ABSTRACT

Under consideration is a nonclassical stationary problem on heat conduction in a body with the pre-set surface temperature and heat flow. The body contains inclusions at unknown locations and with unknown boundaries. The body and inclusions have different constant thermal conductivities. The author explores the possibility of locating inclusions. The article presents an integral criterion based on which a few statements on identification of inclusions in a body are proved.

Keywords: Heat Conduction; Inclusions; Defect; Heterogeneous Medium; Inverse Problem

1. Introduction

Under analysis is a nonclassical problem on heat conduction, with the known body’s surface temperature and surface heat flow. It is assumed that the medium is heterogeneous and contains inclusions. Coefficients of heat conductivity of the medium and inclusions are assumed by constants and differ among themselves. The problem is formulated as the problem on finding heterogeneities (flaw detection problem) by overdetermined surface conditions. On the boundary surface, the condition of continuity of the temperature and heat flow is fulfilled. Such problems belong to the nonclassical problems of mathematical physics.

Stationary problems of heat conduction are described by Equation, where is the temperature of a body and is the Laplace operator. On an interface of two mediums conditions are satisfied

(1)

where is the heat flow; is the coefficient of the thermal conductivity.

Hereinafter is the coefficient of the host medium. It follows from (1) that the normal derivative of becomes discontinuous at the boundary of two surface.

Essentially overdefined condition for the Laplace equation mean assignment of values of and on, or, in our case, the values of and on.

For a homogeneous medium, evidently, the essentially overdetermined conditions cannot be arbitrary, i.e., and are functionally connected on. Let’s receive conditions of coordination for the last and consequences following from them.

Consider a body with volume and surface. Let the functions

(2)

be potentials of the simple layer and double layer, respectively. Here, is the fundamental solution of the Laplace equation [1]; and are the densities of the layers, and n is a vector of external normal to. It is assumed that is piecewise-smooth according to Lyapunov and and fulfill the Holder condition. Let, then

(3)

Here, and are, respectively, the exterior and interior of, the boundary surface not included. It is stated that the densities and are concordant if is continuous in and is equal to on. The consistency of the densities can be interpreted as correspondence of and to the values of a harmonic potential and its normal derivative on in the homogeneous medium. By the uniqueness of the Dirichlet and Neumann problem, the use of the densities and unequivocally recovers the concordant and, respectively. Hereinafter, the set of the concordant densities is denoted in terms of the class.

Statement 1. The densities b and a are concordant, i.e. when and only when the equality below holds true

(4)

where

.

Proof. The necessity follows apparently from Green’s formula for harmonic functions [1]. Proving of the sufficiency uses equivalent of Sokhotsky-Plemeli’s formula

(5)

where, and and are the internal and external limit points relative to, respectively. It follows from (4) that from whence, considering (5), appears which is to be proved.

Statement 1 is similar to the theorem on boundary values of analytic function in the complex-variable function theory where conditions of continuous extension of analytical function from the closed contour to a domain are defined.

It is worth pointing at one property of the functions belonging to the class. Let and and be the values of densities, found from (3), at the boundary of an area on; then. In addition, the condition of the Neumann problem resolvability is fulfilled at and, which means that the heat flows through and are zero.

Subsequently, the potentials and the flow on are assumed known. Then, the concordance conditions (4) are written as For, we also use.

Let’s prove the statement following from the statement 1.

Statement 2. Let on the boundary of the domain with the coefficient of the thermal conductivity coefficient the temperature and the heat flow (on) be assigned such that Then for any it has to be executed.

Proof. On a function is introduced such that and. Assume, that, i.e., in accord with (4)

. (6)

In the same way, from the condition we have

(7)

The flow condition on yields

.

Placing the expression above in (6) and, then, its deduction from (7), considering on, produces

For the simple layer potential, it appears that the external normal derivative. Then, according to [1], i.e. we come to a contradiction with a condition the statement. That is to say, the statement has been proved.

Statement 3 (The theoremof the coefficient problem uniqueness). Let and be assigned on (on). Then of the medium is uniquely found from the condition.

Proof. Inasmuch as, the uniform medium concordance condition (4) can be written as

(8)

Let there exist for which the condition (8) holds true, too. Rewrite (8) for and diminish then (8) by

For the simple layer potential, it appears that the external normal derivative. Then, according to [1], , i.e. we come to a contradiction with a condition the statement. That is to say, the statement has been proved.

Consequence 1. If and are assigned on and and hold true, then.

The condition (8) produces the formula for the coefficient of the thermal conductivity coefficient

.

Let's notice that in [3] conditions for determination of the thermal conductivity coefficient for a non-stationary problem of heat conductivity are received.

Based on the introduced definitions and statements, there are a few inferences for a heterogeneous medium. A heterogeneous medium is understood to be the medium containing inclusions (defects), with the conjugacy condition (1) satisfied at their boundaries. Solving the problem on an extent from to an inclusion boundary defines T and on. . The problem on the extent from belongs to the known problems on the harmonic extension, i.e., Cauchy problem for the Laplace equation [2]. These problems belong to conditionally correct problems of mathematical physics and have the unique solution. Geophysics has many methods of solving such problems. One of methods is offered in [3-6].

It is assumed that the condition is fulfilled on the boundary of the inclusion with. This condition is assumed to be the condition for the inclusion, which means continuity of the solution inside the inclusion, i.e., the inclusion is considered as a homogenous medium. Let us prove the following statement.

Statement 4 (condition of existence of defect in a body). Let and be assigned on . If the body contains an inclusion with the thermal conductivity coefficient (is the coefficient of the thermal conductivity coefficient of the host medium), then.

Proof. Assume. Continuing the decision from to on we will find T, and, consequently,. According to the above mentioned property of on, the condition or, which is the same kind of thing, is to be fulfilled. On the other hand, the solution in the inclusion is continuous, i.e.. Thus, we have that and on. Under consequence 1, this is only possible when, which is a contradiction a statement condition. So, the statement has been proved.

As follows from Statement 4, an inclusion as though initiates features of a field; this means that in construction of the solution in via the extent from, the potential is not expressed in terms of finite functions.

The introduced definitions and proved statements allow stating the uniqueness of finding the inclusion boundary and the heat conductivity coefficient under fulfillment of the conjugacy condition (1).

Statement 5 (Theorem of the unique definition of inclusion boundary). Let in the medium with, and on be known (on) and let contain an inclusion with . Let the solution of the problem on the extent from to, i.e. define and on. Then the condition uniquely defines the boundary of the inclusion.

Proof. We extent the solution from in. Let there be two surfaces and in, and and hold true at these surfaces. Below we consider three cases.

1) Let, Figure 1. Assign an arbitrary function on and. Use the value of on to plot a harmonic function in and find on for this function. Likewise, assign at and find at.

For the harmonic and T we will write down Green’s formula

(9)

Likewise, write Green’s function for

(10)

Summing up (9) and (10) yields a Green formula for the domain

(11)

On the other hand, once the solution is continuously extendable from in the domain, i.e., then this solution has its Green’s formula, too

(12)

Diminution of (12) by (11) produces

(13)

The integral (13) equals zero for the arbitrary function whence it follows that and therefore on, then in the domain. It follow from the harmonicity of in that on, i.e. on. Thus and so, we arrive at contradiction with the condition of our statement.

2) Let, and, Figure 2. It can readily be understood that. Then, inasmuch as, it is evident that, i.e., there is an inclusion inside. Let this domain be denoted as. For the inclusion we have and. Then, , on the one hand, and, on the other hand; besides, , which agrees with the conditions of paragraph 1 of Statement 5. Thus we come to a contradiction.

3) Let, Figure 3. The domain is conditionally divided into two subdomains, one containing, the other containing. The domains are denoted by and, respectively. Let an inclusion be inside. Then. In this case,. Whereupon is to be fulfilled alongside with on, which contradicts consequence 1. In case that the inclusion is inside, the relevant considerations will result in the same contradiction.

With the known thermal conductivity coefficient of the host medium, it is possible to find the thermal conductivity coefficient of the inclusion.

Statement 6 (Theory of the unique definition of the thermal conductivity coefficient of inclusion). Let and be pre-set on (at). Let the medium with the thermal conductivity coefficient

contain an inclusion with the thermal conductivity coefficient and (on). Then the condition uniquely defines for the inclusion.

Proof. Assume that contains two surfaces and where the settings of the theorem are fulfilled and, ,. Likewise Statement 5, a few cases are considered below.

1) Let and. But if the settings are fulfilled on and, then, according to consequenceСписок сокращений 1, it must be that. We have arrived at the contradiction.

2) Let Figure 3. The domain is conditionally divided into two subdomains, one containing, the other containing. The domains are denoted by and, respectively. Let an inclusion be inside

. Accordingly, the domain belongs to. Under the theorem settings, , then, since, we have that. On the other hand, the condition holds true at, too. Then, in pursuance to consequence 1, we get, which is the contradiction. In case that the inclusion is inside, the relevant considerations will reach to the same contradiction.

3) Let or Figure 1. Then and. In other words, we consider the domain with the inclusion and. According to the statement 4 we come to a contradiction.

4) Let, and, Figure 2. Continuing the decision with the on we have. Then according to the statement 4 inclusion is in. Denoted this domain by. The domain belongs to and. It follows whereof that and. According to consequence 1, we arrive at, which is the contradiction. The statement has been proved.

The credibility of the criterion was tested in the two-dimensional calculations. At the side of a unit square, the values of and in a medium enclosing a circular inclusion were pre-set. The field of the inclusion was modeled by the potentials in the form of and. The field inside the inclusion was described by orwhere, are coordinates of the inclusion; is angle between the vector and axis. The constant were found from the conjugacy condition (1). The calculations used

, , , , where and were the thermal conductivity coefficients of the host medium and inclusion, respectively, and was the inclusion radius. In a figure 4 calculations for inclusion in the field and in a figure 5 in the field are presented.

Figures 4 and 5 show the curves for, , i.e. the point moves over the square side at distance 0.015. Curves 1 and 2 correspond to the values of at the distances and. Values H corresponds to a defect depth under the square side.

As the calculations by the criterion showed, the inclusion at the occurrence depth was not revealed; whereas at the inclusion was located at high reliability.

2. Conclusions

1) The criterion allows locating inclusions in a body upon the conjugacy condition (1) at the boundary surface.

2) Based on the criterion, both the boundary of the inclusion and its thermal conductivity are uniquely defined.

3) The criterion is reliable for near-surface inclusions.

The study was supported by the Russian Foundation for Basic Research, Project No. 11-01-00522.

REFERENCES

[1] L. S. Sobolev, “Equations of Mathematical Physics,” Nauka, Moscow, 1966.

[2] D. Lesnic, “The Determination of the Thermal Properties of Homogeneous Heat Conductors,” International Journal of Computational Methods, Vol. 1, No. 3, 2004, pp. 431-443. http://dx.doi.org/10.1142/S0219876204000228

[3] M. M. Lavrentiev, “Ill-Posed Problems in Mathematical Physics,” Novosibirsk, 1962.

[4] D. Lesnic, J. R. Berger and P. A. Martin, “A Boundary Element Regularization Method for the Boundary Determination in Potential Corrosion Damage,” Inverse Problems in Engineering, Vol. 10, No. 2, pp. 163-182.

[5] A. A. Schwab, “Computer Tomography Problem Based on the Hologram Interferometry Method,” In: Studies into Conditionally Ill-Posed Problems of Mathematical Physics, Institute of Mathematics SB AS USSR, Novosibirsk, pp. 157-162.

[6] A. A. Schwab, “Boundary Integral Equations for Inverse Problems in Elasticity Theory,” Journal of Elasticity, Vol. 41, No. 3, 1995, pp. 151-160. http://dx.doi.org/10.1007/BF00041872

NOTES

^*Corresponding author.