A Periodic Solution of the Generalized Forced Liénard Equation

and Applied Analysis 3 Proof. According to the definition ofΦ and (17),Φ : Ω → X is continuous. Hence, in order to use Schauder’s fixed point theorem, we need to prove that Φ is compact and maps Ω into itself. For y ∈ Ω and t ∈ [0, T], we have 󵄨󵄨󵄨Φ [y] (t) 󵄨󵄨󵄨 ≤ T max [R1 ,R2] 󵄨󵄨󵄨󵄨 g 󸀠 (x) 󵄨󵄨󵄨󵄨 φ −1 p (B − A) + |D| , (20) because −B ≤ −S [y] (t) + C ≤ −A, (21) so 0 ≤ −S [y] (t) + C + B ≤ B − A; (22) hence, −S [y] (t) + C ≤ −S [y] (t) + C + B ≤ B − A. (23) Now, by (18) for all t ∈ [0, T], g (R1) − s ≤ Φ [y] (t) ≤ g (R2) − s. (24) So Φ : Ω → Ω. We recall a given sequence {φn(t)} of functions from [a, b] to R, which is called equicontinuous if for every ε > 0 there exists δ > 0 such that for all n ∈ N and for all t1, t2 ∈ [a, b] if |t1 − t2| < δ then |φn(t1) − φn(t2)| < ε. Now, we recall the Ascoli-Arzela theorem as follows. Theorem 2 (Ascoli-Arzela). Let φn(t) be a sequence of functions from [a, b] toRwhich is uniformly bounded and equicontinuous. Then, φn(t) has a uniformly convergent subsequence. In the sequel, we show that Φ is a compact operator on Ω. For this, we show that each bounded sequence {yn} in Ω has a convergent subsequence {yn i } in Ω. Suppose that {yn} is a sequence in Ω. It is clear that {yn} is bounded. Set max[R 1 ,R 2 ]|g (x)|φ p (B − A) = N, for given ε > 0; then, there exists δ > 0 such that 0 < Nδ < ε and, for every t1, t2 ∈ [0, T], if |t1 − t2| < δ, then 󵄨󵄨󵄨Φ [yn] (t1) − Φ [yn] (t2) 󵄨󵄨󵄨 ≤ N 󵄨󵄨󵄨t1 − t2 󵄨󵄨󵄨 < ε. (25) So, 󵄩󵄩󵄩Φ [yn] (t1) − Φ [yn] (t2) 󵄩󵄩󵄩 < ε. (26) Therefore, {Φ[yn](t)} is an equicontinuous sequence on [0, T]. By the Ascoli-Arzela theorem, there exists a subsequence {yn i (t)} of {yn(t)} such that {Φ[yn i ](t)} is uniformly convergent on [0, T]. Hence, Φ is a compact operator. Now, we recall Schauder’s fixed point theorem as follows. Theorem 3 (Schauder’s fixed point theorem). Let X be a Banach space and Ω a closed, bounded, and convex subset of X. If S : Ω → Ω is a compact operator, then S has at least one fixed point on Ω. Therefore, by Schauder’s fixed point theorem, there exists y ∈ Ω such that Φ(y) = y. This means that x = g(y + s) is a solution of (1). When g(x) < 0, Theorem 1 is satisfied by the following changes: Ω = {y ∈ X : y (t) ∈ [g (R2) − s, g (R1) − s]} . (27) Suppose that A, B are defined by A := − 󵄨󵄨󵄨󵄨 T ( 󵄨󵄨󵄨g (R1) − s 󵄨󵄨󵄨 + M0 +M1r +M3r 2 ) + C 󵄨󵄨󵄨󵄨 , B := 󵄨󵄨󵄨󵄨 T ( 󵄨󵄨󵄨g (R1) − s 󵄨󵄨󵄨 + M0 +M1r +M3r 2 ) − C 󵄨󵄨󵄨󵄨 , (28) where r is defined by r = max[0,T]|x (t)|. Now, we have the next corollary. Corollary 4. Suppose that f, k, and g are real functions on R which are locally Lipschitz such that g ∈ C1([R1, R2],R) and g(x) < 0 for real numbers R1 and R2 in which R1 < R2, p is a nonconstant, continuous, and T-periodic real function on [0, T] for T > 0, and φp : R → R is an increasing homeomorphism with φp(0) = 0, and s ∈ R. Let Φ, Ω, A, and B be defined by (10), (27), and (28), respectively. If 2(T max x∈[R1 ,R2] 󵄨󵄨󵄨󵄨 g 󸀠 (x) 󵄨󵄨󵄨󵄨 φ −1 p (B − A) + |D|) ≤ 󵄨󵄨󵄨g (R1) − g (R2) 󵄨󵄨󵄨 , s ∈ [g (R2) + T max [R1,R2] 󵄨󵄨󵄨󵄨 g 󸀠 (x) 󵄨󵄨󵄨󵄨 φ −1 p (B − A) + |D| , g (R1) − T max [R1,R2] 󵄨󵄨󵄨󵄨 g 󸀠 (x) 󵄨󵄨󵄨󵄨 φ −1 p (B − A) − |D|] , (29) then (3) has at least one T-periodic solution on Ω. Proof. Set Ω = {y ∈ X : y (t) ∈ [g (R2) − s, g (R1) − s]} . (30) Now, one can prove the existence of at least one T-periodic solution of (3) by the same argument in the proof of Theorem 1. Now, we consider a special case φ(x) = x and prove the existence of at least one T-periodic solution of the equation x 󸀠󸀠 + (f (x) + k (x) x 󸀠 ) x 󸀠 + g (x) = p (t) + s, (31) where f, k, and g are real functions on R such that g ∈ C1([R1, R2],R), p is a T-periodic real function on [0, T], T > 0, and s ∈ R. Consider that f, k, and g are real functions on R such that g ∈ C1[R1, R2] and g (x) > 0. Suppose that M0 is the maximum value of |p| on [0, T] and M1, M2, and M3 are the maximum values of |f|, |g|, and |k| on [R1, R2]. Also, suppose that φ is defined by (11). In the following, it is shown that (31) has at least one T-periodic solution. 4 Abstract and Applied Analysis Define a new variable y = g(x) − s; thus, (31) can be rewritten as follows: ( y g (g (y + s)) ) 󸀠 + (f (g −1 (y + s)) + k (g −1 (y + s)) y g (g (y + s)) ) × y g (g (y + s)) + y (t) = p (t) . (32) Then, y g (g (y + s))


Introduction
Liénard equation is a kind of differential equations which has a broad set of applications in physics, engineering, and so forth.The existence of at least one periodic solution of this equation has been studied by a number of authors (see, e.g., [1][2][3][4][5][6]).Forced Liénard equation can be considered as an important generalization of Liénard equations.This one appears in a number of physical models such as fluid mechanics and nonlinear elastic mechanical phenomena.

Existence of a Periodic Solution
In this section, we state and prove the existence of at least one periodic solution of the generalized forced Liénard equation by using Schauder's fixed point theorem (see [8][9][10]).In order to do this, at first consider the following generalized forced Liénard equation: + ( () +  ()  Similar to [5], choose a new variable  = () − ; then, (3) can be written as follows: We now integrate (4) to have where  is a constant.In the next step, for all  ∈ [0, ], we define the operator  by So by ( 6), ( 5) can be written as follows: One now integrates (7) to get where  is a constant.Suppose that  = C([0, ], R), where C stands for periodic continuous functions with zero mean value and Note that (, ‖ ⋅ ‖) is a Banach space.Now, define the set Ω as follows: One can show that Ω is a closed, bounded, and convex subspace of .Define the operator Φ : Ω →  by First, by Lemma 1 of [5], there exists a unique choice of  and , such that Φ[]() ∈ .Now, we prove that Φ : Ω → Ω.
In order to show this, let  0 be the maximum value of || on [0, ], and let Hence, for each  ∈ Ω and  ∈ [0, ], we have Using the same method of [5], one can prove the existence of at least one -periodic solution of (1) as follows. Theorem So Φ : Ω → Ω.
We recall a given sequence {  ()} of functions from [, ] to R, which is called equicontinuous if for every  > 0 there exists  > 0 such that for all  ∈ N and for all we recall the Ascoli-Arzela theorem as follows.
In the sequel, we show that Φ is a compact operator on Ω.For this, we show that each bounded sequence {  } in Ω has a convergent subsequence {   } in Ω. Suppose that ( − ) = , for given  > 0; then, there exists  > 0 such that 0 <  <  and, for every So, Therefore, by Schauder's fixed point theorem, there exists  ∈ Ω such that Φ() = .This means that  =  −1 ( + ) is a solution of (1).
When   () < 0, Theorem 1 is satisfied by the following changes: