CC BY
24
MSC 34A55, 65M32, 65L09 DOI: 10.14529/mmp190202
INVERSE PROBLEM FOR SOBOLEV TYPE MATHEMATICAL MODELS
A.A. Zamyshliaeva1, A.V. Lut1
1 South Ural State University, Chelyabinsk, Russian Federation
E-mails: [email protected], [email protected]
The work is devoted to the study of an inverse problem for the linear Sobolev type equation of higher order with an unknown coefficient depending on time. Since the equation might be degenerate the phase space method is used. It consists in construction of projectors splitting initial spaces into a direct sum of subspaces. Actions of operators also split. Therefore, the initial problem is reduced to two problems: regular and singular. The regular one is reduced to the first order nondegenerate problem which is solved via approximations. The needed smoothness of the solution is obtained. Then it is substituted into the singular problem which is solved using the methods of relatively polynomially bounded operator pencils theory. The main result of the work contains sufficient conditions for the existence and uniqueness of the solution to the inverse problem for a complete Sobolev type model of the second order. This technique can be used to investigate inverse problems of the considered type for Boussinesq-Love mathematical model.
Keywords: Sobolev type equation; inverse problem; mathematical models; equation of second order.
Dedicated to the 60-th birthday of outstanding mathematician Jacek Banasiak.
Introduction
Let U, F, Y be Banach spaces, operators A E L(U; F), ker A = {0}, B1, B0 E Cl(U; F), C E L(U; Y), the functions x : [0,T] ^ L(Y; F), f : [0,T] ^ F, tf : [0,T] ^ Y. Consider the following problem
Av''(t) = Biv'(t) + Bov(t) + x(t)q(t) + f (t), t E [0,T], (1)
v(0) = vo, (2)
v'(0) = vi, (3)
Cv(t) = tf(t). (4) One of the prototypes of equation (1) is
(A - A)vtt = a(A - A')vt + /3(A - A")v + g,
which models, among others, the incompressible fluid free surface perturbation under the assumption of motion potentiality and conservation of mass in a layer [1], longitudinal vibrations of an elastic rod [2] and wave processes in smectic and plasma [3]. The problem
of finding a pair of functions v(t) G C2([0,T];U) and q(t) G C1([0,T]; Y) from relations (1) - (4) is called the inverse problem.
Equation (1) refers to Sobolev type equations, since the operator A can be degenerate. Equations of this type of the first order are well studied. For example, there were proved the existence and uniqueness of classical solutions in the case of relatively sectorial operators [4], and [5] presented sufficient conditions for the unique solvability of the inverse problem using semigroups theory. In addition, there are articles devoted to the study of initialfinal problems [6,7], as well as to a numerical study of equations on the graphs [8], where the solution is sought in the form of the Galerkin sum. Sobolev type equations find their application in solving various technical problems, such as in [9]. Higher order equations were studied in [10] and [11] in the case of relatively bounded operator pencils. Semigroups theory finds application also in mathematical bioscience [12].
Problem (1) - (4) is inverse since, in addition to finding the function v(t), we need to restore the function q(t). Inverse problems were studied for the heat equation [13], for the second-order hyperbolic equations [14,15], and also for the multidimensional wave equation of the second order [16].
1. Preliminary Information and Reduction of Initial Inverse Problem
To find a pair of functions v(t) and q(t) we use the results obtained in the research of higher order Sobolev type equations [17]. Denote by B the pencil of operators B1,B0.
Definition 1. The pencil B is called polynomially A-bounded if
3 a G R+ fy G C (H > a) ^ R(B) G L(F;U)). Introduce an important condition:
JrA (B)dp = O, (A)
Y
where y = G C : = r > a}.
Lemma 1. Let the pencil B be polynomially A-bounded and condition (A) be fulfilled. Then the operators
are projectors.
Put U0 = ker P, F0 = ker Q, U1 = im P, F1 = im Q. From the previous Lemma it follows that U = U0 ®U1, F = F0 ©F1. Let Ak(Bk) denote the restriction of the operator A(Bi) onto Uk, k = 0,1; l = 0,1.
Theorem 1. Let the pencil B be polynomially A-bounded and condition (Â) be fulfilled. Then the actions of the operators split:
(i) Ak E L(Uk; Fk), k = 0,1;
(ii) Bk E L(Uk; Fk), k = 0,1, l = 0,1;
(iii) there exists an operator (A1 )-1 E L(F1;U1);
(iv) there exists an operator (B0)-1 E L(F0;U0).
Definition 2. Define the family of operators {Kj, Kj} as follows: K = Ho, Kj = -H1,
K1+1 = Kq2Ho, Kq2+1 = K1 - Kq2H1, q = 1, 2, ... where Ho = (B°)-1A0; H = (B0)-1B0.
Definition 3. The point to is called
(i) a removable singular point of the А-resolvent of pencil B, if Kj = O, Kj = O;
(ii) a pole of order p E N of the A-resolvent of pencil B, if Kp = O or Kj = O, but K1+1 = O, Kp+1 = O; ^
(iii) an essentially singular point of the А-resolvent of the pencil B, if Kj = O for any k E N.
Let the pencil B be polynomially A-bounded and condition (A) be fulfilled, then v(t) can be represented as v(t) = Pv(t) + (I - P)v(t) = u(t) + u(t). Suppose that U0 C ker C. Then, by virtue of Theorem 1 and Lemma 1, problem (1) - (4) is equivalent to the problem of finding the functions u E C2([0,T];U1), u E C2([0,T];U0), q E C 1([0,T]; Y) from the relations
u''(t) = S1u'(t) + S0u(t) + (A1)-1 Qx(t)q(t) + (A1 )-1Qf (t), (5)
u(0) = u0, u'(0) = u1, (6)
Cu(t) = tf(t) = Cv(t), (7)
H0U''(t) = H1u'(t) + u(t) + (B0)-1(1 - Q)x(t)q(t) + (B00)-1(1 - Q)f (t), (8)
u(0) = u0, u'(0) = u1, (9)
where S1 = (A1)-1B1, S0 = (A1)-1B01, u0 = Pv0, u1 = Pvb U0 = (I - P)v0, u1 = (I - P)v1, t E [0,T]. The inverse problem (5) - (7) is called regular, and problem (8), (9) is called singular.
2. Solution of the Regular Inverse Problem
Rewrite problem (5) - (7) in the notation [18]. Let X, Y be Banach spaces, operators S1, S0 E Cl(X),C E L(X, Y), operator-function $ : [0,T] ^ L(Y; X), functions
h : [0, T] ^ X, tf : [0,T] ^ Y
u''(t) = S1u'(t) + S0u(t) + $(t)q(t) + h(t), t E [0,T], (10)
u(0) = u0, u'(0) = u1, (11)
Cu(t) = tf(t). (12)
Theorem 2. Let the pencil B be polynomially A-bounded and condition (А) be fulfilled, moreover, C E L(X; Y), $ E C1 ([0,T]; L(Y; X)), h E C 1([0,T]; X), tf E C3([0,T]; Y), for any t E [0,T] the operator C$(t) be invertible and (C$)-1 E C1([0,T]; L(Y)).
If the compatibility condition Cu1 = ^'(0) is satisfied, then the solution to the inverse problem (10) - (12) exists and is unique in the class of functions q e C 1([0, T]; Y),u e C2([0, T]; X).
Proof. Reduce problem (10) - (12) to the problem for the first order equation
v'(t) = Av(t) + Q(t)q(t) + F(t), t e [0,T], (13)
v(0) = vo, (14)
Bv(t) = Z (t), (15)
where v(t) = ( U« ) , .4 = ( £ /J , Q(f) = ( ) ) ' F(t) = ( h0) ) '
v<°>=( U'(00)) ' vo = (_U0) = ( 0 C ) Z« = *'«•
Put R(t) = —(C$(t)) 1. Therefore, all the conditions of Theorem 6.2.3 from [18], are satisfied, and the function q(t) satisfies the integral equation
q(t) = qo(t) + R(t) ^CS0 J Vi2(t — s)$(s)q(s)ds + CSi J V^(t — s)$(s)q(s)ds j , (16) where
qo(t) = — R(t) (V(t) — CSoVn(t)uo — CS1V21 (t)uo — CS0V12(t)ui — CS1V22(t)ui —
t t N —CSoJ V12(t — s)h(s)ds — CS1 f V22(t — s)h(s)ds — Ch(t)
Thus, the solution to the inverse problem (13) - (15) exists and is unique in the class of functions q e C1([0,T]; Y), v e C 1([0,T]; X2). Thus, we obtain that the solution to the regular inverse problem (10) - (12) exists and is unique, with q e C1 ([0, T]; Y), u e C2([0, T]; X).
□
In order to obtain a solution to a singular problem, we need a greater smoothness of the function q from the solution of a regular problem.
Lemma 2. Let n e N, V e Cn-1([0,T]; L(X)),g e Cn([0,T]; X). Then
\ (n) n-1 t V(t — s)g(s)dsl =^2 V(n-fc"1)(t)g(fc)(0) +f V(t — s)g(n)(s)ds. (17)
k=o
Proof. Apply the method of mathematical induction. First we show that for n =1 formula (17) is true. Find the first derivative with respect to t
t v t
d
J V(t - s)g(s)ds = V(0)g(t) + / | - s)] =
o Jt o
t t (18)
- / 5 M* - s)] = + I V(t- s)g'(s)ds.
Comparing (17) and (18) we conclude that formula (17) for n = 1 is true. Now show that if (17) is true for n, then it is also true for n + 1:
t \(n)\ ' n-1 t
/V(t - s)g(s)ds = E V(n-k)(t)g(k)(0) + V(t)g(n)(0) + / V(t - s)g(n+1)(s)ds =
0 / J k=0 0
n t ft \ ("+1)
= £ V(n-k)(t)g(k)(0) + f V(t - s)g(n+1)(s)ds = /V(t - s)g(s)ds .
k=0 0 0
□
The following theorem provides sufficient conditions for the existence of a more smooth (as p E N) solution q E Cp+2([0, T], Y) of a regular problem.
Theorem 3. Let the pencil B be polynomially A-bounded and condition (A) be fulfilled, moreover, C E L(X; Y), p E N0, $ E Cp+2([0,T]; L(Y; X)), h E Cp+2([0,T]; X), tf E Cp+4([0,T]; Y), for any t E [0,T] operator C$(t) be mvertible, with (C$)-1 E Cp+2([0,T]; L(Y)) and the compatibility condition Cu1 = tf'(0) be satisfied for some u1 E U. Then there exists and a unique solution of (10) - (12), and q E Cp+2([0,T]; Y).
Proof Denote by Vn = f (/u2A - ¡iBx - Б0)_1(М - BJe&dfi,
^12 = 2hi - ¡lBx -
^21 = 2hi ~ ^ ~ ВоУ'Вое^Ч/л,
Y
V22 = zhl(n2A - 11B! - BoY^Ae^d»
y
the propagators of the homogeneous equation (10). Earlier, in the proof of Theorem 2, it was established that the function q(t) satisfies the integral equation (16). Take the natural number n < p + 2. Assuming that q E Cn([0, T]; Y) by Lemma 2 we obtain the equality
, . n—i n—k—i
q(n)(t) = ¿n)(t) + E cnR(k)(t)cs0 E ^(2га-к-т-1)(t)^q)(m)(0)+
, ) n—i n— k— i
k=0 m=0
n n—k t
+ E E Cn>mR(k)(t)CS^ Vi2(t - з)Ф(т) (s)q(n-k-m)(s)ds+
k=0 m=0 0
n- i n- k - i
+ E cnR(k)(t)cSi E V2(n—k—m— i) (t)(^q)(m) (0)+
k=0 m=0
n n—k t
+ E E Cn>mR(k)(t)CS^ V22(t - 5)Ф(т) (s)q(n—k—m)(s)ds,
where = fci^fcji, Ck'm = fc!m!(rara_;fc_m)! and
q0n) (t) = - E CnR(k) (t)(V—k+2) (t) - CS0Vi(rk)(i)u0-k=0
-CSiV2(1n—k)(t)u0 - CS0V1(2n—k)(t)ui - CSiV2(2n—k)(t)ui-
t
-CSo / V12(t - s)h(n—k) (s)ds - CS^ V22(t - s)h(n—k)(s)ds-
0 o
)n—1 n—fc—1
+ E enR(k)(t)cSo E V1(rfc-m-1)(t)h(m)(0)+
fc=0 m=0
n
fc=0 m=0
n—1 n—fc—1
+ E enR(fc)(t)es1 £ v2(2n—k—m—1)(t)h(m)(0)
n
fc=0 m=0
exists from the conditions of this theorem for n = 0,1, + 2.
Show that q e Cp+2([0,T], Y), for this purpose denote ro = qo(0), and for n = 1, 2, + 2 determine the following values
, n—1 n—fc—1 m
(n)
rn = q0n)(0) + E enR(k)(0)es0 E V1(2n—k—m—1)(0)E em$(1) (0)rm—i+
fc=0 m=0 i=0
n—1 n—fc—1 , . m
+ E enR(k)(0)es1 E V2(2n—k—m—1)(0^ em$(i)(0)rm—,.
fc=0 m=0 i=0
Consider the system of integral equations
ft t <?0(t) = q0(t) + R(t) eS^ V12(t - s)$(s)®,(s)ds + eSj V22(t - s)$(s)®,(s)ds
00 n—1 n—fc—1 m
9n(t) = q0n)(t) + E enR(k)(t)es0 E V?^m—^(t) eem$(i)(0)rm—i +
n—1 n—fc—1 _ m
+ E enR(fc)(t)es^ E k—m—^(t) E em$(i)(0)rm—i +
fc=0 m=0 i=0
(19)
n n—fc t
+ E E en'mR(fc)(t)eS^ V12(t - s)$(m)(s)qn—fc—m(s)ds +
fc=0 m=0 0
n n—fc
+ E E en'mR(fc)(t)eSj V>2(t - s)$(m)(s)qn—fc—m(s)ds, n =1,...,p + 2.
fc=0 m=0
Reduce (19) to the Volterra equation of the second kind
t
g(t) = g0(t) + y K (t,s)g (s)ds 0
on the space (e ([0, T]; Y ))p+3 with a matrix operator function K (t, s) given on the triangle A = {(t, s) G R2 : 0 < t < T, 0 < s < t}. By virtue of the continuity of all data of system (19), it has a unique solution
(ç0,q~1,.-,çP+2) G (e ([0, T ]; Y ))p+3-
This solution will be the limit of the sequence of approximations
t t <?0,i(t) = ?0(t) + R(t) ^es^ V12(t - s)$(s)<?0;i—1 (s)ds + es^ V22(t - s)$(s)<?0;i—1 (s)ds
00
30 Bulletin of the South Ural State University. Ser. Mathematical Modelling, Programming
& Computer Software (Bulletin SUSU MMCS), 2019, vol. 12, no. 2, pp. 25-36
t
, ч n—1 n— k—1 , „, m
Jn),
.. I </ _L 10 гъ ± , , , »(</
qn,i(t) = q0n) (t) + E CkR(k)(t)CS0 E V(2n—k—m—cmФ«(0)гт—1+
k=0 m=0 ¿=0
n— 1 n— k— 1 m
+ E CkR(k)(t)CS^ V2(2n—k—m—1)(t)E cmФ(1)(0)rm—1 + (20)
n 1 22 m
k=0 m=0 ¿=0
n n—k t
+ E E CkmR(k)(t)CS^ V12(t - s^(m) (s)qn,—k—m,i—1 (s)ds+
k=0 m=0 0
n n—k t
+ E E Cn>mR(k)(t)CSj V22 (t - s^(m) (s)qfn—k—m,i—1(s)ds, n =1,...,p + 2, i G N,
k=0 m=0 0
which for i —^ to on the interval [0,T] converge uniformly to the functions qn, n = 0, + 2. Set the initial approximation qn>o = 0, n = 0, + 2, then qn+1>o = qn,o, n = 0, + 1. In addition, from (20) it follows that
M0) = r„, n = 0, + 2, i e N. (21)
Assume that for all j = 1, •••,i the equalities qn+1;j(t) = q^j(t), n = 0, + 1 are true. Then, using Lemma 2 and equalities (21), we obtain
(n n—k t
E E C%mR(k\t)CS0JV12(t-s)&m\s)qn-k-m>i(s)ds
k=o m=o o
n+1n—k+1 t
= E E Cn—1>mR(k)(t)CSo/ V12(t — s)$(m) (s)^n—k—m+1,i(s)ds + k=1 m=o o
n n— k
+ E cnR(k)(t)CS0V12(t) E cm—kФМ(0)rn—k—m+
k=0 m=0
n n—k+1 t
+ E E cn'm—1R(k)(t)CS^ V12(t - s^(m) (s)qn—k—m+1,i(s)ds+
k=0 m=1 0
n n—k t
+ E E CkmR(k)(t)CS^ V12(t - s^(m)(S)qin—k—m+1,i(s)ds.
k=0 m=0 0
Denote by ak;TO some values in (22). Taking into account the equalities
Ck + Ck—1 _ Ck Ckim + Ck—1,m + Ck,m—1 _ ckim
Cn + Cn Cn+1, Cn + Cn + Cn Cn+1
we get
n n—k n+1 n—k+1 n n—k+1
E E + Cra '^^k.m +
k=o m=o k=1 m=o k=o m=1
—k
Cn+1ak,m + Cn+1ak,o +2^ Cn+1ao,m + Cn+1ak,n— k+1 + k=1 m=1 k=1 m=1 k=1
n+1n—k+1 k
+ Cn' ao,o + Cn'nao,n+1 + Cn' an+1,o = E E Cn+1ak,m.
k=o m=o
(22)
(23)
From (22) and (23) it follows that
n n— k
i EE Cknmrtk\t)CSQ$Vl2{t-s)&m\s)qn-k-m,{s)ds] =
k=0 m=0 0
n+1 n— k+1 t
= E E Cn+"1R(k)(t)CS^ V12(t - s)$(m) (s)^n—k—m+1,i(s)ds +
k=0 m=0 0
n n— k
+ E CnR(k)(t)CS0V^(t) E Cm—k$(m) (0)rn—k—m.
k=0 m=0
Similarly, we can get that
n n—k
i EE C%mR{k\t)CS1$Vn{t-s)&m\s)qn-k-mt{s)ds ] =
k=0 m=0 0
n+1 n—k+1 t
= E E C„k+"1R(k)(t)CSj V22(t - s)$(m) (s)^n—k—m+1,i(s)ds +
k=0 m=0 0
n n— k
+ E CnR(k)(t)CS1 V22(t) E Cm—k$(m) (0)rn—k—m.
k=0 m=0
Changing the summation indices and re-grading the sums we get
n— 1 n— k — 1 m
I T,ckR^(t)cs0 E ^'(ilE^ioiv,
k=0 m=0 1=0
(24)
(25)
n n— k m
= E Cnk+1R(k)(t)CS^ V^Tk—m)(t) E Cm$(1) (0)rm—i- (26)
k=0 m=0 1=0 v 7
n n— k
- E CnR(k)(t)CS0V12(t) E Cm—k$(m)(0)rn—k—m.
k=0 m=0
Similarly, we get that
(n—1 n—k—1 . . m
ZcZRWvcSi E ^'(ilE^io)^,
k=0 m=0 1=0
n n— k m
= E Cnk+1R(k)(t)CS^ V2(2n—k—m)(t^ Cm$(1) (0)rm—1- (27)
k=0 m=0 1=0 v '
n n— k
- E Cnk R(k)(t)CS1V22 (t) £ Cm—k $(m) (0)rn—k—m.
k=0 m=0
Differentiating (20), and also using (24) - (27), we obtain the equalities q^m = qn+1,i+1, n = 0, ...,p + 1. Thus, the sequence q^ converges as i ^ to to the function q0 uniformly on the interval [0,T], and the sequence q0i = 9m converges as i ^ to to the function q1 uniformly on the segment [0,T]. Therefore, the function q0 is continuously differentiable and q0 = 91. The equalities of qn, = 9n+1, n = 1,...,p + 1, are proved in the same way, which implies that q0 = q E Cp+2([0,T]; Y) and, therefore, q(n) = qn, n = 1, ...,p + 2.
□
3. Solvability of the Original Inverse Problem
Now we can formulate the solvability conditions for the original problem (1) - (4).
Theorem 4. Let the pencil B be polynomially A-bounded and condition (A) be fulfilled, moreover, the to be a pole of order p e No of the A-resolvent of the pencil B, operator C e L(U; Y), Uo C ker C, x e Cp+2([0,T]; L(Y; F)), f e Cp+2([0, T]; F), tf e Cp+4([0,T]; Y), for any t e [0,T] operator C(A1)—1Qx be invertible, with (C(A1)—1Qx)—1 e Cp+2([0,T]; L(Y)), the condition Cu 1 = ^'(0) be satisfied at some initial value u = V1 e U, and the initial values wk = (I — P)vk e Uo satisfy
и,к = -±кЖг1^
j=0
(I - Q)(x(0)q(0) + f (0))
к = 0,1.
Then there exists a unique solution (v,q) of inverse problem (1) - (4), where q G Cp+2([0,T]; Y), v = u + w, whence u G e2([0,T];U1) is the solution of (5) - (7) and the function w G e2([0,T]; U0) is a solution of (8), (9) given by
w(t) = ^ K2(B° )-1
j=0
(I - Q)(x(t)q(t) + f (t))
(28)
Proof. The conditions of Theorems 2 and 3 are satisfied, and therefore there exists a unique solution (q, u) of problem (5) - (7), where q G Cp+2([0,T]; Y), u G c2([0,T];U1).
Using the result of [17] and the required smoothness of the function q, we obtain that there exists a unique solution w G C2([0,T];U0) to (8), (9) given by (28).
□
p
Acknowledgements. The work was supported by Act 211 Government of the Russian Federation, contract no. 02.A03.21.0011.
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Received November 1, 2018
УДК 517.9 DOI: 10.14529/mmp190202
ОБРАТНАЯ ЗАДАЧА ДЛЯ МАТЕМАТИЧЕСКИХ МОДЕЛЕЙ СОБОЛЕВСКОГО ТИПА
А.А. Замышляева1, А.В. Лут1
1 Южно-Уральский государственный университет, г. Челябинск,
Российская Федерация
Работа посвящена исследованию обратной задачи для линейного уравнения соболевского типа высокого порядка с неизвестным коэффициентом, зависящим от времени. Поскольку уравнение может быть вырожденным, используется метод фазового пространства. Он состоит в построении проекторов, расщепляющих исходные пространства в прямую сумму подпространств. Действия операторов также расщепляются. Таким образом, исходная модель сводится к двум задачам: регулярной и сингулярной. Регулярная редуцируется к невырожденной задаче первого порядка, которая решается с помощью аппроксимаций. Получена необходимая гладкость решения. Затем оно подставляется в сингулярную задачу, которая решается с использованием методов теории относительно полиномиально ограниченных пучков операторов. Основной результат работы содержит достаточные условия существования и единственности решения обратной задачи для математической модели соболевского типа второго порядка. Данная методика может быть использована при исследовании обратных задач рассматриваемого типа для математической модели Буссинеска - Лява.
Ключевые слова: уравнения соболевского типа; обратная задача; математические модели; уравнения второго порядка.
Литература
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D.G. Orlovsky, I.A. Vasin. - New York: Marcel Dekker, 2000.
Статья выполнена при поддержке Правительства РФ (Постановление № 211 от 16.03.2013 г.), соглашение № 02.А03.21.0011.
Алена Александровна Замышляева, доктор физико-математических наук, доцент, кафедра «Прикладная математика и программирование:», ЮжноУральский государственный университет (г. Челябинск, Российская Федерация), [email protected].
Александр Валерьевич Лут, аспирант, кафедра «Прикладная математика и программирование>, Южно-Уральский государственный университет (г. Челябинск, Российская Федерация), [email protected].
Поступила в редакцию 1 ноября 201S г.
