# Bilinear Minimax Optimal Control Problems for a von Kárman System with Long Memory

Authors: Jin-soo Hwang
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In the present article, we consider a von Kárman equation with long memory. The goal is to study a quadratic cost minimax optimal control problems for the control system governed by the equation. First, we show that the solution map is continuous under a weak assumption on the data. Then, we formulate the minimax optimal control problem. We show the first and twice Fréchet differentiabilities of the nonlinear solution map from a bilinear input term to the weak solution of the equation. With the Fréchet differentiabilities of the control to solution mapping, we prove the uniqueness and existence of an optimal pair and find its necessary optimality condition.


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## Body

## 1. Introduction

LetΩ be an open bounded domain in R2 with a sufficiently smooth boundary ∂Ω. We set Q=0,T×Ω,Σ=0,T×∂Ω. We consider the following von Kárman system with long memory and the hinged boundary condition in the variables u and v, representing the deflection of the plate and the Airy’s stress, respectively:
(1)u″−γΔu″+Δ2u+∫0tkt−sΔ2usds=u,υ+qu+finQ,Δ2υ=−u,uinQ,u=Δu=υ=∂υ∂v=0onΣ,u0,x=u0x,u′0,x=u1xinΩ,where ′=∂/∂t, the vector ν denotes an outward normal, γ>0 means a constant related to the rotational inertia, k·∈C10,T is a memory kernel, f is a forcing function, and [·,·] is the von Kárman bracket given by
(2)υ,o=∂2υ∂x12∂2o∂x22+∂2o∂x12∂2υ∂x22−2∂2υ∂x1∂x2∂2o∂x1∂x2.The termqu in Equation (1) represents the reset force of the elastic plate in the system. This physical situation naturally leads to the consideration of the bilinear control problem for the control function q, which is used as a force to make the state close to a desired state taking into account. In this motivation, Bradley and Lenhart [1] studied the bilinear optimal control problem for a Kirchhoff plate equation (cf. [2]). And it has been studied in [3] the bilinear optimal control problem of velocity term in a Kirchhoff plate equation.Motivated by [1, 3] with the above physical background, we study here the bilinear minimax control problem for Equation (1) with the control function q based on the Fréchet differentiabilities of the nonlinear solution map. More detailed explanations are given as follows:In our previous study [4], we considered the Dirichlet boundary value problems of Equation (1) without the term qu and studied the optimal control problems for the external forcing control system by the frameworks in Lions [5]. In [4], we proved and used the Gâteaux differentiability of the nonlinear solution map to present the necessary optimality conditions for the optimal controls of the specific observation cases.In this paper, we show the Fréchet differentiability of the solution mapq⟶u from the bilinear control input terms to the solutions of Equation (1). In most cases, the Gâteaux differentiability may be enough to solve a quadratic cost optimal control problem. However, the Fréchet differentiability of a solution map is more desirable for studying the problem with more general cost function like nonquadratic or nonconvex functions. So, this study is an improvement on a previous study [4]. Based on the result, we constructed and solved the bilinear minimax optimal control problems in Equation (1). The minimax control strategies have been used by many researchers for various control problems (see Lasiecka and Triggiani [6] and Li and Yong [7]). As explained in [8], the minimax control framework is employed to take into account of the undesirable effects of system disturbance (or noise) in control inputs such that a cost function achieves its minimum even in the worst disturbances of the system. For the purpose, we replace the bilinear multiplier q in Equation (1) by c+η, where c is a control variable that belongs to the admissible control set Cad, and η is a disturbance (or noise) that belongs to the admissible disturbance set Dad. We also introduce the following cost function to be minimized within Cad and maximized within Dad:
(3)Jc,η=12u−zdL2Q2+α2cL2Q2−β2ηL2Q2,where uis a solution of Equation (1), zd∈L2Q is desired value, and the positive constants α and β are the relative weights of the second and third terms on the right hand side of (3).Our goal of this paper is to find and characterize the optimal control of the cost function (3) for the worst disturbance through control input in Equation (1).This leads to the problem of finding and characterizing the saddle point (c∗,η∗) ∈Cad×Dad satisfying
(4)Jc∗,η≤Jc∗,η∗≤Jc,η∗,∀c,η∈Cad×Dad.In this paper, we use the terminology optimal pair for such a saddle point (c∗,η∗) in (4). For the study of the existence of an optimal pair (c∗,η∗) satisfying (4), we can find results in [8]. In that paper, the author used the minimax theorem in infinite dimensions given in Barbu and Precupanu [9]. And in [10], we extended the result to a quasilinear PDE.On the other hand, in this paper, we use the method given in [11] to obtain the uniqueness as well as the existence of an optimal pair. That is to say, we use the strict convexity (or concavity) arguments of [12] by proving twice (Fréchet) differentiability of the solution map. Also, as we will see later, this method can suggest another condition that ensure strict convexity (or concavity) of the map from control (or noise) to the quadratic cost function (3).Next, we derive an optimality condition for such a (c∗,η∗) in (4). To derive the condition, we refer to the studies on bilinear optimal control problems where the state equations are linear partial differential equations such as the reaction diffusion equation or Kirchhoff plate equation (see [1, 3, 8, 13] and references therein).We now explain the content of this paper. In Section2, we present notations and some necessary lemmas. In Section 3, we prove the well-posedness of Equation (1) with respect to u in the Hadamard sense using some previous results. To name just a few, we can refer to [14–16], and references therein. Especially, in order to prove the local Lipschitz continuity of the nonlinear solution map, we employ the energy equality of Volterra-type integro-differential equation which is proved in [17]. In Section 4, we shall study the minimax optimal control problems: at first, we shall show that the solution map of Equation (1): q⟶u is the first and twice Fréchet differentiable; By using twice Fréchet differentiability of the solution maps c⟶u and η⟶u, we prove that the maps c⟶J and η⟶J are strictly convex and concave, respectively, under the assumptions that α, β are sufficiently large or T>0 is sufficiently small. And we also prove that the maps c⟶J and η⟶J are lower and upper semicontinuous, respectively. Consequently, we can prove the uniqueness and existence of an optimal pair. Next, we derive the necessary optimality condition of an optimal pair for the observation case associated with the cost (3).


## 2. Notations and Preliminaries

Throughout this paper, we useC as a generic constant and omit the integral variables in any definite integrals without confusion.IfX is a Banach space, we denote by X′ its topological dual, and by ⋅,⋅X′,X the duality pairing between X′ and X. We introduce the following abbreviations:
(5)Lp=LpΩ,Wk,p=Wk,pΩ,⋅p=⋅Lp,where p≥1 and Wk,p is the Lp-based Sobolev spaces for k≥1. We denote by Hk, the standard Sobolev spaces Wk,2 for k≥1. And H0k means the completions of C0∞Ω in Hk for k≥1. The duality pairs between H0k and H−kk=1,2 are abbreviated by ⋅,⋅k,−k. The scalar product and norm on L2 are denoted by ·,·2 and ⋅2, respectively. Then, based on the Poincaré inequality and the well-known regularity theory for elliptic boundary value problems (Temam [18] p. 150), the scalar products on H0k (k=1,2) can be given as follows:
(6)υ,oH01=∇υ,∇o2,∀υ,o∈H01,υ,oH02=Δυ,Δo2,∀υ,o∈H02.Then obviously,(7)υH01=∇υ2,∀υ∈H01,oH02=Δo2,∀o∈H02.We define the operatorA which stands for the following:
(8)Au=−Δu,V≡DA=H2∩H01,and consider the operator Mγ=I+γA. We also define the operator A as follows:
(9)Au=Δ2u,DA=u∈H4∩H01Δu=0onΓ.We note that(10)DA1/2=V.By using again the well-known elliptic regularity theory (Temam [18] p. 150), we can obtain
(11)υH2≤CΔυ2,∀υ∈V,oH4≤CΔ2o2,∀o∈DA.Therefore, we can employ(12)υV=Δυ2,∀υ∈V,oDA=Δ2o2,∀o∈DA.It becomes apparent that each topological imbedding(13)V⟶H01⟶L2⟶H−1⟶V′,is continuous and compact. According to Adams [19], we know that when n≤3, the imbedding
(14)V⟶C0Ω¯,is compact.It is well known that the biharmonic operator(15)Δ2:H4∩H02⟶L2,is bijective, and it admits an isometric extension
(16)Δ2:H02⟶H−2.Thus, we can define an operatorG∈ℒL2,H4∩H02orℒH−2,H02by
(17)Gf=giffΔ2g=finΩ,g=∂g∂v=0on∂Ω.Therefore, from Equation (1), one can also note that
(18)υ=−Gu,u∀u∈H02.We collect below some results for the Airy stress function and von Kárman bracket.Lemma 1.
The trilinear formb: V×V×V⟶R given by
(19)bυ,o,φ≡υ,o,φ2,satisfies the property
(20)bυ,o,φ=bυ,φ,o,∀υ,o,φ∈V.Proof.
See ([20], Proposition 1.4.2).Lemma 2.
The bilinear formsυ,o⟶Gυ,o from H2×H2 into W2,∞ and υ,o⟶υ,o from H1×H2 into H−2 are continuous. We also have the following estimates:
(21)Gυ,oW2,∞≤CυH2oH2,∀υ,o∈H2,υ,oH−2≤CυH1oH2,∀υ∈H1,∀o∈H2.
Consequently,(22)φ,Gυ,o2≤CφH2υH2oH2,∀φ,υ,o∈H2.Proof.
See [15, 20].


## 3. Well Posedness of a von Kárman Equation with Long Memory

We introduce the Hilbert spaceW0,T of the weak solutions of Equation (1) given by
(23)W0,T=uu∈L20,T;V,u′∈L20,T;H01,u″∈L2Q,with the norm
(24)uw0,T=uL20,T;V2+u′L20,T;H012+u″L2Q21/2.Definition 3.
Functionu∈W0,T is called a weak solution of Equation (1), if it satisfies
(25)Mγu″⋅,oV′,V+Δu⋅+k∗Δu⋅,Δo2=u⋅,υ⋅+q⋅u⋅+f⋅,o2,Δυ⋅,Δφ2=−u⋅,u⋅,φ2,forallo,φ∈V×H02inthesenseofD′0,T,u0=u0,u′0=u1,where D′0,T is the space of distributions on 0,T.
As indicated in [14], von Kárman nonlinearity is subcritical; thus, the issues of well-posedness and regularity of weak solutions are standard.Theorem 4.
Ifu0,u1∈V×H01,k∈C10,T,q∈L∞Q, and f∈L2Q, then a weak solution u of Equation (1) exists and satisfies:
(26)u∈W0,T∩L∞0,T;V∩W1,∞0,T;H01.
To show the regularity of a weak solutions of Equation (1), we need the following lemma.Lemma 5.
LetX,Y be two Banach spaces, X⊂Y with dense, and X being reflexive. Set
(27)Cw0,T;Y=f∈L∞0,T;Yf⋅,ξY,Y′∈C0,T,∀ξ∈Y′.
Then(28)L∞0,T;X∩Cw0,T;Y=Cw0,T;X.Proof.
See ([21], p. 275).Corollary 6.
Assume thatu is a weak solution of Equation (1). Then, we can assert (after possibly a modification on a set of measure zero) that
(29)u∈Cw0,T;V,u′∈Cw0,T;H01.Proof.
From Dautray and Lions ([22], p. 480), it is clear that W0,T↪C0,T;H01∩C10,T;L2. Therefore, since u∈W0,T∩L∞0,T;V∩W1,∞0,T;H01, the proof is the immediate consequence of Lemma 5 obtained by setting X=V,Y=H01 to have u∈Cw0,T;V and by setting X=H01,Y=L2 to have u′∈Cw0,T;H01.In the sequel, we give the important energy equality of weak solutions of Equation (1). It is used to prove the improved regularity of weak solutions of Equation (1) and used in all estimations in this paper.Lemma 7.
Assume thatu is a weak solution of Equation (1). Then, for each t∈0,T, we have the energy equality
(30)u′t22+γ∇u′t22+Δut22+12Δvt22+2k∗Δut,Δut2=2∫0tk′∗Δu,Δu2ds+2∫0tk0Δu22ds+2∫0tqu,u′2ds+2∫0tf,u′2ds+u122+γ∇u122+Δu022+12Δv022,where Δv0=−Δ−1u0,u0.Proof.
By Corollary6 and the uniform boundedness theorem, we have ut∈V and u′t∈H01 for all t∈0,T. Thus, every function in (30) has meaning for all t∈0,T. Then, we can proceed the proof as in ([17], Proposition 2.1). By regarding f in ([17], Proposition 2.1) as u,v+qu+f in Equation (1), we can deduce that the weak solution u of Equation (1) satisfies
(31)u′t22+γ∇u′t22+Δut22+2k∗Δut,Δut2=2∫0tk′∗Δu,Δu2ds+2∫0tk0Δu22ds+2∫0tu,v+qu+f,u′2ds+u122+γ∇u122+Δu022.
From [4], we can have
(32)2∫0tu,v,u′2ds=−12Δvt22+12Δv022.
Thus, we have (30).
This proves the lemma.From the energy equalities (30) or (31) together with the following well-known Gronwall's lemma, we can prove uniqueness and regularity of weak solutions of Equation (1).Lemma 8.
Letξ· be a nonnegative, absolutely continuous function on 0,T, which satisfies the differentiable inequality for a.e.t∈0,T:(33)ξ′t≤ψtξt+ϕt,where ψ and ϕ are nonnegative, summable functions on 0,T. Then
(34)ξt≤e∫0tψdsξ0+∫0tϕds,for all 0≤t≤T.Proof.
See ([23], p.624).Here, we can state the following theorem.Theorem 9.
Assume thatu0,u1∈V×H01,k∈C10,T,q∈L∞Q, and f∈L2Q. Then Equation (1) has a unique weak solution u in S0,T≡W0,T∩C0,T;V∩C10,T;H01. Moreover, the solution mapping p=u0,u1,q,f⟶up,u′p,vp of P≡V×H01×L∞Q×L2Q into C0,T;V×C0,T;H01×C0,T;W2,∞ is locally Lipschitz continuous.
Indeed, letp1=u01,u11,q1,f1∈P and p2=u02,u12,q2,f2∈P, we prove this theorem by showing the following inequality
(35)up1−up2S0,T+vp1−vp2C0,T;W2,∞≤Cp1−p2P,where C>0 is a constant depending on the data and
(36)p1−p2P=u01−u02V2+u11−u12H012+q1−q2L∞Q2+f1−f2L2Q21/2.Proof of Theorem 9.
Lemma7 allows us to show the regularity of u. It is verified from the data conditions that the right hand side of (30) is continuous in t. Hence, we have that
(37)t⟶∇u′t22+Δut22,is continuous on 0,T. Indeed, u∈C0,T;V∩C10,T;H01.Therefore, considering results in [15, 16] and [14], we can deduce that Equation (1) possesses a unique weak solution u∈S0,T under the data condition u0,u1,f∈H≡V×H01×L2Q such that
(38)uS0,T≤Cu0,u1,fH.Based on the above result, we prove the inequality (35). For the purpose, we denote u1−u2≡up1−up2 by ϕ and v1−v2≡vp1−vp2 by Y. Then, we can know from Equation (1) that ϕ and Y satisfy the following equation in the weak sense:
(39)Mγϕ″+Δ2ϕ+k∗Δ2ϕ=ϕ,v1+u2,Y+q1ϕ+q1−q2u2+f1−f2,Δ2Y=−ϕ,u1+u2inQ,ϕ=Δϕ=Y=∂Y∂ν=0onΣ,ϕ0=u01−u02,ϕ′0=u11−u12inΩ.We note that(40)u2,Y=u2,−Gϕ,u1+u2.Just as deriving the equality (31) from Equation (1), we can know that the weak solution ϕ of Equation (39) satisfies
(41)ϕ′t22+γ∇ϕ′t22+Δϕt22=−2k∗Δϕt,Δϕt2+2∫0tk′∗Δϕ,Δϕ2ds+2∫0tk0Δϕ22ds+2∫0tϕ,v1+u2,Y+q1ϕ+q1−q2u2,ϕ′2ds+2∫0tf1−f2,ϕ′2ds+ϕ′022+γ∇ϕ′022+Δϕ022.At first, we note by (13) that:
(42)2∫0tq1ϕ,ϕ′2ds≤2q1L∞Q∫0tϕ2ϕ′2ds≤C∫0tΔϕ2ϕ′2ds≤C∫0tΔϕ22+ϕ′22ds,(43)2∫0tq1−q2u2,ϕ′2ds≤2∫0tq1−q2u22ϕ′2ds≤q1−q2u2L2Q2+∫0tϕ′22ds≤q1−q2L∞Q2u2L2Q2+∫0tϕ′22ds.Byu2∈S0,T↪L2Q and (38), we can get from (43) that
(44)2∫0tq1−q2u2,ϕ′2ds≤Cq1−q2L∞Q2+∫0tϕ′22ds.For other estimates of the remaining terms on the right hand side of (41), we can refer to the previous results in [4] and obtain with (41) and (42)-(44) the following:
(45)ϕ′t22+∇ϕ′t22+Δϕt22≤Cϕ′022+∇ϕ′022+Δϕ022+q1−q2L∞Q2+f1−f2L2Q2+∫0tΔϕ22+ϕ′22ds≤Cp1−p2P2+∫0tΔϕ22+ϕ′22ds.By applying Lemma8 to (45), we have
(46)∇ϕ′t22+Δϕt22≤Cp1−p2P2.And also, for almostt∈0,T, we obtain by Lemma 2 and (11) that
(47)YtW2,∞2=−Gϕt,u1t+u2tW2,∞2≤CΔϕt22Δu1t+u2t22≤Cu1S0,T2+u2S0,T2Δϕt22.By (38), (46), and (47), we can obtain
(48)YtW2,∞2≤Cp1−p2P2.This implies with (46) that
(49)ϕC0,T;V∩C10,T;H01+YC0,T;W2,∞≤Cp1−p2P.SinceMγ−1A1/2∈LL2 and Mγ−1∈LL2,V, by conducting similar estimations in Equation (39), we can obtain from (49) that
(50)ϕ″L2Q≤Cp1−p2P.Hence, by (49) and (50) we can prove (35).This completes the proof.


## 4. Quadratic Cost Minimax Control Problems

Let the following be the set of the admissible controls:(51)Cad=c∈L∞Qc¯≤c≤c¯a.e.inQ,where c¯and c¯ are given constants, representing lower and upper bounds of the admissible control variables, respectively. Let the following be the set of the admissible disturbances or noises:
(52)Dad=η∈L∞Qη¯≤η≤η¯a.e.inQ,where η¯ and η¯ are given constants, representing lower and upper bounds of admissible disturbance variables, respectively. For variational analysis, we use the L2Q norm on Cad and Dad. For simplicity, we denote by Bad=Cad×Dad. From Theorem 9, we can uniquely define the solution mapBad⟶S0,T, which maps from q=c,η∈Bad to the weak solution uq∈S0,T, where uq satisfies the following equation:
(53)Mγu″q+Δ2uq+∫0tkt−sΔ2uq;sds=uq,vq+c+ηuq+f,Δ2vq=−uq,uqinQ,uq=Δuq=vq=∂vq∂ν=0onΣ,uq;0,x=u0x,u′q;0,x=u1xinΩ.The weak solutionuq of Equation (53) is called the state of the control system Equation (53).To study the quadratic cost minimax optimal control problems for Equation (53), we introduce the following quadratic cost function
(54)Jc,η=12uq−zdL2Q2+α2cL2Q2−β2ηL2Q2,where zd∈L2Q is the desired value, and the positive constants α and β are the relative weights of the second and third terms on the right hand side of (54).As indicated in the introduction, we shall study the minimax optimal control problem as follows: we prove the uniqueness as well as existence of a controlc∗∈Cad and a disturbance (or noise) η∗∈Dad such that c∗,η∗ is a saddle point of the functional J·,· of (54). That is,
(55)Jc∗,η≤Jc∗,η∗≤Jc,η∗,∀c,η∈Bad.Here, we call suchc∗,η∗ in (55) to be an optimal pair for the minimax optimal control problem with the cost (54). And we need to characterize c∗,η∗ in (55) by giving the necessary optimality condition through adjoint equation related to Equation (53) and the cost (54). For this purpose, we have to show the differentiabilities of the control to state mapping.


### 4.1. Differentiabilities of the Nonlinear Solution Map

We study the Fréchet differentiability of the nonlinear solution map, which is an improvement of the previous results in [4] and is more desirable for many applications. From Theorem 9, for fixed u0,u1,f∈V×H01×L2Qin Equation (53), we know that the solution map L∞Q⟶S0,T from q(=c+η in Equation (53)) ∈L∞ (Q) to uq∈S0,T is well defined and continuous.For our study, we present the following definitions.Definition 10.
The solution mapq⟶uq of L∞Q into S0,T is said to be Fréchet differentiable on L∞Q if for any q∈L∞Q, there exists a Tq∈LL∞Q,S0,T such that, for any w∈L∞Q,(56)limwL∞Q⟶0uq+w−uq−TqwS0,TwL∞Q=0.
The operatorTq is called the Fréchet derivative of u at q, which we denote by Duq.Tqw=Duqw∈S0,T is called the Fréchet derivative of u at q in the direction of w∈L∞Q.Definition 11.
LetU be a subset of L∞Q and q,q∗∈U. The solution map q⟶uq of U into S0,T is said to be Gâteaux differentiable at q∗ in the direction q−q∗ if there exists a function Duq∗;q−q∗∈S0,T such that
(57)limλ⟶0uq∗+λq−q∗−uq∗λ−Duq∗;q−q∗S0,T=0.Theorem 12.
The solution mapq⟶uq of L∞Q into S0,T is Fréchet differentiable on L∞Q and the Fréchet derivative of uq at q in the direction w∈L∞Q, that is to say z=Duqw, is the weak solution of
(58)Mγz″+Δ2z+∫0tkt−sΔ2zds=z,−Guq,uq+2uq,−Gz,uq+qz+wuqinQ,z=Δz=0onΣ,z0=0,z′0=0inΩ.
We prove this theorem by two steps.(i)
For anyw∈L∞Q, Equation (58) admits a unique weak solution z∈S0,T, namely, there exists an operator T∈LL∞Q,S0,T satisfying Tw=z=zw(ii)
We show thatuq+w−uq−zS0,T≤CwL∞Q2.Proof.
(i) Let(59)Guq,z≔z,−Guq,uq+2uq,−Gz,uq.
Then, we can estimate the right hand side of (59) as follows. By (11) and Lemma 2 we have
(60)z,−Guq,uq2≤CzH2uqH22≤CuqS0,T2Δz2.
This implies with (38) that
(61)z,−Guq,uq2≤CΔz2.
Similarly, we have(62)2uq,−Gz,uq2≤CΔz2.
Hence, by (61) and (62), we note that
(63)Guq,·∈LV,L2.
Taking into accountwuq∈L2Q and (63), we can employ the linear theory in [17] (cf. [22]) to verify that Equation (58) admits a unique weak solution z∈S0,T. And also by using the energy equality to be satisfied by z as in (31) and following similar estimations in Theorem 9, we can know by (38) that the weak solution z=zw of Equation (58) satisfies
(64)zwS0,T≤CwuqL2Q≤CwL∞QuqL2Q≤CuqS0,TwL∞Q≤CwL∞Q.
Hence, from (64), the mapping w∈L∞Q↦zw∈S0,T is linear and bounded. We can thus infer that there exists a T∈LL∞Q,S0,T such that Tw=zw for each w∈L∞Q.
(ii) We set the differenceuq+w−uq−z=θ. Then, by noting the following:
(65)Mγθ″+Δ2θ+k∗Δ2θ=uq+w,vq+w−uq,vq−z,vq−2uq,−Gz,uq+q+wuq+w−quq−qz−wuq=uq+w,vq+w−vq+θ,vq−2uq,−Gz,uq+qθ+wuq+w−uq≡uq+w,−Guq+w,uq+w+Guq,uq+θ,vq−2uq,−Gz,uq+qθ+wuq+w−uq=uq+w,−Guq+w−uq,uq+w+uq+θ,vq−2uq,−Gz,uq+q+wθ+wz=uq+w,−Gθ,uq+w+uq+uq+w,−Gz,uq+w+uq+θ,vq−2uq,−Gz,uq+q+wθ+wz=uq+w,−Gθ,uq+w+uq+uq+w−uq,−Gz,uq+w+uq+uq,−Gz,uq+w+uq+θ,vq−2uq,−Gz,uq+q+wθ+wz=uq+w,−Gθ,uq+w+uq+uq+w−uq,−Gz,uq+w+uq+uq,−Gz,uq+w−uq+θ,vq+q+wθ+wzinQ,we know that θ satisfies
(66)Mγθ″+Δ2θ+k∗Δ2θ=Ξuq,uq+w,θ+q+wθ+wz+I1+I2inQ,θ=Δθ=0onΣ,θ0,x=0,θ′0,x=0inΩ,in the weak sense, where
(67)Ξuq,uq+w,θ=uq+w,−Gθ,uq+w+uq+θ,vq,I1=uq+w−uq,−Gz,uq+w+uq,I2=uq,−Gz,uq+w−uq.
In a similar argument to (63), we know that
(68)Ξuq,uq+w,·∈LV,L2.
Thus with (68), we can apply the energy equality like (31) to (66) and follow similar estimations as in the Proof of Theorem 9, to obtain
(69)θS0,T≤Cwz+I1+I2L2Q.
By Theorem9 and (64), we can deduce as follows.
(70)wzL2Q≤wL∞QzL2Q≤CwL∞QzS0,T≤CwL∞Q2.
By Lemma2 and (11), we have
(71)I1L2Q≤Cuq+w−uqC0,T;VzC0,T;V×uq+w+uqL20,T;V≤Cuq+w−uqS0,TzS0,T×uq+wS0,T+uqS0,T.
From (38), (64) and Theorem 9, we can obtain with (71) that
(72)I1L2Q≤CwL∞Qq+w−qL∞Q=CwL∞Q2.
By analogy with (71) and (72), we can deduce
(73)I2L2Q≤CwL∞Q2.
Hence, from (69) to (73), we can obtain
(74)θS0,T≤Cwz+I1+I2L2Q≤CwzL2Q+I1L2Q+I2L2Q≤CwL∞Q2.This completes the proof.To show the uniqueness as well as existence of an optimal pair, we are going to use the strict convexity arguments in [12]. To this end, we consider the following results.Theorem 13.
The mapq⟶uq of L∞Q into S0,T is twice Fréchet differentiable at q and the twice Fréchet derivative of u at q in the direction w∈L∞Q, say ρ=D2uqw,w, is a unique solution of the following problem
(75)Mγρ″+Δ2ρ+∫0tkt−sΔ2ρds=Guq,ρ+Fz,uq+qρ+2wzinQ,ρ=Δρ=0onΣ,ρ0=0,ρ′0=0inΩ,where z is the solution of Equation (58), Guq,· is given in (59), and
(76)Fz,uq=4z,−Gz,uq+2uq,−Gz,z.
To prove Theorem13, it is sufficient to show the following:
(i)
ρS0,T≤CwL∞Q2;(ii)
Duq+ww−Duqw−ρS0,T=owL∞QaswL∞Q⟶0.Proof.
(i) By (38) and (64), we can have the following estimate
(77)Fz,uq2≤CuqH2zH22≤CuqS0,TzS0,T2≤CwL∞Q2.
Thus by similar arguments in the proof of (i) of Theorem12, we can show that the weak solution ρ of Equation (75) can be estimated as follows:
(78)ρS0,T≤CFz,uq+2wzL2Q.
By (70) and (77), we know by (78) that
(79)ρS0,T≤CwL∞Q2.
(ii) From Equation (58), we can deduce that κ=Duq+ww is the weak solution of the following equation:
(80)Mγκ″+Δ2κ+∫0tkt−sΔ2κds=Guq+w,κ+q+wκ+wuq+winQ,κ=Δκ=0onΣ,κ0=0,κ′0=0inΩ.
By previous result, we can verify the following(81)κS0,T≤CwL∞Q.
From Equation (58), Equation (75) and Equation (80), χ=κ−z−ρ satisfies the following equation
(82)Mγχ″+Δ2χ+∫0tkt−sΔ2χds=Guq,χ+qχ+∑i=15IiinQ,χ=Δχ=0onΣ,χ0=0,χ′0=0inΩ,in the weak sense, where
(83)I1=κ,−Guq+w−uq,uq+w+uq,I2=2uq+w−uq,−Gκ,uq+w,I3=2uq,−Gκ,uq+w−uq,I4=Fz,uq=4z,−Gz,uq+2uq,−Gz,z,I5=wκ+wθ−wz,where θ=uq+w−uq−z. By Theorem 9, (38) and (81), we can have
(84)I12≤CκH2uq+w−uqH2uq+w+uqH2≤Cuq+wS0,T+uqS0,T×κS0,Tuq+w−uqS0,T≤CwL∞Q2.
By analogy with (84), we obtain
(85)I22≤CwL∞Q2.
Also, by analogy with (84), we have
(86)I32≤CwL∞Q2.
It holds by (77) that
(87)I42≤CwL∞Q2.
By (81) and Theorem 12, we can have
(88)I52≤CwL∞QκS0,T+θS0,T+zS0,T≤CwL∞Q2+wL∞Q3.
By similar arguments to those in the Proof of Theorem9, we can deduce that the weak solution χ of Equation (82) satisfies
(89)χS0,T≤C∑i=15IiL2Q.
From (84) to (89), we can deduce
(90)χS0,T≤CwL∞Q3+wL∞Q2,which implies
(91)χS0,T=owL∞QaswL∞Q⟶0.This completes the proof.Corollary 14.
The mapq⟶uq of U⊂L∞Q into S0,T is twice Gâteaux differentiable at q∗ and such the twice Gâteaux derivative of u in the direction q−q∗, say ρ=D2uq∗q−q∗,q−q∗, is a unique solution of Equation (75) in which q and w are replaced by q∗ and q−q∗, respectively.Proof.
The proof is immediately followed by Theorem13.Lemma 15.
The weak solutionρ of Equation (75) satisfies
(92)ρS0,T≤CwL2Q2.Proof.
Letzw be the weak solution of Equation (58). Then, using S0,T↪C0Q¯, we can get from (64) the following:
(93)zwS0,T≤CwuqL2Q≤CuqC0Q¯wL2Q≤CuqS0,TwL2Q≤CwL2Q.
Letρ be the weak solution of Equation (75). From the second inequality in (77) and (78), we can deduce with (38) and (93) the following:
(94)ρS0,T≤CFz,uqL2Q+wzL2Q≤CuqS0,TzS0,T2+wL2QzS0,T≤CwL2Q2.This completes the proof.


### 4.2. Uniqueness and Existence of an Optimal Pair

To study the existence of an optimal pair, we present the following results.Proposition 16.
The solution mappingq⟶uq from Bad to W0,T of Equation (53) is continuous from the weakly-star topology of Bad to the weak topology of W0,T.Before we prove Proposition16, we need the following compactness lemma.Lemma 17.
LetX,Y, and Z be Banach spaces such that the embeddings X⊂Y⊂Z are continuous and the embedding X⊂Y is compact. Then, a bounded set of W1,∞0,T;X,Z=g∣g∈L∞0,T;X,g′∈L∞0,T;Z is relatively compact in C0,T;Y.Proof.
See Simon [24].Proof of Proposition 16.
Letq=c,η∈Bad and let qn=cn,ηn∈Bad be a sequence such that
(95)qn⟶qweakly−starinBadasn⟶∞.
From now on, each stateun=uqn is the weak solution of
(96)Mγun″+Δ2un+k∗Δ2un=un,vn+cn+ηnun+f,Δ2vn=−un,uninQ,un=Δun=vn=∂vn∂ν=0onΣ,un0,x=u0x,u′n0,x=u1xinΩ.
If we letpn=u0,u1,f,cn+ηn and 0=0,0,0,0, then from Theorem 9, we can know that upn=un in Equation (96) and u0=0 in S0,T and deduce that the following is fulfilled:
(97)upn−u0S0,T=unS0,T≤Cu0,u1,f,cn+ηnP.
Hence, we can deduce from (97) that un and u′n remain in the bounded sets of L∞0,T;V∩W0,T and L∞0,T;H01, respectively. Therefore, by using Rellich's extraction theorem, we can find a subsequence of un, say again un, and find u∈W0,T such that
(98)un⟶uweaklyinW0,Tasn⟶∞,(99)un⟶uweakly−starinL∞0,T;Vasn⟶∞.(100)u′n⟶u′weakly−starinL∞0,T;H01asn⟶∞.
SinceV↪H01 is compact, we can apply Lemma 17 to (99) and (100) with X=V and Y=Z=H01 to verify that
(101)unispre−compactinC0,T;H01.
Hence, we may extract a subsequence, denoted again byun, such that
(102)un⟶ustronglyinC0,T;H01asn⟶∞.
By Lemma2, (11) and (38), we can note for almost t∈0,T that
(103)unt,unt−ut,utH−2=unt−ut,unt+utH−2≤Cunt−utH1unt+utH2≤CunS0,T+uS0,Tun−uC0,T;H01≤Cun−uC0,T;H01.
Therefore, by (102) and (103), we may extract a subsequence, denoted again by un,un, such that
(104)un,un⟶u,ustronglyinC0,T;H−2asn⟶∞.
SinceG∈LH−2,H02, we can deduce with (104) that
(105)vn=−Gun,un⟶v=−Gu,ustronglyinC0,T;H02
asn⟶∞. For any ϕ∈V, we consider for almost all t∈0,T that
(106)unt,vnt−ut,vt,ϕ2≤unt−ut,vnt,ϕ2+ut,vnt−vt,ϕ2.
By Lemma1, we note that the right hand side of (106) equals
(107)unt−ut,vnt,ϕ2+vnt−vt,ut,ϕ2.
By consideringut,ϕ∈V, we have ut,ϕ∈L1. And through (14) we can know that
(108)H02↪L∞,and therefore
(109)L1↪H−2.
Thus, by Lemma2, (109), (11), and (38), we can have as follows:
(110)righthandsideof4.54≤CuntH22ϕH2unt−ut2+vnt−vtH02ut,ϕH−2≤CunC0,T;V2Δϕ2unt−ut2+vnt−vtH02uC0,T;H01Δϕ2≤CunS0,T2+uS0,Tunt−ut2+vnt−vtH02Δϕ2≤Cun−uC0,T;L2+vn−vC0,T;H02Δϕ2.
Thus, from (102), (105), and (110), we may extract a subsequence, if necessary, denoted again by un,vn, such that
(111)un,vn⟶u,vstronglyinC0,T;V′asn⟶∞.
From (95) and (101), we can obtain
(112)cn+ηnun⟶c+ηuweaklyinL2Qasn⟶∞.
We replaceun by unk in Equation (96), if necessary, and take k⟶∞. Then, by the standard arguments in Dautray and Lions ([22], pp.561-565), we conclude that the limit y is a weak solution of
(113)Mγu″+Δ2u+k∗Δ2u=u,v+c+ηu+f,Δ2v=−u,uinQ,u=Δu=v=∂v∂ν=0onΣ,u0,x=u0x,u′0,x=u1xinΩ.
Moreover, from the uniqueness of the weak solutions, we conclude thatu=uq in W0,T, which implies that uqn⟶uq weakly in W0,T.This completes the proof.We now study the uniqueness and existence of an optimal pair.Theorem 18.
For sufficiently largeα and β in (54) there exists unique c∗,η∗∈Bad such that c∗,η∗ satisfies (55).
We prove this Theorem by showing:(i)
For sufficiently largeα and β in (54), we prove the maps c⟶Jc,η and η⟶Jc,η are strictly convex and strictly concave, respectively(ii)
We prove the existence of an optimal pair by showing the mapsc⟶Jc,η and η⟶Jc,η are lower semicontinuous and upper semicontinuous, respectivelyProof.
(i) LetPη be the map c⟶Jc,η and let Qc be the map η⟶Jc,η. To obtain the unique existence of an optimal pair in the minimax optimal control problem, we show the map Pη is strictly convex and lower semicontinuous for all η∈Dad and the map Qc is strictly concave and upper semicontinuous for all c∈Cad. As in [11] (cf. [12]), to show the strict convexity and the strict concavity of each map, we verify the following second Gâteaux derivative conditions
(114)D2Pηc1+ξc2−c1c2−c1,c2−c1>0,∀c1≠c2∈Cad,(115)D2Qcη1+ξη2−η1η2−η1,η2−η1<0,∀η1≠η2∈Dad,
where0<ξ<1. For sufficiently large α in (54), we first prove the convexity of Pη by showing (114). For simplicity, we denote uc1+ξc2−c1,zc1+ξc2−c1 and ρc1+ξc2−c1 by uξ,zξ, and ρξ, respectively. We calculate
(116)DPηc1+ξc2−c1c2−c1=liml⟶0Jc1+ξ+lc2−c1,η−Jc1+ξc2−c1,ηl=DcJc1+ξc2−c1,ηc2−c1=∫0Tuξ−zd,zξ2ds+α∫0Tc1+ξc2−c1,c2−c12dt.
From Corollary14 we know that the map q⟶uq is twice Gâteaux differentiable at q=c1+ξc2−c1 in the direction c2−c1. Thus from (116), we can obtain the second Gâteaux derivative of Pη as follows:
(117)D2Pηc1+ξc2−c1c2−c1,c2−c1=limμ⟶0DcJc1+ξ+μc2−c1,ηc2−c1−DcJc1+ξc2−c1,ηc2−c1μ=∫0Tuξ−zd,ρξ2ds+zξL2Q2+αc2−c1L2Q2,where ρξ is the weak solution of Equation (75) in which q and w are replaced by c1+ξc2−c1 and c2−c1, respectively. Then, by Lemma 15 and (117), we can deduce that
(118)D2Pηc1+ξc2−c1c2−c1,c2−c1≥−ρξC0,T;L2∫0Tuξ−zd2ds+zξL2Q2+αc2−c1L2Q2≥−CTρξS0,Tuξ−zdL2Q+zξL2Q2+αc2−c1L2Q2≥α−CTuξ−zdL2Qc2−c1L2Q2+zξL2Q2.
Hence, we can verify that there exists sufficiently largeαlT,P,Bad,zd>0 such that (114) is satisfied for any α>αlT,P,Bad,zd. Therefore, the map Pη is strictly convex for sufficiently large α>0.
Similarly, we can show that there exist a sufficiently largeβlT,P,Bad,zd>0 such that (115) is satisfied for any β>βlT,P,Bad,zd. This also indicates the strict concavity of Qc.
(ii) Next, we prove the existence of an optimal pairc∗,η∗∈Bad by verifying that Pη is lower semicontinuous for all η∈Dad and Qc is upper semicontinuous for all c∈Cad. Let cn⊂Cad be a minimizing sequence of J. Thus
(119)liminfn⟶∞Jcn,η=infc∈CadJc,η.
Sincecn is bounded in L∞Q⊂L2Q, we can extract a subsequence cnk⊂cn such that
(120)cnk⟶c∗weaklyinL2Qask⟶∞.
Then, by Proposition16, we obtain
(121)ucnk,η⟶uc∗,ηweaklyinW0,Task⟶∞,
∀η∈Dad. Since we know from Dautray-Lions ([22], p.480) that W0,T↪C0,T;H01∩C10,T;L2 and the embedding H01↪L2 is compact, we can use Lemma 17 in which X=H01,Y=Z=L2 to have from (121) that the sequence ucnk,η is relatively compact in C0,T;L2. Thus, we can choose a subsequence of cnk, if necessary, still denoted by cnk such that
(122)ucnk,η⟶uc∗,ηstronglyinC0,T;L2ask⟶∞,
∀η∈Dad. Therefore, we can deduce for the same subsequence cnk given above that
(123)ucnk,η⟶uc∗,ηstronglyinL2Qask⟶∞,
∀η∈Dad. Since the norm is weakly lower semicontinuous, we can verify by (120) and (123) that the map Pη:c⟶Jc,η is lower semicontinuous for all η∈Dad. Hence, we know that
(124)J0η=liminfn⟶∞Jcn,η≥Jc∗,η,∀η∈Dad.
But sinceJ0η≤Jc∗,η, we have
(125)J0η=Jc∗,η=infc∈CadJc,η,∀η∈Dad.
By similar arguments, we can prove thatQc is upper semicontinuous for all c∈Cad. Thus, we can also know that there exists η∗∈Dad such that
(126)J0η∗=supη∈DadJ0η.
From (125) and (126), we can conclude that c∗,η∗∈Bad is an optimal pair for the cost (54).This completes the proof.Remark 19.
Assuming thatT>0 is sufficiently small, instead of assuming α and β are large enough, we can also obtain strict convexity and strict concavity of the maps Pη and Qc, respectively.


### 4.3. Necessary Condition of an Optimal Pair

In this subsection, we study the necessary optimality condition to be satisfied by an optimal pair of the minimax optimal control problem with the cost (54).From Theorem12, we know that the solution map q⟶uq from Bad to S0,T is Fréchet differentiable at q∗=c∗,η∗,and the Fréchet derivative of uqat q=q∗ in the direction w=h,l∈L∞Q2,say z=Duq∗w is a unique weak solution of the following equation:
(127)Mγz″+Δ2z+∫0tkt−sΔ2zds=Guq∗,z+c∗+η∗z+h+luq∗inQ,z=Δz=0onΣ,z0=0,z′0=0inΩ,where Guq∗,z is defined in (59). By Equation (127), we introduce the following adjoint equation corresponding to the cost function (54):
(128)Mγp″+Δ2p+∫tTkσ−tΔ2pdσ=Guq∗,p+c∗+η∗p+uq∗−zdinQ,p=Δp=0onΣ,pT=0,p′T=0inΩ.Proposition 20.
Equation (128) admits a unique solution p∈S0,T.Proof.
Changing the variablet to T−t in Equation (128), we can complete the proof by referring to the results in [4, 17].
We now discuss the first-order optimality conditions for the minimax optimal control problem (55) for the quadratic cost function (54).Theorem 21.
Ifα and β in the cost (54) are large enough or T>0 is sufficiently small, then an optimal control c∗∈Cad and a disturbance η∗∈Dad, namely, an optimal pair q∗=c∗,η∗∈Bad satisfying (55) can be given by:
(129)c∗=maxc¯,min−uq∗pα,c¯,η∗=maxη¯,minuq∗pβ,η¯,where p is the weak solution of Equation (128), c¯ and c¯ are constants given in (51), and η¯ and η¯ are constants given in (52).Proof.
Letq∗=c∗,η∗∈Bad be an optimal pair in (55) with the cost (54) and uq∗ be the corresponding weak solution of Equation (53).
From Theorem12, it is clear that the map q=c,η⟶uq is Gâteaux differentiable at q∗=c∗,η∗ in the direction w=h,l∈L∞Q2 with q∗+εw∈Bad for sufficiently small ε>0. Indeed, we have
(130)uq∗+εw−uq∗ε⟶z=zwstronglyinS0,Tasε⟶0+,where z=Duq∗w is a unique weak solution of Equation (127). Therefore, by (130), we can get the Gâteaux derivative of the cost (54) at q=q∗ in the direction w=h,l as follows:
(131)DJc∗,η∗h,l=limε⟶0+Jc∗+εh,η∗+εl−Jc∗,η∗ε=limε⟶0+12∫0Tuq∗+εw+uq∗−2zd,uq∗+εw−uq∗ε2dt+limε⟶0+α2∫0T2c∗,h2+εh22dt−β2∫0T2η∗,l2+εl22dt=∫0Tuq∗−zd,z2dt+α∫0Tc∗,h2dt−β∫0Tη∗,l2dt,where z=Duq∗w is a solution of Equation (127).
Before we proceed to the calculations, by making use of the fact thatG is a self adjoint operator, we can note by Lemma 1 and (109) that
(132)Guq∗,ϕ,φ2=ϕ,−Guq∗,uq∗,φ2+2uq∗,−Gϕ,uq∗,φ2=ϕ,φ,−Guq∗,uq∗2+2−Gϕ,uq∗,φ,uq∗2,−2=ϕ,φ,−Guq∗,uq∗2+2ϕ,uq∗,−Gφ,uq∗−2,2=ϕ,φ,−Guq∗,uq∗2+2ϕ,uq∗,−Gφ,uq∗2=ϕ,Guq∗,φ2,∀φ,ϕ∈V.
We multiply both sides of the weak form of Equation (128) by z which is a solution of Equation (127), and integrate it over 0,T. Then, we have
(133)∫0TMγp″,zV′,Vdt+∫0TΔp,Δz2dt+∫0T∫tTkσ−tΔpdσ,Δz2dt=∫0TGuq∗,p+c∗+η∗p,z2dt+∫0Tuq∗−zd,z2dt.
By Fubini's theorem, we can know that(134)∫0T∫tTkσ−tΔpdσ,Δz2dt=∫0T∫0tkt−sΔzds,Δp2dt=∫0T∫0tkt−sΔ2zds,pV′,Vdt.
From (132), (134) and terminal values of the weak solution p of Equation (128), we can obtain the following by integration by parts of (133):
(135)∫0Tp,Mγz″+Δ2z+∫0tkt−sΔ2zdtV,V′dt=∫0Tp,Guq∗,z+c∗+η∗z2dt+∫0Tuq∗−zd,z2dt.
Sincez is the solution of Equation (127), we can obtain the following from (135):
(136)∫0Tuq∗−zd,z2dt=∫0Th+luq∗,p2dt.
Therefore, we can deduce that (131) and (136) imply
(137)DJc∗,η∗h,l=∫0Tαc∗+uq∗p,h2dt+∫0T−βη∗+uq∗p,l2dt.
Sinceq∗=c∗,η∗∈Bad is an optimal pair in (55), we know
(138)DcJc∗,η∗h≥0,DηJc∗,η∗l≤0,h,l∈L∞Q2.
Therefore, we can obtain the following from (137) and (138):
(139)∫0Tαc∗+uq∗p,h2dt≥0,∫0T−βη∗+uq∗p,l2dt≤0,where h,l∈L∞Q2. By considering the signs of the variations h and l in (139), which depend on c∗ and η∗, respectively, we can deduce from (139) that
(140)c∗=maxc¯,min−uq∗pα,c¯,η∗=maxη¯,minuq∗pβ,η¯.This completes the proof.


## 4.1. Differentiabilities of the Nonlinear Solution Map

We study the Fréchet differentiability of the nonlinear solution map, which is an improvement of the previous results in [4] and is more desirable for many applications. From Theorem 9, for fixed u0,u1,f∈V×H01×L2Qin Equation (53), we know that the solution map L∞Q⟶S0,T from q(=c+η in Equation (53)) ∈L∞ (Q) to uq∈S0,T is well defined and continuous.For our study, we present the following definitions.Definition 10.
The solution mapq⟶uq of L∞Q into S0,T is said to be Fréchet differentiable on L∞Q if for any q∈L∞Q, there exists a Tq∈LL∞Q,S0,T such that, for any w∈L∞Q,(56)limwL∞Q⟶0uq+w−uq−TqwS0,TwL∞Q=0.
The operatorTq is called the Fréchet derivative of u at q, which we denote by Duq.Tqw=Duqw∈S0,T is called the Fréchet derivative of u at q in the direction of w∈L∞Q.Definition 11.
LetU be a subset of L∞Q and q,q∗∈U. The solution map q⟶uq of U into S0,T is said to be Gâteaux differentiable at q∗ in the direction q−q∗ if there exists a function Duq∗;q−q∗∈S0,T such that
(57)limλ⟶0uq∗+λq−q∗−uq∗λ−Duq∗;q−q∗S0,T=0.Theorem 12.
The solution mapq⟶uq of L∞Q into S0,T is Fréchet differentiable on L∞Q and the Fréchet derivative of uq at q in the direction w∈L∞Q, that is to say z=Duqw, is the weak solution of
(58)Mγz″+Δ2z+∫0tkt−sΔ2zds=z,−Guq,uq+2uq,−Gz,uq+qz+wuqinQ,z=Δz=0onΣ,z0=0,z′0=0inΩ.
We prove this theorem by two steps.(i)
For anyw∈L∞Q, Equation (58) admits a unique weak solution z∈S0,T, namely, there exists an operator T∈LL∞Q,S0,T satisfying Tw=z=zw(ii)
We show thatuq+w−uq−zS0,T≤CwL∞Q2.Proof.
(i) Let(59)Guq,z≔z,−Guq,uq+2uq,−Gz,uq.
Then, we can estimate the right hand side of (59) as follows. By (11) and Lemma 2 we have
(60)z,−Guq,uq2≤CzH2uqH22≤CuqS0,T2Δz2.
This implies with (38) that
(61)z,−Guq,uq2≤CΔz2.
Similarly, we have(62)2uq,−Gz,uq2≤CΔz2.
Hence, by (61) and (62), we note that
(63)Guq,·∈LV,L2.
Taking into accountwuq∈L2Q and (63), we can employ the linear theory in [17] (cf. [22]) to verify that Equation (58) admits a unique weak solution z∈S0,T. And also by using the energy equality to be satisfied by z as in (31) and following similar estimations in Theorem 9, we can know by (38) that the weak solution z=zw of Equation (58) satisfies
(64)zwS0,T≤CwuqL2Q≤CwL∞QuqL2Q≤CuqS0,TwL∞Q≤CwL∞Q.
Hence, from (64), the mapping w∈L∞Q↦zw∈S0,T is linear and bounded. We can thus infer that there exists a T∈LL∞Q,S0,T such that Tw=zw for each w∈L∞Q.
(ii) We set the differenceuq+w−uq−z=θ. Then, by noting the following:
(65)Mγθ″+Δ2θ+k∗Δ2θ=uq+w,vq+w−uq,vq−z,vq−2uq,−Gz,uq+q+wuq+w−quq−qz−wuq=uq+w,vq+w−vq+θ,vq−2uq,−Gz,uq+qθ+wuq+w−uq≡uq+w,−Guq+w,uq+w+Guq,uq+θ,vq−2uq,−Gz,uq+qθ+wuq+w−uq=uq+w,−Guq+w−uq,uq+w+uq+θ,vq−2uq,−Gz,uq+q+wθ+wz=uq+w,−Gθ,uq+w+uq+uq+w,−Gz,uq+w+uq+θ,vq−2uq,−Gz,uq+q+wθ+wz=uq+w,−Gθ,uq+w+uq+uq+w−uq,−Gz,uq+w+uq+uq,−Gz,uq+w+uq+θ,vq−2uq,−Gz,uq+q+wθ+wz=uq+w,−Gθ,uq+w+uq+uq+w−uq,−Gz,uq+w+uq+uq,−Gz,uq+w−uq+θ,vq+q+wθ+wzinQ,we know that θ satisfies
(66)Mγθ″+Δ2θ+k∗Δ2θ=Ξuq,uq+w,θ+q+wθ+wz+I1+I2inQ,θ=Δθ=0onΣ,θ0,x=0,θ′0,x=0inΩ,in the weak sense, where
(67)Ξuq,uq+w,θ=uq+w,−Gθ,uq+w+uq+θ,vq,I1=uq+w−uq,−Gz,uq+w+uq,I2=uq,−Gz,uq+w−uq.
In a similar argument to (63), we know that
(68)Ξuq,uq+w,·∈LV,L2.
Thus with (68), we can apply the energy equality like (31) to (66) and follow similar estimations as in the Proof of Theorem 9, to obtain
(69)θS0,T≤Cwz+I1+I2L2Q.
By Theorem9 and (64), we can deduce as follows.
(70)wzL2Q≤wL∞QzL2Q≤CwL∞QzS0,T≤CwL∞Q2.
By Lemma2 and (11), we have
(71)I1L2Q≤Cuq+w−uqC0,T;VzC0,T;V×uq+w+uqL20,T;V≤Cuq+w−uqS0,TzS0,T×uq+wS0,T+uqS0,T.
From (38), (64) and Theorem 9, we can obtain with (71) that
(72)I1L2Q≤CwL∞Qq+w−qL∞Q=CwL∞Q2.
By analogy with (71) and (72), we can deduce
(73)I2L2Q≤CwL∞Q2.
Hence, from (69) to (73), we can obtain
(74)θS0,T≤Cwz+I1+I2L2Q≤CwzL2Q+I1L2Q+I2L2Q≤CwL∞Q2.This completes the proof.To show the uniqueness as well as existence of an optimal pair, we are going to use the strict convexity arguments in [12]. To this end, we consider the following results.Theorem 13.
The mapq⟶uq of L∞Q into S0,T is twice Fréchet differentiable at q and the twice Fréchet derivative of u at q in the direction w∈L∞Q, say ρ=D2uqw,w, is a unique solution of the following problem
(75)Mγρ″+Δ2ρ+∫0tkt−sΔ2ρds=Guq,ρ+Fz,uq+qρ+2wzinQ,ρ=Δρ=0onΣ,ρ0=0,ρ′0=0inΩ,where z is the solution of Equation (58), Guq,· is given in (59), and
(76)Fz,uq=4z,−Gz,uq+2uq,−Gz,z.
To prove Theorem13, it is sufficient to show the following:
(i)
ρS0,T≤CwL∞Q2;(ii)
Duq+ww−Duqw−ρS0,T=owL∞QaswL∞Q⟶0.Proof.
(i) By (38) and (64), we can have the following estimate
(77)Fz,uq2≤CuqH2zH22≤CuqS0,TzS0,T2≤CwL∞Q2.
Thus by similar arguments in the proof of (i) of Theorem12, we can show that the weak solution ρ of Equation (75) can be estimated as follows:
(78)ρS0,T≤CFz,uq+2wzL2Q.
By (70) and (77), we know by (78) that
(79)ρS0,T≤CwL∞Q2.
(ii) From Equation (58), we can deduce that κ=Duq+ww is the weak solution of the following equation:
(80)Mγκ″+Δ2κ+∫0tkt−sΔ2κds=Guq+w,κ+q+wκ+wuq+winQ,κ=Δκ=0onΣ,κ0=0,κ′0=0inΩ.
By previous result, we can verify the following(81)κS0,T≤CwL∞Q.
From Equation (58), Equation (75) and Equation (80), χ=κ−z−ρ satisfies the following equation
(82)Mγχ″+Δ2χ+∫0tkt−sΔ2χds=Guq,χ+qχ+∑i=15IiinQ,χ=Δχ=0onΣ,χ0=0,χ′0=0inΩ,in the weak sense, where
(83)I1=κ,−Guq+w−uq,uq+w+uq,I2=2uq+w−uq,−Gκ,uq+w,I3=2uq,−Gκ,uq+w−uq,I4=Fz,uq=4z,−Gz,uq+2uq,−Gz,z,I5=wκ+wθ−wz,where θ=uq+w−uq−z. By Theorem 9, (38) and (81), we can have
(84)I12≤CκH2uq+w−uqH2uq+w+uqH2≤Cuq+wS0,T+uqS0,T×κS0,Tuq+w−uqS0,T≤CwL∞Q2.
By analogy with (84), we obtain
(85)I22≤CwL∞Q2.
Also, by analogy with (84), we have
(86)I32≤CwL∞Q2.
It holds by (77) that
(87)I42≤CwL∞Q2.
By (81) and Theorem 12, we can have
(88)I52≤CwL∞QκS0,T+θS0,T+zS0,T≤CwL∞Q2+wL∞Q3.
By similar arguments to those in the Proof of Theorem9, we can deduce that the weak solution χ of Equation (82) satisfies
(89)χS0,T≤C∑i=15IiL2Q.
From (84) to (89), we can deduce
(90)χS0,T≤CwL∞Q3+wL∞Q2,which implies
(91)χS0,T=owL∞QaswL∞Q⟶0.This completes the proof.Corollary 14.
The mapq⟶uq of U⊂L∞Q into S0,T is twice Gâteaux differentiable at q∗ and such the twice Gâteaux derivative of u in the direction q−q∗, say ρ=D2uq∗q−q∗,q−q∗, is a unique solution of Equation (75) in which q and w are replaced by q∗ and q−q∗, respectively.Proof.
The proof is immediately followed by Theorem13.Lemma 15.
The weak solutionρ of Equation (75) satisfies
(92)ρS0,T≤CwL2Q2.Proof.
Letzw be the weak solution of Equation (58). Then, using S0,T↪C0Q¯, we can get from (64) the following:
(93)zwS0,T≤CwuqL2Q≤CuqC0Q¯wL2Q≤CuqS0,TwL2Q≤CwL2Q.
Letρ be the weak solution of Equation (75). From the second inequality in (77) and (78), we can deduce with (38) and (93) the following:
(94)ρS0,T≤CFz,uqL2Q+wzL2Q≤CuqS0,TzS0,T2+wL2QzS0,T≤CwL2Q2.This completes the proof.


## 4.2. Uniqueness and Existence of an Optimal Pair

To study the existence of an optimal pair, we present the following results.Proposition 16.
The solution mappingq⟶uq from Bad to W0,T of Equation (53) is continuous from the weakly-star topology of Bad to the weak topology of W0,T.Before we prove Proposition16, we need the following compactness lemma.Lemma 17.
LetX,Y, and Z be Banach spaces such that the embeddings X⊂Y⊂Z are continuous and the embedding X⊂Y is compact. Then, a bounded set of W1,∞0,T;X,Z=g∣g∈L∞0,T;X,g′∈L∞0,T;Z is relatively compact in C0,T;Y.Proof.
See Simon [24].Proof of Proposition 16.
Letq=c,η∈Bad and let qn=cn,ηn∈Bad be a sequence such that
(95)qn⟶qweakly−starinBadasn⟶∞.
From now on, each stateun=uqn is the weak solution of
(96)Mγun″+Δ2un+k∗Δ2un=un,vn+cn+ηnun+f,Δ2vn=−un,uninQ,un=Δun=vn=∂vn∂ν=0onΣ,un0,x=u0x,u′n0,x=u1xinΩ.
If we letpn=u0,u1,f,cn+ηn and 0=0,0,0,0, then from Theorem 9, we can know that upn=un in Equation (96) and u0=0 in S0,T and deduce that the following is fulfilled:
(97)upn−u0S0,T=unS0,T≤Cu0,u1,f,cn+ηnP.
Hence, we can deduce from (97) that un and u′n remain in the bounded sets of L∞0,T;V∩W0,T and L∞0,T;H01, respectively. Therefore, by using Rellich's extraction theorem, we can find a subsequence of un, say again un, and find u∈W0,T such that
(98)un⟶uweaklyinW0,Tasn⟶∞,(99)un⟶uweakly−starinL∞0,T;Vasn⟶∞.(100)u′n⟶u′weakly−starinL∞0,T;H01asn⟶∞.
SinceV↪H01 is compact, we can apply Lemma 17 to (99) and (100) with X=V and Y=Z=H01 to verify that
(101)unispre−compactinC0,T;H01.
Hence, we may extract a subsequence, denoted again byun, such that
(102)un⟶ustronglyinC0,T;H01asn⟶∞.
By Lemma2, (11) and (38), we can note for almost t∈0,T that
(103)unt,unt−ut,utH−2=unt−ut,unt+utH−2≤Cunt−utH1unt+utH2≤CunS0,T+uS0,Tun−uC0,T;H01≤Cun−uC0,T;H01.
Therefore, by (102) and (103), we may extract a subsequence, denoted again by un,un, such that
(104)un,un⟶u,ustronglyinC0,T;H−2asn⟶∞.
SinceG∈LH−2,H02, we can deduce with (104) that
(105)vn=−Gun,un⟶v=−Gu,ustronglyinC0,T;H02
asn⟶∞. For any ϕ∈V, we consider for almost all t∈0,T that
(106)unt,vnt−ut,vt,ϕ2≤unt−ut,vnt,ϕ2+ut,vnt−vt,ϕ2.
By Lemma1, we note that the right hand side of (106) equals
(107)unt−ut,vnt,ϕ2+vnt−vt,ut,ϕ2.
By consideringut,ϕ∈V, we have ut,ϕ∈L1. And through (14) we can know that
(108)H02↪L∞,and therefore
(109)L1↪H−2.
Thus, by Lemma2, (109), (11), and (38), we can have as follows:
(110)righthandsideof4.54≤CuntH22ϕH2unt−ut2+vnt−vtH02ut,ϕH−2≤CunC0,T;V2Δϕ2unt−ut2+vnt−vtH02uC0,T;H01Δϕ2≤CunS0,T2+uS0,Tunt−ut2+vnt−vtH02Δϕ2≤Cun−uC0,T;L2+vn−vC0,T;H02Δϕ2.
Thus, from (102), (105), and (110), we may extract a subsequence, if necessary, denoted again by un,vn, such that
(111)un,vn⟶u,vstronglyinC0,T;V′asn⟶∞.
From (95) and (101), we can obtain
(112)cn+ηnun⟶c+ηuweaklyinL2Qasn⟶∞.
We replaceun by unk in Equation (96), if necessary, and take k⟶∞. Then, by the standard arguments in Dautray and Lions ([22], pp.561-565), we conclude that the limit y is a weak solution of
(113)Mγu″+Δ2u+k∗Δ2u=u,v+c+ηu+f,Δ2v=−u,uinQ,u=Δu=v=∂v∂ν=0onΣ,u0,x=u0x,u′0,x=u1xinΩ.
Moreover, from the uniqueness of the weak solutions, we conclude thatu=uq in W0,T, which implies that uqn⟶uq weakly in W0,T.This completes the proof.We now study the uniqueness and existence of an optimal pair.Theorem 18.
For sufficiently largeα and β in (54) there exists unique c∗,η∗∈Bad such that c∗,η∗ satisfies (55).
We prove this Theorem by showing:(i)
For sufficiently largeα and β in (54), we prove the maps c⟶Jc,η and η⟶Jc,η are strictly convex and strictly concave, respectively(ii)
We prove the existence of an optimal pair by showing the mapsc⟶Jc,η and η⟶Jc,η are lower semicontinuous and upper semicontinuous, respectivelyProof.
(i) LetPη be the map c⟶Jc,η and let Qc be the map η⟶Jc,η. To obtain the unique existence of an optimal pair in the minimax optimal control problem, we show the map Pη is strictly convex and lower semicontinuous for all η∈Dad and the map Qc is strictly concave and upper semicontinuous for all c∈Cad. As in [11] (cf. [12]), to show the strict convexity and the strict concavity of each map, we verify the following second Gâteaux derivative conditions
(114)D2Pηc1+ξc2−c1c2−c1,c2−c1>0,∀c1≠c2∈Cad,(115)D2Qcη1+ξη2−η1η2−η1,η2−η1<0,∀η1≠η2∈Dad,
where0<ξ<1. For sufficiently large α in (54), we first prove the convexity of Pη by showing (114). For simplicity, we denote uc1+ξc2−c1,zc1+ξc2−c1 and ρc1+ξc2−c1 by uξ,zξ, and ρξ, respectively. We calculate
(116)DPηc1+ξc2−c1c2−c1=liml⟶0Jc1+ξ+lc2−c1,η−Jc1+ξc2−c1,ηl=DcJc1+ξc2−c1,ηc2−c1=∫0Tuξ−zd,zξ2ds+α∫0Tc1+ξc2−c1,c2−c12dt.
From Corollary14 we know that the map q⟶uq is twice Gâteaux differentiable at q=c1+ξc2−c1 in the direction c2−c1. Thus from (116), we can obtain the second Gâteaux derivative of Pη as follows:
(117)D2Pηc1+ξc2−c1c2−c1,c2−c1=limμ⟶0DcJc1+ξ+μc2−c1,ηc2−c1−DcJc1+ξc2−c1,ηc2−c1μ=∫0Tuξ−zd,ρξ2ds+zξL2Q2+αc2−c1L2Q2,where ρξ is the weak solution of Equation (75) in which q and w are replaced by c1+ξc2−c1 and c2−c1, respectively. Then, by Lemma 15 and (117), we can deduce that
(118)D2Pηc1+ξc2−c1c2−c1,c2−c1≥−ρξC0,T;L2∫0Tuξ−zd2ds+zξL2Q2+αc2−c1L2Q2≥−CTρξS0,Tuξ−zdL2Q+zξL2Q2+αc2−c1L2Q2≥α−CTuξ−zdL2Qc2−c1L2Q2+zξL2Q2.
Hence, we can verify that there exists sufficiently largeαlT,P,Bad,zd>0 such that (114) is satisfied for any α>αlT,P,Bad,zd. Therefore, the map Pη is strictly convex for sufficiently large α>0.
Similarly, we can show that there exist a sufficiently largeβlT,P,Bad,zd>0 such that (115) is satisfied for any β>βlT,P,Bad,zd. This also indicates the strict concavity of Qc.
(ii) Next, we prove the existence of an optimal pairc∗,η∗∈Bad by verifying that Pη is lower semicontinuous for all η∈Dad and Qc is upper semicontinuous for all c∈Cad. Let cn⊂Cad be a minimizing sequence of J. Thus
(119)liminfn⟶∞Jcn,η=infc∈CadJc,η.
Sincecn is bounded in L∞Q⊂L2Q, we can extract a subsequence cnk⊂cn such that
(120)cnk⟶c∗weaklyinL2Qask⟶∞.
Then, by Proposition16, we obtain
(121)ucnk,η⟶uc∗,ηweaklyinW0,Task⟶∞,
∀η∈Dad. Since we know from Dautray-Lions ([22], p.480) that W0,T↪C0,T;H01∩C10,T;L2 and the embedding H01↪L2 is compact, we can use Lemma 17 in which X=H01,Y=Z=L2 to have from (121) that the sequence ucnk,η is relatively compact in C0,T;L2. Thus, we can choose a subsequence of cnk, if necessary, still denoted by cnk such that
(122)ucnk,η⟶uc∗,ηstronglyinC0,T;L2ask⟶∞,
∀η∈Dad. Therefore, we can deduce for the same subsequence cnk given above that
(123)ucnk,η⟶uc∗,ηstronglyinL2Qask⟶∞,
∀η∈Dad. Since the norm is weakly lower semicontinuous, we can verify by (120) and (123) that the map Pη:c⟶Jc,η is lower semicontinuous for all η∈Dad. Hence, we know that
(124)J0η=liminfn⟶∞Jcn,η≥Jc∗,η,∀η∈Dad.
But sinceJ0η≤Jc∗,η, we have
(125)J0η=Jc∗,η=infc∈CadJc,η,∀η∈Dad.
By similar arguments, we can prove thatQc is upper semicontinuous for all c∈Cad. Thus, we can also know that there exists η∗∈Dad such that
(126)J0η∗=supη∈DadJ0η.
From (125) and (126), we can conclude that c∗,η∗∈Bad is an optimal pair for the cost (54).This completes the proof.Remark 19.
Assuming thatT>0 is sufficiently small, instead of assuming α and β are large enough, we can also obtain strict convexity and strict concavity of the maps Pη and Qc, respectively.


## 4.3. Necessary Condition of an Optimal Pair

In this subsection, we study the necessary optimality condition to be satisfied by an optimal pair of the minimax optimal control problem with the cost (54).From Theorem12, we know that the solution map q⟶uq from Bad to S0,T is Fréchet differentiable at q∗=c∗,η∗,and the Fréchet derivative of uqat q=q∗ in the direction w=h,l∈L∞Q2,say z=Duq∗w is a unique weak solution of the following equation:
(127)Mγz″+Δ2z+∫0tkt−sΔ2zds=Guq∗,z+c∗+η∗z+h+luq∗inQ,z=Δz=0onΣ,z0=0,z′0=0inΩ,where Guq∗,z is defined in (59). By Equation (127), we introduce the following adjoint equation corresponding to the cost function (54):
(128)Mγp″+Δ2p+∫tTkσ−tΔ2pdσ=Guq∗,p+c∗+η∗p+uq∗−zdinQ,p=Δp=0onΣ,pT=0,p′T=0inΩ.Proposition 20.
Equation (128) admits a unique solution p∈S0,T.Proof.
Changing the variablet to T−t in Equation (128), we can complete the proof by referring to the results in [4, 17].
We now discuss the first-order optimality conditions for the minimax optimal control problem (55) for the quadratic cost function (54).Theorem 21.
Ifα and β in the cost (54) are large enough or T>0 is sufficiently small, then an optimal control c∗∈Cad and a disturbance η∗∈Dad, namely, an optimal pair q∗=c∗,η∗∈Bad satisfying (55) can be given by:
(129)c∗=maxc¯,min−uq∗pα,c¯,η∗=maxη¯,minuq∗pβ,η¯,where p is the weak solution of Equation (128), c¯ and c¯ are constants given in (51), and η¯ and η¯ are constants given in (52).Proof.
Letq∗=c∗,η∗∈Bad be an optimal pair in (55) with the cost (54) and uq∗ be the corresponding weak solution of Equation (53).
From Theorem12, it is clear that the map q=c,η⟶uq is Gâteaux differentiable at q∗=c∗,η∗ in the direction w=h,l∈L∞Q2 with q∗+εw∈Bad for sufficiently small ε>0. Indeed, we have
(130)uq∗+εw−uq∗ε⟶z=zwstronglyinS0,Tasε⟶0+,where z=Duq∗w is a unique weak solution of Equation (127). Therefore, by (130), we can get the Gâteaux derivative of the cost (54) at q=q∗ in the direction w=h,l as follows:
(131)DJc∗,η∗h,l=limε⟶0+Jc∗+εh,η∗+εl−Jc∗,η∗ε=limε⟶0+12∫0Tuq∗+εw+uq∗−2zd,uq∗+εw−uq∗ε2dt+limε⟶0+α2∫0T2c∗,h2+εh22dt−β2∫0T2η∗,l2+εl22dt=∫0Tuq∗−zd,z2dt+α∫0Tc∗,h2dt−β∫0Tη∗,l2dt,where z=Duq∗w is a solution of Equation (127).
Before we proceed to the calculations, by making use of the fact thatG is a self adjoint operator, we can note by Lemma 1 and (109) that
(132)Guq∗,ϕ,φ2=ϕ,−Guq∗,uq∗,φ2+2uq∗,−Gϕ,uq∗,φ2=ϕ,φ,−Guq∗,uq∗2+2−Gϕ,uq∗,φ,uq∗2,−2=ϕ,φ,−Guq∗,uq∗2+2ϕ,uq∗,−Gφ,uq∗−2,2=ϕ,φ,−Guq∗,uq∗2+2ϕ,uq∗,−Gφ,uq∗2=ϕ,Guq∗,φ2,∀φ,ϕ∈V.
We multiply both sides of the weak form of Equation (128) by z which is a solution of Equation (127), and integrate it over 0,T. Then, we have
(133)∫0TMγp″,zV′,Vdt+∫0TΔp,Δz2dt+∫0T∫tTkσ−tΔpdσ,Δz2dt=∫0TGuq∗,p+c∗+η∗p,z2dt+∫0Tuq∗−zd,z2dt.
By Fubini's theorem, we can know that(134)∫0T∫tTkσ−tΔpdσ,Δz2dt=∫0T∫0tkt−sΔzds,Δp2dt=∫0T∫0tkt−sΔ2zds,pV′,Vdt.
From (132), (134) and terminal values of the weak solution p of Equation (128), we can obtain the following by integration by parts of (133):
(135)∫0Tp,Mγz″+Δ2z+∫0tkt−sΔ2zdtV,V′dt=∫0Tp,Guq∗,z+c∗+η∗z2dt+∫0Tuq∗−zd,z2dt.
Sincez is the solution of Equation (127), we can obtain the following from (135):
(136)∫0Tuq∗−zd,z2dt=∫0Th+luq∗,p2dt.
Therefore, we can deduce that (131) and (136) imply
(137)DJc∗,η∗h,l=∫0Tαc∗+uq∗p,h2dt+∫0T−βη∗+uq∗p,l2dt.
Sinceq∗=c∗,η∗∈Bad is an optimal pair in (55), we know
(138)DcJc∗,η∗h≥0,DηJc∗,η∗l≤0,h,l∈L∞Q2.
Therefore, we can obtain the following from (137) and (138):
(139)∫0Tαc∗+uq∗p,h2dt≥0,∫0T−βη∗+uq∗p,l2dt≤0,where h,l∈L∞Q2. By considering the signs of the variations h and l in (139), which depend on c∗ and η∗, respectively, we can deduce from (139) that
(140)c∗=maxc¯,min−uq∗pα,c¯,η∗=maxη¯,minuq∗pβ,η¯.This completes the proof.



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*Source: 1859736-2020-06-16.xml*