Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2007, Article ID 60239, 9 pages doi:10.1155/2007/60239 Research Article Slowly Oscillating Solutions for Differential Equations with Strictly Monotone Operator Chuanyi Zhang and Yali Guo Received 2 August 2006; Accepted 28 February 2007 Recommended by Ondrej Dosly The authors discuss necessary and sufficient conditions for the existence and uniqueness of slowly oscillating solutions for the differential equation u  + F(u) = h(t)withstrictly monotone operator. Particularly, the authors give necessary and sufficient conditions for the existence and uniqueness of slowly oscillating solutions for the differential equation u  + ∇Φ(u) = h(t), where ∇Φ denotes the gradient of the convex function Φ on R N . Copyright © 2007 C. Zhang and Y. Guo. 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. 1. Introduction In this paper, we will consider the following differential equation: u  + F(u) = h(t), (1.1) where the maps h : R → R N and F : R N → R N are continuous. A special class, of the dis- sipative equation (1.1), is the case where the field F is derived from a convex potential Φ: u  + Φ(u) = h(t). (1.2) For the dissipative equation (1.1), Biroli [1], Dafermos [2], Haraux [3], Huang [4], and Ishii [5] have given important contr ibutions to the question of almost periodic solu- tions which are valid even for the abstract evolution equations. In [6], Philippe Cieutat gives necessary and sufficient conditions for the existence and uniqueness of the bounded (resp., almost periodic) solution of (1.2) when the forcing term h(t) is bounded (resp., almost periodic). In the scalar case N = 1, Slyusarchuk established similar results in [7]. But the conditions which are established in [6]for(1.2) do not hold for (1.1), even in 2 Journal of Inequalities and Applications the linear case. So in [6] Cieutat also gives a sufficient condition, then a necessary condi- tion, for the existence and uniqueness of the bounded (resp., almost periodic) solution of (1.1). The numerical space R N is endowed with its standard inner product  N k =1 x k y k , |·| denotes the associated Euclidian norm. We denote by BC(R N ) the Banach space of con- tinuous bounded functions from R to R N endowed with the norm u ∞ := sup t∈R |u(t)|. When k is a positive integer, BC k (R N ) is the space of functions in BC(R N )  C k (R N )such that all their derivatives, up to order k, are bounded functions. When u ∈ BC 1 (R N ), we set u c 1 =u ∞ + u   ∞ and when u ∈ BC 2 (R N ), we set u c 2 =u ∞ + u   ∞ + u   ∞ . In 1984, Sarason in [8] extended almost periodic functions and introduced the defi- nition of remotely periodic functions. The space of remotely periodic functions, as a C ∗ - subalgebra of BC( R N ), is generated by almost periodic functions and slowly oscillating functions which are defined as following. Definit ion 1.1 [8]. A function f ∈ BC(R N ) is said to be slowly oscillating if lim |t|→+∞   f (t + a) − f (t)   = 0, for each a ∈ R, (1.3) the set of all these functions is denoted by SO( R N ). Comparing with the space AP( R) of almost periodic functions, the space of slowly oscillating functions is quite large. In fact, AP( R) = span{e iλt : λ ∈ R}, where the closure is taken in BC( R)(e.g.,see[9] for details). SO(R) not only contains such space as C 0 (R) which consists of all the functions f such that f (t) → 0as|t|→∞,butalsoproperly contains X = span{e iλt α : λ ∈ R,0<α<1} (see [10–12] for details). The only functions in AP( R) ∩ SO(R) are the constant functions on R.Wealsopointoutthattheslowly oscillating functions SO( R N ) studied here form a strict subset of the slowly oscillating functions studied on [13, page 250, Definition 4.2.1]. Thus, all functions in SO( R N )are uniformly continuous. To our knowledge, nobody has investigated the existence and uniqueness of slowly os- cillating solutions for the differential equation (1.1). So in this paper, we g ive a sufficient, then a necessary condition for the existence and uniqueness of slowly oscillating solutions for the differential equation (1.1). We will give sufficient and necessary conditions for the existence and uniqueness of slowly oscillating solutions for differential equation (1.2). To show t he main results of the paper, we need the following definition and lemma. Definit ion 1.2. A function F : R N → R N is said to be strictly monotone on R N if (F(x 1 ) − F(x 2 ),x 1 − x 2 ) > 0forallx 1 ,x 2 ∈ R N such that x 1 = x 2 . Lemma 1.3 [6]. Let F : R N → R N be a continuous and strictly monotone map. Then for every compact subset K of R N and for every ε>0,thereexistsc>0 such that  F  x 1  − F  x 2  ,x 1 − x 2  >c   x 1 − x 2   2 (1.4) for all x 1 ,x 2 ∈ K such that   x 1 − x 2   ≥ ε. (1.5) C. Zhang and Y. Guo 3 2. Main results For each u ∈ BC 1 (R N ), the function t → u  (t)+F(u(t)) belongs to BC(R N ), so we can define the operator Ᏺ 1 : BC 1 (R N ) → BC 0 (R N )withᏲ 1 (u)(t):= u  (t)+F(u(t)) for all u ∈ BC 1 (R N )andt ∈ R.LetSO 1 (R N ) = SO(R N )  C 1 (R N )andu c 1 =u ∞ + u   ∞ for u ∈ SO 1 (R N ). Set Ᏺ 2 = Ᏺ 1 | SO 1 (R N ) . Consider the following assertions: (A) F is a strictly monotone map on R N such that lim |x|→∞  F(x),x  |x| = +∞; (2.1) (B) Ᏺ 2 :(SO 1 (R N ),· C 1 ) → (SO(R N ),· ∞ ) is a homeomorphism; (C) F :( R N ,|·|) → (R N ,|·|) is a homeomorphism. Theorem 2.1. Le t F : R N → R N be a continuous map. Then the following implications hold: (A) ⇒(B)⇒(C). Proof. By [6], (A) implies that Ᏺ 1 :(BC 1 (R N ),· C 1 ) → (BC(R N ),· ∞ )isahome- omorphism. That is, (1.1)hasforeachh from SO( R N ) a unique solution u( t)from SO 1 (R N ), which depends continuously on h. To show (A)⇒(B), it remains to show u ∈ SO(R N )ifh ∈ SO(R N ). Suppose, by the way of contradiction, u(t) ∈ SO(R N ). Then there exist a 0 ,ε 0 > 0and sequence t n →∞such that   u  t n + a 0  − u  t n    ≥ ε 0 . (2.2) Without loss of generality, we can assume t n − t n−1 → +∞. Since u(t) ∈ BC 1 (R N ), u(t)isuniformlycontinuousonR. Then there exists δ>0such that   u  t + a 0  − u(t)   ≥ ε 0 2 , ∀t ∈ p  n , (2.3) where P  n = (t n − δ,t n + δ). Set P n =  t n − δ,t n  , I n =  t n−1 ,t n  , C n =  s ∈ I n :   u  s + a 0  − u(s)   ≥ ε 0 2  , C  n =  s ∈ I n :   u  s + a 0  − u(s)   < ε 0 2  (2.4) and put Φ(t) = u  t + a 0  − u(t). (2.5) Obviously P n ⊂ C n . 4 Journal of Inequalities and Applications Let K = u(R). Note that u(t)isboundedonR N and therefore, K is a compact subset of R N .ByLemma 1.3 there exists c o such that  F  u  s + a o  − F  u(s)  ,u  s + a o  − u(s)  ≥ c o   u  s + a o  − u(s)   2 (2.6) for each s ∈ C n . Note  t n t n −1 d ds  1 2   Φ(s)   2 e 2c 0 s  ds = 1 2   Φ  t n    2 · e 2c 0 t n − 1 2   Φ  t n−1    2 · e 2c 0 t n−1 . (2.7) At the same time, we also have  t n t n −1 d ds  1 2   Φ(s)   2 e 2c 0 s  ds =  C n d ds  1 2   Φ(s)   2 e 2c 0 s  ds+  C  n d ds  1 2   Φ(s)   2 e 2c 0 s  ds. (2.8) Case 1. s ∈ C n , that is,   u  s + a 0  − u(s)   ≥ ε 0 2 . (2.9) We can ge t d ds  1 2   Φ(s)   2  =  Φ(s)  ,Φ(s)  =  h  s + a 0  − h(s),Φ(s)  −  F  u  s + a 0  − F  u(s)  ,Φ(s)  . (2.10) By (2.6), we can obtain d ds  1 2   Φ(s)   2  ≤   h  s + a o  − h(s)   ·   Φ(s)   − c 0   Φ(s)   2 . (2.11) Also we can see d ds  1 2   Φ(s)   2 e 2c 0 s  =  Φ  (s),Φ(s)  · e 2c 0 s + c 0 e 2c 0 s ·   Φ(s)   2 . (2.12) By (2.11), we deduce d ds  1 2   Φ(s)   2 e 2c 0 s  ≤   h  s + a o  − h(s)   ·   Φ(s)   · e 2c 0 s . (2.13) Case 2. s ∈ C  n , that is,   u  s + a 0  − u(s)   < ε 0 2 . (2.14) Moreover , one has d ds  1 2   Φ(s)   2 e 2c 0 s  =  Φ  (s),Φ(s)  · e 2c 0 s + c 0 e 2c 0 s ·   Φ(s)   2 , (2.15)  Φ(s)  ,Φ(s)  =  h  s + a 0  − h(s),Φ(s)  −  F  u  s + a 0  − F  u(s)  ,Φ(s)  . (2.16) C. Zhang and Y. Guo 5 For F is strictly monotone on R N , we can deduce  F  u  s + a 0  − F  u(s)  ,Φ(s)  > 0. (2.17) Moreover , by (2.17)and(2.16)onehas  Φ(s)  ,Φ(s)  <   Φ(s)   ·   h  s + a 0  − h(s)   < ε 0 2   h  s + a 0  − h(s)   . (2.18) By (2.15)and(2.18), we can get d ds  1 2   Φ(s)   2 e 2c 0 s  < ε 0 2   h  s + a 0  − h(s)   · e 2c 0 s + ε 2 0 4 c 0 e 2c 0 s . (2.19) Considering the above two cases, one has 1 2   Φ  t n    2 · e 2c 0 t n − 1 2   Φ  t n−1    2 · e 2c 0 t n−1 =  t n t n−1 d ds  1 2   Φ(s)   2 e 2c 0 s  ds <  C n   h  s + a o  − h(s)   ·   Φ(s)   · e 2c 0 s ds +  C  n  ε 0 2   h  s + a 0  − h(s)   + ε 2 0 4 c 0  · e 2c 0 s ds ≤ sup t∈I n   h  t + a 0  − h(t)   · sup t∈I n   Φ(t)   ·  C n e 2c 0 s ds +  ε 0 2 sup t∈I n   h  t + a 0  − h(t)   + ε 2 0 4 c 0  ·  C  n e 2c 0 s ds. (2.20) Since  C n e 2c 0 s ds ≤  t n t n−1 e 2c 0 s ds,  C  n e 2c 0 s ds ≤  t n t n−1 e 2c 0 s ds−  t n t n −δ e 2c 0 s ds, (2.21) one has 1 2   Φ  t n    2 · e 2c 0 t n − 1 2   Φ  t n−1    2 · e 2c 0 t n−1 ≤ sup t∈I n   h  t + a 0  − h(t)   · sup t∈I n   Φ(t)   ·  t n t n−1 e 2c 0 s ds +  ε 0 2 sup t∈I n   h  t + a 0  − h(t)   + ε 2 0 4 c 0  ·   t n t n−1 e 2c 0 s ds−  t n t n −δ e 2c 0 s ds  . (2.22) 6 Journal of Inequalities and Applications So, we can get 1 2   Φ  t n    2 · e 2c 0 t n − 1 2   Φ  t n−1    2 · e 2c 0 t n−1 ≤ sup t∈I n   h  t + a 0  − h(t)   · sup t∈I n   Φ(t)   · 1 2c 0  e 2c 0 t n − e 2c 0 t n−1  +  ε 0 2 sup t∈I n   h  t+a 0  − h(t)   + ε 2 0 4 c 0  ·  1 2c 0  e 2c 0 t n −e 2c 0 t n−1  − e 2c 0 (t n −δ) ·  e 2c 0 δ −1  · 1 2c 0  . (2.23) That is, 1 2   Φ  t n    2 · e 2c 0 t n − 1 2   Φ  t n−1    2 · e 2c 0 t n−1 ≤ sup t∈I n   h  t + a 0  − h(t)   ·  e 2c 0 t n − e 2c 0 t n−1 2c 0 · sup t∈I n   Φ(t)   + ε 0 2 ·  e 2c 0 t n − e 2c 0 t n−1 2c 0 −  e 2c 0 δ − 1  2c 0 · e 2c 0 (t n −δ)  +  ε 2 0  e 2c 0 t n − e 2c 0 t n−1  8 − ε 2 0  e 2c 0 δ − 1  8 · e 2c 0 (t n −δ)  . (2.24) Thus, sup t∈I n   h  t + a 0  − h(t)   ≥ 1 2   Φ  t n    2 ·e 2c 0 t n − 1 2   Φ  t n−1    2 ·e 2c 0 t n−1 − ε 2 0  e 2c 0 t n −e 2c 0 t n−1  8 + ε 2 0  e 2c 0 δ − 1  8 ·e 2c 0 (t n −δ) sup t∈I n   Φ(t)   · 1 2c 0  e 2c 0 t n −e 2c 0 t n−1  + ε 0 2 ·  1 2c 0  e 2c 0 t n −e 2c 0 t n−1  − e 2c 0 (t n −δ) ·  e 2c 0 δ −1  · 1 2c 0  =  4c 0   Φ  t n    2 −ε 2 0 c 0  · e 2c 0 t n −  4c 0   Φ  t n−1    2 −ε 2 0 c 0  · e 2c 0 t n−1 + ε 2 0 c 0  e 2c 0 δ − 1  · e 2c 0 (t n −δ)  4sup t∈I n   Φ(t)   +2ε 0  ·  e 2c 0 t n − e 2c 0 t n−1  − 2ε 0  e 2c 0 δ − 1  · e 2c 0 (t n −δ) ≥  4c 0   Φ  t n    2 −ε 2 0 c 0  · e 2c 0 t n −  4c 0   Φ  t n−1  | 2 −ε 2 0 c 0  · e 2c 0 t n−1 + ε 2 0 c 0  e 2c 0 δ − 1  · e 2c 0 (t n −δ)  4sup t∈I n   Φ(t)   +2ε 0  ·  e 2c 0 t n − e 2c 0 t n−1  =  4c 0   Φ  t n    2 − ε 2 0 c 0  −  4c 0   Φ  t n−1    2 − ε 2 0 c 0  · e 2c 0 (t n−1 −t n ) + ε 2 0 c 0  e 2c 0 δ − 1  · e −2c 0 δ  4sup t∈I n   Φ(t)   +2ε 0  ·  1 − e 2c 0 (t n−1 −t n )  . (2.25) Since Φ(t) = u(t + a 0 ) − u(t) and the solution u(t)isbounded,thenwecanassume ∃M>0, such that   Φ(t)   <M,foreacht ∈ R. (2.26) C. Zhang and Y. Guo 7 Also we have   Φ  t n    2 ≥ ε 2 0 4 ,fort n ∈ C n . (2.27) Then sup t∈I n   h  t + a 0  − h(t)   ≥ −  4c 0 M 2 − ε 2 0 c 0  · e 2c 0 (t n−1 −t n ) + ε 2 0 c 0  e 2c 0 δ − 1  · e −2c 0 δ  4M +2ε 0  ·  1 − e 2c 0 (t n−1 −t n )  . (2.28) When n → +∞,onehas t n − t n−1 −→ +∞, e 2c 0 (t n−1 −t n ) −→ 0. (2.29) So lim t→+∞   h  t + a 0  − h(t)   > 1 2 · ε 2 0 c 0  e 2c 0 δ − 1  · e −2c 0 δ 4M +2ε 0 = ε 2 0 c 0  e 2c 0 δ − 1  2e 2c 0 δ  4M +2ε 0  > 0. (2.30) This contradicts the fact h(t) ∈ SO(R N ). We must have u(t) ∈ SO(R N ). Finally we show that (B) =⇒ (C). (2.31) If we denote by Ꮿ the set of constant mapping from R to R N ,onehasᏯ ⊂ SO 1 (R N ) and for u ∈ Ꮿ, the function Ᏺ 3 (u) ∈ SO(R N )(Ᏺ 3 (u) = F(u(0)), for all t ∈ R), so we can define the restriction operator of Ᏺ 3 to Ꮿ by Ᏺ 4 : Ꮿ → Ꮿ with Ᏺ 4 (u) = F(u(0)) for all u ∈ Ꮿ and all t ∈ R.Foru ∈ Ꮿ,onehasu C 1 =|u(0)| and Ᏺ 3 (u) ∞ =|F(u(0))|;then it is equivalent to prove Ᏺ 4 or F is a homeomorphism. It remains to prove that Ᏺ 4 is surjective. Let h ∈ Ꮿ. By hypothesis, there exists u ∈ SO 1 (R N )suchthatᏲ 3 (u) = h.we want to prove that u ∈ Ꮿ. For that we denote by u a (t) = u(t + a)forallt and a ∈ R.Note that Ᏺ 3 (u a ) = h for all a ∈ R.ByinjectivityofᏲ 3 ,wededucethatu a (t) = u(t)forall a ∈ R, therefore u ∈ Ꮿ. Remark 2.2. The following example constructed in [6] can be used to show that Asser- tion (A) is not a necessary condition for the existence or the uniqueness of a bounded or slowly oscillating solution of (1.1). Consider the map F : R 2 → R 2 defined by F(x 1 ,x 2 ) = Bx = (−x 2 ,x 1 + x 2 ). The map F is monotone and does not satisfy (A). However, the eigenvalues of B are conjugate and their real parts are equal to 1/2, therefore the lin- ear system u  + Bu = 0 has an exponential dichotomy: namely, there exists k>0suchthat exp(−Bt) L(R) ≤ k exp(−t/2) for all t ≥ 0. As a consequence, the system u  + Bu = 0has precisely one bounded solution on R: u = 0; this implies the injectivity of Ᏺ 1 .Moreover, the following function u(t): =  t −∞ exp(−B(t − s))h(s)ds isasolutionofu  + Bu = h for h ∈ BC(R N ) and satisfies |u(t)|≤2kh ∞ for all t ∈ R; this implies the surjectivity of Ᏺ 1 .SinceᏲ 1 is a bounded linear map between Banach spaces, which is bijective, then Ᏺ 1 is an isomorphism between BC 1 (R 2 )andBC(R N ). To show that (B) holds in this case, it 8 Journal of Inequalities and Applications remains to show that u ∈ SO(R 2 )ifh ∈ SO(R 2 ). In fact, for a ∈ R   u(t + a) − u(t)   =      t+a −∞ exp  − B(t + a − s)  h(s)ds−  t −∞ exp  − B(t − s)  h(s)ds     ≤  t −∞ exp  − B(t − s)    h(s + a) − h(s)   ds. (2.32) It follows that |u(t + a) − u(t)|→0ast →−∞.Sinceh ∈ SO(R 2 ), for ε>0 there exists t 0 > 0suchthat|h(t + a) − h(t)| <εfor all t>t 0 .Now   u(t + a) − u(t)   ≤   t 0 −∞ +  t t 0  exp  − B(t − s)    h(s + a) − h(s)   ds ≤ 2h ∞  t 0 −∞ exp  − B(t − s)  ds+ ε  t t 0 exp  − B(t − s)  ds (2.33) and therefore, u(t + a) − u(t) → 0ast → +∞. This shows that u ∈ SO(R 2 ). Remark 2.3. The following example constructed also in [6] can be used to show that (C) is not a sufficient condition for the existence of slowly oscillating solution of (1.1)even when F is a linear monotone map. Consider the map F : R 2 → R 2 defined by F(x 1 ,x 2 ):= Ax = (−x 2 ,x 1 ). F is a homeomorphism and a monotone map. Let v = (sint,cost). Then v  + Av = 0. Let f be any continuously differentiable function on R such that f (t) = t 1/3 for |t| > 1andleth(t) = f  (t)v(t). Since h(t) → 0as|t|→∞, h ∈ SO(R 2 ). The equation u  + Au = h has no bounded solution, because u(t) = f (t)v(t) is an unbounded solution, therefore Ᏺ 2 is not surjective. Nevertheless, for (1.2) we have the following result. Theorem 2.4. Let Φ beaconvexandcontinuouslydifferentiable function on R N . Assume that F = Φ. Then (A), (B), and (C) are equivalent. We have already shown that (A) ⇒(B)⇒(C) in Theorem 2.1. The equivalence of (A) and (C) is [6, Theorem 1.1].  Acknowledgments The authors would like to thank the referee for the valuable comments. The authors also would like to t hank Professor Ondrej Dosly for his regards. The research is supported by the NSF of China (no. 10671046). 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Applications Volume 2007, Article ID 60239, 9 pages doi:10.1155/2007/60239 Research Article Slowly Oscillating Solutions for Differential Equations with Strictly Monotone Operator Chuanyi Zhang. necessary and sufficient conditions for the existence and uniqueness of slowly oscillating solutions for the differential equation u  + F(u) = h(t)withstrictly monotone operator. Particularly, the authors. of slowly os- cillating solutions for the differential equation (1.1). So in this paper, we g ive a sufficient, then a necessary condition for the existence and uniqueness of slowly oscillating solutions for