# Quadratic Cost Minimax Optimal Control Problems for a Semilinear Viscoelastic Equation with Long Memory

Authors: Jin-soo Hwang
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we study the quadratic cost minimax optimal control problems for a semilinear viscoelastic equation with long memory. A global well-posedness theorem regarding the solutions to its Cauchy problem is given. We formulate the minimax control problem with bilinear control inputs and corresponding disturbances. Under some assumptions, we prove the existence of optimal pairs and give necessary optimality conditions for optimal pairs in some observation cases.


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## Body

## 1. Introduction

We study the minimax optimal control problems for the following viscoelastic equation with long memory:(1)∂2y∂t2−Δy−∫0tkt−sΔysds+yγy=uy+f.As is widely known, equation (1) is considered a typical model of viscoelastic waves with a long-memory damping. By analyzing the above equation with various boundary conditions or additional linear or nonlinear terms, one can study the eventual properties of the viscoelastic dynamics, such as exponential decay or blowup. We can find many research articles in this direction, to name just a few, refer to Cavalcanti et al. [1, 2], Xiao and Liang [3], and Messaoudi [4] and references therein. However, the research on other applications including optimal control or identification problems for the above state equation is few.In this paper, we study the quadratic cost bilinear minimax control problems for equation (1) with the control or disturbance function u. For the bilinear control problems, we refer to Bradley and Lenhart [5–7] in which they studied the bilinear control problems on linear hyperbolic state equations. Recently, Belmiloudi [8] employed the bilinear minimax control framework to study robust control in a linear parabolic state equation. In [9], we extended the results in [8] to a quasilinear beam equation by proving the Fréchet differentiability of the nonlinear solution map.The minimax control methods have been employed by many mathematicians for various control problems (see Arada and Raymond [10], Lasiecka and Triggiani [11], and Li and Yong [12]). As explained in [8, 9], in this paper, the minimax control framework is employed to consider the effects of disturbance (or noise) in control inputs such that a cost function achieves its optimum (minimum) even in the case of the worst disturbances of the system. For the purpose, we replace the bilinear multiplier u in equation (1) by c+η, where c is a control variable that belongs to the admissible control set Uad and η is a disturbance (or noise) that belongs to the admissible disturbance set Vad. We also introduce the following cost function to be minimized within Uad and maximized within Vad:(2)Jc,η=12Cyc+η−Ydℳ2+α2cL20,T2−β2ηL20,T2,where yc+η is the solution of equation (1) with u=c+η, ℳ is a Hilbert space of observation variables, C is a continuous observation operator, Yd∈ℳ is a 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 (2).In this paper, our goal is to find and characterize the optimal controls of the cost function (2) even in the worst disturbances through the control input in equation (1). This leads to the problem of proving the existence of an admissible control c∗∈Uad and disturbance (or noise) η∗∈Vad such that c∗,η∗ is a saddle point of the functional Jc,η of (2), that is,(3)Jc∗,η≤Jc∗,η∗≤Jc,η∗,∀c,η∈Uad×Vad.Then, we derive optimality conditions for suchc∗,η∗ in (3) corresponding to some observation cases. As in [9], we use the term optimal pair to indicate such a saddle point c∗,η∗ in (3).As a main tool to prove the existence of an optimal pairc∗,η∗ satisfying (3), we use the minimax theorem in infinite dimensions given in Barbu and Precupanu [13]. For this, we prove the Fréchet differentiability of the nonlinear solution map and its local Lipschitz continuity.Next, we derive the necessary optimal conditions for some observation cases that should be satisfied by the optimal pairs in these observation cases. To derive these conditions, we refer to the previous results in [5–7] dealing with linear problems. Especially, we deduce the necessary optimality condition for the velocity distribution observation case which is physically meaningful. We propose an appropriate adjoint system for the velocity observation case. To author’s knowledge, this is a newly developed adjoint system to deal with the second-order Volterra-type state equation. Finally, when deducing the optimality condition for the velocity observation case, we use the regularization method proposed by Lions [14] (cf. [15]) to overcome some difficulties. This is another novelty of the paper.


## 2. Preliminaries

### 2.1. Notations and Assumptions

LetΩ be an open, connected, and bounded set of Rnn≤3 with the smooth boundary Γ. We set Q=0,T×Ω and Σ=0,T×Γ for T>0. We consider the following viscoelastic equation with long memory with the Dirichlet boundary condition:(4)∂2y∂t2−Δy−∫0tkt−sΔysds+yγy=uy+f,inQ,y=0,onΣ,y0,x=y0x,∂y∂t0,x=y1x,inΩ,where y0x and y1x are given initial values,f is a given forcing function, k is a memory kernel function, and u is a bilinear forcing control function applied to the system together with displacement y.We denote⋅p=⋅LpΩ,H=L2Ω,andV=H01Ω, where p≥1. We endow the space V with the usual scalar product and norm:(5)ϕ,ψV=∇ϕ,∇ψ2,ϕV=∇ϕ2,∀ϕ,ψ∈V,where ⋅,⋅2 means the inner product on H. Let us denote the topological dual space of V by V′=H−1Ω and the dual pairing between V and V′ by ⋅,⋅V,V′. Then, operator A defined by(6)Aϕ,ψV′,V=ϕ,ψV,∀ϕ,ψ∈V,is a nonnegative self-adjoint operator with domain DA=H2Ω∩V. Also, it is obvious that each natural topological imbedding(7)DA↪V↪H↪V′is continuous and compact.According to Adams [16], we know the following embeddings:(8)H1Ω↪LpΩ,∀p<∞ifn=1,2,p=6ifn=3,(9)H1Ω↪CΩ¯ifn=1,(10)H2Ω↪CΩ¯ifn=2,3.Throughout this paper, the following assumptions will be effective:H1 For the exponent γ in equation (4), we assume that(11)0<γ<∞ifn=1,2,0<γ≤2ifn=3.For the aboveγ, we define the nonlinear function g:R⟶R by gs=sγs. Then, it is easily verified that g∈C1R and g′s=γ+1sγ. For given ϕ∈V, we can deduce from (8) and (11) that the nonlinear operator g⋅:V⟶H, defined by gϕx=gϕx, is well defined.H2 For the memory kernel function k⋅ in equation (4), we assume that(12)k∈C10,T.Now, equation (4) can be rewritten as the abstract initial value problem described by the following second-order semilinear Volterra integrodifferential system:(13)y′′+Ay+∫0tkt−sAyds+gy=uy+f,in0,T,y0=y0,y′0=y1.


### 2.2. Well-Posedness and Continuity of the Solution Map

The Hilbert spaceW0,T of the weak solutions of equation (13) is defined by(14)W0,T=ϕ|ϕ∈L20,T;V,ϕ′∈L20,T;H,ϕ″∈L20,T;V′,equipped with the norm(15)ϕW0,T=ϕL20,T;V2+ϕ′L20,T;H2+ϕ″L20,T;V′21/2,where ϕ′ and ϕ″ denote the first- and second-order time derivatives of ϕ in the sense of distribution.Definition 1.
A functiony is said to be a weak solution of equation (13) if y∈W0,T and y satisfies(16)y′′⋅,φV′,V+y⋅+k∗y⋅,φV+gy⋅,φ2=u⋅y⋅+f⋅,φ2,for allφ∈Vin the sense ofD′0,T,y0=y0,y′0=y1,where D′0,T is the space of distributions in 0,T.Remark 1.
For the existence and uniqueness of weak or strong solutions of equation (13) without the integral term, one can refer to the results of an abstract semilinear wave equation given in Teman ([17], pp. 212–214). In addition, by referring to the results in [18] dealing with the Volterra-type semilinear evolution equations, one can verify the well-posedness of equation (13).
LetX be a Banach space. Set(17)Cw0,T;X=f∈L∞0,T;X|f⋅,ξX,X′∈C0,T,∀ξ∈X′.Lemma 1.
Assume that(18)y0,y1,u,f∈V×H×L20,T×L20,T;Hand y is a corresponding weak solution of equation (13). Then, we can verify that(19)y∈Cw0,T;V,y′∈Cw0,T;H.Proof.
By regarding−gy+uy+f in equation (13) as ft,y,y′ in [18] and noting (18), we can use the result of [18] to obtain(20)y∈W0,T∩L∞0,T;V∩W1,∞0,T;H.
Then, as shown in [18] (cf. Lions and Magenes [19], pp. 275–278), we can obtain (19) from (20). This completes the proof.
The following lemma is used frequently and importantly in this paper.Lemma 2.
Lety be a weak solution of equation (13) with condition (18). Then, for each t∈0,T, we have the following energy equality:(21)y′t22+ytV2+2γ+2ytγ+2γ+2+2k∗yt,ytV=2∫0tk′∗y,yVds+2∫0tk0yV2ds+2∫0tuy,y′2ds+2∫0tf,y′2ds+y122+y0V2+2γ+2y0γ+2γ+2.Proof.
By (19) and the uniform boundedness theorem, yt∈V and y′t∈H for each t∈0,T. Thus, we know that all terms in (21) are meaningful. From (18) and (20), we know that −gy+uy+f∈L20,T;H. Thus, by applying the result in Proposition 2.1 of [18] to equation (13), we obtain(22)y′t22+ytV2+2k∗yt,ytV=2∫0tk′∗y,yVds+2∫0tk0yV2ds−2∫0tgy,y′2ds+2∫0tuy,y′2ds+2∫0tf,y′2ds+y122+y0V2.
Since(23)∫0tgyy′2ds=1γ+2ddtytγ+2γ+2,we can combine (22) and (23) to obtain (21).
This completes the proof.
With the help of (21) or (22), we can address the following theorem.
Throughout this paper, we useC as a generic constant and omit the integral variables in any definite integrals without confusion.Theorem 1.
For given condition (18), let y be the corresponding weak solution of equation (13). Then, we can assure that y∈C0,T;V∩C10,T;H. Moreover, the solution mapping p=y0,y1,f,u⟶yp of P≡V×H×L20,T;H×L20,T into S0,T≡W0,T∩C0,T;V∩C10,T;H is locally Lipschitz continuous. Indeed, for each p1=y01,y11,f1,u1∈P and p2=y02,y12,f2,u2∈P, the following is satisfied:(24)yp1−yp2S0,T≤Cy01−y02V2+y11−y1222+f1−f2L20,T;H2+u1−u2L20,T21/2≡Cp1−p2P,where C>0 is a constant depending on the data.Proof.
Referring to Dautray and Lions ([20], pp. 578–581), we can make use of (22) to show that t⟶y′t22+ytV2 is continuous, which implies the following:(25)y∈C0,T;V∩C10,T;H.
Thus, every weak solutiony of equation (13) with data condition (18) exists in S0,T. From this result, we show inequality (24): we denote y1−y2≡yp1−yp2 by ϕ. Then, from equation (13), we can know that ϕ satisfies the following in the weak sense:(26)ϕ″+Aϕ+∫0tkt−sAϕds+γ+1By1,y2ϕ=u1ϕ+u1−u2y2+f1−f2,in0,T,ϕ0=y01−y02,ϕ′0=y11−y12,where(27)By1,y2=∫01θy1+1−θy2γdθ.
By usingyi∈S0,Ti=1,2, (8), (11), and the Hölder inequality, we note that(28)γ+1By1,y2ϕ2≤Cy1+y2γϕ2≤Cy1V+y2VγϕV≤Cy1C0,T;V+y2C0,T;VγϕV≤Cy1S0,T+y2S0,TγϕV.
To obtain the estimation forϕ in equation (13), we apply energy equality (22) to equation (26). Then, the energy equality for ϕ can be given by(29)ϕ′t22+ϕtV2=−2k∗ϕt,ϕtV+2∫0tk′∗ϕ,ϕVds+2∫0tk0ϕV2ds−2γ+1∫0tBy1,y2ϕ,ϕ′2ds+2∫0tu1ϕ+u1−u2y2,ϕ′2ds+2∫0tf1−f2,ϕ′2ds+ϕ122+ϕ0V2.
By (28), we can verify that(30)2γ+1∫0tBy1,y2ϕ,ϕ′2ds≤Cy1S0,T+y2S0,Tγ∫0tϕVϕ′2ds≤C∫0tϕV2+ϕ′22ds.
By estimating other terms in the right-hand side of (29) as in [18], we can get(31)ϕ′t22+ϕtV2≤C∫0tu12+1ϕ′22+ϕV2ds+ϕ′022+ϕ0V2+f1−f2L20,T;H2+u1−u2y2L20,T;H2.
Here, we note that(32)u1−u2y2L20,T;H2≤u1−u2L20,T2y2C0,T;H2≤Cy2S0,T2u1−u2L20,T2.
Thus, plugging (32) to (31) and applying Gronwall’s lemma, we have the following estimation:(33)ϕ′t22+ϕtV2≤CexpCu1L20,T2+1ϕ′022+ϕ0V2+f1−f2L20,T;H2+u1−u2L20,T2≡Cp1−p2P2,which immediately implies(34)ϕC0,T;V∩C10,T;H≤Cp1−p2P.
SinceA∈ℒV,V′ is an isomorphism, by conducting similar estimations in equation (26), we can obtain from (34) that(35)ϕ″L20,T;V′≤Cp1−p2P.
Hence, by (34) and (35), we can prove (24).
This completes the proof.Remark 2.
From Theorem1, for fixed y0,y1,f∈V×H×L20,T;H, we can define the continuous solution map from the term u∈L20,T to yp∈S0,T satisfying equation (13). Indeed, for each pi=y0,y1,f,ui∈Pi=1,2, we have(36)yp1−yp2S0,T≤Cu1−u2L20,T.
Thus, from now on, to emphasize the fact that the only varying variable of the weak solution of equation (37) is the bilinear multiplier u, we use the notation yu where yu satisfies(37)y″u+Ayu+∫0tkt−sAyuds+gyu=uyu+f,in0,T,yu;0=y0,y′u;0=y1.
Furthermore, we present the following weakly continuous results to study the existence of the optimal pair in the subsequent section. For this, we need the following lemma.Lemma 3.
(see Simon [21]). Let X,Y, and Z be Banach spaces such that each embedding X⊂Y⊂Z is 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.Proposition 1.
For fixedy0,y1,f∈V×H×L20,T;H, the solution map from u∈L20,T to yu∈W0,T of equation (37) is weakly continuous.Proof.
Letu∈L20,T, and let un⊂L20,T be a bounded sequence such that(38)un⟶uweakly inL20,T,asn⟶∞.
From now on, each stateyn=yun is a solution of(39)yn″+Ayn+∫0tkt−sAynds+gyn=unyn+f,in0,T,yn0=y0,yn′0=y1.
From Theorem1, we have(40)ynS0,T≤Cy0,y1,f,unP,which implies that yn remains in a bounded set of W0,T∩W1,∞0,T;V,H. Therefore, we can find a subsequence of yn, say again yn, and find y∈W0,T∩W1,∞0,T;V,H such that(41)yn⟶yweakly inW0,T,asn⟶∞,(42)yn⟶yweakly star inL∞0,T;V,asn⟶∞,(43)yn′⟶y′weakly star inL∞0,T;H,asn⟶∞.
Since the embeddingV↪H is compact, we can apply Lemma 3 to (42) and (43) with X=VandY=Z=H to verify that(44)ynis a precompact set inC0,T;H.
Thus, we can find a subsequence ofyn, if necessary, still denoted by yn, such that(45)yn⟶ystrongly inC0,T;H,asn⟶∞.
For any givenϕ∈L20,T;V, using notation (27), we can deduce(46)gyn−gy,ϕ2=γ+1Byn,yyn−y,ϕ2≤γ+1yn+yγyn−y,ϕ2≤γ+1yn−y2yn+yγϕ2≤Cyn−y2yn+y3γγϕ6≤Cyn−y2ynV+yVγϕV≤CynC0,T;V+yL∞0,T;Vγyn−y2ϕV.
Therefore, from (45) and (46), we can readily find a subsequence of yn, if necessary, still denoted by itself, such that(47)gyn⟶gystrongly inL20,T;V′,asn⟶∞.
From (38) and (45), we may extract a subsequence of un, denoted again by un, such that(48)unyn⟶uyweakly inL20,T;H.
Replacingyn by ynk in equation (39), if necessary, and letting k⟶∞, we can conclude by the standard arguments as in Dautray and Lions ([20], pp.561–565) that the limit y is a weak solution of(49)y″+Ay+∫0tkt−sAyds+gy=uy+f,in0,T,y0=y0,y′0=y1.
Moreover, by the uniqueness of the weak solutions, we can conclude thaty=yu in W0,T, which implies that yun⟶yu weakly in W0,T.
This completes the proof.


## 2.1. Notations and Assumptions

LetΩ be an open, connected, and bounded set of Rnn≤3 with the smooth boundary Γ. We set Q=0,T×Ω and Σ=0,T×Γ for T>0. We consider the following viscoelastic equation with long memory with the Dirichlet boundary condition:(4)∂2y∂t2−Δy−∫0tkt−sΔysds+yγy=uy+f,inQ,y=0,onΣ,y0,x=y0x,∂y∂t0,x=y1x,inΩ,where y0x and y1x are given initial values,f is a given forcing function, k is a memory kernel function, and u is a bilinear forcing control function applied to the system together with displacement y.We denote⋅p=⋅LpΩ,H=L2Ω,andV=H01Ω, where p≥1. We endow the space V with the usual scalar product and norm:(5)ϕ,ψV=∇ϕ,∇ψ2,ϕV=∇ϕ2,∀ϕ,ψ∈V,where ⋅,⋅2 means the inner product on H. Let us denote the topological dual space of V by V′=H−1Ω and the dual pairing between V and V′ by ⋅,⋅V,V′. Then, operator A defined by(6)Aϕ,ψV′,V=ϕ,ψV,∀ϕ,ψ∈V,is a nonnegative self-adjoint operator with domain DA=H2Ω∩V. Also, it is obvious that each natural topological imbedding(7)DA↪V↪H↪V′is continuous and compact.According to Adams [16], we know the following embeddings:(8)H1Ω↪LpΩ,∀p<∞ifn=1,2,p=6ifn=3,(9)H1Ω↪CΩ¯ifn=1,(10)H2Ω↪CΩ¯ifn=2,3.Throughout this paper, the following assumptions will be effective:H1 For the exponent γ in equation (4), we assume that(11)0<γ<∞ifn=1,2,0<γ≤2ifn=3.For the aboveγ, we define the nonlinear function g:R⟶R by gs=sγs. Then, it is easily verified that g∈C1R and g′s=γ+1sγ. For given ϕ∈V, we can deduce from (8) and (11) that the nonlinear operator g⋅:V⟶H, defined by gϕx=gϕx, is well defined.H2 For the memory kernel function k⋅ in equation (4), we assume that(12)k∈C10,T.Now, equation (4) can be rewritten as the abstract initial value problem described by the following second-order semilinear Volterra integrodifferential system:(13)y′′+Ay+∫0tkt−sAyds+gy=uy+f,in0,T,y0=y0,y′0=y1.


## 2.2. Well-Posedness and Continuity of the Solution Map

The Hilbert spaceW0,T of the weak solutions of equation (13) is defined by(14)W0,T=ϕ|ϕ∈L20,T;V,ϕ′∈L20,T;H,ϕ″∈L20,T;V′,equipped with the norm(15)ϕW0,T=ϕL20,T;V2+ϕ′L20,T;H2+ϕ″L20,T;V′21/2,where ϕ′ and ϕ″ denote the first- and second-order time derivatives of ϕ in the sense of distribution.Definition 1.
A functiony is said to be a weak solution of equation (13) if y∈W0,T and y satisfies(16)y′′⋅,φV′,V+y⋅+k∗y⋅,φV+gy⋅,φ2=u⋅y⋅+f⋅,φ2,for allφ∈Vin the sense ofD′0,T,y0=y0,y′0=y1,where D′0,T is the space of distributions in 0,T.Remark 1.
For the existence and uniqueness of weak or strong solutions of equation (13) without the integral term, one can refer to the results of an abstract semilinear wave equation given in Teman ([17], pp. 212–214). In addition, by referring to the results in [18] dealing with the Volterra-type semilinear evolution equations, one can verify the well-posedness of equation (13).
LetX be a Banach space. Set(17)Cw0,T;X=f∈L∞0,T;X|f⋅,ξX,X′∈C0,T,∀ξ∈X′.Lemma 1.
Assume that(18)y0,y1,u,f∈V×H×L20,T×L20,T;Hand y is a corresponding weak solution of equation (13). Then, we can verify that(19)y∈Cw0,T;V,y′∈Cw0,T;H.Proof.
By regarding−gy+uy+f in equation (13) as ft,y,y′ in [18] and noting (18), we can use the result of [18] to obtain(20)y∈W0,T∩L∞0,T;V∩W1,∞0,T;H.
Then, as shown in [18] (cf. Lions and Magenes [19], pp. 275–278), we can obtain (19) from (20). This completes the proof.
The following lemma is used frequently and importantly in this paper.Lemma 2.
Lety be a weak solution of equation (13) with condition (18). Then, for each t∈0,T, we have the following energy equality:(21)y′t22+ytV2+2γ+2ytγ+2γ+2+2k∗yt,ytV=2∫0tk′∗y,yVds+2∫0tk0yV2ds+2∫0tuy,y′2ds+2∫0tf,y′2ds+y122+y0V2+2γ+2y0γ+2γ+2.Proof.
By (19) and the uniform boundedness theorem, yt∈V and y′t∈H for each t∈0,T. Thus, we know that all terms in (21) are meaningful. From (18) and (20), we know that −gy+uy+f∈L20,T;H. Thus, by applying the result in Proposition 2.1 of [18] to equation (13), we obtain(22)y′t22+ytV2+2k∗yt,ytV=2∫0tk′∗y,yVds+2∫0tk0yV2ds−2∫0tgy,y′2ds+2∫0tuy,y′2ds+2∫0tf,y′2ds+y122+y0V2.
Since(23)∫0tgyy′2ds=1γ+2ddtytγ+2γ+2,we can combine (22) and (23) to obtain (21).
This completes the proof.
With the help of (21) or (22), we can address the following theorem.
Throughout this paper, we useC as a generic constant and omit the integral variables in any definite integrals without confusion.Theorem 1.
For given condition (18), let y be the corresponding weak solution of equation (13). Then, we can assure that y∈C0,T;V∩C10,T;H. Moreover, the solution mapping p=y0,y1,f,u⟶yp of P≡V×H×L20,T;H×L20,T into S0,T≡W0,T∩C0,T;V∩C10,T;H is locally Lipschitz continuous. Indeed, for each p1=y01,y11,f1,u1∈P and p2=y02,y12,f2,u2∈P, the following is satisfied:(24)yp1−yp2S0,T≤Cy01−y02V2+y11−y1222+f1−f2L20,T;H2+u1−u2L20,T21/2≡Cp1−p2P,where C>0 is a constant depending on the data.Proof.
Referring to Dautray and Lions ([20], pp. 578–581), we can make use of (22) to show that t⟶y′t22+ytV2 is continuous, which implies the following:(25)y∈C0,T;V∩C10,T;H.
Thus, every weak solutiony of equation (13) with data condition (18) exists in S0,T. From this result, we show inequality (24): we denote y1−y2≡yp1−yp2 by ϕ. Then, from equation (13), we can know that ϕ satisfies the following in the weak sense:(26)ϕ″+Aϕ+∫0tkt−sAϕds+γ+1By1,y2ϕ=u1ϕ+u1−u2y2+f1−f2,in0,T,ϕ0=y01−y02,ϕ′0=y11−y12,where(27)By1,y2=∫01θy1+1−θy2γdθ.
By usingyi∈S0,Ti=1,2, (8), (11), and the Hölder inequality, we note that(28)γ+1By1,y2ϕ2≤Cy1+y2γϕ2≤Cy1V+y2VγϕV≤Cy1C0,T;V+y2C0,T;VγϕV≤Cy1S0,T+y2S0,TγϕV.
To obtain the estimation forϕ in equation (13), we apply energy equality (22) to equation (26). Then, the energy equality for ϕ can be given by(29)ϕ′t22+ϕtV2=−2k∗ϕt,ϕtV+2∫0tk′∗ϕ,ϕVds+2∫0tk0ϕV2ds−2γ+1∫0tBy1,y2ϕ,ϕ′2ds+2∫0tu1ϕ+u1−u2y2,ϕ′2ds+2∫0tf1−f2,ϕ′2ds+ϕ122+ϕ0V2.
By (28), we can verify that(30)2γ+1∫0tBy1,y2ϕ,ϕ′2ds≤Cy1S0,T+y2S0,Tγ∫0tϕVϕ′2ds≤C∫0tϕV2+ϕ′22ds.
By estimating other terms in the right-hand side of (29) as in [18], we can get(31)ϕ′t22+ϕtV2≤C∫0tu12+1ϕ′22+ϕV2ds+ϕ′022+ϕ0V2+f1−f2L20,T;H2+u1−u2y2L20,T;H2.
Here, we note that(32)u1−u2y2L20,T;H2≤u1−u2L20,T2y2C0,T;H2≤Cy2S0,T2u1−u2L20,T2.
Thus, plugging (32) to (31) and applying Gronwall’s lemma, we have the following estimation:(33)ϕ′t22+ϕtV2≤CexpCu1L20,T2+1ϕ′022+ϕ0V2+f1−f2L20,T;H2+u1−u2L20,T2≡Cp1−p2P2,which immediately implies(34)ϕC0,T;V∩C10,T;H≤Cp1−p2P.
SinceA∈ℒV,V′ is an isomorphism, by conducting similar estimations in equation (26), we can obtain from (34) that(35)ϕ″L20,T;V′≤Cp1−p2P.
Hence, by (34) and (35), we can prove (24).
This completes the proof.Remark 2.
From Theorem1, for fixed y0,y1,f∈V×H×L20,T;H, we can define the continuous solution map from the term u∈L20,T to yp∈S0,T satisfying equation (13). Indeed, for each pi=y0,y1,f,ui∈Pi=1,2, we have(36)yp1−yp2S0,T≤Cu1−u2L20,T.
Thus, from now on, to emphasize the fact that the only varying variable of the weak solution of equation (37) is the bilinear multiplier u, we use the notation yu where yu satisfies(37)y″u+Ayu+∫0tkt−sAyuds+gyu=uyu+f,in0,T,yu;0=y0,y′u;0=y1.
Furthermore, we present the following weakly continuous results to study the existence of the optimal pair in the subsequent section. For this, we need the following lemma.Lemma 3.
(see Simon [21]). Let X,Y, and Z be Banach spaces such that each embedding X⊂Y⊂Z is 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.Proposition 1.
For fixedy0,y1,f∈V×H×L20,T;H, the solution map from u∈L20,T to yu∈W0,T of equation (37) is weakly continuous.Proof.
Letu∈L20,T, and let un⊂L20,T be a bounded sequence such that(38)un⟶uweakly inL20,T,asn⟶∞.
From now on, each stateyn=yun is a solution of(39)yn″+Ayn+∫0tkt−sAynds+gyn=unyn+f,in0,T,yn0=y0,yn′0=y1.
From Theorem1, we have(40)ynS0,T≤Cy0,y1,f,unP,which implies that yn remains in a bounded set of W0,T∩W1,∞0,T;V,H. Therefore, we can find a subsequence of yn, say again yn, and find y∈W0,T∩W1,∞0,T;V,H such that(41)yn⟶yweakly inW0,T,asn⟶∞,(42)yn⟶yweakly star inL∞0,T;V,asn⟶∞,(43)yn′⟶y′weakly star inL∞0,T;H,asn⟶∞.
Since the embeddingV↪H is compact, we can apply Lemma 3 to (42) and (43) with X=VandY=Z=H to verify that(44)ynis a precompact set inC0,T;H.
Thus, we can find a subsequence ofyn, if necessary, still denoted by yn, such that(45)yn⟶ystrongly inC0,T;H,asn⟶∞.
For any givenϕ∈L20,T;V, using notation (27), we can deduce(46)gyn−gy,ϕ2=γ+1Byn,yyn−y,ϕ2≤γ+1yn+yγyn−y,ϕ2≤γ+1yn−y2yn+yγϕ2≤Cyn−y2yn+y3γγϕ6≤Cyn−y2ynV+yVγϕV≤CynC0,T;V+yL∞0,T;Vγyn−y2ϕV.
Therefore, from (45) and (46), we can readily find a subsequence of yn, if necessary, still denoted by itself, such that(47)gyn⟶gystrongly inL20,T;V′,asn⟶∞.
From (38) and (45), we may extract a subsequence of un, denoted again by un, such that(48)unyn⟶uyweakly inL20,T;H.
Replacingyn by ynk in equation (39), if necessary, and letting k⟶∞, we can conclude by the standard arguments as in Dautray and Lions ([20], pp.561–565) that the limit y is a weak solution of(49)y″+Ay+∫0tkt−sAyds+gy=uy+f,in0,T,y0=y0,y′0=y1.
Moreover, by the uniqueness of the weak solutions, we can conclude thaty=yu in W0,T, which implies that yun⟶yu weakly in W0,T.
This completes the proof.


## 3. Quadratic Cost Minimax Control Problems

In this section, we study the quadratic cost minimax optimal control problems for equation (52). Let the following be the set of the admissible controls:(50)Uad=c∈L20,T|cl≤c≤cua.e.in0,T,where cl and cu are lower and upper bounds of the control variables, respectively. Let the following be the set of the admissible disturbance or noises:(51)Vad=η∈L20,T|ηl≤η≤ηua.e.in0,T,where ηl and ηu are lower and upper bounds of the disturbance (noise) variables, respectively. For simplicity, let ℱad be a product space defined by ℱad=Uad×Vad.Using Theorem1, for fixed y0,y1,f∈V×H×L20,T;H, we can uniquely define the solution map ℱad⟶S0,T, which maps from c,η∈ℱad via q=c+η to the weak solution yq∈S0,T, where yq satisfies(52)y″q+Ayq+∫0tkt−sAyqds+gyq=c+ηyq+f,in0,T,yq;0=y0,y′q;0=y1.The weak solutionyq is called to be the state of the control system.The quadratic cost function associated with control system (52) is given by(53)Jc,η=12Cyq−Ydℳ2+α2cL20,T2−β2ηL20,T2,where ℳ is a Hilbert space of observation variables, C∈ℒW0,T,ℳ is a continuous linear operator, that is to say, observer, Yd∈ℳ is a 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 (53).As indicated in the introduction, the minimax optimal control problem can be summarized as follows:(i)
Find an admissible controlc∗∈Uad and a disturbance (or noise) η∗∈Vad such that c∗,η∗ is a saddle point of the functional Jc,η of (53), that is,(54)Jc∗,η≤Jc∗,η∗≤Jc,η∗,∀c,η∈ℱad.In this paper, we call such a pairc∗,η∗ in (54) as the optimal pair for the minimax optimal control problem with cost (53). To show the existence of the above saddle point c∗,η∗ (optimal pair), we need to show the solution map is differentiable in some sense. Then, with the assumptions on the weight constants α,β and the exponent γ, we utilize the arguments in Barbu and Precupanu [13] to show the existence of the optimal pairs.(ii)
Characterizec∗,η∗ (optimality condition): in characterizing these optimal pairs, we introduce an appropriate adjoint system corresponding to the observed case and deduce the necessary optimal conditions through the variational inequality.


### 3.1. Differentiability of the Nonlinear Solution Map

In this section, we address the Fréchet differentiability of the nonlinear solution map, which is desirable for many applications.For our study, we define the Fréchet differentiability of the nonlinear solution map as follows.Definition 2.
The solution mapu⟶yu of L20,T into S0,T is said to be Fréchet differentiable on L20,T if for any u∈L20,T, there exists Tu∈ℒL20,T,S0,T such that, for any w∈L20,T,(55)yu+w−yu−TuwS0,TwL20,T⟶0,aswL20,T⟶0.
The operatorTu is called the Fréchet derivative of y at u, which we denote by Dyu. Tuw=Dyuw∈S0,T is called the Fréchet derivative of y at u in the direction of w∈L20,T.Theorem 2.
The solution mapu⟶yu is Fréchet differentiable on L20,T, and the Fréchet derivative of yu at u in the direction w∈L20,T, that is to say z=Dyuw, is given by the solution of(56)z″+Az+∫0tkt−sAzds+γ+1yuγz=uz+wyu,in0,T,z0=0,z′0=0.Proof.
Let(57)Byuz≔yuγz.
Then, by (11) and Theorem 1, we can verify(58)Byuz2≤yu3γγz6≤CyuVγzV≤CyuC0,T;VγzV≤CyuS0,TγzV≤Cy0,y1,f,uPγzV.
Hence, from (58), we see that(59)Byu∈ℒV,H∩ℒH,V′.
Sincew∈L20,T and yu∈S0,T, we know that wyu∈L20,T;H. By (59), we can use the results in [18] to verify that equation (56) admits a unique weak solution z∈S0,T. And by Theorem 1, we can know that the weak solution z=zw of equation (56) satisfies(60)zwS0,T≤CwyuL20,T;H≤CwL20,TyuC0,T;H≤CyuS0,TwL20,T≤Cy0,y1,f,uPwL20,T.
Hence, from (60), the mapping w∈L20,T↦zw∈S0,T is linear and bounded.
We setφ=yu+w−yu−z. Then, using notation (27) and noting(61)gyu+w−gyu−γ+1yuγz=γ+1Byu+w,yuyu+w−yu−γ+1yuγz=γ+1Byu+w,yuφ+γ+1Byu+w,yu−yuγz,we know that φ satisfies(62)φ″+Aφ+∫0tkt−sAφds+γ+1Byu+w,yuφ=u+wφ+wz+I,in0,T,φ0=0,φ′0=0,in the weak sense, where(63)I=−γ+1Byu+w,yu−yuγz.
Arguing as in the proof of Theorem1, we can know that the weak solution φ of equation (62) exists, and the following is fulfilled:(64)φS0,T≤Cwz+IL20,T;H.
By (60), we can deduce(65)wzL20,T;H≤wL20,TzC0,T;H≤CwL20,TzS0,T≤CwL20,T2,Byu+w,yu−yuγz2≤Byu+w,yu−yuγ3z6≤CByu+w,yu−yuγ3zV≤CByu+w,yu−yuγ3zC0,T;V≤CzS0,TByu+w,yu−yuγ3≤CwL20,TByu+w,yu−yuγ3.
By Theorem1, we can see that(66)yu+w−yuS0,T≤CwL20,T.
Thus, ifwL20,T⟶0, then by (66), we can extract a subsequence, still denoted by yu+w, such that(67)Byu+w,yu−yuγ⟶0,a.e.inQ.
And from (11), we know that(68)Byu+w,yu−yuγ3≤8yu+w+yu3γ∈L1Q.
Thus, by the Lebesgue dominated convergence theorem through (67) and (68), we can obtain(69)Byu+w,yu−yuγL3Q⟶0aswL20,T⟶0.
Hence, from (64) to (69), we can obtain(70)φS0,T≤Cwz+IL20,T;H≤CwzL20,T;H+IL20,T;H≤CwL20,TwL20,T+Byu+w,yu−yuγL3Q,which immediately implies that φS0,T=owL20,T as wL20,T⟶0.
This completes the proof.
As before, for simplicity, we denotec+η and c∗+η∗ by q and q∗, respectively. From Theorem 2, we know that the map c,η⟶yq of ℱad to S0,T is Fréchet differentiable at c∗,η∗, and the Fréchet derivative of yq at c∗,η∗ in the direction w=h,l∈L20,T2, say z=Dyq∗w, is a unique weak solution of the following problem:(71)z″+Az+∫0tkt−sAzds+γ+1yq∗γz=q∗z+h+lyq∗,in0,T,z0=0,z′0=0.
The following results will be used in showing the existence of optimal pairs.Proposition 2.
Assume that the exponentγ≥1. For given w∈L20,T, the Fréchet derivative Dyuw is locally Lipschitz continuous on L20,T. Indeed, it is satisfied that(72)Dyu1w−Dyu2wS0,T≤Cu1−u2L20,TwL20,T,where C>0 is a constant depending on the data.Proof.
Letzi=Dyuiwi=1,2 be the weak solutions of equation (56) corresponding to uii=1,2. If we set ϕ=z1−z2, then we can know that ϕ satisfies(73)ϕ″+Aϕ+∫0tkt−sAϕds+γ+1yu1γϕ=u1ϕ+I1+I2,in0,T,ϕ0=0,ϕ′0=0,in the weak sense, where(74)I1=−γγ+1∫01θyu1+1−θyu2γ−1dθyu1−yu2z2,I2=u1−u2z2+wyu1−yu2.
As before, arguing as in the proof of Theorem1, we can deduce that(75)ϕS0,T≤CI1+I2L20,T;H.
In estimatingI1, the case of γ=1 will be clarified as we will see later. Thus, we restrict to the case of γ>1 to estimate I1. By (11), Theorem 1, and (60), we can obtain(76)I12≤γγ+1yu1+yu2γ−1yu1−yu2z22≤Cyu1+yu2γ−16yu1−yu26z26≤Cyu1V+yu2Vγ−1yu1−yu2Vz2V≤Cyu1C0,T;V+yu2C0,T;Vγ−1yu1−yu2C0,T;Vz2C0,T;V≤Cu1−u2L20,TwL20,T,(77)I2L20,T;H≤u1−u2z2L20,T;H+wyu1−yu2L20,T;H≤u1−u2L20,Tz2C0,T;H+wL20,Tyu1−yu2C0,T;H≤Cu1−u2L20,Tz2S0,T+wL20,Tyu1−yu2S0,T≤Cu1−u2L20,TwL20,T.
From (75) to (77), we can get the following:(78)ϕS0,T≤Cu1−u2L20,TwL20,T.
This completes the proof.


### 3.2. Existence of the Optimal Pairs

By the minimax theorem in infinite dimensions in Barbu and Precupanu [13], we study the existence of the optimal pairs in the underlying control system with quadratic cost function (53).Theorem 3.
Assume that the exponentγ≥1 in control system equation (52). Then, for sufficiently large α and β in (53), there exists c∗,η∗∈ℱad satisfying (54).
In order to prove this theorem, we mainly illustrate the following as in [8] (cf. [9]):(i)
For sufficiently largeα and β in (53), the maps c⟶Jc,η and η⟶Jc,η are convex and concave for all η∈Vad and for all c∈Uad, respectively(ii)
The mapsc⟶Jc,η and η⟶Jc,η are lower and upper semicontinuous for all η∈Vad and for all c∈Uad, respectivelyProof.
(i)
Since the mapc⟶yc+η where η is fixed is Fréchet differentiable, the map c⟶Jc,η is also Fréchet differentiable. Let DcJc1,ηc1−c2 and DcJc2,ηc1−c2 be the Fréchet derivatives of the map c⟶Jc,η at c1 and c2 in the direction c1−c2, respectively, where η holds fixed. To verify the convexity of the map c⟶Jc,η, it is sufficient to show that(79)DcJc1,η−DcJc2,ηc1−c2≥0,∀c1,c2∈Uad,for any fixedη∈Vad. (79) means(80)Cyc1+η−Yd,CDcyc1+ηc1−c2ℳ+αc1,c1−c2L20,T−Cyc2+η−Yd,CDcyc2+ηc1−c2ℳ−αc2,c1−c2L20,T≥0,∀c1,c2∈Uad,whereDcyci+ηc1−c2i=1,2 are weak solutions of equation (71), in which q∗z+h+lyp is replaced by ci+ηz+c1−c2yci+ηi=1,2, respectively. We can deduce that (80) is equivalent to(81)Cyc1+η−yc2+η,CDcyc1+ηc1−c2ℳ+Cyc2+η−Yd,CDcyc1+ηc1−c2−Dcyc2+ηc1−c2ℳ+αc1−c2L20,T2≥0,∀c1,c2∈Uad.By Theorem1, (60), and Proposition 2, we can have the following estimations:(82)Cyc1+η−yc2+η,CDcyc1,ηc1−c2ℳ≤Cyc1+η−yc2+ηℳCDcyc1+ηc1−c2ℳ≤Cyc1+η−yc2+ηS0,TDcyc1+ηc1−c2S0,T≤Cc1−c2L20,T2,(83)Cyc2+η−Yd,CDcyc1+ηc1−c2−Dcyc2+ηc1−c2ℳ≤Cyc2+η−YdℳCDcyc1+ηc1−c2−Dcyc2+ηc1−c2ℳ≤CCyc2+ηℳ+Ydℳ×Dcyc1+ηc1−c2−Dcyc2+ηc1−c2S0,T≤CCyc2+ηℳ+Ydℳc1−c2L20,T2≤Cc1−c2L20,T2.From (81) to (83), we can verify that there exists sufficiently large αlP,ℱad,Yd,C,ℳ such that, for any α>αlP,ℱad,Yd,C,ℳ, inequality (79) holds. Thus, the map c⟶Jc,η is convex.Similarly, we can show that there exists sufficiently largeβlP,ℱad,Yd,C,ℳ such that the following inequality(84)DηJc,η1−DηJc,η2η1−η2≤0,∀η1,η2∈Vad,is satisfied for any fixedc∈Uad and β>βlP,ℱad,Yd,C,ℳ. This also indicates the concavity of the map η⟶Jc,η.(ii)
Sincecn is bounded in L20,T, we can extract a subsequence cnk⊂cn such that(85)cnk⟶c∗weakly inL20,T,ask⟶∞.
Then, sinceC is a continuous linear operator on W0,T, by Proposition 1, we obtain(86)Cycnk+η⟶Cyc∗+ηweakly inℳ,ask⟶∞,∀η∈Vad. Since the norm is weakly lower semicontinuous, we can see from (85) and (86) that the map c⟶Jc,η is lower semicontinuous for all η∈Vad. By similar arguments, we can prove that the map η⟶Jc,η is upper semicontinuous for all c∈Uad.
Next, we prove the existence of an optimal pairc∗,η∗∈ℱad: let cn⊂Uad be a minimizing sequence of Jc,η where η∈Vad is fixed. Thus,(87)liminfn⟶∞Jcn,η=infc∈UadJc,η.
Then, by (i) and (ii), we know that(88)J0η=liminfn⟶∞Jcn,η≥Jc∗,η,∀η∈Vad,but since J0η≤Jc∗,η, we have(89)J0η=Jc∗,η=infc∈UadJc,η,∀η∈Vad.
Similarly, we also know that there existsη∗∈Vad such that(90)J0η∗=supη∈VadJ0η.
From (89) and (90), we can conclude that c∗,η∗∈ℱad is an optimal pair for cost (53).
This completes the proof.


### 3.3. Necessary Conditions of Optimal Pairs

In this section, we study the necessary optimality conditions that have to be satisfied by each optimal pair of the minimax optimal control problem with cost (53) in which the following two types of observations are considered:(1)
We takeℳ1=L20,T;H×H and C1∈ℒW0,T,ℳ1 and observe C1yq=yq,yq;T∈L20,T;H×H(2)
We takeℳ2=L20,T;H and C2∈ℒW0,T,ℳ2 and observe C2yq=y′q∈L20,T;HSinceyq∈W0,T⊂C0,T;H, the above observations are meaningful.


#### 3.3.1. Case of Distributive and Terminal Values’ ObservationsC1∈ℒW0,T,ℳ1

We consider the cost functional expressed by(91)Jc,η=12yq−ydL20,T;H2+12yq;T−ydT22+α2cL20,T2−β2ηL20,T2,where yd∈L20,T;H and ydT∈H are desired values. Let c∗,η∗∈L20,T2q∗=c∗+η∗ be an optimal pair subject to equation (52) and quadratic cost (91). Now, we are about to formulate the following adjoint equation corresponding to cost (91) and equation (52) in which c,η with q=c+η is replaced by c∗,η∗ with q∗=c∗+η∗:(92)p″+Ap+∫tTkσ−tApdσ+γ+1yq∗γp=c∗+η∗p+yq∗−yd,in0,T,pT=0,p′T=−yq∗;T+ydT.Proposition 3.
Equation (92) admits a unique weak solution p∈S0,T.Proof.
From the observation conditionsyq∗−yd∈L20,T;H and yq∗;T−ydT∈H and (59), we can refer to the results in [18] (cf. Dautray and Lions [20], pp.655–659) to ensure that equation (92), after reversing the direction of time t⟶T−t, admits a unique weak solution p∈S0,T. This completes the proof.
We now discuss the first-order optimality conditions for minimax optimal control problem (54) for quadratic cost function (91).Theorem 4.
Ifα and β in cost (91) are large enough and the exponent γ≥1, then an optimal control c∗∈Uad and a disturbance η∗∈Vad, namely, an optimal pair c∗,η∗∈ℱad satisfying (54), can be given by(93)c∗=maxcl,min−yq∗,p2α,cu,η∗=maxηl,minyq∗,p2β,ηu,where p is the weak solution of equation (92).Proof.
By the assumptions of Theorem4, we can verify through Theorem 3 that there exists an optimal pair in (54) with cost (91). Let c∗,η∗∈ℱad be an optimal pair in (54) with cost (91) and yq∗ be the corresponding weak solution of equation (52).
Due to Theorem2, we know that the map c,η⟶yq is Fréchet differentiable at c∗,η∗ in the direction w=h,l∈L20,T2 which satisfies for sufficiently small ϵ>0 that c∗+ɛh,η∗+ɛl∈ℱad. Thus, the map c,η⟶yq is also (strongly) Gâteaux differentiable at c∗,η∗ in the direction h,l∈L20,T2. Indeed, we have(94)yq∗+ɛh+l−yq∗ɛ⟶z=zwstrongly inS0,T,as ɛ⟶0+, where z=Dyq∗w is a unique solution of equation (71). Therefore, we can obtain from (94) the Gâteaux derivative of cost (91) at c∗,η∗ in the direction w=h,l as follows:(95)DJc∗,η∗h,l=limɛ⟶0+Jc∗+ɛh,η∗+ɛl−Jc∗,η∗ɛ=limɛ⟶0+12∫0Tyq∗+ɛh+l+yq∗−2yd,yq∗+ɛh+l−yq∗ɛ2dt+limɛ⟶0+12yq∗+ɛh+l;T+yq∗;T−2ydT,yq∗+ɛh+l;T−yq∗;Tɛ2+limɛ⟶0+α22c∗,hL20,T+ɛhL20,T2−β22η∗,lL20,T+ɛlL20,T2=∫0Tyq∗−yd,z2dt+yq∗;T−ydT,zT2+αc∗,hL20,T−βη∗,lL20,T,where z=Dyq∗w is a solution of equation (71).
We multiply both sides of the weak form of equation (92) by z which is a solution of equation (71) and integrate it over 0,T. Then, we obtain(96)∫0Tp″+Ap,zV′,Vdt+∫0T∫tTkσ−tApdσ,zV′,Vdt+∫0Tγ+1yq∗γp−q∗p,z2dt=∫0Tyq∗−yd,z2dt.
By integration by parts, the terminal value of the weak solutionp of equation (92), and Fubini’s theorem, (96) can be written again as(97)∫0Tp,z″+AzV,V′dt−yq∗;T−ydT,zT2+∫0Tp,∫0tkt−sAzdsV,V′dt+∫0Tp,γ+1yq∗γz2dt−∫0Tp,q∗z2dt=∫0Tyq∗−yd,z2dt.
Sincez is the solution of equation (71), by (97), we can deduce the following:(98)∫0Tyq∗−yd,z2dt+yq∗;T−ydT,zT2=h+l,yq∗,p2L20,T.
Therefore, from (95) and (98), we can obtain that(99)DJc∗,η∗h,l=αc∗+yq∗,p2,hL20,T+−βη∗+yq∗,p2,lL20,T.
Sincec∗,η∗∈ℱad is an optimal pair in (54), we know(100)DcJc∗,η∗h≥0,DηJc∗,η∗l≤0,h,l∈L20,T2.
Therefore, we can obtain the following from (99) and (100):(101)αc∗+yq∗,p2,hL20,T≥0,−βη∗+yq∗,p2,lL20,T≤0,where h,l∈L20,T2. By considering the signs of the variations h and l in (101), which depend on c∗ and η∗, respectively, we can deduce from (101) that(102)c∗=maxcl,min−yq∗,p2α,cu,η∗=maxηl,minyq∗,p2β,ηu.
This completes the proof.


#### 3.3.2. Case of Velocity’s Distributive Value ObservationC2∈ℒW0,T,ℳ2

In this case, we consider the following cost functional:(103)Jc,η=12y′q−ydL20,T;H2+α2cL20,T2−β2ηL20,T2,where yd∈L20,T;H is the desired value. Let c∗,η∗∈ℱad be an optimal pair subject to equation (52) and quadratic cost (103). In this case, we introduce the following adjoint equation corresponding to cost (103) and equation (52) in which q=c+η is replaced by q∗=c∗+η∗:(104)p″+Ap+∫tTk0Apds+∫tT∫sTk′σ−sApdσds−γ+1∫tTyq∗γp′d+∫tTq∗p′ds=y′q∗−yd,in0,T,pT=0,p′T=0.Proposition 4.
Equation (104) admits a unique weak solution p∈S0,T.Proof.
By the reversion in timet⟶T−t and the reversion σ⟶T−σ in equation (104), we can obtain the following:(105)p″T−t+ApT−t+∫T−tTk0Apds−∫T−tT∫T−s0k′T−σ−sApT−σdσds−γ+1∫T−tTyq∗γp′ds+∫T−tTq∗p′ds=gT−t,in0,T,pT−0=0,p′T−0=0,where g⋅=y′q∗;⋅−yd⋅. By changing s⟶T−s in equation (105) and then changing every function in the reversed equation like this pT−⋅⟶p⋅, we can have the following initial value problem:(106)p″+Ap+∫0tk0Apds+∫0tk′∗Apds+γ+1∫0tyq∗γp′ds−∫0tq∗p′ds=g,in0,T,p0=0,p′0=0.
Now, we apply the well-known Faedo–Galerkin procedure to equation (106): since V is separable, there exists a basis wmm=1∞ in V such that wmm=1∞ is a complete orthonormal system in H and free and total in V. For each m∈N, we define an approximate solution of equation (106) by(107)pmt=∑j=1mgjmtwj,where pmt satisfies(108)pm″t,wj2+pmt,wjV+∫0tk0pm+k′∗pm,wjVds+∫0tγ+1yq∗γpm′−q∗pm′,wj2ds=g,wj2,in0,T,1≤j≤m,pm0=pm′0=0.
We multiply both sides of equation (108) by gjm′t and sum over j to obtain(109)12ddtpm′t22+pmtV2+∫0tk0pm+k′∗pm,pm′tVds+∫0tγ+1yq∗γpm′−q∗pm′,pm′t2ds=g,pm′t2.
Integrating (109) from 0 to t and by integration by parts, we obtain(110)12pm′t22+pmtV2+∫0tk0pm+k′∗pm,pmtVds−∫0tk0pm+k′∗pm,pmVds+∫0tγ+1yq∗γpm′−q∗pm′,pmt2ds−∫0tγ+1yq∗γpm′−q∗pm′,pm2ds=∫0tg,pm′2ds.
By (59), we can obtain the following estimations:(111)∫0tγ+1yq∗γpm′,pmt2ds≤C∫0tpm′2pmtVds≤ɛpmtV2+Cɛ∫0tpm′22ds,(112)∫0tγ+1yq∗γpm′,pm2ds≤C∫0tpm′2pmtVds≤C∫0tpm′22+pmtV2ds.
And we also have the following estimations:(113)∫0tq∗pm′,pmt2ds≤pmt2∫0tq∗pm′2ds≤CpmtV∫0tq∗pm′2ds≤ɛpmtV2+Cɛ∫0tq∗2pm′22ds,(114)∫0tq∗pm′,pm2ds≤∫0tq∗pm′2pm2ds≤C∫0tq∗pm′2pmVds≤C∫0tq∗2pm′22+pmV2ds.
By (111)–(114) with a properly determined ɛ and estimating other terms in (110) as in the proof of Theorem 1, we can arrive at(115)pm′t22+pmtV2≤CgL20,T;H2+C∫0t1+q∗2pm′22+pmV2ds.
Thus, by Gronwall’s lemma, we see that(116)pm∈W0,T∩L∞0,T;V∩W1,∞0,T;H.
Then, following similar arguments in [18], we can ensure that equation (104) admits a weak solution p∈S0,T. Uniqueness will be proved straightforwardly for g=0.
This completes the proof.
Now, we investigate the first-order optimality conditions for minimax optimal control problem (54) for quadratic cost function (103).Theorem 5.
Ifα and β in cost (103) are large enough and the exponent γ≥1, then an optimal control c∗∈Uad and a disturbance η∗∈Vad, namely, an optimal pair q∗=c∗,η∗∈ℱad satisfying (54), can be given by(117)c∗=maxcl,minyq∗,p′2α,cu,η∗=maxηl,min−yq∗,p′2β,ηu,where p is the weak solution of equation (104).Remark 3.
To get the optimality condition in Theorem5, we need to multiply the weak form of equation (104) by z′. Since we just know that z′∈C0,T;H, it would be a just formal procedure. To overcome this difficulty, we employ the regularization method of Lions ([14], pp. 286–288) which was used to deal with linear hyperbolic problems (cf. [15]).Proof of Theorem5.
By the assumptions of Theorem5, we can verify through Theorem 3 that there exists an optimal pair in (54) with cost (103). Let c∗,η∗∈ℱad be an optimal pair in (54) with cost (103) and yq∗ be the corresponding weak solution of equation (52).
By similar arguments in Theorem4, we can know that cost (103) is also (strongly) Gâteaux differentiable at c∗,η∗ in the direction w=h,l∈L20,T2 which satisfies for sufficiently small ϵ>0 that c∗,η∗+ɛw∈ℱad. Thus, we deduce the following:(118)DJc∗,η∗h,l=limɛ⟶0+Jc∗+ɛh,η∗+ɛl−Jc∗,η∗ɛ=∫0Ty′q∗−yd,z′2dt+αc∗,hL20,T−βη∗,lL20,T,where z=Dyq∗w is a solution of equation (71).
As explained above, we prove this theorem by the regularization method. For this purpose, we extend the time domains of equations (71) and (104) to Rt by introducing ψ and θ as the solutions of(119)ψ″+Aψ+k∗Aψ+γ+1yq∗γψ=q∗ψ+h+lyq∗,in0,T,0,otherwise,ψ0=0,ψ′0=0,(120)θ″+Aθ+∫tTKs,θ+Gq∗,θ′ds=y′q∗−yd,in0,T,0,otherwise,θT=0,θ′T=0,where(121)Ks,θ=k0Aθ+∫sTk′σ−sAθdσ,(122)Gq∗,θ′=−γ+1yq∗γθ′+q∗θ′.
In fact, we have(123)ψ=z,in0,T,θ=p,in0,T,(124)∫0Ty′q∗−yd,z′2dt=∫Rtθ″+Aθ+∫tTKs,θ+Gq∗,θ′ds,ψ′2dt.
For the simplicity, we shall denote the scalar product inH or the antiduality V,V′ by ⋅,⋅. Let ρn be a regularizing sequence on Rt. Then, the right-hand side of (124) becomes(125)limn⟶∞∫Rtθ″+Aθ+∫tTKs,θ+Gq∗,θ′ds∗ρn,ψ′∗ρndt.
Now, we integrate by parts of (125). Then, we put(126)Xn=∫Rt−θ′∗ρn,ψ″∗ρn+Aθ∗ρn,ψ′∗ρn+K⋅,θ+Gq∗,θ′∗ρn,ψ∗ρndt.
Using equation (119), that is, ψ″=−Aψ−k∗Aψ+Gq∗,ψ+h+lyq∗, (126) can be given again by(127)Xn=∫Rt−θ′∗ρn,h+lyq∗∗ρndt+∑i=13Yni,where(128)Yn1=∫Rtθ′∗ρn,Aψ∗ρn+Aθ∗ρn,ψ′∗ρndt,Yn2=∫Rtθ′∗ρn,k∗Aψ∗ρn+K⋅,θ∗ρn,ψ∗ρndt,Yn3=∫Rt−θ′∗ρn,Gq∗,ψ∗ρn+Gq∗,θ′∗ρn,ψ∗ρndt.
We immediately knowYn1=0 by integration by parts. By integration by parts, we can see that(129)Yn2=∫Rt−θ∗ρn,k0Aψ+k′∗Aψ∗ρn+K⋅,θ∗ρn,ψ∗ρndt.
Thus,(130)Yn2⟶∫Rt−θ,k0Aψ+k′∗Aψ+Kt,θ,ψdt,as n⟶∞. From (121) and Fubini’s theorem, we know that the right-hand side of (130) is 0. Therefore, we can obtain Yn2⟶0 as n⟶∞. And we can easily verify that Yn3⟶0 as n⟶∞. Consequently, we can obtain(131)limn⟶∞Xn=∫Rt−θ′,h+lyq∗dt=∫0T−p′,h+lyq∗2dt.
Sincec∗,η∗∈ℱad is an optimal pair in (54), we know(132)DcJc∗,η∗h≥0,DηJc∗,η∗l≤0,h,l∈L20,T2.
From (118), (124), and (131), we know that (132) implies(133)αc∗−yq∗,p′2,hL20,T≥0,−βη∗−yq∗,p′2,lL20,T≤0,where h,l∈L20,T2. By similar arguments in the proof of Theorem 4, we can deduce from (133) that(134)c∗=maxcl,minyq∗,p′2α,cu,η∗=maxηl,min−yq∗,p′2β,ηu.
This completes the proof.


## 3.1. Differentiability of the Nonlinear Solution Map

In this section, we address the Fréchet differentiability of the nonlinear solution map, which is desirable for many applications.For our study, we define the Fréchet differentiability of the nonlinear solution map as follows.Definition 2.
The solution mapu⟶yu of L20,T into S0,T is said to be Fréchet differentiable on L20,T if for any u∈L20,T, there exists Tu∈ℒL20,T,S0,T such that, for any w∈L20,T,(55)yu+w−yu−TuwS0,TwL20,T⟶0,aswL20,T⟶0.
The operatorTu is called the Fréchet derivative of y at u, which we denote by Dyu. Tuw=Dyuw∈S0,T is called the Fréchet derivative of y at u in the direction of w∈L20,T.Theorem 2.
The solution mapu⟶yu is Fréchet differentiable on L20,T, and the Fréchet derivative of yu at u in the direction w∈L20,T, that is to say z=Dyuw, is given by the solution of(56)z″+Az+∫0tkt−sAzds+γ+1yuγz=uz+wyu,in0,T,z0=0,z′0=0.Proof.
Let(57)Byuz≔yuγz.
Then, by (11) and Theorem 1, we can verify(58)Byuz2≤yu3γγz6≤CyuVγzV≤CyuC0,T;VγzV≤CyuS0,TγzV≤Cy0,y1,f,uPγzV.
Hence, from (58), we see that(59)Byu∈ℒV,H∩ℒH,V′.
Sincew∈L20,T and yu∈S0,T, we know that wyu∈L20,T;H. By (59), we can use the results in [18] to verify that equation (56) admits a unique weak solution z∈S0,T. And by Theorem 1, we can know that the weak solution z=zw of equation (56) satisfies(60)zwS0,T≤CwyuL20,T;H≤CwL20,TyuC0,T;H≤CyuS0,TwL20,T≤Cy0,y1,f,uPwL20,T.
Hence, from (60), the mapping w∈L20,T↦zw∈S0,T is linear and bounded.
We setφ=yu+w−yu−z. Then, using notation (27) and noting(61)gyu+w−gyu−γ+1yuγz=γ+1Byu+w,yuyu+w−yu−γ+1yuγz=γ+1Byu+w,yuφ+γ+1Byu+w,yu−yuγz,we know that φ satisfies(62)φ″+Aφ+∫0tkt−sAφds+γ+1Byu+w,yuφ=u+wφ+wz+I,in0,T,φ0=0,φ′0=0,in the weak sense, where(63)I=−γ+1Byu+w,yu−yuγz.
Arguing as in the proof of Theorem1, we can know that the weak solution φ of equation (62) exists, and the following is fulfilled:(64)φS0,T≤Cwz+IL20,T;H.
By (60), we can deduce(65)wzL20,T;H≤wL20,TzC0,T;H≤CwL20,TzS0,T≤CwL20,T2,Byu+w,yu−yuγz2≤Byu+w,yu−yuγ3z6≤CByu+w,yu−yuγ3zV≤CByu+w,yu−yuγ3zC0,T;V≤CzS0,TByu+w,yu−yuγ3≤CwL20,TByu+w,yu−yuγ3.
By Theorem1, we can see that(66)yu+w−yuS0,T≤CwL20,T.
Thus, ifwL20,T⟶0, then by (66), we can extract a subsequence, still denoted by yu+w, such that(67)Byu+w,yu−yuγ⟶0,a.e.inQ.
And from (11), we know that(68)Byu+w,yu−yuγ3≤8yu+w+yu3γ∈L1Q.
Thus, by the Lebesgue dominated convergence theorem through (67) and (68), we can obtain(69)Byu+w,yu−yuγL3Q⟶0aswL20,T⟶0.
Hence, from (64) to (69), we can obtain(70)φS0,T≤Cwz+IL20,T;H≤CwzL20,T;H+IL20,T;H≤CwL20,TwL20,T+Byu+w,yu−yuγL3Q,which immediately implies that φS0,T=owL20,T as wL20,T⟶0.
This completes the proof.
As before, for simplicity, we denotec+η and c∗+η∗ by q and q∗, respectively. From Theorem 2, we know that the map c,η⟶yq of ℱad to S0,T is Fréchet differentiable at c∗,η∗, and the Fréchet derivative of yq at c∗,η∗ in the direction w=h,l∈L20,T2, say z=Dyq∗w, is a unique weak solution of the following problem:(71)z″+Az+∫0tkt−sAzds+γ+1yq∗γz=q∗z+h+lyq∗,in0,T,z0=0,z′0=0.
The following results will be used in showing the existence of optimal pairs.Proposition 2.
Assume that the exponentγ≥1. For given w∈L20,T, the Fréchet derivative Dyuw is locally Lipschitz continuous on L20,T. Indeed, it is satisfied that(72)Dyu1w−Dyu2wS0,T≤Cu1−u2L20,TwL20,T,where C>0 is a constant depending on the data.Proof.
Letzi=Dyuiwi=1,2 be the weak solutions of equation (56) corresponding to uii=1,2. If we set ϕ=z1−z2, then we can know that ϕ satisfies(73)ϕ″+Aϕ+∫0tkt−sAϕds+γ+1yu1γϕ=u1ϕ+I1+I2,in0,T,ϕ0=0,ϕ′0=0,in the weak sense, where(74)I1=−γγ+1∫01θyu1+1−θyu2γ−1dθyu1−yu2z2,I2=u1−u2z2+wyu1−yu2.
As before, arguing as in the proof of Theorem1, we can deduce that(75)ϕS0,T≤CI1+I2L20,T;H.
In estimatingI1, the case of γ=1 will be clarified as we will see later. Thus, we restrict to the case of γ>1 to estimate I1. By (11), Theorem 1, and (60), we can obtain(76)I12≤γγ+1yu1+yu2γ−1yu1−yu2z22≤Cyu1+yu2γ−16yu1−yu26z26≤Cyu1V+yu2Vγ−1yu1−yu2Vz2V≤Cyu1C0,T;V+yu2C0,T;Vγ−1yu1−yu2C0,T;Vz2C0,T;V≤Cu1−u2L20,TwL20,T,(77)I2L20,T;H≤u1−u2z2L20,T;H+wyu1−yu2L20,T;H≤u1−u2L20,Tz2C0,T;H+wL20,Tyu1−yu2C0,T;H≤Cu1−u2L20,Tz2S0,T+wL20,Tyu1−yu2S0,T≤Cu1−u2L20,TwL20,T.
From (75) to (77), we can get the following:(78)ϕS0,T≤Cu1−u2L20,TwL20,T.
This completes the proof.


## 3.2. Existence of the Optimal Pairs

By the minimax theorem in infinite dimensions in Barbu and Precupanu [13], we study the existence of the optimal pairs in the underlying control system with quadratic cost function (53).Theorem 3.
Assume that the exponentγ≥1 in control system equation (52). Then, for sufficiently large α and β in (53), there exists c∗,η∗∈ℱad satisfying (54).
In order to prove this theorem, we mainly illustrate the following as in [8] (cf. [9]):(i)
For sufficiently largeα and β in (53), the maps c⟶Jc,η and η⟶Jc,η are convex and concave for all η∈Vad and for all c∈Uad, respectively(ii)
The mapsc⟶Jc,η and η⟶Jc,η are lower and upper semicontinuous for all η∈Vad and for all c∈Uad, respectivelyProof.
(i)
Since the mapc⟶yc+η where η is fixed is Fréchet differentiable, the map c⟶Jc,η is also Fréchet differentiable. Let DcJc1,ηc1−c2 and DcJc2,ηc1−c2 be the Fréchet derivatives of the map c⟶Jc,η at c1 and c2 in the direction c1−c2, respectively, where η holds fixed. To verify the convexity of the map c⟶Jc,η, it is sufficient to show that(79)DcJc1,η−DcJc2,ηc1−c2≥0,∀c1,c2∈Uad,for any fixedη∈Vad. (79) means(80)Cyc1+η−Yd,CDcyc1+ηc1−c2ℳ+αc1,c1−c2L20,T−Cyc2+η−Yd,CDcyc2+ηc1−c2ℳ−αc2,c1−c2L20,T≥0,∀c1,c2∈Uad,whereDcyci+ηc1−c2i=1,2 are weak solutions of equation (71), in which q∗z+h+lyp is replaced by ci+ηz+c1−c2yci+ηi=1,2, respectively. We can deduce that (80) is equivalent to(81)Cyc1+η−yc2+η,CDcyc1+ηc1−c2ℳ+Cyc2+η−Yd,CDcyc1+ηc1−c2−Dcyc2+ηc1−c2ℳ+αc1−c2L20,T2≥0,∀c1,c2∈Uad.By Theorem1, (60), and Proposition 2, we can have the following estimations:(82)Cyc1+η−yc2+η,CDcyc1,ηc1−c2ℳ≤Cyc1+η−yc2+ηℳCDcyc1+ηc1−c2ℳ≤Cyc1+η−yc2+ηS0,TDcyc1+ηc1−c2S0,T≤Cc1−c2L20,T2,(83)Cyc2+η−Yd,CDcyc1+ηc1−c2−Dcyc2+ηc1−c2ℳ≤Cyc2+η−YdℳCDcyc1+ηc1−c2−Dcyc2+ηc1−c2ℳ≤CCyc2+ηℳ+Ydℳ×Dcyc1+ηc1−c2−Dcyc2+ηc1−c2S0,T≤CCyc2+ηℳ+Ydℳc1−c2L20,T2≤Cc1−c2L20,T2.From (81) to (83), we can verify that there exists sufficiently large αlP,ℱad,Yd,C,ℳ such that, for any α>αlP,ℱad,Yd,C,ℳ, inequality (79) holds. Thus, the map c⟶Jc,η is convex.Similarly, we can show that there exists sufficiently largeβlP,ℱad,Yd,C,ℳ such that the following inequality(84)DηJc,η1−DηJc,η2η1−η2≤0,∀η1,η2∈Vad,is satisfied for any fixedc∈Uad and β>βlP,ℱad,Yd,C,ℳ. This also indicates the concavity of the map η⟶Jc,η.(ii)
Sincecn is bounded in L20,T, we can extract a subsequence cnk⊂cn such that(85)cnk⟶c∗weakly inL20,T,ask⟶∞.
Then, sinceC is a continuous linear operator on W0,T, by Proposition 1, we obtain(86)Cycnk+η⟶Cyc∗+ηweakly inℳ,ask⟶∞,∀η∈Vad. Since the norm is weakly lower semicontinuous, we can see from (85) and (86) that the map c⟶Jc,η is lower semicontinuous for all η∈Vad. By similar arguments, we can prove that the map η⟶Jc,η is upper semicontinuous for all c∈Uad.
Next, we prove the existence of an optimal pairc∗,η∗∈ℱad: let cn⊂Uad be a minimizing sequence of Jc,η where η∈Vad is fixed. Thus,(87)liminfn⟶∞Jcn,η=infc∈UadJc,η.
Then, by (i) and (ii), we know that(88)J0η=liminfn⟶∞Jcn,η≥Jc∗,η,∀η∈Vad,but since J0η≤Jc∗,η, we have(89)J0η=Jc∗,η=infc∈UadJc,η,∀η∈Vad.
Similarly, we also know that there existsη∗∈Vad such that(90)J0η∗=supη∈VadJ0η.
From (89) and (90), we can conclude that c∗,η∗∈ℱad is an optimal pair for cost (53).
This completes the proof.


## 3.3. Necessary Conditions of Optimal Pairs

In this section, we study the necessary optimality conditions that have to be satisfied by each optimal pair of the minimax optimal control problem with cost (53) in which the following two types of observations are considered:(1)
We takeℳ1=L20,T;H×H and C1∈ℒW0,T,ℳ1 and observe C1yq=yq,yq;T∈L20,T;H×H(2)
We takeℳ2=L20,T;H and C2∈ℒW0,T,ℳ2 and observe C2yq=y′q∈L20,T;HSinceyq∈W0,T⊂C0,T;H, the above observations are meaningful.


### 3.3.1. Case of Distributive and Terminal Values’ ObservationsC1∈ℒW0,T,ℳ1

We consider the cost functional expressed by(91)Jc,η=12yq−ydL20,T;H2+12yq;T−ydT22+α2cL20,T2−β2ηL20,T2,where yd∈L20,T;H and ydT∈H are desired values. Let c∗,η∗∈L20,T2q∗=c∗+η∗ be an optimal pair subject to equation (52) and quadratic cost (91). Now, we are about to formulate the following adjoint equation corresponding to cost (91) and equation (52) in which c,η with q=c+η is replaced by c∗,η∗ with q∗=c∗+η∗:(92)p″+Ap+∫tTkσ−tApdσ+γ+1yq∗γp=c∗+η∗p+yq∗−yd,in0,T,pT=0,p′T=−yq∗;T+ydT.Proposition 3.
Equation (92) admits a unique weak solution p∈S0,T.Proof.
From the observation conditionsyq∗−yd∈L20,T;H and yq∗;T−ydT∈H and (59), we can refer to the results in [18] (cf. Dautray and Lions [20], pp.655–659) to ensure that equation (92), after reversing the direction of time t⟶T−t, admits a unique weak solution p∈S0,T. This completes the proof.
We now discuss the first-order optimality conditions for minimax optimal control problem (54) for quadratic cost function (91).Theorem 4.
Ifα and β in cost (91) are large enough and the exponent γ≥1, then an optimal control c∗∈Uad and a disturbance η∗∈Vad, namely, an optimal pair c∗,η∗∈ℱad satisfying (54), can be given by(93)c∗=maxcl,min−yq∗,p2α,cu,η∗=maxηl,minyq∗,p2β,ηu,where p is the weak solution of equation (92).Proof.
By the assumptions of Theorem4, we can verify through Theorem 3 that there exists an optimal pair in (54) with cost (91). Let c∗,η∗∈ℱad be an optimal pair in (54) with cost (91) and yq∗ be the corresponding weak solution of equation (52).
Due to Theorem2, we know that the map c,η⟶yq is Fréchet differentiable at c∗,η∗ in the direction w=h,l∈L20,T2 which satisfies for sufficiently small ϵ>0 that c∗+ɛh,η∗+ɛl∈ℱad. Thus, the map c,η⟶yq is also (strongly) Gâteaux differentiable at c∗,η∗ in the direction h,l∈L20,T2. Indeed, we have(94)yq∗+ɛh+l−yq∗ɛ⟶z=zwstrongly inS0,T,as ɛ⟶0+, where z=Dyq∗w is a unique solution of equation (71). Therefore, we can obtain from (94) the Gâteaux derivative of cost (91) at c∗,η∗ in the direction w=h,l as follows:(95)DJc∗,η∗h,l=limɛ⟶0+Jc∗+ɛh,η∗+ɛl−Jc∗,η∗ɛ=limɛ⟶0+12∫0Tyq∗+ɛh+l+yq∗−2yd,yq∗+ɛh+l−yq∗ɛ2dt+limɛ⟶0+12yq∗+ɛh+l;T+yq∗;T−2ydT,yq∗+ɛh+l;T−yq∗;Tɛ2+limɛ⟶0+α22c∗,hL20,T+ɛhL20,T2−β22η∗,lL20,T+ɛlL20,T2=∫0Tyq∗−yd,z2dt+yq∗;T−ydT,zT2+αc∗,hL20,T−βη∗,lL20,T,where z=Dyq∗w is a solution of equation (71).
We multiply both sides of the weak form of equation (92) by z which is a solution of equation (71) and integrate it over 0,T. Then, we obtain(96)∫0Tp″+Ap,zV′,Vdt+∫0T∫tTkσ−tApdσ,zV′,Vdt+∫0Tγ+1yq∗γp−q∗p,z2dt=∫0Tyq∗−yd,z2dt.
By integration by parts, the terminal value of the weak solutionp of equation (92), and Fubini’s theorem, (96) can be written again as(97)∫0Tp,z″+AzV,V′dt−yq∗;T−ydT,zT2+∫0Tp,∫0tkt−sAzdsV,V′dt+∫0Tp,γ+1yq∗γz2dt−∫0Tp,q∗z2dt=∫0Tyq∗−yd,z2dt.
Sincez is the solution of equation (71), by (97), we can deduce the following:(98)∫0Tyq∗−yd,z2dt+yq∗;T−ydT,zT2=h+l,yq∗,p2L20,T.
Therefore, from (95) and (98), we can obtain that(99)DJc∗,η∗h,l=αc∗+yq∗,p2,hL20,T+−βη∗+yq∗,p2,lL20,T.
Sincec∗,η∗∈ℱad is an optimal pair in (54), we know(100)DcJc∗,η∗h≥0,DηJc∗,η∗l≤0,h,l∈L20,T2.
Therefore, we can obtain the following from (99) and (100):(101)αc∗+yq∗,p2,hL20,T≥0,−βη∗+yq∗,p2,lL20,T≤0,where h,l∈L20,T2. By considering the signs of the variations h and l in (101), which depend on c∗ and η∗, respectively, we can deduce from (101) that(102)c∗=maxcl,min−yq∗,p2α,cu,η∗=maxηl,minyq∗,p2β,ηu.
This completes the proof.


### 3.3.2. Case of Velocity’s Distributive Value ObservationC2∈ℒW0,T,ℳ2

In this case, we consider the following cost functional:(103)Jc,η=12y′q−ydL20,T;H2+α2cL20,T2−β2ηL20,T2,where yd∈L20,T;H is the desired value. Let c∗,η∗∈ℱad be an optimal pair subject to equation (52) and quadratic cost (103). In this case, we introduce the following adjoint equation corresponding to cost (103) and equation (52) in which q=c+η is replaced by q∗=c∗+η∗:(104)p″+Ap+∫tTk0Apds+∫tT∫sTk′σ−sApdσds−γ+1∫tTyq∗γp′d+∫tTq∗p′ds=y′q∗−yd,in0,T,pT=0,p′T=0.Proposition 4.
Equation (104) admits a unique weak solution p∈S0,T.Proof.
By the reversion in timet⟶T−t and the reversion σ⟶T−σ in equation (104), we can obtain the following:(105)p″T−t+ApT−t+∫T−tTk0Apds−∫T−tT∫T−s0k′T−σ−sApT−σdσds−γ+1∫T−tTyq∗γp′ds+∫T−tTq∗p′ds=gT−t,in0,T,pT−0=0,p′T−0=0,where g⋅=y′q∗;⋅−yd⋅. By changing s⟶T−s in equation (105) and then changing every function in the reversed equation like this pT−⋅⟶p⋅, we can have the following initial value problem:(106)p″+Ap+∫0tk0Apds+∫0tk′∗Apds+γ+1∫0tyq∗γp′ds−∫0tq∗p′ds=g,in0,T,p0=0,p′0=0.
Now, we apply the well-known Faedo–Galerkin procedure to equation (106): since V is separable, there exists a basis wmm=1∞ in V such that wmm=1∞ is a complete orthonormal system in H and free and total in V. For each m∈N, we define an approximate solution of equation (106) by(107)pmt=∑j=1mgjmtwj,where pmt satisfies(108)pm″t,wj2+pmt,wjV+∫0tk0pm+k′∗pm,wjVds+∫0tγ+1yq∗γpm′−q∗pm′,wj2ds=g,wj2,in0,T,1≤j≤m,pm0=pm′0=0.
We multiply both sides of equation (108) by gjm′t and sum over j to obtain(109)12ddtpm′t22+pmtV2+∫0tk0pm+k′∗pm,pm′tVds+∫0tγ+1yq∗γpm′−q∗pm′,pm′t2ds=g,pm′t2.
Integrating (109) from 0 to t and by integration by parts, we obtain(110)12pm′t22+pmtV2+∫0tk0pm+k′∗pm,pmtVds−∫0tk0pm+k′∗pm,pmVds+∫0tγ+1yq∗γpm′−q∗pm′,pmt2ds−∫0tγ+1yq∗γpm′−q∗pm′,pm2ds=∫0tg,pm′2ds.
By (59), we can obtain the following estimations:(111)∫0tγ+1yq∗γpm′,pmt2ds≤C∫0tpm′2pmtVds≤ɛpmtV2+Cɛ∫0tpm′22ds,(112)∫0tγ+1yq∗γpm′,pm2ds≤C∫0tpm′2pmtVds≤C∫0tpm′22+pmtV2ds.
And we also have the following estimations:(113)∫0tq∗pm′,pmt2ds≤pmt2∫0tq∗pm′2ds≤CpmtV∫0tq∗pm′2ds≤ɛpmtV2+Cɛ∫0tq∗2pm′22ds,(114)∫0tq∗pm′,pm2ds≤∫0tq∗pm′2pm2ds≤C∫0tq∗pm′2pmVds≤C∫0tq∗2pm′22+pmV2ds.
By (111)–(114) with a properly determined ɛ and estimating other terms in (110) as in the proof of Theorem 1, we can arrive at(115)pm′t22+pmtV2≤CgL20,T;H2+C∫0t1+q∗2pm′22+pmV2ds.
Thus, by Gronwall’s lemma, we see that(116)pm∈W0,T∩L∞0,T;V∩W1,∞0,T;H.
Then, following similar arguments in [18], we can ensure that equation (104) admits a weak solution p∈S0,T. Uniqueness will be proved straightforwardly for g=0.
This completes the proof.
Now, we investigate the first-order optimality conditions for minimax optimal control problem (54) for quadratic cost function (103).Theorem 5.
Ifα and β in cost (103) are large enough and the exponent γ≥1, then an optimal control c∗∈Uad and a disturbance η∗∈Vad, namely, an optimal pair q∗=c∗,η∗∈ℱad satisfying (54), can be given by(117)c∗=maxcl,minyq∗,p′2α,cu,η∗=maxηl,min−yq∗,p′2β,ηu,where p is the weak solution of equation (104).Remark 3.
To get the optimality condition in Theorem5, we need to multiply the weak form of equation (104) by z′. Since we just know that z′∈C0,T;H, it would be a just formal procedure. To overcome this difficulty, we employ the regularization method of Lions ([14], pp. 286–288) which was used to deal with linear hyperbolic problems (cf. [15]).Proof of Theorem5.
By the assumptions of Theorem5, we can verify through Theorem 3 that there exists an optimal pair in (54) with cost (103). Let c∗,η∗∈ℱad be an optimal pair in (54) with cost (103) and yq∗ be the corresponding weak solution of equation (52).
By similar arguments in Theorem4, we can know that cost (103) is also (strongly) Gâteaux differentiable at c∗,η∗ in the direction w=h,l∈L20,T2 which satisfies for sufficiently small ϵ>0 that c∗,η∗+ɛw∈ℱad. Thus, we deduce the following:(118)DJc∗,η∗h,l=limɛ⟶0+Jc∗+ɛh,η∗+ɛl−Jc∗,η∗ɛ=∫0Ty′q∗−yd,z′2dt+αc∗,hL20,T−βη∗,lL20,T,where z=Dyq∗w is a solution of equation (71).
As explained above, we prove this theorem by the regularization method. For this purpose, we extend the time domains of equations (71) and (104) to Rt by introducing ψ and θ as the solutions of(119)ψ″+Aψ+k∗Aψ+γ+1yq∗γψ=q∗ψ+h+lyq∗,in0,T,0,otherwise,ψ0=0,ψ′0=0,(120)θ″+Aθ+∫tTKs,θ+Gq∗,θ′ds=y′q∗−yd,in0,T,0,otherwise,θT=0,θ′T=0,where(121)Ks,θ=k0Aθ+∫sTk′σ−sAθdσ,(122)Gq∗,θ′=−γ+1yq∗γθ′+q∗θ′.
In fact, we have(123)ψ=z,in0,T,θ=p,in0,T,(124)∫0Ty′q∗−yd,z′2dt=∫Rtθ″+Aθ+∫tTKs,θ+Gq∗,θ′ds,ψ′2dt.
For the simplicity, we shall denote the scalar product inH or the antiduality V,V′ by ⋅,⋅. Let ρn be a regularizing sequence on Rt. Then, the right-hand side of (124) becomes(125)limn⟶∞∫Rtθ″+Aθ+∫tTKs,θ+Gq∗,θ′ds∗ρn,ψ′∗ρndt.
Now, we integrate by parts of (125). Then, we put(126)Xn=∫Rt−θ′∗ρn,ψ″∗ρn+Aθ∗ρn,ψ′∗ρn+K⋅,θ+Gq∗,θ′∗ρn,ψ∗ρndt.
Using equation (119), that is, ψ″=−Aψ−k∗Aψ+Gq∗,ψ+h+lyq∗, (126) can be given again by(127)Xn=∫Rt−θ′∗ρn,h+lyq∗∗ρndt+∑i=13Yni,where(128)Yn1=∫Rtθ′∗ρn,Aψ∗ρn+Aθ∗ρn,ψ′∗ρndt,Yn2=∫Rtθ′∗ρn,k∗Aψ∗ρn+K⋅,θ∗ρn,ψ∗ρndt,Yn3=∫Rt−θ′∗ρn,Gq∗,ψ∗ρn+Gq∗,θ′∗ρn,ψ∗ρndt.
We immediately knowYn1=0 by integration by parts. By integration by parts, we can see that(129)Yn2=∫Rt−θ∗ρn,k0Aψ+k′∗Aψ∗ρn+K⋅,θ∗ρn,ψ∗ρndt.
Thus,(130)Yn2⟶∫Rt−θ,k0Aψ+k′∗Aψ+Kt,θ,ψdt,as n⟶∞. From (121) and Fubini’s theorem, we know that the right-hand side of (130) is 0. Therefore, we can obtain Yn2⟶0 as n⟶∞. And we can easily verify that Yn3⟶0 as n⟶∞. Consequently, we can obtain(131)limn⟶∞Xn=∫Rt−θ′,h+lyq∗dt=∫0T−p′,h+lyq∗2dt.
Sincec∗,η∗∈ℱad is an optimal pair in (54), we know(132)DcJc∗,η∗h≥0,DηJc∗,η∗l≤0,h,l∈L20,T2.
From (118), (124), and (131), we know that (132) implies(133)αc∗−yq∗,p′2,hL20,T≥0,−βη∗−yq∗,p′2,lL20,T≤0,where h,l∈L20,T2. By similar arguments in the proof of Theorem 4, we can deduce from (133) that(134)c∗=maxcl,minyq∗,p′2α,cu,η∗=maxηl,min−yq∗,p′2β,ηu.
This completes the proof.


## 3.3.1. Case of Distributive and Terminal Values’ ObservationsC1∈ℒW0,T,ℳ1

We consider the cost functional expressed by(91)Jc,η=12yq−ydL20,T;H2+12yq;T−ydT22+α2cL20,T2−β2ηL20,T2,where yd∈L20,T;H and ydT∈H are desired values. Let c∗,η∗∈L20,T2q∗=c∗+η∗ be an optimal pair subject to equation (52) and quadratic cost (91). Now, we are about to formulate the following adjoint equation corresponding to cost (91) and equation (52) in which c,η with q=c+η is replaced by c∗,η∗ with q∗=c∗+η∗:(92)p″+Ap+∫tTkσ−tApdσ+γ+1yq∗γp=c∗+η∗p+yq∗−yd,in0,T,pT=0,p′T=−yq∗;T+ydT.Proposition 3.
Equation (92) admits a unique weak solution p∈S0,T.Proof.
From the observation conditionsyq∗−yd∈L20,T;H and yq∗;T−ydT∈H and (59), we can refer to the results in [18] (cf. Dautray and Lions [20], pp.655–659) to ensure that equation (92), after reversing the direction of time t⟶T−t, admits a unique weak solution p∈S0,T. This completes the proof.
We now discuss the first-order optimality conditions for minimax optimal control problem (54) for quadratic cost function (91).Theorem 4.
Ifα and β in cost (91) are large enough and the exponent γ≥1, then an optimal control c∗∈Uad and a disturbance η∗∈Vad, namely, an optimal pair c∗,η∗∈ℱad satisfying (54), can be given by(93)c∗=maxcl,min−yq∗,p2α,cu,η∗=maxηl,minyq∗,p2β,ηu,where p is the weak solution of equation (92).Proof.
By the assumptions of Theorem4, we can verify through Theorem 3 that there exists an optimal pair in (54) with cost (91). Let c∗,η∗∈ℱad be an optimal pair in (54) with cost (91) and yq∗ be the corresponding weak solution of equation (52).
Due to Theorem2, we know that the map c,η⟶yq is Fréchet differentiable at c∗,η∗ in the direction w=h,l∈L20,T2 which satisfies for sufficiently small ϵ>0 that c∗+ɛh,η∗+ɛl∈ℱad. Thus, the map c,η⟶yq is also (strongly) Gâteaux differentiable at c∗,η∗ in the direction h,l∈L20,T2. Indeed, we have(94)yq∗+ɛh+l−yq∗ɛ⟶z=zwstrongly inS0,T,as ɛ⟶0+, where z=Dyq∗w is a unique solution of equation (71). Therefore, we can obtain from (94) the Gâteaux derivative of cost (91) at c∗,η∗ in the direction w=h,l as follows:(95)DJc∗,η∗h,l=limɛ⟶0+Jc∗+ɛh,η∗+ɛl−Jc∗,η∗ɛ=limɛ⟶0+12∫0Tyq∗+ɛh+l+yq∗−2yd,yq∗+ɛh+l−yq∗ɛ2dt+limɛ⟶0+12yq∗+ɛh+l;T+yq∗;T−2ydT,yq∗+ɛh+l;T−yq∗;Tɛ2+limɛ⟶0+α22c∗,hL20,T+ɛhL20,T2−β22η∗,lL20,T+ɛlL20,T2=∫0Tyq∗−yd,z2dt+yq∗;T−ydT,zT2+αc∗,hL20,T−βη∗,lL20,T,where z=Dyq∗w is a solution of equation (71).
We multiply both sides of the weak form of equation (92) by z which is a solution of equation (71) and integrate it over 0,T. Then, we obtain(96)∫0Tp″+Ap,zV′,Vdt+∫0T∫tTkσ−tApdσ,zV′,Vdt+∫0Tγ+1yq∗γp−q∗p,z2dt=∫0Tyq∗−yd,z2dt.
By integration by parts, the terminal value of the weak solutionp of equation (92), and Fubini’s theorem, (96) can be written again as(97)∫0Tp,z″+AzV,V′dt−yq∗;T−ydT,zT2+∫0Tp,∫0tkt−sAzdsV,V′dt+∫0Tp,γ+1yq∗γz2dt−∫0Tp,q∗z2dt=∫0Tyq∗−yd,z2dt.
Sincez is the solution of equation (71), by (97), we can deduce the following:(98)∫0Tyq∗−yd,z2dt+yq∗;T−ydT,zT2=h+l,yq∗,p2L20,T.
Therefore, from (95) and (98), we can obtain that(99)DJc∗,η∗h,l=αc∗+yq∗,p2,hL20,T+−βη∗+yq∗,p2,lL20,T.
Sincec∗,η∗∈ℱad is an optimal pair in (54), we know(100)DcJc∗,η∗h≥0,DηJc∗,η∗l≤0,h,l∈L20,T2.
Therefore, we can obtain the following from (99) and (100):(101)αc∗+yq∗,p2,hL20,T≥0,−βη∗+yq∗,p2,lL20,T≤0,where h,l∈L20,T2. By considering the signs of the variations h and l in (101), which depend on c∗ and η∗, respectively, we can deduce from (101) that(102)c∗=maxcl,min−yq∗,p2α,cu,η∗=maxηl,minyq∗,p2β,ηu.
This completes the proof.


## 3.3.2. Case of Velocity’s Distributive Value ObservationC2∈ℒW0,T,ℳ2

In this case, we consider the following cost functional:(103)Jc,η=12y′q−ydL20,T;H2+α2cL20,T2−β2ηL20,T2,where yd∈L20,T;H is the desired value. Let c∗,η∗∈ℱad be an optimal pair subject to equation (52) and quadratic cost (103). In this case, we introduce the following adjoint equation corresponding to cost (103) and equation (52) in which q=c+η is replaced by q∗=c∗+η∗:(104)p″+Ap+∫tTk0Apds+∫tT∫sTk′σ−sApdσds−γ+1∫tTyq∗γp′d+∫tTq∗p′ds=y′q∗−yd,in0,T,pT=0,p′T=0.Proposition 4.
Equation (104) admits a unique weak solution p∈S0,T.Proof.
By the reversion in timet⟶T−t and the reversion σ⟶T−σ in equation (104), we can obtain the following:(105)p″T−t+ApT−t+∫T−tTk0Apds−∫T−tT∫T−s0k′T−σ−sApT−σdσds−γ+1∫T−tTyq∗γp′ds+∫T−tTq∗p′ds=gT−t,in0,T,pT−0=0,p′T−0=0,where g⋅=y′q∗;⋅−yd⋅. By changing s⟶T−s in equation (105) and then changing every function in the reversed equation like this pT−⋅⟶p⋅, we can have the following initial value problem:(106)p″+Ap+∫0tk0Apds+∫0tk′∗Apds+γ+1∫0tyq∗γp′ds−∫0tq∗p′ds=g,in0,T,p0=0,p′0=0.
Now, we apply the well-known Faedo–Galerkin procedure to equation (106): since V is separable, there exists a basis wmm=1∞ in V such that wmm=1∞ is a complete orthonormal system in H and free and total in V. For each m∈N, we define an approximate solution of equation (106) by(107)pmt=∑j=1mgjmtwj,where pmt satisfies(108)pm″t,wj2+pmt,wjV+∫0tk0pm+k′∗pm,wjVds+∫0tγ+1yq∗γpm′−q∗pm′,wj2ds=g,wj2,in0,T,1≤j≤m,pm0=pm′0=0.
We multiply both sides of equation (108) by gjm′t and sum over j to obtain(109)12ddtpm′t22+pmtV2+∫0tk0pm+k′∗pm,pm′tVds+∫0tγ+1yq∗γpm′−q∗pm′,pm′t2ds=g,pm′t2.
Integrating (109) from 0 to t and by integration by parts, we obtain(110)12pm′t22+pmtV2+∫0tk0pm+k′∗pm,pmtVds−∫0tk0pm+k′∗pm,pmVds+∫0tγ+1yq∗γpm′−q∗pm′,pmt2ds−∫0tγ+1yq∗γpm′−q∗pm′,pm2ds=∫0tg,pm′2ds.
By (59), we can obtain the following estimations:(111)∫0tγ+1yq∗γpm′,pmt2ds≤C∫0tpm′2pmtVds≤ɛpmtV2+Cɛ∫0tpm′22ds,(112)∫0tγ+1yq∗γpm′,pm2ds≤C∫0tpm′2pmtVds≤C∫0tpm′22+pmtV2ds.
And we also have the following estimations:(113)∫0tq∗pm′,pmt2ds≤pmt2∫0tq∗pm′2ds≤CpmtV∫0tq∗pm′2ds≤ɛpmtV2+Cɛ∫0tq∗2pm′22ds,(114)∫0tq∗pm′,pm2ds≤∫0tq∗pm′2pm2ds≤C∫0tq∗pm′2pmVds≤C∫0tq∗2pm′22+pmV2ds.
By (111)–(114) with a properly determined ɛ and estimating other terms in (110) as in the proof of Theorem 1, we can arrive at(115)pm′t22+pmtV2≤CgL20,T;H2+C∫0t1+q∗2pm′22+pmV2ds.
Thus, by Gronwall’s lemma, we see that(116)pm∈W0,T∩L∞0,T;V∩W1,∞0,T;H.
Then, following similar arguments in [18], we can ensure that equation (104) admits a weak solution p∈S0,T. Uniqueness will be proved straightforwardly for g=0.
This completes the proof.
Now, we investigate the first-order optimality conditions for minimax optimal control problem (54) for quadratic cost function (103).Theorem 5.
Ifα and β in cost (103) are large enough and the exponent γ≥1, then an optimal control c∗∈Uad and a disturbance η∗∈Vad, namely, an optimal pair q∗=c∗,η∗∈ℱad satisfying (54), can be given by(117)c∗=maxcl,minyq∗,p′2α,cu,η∗=maxηl,min−yq∗,p′2β,ηu,where p is the weak solution of equation (104).Remark 3.
To get the optimality condition in Theorem5, we need to multiply the weak form of equation (104) by z′. Since we just know that z′∈C0,T;H, it would be a just formal procedure. To overcome this difficulty, we employ the regularization method of Lions ([14], pp. 286–288) which was used to deal with linear hyperbolic problems (cf. [15]).Proof of Theorem5.
By the assumptions of Theorem5, we can verify through Theorem 3 that there exists an optimal pair in (54) with cost (103). Let c∗,η∗∈ℱad be an optimal pair in (54) with cost (103) and yq∗ be the corresponding weak solution of equation (52).
By similar arguments in Theorem4, we can know that cost (103) is also (strongly) Gâteaux differentiable at c∗,η∗ in the direction w=h,l∈L20,T2 which satisfies for sufficiently small ϵ>0 that c∗,η∗+ɛw∈ℱad. Thus, we deduce the following:(118)DJc∗,η∗h,l=limɛ⟶0+Jc∗+ɛh,η∗+ɛl−Jc∗,η∗ɛ=∫0Ty′q∗−yd,z′2dt+αc∗,hL20,T−βη∗,lL20,T,where z=Dyq∗w is a solution of equation (71).
As explained above, we prove this theorem by the regularization method. For this purpose, we extend the time domains of equations (71) and (104) to Rt by introducing ψ and θ as the solutions of(119)ψ″+Aψ+k∗Aψ+γ+1yq∗γψ=q∗ψ+h+lyq∗,in0,T,0,otherwise,ψ0=0,ψ′0=0,(120)θ″+Aθ+∫tTKs,θ+Gq∗,θ′ds=y′q∗−yd,in0,T,0,otherwise,θT=0,θ′T=0,where(121)Ks,θ=k0Aθ+∫sTk′σ−sAθdσ,(122)Gq∗,θ′=−γ+1yq∗γθ′+q∗θ′.
In fact, we have(123)ψ=z,in0,T,θ=p,in0,T,(124)∫0Ty′q∗−yd,z′2dt=∫Rtθ″+Aθ+∫tTKs,θ+Gq∗,θ′ds,ψ′2dt.
For the simplicity, we shall denote the scalar product inH or the antiduality V,V′ by ⋅,⋅. Let ρn be a regularizing sequence on Rt. Then, the right-hand side of (124) becomes(125)limn⟶∞∫Rtθ″+Aθ+∫tTKs,θ+Gq∗,θ′ds∗ρn,ψ′∗ρndt.
Now, we integrate by parts of (125). Then, we put(126)Xn=∫Rt−θ′∗ρn,ψ″∗ρn+Aθ∗ρn,ψ′∗ρn+K⋅,θ+Gq∗,θ′∗ρn,ψ∗ρndt.
Using equation (119), that is, ψ″=−Aψ−k∗Aψ+Gq∗,ψ+h+lyq∗, (126) can be given again by(127)Xn=∫Rt−θ′∗ρn,h+lyq∗∗ρndt+∑i=13Yni,where(128)Yn1=∫Rtθ′∗ρn,Aψ∗ρn+Aθ∗ρn,ψ′∗ρndt,Yn2=∫Rtθ′∗ρn,k∗Aψ∗ρn+K⋅,θ∗ρn,ψ∗ρndt,Yn3=∫Rt−θ′∗ρn,Gq∗,ψ∗ρn+Gq∗,θ′∗ρn,ψ∗ρndt.
We immediately knowYn1=0 by integration by parts. By integration by parts, we can see that(129)Yn2=∫Rt−θ∗ρn,k0Aψ+k′∗Aψ∗ρn+K⋅,θ∗ρn,ψ∗ρndt.
Thus,(130)Yn2⟶∫Rt−θ,k0Aψ+k′∗Aψ+Kt,θ,ψdt,as n⟶∞. From (121) and Fubini’s theorem, we know that the right-hand side of (130) is 0. Therefore, we can obtain Yn2⟶0 as n⟶∞. And we can easily verify that Yn3⟶0 as n⟶∞. Consequently, we can obtain(131)limn⟶∞Xn=∫Rt−θ′,h+lyq∗dt=∫0T−p′,h+lyq∗2dt.
Sincec∗,η∗∈ℱad is an optimal pair in (54), we know(132)DcJc∗,η∗h≥0,DηJc∗,η∗l≤0,h,l∈L20,T2.
From (118), (124), and (131), we know that (132) implies(133)αc∗−yq∗,p′2,hL20,T≥0,−βη∗−yq∗,p′2,lL20,T≤0,where h,l∈L20,T2. By similar arguments in the proof of Theorem 4, we can deduce from (133) that(134)c∗=maxcl,minyq∗,p′2α,cu,η∗=maxηl,min−yq∗,p′2β,ηu.
This completes the proof.


## 4. Conclusion

In this paper, we study the bilinear minimax optimal control problems for a semilinear viscoelastic equation with long memory. A global well-posedness theorem regarding the solutions to its Cauchy problem is given. We show the Fréchet differentiability of the nonlinear solution map from the bilinear control input to the solution of the state equation given by the above equation. We formulate the minimax optimal control problem for the state equation. By using and analyzing the properties of the Fréchet derivative of the nonlinear solution map, we show the existence of optimal pairs and find their necessary optimality conditions corresponding to the practical observation cases.



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