Localisation Inverse Problem and Dirichlet-to-Neumann Operator for Absorbing Laplacian Transport

  We study Laplacian transport by the Dirichlet-to-Neumann formalism in isotropic media I   , I     , I  . Our main results concern the solution of the localisation inverse problem of absorbing domains and its relative Dirichlet-to-Neumann operator . In this paper, we define explicitly operator    , and we show that Green-Ostrogradski theorem is adopted to this type of problem in three dimensional case.


Laplacian Transport and Dirichlet-to-Neumann Operators
The theory of Dirichlet-to-Neumann operators is the basis of many research domains in analysis, particularly, those concerning Laplacian transports.It is also very important in mathematical-physics, geophysics, electrochemistry.Moreover, it is very useful in medical diagnosis, such as electrical impedance tomography: In 1989, J. Lee and G. Uhlmann have introduced an example on the determination of conductivity matrix field in a bounded open domain, see e.g.[1].This example is related to measuring the elliptic Dirichlet-to-Neumann map for associated conductivity equation, see e.g.[1].
The problem of electrical current flux is an example of so-called diffusive Laplacian transport.Besides the voltage-to-current problem, the motivation to study this kind of transport comes for instance, from the transfer across biological membranes, see e.g.[2,3].
Let some species of concentration , , diffuse stationary in the isotropic bulk   , where they disappear at a given rate .Then the steady field of concentrations (Laplacian transport with a diffusion coefficient ) obeys the set of equations: , the concentration at the source , ( ), on the interface .
, is a constant concentration of the species inside the cell B B  .This example motivates the following abstract stationary diffusive Laplacian transport problem with absorption on the surface : , .:

 
, dom The advantage of this approach is that as soon as the operator ( 1) is defined, one can apply it to study the mixed boundary value problem (P2).This gives, in particularly, the value of the particle flux due to Laplacian transport across the membrane .Moreover, the total current across the boundary   can be defined (for given f ) in term of Dirihlet-to-Neumann operator (1) as follows: , : d , where  designed the differential element relative to . f There are at least two inverse problems derived from problem (P2): a) geometrical inverse problem: given Dirichlet data and the corresponding (measured) Neumann data g , in (1), on the accessible outer boundary , to reconstruct the shape of the interior boundary , see [5].
 B  b) localisation inverse problem: concerns to localisate of the domain (cell) B with a given shape and the fixed parameters  and , see [6].In Section 2, we introduce the existence and uniqueness for the solution of problem (P2).In Section 3, we introduce our first main result concerning the study of spherical case of problem (P1), whose we give a general method to resolve the type of partial derivative system like (P1), see proposition 3.2.Indeed, we allow an explicit calculations, based on Green-Ostrogradski theorem, for the solution of this problem.
In Section 4, it is our second main result which consist in showing that total current across the external boundary , involving Dirihlet-to-Neumann operator (1), can resolve the localisation inverse problem in three dimensional case, when the compact .

Uniqueness of the Problem (P2)
We suppose that  and be open bounded domains in  with -smooth disjoint boundaries


Then the unit outer-normal to the boundary is well-defined, and we consider the normal derivative in (P2) as the interior limit: The existence of the limit (3) as well as the restriction is insured since has to be harmonic solution of problem (P2) for -smooth boundaries   B    [7].Now, we introduce some indispensable standard notations and definitions, see [8].Let be Hilbert space the Sobolev space of -functions, whosederivatives are also in , and similar, is the .Proof.For existence we refer to [7].To prove the uniqueness, we consider the problem (P2) for  and 0 c   .Then by Gauss-Ostrogradsky theorem, one gets that the corresponding solution yields: The estimate (4) implies that   Const Hence by the boundary condition one gets , and from and by condi (P2) has two solutio tion (5), the problem ns for identical external (on  ) and internal (on Robin boundary c itions.Then by the ndard a uments based on the Holmgren uniqueness theorem [9] for harmonic functions on 2   , one obtains that


Absorbing Laplacian Transport re, we consider the spherical shell of the p so that and the absorbing cell is also a ball B  , whose we denote by 0 . ce tw Hereafter, we denote the pr othesis spherical case.
In the sequel, we resolve the problem (P1) in order to calculate explicitly Dirichlet-to-Neumann operator relative to this case.
Before resolving problem (P1), we need the following theorem which the key of the solution: whose designated the divergence of field vector . and d designated respectively the differen- tial elements relative to  and  .M n  designated the unit outer-normal vector on at arbitrary point M .

Remark 1. Let the orthonormal reference with origin and axis Y Y O
 , which is keen on the line in the sense of the vector .
On the other hand, since for all , spherical harmonic function is harmonic function, then it takes the following form, see [10]: Therefore, we need to calculate the coefficients of ( 7) from the condition boundaries.Indeed, since the radius of points of  are equal to constant 0 , then the condition boundary on r B  implies easily, by identification, the following system: , the radius aren't equal, and d e But, on the boundary  l   , a system of two equations with two unknowns l a and op epend of spherical angl  .Then, for this reason, we use Gauss-Ostrogradski the rem's, whose we show that it is useful to find another relation between the coefficients of (7) where: , we have: On the other hand,  0 l div V can be calculated from (9) by:  

Radius varies between and :
Angle varies between 0 and 2π.
of multiple integrals, we obtain: Moreover, condition boundary on  implies that: , , , Bu that: Indeed: from (7), we have that , .
, we the previous equation, and integrating it on do e other hand, spherical harmonic functions form basis for th Consequently, we deduce, since , , , Here, 0 0 l  designed Dirac funct So, by insert ion.ing (15) in ( 14), we deduce above equality (13) as follows: We continue the proof by inserting (13) in (12): Indeed: unit outer-normal vector M n relative to domain point 2 Showing that: the shape implies that unit 0 , , l Indeed: the symmetry of is below in plan generated by the two vectors r e and  e which are orthogonal to field vector V directed 0 l by  e .So, we obtain: Then, by inserting ( 18) and ( 19) in ( 9), we deduce that:

Boundary Equation
Finally, by inserting ( 16) and ( 20) in (10), we obtain that: The previous equation ends the proof since it is true y 0 l   . osi for an Prop tion 3.3.
problem (P1) have unique solution with e form (7), whose the coefficients are given by: where,   Here, the coef e given by proposi t is given by: and consequently, this implies: .
ation by its value given in (7). Remark 2. For general properties of Dirichlet-to-Neumann operators, mainly existence and uniqueness, the prev s equ we refer to [10], chapter 4.
ii) Second, we aim to find an equation involving the distance On the other hand, by Gauss-O gradsky heorem, one gets: ere dV is the volume differen verifies the following equation: given in proposition r to substitute 0 b , after source localised on the closed boundary  B towards a semipermeable compact interface  of the cell B   0 W  0 D  , one recovers respectively the Neumann and the Dirichlet boundary conditions.Now, we can associate with the problem (P2) a Dirihlet-to-Neumann operator , : h  The main question in this context is to find sufficient conditions insuring that the localization inverse problem is uniquely soluble.Indeed: First, we relate the above problems a) and b) with the Dirichlet-to-Neumann operator (1) by defining explicitly this operator, whose can define the local and total current across the external boundary , which are useful to resolve a) and b).Second, we study the localisation inverse problem in the framework of application outlined in the problem (P2), which consist in finding sufficient (Dirichlet-to-Neumann) conditions to localise the position of the cell B  3    from the experimentally measurable macroscopic response parameters.
. Then the Dirichlet-Neumann problem (P2) has a unique (harmonic) solution in domain B   0 f

3 . 4 .
It is enough to reso    , the et l b given by the two boundary conditions (8) and   f S . Since the solution of problem (P1) is given from proposition 3.3, then we can deduce its relative Dirichletto-Neumann operator: Corollary The Dirichlet-to-Neumann operator (1) is defined by system of two unknowns l a

O center of cell Remark 3 .
Notice that definition of Dirichlet-to-Neumann operator (21) implies that it has as eigenfunctions the spherical harm It is enough to insert (22) in the expression of replacing   r  by its value in term of 0 d : -Neumann hypothesis of problem (P1) on the external boundary, and we can found them from an experimental measures.To summarize, we have found an equation for 0 d which is the distance between the center O of the cell Consequently, the fact that   ends ly unknown d involving the proof. 5. Conclusions 23) is an equation of the on the p ( B and the center O of  , so it remains to find the 0 arameters J  and : f v   , which are the Diri-0 position of the center O .In fact: o  whose the norm of th l current j pectively its maximum and m nimum values, see .Then, from the symmetry o the shape, inver problem for (P1), and we can conclude that 2), involving Dirihlet-to-Neumann fficient to resolve the localisation ough in problem like geometrical inverse problem, see [5].

Figure 1 .
Figure 1.Position of the cell overline B.

Pr osition 3.2. The condition boundary on  im o
like   l b , which it is sufficient to calculate l a and l b : f S .Consequently, we get for each pl