Inverse problem for a linearized quasi-stationary phase field model with degeneracy

Preface

Let Q C R ⁿ be a bounded domain with a boundary д Q оf C^ class. T >  0. в, 5 € R-Consider the initial-boundary value problem

( в + A)( v ( x, 0) — v 0( x )) = 0 , x € Q ,

(1 — 5)v + 5—(x, t) = (1 — 5)w + 5—— (x, t) = 0, (x, t) € дQ x [0, T], ∂n                ∂n for the system of equations vt(x, t) = Av(x, t) — Aw(x, t) + bi(x, t)u(t), (x, t) € Q x [0, T],

0 = v + (в + A)w + b2(x, t)u(t),   (x, t) € Q x [0, T], with overdetermination condition on the subspace of degeneracy jK (У) w (y,t) dy = ф (t), (x,t) € Q x [0 ,T ].

Ω

Up to a linear change of functions v ( x,t ), w ( x,t ), the system coincides with the linearization of the quasistationary phase-field model, describing phase transitions of the first kind in terms of the mesoscopic theory. The unknown functions of the inverse problem (l)-(5) are v ( x,t ), w ( x,t ), u ( t ). The problem is investigated within the framework of a linear inverse problem for an abstract differential equation with a degenerate operator at the derivative, i.e. the Sobolev type equation.

Linear inverse problems for the Sobolev type equations were studied in [1, 2] with an unknown time-independent element u. Problems with an unknown time-dependent element u were considered in linear case in [3], and in nonlinear case in [4, 5]. However, an overdetermination operator, herewith, acted on the resolving Sobolev type equations semigroup image. In the present paper this operator acts on the semigroup kernel.

1.    Statement of the abstract problem

Let X. Y and U be Banach spaces. Consider operators L E L ( X ; Y ) (i. e. linear and continuous) with ker L = { 0 }. M E Cl ( X ; Y ) (linear, closed and densely defined). Ф E L ( X ; U ). B E C 1([0 ,T ]; L ( U ; Y ))■ funeLions y E C 1([0 ,T ]; Y )• Ф E C 1([0 ,T ]; U ) and an element x 0 E Dm- Here Dm is a domain о £ the operator M. endowed with the graph norm ||x 0 || d _m = ||x о Цд - + ll ^Mx о || y-

Theorem 1. [6] Let p E { 0 }U N, the operator M be strongly ( L,p^-radial. Then

• (1)    X = X ⁰ ®X ¹ , Y = Y ⁰ ®Y 1;

• (ii)    a projector along X ⁰ on X ¹ l,.along Y ⁰ оn Y 1) has the form

P = s- lim ( ^RL ( M )) ^{p +1} ,   ( Q = s- lim ( ^LL ( M )) ^{p +1});

• ц^ + ^   ^ц                  ц^ +oo    ^ц

(hi) QL = LP. QMx = MPx for all x E DM ;

• (IV) Lk = L\_xk E L ( X^k ; Y^k ) , Mk = M \ d _m ^k E Cl ( X^k ; Y^k )■ k = 0 , 1;

• (v)    operators M— ¹ E L ( Y ⁰; X ⁰) m id L— ¹ E L ( Y 1; X 1) exist:

• (vi)    the operator H = M— ¹ L 0 is nilpotent of a degree not greater, than p ;

• (vii)    there is a strongly continuous operators semigroup {V ( t ) E L ( X ) : t >  0 }, that resolves the equation. Lx(t ) = Mx ( t ).

While the operator M is strongly ( L,p )-radial, let us consider the inverse problem

Lx(t ) = Mx ( t ) + Bu ( t ) + y ( t ) , t E [0 , T ] ,

Px (0) = x 0 ,

Ф x ( t ) = ф( t ) , t e [0 ,T ] ,

which is to find a pair of functions x E C ¹ ([0 , T ]; X ) O C ([0 , T ]; Dm ) and u E C ¹ ([0 , T ]; U ), named as a solution.

Theorem 2. Let the operator M be strongly ( L,p)-radial, Ф E L ( X ; U ), Ф H = O. X ¹ C кегФ. B E C 1([0 ,T ]; L ( U ; Y ))■ У E C 1([0 ,T ]; Y )■ ( I — Q ) У E C^{p +1}([0 ,T ]; Y )■ Ф E C 1([0 ,T ]; U ). the inverse operator (Ф M 0 1( I — Q ) B ( t )) ^{— 1} exists for all t E [0 , T ] and (Ф M 0 1( I — Q ) B ) ^{— 1} E C 1([0 , T ]; L ( U )), x 0 E Dm OX 1. Then there is a unique solution ( x ; u ) of the problem (6)-(8). It

has the form

t x (t) = V (t) x 0 + У K (t —

s ) L- ¹ Q ( B ( s ) u ( s ) + y ( s )) ds—

—M— 1( I — Q ) B ( t ) u ( t ) — ]^ H^kM— ¹ (( I — Q ) y ( t ))^{( k} ) ,                  (9)

k =0

u ( t ) = — (Ф M— 1( I — Q ) B ( t )) - ¹

( Ф( t ) + ^^ Ф H^kM^— 1(( I — k =0

Q ) y )^{( k )}( t )j

and satisfies the following conditions:

^|x|C !([0 ,T ]; д ) ^ ^c ( ^|Px 0 ^ D m ^{+ 1} Ф li e !([0 ,T ]; U ) ^{+ |y|}CP ⁺¹([0 ,T ]; У )) ,               ⁽¹¹⁾

^|u|C 1([0 ,T ]; U ) ^ ^c ( ^{1 Ф}li e 1([0 ,T ]; U ) ^{+ |y|}e p ⁺¹([0 ,T ]; У )) ,                        ⁽¹²⁾

where a constant c >  0 does not ilepend on x 0. y. Ф.

Proof. Act with the operator Ф on the solution x of the direct problem (6), (7) with the known element u. Then

Ф x ( t ) = Ф( I - P ) x ( t ) = - Ф M- 1( I - Q ) B ( t ) u ( t ) - £ Ф H^kM- ¹ (( I - Q ) y )^{( k} )( t ) = Ф( t ) (13) k =0

lor all t E [0 , T ] due I,о the overdetermiimtion condition (8). the conditions X ¹ С кегФ. Ф H = O and the form of x (see [6, 7]). Thus, the formula (10) holds.

The formula (9) is obtained in [6], when u is unknown. At the same time the equality (13) is taken into account. Estimates (11), (12) follow from (9), (10) and an operator semigroup {E ( t ) E L ( X ) : t >  0 } exponential growth.

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2.    Inverse problem for a linearized quasi-stationary phase field model

Reduce the problem (l)-(5) to (6)-(8). To do this,let us assume X = Y = ( L ₂(Q))² , U = R ,

L =(O O) - ^M =( II    - +а) - B ⁽ t ⁾=( l 2( ::t )) - " t )= * ⁽ t ) ,

H 2(Q) = hh E H²(H) : QdL + (1 - 5 )) h ( x ) = 0 , x e d Q} ,

DM = ( H 2(Q))². Thereby. L E L ( X ) , M E Cl ( X ) , ker L = { 0 }.

Denote Aw = Aw, Da = H2(Q) С L2(Q). Let {pk : k E N} be orthonormal (in the sense of the scalar product (•, •) in L2(Q)) eigenfunctions of the operator A, numbered in decreasing eigenvalues {Xk : k E N}, counting multiplicities. Let -в E a (A). de fine 5k = (в + Xk) 1( в + 1 + Xk)Xk. f<)r Xk = -в- Using the exparision in the basis {pk : k E N} in a space L2(Q). determine operators

( rL ( M ))² =

E

^A k = -

b,V k iV k ⁽ Г '^ k )2

E

^ k = ^{— в}

T ^V k /^V k

^{( в +} ^ k ⁾⁽ Ц-S k ⁾²

O

O

( ^E   U V k lV k

A k = -g ⁽ ^^{-5 k )2} O

E

^A k = ^{— в}

A k V,W }V k

^{( в + A} k ⁾⁽ Ц-S k ⁾²

O

Hence, considering the Hilbert spaces X, Y , we obtain the strong ( L, 1)-radiality of the operator

M [6]. By formulas P = s lim ( ^RL ( M ))², Q = s- lim ( ^LL ( M ))² we derive the projectors Ц^ + ^   ^           ц^ + ^   ^

P =

E (•^,Uk^Pk

A k = -в

^- E

^A k = ^{— в}

(•,V k )V k ^{в + A} k

^O           / ^E (•-Pk ^Pk

I , ^Q =     ^A k = -g

OO

E

^A k = ^{— в}

A k (^,V k ^V k ^{в + A} k

O

Theorem 3. Let. -в E a ( A ). K E L 2(H). (K,p_k) = 0 jtrr X_k = -в- bi E C 1([0 ,T ]; L 2(H)). i = 1 , 2. and (b 1 ( •-t ) ,pk) = 0 Jtrr Xk = -в- (K,b 2( •-t ) ) = 0 for all t E [0 ,T ]. * E C ¹ [0 ,T ]. v 0 E Hl2(QY Then there exists a unique solution of the problem (l)-(5).

Proof. To prove this theorem it is sufficient to verify the conditions of Theorem 2.

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