Math Nachr 278, No 4, 356 – 362 (2005) / DOI 10.1002/mana.200410245 Anti-periodic solutions for evolution equations with mappings in the class (S+ ) Yu Qing Chen∗1 , Yeol Je Cho∗∗2 , and Donal O’Regan∗∗∗3 Department of Mathematics, Foshan University, Foshan, Guangdong 528000, P R China Department of Mathematics Education and the RINS, College of Education, Gyeongsang National University, Chinju 660-701, Korea Department of Mathematics, National University of Ireland, Galway, Ireland Received June 2004, accepted September 2004 Published online 31 January 2005 Key words Evolution equation, anti-periodic solution, a mapping of class (S+ ) MSC (2000) Primary: 34C25, 34G20; Secondary: 47H05, 47H10 In this paper, we study the existence of anti-periodic solutions for the first order evolution equation u (t) + ∂Gu(t) + f (t) = , t ∈ R , u(t + T ) = −u(t) , t ∈ R, in a Hilbert space H, where G : H → R is an even function such that ∂G is a mapping of class (S+ ) and f : R → R satisfies f (t + T ) = −f (t) for any t ∈ R with f (·) ∈ L2 (0, T ; H) c 2005 WILEY-VCH Verlag GmbH & Co KGaA, Weinheim Introduction Okochi [14], [15] initiated the study of anti-periodic solutions for the following nonlinear evolution equation u (t) + ∂φ(u(t)) f (t) , a e t ∈ R , u(t + T ) = −u(t) , t ∈ R, (E 1.1) in Hilbert spaces, where ∂φ is the subdifferential of an even function φ on a real Hilbert space H and f is a T -anti-periodic function It was shown in [15], by applying a fixed point theorem for a non-expansive mapping, that the problem (E 1.1) has a solution Following Okochi’s work, Haraux [13] proved some existence and uniqueness theorems for anti-periodic solutions of gradient type equations by using Brouwer’s and Schauder’s fixed point theorem Later, Aftabizadeh, Aizicovici and Pavel [1], [2], [3] studied the anti-periodic solutions for first and second order evolution equations in Hilbert and Banach spaces by using maximal monotone or m-accretive operator theory We refer the reader to Aizicovici, MacKibben and Reich [4], Chen, Cho [8], [9], [12], Souplet [19], [20] for other works on anti-periodic solutions In this paper, we will study the existence problems of anti-periodic solution for the first order evolution equation u (t) + ∂Gu + f (t) = , a e t ∈ R , u(t + T ) = −u(t) , t ∈ R, (E 1.2) in a real separable Hilbert space H, where G : H → R is an even function such that ∂G is a mapping of class (S+ ) and f : R → R satisfies f (t + T ) = −f (t) for t ∈ R and f (·) ∈ L2 (0, T ; H) This equation is still of gradient type, but we not require any Lipschitz condition on ∂G which is required in [9] and [13] ∗ ∗∗ ∗∗∗ e-mail: yqchen@foshan.net Corresponding author: e-mail: yjcho@gsnu.ac.kr e-mail: donal.oregan@nuigalway.ie c 2005 WILEY-VCH Verlag GmbH & Co KGaA, Weinheim Math Nachr 278, No (2005) / www.mn-journal.com 357 The existence results In this section, assume that H is a real Hilbert space, G : D(G) ⊆ H → R is an even function, and f (t) : R → H is a function satisfies f (t + T ) = −f (t) Consider the following anti-periodic problem u (t) + ∂Gu(t) + f (t) = , a e t ∈ R , u(t) = −u(t + T ) , t ∈ R (E 2.1) Definition 2.1 A function u(·) is called a weak anti-periodic solution of the problem (E 2.1) if u(t + T ) = −u(t) for t ∈ R and the weak derivative u (t) satisfies u (t) + ∂Gu(t) + f (t) = for almost all t ∈ R Lemma 2.2 ([12]) If u, u ∈ L2 (0, T ; H) and u(t + T ) = −u(t) for any t ∈ R, then |u|∞ √ T ≤ T |u (s)| ds In this paper, we shall use different assumptions and different methods from those papers mentioned in Section For completeness, let us recall the definition of mappings in the class (S+ ) in a Hilbert space Definition 2.3 Let T : D(T ) ⊆ H → H be a mapping x0 and lim supn→∞ (T xn , xn − x0 ) ≤ imply that xn → x0 , then we call T a (1) If {xn } ⊂ D(T ), xn mapping of class (S+ ), (2) T is said to be demi-continuous if xn → x0 implies that T xn T x0 For mappings of class (S+ ) in reflexive Banach spaces and their applications, we refer the reader to [5]–[7], [9]–[11], [17] and [18] Theorem 2.4 Let H be a real separable Hilbert space and let G : H → R be even and Fr´echet differentiable If ∂G is a demi-continuous bounded mapping of class (S+ ) and f : R → H is in L2 (0, T ; H) and satisfies f (t + T ) = −f (t) for all t ∈ R, then the equation u + ∂Gu(t) + f (t) = , a e t ∈ R , u(t + T ) = −u(t) , t ∈ R, (E 2.2) has a weak solution P r o o f Since H is separable, there exists an orthogonal basis {e1 , e2 , } of H Set Hn = Span {e1 , e2 , , en } for n = 1, 2, and let Pn : H → Hn be the projection We consider the equation u (t) + Pn ∂Gu(t) + Pn f (t) = , a e t ∈ R , u(t + T ) = −u(t) , t ∈ R (E 2.3) For each n = 1, 2, , set Wn = {u : R → Hn is continuous, u(t + T ) = −u(t)} , and W 1,2 = n u ∈ Wn : T |u (t)|2 dt < ∞ c 2005 WILEY-VCH Verlag GmbH & Co KGaA, Weinheim 358 Chen, Cho, and O’Regan: Anti-periodic solutions for evolution equations Then Wn is a Banach space under the norm |u|∞ = maxt∈[0,T ] |u(t)|, and we may define a norm on W 1,2 n by u W 1,2 T = |u|∞ + 2 |u (t)| dt , where | · | is the norm in H For each v(·) ∈ Wn , consider the following equation u (t) + Pn ∂Gv(t) + Pn f (t) = , a e t ∈ R , u(t) = −u(t + T ) , t ∈ R (E 2.4) It is easy to check that u(t) = − t [Pn ∂Gv(s) + Pn f (s)] ds + T [Pn ∂Gv(s) + Pn f (s)] ds is the unique solution of the problem (E 2.4) Next, we define a mapping L : Wn → Wn as follows: For each v(·) ∈ Wn , Lv is the solution of the problem (E 2.4) We prove that L is continuous Suppose vj (·) ∈ Wn and vj (·) → v0 (·) in Wn Then |vj (·) − v(·)|∞ → as j → ∞ Since (Lvj (t)) − (Lv0 (t)) + (Pn ∂Gvj (t) − Pn ∂Gv0 (t)) = , a e t ∈ R , (2.1) if we multiply both sides of (2.1) by (Lvj (t) − Lv0 (t)) and integrate over (0, T ), we get T T |(Lvj (t) − Lv0 (t)) |2 dt + (Pn ∂Gvj (t) − Pn ∂Gv0 (t))(Lvj (t) − Lv0 (t)) ) dt = Since Pn ∂G is continuous in Hn , we have T 2 |(Lvj (t) − Lv0 (t)) | dt ≤ √ T |Pn ∂Gvj (·) − Pn ∂Gv0 (·)|∞ −→ as j −→ ∞ This with Lemma 2.2 guarantees that L is continuous For each v(·) ∈ Wn , again by the problem (E 2.4), we get T T |(Lv(t)) |2 dt + T (Pn ∂Gv(t), (Lv(t)) ) dt + (Pn f (t), (Lv(t)) ) dt = Thus it follows that T 2 |(Lv(t)) | dt ≤ T |Pn Gv(t)| dt T + |f (t)| dt (2.2) From (2.2), the continuity of Pn ∂G in Hn and Lemma 2.2, we know that L maps bounded sets of Wn to bounded 1,2 sets in Wn The compact embedding of W 1,2 n into Wn together with L : Wn → W n continuous guarantees that L : Wn → Wn is a compact mapping Now, we prove that Lv(·) = λv(·) for v(·) ∈ Wn with √ |v(·)|∞ > T T |f (t)|2 dt and λ ≥ If this is not true, then there exist λ0 ≥ 1, v0 (·) ∈ Wn with |v(·)|∞ > √ T T |f (t)|2 dt such that Lv0 (·) = λ0 v0 (·), i.e., λ0 v0 (t) + Pn ∂Gv0 (t) + Pn f (t) = (2.3) c 2005 WILEY-VCH Verlag GmbH & Co KGaA, Weinheim Math Nachr 278, No (2005) / www.mn-journal.com 359 Multiply both sides of (2.3) by v0 (t) and integrate over [0, T ] to obtain T (2.4) (λ0 v0 (t) + Pn ∂Gv0 (t) + Pn f (t), v0 (t)) dt = Notice that T T (Pn ∂Gv0 (t), v0 (t)) dt = (∂Gv0 (t), v0 (t)) dt = 0 and so we have T T |v0 (t)|2 dt ≤ |f (t)|2 dt , which implies from Lemma 2.2 that √ |v(·)|∞ ≤ which is a contradiction T √ T |f (t)|2 dt T Now, if we take r0 > 2T |f (t)|2 dt the Leray-Schauder degree, we know that , , by the above argument and the homotopy invariance property of deg(I − L, B(0, r0 ), 0) = deg(I, B(0, r0 ), 0) = , where B(0, r0 ) is the open ball centered at with radius r0 in Wn Therefore, L has a fixed point in B(0, r0 ), i.e., there exists (·) ∈ Wn such that Lvn (·) = (·) Hence problem (E 2.3) has a solution (·) It is easy to show that T |vn (t)|2 T dt ≤ |Pn f (t)| dt (2.5) From (2.5) and Lemma 2.2, it follows that |vn (·)|∞ √ T ≤ T |Pn f (t)| dt (2.6) From (2.5), we may assume that vk (·) y(·) ∈ L2 (0, T ; H) and, in view of (2.6), we may also assume that t (0) v0 ∈ H (otherwise, take a subsequence) Since (t) = (0)+ (s) ds, we have (t) v0 (t) = t t v0 + y(s) ds, where y(s) ds is the weak integral Multiply (E 2.3) by (t) − v0 (t) and integrate over [0, T ] to get T [(vn (t), (t) − v0 (t)) + (∂Gvn (t), (t) − Pn v0 (t)) + (Pn f (t), (t) − v0 (t))] dt = Let n → ∞ to obtain T lim n→∞ (∂Gvn (t), (t) − Pn v0 (t)) dt = We claim that limn→∞ T (∂Gvn (t), v0 (t) − Pn v0 (t)) dt = In fact, we have |(∂Gvn (t), v0 (t) − Pn v0 (t))| ≤ |∂Gvn (t)||v0 (t) − Pn v0 (t)| −→ as n −→ ∞ c 2005 WILEY-VCH Verlag GmbH & Co KGaA, Weinheim 360 Chen, Cho, and O’Regan: Anti-periodic solutions for evolution equations for t ∈ [0, T ], and it follows from (2.6) and the boundedness of ∂G that |∂Gvn (t)||v0 (t) − Pn v0 (t)| is bounded by some integrable function, thus by Lebesgue’s dominated convergence theorem it follows that T lim n→∞ (∂Gvn (t), v0 (t) − Pn v0 (t)) dt = Thus we have T lim n→∞ (∂Gvn (t), (t) − v0 (t)) dt = (2.7) However ∂G is a mapping of class (S+ ), and we claim that lim inf (∂Gvn (t), (t) − v0 (t)) ≥ , n→∞ t ∈ [0, T ] Assume that it is false Then there exists t0 ∈ [0, T ] such that lim inf (∂Gvn (t0 ), (t0 ) − v0 (t0 )) < , n→∞ but ∂G is a mapping of class (S+ ), so we have (t0 ) → v0 (t0 ), and this with the demi-continuity of ∂G implies that (∂Gvn (t0 ), (t0 ) − v0 (t0 )) −→ as n → ∞, which is a contradiction This and (2.7) imply that {(∂Gvn (t), (t)−v0 (t))}∞ n=1 converges to zero in measure as ∞ n → ∞ and so {(∂Gvn (t), (t) − v0 (t))}∞ n=1 has a subsequence {(∂Gvnk (t), vnk (t) − v0 (t))}k=1 converging to zero as k → ∞ for almost all t ∈ [0, T ] Since ∂G is a mapping of class (S+ ), we get vnk (t) → v0 (t) for almost all t ∈ [0, T ] From the demi-continuity of ∂G, we have ∂Gvnk (t) ∂Gv0 (t) for almost all t ∈ [0, T ] It is obvious that the weak derivative v0 (t) = ∂Gv0 (t) + f (t) Therefore, equation (E 2.2) has a weak solution An example In this section, we give an application of our results to anti-periodic solutions for partial differential equations Example 3.1 Let a : R → R be a function satisfying the following conditions: (i) a is continuous and |a (x)| ≤ L |x| + C for x ∈ R, where L > 0, C > are constants, (ii) (a (x) − a (y))(x − y) ≥ α(x − y)2 for x, y ∈ R, where α > is a constant Consider the anti-periodic problem ⎧ ⎪ ⎨ut (t, x) = a (ux (t, x))uxx (t, x) + (1 + x ) sin t , (E 3.1) u(t + π, x) = −u(t, x) , ⎪ ⎩ u(t, 0) = , u(t, 1) = for any (t, x) ∈ R × (0, 1) and t ∈ R We call u(t, x) as a generalized solution of (E 3.1) if u(t + π) = −u(t) and for all v(·) ∈ Put ut (t, x)v(x) dx = − H01 ((0, 1)) G(u) = 1 a (ux (t, x))v (x) + (1 + x4 ) sin3 tv(x) dx and almost all t ∈ R |a(u (x))|2 dx for u(·) ∈ H01 ((0, 1)) and c 2005 WILEY-VCH Verlag GmbH & Co KGaA, Weinheim Math Nachr 278, No (2005) / www.mn-journal.com f (t, x) = + x4 sin3 t, 361 (t, x) ∈ R × (0, 1) Then it is easy to check that (∂Gu, v) = a (u (x))v (x) dx , u(·), v(·) ∈ H01 ((0, 1)) (3.1) From assumption (i) and (3.1), we know that ∂G is continuous and bounded on H01 ((0, 1)) and it is obvious that f (t + π, x) = −f (t, x) for any (t, x) ∈ R × (0, 1) Therefore, (E 3.1) is equivalent to u (t) = −∂Gu(t) + f 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