# Inverse Problems for a Parabolic Integrodifferential Equation in a Convolutional Weak Form

Authors: Kairi Kasemets; Jaan Janno
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We deduce formulas for the Fréchet derivatives of cost functionals of several inverse problems for a parabolic integrodifferential equation in a weak formulation. The method consists in the application of an integrated convolutional form of the weak problem and all computations are implemented in regular Sobolev spaces.


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## Body

## 1. Introduction

Many methods to solve inverse problems (e.g., the Landweber iteration, conjugate gradient method) use the Fréchet derivatives of the cost functionals of these problems [1]. The explicit formula for the Fréchet derivative in terms of the variation of the unknowns of the inverse problem contains the solution of an adjoint problem.The derivation of the explicit formula for such a Fréchet derivative includes testing the direct problem with the solution of the adjoint problem and vice versa: testing the adjoint problem with the solution of the direct problem. In the case of the parabolic weak problem, such a procedure is cumbersome, because of the asymmetry of the properties of the solution and the test function. In the classical formulation of the parabolic weak problem (see, e.g., [2] and also (19) below), the test function must have higher time regularity than the weak solution. This means that in case of nonsmooth coefficients neither the solution of the direct problem nor the solution of the adjoint problem can be used as a test function. Another formulation of the weak parabolic problem consists in reducing the problem to an abstract Cauchy problem over the time variable (see, e.g., [3]). In such a case, a partial integration over the time has to be implemented within singular distributions in the derivation procedure.In this paper, we present a new method that enables the deduction of the formulas for the Fréchet derivatives for cost functionals of inverse problems related to weak solutions of parabolic problems. The method is based on an integrated convolutional form of the weak direct problem. The requirements to the test function are weaker than in the classical case and coincide with the properties of the solution of the direct problem. All computations in the deduction procedure can be implemented within usual regular Sobolev spaces.More precisely, we will consider inverse problems related to a parabolic integrodifferential equation that occur in heat flow with memory [4–6]. This equation contains a time convolution. Therefore, the convolutional form of the weak problem is especially suitable. Supposedly, the proposed method can be generalised to parabolic systems, as well.


## 2. Formal Direct Problem: Notation

LetΩ be an  n-dimensional domain, where n≥1, and Γ be the boundary of Ω. Let Γ=Γ1∪Γ2 where either Γ1 or Γ2 is allowed to be an empty set. In case n≥2, we assume that Γ is sufficiently smooth, meas Γ1∩Γ2=0, and for any j∈{1;2} it holds either Γj=∅ or meas Γj>0. Denote
(1)Ωt=Ω×(0,t),Γ1,t=Γ1×(0,t),Γ2,t=Γ2×(0,t),
for t≥0. Consider the problem (direct problem) to find u(x,t):ΩT→ℝ such that
(2)ut=Au-m*Au+f+∇·ϕinΩT,(3)u=u0inΩ×{0},(4)u=ginΓ1,T,(5)-νA·∇u+m*νA·∇u=ϑu+h+ν·ϕinΓ2,T,
where T>0 is a fixed number,
(6)Av=∑i,j=1n(aijvxj)xi+av,νA=(∑j=1n‍aijνj∣i=1,…,n),ν=(ν1,…,νn)-outernormalofΓ2,aij,a,u0:Ω→ℝ,  f:ΩT→ℝ,  ϕ:ΩT→ℝn,  g:Γ1,T→ℝ,   ϑ:Γ2→ℝ, h:Γ2,T→ℝ, m:(0,T)→ℝ are given functions, the subscripts  t,  xj,  xi denote the partial derivatives and
(7)m*w(t)=∫0tm(t-τ)w(τ)dτ
denotes the time convolution. In case Γ1=∅ (Γ2=∅), the boundary condition (4) and (5) is dropped.The problem (2)–(5) describes the heat flow in the body Ω with the thermal memory. Concerning the physical background, we refer the reader to [4, 6, 7]. The solution u is the temperature of the body and m is the heat flux relaxation (or memory) kernel. The boundary condition (5) is of the third kind where the term  -νA·∇u+m*νA·∇u  equals the heat flux in the direction of the conormal vector.Let us introduce some additional notations. Lett>0. We use the Sobolev spaces
(8)W2l(Ω)={[∑|α|≤l‍∥Dxαv∥L2(Ω)2]1/2v:Ω→ℝ:∥v∥W2l(Ω)∶=[∑|α|≤l‍∥Dxαv∥L2(Ω)2]1/2<∞}.
Here, l=0,1,2,…, α=(α1,…,αn) is the multiindex, |α|=α1+⋯+αn and Dxαv=∂|α|v/∂x1α1⋯∂xnαn. Further, let X be a Banach space. We denote by C([0,t];X) the space of abstract continuous functions from [0,t] to X endowed with the usual maximum norm  ∥v∥C([0,t];X):=maxs∈[0,t]∥v(s)∥.  Moreover, let
(9)Lp((0,t);X)≔{[∑|α|≤l‍∥Dxαv∥L2(Ω)2]1/2v:(0,t)→X:∥v∥Lp((0,t);X)≥[∫0t‍∥v(s)∥pds]1/p<∞[∑|α|≤l‍∥Dxαv∥L2(Ω)2]1/2}for1<p<∞,L∞((0,t);X)∶={supt∈(0,T)v:(0,t)→X:∥v∥L∞((0,t);X)=esssupt∈(0,T)∥v(s)∥<∞}.
By means of these spaces, we define the following important functional spaces:
(10)𝒰(Ωt)=C([0,t];L2(Ω))∩L2((0,t);W21(Ω)),𝒰0(Ωt)={η∈𝒰(Ωt):η|Γ1,t=0incaseΓ1≠∅}.Convention. In case n=1, the integrals ∫Γj‍v(x)dΓ,  j=1,2 are equal to ∑k=1Kv(xk), where xk∈Γj and K is the number of points in Γj,  and Lp(Γj) is simply ℝK.


## 3. Weak Direct Problem and Its Convolutional Form

Let us return to the direct problem (2)–(5). Throughout the paper we assume the following basic regularity conditions on the coefficients, the kernel, and the initial and boundary functions: (11)aij∈L∞(Ω),aij=aji,ϑ∈C(Γ¯2),ϑ≥0,(12)a∈Lq1(Ω),whereq1=1ifn=1,q1>n2ifn≥2(13)m∈L1(0,T),g∈L2((0,T);W21(Ω)),gt∈L2(ΩT),(14)f∈L2((0,T);Lq2(Ω)),whereq2=1ifn=1,q2∈(1,q1)ifn=2,q2=2nn+2ifn≥3,(15)ϕ=(ϕ1,…,ϕn)∈(L2(ΩT))n(16)u0∈L2(Ω),h∈L2(Γ2,T)
and the ellipticity condition
(17)∑i,j=1n‍aijλiλi≥ϵ|λ|2,x∈Ω¯,λ∈ℝnwithsomeϵ>0,
(for the sake of simplicity we introduced an assumption for the extension of g onto  ΩT).The first aim is to reformulate the problem (2)–(5) in a weak form. Let us suppose that aij∈W21(Ω),   (∂/∂xi)ϕi∈L2(ΩT),  i=1,…,n and (2)–(5) has a classical solution u∈W22,1(ΩT). Then, we multiply (2) with a test function η from the space
(18)𝒯(ΩT)={η∈L2((0,T);W21(Ω)):ηt∈L2((0,T);L2(Ω)),η|Γ1,T=0incaseΓ1≠∅W21}
and integrate by parts with respect to time and space variables. We obtain the following relation:
(19)0=∫Ω[u(x,T)η(x,T)-u0(x)η(x,0)]dx-∬ΩTuηtdxdt+∬ΩT[∑i,j=1naij(uxj-m*uxj)ηxi-a(u-m*u)η]dxdt+∬Γ2,T(ϑu+h)ηdΓdt-∬ΩT(fη-ϕ·∇η)dxdt.
This relation makes sense also in a more general case when aij, ϕ satisfies only (11) and (15) and  u does not have regular first-order time and second-order spatial derivatives.We call aweak solution of the problem (2)–(5) a function from the space 𝒰(ΩT) that satisfies the relation (19) for any η∈𝒯(ΩT) and in case Γ1≠∅ fulfills the boundary condition (4).Lemma 1.
The following assertions are valid:(i)
𝒰(ΩT)↪L2((0,T);Lq3(Ω)) where  q3=∞  if  n=1, q3∈(q1q2/(q1-q2),∞) if n=2 and  q3=2n/(n-2) if n>2, where  q1 and  q2 are given in (12) and (14), respectively;(ii)
for anyu∈L2((0,T);Lq3(Ω)) it holds  au∈L2((0,T);Lq2(Ω)) and ∥au∥L2((0,T);Lq2(Ω))≤C∥a∥Lq1(Ω)∥u∥L2((0,T);Lq3(Ω)), where C  is a constant.Proof.
Since𝒰(ΩT)↪L2((0,T);W21(Ω)), the assertion (i) follows from the continuous embedding of W21(Ω) in Lq3(Ω). The assertion (ii) can be proved by means of Hölder's inequality.Theorem 2.
The problem (2)–(5) has a unique weak solution. This solution satisfies the estimate
(20)∥u∥𝒰(ΩT)≤C0[∫∥u0∥L2(Ω)+∥f∥L2((0,T);Lq2(Ω))+∥ϕ∥(L2(ΩT))n+θ1{∥g∥L2((0,T);W21(Ω))+∥gt∥L2(ΩT)}+θ2∥h∥L2(Γ2,T)∫],
where θ1=0 in case Γ1=∅, θ2=0 in case Γ2=∅ and C0 is a constant depending on Ω,  Γj,  aij,  a,  ϑ  and  m.Proof.
The assertion of the theorem in casem=0 is well known from the theory of parabolic equations (see, e.g., [2]). Let 𝒵 be the operator that assigns to the data vector d:=(u0,f,ϕ,g,h) the weak solution of the problem (2)–(5) in case m=0. Then it holds
(21)∥𝒵d∥𝒰(ΩT)≤RHS,
where RHS is the right-hand side of (20).
Further, let us formulate the problem for the differencev=u-𝒵d. Introducing the linear operator 𝒜  by the formula
(22)𝒜w=𝒵(0,-am*w,-∑j=1naijm*wxj,0,0),
the weak problem (2)–(5) for the solution u∈𝒰(ΩT) equivalent to the following operator equation for the quantity v:
(23)v=𝒜v+𝒜𝒵d.
We have to estimate 𝒜. For this purpose, we firstly prove the following auxiliary inequality:
(24)∥m*y∥L2((0,t);Lp(Ω))≤∫0t|m(t-τ)|∥y∥L2((0,τ);Lp(Ω))dτ,t∈[0,T],
for any p≥1 and y∈L2((0,t);Lp(Ω)).
Denotingy-(t)=y(·,t), ∥y-(t)∥=∥y-(t)∥Lp(Ω), making use of the following property of the Bochner integral: ∥∫0sw(s,τ)dτ∥Lp(Ω)≤∫0s∥w(s,τ)∥Lp(Ω)dτ for functions w∈L1((0,T);Lp(Ω)) and the Cauchy's inequality, the relation (24) can be deduced by means of the following computations:
(25)∥m*y∥L2((0,t);Lp(Ω))=[∫0t∥∫0s‍m(τ)y-(s-τ)dτ∥Lp(Ω)2ds]1/2≤I,whereI=[∫0t(∫0s|m(τ)|∥y-(s-τ)∥dτ)2ds]1/2I=[∫0t|m(τ)|∫τt∥y-(s-τ)∥∫0s|m(z)|∥y-(s-z)∥dzdsdτ]1/2I≤[∫0t‍|m(τ)|[∫τt‍∥y-(s-τ)∥2ds]1/2{[∫τt‍(∫0s‍|m(z)|∥y-(s-z)∥dz)2ds]1/2dτ}×[∫τt‍(∫0s‍|m(z)|∥y-(s-z)∥dz)2ds]1/2dτ]1/2I≤[∫0t‍|m(τ)|[∫0t-τ‍∥y-(s)∥2ds]1/2dτ]1/2I×[∫0t‍(∫0s‍|m(z)|∥y-(s-z)∥dz)2ds]1/4I=[∫0t‍|m(t-τ)|∥y∥L2((0,τ);Lp(Ω))dτ]1/2×I1/2.
Next, lett∈[0,T] and introduce the operator(26)Ptw={winΩt0inΩT∖Ωt.
Due to the causality we have 𝒵(0,Ptf,Ptϕ,0,0)(x,t)=𝒵(0,f,ϕ,0,0)(x,t) for any (x,t)∈Ωt. Using these relations, the continuity of the linear operator 𝒵, the inequality (24), and the boundedness of  aij, we compute the following:
(27)∥𝒜w∥𝒰(Ωt)=∥𝒵(0,-am*w,-∑j=1naijm*wxj,0,0)∥𝒰(Ωt)=∥𝒵(0,-Pt[am*w],-Pt[∑j=1naijm*wxj],0,0)∥𝒰(Ωt)≤∥𝒵(0,-Pt[am*w],-Pt[∑j=1naijm*wxj],0,0)∥𝒰(ΩT)≤C^[∥Pt[am*w]∥L2((0,T);Lq2(Ω))+∑i=1n∥Pt[aijm*wxj]∥L2(ΩT)]=C^[∥am*w∥L2((0,t);Lq2(Ω))+∑i=1n∥aijm*wxj∥L2(Ωt)]≤C~∫0t‍|m(t-τ)|(∥aw∥L2((0,τ);Lq2(Ω))+∥∇w∥L2(Ωτ))dτ,
with some constants C^  and C~  depending on  Ω, Γj, aij, a, ϑ.  Using Lemma 1, we obtain
(28)∥aw∥L2((0,τ);Lq2(Ω))≤C1∥a∥Lq1(Ω)∥w∥L2((0,τ);W21(Ω)).
Using this relation in (27), we arrive at the following basic estimate for 𝒜:
(29)∥𝒜w∥𝒰(Ωt)≤C2∫0t|m(t-τ)|∥w∥𝒰(Ωτ)dτ,t∈[0,T],
where C2 is a constant depending on Ω, Γj, aij, a, ϑ. Let us define the weighted norms in 𝒰(ΩT): ∥v∥σ=sup0<t<Te-σt∥v∥𝒰(Ωt) where σ≥0. The estimate (29) implies the further estimate
(30)∥𝒜w∥σ≤C2sup0<t<Te-σt∫0t|m(t-τ)|∥w∥𝒰(Ωτ)dτ=C2sup0<t<T∫0te-σ(t-τ)|m(t-τ)|e-στ∥w∥𝒰(Ωτ)dτ≤C2∫0Te-σs|m(s)|ds∥w∥σ.
Since ∫0Te-σs|m(s)|ds→0 as σ→∞, there exists σ0, depending on C2 and m, such that C2∫0Te-σ0s|m(s)|ds≤1/2. Thus, ∥𝒜w∥σ0≤(1/2)∥w∥σ0. The operator 𝒜 is a contraction in 𝒰(ΩT). This implies the existence and uniqueness assertions of the theorem.
To prove the estimate (20), we firstly deduce from (23) the inequality ∥v∥σ0≤∥𝒜v∥σ0+∥𝒜𝒵d∥σ0≤(1/2)[∥v∥σ0+∥𝒵d∥σ0]. This implies ∥v∥σ0≤∥𝒵d∥σ0. Using the equivalence relations e-σ0T∥·∥0≤∥·∥σ0≤∥·∥0, where ∥·∥0=∥·∥𝒰(ΩT) and (21), we reach (20).We note the upper integration boundT in (19) can be released to be any number t from the interval [0,T]. Indeed, (19) is equivalent to the following problem:
(31)0=∫Ω[u(x,t)η(x,t)-u0(x)η(x,0)]dx-∬Ωtuηtdxdt+∬Ωt[∑i,j=1naij(uxj-m*uxj)ηxi-a(u-m*u)η∑i,j=1n]dxdt+∬Γ2,t(ϑu+h)ηdΓdt-∬Ωt(fη-ϕ·∇η)dxdt,t∈[0,T],
for any η∈𝒯(ΩT). This assertion can be proved using the standard technique defining the test function as follows:
(32)ηϵ(x,t)={η(x,τ)forτ∈[0,t],η(x,τ)(1-τ-tϵ)forτ∈(t,t+ϵ),0forτ∈[t+ϵ,T],
and letting the parameter ϵ to approach 0.Next we transform the weak direct problem (31) to a form that does not contain a time derivative of the test function η. This form enables the extension of the test space. This is useful for treatment of problems for adjoint states of quasisolutions of inverse problems in next sections.Theorem 3.
The functionu∈𝒰(ΩT) satisfies the relation (19) for any η∈𝒯(ΩT) if and only if it satisfies the following relation:
(33)0=∫Ωu*ηdx-∫Ω∫0tu0(x)η(x,τ)dτdx+∫Ω1*[∑i,j=1naij(uxj-m*uxj)*ηxi-a(u-m*u)*η∑i,j=1n]dx+∫Γ21*(ϑu+h)*ηdΓ-∫Ω1*(f*η-∑i=1nϕi*ηxi)dx,t∈[0,T],
for any η∈𝒰0(ΩT).Here, according to the definition of the time convolution in the previous section,1*w(t)=∫0tw(τ)dτ.Proof.
It is sufficient to prove thatu∈𝒰(ΩT) satisfies (31) for any η∈𝒯(ΩT) if and only if it satisfies (33) for any  η∈𝒰0(ΩT). Suppose that u∈𝒰(ΩT) satisfies (31) and choose an arbitrary η∈𝒯(ΩT). Let t1 be an arbitrary number on the interval [0,T] and choose some function ξt1∈𝒯(ΩT) such that the relation
(34)ξt1(x,t)=η(x,t1-t)fort∈[0,t1]
is valid. For instance, it is possible to define  ξt1  as a periodic function with respect to t, that is,  ξt1(x,t)=η(x,t1-t) for t∈[0,t1], ξt1(x,t)=η(x,t-t1) for t∈[t1,2t1], ξt1(x,t)=η(x,3t1-t) for t∈[2t1,3t1] and so on. Using the relation (31) with η replaced by ξt1 and setting there t=t1 we obtain the equality
(35)0=K1(t1)+K2(t1),
where
(36)K1(t)=∫Ω[u(x,t)η(x,0)-u0(x)η(x,t)]dx+∫Ω∫0tu(x,τ)ηt(x,t-τ)dτdx,(37)K2(t)=∫Ω[∑i,j=1naij(uxj-m*uxj)*ηxi-a(u-m*u)*η∑i,j=1n]dx+∫Γ2(ϑu+h)*ηdΓ-∫Ω(f*η-∑i=1nϕi*ηxi)dx.
Note that the time derivative of η can be removed from K1 by integration. Indeed, let t2∈[0,T]. Then
(38)∫0t2K1(t1)dt1=∫0t2∫Ωu(x,t1)η(x,0)dxdt1-∫0t2∫Ωu0(x)η(x,t1)dxdt1+∫0t2∫Ω∫0t1u(x,τ)ηt1(x,t1-τ)dτdxdt1.
Changing the order of the integrals over τ and t1 in the last term, we easily obtain
(39)∫0t2K1(t1)dt1=∫Ω∫0t2u(x,τ)η(x,t2-τ)dτdx-∫0t2∫Ωu0(x)η(x,t1)dxdt1.
Integrating now the whole equality (35) over t1 from 0 to t2, observing (37) and (39), and finally redenoting t2 by t, we reach the desired relation (33). Summing up, we have proved that (33) holds for any η∈𝒯(ΩT). But all terms in the right-hand side of (33) are well defined for η∈𝒰0(ΩT), too. Since 𝒯(ΩT) is densely embedded in  𝒰0(ΩT), we conclude that (33) holds for any η∈𝒰0(ΩT).
It remains to show that (33) implies (31). Suppose that u∈𝒰(ΩT) satisfies (33) and choose an arbitrary η∈𝒯(ΩT) and t1∈[0,T]. Again, let ξt1 be a function from 𝒯(ΩT) such (34) is valid. Inserting ξt1 instead of η into (33), differentiating with respect to t and setting t=t1 we come to the relation (31). Theorem is proved.Corollary 4.
A functionu∈𝒰(ΩT) is a weak solution of (2)–(5) if and only if it satisfies the relation (33) for any η∈𝒰0(ΩT) and in case Γ1≠∅ fulfills the boundary condition (4).


## 4. Inverse Problems and Quasisolutions

In the sequel, let us pose some inverse problems for the weak solution of (2)–(5). These problems are selected in order to demonstrate the wide possibilities of the method that we will introduce in Section 5.Firstly, we suppose that (2)–(5) has the following specific form:
(40)ut=Au-m*Au+f0+∇·ϕ+∑j=1Nγj(t)ωj(x)inΩT,u=u0inΩ×{0},u=ginΓ1,T,-νA·∇u+m*νA·∇u=ϑu+h+ν·ϕinΓ2,T,
where ω=(ω1,…,ωN) is unknown. The coefficients and other given functions f0, ϕ, u0, g, h are assumed to satisfy (11)–(17). Moreover, γ∈(L2(0,T))N is prescribed.IP1. Find the vector ω∈(L2(Ω))N such that the weak solution of (40) satisfies the following instant additional conditions:
(41)u(x,Ti)=uTi(x),x∈Ω,i=1,2,…,N,
where 0<T1<T2<⋯<TN≤T and uTi∈L2(Ω), i=1,…,N are given functions (observations of u).Since∑i=1Nγjωj∈L2((0,T);L2(Ω))⊂L2((0,T);Lq2(Ω)) for ω∈(L2(Ω))N, the weak solution u of (40) exists in 𝒰(ΩT); hence, it has traces u(·,Ti)∈L2(Ω), i=1,…,N. In practice, the term ∑j=1Nγjωj may represent an approximation of a more general function F(x,t)∈L2(ΩT), where γj, j=1,2,… form a basis in L2(0,T).Further, letu0 also be unknown.IP2. Find the vector ω∈(L2(Ω))N and u0∈L2(Ω) such that the weak solution of (40) satisfies the following integral additional conditions:
(42)∫0Tκi(x,t)u(x,t)dt=vi(x),x∈Ω,i=1,2,…,N+1,
where vi∈L2(Ω), i=1,…,N+1 are given observation functions and κi, i=1,…,N are given weights that satisfy the following condition:
(43)|κi(x,t)|≤κ^(t)inΩT,i=1,…,N+1withsomeκ^∈L2(0,T).
Note that the integral ∫0Tκi(·,t)u(·,t)dt in (42) belongs to L2(Ω) for any ω∈(L2(Ω))N and u0∈L2(Ω). Indeed, for such ω and u0 it holds u∈𝒰(ΩT)⊂L2(ΩT), which implies
(44)∥∫0Tκi(x,t)u(x,t)dt∥L2(Ω)≤∥κ^∥L2(0,T)∥u∥L2(ΩT)<∞.
In practice, the weights κi are usually concentrated in neighborhoods of some fixed values of time t=Ti.Finally, let us pose a nonlinear inverse problem for the coefficienta and the kernel m. Assume that n∈{1;2;3}. Then any coefficient a that belongs to L2(Ω)  satisfies (12). Moreover, let us set q1=2 if n=2 and Γ2≢∅. The other coefficients and the given functions u0, f, ϕ, g, h are assumed to satisfy (11)–(17).IP3. Find a∈L2(Ω) and m∈L1(0,T) such that the weak solution of (2)–(5) satisfies the following integral additional conditions:
(45)u(x,T)=uT(x),x∈Ω,∫Γ2‍κ(x,t)u(x,t)dΓ=v(t),t∈(0,T),
where uT∈L2(Ω),   v∈L2(0,T) are given observation functions and κ is a given weight function such that κ∈L∞((0,T);L2(Γ2)).As in IP1, we can show that the traceu(·,T) belongs to L2(Ω). Moreover, using the property u∈𝒰(ΩT), the embedding of W21(Ω) in L2(Γ2) and Hölder’s inequality, one can immediately check that the term  ∫Γ2‍κ(x,·)u(x,·)dΓ in (45) belongs to L2(0,T).Available existence, uniqueness, and stability results for IP1–IP3 require stronger smoothness of the data than imposed in the present paper. Let us cite some of these results.In caseN=1, the well posedness of IP1 was proved in [8]. Partial results were deduced earlier in [9]. A more general problem involving both IP1 and IP2 without the unknown u0 in case N=1 was studied in [10] by means of different techniques. IP1 and IP2 in case m=0 and N=1 were treated in many papers, for example,  [11–14]. The case N>1 is open even if  m=0. Inverse problems to determine  m with given a were studied in a number of papers, for example,  [7, 15–23]. The problem for a with given m was treated in [8].In the present paper, we will deal with quasisolutions of IP1–IP3 and related cost functionals. Denote𝒵1=(L2(Ω))N. Let M⊆𝒵1. The quasi-solution of IP1 in the set M is an element ω*∈argminω∈MJ1(ω), where J1 is the following cost functional
(46)J1(ω)=∑i=1N∥u(x,Ti;ω)-uTi(x)∥L2(Ω)2,
and u(x,t;ω) is the solution of (40) that corresponds to a fixed element ω.Similarly, letM⊆𝒵2:=(L2(Ω))N+1. The quasi-solution of IP2 in the set M is z*∈argminz∈MJ2(z), where J2 is the cost functional
(47)J2(z)=∑i=1N+1∥∫0Tκi(x,t)u(x,t;z)dt-vi(x)∥L2(Ω)2,
and u(x,t;z) is the weak solution of (40) that corresponds to a given vector  z=(ω,u0).Finally, definingM⊆𝒵3:=L2(Ω)×L2(0,T), the quasi-solution of IP3 in M is an element   z*∈argminz∈MJ3(z), where J3 is the cost functional
(48)J3(z)=∥u(x,T;z)-uT(x)∥L2(Ω)2+∥∫Γ2κ(x,t)u(x,t;z)dΓ-v(t)∥L2(0,T)2,
and u(x,t;z) is the weak solution of the direct problem (2)–(5) corresponding to given z=(a,m). Here, we restricted the space for the unknown m to L2(0,T), because we will seek for the Fréchet derivative of J3 in a Hilbert space. Moreover, the kernel of the second addend corresponding to m in the representation formula of J3′ (90) is an element of L2(0,T) and in general does not belong to the adjoint space L∞(0,T).According to the above-mentioned arguments, the functionalsJk, k=1,2,3, are well-defined in  𝒵1, 𝒵2, and 𝒵3, respectively.


## 5. The Fréchet Derivatives of Cost Functionals of Inverse Problems

### 5.1. General Procedure

Suppose that the solutionu of the direct problem depends on a vector of parameters p that has to be determined in an inverse problem making use of certain measurements of u. Let the quasi-solution of the inverse problem be sought by a method involving the Fréchet derivative of the cost functional (i.e., some gradient-type method). Usually in practice, a solution of a proper adjoint problem is used to represent the Fréchet derivative.We introduce a general procedure to deduce such adjoint problems. Assume thatΔu is the difference of solutions of the direct problem corresponding to a difference of the vector of the parameters  Δp. More precisely, we suppose that Δu is a solution of the following problem:
(49)Δut=AΔu-m*AΔu+f†+∇·ϕ†inΩT,(50)Δu=Δu0inΩ×{0},(51)Δu=0inΓ1,T,(52)-νA·∇Δu+m*νA·∇Δu=ϑΔu+h†+ν·ϕ†inΓ2,T,
with some data f†, ϕ†, Δu0, h† depending on Δp. We restrict ourselves to the case when the Dirichlet boundary condition of the state u is independent of p. Therefore, the condition (51) for  Δu  is homogeneous.In practice, the adjoint parabolic problems are usually formulated as backward problems. In our context, it is better to pose adjoint problems in the forward form. The involved memory term withm is defined via a forward convolution and from the practical viewpoint, it is preferable to have the direct and adjoint problems in a similar form (e.g., to simplify parallelisation of computations).More precisely, let the adjoint stateψ be a solution of the following problem:
(53)ψt=Aψ-m*Aψ+f∘+∇·ϕ∘inΩT,ψ=u∘inΩ×{0},ψ=0inΓ1,T,-νA·∇ψ+m*νA·∇ψ=ϑψ+h∘+ν·ϕ∘inΓ2,T,
where f∘, ϕ∘, u∘, and h∘ are some data depending on Δu and the cost functional under consideration.Assume that the quadrupletsf†, ϕ†, Δu0, h†, and f∘, ϕ∘, u∘, h∘ satisfy the conditions (14)–(16). Then, due to Theorem 2, the problems (49)–(52) and (53) have unique weak solutions in the space 𝒰(ΩT). Actually, we have Δu,ψ∈𝒰0(ΩT) because of the homogeneous boundary conditions on Γ1,T.Let us write the relation (33) for Δu and use the test function η=ψ. Then we obtain for any t∈[0,T](54)0=∫ΩΔu*ψdx-∫Ω∫0tΔu0(x)ψ(x,τ)dτdx+∫Ω1*[∑i,j=1naij(Δuxj-m*Δuxj)*ψxi-a(Δu-m*Δu)*ψη∑i,j=1n]dx+∫Γ21*(ϑΔu+h†)*ψdΓ-∫Ω1*(f†*ψ-∑i=1nϕi†*ψxi)dx.
Secondly, let us write this relation for ψ and use the test function η=Δu. Then we have for any  t∈[0,T](55)0=∫Ωψ*Δudx-∫Ω∫0tu∘(x)Δu(x,τ)dτdx+∫Ω1*[∑i,j=1naij(ψxj-m*ψxj)*Δuxi-a(ψ-m*ψ)*Δu∑i,j=1n]dx+∫Γ21*(ϑψ+h∘)*ΔudΓ-∫Ω1*(f∘*Δu-∑i=1nϕi∘*Δuxi)dx.
Subtracting (54) from (55), using the commutativity of the convolution, the symmetricity relations aij=aji and differentiating with respect to t, we arrive at the following basic equality that can be used in various inverse problems:
(56)∫Ωu∘(x)Δu(x,t)dx-∫Γ2h∘*ΔudΓ+∫Ω(f∘*Δu-∑i=1nϕi∘*Δuxi)dx=∫ΩΔu0(x)ψ(x,t)dx-∫Γ2h†*ψdΓ+∫Ω(f†*ψ-∑i=1nϕi†*ψxi)dx,t∈[0,T].


### 5.2. Derivative ofJ1

Theorem 5.
The functionalJ1  is the Fréchet differentiable in (L2(Ω))N and
(57)J1′(ω)Δω=∑j=1N∫Ω∑i=1N∫0Tiγj(t)ψi(x,Ti-t;ω)dtΔωj(x)dx,
where ψi∈𝒰(ΩTi), i=1,…,N, are the unique ω-dependent weak solutions of the following problems:
(58)ψi,t=Aψi-m*AψiinΩTi,ψi=2[u(x,Ti;ω)-uTi(x)]inΩ×{0},ψi=0inΓ1,Ti,-νA·∇ψi+m*νA·∇ψi=ϑψiinΓ2,Ti,i=1,…,N.Proof.
Let us fix someω,Δω∈(L2(Ω))N. One can immediately check that it holds
(59)J1(ω+Δω)-J1(ω)=2∑i=1N∫Ω[u(x,Ti;ω)-uTi(x)]Δu(x,Ti;ω)dx+∑i=1N∫ΩΔu(x,Ti;ω)2dx,
where Δu(x,t;ω)=u(x,t;ω+Δω)-u(x,t;ω)∈𝒰0(ΩT) is the weak solution of the following problem:
(60)Δut=AΔu-m*AΔu+∑j=1NγjΔωjinΩT,Δu=0inΩ×{0},Δu=0inΓ1,T,-νA·∇Δu+m*νA·∇Δu=ϑΔuinΓ2,T.
Applying the estimate (20) to the solution of this problem we deduce the following estimate for the second term in the right-hand side of (59):
(61)∑i=1N∫ΩΔu(x,Ti;ω)2dx≤n∥Δu∥𝒰(ΩT)2≤C02n∥∑j=1NγjΔωj∥L2((0,T);Lq2(Ω))2≤C4∥ω∥(L2(Ω))N2,
with some constant  C4. This implies that  J1  is the Fréchet differentiable and the first term in the right-hand side of (59) represents the Fréchet derivative, that is,
(62)J1′(ω)Δω=∑i=1Nσiwithσi=2∫Ω[u(x,Ti;ω)-uTi(x)]Δu(x,Ti;ω)dx.
Further, let us use the method presented in Section5.1 to deduce the proper adjoint problems. Comparing (60) with (49)–(52) we see that f†=∑j=1NγjΔωj, ϕ†=Δu0=h†=0. Therefore, the relation (56) has the form
(63)∫Ωu∘(x)Δu(x,t)dx-∫Γ2h∘*ΔudΓ+∫Ω(f∘*Δu-∑i=1nϕi∘*Δuxi)dx=∑j=1N∫ΩγjΔωj*ψdx,t∈[0,T].
In order to deduce a formula for the component σi in the quantity J1′(ω)Δω, we set u∘=ui∘=2[u(x,Ti;ω)-uTi(x)], h∘=f∘=ϕ∘=0 and t=Ti in (63). Then we immediately have
(64)σi=∑j=1N∫ΩγjΔωj*ψidx∣t=Ti,
where according to (53) and the definition of ui∘,h∘,ϕ∘, the function ψi is the weak solution of the problem (58) in the domain ΩT instead of ΩTi. Due to Theorem 2, this problem has a unique solution. From (62) and (64) we obtain (57). The latter formula contains the values of ψi in ΩTi. Therefore, we can restrict the problem (58) from ΩT to ΩTi.To use the formula (57) one has to solve N weak problems for the functions ψi in domains ΩTi. In the following theorem, we will show that computational work related to the evaluation of the Fréchet derivative can be considerably reduced. Actually, it is sufficient to solve N weak problems in the smaller domains ΩTi-Ti-1,  i=1,…,N. Here,  T0=0.Theorem 6.
The Fréchet derivative of the functionalJ1 can also be written in the form
(65)J1′(ω)Δω=∑j=1N∫Ω∑l=1N∫Tl-1Tlγj(t)βl(x,Tl-t,ω)dtΔωj(x)dx,
where βl∈𝒰0(ΩTl-Tl-1) are the unique ω-dependent weak solutions of the following sequence of recursive problems in the domains ΩTl-Tl-1:
(66)βl,t=Aβl-m*Aβl-afl-∇·ΦlinΩTl-Tl-1,βl=u0linΩ×{0},βl=0inΓ1,Tl-Tl-1,-νA·∇βl+m*νA·∇βl=ϑβl-ν·ΦlinΓ2,Tl-Tl-1,
where l=N,N-1,…,2,1. Here,
(67)u0l(x)=2[u(x,Tl;ω)-uTl(x)]+Θlβl+1(x,Tl+1-Tl)
and the function fl and the vector  Φl are defined via  βN,βN-1,…,βl+1 as follows:
(68)fl=Θl∑k=lN-1‍∫0Tk+1-Tk‍m(Tk-Tl+t+τ)βk+1×(x,Tk+1-Tk-τ)dτ,Φl=(Φ1l,…,Φnl), Φil=∑j=1n‍aij(∂/∂xj)fl  and  ΘN=0, Θl=1  for  l<N.Proof.
Firstly, let us check that (66) indeed have unique weak solutions βl in 𝒰0(ΩTl-Tl-1). To this end we can use Theorem 2. For the problem βN this is immediate, because the initial condition of the problem for  βN belongs to  L2(Ω)  and other equations in this problem are homogeneous. Further, we use the induction. Choose some l in the range N>l≥1 and suppose that βk+1∈𝒰0(ΩTk+1-Tk) for all k such that  N-1≥k≥l. The aim is to us to show that then the problem for βl has a unique weak solution in 𝒰0(ΩTl-Tl-1). Let us represent the kth addend in (68) in the form
(69)Ik=∫0Tk+1-Tkm(Tk-Tl+t+τ)βk+1(x,Tk+1-Tk-τ)dτ=∫tTk+1-Tk+tm(Tk-Tl+τ)βk+1(x,Tk+1-Tk+t-τ)dτ.
For any  k in the range  N-1≥k≥l we have
(70)∥Ik∥L2((0,Tl-Tl-1);W21(Ω))2≤∑|α|≤1∫0Tl-Tl-1[∫tTk+1-Tk+t|m(Tk-Tl+τ)|·∥Dxαβk+1(·,Tk+1-Tk+t-τ)∥L2(Ω)dτ∫tTk+1-Tk+t]2dt=∑|α|≤1∫0Tl-Tl-1[∫tTk+1-Tk+t∫0Tk+1-Tk+t|m(Tk-Tl+τ)|zk,α·(Tk+1-Tk+t-τ)dτ∫tTk+1-Tk+t]2dt=∑|α|≤1∫Tk+1-TkTl-Tl-1+Tk+1-Tk[∫0tmk(τ)zk,α(t-τ)dτ]2dt,
where zk,α(t)=∥Dxαβk+1(·,t)∥L2(Ω)  for  t∈[0,Tk+1-Tk], zk,α(t)=0 for t∉[0,Tk+1-Tk] and mk(t)=|m(Tk-Tl+t)|. Since m∈L1(0,T) and βk+1∈L2((0,Tk+1-Tk);W21(Ω)), we have mk∈L1(0,Tl-Tl-1+Tk+1-Tk) and  zk,α∈L2(0,Tl-Tl-1+Tk+1-Tk). Due to the Young’s theorem for convolutions, we get mk*zk,α∈L2(0,Tl-Tl-1+Tk+1-Tk). Therefore,  ∥Ik∥L2((0,Tl-Tl-1);W21(Ω))2<∞. This implies that fl belongs to L2((0,Tl-Tl-1);W21(Ω)). From the latter relation and aij∈L∞(Ω) we immediately have Φl∈(L2(ΩTl-Tl-1))n. Using the embedding theorem and Lemma 1 we see that afl satisfies the property (14). Finally, the initial condition  u0l belongs to L2(Ω), because u-uTl,  βl+1∈C([0,Tl+1-Tl],L2(Ω)). All assumptions of Theorem 2 are satisfied for the problem for βl. Consequently, it possesses a unique weak solution in  𝒰0(ΩTl-Tl-1).
Secondly, let us define the functions(71)βl*(x,t)=∑i=lNψi(x,Ti-Tl+t)for(x,t)∈ΩTl-Tl-1,
where l=1…,N and  ψi  are the solutions of  (58). We are going to show that βl=βl*, l=1,…,N. From the definition of  βl* using the value of ψl(x,0) and simple computations, we immediately get
(72)βl*(x,0)=2[u(x,Tl;ω)-uTl(x)]+Θlβl+1*(x,Tl+1-Tl),l=1,…,N.
Let us fix l=1,…,N and choose some η∈𝒯(ΩTl-Tl-1). We continue η by the formulae η(x,t)=η(x,Tl-Tl-1) for  t>Tl-Tl-1 and  η(x,t)=η(x,0)  for t<0. Further, let us define ηi(x,t)=η(x,Tl-Ti+t) where i=l,…,N. By the definition, it holds ηi∈𝒯(ΩTi).
Let us write down the weak form (31) for the problem for ψi (58) with the test function ηi. We fix some t∈[0,Tl-Tl-1] and compute the difference of this weak problem with t replaced by Ti-Tl+t and t replaced by Ti-Tl and take the sum over  i=l,…,N.  This results in the following expression:
(73)0=Z1+Z2+Z3+Z4,
where
(74)Z1=∑i=lN∫Ω[ψi(x,Ti-Tl+t)ηi(x,Ti-Tl+t)-ψi(x,Ti-Tl)ηi(x,Ti-Tl)]dx,Z2=∑i=lN∫∫ΩTi-Tl+t∖ΩTi-Tl[∑s,j=1n-ψiηi,t+∑s,j=1nasjψi,xjηi,xs-aψiηi]dxdt+∑i=lN∫∫Γ2,Ti-Tl+t∖Γ2,Ti-TlϑψiηidΓdt,Z3=∑i=lN∫∫ΩTi-Tl+t∖ΩTi-Tlam*ψiηidxdt,Z4=-∑i=lN∫∫ΩTi-Tl+t∖ΩTi-Tl∑s,j=1nasjm*ψi,xjηi,xsdxdt.
Using the definitions of  η and  βl* and the formula (72), we have
(75)Z1=∫Ω[βl*(x,t)η(x,t)-βl*(x,0)η(x,0)]dx=∫Ω[∫βl*(x,t)η(x,t)-{2[u(x,Tl;ω)-uTl(x)]+Θlβl+1*(x,Tl+1-Tl)}×η(x,0)∫]dx.
Similarly, using the definitions of η and βl* and changing the variable of integration in  Z2, we deduce
(76)Z2=∫∫Ωt[-βl*ηt+∑s,j=1nasjβl*ηxs-aβl*η]dxdt+∫∫Γ2,tϑβl*ηdΓdt.
By the change of variable, the quantity Z3 is transformed to
(77)Z3=∫∫Ωta(x)∑i=lN(m*ψi)(x,Ti-Tl+t)η(x,t)dxdt.
Let us consider the term ∑i=lN(m*ψi)(x,Ti-Tl+t) in the latter formula. We compute
(78)∑i=lN(m*ψi)(x,Ti-Tl+t)=∑i=lN∫0Ti-Tl+tm(τ)ψi(x,Ti-Tl+t-τ)dτ=∫0tm(τ)∑i=lNψi(x,Ti-Tl+t-τ)dτ+∑i=lN∫tTi-Tl+tm(τ)ψi(x,Ti-Tl+t-τ)dτ=∫0tm(τ)βl*(x,t-τ)dτ+∑i=lN∑k=li-1∫0Tk+1-Tkm(Tk-Tl+t+τ)ψi(x,Ti-Tk-τ)dτ=(m*βl*)(x,t)+∑k=lN-1∫0Tk+1-Tkm(Tk-Tl+t+τ)×∑i=k+1Nψi(x,Ti-Tk-τ)dτ=(m*βl*)(x,t)+∑k=lN-1∫0Tk+1-Tkm(Tk-Tl+t+τ)×βk+1*(x,Tk+1-Tk-τ)dτ.
Thus, (77) reads
(79)Z3=∫∫Ωtam*βl*ηdxdt+∫∫Ωta[∑k=lN-1∫0Tk+1-Tkm(Tk-Tl+t+τ)×βk+1*(x,Tk+1-Tk-τ)dτ∑k=lN-1]ηdxdt.
Using similar computations, we obtain
(80)Z4=-∫∫Ωt∑s,j=1nasjm*βl,xj*ηxsdxdt-∫∫Ωt∑s,j=1nasj×[∑k=lN-1∫0Tk+1-Tkm(Tk-Tl+t+τ)×βk+1,xj*(x,Tk+1-Tk-τ)dτ∑k=lN-1]ηxsdxdt.
Plugging (75), (76), (79), and (80) into (73), we arrive at a certain weak problem for βl* that coincides with the weak problem for βl. Moreover, since ψi∈𝒰0(ΩTi), from (71) we see that βl*∈𝒰0(ΩTl-Tl-1). But we have shown the uniqueness of the weak solutions of the problems for βl in 𝒰0(ΩTl-Tl-1). This implies βl*=βl.
Finally, from (57), we have
(81)J1′(ω)Δω=∑j=1N∫Ω∑i=1N∫0Tiγj(t)ψi(x,Ti-t)dtΔωj(x)dx=∑j=1N∫Ω∑i=1N∑l=1i∫Tl-1Tlγj(t)ψi(x,Ti-t)dtΔωj(x)dx.
Changing here the order of sums over i and l and observing (71) with βl* replaced by βl, we obtain (65). The proof is complete.


### 5.3. Derivative ofJ2

Theorem 7.
The functionalJ2 is the Fréchet differentiable in (L2(Ω))N+1 and
(82)J2′(z)Δz=∑j=1N∫Ω∫0Tγj(t)ψ(x,T-t;z)dtΔωj(x)dx+∫Ωψ(x,T;z)Δu0(x)dx,
where ψ∈𝒰(ΩT) is the unique z-dependent weak solution of the following problem:
(83)ψt=Aψ-m*Aψ+2∑i=1N+1κi(x,T-t)×[∫0Tκi(x,τ)u(x,τ;z)dτ-vi(x)]inΩT,ψ=0inΩ×{0},ψ=0inΓ1,T,-νA·∇ψ+m*νA·∇ψ=ϑψinΓ2,T.Proof.
Let us fix someΔz=(Δω,Δu0)∈(L2(Ω))N+1. It holds
(84)J2(z+Δz)-J2(z)=2∑i=1N+1∫Ω∫0Tκi(x,t)×[∫0Tκi(x,τ)u(x,τ;z)dτ-vi(x)]×Δu(x,t;z)dtdx+∑i=1N+1∫Ω[∫0Tκi(x,t)Δu(x,t;z)dt]2dx,
where Δu(x,t;z)=u(x,t;z+Δz)-u(x,t;z)∈𝒰0(ΩT) is the weak solution of the following problem:
(85)Δut=AΔu-m*AΔu+∑j=1NγjΔωjinΩT,Δu=Δu0inΩ×{0},Δu=0inΓ1,T,-νA·∇Δu+m*νA·∇Δu=ϑΔuinΓ2,T.
Using (43), the Cauchy inequality and estimate (20) from Theorem 2 for the problem of Δu(x,t;z), we come to the estimate
(86)|∑i=1N+1∫Ω[∫0Tκi(x,t)Δu(x,t;z)dt]2dx|≤(N+1)∥κ^∥L2(0,T)2∥Δu∥L2(ΩT)2≤C5(N+1)∥κ^∥L2(0,T)2∥Δu∥𝒰(ΩT)2≤C6∥Δz∥(L2(Ω))N+12,
with some constants C5 and C6. Therefore, J2 is the Fréchet differentiable and the first term in the right-hand side of (84) represents the Fréchet derivative, that is,
(87)J2′(z)Δz=2∑i=1N+1∫Ω∫0Tκi(x,t)×[∫0Tκi(x,τ)u(x,τ;z)dτ-vi(x)]×Δu(x,t;z)dtdx.
Comparing (85) with (49)–(52), we see that f†=∑j=1NγjΔωj, ϕ†=h†=0. Consequently, the relation (56) has the form
(88)∫Ωu∘(x)Δu(x,t)dx-∫Γ2h∘*ΔudΓ+∫Ω(f∘*Δu-∑i=1nϕi∘*Δuxi)dx=∫ΩΔu0(x)ψ(x,t)dx+∑j=1N∫ΩγjΔωj*ψdx,t∈[0,T].
To deduce a formula for  J2′(z)Δz, we define
(89)f∘=2∑i=1N+1κi(x,T-t)[∫0Tκi(x,τ)u(x,τ;z)dτ-vi(x)],u∘=h∘=ϕ∘=0 and t=T in (88). Then from (87) and (88), we obtain (82), where due to (53), ψi is the weak solution of the problem (83). In view of Theorem 2, this problem has a unique solution in 𝒰(ΩT).


### 5.4. Derivative ofJ3

Theorem 8.
The functionalJ3 is the Fréchet differentiable in L2(Ω)×L2(0,T) and
(90)J3′(z)Δz=∫Ω[(u-m*u)*ψ](x,T;z)Δa(x)dx+∫0T∫Ω[∑i,j=1naijuxj*ψxi-au*ψ]×(x,T-t;z)dxΔm(t)dt,
where ψ∈𝒰0(ΩT) is the unique z-dependent weak solution of the problem
(91)ψt=Aψ-m*AψinΩT,ψ=2[u(x,T;z)-uT(x)]inΩ×{0},ψ=0inΓ1,T,-νA·∇ψ+m*νA·∇ψ=ϑψ-2κ(x,T-t)×[∫Γ2κ(y,T-t)u(y,T-t;z)dΓ-v(T-t)]inΓ2,T.Proof.
Due tou(x,t;z)∈𝒰(ΩT),  κ∈L∞((0,T);L2(Γ2)), v∈L2(0,T), and uT∈L2(Ω), the problem (91) satisfies the assumptions of Theorem 2. Therefore, it has a unique weak solution in 𝒰0(ΩT).
LetΔz=(Δa,Δm)∈L2(Ω)×L2(0,T) and define Δ~u=u(x,t;z+Δz)-u(x,t;z). We split Δ~u as follows: Δ~u=Δu+Δ^u, where Δu is the weak solution of the following problem:
(92)Δut=AΔu-m*AΔu+Δa[u-m*u]-Δm*au-∇·[Δm*∑j=1n‍aijuxj]inΩT,Δu=0inΩ×{0},Δu=0inΓ1,T,-νA·∇Δu+m*νA·∇Δu=ϑΔu-ν·[Δm*∑j=1n‍aijuxj]inΓ2,T.
In view of Lemma 1(i), u∈𝒰(ΩT), m∈L1(0,T), and the Young’s theorem, it holds u-m*u∈L2((0,T);Lq3(Ω)). Therefore, Lemma 1(ii) implies
(93)∥Δa[u-m*u]∥L2((0,T);Lq2(Ω))≤C8(u,m)∥Δa∥Lq1(Ω)≤C9(u,m)∥Δa∥L2(Ω),
where C8 and C9 are some constants depending on u,m. Moreover, since u∈L2((0,T);W21(Ω)), by Young’s inequality we have also
(94)∥Δm*∑j=1n‍aijuxj∥(L2(ΩT))n≤C10(u)∥Δm∥L1(0,T)≤C11(u)∥Δm∥L2(0,T),
with some constants C10 and C11 depending on u. The obtained estimates show that assumptions of Theorem 2 are satisfied for the problem (92) and it indeed has a unique weak solution Δu∈𝒰(ΩT). Moreover, applying the relation (20) from Theorem 2, we get
(95)∥Δu∥𝒰(ΩT)≤C12(m,u)[∥Δa∥L2(Ω)+∥Δm∥L2(0,T)]=C12(m,u)∥Δz∥,
where C12(m,u) is a constant depending on  m,u.
Further, writing the problem forΔ~u and subtracting the problem for Δu, we obtain the following problem for  Δ^u:
(96)Δ^ut=AΔ^u-m*AΔ^u+f+f^+∇·ϕ+∇·ϕ^inΩT,Δ^u=0inΩ×{0},Δ^u=0inΓ1,T,-νA·∇Δ^u+m*νA·∇Δ^u=ϑΔ^u+ν·ϕ+ν·ϕ^inΓ2,T,
where
(97)f=ΔaΔu-(m+Δm)*ΔaΔu-Δm*Δau-Δm*aΔu,f^=ΔaΔ^u-(m+Δm)*ΔaΔ^u-Δm*aΔ^u,ϕ=-Δm*∑j=1naijΔuxj,ϕ^=-Δm*∑j=1naijΔ^uxj.
Using again Lemma1 and the Young’s inequality, we deduce the estimates
(98)∥f∥L2((0,T);Lq2(Ω))≤C13{[∥Δa∥L2(Ω)+{∥m∥L2(0,T)+∥Δm∥L2(0,T)}∥Δa∥L2(Ω)+∥Δm∥L2(0,T)∥a∥L2(Ω)]∥Δu∥𝒰(ΩT)+∥u∥𝒰(ΩT)∥Δa∥L2(Ω)∥Δm∥L2(0,T)}≤C14(z,u){[∥Δz∥+∥Δz∥2]∥Δu∥𝒰(ΩT)+∥Δz∥2},∥f^∥L2((0,T);Lq2(Ω))≤C15(z)[∥Δz∥+∥Δz∥2]∥Δ^u∥𝒰(ΩT),∥ϕ∥(L2(ΩT))n≤C16∥Δm∥L2(0,T)∥Δu∥𝒰(ΩT)≤C16∥Δz∥∥Δu∥𝒰(ΩT),∥ϕ^∥(L2(ΩT))n≤C17∥Δz∥∥Δ^u∥𝒰(ΩT),
with some constants C13⋯C17. Therefore, applying the relation (20) to the solution of the problem (96) we obtain
(99)∥Δ^u∥𝒰(ΩT)≤C18(z,u)×{[∥Δz∥+∥Δz∥2]{∥Δu∥𝒰(ΩT)+∥Δ^u∥𝒰(ΩT)}+∥Δz∥2},
with some constant C18. In case ∥Δz∥ is small enough, that is,
(100)∥Δz∥+∥Δz∥2≤12C18(z,u),
we have
(101)∥Δ^u∥𝒰(ΩT)≤2C18(z,u)×{[∥Δz∥+∥Δz∥2]∥Δu∥𝒰(ΩT)+∥Δz∥2}.
In view of (95), this implies
(102)∥Δ^u∥𝒰(ΩT)≤C19(z,u)[∥Δz∥2+∥Δz∥3],
with a constant C19.
Similarly, for the solution of the problem (92), we deduce the estimate
(103)∥Δu∥𝒰(ΩT)≤C20(z,u)∥Δz∥,
with a constant C20.
Now, we write the difference ofJ3  in the following form:
(104)J3(z+Δz)-J3(z)=2∫Ω[u(x,T;z)-uT(x)]Δu(x,T;z)dx+2∫0T∫Γ2κ(x,t)[∫Γ2κ(y,t)u(y,t;z)dΓ-v(t)]×Δu(x,t;z)dΓdt+Θ,
where
(105)Θ=2∫Ω[u(x,T;z)-uT(x)]Δ^u(x,T;z)dx+2∫0T∫Γ2κ(x,t)[∫Γ2‍κ(y,t)u(y,t;z)dΓ-v(t)]×Δ^u(x,t;z)dΓdt+∫Ω{(Δu+Δ^u)(x,T;z)}2dx+∫0T{∫Γ2κ(x,t)(Δu+Δ^u)(x,t;z)dΓ}2dt.
Using (102), (103), and the property κ∈L∞((0,T);L2(Γ2)), we obtain the estimate |Θ|≤C21(z,u)∑j=26∥Δz∥j in case (100). This shows that J3 is the Fréchet differentiable and
(106)J3′(z)Δz=2∫Ω[u(x,T;z)-uT(x)]Δu(x,T;z)dx+2∫0T∫Γ2κ(x,t)×[∫Γ2κ(y,t)u(y,t;z)dΓ-v(t)]×Δu(x,t;z)dΓdt.
Finally, let us prove (90) and (91). Comparing (92) with (49)–(52), we see that f†=Δa[u-m*u]-Δm*au, ϕi†=-Δm*∑j=1naijuxj and Δu0=h†=0. Thus, (56) reads
(107)∫Ωu∘(x)Δu(x,t)dx-∫Γ2h∘*ΔudΓ+∫Ω(f∘*Δu-∑i=1nϕi∘*Δuxi)dx=∫Ω(∑i,j=1n{Δa[u-m*u]-Δm*au}*ψ+Δm*∑i,j=1naijuxj*ψxi)dx,t∈[0,T].
In order to obtain a formula for the right-hand side in (106), we set u∘=2[u(x,T;z)-uT(x)],
(108)h∘(x,t)=-2κ(x,T-t)×[∫Γ2κ(y,T-t)u(y,T-t;z)dΓ-v(T-t)],f∘=ϕ∘=0 and t=T. Then, we obtain (90), where in view of (53) the function ψ is the weak solution of (91).


## 5.1. General Procedure

Suppose that the solutionu of the direct problem depends on a vector of parameters p that has to be determined in an inverse problem making use of certain measurements of u. Let the quasi-solution of the inverse problem be sought by a method involving the Fréchet derivative of the cost functional (i.e., some gradient-type method). Usually in practice, a solution of a proper adjoint problem is used to represent the Fréchet derivative.We introduce a general procedure to deduce such adjoint problems. Assume thatΔu is the difference of solutions of the direct problem corresponding to a difference of the vector of the parameters  Δp. More precisely, we suppose that Δu is a solution of the following problem:
(49)Δut=AΔu-m*AΔu+f†+∇·ϕ†inΩT,(50)Δu=Δu0inΩ×{0},(51)Δu=0inΓ1,T,(52)-νA·∇Δu+m*νA·∇Δu=ϑΔu+h†+ν·ϕ†inΓ2,T,
with some data f†, ϕ†, Δu0, h† depending on Δp. We restrict ourselves to the case when the Dirichlet boundary condition of the state u is independent of p. Therefore, the condition (51) for  Δu  is homogeneous.In practice, the adjoint parabolic problems are usually formulated as backward problems. In our context, it is better to pose adjoint problems in the forward form. The involved memory term withm is defined via a forward convolution and from the practical viewpoint, it is preferable to have the direct and adjoint problems in a similar form (e.g., to simplify parallelisation of computations).More precisely, let the adjoint stateψ be a solution of the following problem:
(53)ψt=Aψ-m*Aψ+f∘+∇·ϕ∘inΩT,ψ=u∘inΩ×{0},ψ=0inΓ1,T,-νA·∇ψ+m*νA·∇ψ=ϑψ+h∘+ν·ϕ∘inΓ2,T,
where f∘, ϕ∘, u∘, and h∘ are some data depending on Δu and the cost functional under consideration.Assume that the quadrupletsf†, ϕ†, Δu0, h†, and f∘, ϕ∘, u∘, h∘ satisfy the conditions (14)–(16). Then, due to Theorem 2, the problems (49)–(52) and (53) have unique weak solutions in the space 𝒰(ΩT). Actually, we have Δu,ψ∈𝒰0(ΩT) because of the homogeneous boundary conditions on Γ1,T.Let us write the relation (33) for Δu and use the test function η=ψ. Then we obtain for any t∈[0,T](54)0=∫ΩΔu*ψdx-∫Ω∫0tΔu0(x)ψ(x,τ)dτdx+∫Ω1*[∑i,j=1naij(Δuxj-m*Δuxj)*ψxi-a(Δu-m*Δu)*ψη∑i,j=1n]dx+∫Γ21*(ϑΔu+h†)*ψdΓ-∫Ω1*(f†*ψ-∑i=1nϕi†*ψxi)dx.
Secondly, let us write this relation for ψ and use the test function η=Δu. Then we have for any  t∈[0,T](55)0=∫Ωψ*Δudx-∫Ω∫0tu∘(x)Δu(x,τ)dτdx+∫Ω1*[∑i,j=1naij(ψxj-m*ψxj)*Δuxi-a(ψ-m*ψ)*Δu∑i,j=1n]dx+∫Γ21*(ϑψ+h∘)*ΔudΓ-∫Ω1*(f∘*Δu-∑i=1nϕi∘*Δuxi)dx.
Subtracting (54) from (55), using the commutativity of the convolution, the symmetricity relations aij=aji and differentiating with respect to t, we arrive at the following basic equality that can be used in various inverse problems:
(56)∫Ωu∘(x)Δu(x,t)dx-∫Γ2h∘*ΔudΓ+∫Ω(f∘*Δu-∑i=1nϕi∘*Δuxi)dx=∫ΩΔu0(x)ψ(x,t)dx-∫Γ2h†*ψdΓ+∫Ω(f†*ψ-∑i=1nϕi†*ψxi)dx,t∈[0,T].


## 5.2. Derivative ofJ1

Theorem 5.
The functionalJ1  is the Fréchet differentiable in (L2(Ω))N and
(57)J1′(ω)Δω=∑j=1N∫Ω∑i=1N∫0Tiγj(t)ψi(x,Ti-t;ω)dtΔωj(x)dx,
where ψi∈𝒰(ΩTi), i=1,…,N, are the unique ω-dependent weak solutions of the following problems:
(58)ψi,t=Aψi-m*AψiinΩTi,ψi=2[u(x,Ti;ω)-uTi(x)]inΩ×{0},ψi=0inΓ1,Ti,-νA·∇ψi+m*νA·∇ψi=ϑψiinΓ2,Ti,i=1,…,N.Proof.
Let us fix someω,Δω∈(L2(Ω))N. One can immediately check that it holds
(59)J1(ω+Δω)-J1(ω)=2∑i=1N∫Ω[u(x,Ti;ω)-uTi(x)]Δu(x,Ti;ω)dx+∑i=1N∫ΩΔu(x,Ti;ω)2dx,
where Δu(x,t;ω)=u(x,t;ω+Δω)-u(x,t;ω)∈𝒰0(ΩT) is the weak solution of the following problem:
(60)Δut=AΔu-m*AΔu+∑j=1NγjΔωjinΩT,Δu=0inΩ×{0},Δu=0inΓ1,T,-νA·∇Δu+m*νA·∇Δu=ϑΔuinΓ2,T.
Applying the estimate (20) to the solution of this problem we deduce the following estimate for the second term in the right-hand side of (59):
(61)∑i=1N∫ΩΔu(x,Ti;ω)2dx≤n∥Δu∥𝒰(ΩT)2≤C02n∥∑j=1NγjΔωj∥L2((0,T);Lq2(Ω))2≤C4∥ω∥(L2(Ω))N2,
with some constant  C4. This implies that  J1  is the Fréchet differentiable and the first term in the right-hand side of (59) represents the Fréchet derivative, that is,
(62)J1′(ω)Δω=∑i=1Nσiwithσi=2∫Ω[u(x,Ti;ω)-uTi(x)]Δu(x,Ti;ω)dx.
Further, let us use the method presented in Section5.1 to deduce the proper adjoint problems. Comparing (60) with (49)–(52) we see that f†=∑j=1NγjΔωj, ϕ†=Δu0=h†=0. Therefore, the relation (56) has the form
(63)∫Ωu∘(x)Δu(x,t)dx-∫Γ2h∘*ΔudΓ+∫Ω(f∘*Δu-∑i=1nϕi∘*Δuxi)dx=∑j=1N∫ΩγjΔωj*ψdx,t∈[0,T].
In order to deduce a formula for the component σi in the quantity J1′(ω)Δω, we set u∘=ui∘=2[u(x,Ti;ω)-uTi(x)], h∘=f∘=ϕ∘=0 and t=Ti in (63). Then we immediately have
(64)σi=∑j=1N∫ΩγjΔωj*ψidx∣t=Ti,
where according to (53) and the definition of ui∘,h∘,ϕ∘, the function ψi is the weak solution of the problem (58) in the domain ΩT instead of ΩTi. Due to Theorem 2, this problem has a unique solution. From (62) and (64) we obtain (57). The latter formula contains the values of ψi in ΩTi. Therefore, we can restrict the problem (58) from ΩT to ΩTi.To use the formula (57) one has to solve N weak problems for the functions ψi in domains ΩTi. In the following theorem, we will show that computational work related to the evaluation of the Fréchet derivative can be considerably reduced. Actually, it is sufficient to solve N weak problems in the smaller domains ΩTi-Ti-1,  i=1,…,N. Here,  T0=0.Theorem 6.
The Fréchet derivative of the functionalJ1 can also be written in the form
(65)J1′(ω)Δω=∑j=1N∫Ω∑l=1N∫Tl-1Tlγj(t)βl(x,Tl-t,ω)dtΔωj(x)dx,
where βl∈𝒰0(ΩTl-Tl-1) are the unique ω-dependent weak solutions of the following sequence of recursive problems in the domains ΩTl-Tl-1:
(66)βl,t=Aβl-m*Aβl-afl-∇·ΦlinΩTl-Tl-1,βl=u0linΩ×{0},βl=0inΓ1,Tl-Tl-1,-νA·∇βl+m*νA·∇βl=ϑβl-ν·ΦlinΓ2,Tl-Tl-1,
where l=N,N-1,…,2,1. Here,
(67)u0l(x)=2[u(x,Tl;ω)-uTl(x)]+Θlβl+1(x,Tl+1-Tl)
and the function fl and the vector  Φl are defined via  βN,βN-1,…,βl+1 as follows:
(68)fl=Θl∑k=lN-1‍∫0Tk+1-Tk‍m(Tk-Tl+t+τ)βk+1×(x,Tk+1-Tk-τ)dτ,Φl=(Φ1l,…,Φnl), Φil=∑j=1n‍aij(∂/∂xj)fl  and  ΘN=0, Θl=1  for  l<N.Proof.
Firstly, let us check that (66) indeed have unique weak solutions βl in 𝒰0(ΩTl-Tl-1). To this end we can use Theorem 2. For the problem βN this is immediate, because the initial condition of the problem for  βN belongs to  L2(Ω)  and other equations in this problem are homogeneous. Further, we use the induction. Choose some l in the range N>l≥1 and suppose that βk+1∈𝒰0(ΩTk+1-Tk) for all k such that  N-1≥k≥l. The aim is to us to show that then the problem for βl has a unique weak solution in 𝒰0(ΩTl-Tl-1). Let us represent the kth addend in (68) in the form
(69)Ik=∫0Tk+1-Tkm(Tk-Tl+t+τ)βk+1(x,Tk+1-Tk-τ)dτ=∫tTk+1-Tk+tm(Tk-Tl+τ)βk+1(x,Tk+1-Tk+t-τ)dτ.
For any  k in the range  N-1≥k≥l we have
(70)∥Ik∥L2((0,Tl-Tl-1);W21(Ω))2≤∑|α|≤1∫0Tl-Tl-1[∫tTk+1-Tk+t|m(Tk-Tl+τ)|·∥Dxαβk+1(·,Tk+1-Tk+t-τ)∥L2(Ω)dτ∫tTk+1-Tk+t]2dt=∑|α|≤1∫0Tl-Tl-1[∫tTk+1-Tk+t∫0Tk+1-Tk+t|m(Tk-Tl+τ)|zk,α·(Tk+1-Tk+t-τ)dτ∫tTk+1-Tk+t]2dt=∑|α|≤1∫Tk+1-TkTl-Tl-1+Tk+1-Tk[∫0tmk(τ)zk,α(t-τ)dτ]2dt,
where zk,α(t)=∥Dxαβk+1(·,t)∥L2(Ω)  for  t∈[0,Tk+1-Tk], zk,α(t)=0 for t∉[0,Tk+1-Tk] and mk(t)=|m(Tk-Tl+t)|. Since m∈L1(0,T) and βk+1∈L2((0,Tk+1-Tk);W21(Ω)), we have mk∈L1(0,Tl-Tl-1+Tk+1-Tk) and  zk,α∈L2(0,Tl-Tl-1+Tk+1-Tk). Due to the Young’s theorem for convolutions, we get mk*zk,α∈L2(0,Tl-Tl-1+Tk+1-Tk). Therefore,  ∥Ik∥L2((0,Tl-Tl-1);W21(Ω))2<∞. This implies that fl belongs to L2((0,Tl-Tl-1);W21(Ω)). From the latter relation and aij∈L∞(Ω) we immediately have Φl∈(L2(ΩTl-Tl-1))n. Using the embedding theorem and Lemma 1 we see that afl satisfies the property (14). Finally, the initial condition  u0l belongs to L2(Ω), because u-uTl,  βl+1∈C([0,Tl+1-Tl],L2(Ω)). All assumptions of Theorem 2 are satisfied for the problem for βl. Consequently, it possesses a unique weak solution in  𝒰0(ΩTl-Tl-1).
Secondly, let us define the functions(71)βl*(x,t)=∑i=lNψi(x,Ti-Tl+t)for(x,t)∈ΩTl-Tl-1,
where l=1…,N and  ψi  are the solutions of  (58). We are going to show that βl=βl*, l=1,…,N. From the definition of  βl* using the value of ψl(x,0) and simple computations, we immediately get
(72)βl*(x,0)=2[u(x,Tl;ω)-uTl(x)]+Θlβl+1*(x,Tl+1-Tl),l=1,…,N.
Let us fix l=1,…,N and choose some η∈𝒯(ΩTl-Tl-1). We continue η by the formulae η(x,t)=η(x,Tl-Tl-1) for  t>Tl-Tl-1 and  η(x,t)=η(x,0)  for t<0. Further, let us define ηi(x,t)=η(x,Tl-Ti+t) where i=l,…,N. By the definition, it holds ηi∈𝒯(ΩTi).
Let us write down the weak form (31) for the problem for ψi (58) with the test function ηi. We fix some t∈[0,Tl-Tl-1] and compute the difference of this weak problem with t replaced by Ti-Tl+t and t replaced by Ti-Tl and take the sum over  i=l,…,N.  This results in the following expression:
(73)0=Z1+Z2+Z3+Z4,
where
(74)Z1=∑i=lN∫Ω[ψi(x,Ti-Tl+t)ηi(x,Ti-Tl+t)-ψi(x,Ti-Tl)ηi(x,Ti-Tl)]dx,Z2=∑i=lN∫∫ΩTi-Tl+t∖ΩTi-Tl[∑s,j=1n-ψiηi,t+∑s,j=1nasjψi,xjηi,xs-aψiηi]dxdt+∑i=lN∫∫Γ2,Ti-Tl+t∖Γ2,Ti-TlϑψiηidΓdt,Z3=∑i=lN∫∫ΩTi-Tl+t∖ΩTi-Tlam*ψiηidxdt,Z4=-∑i=lN∫∫ΩTi-Tl+t∖ΩTi-Tl∑s,j=1nasjm*ψi,xjηi,xsdxdt.
Using the definitions of  η and  βl* and the formula (72), we have
(75)Z1=∫Ω[βl*(x,t)η(x,t)-βl*(x,0)η(x,0)]dx=∫Ω[∫βl*(x,t)η(x,t)-{2[u(x,Tl;ω)-uTl(x)]+Θlβl+1*(x,Tl+1-Tl)}×η(x,0)∫]dx.
Similarly, using the definitions of η and βl* and changing the variable of integration in  Z2, we deduce
(76)Z2=∫∫Ωt[-βl*ηt+∑s,j=1nasjβl*ηxs-aβl*η]dxdt+∫∫Γ2,tϑβl*ηdΓdt.
By the change of variable, the quantity Z3 is transformed to
(77)Z3=∫∫Ωta(x)∑i=lN(m*ψi)(x,Ti-Tl+t)η(x,t)dxdt.
Let us consider the term ∑i=lN(m*ψi)(x,Ti-Tl+t) in the latter formula. We compute
(78)∑i=lN(m*ψi)(x,Ti-Tl+t)=∑i=lN∫0Ti-Tl+tm(τ)ψi(x,Ti-Tl+t-τ)dτ=∫0tm(τ)∑i=lNψi(x,Ti-Tl+t-τ)dτ+∑i=lN∫tTi-Tl+tm(τ)ψi(x,Ti-Tl+t-τ)dτ=∫0tm(τ)βl*(x,t-τ)dτ+∑i=lN∑k=li-1∫0Tk+1-Tkm(Tk-Tl+t+τ)ψi(x,Ti-Tk-τ)dτ=(m*βl*)(x,t)+∑k=lN-1∫0Tk+1-Tkm(Tk-Tl+t+τ)×∑i=k+1Nψi(x,Ti-Tk-τ)dτ=(m*βl*)(x,t)+∑k=lN-1∫0Tk+1-Tkm(Tk-Tl+t+τ)×βk+1*(x,Tk+1-Tk-τ)dτ.
Thus, (77) reads
(79)Z3=∫∫Ωtam*βl*ηdxdt+∫∫Ωta[∑k=lN-1∫0Tk+1-Tkm(Tk-Tl+t+τ)×βk+1*(x,Tk+1-Tk-τ)dτ∑k=lN-1]ηdxdt.
Using similar computations, we obtain
(80)Z4=-∫∫Ωt∑s,j=1nasjm*βl,xj*ηxsdxdt-∫∫Ωt∑s,j=1nasj×[∑k=lN-1∫0Tk+1-Tkm(Tk-Tl+t+τ)×βk+1,xj*(x,Tk+1-Tk-τ)dτ∑k=lN-1]ηxsdxdt.
Plugging (75), (76), (79), and (80) into (73), we arrive at a certain weak problem for βl* that coincides with the weak problem for βl. Moreover, since ψi∈𝒰0(ΩTi), from (71) we see that βl*∈𝒰0(ΩTl-Tl-1). But we have shown the uniqueness of the weak solutions of the problems for βl in 𝒰0(ΩTl-Tl-1). This implies βl*=βl.
Finally, from (57), we have
(81)J1′(ω)Δω=∑j=1N∫Ω∑i=1N∫0Tiγj(t)ψi(x,Ti-t)dtΔωj(x)dx=∑j=1N∫Ω∑i=1N∑l=1i∫Tl-1Tlγj(t)ψi(x,Ti-t)dtΔωj(x)dx.
Changing here the order of sums over i and l and observing (71) with βl* replaced by βl, we obtain (65). The proof is complete.


## 5.3. Derivative ofJ2

Theorem 7.
The functionalJ2 is the Fréchet differentiable in (L2(Ω))N+1 and
(82)J2′(z)Δz=∑j=1N∫Ω∫0Tγj(t)ψ(x,T-t;z)dtΔωj(x)dx+∫Ωψ(x,T;z)Δu0(x)dx,
where ψ∈𝒰(ΩT) is the unique z-dependent weak solution of the following problem:
(83)ψt=Aψ-m*Aψ+2∑i=1N+1κi(x,T-t)×[∫0Tκi(x,τ)u(x,τ;z)dτ-vi(x)]inΩT,ψ=0inΩ×{0},ψ=0inΓ1,T,-νA·∇ψ+m*νA·∇ψ=ϑψinΓ2,T.Proof.
Let us fix someΔz=(Δω,Δu0)∈(L2(Ω))N+1. It holds
(84)J2(z+Δz)-J2(z)=2∑i=1N+1∫Ω∫0Tκi(x,t)×[∫0Tκi(x,τ)u(x,τ;z)dτ-vi(x)]×Δu(x,t;z)dtdx+∑i=1N+1∫Ω[∫0Tκi(x,t)Δu(x,t;z)dt]2dx,
where Δu(x,t;z)=u(x,t;z+Δz)-u(x,t;z)∈𝒰0(ΩT) is the weak solution of the following problem:
(85)Δut=AΔu-m*AΔu+∑j=1NγjΔωjinΩT,Δu=Δu0inΩ×{0},Δu=0inΓ1,T,-νA·∇Δu+m*νA·∇Δu=ϑΔuinΓ2,T.
Using (43), the Cauchy inequality and estimate (20) from Theorem 2 for the problem of Δu(x,t;z), we come to the estimate
(86)|∑i=1N+1∫Ω[∫0Tκi(x,t)Δu(x,t;z)dt]2dx|≤(N+1)∥κ^∥L2(0,T)2∥Δu∥L2(ΩT)2≤C5(N+1)∥κ^∥L2(0,T)2∥Δu∥𝒰(ΩT)2≤C6∥Δz∥(L2(Ω))N+12,
with some constants C5 and C6. Therefore, J2 is the Fréchet differentiable and the first term in the right-hand side of (84) represents the Fréchet derivative, that is,
(87)J2′(z)Δz=2∑i=1N+1∫Ω∫0Tκi(x,t)×[∫0Tκi(x,τ)u(x,τ;z)dτ-vi(x)]×Δu(x,t;z)dtdx.
Comparing (85) with (49)–(52), we see that f†=∑j=1NγjΔωj, ϕ†=h†=0. Consequently, the relation (56) has the form
(88)∫Ωu∘(x)Δu(x,t)dx-∫Γ2h∘*ΔudΓ+∫Ω(f∘*Δu-∑i=1nϕi∘*Δuxi)dx=∫ΩΔu0(x)ψ(x,t)dx+∑j=1N∫ΩγjΔωj*ψdx,t∈[0,T].
To deduce a formula for  J2′(z)Δz, we define
(89)f∘=2∑i=1N+1κi(x,T-t)[∫0Tκi(x,τ)u(x,τ;z)dτ-vi(x)],u∘=h∘=ϕ∘=0 and t=T in (88). Then from (87) and (88), we obtain (82), where due to (53), ψi is the weak solution of the problem (83). In view of Theorem 2, this problem has a unique solution in 𝒰(ΩT).


## 5.4. Derivative ofJ3

Theorem 8.
The functionalJ3 is the Fréchet differentiable in L2(Ω)×L2(0,T) and
(90)J3′(z)Δz=∫Ω[(u-m*u)*ψ](x,T;z)Δa(x)dx+∫0T∫Ω[∑i,j=1naijuxj*ψxi-au*ψ]×(x,T-t;z)dxΔm(t)dt,
where ψ∈𝒰0(ΩT) is the unique z-dependent weak solution of the problem
(91)ψt=Aψ-m*AψinΩT,ψ=2[u(x,T;z)-uT(x)]inΩ×{0},ψ=0inΓ1,T,-νA·∇ψ+m*νA·∇ψ=ϑψ-2κ(x,T-t)×[∫Γ2κ(y,T-t)u(y,T-t;z)dΓ-v(T-t)]inΓ2,T.Proof.
Due tou(x,t;z)∈𝒰(ΩT),  κ∈L∞((0,T);L2(Γ2)), v∈L2(0,T), and uT∈L2(Ω), the problem (91) satisfies the assumptions of Theorem 2. Therefore, it has a unique weak solution in 𝒰0(ΩT).
LetΔz=(Δa,Δm)∈L2(Ω)×L2(0,T) and define Δ~u=u(x,t;z+Δz)-u(x,t;z). We split Δ~u as follows: Δ~u=Δu+Δ^u, where Δu is the weak solution of the following problem:
(92)Δut=AΔu-m*AΔu+Δa[u-m*u]-Δm*au-∇·[Δm*∑j=1n‍aijuxj]inΩT,Δu=0inΩ×{0},Δu=0inΓ1,T,-νA·∇Δu+m*νA·∇Δu=ϑΔu-ν·[Δm*∑j=1n‍aijuxj]inΓ2,T.
In view of Lemma 1(i), u∈𝒰(ΩT), m∈L1(0,T), and the Young’s theorem, it holds u-m*u∈L2((0,T);Lq3(Ω)). Therefore, Lemma 1(ii) implies
(93)∥Δa[u-m*u]∥L2((0,T);Lq2(Ω))≤C8(u,m)∥Δa∥Lq1(Ω)≤C9(u,m)∥Δa∥L2(Ω),
where C8 and C9 are some constants depending on u,m. Moreover, since u∈L2((0,T);W21(Ω)), by Young’s inequality we have also
(94)∥Δm*∑j=1n‍aijuxj∥(L2(ΩT))n≤C10(u)∥Δm∥L1(0,T)≤C11(u)∥Δm∥L2(0,T),
with some constants C10 and C11 depending on u. The obtained estimates show that assumptions of Theorem 2 are satisfied for the problem (92) and it indeed has a unique weak solution Δu∈𝒰(ΩT). Moreover, applying the relation (20) from Theorem 2, we get
(95)∥Δu∥𝒰(ΩT)≤C12(m,u)[∥Δa∥L2(Ω)+∥Δm∥L2(0,T)]=C12(m,u)∥Δz∥,
where C12(m,u) is a constant depending on  m,u.
Further, writing the problem forΔ~u and subtracting the problem for Δu, we obtain the following problem for  Δ^u:
(96)Δ^ut=AΔ^u-m*AΔ^u+f+f^+∇·ϕ+∇·ϕ^inΩT,Δ^u=0inΩ×{0},Δ^u=0inΓ1,T,-νA·∇Δ^u+m*νA·∇Δ^u=ϑΔ^u+ν·ϕ+ν·ϕ^inΓ2,T,
where
(97)f=ΔaΔu-(m+Δm)*ΔaΔu-Δm*Δau-Δm*aΔu,f^=ΔaΔ^u-(m+Δm)*ΔaΔ^u-Δm*aΔ^u,ϕ=-Δm*∑j=1naijΔuxj,ϕ^=-Δm*∑j=1naijΔ^uxj.
Using again Lemma1 and the Young’s inequality, we deduce the estimates
(98)∥f∥L2((0,T);Lq2(Ω))≤C13{[∥Δa∥L2(Ω)+{∥m∥L2(0,T)+∥Δm∥L2(0,T)}∥Δa∥L2(Ω)+∥Δm∥L2(0,T)∥a∥L2(Ω)]∥Δu∥𝒰(ΩT)+∥u∥𝒰(ΩT)∥Δa∥L2(Ω)∥Δm∥L2(0,T)}≤C14(z,u){[∥Δz∥+∥Δz∥2]∥Δu∥𝒰(ΩT)+∥Δz∥2},∥f^∥L2((0,T);Lq2(Ω))≤C15(z)[∥Δz∥+∥Δz∥2]∥Δ^u∥𝒰(ΩT),∥ϕ∥(L2(ΩT))n≤C16∥Δm∥L2(0,T)∥Δu∥𝒰(ΩT)≤C16∥Δz∥∥Δu∥𝒰(ΩT),∥ϕ^∥(L2(ΩT))n≤C17∥Δz∥∥Δ^u∥𝒰(ΩT),
with some constants C13⋯C17. Therefore, applying the relation (20) to the solution of the problem (96) we obtain
(99)∥Δ^u∥𝒰(ΩT)≤C18(z,u)×{[∥Δz∥+∥Δz∥2]{∥Δu∥𝒰(ΩT)+∥Δ^u∥𝒰(ΩT)}+∥Δz∥2},
with some constant C18. In case ∥Δz∥ is small enough, that is,
(100)∥Δz∥+∥Δz∥2≤12C18(z,u),
we have
(101)∥Δ^u∥𝒰(ΩT)≤2C18(z,u)×{[∥Δz∥+∥Δz∥2]∥Δu∥𝒰(ΩT)+∥Δz∥2}.
In view of (95), this implies
(102)∥Δ^u∥𝒰(ΩT)≤C19(z,u)[∥Δz∥2+∥Δz∥3],
with a constant C19.
Similarly, for the solution of the problem (92), we deduce the estimate
(103)∥Δu∥𝒰(ΩT)≤C20(z,u)∥Δz∥,
with a constant C20.
Now, we write the difference ofJ3  in the following form:
(104)J3(z+Δz)-J3(z)=2∫Ω[u(x,T;z)-uT(x)]Δu(x,T;z)dx+2∫0T∫Γ2κ(x,t)[∫Γ2κ(y,t)u(y,t;z)dΓ-v(t)]×Δu(x,t;z)dΓdt+Θ,
where
(105)Θ=2∫Ω[u(x,T;z)-uT(x)]Δ^u(x,T;z)dx+2∫0T∫Γ2κ(x,t)[∫Γ2‍κ(y,t)u(y,t;z)dΓ-v(t)]×Δ^u(x,t;z)dΓdt+∫Ω{(Δu+Δ^u)(x,T;z)}2dx+∫0T{∫Γ2κ(x,t)(Δu+Δ^u)(x,t;z)dΓ}2dt.
Using (102), (103), and the property κ∈L∞((0,T);L2(Γ2)), we obtain the estimate |Θ|≤C21(z,u)∑j=26∥Δz∥j in case (100). This shows that J3 is the Fréchet differentiable and
(106)J3′(z)Δz=2∫Ω[u(x,T;z)-uT(x)]Δu(x,T;z)dx+2∫0T∫Γ2κ(x,t)×[∫Γ2κ(y,t)u(y,t;z)dΓ-v(t)]×Δu(x,t;z)dΓdt.
Finally, let us prove (90) and (91). Comparing (92) with (49)–(52), we see that f†=Δa[u-m*u]-Δm*au, ϕi†=-Δm*∑j=1naijuxj and Δu0=h†=0. Thus, (56) reads
(107)∫Ωu∘(x)Δu(x,t)dx-∫Γ2h∘*ΔudΓ+∫Ω(f∘*Δu-∑i=1nϕi∘*Δuxi)dx=∫Ω(∑i,j=1n{Δa[u-m*u]-Δm*au}*ψ+Δm*∑i,j=1naijuxj*ψxi)dx,t∈[0,T].
In order to obtain a formula for the right-hand side in (106), we set u∘=2[u(x,T;z)-uT(x)],
(108)h∘(x,t)=-2κ(x,T-t)×[∫Γ2κ(y,T-t)u(y,T-t;z)dΓ-v(T-t)],f∘=ϕ∘=0 and t=T. Then, we obtain (90), where in view of (53) the function ψ is the weak solution of (91).


## 6. Further Aspects of Minimisation

### 6.1. Existence of Quasisolutions

For the convenience, we will use also the symbolz  to denote the argument ω of J1.Theorem 9.
(i)  Letk∈{1;2} and M⊂𝒵k be bounded, closed, and convex. Then,  IPk has a quasi-solution in M. The set of quasisolutions is closed and convex.
(ii) Letk∈{1;2;3} and M⊂𝒵3 be compact. Then  IPk has a quasi-solution in M.Proof.
Let us prove (i). The existence assertion follows from Weierstrass existence theorem (see [24, Section 2.5, Theorem  2D]) once we have proved that Jk is weakly sequentially lower semicontinuous in ℱ, that is,
(109)Jk(z)≤liminfn→∞Jk(zn)aszn⇀zin𝒵k.
But (109) follows from the continuity and convexity of  Jk [24]. The convexity of Jk can be immediately deduced making use of the linearity of the ingredient u(x,t;z) with respect to z inside the quadratic functional Jk (for similar computations see [25, Theorem  2]). The closedness of the set of quasisolutions is again a direct consequence of the continuity of Jk. The convexity of the set of solutions follows from the convexity of Jk.
Next, we prove (ii). Letm=infz∈MJk(z) and zl∈M be the minimising sequence, that is,  limJk(zl)=m. By the compactness, there exists a subsequence zlj∈M such that limzlj=z*∈M. Due to the continuity of Jk we have limJk(zlj)=Jk(z*). Thus, Jk(z*)=m. This proves (ii).In practice, the compact setM may be a bounded and closed finite-dimensional subset of  𝒵k. The proof of weak lower semicontinuity of J3 may be harder because this functional is not convex.


### 6.2. Discretisation and Minimisation

Let us consider the penalised discrete problems(110)z†∈argminz∈𝒵k,LΦk,L(z),Φk,L=ΠL(z)+Jk(z),
where k∈{1;2;3},  𝒵k,L is an L-dimensional subspace of Zk (L∈{1,2,…}) and ΠL is a penalty function related to the set  ML=PLM with PL  being the orthogonal projection onto 𝒵k,L. The general assumptions for ΠL are
(111)ΠL-accretive,convex,Fréchetdifferentiable,ΠL′-uniformlyLipschitzcontinuousin𝒵k,L.Theorem 10.
The problem (110) has a solution.Proof.
The proof repeats the proof of the statement (ii) of Theorem9, because in view of the accretivity of Φk,L, a minimizing sequence is bounded and in a finite-dimensional space any bounded sequence is compact.The Fréchet derivative ofΦk,L, that is,  Φk,L′(z)=ΠL′(z)+Jk′(z)∈𝒵k,L*=𝒵k,L can be identified by a certain element in  𝒵k,L, that is,
(112)Φk,L′(z)Δz=〈Φk,L′(z),Δz〉𝒵k∀Δz∈𝒵k,L,
where 〈·,·〉𝒵k is the inner product of 𝒵k. In particular, the addend Jk′(z) is identical to the element PLwk(z) where wk(z) is the kernel of the functional Jk′(z). Thus, by virtue of (57), (65), (82), and (90), it holds
(113)w1(z)=(∑i=1N∫0Tiγj(t)ψi(·,Ti-t;z)dt∣j=1,…,N)=(∑l=1N∫Tl-1Tlγj(t)βl(·,Tl-t;z)dt∣j=1,…,N),w2(z)=(∫0Tγj(t)ψ(·,T-t;z)dt∣j=1,…,N,ψ(·,T;z)),w3(z)=(∑i,j=1n(u-m*u)*ψ(·,T),∫Ω[∑i,j=1naijuxj*ψxi-au*ψ](x,T-·)dx).
In w1, the functions ψi and βl are the z- (or, equivalently, ω-) dependent weak solutions of the problems (58) and (66), respectively. In w2 the function ψ is the weak solution of (83) and in w3 the functions u and ψ are the z-dependent weak solutions of (2)–(5) and (91), respectively.Example 11.
Consider the casek=1. Let M={z∈𝒵1:∥z∥≤ρ}, where ρ>0. Further, let ξj, j=1,2,…, be an orthonormal basis in L2(Ω) and 𝒵1,L=(span(ξ1,…,ξL))N. Then J1′(z) is in 𝒵1,L identical to the element
(114)PLw1(z)=(∑i=1L∫Ω∑l=1N∫Tl-1Tlγj(t)βl(x,Tl-t;z)dt×ξi(x)dxξi∣j=1,…,N∑i,j=1n).
Moreover, it holds ML={z∈𝒵1,L:∥z∥≤ρ}. Define a convex penalty function ΠL∈C∞[0,∞) such that ΠL(z)=0 for ∥z∥≤ρ and ΠL(z)=d(∥z∥2-ρ2) for ∥z∥≥ρ+ε with some d,ε>0. Then ΠL  satisfies (111).Letk∈{1;2;3}. Choose some initial guess z0∈𝒵k,L. Compute the approximate solutions by the gradient method
(115)zs+1=zs-csΦk,L′(zs),
where s=0,1,2,…  and  cs>0.Theorem 12.
Letk∈{1;2} and cs be chosen by the rule
(116)infc>0Φk,L(zs-cΦk,L′(zs))≤Φk,L(zs-csΦk,L′(zs))≤infc>0Φk,L(zs-cΦk,L′(zs))+δs,
where δs≥0,   ∑s=0∞δs=:δ<∞. Then it holds dist(zs,S)→0 as s→∞, where S is the set of solutions of (110).Proof.
The assertion follows from Theorem5.1.2 of [26] once we have proved that Φk,L′ is uniformly Lipschitz continuous, Φk,L is convex, and the set M(z0)={z∈𝒵k,L:Φk,L(z)≤Φk,L(z0)+δ} is bounded. The convexity of Φk,L follows from the convexity of its addends ΠL and Jk. The boundedness of M(z0) is a direct consequence of the accretivity of Φk,L following from the accretivity of the addend ΠL.
It remains to show the uniform Lipschitz continuity ofJk′ in 𝒵k,L (such a property for ΠL′  is assumed in (111)). Let k=1. Then by (113) and Jk′(z)=PLw1(z) for any z,z~∈𝒵k,L, we have
(117)∥J1′(z~)-J1′(z)∥≤∥PL∥∥w1(z~)-w1(z)∥≤C22∑i=1N∥ψi(·,·;z~)-ψi(·,·;z)∥𝒰(ΩT),
where C22 is a constant independent of z and z~. Further, observing (58) and (40), the estimate (20) of Theorem 2 and z=ω, we deduce
(118)∥J1′(z~)-J1′(z)∥≤2C22C0∑i=1N∥u(·,Ti;z~)-u(·,Ti;z)∥L2(Ω)≤C23∥u(·,·;z~)-u(·,·;z)∥𝒰(ΩT)≤C24∥z~-z∥,
where C23, C24 are independent of z and z~. This proves the uniform Lipschitz continuity of J1′. Such a property of J2′ can be proved in a similar manner.The convergence ofzs in case k=3 is an open issue. This case is more complex because IP3 is nonlinear and the Fréchet derivative of J3 is not uniformly Lipschitz continuous.The quasisolutions of IP1–IP3 are not expected to be stable with respect to the noise of the data, that is, the problems under consideration may be ill posed. Nevertheless, from the intuitive viewpoint, a discretisation should regularise an ill-posed problem. Such a property of the discretisation has been proved in many cases [27, 28]. Alternatively, the index s of the gradient method could be used as a regularization parameter (see [29, 30]). Moreover, the addend ΠL can be defined to be the stabilizing term of the Tikhonov's method instead of the penalty function, that is, ΠL=α∥z∥2, where α>0 is the regularisation parameter. Such a  ΠL satisfies (111).


## 6.1. Existence of Quasisolutions

For the convenience, we will use also the symbolz  to denote the argument ω of J1.Theorem 9.
(i)  Letk∈{1;2} and M⊂𝒵k be bounded, closed, and convex. Then,  IPk has a quasi-solution in M. The set of quasisolutions is closed and convex.
(ii) Letk∈{1;2;3} and M⊂𝒵3 be compact. Then  IPk has a quasi-solution in M.Proof.
Let us prove (i). The existence assertion follows from Weierstrass existence theorem (see [24, Section 2.5, Theorem  2D]) once we have proved that Jk is weakly sequentially lower semicontinuous in ℱ, that is,
(109)Jk(z)≤liminfn→∞Jk(zn)aszn⇀zin𝒵k.
But (109) follows from the continuity and convexity of  Jk [24]. The convexity of Jk can be immediately deduced making use of the linearity of the ingredient u(x,t;z) with respect to z inside the quadratic functional Jk (for similar computations see [25, Theorem  2]). The closedness of the set of quasisolutions is again a direct consequence of the continuity of Jk. The convexity of the set of solutions follows from the convexity of Jk.
Next, we prove (ii). Letm=infz∈MJk(z) and zl∈M be the minimising sequence, that is,  limJk(zl)=m. By the compactness, there exists a subsequence zlj∈M such that limzlj=z*∈M. Due to the continuity of Jk we have limJk(zlj)=Jk(z*). Thus, Jk(z*)=m. This proves (ii).In practice, the compact setM may be a bounded and closed finite-dimensional subset of  𝒵k. The proof of weak lower semicontinuity of J3 may be harder because this functional is not convex.


## 6.2. Discretisation and Minimisation

Let us consider the penalised discrete problems(110)z†∈argminz∈𝒵k,LΦk,L(z),Φk,L=ΠL(z)+Jk(z),
where k∈{1;2;3},  𝒵k,L is an L-dimensional subspace of Zk (L∈{1,2,…}) and ΠL is a penalty function related to the set  ML=PLM with PL  being the orthogonal projection onto 𝒵k,L. The general assumptions for ΠL are
(111)ΠL-accretive,convex,Fréchetdifferentiable,ΠL′-uniformlyLipschitzcontinuousin𝒵k,L.Theorem 10.
The problem (110) has a solution.Proof.
The proof repeats the proof of the statement (ii) of Theorem9, because in view of the accretivity of Φk,L, a minimizing sequence is bounded and in a finite-dimensional space any bounded sequence is compact.The Fréchet derivative ofΦk,L, that is,  Φk,L′(z)=ΠL′(z)+Jk′(z)∈𝒵k,L*=𝒵k,L can be identified by a certain element in  𝒵k,L, that is,
(112)Φk,L′(z)Δz=〈Φk,L′(z),Δz〉𝒵k∀Δz∈𝒵k,L,
where 〈·,·〉𝒵k is the inner product of 𝒵k. In particular, the addend Jk′(z) is identical to the element PLwk(z) where wk(z) is the kernel of the functional Jk′(z). Thus, by virtue of (57), (65), (82), and (90), it holds
(113)w1(z)=(∑i=1N∫0Tiγj(t)ψi(·,Ti-t;z)dt∣j=1,…,N)=(∑l=1N∫Tl-1Tlγj(t)βl(·,Tl-t;z)dt∣j=1,…,N),w2(z)=(∫0Tγj(t)ψ(·,T-t;z)dt∣j=1,…,N,ψ(·,T;z)),w3(z)=(∑i,j=1n(u-m*u)*ψ(·,T),∫Ω[∑i,j=1naijuxj*ψxi-au*ψ](x,T-·)dx).
In w1, the functions ψi and βl are the z- (or, equivalently, ω-) dependent weak solutions of the problems (58) and (66), respectively. In w2 the function ψ is the weak solution of (83) and in w3 the functions u and ψ are the z-dependent weak solutions of (2)–(5) and (91), respectively.Example 11.
Consider the casek=1. Let M={z∈𝒵1:∥z∥≤ρ}, where ρ>0. Further, let ξj, j=1,2,…, be an orthonormal basis in L2(Ω) and 𝒵1,L=(span(ξ1,…,ξL))N. Then J1′(z) is in 𝒵1,L identical to the element
(114)PLw1(z)=(∑i=1L∫Ω∑l=1N∫Tl-1Tlγj(t)βl(x,Tl-t;z)dt×ξi(x)dxξi∣j=1,…,N∑i,j=1n).
Moreover, it holds ML={z∈𝒵1,L:∥z∥≤ρ}. Define a convex penalty function ΠL∈C∞[0,∞) such that ΠL(z)=0 for ∥z∥≤ρ and ΠL(z)=d(∥z∥2-ρ2) for ∥z∥≥ρ+ε with some d,ε>0. Then ΠL  satisfies (111).Letk∈{1;2;3}. Choose some initial guess z0∈𝒵k,L. Compute the approximate solutions by the gradient method
(115)zs+1=zs-csΦk,L′(zs),
where s=0,1,2,…  and  cs>0.Theorem 12.
Letk∈{1;2} and cs be chosen by the rule
(116)infc>0Φk,L(zs-cΦk,L′(zs))≤Φk,L(zs-csΦk,L′(zs))≤infc>0Φk,L(zs-cΦk,L′(zs))+δs,
where δs≥0,   ∑s=0∞δs=:δ<∞. Then it holds dist(zs,S)→0 as s→∞, where S is the set of solutions of (110).Proof.
The assertion follows from Theorem5.1.2 of [26] once we have proved that Φk,L′ is uniformly Lipschitz continuous, Φk,L is convex, and the set M(z0)={z∈𝒵k,L:Φk,L(z)≤Φk,L(z0)+δ} is bounded. The convexity of Φk,L follows from the convexity of its addends ΠL and Jk. The boundedness of M(z0) is a direct consequence of the accretivity of Φk,L following from the accretivity of the addend ΠL.
It remains to show the uniform Lipschitz continuity ofJk′ in 𝒵k,L (such a property for ΠL′  is assumed in (111)). Let k=1. Then by (113) and Jk′(z)=PLw1(z) for any z,z~∈𝒵k,L, we have
(117)∥J1′(z~)-J1′(z)∥≤∥PL∥∥w1(z~)-w1(z)∥≤C22∑i=1N∥ψi(·,·;z~)-ψi(·,·;z)∥𝒰(ΩT),
where C22 is a constant independent of z and z~. Further, observing (58) and (40), the estimate (20) of Theorem 2 and z=ω, we deduce
(118)∥J1′(z~)-J1′(z)∥≤2C22C0∑i=1N∥u(·,Ti;z~)-u(·,Ti;z)∥L2(Ω)≤C23∥u(·,·;z~)-u(·,·;z)∥𝒰(ΩT)≤C24∥z~-z∥,
where C23, C24 are independent of z and z~. This proves the uniform Lipschitz continuity of J1′. Such a property of J2′ can be proved in a similar manner.The convergence ofzs in case k=3 is an open issue. This case is more complex because IP3 is nonlinear and the Fréchet derivative of J3 is not uniformly Lipschitz continuous.The quasisolutions of IP1–IP3 are not expected to be stable with respect to the noise of the data, that is, the problems under consideration may be ill posed. Nevertheless, from the intuitive viewpoint, a discretisation should regularise an ill-posed problem. Such a property of the discretisation has been proved in many cases [27, 28]. Alternatively, the index s of the gradient method could be used as a regularization parameter (see [29, 30]). Moreover, the addend ΠL can be defined to be the stabilizing term of the Tikhonov's method instead of the penalty function, that is, ΠL=α∥z∥2, where α>0 is the regularisation parameter. Such a  ΠL satisfies (111).



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*Source: 297104-2013-03-21.xml*