Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2013, Article ID 724190, 10 pages http://dx.doi.org/10.1155/2013/724190 Research Article Optimal Control Problems for Nonlinear Variational Evolution Inequalities Eun-Young Ju and Jin-Mun Jeong Department of Applied Mathematics, Pukyong National University, Busan 608-737, Republic of Korea Correspondence should be addressed to Jin-Mun Jeong; jmjeong@pknu.ac.kr Received 12 November 2012; Revised 4 January 2013; Accepted 6 January 2013 Academic Editor: Ryan Loxton Copyright © 2013 E.-Y. Ju and J.-M. Jeong. 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 deal with optimal control problems governed by semilinear parabolic type equations and in particular described by variational inequalities. We will also characterize the optimal controls by giving necessary conditions for optimality by proving the GaΜteaux differentiability of solution mapping on control variables. 1. Introduction In this paper, we deal with optimal control problems governed by the following variational inequality in a Hilbert space π»: (π₯σΈ (π‘) + π΄π₯ (π‘) , π₯ (π‘) − π§) + π (π₯ (π‘)) − π (π§) ≤ (π (π‘, π₯ (π‘)) + π΅π’ (π‘) , π₯ (π‘) − π§) , a.e., 0 < π‘ ≤ π, π§ ∈ π, (1) π₯ (0) = π₯0 . Here, π΄ is a continuous linear operator from π into π∗ which is assumed to satisfy GaΜrding’s inequality, where π is dense subspace in π». Let π : π → (−∞, +∞] be a lower semicontinuous, proper convex function. Let U be a Hilbert space of control variables, and let π΅ be a bounded linear operator from U into πΏ2 (0, π; π»). Let Uad be a closed convex subset of U, which is called the admissible set. Let π½ = π½(π£) be a given quadratic cost function (see (61) or (103)). Then we will find an element π’ ∈ Uad which attains minimum of π½(π£) over Uad subject to (1). Recently, initial and boundary value problems for permanent magnet technologies have been introduced via variational inequalities in [1, 2] and nonlinear variational inequalities of semilinear parabolic type in [3, 4]. The papers treating the variational inequalities with nonlinear perturbations are not many. First of all, we deal with the existence and a variation of constant formula for solutions of the nonlinear functional differential equation (1) governed by the variational inequality in Hilbert spaces in Section 2. Based on the regularity results for solution of (1), we intend to establish the optimal control problem for the cost problems in Section 3. For the optimal control problem of systems governed by variational inequalities, see [1, 5]. We refer to [6, 7] to see the applications of nonlinear variational inequalities. Necessary conditions for state constraint optimal control problems governed by semilinear elliptic problems have been obtained by Bonnans and Tiba [8] using methods of convex analysis (see also [9]). Let π₯π’ stand for solution of (1) associated with the control π’ ∈ U. When the nonlinear mapping π is Lipschitz continuous from R × π into π», we will obtain the regularity for solutions of (1) and the norm estimate of a solution of the above nonlinear equation on desired solution space. Consequently, in view of the monotonicity of ππ, we show that the mapping π’ σ³¨→ π₯π’ is continuous in order to establish the necessary conditions of optimality of optimal controls for various observation cases. In Section 4, we will characterize the optimal controls by giving necessary conditions for optimality. For this, it is necessary to write down the necessary optimal condition due to the theory of Lions [9]. The most important objective of such a treatment is to derive necessary optimality conditions 2 Abstract and Applied Analysis that are able to give complete information on the optimal control. Since the optimal control problems governed by nonlinear equations are nonsmooth and nonconvex, the standard methods of deriving necessary conditions of optimality are inapplicable here. So we approximate the given problem by a family of smooth optimization problems and afterwards tend to consider the limit in the corresponding optimal control problems. An attractive feature of this approach is that it allows the treatment of optimal control problems governed by a large class of nonlinear systems with general cost criteria. If π» is identified with its dual space we may write π ⊂ π» ⊂ π∗ densely and the corresponding injections are continuous. The norm on π, π», and π∗ will be denoted by || ⋅ ||, | ⋅ |, and || ⋅ ||∗ , respectively. The duality pairing between the element π£1 of π∗ and the element π£2 of π is denoted by (π£1 , π£2 ), which is the ordinary inner product in π» if π£1 , π£2 ∈ π». For π ∈ π∗ we denote (π, π£) by the value π(π£) of π at π£ ∈ π. The norm of π as element of π∗ is given by π£∈π |(π, π£)| . βπ£β (2) Therefore, we assume that π has a stronger topology than π» and, for brevity, we may regard that βπ’β∗ ≤ |π’| ≤ βπ’β , ∀π’ ∈ π. (3) Let π(⋅, ⋅) be a bounded sesquilinear form defined in π×π and satisfying GaΜrding’s inequality Re π (π’, π’) ≥ π1 βπ’β2 − π2 |π’|2 , (4) where π1 > 0 and π2 is a real number. Let π΄ be the operator associated with this sesquilinear form: (π΄π’, π£) = π (π’, π£) , π’, π£ ∈ π. (5) Then −π΄ is a bounded linear operator from π to π∗ by the Lax-Milgram Theorem. The realization of π΄ in π» which is the restriction of π΄ to π· (π΄) = {π’ ∈ π : π΄π’ ∈ π»} (6) is also denoted by π΄. From the following inequalities π1 βπ’β2 ≤ Re π (π’, π’) + π2 |π’|2 ≤ πΆ |π΄π’| |π’| + π2 |π’|2 ≤ (πΆ |π΄π’| + π2 |π’|) |π’| ≤ max {πΆ, π2 } βπ’βπ·(π΄) |π’| , (7) (10) where each space is dense in the next one with continuous injection. Lemma 1. With the notations (9) and (10), we have (π, π∗ )1/2,2 = π», (π·(π΄), π»)1/2,2 = π, (11) 2 1/2 βπ’βπ·(π΄) = (|π΄π’| + |π’| ) (8) is the graph norm of π·(π΄), it follows that there exists a constant πΆ0 > 0 such that 1/2 βπ’β ≤ πΆ0 βπ’β1/2 π·(π΄) |π’| . It is also well known that π΄ generates an analytic semigroup π(π‘) in both π» and π∗ . For the sake of simplicity we assume that π2 = 0 and hence the closed half plane {π : Re π ≥ 0} is contained in the resolvent set of π΄. If π is a Banach space, πΏ2 (0, π; π) is the collection of all strongly measurable square integrable functions from (0, π) into π and π1,2 (0, π; π) is the set of all absolutely continuous functions on [0, π] such that their derivative belongs to πΏ2 (0, π; π). πΆ([0, π]; π) will denote the set of all continuously functions from [0, π] into π with the supremum norm. If π and π are two Banach spaces, L(π, π) is the collection of all bounded linear operators from π into π, and L(π, π) is simply written as L(π). Here, we note that by using interpolation theory we have πΏ2 (0, π; π) ∩ π1,2 (0, π; π∗ ) ⊂ πΆ ([0, π] ; π») . (9) (12) First of all, consider the following linear system: π₯σΈ (π‘) + π΄π₯ (π‘) = π (π‘) , π₯ (0) = π₯0 . (13) By virtue of Theorem 3.3 of [11] (or Theorem 3.1 of [12, 13]), we have the following result on the corresponding linear equation of (13). Lemma 2. Suppose that the assumptions for the principal operator π΄ stated above are satisfied. Then the following properties hold. (1) For π₯0 ∈ π = (π·(π΄), π»)1/2,2 (see Lemma 1) and π ∈ πΏ2 (0, π; π»), π > 0, there exists a unique solution π₯ of (13) belonging to πΏ2 (0, π; π· (π΄)) ∩ π1,2 (0, π; π») ⊂ πΆ ([0, π] ; π) (14) and satisfying σ΅© σ΅© βπ₯βπΏ2 (0,π;π·(π΄))∩π1,2 (0,π;π») ≤ πΆ1 (σ΅©σ΅©σ΅©π₯0 σ΅©σ΅©σ΅© + βπβπΏ2 (0,π;π») ) , where 2 π· (π΄) ⊂ π ⊂ π» ⊂ π∗ ⊂ π·(π΄)∗ , where (π, π∗ )1/2,2 denotes the real interpolation space between π and π∗ (Section 1.3.3 of [10]). 2. Regularity for Solutions βπβ∗ = sup Thus we have the following sequence (15) where πΆ1 is a constant depending on π. (2) Let π₯0 ∈ π» and π ∈ πΏ2 (0, π; π∗ ), π > 0. Then there exists a unique solution π₯ of (13) belonging to πΏ2 (0, π; π) ∩ π1,2 (0, π; π∗ ) ⊂ πΆ ([0, π] ; π») (16) Abstract and Applied Analysis 3 and satisfying βπ₯βπΏ2 (0,π;π)∩π1,2 (0,π;π∗ ) (2) We assume the following. σ΅¨ σ΅¨ ≤ πΆ1 (σ΅¨σ΅¨σ΅¨π₯0 σ΅¨σ΅¨σ΅¨ + βπβπΏ2 (0,π;π∗ ) ) , (17) where πΆ1 is a constant depending on π. Let π be a nonlinear single valued mapping from [0, ∞)× π into π». (F) We assume that σ΅© σ΅¨σ΅¨ σ΅¨ σ΅© σ΅¨σ΅¨π (π‘, π₯1 ) − π (π‘, π₯2 )σ΅¨σ΅¨σ΅¨ ≤ πΏ σ΅©σ΅©σ΅©π₯1 − π₯2 σ΅©σ΅©σ΅© , (18) (A) π΄ is symmetric and there exists β ∈ π» such that for every π > 0 and any π¦ ∈ π·(π) π½π (π¦ + πβ) ∈ π· (π) , π (π½π (π¦ + πβ)) ≤ π (π¦) , (26) where π½π = (πΌ + ππ΄)−1 . Then for π’ ∈ πΏ2 (0, π; π), π΅ ∈ L(π, π»), and π₯0 ∈ π·(π) ∩ π (1) has a unique solution π₯ ∈ πΏ2 (0, π; π· (π΄)) ∩ π1,2 (0, π; π») ∩ πΆ ([0, π] ; π») , (27) for every π₯1 , π₯2 ∈ π. Let π be another Hilbert space of control variables and take U = πΏ2 (0, π; π) as stated in the Introduction. Choose a bounded subset π of π and call it a control set. Let us define an admissible control Uad as Uad = {π’ ∈ πΏ2 (0, π; π) : π’ is strongly measurable function satisfying π’ (t) ∈ π for almost all π‘} . (19) Noting that the subdifferential operator ππ is defined by ππ (π₯) = {π₯∗ ∈ π∗ ; π (π₯) ≤ π (π¦) + (π₯∗ , π₯ − π¦) , π¦ ∈ π} , (20) the problem (1) is represented by the following nonlinear functional differential problem on π»: which satisfies σ΅© σ΅© βπ₯βπΏ2 ∩π1,2 ∩πΆ ≤ πΆ2 (1 + σ΅©σ΅©σ΅©π₯0 σ΅©σ΅©σ΅© + βπ΅π’βπΏ2 (0,π;π») ) . Remark 4. In terms of Lemma 1, the following inclusion πΏ2 (0, π; π) ∩ π1,2 (0, π; π∗ ) ⊂ πΆ ([0, π] ; π») The following Lemma is from BreΜzis [14, Lemma A.5]. Lemma 5. Let π ∈ πΏ1 (0, π; R) satisfying π(π‘) ≥ 0 for all π‘ ∈ (0, π) and π ≥ 0 be a constant. Let π be a continuous function on [0, π] ⊂ R satisfying the following inequality: π‘ 1 1 2 π (π‘) ≤ π2 + ∫ π (π ) π (π ) ππ , 2 2 0 (21) π₯ (0) = π₯0 . (29) is well known as seen in (9) and is an easy consequence of the definition of real interpolation spaces by the trace method (see [4, 13]). π₯σΈ (π‘) + π΄π₯ (π‘) + ππ (π₯ (π‘)) ∋ π (π‘, π₯ (π‘)) + π΅π’ (π‘) , 0 < π‘ ≤ π, (28) π‘ ∈ [0, π] . (30) Then, Referring to Theorem 3.1 of [3], we establish the following results on the solvability of (1). Proposition 3. (1) Let the assumption (F) be satisfied. Assume that π’ ∈ πΏ2 (0, π; π), π΅ ∈ L(π, π∗ ), and π₯0 ∈ π·(π) where π·(π) is the closure in π» of the set π·(π) = {π’ ∈ π : π(π’) < ∞}. Then, (1) has a unique solution π₯ ∈ πΏ2 (0, π; π) ∩ πΆ ([0, π] ; π») (22) which satisfies 0 π₯σΈ (π‘) = π΅π’ (π‘) − π΄π₯ (π‘) − (ππ) (π₯ (π‘)) + π (π‘, π₯ (π‘)) , (23) 0 where (ππ) : π» → π» is the minimum element of ππ and there exists a constant πΆ2 depending on π such that σ΅¨ σ΅¨ βπ₯βπΏ2 ∩πΆ ≤ πΆ2 (1 + σ΅¨σ΅¨σ΅¨π₯0 σ΅¨σ΅¨σ΅¨ + βπ΅π’βπΏ2 (0,π;π∗ ) ) , (24) where πΆ2 is some positive constant and πΏ2 ∩ πΆ = πΏ2 (0, π; π) ∩ πΆ([0, π]; π»). Furthermore, if π΅ ∈ L(π, π») then the solution π₯ belongs to π1,2 (0, π; π») and satisfies σ΅¨ σ΅¨ βπ₯βπ1,2 (0,π;π») ≤ πΆ2 (1 + σ΅¨σ΅¨σ΅¨π₯0 σ΅¨σ΅¨σ΅¨ + βπ΅π’βπΏ2 (0,π;π») ) . (25) π‘ |π (π‘)| ≤ π + ∫ π (π ) ππ , 0 π‘ ∈ [0, π] . (31) For each (π₯0 , π’) ∈ π» × πΏ2 (0, π; π), we can define the continuous solution mapping (π₯0 , π’) σ³¨→ π₯. Now, we can state the following theorem. Theorem 6. (1) Let the assumption (F) be satisfied, π₯0 ∈ π», and π΅ ∈ L(π, π∗ ). Then the solution π₯ of (1) belongs to π₯ ∈ πΏ2 (0, π; π) ∩ πΆ([0, π]; π») and the mapping π» × πΏ2 (0, π; π) ∋ (π₯0 , π’) σ³¨σ³¨→ π₯ ∈ πΏ2 (0, π; π) ∩ πΆ ([0, π] ; π») (32) is Lipschtz continuous; that is, suppose that (π₯0π , π’π ) ∈ π» × πΏ2 (0, π; π) and π₯π be the solution of (1) with (π₯0π , π’π ) in place of (π₯0 , π’) for π = 1, 2, σ΅© σ΅©σ΅© σ΅©σ΅©π₯1 − π₯2 σ΅©σ΅©σ΅©πΏ2 (0,π;π)∩πΆ([0,π];π») σ΅¨ σ΅© σ΅© σ΅¨ ≤ πΆ {σ΅¨σ΅¨σ΅¨π₯01 − π₯02 σ΅¨σ΅¨σ΅¨ + σ΅©σ΅©σ΅©π’1 − π’2 σ΅©σ΅©σ΅©πΏ2 (0,π;π) } , where πΆ is a constant. (33) 4 Abstract and Applied Analysis (2) Let the assumptions (A) and (F) be satisfied and let π΅ ∈ L(π, π») and π₯0 ∈ π·(π) ∩ π. Then π₯ ∈ πΏ2 (0, π; π·(π΄)) ∩ π1,2 (0, π; π»), and the mapping integrating the above inequality over (0, π‘), we have π‘ π−2π2 π‘ ∫ |π(π )|2 ππ 0 π × πΏ2 (0, π; π) ∋ (π₯0 , π’) 2 1,2 σ³¨σ³¨→ π₯ ∈ πΏ (0, π; π· (π΄)) ∩ π (0, π; π») π‘ π 1σ΅¨ σ΅¨2 ≤ 2 ∫ π−2π2 π { σ΅¨σ΅¨σ΅¨π₯01 − π₯02 σ΅¨σ΅¨σ΅¨ + ∫ π» (π ) ππ } ππ 2 0 0 (34) is continuous. Proof. (1) Due to Proposition 3, we can infer that (1) possesses a unique solution π₯ ∈ πΏ2 (0, π; π) ∩ πΆ([0, π]; π») with the data condition (π₯0 , π’) ∈ π» × πΏ2 (0, π; π). Now, we will prove the inequality (33). For that purpose, we denote π₯1 − π₯2 by π. Then + π΅ (π’1 (π‘) − π’2 (π‘)) , 0 < π‘ ≤ π, π‘ π‘ 1 − π−2π2 π‘ σ΅¨σ΅¨ σ΅¨2 −2π π σ΅¨σ΅¨π₯01 − π₯02 σ΅¨σ΅¨σ΅¨ + 2 ∫ ∫ π 2 πππ» (π ) ππ 2π2 0 π = 1 − π−2π2 π‘ σ΅¨σ΅¨ σ΅¨2 σ΅¨π₯ − π₯02 σ΅¨σ΅¨σ΅¨ 2π2 σ΅¨ 01 1 π‘ −2π2 π −2π2 π‘ −π ) π» (π ) ππ . ∫ (π π2 0 + πσΈ (π‘) + π΄π (π‘) + ππ (π₯1 (π‘)) − ππ (π₯2 (π‘)) ∋ π (π‘, π₯1 (π‘)) − π (π‘, π₯2 (π‘)) = (35) Thus, we get π (0) = π₯01 − π₯02 . π‘ π2 ∫ |π(π )|2 ππ ≤ Multiplying on the above equation by π(π‘), we have 0 π‘ 1 π |π (π‘)|2 + π1 βπ (π‘)β2 2 ππ‘ 2π2 (π‘−π ) + ∫ (π 0 ≤ π2 |π (π‘)|2 σ΅¨ σ΅¨ σ΅¨ σ΅¨ + {σ΅¨σ΅¨σ΅¨π (π‘, π₯1 (π‘)) − π (π‘, π₯2 (π‘))σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨π΅ (π’1 (π‘) − π’2 (π‘))σ΅¨σ΅¨σ΅¨} × |π (π‘)| . (36) 1 2π2 π‘ σ΅¨ σ΅¨2 − 1) σ΅¨σ΅¨σ΅¨π₯01 − π₯02 σ΅¨σ΅¨σ΅¨ (π 2 (41) − 1) π» (π ) ππ . Combining this with (38) it holds that π‘ 1 1 σ΅¨ σ΅¨2 |π (π‘)|2 + π1 ∫ βπ (π )β2 ππ ≤ π2π2 π‘ σ΅¨σ΅¨σ΅¨π₯01 − π₯02 σ΅¨σ΅¨σ΅¨ 2 2 0 π‘ 2π2 (π‘−π ) +∫ π Put σ΅¨ σ΅¨ π» (π‘) = (πΏ βπ (π‘)β + σ΅¨σ΅¨σ΅¨π΅ (π’1 (π‘) − π’2 (π‘))σ΅¨σ΅¨σ΅¨) |π (π‘)| . (40) 0 (42) π» (π ) ππ . (37) By Lemma 5, the following inequality By integrating the above inequality over [0, π‘], we have π‘ 2 1 −π2 π‘ (π |π (π‘)|) + π1 π−2π2 π‘ ∫ βπ(π )β2 ππ 2 0 π‘ 1 |π(π‘)|2 + π1 ∫ βπ (π )β2 ππ 2 0 π‘ π‘ 1σ΅¨ σ΅¨2 ≤ σ΅¨σ΅¨σ΅¨π₯01 − π₯02 σ΅¨σ΅¨σ΅¨ + π2 ∫ |π (π )|2 ππ + ∫ π» (π ) ππ . 2 0 0 1σ΅¨ σ΅¨2 ≤ σ΅¨σ΅¨σ΅¨π₯01 − π₯02 σ΅¨σ΅¨σ΅¨ 2 (38) π‘ σ΅¨ σ΅¨ + ∫ π−π2 π (πΏ βπ (π )β + σ΅¨σ΅¨σ΅¨π΅ (π’1 (π ) − π’2 (π ))σ΅¨σ΅¨σ΅¨) π−π2 π |π (π )| ππ 0 Note that π −2π2 π‘ π‘ {π ∫ |π (π )|2 ππ } ππ‘ 0 (43) implies that π‘ 1 ≤ 2π−2π2 π‘ { |π (π‘)|2 − π2 ∫ |π (π )|2 ππ } 2 0 π‘ 1σ΅¨ σ΅¨2 ≤ 2π−2π2 π‘ { σ΅¨σ΅¨σ΅¨π₯01 − π₯02 σ΅¨σ΅¨σ΅¨ + ∫ π» (π ) ππ } , 2 0 (39) σ΅¨ σ΅¨ π−π2 π‘ |π (π‘)| ≤ σ΅¨σ΅¨σ΅¨π₯01 − π₯02 σ΅¨σ΅¨σ΅¨ π‘ −π2 π +∫ π 0 σ΅¨ σ΅¨ (πΏ βπ (π )β + σ΅¨σ΅¨σ΅¨π΅ (π’1 (π ) − π’2 (π ))σ΅¨σ΅¨σ΅¨) ππ . (44) Abstract and Applied Analysis 5 From (42) and (44) it follows that π‘ 1 1 σ΅¨ σ΅¨2 |π(π‘)|2 + π1 ∫ βπ(π )β2 ππ ≤ π2π2 π‘ σ΅¨σ΅¨σ΅¨π₯01 − π₯02 σ΅¨σ΅¨σ΅¨ 2 2 0 π‘ σ΅¨ σ΅¨ + ∫ π2π2 (π‘−π ) (πΏ βπ (π )β + σ΅¨σ΅¨σ΅¨π΅ (π’1 (π ) − π’2 (π ))σ΅¨σ΅¨σ΅¨) ππ2 π 0 σ΅¨ σ΅¨ × σ΅¨σ΅¨σ΅¨π₯01 − π₯02 σ΅¨σ΅¨σ΅¨ ππ π‘ 2π2 (π‘−π ) +∫ π 0 σ΅¨ σ΅¨ (πΏ βπ (π )β + σ΅¨σ΅¨σ΅¨π΅ (π’1 (π ) − π’2 (π ))σ΅¨σ΅¨σ΅¨) (45) π σ΅¨ σ΅¨ × ∫ ππ2 (π −π) (πΏ βπ (π)β + σ΅¨σ΅¨σ΅¨π΅ (π’1 (π) − π’2 (π))σ΅¨σ΅¨σ΅¨) ππππ 0 = πΌ + πΌπΌ + πΌπΌπΌ. Putting σ΅¨ σ΅¨ πΊ (π ) = βπ (π )β + σ΅¨σ΅¨σ΅¨π΅ (π’1 (π ) − π’2 (π ))σ΅¨σ΅¨σ΅¨ . (46) The third term of the right hand side of (45) is estimated as π‘ π = πΏ2 π2π2 π‘ ∫ π‘ 0 π₯1 (π‘) − π₯2 (π‘) = π₯01 − π₯02 + ∫ (π₯1Μ (π ) − π₯2Μ (π )) ππ , 0 0 2 π 1 π {∫ π−π2 π βπΊ(π)β ππ} ππ 2 ππ 0 1 2 2π2 π‘ π‘ −π2 π πΏ π {∫ π βπΊ(π)β ππ} 2 0 ≤ 1 2 2π2 π‘ 1 − π−2π2 π‘ π‘ πΏπ ∫ βπΊ (π)β2 ππ 2 2π2 0 = ≤ + σ΅© σ΅©1/2 ≤ πΆ0 σ΅©σ΅©σ΅©π₯1 − π₯2 σ΅©σ΅©σ΅©πΏ2 (0,π;π·(π΄)) π‘ σ΅¨2 σ΅¨ ∫ (βπ (π )β2 + σ΅¨σ΅¨σ΅¨π΅ (π’1 (π ) − π’2 (π ))σ΅¨σ΅¨σ΅¨ ) ππ . 0 (47) The second term of the right hand side of (45) is estimated as π‘ σ΅¨ σ΅¨ πΌπΌ = π2π2 π‘ ∫ π−π2 π (πΏ βπ (π )β + σ΅¨σ΅¨σ΅¨π΅ (π’1 (π ) − π’2 (π ))σ΅¨σ΅¨σ΅¨) ππ 0 1 2π2 π‘ 2 π‘ σ΅¨2 σ΅¨ π πΏ ∫ (βπ (π )β2 + σ΅¨σ΅¨σ΅¨π΅ (π’1 (π ) − π’2 (π ))σ΅¨σ΅¨σ΅¨ ) ππ 2 0 (48) 1 σ΅¨ σ΅¨2 + π2π2 π‘ σ΅¨σ΅¨σ΅¨π₯01 − π₯02 σ΅¨σ΅¨σ΅¨ . 2 Thus, from (47) and (48), we apply Gronwall’s inequality to (15), and we arrive at π‘ 1 |π (π‘)|2 + π1 ∫ βπ (π )β2 ππ 2 0 π1 π 1/2 σ΅© σ΅© σ΅©1/2 σ΅©1/2 × {π1/4 σ΅©σ΅©σ΅©π₯01 − π₯02 σ΅©σ΅©σ΅© + ( ) σ΅©σ΅©σ΅©π₯1 − π₯2 σ΅©σ΅©σ΅©π1,2 (0,π;π») } √2 σ΅© σ΅©1/2 σ΅© σ΅©1/2 ≤ πΆ0 π1/4 σ΅©σ΅©σ΅©π₯01 − π₯02 σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π₯1 − π₯2 σ΅©σ΅©σ΅©πΏ2 (0,π;π·(π΄)) π 1/2 σ΅©σ΅© σ΅© ) σ΅©σ΅©π₯1 − π₯2 σ΅©σ΅©σ΅©πΏ2 (0,π;π·(π΄))∩π1,2 (0,π;π») √2 σ΅© σ΅© ≤ 2−7/4 πΆ0 σ΅©σ΅©σ΅©π₯01 − π₯02 σ΅©σ΅©σ΅© + πΆ0 ( σ΅¨ σ΅¨ × σ΅¨σ΅¨σ΅¨π₯01 − π₯02 σ΅¨σ΅¨σ΅¨ ≤ (52) σ΅©σ΅© σ΅© σ΅© σ΅©1/2 σ΅© σ΅©1/2 σ΅©σ΅©π₯1 − π₯2 σ΅©σ΅©σ΅©πΏ2 (0,π;π) ≤ πΆ0 σ΅©σ΅©σ΅©π₯1 − π₯2 σ΅©σ΅©σ΅©πΏ2 (0,π;π·(π΄)) σ΅©σ΅©σ΅©π₯1 − π₯2 σ΅©σ΅©σ΅©πΏ2 (0,π;π») πΏ (π2π2 π‘ − 1) ∫ βπΊ (π )β2 ππ 4π2 0 2π2 π σ΅©σ΅© σ΅© σ΅©π₯ − π₯2 σ΅©σ΅©σ΅©π1,2 (0,π;π») . √2 σ΅© 1 Hence arguing as in (9) we get π‘ πΏ2 (π2π2 π‘ − 1) (51) we get, noting that | ⋅ | ≤ || ⋅ ||, σ΅© σ΅© σ΅© σ΅©σ΅© σ΅©σ΅©π₯1 − π₯2 σ΅©σ΅©σ΅©πΏ2 (0,π;π») ≤ √π σ΅©σ΅©σ΅©π₯01 − π₯02 σ΅©σ΅©σ΅© 2 = 2 Since π‘ πΌπΌπΌ = πΏ2 π2π2 π‘ ∫ π−π2 π βπΊ (π )β ∫ π−π2 π βπΊ (π)β ππππ 0 where πΆ > 0 is a constant. Suppose (π₯0π , π’π ) → (π₯0 , π’) in π»× πΏ2 (0, π; π), and let π₯π and π₯ be the solutions (1) with (π₯0π , π’π ) and (π₯0 , π’), respectively. Then, by virtue of (49), we see that π₯π → π₯ in πΏ2 (0, π, π) ∩ πΆ([0, π]; π»). (2) It is easy to show that if π₯0 ∈ π and π΅ ∈ L(π, π»), then π₯ belongs to πΏ2 (0, π; π·(π΄))∩π1,2 (0, π; π»). Let (π₯0π , π’π ) ∈ π × πΏ2 (0, π; π»), and π₯π be the solution of (1) with (π₯0π , π’π ) in place of (π₯0 , π’) for π = 1, 2. Then in view of Lemma 2 and assumption (F), we have σ΅© σ΅©σ΅© σ΅©σ΅©π₯1 − π₯2 σ΅©σ΅©σ΅©πΏ2 (0,π;π·(π΄))∩π1,2 (0,π;π») σ΅© σ΅© σ΅© σ΅© ≤ πΆ1 {σ΅©σ΅©σ΅©π₯01 − π₯02 σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π (⋅, π₯1 ) − π(⋅, π₯2 )σ΅©σ΅©σ΅©πΏ2 (0,π;π») σ΅© σ΅© +σ΅©σ΅©σ΅©π΅ (π’1 − π’2 )σ΅©σ΅©σ΅©πΏ2 (0,π;π») } (50) σ΅© σ΅© σ΅© σ΅© ≤ πΆ1 {σ΅©σ΅©σ΅©π₯01 − π₯02 σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅ (π’1 − π’2 )σ΅©σ΅©σ΅©πΏ2 (0,π;π») σ΅© σ΅© +πΏσ΅©σ΅©σ΅©π₯1 − π₯2 σ΅©σ΅©σ΅©πΏ2 (0,π:π) } . σ΅¨2 σ΅¨ σ΅¨ σ΅¨2 ≤ πΆ (σ΅¨σ΅¨σ΅¨π₯01 − π₯02 σ΅¨σ΅¨σ΅¨ + ∫ σ΅¨σ΅¨σ΅¨π΅ (π’1 (π ) − π’2 (π ))σ΅¨σ΅¨σ΅¨ ππ ) , 0 (49) + 2πΆ0 ( π 1/2 σ΅©σ΅© σ΅© ) σ΅©σ΅©π₯1 − π₯2 σ΅©σ΅©σ΅©πΏ2 (0,π;π·(π΄))∩π1,2 (0,π;π») . √2 Combining (50) and (53) we obtain σ΅©σ΅©σ΅©π₯1 − π₯2 σ΅©σ΅©σ΅© 2 σ΅© σ΅©πΏ (0,π;π·(π΄))∩π1,2 (0,π;π») σ΅© σ΅© σ΅© σ΅© ≤ πΆ1 {σ΅©σ΅©σ΅©π₯01 − π₯02 σ΅©σ΅©σ΅©} + σ΅©σ΅©σ΅©π΅π’1 − π΅π’2 σ΅©σ΅©σ΅©πΏ2 (0,π;π») σ΅© σ΅© + 2−7/4 πΆ0 πΆ1 πΏ σ΅©σ΅©σ΅©π₯01 − π₯02 σ΅©σ΅©σ΅© + 2πΆ0 πΆ1 ( π 1/2 σ΅©σ΅© σ΅© ) πΏσ΅©σ΅©π₯1 − π₯2 σ΅©σ΅©σ΅©πΏ2 (0,π;π·(π΄))∩π1,2 (0,π;π») . √2 (53) (54) 6 Abstract and Applied Analysis Suppose that (π₯0π , π’π ) σ³¨σ³¨→ (π₯0 , π’) ∈ π × πΏ2 (0, π; π) , (55) and let π₯π and π₯ be the solutions (1) with (π₯0π , π’π ) and (π₯0 , π’), respectively. Let 0 < π1 ≤ π be such that 2πΆ0 πΆ1 ( π1 1/2 ) πΏ < 1. √2 (56) Then by virtue of (54) with π replaced by π1 we see that π₯π σ³¨→ π₯ ∈ πΏ2 (0, π1 ; π· (π΄)) ∩ π1,2 (0, π1 ; π») . (57) This implies that (π₯π (π1 ), (π₯π )π1 ) σ³¨→ (π₯(π1 ), π₯π1 ) in π × πΏ2 (0, π; π·(π΄)). Hence the same argument shows that π₯π σ³¨→ π₯ in πΏ2 (π1 , min {2π1 , π} ; π· (π΄)) ∩ π1,2 (π1 , min {2π1 , π} ; π») . (58) Repeating this process we conclude that π₯π πΏ2 (0, π; π·(π΄)) ∩ π1,2 (0, π; π»). σ³¨→ π₯ in In this section we study the optimal control problems for the quadratic cost function in the framework of Lions [9]. In what follows we assume that the embedding π·(π΄) ⊂ π ⊂ π» is compact. Let π be another Hilbert space of control variables, and π΅ be a bounded linear operator from π into π»; that is, (59) which is called a controller. By virtue of Theorem 6, we can define uniquely the solution map π’ σ³¨→ π₯(π’) of πΏ2 (0, π; π) into πΏ2 (0, π; π) ∩ πΆ([0, π]; π»). We will call the solution π₯(π’) the state of the control system (1). Let π be a Hilbert space of observation variables. The observation of state is assumed to be given by π§ (π’) = πΊπ₯ (π’) , ∗ πΊ ∈ L (πΆ (0, π; π ) , π) , (60) where πΊ is an operator called the observer. The quadratic cost function associated with the control system (1) is given by σ΅©2 σ΅© π½ (π£) = σ΅©σ΅©σ΅©πΊπ₯ (π£) − π§π σ΅©σ΅©σ΅©π + (π π£, π£)πΏ2 (0,π;π) for π£ ∈ πΏ2 (0, π; π) , (61) where π§π ∈ π is a desire value of π₯(π£) and π ∈ L(πΏ2 (0, π; π)) is symmetric and positive; that is, (π π£, π£)πΏ2 (0,π;π) = (π£, π π£)πΏ2 (0,π;π) ≥ πβπ£β2πΏ2 (0,π;π) W = πΏ2 (0, π; π) ∩ π1,2 (0, π; π∗ ) ⊂ πΆ ([0, π] ; π») (63) endowed with the norm β⋅βW = max {β⋅βπΏ2 (0,π;π) , β⋅βπ1,2 (0,π;π∗ ) } . (64) Let Ω be an open bounded and connected set of Rπ with smooth boundary. We consider the observation πΊ of distributive and terminal values (see [15, 16]). (1) We take π = πΏ2 ((0, π)×Ω)×πΏ2 (Ω) and πΊ ∈ L(W, π) and observe π§ (π£) = πΊπ₯ (π£) = (π₯ (π£; ⋅) , π₯ (π£, π)) ∈ πΏ2 ((0, π) × Ω) × πΏ2 (Ω) . (65) (2) We take π = πΏ2 ((0, π) × Ω) and πΊ ∈ L(W, π) and observe π§ (π£) = πΊπ₯ (π£) = π¦σΈ (π£; ⋅) ∈ πΏ2 ((0, π) × Ω) . (66) The above observations are meaningful in view of the regularity of (1) by Proposition 3. 3. Optimal Control Problems π΅ ∈ L (π, π») , Remark 7. The solution space W of strong solutions of (1) is defined by (62) for some π > 0. Let Uad be a closed convex subset of πΏ2 (0, π; π), which is called the admissible set. An element π’ ∈ Uad which attains minimum of π½(π£) over Uad is called an optimal control for the cost function (61). Theorem 8. (1) Let the assumption (F) be satisfied. Assume that π΅ ∈ L(π, π∗ ) and π₯0 ∈ π·(π). Let π₯(π’) be the solution of (1) corresponding to π’. Then the mapping π’ σ³¨→ π₯(π’) is compact from πΏ2 (0, π; π) to πΏ2 (0, π; π»). (2) Let the assumptions (A) and (F) be satisfied. If π΅ ∈ L(π, π») and π₯0 ∈ π·(π) ∩ π, then the mapping π’ σ³¨→ π₯(π’) is compact from πΏ2 (0, π; π) to πΏ2 (0, π; π). Proof. (1) We define the solution mapping π from πΏ2 (0, π; π) to πΏ2 (0, π; π») by ππ’ = π₯ (π’) , π’ ∈ πΏ2 (0, π; π) . (67) In virtue of Lemma 2, we have βππ’βπΏ2 (0,π;π)∩π1,2 (0,π;π∗ ) σ΅¨ σ΅¨ = βπ₯ (π’)β ≤ πΆ1 {σ΅¨σ΅¨σ΅¨π₯0 σ΅¨σ΅¨σ΅¨ + βπ΅π’βπΏ2 (0,π;π∗ ) } . (68) Hence if π’ is bounded in πΏ2 (0, π; π), then so is π₯(π’) in πΏ2 (0, π; π) ∩ π1,2 (0, π; π∗ ). Since π is compactly embedded in π» by assumption, the embedding πΏ2 (0, π; π) ∩ π1,2 (0, π; π∗ ) ⊂ πΏ2 (0, π; π») is also compact in view of Theorem 2 of Aubin [17]. Hence, the mapping π’ σ³¨→ ππ’ = π₯(π’) is compact from πΏ2 (0, π; π) to πΏ2 (0, π; π»). (2) If π·(π΄) is compactly embedded in π by assumption, the embedding πΏ2 (0, π; π· (π΄)) ∩ π1,2 (0, π; π») ⊂ πΏ2 (0, π; π) (69) is compact. Hence, the proof of (2) is complete. As indicated in the Introduction we need to show the existence of an optimal control and to give the characterizations of them. The existence of an optimal control π’ for the cost function (61) can be stated by the following theorem. Abstract and Applied Analysis 7 Theorem 9. Let the assumptions (A) and (F) be satisfied and π₯0 ∈ π·(π) ∩ π. Then there exists at least one optimal control π’ for the control problem (1) associated with the cost function (61); that is, there exists π’ ∈ Uad such that π½ (π’) = inf π½ (π£) := π½. π₯σΈ (π‘) + π΄π₯ (π‘) + ππ (π₯ (π‘)) ∋ π (π‘, π₯ (π‘)) + π΅π’ (π‘) , (70) π£∈Uad Proof. Since Uad is nonempty, there is a sequence {π’π } ⊂ Uad such that minimizing sequence for the problem (70) satisfies inf π½ (π£) = lim π½ (π’π ) = π. π→∞ π£∈Uad Thus we have proved that π₯(π‘) satisfies a.e. on (0, π) the following equation: (71) Obviously, {π½(π’π )} is bounded. Hence by (62) there is a positive constant πΎ0 such that σ΅© σ΅©2 πσ΅©σ΅©σ΅©π’π σ΅©σ΅©σ΅© ≤ (π π’π , π’π ) ≤ π½ (π’π ) ≤ πΎ0 . π₯πσΈ (π‘) + π΄π₯π (π‘) + ππ (π₯π (π‘)) ∋ π (π‘, π₯π (π‘)) + π΅π’π (π‘) , 0 < π‘ ≤ π, (73) π₯π (0) = π₯0 . (80) π₯ (0) = π₯0 . Since πΊ is continuous and || ⋅ ||π is lower semicontinuous, it holds that σ΅©σ΅©σ΅©πΊπ₯(π’) − π§π σ΅©σ΅©σ΅© ≤ lim inf σ΅©σ΅©σ΅©πΊπ₯ (π’π ) − π§π σ΅©σ΅©σ΅© . (81) σ΅©π σ΅©π σ΅© π→∞ σ΅© It is also clear from lim inf π → ∞ βπ 1/2 π’π βπΏ2 (0,π;π) βπ 1/2 π’βπΏ2 (0,π;π) that (72) This shows that {π’π } is bounded in Uad . So we can extract a subsequence (denoted again by {π’π }) of {π’π } and find a π’ ∈ Uad such that π€ − lim π’π = π’ in π. Let π₯π = π₯(π’π ) be the solution of the following equation corresponding to π’π : a.e., 0 < π‘ ≤ π, ≥ lim inf (π π’π , π’π )πΏ2 (0,π;π) ≥ (π π’, π’)πΏ2 (0,π;π) . (82) π = lim π½ (π’π ) ≥ π½ (π’) . (83) π→∞ Thus, π→∞ But since π½(π’) ≥ π by definition, we conclude π’ ∈ Uad is a desired optimal control. 4. Necessary Conditions for Optimality By (15) and (17) we know that {π₯π } and {π₯πσΈ } are bounded in πΏ2 (0, π; π) and πΏ2 (0, π; π∗ ), respectively. Therefore, by the extraction theorem of Rellich’s, we can find a subsequence of {π₯π }, say again {π₯π }, and find π₯ such that 2 π₯π (⋅) σ³¨→ π₯ (⋅) weakly in πΏ (0, π; π) ∩ πΆ ([0, π] ; π») , π₯πσΈ σ³¨→ π₯σΈ , weakly in πΏ2 (0, π; π∗ ) . (74) However, by Theorem 8, we know that π₯π (⋅) σ³¨→ π₯ (⋅) , strongly in πΏ2 (0, π; π) . (75) (76) By the boundedness of π΄ we have π΄π₯π σ³¨→ π΄π₯, strongly in πΏ2 (0, π; π∗ ) . (77) Since ππ(π₯π ) are uniformly bounded from (73)–(77) it follows that ππ (π₯π ) σ³¨→ π (⋅, π₯) + π΅π’ − π₯σΈ − π΄π₯, weakly in πΏ2 (0, π; π∗ ) , (78) and noting that ππ is demiclosed, we have that π (⋅, π₯) + π΅π’ − π₯σΈ − π΄π₯ ∈ ππ (π₯) in πΏ2 (0, π; π∗ ) . π·π½ (π’) (π£ − π’) ≥ 0, π£ ∈ Uad (84) and to analyze (84) in view of the proper adjoint state system, where π·π½(π’) denote the GaΜteaux derivative of π½(π£) at π£ = π’. Therefore, we have to prove that the solution mapping π£ σ³¨→ π₯(π£) is GaΜteaux differentiable at π£ = π’. Here we note that from Theorem 6 it follows immediately that lim π₯ (π’ + ππ€) π→0 = π₯ (π’) , strongly in πΏ2 (0, π; π) ∩ πΆ ([0, π] ; π») . From (F) it follows that π (⋅, π₯π ) σ³¨→ π (⋅, π₯) , strongly in πΏ2 (0, π; π») . In this section we will characterize the optimal controls by giving necessary conditions for optimality. For this it is necessary to write down the necessary optimal condition (79) (85) The solution map π£ σ³¨→ π₯(π£) of πΏ2 (0, π; π) into πΏ2 (0, π; π) ∩ πΆ([0, π]; π») is said to be GaΜteaux differentiable at π£ = π’ if for any π€ ∈ πΏ2 (0, π; π) there exists a π·π₯(π’) ∈ L(πΏ2 (0, π; π), πΏ2 (0, π; π) ∩ πΆ([0, π]; π») such that σ΅©σ΅© 1 σ΅© σ΅©σ΅© (π₯ (π’ + ππ€) − π₯ (π’)) − π·π₯ (π’) π€σ΅©σ΅©σ΅© σ³¨→ 0 σ΅©σ΅© σ΅©σ΅© σ΅©π σ΅© as π σ³¨→ 0. (86) The operator π·π₯(π’) denotes the GaΜteaux derivative of π₯(π’) at π£ = π’ and the function π·π₯(π’)π€ ∈ πΏ2 (0, π; π)∩πΆ([0, π]; π»)) is called the GaΜteaux derivative in the direction π€ ∈ πΏ2 (0, π; π), which plays an important part in the nonlinear optimal control problems. First, as is seen in Corollary 2.2 of Chapter II of [18], let us introduce the regularization of π as follows. 8 Abstract and Applied Analysis Lemma 10. For every π > 0, define and, by the relation (12), σ΅©σ΅© σ΅©2 σ΅©π₯ − π½π π₯σ΅©σ΅©σ΅©∗ + π (π½π π₯) : ∀π > 0, π₯ ∈ π»} , ππ (π₯) = { σ΅© 2π (87) where π½π = (πΌ + ππ)−1 . Then the function ππ is FreΜchet differentiable on π» and its FrecΜhet differential πππ is Lipschitz continuous on π» with Lipschitz constant π−1 . In addition, lim ππ (π₯) = π (π₯) , ∀π₯ ∈ π», π→0 π (π½π π₯) ≤ ππ (π₯) ≤ π (π₯) , 0 lim π ππ (π₯) = (ππ) (π₯) , (88) ∀π₯ ∈ π», where (ππ)0 (π₯) is the element of minimum norm in the set ππ(π₯). Now, we introduce the smoothing system corresponding to (1) as follows. π₯σΈ (π‘) + π΄π₯ (π‘) + πππ (π₯ (π‘)) = π (π‘, π₯ (π‘)) + π΅π’ (π‘) , 0 < π‘ ≤ π, (89) π₯ (0) = π₯0 . Lemma 11. Let the assumption (F) be satisfied. Then the solution map π£ σ³¨→ π₯(π£) of πΏ2 (0, π; π) into πΏ2 (0, π; π)∩πΆ([0, π]; π») is Lipschtz continuous. Moreover, let us assume the condition (A) in Proposition 3. Then the map π£ σ³¨→ πππ (π₯(π£)) of πΏ2 (0, π; π) into πΏ2 (0, π; π») ∩ πΆ([0, π]; π∗ ) is also Lipschtz continuous. Proof. We set π€ = π£ − π’. From Theorem 6, it follows immediately that βπ₯ (π’ + ππ€) − π₯ (π’)βπΆ([0,π];π») ≤ const. |π| βπ€βπΏ2 (0,π;π) , 2 (90) 2 so the solution map π£ σ³¨→ π₯(π£) of πΏ (0, π; π) into πΏ (0, π; π) ∩ πΆ([0, π]; π») is Lipschtz continuous. Moreover, since πππ (π₯ (π’; π‘)) − πππ (π₯ (π’ + ππ€; π‘)) = π₯σΈ (π’ + ππ€; π‘) − π₯σΈ (π’; π‘) + π΄ (π₯ (π’ + ππ€; π‘) − π₯ (π’; π‘)) − {π (π‘, π₯ (π’ + ππ€; π‘)) − π (π‘, π₯ (π’; π‘))} − ππ΅π€ (π‘) , (91) by the assumption (A) and (2) of Theorem 6, it holds σ΅©σ΅©σ΅©πππ (π₯ (π’ + ππ€)) − πππ (π₯ (π’))σ΅©σ΅©σ΅© 2 σ΅© σ΅©πΏ (0,π;π») σ΅© σ΅©σ΅© σΈ ≤ σ΅©σ΅©σ΅©π₯ (π’ + ππ€) − π₯σΈ (π’)σ΅©σ΅©σ΅©σ΅©πΏ2 (0,π;π») + βπ₯ (π’ + ππ€) − π₯ (π’)βπΏ2 (0,π;π·(π΄)) + πΏβπ₯ (π’ + ππ€) − π₯ (π’)βπΏ2 (0,π;π) + |π|βπ΅ββπ€βπΏ2 (0,π;π) ≤ const. |π|βπ€βπΏ2 (0,π;π) + βπ΄βL(π,π∗ ) β(π₯ (π’ + ππ€; π‘) − π₯ (π’; π‘))β (93) + πΏ βπ₯ (π’ + ππ€; π‘) − π₯ (π’; π‘)β + |π| βπ΅β|π€ (π‘)| ≤ const. |π|βπ€βπΏ2 (0,π;π) . ∀π > 0, π₯ ∈ π», π→0 σ΅© σ΅©σ΅© σ΅©σ΅©πππ (π₯ (π’ + ππ€; π‘)) − πππ (π₯ (π’; π‘))σ΅©σ΅©σ΅©∗ σ΅© σ΅© ≤ σ΅©σ΅©σ΅©σ΅©π₯σΈ (π’ + ππ€; π‘) − π₯σΈ (π’; π‘)σ΅©σ΅©σ΅©σ΅©∗ So we know that the map π£ σ³¨→ πππ (π₯(π£)) of πΏ2 (0, π; π) into πΏ2 (0, π; π») ∩ πΆ([0, π]; π∗ ) is also Lipschtz continuous. Let the solution space W1 of (1) of strong solutions is defined by W1 = πΏ2 (0, π; π· (π΄)) ∩ π1,2 (0, π; π») (94) as stated in Remark 7. In order to obtain the optimality conditions, we require the following assumptions. (F1) The GaΜteaux derivative π2 π(π‘, π₯) in the second argument for (π‘, π₯) ∈ (0, π) × π is measurable in π‘ ∈ (0, π) for π₯ ∈ π and continuous in π₯ ∈ π for a.e. π‘ ∈ (0, π), and there exist functions π1 , π2 ∈ πΏ2 (R+ ; R) such that σ΅©σ΅© σ΅© σ΅©σ΅©π2 π (π‘, π₯)σ΅©σ΅©σ΅©∗ ≤ π1 (π‘) + π2 (βπ₯β) , ∀ (π‘, π₯) ∈ (0, π) × π. (95) (F2) The map π₯ → πππ (π₯) is GaΜteaux differentiable, and the value π·πππ (π₯)π·π₯(π’) is the GaΜteaux derivative of πππ (π₯)π₯(π’) at π’ ∈ πΏ2 (0, π; π) such that there exist functions π3 , π4 ∈ πΏ2 (R+ ; R) such that σ΅© σ΅©σ΅© σ΅©σ΅©π·πππ (π₯) π·π₯ (π’)σ΅©σ΅©σ΅©∗ ≤ π3 (π‘) + π4 (βπ’βπΏ2 (0,π;π) ) , ∀π’ ∈ πΏ2 (0, π; π) . (96) Theorem 12. Let the assumptions (A), (F1), and (F2) be satisfied. Let π’ ∈ Uad be an optimal control for the cost function π½ in (61). Then the following inequality holds: (πΆ∗ Λ π (πΆπ₯ (π’) − π§π ) , π¦)W1 + (π π’, π£ − π’)πΏ2 (0,π;π) ≥ 0, ∀π£ ∈ Uad , (97) where π¦ = π·π₯(π’)(π£ − π’) ∈ πΆ([0, π]; π∗ ) is a unique solution of the following equation: 0 π¦σΈ (π‘) + π΄π¦ (π‘) + π·(ππ) (π₯) (π¦ (π‘)) = π2 π (π‘, π₯) π¦ (π‘) + π΅π€ (π‘) , (92) 0 < π‘ ≤ π, (98) π¦ (0) = 0. Proof. We set π€ = π£ − π’. Let π ∈ (−1, 1), π =ΜΈ 0. We set π¦ = lim π−1 (π₯ (π’ + ππ€) − π₯ (π’)) = π·π₯ (π’) π€. π→0 (99) Abstract and Applied Analysis 9 Proof. Let π₯(π‘) = π₯0 (π‘) be a solution of (1) associated with the control 0. Then it holds that From (89), we have σΈ σΈ π₯ (π’ + ππ€) − π₯ (π’) + π΄ (π₯ (π’ + ππ€) − π₯ (π’)) + πππ (π₯ (π’ + ππ€)) − πππ (π₯ (π’)) (100) = π (⋅, π₯ (π’ + ππ€)) − π (⋅, π₯ (π’)) + ππ΅π€. π π σ΅©2 σ΅© = ∫ σ΅©σ΅©σ΅©πΆ (π₯π£ (π‘) − π₯ (π‘)) + πΆπ₯ (π‘) − π§π (π‘)σ΅©σ΅©σ΅©π ππ‘ 0 Then as an immediate consequence of Lemma 11 one obtains lim 1 π→0π {πππ (π₯ (π’ + ππ€; π‘)) − πππ (π₯ (π’; π‘))} = π·πππ (π₯) π¦ (π‘) , π + ∫ (π π£ (π‘) , π£ (π‘)) ππ‘ 0 1 {π (π‘, π₯ (π’ + ππ€; π‘)) − π (π‘, π₯ (π’; π‘))} = π2 π (π‘, π₯)π¦ (π‘) , π→0π (101) lim thus, in the sense of (F2), we have that π¦ = π·π₯(π’)(π£ − π’) satisfies (98) and the cost π½(π£) is GaΜteaux differentiable at π’ in the direction π€ = π£ − π’. The optimal condition (84) is rewritten as (πΆπ₯ (π’) − π§π , π¦)π + (π π’, π£ − π’)πΏ2 (0,π;π) = (πΆ∗ Λ π (πΆπ₯ (π’) − π§π ) , π¦)W1 + (π π’, π£ − π’)πΏ2 (0,π;π) ≥ 0, π σ΅© σ΅©2 π½1 (π£) = ∫ σ΅©σ΅©σ΅©πΆπ₯π£ (π‘) − π§π (π‘)σ΅©σ΅©σ΅©π ππ‘ + ∫ (π π£ (π‘) , π£ (π‘)) ππ‘ 0 0 π σ΅©2 σ΅© = π (π£, π£) − 2πΏ (π£) + ∫ σ΅©σ΅©σ΅©π§π (π‘) − πΆπ₯ (π‘)σ΅©σ΅©σ΅©π ππ‘, 0 (108) where π π (π’, π£) = ∫ (πΆ (π₯π’ (π‘) − π₯ (π‘)) , πΆ (π₯π£ (π‘) − π₯ (π‘)))π ππ‘ 0 (102) π + ∫ (π π’ (π‘) , π£ (π‘)) ππ‘, ∀π£ ∈ Uad . 0 π With every control π’ ∈ πΏ2 (0, π; π), we consider the following distributional cost function expressed by π 0 (109) π σ΅©2 σ΅© π½1 (π’) = ∫ σ΅©σ΅©σ΅©πΆπ₯π’ (π‘) − π§π (π‘)σ΅©σ΅©σ΅©π ππ‘ + ∫ (π π’ (π‘) , π’ (π‘)) ππ‘, 0 0 (103) where the operator πΆ is bounded from π» to another Hilbert space π and π§π ∈ πΏ2 (0, π; π). Finally we are given that π is a self adjoint and positive definite: π ∈ L (π) , πΏ (π£) = ∫ (π§π (π‘) − πΆπ₯ (π‘) , πΆ (π₯π£ (π‘) − π₯ (π‘)))π ππ‘. (π π’, π’) ≥ π βπ’β , π > 0. (104) Let π₯π’ (π‘) stand for solution of (1) associated with the control π’ ∈ πΏ2 (0, π; π). Let Uad be a closed convex subset of πΏ2 (0, π; π). Theorem 13. Let the assumptions in Theorem 12 be satisfied and let the operators πΆ and π satisfy the conditions mentioned above. Then there exists an element π’ ∈ Uad such that π½1 (π’) = inf π½1 (π£) . 0 π£ ∈ πΏ2 (0, π; π) . (110) If π’ is an optimal control, similarly for (97), (84) is equivalent to π ∫ (πΆ∗ Λ π (πΆπ₯π’ (π‘) − π§π (π‘)) , π¦ (π‘)) ππ‘ 0 (111) π + ∫ (π π’ (π‘) , (π£ − π’) (π‘)) ππ‘ ≥ 0. 0 Now we formulate the adjoint system to describe the optimal condition: Furthermore, the following inequality holds: π π (π£, π£) ≥ πβπ£β2 , (105) π£∈Uad ∗ ∫ (Λ−1 π π΅ ππ’ (π‘) + π π’ (π‘) , (π£ − π’) (π‘)) ππ‘ ≥ 0, The form π(π’, π£) is a continuous form in πΏ2 (0, π; π) × πΏ2 (0, π; π) and from assumption of the positive definite of the operator π , we have ∀π£ ∈ Uad , (106) ∗ holds, where Λ π is the canonical isomorphism π onto π and ππ’ satisfies the following equation: ππ’σΈ (π‘) − π΄∗ ππ’ (π‘) − π·πππ (π₯)∗ ππ’ (π‘) + π2 π(π‘, π₯)∗ ππ’ (π‘) = − (πΆ∗ Λ π πΆπ₯π’ (π‘) − π§π (π‘)) , for 0 < π‘ ≤ π, (112) ππ’ (π) = 0. 0 ππ’σΈ (π‘) − π΄∗ ππ’ (π‘) − π·(ππ) (π₯)∗ ππ’ (π‘) + π2 π(π‘, π₯)∗ ππ’ (π‘) ∗ = −πΆ Λ π (πΆπ₯π’ (π‘) − π§π (π‘)) , for 0 < π‘ ≤ π, ππ’ (π) = 0. (107) Taking into account the regularity result of Proposition 3 and the observation conditions, we can assert that (112) admits a unique weak solution ππ’ reversing the direction of time π‘ → π − π‘ by referring to the well-posedness result of Dautray and Lions [19, pages 558–570]. 10 Abstract and Applied Analysis We multiply both sides of (112) by π¦(π‘) of (98) and integrate it over [0, π]. Then we have π ∫ (πΆ∗ Λ π (πΆπ₯π’ (π‘) − π§π (π‘)) , π¦ (π‘)) ππ‘ 0 π π 0 0 = − ∫ (ππ’σΈ (π‘) , π¦ (π‘)) ππ‘ + ∫ (π΄∗ ππ’ (π‘) , π¦ (π‘)) ππ‘ π (113) ∗ + ∫ (π·πππ (π₯) ππ’ (π‘) , π¦ (π‘)) ππ‘ 0 π − ∫ (π2 π(π‘, π₯)∗ ππ’ (π‘) , π¦ (π‘)) ππ‘. 0 By the initial value condition of π¦ and the terminal value condition of ππ’ , the left hand side of (113) yields − (ππ’ (π) , π¦ (π)) + (ππ’ (0) , π¦ (0)) π π 0 0 + ∫ (ππ’ (π‘) , π¦σΈ (π‘)) ππ‘ + ∫ (ππ’ (π‘) , π΄π¦ (π‘)) ππ‘ π + ∫ (ππ’ (π‘) , π·πππ (π₯) π¦ (π‘)) ππ‘ 0 (114) π − ∫ (ππ’ (π‘) , π2 π (π‘, π₯) π¦ (π‘)) ππ‘ 0 π = ∫ (ππ’ (π‘) , π΅ (π£ − π’) (π‘)) ππ‘. 0 Let π’ be the optimal control subject to (103). Then (111) is represented by π ∫ (ππ’ (π‘) , π΅ (π£ − π’) (π‘)) ππ‘ + ∫ (π π’ (π‘) , (π£ − π’) (π‘)) ππ‘ ≥ 0, Ω 0 (115) which is rewritten by (106). Note that πΆ∗ ∈ π΅(π∗ , π») and for π and π in π» we have (πΆ∗ Λ π πΆπ, π) = β¨πΆπ, πΆπβ©π , where duality pairing is also denoted by (⋅, ⋅). Remark 14. Identifying the antidual π with π we need not use the canonical isomorphism Λ π . However, in case where π ⊂ π∗ this leads to difficulties since π» has already been identified with its dual. Acknowledgment This research was supported by Basic Science Research Program through the National research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (2012-0007560). References [1] V. Barbu, Analysis and Control of Nonlinear Infinite-Dimensional Systems, vol. 190, Academic Press, Boston, Mass, USA, 1993. [2] G. Bertotti and I. Mayergoyz, The Science of Hysteresis, Academic Press, Elsevier, 2006. [3] J.-M. Jeong and J.-Y. Park, “Nonlinear variational inequalities of semilinear parabolic type,” Journal of Inequalities and Applications, vol. 6, no. 2, pp. 227–245, 2001. [4] J.-L. Lions, “Quelques probleΜmes de la theΜorie des eΜquations non lineΜaires d’eΜvolution,” in Problems in Non-Linear Analysis, pp. 189–342, Edizioni Cremonese, Rome, Italy, 1971. 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