On some inverse problem for a parabolic equation with a parameter Текст научной статьи по специальности «Математика»

Journal of Siberian Federal University. Mathematics & Physics 2015, 8(3), 281—290

УДК 517.95

On some Inverse Problem for a Parabolic Equation with a Parameter

Kirill V. Korshun*

Institute of Mathematics and Computer Science Siberian Federal University Svobodny, 79, Krasnoyarsk, 660041

Russia

Received 10.05.2015, received in revised form 01.06.2015, accepted 20.06.2015 An inverse boundary-value problem for n-dimensional parabolic equation with a parameter is considered. Sufficient conditions for existence and uniqueness of solution in continuously differentiable class are obtained.

Keywords: differential equation, boundary-value problem, method of weak approximation. DOI: 10.17516/1997-1397-2015-8-3-281-290

Today inverse problems of mathematical physics play an important role in science and applications [1]. Coefficient inverse problems for parabolic equations are problems of finding solutions of differential equation with one (or more) unknown coefficients.

An inverse problem for a parabolic equation with a parameter is investigated. The parameter has the same dimension as the spatial variable.

Inverse problems with unknown parameter arise in various problems: in studying boundary-value problems for mixed-type equations and equation systems [2,3]; in solving various inverse problems [4-7]; in studying boundary-value problems for equation systems with small parameters [8,9].

1. Problem formulation

We consider the boundary-value problem

= AAxu(t, x, y) + ^(i, y)f (t, x, y), (1)

u(0,x,y) = uo(x, y), (2)

u(t,x,y)|x£dn = 0, (3)

u(t,x,y)|x=y = ^(t,y), (t,x,y) e Qt, (4)

where

Qt = {(t,x,y)|t e [0,T], x e n, y e D},

T > 0, n is a rectangular cuboid [0, l1] x [0,l2] x • • • x [0,ln] in Rn, D is a compact subset of

n 32

n with smooth boundary 3D, Ax = ^^ ■—2 is the Laplace operator, u(t,x,y) and y(t,y) are

^ dx2

i=i 1

unknown functions. Functions f (t, x,y), u0(x,y) are given.

* [email protected] © Siberian Federal University. All rights reserved

We use following notation:

Da =

dV

x daix\ ...d°nxn

is partial differential operator with respect to spatial variables xi... xn, where a is multi-index notation (a = (ai,. .., an), |a| = ai + • • • + an, aj ^ 0, aj £ Z);

Di

glil

y dilyi ...dinyn is partial differential operator with respect to parameter y;

K > 0, i £ N,

are nonnegative constants that depends only on initial conditions of problem (1)-(4);

Zp(Q) = {(u(t,x,y),^(t,y)) |D>(t,x,y) £ C([0,T] x Q x D), |D>(t,x,y)| < K, ^(t,y) £ C([0,T] x D), |a| < p - 2}

is the class of continuous functions.

Let us assume that the following conditions are fulfilled:

|f (t, y,y)| > Ki > 0, y £ D, |DaD£uo(x,y)| < K2, |a| < p, |£| < 1, (t,x,y) £ Qt, p > 6;

DaDi f (t,x,y) x y f (t, y, y)

< K3, |Di<Mt,y)| < K4,

dk .

dxk uo(xi,...,xj dk

—T f (t, xi,..., x dxk v

,xn,y)| xi=0,xi=li = 0,

■,xn,y)|xi=o,xi=ii = 0, i = 1, ...,n, k = 0, 2,4, 6.

(5)

(6) (7)

We prove the following statements:

Theorem 1.1. Lei us assume that initial data of problem (1)-(4) satisfy (5)-(7) for some p. Then the problem has a solution of class Zp.

Theorem 1.2. The solution of problem (1)-(4) of class Zp is unique. Theorem 1.3. Let us consider the Cauchy problem (1), (2), (4) in domain

E = {(t, x, y)|t £ [0,T],x £ Rn,y £ D}.

a. This problem has a solution of class Zp(Rn) if conditions (5) are fulfilled in domain E.

b. The solution of the problem is unique.

2. Proof of existence

The proof of Theorem 1.1 is based on reduction of boundary-value problem to Cauchy problem. We construct an extension of functions u0, f from set QT to E in n steps. At the first step we extend functions u0, f to R with respect to variable xi as follows:

uo(-xi, x2,... ,xn,y) = -uo(xi,x2,..., xn,y),

f (t, -xi, x2,..., xn, y) = -f (t, xi, x2,..., xn, y), xi e [0, li];

uo(xi + 2kli, x2,..., xn, y) = uo(xi, x2,..., xn, y),

f (t, xi + 2kli,x2,... ,xn,y) = f (t,xi,x2,... ,xn,y), k e Z, xi e [0, li].

At i-th step (2 < i < n) we extend functions u0, f from [0, lj] to R with respect to variable xj in the same way. We denote the extensions of functions u0, f as u0, f *, respectively.

By (5), (6), functions u*, f* have continuous partial derivatives with respect to variables

xi, ..., xn up to p-th order on whole set Rn. One should note that functions u*, f * are odd

x0

„„+. TD>n r\ J-„ + f t- „ u0

and periodic with respect to variables xj with period 2lj. By this, the following conditions are fulfilled:

u*(xi,..., xj,..., xn, y) + u*(xi,..., -xj,..., xn, y) = 0, (8)

u*(xi,..., lj + xj,..., xn,y) + u*(xi,..., lj - xj,..., xn,y) = 0, (9)

f *(t, xi,..., xj,..., xn, y) + f *(t, xi,..., -xj,..., xn, y) = 0, (10)

f *(t, xi,. . . ,lj + xj,...,xn ,y) + f * (t,xi,...,lj - xj,...,xn,y) =0. (11)

**

We use u*, f* as the initial data for the Cauchy problem

du(t, x, y)

AAxu(t, x, y) + M(t, y)f *(t, x, y), (12)

u(0, x,y)= u*(x,y), (13)

dt

y) = 0

u(t, x, y) |x=y = ^(t,y), (14)

for t e [0, T], x e Rn, y e D c Rn.

After substitution x = y, y e D to (12) one can find ^(t,y):

M(t,y) = 7^7"-r(&(t,y) - AAxu(t,y,y)), y e D. (15)

f *(t,y,y)

Using (15), we reduce problem (12)-(14) to auxiliary Cauchy problem for nonclassic partial differential equation

du(tdtx,y) = AAxu(t, x, y) + f* (t"y y) (&(t, y) - AAxu(t, y, y)) f*(t, x, y), (16)

u(0, x, y) = u*(x, y), t e [0,T], x e Rn, y e D. (17)

Existence of solution of problem (16)-(17) is proved with the use of the method of weak approximation (MWA, see [10-12]). We split the problem into two fractional steps and make time shift by T/2 in the trace of unknown function:

duT(t,x,y) =2AAxuT(t,x,y), t e (kr, (k +i /2)t], (18) dt

duT (^;x,y) =2 f^j (^t(t,y) - AAxuT (t T /2, y, y)), t e ((k +i /2 )t, (k +l)r ], (19)

uT (0, x, y) = u* (x, y), k = 0, ...,N - l, Nt = T, x e Rn, y e D. (20)

We prove (see Appendix 3.) that functions

d d d

-D>T(t,x,y), — DauT(t, x, y), — D>T(t, x, y), |a| < p - 2 (21)

are uniformly (with respect to t) bounded in domain E. This implies uniform boundedness and uniform equicontinuity of function sets {DauT}, |a| < p — 2 in compact subset

nM = {(t, x, y) |t £ [0,T], |x,| < M, y £ D, i = 1,.. .,n}.

Applying Arzela-Ascoli theorem about compactness, we show the existence of the subsequence uTk (t,x,y) of sequence uT(t,x, y), which converges to some function u(t,x,y) with its partial derivatives {DauT}, |a| ^ p — 2. It follows from the theorem on convergence of MWA that function u(t,x,y) is a solution to (16)-(17) in nM and

||D>T — Dau||c(nM) - 0, |a| < p — 2

for t — 0. Since M is an arbitrary constant, function u(t, x, y) is a solution to (16)-(17) in whole domain E.

We prove that pair of functions (u(t,x,y), ^(t,y)) (where ^(t,y) is given by (15)) is solution to (12)-(14). Because u(t, x,y) is a solution to (16), (17) substitution of (u(t, x,y), ^(t, y)) to (12), (13) gives us identity (16), (17). After substitution x = y, y £ D to (16), (17) we show that u(t, x, y) satisfies

du(t,y,y) , n _ n

—dt— = y), y£ D.

We assume that ^(t, y) satisfies initial data:

uo(y,y) = ^(0, y), y £ D. Under this assumption function ^(t) = u(t, y, y) — ^>(t, y) is a solution to Cauchy problem

V>'(t) = 0,

^(0) = 0.

Thus ^(t) = 0 and (14) is fulfilled.

Remark. If we assume that u*, f * are arbitrary functions satisfying (5) in domain E then we prove Theorem 1.3 a.

We prove that the solution of Cauchy problem u(t, x,y) satisfies boundary conditions (3). Solution uT of split problem (18)-(20) satisfies

uT(t, xi, .. ., xj, ..., xn, y) + uT(t, xi, .. ., —xj, .. ., xn, y) = 0, (22)

uT(t, xi, .. ., lj + xj,. .., xn, y) + uT(t, xi, ..., lj — xj,. .., xn, y) = 0 (23)

for any t > 0, as it is proved in Appendix 3.. Because uT(t, x, y) converges to u(t,x,y) in nM for any M > 0 we can set Mo > max(li,..., ln). Then we have QT C nMo.

Relations (22)-(23) have a limit as t — 0. We assume t — 0 and xj = 0 in (22)-(23) and obtain (3). Solution of Cauchy problem (12)-(14) satisfies (1), (2), (4) in QT and (3) is fulfilled. This proves Theorem 1.1.

3. Proof of uniqueness

Let us assume that (ui(t,x,y),y)), (u2(t,x,y),(t, y)) are two arbitrary solutions of problem (1)-(4) of class We denote u* = u1 — u2, = — Functions u*, satisfy the

following problem

du * (t, x, y)

dt

u * (0, x, y) =0, u* (t,x,y)|xeôn = 0,

AAx u* (t,x,y) + (t,y)/ (t,x,y),

u* (t,x,y)|a

0,

(24)

(25)

(26) (27)

for (t, x, y) e Qt.

After substitution x = y, y e D into (24) one can find ^ * (t, y) using (15) with ^(t) = 0. Next we substitute ^ * (t, y) into (24). Function u * satisfies the following problem

du * (t,x,y^ A • /(t,x,y) • Axu* (t,y,y)

■ = AAxu (t, x, y)--

dt

u * (0, x, y) = 0, u * (t,x,y)|xeôn = 0,

/(t, y, y)

(28)

(29)

(30)

d2î

for (t, x, y) e Qt.

We differentiate twice relations (28)-(30) with respect to xj. Then " " is a solution to

dx2

second-order parabolic boundary-value problem

__ u (t,x,y) = AAx ¿2 u* (t,x,y) —

dtd d2

u (0, x,y) = 0,

d2 *

u*(t, x,y)|xeôn = 0,

A • dx2 /(t,x,y) • Axu * (t,y,y)

/ (t, y, y)

(31)

(32)

(33)

for i = 1,

We apply the maximum principle to (31)-(33) and obtain

^ K3t sup |Axu *(t, x, y)|, i = l,..., n.

d 2

dx2u (t,x,y)

Summation of these inequalities for i = 1,..., n gives

E

d2

u (t,x,y)

^ K3nt sup |Axu * (t, x, y)| ^ K3nt^^ sup

x£l"

d2

u (t,x,y)

One can set £ so as K3n£ < 1 and obtain

supn

x£l"

d2

u (t,x,y)

= 0, t G [0, £].

n.

x

x

This proves that the right-hand side of (28) is equal to zero. By the maximum principle u* (t, x, y) = 0 for t £ [0,£].

Let us consider problem (28), (30) for t £ [£, T] with initial data u*(£, x, y) = 0. Using the same reasoning we prove that u*(t, x, y) = 0 for t £ [£, 2£]. After finite number of steps we prove that u*(t, x, y) = 0 for t £ [0, T]. With u* =0 in (24) we have

M*(t,y)f (t,x,y) =0, Vx £ Q, Vy £ D.

Since f (t, x, y) = 0 for x = y we have M(t, y) = 0. This proves Theorem 1.2.

Note. Let us assume that (ui,Mi), (u2,m2) are two arbitrary solutions of the Cauchy problem (1), (2), (4) in domain E and formulate the following Cauchy problem for u* = ui — u2, m* = Mi — M2-'

(dtx,y) = AAxu*(t,x,y) + M*(t, y)f (t,x,y), (34)

u*(0, x, y) = 0, (35)

u*(t, y, y) = 0, (36)

for (t, x, y) £ E.

One can prove in exactly the same way as we did it for (24)-(27) that u* = 0 and m* = 0. This proves Theorem 1.3 b.

Appendix

A. Proof of statement (21)

Split-problem (18)-(20) is n-dimensional Cauchy problem for parabolic equation (18), (20) at the first fractional step and the Cauchy problem for ordinary differential equation (19), (20) at the second fractional step. Note that the initial data of split-problem satisfies (5). We use the following notation:

(t)= sup sup |D^Dß(£,x,y)|, UT(t) =

= EE ua,ß (t), (t) = E ua,o(t), (37)

M<P IßK1 H<P

are nonnegative increasing functions. They are bounds of uT and its partial derivatives.

Zeroth whole step (k = 0) is considered. At the first fractional step we differentiate (18), (20) up to p times with respect to Xj and once with respect to yj and obtain

d

— D^DßuT(t, x, y) = 2AAKD^uT(t, x, y), t G (0,T /2], x G Rn, y G D. dt " "

The application of the maximum principle to this equation gives

| D^DßuT (t, x, y) | < K2, t G [0,T /2]. (38)

Solution of problem (18), (20) at the second fractional step can be expressed in explicit form:

uT (t, x, y) = uT (T/2, x, y^T f (^,x,y) y) - AAxuT (e T /2, y, y)) d£, t e [T/2, t ].

A/2 j(e,y,y)

Upon differentiating this identity with respect to xj, we obtain bounds for respective partial derivatives:

|D>T (t,x,y)| < |D>T (T/2,x,y)| + KstI 1+ sup |AxuT(e T /2,y,y)|) , t e [T / 2, t ]. (39)

V «e[T /2 ,t ] J

Using (37), inequalities (38)-(39) can be expressed in the following form: UT,0(t) < ua,0(0), t e [0,T /2],

UT,0(t) < UT,0(0) + K5T ^1+ £ uT,0(0)j < UT,0(0) + K5T (1 + c/T(0)), t e [0,t].

The same technique can be applied on the first and subsequent whole steps. At the first whole step (k = l) bounds for uT and its partial derivatives are

UT,0(t) < ut,0(0) + K5T (1 + UT(0)), t e [t,3t /2], at the first fractional step and

|D>T(t,x,y)| <|D>T(3t/2,x,y)| + K5T (1+ sup |AxuT(e T /2,y,y)|J , t e [3t/2, 2t],

V «£[3t/2,2t] J

at the second fractional step. Hence

UT,0(t) < UT,0(0) + K5T (1 + C/T(0) + 1 + C/T(t)), t e [t, 2t].

After applying this technique k times, we obtain

k

UT,0(t) < UT,0(0) + K5^ (1 + f/T((j - 1)t)), t e [0,kT], k = 1,..., N. (40) j=i

Then we sum (40) over all a, |a| < p and prove that

k

C/T(t) < UT(0)+ K6t£ (1 + UT((j - 1)t)) <

j=i

< (1 + U7T(0))(1 + K6t)k - 1 < K7, t e [0,kT], k = 1,..., N. (41)

Because

(1 + K6T)k < (1 + K6t)N < (eK«T)N = eK«NT = eK«T,

K7 does not depend on t and (41) is the uniform bound.

d

Consider first-order partial derivatives —— D?uT. The partial derivatives can be estimated

dyj x

by (38) with | = 1 at every first fractional step. At second fractional steps we first differentiate

the explicit solution of (19), (20) with respect to xj and then with respect to yj (considering uT (f, y, y) as composite function of y):

4DauT (t,x,y)=£DauT (T/2,x,y) -+ /1 (2Da £ §Sy)x

x y) — AAxuT(e —t /2, y, y)) de+/' ^aff^) x

/ d d d M dy:^y) - AdX:AxuT(e -T /2,y, y) - a^A*wT(e — /2, y, y) ) ^

Because every partial derivative DauT is bounded by (41) the following inequalities are true: d

—D>T (t,x,y) dy:

<

d

— d>t (t/2,x,y)

+ t (k • (K + AK7) + K3• (K4 + AK7+

+A sup

«e[0,T/2]

d

—AxuT(e T /2,y,y) dy:

UT,i(t) < UT,i(0) + K8t I 1 + UT,i(0) ) < UT,i(0) + Kst (1 + UT(0)), t £ [0, t]. (42)

|a|=2

Using the same line of reasoning on every whole step, we obtain

k

UT,i(t) < UT,i(0) + K8^ (1 + UT ((j — 1)t )), t £ [0,kT], k = 1,...,N. (43)

j=i

Then we sum (43) over all a, p, |a| < p, |p| = 1 and obtain

k

UT(t) < UT(0) + (1 + UT((j - 1)t)) < (44)

j=i ( ) < (1 + UT(0)) (1 + K9T)k - 1 < K10, t G [0, kr], k = 1,..., N.

Inequality (44) shows uniform (with respect to t) boundedness of partial derivatives

DD uT (t,x,y).

We differentiate (18), (19) with respect to up to p - 2 times. Because the right-hand side contains uniformly bounded functions then the left-hand side

d

-D>T(t,x,y), |a| < p - 2, is also uniformly bounded. This proves statement (21).

B. Proof of relations (22) and (23)

We prove relations (22) and (23) with the use of the method of fractional steps.

At t = 0 relations (22), (23) are fulfilled. It follows from (8) and (9). At the first fractional step uT satisfies the Cauchy problem (18), (20). The solution of this problem is of the form (see [13])

uT(t,xi,x2,...,xn,y) = / u *(ei,e2,...,en,y)w(x,e,t,0)de^...den, (45)

■J Rn

W (xi,x2, ...,!„ ,Ci,C2, . . . ,Cn,t, z)

4n(t -

: exp

/ v (xj-gj)2 \ 2Â

i=i

4(t - z)

(46)

We substitute this solution into (22) and (23) and obtain

"(t,Xl, . . . , Ci +

Ci +j- Xi ..... Xn

, y) + uT (t, Xl,...,Ci - Xi, ...,x„,y) =

[ ^^ • • •

/Kn 4nt(2A)n/2

/ (Ci + Xi - Ci)2 + £ (Xj - Cj )

exp

j=i

8 At

+

/ (Ci - Xi - Ci)2 + E (Xj - Cj)

+ exp

j=i

8 At

dCi ... dCn

u 0 (Ci,...,Ci - Ci....,Cn

( E(Xj - Cj)

4nt(2A)n/2

exp

j=i

8 At

exp -

(Xi + Ci)2

8 At

+ exp -

(Xi - Ci)2

8 At

dCi .. . dCn, i =l,...,n, Ci =0,1i, t G (0,T /2].

Note that all integrands are odd functions with respect to ej, hence all integrals are equal to zero.

At the second fractional step, uT have the following form:

/ (C,X,y)

U(t, X, y) = uT(T/2, X, y) + 2 f f (C,X,y)) (&(C, y) - AAxUT(C T /2, y, y)) dC, t G [T/2, t]

Jt/2 f (C,y,y)

We substitute this expression into (22) and (23) and obtain

u (t7 X i 5 . . . , Ci + Xi 5 ... 5 Xn 5

' T

y) + uT(t, Xi, . .., Ci - Xi, .. ., Xn,y) =

u ^2 ,Xi,...,Ci + Xi,..., Xn , y J + u ^ ,Xi ,...,Ci Xi,... , Xn, y J +

+2 J (/ 0 (C,Xi, . . . , Ci + Xi, . . . ,X„,y) + f (C, Xi,.. ., Ci - Xi, . . . ,X„,y)) • ...dC = 0,

where i = l,..., n, Ci = 0,1i. All terms in this identity are equal to zero by statements proved earlier.

Thus, relations (22) and (23) are fulfilled for t G [0, t]. Using the same line of reasoning k times, we prove that (22) and (23) are fulfilled for t G [0, kT] and, therefore, for all t G [0, T].

l

u

2

2

2

x

R

x

References

[1] Yu.Ya.Belov, Inverse problems for partial differential equations, Utrecht etc.,VSP, 2002.

[2] Yu.Ya.Belov, A linear stationary problem of ocean dynamics, Math. Notes, 26(1979), no. 1, 513-517.

[3] Yu.Ya.Belov, Splitting in a degenerate quasilinear parabolic equation, Math. Notes, 46(1989), no. 6, 906-909.

[4] S.S.Akhtamova, On some inverse problems for parabolic equations, Doklady AN SSSR, 316(1991), no 4, 791-795 (in Russian).

[5] Yu.Ya.Belov, On some inverse problem for semilinear parabolic equation, Doklady AN SSSR, 316(1991), no. 5, 1034-1038 (in Russian).

[6] Yu.Ya.Belov, K.V.Korshun, An Identification Problem of Source Function in the Burgers-type Equation, Journal of Siberian Federal Universtity. Mathematics & Physics, 5(2012), no. 4, 497-506 (in Russian).

[7] Yu.Ya.Belov, K.V.Korshun, On some inverse problem for the Burgers-type Equation, Sibirskiy Zhurnal Industrailnoy Matematiki, 16(2013), no. 3, 28-40 (in Russian).

[8] Yu.Ya.Belov, On some linear partial differential systems with small parameter, Dokl. Akad. Nauk SSSR, 231(1976), no. 5, 1037-1040 (in Russian).

[9] A.P.Oskolkov, On some quasilinear parabolic system with small parameter approximating Navier-Stokes equation system, Zapiski nauchnykh seminarov LOMI AN SSSR, 32(1971), 78-103 (in Russian).

[10] N.N.Yanenko, On the weak approximation of the differential equations systems, Siberian Mat. Zh., 5(1964), no. 6, 1431-1434 (in Russian).

[11] N.N.Yanenko, G.V.Demidov, The research of a Cauchy problem by method of weak approximation, Dokl. Akad. Nauk SSSR, 6(1966), 1242-1244 (in Russian).

[12] Yu.Ya.Belov, S.A.Cantor, Weak approximation method, Krasnoyarsky gos. univ., Krasnoyarsk, 1999 (in Russian).

[13] A.D.Polyanin, Reference book to linear equations of mathematical physics, Nauka, Moscow, 2001 (in Russian).

Об одной обратной задаче для параболического уравнения с параметром

Кирилл В. Коршун

В статье рассмотрена краевая обратная задача для n-мерного параболического уравнения с параметром. Получены достаточные условия на входные данные, обеспечивающие однозначную разрешимость задачи в классе гладких функций.

Ключевые слова: дифференциальные уравнения, краевая задача, метод слабой аппроксимации.