On uniqueness in the problems of determining point sources in mathematical models of heat and mass transfer

Under consideration is the inverse problem of recovering the point sources in the model ut + Lu = ^Ni (t)5(x-xi) + fо(t,x),Lu = -Nu + ^ai (x)ux + a0 (x)u,(1)

i=1                                                                   i=1

where (x, t )e Q = ( 0, T )x G, G is a domain in >n (n = 2,3) with boundary Ге C2 . The unknowns are the functions Ni (t). The equation (1) is furnished with the initial and boundary conditions

Bus = g, ut=0 = Uo (x), S = ( 0, T )хГ,(2)

,          ..       „       Su                  „                 .    ,                 ,                    ,        „ x „ ,             „.

where either Bu =   + σu , or Bu = u ( ν is the outward unit normal to Γ ), and the overdetermination

Sv conditions

u ( y j ,t ) = ^ j ( t ) , j = ^1,2, ™ ^, s. ⁽³⁾

These problems arise in mathematical modelling of heat and mass transfer processes, diffusion, filtration, and in many other fields (see [1–3]). In the theory of heat and mass transfer, the function u is the concentration of a transferred substance and the right part characterizes sources (sinks) [1]. In the most general formulation of the problem (1)-(3), the intensities Ni (t) of point sources, their locations xi and the number m are quantities to be determined. Some descriptions of models of this type can be found, for example, in [1]. A lot of articles are devoted to solving these inverse problems. The main results are connected with numerical methods of solving the problem and many of them are far from justified (see [4–16]). The problem is ill-posed and examples when the problem is not solvable or has many solutions are easily constructed. Very often the methods rely on reducing the problem to an optimal control problem and minimization of the corresponding objective functional [2, 4, 5, 9, 16]. However, it is possible that the corresponding functionals can have many local minima. Some theoretical results devoted to the problem (1)–(3) are available in [17–21]. The stationary case is treated in [20], where the Dirichlet data are complemented with the Neumann data and these data allow to solve the problem on recovering the number of sources, their locations, and intensities using test functions and a Prony-type algorithm. The model problem (1)–(3) (G= Rn ) is considered in [21], where the explicit representation of solutions to the direct problem (the Poisson formula) and auxiliary variational problem are employed to determine numerically the quantities ^ Nirj (here Ni (t) = const for all i and rj = I xi - yj I )• The quantities found allow to determine the points {xi} and intensities Ni (see Theorem 2 and the corresponding algorithm in [21]). So the results of [21], for instance, say that the problem (1)–(3) and more general problem of simultaneous recovering points {xi} and intensities {Ni} in some model situations is uniquely solvable. In the one-dimensional case uniqueness theorem for solutions to the problem (1)–(3) with n = 1, m = 1 is stated in [17]. Similar results are presented also in [22].

In this article the main attention is paid to uniqueness questions of solutions to the problem in some model cases and the general case as well. Examples showing the accuracy of the results obtained are displayed. The constructions can be used when developing numerical algorithms. The results are based on asymptotic representations of the Green functions of the corresponding elliptic problems (see [23]).

Preliminaries

First, we describe our conditions on the data and some corollaries of the results in [23]. Let G be a domain in Mn . The symbols Lp (G) and Wp (G) (1 < p <^) stand for the Lebesgue and Sobolev spaces [24]. We also use the spaces Ck (G) of k times differentiable functions (see the definitions in [24]). If r,S are some sets then the symbol p(F,S) stands for the distance between these sets. The symbol D (L) stands for the domain of an operator L. Denote by Br (x0) the ball of radius r centered at x0 . Let a =( a1, a2) for n = 2 and a =( a1, a2, a3) for n = 3. The brackets (•, p denote the inner prod- uct in Rn . Let

V ^{( x} ) = 2

j ( a ( x ₀ + т ( x - x ₀ )), ( x - x ₀ )) d t .

The coefficients in (1) are assumed to be real-valued and a e W2(G)(i = 1,...,n),V^,Av,a0 eL^G),a e C1 (Г),(4)

Consider the problem

n

-Au + ^ aiux + a0u + Xu = 5(x - x0), x e G c >n,(5)

i=1

Bu\T = 0.(6)

For the reader’s convenience, here we present some results the article [23] (see Theorem 3.5, 3.9, 3.11, 3.12). We consider compact K c G , containing x ₀ , with properties: if Bu = u and G is a domain with compact boundary then the convex hull of K is contained in G ; if G is a domain with compact boundary and Bu ^ u then K c B^ p^ ( x ₀) ; if G = >ⁿ or G = R+ , then K is an arbitrary compact.

Theorem 1. [23]. Assume that the conditions (4) hold, K is a compact with the above properties, and if G = R+ , then, in case Bu ^ u , a = 0 (i.e., Bu = ux ) and ai = 0 for i = 1,2,...,n . Then there exists X1 > 0 such that for all X > X1 a solution un (x) (n = 2,3) to the problem (5), (6) in every domain {y e K :0 < s < | y - xi |, i = 1,2,_, m} admits the representation u 2 ( x ) =

¹          e ^ ⁽ x ^)- ^ x ^- x " (1 + O (-1=));

1/4                     V- u3 (x) =-----1-----t^x)-~ x-xo| (1 + O(^)).

^{3 (} )   4 n | x - x ₀|             ^V      4^;

Next theorem deals with solvability of the direct problem (1), (2). Let u0 ( x) g W2 (G), u0 (x)|  = g (x,0) if Bu = u.

We also suppose that f0 g L2 (Q),g(x,t) g W23/4,3/2 (S) if Bu = u,g(x,t) g W21/4J/2 (S) if Bu * u.

Consider auxiliary problems ut + Lu = f) (t, x), Bus = g, ut=0 = u0 ( x ), wt + Lw = ^T Ni (t )б (x - x), BwS = 0, wt=0 = 0.                    (12)

i = 1

Let Wp,B (G) be a space of functions u g Wp (G) satisfying the homogeneous Dirichlet condition whenever Bu = u and Wp,B (G) = Wp (G) if Bu * u . Denote by W-B (G) the dual space to WB (G) (the duality is defined by the inner product in L2 (G), see [25]).

The following theorem follows from [26], theorem 2 and [27], theorem 8.2.

Theorem 2. Let T <  ro and let p g ( 1, n / ( n - 1 ) ) . Assume that the conditions (9), (10) hold, a i g L _ro( G ) ( i = 0,1,..., n ) , and N i g L ₂ ( 0, T ) ( i = 1,2,..., m ) . Then there exists a unique solution to the problem (1), (2) such that u = w₀ +w , where w ₀ g W ₂ ^1,2 ( Q ) is a solution to the problem (11), w is a solution to the problem (12), w g L ₂ ( 0, to ; W p , _B ( G ) ) , w t g L ₂ ( 0, to ; W - B ( G ) ) and w g W ^1,2 ( Q _£ ) with Q _£ = { ( x , t ) g Q :| x - x i | > £ V i <  m } for all £ >  0.

Main results

Here we present our uniqueness theorem for solutions to the problem (1)–(3). We introduce the functions

1 r_

P j ( ^x ) = ^-- J ^{( a (} y j ^{+ t ( x -} y j )), ^{( x -} y j ^{)) d T} .

2 0

Let б, = min, r.■, j = 1,2,^, s, where r;= | x - y, |. Let A be the matrix with entries aH=e’j (i) j            i ij                                         ij        i       jJi if | x - y= |=5,- and ah = 0 otherwise. We assume, that: i        j         jji det A0 ^ 0(13)

Condition (4) is rewritten as follows: the coefficients of L are real-valued and ai g WTO (G)(i = 1,^,n),V^j, A^j,a0 g Lro (G)(j

Firstly, we justify uniqueness in the inverse problem (1)–(3) of recovering a solutions u and intensities N i ( i = 1, ^ , m ) . Points { x i } and their number are assumed to be known.

Theorem 3. Assume that T <  to , m = s , and the conditions (13), (14) hold. Then a solution ( u , N ) to the problem (1)-(3) such that u belongs to the class described in Theorem 2 and N i g L ₂ ( 0, T ) is unique.

Proof. It suffices to demonstrate that a solution to the problem (1)–(3) with homogeneous data is zero. In this case the auxiliary function w0 = 0. Let a function u such that u g W21,2 (Q£) for any £ > 0, u g L2 (0, T; Wp,B ( G)), ut g L2 (0, T; Wp-B (G)) be a solution to the problem ut + Lu = ^Nt (t)3(x - xi),(15)

i = 1

BuS = 0, ut=0 = 0,(16)

u (yj, t ) = 0, j = 1,2,..., 5.(17)

We integrate the equation (15) with respect to t and make the change of variables w = J ^u ( r ) d r . This function is a solution to the problem

Wt + Lw = £s, (t3 (x - x), Si = J N (r)dr e W2 (0, T), Si (0) = 0,(18)

i=10

BwS = 0, wt=o = 0,(19)

w (yj, t ) = 0, j = 1,2,..., s.(20)

Put w = eXtv, where X e R . This function satisfies the equation vt+Lv + Xv = £ si(t)e ,y(x-xi).(21)

i=1

Let Vj (x, X) be a solution to the problem **.*_/           **

Lvj+ 2vj = 3(x-yj),B vj|r= 0(22)

where L* - formally adjoint operator to operator L , B v = v , if Bu = u and B*v = ^^V + ( ^ + ( a,v ) ) v otherwise. The problem (22) is the adjoint problem to the problem

Lv + 2v = 3( x - yj ), Bv|r = 0.(23)

Multiplying the equation (21) by vj , integrating the result over Q , and using (20), we obtain the equalities mT

( v ( T , x ) , v j ( T , x ) ) = J v ( T , x ) v j ( T , x ) dx = У J s i ( t ) e ddtVj ( x i ) .

i = 1 0

G

The equality (24) can be rewritten as

A (X) S = F,(25)

T where the vectors S, F have the coordinates Si = J si (t) e 2 dt and Fi = 4n3j (v (T, x), v j (x)) e for n = 3 and Fi = 2^23jn21/4 (v (T, x), vj (T, x)) e^3jj for n = 2. Transform the representation fj = ( v ( T, x ), vj ( T, x )) = e-2T ( w(T, x), (2 + L* )-13 (x - yj )) = e -XT ((2 + L )-1 w(T, x ),3 (x - y)) = e ~XT (2 + L )-1 w(T, x )|    , x = yi

Note that the last expression makes sense and these formal transformations are justified. Indicate that w , w_t e L ₂ ( 0, T ; W p1, _B ( G ) ) . In particular, we infer w e C ( [ 0, T ] ; W p1 ( G ) ) after a possible change on a set of zero measure. By embedding theorems, w e C ( [ 0, T ] ; L q ( G ) ) с q <  3 p / ( 3 - p ) for n = 2,3. In this case the expression ( 2 + L ) ^{- 1} w ( T , x ) e W^(G ) is well-defined if the parameter X is sufficiently large, say X >  X 0 >  0 for some 2 0 . However, W q ( G ) c C ( G ) when n = 2,3 and q >  3/2 . Thus, we can consider the value ( 2 + L ) ^{- 1} w ( T , x )|^_ . There is the estimate

( X + L ) ^{- 1} w(T , y j ) < X + L ) ^{- 1} w(T , x )\\ c ( g ) <  C 0 \|( X + L ) ^{- 1} w(T , x )\\^ _(G)< q || w(T , x )|| _ц ( _G ) ,  (26)

where the constants do not depend on the parameter X >  X 0 and we use resolvent estimates for the elliptic operators (see. [27, Ch.2]). As a consequence, we obtain the estimate

| Fj |< c2XeTXeX5j VX > X0, where is the constant c2 does not depend on X and у = 0 when n = 3 and у = 1/4 when n = 2 . Fix an arbitrary 8 g (0,T). The above estimate implies that there exists a constant C0 (8) > 0 such that

C      ^-( T^-8)X

\ f । <  CW---- ^j

By Theorem 1, the entries b jj ( X ) of A ( X ) are representable as

/ x             * /   \ XX 8 •

I I 1 = a. 1 + О -т=

ji

bjj (X)=4n8jvj(xj)e j for n = 3 and b,( X) = 2^25^* (Xj) X1'4 e-8

ji

for n = 2 . Under the condition (13), we can assume that the matrix A ( X ) is invertible for X >  X 0 and the elements of the inverse matrix A ^{- 1} = { s j } are bounded by a constant independent of X ; otherwise, we increase the parameter X ₀ . Therefore, we have

m

Sj (X) = Ssjj(X)F, (X)

j = 1

and estimate (27) ensures that

Consider the functions S_j ( X 0 + z ) , where z is a complex parameter, Re z >  0 . The function S_j ( X 0 + z ) = J_o s_j ( t ) e ^Xt e"^ztdt is the Laplace transform of the function s ( t ) = s _j ( t ) e ^Xt for t <  T and s ( t ) = 0 for t > T . Introduce an additional function W ( z ) = ze z ^{( T} - ⁸ ) S _j ( X 0 + z ) . It is analytic in the right half-plane and is bounded by a constant C ₁ on the real semi-axis ⁺ . Estimate this function on the on the imaginary axis. Integrating by parts, we have

For z = y we have the estimate

s _j ( T ) e'^ ^ Te"^zT + J s_j ‘ ( t ) e"^A 0^t e ^{- zt}dt .

0                  J

\ W ( z ) | <^ \ S j ( T ) \ + \\ S j '|\_ii( 0, _T )

= C₃ V z = jy , y g R.

In each of the sectors 0 <  arg z <  n /2 , - n /2 <  arg z <  0 the function W ( z ) admits the estimate

W ( z ) \ < e^{|z | ( T -8} )

| s^{j ( T} ) I ⁺ || s^j H L ( 0, T )

V Re z >  0.

Applying the Fragment–Lindelef Theorem (see theorem 5.6.1 in [28]) we obtain that in each of the sectors 0 < arg z < n /2 , -n /2 < arg z < 0 the function W (z) admits the estimate W (z )< max (C1, C3 ) = C4 V Re z > 0.

Therefore,

| S i ( Я + 2 ) | = | I ( S i ( t ) ) ( z ) | <  C 4 ( e ) e ^{-( T -e} ) Re ^z /1 z | V Re z >  0.                (32)

We have equality ( ст > Я ₀, p = ст + i^ ) . с + i да                              . да

Si (t) = z— J epL(Si)(p)dp = V eeeieLL(Si)(с + i^)d^. 2ni                    2n с-i да                             —да and, thereby, да

S i ( t ) e -< ‘ - T^-e » = ± J       T- > I ( s i )( c + И ) d ^ _

2 n

—да

The Parseval identity yields

|| S i ( t ) e ^M' - T - > ■» ||² (_, да ) = ' ^да e ^M T^— = ) | I ( s , )( c + , _fl|² d ^ <  C / ■ ) ^да           d ^ <  Д И .

2 П                            2 П                2 M

-да                                                  -да        7

Since this inequality is true for all с >  0, s i ( t ) = 0 for t <  T - e . Since the parameter e is arbitrary, s i ( t ) = 0 for t <  T . We infer N i ( t ) = 0 for t <  T and every i and, therefore, the right-hand side of (15) vanishes which implies that u = 0 .                                                        □

We note that the following condition is actually a necessary condition for the uniqueness of solutions to the problem (1)–(3). If it fails then any number of the points { y_i } does not ensure uniqueness of solutions (see examples below).

Condition (A). For n = 2, any three points { y_t } do not lie on the same straight line and, for n = 3 any four points { y_i } do not lie on the same plane.

Next, we describe some model situation in which Lu = -A u + Я о u , Я >  0, G = R ⁿ and functions N_i on the right-hand side of (1) are real constants.

Theorem 4. Let u1,u2 be two solutions to the problem (1)–(3) from class described in the theorem rj

1 with the right-hand sides in (1) of the form Z N ii 5 ( x - x i ) ( N j = const, j = 1,2 ) , the condition (A) i = 1

holds, and s >  2 r + 1 in the case n = 2 and s >  3r + 1 in the case n = 3, where r >  max ( r 1 , r₂ ) (i. e., there is the upper bound for the number max ( r 1 , r ₂ ) ). Then u 1 = u ₂ , r 1 = r ₂ , and N 1 ¹ = № for all i , i. e., a solution to the problem of recovering the number m , points x_i , and constants N_i is unique.

Proof. Let the functions u 1 ,u₂ do not coincide and let w = u 1 - u₂ . The function w satisfies the homogeneous initial data and over determination conditions (3) and we have (after renumbering the constants N_i^j and points x_i )

r3

wt + Iw = ^Ni5(x-xi)-^Ci5(x-x*), 2r > r3 + r4,Ni,Ci = const,(33)

i=1

where Ni, Cj > 0 for all i,j. Without loss of generality, we can assume that all the numbers Nt,Ct are not equal to zero and all points xi,xi are distinct. Let, for example, n = 3. For simplicity, take Я0 = 0 . The proof is the same for other values of this parameter. Applying the Laplace transform, we infer

Using (3), we obtain

r ₃

z

\           N      - ^|x - xt | w (x) = > --------i------e       i

^{V }}  “ 4 n | x - x i | Я

r ₄

E      Ci      --'Д |x-x* | e.

i = 1 4 n | x - x i | Я

_____N i _____e-^ yj - x i _ rv_____C i _____e-^ y j' - ^x * ^| / = 12 e                                                        e                              ,         -*-,

4 n | y j - x i | Я              i T 14 n | y j - x *| Я            ^J

, s .

For definiteness, we assume that r₃ >  r ₄ . Let us show that the sets of numbers {r ij =x i - y j |: i = 1,2 - , r ₃}, {r ij = | x * - y j |, i = 1,2 - , r ₄} coincide for all j . Fix the parameter j . Let 5 1 j = min i r jj , 5 1 j = min i r ij . Demonstrate that 5 1 j = 5 1 j . Assume the contrary. Let, for example,

51 j < 51 j . Multiply the system (35) by 4n51 jXe   j and passing to the limit as X ^ +^ we obtain the equality

L   Ni= °-(36)

i ^{:| x} i ^- y j ⁵ j

It is a contradiction, since N i >  °. So, 5 1 j = 5 1 j and multiplying the system (35) by 4 n5 1 j te ^j ^j and passing to the limit as X ^ + ^ we also derive that

L  Ni = J

i ^{:| x} i - y ¹ | = ^{5 1} j         i :| x * - y j | = 5 1 j

So, we can reduce the following sums on the left and on the right in the equalities (35): у         N i     ^- 4^\ y j - x i ^|     у _____ C i_____ - x/^ - х *

i:x-y?|=51j.4n | yj - xi I ^          ’i:x* —yjk/^ I yj - x ^'

Denote 52= min.  . e rti  and 52= min.  . e Гц . Repeating the arguments, we obtain that j           i :rij >°1 j ij                 j           1 :rij >°1 jj

5 2j = 5 2j and, thereby,

L n = L. C i - i ^:| xi ^- y j k ^ j         i ^:| x i * - y j L j

Again, abbreviated equal summands (35), we arrive at the system (35), where the sums on the left and on the right are taken over i : r ij > 5 2j и i : r ij > 5 2j , respectively. It is now obvious by induction that there are pairs of equal numbers 5 _k j , 5 kj k = 1,2, - , ^r ° j <  min ( r ₃, r ₄ ) and

L    N i = L    C i , k = 1,2, - , r ° j ,                      (37)

i:| xi- yj|=5kj         i: | x* - yj |=5kj moreover, the left-hand and right-hand sides of these equalities are positive. So, the sets of numbers {rij = xi -yj-1: i = 1,2-,r3}, {rij = | x* -yj |,i = 1,2-,r4} coincide for all j. In particular, it follows that for any point, for example, x1 and any j , there exists a point xi such that

*

^{| x}1 ^- y j ^| = ^| xij ^- y j ^| , j = ^{1,2, —} , » .

But we have s > 3r +1 and r4 < r is the number of points {x*}. Hence, among the points {x*.} y=1 there are four coinciding points. After renumbering if necessary we can assume that these points are x*, x* , x*, x* . Then the equalities i1, i2, i3, i4                             1

*

| x1 -yj |=| xij -yj |, j =1,2,—,s imply that the points yj with j = 1,2,3,4 lie in the same plane which is perpendicular to the segment ^x1,x^ J, but this fact contradicts to the conditions (A). So, w = °.

The proof in the case of n = 2 is almost the same but we use an asymptotic representation for a fundamental solution i^H°°1)(i4x | x-x° |) defined by the equality (7), where у = ° . As in the case of n = 3, we arrive at contradiction with the condition (A).                                              □

We display the corresponding examples showing the accuracy the results obtained. The following example shows that if the condition (A) fails then the problem of recovering the intensities of sources

(sinks) located at x1, x2 has a nonunique solution. At the same time, it is an example of the nonuniqueness in the problem of recovering the intensity of one source and its location. Note that the problem of determining the location of one source x0 and its intensity N (t) is simple enough and to uniquely recover these parameters we need two measurements in the case of n = 1 [22], three measurements in the case of n = 2 [28] and four measurements (that is s = 4 in (3)) in the case of n = 4 [25]. The smaller number of points does not allow to define the parameters N(t), x0 uniquely. We should also require that the point x0 lie between two measurement points in the case of n = 1 and the condition

• (A) holds in the case of n = 2,3. The numerical solution of the problem of recovering one source is treated in the articles [6, 9–15, 19, 28].

Example 1. First we take n = 3, G = R ⁿ , Lu = -A u . Let u be a solution to the equation (1) satisfying the homogeneous initial conditions with the right-hand side in (1) of the form

The Laplace transform of this solution to the problem (1)–(2) is written as u = NT (X)(----1----e- 2 - x--1----e - 2 -x2|).

• ⁷ 4 n | x - x 1 1             4 n | x - x ₂ |

Let P be the plane perpendicular to the segment [ x 1 , x ₂ ] and passing through its center. We have u ( y , 2 ) = 0 V y e P.

• So,    u ( y , t ) = 0 for all y e P . Precisely the same example can be constructed in the case n = 2 . We take the perpendicular to the segment [ x 1 , x ₂ ] passing through its center rather than the plane P . Thus, if condition (A) fails then any number of measurement points does not allow to determine the intensity and the location of the sources.

Example 2. Consider the case of G = >n , Lu = -Au. Let us show that the conditions (3) with s = 4 in the case of n = 2 and s = 6 in the case of n = 3 does not allow to determine location of two sources and their intensities even if the condition (A) holds. Let u1 , u2 be solutions to the equation (10) satisfying the homogeneous initial conditions in which the right-hand sides are of the form

Let, for example, n = 3. Then the Laplace transforms of u 1 , u ₂ are as follows:

N     - 2 - x |                N    - 2 - x * |.

i = 1^{4 n} | x ^- x i |                         i = 1 4 n | x - x_i|

Here we use explicit representations of the fundamental solution for the Helmgoltz equation (see, for example, in [30, §3.1] or [31, ch. 4, 8]). We take   x1 =( a, a ,0), x* =( a, - a ,0), x2 =(- a, - a ,0), x2 =(- a, a ,0) (a > 0). As is easily seen, the functions u1, u2 coincide at the points yx =( M ,0,0), y2 =(-M ,0,0 ), y3 =( 0, M ,0 ), y 4 =( 0, -M ,0 ), y5 =( 0,0, M ), y6 =( 0,0, -M ),      where

M >  0 and, thus, the problem of recovering the locations of 2 sources and their intensities admits several solutions in the case of s = 6 . It follows from the theorem 2 that in the case of s = 7 points x 1 , x ₂ and the intensities are determined uniquely (if the condition (A) holds and the intensities are constants).

Consider the case of n = 2 . As before, we construct functions u_x, u ₂ whose Laplace transform is of the form

• 2                                      2

u=i^ H01)(i   | x- xj|)s u2=i^ H01) (i   | x- xj|)s where H0 is the Hankel functions [32]. Let us take x1 =( a, a ), x* =( a, - a), x2 =(-a, - a), x* =(-a, a) (a > 0). It is easy to check that uti (yj,Л) = U2 (yj,2)vj = 1,^,4,ЯеГ- (39) where y1 =(M,0),y2 =(-M,0),y3 =(0,M),y4 =(0,-M). It follows from the theorem 2 that the points x1, x2 and intensities are determined uniquely in case s = 5 (if condition (A) holds and the intensities are constants).

Remark 1. The examples show that the number of minima of the corresponding objective functionals introduced if we solve the problem (1)–(3) numerically reducing the problem to an optimal control problem can be large and even can be a manifold.

Remark 2. Relying on asymptotic representations and Theorem 1 in the case of constant values N i , we can construct a numerical algorithm for finding sources { x i } employing the ideas from the article [19]. Some review of the results connected with numerical determining point sources can be found in the article [33] and some results in [34–37].