω (4.6) F (t, x) =. e(t)h(x)

13 NON AUTONONOUS DIFFERENTIAL EQUATIONS OF FIFTH ORDER 15 According to condition (viii) of the theorem and the boundedness of e, we have e(t) e as t, where 1 e e E. If we set (4.7) H( x) =, e h(x) then the conditions on H( x) of Lemma 3 are satisfied. Since all solutions of (3.1) are bounded, it follows from Lemma 3 that every solution of (3.1) approaches the largest semi-invariant set of the system x = H( x) contained in Ω as t. From (4.7); x = H( x) is the system which has the solutions ẋ =, ẏ =, ż =, ẇ = and u = e h(x), x = k 1, y = k 2, z = k 3, w = k 4, and u = k 5 e h(k 1 )(t t ). In order to remain in Ω, the above solutions must satisfy k 2 =, k 3 =, k 4 = and k 5 e h(k 1 )(t t ) = for allt t, which implies k 5 =, h(k 1 ) =, and thus k 1 = k 5 =. Therefore the only solution of x = H( x) remaining in Ω is x =, that is, the largest semi-invariant set of x = H( x) contained in Ω is the point (,,,, ). Consequently we obtain x(t), ẋ(t), ẍ(t),... x (t), x (4) (t) as t. References [1] Abou-El-Ela, A. M. A. and Sadek, A. I., On complete stability of the solutions of non-linear differential equations of the fifth-order, Proc. of Assiut First Intern. Conf. Part V (199), [2] Abou-El-Ela, A. M. A. and Sadek, A. I., On the asymptotic behaviour of solutions of certain non-autonomous differential equations, J. Math. Anal. Appl. 237 (1999), [3] Chukwu, E. N., On the boundedness and stability of some differential equations of the fifth-order, Siam J. Math. Anal. 7(2) (1976), [4] Hara, T., On the stability of solutions of certain third order ordinary differential equation, Proc. Japan Acad. 47 (1971), [5] Swich, K. E., On the boundedness and the stability of solutions of some differential equations of the third order, J. London Math. Soc. 44 (1969), [6] Tiryaki, A. and Tunc, C., On the boundedness and the stability properties for the solutions of certain fifth order differential equations, Hacet. Bull. Nat. Sci. Eng. Ser. B, 25 (1996),

14 16 A. I. SADEK [7] Tunc, C., On the boundedness and the stability results for the solutions of certain fifth order differential equations, Istambul. Univ. Fen Fak. Mat. Derg. 54 (1997), [8] Yamamoto, M., On the stability of solutions of some non-autonomous differential equation of the third order, Proc. Japan Acad. 47 (1971), [9] Yoshizawa, T., Stability theorem by Liapunov second method, The Mathematical Society of Japan, Tokyo (1966).! " # $ %'& %)( *,+ - sadeka1961@hotmail.com