Jul 3, 2016 - We consider the inverse Stefan type free boundary problem, where ... In order to find f(x, t) and g(t) along with u(x, t) and s(t), we must have additional ... Assume that we are able to measure the temperature on our domain and ...
FRECHET DIFFERENTIABILITY IN BESOV SPACES IN THE OPTIMAL CONTROL OF
arXiv:1604.00057v1 [math.AP] 31 Mar 2016
PARABOLIC FREE BOUNDARY PROBLEMS
UGUR G. ABDULLA & JONATHAN GOLDFARB Department of Mathematical Sciences Florida Institute of Technology Melbourne, FL 32901, abdulla@fit.edu Abstract. We consider inverse Stefan type free boundary problem, where information on the boundary heat flux and density of the sources are missing and must be found along with the temperature and free boundary. Following new variational formulation introduced recently in U.G. Abdulla, Inverse Problems and Imaging, 7, 2(2013), 307–340, we pursue optimal control problem where boundary heat flux, density of sources and free boundary are components of the control vector, and the optimality criteria consists of the minimiztion of the L2 -norm declinations of the temperature measurements at the final moment, phase transition temperature and final position of the free boundary. We prove the Frechet differentiability in Besov spaces, and derive the formula for the Frechet differential under minimal regularity assumptions on the data.
1
2
1. Introduction and Statement of Main Results 1.1. Inverse Stefan Problem. Consider the general one-phase Stefan problem: (aux )x + bux + cu − ut = f, in Ω
(1)
u(x, 0) = φ(x), 0 ≤ x ≤ s(0) = s0
(2)
a(0, t)ux (0, t) = g(t), 0 ≤ t ≤ T
(3)
a(s(t), t)ux (s(t), t) + γ(s(t), t)s′ (t) = χ(s(t), t), 0 ≤ t ≤ T
(4)
u(s(t), t) = µ(t), 0 ≤ t ≤ T,
(5) where (6)
Ω = {(x, t) : 0 < x < s(t), 0 < t ≤ T }
where a, b, c, f, φ, g, γ, χ, µ are given functions. Assume that f (x, t) and g(t) are not known, where f (x, t) is the density of heat sources and g(t) is the heat flux at x = 0, the left boundary of our domain. In order to find f (x, t) and g(t) along with u(x, t) and s(t), we must have additional information. Assume that we are able to measure the temperature on our domain and the position of the free boundary at the final moment T . (7)
u(x, T ) = w(x), 0 ≤ x ≤ s(T ) = s∗ .
Under these conditions, we are required to solve an inverse Stefan problem (ISP): find a tuple {u(x, t), s(t), g(t), f (x, t)} that satisfy conditions (1)–(7). ISP is not well posed in the sense of Hadamard: the solution may not exist; if it exists, it may not be unique, and in general it does not exhibit continuous dependence on the data. The main methods available for ISP are based on a variational formulation, Frechet differentiability, and iterative gradient methods. We cite recent papers [1, 2] and the monograph [3] for a list of references. The established variational methods in earlier works fail in general to address two issues: • The solution of ISP does not depend continuously on the phase transition temperature µ(t) from (5). A small perturbation of the phase transition temperature may imply significant change of the solution to the inverse Stefan problem. • In the existing formulation, at each step of the iterative method a Stefan problem must be solved (that is, for instance, the unknown heat flux g is given, and the corresponding u(x, t) and s(t) are calculated) which incurs a high computational cost. A new method developed in [1, 2] addresses both issues with a new variational formulation. The key insight is that the free boundary problem has a similar nature to an inverse problem, so putting them into the same framework gives a conceptually clear formulation of the problem; proving existence, approximation, and differentiability is a resulting challenge. Existence of the optimal control and the convergence of the sequence of discrete optimal control problems to the continuous optimal control problem was proved in [1, 2]. Our goal in this work is to prove the Frechet differentiability and to derive the formula for the Frechet differential under the minimal regularity assumptions on the data. 1.2. Notation. We will use the notation 1I (x) =
(
1, 0,
x∈I x 6∈ I
for the indicator function of the set I, and [r] for the integer part of the real number r. We will require the notions of Sobolev-Slobodeckij or Besov spaces [4, 5, 6, 7]. In this section, assume U is a domain in R and denote by QT = (0, 1) × (0, T ] .
OPTIMAL CONTROL OF PARABOLIC FREE BOUNDARY PROBLEMS
3
• For ℓ ∈ Z+ , Wpℓ (U ) is the Banach space of measurable functions with finite norm !1/p Z X ℓ k p d u k dx U k=0 dx
kukW ℓ (U ) :=
(8)
p
• For ℓ 6∈ Z+ , Bpℓ (U ) is the Banach space of measurable functions with finite norm
(9)
kukBℓ (U ) := kukW [ℓ] (U ) + [u]Bℓ (U ) where [u]pBℓ (U ) := p
p
p
p
Z Z ∂ [ℓ]u(x) − ∂x[ℓ] U
U
p
∂ [ℓ]u(y) ∂x[ℓ]
|x − y|1+p(ℓ−[ℓ])
dx dy
• If ℓ ∈ Z+ , the seminorm [u]Bℓ (U ) is given by p
(10)
[u]pBℓ (U ) p
:=
Z
∞
−∞
Z
p ℓ−1 ∂ ℓ−1 u x+y u(x) 1 ∂ ℓ−1 u(y) ∂ 2 dy dx −2 + ℓ−1 ℓ−1 ℓ−1 |x − y|1+p ∂x ∂x −∞ ∂x ∞
[8, thm. 5, p. 72]. By [5, §18, thm. 9], it follows that p = 2 and ℓ ∈ Z+ , the Bpℓ (U ) norm is equivalent to the Wpℓ (U ) norm (i.e. the two spaces coincide.) • W21,1 (QT ) is the Hilbert space of L2 (QT ) functions with a weak x- and t-derivatives belonging to L2 (QT ). The inner product in W21,1 (QT ) is hu, vi =
Z
[uv + ux vx + ut vt ] dx dt QT
• W22ℓ,ℓ (Ω), ℓ = 1, 2, . . . is the Hilbert space of u ∈ L2 (Ω) for which ∂ 2r+s u ∈ L2 (Ω), 0 ≤ 2r + s ≤ 2ℓ. ∂tr xs The inner product in W22ℓ,ℓ (Ω) is hu, vi =
Z X 2ℓ X ∂ 2r+s u ∂ 2r+s v dx dt r s ∂tr xs Ω j=0 2r+s=j ∂t x
where the summation indices r, s ≥ 0 satisfy 2r + s = j, j = 0, . . . , 2ℓ. ℓ1 ,ℓ2 • Let 1 ≤ p < ∞, ℓ1 , ℓ2 > 0. The Besov space Bp,x,t (QT ) is defined as the closure of the set of smooth functions under the norm 1/p Z 1 Z T 1/p ku(x, t)kp ℓ1 dt kukBℓ1 ,ℓ2 (Q ) = ku(x, t)kp ℓ2 dx + p,x,t
T
0
Bp (0,1)
0
Bp [0,T ]
When p = 2, if either ℓ1 or ℓ2 is an integer, the Besov seminorm may be replaced with the corresponding Sobolev seminorm due to equivalence of the norms. α,α/2 • The Hölder space Cx,t (QT ) is the set of continuous functions with [α] x-derivatives and [α/2] tderivatives, and for which the highest order x- and t-derivatives satisfy Hölder conditions of order α − [α] and α/2 − [α/2], respectively. 1.3. Optimal Control Problem. Fix any α > 0. Consider the minimization of the functional Z s(T ) Z T (11) J (v) = β0 |u(x, T ; v) − w(x)|2 dx + β1 |u(s(t), t; v) − µ(t)|2 dt + β2 |s(T ) − s∗ |2 0
0
4
OPTIMAL CONTROL OF PARABOLIC FREE BOUNDARY PROBLEMS
on the control set
n 1/2+α 1,1/4+α [0, T ] × W22 [0, T ]; s(0) = s0 VR = v = (f, g, s) ∈ H := B2,x,t (D) × B2 s′ (0) = 0, g(0) = a(0, 0)φ′ (0), 0 < δ ≤ s(t) ≤ ℓ; o kvkH := max kf kB1,1/4+α (D) , kgkB1/2+α [0,T ] , kskW 2 [0,T ] ≤ R ,
(12)
2,x,t
2
2
where β0 , β1 , β2 ≥ 0 and ℓ, δ, R > 0 are given. Define
D = {(x, t) : 0 ≤ x ≤ ℓ, 0 ≤ t ≤ T } For a given control vector v ∈ VR , the state vector u(x, t; v) is a solution to the Neumann problem (1)–(4). The formulated optimal control problem (11)–(12) will be called Problem I. Since the data appearing in the Neumann problem (1)–(4) are in general non-smooth, the solutions may not exist in the classical sense. The notion of solution must be understood in a weak sense, i.e. for a fixed control vector v ∈ VR u ∈ W22,1 (Ω) is called a solution of the Neumann problem (1)–(4) if it satisfies the equation (1) and conditions (2)–(4) pointwise almost everywhere. 1.4. Statement of Main Results. Let α > 0 is fixed as in (12). The main results are established under the assumptions 0 < a0 ≤ a(x, t) in D
(13)
1/2+α∗ ,1/4+α∗
a, ax , b, c ∈ Cx,t
(14) w∈
(15)
3/2+2α (0, ℓ), B2
χ, γ ∈
(16)
φ∈
(D)
3/2+2α (0, s0 ), B2
3/2+2α∗ ,3/4+α∗ B2,x,t (D) 1/4
µ ∈ B2 [0, T ]
(17)
where α∗ > α is arbitrary, and χ, φ satisfy the compatibility condition χ(s0 , 0) = φ′ (s0 )a(s0 , 0).
(18)
Given a control vector v ∈ VR , under the conditions (13)–(18) there exists a unique pointwise a.e. solution u ∈ W22,1 (Ω) of the Neumann problem (1)–(4) such that the transformed function u e(y, t) = u(ys(t), t) belongs to 5/2+2α,5/4+α (QT ) and solves (49)–(52) ([8]; see Lemma 4 below.) B2,x,t Definition 1. For given v and u = u(x, t; v), ψ ∈ W22,1 (Ω) is a solution to the adjoint problem if (19) aψx x − (bψ)x + cψ + ψt = 0, in Ω ψ(x, T ) = 2β0 (u(x, T ) − w(x)), 0 ≤ x ≤ s(T )
(20)
a(0, t)ψx (0, t) − b(0, t)ψ(0, t) = 0, 0 ≤ t ≤ T
(21) (22)
a(s(t), t)ψx (s(t), t) − (b(s(t), t) + s′ (t))ψ(s(t), t) = 2β1 (u(s(t), t) − µ(t)), 0 ≤ t ≤ T
Theorem 1. Problem I has a solution, and the functional J (v) is differentiable in the sense of Frechet, and the first variation is Z Z T dJ (v) = − ψδf dx dt − ψ(0, t)δg(t) dt+ + (23)
−
Z
Z
Ω T
0 T 0
0
2β1 (u − µ)ux + ψ χx − γx s′ − aux γψ
x=s(t)
x
x=s(t)
δs(t) dt−
δs′ (t) dt + β0 |u(s(T ), T ) − w(s(T ))|2 + 2β2 (s(T ) − s∗ ) δs(T )
where ψ is a solution to the adjoint problem in the sense of definition 1, and δv = (δs, δg, δf ) is a variation of the control vector v ∈ VR such that v + δv ∈ VR .
OPTIMAL CONTROL OF PARABOLIC FREE BOUNDARY PROBLEMS
5
2. Heuristic Derivation of the Frechet Differential To give a first indication of the form of the gradient, we apply the heuristic method of Lagrange-type multipliers; the rigorous proof follows in Section 4. Consider the functional: Z T Z s(t) L(g, f, s, u, ψ) = J (v) + ψ [(aux )x + bux + cu − ut − f ] dx dt. 0
0
Define δv = (δs, δg, δf ), v¯ = v + δv = (s, g¯, f¯). Let u(x, t) = u(x, t, v¯). We will also denote by s˜(t) = s(t) + θ(t)δs(t) where 0 ≤ θ(t) ≤ 1 standing for all functions arising from application of mean value theorem in the region between s(t) and s(t). Define sb = min(s, s),
0 ≤ t ≤ T,
b = {(x, t) : 0 < x < sb(t), 0 < t ≤ T } Ω
b δu(x, t) = u(x, t) − u(x, t) in Ω
The increment δv must be made in such a way that v + δv ∈ VR , so in particular δs(0) = 0, δs′ (0) = 0. Simiarlly, the incremented solution u must satisfy the initial condition (2) and corresponding boundary conditions (3) and (4), so in particular δu(x, 0) = 0,
(24)
a(0, t)δux (0, t) = δg(t).
In what follows, all terms of higher than linear order with respect to δv will be absorbed into the expression o(δv). Partition the time domain as [0, T ] = T1 ∪ T2 where T1 = {t ∈ [0, T ] : δs(t) < 0} ,
T2 = [0, T ] \ T1 = {t ∈ [0, T ] : δs(t) ≥ 0}
Calculate (25)
∆L = ∆J + ∆I
where ∆I =
Z
T
0
(26) (27)
−
Z
Z
s(t)
ψ [(aux )x + bux + cu − ut − f − δf ] dx dt−
0
T
0
Z
s(t) 0
ψ [(aux )x + bux + cu − ut − f ] dx dt
∆J = J (v + δv) − J (v) = J1 + J2 + J3
and the terms J1 -J3 are given by J1 = β0 J2 = β1 (28)
"Z
Z
s(T )
|u(x, T ) − w(x)| dx −
0 T
0
2
|u(s(t), t) − µ(t)|2 dt −
J3 = β2 |s(T ) − s∗ |2 − |s(T ) − s∗ |2
Z
Z
0
T 0
s(T )
|u(x, T ) − w(x)| dx
J22 = β1
Z
Z
#
|u(s(t), t) − µ(t)|2 dt = J21 + J22
where J21 = β1
2
T1
|u(s(t), t) − µ(t)|2 − |u(s(t), t) − µ(t)|2 dt
T2
|u(s(t), t) − µ(t)|2 − |u(s(t), t) − µ(t)|2 dt
6
OPTIMAL CONTROL OF PARABOLIC FREE BOUNDARY PROBLEMS
In J1 , calculate J1 = β0
Z
s b(T ) 0
− β0
Z
|u(x, T ) − w(x)|2 − |u(x, T ) − w(x)|2 dx + β0
s(T ) s b(T )
Z
s(T )
s b(T )
|u(x, T ) − w(x)|2 dx−
|u(x, T ) − w(x)|2 dx
The first term comprises the increment of a quadratic functional, while Z
s(T )
s b(T )
Z
s(T )
s b(T )
|u(x, T ) − w(x)|2 dx = 1T2 (T ) |u(s(T ), T ) − w(s(T ))|2 δs(T ) + o(δv) |u(x, T ) − w(x)|2 dx = −1T1 (T ) |u(s(T ), T ) − w(s(T ))|2 δs(T ) + o(δv)
Hence (29)
J1 = β0
Z
s b(T ) 0
2(u − w)δu t=T dx + β0 |u(s(T ), T ) − w(s(T ))|2 δs(T ) + o(δv)
Using the identity for t ∈ T1
u(s(t), t) = u(s(t), t) + ux (˜ s(t), t)δs(t) + δu(s(t), t) in J21 , it follows that (30)
J21 = β1
Z
T1
2 u(s(t), t) − µ(t) ux (s(t), t)δs(t) dt + β1
Z
T1
2 u(s(t), t) − µ(t) δu(s(t), t) dt + o(δv)
Similarly, use the identity for t ∈ T2 u(s(t), t) = u(s(t), t) + δu(s(t), t) + ux (˜ s(t), t)δs(t) in J22 to derive J22 = β1
(31)
Z
T2
2 u(s(t), t) − µ(t) δu(s(t), t) dt + β1
Z
T2
2 u(s(t), t) − µ(t) ux (s(t), t)δs(t) dt + o(δv)
From (30), (31), it follows that (32)
J2 = β1
Z
T 0
For J3 , we have (33)
2 u(s(t), t) − µ(t) ux (s(t), t)δs(t) dt + β1
Z
T 0
2 u(s(t), t) − µ(t) δu(b s(t), t) dt + o(δv)
J3 = 2β2 (s(T ) − s∗ )δs(T ) + o(δv).
With the help of the PDE (1), from (26) it follows that ∆I = 0 and Z (34) ψ aδux x + bδux + cδu − δut − δf dx dt ∆I = b Ω
OPTIMAL CONTROL OF PARABOLIC FREE BOUNDARY PROBLEMS
7
Integrating by parts with respect to x- and t-variables, derive ∆I =
Z
b Ω
+
=
− Z
Z
Z
b Ω
+ + −
(35)
[(aψx )x − (ψb)x + ψc + ψt ] δu dx dt + T 0
Z
− aψx + b + sb′ ψ δu x=bs(t) dt − Z
T 0
Z
aψδux
x=b s(t)
dt+
ψδf dx dt−
b Ω
Z sb(T ) (aψx − bψ)δu x=0 dt − ψ(x, T )δu(x, T ) dx = 0 0 0 Z [(aψx )x − (ψb)x + ψc + ψt ] δu dx dt + aψδux x=s(t) dt+
Z
Z
Z
T
ψ(0, t)δg(t) dt +
T
T1
T
0
T2
s(t), t) dt − −aψx + b + s′ ψ x=s(t) δu(b
aψδux
s b(T )
dt −
x=s(t)
Z
T
ψ(0, t)δg(t) dt +
0
Z
Z
0
ψδf dx dt+
Ω T
(aψx − bψ)δu x=0 dt−
ψ(x, T )δu(x, T ) dx + o(δv) 0
Using the boundary conditions for u on the moving boundary s (4) and mean value theorem, it follows that for t ∈ T1 , a(s(t), t)δux (s(t), t) = χ(s(t), t) − γ(s(t), t)s′ (t) − a(s(t), t)ux (s(t), t) − − χ(s(t), t) − γ(s(t), t)s′ (t) − a(s(t), t)ux (s(t), t) δs(t) = χx (˜ s(t), t)δs(t) − γx (˜ s(t), t)δs(t)s′ (t) − γ(s(t), t)δs′ (t) − (aux )x
(36)
x=˜ s(t)
Using (36) it follows that
Z i h ψ(s(t), t) χx δs − γx δss′ − (aux )x δs x=˜s(t) − γ(s(t), t)δs′ (t) dt [ψaδux ]x=s(t) dt = T T1 Z h 1 i = ψ χx δs − γx δss′ − γδs′ − (aux )x δs dt + o(δv)
Z (37)
x=s(t)
T1
Applying the boundary condition for u on the moving boundary s (4) similarly to the derivation of (36)–(37), it follows that, for t ∈ T2 , (38)
Z
T2
[ψaδux ]x=s(t) dt =
Z
T2
h i ψ χx δs − γx s′ δs − γδs′ − (aux )x δs
x=s(t)
dt + o(δv)
Using (37) and (38) in (35), it follows that ∆I =
Z
b Ω
− (39)
−
[(aψx )x − (ψb)x + ψc + ψt ] δu dx dt +
Z
ψδf dx dt +
Ω Z T 0
Z
T 0
h
ψ(0, t)δg(t) dt +
′
Z
0
T
s(t), t) dt− −aψx + b + s′ ψ x=s(t) δu(b
ψ χx δs − γx s δs − γδs′ − (aux )x δs Z
0
T
(aψx − bψ)δu x=0 dt −
Z
s b(T )
i
x=s(t)
dt−
ψ(x, T )δu(x, T ) dx + o(δv) 0
8
OPTIMAL CONTROL OF PARABOLIC FREE BOUNDARY PROBLEMS
Taking the sum of ∆I and ∆J using (28) and (29), (32), (33) gives Z sb(T ) h Z T i ∆L = 2β0 (u − w) − ψ δu dx + s(t), t) dt+ −aψx + b + s′ ψ + 2β1 (u − µ) x=s(t) δu(b t=T
0
+
+ − (40)
−
Z
Z
Z
T
0
b Ω
2β1 (u − µ)ux
x=s(t)
0
δs(t) dt + [β0 (u(s(T ), T ) − w(s(T )))2 + 2β2 (s(T ) − s∗ )]δs(T )+
[(aψx )x − (ψb)x + ψc + ψt ] δu dx dt + ψδf dx dt +
Ω Z T
Z
0
T
Z
T 0
(aψx − bψ)δu x=0 dt+
h i ψ χx δs − γx s′ δs − γδs′ − (aux )x δs
x=s(t)
dt−
ψ(0, t)δg(t) dt + o(δv)
0
Due to arbitraricity of the the incremented variables δu, δs, etc. all of the coefficients on these variables must be zero. In particular, it follows that ψ should satisfy (19)–(22) in a pointwise a.e. sense. All of the remaining terms depend linearly on the increment δv, so the Frechet differential dJ is (23). In this form, the adjoint problem (19)–(22) plays the role of the Lagrange multiplier corresponding to the PDE “constraint” in this setting. 3. Preliminary Results 3.1. Existence Results and Energy Estimates. 2ℓ,ℓ 3.1.1. Bp,x,t (QT )-Solutions. Consider the problem
auxx + bux + cu − ut = f in QT
(41)
a(0, t)ux (0, t) = χ1 (t), 0 ≤ t ≤ T
(42)
a(1, t)ux (1, t) = χ2 (t), 0 ≤ t ≤ T
(43)
u(x, 0) = φ(x), 0 ≤ x ≤ 1
(44)
Let ℓ > 1 be fixed, p > 1. The following key result is due to Solonnikov [8] Lemma 1. [8, §7, thm. 17] Suppose that ∗
2ℓ −2,ℓ a, b, c ∈ Cx,t
(45)
∗
−1
(QT ), arbitrary ℓ∗ > ℓ
2ℓ− 2
1− 1 ℓ− 2 2p
2ℓ−2,ℓ−1 f ∈ Bp,x,t (QT ), φ ∈ Bp p (0, 1), χ1 , χ2 ∈ Bp i h 3 − 12 holds; that is, and the consistency condition of order k = ℓ − 2p
(46)
∂ j (aux ) ∂ j (aux ) d j χ1 d j χ2 (0, 0) = (0), (1, 0) = (0), j j j ∂x dt ∂x dtj Then the solution u of (41)–(44) satisfies the energy estimate h (47) kukB2ℓ,ℓ (Q ) ≤ C kf kB2ℓ−1,ℓ−1 (Q ) + kφkB2ℓ−2/p (0,1) + kχ1 k ℓ− 1 − p,x,t
p,x,t
T
p
Bp
2
j = 0, . . . , k
1 2p
(0,T )
+ kχ2 k
ℓ− 1 − 1 2 2p Bp (0,T )
i
3 2p
6∈ Z+ , and when ℓ ∈ Z+ , h kukW 2ℓ,ℓ (Q ) ≤ C kf kW 2ℓ−1,ℓ−1 (Q
when ℓ, ℓ − (48)
T
(0, T )
p,x,t
T
p,x,t
T
+ kφkB2ℓ−2/p (0,1) + kχ1 k ) p
ℓ− 1 − 1 2 2p (0,T ) Bp
+ kχ2 k
ℓ− 1 − 1 2 2p Bp (0,T )
i
In particular, energy estimates (47), (48) imply the existence and uniqueness of the solution in respective spaces 2ℓ,ℓ 2ℓ,ℓ Bp,x,t (QT ) or Wp,x,t (QT ). Note that when k = 0, the consistency condition of order k is the condition of continuity of the boundary functions: (aux )(0, 0) = a(0, 0)φ′ (0) = χ1 (0),
(aux )(1, 0) = a(1, 0)φ′ (1) = χ2 (0)
OPTIMAL CONTROL OF PARABOLIC FREE BOUNDARY PROBLEMS
9
3.2. Traces and Embeddings of Besov Functions. For functions u ∈ W22,1 (Ω), the applicability of the boundary conditions are justified by the following trace and regularity results. Lemma 2. [10, lem. II.3.3] If u ∈ W22,1 (Ω), then u has a Hölder continuous representative in Ω; in particular, 1/2,1/4 u ∈ Cx,t (Ω). Moreover [10, lem. II.3.4], the following bounded embeddings of traces hold: 1/4 3/4 u s(t), t , u(0, t) ∈ B2 (0, T ), ux s(t), t , ux (0, t) ∈ B2 (0, T )
and for any fixed 0 ≤ t¯ ≤ T ,
u(·, t¯) ∈ W21 (0, s(t¯)) Lemma 3. [8, §4, thm. 9] For a function u ∈ B22ℓ,ℓ (QT ), the following bounded embeddings of traces hold: for any fixed 0 ≤ t ≤ T, u(·, t) ∈ B22ℓ−1 [0, 1] when ℓ > 1/2
For any fixed 0 ≤ x ≤ 1, ℓ−1/4
u(x, ·) ∈ B2 ux (x, ·) ∈ uxx (x, ·) ∈
[0, T ] when ℓ > 1/4
ℓ−3/4 [0, T ] B2 ℓ−5/4 [0, T ] B2
when ℓ > 3/4 when ℓ > 5/4
3.3. Consequences of Energy Estimates and Embeddings. For given v = (f, s, g) ∈ VR transform the domain Ω to the cylindrical domain QT by the change of variables y = x/s(t). Let d = d(x, t), (x, t) ∈ Ω stand for any of a, b, c, f, γ, χ, define the function d˜ by ˜ t) = d xs(t), t , and φ(x) ˜ d(x, = φ xs(t) The transformed function u e is a pointwise a.e. solution of the Neumann problem 1 1 a ˜u ey y + ˜b + ys′ (t) u (49) ey + c˜u e−u et = f˜, in QT s2 s ˜ (50) u e(x, 0) = φ(x), 0≤x≤1 a ˜(0, t)e uy (0, t) = g(t)s(t), 0 ≤ t ≤ T
(51)
(52)
a ˜(1, t)e uy (1, t) = χ(1, ˜ t)s(t) − γ˜ (1, t)s′ (t)s(t), 0 ≤ t ≤ T
Lemma 4. For fixed v ∈ VR , there exists a unique solution u ∈ W22,1 (Ω) of the Neumann problem (1)–(4) for which the 5/2+2α,5/4+α (QT ) solves (49)–(52) and satisfies the following energy estimate transformed function u e ∈ B2,x,t ke ukB5/2+2α,5/4+α (Q ) ≤ C kf kB1,1/4+α (D) + kφkB3/2+2α (0,s ) + 0 T 2 2,x,t 2,x,t + kgkB1/2+α [0,T ] + kχkB3/2+2α∗ ,3/4+α∗ (D) + kγkB3/2+2α∗ ,3/4+α∗ (D) (53) 2
2,x,t
2,x,t
where α∗ > α is arbitrary.
Proof. Assumptions (13)–(18) imply the applicability of Lemma 1 with p = 2, ℓ = 5/4 + α to the Neumann problem (49)–(52); by energy estimate (47),
h
ke ukB5/2+2α,5/4+α (Q ) ≤ C f˜ 1/2+2α,1/4+α + φ˜ 3/2+2α + kg(t)s(t)kB1/2+α (0,T ) + T 2,x,t B2,x,t (QT ) B2 (0,1) 2 i
′ + χ(1, ˜ t)s(t) − γ˜ (1, t)s (t)s(t) 1/2+α B2
(0,T )
3/2+2α∗ ,3/4+α∗ B2,x,t (QT ),
By Sobolev trace embedding recalled in Lemma 3 for functions in
χ(1, ˜ t)s(t) − γ ˜ (1, t)s′ (t)s(t) B1/2+α (0,T ) ≤ C kχk ˜ B3/2+2α∗ ,3/4+α∗ (Q 2
2,x,t
T)
we have
+ k˜ γ kB3/2+2α∗ ,3/4+α∗ (Q 2,x,t
T)
10
OPTIMAL CONTROL OF PARABOLIC FREE BOUNDARY PROBLEMS
Considering the term f˜
˜
f
1/2+2α,1/4+α
B2,x,t
1/2+2α,1/4+α
B2,x,t
(QT )
= f˜
, by definition
L2 (QT )
Z
+
(QT )
T 0
Z
+
T 0
Z
Z
1
T 0
Z
1 0
Z
1
|f (xs(t), t) − f (ys(t), t)|2 |t − τ |1+2(1/4+α)
0
|f (ys(t), t) − f (ys(τ ), τ )|2 |t − τ |1+2(1/4+α)
0
dy dt dτ
dy dx dτ
!1/2
!1/2
+
The first two terms are estimated in a straightforward way in terms of kf kB1/2+2α,1/4+α (D) ; the last term can be 2,x,t
estimated through the sum of two terms I1 + I2 where !1/2 Z TZ TZ 1 |f (ys(τ ), t) − f (ys(τ ), τ )|2 dy dt dτ , I1 = |t − τ |1+2(1/4+α) 0 0 0 !1/2 Z TZ TZ 1 |f (ys(t), t) − f (ys(τ ), t)|2 I2 = dy dt dτ |t − τ |1+2(1/4+α) 0 0 0
I1 can also be estimated in terms of kf kB1/2+2α,1/4+α (D) ; the last requires increased regularity of f with respect to 2,x,t
the space variable: by CBS inequality, I2 ≤
Z
0
T
Z
T 0
Z
1 0
Z
1 0
fx y(θs(t) + (1 − θ)s(τ )), t 2 |s(t) − s(τ )|2 |t − τ |1+2(1/4+α)
dθ dy dt dτ
!1/2
By mean value theorem, Morrey’s inequality, and condition (12) on s,
Similarly,
1 I2 ≤ C s′ W 1 [0,T ] T 3/4−α √ kfx kL2 (D) 2 δ kχk ˜ B3/2+2α∗ ,3/4+α∗ (Q
≤ C kχkB3/2+2α∗ ,3/4+α∗ (D)
k˜ γ kB3/2+2α∗ ,3/4+α∗ (Q
≤ C kγkB3/2+2α∗ ,3/4+α∗ (D)
T)
2,x,t
T)
2,x,t
2,x,t
2,x,t
Estimate (53) follows.
By the trace embedding result of Lemma 2, it follows that the functional J (v) of (11) is well defined for v ∈ VR . Lemma 5. For fixed v ∈ VR , given the corresponding state vector u = u(x, t; v), there exists a unique solution ψ ∈ W22,1 (Ω) of the adjoint problem (19)–(22) and the following energy estimate is valid kψkW 2,1 (Ω) ≤ C kf kL2 (Ω) + kφkW 1 (0,s0 ) + kgkB1/2+α [0,T ] + kχk 1+2α, 1 +α + 2
2
B2,x,t 2
2
(54)
+ kskW 2 [0,T ] kγk 2
1
1+2α, +α B2,x,t 2 (D)
+ kwkW 1 [0,s(T )] + kµkB1/4 (0,T ) 2
2
(D)
Proof. By Lemma 4, there exists a solution u of the problem (1)–(4). In particular, W22,1 -norm of the solution satisfies the following energy estimate
(55) kukW 2,1 (Ω) ≤ C kf kL2 (Ω) + kφkW 1 [0,s0 ] + kgkB1/4 (0,T ) + χ|x=s(t) + γ|x=s(t) s′ (t) B1/4 (0,T ) 2
2
2
2
Indeed, by applying (48) to the transformed solution u e, and due to equivalence of W21 [0, ℓ] and B21 [0, ℓ] norms, (55) follows. From the trace embedding result of Lemma 3 it follows that + kγk 1+2α, 1 +α kχ(s(t), t)kB1/4 (0,T ) + kγ(s(t), t)kB1/4 (0,T ) ≤ C kχk 1+2α, 1 +α 2
2
B2,x,t 2
(D)
B2,x,t 2
(D)
OPTIMAL CONTROL OF PARABOLIC FREE BOUNDARY PROBLEMS
11
and therefore from (55) we get kukW 2,1 (Ω) ≤ C 2
kf kL2 (Ω) + kφkW 1 [0,s0 ] + kgkB1/4 (0,T ) + kχk 2
+ kskW 2 [0,T ] kγk
(56)
2
1+2α, 1 +α (D) B2,x,t 2
1
1+2α, +α B2,x,t 2 (D)
2
+
From [10, thm. IV.9.1] and [11, thm. 5.4] it follows that there exists a unique solution ψ ∈ W22,1 (Ω) of the third type adjoint initial boundary value problem (19)–(22) and the following energy estimate is satisfied: kψkW 2,1 (Ω) ≤ C ku(x, T ; v) − w(x)kW 1 [0,s(T )] + ku(s(t), t) − µ(t)kB1/4 (0,T ) 2
2
2
By embedding theorem recalled in Lemma 2 [10, lem. II.3.3] it follows kψkW 2,1 (Ω) ≤ C kukW 2,1 (Ω) + kwkW 1 [0,s(T )] + kµkB1/4 (0,T ) 2
2
2
2
Applying energy estimate (3.3), (54) follows. Lemma is proved.
Lemma 6. Problem I has a solution, i.e. V∗ := v ∈ VR : J (v) = J∗ =: inf J (v) 6= 0 v∈VR
The proof proceeds in a similar way to [1, thm. 1.4]. Proof. Let {vn = (sn , gn , fn )} ⊆ VR be a minimizing sequence, i.e. J (vn ) → J∗ Since VR is a bounded and closed subset of an Hilbert space H, {vn } is weakly precompact (see e.g. [12, ch. V, § 2, p.126]); that is, there exists a subsequence vnk which converges weakly. Assume the whole sequence converges weakly to v∗ = (s∗ , g∗ , f∗ ) ∈ VR . Compact Sobolev embedding theorem implies vn → v strongly in W21 [0, T ] × L2 [0, T ] × L2 (D). The corresponding solutions un ∈ W22,1 (Ω) and transformed solutions u en ∈ W22,1 (QT ) are uniformly bounded by (55); that is, there exists C > 0 such that ke un kW 2,1 (QT ) ≤ C 2
It follows that there exists an element v˜ ∈
W22,1 (QT )
such that for some subsequence nk ,
u enk → v˜ weakly in W22,1 (QT )
Multiplying (1) written for u enk by an arbitrary test function Φ ∈ L2 (QT ) and passing to the limit as nk → ∞, it follows that v˜ is a W22,1 (QT ) weak solution of (49)–(52). Since all weak limit points of {e un } are W22,1 (QT ) weak solutions of the same equation, by uniqueness of the weak solution, it follows that v˜ = u e∗ and the whole sequence u en → u e∗ weakly in W22,1 (QT ). Sobolev trace theorem [4] then implies that u en (y, T ) → u e∗ (y, T ) in C[0, 1]
and hence
un (ysn (T ), T ) → u∗ (ys∗ (T ), T ) in C[0, 1] Together with the convergence of sn (t) → s∗ (t) uniformly on [0, T ], it follows that Z
sn (T ) 0
|un (x, T ) − w(x)|2 dx −
Z
s∗ (T ) 0
|u∗ (x, T ) − w(x)|2 dx → 0
12
OPTIMAL CONTROL OF PARABOLIC FREE BOUNDARY PROBLEMS
The other two terms in J (v) are handled similarly, so it follows that lim J (vn ) = J (v∗ ) = J∗
n→∞
Lemma is proved.
The following technical Lemma plays a key role in showing that remainder terms appearing in the derivation of formula (40) for the increment are of higher than linear order with respect to δv, and hence in completing the rigorous proof of the main result. Take u e, a ˜, etc. as in the proof of Lemma (4), transform the domain Ωs¯ to QT by e e taking y¯ = x/s(t), etc. in a similar way, and define u, a, etc. For d standing for any of u, f , a, b, c, γ, χ, denote by ˜ t), (y, t) ∈ QT ˜ t) = e d(y, t) − d(y, ∆d(y,
Lemma 7. If a, b, c satisfy (13), (14) and χ, γ satisfy (16), then k∆e ukB5/2+2α,5/4+α (Q
(57)
T)
2,x,t
→ 0 as δv → 0 in H.
Proof. A straightforward calculation shows that ∆e u solves e e e 1 ay e a ˜ a a ′ ˜ e ∆e u + u eyy + ∆e u + + b + y¯ s c∆e u − ∆e u = ∆ f + − y yy t s¯2 s¯ s¯ s2 s¯2 ! e ˜b e s′ s¯′ a ˜y ay b (58) u ey − ∆˜ cu e + − + − + − s2 s¯2 s s¯ s s¯ in QT ,
∆e u(x, 0) = 0, 0 ≤ x ≤ 1
(59)
e a(0, t)∆e uy (0, t) = δs(t)g(t) + s(t)δg(t) + δs(t)δg(t), 0 ≤ t ≤ T
(60) and
(61)
e t) − γ(1, e t)¯ e a(1, t)∆e uy (1, t) = −∆˜ a(1, t)e uy (1, t) + χ(1, s′ (t) δs(t)+ e t)δs′ (t) − ∆˜ + ∆χ(1, ˜ t) − γ(1, γ (1, t)s′ (t) s(t) 5/2+2α,5/4+α
, it follows from (58)–(61), and Minkowski inequality that By the energy estimate (47) for functions in B2,x,t X 6
a(0, t)∆e uy (0, t) 1/2+α + kΓi kB1/2+2α,1/4+α (Q ) + e k∆e ukB5/2+2α,5/4+α (Q ) ≤ C T
2,x,t
(62)
where
1/2+α
B2
[0,T ]
Γ2 =
B2
[0,T ]
e e a ˜ a ˜y a ay − u e , Γ = − u ey yy 3 s2 s¯2 s2 s¯2 ! ′ ˜b e s s¯′ b u ey , Γ5 = − − u ey , Γ6 = ∆˜ cu e s s¯ s s¯
Γ1 = ∆f˜, Γ4 =
T
2,x,t
i=1
+ e a(1, t)∆e uy (1, t)
The last two terms in (62) converge to zero by (60), (61). The remaining terms are all estimated in a similar way; we demonstrate the estimation for one such term. Since s → s uniformly as δv → 0, to prove that kΓ2 k → 0, due to Lebesgue’s dominated convergence theorem, it follows that it is sufficient to show that the integrands in
a
e
˜u
au
e e , yy yy
s2
1/2+2α,1/4+α
s2
1/2+2α,1/4+α (QT ) B (QT ) B 2,x,t
2,x,t
OPTIMAL CONTROL OF PARABOLIC FREE BOUNDARY PROBLEMS
13
are bounded above by an integrable function. We demonstrate how the first term is estimated; the second is estimated in a nearly identical way. By definition,
Z T Z 1 1/2 Z T Z 1 Z 1 1/2
a
˜u
e = I dy dt + I dy dx dt + yy 1 2
s2
1/2+2α,1/4+α 0 0 0 0 0 (QT ) B2,x,t 1/2 Z T Z T Z 1 I3 dy dt dτ + (63) 0
0
0
where 2 a(ys(t),t) 2 eyy (y, t) − a(xs(t),t) u eyy (x, t) s2 (t) u 1 s2 (t) uyy (y, t) , I2 = I1 = 2 a(ys(t), t)e s (t) |x − y|1+2(1/2+2α) 2 a(ys(t),t) ),τ ) eyy (y, t) − a(ys(τ u eyy (y, τ ) s2 (t) u s2 (τ ) I3 = |t − τ |1+2(1/4+α)
Estimate in I1 using the condition (12) on s and condition (14) on a, and (53) to derive I1 ≤
(64)
1 kak2C(D) |e uyy (y, t)|2 δ2
By (12), (14), I2 ≤
(65)
2 δ2
ℓ2 kax k2C(D) |e uyy (y, t)|2 |x − y|−4α + kak2C(D) 5/2+2α,5/4+α
Both I1 and I2 are integrable by definition of the norm in B2,x,t
|e uyy (y, t) − u eyy (x, t)|2 |x − y|2+4α
(QT ).In a similar way, derive
I3 ≤ 4 (I31 + I32 + I33 + I34 ) where I31 = I33 =
|a(ys(t), t) − a(ys(τ ), t)|2 |e uyy (y, t)|2 3/2+2α
,
|t − τ | s4 (t) 2 2 2 |a(ys(τ ), τ )| s (τ ) − s2 (t) |e uyy (y, t)|2 |t − τ |3/2+2α |s2 (t)s2 (τ )|2
I32 = ,
|a(ys(τ ), t) − a(ys(τ ), τ )|2 |e uyy (y, t)|2
I34 =
|t − τ |3/2+2α s4 (t)
|a(ys(τ ), τ )|2 |e uyy (y, t) − u eyy (y, τ )|2 |t − τ |3/2+2α s4 (τ )
By estimate (12) on s, assumption (14), mean value theorem and Morrey inequality, I31 + I32 + I33 ≤ +C
C ks′ k2
W21 [0,T ]
kak2C(D)
T 1/2−2α kax k2C(D) δ4
2 4ℓ2 ks′ kW 1 [0,T ] 2
δ4
T 1/2−2α
5/2+2α,5/4+α
+
kak2C 0,1/4+α∗ (D) x,t
|t − τ |1−2(α
∗ −α)
δ4
|e uyy (y, t)|2
which is integrable by definition of norm in B2,x,t (QT ) and integrability of the singularities on the curve t = τ in R2 . 5/2+2α,5/4+α The term I34 requires increased time regularity of u eyy , which, for arbitrary functions in B2,x,t (QT ), does not follow from the definition of the norm. However, since u e is a pointwise a.e. solution of (49), estimate (12) on s, and assumption (14) (a ∈ C(D)), and triangle inequality, it follows that I34 ≤
4 kak2C(D) δ4
1 2 3 4 I34 + I34 + I34 + I34
14
OPTIMAL CONTROL OF PARABOLIC FREE BOUNDARY PROBLEMS
where
1 I34 =
2 I34 =
3 I34 =
4 I34 =
h a˜ +s˜b+ys′ s it 2 x u e y a ˜ τ
|t − τ |3/2+2α 2 a(xs(τ ), τ )s2 (t)˜ c(x, t)e u(x, t) − a(xs(t), t)s2 (τ )˜ c(x, τ )e u(x, τ ) |t − τ |3/2+2α |a(xs(τ ), τ )a(xs(t), t)|2 2 a(xs(τ ), τ )s2 (t)e ut (x, t) − a(xs(t), t)s2 (τ )e ut (x, τ )
|t − τ |3/2+2α |a(xs(τ ), τ )a(xs(t), t)|2 2 a(xs(τ ), τ )s2 (t)f˜(x, t) − a(xs(t), t)s2 (τ )f˜(x, τ ) |t − τ |3/2+2α |a(xs(τ ), τ )a(xs(t), t)|2
Each term is now easily estimated using Sobolev embedding; we demonstrate the estimation of the highest regularity 3 term I34 . By triangle inequality, By triangle inequality, condition (13) on a and condition (12) on s, assumption (14) 0,1/4+α∗ (D),) mean value theorem, and Morrey’s inequality, (ax ∈ C(D) and a ∈ Cx,t 3 I34 ≤
(66)
2 ℓ4 |e ut (x, t)|2 ℓ4 kax k2C(D) s′ W 1 [0,T ] |e + ut (y, t)|2 T 1/2−2α + C 4 kak2C 0,1/4+α (D) ∗ 4 2 a0 a0 x,t |t − τ |1−2(α −α) ℓ4 |e ut (x, t) − u et (x, τ )|2 4ℓ2 2 ut (y, t)|2 + 2 + 2 s′ W 1 [0,T ] T 1/2−2α |e 2 a0 a0 |t − τ |3/2+2α 5/2+2α,5/4+α
which is integrable by the definition of the norm in B2,x,t
(QT ).
4. Proof of Main Result In this section, we give a rigorous justification for the formula given for the first variation of the functional J in Section 2; repeated calculations will now be omitted with a reference to the previous equation. Note that the heuristic derivation using the formulated Lagrangian is invalid since the adjoint ψ is only defined in Ω; instead, the functional, direct, and adjoint equations are transformed explicitly. Lemma 6 establishes the existence of the solution; the formula for the gradient is to be established. The important facts now used are 5/2+2α,5/4+α
• Under the assumptions (13)–(18), and given v ∈ VR , there exists a unique solution u ∈ B2,x,t of (1)–(4) (Lemma 4), • Subsequently, the corresponding adjoint ψ ∈ W22,1 (Lemma 5) also exists. • Then, precise embedding results (Lemmas 2 and 3) apply to guarantee that all of the traces appearing in the derivation exist as well. Proceed just as in Section 2, partitioning the time domain as [0, T ] = T1 ∪ T2 as before, and consider first the increment of the functional, ∆J . Decomposing the increment as in (28), formula (29) is valid with o(δv) replaced by R1 + · · · + R5 where Z s(T ) Z s(T ) R1 = 1T1 (T )β0 |δu(x, T )|2 dx |δu(x, T )|2 dx, R2 = 1T2 (T )β0 0 0 s(T ), T ) − w(˜ s(T ))|2 − |u(s(T ), T ) − w(s(T ))|2 δs(T ) R3 = 1T1 (T )β0 |u(˜ R4 = 1T2 (T )β0 |u(˜ s(T ), T ) − w(˜ s(T ))|2 − |u(s(T ), T ) − w(s(T ))|2 δs(T ) R5 = 1T2 (T )β0 |u(s(T ), T ) − w(s(T ))|2 − |u(s(T ), T ) − w(s(T ))|2 δs(T )
OPTIMAL CONTROL OF PARABOLIC FREE BOUNDARY PROBLEMS
15
Similarly, formula (32) is valid with o(δv) replaced by R6 + · · · + R10 where Z Z s(t), t)δs(t)|2 dt β1 |δu(s(t), t) + ux (˜ R6 = β1 |ux (˜ s(t), t)δs(t) + δu(s(t), t)|2 dt, R7 = T2 T1 Z Z x=s(t) x=˜ s(t) R8 = 2β1 u − µ x=s(t) [ux ]x=s(t) δs(t) dt 2β1 u − µ x=s(t) [ux ]x=˜s(t) δs(t) dt, R9 = T2 T1 Z R10 = 2β1 u(s(t), t) − µ(t) (ux (s(t), t) − ux (s(t), t)) δs(t) dt T2
Lastly,
J3 = 2β2 (s(T ) − s∗ )δs(T ) + R11 where R11 = β2 |δs(T )|2
(67)
Proceeding as before with the term ∆I, formula (35) is valid with o(δv) replaced by R12 + R13 where Z Z s(t) Z x=s(t) R12 = ψδf dx dt, R13 = − aψx + bψ δu(s(t), t) dt T1
s(t)
x=s(t)
T1
Using the boundary condition (4) for u on the moving boundary s, as in (36), (37) it follows that Z Z h 18 X i ψaδux (68) Ri dt + dt = ψ χx δs − γx δss′ − γδs′ − (aux )x δ x=s(t)
T1
where
x=s(t)
T1
i=14
Z
Z h i h i x=s(t) x=s(t) [ψ]x=s(t) χx δs − γx δss′ − γδs′ δs(t) dt dt, R15 = [ψ]x=s(t) (aux )x x=˜ s(t) x=s(t) T1 T1 Z i h R16 = ψ(s(t), t) χx (˜ s(t), t) − χx (s(t), t) − (γx (˜ s(t), t) + γx (s(t), t)) s(t) δs(t) dt T Z 1 Z x=˜s(t) R17 = − ψ(s(t), t)γx (s(t), t)δs(t)δs′ (t) dt, R18 = − ψ(s(t), t)(aux )x δs(t) dt
R14 =
T1
x=s(t)
T1
Applying the boundary condition (4) for u on the moving boundary s, as in (38), it follows that Z Z h 22 X i dt + (69) Ri ψaδux dt = ψ χx δs − γx δss′ − γδs′ − (aux )x δ x=s(t)
T2
where
i=19
Z
i h ψ(s(t), t) χx (˜ s(t), t) − χx (s(t), t) − γx (˜ s(t), t) − γx (s(t), t) s′ (t) δs(t) dt T Z 2 Z h ix=˜s(t) =− ψ(s(t), t)γx (s(t), t)δs′ (t)δs(t), R21 = − ψ(s(t), t) (aux )x δs(t) dt x=s(t) T2 T2 Z h i δs(t) dt R22 = − ψ(s(t), t) aux x − aux x
R19 = R20
x=s(t)
T1
x=s(t)
T2
Therefore, taking the sum of ∆J using decomposition (28) with (29), (32), and (67) and ∆I using identities (68) and (69), as in formula (40) the the following formula for the increment of J is valid ∆J (v) = dJ (v) +
22 X
Ri .
i=1
Denote by C a constant depending on the fixed data Ω, D, VR , a, etc. For R22 , applying Morrey’s inequality and Sobolev embedding of Lemma 2, derive Z h i |R22 | ≤ C kψkW 2,1 (Ω) kδskW 1 [0,T ] aux x − aux x x=s(t) dt 2 2 T2
16
OPTIMAL CONTROL OF PARABOLIC FREE BOUNDARY PROBLEMS
Expand on the right-hand side as ≤ C kψkW 2,1 (Ω) kδskW 1 [0,T ] 2
2
Z
T2
h i dt ax ux − ax ux + auxx − auxx x=s(t)
By Minkowski inequality and CBS inequality, it follows that √ |R22 | = kψkW 2,1 (Ω) kδskW 1 [0,T ] T [R22,1 + R22,2 ] 2
2
where R22,1 =
Z
T2
2 !1/2 dt , x=s(t)
h i ax ux − ax ux
R22,2 =
Z
T2
h 2 !1/2 i auxx − auxx dt x=s(t)
Take the transformation y = x/s(t) and apply Minkowski inequality to derive "Z 1/2 Z 1/2 # q 2 2 |∆e uxx (1, t)| dt |e uxx (1, t) − u exx (s(t)/s(t), t)| dt + R22,2 ≤ kakC T2
T2
By trace embedding of Theorem 3 with ℓ = 5/4 + α and Lemma 7, it follows that Z 1/2 2 |∆e uxx (1, t)| dt ≤ k∆e ukB5/2+2α,5/4+α (Q ) → 0 as δv → 0 T2
2,x,t
T
5/2+2α,5/4+α
Since u e ∈ B2,x,t (QT ), the trace u exx exists; by uniform convergence of s to s as δv → 0, it follows that s/s → 1 as δv → 0. Then continuity of L2 norm with respect to shift implies Z 1/2 |e uxx (1, t) − u exx (s(t)/s(t), t)|2 dt → 0 as δv → 0 T2
Hence R22,2 → 0 as δv → 0. The proof that R22,1 → 0 as δv → 0 is similar. It follows that R22 = o(δv). Each remainder term Ri is shown to be of higher than linear order with respect to δv using a similar application of the energy estimates from Lemmas 4, 5, 7, trace embedding theorem [8, §3, thm. 9], and the continuity of L2 with respect to shift. Theorem is proved. Acknowledgement. This research was funded by National Science Foundation: grant #1359074–REU Site: Partial Differential Equations and Dynamical Systems at Florida Instiute of Technology. Two REU students Jessica Pillow and Dylanger Pittman worked on part of the project restricted to heuristic derivation given in Section 2 with the intention to implement gradient type method for numerical analysis. References [1] Abdulla, U.: 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. Inverse Problems and Imaging 7(2), 307–340 (2013). DOI 10.3934/ipi.2013.7.307 [2] Abdulla, U.: On the optimal control of the free boundary problems for the second order parabolic equations. II. Convergence of the method of finite differences. Inverse Problems and Imaging (To Appear) (2016). URL http://arxiv.org/abs/1506.02341 [3] Gol’dman, N.: Inverse Stefan Problems. Kluwer Academic Publishers Group, Dodrecht (1997) [4] Besov, O.V., Ilin, V.P., Nikolskii, S.M.: Integral Representations of Functions and Imbedding Theorems, vol. 1. John Wiley & Sons (1979) [5] Besov, O.V., Ilin, V.P., Nikolskii, S.M.: Integral Representations of Functions and Imbedding Theorems, vol. 2. John Wiley & Sons (1979) [6] Kufner, A., John, O., Fuˇcik, S.: Function Spaces. Noordhoff International Publishing, Leyden, The Netherlands (1977) [7] Nikol’skii, S.M.: Approximation of Functions of Several Variables and Imbedding Theorems. Springer-Verlag, New York-Heidelberg (1975) [8] Solonnikov, V.A.: A-priori estimates for solutions of second-order equations of parabolic type, Trudy Matematischeskogo instituta im. V. A. Steklova, vol. 70. Nauka, Moscow-Leningrad (1964) [9] Evans, L.C., Gariepy, R.F.: Measure Theory and Fine Properties of Functions. Studies in Advanced Mathematics. CRC Press (1992) [10] Ladyzhenskaya, O.A., Solonnikov, V.A., Uraltseva, N.N.: Linear and Quasilinear Equations of Parabolic Type, Translations of Mathematical Monographs, vol. 23. American Mathematical Society, Providence, R. I. (1968) [11] Solonnikov, V.A.: On boundary value problems for linear parabolic systems of differential equations in general form. Proceedings of the Steklov Institute of Mathematics 83, 1–184 (1965) [12] Yosida, K.: Functional Analysis. Classics in Mathematics. Springer (1996)
