Convergence of solutions of certain non-homogeneous third order ...

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Convergence of solutions of certain non-homogeneous third order ...

Kragujevac J. Math. 31 (2008) 5–16.CONVERGENCE OF SOLUTIONS OF CERTAINNON-HOMOGENEOUS THIRD ORDER ORDINARYDIFFERENTIAL EQUATIONSA. U. Afuwapẹ 1 and M.O. Omeike 21 Departmento de Matemáticas, Universidad de Antioquia, Calle 67, No. 53-108,AA1226 Medellín, Colombia.(e-mail: aafuwape@yahoo.co.uk)2 Department of Mathematics, University of Agriculture, Abeokuta, Nigeria.(e-mail: moomeike@yahoo.com)(Received November 1, 2006)Abstract. This paper is concerned with differential equations of the form...x +aẍ + g(ẋ) + h(x) = p(t, x, ẋ, ẍ)where a is a positive constant and g,h and p are continuous in their respective arguments,with functions g and h not necessarily differentiable. By introducing a complete Lyapunovfunction, as well as restricting the incrementary ratio η −1 {h(ξ + η) − h(ξ)}, (η ≠ 0), of h toa closed sub-interval of the Routh-Hurwitz interval, we prove the convergence of solutionsfor this equation. This generalizes earlier results.1 Research supported by Universidad de Antioquia grant from SUI No. IN10095CE


61. INTRODUCTIONThis paper considers the convergence of solutions of the third-order nonlineardifferential equations of the form...x +aẍ + g(ẋ) + h(x) = p(t, x, ẋ, ẍ) (1)in which a > 0 is a constant, functions g, h and p are continuous in their respectivearguments.Any two solutions x 1 (t),x 2 (t) of (1) are said to converge ifx 2 (t) − x 1 (t) −→ 0, ẋ 2 (t) − ẋ 1 (t) −→ 0, ẍ 2 (t) − ẍ 1 (t) −→ 0 (2)as t → ∞. If the relations (2) are true of each (arbitrary) pair of solutions of (1) weshall describe this by saying that all solutions of (1) converge.Many results have been obtained on the convergence of solutions of third-order differentialequations ([2, 5, 7, 9, 10] ). In [10], Tejumola established the convergenceof solutions of (1) assuming that h (of class C ′ ) and g satisfies the Routh-Hurwitzconditionh ′ (x) ≤ c ≤ ab,for some constants b, c and δ, andh(x 2 ) − h(x 1 )x 2 − x 1≥ δ > 0, (x 1 ≠ x 2 ), (3)0 < b ≤ g(y 2) − g(y 1 )y 2 − y 1≤ b 0 < ∞, (y 1 ≠ y 1 ) (4)respectively, with additional conditions thatand{h(x 2 − x 1 ) − h(x 2 ) + h(x 1 )} 2 ≤ B(x 2 − x 1 ) 2 , (5){g(y 2 − y 1 ) − g(y 2 ) + g(y 1 )} 2 ≤ C(y 2 − y 1 ) 2 (6)with constants B, C sufficiently small.In this work, we consider a somewhat different approach to [5] in that we assumeg(ẋ), h(x) are not necessarily differentiable but satisfy (4) and0 < δ ≤ h(x 2) − h(x 1 )x 2 − x 1≤ kab, (x 1 ≠ x 2 ), (7)


7respectively, where k < 1 is a positive constant whose estimate is given later. Moreover,we do not need additional conditions (5) and (6).2. MAIN RESULTSThe main results in this paper, which are in some respects generalizations of [4],are the following :Theorem 1. Suppose that g(0) = h(0) and that(i) there are constants b 0 > 0, b > 0 such that g(y) satisfies inequalities (4) ;(ii) there are constants δ > 0, k < 1 such that for any ξ, η, (η ≠ 0),the incrementaryratio for h satisfiesη −1 {h(ξ + η) − h(ξ)} lies in I 0 (8)with I 0 = [δ, kab];(iii) there is a continuous function φ(t) such that|p(t, x 2 , y 2 , z 2 ) − p(t, x 1 , y 1 , z 1 )|≤ φ(t){|x 2 − x 1 | + |y 2 − y 1 | + |z 2 − z 1 |} (9)holds for arbitrary t, x 1 , y 1 , z 1 , x 2 , y 2 and z 2 .Then, there exists a constant D 1 such that if∫ t0φ ν (τ)dτ ≤ D 1 t (10)for some ν, in the range 1 ≤ ν ≤ 2, then all solutions of (1) converge.A very important step in the proof of Theorem 1 will be to give estimate for anytwo solutions of (1). This in itself , being of independent interest, is giving as:


8Theorem 2. Let x 1 (t), x 2 (t) be any two solutions of (1). Suppose that all theconditions of Theorem 1 are satisfied , then for each fixed ν, in the range 1 ≤ ν ≤ 2,there exists constants D 2 , D 3 and D 4 such that for t 2 ≥ t 1 ,whereS(t 2 ) ≤ D 2 S(t 1 )exp{−D 3 (t 2 − t 1 ) + D 4∫ t2t 1φ ν (τ)dτ} (11)S(t) = {[x 2 (t) − x 1 (t)] 2 + [ẋ 2 (t) − ẋ 1 (t)] 2 + [ẍ 2 (t) − ẍ 1 (t)] 2 }. (12)If we put x 1 (t) = 0 and t 1 = 0, we immediately obtain:Corollary 1. If p = 0 and hypotheses (i) and (ii) of Theorem 1 hold, then thetrivial solution of (1) is exponentially stable in the large.Further, if we put ξ = 0 in (2.1) with η(η ≠ 0) arbitrary, we obtain :Corollary 2. If p ≠ 0 and hypotheses (i) and (ii) of Theorem 1 hold for arbitraryη(η ≠ 0) , and ξ = 0, then there exists a constant D 5 > 0 such that every solutionx(t) of (1) satisfies|x(t)| ≤ D 5 ; |ẋ(t)| ≤ D 5 ; |ẍ(t)| ≤ D 5 . (13)and3. PRELIMINARY RESULTSBy setting ẋ = y, ẏ = z , the equation (1) may be replaced with the systemẋ = y, ẏ = z, ż = −az − g(y) − h(x) + p(t, x, y, z) (14)Let (x i (t), y i (t), z i (t)), (i = 1, 2), be any two solutions of (3.1) such thatb ≤ g(y 2) − g(y 1 )y 2 − y 1≤ b 0 (y 2 ≠ y 1 ); (15)δ ≤ h(x 2) − h(x 1 )x 2 − x 1≤ kab, (x 2 ≠ x 1 ), (16)


10where θ = p(t, x 2 , y 2 , z 2 ) − p(t, x 1 , y 1 , z 1 ).Proof of Lemma 2. On using (3.1), a direct computation of dW dtgives aftersimplificationwheredWdt= −W 1 − W 2 − W 3 − W 4 − W 5 + W 6 (21)W 1 = 1 4 b(1 − β)H(x 2, x 1 )(x 2 − x 1 ) 2 + 1 4 α(z 2 − z 1 ) 2+ 1 4 a[G(y 2, y 1 ) − b(1 − β)](y 2 − y 1 ) 2 ;W 2 = 1 4 b(1 − β)H(x 2, x 1 )(x 2 − x 1 ) 2 + 1 4 α(z 2 − z 1 ) 2+(1 + αa −1 )H(x 2 , x 1 )(x 2 − x 1 )(z 2 − z 1 );W 3 = 1 4 b(1 − β)H(x 2, x 1 )(x 2 − x 1 ) 2+ 1 4 a[G(y 2, y 1 ) − b(1 − β)](y 2 − y 1 ) 2+aH(x 2 , x 1 )(x 2 − x 1 )(y 2 − y 1 );W 4 = 1 4 b(1 − β)H(x 2, x 1 )(x 2 − x 1 ) 2+ 1 4 a[G(y 2, y 1 ) − b(1 − β)](y 2 − y 1 ) 2+b(1 − β)[G(y 2 , y 1 ) − b](x 2 − x 1 )(y 2 − y 1 );W 5 = 1 4 a[G(y 2, y 1 ) − b(1 − β)](y 2 − y 1 ) 2 + 1 4 α(z 2 − z 1 ) 2+(αa −1 + 1)[G(y 2 , y 1 ) − b](y 2 − y 1 )(z 2 − z 1 );W 6 = {b(1 − β)(x 2 − x 1 ) + a(y 2 − y 1 ) + (1 + αa −1 )(z 2 − z 1 )}θ(t);withG(y 2 , y 1 ) = g(y 2) − g(y 1 )y 2 − y 1, (y 2 ≠ y 1 ); (22)


11andhaveH(x 2 , x 1 ) = h(x 2) − h(x 1 )x 2 − x 1, (x 2 ≠ x 1 ). (23)Further, let us denote H(x 2 , x 1 ) and G(y 2 , y 1 ) simply by H and G, respectively.For strictly positive constants k 1 , k 2 , k 3 , and k 4 conveniently chosen later, we(αa −1 + 1)H(x 2 − x 1 )(z 2 − z 1 )= {k 1 (αa −1 + 1) 1 12 H 2 (x2 − x 1 ) + 1 2 k−1 1 (αa −1 + 1) 1 12 H 2 (z2 − z 1 )} 2−k 2 1(αa −1 + 1)H(x 2 − x 1 ) 2 − 1 4 k−2 1 (αa −1 + 1)H(z 2 − z 1 ) 2 ;aH(x 2 − x 1 )(y 2 − y 1 ) = {k 2 a 1 12 H 2 (x2 − x 1 ) + 1 2 k−1 2 a 1 12 H 2 (y2 − y 1 )} 2b(1 − β)(G − b)(x 2 − x 1 )(y 2 − y 1 )−k 2 2aH(x 2 − x 1 ) 2 − 1 4 k−2 2 aH(y 2 − y 1 ) 2 ;= {k 3 b 1 2 (1 − β) 1 2 (G − b) 1 2 (x 2 − x 1 ) + 1 2 k−1 3 b 1 2 (1 − β) 1 2 (G − b) 1 2 (y 2 − y 1 )} 2−k 2 3b(1 − β)(G − b)(x 2 − x 1 ) 2 − 1 4 k−2 3 b(1 − β)(G − b)(y 2 − y 1 ) 2 ;(αa −1 + 1)(G − b)(y 2 − y 1 )(z 2 − z 1 )= {k 4 (αa −1 + 1) 1 2 (G − b) 1 2 (y 2 − y 1 )+ 1 2 k−1 4 (αa −1 + 1) 1 2 (G − b) 1 2 (z 2 − z 1 )} 2Thus,W 2−k 2 4(αa −1 + 1)(G − b)(y 2 − y 1 ) 2 − 1 4 k−2 1 (αa −1 + 1)(G − b)(z 2 − z 1 ) 2 ;= {k 1 (αa −1 + 1) 1/2 H 1/2 (x 2 − x 1 ) + 1 2 k−1 1 (αa −11 + 1) 1/2 H 1/2 (z 2 − z 1 )} 2+{ 1 4 b(1 − β)H − k2 1(αa −1 + 1)H}(x 2 − x 1 ) 2+{ 1 4 α − 1 4 k−2 1 (αa −1 + 1)H}(z 2 − z 1 ) 2 ;


12W 3 = {k 2 a 1/2 H 1/2 (x 2 − x 1 ) + 1 2 k−1 2 a 1/2 H 1/2 (y 2 − y 1 )} 2+{ 1 4 a[G − b(1 − β)] − 1 4 k−2 2 aH}(y 2 − y 1 ) 2+{ 1 4 b(1 − β)H − k2 2aH}(x 2 − x 1 ) 2 ;W 4 = {k 3 b 1/2 (1 − β) 1/2 (G − b) 1/2 (x 2 − x 1 )+ 1 2 k−1 3 b 1/2 (1 − β) 1/2 (G − b) 1/2 (y 2 − y 1 )} 2+{ 1 4 b(1 − β)H − k2 3b(1 − β)(G − b)}(x 2 − x 1 ) 2+{ 1 4 a[G − b(1 − β)] − 1 4 k−2 3 b(1 − β)(G − b)}(y 2 − y 1 ) 2 ;W 5 = {k 1 (αa −1 + 1) 1/2 (G − b) 1/2 (y 2 − y 1 )+ 1 2 k−1 4 (αa −11 + 1) 1/2 (G − b) 1/2 (z 2 − z 1 )} 2+{ 1 4 a[G − b(1 − β)] − k2 4(αa −1 + 1)(G − b)}(y 2 − y 1 ) 2+{ 1 4 α − 1 4 k−2 4 (αa −1 + 1)(G − b)}(z 2 − z 1 ) 2 .Furthermore, by using (15), we obtain for all x i , z i (i=1,2) in IR,W 2 ≥ 0, (24)ifk 2 1 ≤(1 − β)ab4(a + α)and for all x i , y i (i=1,2) in IR,withH ≤ α(1 − β)a2 b16(a + α) 2 ,ifW 3 ≥ 0 (25)k 2 2 ≤b(1 − β)4awithH ≤β(1 − β)b2.4aCombining all the inequalities in (24) and (25), we have for all x i , y i , z i (i=1,2) inIR ,W 2 ≥ 0 and W 3 ≥ 0


13ifα(1 − β)ab β(1 − β)bH ≤ kab with k = min{ ; } < 1. (26)16(a + α)24a 2Also, for all x i , y i (i=1,2) in IR,W 4 ≥ 0 (27)ifand for all y i , z i (i=1,2) in IR,(1 − β)(b 0 − b)aβ≤ k 2 3 ≤δb 0 − b ,W 5 ≥ 0, (28)if(a + α)(b 0 − b)aα≤ k 2 4 ≤a 2 bβ4(a + α)(b 0 − b) .We are now left with estimates W 1 and W 6 . There exists a positive constant D 10such thatW 1 ≥ D 10 {(x 2 − x 1 ) 2 + (y 2 − y 1 ) 2 + (z 2 − z 1 ) 2 } (29)where D 10 = 1 min{bδ(1 − β); abβ; α}4whileW 6 ≤ D 11 {(x 2 − x 1 ) 2 + (y 2 − y 1 ) 2 + (z 2 − z 1 ) 2 } 1/2 |θ(t)| (30)whereD 11 = 2 max{b(1 − β); a; (1 + αa −1 )}.Thus, combining (24), (25), (27) in (21) and using (19) we have thatdWdt≤ −D 10 S(t) + D 11 S 1 2 (t)|θ(t)|. (31)This completes the proof of Lemma 2.✷


144. PROOF OF THEOREM 2This follows directly from [5], on using inequality (31) . Let ν be any constantin the range 1 ≤ ν ≤ 2. Set 2µ = 2 − ν , so that 0 ≤ 2µ ≤ 1. We re-write (31) inthe formwhereConsidering the two cases(i) |θ| ≤ D 10 D −111 S 1 2 and(ii) |θ| > D 10 D −111 S 1 2dWdt + D 10S ≤ D 11 S µ W ∗ (32)W ∗ = (|θ| − D 10 D −111 S 1 2 )S12 −µ .separately, we find that in either case, there exists some constant D 12W ∗ ≤ D 12 |θ| 2(1−µ) . Thus, using (9), inequality (32) becomessuch thatdWdt + D 10S ≤ D 13 S µ φ 2(1−µ) S 1−µ (33)whereD 13 ≥ 2D 11 D 12 . This immediately givesdWdt + (D 14 − D 15 φ ν (t))W ≤ 0 (34)after using Lemma 1 on W , with D 14 and D 15 as some positive constants. On integrating(34) from t 1 to t 2 , (t 2 ≥ t 1 ), we obtainW (t 2 ) ≤ W (t 1 ) exp{−D 14 (t 2 − t 1 ) + D 15∫ t2t 1φ ν (τ)dτ}. (35)Again, using Lemma 1, we obtain (11), with D 2 = D 7 D −16 ,D 3 = D 14 and D 4 = D 15 .This completes the proof of Theorem 2.✷


155. PROOF OF THEOREM 1This follows from the estimate (11) and the condition (10) on φ(t). ChooseD 1 = D 3 D −14 in (10). Then, as t = (t 2 − t 1 ) −→ ∞, S(t) −→ 0, which proves thatas t −→ ∞,x 2 (t) − x 1 (t) −→ 0, ẋ 2 (t) − ẋ 1 (t) −→ 0, ẍ 2 (t) − ẍ 1 (t) −→ 0.This completes the proof of Theorem 1.✷Remark: As remarked in [5], the results remain valid if we replace φ(t) in (10) bya constant D 16 > 0.References[1] A. U. Afuwape, M. O. Omeike, Further Ultimate Boundedness of Solutions ofsome System of Third Order Nonlinear Ordinary Differential Equations,ActaUniv. Palacki. Olomuc., Fac. rer. nat., Mathematica 43 (2004) 7-20. MR. 2006e:34115.[2] A. U. Afuwape, M. O. Omeike, Convergence of solutions of certain system ofthird-order nonlinear ordinary differential equations, Ann. of Diff. Eqs. 21: 4(2005), 533-540.[3] J. Andres, Recent Results On Third Order Non-Linear ODEs, J. Nig. Math. Soc.Vol. 14/15, (1995/96), 41-66. MR.2001c:34001[4] J. O. C. Ezelio, A note on the convergence of solutions of certain second orderdifferential equations, Portugaliae Mathematica, 24 (1965), 49-58. MR. 34 #4597.


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