on the optimal control of the free boundary

doi:10.3934/ipi.2013.7.307

Inverse Problems and Imaging Volume 7, No. 2, 2013, 307–340

ON THE OPTIMAL CONTROL OF THE FREE BOUNDARY PROBLEMS FOR THE SECOND ORDER PARABOLIC EQUATIONS. I. WELL-POSEDNESS AND CONVERGENCE OF THE METHOD OF LINES

Ugur G. Abdulla Department of Mathematics, Florida Institute of Technology Melbourne, Florida 32901, USA

(Communicated by Antonin Chambolle) Abstract. We develop a new variational formulation of the inverse Stefan problem, where information on the heat flux on the fixed boundary is missing and must be found along with the temperature and free boundary. We employ optimal control framework, where boundary heat flux and free boundary are components of the control vector, and optimality criteria consists of the minimization of the sum of L2 -norm declinations from the available measurement of the temperature flux on the fixed boundary and available information on the phase transition temperature on the free boundary. This approach allows one to tackle situations when the phase transition temperature is not known explicitly, and is available through measurement with possible error. It also allows for the development of iterative numerical methods of least computational cost due to the fact that for every given control vector, the parabolic PDE is solved in a fixed region instead of full free boundary problem. We prove well-posedness in Sobolev spaces framework and convergence of discrete optimal control problems to the original problem both with respect to cost functional and control.

1. Description of main results. 1.1. Introduction and motivation. Consider the general one-phase Stefan problem ([14, 25]): find the temperature function u(x, t) and the free boundary x = s(t) from the following conditions (1) (2)

(a(x, t)ux )x + b(x, t)ux + c(x, t)u − ut = f (x, t),

for (x, t) ∈ Ω

0 ≤ x ≤ s(0) = s0

u(x, 0) = φ(x),

0≤t≤T

(3)

a(0, t)ux (0, t) = g(t),

(4)

a(s(t), t)ux (s(t), t) + γ(s(t), t)s0 (t) = χ(s(t), t),

(5)

u(s(t), t) = µ(t),

0≤t≤T

0≤t≤T

2010 Mathematics Subject Classification. Primary: 35R30, 35R35, 35K20, 35Q93; Secondary: 65M32, 65N21. Key words and phrases. Inverse Stefan problem, optimal control, second order parabolic PDE, Sobolev spaces, energy estimate, embedding theorems, traces of Sobolev functions, method of lines, discrete optimal control problem, convergence in functional, convergence in control. 307

c

2013 American Institute of Mathematical Sciences

308

Ugur G. Abdulla

where a, b, c, f , φ, g, γ, χ, µ are known functions and a(x, t) ≥ a0 > 0,

(6)

s0 > 0

Ω = {(x, t) : 0 < x < s(t), 0 < t ≤ T } In the physical context, f characterizes the density of the sources, φ is the initial temperature, g is the heat flux on the fixed boundary and µ is the phase transition temperature. Assume now that some of the data is not available, or involves some measurement error. For example, assume that the heat flux g(t) on the fixed boundary x = 0 is not known and must be found along with the temperature u(x, t) and the free boundary s(t). In order to do that, some additional information is needed. Assume that this additional information is given in the form of the temperature measurement along the boundary x = 0: (7)

u(0, t) = ν(t),

for 0 ≤ t ≤ T

Inverse Stefan Problem (ISP): Find the functions u(x, t) and s(t) and the boundary heat flux g(t) satisfying conditions (1)-(7). Motivation for this type of inverse problem arose, in particular, in modeling of the bioengineering problem on the laser ablation of biological tissues through Stefan problem (1)-(6), where s(t) is the ablation depth at the moment t. The boundary temperature measurement u(0, t) contains an error, which makes it impossible to get reliable measurements of the boundary heat flux g(t), and the ISP must be solved for its identification. This approach allows to regularize an error contained in a measurement ν(t). Another advantage of this approach is that, in fact, condition (5) can be treated as a measurement of the temperature on the ablation front, and our approach allows us to regularize an error contained in temperature measurement µ(t) on the ablation front. Still another important motivation arises in optimal control of Stefan problem, where controlling g(t) is equivalent of controlling external temperature along the fixed boundary. It should be pointed out that the method of this paper can be applied to different type of inverse problems. For example, (7) can be replaced with u(x, T ) = w(x),

for 0 ≤ x ≤ s(T ),

meaning that measurement is taken for the final temperature distribution w(x) and final ablation depth s(T ). ISP is not well posed in the sense of Hadamard. If there is no coordination between the input data, the exact solution may not exist. Even if it exists, it might be not unique, and most importantly, there is no continuous dependence of the solution on the data. Inverse Stefan problem was first mentioned in [9], in the form of finding a heat flux on the fixed boundary which provides a desired free boundary. This problem is similar to non-characteristic Cauchy problem for the heat equation. The variational approach for solving this ill-posed inverse Stefan problem was performed in [6, 7]. First result on the optimal control of the Stefan problem appeared in [35]. It consists of finding optimal value of the external temperature along the fixed boundary, in order to ensure that the solutions of the Stefan problem are close to the measurements taken at the final moment. In [35] existence result was proved. In [37] the Frechet differentiability and the convergence of the difference schemes was proved for the same problem and Tikhonov regularization was suggested. Later development of the inverse Stefan problem was along these two lines: Inverse Stefan problems with given phase boundaries were Inverse Problems and Imaging

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considered in [1, 3, 5, 8, 10, 11, 12, 17, 31, 15]; optimal control of Stefan problems, or equivalently inverse problems with unknown phase boundaries were investigated in [2, 13, 18, 19, 20, 21, 22, 24, 28, 26, 29, 30, 34, 15]. We refer to monography [15] for a complete list of references of both types of inverse Stefan problems, both for linear and quasilinear parabolic equations. The main methods used to solve inverse Stefan problem are based on variational formulation, method of quasisolutions or Tikhonov regularization which takes into account ill-posedness in terms of the dependence of the solution on the inaccuracy involved in the measurement (7), Frechet differentiability and iterative conjugate gradient methods for numerical solution. Despite its effectiveness, this approach has some deficiencies in many practical applications: • Solution of the inverse Stefan problem is not continuously dependent on the phase transition temperature µ(t): small perturbation of the phase transition temperature may imply significant change of the solution to the inverse Stefan problem. Accordingly, any regularization which equally takes into account instability with respect to both ν(t) from measurement (7), and the phase transition temperature µ(t) from (5) will be preferred. It should be also mentioned that in many applications the phase transition temperature is not known explicitly. In many processes the melting temperature of pure material at a given external action depends on the process evolution. For example, gallium (Ga, atomic number 31) may remain in the liquid phase at temperatures well below its mean melting temperature ([25]). • Numerical implementation of the iterative gradient type methods within the existing approach requires to solve full free boundary problem at every step of the iteration, and accordingly requires quite high computational cost. Iterative gradient method which requires at every step solution of the boundary value problem in a fixed region would definitely be much more effective in terms of the computational cost. The main goal of this project is to develop a new variational approach based on the optimal control theory which is capable of addressing both of the mentioned issues and allows the inverse Stefan problem to be solved numerically with least computational cost by using conjugate gradient methods in Hilbert spaces. In this paper we prove the existence of the optimal control and convergence of the family of time-discretized optimal control problems to the continuous problem both with respect to cost functional and control. We employ Sobolev spaces framework which allows to reduce the reguarity and structural requirements on the data. We address the problems of convergence of the fully discretized family of optimal control problems, Frechet differentiability and iterative conjugate gradient methods in Hilbert spaces in an upcoming paper. Throughout the paper we use usual notation for Sobolev spaces according to references [23, 4, 27, 32, 33]. For convenience, we describe in the next section the notations for the Sobolev spaces used throughout the paper.

1.2. Notation of Sobolev spaces. L2 [0, T ] – Hilbert space with scalar product Z (u, v) =

T

uvdt 0

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W2k [0, T ], k = 1, 2 – Hilbert space of all elements of L2 [0, T ] whose weak derivatives up to order k belong to L2 [0, T ] and scalar product is defined as Z TX k ds u ds v (u, v) = dt s s 0 s=0 dt dt 1

W24 [0, T ] – Banach space of all elements of L2 [0, T ] with finite norm Z T Z T |u(t) − u(τ )|2 21 2 dt kuk 14 = kukL2 [0,T ] + dτ 3 W2 [0,T ] |t − τ | 2 0 0 L2 (Ω) – Hilbert space with scalar product Z (u, v) = uvdxdt Ω

W21,0 (Ω) – Hilbert space of all elements of L2 (Ω) whose weak derivative to L2 (Ω), and scalar product is defined as Z ∂u ∂v (u, v) = uv + dxdt ∂x ∂x Ω

∂u ∂x

belongs

W21,1 (Ω) – Hilbert space of all elements of L2 (Ω) whose weak derivatives belong to L2 (Ω), and scalar product is defined as Z ∂u ∂v ∂u ∂v (u, v) = uv + + dxdt ∂x ∂x ∂t ∂t Ω

∂u ∂u ∂x , ∂t

V2 (Ω) – Banach space of all elements of W21,0 (Ω) with finite norm

2 21

∂u 2

kukV2 (Ω) = esssup0≤t≤T ku(x, t)kL2 [0,s(t)] +

∂x L2 (Ω) V21,0 (Ω) – Banach space which is a completion of W21,1 (Ω) in the norm of V2 (Ω). It consists of all elements of V2 (Ω), continuous with respect to t in norm of L2 [0, s(t)] and with finite norm

2 12

∂u 2

1,0 kukV (Ω) = max ku(x, t)kL2 (0,s(t)) +

2 0≤t≤T ∂x L2 (Ω) W22,1 (Ω) – Banach space of all elements of L2 (Ω) whose weak derivatives ∂u ∂x , ∂2u ∂x2 belong to L2 (Ω), and the norm is defined as

2

2 2

2 12

∂u

∂u

∂ u 2

2,1 kukW (Ω) = kukL2 (Ω) + + +

2 ∂x L2 (Ω) ∂t L2 (Ω) ∂x2 L2 (Ω)

∂u ∂t ,

1.3. Optimal control problem. In this section we formulate a new variational formulation of the inverse problem which takes into account the deficiencies described in Section 1.1. Consider a minimization of the cost functional (8)

J (v) = β0 ku(0, t) − ν(t)k2L2 [0,T ] + β1 ku(s(t), t) − µ(t)k2L2 [0,T ]

on the control set n VR = v = (s, g) ∈ W22 [0, T ] × W21 [0, T ] : δ ≤ s(t) ≤ l, o s(0) = s0 , max( kskW22 ; kgkW21 ) ≤ R Inverse Problems and Imaging

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where δ, l, R, β0 , β1 are given positive numbers, and u = u(x, t; v) be a solution of the Neumann problem (1)-(4). Definition 1.1. The function u ∈ W21,1 (Ω) is called a weak solution of the problem (1)-(4) if u(x, 0) = φ(x) ∈ W21 [0, s0 ] and Z T Z s(t) [aux Φx − bux Φ − cuΦ + ut Φ + f Φ] dx dt 0= 0

Z (9)

+

0

T

[γ(s(t), t)s0 (t) − χ(s(t), t)]Φ(s(t), t) dt +

Z

T

g(t)Φ(0, t) dt 0

0

for arbitrary Φ ∈ W21,1 (Ω) We also need a notion of weak solution from V2 (Ω) of the Neumann problem: Definition 1.2. The function u ∈ V2 (Ω) is called a weak solution of (1)-(4) if Z T Z s(t) Z s0 0= [aux Φx − bux Φ − cuΦ − uΦt + f Φ] dx dt − φ(x)Φ(x, 0) dx+ Z

0 T

0

0

Z g(t)Φ(0, t) dt +

(10) 0

T

[γ(s(t), t)s0 (t) − u(s(t), t)s0 (t) − χ(s(t), t)]Φ(s(t), t) dt

0

for arbitrary Φ ∈ W21,1 (Ω) such that Φ|t=T = 0. Formally, (10) follows from (9) by applying integration by parts to the term ut Φ. By standard methods, one can prove that any weak solution from W21,1 (Ω) is a weak solution from V2 (Ω), and any weak solution from V2 (Ω) is a weak solution from W21,1 (Ω) if its weak derivative ut belongs to L2 (Ω). If u is a weak solution either from V2 (Ω) (or W21,1 (Ω)), then traces u|x=0 and u|x=s(t) are elements of L2 [0, T ], when s ∈ W22 [0, T ] ([27, 23]) and cost functional J (v) is well defined. Furthermore, formulated optimal control problem will be called Problem I. 1.4. Discrete optimal control problem. Let ωτ = {tj = j · τ, j = 0, 1, . . . , n} be a grid on [0, T ] and τ = Tn . Consider a discretized control set n VRn = [v]n = ([s]n , [g]n ) ∈ R2n+2 : 0 < δ ≤ sk ≤ l, o max(k[s]n k2w2 ; k[g]n k2w1 ) ≤ R2 2

2

where, [s]n = (s0 , s1 , ..., sn ) ∈ Rn+1 , [g]n = (g0 , g1 , ..., gn ) ∈ Rn+1 n−1 n−1 n n−1 n X X X X X 2 k[s]n k2w2 = τ s2k + τ s2t,k + τ s2tt,k , k[g]n k2w1 = τ gk2 + τ gt,k . 2

2

k=0

k=1

k=1

k=0

k=1

under the standard notation for the finite differences: sk − sk−1 sk+1 − sk 2 sk+1 − 2sk + sk−1 st,k = , st,k = , stt,k = . τ τ τ2 Introduce two mappings Qn and Pn between continuous and discrete control sets: Qn (v) = [v]n = ([s]n , [g]n ),

for v ∈ VR

where sk = s(tk ), gk = g(tk ), k = 0, 1, ..., n. Pn ([v]n ) = v n = (sn , g n ) ∈ W22 [0, T ] × W21 [0, T ] Inverse Problems and Imaging

for [v]n ∈ VRn ,

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where (11) n

s (t) =

2

t s0 + 2τ st,1 0 ≤ t ≤ τ, sk−1 + (t − tk−1 − τ2 )st,k−1 + 12 (t − tk−1 )2 stt,k−1 , tk−1 ≤ t ≤ tk ,

for k = 2, n and gk − gk−1 (t − tk−1 ), tk−1 ≤ t ≤ tk , k = 1, n. τ Introduce Steklov averages Z Z 1 tk 1 tk dk (x) = d(x, t) dt, hk = h(t) dt, τ tk−1 τ tk−1 g n (t) = gk−1 +

where d stands for any of the functions a, b, c, f , and h stands for any of the functions ν, µ, g or g n . Given v = (s, g) ∈ VR we define Steklov averages of traces Z Z 1 tk 1 tk (12) χks = χ(s(t), t) dt, (γs s0 )k = γ(s(t), t)s0 (t) dt. τ tk−1 τ tk−1 Given [v]n = ([s]n , [g]n ) ∈ VRn we define Steklov averages χksn and (γsn (sn )0 )k through (12) with s replaced by sn from (11). Next we define a discrete state vector through time-discretization of the integral identity (9) Definition 1.3. Given discrete control vector [v]n , the vector function [u([v]n )]n = (u(x; 0), u(x; 1), ..., u(x; n)) is called a discrete state vector if (a) u(x; 0) = φ(x) ∈ W21 [0, s0 ]; (b) For arbitrary k = 1, 2, . . . , n, u(x; k) ∈ W21 [0, sk ] satisfy the integral identity Z sk du(x; k) du(x; k) 0 η (x) − bk η(x) − ck (x)u(x; k)η(x) + fk (x)η(x) ak (x) dx dx 0 + ut (x; k)η(x) dx + (γsn (sn )0 )k − χksn η(sk ) + gkn η(0) = 0, (13) for arbitrary η ∈ W21 [0, sk ], where ut (x; k) =

u(x; k) − u(x; k − 1) . τ

(c) For arbitrary k = 0, 1, ..., n, u(x; k) ∈ W21 [0, sk ] iteratively continued to [0, l] as (14)

u(x; k) = u(2n sk − x; k), 2n−1 sk ≤ x ≤ 2n sk , n = 1, nk , nk ≤ N h i where N = 1 + log2 δl and [r] means integer part of r.

Consider a discrete optimal control problem of minimization of the cost functional n n 2 2 X X (15) In ([v]n ) = β0 τ u(0; k) − νk + β1 τ u(sk ; k) − µk k=1

k=1

on a set VRn subject to the state vector defined in Definition 1.3. Furthermore, formulated discrete optimal control problem will be called Problem In . Inverse Problems and Imaging

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Throughout we use piecewise constant and piecewise linear interpolations of the discrete state vector: given discrete state vector [u([v]n )]n = (u(x; 0), u(x; 1), ..., u(x; n)), let uτ (x, t) = u(x; k), if tk−1 < t ≤ tk , 0 ≤ x ≤ l, k = 0, n, u ˆτ (x, t) = u(x; k − 1) + ut (x; k)(t − tk−1 ), τ

u ˆ (x, t) = u(x; n),

if tk−1 < t ≤ tk , 0 ≤ x ≤ l, k = 1, n,

if t ≥ T, 0 ≤ x ≤ l.

Obviously, we have uτ ∈ V2 (D), u ˆτ ∈ W21,1 (D) 1.5. Formulation of the main result. Let D = {(x, t) : 0 < x < l, 0 < t ≤ T } Throughout the whole paper we assume the following conditions are satisfied by the data: a, b, c ∈ L∞ (D), f ∈ L2 (D), φ ∈ W21 [0, s0 ], γ, χ ∈ W21,1 (D), µ, ν ∈ L2 [0, T ], the coefficient a satisfies (6) almost everywhere on D, the generalized derivatives ∂a ∂a ∂t , ∂x exists and Z T ∂a ∂a ∈ L∞ (D), (16) esssup0≤x≤l dt < +∞. ∂x ∂t 0 Our main theorems read: Theorem 1.4. The Problem I has a solution, i.e. V∗ = {v ∈ VR : J (v) = J∗ ≡ inf J (v)} = 6 ∅ v∈VR

Theorem 1.5. Sequence of discrete optimal control problems In approximates the optimal control problem I with respect to functional, i.e. lim In∗ = J∗ ,

(17)

n→+∞

where In∗ = infn In ([v]n ), n = 1, 2, ... VR

If [v]nε ∈

VRn

is chosen such that In∗ ≤ In ([v]nε ) ≤ In∗ + εn , εn ↓ 0, n

then the sequence v = (sn , g n ) = Pn ([v]nε ) converges to some element v∗ = (s∗ , g∗ ) ∈ V∗ weakly in W22 [0, T ] × W21 [0, T ], and strongly in W21 [0, T ] × L2 [0, T ]. In particular sn converges to s∗ uniformly on [0, T ]. Moreover, piecewise linear interpolation u ˆτ of the discrete state vector [u[v]nε ]n converges to the solution 1,1 u(x, t; v∗ ) ∈ W2 (Ω∗ ) of the Neumann problem (1)-(4) weakly in W21,1 (Ω∗ ). Strategy of proofs. Proofs of both Theorem 1.4 and Theorem 1.5 are significantly based on two energy estimates for the discrete state vector. In Theorem 3.1 we prove the first energy estimate. By using it in Theorem 3.4 we prove that the sequence of interpolated solutions is weakly compact in W21,0 (Ω) and the weak limit is the solution u ∈ V21,0 (Ω) of the problem (1)-(4). In Theorem 3.5 the second energy estimate is proved, which allows us to prove in Theorem 3.8 that the sequence of interpolated solutins is weakly compact in W21,1 (Ω) and the weak limit Inverse Problems and Imaging

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is the solution u ∈ W21,1 (Ω) of the problem (1)-(4). By applying the first and second energy estimates, it is proved in Section 3.3 that functional J (v) is weakly continuous in W22 [0, T ] × W21 [0, T ]. Since VR is weakly compact, the assertion of Theorem 1.4 follows from the Weierstrass theorem in weak topology. The proof of Theorem 1.5 is more involved and requires more delicate analysis of the relation between discrete and continuous functionals and state vectors. To make it easy for the reader, we formulate in Lemma 2.2 a known necessary and sufficient condition for convergence. By applying the second energy estimate, it is proved in Section 3.4 that the conditions of the general criteria are satisfied. 2. Preliminary results. In a Lemma 2.1 below we prove existence and uniqueness of the discrete state vector [u([v]n )]n (see Definition 1.3) for arbitrary discrete control vector [v]n ∈ VRn . In a Lemma 2.2 we remind a general approximation criteria for the optimal control problems from ([36]). In a Lemma 2.3 we prove some properties of the mappings Qn and Pn between continuous and discrete control sets. Lemma 2.1. For sufficiently small time step τ , there exists a unique discrete state vector [u([v]n )]n for arbitrary discrete control vector [v]n ∈ VRn . Proof. To prove uniqueness, it is enough to show that if u(x; k − 1) ≡ 0, (γsn (sn )0 )k = 0, χksn = 0, gkn = 0, fk (x) ≡ 0 then u(x; k) which solves (13) vanishes identically. Under these assumptions by choosing η(x) = u(x; k) in (13) we have (18) Z sk du(x; k) 2 du(x; k) 1 − bk u(x; k) − ck (x)u2 (x; k) + u2 (x; k) dx = 0. ak (x) dx dx τ 0 Using (6) and Cauchy inequality with ε > 0 we derive that Z sk Z du(x; k) 2 1 sk 2 a0 dx + u (x; k) dx ≤ dx τ 0 Z sk0 M Z sk du(x; k) 2 εM dx + +M (19) u2 (x; k) dx, 2 0 dx 2ε 0 where M = max ||a||L∞ (D) ; ||b||L∞ (D) ; ||c||L∞ (D) . By choosing ε = a0 /M in (19) we have Z Z 1 1 sk 2 a0 sk du(x; k) 2 dx + − u (x; k) dx ≤ 0, (20) 2 0 dx τ τ0 0 where τ0 =

M2

+M

−1

. 2a0 From (20) it follows that u(x; k) ≡ 0 if τ < τ0 . To prove an existence we apply Galerkin method. Consider an approximate solution N X uN (x) = di ψi (x) i=1 Inverse Problems and Imaging

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where {ψi } is a fundamental system in W21 [0, sk ] and the coefficients {di } solve the following system Z sk h duN 0 duN ak (x) ψi (x) − bk (x) ψi (x) − ck (x)uN (x)ψi (x) dx dx 0 Z i 1 1 sk u(x; k − 1)ψi (x) dx + uN (x)ψi (x) + fk (x)ψi (x) dx = τ τ 0 − (γsn (sn )0 )k − χksn ψi (sk ) − gkn ψi (0), i = 1, . . . , N (21) which is equivalent to N Z sk h X ak (x)ψj0 (x)ψi0 (x) − bk (x)ψj0 (x)ψi (x) − ck (x)ψj (x)ψi (x) 0

j=1

(22)

Z sk h i i 1 1 − fk (x)ψi (x) + u(x; k − 1)ψi (x) dx + ψj (x)ψi (x) dx dj = τ τ 0 n 0 k k n n − (γs (s ) ) − χsn ψi (sk ) − gk ψi (0), i = 1, . . . , N.

Homogeneous system corresponding to (22) is N Z sk h X ak (x)ψj0 (x)ψi0 (x) − bk (x)ψj0 (x)ψi (x) − ck (x)ψj (x)ψi (x) 0

j=1

i 1 + ψj (x)ψi (x) dx dj = 0, i = 1, . . . , N τ Let us multiply each equation in (23) by di and add with respect to i: 2 Z sk h duN (x) duN (x) uN (x)dx ak (x) − bk (x) dx dx 0 i 1 −ck (x)u2N (x) + u2N (x)) dx = 0 (24) τ As before, from (24) it follows that uN ≡ 0, and therefore the homogeneous system (23) has only the trivial solution. This proves the uniqueness of the approximate solution uN (x). Let us now prove uniform estimation of the sequence {uN (x)}. Multiply (21) by di and add with respect to i = 1, . . . , N : Z sk h du 2 duN N uN (x) − ck (x)u2N (x) ak (x) − bk (x) dx dx 0 Z i 1 2 1 sk + uN (x) + fk (x)uN (x) dx = u(x; k − 1)uN (x) dx τ τ 0 (25) − (γsn (sn )0 )k − χksn uN (sk ) + gkn uN (0). (23)

We estimate the four integrals on the left-hand side of (25) as we did before to prove (20) and derive Z Z sk h a0 sk duN (x) 2 1 u2N (x) dx ≤ |gkn ||uN (0)| + |(γsn (sn )0 )k | dx + 2 0 dx 2τ 0 Z sk i 1 (26) +|χksn | |uN (sk )| + |fk (x)| + |u(x; k − 1)| |uN (x)| dx τ 0 for all τ ≤ (27)

τ0 2 .

By Morrey’s inequality we have

max{|uN (0)|; |uN (sk )|} ≤ kuN kC[0,sk ] ≤ CkuN kW21 [0,sk ] ,


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where the constant C is independent of N and τ . By using Cauchy inequalities with appropriately chosen ε > 0, from (26) and (27) it easily follows that h kuN k2W 1 [0,sk ] ≤ C ku(x; k − 1)k2L2 [0,sk ] + kfk k2L2 (0,sk ) 2 i 2 + (γsn (sn )0 )k + |χksn |2 + |gkn |2 , (28) where C does not depend on N , but depends on the time step τ . From (28) it follows that {uN } is weakly compact in W21 [0, sk ]. Let v(x) be its weak limit point in W21 [0, sk ]. Passing to the limit in (21) it follows that v(x) satisfies (13) for η(x) = ψi (x). Since {ψi } is a fundamental system in W21 [0, sk ], it follows that v(x) satisfies (13) for every η(x) ∈ W21 [0, sk ]. Hence v(x) = u(x; k) is a solution of (13) and in view of uniqueness the whole sequence uN converges weakly in W21 [0, sk ] to u(x; k). Lemma is proved. The following known criteria will be used in the proof of Theorem 1.5. Lemma 2.2. [36] Sequence of discrete optimal control problems In approximates the continuous optimal control problem I if and only if the following conditions are satisfied: (1) for arbitrary sufficiently small ε > 0 there exists number N1 = N1 (ε) such that QN (v) ∈ VRn for all v ∈ VR−ε and N ≥ N1 ; and for any fixed ε > 0 and for all v ∈ VR−ε the following inequality is satisfied: (29) lim sup IN (QN (v)) − J (v) ≤ 0. N →∞

(2) for arbitrary sufficiently small ε > 0 there exists number N2 = N2 (ε) such that PN ([v]N ) ∈ VR+ε for all [v]N ∈ VRN and N ≥ N2 ; and for all [v]N ∈ VRN , N ≥ 1 the following inequality is satisfied: (30) lim sup J (PN ([v]N )) − IN ([v]N ) ≤ 0. N →∞

(3) the following inequalities are satisfied: lim sup J∗ (ε) ≥ J∗ , lim inf J∗ (−ε) ≤ J∗ ,

(31)

ε→0

ε→0

where J∗ (±ε) = inf J (u). VR±ε

In the next lemma we prove that the mappings Qn and Pn introduced in Section 1.4 satisfy the conditions of Lemma 2.2. Lemma 2.3. For arbitrary sufficiently small ε > 0 there exists nε such that (32)

Qn (v) ∈ VRn ,

(33)

Pn ([v]n ) ∈ VR+ε ,

for all v ∈ VR−ε

and n > nε .

for all [v]n ∈ VRn

and n > nε .

Proof. Let 0 < ε nε =

h

RT ε

i

+ 1. Hence, (32) is proved.

Let us know choose [v]n ∈ VRn . We simplify the notation and assume v = (s, g) = Pn ([v]n ). Through direct calculations we derive (38)

ksk2W 2 [0,T ] ≤

n−1 X

2

τ s2k +

k=0

n−1 X

τ s2t,k +

n−1 X

1 2 1 τ s2tt,k + τ st,1 + s2t,1 + Cτ, 3 τ

k=1

k=1

where C is independent of τ . By using CBS inequality we have 2 τ st,1

Zτ ≤

1 2 1 |s (t)| dt, st,1 = 3 τ τ 0

2

0

(39)

Zτ Zt 0

1 2τ

Zτ Zt 0

1 2

|s00 (ξ)|2 dξdt ≤

Zτ

0

2 s00 (ξ)dξdt ≤

0

|s00 (t)|2 dt.

0

Since [v]n ∈ VRn , from (38),(39) it follows that ksk2W 2 [0,T ] ≤ C1 ,

(40)

2

where C1 is independent of τ . This implies that lim kskW22 [0,τ ] = 0.

(41)

τ →0

In a similar way we calculate (42)

kgk2W 1 [0,T ] ≤

n−1 X

2

k=0

τ gk2 +

n X

2 τ gt,k + Cτ.

k=1

Hence, from (38),(39) and (42) it follows that max ksk2W 2 [0,T ] , kgk2W 1 [0,T ] ≤ max k[s]n k2w2 , k[g]n k2w1 2

2

2

2

1 1 (43) +Cτ + ks0 k2W 1 [0,τ ] ≤ R2 + Cτ + ks0 k2W 1 [0,τ ] , 2 2 2 2 From (41) it follows that given ε > 0 we can choose nε such that for any n > nε 1 (44) R2 + Cτ + ks0 k2W 1 [0,τ ] ≤ (R + ε)2 . 2 2 Inverse Problems and Imaging

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Ugur G. Abdulla

From (43) and (44), (33) follows. Lemma is proved. Corollary 1. Let either [v]n ∈

VRn

or [v]n = Qn (v) for v ∈ VR . Then

|sk − sk−1 | ≤ Cτ, k = 1, 2, · · · , n

(45)

where C is independent of n. Indeed, if v ∈ VR , then s0 ∈ W21 [0, T ] and by Morrey inequality ks0 kC[0,T ] ≤ C1 ks0 kW21 [0,T ] ≤ C1 R

(46)

and hence for the first component [s]n of [v]n = Qn (v) we have (45). Also, if [v]n ∈ VRn , then the sequence v n = Pn ([v]n ) belongs to VR+1 by Lemma 2.3 and the component sn of v n satisfies (46). Since, sk + sk−1 sn (0) = s0 , sn (tk ) = , k = 1, · · · , n 2 from (46), (45) easily follows. 3. Proofs of the main results. 3.1. First energy estimate and its consequences. The main goal of this section to prove the following energy estimation for the discrete state vector. Theorem 3.1. For all sufficiently small τ discrete state vector [u([v]n )]n satisfies the following stability estimations: Z l n Z l X du(x; k) 2 max u2 (x; k) dx + τ dx ≤ 0≤k≤n 0 dx k=1 0 C kφk2L2 (0,s0 ) + kg n k2L2 (0,T ) + kf k2L2 (D) + kγ(sn (t), t)(sn )0 (t)k2L2 (0,T ) Z sk+1 n−1 X n 2 +kχ(s (t), t)kL2 (0,T ) + 1+ (sk+1 − sk ) (47) u2 (x; k)dx , sk

k=1

Z max

0≤k≤n

C (48)

l

u2 (x; k) dx + τ

0

n Z l n Z l X X du(x; k) 2 ut2 (x; k)dx ≤ dx + τ 2 dx 0 0 k=1

k=0

+ kγ(s (t), t)(sn )0 (t)k2L2 (0,T ) Z sk+1 n−1 X +kχ(sn (t), t)k2L2 (0,T ) + 1+ (sk+1 − sk ) u2 (x; k)dx ,

kφk2W 1 (0,s0 ) 2

+

kg n k2L2 (0,T )

+

kf k2L2 (D)

k=1

n

sk

where C is independent of τ and 1+ be an indicator function of the positive semiaxis. We split the proof into two Lemmas. Lemma 3.2. For all sufficiently small τ , discrete state vector [u([v]n )]n satisfies the following estimation: Z sk n Z sk n Z sk X X du(x; k) 2 2 2 ut2 (x; k)dx ≤ u (x; k) dx + τ max dx + τ 1≤k≤n 0 dx 0 0 k=1 k=1 2 n 2 2 n C kφkL2 (0,s0 ) + kg kL2 (0,T ) + kf kL2 (D) + kγ(s (t), t)(sn )0 (t)k2L2 (0,T ) Z sk+1 n−1 X 1+ (sk+1 − sk ) (49) +kχ(sn (t), t)k2L2 (0,T ) + u2 (x; k)dx , k=1 Inverse Problems and Imaging

sk

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319

where C is independent of τ . Proof. By choosing η(x) = 2τ u(x; k) in (13) and by using the equality 2τ ut (x; k)u(x; k) = u2 (x; k) − u2 (x; k − 1) + τ 2 u2t (x; k) we have Z sk u2 (x; k − 1)dx + τ 2 ut2 (x; k)dx 0 0 0 Z sk h Z sk du(x; k) 2 du(x; k) bk (x) u(x; k) + ck (x)u2 (x; k) +2τ ak (x) dx = 2τ dx dx 0 0 i −fk (x)u(x; k) dx − 2τ (γsn (sn )0 )k − χksn u(sk ; k) − 2τ gkn u(0; k). (50) Z

sk

u2 (x; k)dx −

Z

sk

Using (6), Cauchy inequalities with appropriately chosen ε > 0, and Morrey inequality (27) from (50) we derive that Z sk Z sk Z sk du(x; k) 2 2 2 u (x; k)dx − u (x; k − 1)dx + a0 τ dx dx 0 0 0 Z sk h +τ 2 u2t (x; k)dx ≤ C1 τ |(γsn (sn )0 )k |2 + |χksn |2 0 Z sk Z sk i +|gkn |2 + fk2 (x)dx + (51) u2 (x; k)dx , 0

0

where C1 is independent of τ . Assuming that τ < C1 , from (51) it follows that Z sk Z sk−1 2 (1 − C1 τ ) u (x; k)dx ≤ u2 (x; k − 1)dx 0 Z sk 0 +1+ (sk − sk−1 ) u2 (x; k − 1)dx sk−1

Z n 0 k 2 k 2 n 2 +C1 τ |(γsn (s ) ) | + |χsn | + |gk | +

(52)

sk

fk2 (x)dx

,

0

By induction we have Z sk

2

u (x; k)dx ≤ (1 − C1 τ ) 0

−k

Z

s0

φ2 (x)dx

0

Z k n h X j 2 −k+j−1 n 0 j 2 n 2 + (1 − C1 τ ) C1 τ |(γsn (s ) ) | + |χsn | + |gj | +

i fj2 (x)dx

0

j=1

Z (53)

sj

sj

+1+ (sj − sj−1 )

o u2 (x; j − 1)dx .

sj−1

For arbitrary 1 ≤ j ≤ k ≤ n we have (54)

C1 T −n → eC 1 T , (1 − C1 τ )−k+j−1 ≤ (1 − C1 τ )−k ≤ (1 − C1 τ )−n = 1 − n

as τ → 0. Accordingly for sufficiently small τ we have (55)

(1 − C1 τ )−k+j−1 ≤ 2eC1 T


for 1 ≤ j ≤ k ≤ n, Volume 7, No. 2 (2013), 307–340

320

Ugur G. Abdulla

By applying CBS inequality from (53)-(55) it follows that sk

Z max

1≤k≤n

0

u2 (x; k) dx ≤ C2 kφk2L2 (0,s0 ) + kg n k2L2 (0,T )

+kγ(sn (t), t)(sn )0 (t)k2L2 (0,T ) + kχ(sn (t), t)k2L2 (0,T ) Z sk+1 n−1 X 2 +kf kL2 (D) + 1+ (sk+1 − sk ) u2 (x; k)dx .

(56)

sk

k=1

where C2 is independent of τ . Having (56), we perform summation of (51) with respect to k from 1 to n and derive sn

Z

u2 (x; n) dx + a0 τ

0

n Z X k=1

2kφk2L2 (0,s0 )

sk

0

n Z sk du(x; k) 2 X ut2 (x; k)dx ≤ dx + τ 2 dx 0 k=1

+ kγ(s (t), t)(sn )0 (t)k2L2 (0,T ) n Z sk X n 2 +kχ(s (t), t)kL2 (0,T ) + τ u2 (x; k)dx

+

C1 kg n k2L2 (0,T )

+

kf k2L2 (D)

k=1

(57)

+

n−1 X

Z

n

0

sk+1

u2 (x; k)dx,

1+ (sk+1 − sk ) sk

k=1

From (56) and (57), (49) follows. Lemma is proved. In the next lemma we prove a nice property of the extension introduced in the Definition 1.3, which allows to extend the estimation (49) to (47) and (48). Lemma 3.3. Given discrete control vector [v]n ∈ VRn , a discrete state vector [u([v]n )]n satisfies the inequalty l

Z max

1≤k≤n

n Z l n Z l X X du(x; k) 2 ut2 (x; k)dx ≤ u (x; k) dx + τ dx + τ dx 0 0 2

0

k=1

k=0

(58) C

Z max

1≤k≤n

sk

u2 (x; k) dx + τ

0

n Z X k=0

0

sk

n Z sk du(x; k) 2 X ut2 (x; k)dx , dx + τ dx 0 k=1

where C is independent of τ . Proof. By induction it follows that the first two terms on the left hand side are estimated by the first two terms on the right hand side with the constant C = 2N , where N is defined in (14). Define a family of functions {˜ u(y; k), k = 0, ..., n} as u ˜(y; 0) = φ(ys0 ), u ˜(y; k) = u(ysk ; k), 0 ≤ y ≤ 1, k = 1, ..., n. As before, assume they are all continued by induction to semiaxis {y ≥ 0} as u ˜(y; k) = u ˜(2n − y; k), Inverse Problems and Imaging

for 2n−1 ≤ y ≤ 2n . Volume 7, No. 2 (2013), 307–340


321

We have N

2Z sk Zl n n X X 2 u2t (x; k)dx = τ ut (x; k)dx ≤ τ k=1

k=1

0

0

N

n X

2Z sk

τ

k=1

hu ˜(x/sk ; k) − u ˜(x/sk−1 ; k − 1) i2 dx = τ

0 N

Z2 h n X u ˜(y; k) − u ˜(ysk /sk−1 ; k − 1) i2 dy ≤ I1 + I2 τ sk τ

(59)

k=1

0

where N

I1 = 2

n X k=1

Z2 h u ˜(y; k) − u ˜(y; k − 1) i2 dy = · · · = τ sk τ 0 N +1

2

n X

Z1 τ sk

k=1

u ˜t2 (y; k)dy =

0

Zsk h n X u(x; k) − u(xsk−1 /sk ; k − 1) i2 N +1 2 τ dx ≤ τ k=1

0

2

N +2

Zsk n X τ u2t (x; k)dx+ k=1

2N +2

(60)

n X

τ

k=1

Zsk h

0

u(x; k − 1) − u(xsk−1 /sk ; k − 1) i2 dx τ

0

N

(61)

I2 = 2

n X k=1

Z2 h u ˜(y; k − 1) − u ˜(ysk /sk−1 ; k − 1) i2 dy. τ sk τ 0

By using CBS inequality, Fubini’s theorem and Corollary 1 we have Zsk h n X u(x; k − 1) − u(xsk−1 /sk ; k − 1) i2 τ dx = τ

k=1

(62)

0

Zsk Zx n X 1 τ

k=1

0

x

sk−1 sk


n−1 Z l du(ξ; k − 1) 2 C12 l X du(x; k) 2 dξ dx ≤ τ dx, dξ δ dx k=0

0

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322

Ugur G. Abdulla N

N Z2 n u(x; k − 1) 2 22N +1 C12 N 2 X d˜ τ I2 ≤ dx = δ dx k=1

0

n−1 Z1 23N +1 C12 N 3 X u(x; k) 2 d˜ τ dx ≤ δ dx k=0

0

3 n−1 X

23N +1 C12 N l δ

(63)

k=0

τ

Zsk du(x; k) 2 dx dx 0

Hence, from (59)-(63) it follows that

(64)

sk Zl Zsk n n n−1 X X Z du(x; k) 2 X 2 τ ut (x; k)dx ≤ C τ τ u2t (x; k)dx dx + dx

k=1

k=0

0

0

k=1

0

where C is independent of τ . From (64),(58) follows. Lemma is proved. It can be easily seen that Theorem 3.1 follows from Lemma 3.2 and Lemma 3.3. Let [v]n ∈ VRn , n = 1, 2, ... be a sequence of discrete controls. From Lemma 2.3 it follows that the sequence {Pn ([v]n )} is weakly precompact in W22 [0, T ] × W21 [0, T ]. Assume that the whole sequence converges to v = (s, g) weakly in W22 [0, T ] × W21 [0, T ]. This implies the strong convegence in W21 [0, T ] × L2 [0, T ]. Conversely, given control v = (s, g) ∈ VR we can choose sequence of discrete controls [v]n = Qn (v). Appplying Lemma 2.3 twice one can easily establish that the sequence {Pn ([v]n } converges to v = (s, g) weakly in W22 [0, T ] × W21 [0, T ], and strongly in W21 [0, T ] × L2 [0, T ]. In the next theorem we prove the continuous dependence of the family of interpolarions {uτ } on this convergence. Theorem 3.4. Let [v]n ∈ VRn , n = 1, 2, ... be a sequence of discrete controls and the sequence {Pn ([v]n } converges strongly in W21 [0, T ] × L2 [0, T ] to v = (s, g). Then the sequence {uτ } converges as τ → 0 weakly in W21,0 (Ω) to weak solution u ∈ V21,0 (Ω) of the problem (1)-(4), i.e. to the solution of the integral identity (10). Moreover, u satisfies the energy estimate (65) kuk2V 1,0 (D) ≤ C kφk2L2 (0,s0 ) +kgk2L2 (0,T ) +kf k2L2 (D) +kγk2W 1,0 (D) +kχk2W 1,0 (D) 2

2

2

Proof. In addition to quadratic interpolation of [s]n from (11), consider two linear interpolations: sñ (t) = sk−1 +

sk − sk−1 (t − tk−1 ), tk−1 ≤ t ≤ tk , k = 1, n; sñ (t) ≡ sn , t ≥ T ; τ sñ1 (t) = sñ (t + τ ), 0 ≤ t ≤ T.

It can be easily proved that both sequences sñ and sñ1 are equivalent to the sequence sn in W21 [0, T ] and converge to s strongly in W21 [0, T ]. In particular, (66)

sup k˜ sn1 kW21 [0,T ] < C∗ n


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323

where C∗ is independent of n. We estimate the last term on the right-hand side of (47) as follows: Z sk+1 n−1 X 1+ (sk+1 − sk ) u2 (x; k)dx = sk

k=1 n−1 X

tk

Z

tk+1

1+ (sk+1 − sk )

2 0 (˜ sn ) (t) uτ (˜ sn (t), t − τ ) dt =

tk

k=1 n−1 X

(67)

0

(˜ sn ) (t)u2 (˜ sn (t); k)dt =

1+ (sk+1 − sk )

k=1 n−1 X

tk+1

Z

Z

tk

1+ (sk+1 − sk )

2 0 (˜ sn1 ) (t) uτ (˜ sn1 (t), t) dt.

tk−1

k=1

By applying CBS inequality we have Z sk+1 n−1 X 1+ (sk+1 − sk ) (68) u2 (x; k)dx ≤ k(˜ sn1 )0 kL2 [0,T ] kuτ (˜ sn1 (t), t)k2L4 [0,T ] . sk

k=1

From the results on traces of the elements of space V2 (D) ([23, 4, 27]) it follows that for arbitrary u ∈ V2 (D) the following inequality is valid ˜ (69) ku(˜ sn1 (t), t)kL [0,T ] ≤ Ckuk V (D) , 4

2

with the constant C˜ being independent of u as well as n. From (66),(68) and (69) it follows that Z sk+1 n−1 X ˜ τ k2 1+ (sk+1 − sk ) (70) u2 (x; k)dx ≤ C∗ Cku V2 (D) . sk

k=1

If the constant C∗ from (66) satisfies the condition ˜ −1 (71) C∗ < (C C) then from (47) and (70) it follows that kuτ k2V2 (D) ≤ C kφk2L2 (0,s0 ) + kg n k2L2 (0,T ) + kf k2L2 (D) + (72) kγ(sn (t), t)(sn )0 (t)k2L2 (0,T ) + kχ(sn (t), t)k2L2 (0,T ) , where C is another constant independent of n. By applying the results on the traces of elements of W21,0 (D) ([4, 27]) on smooth curve x = sn (t), Morrey inequality for (sn )0 and (33) we have kγ(sn (t), t)(sn )0 (t)kL2 (0,T ) ≤ k(sn )0 kC[0,T ] kγ(sn (t), t)kL2 [0,T ] ≤ C3 kγkW 1,0 (D) 2

(73)

kχ(sn (t), t)kL2 [0,T ] ≤ C3 kχkW 1,0 (D) , 2

where C3 is independent of γ, χ and n. Hence, from (72),(73) and strong convergence of g n to g in L2 [0, T ] it follows the estimation (74) kuτ k2V2 (D) ≤ C kφk2L2 (0,s0 ) + kgk2L2 (0,T ) + kf k2L2 (D) + kγk2W 1,0 (D) + kχk2W 1,0 (D) , 2

2

with C being independent of n. If (71) is not satisfied, then we can partition [0, T ] into finitely many segments [tnj−1 , tnj ], j = 1, q with tn0 = 0, tnq = T in such a way that by replacing [0, T ] Inverse Problems and Imaging

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324

Ugur G. Abdulla

with any of the subsegments [tnj−1 , tnj ] (66) will be satisfied with C∗ small enough to obey (71). Hence, we divide D into finitely many subsets Dj = D ∩ {tnj−1 < t ≤ tnj } such that every norm kuτ k2V2 (Dj ) is uniformly bounded through the right-hand side of (74). Summation with j = 1, . . . , q implies (74). From (74) it follows that the sequence {uτ } is weakly precompact in W21,0 (D). Let u ∈ W21,0 (D) be a weak limit point of uτ in W21,0 (D), and assume that whole sequence {uτ } converges to u weakly in W21,0 (D). Let us prove that in fact u satisfies the integral identity (10) for arbitrary test function Φ ∈ W21,1 (Ω) such that Φ|t=T = 0. Due to density of C 1 (Ω) in W21,1 (Ω) it is enough to assume Φ ∈ C 1 (Ω). Without loss of generality we can also assume that Φ ∈ C 1 (DT +τ ), Φ ≡ 0, for T ≤ t ≤ T + τ , where DT +τ = {(x, t) : 0 < x < l, 0 < t ≤ T + τ } Otherwise, we can continue Φ to DT +τ with the described properties. Let Φ(x; k) = Φ(x, kτ ),

Φt (x; k) =

Φ(x; k + 1) − Φ(x; k) τ

As before, we construct piecewise constant interpolations Φτ , Φτt . Obviously, the τ ∂Φ τ sequences {Φτ }, { ∂Φ ∂x } and {Φt } converge as τ → 0 uniformly in D to Φ, ∂x and ∂Φ ∂t respectively. By choosing in (13) η(x) = τ Φ(x; k), after summation with respect to k = 1, n and transformation of the time difference term as follows

τ

n Z X k=1

sk

ut¯(x; k)Φ(x; k) dx = −τ

0 s1

φ(x)Φ(x; 1) dx − 0 n Z X

tk

Z

tk−1

k=1

n−1 X Z tk+1 k=1

Z

T

sk+1

uτ Φτt dx dt −

s1

Z

0

φ(x)Φτ (x, τ ) dx−

0

(˜ sn )0 (t)uτ (˜ sn (t), t − τ )Φτ (˜ sn (t), t − τ ) dt =

Z

0

(75)

u(x; k)Φ(x; k) dx =

sk

tk

− Z

n−1 X Z sk+1 k=1

u(x; k)Φt (x; k) dx−

0

k=1

Z

−

n−1 X Z sk+1

0

s(t)

uτ Φτt dx dt −

Z

s1

φ(x)Φτ (x, τ ) dx−

0

T −τ

(˜ sn1 )0 (t)uτ ((˜ sn1 )(t), t)Φτ ((˜ sn1 )(t), t) dt −

0

n−1 X Z tk k=1

tk−1

Z

sk+1

uτ Φτt dx dt

s(t)

we derive that Inverse Problems and Imaging

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T

Z

s(t)

Z 0

0

∂uτ ∂Φτ ∂uτ τ a −b Φ − cuτ Φτ + f Φτ − uτ Φτt ∂x ∂x ∂x Z s0 Z T φ(x)Φτ (x, τ ) dx + g n (t)Φτ (0, t) dt−

0

325

dx dt−

0 T −τ

Z

(˜ sn1 )0 (t)uτ ((˜ sn1 )(t), t)Φτ ((˜ sn1 )(t), t) dt+

0 T

Z (76)

h

i γ(sn (t), t)(sn )0 (t) − χ(sn (t), t)) Φτ (sn (t), t) dt − R = 0

0

where n Z X

tk

Z

k=1 tk

Z

sk

∂uτ ∂Φτ ∂uτ τ τ τ τ R= a −b Φ − cu Φ + f Φ dx dt− ∂x ∂x ∂x k=1 tk−1 s(t) Z s1 n−1 X Z tk Z sk+1 uτ Φτt dx dt + φ(x)Φτ (x, τ ) dx +

n Z X

tk−1

k=1

Let ∆=

n [

tk−1

s(t)

s0

sk

h i ∂Φτ dx dt+ γ(sn (t), t)(sn )0 (t) − χ(sn (t), t)) ∂x sn (t)

{(x, t) : tk−1 < t < tk , min(s(t), sk ) < x < max(s(t), sk )}

k=1

|∆| denotes the Lebesgue measure of ∆. Since √ n Z tk Z tk X 2 T 0 0 |∆| ≤ ks kL2 (0,T ) τ → 0 |s (τ )| dτ dt ≤ 3 tk−1 t

as τ → 0

k=1

and all of the integrands are uniformly bounded in L1 (D), it follows that the first term in the expression of R converges to zero as τ → 0. In a similar way one can see that the second and fourth terms also converge to zero as τ → 0. The third term in the expression of R converges to zero due to Corollary 1 and uniform convergence of {Φτ } in D. Hence, we have (77)

lim R = 0

τ →0

Due to weak convergence of uτ to u in W21,0 (D) and uniform convergence of the τ ∂Φ ∂Φ τ sequences {Φτ }, { ∂Φ ∂x } and {Φt } to Φ, ∂x and ∂t respectively, passing to limit as τ → 0, it follows that first, second and third integrals on the left-hand side of (76) converge to similar integrals with uτ , Φτ , Φτt , Φτ (x, τ ) and Φτ (0, t) replaced n by u,Φ, ∂Φ ∂t , Φ(x, 0) and Φ(0, t) respectively. Since s converges to s strongly in 1 n n W2 [0, T ], the traces γ(s (t), (t)), χ(s (t), t) converge strongly in L2 [0, T ] to traces γ(s(t), (t)), χ(s(t), t) respectively. Since Φτ (sn (t), t) converge uniformly on [0, T ] to Φ(s(t), t), passing to the limit as τ → 0, the last integral on the left-hand side of (76) converge to similar integral with sn and Φτ replaced by s and Φ. It only remains to prove that Z T −τ Z T (78) lim (˜ sn1 )0 (t)uτ (˜ sn1 (t), t)Φτ (˜ sn1 (t), t) dt = s0 (t)u(s(t), t)Φ(s(t), t) τ →0

0


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Ugur G. Abdulla

Since {˜ sn1 } converges to s strongly in W21 [0, T ], from (74) it follows that {uτ (˜ sn1 (t), t)} is uniformly bounded in L2 [0, T ] and kuτ (˜ sn1 (t), t) − uτ (s(t), t)kL2 [0,T ] → 0

(79)

as τ → 0

Since {uτ } converges to u weakly in W21,0 (D), it follows that uτ (s(t), t) → u(s(t), t),

(80)

weakly in L2 [0, T ]

Since {Φτ (˜ sn1 (t), t)} converges to Φ(s(t), t) uniformly in [0, T ], from (79),(80), (78) easily follows. Passing to the limit as τ → 0, from (76) it follows that u satisfies integral identity (10), i.e. it is a weak solution of the problem (1)-(4). Since this solution is unique ([23]) it follows that indeed the whole sequence {uτ } converges to u ∈ V21,0 (Ω) weakly in W21,0 (Ω). From the property of weak convergence and (74),(65) follows. Theorem is proved. In particular, Theorem 3.4 implies the following well-known existence result ([23]): Corollary 2. For arbitrary v = (s, g) ∈ VR there exists a weak solution u ∈ V21,0 (Ω) of the problem (1)-(4) which satisfy the energy estimate (65) Remark. All the proofs in this section are performed by using only assumptions φ ∈ L2 [0, l], γ, χ ∈ W21,0 (D), a ∈ L∞ (D), and (6) instead of conditions imposed in Section 1.5. 3.2. Second energy estimate and its consequences. The main goal of this section to prove the following energy estimation for the discrete state vector. Theorem 3.5. For all sufficiently small τ discrete state vector [u([v]n )]n satisfies the following stability estimation: Z l n Z l X du(x; k) 2 dx + τ max u2t¯ (x; k) dx ≤ 1≤k≤n 0 dx 0 k=1 2 2 2 2 C kφkW 1 [0,l] + kg n k 41 + kf kL2 (D) + kγ(sn (t), t)(sn )0 (t)k 41 + 2

(81)

W2 [0,T ]

W2 [0,T ]

n

2

kχ(s (t), t)k

1

W24 [0,T ]

+

n−1 X

Z

sk+1

1+ (sk+1 − sk )

k=1

u (x; k)dx , 2

sk

We split the proof into two lemmas. Lemma 3.6. Let given discrete control vector [v]n , along with discrete state vector [u([v]n )]n , the vector function [˜ u([v]n )]n = (˜ u(x; 0), u ˜(x; 1), ..., u ˜(x; n)) is defined as u ˜(x; k) = Inverse Problems and Imaging

u(x; k) 0 ≤ x ≤ sk , u(sk ; k) sk ≤ x ≤ l, k = 0, n. Volume 7, No. 2 (2013), 307–340


327

Then for all sufficiently small τ , [˜ u([v]n )]n satisfies the following estimation: 2 m Z sk X d˜ u(x; k) u ˜2t¯ (x; k) dx+ max dx dx + τ 1≤k≤n 0 0 k=1 2 m Z sk X d˜ u (x; k) 2 2 τ2 dx ≤ C kφkW 1 [0,l] + kg n k 41 + 2 W2 [0,T ] dx ¯ 0 t Z

sk

k=1

2

kγ(sn (t), t)(sn )0 (t)k

2

1

W24 [0,T ]

2 kf kL2 (D)

(82)

+

n−1 X

+ kχ(sn (t), t)k

1

W24 [0,T ]

Z

sk+1

1+ (sk+1 − sk )

+

u (x; k)dx , 2

sk

k=1

Proof. By choosing η(x) = 2τ u ˜t (x; k) in (13) and by using the following identity 2τ ak (x)

2 2 d˜ u(x; k) d˜ u(x; k) d˜ u(x; k − 1) = ak (x) − ak−1 (x) dx dx dx t¯ 2 2 d˜ u(x; k) d˜ u(x; k − 1) + τ 2 ak (x) , −τ akt¯(x) dx dx t¯

d˜ u(x; k) dx

(83)

we have 2 d˜ u(x; k − 1) dx − dx+ ak−1 (x) ak (x) dx 0 0 2 Z sk Z sk d˜ u(x; k) 2τ ≤ (˜ ut¯(x; k))2 dx + τ 2 ak (x) dx 0 0 t¯ 2 Z sk Z sk d˜ u(x; k − 1) d˜ u(x; k) τ dx + 2τ u ˜t¯(x; k) dx+ akt¯(x) bk (x) dx dx 0 0 Z sk Z sk 2τ ck (x)˜ u(x; k)˜ ut¯(x; k) dx − 2τ fk (x)ut¯(x; k) dx− 0 0 ˜t¯(0; k) 2τ (γsn (sn )0 )k − χksn u ˜t¯(sk ; k) − 2τ gkn u Z

(84)

sk

d˜ u(x; k) dx

2

Z

sk−1

By adding inequalities (84) with respect to k from 1 to arbitrary m ≤ n we derive 2 m Z sk X d˜ u(x; m) am (x) u ˜2t¯ (x; k) dx+ dx + 2τ dx 0 0 k=1 2 2 m Z sk m Z sk X X d˜ u(x; k − 1) d˜ u (x; k) 2 τ ak (x) dx ≤ τ akt¯(x) dx+ dx dx t¯ k=1 0 k=1 0 m Z sk m Z sk X X d˜ u(x; k) 2τ bk (x) u ˜t¯(x; k) dx + 2τ ck (x)˜ u(x; k)˜ ut¯(x; k) dx− dx k=1 0 k=1 0 2 Z s0 m Z sk X dφ 2τ fk (x)˜ ut¯(x; k) dx + a0 (x) dx− dx 0 0 Z

sm

k=1

(85)

m m X X 2τ (γsn (sn )0 )k − χksn u ˜t¯(sk ; k) − 2τ gkn u ˜t¯(0; k) k=1


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By using (6),(16) and by applying Cauchy inequalities with appropriately chosen ε > 0, from (85) it follows that 2 Z sm m Z sk X d˜ u(x; k) u ˜2t¯ (x; k) dx+ a0 dx dx + τ 0 0 k=1 2 m Z sk m Z sk X X d˜ u (x; k) 2 a0 τ dx ≤ Cτ u2 (x; k) dx+ dx ¯ 0 0 t k=1 k=1 Z sk Z sk Z s0 2 du(x; k) 2 dφ 2 dx + dx− f (x) dx + C k dx dx 0 0 0 m m X X 2τ (γsn (sn )0 )k − χksn u ˜t¯(sk ; k) − 2τ gkn u (86) ˜t¯(0; k) dx k=1

k=1

where C is independent of n. Note that we replaced u ˜ with u in first two integrals on the right-hand side of (86). Since γ, χ ∈ W21,1 (D) we have γ(sn (t), t), χ(sn (t), t) ∈ 1

W24 [0, T ] ([27, 4, 23]) and (87)

kγ(sn (t), t)k

1

W24 [0,T ]

≤ CkγkW 1,1 (D) , kχ(sn (t), t)k

1

W24 [0,T ]

2

≤ CkχkW 1,1 (D) , 2

where C is independent of n. According to Lemma 2.3 Pn ([v]n ) ∈ VR+1 . By apply1

ing Morrey inequality to (sn )0 we easily deduce that γ(sn (t), t)(sn )0 (t) ∈ W24 [0, T ] and moreover, kγ(sn (t), t)(sn )0 (t)k

1

W24 [0,T ]

≤ C1 kγ(sn (t), t)k

1

W24 [0,T ]

ksn kW22 [0,T ]

≤ CkγkW 1,1 (D) ,

(88)

2

where C is independent of n. Let w(x, t) be a function in W22,1 (D) such that (89)

w(x, 0) = φ(x)

for x ∈ [0, s0 ], a(0, t)wx (0, t) = g n (t)

for a.e. t ∈ [0, T ]

(90)

a(sn (t), t)wx (sn (t), t) = γ(sn (t), t)(sn )0 (t) − χ(sn (t), t)

for a.e. t ∈ [0, T ]

and h kwkW 2,1 (D) ≤ C kg n k

+ kφ(x)kW 1 [0,s0 ] 2 i + kγ(sn (t), t)(sn )0 (t) − χ(sn (t), t)k 41 2

(91)

1

W24 [0,T ]

W2 [0,T ]

The existence of w follows from the result on traces of Sobolev functions [4, 27]. For example, w can be constructed as a solution from W22,1 (Ωn ) of the heat equation in Ωn = {0 < x < sn (t), 0 < t < T } under initial-boundary conditions (89),(90) with subsequent continuation to W22,1 (D) with norm preservation [32, 33]. Hence, by replacing in the original problem (1)-(4) s, g and u with sn , g n and u − w respectively, we can derive modified (86) without the last three terms on the right-hand side and with fk (x), replaced by Steklov average Fk (x) of (92)

F = f + wt − (awx )x − bwx − cw ∈ L2 (D).

By using the stability estimation (49), from modified (86),(91) and (92), (82) follows. Lemma is proved. Inverse Problems and Imaging

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329

In the next lemma we prove (81) with l being replaced with sk on the left-hand side. Lemma 3.7. For all sufficiently small τ , discrete state vector [u([v]n )]n satisfies the following estimation: n Z sk X du(x; k) 2 dx + τ u2t¯ (x; k) dx ≤ dx 0

sk

Z max

1≤k≤n

0

2

k=1

2

C kφkW 1 [0,l] + kg n k 2

kχ(sn (t), t)k

(93)

1 W24

2

2

1

W24 [0,T ]

2

+ [0,T ]

+ kf kL2 (D) + kγ(sn (t), t)(sn )0 (t)k

1

W24 [0,T ]

n−1 X

sk+1

Z 1+ (sk+1 − sk )

+

u2 (x; k)dx ,

sk

k=1

Proof. Obviously, we can equivalently replace u ˜ with u in the first term on the left-hand side of (82). We can do so also in the second term provided sk−1 ≥ sk for all k = 1, m. Hence, we only need to estimate Zsk

u2t¯ (x; k) dx, sk−1 < sk .

0

By using (45) we have Zsk

u2t¯ (x; k)dx

0

sZk−1

u ˜2t¯ (x; k)dx +

=

Zsk

u2t¯ (x; k)dx,

sk−1

0

Zsk Zsk u(x; k) − u(2sk−1 − x; k) 2 u(x; k) − u(x; k − 1) 2 dx ≤ 2 dx τ τ sk−1

sk−1

Zsk +2

u(2s 2 k−1 − x; k) − u(2sk−1 − x; k − 1) dx ≤ τ

sk−1

Zsk 1 2 τ

2sk−1 −x

sk−1

2 τ2

Zsk

Zx

Zx

sZk−1

du(y; k) 2 dx + 2 dy

sk−1 −(sk −sk−1 )

du(y; k) 2 dy2(x − sk−1 )dx + 2 dy

sk−1 2sk−1 −x

sZk−1

u ˜2t¯ (x; k)dx ≤

sk−1 −Cτ

Zsk (94)

u ˜2t¯ (x; k)dx ≤

2 sk−1 −Cτ


d˜ u(x; k) 2 dx + 2 dx

sZk−1

u ˜2t¯ (x; k)dx.

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Hence, for sufficiently small τ we have Zsk

u2t¯ (x; k)dx

≤2 sk−1 −Cτ

0 sZk−1

(95)

sZk−1 d˜ u(x; k) 2 u ˜2t¯ (x; k)dx dx + dx

Zsk

+2

u ˜2t¯ (x; k)dx

Zsk ≤2

sk−1 −Cτ

0 sZk−1 d˜ u(x; k) 2 u ˜2t¯ (x; k)dx. dx + 3 dx 0

0

From (82) and (95), (93) follows. Lemma is proved. It can be easily seen that Theorem 3.5 follows from Lemma 3.7 and extension Lemma 3.3. Second energy estimate (81) allows to strengthen the result of Theorem 3.4. Theorem 3.8. Let [v]n ∈ VRn , n = 1, 2, ... be a sequence of discrete controls and the sequence {Pn ([v]n } converges strongly in W21 [0, T ] × L2 [0, T ] to v = (s, g). Then the sequence {ˆ uτ } converges as τ → 0 weakly in W21,1 (Ω) to weak solution u ∈ W21,1 (Ω) of the problem (1)-(4), i.e. to the solution of the integral identity (9). Moreover, u satisfies the energy estimate + kf k2L2 (D) + kuk2W 1,1 (D) ≤ C kφk2W 1 (0,s0 ) + kgk2 1 2 2 W24 [0,T ] (96) kγk2W 1,1 (D) + kχk2W 1,1 (D) 2

2

Proof. The last term on the right-hand side of the second energy estimate (81) is estimated in Theorem 3.4 along (66)-(70). By using Theorems 3.1 and 3.4, from (81),(87),(88) it follows that the sequence {ˆ uτ } satisfies the estimate kˆ uτ k2W 1,1 (D) ≤ C kφk2W 1 (0,s0 ) + kg n k2 1 + kf k2L2 (D) + 2 2 W24 [0,T ] (97) kγk2W 1,1 (D) + kχk2W 1,1 (D) 2

1 4

2

n

W21 -norm

Note that W2 -norm of g is bounded by which is uniformly bounded by τ Lemma 2.3. Hence, {ˆ u } is weakly precompact in W21,1 (D). It follows that it is strongly precompact in L2 (D). Let u be a weak limit point of {ˆ uτ } in W21,1 (D), and therefore a strong limit point in L2 (D). From (81) it follows that n

τ

kˆ u −

uτ k2L2 (D)

1 X = τ3 3

Zl

u2t¯ (x; k)dx → 0,

as τ → 0.

k=1 0

Therefore, u is a strong limit point of the sequence {uτ } in L2 (D). By Theorem 3.4 whole sequence {uτ } converges weakly in W21,0 (Ω) to the unique weak solution from V21,0 (Ω) of the problem (1)-(4). Hence, u is a weak solution of the problem (1)-(4) and we conclude that whole sequence {ˆ uτ } converges weakly in W21,1 (D) 1,1 to u ∈ W2 (D) which is a weak solution of the problem (1)-(4) from W21,1 (Ω). From the property of weak convergence it follows that u satisfies (96). Theorem is proved. In particular, Theorem 3.8 implies the following existence result: Corollary 3. For arbitrary v = (s, g) ∈ VR there exists a weak solution u ∈ W21,1 (Ω) of the problem (1)-(4) which satisfy the energy estimate (96) Inverse Problems and Imaging

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Remark. In fact, we proved slightly higher regularity of u, and both in Theorem 3.8 and Corollary 3 W21,1 (D)-norm on the left-hand side of (96) can be replaced with the norm kuk = max ku(x, t)kW21 [0,l] + kut kL2 (D) 0≤t≤T

3.3. Proof of Theorem 1.4. Let {vn } ∈ VR be a minimizing sequence lim J (vn ) = J∗

n→∞

Sequence vn = (sn , gn ) is weakly precompact in W22 [0, T ] × W21 [0, T ]. Assume that the whole sequence vn = (sn , gn ) converge to some limit function v = (s, g) ∈ VR weakly in W22 [0, T ] × W21 [0, T ], and hence, strongly in W21 [0, T ] × L2 [0, T ]. Let un = u(x, t; vn ), u = u(x, t; v) ∈ W21,1 (D) are weak solutions of (1)-(4) in W21,1 (Ωn ) and W21,1 (Ω) respectively. By Corollary 3, both satisfy energy estimation (96) with gn and g on the right-hand side respectively. Since vn ∈ VR , kun kW 1,1 (D) is 2 uniformly bounded. Hence, the sequence ∆u = un − u satisfies k∆ukW 1,1 (D) ≤ C

(98)

2

uniformly with respect to n. Accordingly, {∆u} is weakly precompact in W21,1 (D). Without loss of generality assume that the whole sequence un − u converges weakly in W21,1 (D) to some function w ∈ W21,1 (D). Let us subtract integral identities (9) for un and u, by assuming that the fixed test function Φ belongs to W21,1 (D). Indeed, otherwise Φ can be continued to D as an element of W21,1 (D). Z T Z s(t) a∆ux Φx − b∆ux Φ − c∆uΦ + ∆ut Φx dx dt+ 0

0

Z

T

(gn (t) − g(t)) Φ(0, t) dt+ 0 T

Z

[γ(sn (t), t)s0n (t) − γ(s(t), t)s0 (t) − χ(sn (t), t) + χ(s(t), t)] Φ(s(t), t) dt

+ 0

T

Z

Z

sn (t)

{a(un )x Φx − b(un )x Φ − cun Φ + (un )t Φ + f Φ} dx dt

+ 0

Z (99)

+

s(t)

T

[γ(sn (t), t)s0n (t) − χ(sn (t), t)] [Φ(sn (t), t) − Φ(s(t), t)] dt = 0

0

By using energy estimate (96), and continuity of traces γ(s(t), t), χ(s(t), t) of elements γ, χ ∈ W21,1 (D), strongly in L2 [0, T ] with respect to s ∈ W21 [0, T ], passing to the limit as n → +∞, from (99) it follows that the weak limit function w satisfies the integral identity Z T Z s(t) (100) {awx Φx − bwx Φ − cwΦ + wt Φ} dx dt = 0 0

0

for arbitrary Φ ∈ W21,1 (D). Since, any element Φ ∈ W21,1 (Ω) can be continued to D as element of W21,1 (D), (100) is valid for arbitrary Φ ∈ W21,1 (Ω). Hence, w is a weak solution from W21,1 (Ω) of the problem (1)-(4) with f = g = γ = χ = 0. From (96) and uniqueness it follows that w = 0. Thus un converges to u weakly in W21,1 (D). From Sobolev trace theorem ([4, 27]) it follows that kun (0, t) − u(0, t)kL2 [0,T ] → 0, kun (s(t), t) − u(s(t), t))kL2 [0,T ] → 0 Inverse Problems and Imaging

as n → ∞,

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Ugur G. Abdulla

kun (sn (t), t) − u(s(t), t))kL2 [0,T ] ≤ kun (sn (t), t) − un (s(t), t)kL2 [0,T ] +kun (s(t), t) − u(s(t), t))kL2 [0,T ] → 0

as n → ∞.

Hence, we have J (v) = lim J (vn ) = J∗ n→∞

and v is a solution of the Problem I. Theorem is proved. Remark. By applying first and second energy estimates we proved that functional J (v) is weakly continuous in W22 [0, T ] × W21 [0, T ]. Since VR is weakly compact existence of the optimal control follows from Weierstrass theorem in weak topology. 3.4. Proof of Theorem 1.5. We split the remainder of the proof into three lemmas. Lemma 3.9. Let J∗ (±) = inf J (v), ε > 0. Then VR±

lim J∗ () = J∗ = lim J∗ (−)

(101)

→0

→0

Proof. Note that for 0 < 1 < 2 we have J∗ (2 ) ≤ J∗ (1 ) ≤ J∗ ≤ J∗ (−1 ) ≤ J∗ (−2 ) Therefore lim J∗ () ≤ J∗ and lim J∗ (−) ≥ J∗ exist. Let us choose v ∈ VR+ such →0

→0

that lim J (v ) − J∗ () = 0

(102)

→0

Since v = (s , g ) is weakly precompact in W22 [0, T ] × W21 [0, T ], there exists some subsequence 0 such that s0 → s∗ weakly in W22 [0, T ], g0 → g∗ weakly in W21 [0, T ] as 0 → 0 Moreover, we have v∗ = (s∗ , g∗ ) ∈ VR . Since J (v) is weakly continuous it follows that (103)

lim J (v0 ) = J (v∗ ).

0 →0

From (102),(103) it follows that lim J∗ (0 ) = J∗

0 →0

which implies the first relation in (101). To prove the second relation in (101), take 0 > 0 and v˜ = (˜ s, g˜) ∈ VR−0 . Let {αk } be a real sequence with 0 < αk < 1, lim αk = 0 and set k→+∞

vk = (sk , gk ) = αk v˜ + (1 − αk )v∗ where J (v∗ ) = J∗ . We have vk ∈ VR−αk 0 and vk converges to v∗ strongly in W22 [0, T ] × W21 [0, T ]. Since J (v) is continuous, vk is a minimizing sequence: lim J (vk ) = J∗

k→∞

For fixed k choose arbitrary such that 0 < < 0 αk . We obviously have J∗ (−) ≤ J (vk ), Passing to the limit as → 0 we have

0 < < 0 αk

lim J∗ (−) ≤ J (vk )

→0 Inverse Problems and Imaging

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333

Now we pass to the limit as k → +∞ and get lim J∗ (−) ≤ J∗

→0

Since the opposite inequality is obvious, (101) follows. Lemma is proved.

Lemma 3.10. For arbitrary v = (s, g) ∈ VR , lim In (Qn (v)) = J (v)

(104)

n→∞

Proof. Let v ∈ VR , u = u(x, t; v), Qn (v) = [v]n and [u([v]n )]n be a corresponding discrete state vector. In Theorem 3.8 it is proved that the sequence {ˆ uτ } con1,1 verges to u weakly in W2 (Ω). This implies that the sequences of traces {ˆ uτ (0, t)} τ and {ˆ u (s(t), t)} converge strongly in L2 [0, T ] to corresponding traces u(0, t) and u(s(t), t). Let us prove that that the sequences of traces {uτ (0, t)} and {uτ (s(t), t)} converge strongly in L2 [0, T ] to traces u(0, t) and u(s(t), t) respectively. By Sobolev embedding theorem ([4, 27]) it is enough to prove that the sequences {uτ } and {ˆ uτ } 1,0 are equivalent in strong topology of W2 (Ω). In Theorem 3.8 it is proved that they are equivalent in strong topology of L2 (D). It remains only to demonstrate that the uτx } are equivalent in strong topology of L2 (Ω). sequences of derivatives {uτx } and {ˆ We have min(s Zk−1 ;sk ) n d˜ u(x; k) 2 1X 3 τ τ 2 τ dx + kuτx − u ˆτx k2L2 (Γn ) , (105) kux − u ˆx kL2 (Ω) ≤ 3 dx t¯ k=1

n

0

n

where sk = s (tk ), s is the first component of Pn ([v]n ) and Γn = ∪nk=1 {tk−1 < t ≤ tk , min(sk−1 ; sk ) < x < s(t)} Since sn converges to s uniformly on [0, T ], it follows that the Lebesgue measure of Γn converges to zero as n → +∞. By Theorems 3.4 and 3.8 the integrand is uniformly bounded in L2 (D). Therefore, the second term on the right-hand side of (105) converges to zero as n → +∞. First term on the right-hand side of (105) converges to zero due to stability estimation (82) and the claim is proved. Let ν τ (t) = ν k , µτ (t) = µk , if tk−1 < t ≤ tk , k = 1, . . . , n. We have kν τ − νkL2 [0,T ] → 0, kµτ − µkL2 [0,T ] → 0 as τ → 0

(106)

We estimate the first term in In (Qn (v)) as follows Z T n X (107) β0 τ |u(0; k) − ν k |2 = β0 |uτ (0, t) − ν τ (t)|2 dt 0

j=1

From (106) it follows that n X (108) lim β0 τ |u(0; k) − ν k |2 = β0 ku(0, t) − ν(t)k2L2 [0,T ] n→∞

k=1

We estimate the second term in In (Qn (v)) as follows n n Z tk Z sk X X ∂uτ τ (u (s(t), t) − µτ (t)) dx dt β1 τ |u(sk ; k) − µk |2 = 2β1 ∂x t s(t) k=1 k=1 k−1 !2 Z Z sk n n Z tk t X k X ∂uτ k 2 +β1 |u(s(t); k) − µ | dt + β1 dx dt = tk−1 tk−1 s(t) ∂x k=1

(109) Inverse Problems and Imaging

k=1

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Ugur G. Abdulla

We have T

Z (110)

n→∞

n→∞

Z

|uτ (s(t), t) − µτ (t)|2 dt = β1

lim I2 = lim β1 0

τ

T

|u(s(t), t) − µ(t)|2 dt

0

τ

τ

Since k(u )x kL2 (D) and ku (s(t), t) − µ kL2 [0,T ] are uniformly bounded, and {sn } converges to s uniformly on [0, T ], by applying CBS inequality it easily follows that (111)

lim I1 = 0, lim I3 = 0

n→∞

n→∞

From (109)-(111) it follows that (112)

lim β1 τ

τ →0

Z n X u(sk ; k) − µk 2 = β1

T

2

|u(s(t), t) − µ(t)| dt

0

k=1

Therefore, from (107) and (112), (104) follows. Lemma is proved.

Lemma 3.11. For arbitrary [v]n ∈ VRn (113) lim J (Pn ([v]n )) − In ([v]n ) = 0 n→∞

and v n = (sn , g n ) = Pn ([v]n ). From Lemma 2.3 it follows Proof. Let [v]n ∈ that the sequence {Pn ([v]n } is weakly precompact in W22 [0, T ] × W21 [0, T ]. Assume that the whole sequence converges to v˜ = (˜ s, g˜) weakly in W22 [0, T ] × W21 [0, T ]. 1 This implies the strong convegence in W2 [0, T ] × L2 [0, T ]. From the well-known property of weak convergence it follows that v˜ ∈ VR . In particular sn converges to s˜ uniformly on [0, T ] and we have VRn

lim max |sn (ti ) − s˜(ti )| = 0

(114)

n→∞ 0≤i≤n

Let Qn (˜ v ) = [˜ v ]n We have (115) In [v]n − J (v n ) = In [v]n − In [˜ v ]n + In [˜ v ]n − J (˜ v ) + J (˜ v ) − J (v n ) In Section 3.3 we proved the weak continuity of the functional J (v), i.e. lim (J (˜ v ) − J (v n )) = 0.

n→∞

From Lemma 3.10 it follows that lim In [˜ v ]n − J (˜ v ) = 0.

n→∞

Hence, we only need to prove that lim In [v]n − In [˜ v ]n = 0

(116)

n→∞

Let [u([v]n )]n = (un (x; 0), un (x; 1), ..., un (x; n)), [u([˜ v ]n )]n = (˜ u(x; 0), u ˜(x; 1), ..., u ˜(x; n)) are corresponding discrete state vectors according to Definition 1.3. Let sk = sn (tk ), s˜k = s˜(tk ) and ∆u(x; k) = un (x; k) − u ˜(x; k) Inverse Problems and Imaging

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335

We have In ([v]n ) − In ([˜ v ] n ) = β0

n X

τ un (0; k) − f0k

2

+ β1

k=1

β0 β0

n X

τ un (sk ; k) − f1k

2

−

k=1

τ u ˜(0; k) − f0k

k=1 n X

n X

2

− β1

n X

τ u ˜(˜ sk ; k) − f1k

2

=

k=1 2

τ (∆u(0; k)) + 2β0

k=1

n X

τ ∆u(0; k) u ˜(0; k) − f0k +

k=1

β1

n X

2

τ (∆u(sk ; k) + u ˜(sk ; k) − u ˜(˜ sk ; k)) +

k=1

(117)

2β1

n X

τ (∆u(sk ; k) + u ˜(sk ; k) − u ˜(˜ sk ; k)) u ˜(˜ sk ; k) − f1k

k=1

From the estimations of Sections 3.1 and 3.2 it follows that the sequences {un (x; k)}, {˜ u(x; k)} are uniformly bounded in W21 [0, l]. From (114) it follows that 2 Z n n s˜k d˜ X X u(x; k) 2 dx τ (˜ u(sk ; k) − u ˜(˜ sk ; k)) = τ sk dx k=1 k=1 2 n Z s˜k ∂ u X ˜(x; k) dx → 0, as n → +∞. (118) τ ≤ max |sk − s˜k | sk 1≤k≤n ∂x k=1

From (117) and (118) it follows that in order to prove (116) it is enough to prove that n h X 2 2 i (119) R= τ ∆u(0; k) + ∆u(sk ; k) → 0 as n → +∞ k=1

By the Morrey inequality we have "Z Z n sk X ∆u(x; k) 2 dx + (120) R≤C τ k=1

0

0

sk

# d∆u(x; k) 2 dx dx

where C is independent of n. Let us subtract integral identities (13) for un (x; k) and u ˜(x; k), by assuming that the fixed test function η belongs to W21 [0, l]. Indeed, otherwise η can be continued to [0, l] as a element of W21 [0, l]: Z sk d∆u dη d∆u ak (x) − bk (x) η(x) − ck (x)∆uη + ∆ut¯η dx+ dx dx dx 0 Z s˜k d˜ u dη d˜ u ak (x) − bk (x) η − ck (x)˜ uη + fk (x)η + u ˜t¯η dx+ dx dx dx sk n 0 k 0 k (γsn (s ) ) − (γs˜s˜ ) η(sk ) + (γs˜s˜0 )k [η(sk ) − η(˜ sk )] − k k k n n −χs˜ [η(sk ) − η(˜ (121) sk )] − χsn − χs˜ η(sk ) + (gk − g˜k )η(0) = 0 Our goal now is to derive from (121) that the right-hand side of (120) converges to zero as n → +∞. The proof goes along the same lines as the derivation of the first energy estimate in Lemma 3.2. By choosing η(x) = 2τ ∆u(x; k) in (121), and Inverse Problems and Imaging

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by using (6), Cauchy inequalities with appropriately chosen ε > 0, and Morrey inequality (27) we derive similar to (51): Z sk Z sk d∆u(x; k) 2 2 dx + ∆u a0 τ (x; k) dx − ∆u2 (s; k − 1) dx+ dx 0 0 0 Z sk Z sk 2 2 n n 2 2 ∆ut¯ (x; k)dx ≤ C1 τ ∆u (x; k) dx + |gk − g˜k | − τ Z

sk

0

0

s˜k

d˜ u d∆u(x; k) d˜ u ak (x) 2τ − bk (x) ∆u(x; k) − ck (x)˜ u∆u(x; k)+ dx dx dx sk Z tk Z sk d∆u(x; k) fk (x)∆u(x; k) + u ˜t¯∆u(x; k) dx − 2 γ(˜ s(t), t)˜ s0 (t) dx dt− dx tk−1 s˜k Z tk 2 (γ(sn (t), t)(sn )0 (t) − γ(˜ s(t), t)˜ s0 (t)) dt ∆u(sk ; k)+ Z

tk−1 tk

Z

(χ(sn (t), t) − χ(˜ s(t), t)) dt∆u(sk ; k)+

2 tk−1

tk

Z (122)

sk

Z

2

χ(˜ s(t), t) tk−1

s˜k

d∆u(x; k) dx dt dx

By applying the technique along with (52)-(56) from (122) it follows that for all sufficiently small τ sk

Z

∆u2 (x; k) dx ≤ C kg n − g˜k2L2 [0,T ] +

max

1≤k≤n n−1 X

(123)

0

Z

sj+1

1+ (sj+1 − sj )

∆u2 (x; j)dx +

sj

j=1

n X

|Lj |

j=1

where C is independent of τ and Z

s˜j

Lj = τ

aj (x)

sj

Z

tj

d˜ u d∆u(x; j) d˜ u − bj (x) ∆u(x; j) − cj (x)˜ u∆u(x; j)+ dx dx dx fj (x)∆u(x; j) + u ˜t¯∆u(x; j) dx+

[γ(sn (t), t)(sn )0 (t) − γ(˜ s(t), t)˜ s0 (t)] dt ∆u(sj ; j)+

tj−1

Z Z

tj

Z

tj−1 tj

sj

γ(˜ s(t), t)˜ s0 (t)

s˜j

d∆u(x; j) dx dt+ dx

(χ(sn (t), t) − χ(˜ s(t), t)) dt ∆u(sj ; j)+

tj−1

Z

tj

Z

sj

(124)

χ(˜ s(t), t) tj−1


s˜j

d∆u(x; j) dx dt dx Volume 7, No. 2 (2013), 307–340


337

Having (123) we perform summation in (122) with respect to k from 1 to n and derive Z sk Z sk n X d∆u(x; k) 2 2 dx+ max ∆u (x; k) dx + τ 1≤k≤n 0 dx 0 k=1 Z sk n X τ2 ∆u2t¯ (x; k) dx ≤ C1 kg n − g˜k2L2 [0,T ] + 0

k=1 n−1 X

(125)

Z

sj+1

1+ (sj+1 − sj )

∆u2 (x; j)dx +

sj

j=1

n X

|Lj |

j=1

where C1 is independent of τ . Our next goal is to absorb the first term on the righthand side of (125) into the left-hand side. We apply the same method used in the proof of Theorem 3.4 (see (66)-(72)). The only difference is that in the estimations (69) and (70) we replace D with Ω1n = {0 < t < T, 0 < x < sñ1 (t)} Let us also introduce the region n [ Ωn = {tk−1 < t ≤ tk , 0 < x < sk } k=1

Note that k∆uτ k2V2 (Ωn ) = max

1≤k≤n

Z

sk

∆u2 (x; k) dx +

0

Z n X τ k=1

sk

0

d∆u(x; k) 2 dx dx

Hence, by applying the method used in Theorem 3.4 we derive from (125) the following estimate: n X (126) k∆uτ k2V2 (Ωn ) ≤ C2 kg n − g˜k2L2 [0,T ] + k∆uτ k2V2 (Ω1n −Ωn ) + |Lj | j=1

{˜ sn1 }

n

where C2 is independent of τ . Since the sequences and {s } are equivalent in strong topology of W21 [0, T ], the second term on the right-hand side of (126) converges to zero as n → +∞. Due to convergence of g n to g˜ in L2 [0, T ], it only remains to prove that n X (127) lim |Lj | = 0. n→+∞

j=1

We have Z s˜j n Z s˜j n Z X d˜ u(x; j) d∆u(x; j) X tj d˜ uτ d∆uτ aj (x) dx = a(x, t) dx τ sj dx dx dx dx tj−1 sj j=1 j=1

τ

∂u

∂∆uτ ˜

(128) ≤C

∂x ˜ ∂x ˜ L2 (∆) L2 (∆) where ˜ = ∆

n [

{(x, t) : tj−1 < t < tj , min(sj , s˜j ) < x < max(sj , s˜j )}

j=1

˜ converges to zero as n → ∞. From (114) it follows that the Lebesgue measure of ∆ 1,0 Since by the first energy estimate W2 (D) norm of u ˜τ and ∆uτ are uniformly Inverse Problems and Imaging

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338

Ugur G. Abdulla

bounded, the right-hand side of (128) converges P to zero as n → ∞. For the same n reason, the next three terms in the expression of j=1 |Lj | also converge to zero as n → ∞. We have  21  n Z n Z s˜j l X X u ˜2t¯ (x; j) dx k∆uτ kL2 (∆) τ (129) u ˜¯(x; j)∆u(x; j) dx ≤ τ ˜ sj t 0 j=1

j=1

W21,1 (D)

ˆ˜τ is uniformly bounded. AccordBy the second energy estimate norm of u ingly, the right-hand side of (129) converges to zero as n → +∞. We have n Z tj X s(t), t)˜ s0 (t) ∆u(sj , j) dt ≤ γ(sn (t), t)(sn )0 (t) − γ(˜ j=1

tj−1

n Z X j=1

tj

tj−1

s(t), t)˜ s0 (t) × γ(sn (t), t)(sn )0 (t) − γ(˜ Z

sj

∆u(˜ s(t); j) + s˜(t)

∂∆uτ (x, t) dx dt ≤ ∂x

kγ(sn (t), t) − γ(˜ s(t), t)kL2 [0,T ] k(sn )0 kC[0,T ] k∆uτ (˜ s(t), t)kL2 [0,T ] + k(sn )0 (t) − s˜0 (t)kL2 [0,T ] kγ(˜ s(t), t)kL4 [0,T ] k∆uτ (˜ s(t), t)kL4 [0,T ] + ! 12 Z tj n X n n 0 0 2 |γ(s (t), t)(s ) (t) − γ(˜ s(t), t)˜ s (t)| dt × j=1

tj−1

Z 2  21 sj ∂∆uτ  dx dt ≤ Ckγ(sn (t), t) − γ(˜ s(t), t)kL2 [0,T ] × tj−1 s˜(t) ∂x 

Z

tj

k(sn )0 kW21 [0,T ] k∆uτ kW 1,0 (D) + Ck(sn )0 (t) − s˜0 (t)kL2 [0,T ] kγkW 1,1 (D) × 2

(130)

2

k∆uτ kV2 (D) + kγ(sn (t), t)(sn )0 (t) − γ(˜ s(t), t)˜ s0 (t)kL2 [0,T ] ×

21

∂∆uτ n

max |˜ s (t) − s (t)|

∂x ˜ 0≤t≤T L2 (∆)

and all three terms on the right-hand side converge to zero as P n → +∞. Similarly n one can prove that all the last three terms in the expression of j=1 |Lj | converges to zero as n → ∞. Hence, (127) is proved. From (126) and (127), (119) follows. Lemma is proved. Having Lemmas 3.9, 3.10 and 3.11, Theorem 1.5 follows from Lemma 2.2 REFERENCES [1] O. M. Alifanov, “Inverse Heat Transfer Problems,” Springer-Verlag Telos, 1995. [2] J. Baumeister, Zur optimal Steuerung von frien Randwertausgaben, ZAMM, 60 (1980), 335– 339. [3] J. B. Bell, The non-characteristic Cauchy problem for a class of equations with time dependence. I. Problem in one space dimension, SIAM J. Math. Anal., 12 (1981), 759–777. [4] O. V. Besov, V. P. Il’in and S. M. Nikol’skii, “Integral Representations of Functions and Embedding Theorems,” Winston & Sons, Washington, D.C.; John Wiley & Sons, 1978. [5] B. M. Budak and V. N. Vasil’eva, On the solution of the inverse Stefan problem, Soviet Math. Dokl, 13 (1972), 811–815. [6] B. M. Budak and V. N. Vasil’eva, The solution of the inverse Stefan problem, USSR Comput. Maths. Math. Phys, 13 (1973), 130–151. Inverse Problems and Imaging

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[36] F. P. Vasil’ev, “Methods for Solving Extremal Problems. Minimization Problems in Function Spaces, Regularization, Approximation,” (in Russian), Moscow, Nauka, 1981. [37] A. D. Yurii, On an optimal Stefan problem, Dokl. Akad. Nauk SSSR, 251 (1980), 1317–1321.

Received March 2012; revised October 2012. E-mail address: [email protected]


Volume 7, No. 2 (2013), 307–340