An Identification Problem for a Linear Evolution Equation in a Banach Space

DISCRETE AND CONTINUOUS doi:10.3934/dcdss.2020081 DYNAMICAL SYSTEMS SERIES S Volume 13, Number 5, May 2020 pp. 1429–1440

AN IDENTIFICATION PROBLEM FOR A LINEAR EVOLUTION EQUATION IN A BANACH SPACE

Viorel Barbu Octav Mayer Institute of Mathematics of the Romanian Academy Bdul Carol I, 8, Iasi, Romania Gabriela Marinoschi∗ Gheorghe Mihoc-Caius Iacob Institute of Mathematical Statistics and Applied Mathematics of the Romanian Academy Calea 13 Septembrie 13, 050711 Bucharest, Romania

Abstract. We study a problem of a parameter identification related to a linear evolution equation in a Banach space, using an additional information about the solution. For sufficiently regular data we provide an exact solution given by a Volterra integral equation, while for less regular data we obtain an approximating solution by an optimal control approach. Under certain hypotheses, the characterization of the limit of the sequence of the approximating solutions reveals that it is a solution to the original identification problem. An application to an inverse problem arising in population dynamics is presented.

1. Problem statement. The problem of parameter identiﬁcation in a dynamic mathematical model describing a real process is of particular interest in a large range of sciences, from engineering to economy, biology, medicine or environmental sciences. Its mathematical approach may be challenging and the problem is not one of a particular estimation technique. Of course, the ideal solution is to obtain a close identiﬁcation formula for the parameter, but generally this can be done under strong hypotheses. Let us explain this by using an abstract formulation, considering the evolution equation y0(t) = Ay(t) + u(t)z(t), t ∈ (0,T ), (1)

y(0) = y0, (2) where A is an infinitesimal operator of a C0-semigroup on a Banach space X, and z : [0,T ] → X. Given the demand (1)-(2), the requirement is to recover the unknown function u : [0,T ] → R, from the supply φ(y(t)) = g(t), t ∈ [0,T ], (3) where g : [0,T ] → R is known and φ ∈ X∗, the dual of X. A large class of identification problems for linear evolution equations of first-order in the hyperbolic and parabolic cases and of second-order has been studied in papers by Angelo Favini and co-authors. A particular attention has been given to strongly

2010 Mathematics Subject Classiﬁcation. Primary: 35R30, 47Dxx, 49J27, 49K15, 49K20, 65N21, 92D25. Key words and phrases. Identiﬁcation problems, linear evolution equations, semigroup, optimal control, population dynamics. ∗ Corresponding author: Gabriela Marinoschi.

1429 1430 VIOREL BARBU AND GABRIELA MARINOSCHI degenerate equations and recently to integro-differential equations. Sufficient conditions for the existence and uniqueness of the solution to the identification problem have been found. They include the condition φ(z(t)) 6= 0. Without exhausting the list we can cite a few results: [1], [2], [3], [4], [5], [8], [9], [10], [11], [12]. In this paper, we shall discuss the possibility of solving such a problem in two situations. The general assumptions, besides the fact that A is a generator of a C0-semigroup on the Banach space X, are:

(i) y0 ∈ X (ii) z ∈ L2(0,T ; X) (iii) g ∈ L2(0,T ; R). This paper addresses the reconstruction of u, and consequently of y, as a solution to (1)–(2), from the observation (3). We shall call it problem (IP ). A ﬁrst result is to solve this problem in an exact form, under stronger conditions, in Section 2. For more relaxed hypotheses, including also φ(z(t)) = 0 on some intervals, we shall introduce an optimal control approach providing an approximating solution to (IP ). The sequence of these solutions will tend to the solution to (IP ) if this exists in this case. These will be detailed in Section 3. Finally, in Section 4, we shall give an example of application to an inverse problem related to population dynamics.

2. A direct approach. Deﬁnition 2.1. Let u ∈ C[0,T ] and z ∈ C([0,T ]; X). A strong solution to (1)–(2) is a function y ∈ C1([0,T ]; X) T C([0,T ]; D(A)) which satisﬁes (1) for all t ∈ [0,T ]. Theorem 2.2. Let us assume that

y0 ∈ D(A), z ∈ C([0,T ]; D(A)), (4) φ(z(t)) 6= 0, for all t ∈ [0,T ], (5) 1 g ∈ C ([0,T ]; R). (6) Then, problem (IP ) has a unique solution (u, y) ∈ C[0,T ] × C1([0,T ]; X) ∩ C([0,T ]; D(A)).

Proof. Since A is a generator of a C0-semigroup on X, the Cauchy problem (1)–(2) has a unique mild solution y ∈ C([0,T ]; X) given by Z t At A(t−s) y(t) = e y0 + e z(s)u(s)ds, for all t ∈ [0,T ]. (7) 0 By (4) this turns out to be actually a strong solution. Applying φ we get Z t At A(t−s) φ(y(t)) = φ(e y0) + φ(e z(s))u(s)ds. (8) 0 Denoting G(t, s) := φ(eA(t−s)z(s)), 0 ≤ s ≤ t ≤ T, Z t Γu(t) := G(t, s)u(s)ds 0 and recalling (3), we can write (8) as At Γu(t) = g(t) − φ(e y0). (9) We note that (t, s) → G(t, s) is differentiable on [0,T ] × [0,T ] and the right-hand side of (9) is differentiable on [0,T ]. This is an integral equation of the first kind AN IDENTIFICATION PROBLEM 1431 which can be reduced to a linear Volterra equation of second kind by a standard way. Namely, by differentiating (9) with respect to t we obtain Z t At 0 G(t, t)u(t) − G(t, 0)u(0) + Gt(t, s)u(s)ds + φ(e Ay0) = g (t). 0 Next, writing

0 At At −1 f(t) := (g (t) − φ(e Ay0) + φ(e z(0))u(0))(G(t, t)) , and observing that G(t, t) = φ(z(t)) 6= 0 for all t ∈ [0,T ], we get Z t u(t) = K(t, s)u(s)ds + f(t), for all t ∈ [0,T ], (10) 0 where G (t, s) K(t, s) = − t ,G (t, s) = φ(eA(t−s)Az(s)), 0 ≤ s ≤ t ≤ T. G(t, t) t For the continuity of u we should choose u(0) = 0. Indeed, by (10) we have

0 −1 u(0) = f(0) = (g (0) − φ(Ay0) + φ(z(0))u(0))(φ(z(0))) , 0 which yields g (0) = φ(Ay0), and diﬀerentiating (3) we get u(0) = 0, because 0 g (0) = φ(Ay0) + u(0)φ(z(0)).

Under our hypotheses f ∈ C[0,T ], R) and so the Volterra integral equation of second kind (10) has a unique solution u ∈ C[0,T ], given by Z t u(t) = f(t) + K(t, s)f(s)ds, for all t ∈ [0,T ], (11) 0 with the resolvent kernel ∞ X K(t, s) = Kj(t, s), for all 0 ≤ s ≤ T, 0 ≤ t ≤ T, j−0 where Z t Kn(t, s) = K(t, τ)Kn−1(τ, s)dτ, K0(t, s) = K(t, s). s As also known, u(t) can be iteratively obtained as Z t un+1(t) = K(t, s)un(s)ds + f(t), u0(t) = u(0), (12) 0 and the sequence un(t)n≥1 converges strongly to u(t), the solution to (10), as n → ∞. By (7) it follows that the corresponding solution y(t) is unique, hence problem (IP ) has a unique strong solution (u, y), as claimed.

The following results is easily observed by (12). Corollary 2.3. Under the hypotheses of Theorem 2.2, if u(0) ≥ 0, f(t) ≥ 0 for all t ∈ [0,T ] and K(t, s) ≥ 0 for all t, s ∈ [0,T ], it follows that u(t) ≥ 0 for all t ∈ [0,T ]. Moreover, if u(0) ≤ 0, f(t) ≤ 0 for all t ∈ [0,T ] and K(t, s) ≤ 0 for all t, s ∈ [0,T ], it follows that u(t) ≤ 0 for all t ∈ [0,T ]. 1432 VIOREL BARBU AND GABRIELA MARINOSCHI

3. An optimal control approach. An alternative for solving problem (IP ) is to use an optimal control approach by considering the minimization problem ( ) 1 Z T Min J(u) = (φ(y(t)) − g(t))2dt; u ∈ C[0,T ] (P ) 2 0 for all (u, y) satisfying (1)–(2). It is obvious that if (1)–(2) has a unique solution, this turns out to be the unique solution to (P ), but the converse assertion is not generally true. It is not clear if (P ) may have a solution, especially if we do not assume that φ(z(t)) 6= 0. The way to get however a solution is to use an approximating control problem (Pλ) which ∗ ∗ provides an approximating solution (uλ, yλ) and to check if this could tend to the solution to (P ), if the latter has one. In this approximating problem we shall require less regularity for the data than in the ﬁrst case of Section 2. The hypotheses used in this part are (i)–(iii) and X is a reﬂexive Banach space. We stress that we do not require that φ(z(t)) is nonzero. Let λ be positive and let us introduce the family of optimal control problems ( Z T ) 1 2 λ 2 Min Jλ(u) = (φ(y(t)) − g(t)) dt + kukL2(0,T ) (Pλ) 2 0 2 subject to (1)–(2), where 2 U = {u ∈ L (0,T ); u(t) ∈ [a, b] ⊂ R, a.e. t ∈ (0,T )}. Here, u acts as a control.

∗ ∗ ∗ Proposition 3.1. Problem (Pλ) has a unique solution (uλ, yλ), where yλ is the mild solution to (1)–(2).

Proof. Since Jλ(u) ≥ 0 it follows that its inﬁmum dλ exists and it is nonnegative. n n n Let us consider a minimizing sequence (uλ, yλ ) with uλ ∈ U and so 1 d ≤ J(un, yn) ≤ d + , for n ≥ 1. (13) λ λ λ λ n It follows that we can select a subsequence, denoted still by n such that n ∗ ∞ uλ → uλ weak-* in L (0,T ), ∗ n and uλ is in U. For u = uλ problem (1)–(2) has a unique mild solution (since n 1 uλz ∈ L (0,T ; X) and y0 ∈ X), that is Z t n At A(t−s) n yλ (t) = e y0 + e z(s)uλ(s)ds, for all t ∈ [0,T ], 0 whence we deduce that Z t n ∗ At A(t−s) ∗ yλ (t) → yλ(t) = e y0 + e z(s)uλ(s)ds 0 ∗ weakly in X for all t ∈ [0,T ]. Thus, yλ follows to be the solution to (1)–(2). Next, since φ ∈ X∗ we have n ∗ φ(yλ (t)) → φ(yλ(t)), for all t ∈ [0,T ] ∗ and passing to the limit in (13) we get Jλ(uλ) = dλ. We have got that (Pλ) has a solution. Finally, we note that the functional is strictly convex, so that the minimum ∗ uλ is unique. AN IDENTIFICATION PROBLEM 1433

∗ Let us denote by h·, ·iX∗,X the pairing between the dual spaces X and X and by P[a,b](f) the projection of f on [a, b]. ∗ ∗ Proposition 3.2. Let (uλ, yλ) be the solution to (Pλ). Then, the ﬁrst order neces- sary conditions of optimality are 1 u∗ (t) = P hp (t), z(t)i , a.e. t ∈ (0,T ), (14) λ [a,b] λ λ X∗,X where pλ is the solution to the dual backward problem dp − λ = A∗p (t) − (φ(y∗(t)) − g(t))φ, t ∈ (0,T ), (15) dt λ λ pλ(T ) = 0, and A∗ is the adjoint of A.

∗ ∗ Proof. Let (uλ, yλ) be the solution to (Pλ) and for σ > 0 let us denote σ ∗ ∗ uλ = uλ + σw, with w = v − uλ and v ∈ U. Now, we consider the equation in variations dY (t) = AY (t) + w(t)z(t), t ∈ (0,T ), (16) dt Y (0) = 0, which has a unique mild solution Y ∈ C([0,T ]; X). It is easily seen by (1)–(2) that the function σ ∗ yu − yu Y := lim strongly in C([0,T ; X) σ→0 σ σ is the unique mild solution to the Cauchy problem (16). We specify that yu and σ yu are the solutions to (1)–(2) corresponding to uσ and u∗, respectively. ∗ ∗ Now, since (uλ, yλ) is optimal in (Pλ) it satisﬁes ∗ σ Jλ(uλ) ≤ Jλ(uλ), whence by performing a short calculation, dividing by λ and passing to the limit as λ → 0, we get the optimality relation Z T Z T ∗ ∗ (φ(yλ(t)) − g(t))φ(Y (t))dt + λ uλ(t)w(t)dt ≥ 0. (17) 0 0 ∗ ∗ ∗ We recall that, since X is reﬂexive, A is generating a C0-semigroup on X (see [14], p. 41). Then, since φ ∈ X∗ equation (15) has a unique mild solution ∗ pλ ∈ C([0,T ]; X ), Z T A∗(s−t) ∗ pλ(t) = e (φ(yλ(s)) − g(s))φds. t ∗ Now, if pλ and Y would be strong solutions (if e.g. φ ∈ D(A ) and wz ∈ D(A)), we could multiply (16) by pλ and to integrate by parts to get Z T hpλ(T ),Y (T )iX∗,X − hpλ(0),Y (0)iX∗,X − h(pλ)t(t),Y (t)iX∗,X dt 0 Z T Z T = hpλ(t), AY (t)iX∗,X dt + hpλ(t), w(t)z(t)iX∗,X dt. 0 0 1434 VIOREL BARBU AND GABRIELA MARINOSCHI

Since in our case the data are not regular, the same result of integration by parts can actually follow by passing to the limit in regularizing problems for Y and pλ, corresponding to regular data. Next, by some calculations taking into account the initial and ﬁnal conditions in (16) and (15) we get Z T Z T ∗ h−(pλ)t(t) − A pλ(t),Y (t)iX∗,X dt = hpλ(t), z(t)iX∗,X w(t)dt, 0 0 which implies by (15) Z T Z T ∗ − φ(yλ(t) − g(t))φ(Y (t))dt = hpλ(t), z(t)iX∗,X w(t)dt. 0 0 By comparison with (17) we ﬁnd Z T ∗ (− hpλ(t), z(t)iX∗,X + λuλ(t))w(t)dt ≥ 0, 0 ∗ whence taking into account that w = v − uλ, we get Z T ∗ ∗ (hpλ(t), z(t)iX∗,X − λuλ(t))(uλ(t) − v(t))dt ≥ 0, for any v ∈ U. 0 Thus, we deduce that ∗ ∗ ∗ hpλ(t), z(t)iX∗,X − λuλ(t) ∈ N[a,b](uλ(t)) = ∂I[a,b](uλ(t)), a.e. t ∈ (0,T ),

∗ ∗ R where N[a,b](uλ(t)) is the normal cone to [a, b] at uλ(t) and ∂I[a,b] : R → 2 is the subdiﬀerential of the indicator function of [a, b]. Then, 1 1 u∗ (t) + ∂I (u∗ (t)) 3 hp (t), z(t)i λ λ [a,b] λ λ λ X∗,X which implies (14), as claimed.

Let M := {u ∈ L2(0,T ); u(t) ∈ [a, b] a.e., (u, y) solves (1)–(3)}.

It is easily seen that M a convex and close subset. Let us denote by PrM(0) the projection of 0 on M.

∗ ∗ Theorem 3.3. Let (uλ, yλ) be a solution to (Pλ). If M 6= ∅, then for λ → 0 we have ∗ 2 uλ → ue strongly in L (0,T ), (18) ∗ yλ → ye strongly in C([0,T ]; X), (19) where (u,e ye) is a solution to the identiﬁcation problem (1)-(3). Moreover, ue = PrM(0). ∗ ∗ Proof. Let (uλ, yλ) be a solution to (Pλ). By the optimality condition we have Z T ∗ 1 ∗ 2 λ ∗ 2 Jλ(uλ) = (φ(yλ(t)) − g(t)) dt + kuλkL2(0,T ) 2 0 2 Z T 1 2 λ 2 ≤ Jλ(u) = (φ(y(t)) − g(t)) dt + kukL2(0,T ) , 2 0 2 AN IDENTIFICATION PROBLEM 1435 for all u ∈ U. In particular, let us u = u∗ and y = y∗ where (u∗, y∗) is any solution to (1)-(3), if M= 6 ∅. Hence φ(y∗(t)) = g(t), a.e. t ∈ (0,T ) and we have u∗ ≤ ku∗k . (20) λn L2(0,T ) L2(0,T ) ∗ 2 Hence, on a subsequence λn → 0, we get u → u weakly in L (0,T ). Consequently, λn e y∗ (t) → y(t) weakly in X, for all t ∈ [0,T ], where λn e Z t At A(t−s) ye(t) = e y0 + e z(s)ue(s)ds, t ∈ [0,T ] 0 and so ye(t) is a solution to (1)–(2). Moreover, φ(y∗ ) → φ(y) as λ → 0 λn e and so (u,e ye) is a mild solution to (1)–(3), that is ue ∈ M. On the one hand, by (20) we have kuk ≤ lim inf u∗ ≤ ku∗k L2(0,T ) λn L2(0,T ) L2(0,T ) e λn→0 ∗ for any u ∈ M and so we deduce that the distance from ue to 0 is the smallest, that is ue = PrM(0). (21) Namely, ue = 0 if 0 ∈ M and ue ∈ ∂M, if 0 ∈/ M. On the other hand, it follows that u∗ → u strongly in L2(0,T ) (22) λn e ∗ ∗ ∗ (see [6], p. 78), because, by (20), lim sup u ≤ ku k 2 and u → u λn L2(0,T ) L (0,T ) λn e λn→0 2 weakly in L (0,T ) (which is strictly convex). Finally, by the uniqueness of ue deﬁned by (21) it follows that (22) holds for all {λn} → 0 and so (18) follows, as claimed. Moreover, this implies (19), too. This ends the proof.

4. Application to population dynamics control. In this section, we present an application to a problem arising in the control of population dynamics. Let us consider a population structured with respect to the age a, diﬀusing in a habitat Ω, whose dynamics is governed by a supplementary mortality rate µ(a, x), a fertility rate β(a) and a free term u(t)z(a, x) showing a change of population, possibly due to demographic events. Both these rates are nonnegative and essentially bounded, + + 0 ≤ β(a) ≤ β+, a.e. a ∈ (0, a ), 0 ≤ µ(a, x) ≤ µ+, a.e. (a, x) ∈ (0, a ) × Ω. The equations describing the dynamics of the population of density y are: + yt + ya − ∆y + µ(a, x)y = u(t)z(a, x), in (0,T ) × (0, a ) × Ω, (23) Z a+ y(t, 0, x) = β(a)y(t, a, x)da, in (0,T ) × Ω, (24) 0 − ∇y · ν = 0, on (0,T ) × (0, a+) × ∂Ω, (25) + y(0, a, x) = y0, in (0, a ) × Ω, (26) where a+ is the maximum life age, ν is the outer unit normal to the boundary of Ω and (24) is the renewal equation in population dynamics, providing the number of newborns. The problem is to ﬁnd the rate u(t) such that to ensure the value g(t) + of a prescribed total population with ages in a certain age range [a1, a2] ⊂ [0, a ], Z a2 Z φ(y(t)) = y(t, a, x)dxda = g(t), a.e. t ∈ (0,T ). (27) a1 Ω 1436 VIOREL BARBU AND GABRIELA MARINOSCHI

We shall study the problem in the following functional framework. Let us denote H := L2(Ω),V = H1(Ω),V 0 = (H1(Ω))0 H = L2(0, a+; H), V = L2(0, a+; V ), V0 = L2(0, a+; V 0). 0 We introduce B0 : V → V by Z a+ Z hB0y, ψiV0,V = (∇y · ∇ψ + µ(a, x)yψ)dxda, for all ψ ∈ V, 0 Ω and B : D(B) ⊂ H → H, by By = ya + B0y for all y ∈ D(B), where ( Z a+ ) 0 D(B) = y ∈ V; ya ∈ V , By ∈ H, y(0, x) = β(a)y(a, x)da . 0 Then, problem (23)–(26) can be written as dy (t) = Ay(t) + u(t)z(a, x), a.e. t ∈ (0,T ), (28) dt y(0) = y0, where A = −B. We have Proposition 4.1. Let 1 y0 ∈ D(B), z ∈ H, g ∈ C ([0,T ]; R), Z a2 Z z(a, x)dxda 6= 0. a1 Ω Then, the identiﬁcation problem (23)–(27) has a unique strong solution given by (11), where R a2 R eA(t−s)Azdxda K(t, s) = − a1 Ω , R a2 R z(a, x)dxda a1 Ω

0 R a2 R At R a2 R At g (t) − e Ay0dxda + e zu(0)dxda f(t) = a1 Ω a1 Ω , R a2 R z(a, x)dxda a1 Ω u(0) = 0. Proof. One applies the results in Section 2, for the operator A = −B, after showing that A generates a C0-semigroup on the Hilbert space X = H. Equivalently, we show that B = −A is quasi m-accretive on H. Indeed, B is quasi monotone Z 0 0 2 1 2 + hλ I + B)y, yiV0,V = λ kykH + y (a , x)dx 2 Ω + !2 1 Z Z a − β(a)y(a, x)da dx 2 Ω 0 Z a+ Z 2 2 + k∇ykH + µ(a, x)y dxda 0 Ω 1 1 ≥ λ0 − β2 a+ kyk2 ≥ 0, for λ0 > β2 a+. 2 + H 2 + Moreover, B is quasi m-accretive because Range(λ0I + B) = H. To prove this we consider the equation 0 + λ y + ya − ∆y + µ(a, x)y = h, in (0, a ) × Ω, (29) AN IDENTIFICATION PROBLEM 1437 with the boundary conditions Z a+ ∇y · v = 0 on (0, a+) × ∂Ω, y(0, x) = β(a)y(a, x)da (30) 0 and h ∈ H and show that it has a solution y ∈ D(B), by the Banach fixed point theorem. Namely, let us fix ζ ∈ H and consider the equation (29) with the boundary conditions Z a+ ∇y · v = 0 on (0, a+) × ∂Ω, y(0, x) = β(a)ζ(a, x)da. (31) 0 By the general results concerning parabolic boundary-value systems, e.g., by Lions’ theorem, problem (29), (31) has a unique solution yζ ∈ C([0, a+]; H)∩L2(0, a+; V )∩ W 1,2(0, a+; V 0). Then, we define the mapping Φ : H → H by Φ(ζ) = yζ the solution to (29), (31). By a direct computation, one can prove that Φ is a contraction on H for λ0 large enough, hence Φ(ζ) = yζ = ζ and so we can replace ζ in (31) by yζ . Indeed, considering two solutions y1 and y2 corresponding to ζ1 and ζ2 we multiply + the difference equation for y1 − y2 by y1 − y2 and integrate over (0, a ) × Ω. We get 2 Z Z a+ ! 1 + 2 0 2 1 (y1 − y2)(a ) H + λ ky1 − y2kH ≤ β(a)(ζ1 − ζ2)(a, x)da dx, 2 2 Ω 0 whence 1 λ0 ky − y k2 ≤ β2 a+ kζ − ζ k2 , 1 2 H 2 + 1 2 H showing that for λ0 large enough, Φ is a contraction on H. It follows that system (29), (30) has a unique solution in the previous indicated spaces. Moreover, By = h − λ0y ∈ H, hence y ∈ D(B) for each h ∈ H. Thus, A quasi m-accretive and so A = −B generates a C0-semigroup on H, so that Theorem 2.2 can be applied. For weaker regularity of the problem data we resort to the optimal control approach by solving the minimization problem  2  Z T Z a+ Z ! Z T  1 λ 2  Min Jλ(u) = y(t, a, x)dxda − g(t) dt + u (t)dt (Peλ)  2 0 0 Ω 2 0  for all u ∈ U = {v ∈ L2(0,T ); v(t) ∈ [a, b] ⊂ R a.e. t ∈ (0,T )}, and applying the results in Section 3.

2 Proposition 4.2. Let y0 ∈ H, z ∈ H, g ∈ L (0,T ; R). Then, problem (Peλ) has a ∗ unique solution uλ given by ∗ 1 u (t) = P hp(t), z(t)i 0 , a.e. t ∈ (0,T ), λ [a,b] λ V ,V where p is the solution to the dual backward problem + pt + pa + ∆y − µ(a, x)p + β(a)p(0, x) = F (t), in (0,T ) × (0, a ) × Ω, p(t, a+, x) = 0, in (0,T ) × Ω, ∇p · ν = 0, on (0,T ) × (0, a+) × ∂Ω, p(T, a, x) = 0, in (0, a+) × Ω,

Z a+ Z ∗ F (t) = yλ(t, a, x)dxda − g(t) 0 Ω 1438 VIOREL BARBU AND GABRIELA MARINOSCHI

∗ ∗ ∗ and yλ is the solution to (28) corresponding to uλ. If M= 6 ∅, then uλ → ue strongly 2 in L (0,T ) and (u,e ye) is a solution to (23)–(27). Moreover, ue = PrM(0). Proof. The state system (28) has a unique solution y ∈ C([0,T ]; H) ∩ L2(0,T ; V) ∩ C([0, a+]; L2(0,T ; H)). The proof can be led as in [7], [13]. The existence of the control follows by Proposition 3.1. In Proposition 3.2, the optimality relation (17) reads now Z T Z a+ Z Z T ∗ F (t)Y (t, a, x)dxdadt + λ uλ(t)w(t)dt ≥ 0 0 0 Ω 0 and the system in variations is dY (t) = AY (t) + w(t)z(a, x), t ∈ (0,T ), dt Y (0) = 0, with A = −B previously deﬁned in this section. Like the state system, it has a unique solution Y ∈ C([0,T ]; H) ∩ L2(0,T ; V) ∩ C([0, a+]; L2(0,T ; H)).

We set Ape = pa + ∆p − µ(a, x)p + β(a)p(0, ·), Ae : D(Ae) ⊂ H → H with

n 0 + + o D(Ae) = p ∈ V; pa ∈ V , Ape ∈ H, p(a , x) = 0, ∇p · ν = 0 on (0, a ) × Ω .

First, we prove that Ae is actually A∗, the dual of A. Indeed, we easily see that D E Ap,e y = hp, AyiV,V0 , ∀p ∈ D(Ae), y ∈ D(A), V0,V but this is not suﬃcient to ensure that Ae is the adjoint of A, because A∗ may be an extension of A.e It suﬃces however to prove that Be = −Ae is quasi m-accretive on H and so, Be being maximal it follows that Be = B∗ which implies that Ae = A∗. We have Z Z D 0 E 0 2 1 2 + 1 2 2 λ I + Be)q, q = λ kpkH − p (a , x)dx + p (0, x)dx + k∇pkH V0,V 2 Ω 2 Ω Z a+ Z Z a+ Z + µp2dxda − β(a)p(0, x)pdxda 0 Ω 0 Ω Z 0 2 + 2 1 2 0 2 + ≥ (λ − 4β+a ) kpkH + p (0, x)dx > 0 for λ > 4β+a . 4 Ω For the quasi m-accretiveness, we consider the system

0 + λ p − pa − ∆y + µ(a, x)p − β(a)p(0, x) = h, in (0, a ) × Ω, p(a+, x) = 0, in Ω, ∇p · ν = 0, on (0, a+) × ∂Ω, in which we make the transformation a0 = a+ − a, p(a+ − a, x) = q(a0, x), getting

0 0 0 + + λ q + qa0 − ∆q + µe(a , x)q − βe(a )q(a , x) = h, in (0, a ) × Ω, q(0, x) = 0, in Ω, (32) ∇q · ν = 0, on (0, a+) × ∂Ω. AN IDENTIFICATION PROBLEM 1439

0 + 0 0 + 0 Here, βe(a ) = β(a − a ), µe(a , x) = µ(a − a , x). For h ∈ H we should have a solution q ∈ D(Be). We proceed again by a ﬁxed point theorem, by taking v ∈ C([0, a+]; H), ﬁxing v(a+, x) in the equation

0 0 0 + λ q + qa0 − ∆q + µ(a , x)q = h + βe(a )v(a , x) ∈ H (33) and showing that the function Ψ(v(a+, ·)) = q(a+, ·) maps H into H and it is a contraction. The equation (33) together with the conditions in (32) has a unique solution q ∈ C([0, a+]; H) ∩ L2(0, a+; H2(Ω)) ∩ W 1,2(0, a+; V 0). Now, writing the diﬀerence ω of two solutions q and q, corresponding to v and v we have

0 0 + ωa0 − ∆ω + µω + λ ω = βe(a )vb(a , x), + + + with ∇ω · ν = 0 and ω(0, x) = 0 in Ω, vb(a , x) = v(a , x) − v(a , x). We multiply the equation by ω and integrate over (0, a+) × Ω. We get

+ + 1 Z a+ Z a Z ε Z a Z ω2(a+, x)dx+λ0 kωk2 ≤ β2(a0)ω2dxda+ v2(a+, x)dxda, H e + b 2 Ω ε 0 Ω a 0 Ω which yields Z + Z 1 2 + 0 a 2 2 2 + ω (a , x)dx + λ − β+ kωkH ≤ ε vb (a , x)dx. 2 Ω ε Ω For 2ε < 1 we get that Ψ is a contraction on H. Moreover, it is obvious that the solution q ∈ D(Be) and the proof of the quasi m-accretiveness is ended. Thus, the adjoint system dp − (t) = Ape (t) + F (t), t ∈ (0,T ), dt p(T ) = 0 has a unique solution p ∈ C([0,T ]; H) ∩ L2(0,T ; V) ∩ C([0, a+]; L2(0,T ; H)). Now, we can apply Proposition 3.2 and Theorem 3.3 to get the assertions in Proposition 4.2. Finally, as a remark, since y is a density it should be positive. Assuming that y0 ≥ 0 and that the free term in (23) is nonnegative, then by multiplying (23) by the negative part (y)−, making some computations and applying the Stampac- chia lemma, we get that the solution is positive. Taking into account the expres- ∗ sion (14) for uλ, in order to have a nonnegative free term in (23) we must have z(a, x) hp(t), z(t)iV0 ,V ≥ 0 for a.e. t ∈ (0,T ).

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