# Constraints Optimal Control Governing by Triple Nonlinear Hyperbolic Boundary Value Problem

Authors: Jamil A. Ali Al-Hawasy; Lamyaa H. Ali
Journal: Journal of Applied Mathematics
 (2020)

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## Abstract

The focus of this work lies on proving the existence theorem of a unique state vector solution (Stvs) of the triple nonlinear hyperbolic boundary value problem (TNHBVP) when the classical continuous control vector (CCCVE) is fixed by using the Galerkin method (Galm), proving the existence theorem of a unique constraints classical continuous optimal control vector (CCCOCVE) with vector state constraints (equality EQVC and inequality INEQVC). Also, it consists of studying for the existence and uniqueness adjoint vector solution (Advs) of the triple adjoint vector equations (TAEqs) associated with the considered triple state equations (Tsteqs). The Fréchet Derivative (Frde.) of the Hamiltonian (HAM) is found. At the end, the theorems for the necessary conditions and the sufficient conditions of optimality (Necoop and Sucoop) are achieved.


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## Body

## 1. Introduction

The subject of optimal control problem (OCP) plays a basic role in many real life problems in different branches of sciences; for example, in, medicine [1], engineering and social sciences [2], biology [3], ecology [4], electric power [5], aerospace [6], and many other branches.This role encouraged many researchers to go deeply into studying the OCPS governed by differential equations (deqs). Such OCP problems are studied at the beginning for the systems which are controlled by nonlinear ordinary deqs (nodeqs) [7] or by linear deqs (lpdeqs) [8]. Later great interests have been made to study this subject but for systems which are controlled by pdeqs of elliptic type (ET) [9], or of hyperbolic type (HT) [10], or of parabolic type (PT) [11], or by couple of npdeqs of ET [12], or of PT [13], or of HT [14].Recently, the attention in this subject is magnified to deal with more general types as the studying the CCCOPCVE controlling by TBVP of ET [15], and of PT [16]. All these studies motivated us to look deep inside the CCCOPCVE controlled by TNHBVP.In this work and at first, we give a mathematical description for the CCCOPCVE, and then the TNHBVP is written in its weak form (wkf), and the existence and uniqueness theorem of the Stvs for the TNHBVP using the Galm with the Aubin compactness theorem is proved under appropriate hypotheses when the CCCVE is given. Under reasonable hypotheses, the objective function and the EQVC and INEQVC are proved continuous. The proof of the existence theorem of a CCCOPCVE governed by the TNHBVP is achieved. Under a certain hypotheses, the study of the existence theorem for a unique Advs of the TAEqs associated with the considered Tsteqs is done. The Fréchet Derivative (Frde.) of the HAM is found. Finally, the Necoop and the Sucoop theorems for the CCCOPCVE are proved.


### 1.1. Problem Description

LetI=0,T, T<∞, E be a bounded and open region in ℝ2 with Lipschitz (Lip.) boundary ∂E, Π=E×I, and ∂Π=∂E×I. The considered CCOPCP consists of the Steq which is given by the TNLHPDEqs:(1)ψ1tt−Δψ1+ψ1−ψ2−ψ3=k1x,t,ψ1,ω1,inΠ,(2)ψ2tt−Δψ2+ψ2+ψ3+ψ1=k2x,t,ψ2,ω2,inΠ,(3)ψ3tt−Δψ3+ψ3+ψ1−ψ2=k3x,t,ψ3,ω3,inΠ,with the BCs and ICs.(4)ψ1x,t=0,ψ2x,t=0,ψ3x,t=0,on∂Π,(5)ψ1x,0=ψ10x,ψ2x,0=ψ20x,ψ3x,0=ψ30x,onE,(6)ψ1tx,0=ψ11x,ψ1tx,0=ψ21x,ψ3tx,0=ψ31x,onE,where ψ⟶=ψ1,ψ2,ψ3∈H1E3 is the Stvs, corresponding to the CCCVE ω⟶=ω1,ω2,ω3∈L2Π3 and k1,k2,k3∈L2Π3 is a function defined on Π×ℛ×∁1×Π×ℛ×∁2×Π×ℛ×∁3 with ∁i⊂ℛ, for i=1,2,3.The controls set areω⟶∈W^⟶,W^⟶⊂L2Π3 with W^⟶=W^⟶∈L2Π3W^⟶∈∁⟶,a.e.inΠ, with ∁⟶⊂ℛ3.The cost function is(7)M0ω⟶=∑i=13∫Πm0ix,t,ψi,ωidxdt.The EQVC and INEQVC on the state vectors are(8)Mrω⟶=∑i=13∫Πmrix,t,ψi,ωidxdt=0,1≤r≤p,(9)Mrω⟶=∑i=13∫Πmrix,t,ψi,ωidxdt≤0,p+1≤r≤q.The set of admissible control vector isW^⟶A=ω⟶∈W^⟶Mrω⟶=0,Mr+pω⟶≤0,1≤r≤p.The continuous optimal control problem is to findω⟶∈W^⟶A such that M0ω⟶=w^⟶∈W^A⟶minM0w^⟶.Letϒ⟶=.ϒ1×.ϒ2×.ϒ3=υ⟶:υ⟶∈H1Ω3,withυ1=υ2=υ2=0on∂E, υ⟶=υ1,υ2,υ3.We denote byυ,υ, υ,υ1, and υ⟶,υ⟶1=∑i=13υi,υi1 to the inner products in L2E, H1E, and ϒ⟶ respectively while the norms in these spaces is denoted by υ0, υ1, and υ⟶12=∑i=13υi12, ϒ⟶∗ is denoted the dual of ϒ⟶. Also, the symbol–will be used to indicate that the convergence of a sequence is weak, while the strong convergence of a sequence will be indicated by ⟶.The wkf of problem (1)–(6) when ψ⟶∈H01E3 is given almost everywhere on I for each υ1∈ϒ1, υ2∈ϒ2, and υ3∈ϒ3:(10)ψ1tt,υ1+∇ψ1,∇υ1+ψ1−ψ2−ψ3,υ1=k1ψ1,ω1,υ1,ψ1⋅,t∈Y1,(11)ψ10,υ1=ψ10,υ1,ψ11,υ1=ψ1t0,υ1,(12)ψ2tt,υ2+∇ψ2,∇υ2+ψ2+ψ1+ψ3,υ2=k2ψ2,ω2,υ2,ψ2⋅,t∈Y2,(13)ψ20,υ2=ψ20,υ2,ψ21,v2=ψ2t0,υ2,(14)ψ3tt,υ3+∇ψ3,∇υ3+ψ3+ψ1−ψ2,υ3=k3ψ3,ω3,υ3,ψ3⋅,t∈Y3,(15)ψ30,υ3=ψ30,υ3,ψ31,υ3=ψ3t0,υ3.The following assumptions (Assums.) are needed to investigate the classical continuous optimal control problem (CCOPCP).Assumption A.
k
i is of the Carathéodory type (Caraty.) on Π×ℛ×∁i and satisfies the following conditions for x,t∈Π and ∀i=1,2,3:(16)kix,t,ψi,ωi≤Fix,t+βiψi,whereψi,ωi∈ℛ,βi>0,Fi∈L2Π,kix,t,ψi,ωi−kix,t,ψ¯i,ωi≤Liψi−ψ¯,whereψi,ψ¯i,ωi∈ℛ,Li>0.


### 1.2. The Solution of the State Equations

In this part, the existence theorem of a unique solution for triple nonlinear hyperbolic partial differential equations (TNLHPDEQs) under AssumptionA is proved when the control vector is given, and the following proposition will be needed.Proposition 1 (see [17]).
SupposeD⊂ℛss=2,3, k:D×ℛn⟶ℛm is of Caraty. It satisfies kυ,x≤αυ+βυxa, for each υ,x∈D×ℛn, where x∈LbD,ℛn, α∈L1D,R, β∈Lb/b−aD,R,a∈0,b, if b≠∞, α=0 if b=∞. Then, the functional Kx=∫Dkυ,xυdv is cont.Theorem 1.
Existence and Uniqueness of the Stvs: with AssumptionA, for any given ω⟶∈L2Π3, the wkf of (10)–(15) has a unique solution ψ⟶=ψ1,ψ2,ψ3 s.t. ψ⟶∈L2I,ϒ3, ψ⟶t=ψ1t,ψ2t,ψ3t∈L2Π3, and ψ⟶tt=ψ1t,ψ2t,ψ3t∈L2I,ϒ∗3.Proof.
Letϒ⟶n=ϒn×ϒn×ϒn⊂ϒ⟶ (for each n) be the set of cont. and piecewise affine function in E. ϒ⟶nn=1∞ be a sequence of subspaces of ϒ⟶, such that ∀υ⟶=υ1,υ2,υ3∈ϒ⟶, there exists a sequence v⟶n with υ⟶n=υ1n,υ2n,υ3n∈ϒ⟶n,∀n, and υ⟶n⟶υ⟶ in Y⟶ ⟹ υ⟶n⟹υ⟶ in L2E3. υ⟶j=υ1j,υ2j,υ3j:j=1,2,…,n be a finite basis of ϒ⟶n (where υ⟶j is cont. function in E, with υ⟶jx=0 on the boundary ∂E) and let ψ⟶n=ψ1n,ψ2n,ψ3n be the Galerkin approximate solution (Galso) to the exact solution ψ⟶=ψ1,ψ2,ψ3 such that(17)ψin=∑j=1ncijtυijx,where cijt are unknown functions of t, ∀j=1,2,…,n, and. ∀i=1,2,3.
The wkf of (10)–(15) is approximated with respect to x using the Galm, and substituting ψint=ζini=1,2,3 in the obtained equations, they become for each υ1,υ2,υ3∈ϒn:(18)ζ1nt,υ1+∇ψ1n,∇υ1+ψ1n,υ1−ψ2n,υ1−ψ3n,υ1=k1ψ1n,ω1,υ1,(19)ψ1n0,υ1=ψ10,υ1,ψ1n1,υ1=ψ11,υ1,(20)ζ2nt,υ+∇ψ2n,∇υ2+ψ2n,υ2+ψ3n,υ2+ψ1n,υ2=k2ψ2n,ω2,υ2,(21)ψ2n0,υ2=ψ20,υ2,ψ2n1,v2=ψ21,υ2,(22)ζ3nt,υ3+∇ψ3n,∇υ3+ψ3n,υ3+ψ1n,υ3−ψ2n,υ3=k3ψ3n,ω3,υ3,(23)ψ3n0,υ3=ψ30,υ3,ψ3n1,υ3=ψ31,υ3,where ψin0=ψin0x=ψinx,0∈ϒn (respectively, ζin0=ψin1=ψin1x=ψintx,0∈L2E) be the projection of yi0 onto V (be the projection of ψi1=ψit onto L2E), ∀i=1,2,3, i.e.,(24)ψin0⟶ψi0inV,withψ⟶n01≤b0andψ⟶n00≤b0,(25)ψin1⟶ψi1,inL2Eandψ⟶n10≤b1.
Substituting (17) for each i=1,2,3, respectively, in the pairs (18) and (19), (20) and (21), and in (22) and (23), setting υ1=υ1i, υ2=υ2i, υ3=υ3i, the obtained equations will be written as the following 1st order nodeqs with their ICs and have a unique solution ψ⟶n∈CI,ϒ⟶, i.e., for each l=1,2,3 and m¯=0,1:(26)A1D1′t+B1C1t−E1C2t−H1C3t=b1Y¯1TxC1t,A1C10=b10,A1D10=b11,A2D2′t+B2C2t+E2C1t+H2C3t=b2Y¯2TxC2t,A2C20=b20,A2D20=b21,A3D3′t+B3C3t+E3C1t−H3C3t=b3Y¯3TxC3t,A3C30=b30,A3D30=b31,where Clt=cljtn×1, C2′t=c2j′tn×1, Dl′t=dlj′tn×1, Dlt=dljtn×1, bl=blin×1, bli=klVlTclt,ωl,υli, blm¯=bljm¯, bljm¯=ψlm¯,vlj, Al=alijn×n, alij=υlj,υli, E1=e1ijn×n, e1ij=υ2j,υ1i, H1=f1ijn×n f1ij=υ3j,υ1i, E2=e2ijn×n, e2ij=υ1j,υ2i, H2=f2ijn×n, f2ij=υ3j,υ2i, E3=e3ijn×n, e3ij=υ1j,υ3i, H3=f3ijn×n, f3ij=υ2j,υ3i, Bl=blijn×n, and blij=∇υlj,∇υli+υlj,υli.
Then, corresponding to the sequenceϒ⟶n, there exists a sequence of the following approximation problems, i.e., for each υ⟶n=υ1n,υ2n,υ3n⊂ϒ⟶n, and n=1,2,…:(27)ψ1ntt,υ1n+∇ψ1n,∇υ1n+ψ1n−ψ2n−ψ3n,υ1n=k1ψ1n,ω1,υ1n,(28)ψ1n0,υ1n=ψ10,υ1n,ψ1n1,υ1=ψ11,υ1n,(29)ψ2ntt,υ2n+∇ψ2n,∇υ2n+ψ2n+ψ3n+ψ1n,υ2n=k2ψ2n,ω2,υ2n,(30)ψ2n0,υ2n=ψ20,υ2n,ψ2n1,υ2=ψ21,υ2n,(31)ψ3ntt,υ3n+∇ψ3n,∇υ3n+ψ3n+ψ1n−ψ2n,υ3n=k3ψ3n,ω3,υ3n,(32)ψ3n0,υ3n=ψ30,υ3n,ψ3n1,υ3=ψ31,υ3n,which has a sequence of unique solution ψ⟶n. Substituting υin=ψint for i=1,2,3 in (25)–(27), respectively, adding the three obtained equations together, and employing Lemma 1.2 in [18] for the first term of the LHS, to get (33) which is given by(33)ddtψ⟶ntt02+ψ⟶n12=2ψ2n,ψ1nt+ψ3n,ψ1nt−ψ1n,ψ2nt−ψ3n,ψ2nt−ψ1n,ψ3nt+ψ2n,ψ3nt+k1ψ1n,ω1,ψ1nt+k2ψ2n,ω2,ψ2nt+k3ψ3n,ω3,ψ3nt.
Or (33) can be rewritten as (34) which is(34)ddtψ⟶ntt02+ψ⟶n12≤2ψ2n,ψ1nt+ψ3n,ψ1nt+ψ1n,ψ2nt+ψ3n,ψ2nt+ψ1n,ψ3nt+ψ2n,ψ2nt+k1ψ1n,ω1,ψ1nt+k2ψ2n,ω2,ψ2nt+k3ψ3n,ω3,ψ3nt.
Using AssumptionA for the RHS of (34), integrating both sides on 0,t, using ψin0≤ψin1≤ψ⟶n1, and ψint0≤ψ⟶nt0, we get (35)∫0tddtψ⟶ntt02+ψ⟶n12dt≤∫0tψ⟶nt02+ψ⟶n12dt+∫0t∑i=13Fi02dt+β4∫0tψ⟶nt02+ψ⟶n12dt+∫0tψ⟶nt02dt≤∑i=13FiQ2+β6∫0tψ⟶nt02+ψ⟶n12dt≤β8+β7∫0tψ⟶nt02+ψ⟶n12dt,where β4=∑i=13βi, β5=1+β4, β6=2+β4, β7=maxβ5,β6, β8=∑i=13bi′.
Sinceψ⟶n01≤b1, and ψ⟶n10≤b0, with β9=b0+b1+β7, then (35) is reduced to(36)ψ⟶ntt02+ψ⟶nt12≤β9+β7∫0tψ⟶n02+ψ⟶nt12dt.
Appling the Belman–Gronwall (BGin) inequality, the abovementioned inequality gives∀t∈0,T:(37)ψ⟶ntt02+ψ⟶nt12≤β9eβ7=b2c⇛ψ⟶ntt02≤b2c,ψ⟶nt12≤b2c,∀t∈0,T.
Easily, one can obtain thatψ⟶nttQ≤b1candψ⟶ntL2I,Y≤bc.
Then, by applying Alaoglu’s theorem (Algth),ψ⟶nn∈N has a subsequence; it is not loss of generality to say ψ⟶nn∈N such that ψ⟶nt⇀ψ⟶ in L2Π3 and ψ⟶n⇀ψ⟶ in L2I,Y3, and(38)L2ℛ,Y3⊂L2ℛ,E3≅L2ℛ,E∗3⊂L2ℛ,Y∗3.
Then, the Aubin compactness theorem [18] can be applied here to get that ψ⟶n⟶ψ⟶ in L2Π3. Now, multiplying both sides of (27) and (29), and (31) by χit∈C20,T, such that χiT=χi′T=0, χi0≠0, χi′0≠0,∀i=1,2, integrating on, finally integrating by parts twice the 1st term of each one of the obtained three equations, led to(39)−∫0Tddtψ1n,υ1nχ1′tdt+∫0T∇ψ1n,∇υ1nχ1t+ψ1n,υ1n−ψ2n,υ1n−ψ3n,υ1nχ1tdt=∫0Tk1ψ1n,ω1,υ1nχ1tdt+ψ1n1,υ1nχ10,(40)∫0Tψ1n,υ1nχ1″tdt+∫0T∇ψ1n,∇υ1nχ1t+ψ1n,υ1n−ψ2n,υ1n−ψ3n,υ1nχ1tdt=∫0Tk1ψ1n,ω1,υ1nχ1tdt+ψ1n1,υ1nχ10+ψ1n0,υ1nχ1′0,(41)−∫0Tddtψ2n,υ2nχ2′tdt+∫0T∇ψ2n,∇υ2nχ2t+ψ2n,υ2n+ψ3n,υ2n+ψ1n,υ2nχ2tdt=∫0Tk2ψ2n,ω2,υ2nχ2tdt+ψ2n1,υ2nχ20,(42)∫0Tψ2n,υ2nχ2″tdt+∫0T∇ψ2n,∇υ2nχ2t+ψ2n,υ2n+ψ3n,υ2n+ψ1n,υ2nχ2tdt=∫0Tk2ψ2n,ω2,υ2nχ2tdt+ψ2n1,υ2nχ20+ψ2n0,υ2nχ2′0,(43)−∫0Tddtψ3n,υ3nχ3′tdt+∫0T∇ψ3n,∇υ3nχ3t+ψ3n,υ3n+ψ1n,υ3n−ψ2n,υ3nχ3tdt=∫0Tk3ψ3n,ω3,υ3nχ3tdt+ψ3n1,υ3nχ30,(44)∫0Tψ3n,υ3nχ3″tdt+∫0T∇ψ3n,∇υ3nχ3t+ψ3n,υ3n+ψ1n,υ3n−ψ2n,υ3nχ3tdt=∫0Tk3ψ3n,ω3,υ2nχ3tdt+ψ3n1,υ3nχ30+ψ3n0,υ3nχ3′0.
Now, for eachi=1,2,3, we have the following convergences:First, sinceυin⟶υiinY⟶υinχit⟶υiχitυinχi′t⟶υiχi′t,inL2I,Yυinχi0⟶υiχi0,inL2EOn the other hand, sinceυin⟶υiinL2E⟶υinχi′t⟶υiχi′tυinχi″t⟶υiχi″t,inL2Πυinχi′0⟶υiχi′0,inL2ESecond, we haveψint⇀ψit,inL2Πψin⇀ψi,inL2I,Yψin⟶ψi,inL2ΠThird, letwin=υinχi and wi=υiχi, then win⟶wi in L2Π and win is measurable in E, so from Assumption (A–I) and Proposition 1, the integral ∫Πkix,t,ψin,ωiwindxdt is cont. with respect to ψin,ωi,win, then(45)∫0Tkiψin,ωi,υinχitdt⟶∫0Tkiψi,ωi,υiχitdt.Now, from these convergences and (24) and (25), for i=1, we can passage to the limits in (39) and (40) to get(46)−∫0Tψ1t,υ1χ1′tdt+∫0T∇ψ1,∇υ1+ψ1−ψ2−ψ3,υ1χ1tdt=∫0Tk1ψ1,ω1,υ1χ1tdt+ψ11,υ1χ10,(47)∫0Tψ1,υ1χ1″tdt+∫0T∇ψ1,∇υ1+ψ1−ψ2−ψ3,υ1χ1tdt=∫0Tk1ψ1,ω1,υ1χ1tdt+ψ11,υ1χ10+ψ10,υ1χ1′0,Case 1.
Chooseχi∈C20,T, s.t. χi0=χi′0=χiT=χi′T=0. Using these values in (47) (for i=1), using integration by parts twice for the first terms in the LHS of the obtained equation, yields to(48)∫0Tψ1tt,υ1χ1tdt+∫0T∇ψ1,∇υ1+ψ1−ψ2−ψ3,υ1χ1tdt=∫0Tk1ψ1,ω1,υ1χ1tdt,which give that ψ1 is a solution of (10) (a.e. on I).
Similar way can be used fori=2,3, with (41)–(44) respectively to get that ψ2 and ψ3 are solutions of and (12) and (14) respectively (a.e. on I).Case 2.
Chooseχi∈C20,T, such that χiT≠0 and χi0≠0. For i=1, integrating both sides of (10) on 0,T after multiplying it by χ1t, using integrating by parts for the first term in the LHS of the obtained equation, then subtracting the obtained equation from (41), we get ψi1,υiχi0=ψit0,υiχi0.
Also, similar way can be used but fori=2,3 with the pairs (12) and (47) and (14) and (48), respectively, to get the same result.Case 3.
Chooseχi∈C20,T, such that χi0=χiT=χi′T=0, χi′0≠0. For i=1, integrating both sides of (10) on 0,T after multiplying it by χ1t, using integrating by parts for the first term in the LHS of the obtained equation, then subtracting this obtained equation from (47), we get ψi0,υiχi′0=ψi0,υiχi′0. Also, for i=2,3 and by using (12) and (14), we can use a similar way to get the same result.
From the last two cases, easily we can get the ICs (11) and (13), and (15).
Uniqueness of the solution: letψ⟶=ψ1,ψ2,ψ3 and ψ¯⟶=ψ¯1,ψ¯2,ψ¯3 be two solutions of the wkf (10)–(15), i.e., ψi and ψ¯i (for each i=1,2,3) are satisfied the wkf (10)–(15), subtracting each equality from the other and letting υi=ψi−ψ¯i, yields to(49)ψi−ψitt,ψi−ψ¯i+ψi−ψ¯i12=kiψi,ωi−kiψ¯i,ωi,ψi−ψ¯i,ψi−ψ¯i0,ψi−ψ¯i0=0,ψi−ψ¯it0,ψi−ψ¯it0=0.
Adding these three equations, using Lemma 1.2 in ref. [18] on the first term in LHS of the obtained equation which will be positive, integrating both sides from 0 to t, using the initial conditions, the Lipshctiz property on the RHS, and lastly applying the B–G inequality, to get(50)∫0tddtψ⟶−ψ¯⟶tt02+2ψ⟶−ψ¯⟶12dt≤2L∫0tψ⟶−ψ¯⟶tt02+ψ⟶−ψ¯⟶12dtψ⟶−ψ¯⟶tt02+ψ⟶−ψ¯⟶12≤⇛ψ⟶−ψ¯⟶t12=0,∀t∈I⇛ψ⟶−ψ¯⟶tL2I,Υ=0⇛,the solution is unique.Lemma 1.
In addition to AssumptionA, if the functions ki (for each i=1,2,3) is Lip. with respect to yi and ωi, and if the control vector is bounded, then the operator ω⟶↦ψ⟶ω⟶ from L2Π3 into L∞I,L2E3 or in to L2I,Y3 or in to L2Π3 is cont.Proof.
Letω⟶=ω1,ω2,ω2,ω¯⟶=ω¯1,ω¯2,ω¯3∈L2Π3, Δω⟶=ω¯⟶−ω⟶∈L2Π3, then by Theorem 1, ψ⟶=ψ⟶ω⟶=ψ1,ψ2,ψ3 and ψ¯⟶=ψ¯⟶ω⟶=ψ¯1,ψ¯2,ψ¯3 are their corresponding states’ solutions which satisfy the wkf of (10)–(15), setting Δψ⟶=Δψ1,Δψ2,Δψ3=ψ¯⟶−ψ⟶, then(51)Δψ1tt,υ1+∇Δψ1,∇υ1+Δψ1−Δψ2−Δψ3,υ1=k1ψ1+Δψ1,ω1+Δω1−k1ψ1,ω1,υ1,(52)Δψ1x,0=0,Δψ1tx,0=0,(53)Δψ2tt,υ2+∇Δψ2,∇υ2+Δψ2+Δψ1+Δψ3,υ2=k2ψ2+Δψ2,ω2+Δω2−k2ψ2,ω2,υ2,(54)Δψ2x,0=0,Δψ2tx,0=0,(55)Δψ3tt,v3+∇Δψ3,∇v3+Δψ3+Δψ1−Δψ2,v3=k3ψ3+Δψ3,ω3+Δω3−k3ψ3,ω3,υ3,(56)Δψ3x,0=0,Δψ3tx,0=0.
Substitutingυi=Δψit for each i=1,2,3 in (51), (53), and (55), respectively, adding the obtained three equations together, and using the same way that we used to get (32), a similar equation can be obtained but with Δψ⟶ in state of ψ⟶n, and then integration of both sides on 0,t, using the Lip. property on ki, i=1,2,3, with respect to each dependent variable, yields(57)∫0tddtΔψ⟶tt02+Δψ⟶12≤2∫0tΔψ2+Δψ3Δψ1t+Δψ1+Δψ3Δψ2tdt+2∫0tΔψ1+Δψ2Δψ3t+L¯1Δψ1+L¯¯1Δω1Δψ1tdt+2∫0tL¯2Δψ2+L¯¯2Δω2Δψ2t+L¯3Δψ3+L¯¯3Δω3Δψ3tdt.
Using the definitions of the norms and the relations between them, we get(58)Δψt⟶t02+Δψ⟶12≤2∫0tΔψ⟶02+Δψt⟶12dt+L˜1∫0tΔψ⟶02+Δψ⟶t12dt+L¯2∫0TΔω⟶02dt+L¯2∫0tΔψt⟶12dt≤L¯2Δω⟶tΠ2+L1∫0tΔψ⟶02+Δψt⟶12dt,where L˜1=maxL¯1,L¯2,L¯3, L¯2=maxL¯¯1,L¯¯2,L¯¯3, and L1=max2+L˜1,2+L˜1+L¯2.
Applying the BGin, withL2=L¯2eL1, we get(59)Δψ⟶tt02+Δψ⟶12≤L2Δω⟶tΠ2,∀t∈I¯⇛Δψ⟶t12≤L2Δω⟶tΠ2,∀t∈I¯⇛Δψ⟶L∞I,L2Ε≤LΔω⟶Π,Δψ⟶L2I,Υ≤LΔω⟶ΠandΔψ⟶Π≤LΔω⟶Π.From the abovementioned three inequalities, the Lip. continuity of the operatorω⟶↦ψ⟶ easily obtained.


### 1.3. The Existence of a Classical Optimal Control

This section concerned with proving the existence theorem with a CCOPCV satisfying the EQVC and INEQVC is studied. Hence, the following assumption and lemmas will be needed.Assumption B.
Considermri (for r=0,…,q and i=1,2,3) is of Caraty. on Π×ℛ×∁i and satisfies ψi∈ℛ and ωi∈∁i:(60)mrix,t,ψi,ωi≤Mrix,t+ckiψi2,whereMri∈L1Π,∀i=1,2,3,∀r=0,…,q.Lemma 2.
With AssumptionB, the functional ω⟶↦Mrω⟶ is cont. on L2Π3.Proof.
Using AssumptionB and Proposition 1 gives ∫Qmrix,t,ψi,ωidxdt is cont. on L2Π, ∀i=1,2,3, and ∀r=0,…,q; hence, Mlω⟶ is cont. on L2Π3.Lemma 3 (see [13]).
Letm:Q×ℛ2⟶ℛ is of Caraty. on Π×ℛ×ℛ and satisfies mx,ψ,ω≤mx,t+cψ2, where x,t∈L1Π, ω∈∁, c≥0, ∁⊂ℛ is compact. Then, ∫Πmx,ψ,ωdx is cont. on L2Π, with respect to ψ.Theorem 2.
With the AssumptionsA and B, if the set ∁⟶ is convex (cox.) and compact (com), W^⟶A≠∅, and the function ki ∀i=1,2,3 has the form(61)kix,t,ψi,ωi=ki1x,t,ψi+ki2x,tωi,where ki1x,t,ψi≤ηix,t+ciψi, ki2x,t≤ki, ηi∈L2Π, ci≥0.In additionmri is independent of ωi (∀i=1,2,3, and r=1,…,p), mri (∀i=1,2,3, and r=1,…,p) is cox. with respect to ωi for fixed x,t,ψi, there exists a CCOPCV.Proof.
From the Assumption on∁i⊂ℛ ∀i=1,2,3 and Egorov’s theorem, one obtains that W^⟶1×W^⟶2×W^⟶3=W^⟶ is weakly com. Since W^⟶A≠∅, hence there is ω¯⟶∈W⟶A s.t. rω¯⟶=0,1≤r≤p, Mkω¯⟶≤0, for p+1≤r≤q and there is a minimizing sequence ω⟶ρ s.t. ω⟶ρ∈W^⟶A,∀ρ, which satisfies ρ⟶∞limM0ω⟶ρ=ω¯⟶∈W⟶AinfM0ω¯⟶. Since ω⟶ρ∈W^⟶A,∀ρ and W^⟶ is weakly com., then ω⟶ρ has a subsequence say again ω⟶ρ which converges weakly to some ω⟶ in W^⟶, i.e., ω⟶ρ⇀ω in L2Π3 and ω⟶ρΠ≤c,∀ρ. From Theorem 1, for any given control ω⟶ρ, then ψ⟶ρ=ψ⟶ω⟶ρ is a unique solution for the Tsteqs, and ψ⟶ρL2I,Y, ψ⟶ρtL2Π are bounded, then by Algth, the sequences ψ⟶ρ and ψ⟶ρt have a weakly converging subsequences, say for simplicity, ψ⟶ρ and ψ⟶ρt, i.e.,(62)ψ⟶ρ⇀ψ⟶,inL2I,Y3,ψ⟶ρt⇀ψ⟶t,inL2Π3.
Then, by applying the Aubin compactness theorem [18], the sequence ψ⟶ρ has a strongly converging subsequence, say for simplicity, ψ⟶ρ such that ψ⟶ρ⟶ψ⟶ in L2Π3.
Now, for eachρ, substitute the solution ψ1ρ,ψ2ρ,ψ3ρ in the wkf of (18), (20), and (22), then multiply both sides of each one by χit (with χi∈C20,T, such that χiT=χi′T=0, χi0≠0,χi′0≠0, for i=1,2,3). After rewriting the first terms in the LHS in each one of them, integrating both sides on 0,T, and then by applying integration by parts for these first terms, we get(63)∫0Tddtψ1ρt,υ1χ1tdt+∫0T∇ψ1ρ,∇υ1+ψ1ρ,υ1−ψ2ρ,υ1−ψ3ρ,υ1χ1tdt=∫0Tk11x,t,ψ1ρ+k12x,tω1ρ,υχ1tdt,(64)∫0Tddtψ2ρt,υ2χ2tdt+∫0T∇ψ2ρ,∇υ2+ψ2ρ,υ2+ψ1ρ,υ2+ψ3ρ,υ2χ2tdt=∫0Tk21x,t,ψ2ρ+k22x,tω2ρ,υ2χ2tdt,(65)∫0Tddtψ3ρt,υ3χ3tdt+∫0T∇ψ3ρ,∇υ3+ψ3ρ,υ3+ψ1ρ,υ3−ψ2ρ,υ3χ3tdt=∫0Tk31x,t,ψ3ρ+k32x,tω3ρ,υ3χ3tdt.
One can passage the limits in the LHS of (50)–(52) by applying the same manner which is applied in the proof of Theorem 1 to passage the limits in RHS of these equations; we suppose ∀i=1,2,3, υi∈CΕ¯, wi=υiχit, then wi∈CΠ¯∈L∞I,V⊂L2Π, set k¯i1ψ1ρ=ki1ψiρwi, then k¯i1:Π×ℝ⟶ℝ is of Caraty., using Proposition 1, to get the integral ∫Πki1ψiρwidxdt is cont. with respect to ψiρ, but ψiρ⟶ψi in L2Π and ωiρ⇀ωi in L2Π, then(66)∫Πki1ψiρwidxdt⟶∫Πki1ψiwidxdt,∀wi∈CΠ¯fori=1,2,3,(67)∫Πki2x,tωiρwidxdt⟶∫Πki2x,tωiwidxdt,∀wi∈CΠ¯,fori=1,2,3.
Thus, (66) and (67) are holded for every υi∈ϒ, since CE¯ is dense in ϒ; hence, we get the wkf (10), (12), and (14).
Also, the same manner which is applied in the proof of Theorem2 can be used here to passage the limits in the ICs. Hence, ψ1,ψ2,ψ3 is a solution of the wkf of (10)–(15).
On the other hand, sincemri (for i=1,2,3 and r=1,2,…,p) is independent of ωi and cont. with respect to ψi, by Lemma 2, ∫Πmrix,t,ψiρdxdt is cont. with respect to yi, ψ⟶ρ⟶ψ⟶ in L2Π3, then(68)∫Πmrix,t,ψiρdxdt⟶∫Πmrix,t,ψidxdt.Hence,Mrω⟶=ρ⟶∞limMrω⟶ρ=0.
Now, sincemrix,t,ψi,ωi (for each r=0,1+p,2+p,…,q and i=1,2,3) is cont. with respect to ψi,ωi and since Ci is compact, then using Lemma 3, we get(69)∫Πmrix,t,ψiρ,ωiρdxdt⟶∫Πmrix,t,ψi,ωiρdxdt.
On the other hand,∫Πmrix,t,ψi,ωiρdxdt is cox. and cont. with respect to ωi (since mrix,t,ψi,ωi is cox. and cont. with respect to ωi), then ∫Πmrix,t,ψi,ωidxdt Necessary and Sufficient is weakly lower semicont. (welsc) with respect to ωi, i.e.,(70)∫Πmrix,t,ψi,ωidxdt≤ρ⟶∞liminf∫Πmrix,t,ψi,ωiρ−mrix,t,ψiρ,ωiρdxdt+ρ⟶∞liminf∫Πmrix,t,ψiρ,ωiρdxdt≤ρ⟶∞liminf∫Πmrix,t,ψiρ,ωiρdxdt,by55⇛∑i=12∫Πmrix,t,ψi,ωidxdt≤k⟶∞liminf∑i=12∫Πmrix,t,ψik,ωiρdxdt⇛Mrω⟶≤limρ⟶∞infMrω⟶ρ,then Mrω⟶≤0 r=p+1,…,q since ω⟶ρ∈W^⟶A,∀ρ, and we get that (71)M0ω⟶≤ρ⟶∞liminfM0ω⟶ρ=limρ⟶∞M0ω⟶ρ=ω¯⟶∈W⟶AinfM0ω¯⟶=ω¯⟶∈W⟶AMinM0ω¯⟶.
Thus,ω⟶ is a CCOPCV.Assumption C.
Assume forr=0,…,q and i=1,2,3, the functions ki, kiψi, kiωi, mriψi, and mriωi are defined and are of Caraty. on Π×ℛ×∁′ (where ∁′ is an open subset in ∁) and satisfy kiψix,t,ψi,ωi≤Li, kiωix,t,ψi,ωi≤Li′:(72)mriψix,t,ψi,ωi≤Mri5x,t+cri5ψi,mriωix,t,ψi,ωi≤Mri6x,t+cri6ψi,where x,t∈Π, ψi,ωi∈ℝ, Mri5,Mri6∈L2Π. Li,Li′,cri5,cri6≥0.
Note: for simplicity, in the following theorem, we will drop the indexk from the functions mli and Ml. Also, we assume the Assums. (A), (B), and (C) are considered.Theorem 3.
Consider the TAEqsξ⟶=ξ1,ξ2,ξ3 of Tseqs (1)–(6) are defined by(73)ξ1tt−Δξ1+ξ1+ξ2+ξ3=ξ1k1ψ1x,t,ψ1,ω1+m1ψ1x,t,ψ1,ω1,onΠ,ξ1=0,onΣξ1x,T=ξ1tx,T=0,onE,ξ2tt−Δξ2+ξ2−ξ1−ξ3=ξ2k2ψ2x,t,ψ2,ω2+m2ψ2x,t,ψ2,ω2,onΠ,ξ2=0,onΣ,ξ2x,T=ξ2tx,T=0,onE,ξ3tt−Δξ3+ξ3−ξ1+ξ2=ξ13k3ψ3x,t,ψ3,ω3+m3ψ3x,t,ψ3,ω3,onΠ,ξ3=0,onΣξ3x,T=ξ3tx,T=0,onE,and the Ham is given by ℋx,t,ψ⟶,ω⟶,ξ⟶=∑i=13ζix,t,ψi,ωi, where ζi=ξikix,t,ψi,ωi+mix,t,ψi,ωi, for each i=1,2,3.Then, the Frde. ofG is defined by(74)M′ω⟶Δω⟶=∫Πℋω⟶x,t,ψ⟶,ω⟶,ξ⟶Δω⟶dxdt,where ℋω⟶x,t,ψ⟶,ξ⟶,ω⟶=ζ1ω1,ζ1ω1,ζ1ω1, and Δω⟶=Δω1,Δω2,Δω3, with ζiωi=ξikiωix,t,ψi,ωi+miωix,t,ψi,ωi, mi=∑r=0qκrmri and ξi=∑r=0qκrξli, for each i=1,2,3.Proof.
at first, let the wkf of the TAEqs are given∀υ1,υ2,υ3∈Y, by(75)ξ1tt,υ1+∇ξ1,∇υ1+ξ1+ξ2+ξ3,υ1=ξ1k1ψ1+m1ψ1,υ1,∀v1∈Ya.e.onI,(76)ξ1T,υ1=ξ1tT,υ1=0,(77)zξ2t,υ2+∇ξ2,∇υ2+ξ2−ξ1−ξ3,υ2=ξ2k2ψ2+m2ψ2,υ2,∀υ2∈Ya.e.onI,(78)ξ2T,υ2=ξ2tT,v2=0(79)ξ3t,υ3+∇ξ3,∇υ3+ξ3−ξ1+ξ2,υ3=ξ3k3ψ3+m3ψ3,υ3,∀υ2∈Ya.e.onI,(80)ξ3T,υ3=ξ3tT,υ3=0.
From the assumptions and using the same manner which is applied in the proof of Theorem1, once can prove that the wkf (75)–(80) has a unique solution ξ⟶=ξ1,ξ2,ξ3∈L2Π3.
Substitutingυi=Δψi for each i=1,2,3 in (75), (77), and (79) and integrating both sides on 0,T, we get(81)∫0TΔψ1,ξ1tt+∇ξ1,∇Δψ1+ξ1+ξ3+ξ2,Δψ1dt=∫0Tξ1k1ψ1+g1ψ1,Δψ1dt,(82)∫0TΔψ2,ξ2tt+∇ξ2,∇Δψ2+ξ2−ξ1−ξ3,Δψ2dt=∫0Tξ2k2ψ2+g2ψ2,Δψ2dt,(83)∫0TΔψ3,ξ3tt+∇ξ3,∇Δψ3+ξ3−ξ1+ξ2,Δψ3dt=∫0Tξ3k3ψ3+g3ψ3,Δψ3dt.
Now, letω⟶,ω¯⟶∈L2Π3, Δω⟶=ω¯⟶−ω⟶∈L2Q3, and then by Theorem 1, ψ⟶=ψ⟶ω⟶ and ψ¯⟶=ψ⟶ω¯⟶ are their corresponding solutions. Set Δψ⟶=Δψ1,Δψ2,Δψ3=ψ¯⟶−ψ⟶, substitute υi=ξi for each i=1,2,3 in (51), (53), and (55), integrate both sides on 0,T, and then integrate by parts twice the first term in the LHS of each equation. Finding for each i=1,2,3 the Frde. of ki in the RHS of each equation which are exist from the Assumption C, then by Lemma 1, and the inequality of Minkowiski, one has(84)∫0TΔψ1,ξ1tt+∇Δψ1,∇ξ1+Δψ1−Δψ2−Δψ3,ξ1dt=∫0Tk1ψ1Δψ1+k1ω1Δω1,ξ1dt+O1Δω⟶Δω⟶Π,(85)∫0TΔψ2,ξ2tt+∇Δψ2,∇ξ2+Δψ2+Δψ1+Δψ3,ξ2dt=∫0Tk2ψ2δψ2+k2ω2Δω2,ξ2dt+O2Δω⟶Δω⟶Π,(86)∫0TΔψ3,ξ3tt+∇δψ3,∇ξ3+Δψ2ε+Δψ1−Δψ2,ξ3dt=∫0Tk3ψ3Δψ3+k3ω3Δω3,ξ3dt+O3Δω⟶Δω⟶Π.
Subtracting (84)–(86) from, respectively, adding the obtain equations, we get(87)∫0T∑i=13kiωiΔωi,ξidt+O4Δω⟶=∫0T∑i=13miψi,Δψidt,where. O4ε⟶0,asΔω⟶⟶0,withO4Δω⟶=∑i=13OiΔω⟶Δω⟶Π.
On the other hand, from the AssumptionB on mi (for i=1,2,3), the Frde definition, Lemma 1, and by applying the inequality of Minkowiski, we obtain(88)Mω⟶+Δω⟶−Mω⟶=∫Π∑i=13miψiΔψi+miωiΔωidxdt+O5Δω⟶Δω⟶Π,where O5Δω⟶⟶0, as Δω⟶⟶0.
Now, by substituting (87) in (88), one has(89)Mω⟶+Δω⟶−Mω⟶=∫Π∑i=13ζiωix,t,ψi,ωiΔωidxdt+O6Δω⟶Δω⟶Π,where O6Δω⟶=O5Δω⟶+O5Δω⟶⟶0, as Δω⟶⟶0.
Finally, from the Frde. of, we get(90)M′ω⟶Δω⟶=∫Πℋω⟶x,t,ψ⟶,ω⟶,ξ⟶Δω⟶dxdt.


### 1.4. Necessary and Sufficient Conditions for Optimality

This section deals with the theorems for the Necoop necessary under certain hypotheses which are proved as follows:Theorem 4.
Necoop (multipliers theorem):(a)
with Assumptions A, B, and C, ifW^⟶ is cox. and the ω⟶∈W^⟶A is optimal, then there are multipliers κr∈ℝ,r=0,1,…,p,p+1,…,q with κr≥0, for r=0,p+1,…,q,∑r=0qκr=1 such that the following Kuhn-Tucker-Lagrange (K. T. L.) conditions are holded:(91)∫Πℋω⟶x,t,ψ⟶,ω⟶,ξ⟶Δω⟶dxdt≥0,∀ω¯⟶∈W^⟶,Δω⟶=ω¯⟶−ω⟶,(92)κrMrω⟶=0,forr=p+1,…,q.(b)
Inequality (91) is equivalent to the (weak) piecewise minimum principle(93)ℋω⟶x,t,ψ⟶,ξ⟶,ω⟶⋅ω⟶t=ℋω⟶ω¯⟶∈∁⟶minx,t,ψ⟶,ξ⟶,ω⟶⋅ω¯⟶t,a.e.onΠ.Proof.
(a)
From Lemma2, the functional Mrω⟶ (for r=0,1,…,q) is cont., and from Theorem 3, the functional Mr′ (for r=0,1,…,q) is cont. with respect to ω¯⟶−ω⟶ and linear in ω¯⟶−ω⟶, then Mk′ is L− differential for every L, and then applying the K. T. L. theorem [5], there are multipliers κr∈ℝ,r=0,1,…,q with κr≥0, for r=0,p+1,…,q,∑r=0qκr=1, such that (91)–(93) are satisfied, by using Theorem 3, then (91) becomes ∑r=0qκr∫Π∑i=13ζiωix,t,ψi,ωiΔωidxdt≥0, which can be rewritten as(94)∫Πζ⟶ω⟶⋅ω¯⟶k−ω⟶dxdt≥0,∀ω¯⟶∈W^⟶withζ⟶ω⟶=ζ1ω1,ζ2ω2,ζ3ω3.
(b)
Letω¯⟶l be a dense sequence (dse) in W^⟶, m denotes the Lebesgue measure on Π, and Γ⊂Π be a measurable subset with property ω¯⟶x,t=ω¯⟶lx,t,ifx,t∈Γω⟶x,t,ifx,t∉Γ.
Therefore, (94) becomes(95)∫Γζ⟶ω⟶⋅ω¯⟶−ω⟶dxdt≥0,which implies toζ⟶ω⟶⋅ω¯⟶l−ω⟶≥0, a.e.on∏.
This means the inequality is satisfied on the whole regionQ except in a subset Πl such that mΠl=0,∀l, where m represents the Lebesgue measure; thus, the inequality holds on Π except in the union ∪lΠl with m∪lΠl=0, but ω¯⟶l is a dse in W^⟶, then there is ω¯⟶∈W^⟶ such that ∫Πζ⟶ω⟶⋅ω¯⟶l−ω⟶≥0, a.e. on Π,∀ω¯⟶∈W^⟶.
That is, (91) gives (94). The converse is clear.Theorem 5.
Sucoop: besides the Assumptions A, B, and C, supposeW^⟶ is cox., with ∁⟶ cox., ki and mri ∀r=1,2,…,p are affine with respect to ψi,ωi for each x,t, mri ∀r=0,p+1,…,q are cox. with respect to ψi,ωi ∀x,t, for i=1,2,3. Then, the Necoop for Theorem 4 with κ0>0 are sufficient.Proof.
Assumeω⟶∈W^⟶A satisfies the condition (91) and (92). Let Mω⟶=∑r=0qκrMrω⟶, then using Theorem 3, we get(96)M′ω⟶Δω⟶=∑r=0qκrMr′ω⟶Δω⟶=∑r=0qκr∫Π∑i=13ζiωi⋅δωidxdt≥0.
Sincekix,t,ψi,ωi=ki1x,tψi+ki2x,tωi+ki3x,t,i=1,2,3.
Letω=ω1,ω2,ω3 and ω¯⟶=ω¯1,ω¯2,ω¯3 be two given control vectors, then ψ⟶=ψω1,ψω2,ψω3=ψ1,ψ2,ψ3 and ψ¯⟶=ψ¯ω1,ψ¯ω¯2,ψ¯ω¯3=ψ¯1,ψ¯2,ψ¯3 represent the corresponding stats solutions. Substituting ω⟶,y⟶ in (1)–(6), multiplying all the obtained equalities by α∈0,1 once and on the other hand substituting ω¯⟶,y¯⟶ in (1)–(6), multiplying all the obtained equalities by 1−α once again, and lastly adding each pair from the corresponding equalities together, we obtain(97)αψ1+1−αψ¯1tt−Δαψ1+1−αψ¯1+αψ1−ψ2−ψ3+1−αψ¯1−ψ¯2−ψ¯3=αk11x,tψ1+k12x,tω1+1−αk11x,tψ¯1+k12x,tω¯1+k13x,t,(98)αψ1x,t+1−αψ¯1x,0=0,(99)αψ1x,0+1−αψ¯1x,0=ψ10x,αψ1tx,0+1−αψ¯1tx,0=ψ11x,(100)αψ2+1−αψ¯2tt−Δαψ2+1−αψ¯2+αψ2+ψ1+ψ3+1−αψ¯2+ψ¯1+ψ¯3=αk21x,tψ2+k22x,tω2+1−αk21x,tψ¯2+k22x,tω¯2+k23x,t,(101)αψ2x,t+1−αψ¯2x,0=0,(102)αψ2x,0+1−αψ¯2x,0=ψ20x,αψ2tx,0+1−αψ¯2x,0=ψ21x,(103)αψ3+1−αψ¯3tt−Δαψ3+1−αψ¯3+αψ3+ψ1−ψ2+1−αψ¯3+ψ¯1−ψ¯2=αk31x,tψ3+k32x,tω3+1−αk31x,tψ¯3+k32x,tω¯3+k23x,t,(104)αψ3x,t+1−αψ¯3x,0=0,(105)αψ3x,0+1−αψ¯3x,0=ψ30x,αψ3tx,0+1−αψ¯3tx,0=ψ31x.
Equations (97)–(105) show that if the control vector is ω˜⟶=ω˜1,ω˜2,ω˜3 with ω˜⟶=ω⟶+1−αω¯⟶ then its corresponding state vector is ψ˜⟶=ψ˜1,ψ˜2,ψ˜3 with ψ˜i=ψiω˜i=ψiαωi+1−αω¯i=αψi+1−αψ¯i,∀i=1,2,3. This gives the operator ω⟶↦y⟶ω⟶ is convex–linear with respect to ψ⟶,ω⟶ for any x,t∈Π.
On the other hand, the functionMlω⟶ (for k=1,…,p) is convex–linear with respect to ψ⟶,ω⟶ for any x,t∈Π, this backs to the fact that the sum of two affine functions mrix,t,ψi,ωi (for each i=1,2,3, and r=1,…,p with respect to ψi,ωi and ∀x,t∈Π is affine and the operator ω⟶↦ψ⟶ω⟶ is convex-linear.
The functionsMrω⟶,∀r=0,p+1,…,q are cox. with respect to ψ⟶,ω⟶,∀x,t∈Π (from the assumptions on the functions mr1 and mr2, ∀r=0,p+1,…,q). Hence, Mω⟶ is cox. with respect to ψ⟶,ω⟶,∀x,t∈Π in the cox. set W^⟶, and has a cont. Fréd satisfies M′ω⟶⋅ω¯⟶−ω⟶≥0⇛Mω⟶ has a minimum at ω⟶⇛Mω⟶≤Mω¯⟶,∀ω¯⟶∈W^⟶⇛(106)∑r=0pκrMrω⟶+∑r=p+1qκrMrω⟶≤∑r=0pκrMrω¯⟶+∑r=p+1qκrMrω¯⟶,∀ω¯⟶∈W^⟶.
Letω¯⟶∈W^⟶A, with kk≥0 (for r=p+1,…,q), then from (8), (9), and (77), the abovementioned inequality becomes(107)κ0M0ω⟶≤κ0M0ω¯⟶,∀ω¯⟶∈W^⟶⇛G0ω⟶≤G0ω¯⟶,∀ω¯⟶∈W^⟶⇛∴ω⟶is a CCOCVE.


## 1.1. Problem Description

LetI=0,T, T<∞, E be a bounded and open region in ℝ2 with Lipschitz (Lip.) boundary ∂E, Π=E×I, and ∂Π=∂E×I. The considered CCOPCP consists of the Steq which is given by the TNLHPDEqs:(1)ψ1tt−Δψ1+ψ1−ψ2−ψ3=k1x,t,ψ1,ω1,inΠ,(2)ψ2tt−Δψ2+ψ2+ψ3+ψ1=k2x,t,ψ2,ω2,inΠ,(3)ψ3tt−Δψ3+ψ3+ψ1−ψ2=k3x,t,ψ3,ω3,inΠ,with the BCs and ICs.(4)ψ1x,t=0,ψ2x,t=0,ψ3x,t=0,on∂Π,(5)ψ1x,0=ψ10x,ψ2x,0=ψ20x,ψ3x,0=ψ30x,onE,(6)ψ1tx,0=ψ11x,ψ1tx,0=ψ21x,ψ3tx,0=ψ31x,onE,where ψ⟶=ψ1,ψ2,ψ3∈H1E3 is the Stvs, corresponding to the CCCVE ω⟶=ω1,ω2,ω3∈L2Π3 and k1,k2,k3∈L2Π3 is a function defined on Π×ℛ×∁1×Π×ℛ×∁2×Π×ℛ×∁3 with ∁i⊂ℛ, for i=1,2,3.The controls set areω⟶∈W^⟶,W^⟶⊂L2Π3 with W^⟶=W^⟶∈L2Π3W^⟶∈∁⟶,a.e.inΠ, with ∁⟶⊂ℛ3.The cost function is(7)M0ω⟶=∑i=13∫Πm0ix,t,ψi,ωidxdt.The EQVC and INEQVC on the state vectors are(8)Mrω⟶=∑i=13∫Πmrix,t,ψi,ωidxdt=0,1≤r≤p,(9)Mrω⟶=∑i=13∫Πmrix,t,ψi,ωidxdt≤0,p+1≤r≤q.The set of admissible control vector isW^⟶A=ω⟶∈W^⟶Mrω⟶=0,Mr+pω⟶≤0,1≤r≤p.The continuous optimal control problem is to findω⟶∈W^⟶A such that M0ω⟶=w^⟶∈W^A⟶minM0w^⟶.Letϒ⟶=.ϒ1×.ϒ2×.ϒ3=υ⟶:υ⟶∈H1Ω3,withυ1=υ2=υ2=0on∂E, υ⟶=υ1,υ2,υ3.We denote byυ,υ, υ,υ1, and υ⟶,υ⟶1=∑i=13υi,υi1 to the inner products in L2E, H1E, and ϒ⟶ respectively while the norms in these spaces is denoted by υ0, υ1, and υ⟶12=∑i=13υi12, ϒ⟶∗ is denoted the dual of ϒ⟶. Also, the symbol–will be used to indicate that the convergence of a sequence is weak, while the strong convergence of a sequence will be indicated by ⟶.The wkf of problem (1)–(6) when ψ⟶∈H01E3 is given almost everywhere on I for each υ1∈ϒ1, υ2∈ϒ2, and υ3∈ϒ3:(10)ψ1tt,υ1+∇ψ1,∇υ1+ψ1−ψ2−ψ3,υ1=k1ψ1,ω1,υ1,ψ1⋅,t∈Y1,(11)ψ10,υ1=ψ10,υ1,ψ11,υ1=ψ1t0,υ1,(12)ψ2tt,υ2+∇ψ2,∇υ2+ψ2+ψ1+ψ3,υ2=k2ψ2,ω2,υ2,ψ2⋅,t∈Y2,(13)ψ20,υ2=ψ20,υ2,ψ21,v2=ψ2t0,υ2,(14)ψ3tt,υ3+∇ψ3,∇υ3+ψ3+ψ1−ψ2,υ3=k3ψ3,ω3,υ3,ψ3⋅,t∈Y3,(15)ψ30,υ3=ψ30,υ3,ψ31,υ3=ψ3t0,υ3.The following assumptions (Assums.) are needed to investigate the classical continuous optimal control problem (CCOPCP).Assumption A.
k
i is of the Carathéodory type (Caraty.) on Π×ℛ×∁i and satisfies the following conditions for x,t∈Π and ∀i=1,2,3:(16)kix,t,ψi,ωi≤Fix,t+βiψi,whereψi,ωi∈ℛ,βi>0,Fi∈L2Π,kix,t,ψi,ωi−kix,t,ψ¯i,ωi≤Liψi−ψ¯,whereψi,ψ¯i,ωi∈ℛ,Li>0.


## 1.2. The Solution of the State Equations

In this part, the existence theorem of a unique solution for triple nonlinear hyperbolic partial differential equations (TNLHPDEQs) under AssumptionA is proved when the control vector is given, and the following proposition will be needed.Proposition 1 (see [17]).
SupposeD⊂ℛss=2,3, k:D×ℛn⟶ℛm is of Caraty. It satisfies kυ,x≤αυ+βυxa, for each υ,x∈D×ℛn, where x∈LbD,ℛn, α∈L1D,R, β∈Lb/b−aD,R,a∈0,b, if b≠∞, α=0 if b=∞. Then, the functional Kx=∫Dkυ,xυdv is cont.Theorem 1.
Existence and Uniqueness of the Stvs: with AssumptionA, for any given ω⟶∈L2Π3, the wkf of (10)–(15) has a unique solution ψ⟶=ψ1,ψ2,ψ3 s.t. ψ⟶∈L2I,ϒ3, ψ⟶t=ψ1t,ψ2t,ψ3t∈L2Π3, and ψ⟶tt=ψ1t,ψ2t,ψ3t∈L2I,ϒ∗3.Proof.
Letϒ⟶n=ϒn×ϒn×ϒn⊂ϒ⟶ (for each n) be the set of cont. and piecewise affine function in E. ϒ⟶nn=1∞ be a sequence of subspaces of ϒ⟶, such that ∀υ⟶=υ1,υ2,υ3∈ϒ⟶, there exists a sequence v⟶n with υ⟶n=υ1n,υ2n,υ3n∈ϒ⟶n,∀n, and υ⟶n⟶υ⟶ in Y⟶ ⟹ υ⟶n⟹υ⟶ in L2E3. υ⟶j=υ1j,υ2j,υ3j:j=1,2,…,n be a finite basis of ϒ⟶n (where υ⟶j is cont. function in E, with υ⟶jx=0 on the boundary ∂E) and let ψ⟶n=ψ1n,ψ2n,ψ3n be the Galerkin approximate solution (Galso) to the exact solution ψ⟶=ψ1,ψ2,ψ3 such that(17)ψin=∑j=1ncijtυijx,where cijt are unknown functions of t, ∀j=1,2,…,n, and. ∀i=1,2,3.
The wkf of (10)–(15) is approximated with respect to x using the Galm, and substituting ψint=ζini=1,2,3 in the obtained equations, they become for each υ1,υ2,υ3∈ϒn:(18)ζ1nt,υ1+∇ψ1n,∇υ1+ψ1n,υ1−ψ2n,υ1−ψ3n,υ1=k1ψ1n,ω1,υ1,(19)ψ1n0,υ1=ψ10,υ1,ψ1n1,υ1=ψ11,υ1,(20)ζ2nt,υ+∇ψ2n,∇υ2+ψ2n,υ2+ψ3n,υ2+ψ1n,υ2=k2ψ2n,ω2,υ2,(21)ψ2n0,υ2=ψ20,υ2,ψ2n1,v2=ψ21,υ2,(22)ζ3nt,υ3+∇ψ3n,∇υ3+ψ3n,υ3+ψ1n,υ3−ψ2n,υ3=k3ψ3n,ω3,υ3,(23)ψ3n0,υ3=ψ30,υ3,ψ3n1,υ3=ψ31,υ3,where ψin0=ψin0x=ψinx,0∈ϒn (respectively, ζin0=ψin1=ψin1x=ψintx,0∈L2E) be the projection of yi0 onto V (be the projection of ψi1=ψit onto L2E), ∀i=1,2,3, i.e.,(24)ψin0⟶ψi0inV,withψ⟶n01≤b0andψ⟶n00≤b0,(25)ψin1⟶ψi1,inL2Eandψ⟶n10≤b1.
Substituting (17) for each i=1,2,3, respectively, in the pairs (18) and (19), (20) and (21), and in (22) and (23), setting υ1=υ1i, υ2=υ2i, υ3=υ3i, the obtained equations will be written as the following 1st order nodeqs with their ICs and have a unique solution ψ⟶n∈CI,ϒ⟶, i.e., for each l=1,2,3 and m¯=0,1:(26)A1D1′t+B1C1t−E1C2t−H1C3t=b1Y¯1TxC1t,A1C10=b10,A1D10=b11,A2D2′t+B2C2t+E2C1t+H2C3t=b2Y¯2TxC2t,A2C20=b20,A2D20=b21,A3D3′t+B3C3t+E3C1t−H3C3t=b3Y¯3TxC3t,A3C30=b30,A3D30=b31,where Clt=cljtn×1, C2′t=c2j′tn×1, Dl′t=dlj′tn×1, Dlt=dljtn×1, bl=blin×1, bli=klVlTclt,ωl,υli, blm¯=bljm¯, bljm¯=ψlm¯,vlj, Al=alijn×n, alij=υlj,υli, E1=e1ijn×n, e1ij=υ2j,υ1i, H1=f1ijn×n f1ij=υ3j,υ1i, E2=e2ijn×n, e2ij=υ1j,υ2i, H2=f2ijn×n, f2ij=υ3j,υ2i, E3=e3ijn×n, e3ij=υ1j,υ3i, H3=f3ijn×n, f3ij=υ2j,υ3i, Bl=blijn×n, and blij=∇υlj,∇υli+υlj,υli.
Then, corresponding to the sequenceϒ⟶n, there exists a sequence of the following approximation problems, i.e., for each υ⟶n=υ1n,υ2n,υ3n⊂ϒ⟶n, and n=1,2,…:(27)ψ1ntt,υ1n+∇ψ1n,∇υ1n+ψ1n−ψ2n−ψ3n,υ1n=k1ψ1n,ω1,υ1n,(28)ψ1n0,υ1n=ψ10,υ1n,ψ1n1,υ1=ψ11,υ1n,(29)ψ2ntt,υ2n+∇ψ2n,∇υ2n+ψ2n+ψ3n+ψ1n,υ2n=k2ψ2n,ω2,υ2n,(30)ψ2n0,υ2n=ψ20,υ2n,ψ2n1,υ2=ψ21,υ2n,(31)ψ3ntt,υ3n+∇ψ3n,∇υ3n+ψ3n+ψ1n−ψ2n,υ3n=k3ψ3n,ω3,υ3n,(32)ψ3n0,υ3n=ψ30,υ3n,ψ3n1,υ3=ψ31,υ3n,which has a sequence of unique solution ψ⟶n. Substituting υin=ψint for i=1,2,3 in (25)–(27), respectively, adding the three obtained equations together, and employing Lemma 1.2 in [18] for the first term of the LHS, to get (33) which is given by(33)ddtψ⟶ntt02+ψ⟶n12=2ψ2n,ψ1nt+ψ3n,ψ1nt−ψ1n,ψ2nt−ψ3n,ψ2nt−ψ1n,ψ3nt+ψ2n,ψ3nt+k1ψ1n,ω1,ψ1nt+k2ψ2n,ω2,ψ2nt+k3ψ3n,ω3,ψ3nt.
Or (33) can be rewritten as (34) which is(34)ddtψ⟶ntt02+ψ⟶n12≤2ψ2n,ψ1nt+ψ3n,ψ1nt+ψ1n,ψ2nt+ψ3n,ψ2nt+ψ1n,ψ3nt+ψ2n,ψ2nt+k1ψ1n,ω1,ψ1nt+k2ψ2n,ω2,ψ2nt+k3ψ3n,ω3,ψ3nt.
Using AssumptionA for the RHS of (34), integrating both sides on 0,t, using ψin0≤ψin1≤ψ⟶n1, and ψint0≤ψ⟶nt0, we get (35)∫0tddtψ⟶ntt02+ψ⟶n12dt≤∫0tψ⟶nt02+ψ⟶n12dt+∫0t∑i=13Fi02dt+β4∫0tψ⟶nt02+ψ⟶n12dt+∫0tψ⟶nt02dt≤∑i=13FiQ2+β6∫0tψ⟶nt02+ψ⟶n12dt≤β8+β7∫0tψ⟶nt02+ψ⟶n12dt,where β4=∑i=13βi, β5=1+β4, β6=2+β4, β7=maxβ5,β6, β8=∑i=13bi′.
Sinceψ⟶n01≤b1, and ψ⟶n10≤b0, with β9=b0+b1+β7, then (35) is reduced to(36)ψ⟶ntt02+ψ⟶nt12≤β9+β7∫0tψ⟶n02+ψ⟶nt12dt.
Appling the Belman–Gronwall (BGin) inequality, the abovementioned inequality gives∀t∈0,T:(37)ψ⟶ntt02+ψ⟶nt12≤β9eβ7=b2c⇛ψ⟶ntt02≤b2c,ψ⟶nt12≤b2c,∀t∈0,T.
Easily, one can obtain thatψ⟶nttQ≤b1candψ⟶ntL2I,Y≤bc.
Then, by applying Alaoglu’s theorem (Algth),ψ⟶nn∈N has a subsequence; it is not loss of generality to say ψ⟶nn∈N such that ψ⟶nt⇀ψ⟶ in L2Π3 and ψ⟶n⇀ψ⟶ in L2I,Y3, and(38)L2ℛ,Y3⊂L2ℛ,E3≅L2ℛ,E∗3⊂L2ℛ,Y∗3.
Then, the Aubin compactness theorem [18] can be applied here to get that ψ⟶n⟶ψ⟶ in L2Π3. Now, multiplying both sides of (27) and (29), and (31) by χit∈C20,T, such that χiT=χi′T=0, χi0≠0, χi′0≠0,∀i=1,2, integrating on, finally integrating by parts twice the 1st term of each one of the obtained three equations, led to(39)−∫0Tddtψ1n,υ1nχ1′tdt+∫0T∇ψ1n,∇υ1nχ1t+ψ1n,υ1n−ψ2n,υ1n−ψ3n,υ1nχ1tdt=∫0Tk1ψ1n,ω1,υ1nχ1tdt+ψ1n1,υ1nχ10,(40)∫0Tψ1n,υ1nχ1″tdt+∫0T∇ψ1n,∇υ1nχ1t+ψ1n,υ1n−ψ2n,υ1n−ψ3n,υ1nχ1tdt=∫0Tk1ψ1n,ω1,υ1nχ1tdt+ψ1n1,υ1nχ10+ψ1n0,υ1nχ1′0,(41)−∫0Tddtψ2n,υ2nχ2′tdt+∫0T∇ψ2n,∇υ2nχ2t+ψ2n,υ2n+ψ3n,υ2n+ψ1n,υ2nχ2tdt=∫0Tk2ψ2n,ω2,υ2nχ2tdt+ψ2n1,υ2nχ20,(42)∫0Tψ2n,υ2nχ2″tdt+∫0T∇ψ2n,∇υ2nχ2t+ψ2n,υ2n+ψ3n,υ2n+ψ1n,υ2nχ2tdt=∫0Tk2ψ2n,ω2,υ2nχ2tdt+ψ2n1,υ2nχ20+ψ2n0,υ2nχ2′0,(43)−∫0Tddtψ3n,υ3nχ3′tdt+∫0T∇ψ3n,∇υ3nχ3t+ψ3n,υ3n+ψ1n,υ3n−ψ2n,υ3nχ3tdt=∫0Tk3ψ3n,ω3,υ3nχ3tdt+ψ3n1,υ3nχ30,(44)∫0Tψ3n,υ3nχ3″tdt+∫0T∇ψ3n,∇υ3nχ3t+ψ3n,υ3n+ψ1n,υ3n−ψ2n,υ3nχ3tdt=∫0Tk3ψ3n,ω3,υ2nχ3tdt+ψ3n1,υ3nχ30+ψ3n0,υ3nχ3′0.
Now, for eachi=1,2,3, we have the following convergences:First, sinceυin⟶υiinY⟶υinχit⟶υiχitυinχi′t⟶υiχi′t,inL2I,Yυinχi0⟶υiχi0,inL2EOn the other hand, sinceυin⟶υiinL2E⟶υinχi′t⟶υiχi′tυinχi″t⟶υiχi″t,inL2Πυinχi′0⟶υiχi′0,inL2ESecond, we haveψint⇀ψit,inL2Πψin⇀ψi,inL2I,Yψin⟶ψi,inL2ΠThird, letwin=υinχi and wi=υiχi, then win⟶wi in L2Π and win is measurable in E, so from Assumption (A–I) and Proposition 1, the integral ∫Πkix,t,ψin,ωiwindxdt is cont. with respect to ψin,ωi,win, then(45)∫0Tkiψin,ωi,υinχitdt⟶∫0Tkiψi,ωi,υiχitdt.Now, from these convergences and (24) and (25), for i=1, we can passage to the limits in (39) and (40) to get(46)−∫0Tψ1t,υ1χ1′tdt+∫0T∇ψ1,∇υ1+ψ1−ψ2−ψ3,υ1χ1tdt=∫0Tk1ψ1,ω1,υ1χ1tdt+ψ11,υ1χ10,(47)∫0Tψ1,υ1χ1″tdt+∫0T∇ψ1,∇υ1+ψ1−ψ2−ψ3,υ1χ1tdt=∫0Tk1ψ1,ω1,υ1χ1tdt+ψ11,υ1χ10+ψ10,υ1χ1′0,Case 1.
Chooseχi∈C20,T, s.t. χi0=χi′0=χiT=χi′T=0. Using these values in (47) (for i=1), using integration by parts twice for the first terms in the LHS of the obtained equation, yields to(48)∫0Tψ1tt,υ1χ1tdt+∫0T∇ψ1,∇υ1+ψ1−ψ2−ψ3,υ1χ1tdt=∫0Tk1ψ1,ω1,υ1χ1tdt,which give that ψ1 is a solution of (10) (a.e. on I).
Similar way can be used fori=2,3, with (41)–(44) respectively to get that ψ2 and ψ3 are solutions of and (12) and (14) respectively (a.e. on I).Case 2.
Chooseχi∈C20,T, such that χiT≠0 and χi0≠0. For i=1, integrating both sides of (10) on 0,T after multiplying it by χ1t, using integrating by parts for the first term in the LHS of the obtained equation, then subtracting the obtained equation from (41), we get ψi1,υiχi0=ψit0,υiχi0.
Also, similar way can be used but fori=2,3 with the pairs (12) and (47) and (14) and (48), respectively, to get the same result.Case 3.
Chooseχi∈C20,T, such that χi0=χiT=χi′T=0, χi′0≠0. For i=1, integrating both sides of (10) on 0,T after multiplying it by χ1t, using integrating by parts for the first term in the LHS of the obtained equation, then subtracting this obtained equation from (47), we get ψi0,υiχi′0=ψi0,υiχi′0. Also, for i=2,3 and by using (12) and (14), we can use a similar way to get the same result.
From the last two cases, easily we can get the ICs (11) and (13), and (15).
Uniqueness of the solution: letψ⟶=ψ1,ψ2,ψ3 and ψ¯⟶=ψ¯1,ψ¯2,ψ¯3 be two solutions of the wkf (10)–(15), i.e., ψi and ψ¯i (for each i=1,2,3) are satisfied the wkf (10)–(15), subtracting each equality from the other and letting υi=ψi−ψ¯i, yields to(49)ψi−ψitt,ψi−ψ¯i+ψi−ψ¯i12=kiψi,ωi−kiψ¯i,ωi,ψi−ψ¯i,ψi−ψ¯i0,ψi−ψ¯i0=0,ψi−ψ¯it0,ψi−ψ¯it0=0.
Adding these three equations, using Lemma 1.2 in ref. [18] on the first term in LHS of the obtained equation which will be positive, integrating both sides from 0 to t, using the initial conditions, the Lipshctiz property on the RHS, and lastly applying the B–G inequality, to get(50)∫0tddtψ⟶−ψ¯⟶tt02+2ψ⟶−ψ¯⟶12dt≤2L∫0tψ⟶−ψ¯⟶tt02+ψ⟶−ψ¯⟶12dtψ⟶−ψ¯⟶tt02+ψ⟶−ψ¯⟶12≤⇛ψ⟶−ψ¯⟶t12=0,∀t∈I⇛ψ⟶−ψ¯⟶tL2I,Υ=0⇛,the solution is unique.Lemma 1.
In addition to AssumptionA, if the functions ki (for each i=1,2,3) is Lip. with respect to yi and ωi, and if the control vector is bounded, then the operator ω⟶↦ψ⟶ω⟶ from L2Π3 into L∞I,L2E3 or in to L2I,Y3 or in to L2Π3 is cont.Proof.
Letω⟶=ω1,ω2,ω2,ω¯⟶=ω¯1,ω¯2,ω¯3∈L2Π3, Δω⟶=ω¯⟶−ω⟶∈L2Π3, then by Theorem 1, ψ⟶=ψ⟶ω⟶=ψ1,ψ2,ψ3 and ψ¯⟶=ψ¯⟶ω⟶=ψ¯1,ψ¯2,ψ¯3 are their corresponding states’ solutions which satisfy the wkf of (10)–(15), setting Δψ⟶=Δψ1,Δψ2,Δψ3=ψ¯⟶−ψ⟶, then(51)Δψ1tt,υ1+∇Δψ1,∇υ1+Δψ1−Δψ2−Δψ3,υ1=k1ψ1+Δψ1,ω1+Δω1−k1ψ1,ω1,υ1,(52)Δψ1x,0=0,Δψ1tx,0=0,(53)Δψ2tt,υ2+∇Δψ2,∇υ2+Δψ2+Δψ1+Δψ3,υ2=k2ψ2+Δψ2,ω2+Δω2−k2ψ2,ω2,υ2,(54)Δψ2x,0=0,Δψ2tx,0=0,(55)Δψ3tt,v3+∇Δψ3,∇v3+Δψ3+Δψ1−Δψ2,v3=k3ψ3+Δψ3,ω3+Δω3−k3ψ3,ω3,υ3,(56)Δψ3x,0=0,Δψ3tx,0=0.
Substitutingυi=Δψit for each i=1,2,3 in (51), (53), and (55), respectively, adding the obtained three equations together, and using the same way that we used to get (32), a similar equation can be obtained but with Δψ⟶ in state of ψ⟶n, and then integration of both sides on 0,t, using the Lip. property on ki, i=1,2,3, with respect to each dependent variable, yields(57)∫0tddtΔψ⟶tt02+Δψ⟶12≤2∫0tΔψ2+Δψ3Δψ1t+Δψ1+Δψ3Δψ2tdt+2∫0tΔψ1+Δψ2Δψ3t+L¯1Δψ1+L¯¯1Δω1Δψ1tdt+2∫0tL¯2Δψ2+L¯¯2Δω2Δψ2t+L¯3Δψ3+L¯¯3Δω3Δψ3tdt.
Using the definitions of the norms and the relations between them, we get(58)Δψt⟶t02+Δψ⟶12≤2∫0tΔψ⟶02+Δψt⟶12dt+L˜1∫0tΔψ⟶02+Δψ⟶t12dt+L¯2∫0TΔω⟶02dt+L¯2∫0tΔψt⟶12dt≤L¯2Δω⟶tΠ2+L1∫0tΔψ⟶02+Δψt⟶12dt,where L˜1=maxL¯1,L¯2,L¯3, L¯2=maxL¯¯1,L¯¯2,L¯¯3, and L1=max2+L˜1,2+L˜1+L¯2.
Applying the BGin, withL2=L¯2eL1, we get(59)Δψ⟶tt02+Δψ⟶12≤L2Δω⟶tΠ2,∀t∈I¯⇛Δψ⟶t12≤L2Δω⟶tΠ2,∀t∈I¯⇛Δψ⟶L∞I,L2Ε≤LΔω⟶Π,Δψ⟶L2I,Υ≤LΔω⟶ΠandΔψ⟶Π≤LΔω⟶Π.From the abovementioned three inequalities, the Lip. continuity of the operatorω⟶↦ψ⟶ easily obtained.


## 1.3. The Existence of a Classical Optimal Control

This section concerned with proving the existence theorem with a CCOPCV satisfying the EQVC and INEQVC is studied. Hence, the following assumption and lemmas will be needed.Assumption B.
Considermri (for r=0,…,q and i=1,2,3) is of Caraty. on Π×ℛ×∁i and satisfies ψi∈ℛ and ωi∈∁i:(60)mrix,t,ψi,ωi≤Mrix,t+ckiψi2,whereMri∈L1Π,∀i=1,2,3,∀r=0,…,q.Lemma 2.
With AssumptionB, the functional ω⟶↦Mrω⟶ is cont. on L2Π3.Proof.
Using AssumptionB and Proposition 1 gives ∫Qmrix,t,ψi,ωidxdt is cont. on L2Π, ∀i=1,2,3, and ∀r=0,…,q; hence, Mlω⟶ is cont. on L2Π3.Lemma 3 (see [13]).
Letm:Q×ℛ2⟶ℛ is of Caraty. on Π×ℛ×ℛ and satisfies mx,ψ,ω≤mx,t+cψ2, where x,t∈L1Π, ω∈∁, c≥0, ∁⊂ℛ is compact. Then, ∫Πmx,ψ,ωdx is cont. on L2Π, with respect to ψ.Theorem 2.
With the AssumptionsA and B, if the set ∁⟶ is convex (cox.) and compact (com), W^⟶A≠∅, and the function ki ∀i=1,2,3 has the form(61)kix,t,ψi,ωi=ki1x,t,ψi+ki2x,tωi,where ki1x,t,ψi≤ηix,t+ciψi, ki2x,t≤ki, ηi∈L2Π, ci≥0.In additionmri is independent of ωi (∀i=1,2,3, and r=1,…,p), mri (∀i=1,2,3, and r=1,…,p) is cox. with respect to ωi for fixed x,t,ψi, there exists a CCOPCV.Proof.
From the Assumption on∁i⊂ℛ ∀i=1,2,3 and Egorov’s theorem, one obtains that W^⟶1×W^⟶2×W^⟶3=W^⟶ is weakly com. Since W^⟶A≠∅, hence there is ω¯⟶∈W⟶A s.t. rω¯⟶=0,1≤r≤p, Mkω¯⟶≤0, for p+1≤r≤q and there is a minimizing sequence ω⟶ρ s.t. ω⟶ρ∈W^⟶A,∀ρ, which satisfies ρ⟶∞limM0ω⟶ρ=ω¯⟶∈W⟶AinfM0ω¯⟶. Since ω⟶ρ∈W^⟶A,∀ρ and W^⟶ is weakly com., then ω⟶ρ has a subsequence say again ω⟶ρ which converges weakly to some ω⟶ in W^⟶, i.e., ω⟶ρ⇀ω in L2Π3 and ω⟶ρΠ≤c,∀ρ. From Theorem 1, for any given control ω⟶ρ, then ψ⟶ρ=ψ⟶ω⟶ρ is a unique solution for the Tsteqs, and ψ⟶ρL2I,Y, ψ⟶ρtL2Π are bounded, then by Algth, the sequences ψ⟶ρ and ψ⟶ρt have a weakly converging subsequences, say for simplicity, ψ⟶ρ and ψ⟶ρt, i.e.,(62)ψ⟶ρ⇀ψ⟶,inL2I,Y3,ψ⟶ρt⇀ψ⟶t,inL2Π3.
Then, by applying the Aubin compactness theorem [18], the sequence ψ⟶ρ has a strongly converging subsequence, say for simplicity, ψ⟶ρ such that ψ⟶ρ⟶ψ⟶ in L2Π3.
Now, for eachρ, substitute the solution ψ1ρ,ψ2ρ,ψ3ρ in the wkf of (18), (20), and (22), then multiply both sides of each one by χit (with χi∈C20,T, such that χiT=χi′T=0, χi0≠0,χi′0≠0, for i=1,2,3). After rewriting the first terms in the LHS in each one of them, integrating both sides on 0,T, and then by applying integration by parts for these first terms, we get(63)∫0Tddtψ1ρt,υ1χ1tdt+∫0T∇ψ1ρ,∇υ1+ψ1ρ,υ1−ψ2ρ,υ1−ψ3ρ,υ1χ1tdt=∫0Tk11x,t,ψ1ρ+k12x,tω1ρ,υχ1tdt,(64)∫0Tddtψ2ρt,υ2χ2tdt+∫0T∇ψ2ρ,∇υ2+ψ2ρ,υ2+ψ1ρ,υ2+ψ3ρ,υ2χ2tdt=∫0Tk21x,t,ψ2ρ+k22x,tω2ρ,υ2χ2tdt,(65)∫0Tddtψ3ρt,υ3χ3tdt+∫0T∇ψ3ρ,∇υ3+ψ3ρ,υ3+ψ1ρ,υ3−ψ2ρ,υ3χ3tdt=∫0Tk31x,t,ψ3ρ+k32x,tω3ρ,υ3χ3tdt.
One can passage the limits in the LHS of (50)–(52) by applying the same manner which is applied in the proof of Theorem 1 to passage the limits in RHS of these equations; we suppose ∀i=1,2,3, υi∈CΕ¯, wi=υiχit, then wi∈CΠ¯∈L∞I,V⊂L2Π, set k¯i1ψ1ρ=ki1ψiρwi, then k¯i1:Π×ℝ⟶ℝ is of Caraty., using Proposition 1, to get the integral ∫Πki1ψiρwidxdt is cont. with respect to ψiρ, but ψiρ⟶ψi in L2Π and ωiρ⇀ωi in L2Π, then(66)∫Πki1ψiρwidxdt⟶∫Πki1ψiwidxdt,∀wi∈CΠ¯fori=1,2,3,(67)∫Πki2x,tωiρwidxdt⟶∫Πki2x,tωiwidxdt,∀wi∈CΠ¯,fori=1,2,3.
Thus, (66) and (67) are holded for every υi∈ϒ, since CE¯ is dense in ϒ; hence, we get the wkf (10), (12), and (14).
Also, the same manner which is applied in the proof of Theorem2 can be used here to passage the limits in the ICs. Hence, ψ1,ψ2,ψ3 is a solution of the wkf of (10)–(15).
On the other hand, sincemri (for i=1,2,3 and r=1,2,…,p) is independent of ωi and cont. with respect to ψi, by Lemma 2, ∫Πmrix,t,ψiρdxdt is cont. with respect to yi, ψ⟶ρ⟶ψ⟶ in L2Π3, then(68)∫Πmrix,t,ψiρdxdt⟶∫Πmrix,t,ψidxdt.Hence,Mrω⟶=ρ⟶∞limMrω⟶ρ=0.
Now, sincemrix,t,ψi,ωi (for each r=0,1+p,2+p,…,q and i=1,2,3) is cont. with respect to ψi,ωi and since Ci is compact, then using Lemma 3, we get(69)∫Πmrix,t,ψiρ,ωiρdxdt⟶∫Πmrix,t,ψi,ωiρdxdt.
On the other hand,∫Πmrix,t,ψi,ωiρdxdt is cox. and cont. with respect to ωi (since mrix,t,ψi,ωi is cox. and cont. with respect to ωi), then ∫Πmrix,t,ψi,ωidxdt Necessary and Sufficient is weakly lower semicont. (welsc) with respect to ωi, i.e.,(70)∫Πmrix,t,ψi,ωidxdt≤ρ⟶∞liminf∫Πmrix,t,ψi,ωiρ−mrix,t,ψiρ,ωiρdxdt+ρ⟶∞liminf∫Πmrix,t,ψiρ,ωiρdxdt≤ρ⟶∞liminf∫Πmrix,t,ψiρ,ωiρdxdt,by55⇛∑i=12∫Πmrix,t,ψi,ωidxdt≤k⟶∞liminf∑i=12∫Πmrix,t,ψik,ωiρdxdt⇛Mrω⟶≤limρ⟶∞infMrω⟶ρ,then Mrω⟶≤0 r=p+1,…,q since ω⟶ρ∈W^⟶A,∀ρ, and we get that (71)M0ω⟶≤ρ⟶∞liminfM0ω⟶ρ=limρ⟶∞M0ω⟶ρ=ω¯⟶∈W⟶AinfM0ω¯⟶=ω¯⟶∈W⟶AMinM0ω¯⟶.
Thus,ω⟶ is a CCOPCV.Assumption C.
Assume forr=0,…,q and i=1,2,3, the functions ki, kiψi, kiωi, mriψi, and mriωi are defined and are of Caraty. on Π×ℛ×∁′ (where ∁′ is an open subset in ∁) and satisfy kiψix,t,ψi,ωi≤Li, kiωix,t,ψi,ωi≤Li′:(72)mriψix,t,ψi,ωi≤Mri5x,t+cri5ψi,mriωix,t,ψi,ωi≤Mri6x,t+cri6ψi,where x,t∈Π, ψi,ωi∈ℝ, Mri5,Mri6∈L2Π. Li,Li′,cri5,cri6≥0.
Note: for simplicity, in the following theorem, we will drop the indexk from the functions mli and Ml. Also, we assume the Assums. (A), (B), and (C) are considered.Theorem 3.
Consider the TAEqsξ⟶=ξ1,ξ2,ξ3 of Tseqs (1)–(6) are defined by(73)ξ1tt−Δξ1+ξ1+ξ2+ξ3=ξ1k1ψ1x,t,ψ1,ω1+m1ψ1x,t,ψ1,ω1,onΠ,ξ1=0,onΣξ1x,T=ξ1tx,T=0,onE,ξ2tt−Δξ2+ξ2−ξ1−ξ3=ξ2k2ψ2x,t,ψ2,ω2+m2ψ2x,t,ψ2,ω2,onΠ,ξ2=0,onΣ,ξ2x,T=ξ2tx,T=0,onE,ξ3tt−Δξ3+ξ3−ξ1+ξ2=ξ13k3ψ3x,t,ψ3,ω3+m3ψ3x,t,ψ3,ω3,onΠ,ξ3=0,onΣξ3x,T=ξ3tx,T=0,onE,and the Ham is given by ℋx,t,ψ⟶,ω⟶,ξ⟶=∑i=13ζix,t,ψi,ωi, where ζi=ξikix,t,ψi,ωi+mix,t,ψi,ωi, for each i=1,2,3.Then, the Frde. ofG is defined by(74)M′ω⟶Δω⟶=∫Πℋω⟶x,t,ψ⟶,ω⟶,ξ⟶Δω⟶dxdt,where ℋω⟶x,t,ψ⟶,ξ⟶,ω⟶=ζ1ω1,ζ1ω1,ζ1ω1, and Δω⟶=Δω1,Δω2,Δω3, with ζiωi=ξikiωix,t,ψi,ωi+miωix,t,ψi,ωi, mi=∑r=0qκrmri and ξi=∑r=0qκrξli, for each i=1,2,3.Proof.
at first, let the wkf of the TAEqs are given∀υ1,υ2,υ3∈Y, by(75)ξ1tt,υ1+∇ξ1,∇υ1+ξ1+ξ2+ξ3,υ1=ξ1k1ψ1+m1ψ1,υ1,∀v1∈Ya.e.onI,(76)ξ1T,υ1=ξ1tT,υ1=0,(77)zξ2t,υ2+∇ξ2,∇υ2+ξ2−ξ1−ξ3,υ2=ξ2k2ψ2+m2ψ2,υ2,∀υ2∈Ya.e.onI,(78)ξ2T,υ2=ξ2tT,v2=0(79)ξ3t,υ3+∇ξ3,∇υ3+ξ3−ξ1+ξ2,υ3=ξ3k3ψ3+m3ψ3,υ3,∀υ2∈Ya.e.onI,(80)ξ3T,υ3=ξ3tT,υ3=0.
From the assumptions and using the same manner which is applied in the proof of Theorem1, once can prove that the wkf (75)–(80) has a unique solution ξ⟶=ξ1,ξ2,ξ3∈L2Π3.
Substitutingυi=Δψi for each i=1,2,3 in (75), (77), and (79) and integrating both sides on 0,T, we get(81)∫0TΔψ1,ξ1tt+∇ξ1,∇Δψ1+ξ1+ξ3+ξ2,Δψ1dt=∫0Tξ1k1ψ1+g1ψ1,Δψ1dt,(82)∫0TΔψ2,ξ2tt+∇ξ2,∇Δψ2+ξ2−ξ1−ξ3,Δψ2dt=∫0Tξ2k2ψ2+g2ψ2,Δψ2dt,(83)∫0TΔψ3,ξ3tt+∇ξ3,∇Δψ3+ξ3−ξ1+ξ2,Δψ3dt=∫0Tξ3k3ψ3+g3ψ3,Δψ3dt.
Now, letω⟶,ω¯⟶∈L2Π3, Δω⟶=ω¯⟶−ω⟶∈L2Q3, and then by Theorem 1, ψ⟶=ψ⟶ω⟶ and ψ¯⟶=ψ⟶ω¯⟶ are their corresponding solutions. Set Δψ⟶=Δψ1,Δψ2,Δψ3=ψ¯⟶−ψ⟶, substitute υi=ξi for each i=1,2,3 in (51), (53), and (55), integrate both sides on 0,T, and then integrate by parts twice the first term in the LHS of each equation. Finding for each i=1,2,3 the Frde. of ki in the RHS of each equation which are exist from the Assumption C, then by Lemma 1, and the inequality of Minkowiski, one has(84)∫0TΔψ1,ξ1tt+∇Δψ1,∇ξ1+Δψ1−Δψ2−Δψ3,ξ1dt=∫0Tk1ψ1Δψ1+k1ω1Δω1,ξ1dt+O1Δω⟶Δω⟶Π,(85)∫0TΔψ2,ξ2tt+∇Δψ2,∇ξ2+Δψ2+Δψ1+Δψ3,ξ2dt=∫0Tk2ψ2δψ2+k2ω2Δω2,ξ2dt+O2Δω⟶Δω⟶Π,(86)∫0TΔψ3,ξ3tt+∇δψ3,∇ξ3+Δψ2ε+Δψ1−Δψ2,ξ3dt=∫0Tk3ψ3Δψ3+k3ω3Δω3,ξ3dt+O3Δω⟶Δω⟶Π.
Subtracting (84)–(86) from, respectively, adding the obtain equations, we get(87)∫0T∑i=13kiωiΔωi,ξidt+O4Δω⟶=∫0T∑i=13miψi,Δψidt,where. O4ε⟶0,asΔω⟶⟶0,withO4Δω⟶=∑i=13OiΔω⟶Δω⟶Π.
On the other hand, from the AssumptionB on mi (for i=1,2,3), the Frde definition, Lemma 1, and by applying the inequality of Minkowiski, we obtain(88)Mω⟶+Δω⟶−Mω⟶=∫Π∑i=13miψiΔψi+miωiΔωidxdt+O5Δω⟶Δω⟶Π,where O5Δω⟶⟶0, as Δω⟶⟶0.
Now, by substituting (87) in (88), one has(89)Mω⟶+Δω⟶−Mω⟶=∫Π∑i=13ζiωix,t,ψi,ωiΔωidxdt+O6Δω⟶Δω⟶Π,where O6Δω⟶=O5Δω⟶+O5Δω⟶⟶0, as Δω⟶⟶0.
Finally, from the Frde. of, we get(90)M′ω⟶Δω⟶=∫Πℋω⟶x,t,ψ⟶,ω⟶,ξ⟶Δω⟶dxdt.


## 1.4. Necessary and Sufficient Conditions for Optimality

This section deals with the theorems for the Necoop necessary under certain hypotheses which are proved as follows:Theorem 4.
Necoop (multipliers theorem):(a)
with Assumptions A, B, and C, ifW^⟶ is cox. and the ω⟶∈W^⟶A is optimal, then there are multipliers κr∈ℝ,r=0,1,…,p,p+1,…,q with κr≥0, for r=0,p+1,…,q,∑r=0qκr=1 such that the following Kuhn-Tucker-Lagrange (K. T. L.) conditions are holded:(91)∫Πℋω⟶x,t,ψ⟶,ω⟶,ξ⟶Δω⟶dxdt≥0,∀ω¯⟶∈W^⟶,Δω⟶=ω¯⟶−ω⟶,(92)κrMrω⟶=0,forr=p+1,…,q.(b)
Inequality (91) is equivalent to the (weak) piecewise minimum principle(93)ℋω⟶x,t,ψ⟶,ξ⟶,ω⟶⋅ω⟶t=ℋω⟶ω¯⟶∈∁⟶minx,t,ψ⟶,ξ⟶,ω⟶⋅ω¯⟶t,a.e.onΠ.Proof.
(a)
From Lemma2, the functional Mrω⟶ (for r=0,1,…,q) is cont., and from Theorem 3, the functional Mr′ (for r=0,1,…,q) is cont. with respect to ω¯⟶−ω⟶ and linear in ω¯⟶−ω⟶, then Mk′ is L− differential for every L, and then applying the K. T. L. theorem [5], there are multipliers κr∈ℝ,r=0,1,…,q with κr≥0, for r=0,p+1,…,q,∑r=0qκr=1, such that (91)–(93) are satisfied, by using Theorem 3, then (91) becomes ∑r=0qκr∫Π∑i=13ζiωix,t,ψi,ωiΔωidxdt≥0, which can be rewritten as(94)∫Πζ⟶ω⟶⋅ω¯⟶k−ω⟶dxdt≥0,∀ω¯⟶∈W^⟶withζ⟶ω⟶=ζ1ω1,ζ2ω2,ζ3ω3.
(b)
Letω¯⟶l be a dense sequence (dse) in W^⟶, m denotes the Lebesgue measure on Π, and Γ⊂Π be a measurable subset with property ω¯⟶x,t=ω¯⟶lx,t,ifx,t∈Γω⟶x,t,ifx,t∉Γ.
Therefore, (94) becomes(95)∫Γζ⟶ω⟶⋅ω¯⟶−ω⟶dxdt≥0,which implies toζ⟶ω⟶⋅ω¯⟶l−ω⟶≥0, a.e.on∏.
This means the inequality is satisfied on the whole regionQ except in a subset Πl such that mΠl=0,∀l, where m represents the Lebesgue measure; thus, the inequality holds on Π except in the union ∪lΠl with m∪lΠl=0, but ω¯⟶l is a dse in W^⟶, then there is ω¯⟶∈W^⟶ such that ∫Πζ⟶ω⟶⋅ω¯⟶l−ω⟶≥0, a.e. on Π,∀ω¯⟶∈W^⟶.
That is, (91) gives (94). The converse is clear.Theorem 5.
Sucoop: besides the Assumptions A, B, and C, supposeW^⟶ is cox., with ∁⟶ cox., ki and mri ∀r=1,2,…,p are affine with respect to ψi,ωi for each x,t, mri ∀r=0,p+1,…,q are cox. with respect to ψi,ωi ∀x,t, for i=1,2,3. Then, the Necoop for Theorem 4 with κ0>0 are sufficient.Proof.
Assumeω⟶∈W^⟶A satisfies the condition (91) and (92). Let Mω⟶=∑r=0qκrMrω⟶, then using Theorem 3, we get(96)M′ω⟶Δω⟶=∑r=0qκrMr′ω⟶Δω⟶=∑r=0qκr∫Π∑i=13ζiωi⋅δωidxdt≥0.
Sincekix,t,ψi,ωi=ki1x,tψi+ki2x,tωi+ki3x,t,i=1,2,3.
Letω=ω1,ω2,ω3 and ω¯⟶=ω¯1,ω¯2,ω¯3 be two given control vectors, then ψ⟶=ψω1,ψω2,ψω3=ψ1,ψ2,ψ3 and ψ¯⟶=ψ¯ω1,ψ¯ω¯2,ψ¯ω¯3=ψ¯1,ψ¯2,ψ¯3 represent the corresponding stats solutions. Substituting ω⟶,y⟶ in (1)–(6), multiplying all the obtained equalities by α∈0,1 once and on the other hand substituting ω¯⟶,y¯⟶ in (1)–(6), multiplying all the obtained equalities by 1−α once again, and lastly adding each pair from the corresponding equalities together, we obtain(97)αψ1+1−αψ¯1tt−Δαψ1+1−αψ¯1+αψ1−ψ2−ψ3+1−αψ¯1−ψ¯2−ψ¯3=αk11x,tψ1+k12x,tω1+1−αk11x,tψ¯1+k12x,tω¯1+k13x,t,(98)αψ1x,t+1−αψ¯1x,0=0,(99)αψ1x,0+1−αψ¯1x,0=ψ10x,αψ1tx,0+1−αψ¯1tx,0=ψ11x,(100)αψ2+1−αψ¯2tt−Δαψ2+1−αψ¯2+αψ2+ψ1+ψ3+1−αψ¯2+ψ¯1+ψ¯3=αk21x,tψ2+k22x,tω2+1−αk21x,tψ¯2+k22x,tω¯2+k23x,t,(101)αψ2x,t+1−αψ¯2x,0=0,(102)αψ2x,0+1−αψ¯2x,0=ψ20x,αψ2tx,0+1−αψ¯2x,0=ψ21x,(103)αψ3+1−αψ¯3tt−Δαψ3+1−αψ¯3+αψ3+ψ1−ψ2+1−αψ¯3+ψ¯1−ψ¯2=αk31x,tψ3+k32x,tω3+1−αk31x,tψ¯3+k32x,tω¯3+k23x,t,(104)αψ3x,t+1−αψ¯3x,0=0,(105)αψ3x,0+1−αψ¯3x,0=ψ30x,αψ3tx,0+1−αψ¯3tx,0=ψ31x.
Equations (97)–(105) show that if the control vector is ω˜⟶=ω˜1,ω˜2,ω˜3 with ω˜⟶=ω⟶+1−αω¯⟶ then its corresponding state vector is ψ˜⟶=ψ˜1,ψ˜2,ψ˜3 with ψ˜i=ψiω˜i=ψiαωi+1−αω¯i=αψi+1−αψ¯i,∀i=1,2,3. This gives the operator ω⟶↦y⟶ω⟶ is convex–linear with respect to ψ⟶,ω⟶ for any x,t∈Π.
On the other hand, the functionMlω⟶ (for k=1,…,p) is convex–linear with respect to ψ⟶,ω⟶ for any x,t∈Π, this backs to the fact that the sum of two affine functions mrix,t,ψi,ωi (for each i=1,2,3, and r=1,…,p with respect to ψi,ωi and ∀x,t∈Π is affine and the operator ω⟶↦ψ⟶ω⟶ is convex-linear.
The functionsMrω⟶,∀r=0,p+1,…,q are cox. with respect to ψ⟶,ω⟶,∀x,t∈Π (from the assumptions on the functions mr1 and mr2, ∀r=0,p+1,…,q). Hence, Mω⟶ is cox. with respect to ψ⟶,ω⟶,∀x,t∈Π in the cox. set W^⟶, and has a cont. Fréd satisfies M′ω⟶⋅ω¯⟶−ω⟶≥0⇛Mω⟶ has a minimum at ω⟶⇛Mω⟶≤Mω¯⟶,∀ω¯⟶∈W^⟶⇛(106)∑r=0pκrMrω⟶+∑r=p+1qκrMrω⟶≤∑r=0pκrMrω¯⟶+∑r=p+1qκrMrω¯⟶,∀ω¯⟶∈W^⟶.
Letω¯⟶∈W^⟶A, with kk≥0 (for r=p+1,…,q), then from (8), (9), and (77), the abovementioned inequality becomes(107)κ0M0ω⟶≤κ0M0ω¯⟶,∀ω¯⟶∈W^⟶⇛G0ω⟶≤G0ω¯⟶,∀ω¯⟶∈W^⟶⇛∴ω⟶is a CCOCVE.


## 2. Conclusions

The Galm with the Aubin compactness theorem are applied successfully to prove the existence of a unique “continuous state vector” solution for the TNLHPDEqs for a given cont. CCOPCVE. The existence theorem of governing by the considered the TNLHPDEqs with EQVC and INEQVC is proved. The existence of a unique solution of ATEqs associated with the considered Tsteqs is studied. The Frde. of the Ham is derived. The theorems of the Necoop and the Sucoop of the constrained problem are proved.



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*Source: 8021635-2020-04-10.xml*