# On an Invariant Set in the Heat Conductivity Problem with Time Lag

Authors: M. Tukhtasinov; G. I. Ibragimov; N. O. Mamadaliev
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The problems of weak and strong invariance of a constant multivalued mapping with respect to the heat conductivity equation with time lag are studied. Sufficient conditions of weak and strong invariance of a given multivalued mapping are obtained.


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## Body

## 1. Introduction

### 1.1. Related Works

There are many theoretical and practical problems in control problems with distributed parameters where known methods do not work to solve them. The typical examples of such problems are conservation of temperature of a volume within admissible bounds and deviation from undesirable states.Note that the works [1–7] were devoted to the problems of invariance of given sets for the controlled systems. In these works, some results on construction of core of liveness, the maximal weak invariant subset of the given set was obtained for the control system.In the paper [3], a family of trajectories, which is kept to be within the given set until a certain time (viability), is analytically described for control systems given by differential inclusions. The paper [8] deals with the problem of diffusion process control by boundary control.However, all the works mentioned above relate to control systems with concentrated parameters. In the papers [9, 10], weak and strong invariance of the given set with respect to a system with distributed parameters were studied. Alimov [11], Albeverio and Alimov [12] studied interesting applied control problems on heat distribution by convectors in a volume.The present paper deals with the problems of weak and strong invariance of given multivalued mapping for the 3rd heat conductivity boundary value problem with time lag. In the equation of this problem, the control parameter appears on the right hand side. We obtain conditions which can be easily checked to determine the invariance of the given constant multivalued mapping.


### 1.2. Preliminaries

First of all, we recall some definitions. A bounded regionΩ(⊂ℝn) is referred to as the region with piecewise smooth boundary if its boundary Γ=Ω-∖Ω can be represented as follows: Γ=∑j=1NΓ-j, where Γj⊂Γ is an open set with respect to the topology on Γ induced by the topology in ℝn. Moreover, each Γj is a connected surface of class C1; that is, for any point x0∈Γj there exists a ball Uϵ(x0) of radius ϵ>0 such that the piece Γj∩Uϵ(x0) of the surface is given by the equation of the form xk=fk(x1,…,xk-1,xk+1,…,xn), where fk(·)∈C1 and 1≤k≤n.LetΩ be a region in ℝn with piecewise smooth boundary. We will use the letter A to denote the following differential operator [13, 14]:
(1)A(x,D)φ=-∑i,j=1n‍∂∂xi(aij(x)∂φ∂xj),
where the functions aij(x)∈C1(Ω)∩C(Ω-) satisfy the conditions aij(x)=aji(x),x∈Ω, and
(2)∑i,j=1n‍aij(x)ξiξj≥γ∑i=1n‍ξi2,
for all (ξ1,ξ2,…,ξn)∈Rn. The inequality (2) is called the condition of uniform ellipticity of the operator A defined by (1). The domain of the operator A is C2(Ω), which is the space of functions that are twice continuously differentiable in Ω and continuous on Ω∪∂Ω.Define the operatorP by the equation
(3)Pφ=∑i,j=1n‍aij(x)∂φ∂xjcos(l,xi)+k(x)φ,x∈∂Ω,
where l denotes the upward unit normal vector on ∂Ω and k(x) is a given positive continuous function defined on ∂Ω.Definition 1.
The numberλ for which the following boundary value problem:
(4)A(x,D)φ=λφ,x∈Ω,Pφ=0,x∈∂Ω,
has nonzero solution φλ(x), x∈Ω is called eigenvalue of this boundary value problem, and the solution φλ(x) is called eigenfunction of the boundary value problem.Since problem (4) is homogeneous, we assume that
(5)∫Ω‍φλ2dx=1.
As the operator (1) is self-adjoint, then it has a discrete spectrum [13, 14]; that is,(i)
there exist a countable set of eigenvaluesλk of the problem (4) such that
(6)0<λ1≤λ2≤⋯≤λn≤⋯,λn⟶∞,n⟶∞,(ii)
for each eigenvalueλk, there are a finite number of eigenfunctions φk corresponding to it such that ∫Ωφi(x)φj(x)dx=δij, where
(7)δij={1,ifi=j,0,ifi≠j,(iii)
the set of all eigenvalues{φk} is complete (closed) in the space L2(Ω); that is, any function f from the space L2(Ω) can be uniquely represented in the form
(8)f(x)=∑k=1∞‍fkφk(x),where the equality is understood in the following sense:(9)∫Ω‍[f(x)-∑k=1N‍fkφk(x)]2dx⟶0,N⟶∞.The Fourier coefficientsfk in the Fourier series (8) of f(x) are defined by formula
(10)fk=∫Ω‍f(x)φk(x)dx.
Since the operator A is self-adjoint, in certain conditions on the functions aij, φ, and ψ, the Green formula is written in the form
(11)∫Ω‍(ψA(x,D)φ-φA(x,D)ψ)dx=∫∂Ω‍(ψPφ-φPψ)dS.
Using these, we construct the following spaces depending on a parameter. Let r be any nonnegative number. Denote
(12)ℓr={α=(α1,α2,…)∣∑i=1∞‍λirαi2<∞},Hr(Ω)={f∈L2(Ω)∣f=∑i=1∞‍αiφi,α∈ℓr}.We now define inner product and norm in the spacesℓr,Hr=Hr(Ω). Let
(13)α,β∈ℓr,f=∑i=1∞‍αiφi,g=∑i=1∞‍βiφi.
Set
(14)(α,β)r=∑i=1∞‍λirαiβi,∥α∥=(α,α)r1/2,(f,g)r=(α,β)r,∥f∥=∥α∥.
It should be noted that H0(Ω)=L2(Ω) and Hr(Ω)⊂Hs(Ω) for all 0≤s≤r.Denote byC(0,T;Hr(Ω)) and L2(0,T;Hr(Ω)) the spaces of continuous functions and summable quadratically measurable functions defined on [0,T] with the values in Hr(Ω), respectively, where T is a positive number.


### 1.3. Auxiliary Statements

Consider the following heat exchange control problem with lag:(15)∂z(x,t)∂t+A(x,D)z(x,t)=z0(x,t-h)+u(x,t),0<t≤T,
with boundary
(16)Pz(x,t)=0,0≤t≤T,x∈∂Ω,
and initial conditions
(17)z(x,0)=z0(x,0),
where z0(x,·)∈X; X={z(x,·)∣z(x,t)∈Hr,-h≤t≤0}. Here, z(·)=z(x,·) and u(·)=u(x,·) are abstract functions whose values at each t>0 are unique elements of the space Hr(Ω); h is a positive fixed number and T is a positive number.Further, we use the same letterA to denote the bounded operator that maps Hr into Hr+2 and is defined by the following formula
(18)Aφ=A(∑k=1∞‍ckφk)=∑k=1∞‍λkckφk,
where φ=∑k=1∞ckφk. Clearly, we obtain from this that ∥Af∥Hr=∥f∥Hr+2.The problem (15)–(17) is understood in the sense of the theory of generalized functions (the theory of distributions) with the values in Hr. We will look for the solution of the problem (15)–(17), which is continuous with respect to t and its values belong to one of the spaces Hr, and the initial and boundary conditions are considered as equality of elements of these spaces.In the control problem (15)–(17), the control function u(·) is subjected to either constraint
(19)∥u(t)∥Hr≤ρ,0≤t≤T,
or
(20)u(·)∈L2([0,h];Hr),∥u(·)∥L2([0,h];Hr)≤ρ.
Accordingly, in this paper, we consider both of the problems (15)–(19) and (15)–(17), (20). The control function u(·) that satisfies either (19) or (20) is called admissible.Assume that the problem (15)–(17) has a solution z(t), 0≤t≤h, at some admissible control u(·). Then we have
(21)z(t)=∑k=1∞‍zk(t)φk(x);zk(t)=∫Ω‍z(t)φk(x)dx.
It follows from (15) that
(22)∫Ω‍(∂z(t)∂t+Az(x)-z0(t-h)-u(t))φk(x)dx≡0,000000000000000000000000000000000k=1,2,….
Also, by definition of the functions z(t), φk(x) we obtain
(23)Pz(t)=0,0≤t≤h,x∈∂Ω,Pφk=0,Aφk=λkφk.
Therefore, in view of (11) we have
(24)∫Ω‍φk(x)Az(t)dx=λkzk(t).
Denoting
(25)z0k(t-h)=∫Ω‍z0(t-h)φk(x)dx,uk(t)=∫Ω‍u(t)φk(x)dx
and using the fact that
(26)∫Ω‍∂z(t)∂tφk(x)dx=dzk(t)dt,
we deduce from formulas (22) and (24) that
(27)z˙k(t)+λkzk(t)=z0k(t-h)+uk(t),00000000000≤t≤h,k=1,2,….
Since the function (21) must satisfy the initial condition z(0)=z0(0), we have
(28)zk(0)=z0k0=∫Ω‍z0(0)φk(x)dx,k=1,2,….
From this and (27) we see that
(29)zk(t)=e-λktz0k0+∫0t‍e-λk(t-τ)(z0k(τ-h)+uk(τ))dτ,000000000000000000000000000000000k=1,2,….
Thus, (21) and (29) define the formal solution of the problem (15)–(17) on [0,h] which can be written as follows:
(30)z(t)=∑k=1∞‍[e-λktz0k0+∫0t‍e-λk(t-τ)(z0k(τ-h)+uk(τ))dτ]00000×φk(x).The following assumption will be needed throughout the paper.Assumption 2.
Let functionsz(·) of the set X satisfy the following conditions:
(31)∑k=1∞‍λkr∫0t‍|z0k(τ-h)|2dτ≤b2,∑k=1∞‍λkr|z0k(0)|2≤b2,(32)∑k=1∞‍λkr∫0t‍e-λk(t-τ)|z0k(τ-h)|2dτ≤b2∫0t‍e-λ1(t-τ)dτ,0≤t≤h.Proposition 3.
For any functionf(t),∥f(t)∥Hr≤M, 0≤t≤T, the following inequality holds:
(33)∑k=1∞‍λkr(∫0t‍e-λk(t-τ)fk(τ)dτ)2≤a12(t)M2;a1(t)=1-e-λ1tλ1,0≤t≤T,
where λ1,λ2,… are the eigenvalues of the operator A arranged in decreasing order, M is a positive number, and fk(·) is the Fourier coefficient of the function f(t).Proof.
Indeed, for0≤t≤T, using the Cauchy-Schwartz inequality and the fact that
(34)∥f(τ)∥Hr2=∑k=1∞‍λkrfk2(τ)≤M2,0≤τ≤t≤T,
we obtain
(35)∑k=1∞‍λkr(∫0t‍e-λk(t-τ)fk(τ)dτ)2=∑k=1∞‍λkr(∫0t‍e-(λk/2)(t-τ)·e-(λk/2)(t-τ)fk(τ)dτ)2≤∑k=1∞‍λkr∫0t‍e-λ1(t-τ)dτ∫0t‍e-λ1(t-τ)fk2(τ)dτ≤∫0t‍e-λ1(t-τ)dτ∫0t‍e-λ1(t-τ)·∑k=1∞‍λkrfk2(τ)dτ≤a1(t)∫0t‍e-λ1(t-τ)M2dτ=a12(t)M2.Theorem 4.
Letu(t), 0≤t≤h be an admissible control that satisfies the condition (19), and let Assumption 2 be satisfied. Then there exists a unique solution of the problem (15)–(17) in the space C([0,h];Hr(Ω)).Proof.
To prove the theorem we use the formal representation (30) of the solution of the problem (15)–(17). By definition of the norm in Hr(Ω) for 0≤t≤h, we have
(36)∥z(t)∥Hr2=∑k=1∞‍λkr|zk(t)|2=∑k=1∞‍λkr(e-λktz0k0+∫0t‍e-λk(t-τ)fk(τ)dτ)2,
where fk(τ)=z0k(τ-h)+uk(τ).
We use the Cauchy-Schwartz inequality to obtain the following chain of relations:(37)∑k=1∞‍λkr(∫0t‍e-λk(t-τ)fk(τ)dτ)2≤∑k=1∞‍λkr(∫0t‍e-(λk(t-τ))/2e-(λk(t-τ))/2fk(τ)dτ)2≤∑k=1∞‍λkr∫0t‍e-λk(t-τ)dτ∫0t‍e-λk(t-τ)fk2(τ)dτ≤∑k=1∞‍λkr1-e-λktλk∫0t‍e-λk(t-τ)fk2(τ)dτ.
Therefore,
(38)∑k=1∞‍λkr1-e-λktλk∫0t‍e-λk(t-τ)fk2(τ)dτ≤a1(t)∑k=1∞‍λkr∫0t‍e-λk(t-τ)(z0k(τ-h)+uk(τ))2dτ.
Next, we expand the integrand on the right hand side of the last inequality; then we use the Cauchy-Schwartz inequality and after that we use Assumption 2 to obtain
(39)∑k=1∞‍λkr(∫0t‍e-λk(t-τ)fk(τ)dτ)2≤(b+ρ)2a12(t).
Expanding the square in (36), and then using (39) and Assumption 2, we arrive at
(40)∥z(t)∥Hr2≤(e-λ1tb+(b+ρ)a1(t))2.
Hence, z(t)∈Hr(Ω) for each t∈[0,h].
We now show that the functionz(t), 0≤t≤h, is continuous with respect to the norm of the space Hr. We have
(41)∥z(t+δ)-z(t)∥Hr2=∑k=1∞‍λkr|zk(t+δ)-zk(t)|2=∑k=1N‍λkr|zk(t+δ)-zk(t)|2+∑k=N+1∞‍λkr|zk(t+δ)-zk(t)|2,
where the numbers δ and N will be chosen depending on any ε>0.
By (29), We have
(42)zk(t+δ)-zk(t)=z0k0e-λkt(e-λkδ-1)+∫0t‍e-λk(t-τ)(e-λkδ-1)fk(τ)dτ+∫tt+δ‍e-λk(t+δ-τ)fk(τ)dτ.
Combining this with (41), we obtain
(43)∥z(t+δ)-z(t)∥Hr2≤I1+I2+I3;I1=3∑k=1N‍λkr|e-λkδ-1|2(|z0k0|2+|∫0t‍e-λk(t-τ)fk(τ)dτ|2)I2=3∑k=N+1∞‍λkr(|z0k0|2+|∫0t‍e-λk(t-τ)fk(τ)dτ|2)I3=3∑k=1∞‍λkr|∫tt+δ‍e-λk(t+δ-τ)fk(τ)dτ|2.
Let ε be an arbitrary positive number. Since the series in (31) and (39) are convergent, one can choose the number N so that I2<ε2/3. Then, we choose δ so that I1<ε2/3. Further, similar to (39) we have
(44)∑k=1∞‍λkr(∫tt+δ‍e-λk(t+δ-τ)fk(τ)dτ)2≤e-2λ1δ∑k=1∞‍λkr(∫tt+δ‍e-λk(t-τ)fk(τ)dτ)2≤e-2λ1δ((b+ρ)∫tt+δ‍e-λk(t-τ)dτ)2.
It is clear from these inequalities that we can choose δ so that I3<ε2/3. From the estimations for I1,I2,I3 obtained above we conclude according to (43) that ∥z(t+δ)-z(t)∥Hr<ε, meaning that the function z(t), 0≤t≤h, is continuous.
We now show that the problem (15)–(17) has a unique solution. We assume the contrary, that it has two different solutions z′(t), z′′(t), 0≤t≤h for the same functions z0(·)∈X and u(·). Then their difference Z(t)=z′(t)-z′′(t), 0≤t≤h, as a solution of corresponding homogeneous problem (15)–(17), can be represented in the form of the Fourier series (21), and the Fourier coefficients Zk(t), 0≤t≤h, of the function Z(t), 0≤t≤h, are solutions of the following infinite system of differential equations:
(45)z˙k(t)+λkzk(t)=0,zk(0)=0,000000000≤t≤h,k=1,2,….
Therefore, from the representation (29) we obtain Zk(t)=0 on 0≤t≤h. Consequently, Z(t)=0, 0≤t≤h, a contradiction. This proves the theorem.


## 1.1. Related Works

There are many theoretical and practical problems in control problems with distributed parameters where known methods do not work to solve them. The typical examples of such problems are conservation of temperature of a volume within admissible bounds and deviation from undesirable states.Note that the works [1–7] were devoted to the problems of invariance of given sets for the controlled systems. In these works, some results on construction of core of liveness, the maximal weak invariant subset of the given set was obtained for the control system.In the paper [3], a family of trajectories, which is kept to be within the given set until a certain time (viability), is analytically described for control systems given by differential inclusions. The paper [8] deals with the problem of diffusion process control by boundary control.However, all the works mentioned above relate to control systems with concentrated parameters. In the papers [9, 10], weak and strong invariance of the given set with respect to a system with distributed parameters were studied. Alimov [11], Albeverio and Alimov [12] studied interesting applied control problems on heat distribution by convectors in a volume.The present paper deals with the problems of weak and strong invariance of given multivalued mapping for the 3rd heat conductivity boundary value problem with time lag. In the equation of this problem, the control parameter appears on the right hand side. We obtain conditions which can be easily checked to determine the invariance of the given constant multivalued mapping.


## 1.2. Preliminaries

First of all, we recall some definitions. A bounded regionΩ(⊂ℝn) is referred to as the region with piecewise smooth boundary if its boundary Γ=Ω-∖Ω can be represented as follows: Γ=∑j=1NΓ-j, where Γj⊂Γ is an open set with respect to the topology on Γ induced by the topology in ℝn. Moreover, each Γj is a connected surface of class C1; that is, for any point x0∈Γj there exists a ball Uϵ(x0) of radius ϵ>0 such that the piece Γj∩Uϵ(x0) of the surface is given by the equation of the form xk=fk(x1,…,xk-1,xk+1,…,xn), where fk(·)∈C1 and 1≤k≤n.LetΩ be a region in ℝn with piecewise smooth boundary. We will use the letter A to denote the following differential operator [13, 14]:
(1)A(x,D)φ=-∑i,j=1n‍∂∂xi(aij(x)∂φ∂xj),
where the functions aij(x)∈C1(Ω)∩C(Ω-) satisfy the conditions aij(x)=aji(x),x∈Ω, and
(2)∑i,j=1n‍aij(x)ξiξj≥γ∑i=1n‍ξi2,
for all (ξ1,ξ2,…,ξn)∈Rn. The inequality (2) is called the condition of uniform ellipticity of the operator A defined by (1). The domain of the operator A is C2(Ω), which is the space of functions that are twice continuously differentiable in Ω and continuous on Ω∪∂Ω.Define the operatorP by the equation
(3)Pφ=∑i,j=1n‍aij(x)∂φ∂xjcos(l,xi)+k(x)φ,x∈∂Ω,
where l denotes the upward unit normal vector on ∂Ω and k(x) is a given positive continuous function defined on ∂Ω.Definition 1.
The numberλ for which the following boundary value problem:
(4)A(x,D)φ=λφ,x∈Ω,Pφ=0,x∈∂Ω,
has nonzero solution φλ(x), x∈Ω is called eigenvalue of this boundary value problem, and the solution φλ(x) is called eigenfunction of the boundary value problem.Since problem (4) is homogeneous, we assume that
(5)∫Ω‍φλ2dx=1.
As the operator (1) is self-adjoint, then it has a discrete spectrum [13, 14]; that is,(i)
there exist a countable set of eigenvaluesλk of the problem (4) such that
(6)0<λ1≤λ2≤⋯≤λn≤⋯,λn⟶∞,n⟶∞,(ii)
for each eigenvalueλk, there are a finite number of eigenfunctions φk corresponding to it such that ∫Ωφi(x)φj(x)dx=δij, where
(7)δij={1,ifi=j,0,ifi≠j,(iii)
the set of all eigenvalues{φk} is complete (closed) in the space L2(Ω); that is, any function f from the space L2(Ω) can be uniquely represented in the form
(8)f(x)=∑k=1∞‍fkφk(x),where the equality is understood in the following sense:(9)∫Ω‍[f(x)-∑k=1N‍fkφk(x)]2dx⟶0,N⟶∞.The Fourier coefficientsfk in the Fourier series (8) of f(x) are defined by formula
(10)fk=∫Ω‍f(x)φk(x)dx.
Since the operator A is self-adjoint, in certain conditions on the functions aij, φ, and ψ, the Green formula is written in the form
(11)∫Ω‍(ψA(x,D)φ-φA(x,D)ψ)dx=∫∂Ω‍(ψPφ-φPψ)dS.
Using these, we construct the following spaces depending on a parameter. Let r be any nonnegative number. Denote
(12)ℓr={α=(α1,α2,…)∣∑i=1∞‍λirαi2<∞},Hr(Ω)={f∈L2(Ω)∣f=∑i=1∞‍αiφi,α∈ℓr}.We now define inner product and norm in the spacesℓr,Hr=Hr(Ω). Let
(13)α,β∈ℓr,f=∑i=1∞‍αiφi,g=∑i=1∞‍βiφi.
Set
(14)(α,β)r=∑i=1∞‍λirαiβi,∥α∥=(α,α)r1/2,(f,g)r=(α,β)r,∥f∥=∥α∥.
It should be noted that H0(Ω)=L2(Ω) and Hr(Ω)⊂Hs(Ω) for all 0≤s≤r.Denote byC(0,T;Hr(Ω)) and L2(0,T;Hr(Ω)) the spaces of continuous functions and summable quadratically measurable functions defined on [0,T] with the values in Hr(Ω), respectively, where T is a positive number.


## 1.3. Auxiliary Statements

Consider the following heat exchange control problem with lag:(15)∂z(x,t)∂t+A(x,D)z(x,t)=z0(x,t-h)+u(x,t),0<t≤T,
with boundary
(16)Pz(x,t)=0,0≤t≤T,x∈∂Ω,
and initial conditions
(17)z(x,0)=z0(x,0),
where z0(x,·)∈X; X={z(x,·)∣z(x,t)∈Hr,-h≤t≤0}. Here, z(·)=z(x,·) and u(·)=u(x,·) are abstract functions whose values at each t>0 are unique elements of the space Hr(Ω); h is a positive fixed number and T is a positive number.Further, we use the same letterA to denote the bounded operator that maps Hr into Hr+2 and is defined by the following formula
(18)Aφ=A(∑k=1∞‍ckφk)=∑k=1∞‍λkckφk,
where φ=∑k=1∞ckφk. Clearly, we obtain from this that ∥Af∥Hr=∥f∥Hr+2.The problem (15)–(17) is understood in the sense of the theory of generalized functions (the theory of distributions) with the values in Hr. We will look for the solution of the problem (15)–(17), which is continuous with respect to t and its values belong to one of the spaces Hr, and the initial and boundary conditions are considered as equality of elements of these spaces.In the control problem (15)–(17), the control function u(·) is subjected to either constraint
(19)∥u(t)∥Hr≤ρ,0≤t≤T,
or
(20)u(·)∈L2([0,h];Hr),∥u(·)∥L2([0,h];Hr)≤ρ.
Accordingly, in this paper, we consider both of the problems (15)–(19) and (15)–(17), (20). The control function u(·) that satisfies either (19) or (20) is called admissible.Assume that the problem (15)–(17) has a solution z(t), 0≤t≤h, at some admissible control u(·). Then we have
(21)z(t)=∑k=1∞‍zk(t)φk(x);zk(t)=∫Ω‍z(t)φk(x)dx.
It follows from (15) that
(22)∫Ω‍(∂z(t)∂t+Az(x)-z0(t-h)-u(t))φk(x)dx≡0,000000000000000000000000000000000k=1,2,….
Also, by definition of the functions z(t), φk(x) we obtain
(23)Pz(t)=0,0≤t≤h,x∈∂Ω,Pφk=0,Aφk=λkφk.
Therefore, in view of (11) we have
(24)∫Ω‍φk(x)Az(t)dx=λkzk(t).
Denoting
(25)z0k(t-h)=∫Ω‍z0(t-h)φk(x)dx,uk(t)=∫Ω‍u(t)φk(x)dx
and using the fact that
(26)∫Ω‍∂z(t)∂tφk(x)dx=dzk(t)dt,
we deduce from formulas (22) and (24) that
(27)z˙k(t)+λkzk(t)=z0k(t-h)+uk(t),00000000000≤t≤h,k=1,2,….
Since the function (21) must satisfy the initial condition z(0)=z0(0), we have
(28)zk(0)=z0k0=∫Ω‍z0(0)φk(x)dx,k=1,2,….
From this and (27) we see that
(29)zk(t)=e-λktz0k0+∫0t‍e-λk(t-τ)(z0k(τ-h)+uk(τ))dτ,000000000000000000000000000000000k=1,2,….
Thus, (21) and (29) define the formal solution of the problem (15)–(17) on [0,h] which can be written as follows:
(30)z(t)=∑k=1∞‍[e-λktz0k0+∫0t‍e-λk(t-τ)(z0k(τ-h)+uk(τ))dτ]00000×φk(x).The following assumption will be needed throughout the paper.Assumption 2.
Let functionsz(·) of the set X satisfy the following conditions:
(31)∑k=1∞‍λkr∫0t‍|z0k(τ-h)|2dτ≤b2,∑k=1∞‍λkr|z0k(0)|2≤b2,(32)∑k=1∞‍λkr∫0t‍e-λk(t-τ)|z0k(τ-h)|2dτ≤b2∫0t‍e-λ1(t-τ)dτ,0≤t≤h.Proposition 3.
For any functionf(t),∥f(t)∥Hr≤M, 0≤t≤T, the following inequality holds:
(33)∑k=1∞‍λkr(∫0t‍e-λk(t-τ)fk(τ)dτ)2≤a12(t)M2;a1(t)=1-e-λ1tλ1,0≤t≤T,
where λ1,λ2,… are the eigenvalues of the operator A arranged in decreasing order, M is a positive number, and fk(·) is the Fourier coefficient of the function f(t).Proof.
Indeed, for0≤t≤T, using the Cauchy-Schwartz inequality and the fact that
(34)∥f(τ)∥Hr2=∑k=1∞‍λkrfk2(τ)≤M2,0≤τ≤t≤T,
we obtain
(35)∑k=1∞‍λkr(∫0t‍e-λk(t-τ)fk(τ)dτ)2=∑k=1∞‍λkr(∫0t‍e-(λk/2)(t-τ)·e-(λk/2)(t-τ)fk(τ)dτ)2≤∑k=1∞‍λkr∫0t‍e-λ1(t-τ)dτ∫0t‍e-λ1(t-τ)fk2(τ)dτ≤∫0t‍e-λ1(t-τ)dτ∫0t‍e-λ1(t-τ)·∑k=1∞‍λkrfk2(τ)dτ≤a1(t)∫0t‍e-λ1(t-τ)M2dτ=a12(t)M2.Theorem 4.
Letu(t), 0≤t≤h be an admissible control that satisfies the condition (19), and let Assumption 2 be satisfied. Then there exists a unique solution of the problem (15)–(17) in the space C([0,h];Hr(Ω)).Proof.
To prove the theorem we use the formal representation (30) of the solution of the problem (15)–(17). By definition of the norm in Hr(Ω) for 0≤t≤h, we have
(36)∥z(t)∥Hr2=∑k=1∞‍λkr|zk(t)|2=∑k=1∞‍λkr(e-λktz0k0+∫0t‍e-λk(t-τ)fk(τ)dτ)2,
where fk(τ)=z0k(τ-h)+uk(τ).
We use the Cauchy-Schwartz inequality to obtain the following chain of relations:(37)∑k=1∞‍λkr(∫0t‍e-λk(t-τ)fk(τ)dτ)2≤∑k=1∞‍λkr(∫0t‍e-(λk(t-τ))/2e-(λk(t-τ))/2fk(τ)dτ)2≤∑k=1∞‍λkr∫0t‍e-λk(t-τ)dτ∫0t‍e-λk(t-τ)fk2(τ)dτ≤∑k=1∞‍λkr1-e-λktλk∫0t‍e-λk(t-τ)fk2(τ)dτ.
Therefore,
(38)∑k=1∞‍λkr1-e-λktλk∫0t‍e-λk(t-τ)fk2(τ)dτ≤a1(t)∑k=1∞‍λkr∫0t‍e-λk(t-τ)(z0k(τ-h)+uk(τ))2dτ.
Next, we expand the integrand on the right hand side of the last inequality; then we use the Cauchy-Schwartz inequality and after that we use Assumption 2 to obtain
(39)∑k=1∞‍λkr(∫0t‍e-λk(t-τ)fk(τ)dτ)2≤(b+ρ)2a12(t).
Expanding the square in (36), and then using (39) and Assumption 2, we arrive at
(40)∥z(t)∥Hr2≤(e-λ1tb+(b+ρ)a1(t))2.
Hence, z(t)∈Hr(Ω) for each t∈[0,h].
We now show that the functionz(t), 0≤t≤h, is continuous with respect to the norm of the space Hr. We have
(41)∥z(t+δ)-z(t)∥Hr2=∑k=1∞‍λkr|zk(t+δ)-zk(t)|2=∑k=1N‍λkr|zk(t+δ)-zk(t)|2+∑k=N+1∞‍λkr|zk(t+δ)-zk(t)|2,
where the numbers δ and N will be chosen depending on any ε>0.
By (29), We have
(42)zk(t+δ)-zk(t)=z0k0e-λkt(e-λkδ-1)+∫0t‍e-λk(t-τ)(e-λkδ-1)fk(τ)dτ+∫tt+δ‍e-λk(t+δ-τ)fk(τ)dτ.
Combining this with (41), we obtain
(43)∥z(t+δ)-z(t)∥Hr2≤I1+I2+I3;I1=3∑k=1N‍λkr|e-λkδ-1|2(|z0k0|2+|∫0t‍e-λk(t-τ)fk(τ)dτ|2)I2=3∑k=N+1∞‍λkr(|z0k0|2+|∫0t‍e-λk(t-τ)fk(τ)dτ|2)I3=3∑k=1∞‍λkr|∫tt+δ‍e-λk(t+δ-τ)fk(τ)dτ|2.
Let ε be an arbitrary positive number. Since the series in (31) and (39) are convergent, one can choose the number N so that I2<ε2/3. Then, we choose δ so that I1<ε2/3. Further, similar to (39) we have
(44)∑k=1∞‍λkr(∫tt+δ‍e-λk(t+δ-τ)fk(τ)dτ)2≤e-2λ1δ∑k=1∞‍λkr(∫tt+δ‍e-λk(t-τ)fk(τ)dτ)2≤e-2λ1δ((b+ρ)∫tt+δ‍e-λk(t-τ)dτ)2.
It is clear from these inequalities that we can choose δ so that I3<ε2/3. From the estimations for I1,I2,I3 obtained above we conclude according to (43) that ∥z(t+δ)-z(t)∥Hr<ε, meaning that the function z(t), 0≤t≤h, is continuous.
We now show that the problem (15)–(17) has a unique solution. We assume the contrary, that it has two different solutions z′(t), z′′(t), 0≤t≤h for the same functions z0(·)∈X and u(·). Then their difference Z(t)=z′(t)-z′′(t), 0≤t≤h, as a solution of corresponding homogeneous problem (15)–(17), can be represented in the form of the Fourier series (21), and the Fourier coefficients Zk(t), 0≤t≤h, of the function Z(t), 0≤t≤h, are solutions of the following infinite system of differential equations:
(45)z˙k(t)+λkzk(t)=0,zk(0)=0,000000000≤t≤h,k=1,2,….
Therefore, from the representation (29) we obtain Zk(t)=0 on 0≤t≤h. Consequently, Z(t)=0, 0≤t≤h, a contradiction. This proves the theorem.


## 2. Main results

Consider heat exchange control problem with lag (15)–(17).Definition 5.
A setW⊂R1 is referred to as the strong invariant set on the time interval [0,T] with respect to the problem (15)–(19) if for any z0(t), ∥z0(t)∥Hr∈W, and u(t), ∥u(t)∥Hr≤ρ, t∈[0,T], the inclusion ∥z(t)∥Hr∈W holds true on [0,T].Definition 6.
A setW⊂R1 is referred to as the weak invariant set on the time interval [0,T] with respect to the problem (15)–(19) if for any z0(t), ∥z0(t)∥Hr∈W there exists u(t), ∥u(t)∥Hr≤ρ, t∈[0,T], such that the inclusion ∥z(t)∥Hr∈W holds true on [0,T].Further, strong and weak invariant sets of the formW=[0,b] will be investigated, where b is a positive number. We will find relations among the parameters T, b, ρ, and λi in order to guarantee strong or weak invariance of the set W on the time interval [0,T] with respect to the problem (15)–(19) (or (15)–(17), (20)).Theorem 7.
Ifλ1≥1, the set W=[0,b] is strong invariant on [0,T] with respect to the problem (15)–(19) if and only if
(46)ρ≤(λ1-1)b.Proof.
Let (46) hold true and let z0(t), ∥z0(t)∥Hr≤b, -h≤t≤0, and u(t), ∥u(t)∥Hr≤ρ, 0≤t≤h be any functions. Using the same reasoning as in the proof of Theorem 4, we get from (40) that
(47)∥z(t)∥Hr2≤χ2(t);χ(t)=e-λ1tb+a1(t)(b+ρ),0≤t≤h.
Then(48)χ(0)=b,χ′(t)=-λ1e-λ1tb+e-λ1t(b+ρ)=e-λ1t(-λ1b+b+ρ)≤0.
Hence, χ′(t)≤0 meaning that the function χ(t) is nonincreasing. For this reason, ∥z(t)∥Hr≤b, 0≤t≤h.
The same reasoning can be applied for the time interval[h,2h] to obtain ∥z(t)∥Hr≤b. For any positive number T>0, we can continue in this fashion to obtain
(49)∥z(t)∥Hr≤b,0≤t≤T;
that is, W is strong invariant.
We now turn to the proof of the second part of the theorem. Let the setW be strong invariant. Suppose, contrary to our claim, that the inequality (46) fails to hold. Then, clearly, the function χ(t) is increasing. Choose the functions z0(·),u(·) as follows:
(50)z01(t-h)=bλ1r/2,u1(t)=ρλ1r/2,z0k(t-h)=0,uk(t)=0,k≥2,0≤t≤h.
Then
(51)∥z(t)∥Hr2=λ1r(e-λ1tz010+∫0t‍e-λ1(t-τ)(z01(t-τ)+u1(τ))dτ)2=χ2(t).
Since the function χ(t) is increasing on [0,h] and χ(0)=b, therefore t,χ(t)>b; that is, ∥z(t)∥Hr>b, which contradicts the strong invariance of the set W. This completes the proof of Theorem 7.Theorem 8.
LetT be any positive number. If λ1≥1, the set W=[0,b] is weak invariant on [0,T] with respect to the problem (15)–(19).Proof.
Letλ1≥1. Show that, under condition 9, the set W=[0,b] is weak invariant with respect to the problem (15)–(17). Let z0(·) be an arbitrary function from X. Set u(·)=0. Then we obtain from the representation (29) of the solution of the problem (27) that
(52)∥z(t)∥Hr2=∑k=1∞‍λkr(e-λktz0k0+∫0t‍e-λk(t-τ)z0k(t-τ)dτ)2≤e-2λ1t∑k=1∞‍λkr|z0k0|2+2e-λ1t∑k=1∞‍λkr|z0k0|∫0t‍e-λk(t-τ)|z0k(t-τ)|dτ+∑k=1∞‍λkr(∫0t‍e-λk(t-τ)z0k0(t-τ)dτ)2≤e-2λ1tb2+2e-λ1tba1(t)b+a12(t)b2=(e-λ1t+a1(t))2b2.
Here, we used the following inequality:
(53)∑k=1∞‍λkr|z0k0|∫0t‍e-λk(t-τ)|z0k(t-τ)|dτ≤∑k=1∞‍λkr|z0k0|2∑k=1∞‍λkr(∫0t‍e-λk(t-τ)|z0k(t-τ)|dτ)2
and Assumption 2. Since the function
(54)ξ(t)=e-λ1t+1-e-λ1tλ1,0≤t≤h
satisfies
(55)ξ(0)=1,ξ′(t)=(1-λ1)e-λ1t≤0,
we have 0<ξ(t)≤1. Then in view of (52) we conclude that ||z(t)||Hr≤b, 0≤t≤h. Repeated application of this reasoning to the intervals [jh,(j+1)h], j=1,2,… enables us to conclude that ∥z(t)∥Hr≤b, 0≤t≤T, and this completes the proof.We now turn to the problem (15)–(17), (20).Definition 9.
The setW⊂R1 is called weak invariant on [0,T] with respect to the problem (15)–(17), (20) if for any z0(t), ∥z0(t)∥Hr∈W, t∈[-h,0] there exists u(t), ∥u(·)∥L2([0,T];Hr)≤ρ, such that the inclusion ||z(t)||Hr∈W holds for all t∈[0,T].Definition 10.
The setW⊂R1 is called strong invariant on [0,T] with respect to the problem (15)–(17), (20) if for any z0(t), ∥z0(t)∥Hr∈W, t∈[-h,0] and u(t), ∥u(·)∥L2([0,T];Hr)≤ρ the inclusion ∥z(t)∥Hr∈W holds for all t∈[0,T].Proposition 11.
For any functiong(τ), 0≤τ≤t≤T, for which ∥g(·)∥L2([0,T];Hr)≤N, the following inequality holds:
(56)∑k=1∞‍λkr(∫0t‍e-λk(t-τ)gk(τ)dτ)2≤a2(t)N2,00000000000000000000000000≤t≤T;a2(t)=1-e-2λ1t2λ1,
where T a positive number.The proof follows from the following chain of relations(57)∑k=1∞‍λkr(∫0t‍e-λk(t-τ)gk(τ)dτ)2≤∑k=1∞‍λkr∫0t‍e-2λk(t-τ)dτ·∫0t‍gk2(τ)dτ≤a2(t)∑k=1∞‍λkr∫0t‍gk2(τ)dτ=a2(t)∥g(·)∥L2([0,t];Hr)2≤a2(t)N2;
here we used the Cauchy-Schwartz inequality and the relation e-λkt≤e-λ1t, t≥0, k=1,2,….Theorem 12.
Ifρ>0, then the set W=[0,b] is not strong invariant with respect to the problem (15)–(17), (20) on [0,T], where T is any number.Proof.
Letρ>0. To prove the theorem, we use the fact that modulus of the admissible control function can take any big values on small time interval. Consider the function
(58)η(t)=e-λ1tb+1-e-λ1tλ1b+e-λ1ttρ,t≥0.
Note that η(0)=b and η(t)>0 for all t>0. Assuming t>0, we have
(59)η′(t)=e-λ1t(b-λ1b-λ1ρt+ρ2t).
Due to the term e-λ1t(ρ/2t), we obtain that there exists t1, 0<t1<h, such that η′(t)>0 whenever 0<t≤t1.
We now specify the initial function and admissible control function on[0,h] as follows:
(60)z0(τ-h)=bλ1r/2φ1,0≤τ≤h,u(τ)={ρλ1r/2t1φ1,0≤τ<t1,0,t1≤τ≤h.
We observe that
(61)∥z0(τ)∥Hr≤b,-h≤τ≤0,∥u(·)∥L2([0,h];Hr)≤ρ.
Then by the representation (30) we have
(62)z(t)=(e-λ1tbλ1r/2+∫0t‍e-λ1(t-τ)(bλ1r/2+ρλ1r/2t1)dτ)φ1,0000000000000000000000000000000.00000000≤t≤t1.
Hence
(63)∥z(t)∥Hr2=λ1r(e-λ1tbλ1r/2+∫0t‍e-λ1(t-τ)(bλ1r/2+ρλ1r/2t1)dτ)2,000000000000000000000000000000000000≤t≤t1.
This implies that
(64)∥z(t1)∥Hr=e-λ1t1b+b∫0t1‍e-λ1(t1-τ)dτ+∫0t1‍e-λ1(t1-τ)ρt1dτ≥e-λ1t1b+b∫0t1‍e-λ1(t1-τ)dτ+∫0t1‍e-λ1t1ρt1dτ=η(t1).
Sinceη(0)=b and η′(t)>0 for 0<t≤t1, then η(t1)>b. Therefore ∥z(t1)∥Hr>b which shows that the set W=[0,b] is not strong invariant, and the proof of Theorem 12 is complete.Theorem 13.
Ifλ1≥1, then the set W=[0,b] is weak invariant with respect to the problem (15)–(17), (20) on [0,h].The proof of the theorem is similar to that of Theorem8.



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