Research Article Global Existence and Boundedness of Solutions to a Second-Order Nonlinear Differential System

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1 Applied Mathematics Volume 212, Article ID 63783, 12 pages doi:1.1155/212/63783 Research Article Global Existence and Boundedness of Solutions to a Second-Order Nonlinear Differential System Changjian Wu, 1 Shushuang Hao, 2 and Changwei Xu 3 1 Zhujiang College, South China Agricultural University, Guangdong, Conghua 519, China 2 Department of Information Engineering, Huanghe Science and Technology College, Henan, Zhengzhou 4563, China 3 College of Science, Zhongyuan University of Technology, Henan, Zhengzhou 456, China Correspondence should be addressed to Shushuang Hao, hss9811@yahoo.com.cn Received 25 March 212; Accepted 2 August 212 Academic Editor: Carla Roque Copyright q 212 Changjian Wu et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We investigate the global existence and boundedness of solutions to a second-order nonlinear differential system. 1. Introduction In this paper, we study the nonlinear system x 1 [ ( ) ] c y b x, a x y a x h x e t, 1.1 where a : R,, b, c, h : R R, ande : R R are continuous. As a particular case of 1.1 we have well-known Liẽnard equation as follows: x f x x h x e t. 1.2

2 2 Applied Mathematics with a x 1, b x x f s, c x x, x R and the second-order nonlinear differential equation as follows: x ( f x g x x ) x h x e t 1.3 for a x exp x g s, b x x a s f s, c x x, x R. System 1.1 can be regarded as a mathematical model for many phenomena in applied sciences theory of feedback electronic circuits, motion of a mass-spring system. It has been investigated by several authors, compare 1 4 and the citations therein. The purpose of this paper is to present new results on the global existence and boundedness of solutions for the system 1.1. The obtained results improve the recent results in 1, 5. Our paper is divided into two section. In Section 2, we prove the global existence of solutions for 1.1.InSection 3, we get some new results on boundedness of solutions for the system Global Existence In this section, we will present new results on the global existence of solutions to system 1.1 under general conditions on the nonlinearities. Let us first define C ( y ) Then, we have the following. Theorem 2.1. Assume that y c s, i there exists some K, such that H x x a 2 s h s. 2.1 sgn x H x K, x R, sgn ( y ) C ( y ) K, y R, 2.2 ii there exist some N and Q>, such that H x <Q, x >N, ( ) C y <Q, y >N, 2.3 iii lim y sgn y C y Q, lim x 1/ Q sgn x H x sgn x b x, iv there exist two positive functions μ, ω C,K Q,, such that a x c ( y ) min { μ ( sgn x H x K ) ω ( sgn x H x K ), μ ( sgn ( y ) C ( y ) K ) ω ( sgn ( y ) C ( y ) K )}, 2.4 x >N, y >N,

3 Applied Mathematics 3 If v sgn x a x b x h x μ sgn x H x K ω sgn x H x K, x >Nand h x M<, x R. K Q, μ s ω s 2.5 then every solution of 1.1 exists globally. Proof. Due that a : R,, b, c, h : R R and e : R R are continuous, by Peano s Existence Theorem 6, we have that the system 1.1 with any initial data x,y possesses a solution x t,y t on,t for some maximal T>. If T<, one has ( ) lim x t y t. 2.6 t T First, assume that lim t T y t. Since y t is continuous, there exists T <Tsuch that y t >N, t T,T. 2.7 Take V 1 t, x, y sgn y C y K, t R, x, y R.Differentiating V 1 t, x, y with respect to t along solution x t,y t of 1.1, we have dv 1 sgn ( y )[ a x c ( y ) h x a x c ( y ) e t ] h x e t a x c ( y ) M e t [ μ ( sgn ( y ) C ( y ) K ) ω ( sgn ( y ) C ( y ) K )], t T,T. 2.8 Since sgn y t C y t K<Q K, t T,T,weobtain dv 1 t μ V 1 t ω V 1 t M e t, t T,T. 2.9 We denote that V 1 t V 1 t, x t,y t. K Q Since lim y sgn y C y Q, / μ s ω s, y t, C y are continuous, there exists T t 1 <t 2 <Tsuch that V1 t 2 V 1 t 1 T T μ s ω s >M e t. 2.1

4 4 Applied Mathematics Integrating 2.9 on t 1,t 2 with respect to t and using the above relation, we obtain the following contradiction: T T M e t < t2 V 1 t 1 μ s ω s dv 1 t t 1 μ V 1 t ω V 1 t T T M e t M e t. t 1 V1 t 2 t Thus, there exists an M> such that y t M, t,t Second, by the result above, we have lim t T x t. If lim x 1/ Q sgn x H x, that is, lim x sgn x H x Q,weset V 2 ( t, x, y ) sgn x H x K, t R, x,y R Since x t is continuous, there exists T 1 <Tsuch that x t >N, t T 1,T Differentiating V 2 t, x, y with respect to t along solution x t,y t of 1.1, we have dv 2 sgn x [ a x c ( y ) h x a x b x h x ] M 1 [ μ ( sgn x H x K ) ω ( sgn x H x K )], t T 1,T Since sgn x t H x t K<Q K, t T 1,T,weobtain dv 2 t μ V 2 t ω V 2 t M 1, t T 1,T We denote V 2 t V 2 t, x t,y t. Since lim x sgn x H x Q, K Q / μ s ω s, x t,h x are continuous, there exists T 1 t 3 <t 4 <Tsuch that V2 t 4 V 2 t 3 T > M 1. μ s ω s 2.17

5 Applied Mathematics 5 Integrating 2.16 on t 3,t 4 with respect to t and using the above relation, we obtain the contradiction as follows: T M 1 < V2 t 4 t4 V 2 t 3 t4 μ s ω s dv 2 t t 3 μ V 2 t ω V 2 t t 3 M 1 M 1 T So consider lim x 1/ Q sgn x H x <. By iii, we have lim x sgn x b x. Set W ( t, x, y ) x, t R,x,y R Then, along solutions to 1.1 we have dw 1 [ ( ) ] c y b x. 2.2 a x If lim t T x t, we deduce that there exist x 1 and x 2 such that x <x 1 <x 2 and dw <, x 1 x x 2, y M Then, by the continuity of the solution, there exist < t 1 < t 2 < T such that x t 1 x 1, x t 2 x 2. Integrating 2.21 on t 1,t 2, we have W ( t 1,x t 1,y t 1 ) x 1 >x 2 W ( t 2,x t 2,y t 2 ) This contradicts x 1 <x 2. Hence x t is bounded from above. Similarly, if lim t T x t, we can obtain a contradiction by setting W t, x, y x. Thus, it follows that x t is also bounded from above. This forces T and completes the proof of Theorem 2.1. Example 2.2. Consider the following nonlinear system: x y 1 x 2 1 y 2, 1 t 2. 1 x 2 1 x Set a x 1/ 1 x 2, b x, c y 1/ 1 y 2, h x 1, e t t 2. Then we have C ( y ) arctan y, H x arctan x. 2.24

6 6 Applied Mathematics Take K N, Q π/2, and μ θ ω θ cos θ/2, θ,π/2. Notethat a x ( ) c y x 2 1 y min 1 2, 1 1 x 2 1 y 2 min { μ ( sgn x arctan x K ) ω ( sgn x arctan x K ), μ ( sgn ( y ) arctan y K ) ω ( sgn ( y ) arctan y K )} 2.25 sgn x a x b x h x, h x 1and π/2 dθ π/2 μ θ ω θ dθ cos θ dx x Applying Theorem 2.1, we know that every solution of 2.23 exists globally. Observe that the theorem and corollary in 5 cannot be used in the present case. Theorem 2.3. Assume that i there exists some K, such that H x K, x R, C ( y ) K, y R, 2.27 ii there exist some N and Q>, such that H x <Q, x >N, C y < Q, y > N, 2.28 iii lim y C y Q, lim x 1/ Q H x sgn x b x, iv there exist two positive functions μ, ω C,K Q,, such that a x c ( y ) min { μ H x K ω H x K, μ ( C ( y ) K ) ω ( C ( y ) K )}, 2.29 x >N, y >N, v a x b x h x μ H x K ω H x K, x >Nand h x M<, x R. If K Q, μ s ω s 2.3 then every solution of 1.1 exists globally.

7 Applied Mathematics 7 Proof. The proof of Theorem 2.3 is similar to that of Theorem 2.1, so we omit it. Example 2.4. Consider the following nonlinear system: x y 2y ( ) 2y ln( 1 x 2) 1 y 2 2 ( ), 1 y 2 2 2x 1 x 2 1 ln 1 x 2 t 3 1 ln 1 x Set a x 1/ 1 ln 1 x 2, b x, c y 2y/ 1 y 2 2, h x 2x/ 1 x 2, e t t 3. Then we have C y 1 1/ 1 y 2, H x 1 1/ 1 ln 1 x 2. Take K N, Q 1and μ t ω t 1 t /2, t, 1. Notethat a x ( ) c y 1 1 ln 1 x 2 2 y { } ( ) min 1 1 y ln 1 x 2, 1 1 y 2 min { μ H x K ω H x K,μ ( C ( y ) K ) ω ( C ( y ) K )} a x b x h x, h x 2 x / 1 x 2 1and 1 1 μ s ω s s Applying Theorem 2.3, we know that every solution of 2.31 exists globally. Observe that the theorem and corollary in 5 cannot be uses in the present case. 3. Boundedness In this section, we will present some results on the boundedness of solutions to 1.1 under general conditions on the nonlinearities. Theorem 3.1. Assume that i there exist functions f 1, f 2 C R,R such that f 1 t y t a x h x e t f 2 t, x R, t R, 3.1 and f 1 t <, f 2 t <, ii lim x sgn x b x. If the solution x t,y t of 1.1 exists globally, then x t,y t is bounded. Proof. By i, we have f 1 t y t f 2 t, x R, t R. 3.2

8 8 Applied Mathematics Integrating 3.2 on,t with respect to t, we have t f 1 s y t y t f 2 s. 3.3 Since f 1 t <, f 2 t <. Thus, there exists a Y>such that y t Y, t. 3.4 Set W ( t, x, y ) x, t R,x,y R. 3.5 Then, along solutions to 1.1, we have dw 1 [ ( ) ] c y b x. 3.6 a x If lim t x t, we deduce that there exist x 1 and x 2 such that x >x 1 >x 2 and dw <, x 2 x x 1, y Y. 3.7 Then, by the continuity of the solution, we have that there exist <t 1 <t 2 < such that x t 1 x 1 and x t 2 x 2. Integrating 3.7 on t 1,t 2,weget W ( t 1,x t 1,y t 1 ) x 1 > x 2 W ( t 2,x t 2,y t 2 ). 3.8 This contradicts x 1 >x 2. Hence x t is bounded from below. Similarly, if lim t T x t, we can obtain a contradiction by setting W t, x, y x. Thus, it follows that x t is also bounded from above. This completes the proof of Theorem 3.1. Example 3.2. Consider the following nonlinear system: x y 1 x 2 x 1 x 2, 1 y x 2 1 t Set a x 1/ 1 x 2, b x x, c y 1/ 1 y 2, h x 1, e t 1 1/ 1 t 2. Then we have C y arctan y and H x arctan x. Take K N, Q π/2andμ θ ω θ cos θ/2, θ,π/2. Applying Theorem 2.1, we know that every solution of 3.9 exists globally.

9 Applied Mathematics 9 Take f 1 t, f 2 t 1/ 1 t 2, we have f 1 t y t f 2 t 1 1 t 2, x R, t R 3.1 and f 2 t π/2 <, lim x sgn x b x lim x x sgn x. Applying Theorem 3.1, we know that every solution of 3.9 is bounded. Theorem 3.3. Assume that i there exists some K, such that sgn x H x K, x R, sgn ( y ) C ( y ) K, y R, 3.11 ii there exist some N and Q>, such that H x <Q, C ( y ) <Q, x >N, y >N, 3.12 iii lim y sgn y C y Q, lim x 1/ Q sgn x H x sgn x b x, iv there exist two positive functions μ, ω C,K Q,, such that a x c ( y ) min { μ ( sgn x H x K ) ω ( sgn x H x K ), μ ( sgn ( y ) C ( y ) K ) ω ( sgn ( y ) C ( y ) K )}, 3.13 x >N, y >N, v sgn x a x b x h x, x >Nand h x M<, x R, vi E e t <, and K Q / μ s ω s. If there exists g x such that a x < ( ) g x, x,y R c y b x 3.14 and G g x h x dx <, then every solution of 1.1 is bounded. Proof. Let x t,y t be a solution to 1.1 with initial data x,y.bytheorem 2.1, we have that x t,y t exists globally. If x t,y t is unbounded, we have lim t x t or lim y t. t 3.15 First, assume that lim t T y t.

10 1 Applied Mathematics Since y t is continuous, there exists T 1 such that y t >N, t T1, Take V 1 t, x, y sgn y C y K, t R, x, y R. Differentiating V 1 t, x, y with respect to t along solution x t,y t of 1.1, we have dv 1 sgn ( y )[ a x c ( y ) h x a x c ( y ) e t ] h x e t a x c ( y ) h x e t [ μ ( sgn ( y ) C ( y ) K ) ω ( sgn ( y ) C ( y ) K )], t T 1, Since sgn y t C y t K<Q K, t T 1,,weobtain dv 1 t μ V 1 t ω V 1 t h x t e t, t T 1, We denote V 1 t V 1 t, x t,y t. By vi, there exists T 1 t 1 <t 2 < such that V1 t 2 V 1 t 1 >G E. μ s ω s 3.19 Integrating 3.18 on t 1,t 2 with respect to t and using the above relation, we obtain the contradiction as follows: G E< V1 t 2 t2 V 1 t 1 V1 t 2 μ s ω s dv 1 t V 1 t 1 μ V 1 t ω V 1 t t 1 h x t t2 t 1 e t g x h x dx E G E. x t2 h x dx x t 1 x t E 3.2 Thus, there exists a Y>such that y t Y, t, Second assume that lim t T x t. If lim x 1/ Q sgn x H x, that is, lim x sgn x H x Q,weset V 2 ( t, x, y ) sgn x H x K, t R, x,y R. 3.22

11 Applied Mathematics 11 Since x t is continuous, there exists T 2 < such that x t >N, t T 2, Differentiating V 2 t, x, y with respect to t along solution x t,y t of 1.1, we have dv 2 sgn x [ a x c ( y ) h x a x b x h x ] h x [ μ ( sgn x H x K ) ω ( sgn x H x K )], t T 2, Since sgn x t H x t K<Q K, t T 2,,weobtain dv 2 t μ V 2 t ω V 2 t h x t, t T 2, We denote that V 2 t V 2 t, x t,y t. By vi, there exists T 2 t 3 <t 4 <Tsuch that V2 t 4 V 2 t 3 μ s ω s >G Integrating 3.25 on t 3,t 4 with respect to t and using the above relation, we obtain the contradiction as follows: G< V2 t 4 V 2 t 3 x t4 t4 μ s ω s dv 2 t t4 t 3 μ V 2 t ω V 2 t h x t t 3 h x dx x t 3 x t g x h x dx G So consider lim x 1/ Q sgn x H x <. By iii, we have lim x sgn x b x. The proof of this condition is similar to that of Theorem 3.1, so we omit it. Thus, it follows that x t is also bounded from above. Then every solution of 1.1 is bounded. This completes the proof of Theorem 3.3. Theorem 3.4. Assume that i there exists some K, such that H x K, x R, C ( y ) K, y R, 3.28

12 12 Applied Mathematics ii there exist some N and Q>, such that H x <Q, C ( y ) <Q, x >N, y >N, 3.29 iii lim y C y Q, lim x 1/ Q H x sgn x b x, iv there exist two positive functions μ, ω C,K Q,, such that a x c ( y ) min { μ H x K ω H x K, μ ( C ( y ) K ) ω ( C ( y ) K )}, 3.3 x >N, y >N, v sgn x a x b x h x, x >Nand h x M<, x R, vi E e t <, and K Q / μ s ω s. If there exists g x such that a x < ( ) g x, x,y R c y b x 3.31 and G g x h x dx <, then every solution of 1.1 is bounded. Proof. The proof of Theorem 3.4 is similar to that of Theorem 3.3, so we omit it. Acknowledgments The authors thank the helpful suggestions by Professor Zhaoyang YIN and the referees for careful reading. This work is supported by Natural Science Foundation of the Education Department of Henan Province, China no. 211C115. References 1 A. Constantin, A note on a second-order nonlinear differential system, Glasgow Mathematical Journal, vol. 42, no. 2, pp , 2. 2 T. A. Burton, On the equation x f x h x x g x e t, Annali di Matematica Pura ed Applicata, vol. 85, pp , A. Constantin, Global existence of solutions for perturbed differential equations, Annali di Matematica Pura ed Applicata, vol. 168, pp , J. Kato, On a boundedness condition for solutions of a generalized Liénard equation, Differential Equations, vol. 65, no. 2, pp , Z. Yin, Global existence and boundedness of solutions to a second order nonlinear differential system, Studia Scientiarum Mathematicarum Hungarica, vol. 41, no. 4, pp , V. Lakshmikantham and S. Leela, Differential and Integral Inequalities Volume I: Ordinary Differential Equation, Academic Press, London, UK, 1969.

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