Research Article Existence and Uniqueness of Homoclinic Solution for a Class of Nonlinear Second-Order Differential Equations

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1 Applied Mathematics Volume 2012, Article ID , 13 pages doi: /2012/ Research Article Existence and Uniqueness of Homoclinic Solution for a Class of Nonlinear Second-Order Differential Equations Lijuan Chen and Shiping Lu School of Mathematics and Statistics, Nanjing University of Information Science and Technology, Nanjing , China Correspondence should be addressed to Lijuan Chen, cljung@sohu.com Received 11 September 2012; Accepted 13 November 2012 Academic Editor: Julián López-Gómez Copyright q 2012 L. Chen and S. Lu. 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. The authors study the existence and uniqueness of a set with -periodic solutions for a class of second-order differential equations by using Mawhin s continuation theorem and some analysis methods, and then a unique homoclinic orbit is obtained as a limit point of the above set of periodic solutions. 1. Introduction In this paper, we study the existence and uniqueness of homoclinic solutions for the following nonlinear second-order differential equations: u t g ( u t ) h u t f t, 1.1 where u t R, g, h and f are all in C R, R. As usual we say that a nonzero solution u t of 1.1 is homoclinic to 0 if u t 0 and u t 0as t. Equation 1.1 is important in the applied sciences such as nonlinear vibration of masses, see 1 3 and the references therein. But most of the authors in those papers are interested in the study of problems of periodic solutions. Recently, the existence of homoclinic solutions for some second-order ordinary differential equation system has been extensively studied by using critical point theory, see 4 13 and the references therein. For example,

2 2 Applied Mathematics in 9, by using the Mountain Pass theorem, Lv et al. discussed the existence of homoclinic solutions for the following second-order Hamiltonian systems: u t L t u t w t, u t 0, 1.2 and in 13, the authors by means of variational method studied the problem of homoclinic solutions for the forced pendulum equation without the first derivative term. But, as far as we know, there were few papers studying the existence of homoclinic solution for the equation such as 1.1. This is due to the fact that 1.1 contains the first derivative term g u t. This implies that the differential equation is not the Euler Lagrange equation associated with some functional I : W 1,p R. So the method of critical point theory or variational method in 4 13 cannot be applied directly. Although paper 13 discussed the existence of homoclinic solutions for the following equation containing the first derivative term: x t f t x t β t x t g t, x t 0, 1.3 the term containing the first derivative is only linear with respect to x t. In order to investigate the homoclinic solutions to 1.1, firstly, we study the existence of -periodic solutions to the following equation for each k N: u t g ( u t ) h u t f k t, 1.4 where f k : R R is a -periodic function such that f t, t, kt ε 0 f k t f kt ε 0 f f kt ε 0 t kt ε 0, ε 0 t kt ε 0,kT, 1.5 T>0 is a given constant, and ε 0 0,T is a constant independent of k. Then a homoclinic solution to 1.1 is obtained as a limit point of the set {u k t }, where u k t is an arbitrary -periodic solution to 1.4 for each k N. The significance of present paper is that we not only investigate the existence of homoclinic solution to 1.1, but also study the uniqueness of the homoclinic solution and, the existence of -periodic solutions to 1.4 is obtained by using Mawhin s continuation theorem 14, not by using the methods of critical point theory, which is quite different from the approaches of 4 13, 15. Furthermore, the method to obtain the homoclinic solution to 1.1 is also different from the corresponding ones of Main Lemmas For each k N, letc {x x C R, R,x t x t }, C 1 {x x C1 R, R, x t x t }, the norms of C and C 1 are defined by max t,kt x t and x C 1 max{ x, x }, respectively, then C and C 1 are all Banach spaces. Furthermore for x C, x r x t r dt 1/r, where r 1,.

3 Applied Mathematics 3 Lemma 2.1 see 12. Let a>0 and q W 1,p R, R, then for every t R, the following inequality holds: q t 2a 1/μ ( t a t a ) 1/μ ( t a q s μ ds a 2a 1/p q s 1/p ds) p, 2.1 t a where μ, p 1, are constants. Lemma 2.2 see 12. Let q W 1,p R, Rn, then the following inequality holds: q T 1/ν ( ) 1/ν ( kt q s ν ds T p 1 /p q s 1/p ds) p, 2.2 where ν and p are constants with ν>1 and p>1. In order to use Mawhin s continuation theorem for investigating the existence of periodic solutions to 1.4, we give some definitions associated with Mawhin s continuation theorem. Definition 2.3 see 14. LetX and Y be two Banach spaces with norms x X and x Y, respectively. A linear operator L : D L X Y 2.3 is said to be a Fredholm opeartor with index zero provided that 1 Im L is a closed subset of Y; 2 dim Ker L codim Im L<. If L : D L X Y is a Fredholm operator with index zero, then X ker L X 1 and Y Im L Y 1.LetP : X ker L and Q : Y Y 1 be the continuous projectors. Clearly, ker L D L X 1 {0}, thus the restriction L P : L D L X1 is invertible. Denote by K P the inverse of L P. Definition 2.4 see 14. LetX and Y be two Banach spaces with norms x X and x Y, respectively, and the operator L : D L X Y 2.4 is a Fredholm operator with index zero, Ω X is an open bounded set with D L Ω / φ. A continuous operator N : Ω X Y is said to be L-compact in Ω, provided that 1 K P I Q N Ω is a relative compact set of X; 2 QN Ω is a bounded set of Y.

4 4 Applied Mathematics Lemma 2.5 see 14. Suppose that X and Y are two Banach spaces, and L : D L X Y is a Fredholm operator with index zero. Furthermore, Ω X is an open bounded subset and N : Ω Y is L-compact on Ω. If all the following conditions hold: 1 Lx / λnx, for all x Ω D L,λ 0, 1 ; 2 Nx / Im L, for all x Ω Ker L; 3 deg{jqn,ω Ker L, 0} / 0, where J :ImQ Ker L is an isomorphism. Then equation Lx Nx has a solution on Ω D L. Lemma 2.6. Assume that there are positive constants m, m 1, n, l 0, and l 1 with l 0 l 1, such that the following conditions hold. A1 sup t R f t <, R f t l 0 1 /l 0 dt <, R f t l 1 1 /l 1 dt < and R f t 2 dt <. A2 m 1 x l 0 1 xg x m x l 0 1, x R, xh x n x l 1 1, x R. 2.5 A3 h C 1 R, R with h x 0 for all x R. Then for every k N, 1.4 possesses a -periodic solution. Remark 2.7. From 1.5, wesee fk sup f t <, k N, t R ( kt fk l0 1 /l 0 fk s ) l0 / l 0 1 l 0 1 /l 0 ds ( ε0 ( R fk s ) l0 / l 0 1 ( l 0 1 /l 0 ds fk s ) l0 / l 0 1 l 0 1 /l 0 ds kt ε 0 ) l0 / l 0 1 f s l 0 1 /l 0 ds ε l 0/ l sup f t, k N, t R 2.6 which together with assumption A1 yields that f k and f k l0 1 /l 0 independent of k N. are two constants < and f k 2 < are two constants indepen- Similarly, we have that f k l1 1 /l 1 dent of k N.

5 Applied Mathematics 5 Proof. Set X C 1, Y C, L : D L X Y, Lu u, where D L {u u C 2 }, and N : C 1 C, Nu t g ( u t ) h u t f k t. 2.7 Clearly, Ker L R, ImL {y Y : y s ds 0}, which implies that Im L is a closed subset of X, and dim Ker L codim Im L 1 <. SoL is a Fredholm operator with index zero. Let P : X Ker L, Q : Y Y/ Im L 2.8 be defined respectively by Px 1/ 0 x s ds, Qx 1/ 0 x s ds and let L P L X Ker P : X Ker P Im L. 2.9 Then L P has a unique continuous pseudo-inverse L 1 P on Im L defined by L 1 P y t Fy t, where [ ] Fy t G t, s y s ds, 0 G t, s s t, 0 s<t, t s, t s For each open bounded set Ω C, from the above formula, it is easy to see that the mapper N is L-compact on Ω. Step 1. For each k N,letΩ 1 {x C 1 : Lx λnx, λ 0, 1 }, thatis, Ω 1 { x C 1 : x t λg ( x t ) } λh x t λf k t,λ 0, We will show that Ω 1 is bounded in C 1. Suppose that u Ω 1, then u t λg ( u t ) λh u t λf k t, λ 0, Multiplying both sides of 2.12 by u t and integrating on the interval, kt, we have from assumption A2 that m u t l0 1 dt u t g ( u t ) dt f k t u t dt. 2.13

6 6 Applied Mathematics By using hólder inequality, we get u l 0 1 l fk l0 m 1 /l 0 u l0 1, 2.14 which together with the conclusion of Remark 2.7 shows ( ) u 1 l0 1 1/l0 fk l0 m 1 /l 0 : β Clearly, β 1 is a constant independent of k and λ. Multiplying both sides of 2.12 by u t and integrating on the interval, kt,we have u 2 2 λ ( u t ) 2 h u t dt λ f k t u t dt It follows from 2.16 and assumption A3 that which implies u 2 2 fk t u t dt fk 2 u 2, u 2 fk 2 : β By using Lemma 2.2, we have u T 1/ l 0 1 u l0 1 T 1/2 u 2 T 1/ l 0 1 β 1 T 1/2 β : β. Clearly, β is a constant independent of k and λ. On the other hand, multiplying both sides of 2.12 by u t and integrating on the interval, kt, we have u t 2 dt λ h u t u t dt λ g ( u t ) u t dt λ f k t u t dt It follows from assumption A2 that λn u t l 1 1 dt ( u t ) 2 dt λm1 u t l 0 u t dt λ fk t u t dt, 2.21

7 Applied Mathematics 7 which together with 2.19 and l 0 l 1 results in n u t l1 1 dt m 1 u t l0 u t dt fk t u t dt ( m 1 u t ) l1 / l 1 1 ( ) 1/ l1 1 l 0 l 1 1 /l 1 dt u t l1 1 dt ( m 1 ( m 1 β l 0 1 / l 1 1 f k t ) l1 / l 1 1 ( ) 1/ l1 1 l 1 1 /l 1 dt u t l1 1 dt u t l 0 1 u t ) l1 / l 1 1 l 0 l 1 /l 1 dt u l1 1 fk l1 1 /l 1 u l1 1 ( u t ) l1 / l 1 1 l 0 1 dt u l1 1 fk l1 1 /l 1 u l1 1 m 1 β l 0 l 1 / l 1 1 ( u l0 1) l1 l 0 1 / l 1 1 u l1 1 fk l1 1 /l 1 u l 1 1 m 1 β l 0 1 / l 1 1 β l 1 l 0 1 / l u l1 1 f k l1 1 /l 1 u l that is, u l 1 1 l [ m 1 β l 0 l 1 / l 1 1 β l 1 l 0 1 / l 1 1 n 1 ] fk l1 1 /l 1 u l Therefore u l1 1 ( 1 n ) 1/l1[ m 1 β l 0 l 1 / l 1 1 β l 1 l 0 1 / l fk l1 1 /l 1 ] 1/l1 : α 1, 2.24 where α 1 is a constant independent of k and λ. ByusingLemma 2.2 again, we get u T 1/ l 1 1 u l1 1 T l 0/ l 0 1 u l0 1 T 1/ l 1 1 α 1 T l 0/ l 0 1 β 1 : α Obviously, α is a constant independent of k and λ. Therefore, if u Ω 1, then by 2.19 we see that u C 1 max { u, } { } u max α, β sup f t : M. t R 2.26 Clearly, M >0 is a constant independent of k and λ; thatis,ω 1 is uniformly bounded for all k N and λ 0, 1.

8 8 Applied Mathematics Step 2. From assumptions A2 and A3, we see that there must be a constant M 1 > 0 such that h M 1 f k > 0and h M 1 f k < 0. Set Ω 2 {u t C 1 : u C 1 <M}, where M max{m 1, M}. We will show that Nu / Im L, for all u Ω 2 Ker L. In fact, by assumption A2,weseethatg 0 0, and if u Ω 2 Ker L, then u t M or u M.So QN u 1 [ h M fk t ] dt h M f k h M 1 f k > 0, u t M, QN u 1 [ h M fk t ] dt h M f k h M 1 f k < 0, u t M, 2.27 where f k 1/ f k t dt. This implies that Nu / Im L, for all u Ω 2 Ker L. Step 3. Set J :ImQ Ker L, Jx x, we will show deg{jqn,ω 2 Ker L, 0} / 0. Let H x, μ μx 1 μ JQNx, for all x Ω 2 Ker L, when x Ω 2 Ker L, we have x ±M and H ( M, μ ) μm ( 1 μ ) JQN M μm ( 1 μ ) QN M > 0, H ( M, μ ) μm ( 1 μ ) JQN M μm ( 1 μ ) QN M < So for all μ 0, 1, H Ω 2 Ker L,μ / 0, and then deg{jqn,ω 2 Ker L, 0} deg{h, 0, Ω 2 Ker L, 0} deg{h, 1, Ω 2 Ker L, 0} deg{i,ω 2 Ker L, 0} Therefore, by Lemma 2.5, 1.4 has a -periodic solution u k Ω 2. Remark 2.8. Suppose that all the conditions in Lemma 2.6 hold. We see that for each k N, 1.4 has a -periodic solution u k Ω 2. This implies that u k M, u k M Furthermore, as same as the proof of step 1 in Lemma 2.6 with replacing u t by u k t, we have u k l1 1 α 1, u k l0 1 β 1, 2.31 where α 1 and β 1 are two positive constants independent of k N.

9 Applied Mathematics 9 Lemma 2.9 see 12. Let u k C 1 be the -periodic solution for 1.4 and satisfies 2.30 and 2.31 for all k N. Then there exists a function u 0 C 1 R, R such that for each interval c, d R, there is a subsequence {u kj } of {u k } k N with u k j t u 0 t uniformly on c, d. 3. Main Results Theorem 3.1. Suppose that assumptions (A1), (A2), and (A3) in Lemma 2.6 hold. Then 1.1 has a unique homoclinic solution. Proof. Since assumptions A1, A2, onsisting of Kuratowski operations we used following principles and A3 in Lemma 2.6 hold, by using Lemma 2.6, weseethat 1.4 has a periodic solution u k t satisfying 2.30 and 2.31 for each k N. It follows from Lemma 2.9 that there exists a u 0 C 1 R, R such that for each interval c, d R, there is a subsequence {u kj } of {u k } k N satisfying u k j t u 0 t uniformly on c, d. Below, we will show that u 0 t is just a unique homoclinic solution to 1.1. Step 1. We show that u 0 is a solution of 1.1. In view of u kj t being a 2k j T-periodic solution to 1.4, we have ( ) ( ) u k j t g u k j t h u kj t f kj t, for t [ k j T, k j T ], j N. 3.1 Take a, b R such that a<b, there exists j 0 N such that for all j>j 0 ( ) ( ) u k j t g u k j t h u kj t f t, for t a, b. 3.2 Integrating 3.2 from a to t a, b, we have u k j t u k j a t [ ( ) ( ) ] g u k j s h u kj s f s ds, for t a, b. 3.3 a Since Lemma 2.9 shows that u kj u 0 uniformly on a, b and u k j u 0 uniformly on a, b as j,letj in 3.3, weget t u 0 t u 0 a [ ( g u 0 s ) h u 0 s f s ] ds, for t a, b. 3.4 a In view of a and b are arbitrary, 3.4 shows that u 0 t is a solution of 1.1. Step 2. We prove that u 0 t 0, as t ±.

10 10 Applied Mathematics Obviously, for every i N, there exists j i N such that for all j>j i, we have it it [ ukj t l 1 1 u ] k j t l 0 1 dt kj T k j T [ ukj t l 1 1 u ] k j t l 0 1 dt α l β l : M It follows that [ u 0 t l 1 1 u 0 t l 0 1 ] dt it [ lim u 0 t l1 1 u 0 t ] l 0 1 dt i it it lim lim i j it M 2, [ ukj t l 1 1 u ] k j t l 0 1 dt 3.6 and then t r [ u 0 t l 1 1 u 0 t l 0 1 ] dt 0, as r, 3.7 which yields t r u 0 t l 1 1 dt 0, t r u 0 t l 0 1 dt 0, as r. 3.8 By using Lemma 2.1, ast ±, u 0 t 2a 1/ l 1 1 ( t a ) 1/ l1 1 t a a 2a 1/ l 0 1 u 0 s l 1 1 ds ( t a t a ) 1/ l0 1 u 0 s l 0 1 ds So we have u 0 t 0, as t ±. Step 3. We will show that u 0 t 0, as t ±. 3.10

11 Applied Mathematics 11 From the Remark 2.8 and Lemma 2.9, we have u 0 t M, u 0 t M, for all t R, 3.11 which together with 1.1 implies that u 0 g M h M sup f t, t R 3.12 where g M max x M g x and h M max x M h x. Ifu 0 t 0, as t ±, then there exist a ε 0 0, 1/2 and a sequence {t k } such that t 1 < t 2 < t 3 <, t k 1 < t k 1, k N, u 0 t k 2ε0, k N From this, we have for t t k,t k ε 0 / 1 M 1 t u 0 t u 0 t k u 0 s ds u 0 t k t u 0 s ds ε t k t k It follows that u 0 t l0 1 dt k 1 tk ε 0 / 1 M 1 t k u 0 t l 0 1 dt, 3.15 which contradicts 3.6,andso 3.10 holds. Step 4. Finally, we will prove that 1.1 possesses a unique homoclinic solution. In order to do it, let u t u 1 t u 2 t, where u 1 t and u 2 t are two arbitrary homoclinic solutions of 1.1. Then u t 0 as t ± We will show that u t If 3.17 does not hold, then there must be a t R such that u t > or u t <

12 12 Applied Mathematics If u t > 0, then from 3.16, we see that there is a constant X > 0 such that t X, X and u t <u t /2 fort,x X,. Lett X, X such that u t max t X,X u t, then u t u t > u t u t > 0, sup u t, t,x X, that is, u t max t R u t So u t 0andu t 0, and then from 1.1,wesee h u 1 t h u 2 t u 1 t u 2 t u t By using the condition A3, we have that u t u 1 t u 2 t 0, 3.24 which contradicts to This contradiction implies that 3.18 does not hold. Similarly, we can prove that 3.19 does not hold, either. So u t 0. As an application, we consider the following example: u t m ( u t ) 3 n u t e t/2 e t e t, 3.25 where m, n > 0 are constants and, f t e t/2 / e t e t. Corresponding to 1.1, we can chose l 0 3andl 1 1 such that assumptions A2 and A3 hold. Furthermore, by the direct calculation, we can easily obtain that R R f t l 1 1 /l 1 dt f t l 0 1 /l 0 dt f t 2 π dt 4 <, f t 4/3 3 dt 2 < This implies that assumption A1 also holds. So by applying Theorem 3.1, we know that 3.25 possesses a unique homoclinic solution. Acknowledgments The authors are very grateful to the referee for her/his careful reading of the original paper and for her/his valuable suggestions for improving this paper. This work was sponsored by

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