ON AN INVERSE PROBLEM FOR A STATIONARY EQUATION WITH BOUNDARY CONDITION OF THE THIRD KIND Текст научной статьи по специальности «Математика»

DOI: 10.17516/1997-1397-2021-14-5-659-666 УДК 517.95

On an Inverse Problem for a Stationary Equation with Boundary Condition of the Third Kind

Alexander V. Velisevich*

Siberian Federal University Krasnoyarsk, Russian Federation

Received 10.04.2021, received in revised form 10.05.2021, accepted 20.06.2021 Abstract. The identification of an unknown coefficient in the lower term of elliptic second-order differential equation Mu + ku = f with boundary condition of the third kind is considered. The identification of the coefficient is based on integral boundary data. The local existence and uniqueness of the strong solution for the inverse problem is proved.

Keywords: inverse problem for PDE, boundary value problem, second-order elliptic equation, existence and uniqueness theorem.

Citation: A.V. Velisevich, On an Inverse Problem for a Stationary Equation with Boundary Condition of the Third Kind, J. Sib. Fed. Univ. Math. Phys., 2021, 14(5), 659-666. DOI: 10.17516/1997-1397-2021-14-5-659-666.

Introduction

In this paper an inverse problem for some stationary equation is considered. Problem. For given functions f (x), a(x), /3(x), h(x) and constant j find function u(x) and constant k that satisfy the equation

—div(M(x)Vu) + m(x)u + ku = f, (1)

boundary condition

du \

-L + a(x)u) = 8(x), (2)

J dn ' y h W

and the condition of overdetermination

/ uh(x)ds = /л. (3)

JSQ

Here Q C R" is a bounded domain with boundary 8Q, t G (0,T), M(x) = (mj(x)) is a matrix

d

of functions mj (x), i,j = 1, 2,...,n; m(x) is a scalar function, = (M(x)V, n), n is the

dN

unit vector of the outward normal to the boundary dQ.

A main goal of this paper is to establish the existence and uniqueness of the strong solution of inverse problem (1)-(3). The additional integral boundary data similar to condition of overdetermination (3) were considered [1-3]. Following the idea given in [1-3] and using method

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

developed in [4], we prove the existence of the solution by reducing the inverse problem to an operator equation of the second kind for the unknown coefficient. Note that the problem for the same equation with Dirichlet boundary conditions was considered [5].

The study of inverse problems for the elliptic equations goes back to fundamental work of M. M. Lavrentiev [6]. Inverse problems for the elliptic equation with special boundary conditions (non-local conditions, non-classical conditions) were considered [7-9].

Such problems arise in determination of unknown physical properties of a medium. In particular, the lowest coefficient k specifies, for instance, the catabolism of contaminants due to chemical reactions [10] or the absorption in diffusion and acoustic problems [11].

1. The preliminaries

The following notations are used || • ||R, (■, — the norm and the inner product in R"; || ■ ||, (■, ■) — the norm and the inner product in L2(Q); || ■ |j, (■, ■) 1 — the norm in Wj (Q),

o

j = 1,2, and the duality relation between W2 (Q) and W—1 (Q), respectively. The linear operator M : W21(Q) ^ (W21(Q))* of the form

M = —div(M(x)V) + m(x)I,

Is introduced, where I is the identity operator. The notation

(Mv1,v2)M = ((M(xx)Vv1, Vv2)r + m(x)v1v2)dx Jn

is also used for v1,v2 £ W2(Q). The following assumptions hold throughout the paper

I. mj (x), dmj/dxi, i,j,l = 1, 2,...,n,and m(x) are bounded in Q. Operator M is elliptic, that is, there exist positive constants m0 and m1 such that for all v £ W21(Q)

mo||v||2 < aMv,v)M < m1Ml (4)

II. M is self-adjoint, that is, mj(x) = mji(x) for i,j = 1,... ,n.

The existence and uniqueness results for problem (1)-(3) is based on two lemmas for direct problem (1)-(2) with known coefficient k.

Lemma 1.1. Let u be the strong solution of problem (1)-(2). If f > 0, ¡3 > 0, a > 0, k > 0 and assumptions I, II are fulfilled, then u ^ 0 almost everywhere in Q.

Proof. Multiplying (1) by u = min{u, 0} in terms of the inner product in L2(Q) and integrating by parts in first term, we obtain

(Mu,u)1 + k||u||2 + / au2ds — ¡3uds — (f,u) = 0. Jen Jen

Taking into account the lemma conditions, the last equality implies that

m1||u||2 < 0.

So, u = 0 almost everywhere in Q. Lemma is proved. □

Lemma 1.2. Let ui,u2 G W22(Q) are the solutions of the problems

Mui + kiUi = fi,

= ßi

dui

+ au.

dN

dQ

here i = 1, 2.

If 0 ^ ki ^ k2, 0 ^ ¡2 ^ ¡i, 0 ^ f2 ^ fi and a(x) ^ 0 then ui ^ u2 ^ 0 for almost all x G Q.

Proof. By Lemma (1.1), u ^ 0, i = 1, 2, for almost all x G Q. The difference ui — u2 satisfies equation

M (ui — u2 ) + ki(ui — u2 ) = (k2 — ki)u2 + fi — f2, (5)

and boundary condition

———— + a(x)(u1 — u2) dN K 'K '

dQ

= ßl = ß2 •

Taking into account the lemma conditions, the right side of (5) is non-negative and ß1 = ß2 > 0. So, by Lemma (1.1), u1 = u2 > 0 for almost all x G Q. Lemma is proved. □

2. Existence and uniqueness

First of all the solution of the inverse problem should be defined. By the solution of the inverse problem is meant function u G W2(Q) and a positive real number k. They satisfy equation (1) almost everywhere in Q and conditions (2)-(3) almost everywhere on dQ. Now, to formulate the theorem functions a, aT and b are introduced as the solution of the problems

Ma = f (x), MaT + TaT = f, Mb = 0,

da

+ a(x)a dN K '

daT . , T + a(x)a dN '

dQ

ß(x);

dQ

ß(x);

db

+ a(x)b dN K '

dQ

h(x),

(6)

(7)

(8)

where t > 0 is a real number.

Theorem 2.1. Let dQ G C2 and assumptions I, II are fulfilled. Suppose also that

(i) f (x) G L2(Q), 3(x), h(x) G w3/2(dQ,), a(x) G C(dQ);

(ii) f (x) ^ 0 almost everywhere in Q; 3(x) ^ 0, a(x) ^ 0, h(x) ^ 0 for almost all x G dQ and there is a smooth piece r of the boundary dQ and a constant 6 > 0 such that 3 ^ 8 and w ^ 8 almost everywhere on r.

Then problem (1)-(3) has a solution {u,k}. Moreover, the estimates

aT < u < a, 0 < k < t, \\u\\2 < C(t + 1)\\a\\ + \\a\\2 hold with some t > 0, and constant C depends on mesQ,T,m0 and m\. If

mo(a, b)2

0 < n = ^ <

where = f ahds = (f, b), then the solution is unique.

dQ

(9)

Proof. Following the idea given in [4] and the method developed in [1], the original problem is reduced to an equivalent inverse problem with a non-linear operator equation for k. It follows from (1)-(3) that function w = a — u and the constant k satisfy the following relations

Mw + kw = ka, (11) dw \

-¡L + a(x)w) =0, (12)

dN K ' ) da ' v ;

/ whds = / ahds — j. (13)

Jsa Jen

Taking into account (8), (11) and (12), multiplying (9) by b in terms of the inner product in L2(Q) and integrating by parts twice, we obtain

k(u, b) = / ahds + (f,b) — j = ^ — j.

JSQ

i da

Let operator A : R+ ^ R maps every y e R+ into the real number Ay by the rule

Ay =!—S' (14)

(uy,b)

where uy is the solution of direct problem (1)-(2) with y = k. One can show that the original problem is solvable if and only if operator A has a fixed point, i.e., the operator equation Ak = k has a solution.

Now we need to prove that there exists t > 0 such that operator A defined for all k e [0, t], is continuous on [0, t], and maps [0, t] into itself. Indeed, Lemma 1.2 implies that for all 0 < y < t

aT ^ uy ^ a. (15)

Therefore

Ay > 0.

(a,b)

On the other hand, let us introduce the difference between (6) and (7)

M (a — aT) + t (a — aT) = Ta.

Then, multiplying the difference by a — aT in terms of the inner product in L2(il), integrating by parts in the first term and estimating the left-hand side of the result with the help of (4), we obtain

r T 2

m0\\a — aT ||2 + 2 a (a — aT )2ds + 2t ||a — aT ||2 < -\\a\\2.

Jsa mo

This estimate and (15) allows one to obtain the lower bound of (uy, b) in (14)

(uy ,b) > (aT ,b) = (a, b) — (a — aT ,b) > (a,b) — \\a\\\\b\\ > 0. (16)

Im o

Hence _

0 ^ ^ ^ v/mo(a,b)

^ ^ lla

In view of (14) and (16)

^ — j

Ay ^ M-^mjar^ ^T-

Accordingly, the relation Ay ^ t holds for all t > 0 such that

T 2

-_\\a\\\\b\\- t(a,b) + * — 1 < 0. (17)

Vmo

The last inequality is possible, because it follows from the theorem conditions that

D = (a,b)2 = ^N""" > 0.

Jm0

Then (17) is valid for t that obeys the inequality

/m0((a, b) -VD) < < /mQ((a, b) + VD) 2\\a\\\\b\\ ^ T " 2\\a\\\\b\\ '

Thus, the operator A maps the segment into itself.

Now one can obtain the estimate of uy in W2(Q) provided that y G [0, t]. Let wy = a — uy. This function satisfies (11)-(13) with y = k. Multiplying (11) for k = y by wy in terms of the inner product in Ly(Q) and integrating by parts in the first term, we obtain

(Mwy ,wy) + y\\wy\\2 = (Mwy ,wy )m + y\\wy\\2 + / awyds = y(a,Wy).

■Jdn

In view of (15) and the definition of wy, we have

' < t\\a\\y.

Taking into account the ellipticity of operator M, the last two relations implies that

y / awy dx Q

t

\\uy11l — ||aH + \\a\\i. (18)

mo

In accordance with [12], direct problem (11)-(12) has a unique solution wy G Wy(Q) for all y ^ 0. Furthermore, (11) is fulfilled almost everywhere in Q. and Mwy G Ly(il). Multiplying (11) with k = y by Mwy in terms of the inner product in Ly (il) and integrating by parts in the second component, one can obtain the equality

\\Mwy\\y + y(wy, Mwy) M + I ow2y ds = y(aM,wy). (19)

JdQ

In accordance with (4), the second term of (19) is non-negative and

y\(a,Mwy)| < t\\a\\\\MwVy\\ < 1 ty\\a\\y + 2\\Mwy\\y. it follows from the last two relations that

\\Mwy\\y < T2\\a\\2. (20)

In view of the definition of wy and the inequality [12]

II^Ny < Cm(\\Mv\\ + \M\),

valid for all v £ n W22(Q) with the constant CM depending on M and mesQ, relations

(18), (20) imply the estimate

IMI2 < IWyII2 + IMI2 < cm(t +1)yay + ||a||2.

Now one can show that operator A is continuous on segment [0, t]. Let y1 ,y2 £ [0, t] and uyi, uyi are the solutions of problem (11), (12) with y1 = k and y2 = k, respectively. By the definition of operator A, (15) and (16)

A _A 11uy2 - uyiWHK^ - v) . 11uy2 - uyiWHK^ - v) (21)

|Ayi Ay21 ^ (aT,b)2 ^ ((a, b) - ^IMPII)2 ■ (21)

On the other hand, multiplying the difference of equation (1) for k = y1 and k = y2 by uyi — uy2 in terms of the inner product in L2(Q) and integrating by parts in the first term of the resulting equality,we obtain

M (uyi - uy2 ,uyi ))i + a(uyi - uy2)2ds + yiWuyi — uy2 W2 = (y2 — yi)(uy2 ,uyi — uy2). (22) JSQ

In accordance with (4) and the non-negativity of y1, the left side of (22) can be estimated as

M (uyi - uy2 ),uyi )1 + / a(uyi - uy2)2ds + y1IIuyi - uy2f > moWuyi — uy2Wl. JSQ

The right term of (22) is estimated with the use of (15) as

i(y2 - y1 )(uy2 ,uyi— uy2 )i < 2mmoy-y^w^2 + IIuyi- uy2 w?.

Hence, we obtain the relation

IK1 - uV2 111 < — Il«y|y2 - Vil (23)

mo

Then, joining (21) with

2^m0((a, b) -VD)

t=—I-

and (23), we obtain the inequality

to,

\Ay! - Ay2 \ < NN^*"^ y - y2\ (24)

1 yl ^ ((a, b) + 2\/D)

which implies the continuity of operator A. Thus, according to the Brouwer fixed point theorem, operator A has a fixed point k* G [0, to] and the pair {u*, k*}, where function u* satisfies (1)-(2) with k = k*, gives a solution of problem (1)-(3).

It remains to prove that the solution of problem (1)-(3) is unique under assumption (10). In this case, operator A is a contractor on the segment [0, t0] because A satisfies (24) with

q = ||a||||b||(tt - _) < (a, b)2 < 1 (a,b)+2VD ((a, b) + 2%/D)2 '

Let us assume that (u', k') and (u'', k'') are two solutions of problem (1)-(2). Then k', k'' are the fixed points of operator A. By (24)

\k' - k''\ = \Ak' - Ak''\ < q\k' - k''\

whence k' — k" = 0. This in turn implies u' — u" = 0 in view of (23). Theorem is proved. □

Under assumption (10) the solution {u, k} depends continuously on the input data of original problem.

Remark 1. Condition (4) is valid when m(x) ^ m0 > 0 almost everywhere in il, or a(x) > a0 > 0 almost everywhere in dQ, here m0,a0 are some constants. In the last case left inequality holds due to the Friedrichs inequality.

Remark 2. The main theorem is correct for a more general type of operator M:

M = — div(^(x)V) + (mV) + m(x)I,

where m G L^(il) is vector of functions mi(x), i = 1,... ,n.

This work was supported by Russian Foundation of Basic Research [grant no. 20-31-90053].

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Об одной обратной задаче для эллиптического уравнения со смешанными граничными условиями третьего рода

Александр В. Велисевич

Аннотация. В данной работе рассматривается обратная задача для эллиптического уравнения с граничными условием третьего рода и условием интегрального переопределения. Доказано существование и единственность решения, а также непрерывная зависимость решения от входных данных.

Ключевые слова: обратная задача, краевая задача, эллиптическое уравнение, теорема существования и единственности.