On the uniqueness theorem for one nonstationary 3-D inverse heat conductivity problem in a layered domain

In this work, we consider the inverse 3-D heat transfer problem with any fixed n∈ discontinuous heat conductivity coefficients. At first with the help of conservative averaging method, we reduce the considered problem to the 1-D inverse problems. Then we prove the uniqueness theorem for obtained 1-D inverse problem. Key-Words: Heat transfer problem, inverse problem, conservative averaging 1 Mathematical formulation of the initial problem In this paper we consider the 3-D problem for the function ( ) , , , U x y z t in the parallelepiped ( ) [ ] [ ] [ ] { } 3 , , : 0, , 0, , 0, def n D x y z x x y Y z Z ≡ ∈ ∈ ∈ ∈ . Let us the function ( ) , , , U x y z t satisfies the following conditions and heat conductivity equation: ( ) ( ) ( ) 1 0 ( , , , ) ( , , , ), , , 0, , 0, , 0, 1, , 0, t j j j x y z t k U x y z t x x x y Y z Z x j n t U − = ⋅ Δ ∈ ∈ ∈ = = > (1) 0 ( , , , ) 0, t U x y z t = = (2) 0 ( , , , ) 0, x x U x y z t = = (3) [ ] ( , , , ) ( , , , ) ( ), n n x x x U x y z t C k U x y z t t ν = + ⋅ ⋅ = (4) 0 0 ( , , , ) ( , , , ) , 1, 1, j j x x x x U x y z t U x y z t j n − + = = = = − (5) 0 0 1 ( , , , ) ( , , , ) , 1, 1, j j j j x x x x k U x y z t k U x y z t j n + − + = = ⋅ = ⋅ = − (6) 0 ( , , , ) ( , , , ) 0, y y y y Y U x y z t U x y z t = = = = (7) 0 ( , , , ) ( , , , ) 0. z z z z Z U x y z t U x y z t = = = = (8) From the problem (1)-(8) we have to determine the following data: • the number n N ∈ that means a number of discontinuity of heat conduction coefficient; • the discontinuity points ( 1, 1) j x j n = − ; • the heat conduction coefficients ( 1, ) j k j n = ; • the constant C ; • the function ( ) , , , , U x y z t where ( ) , , , 0 x y z D t ∈ > . In addition, it is proposed that we have the some additional data. These data will be formulated in the next section of the present paper. Now let us introduce the function (as in the conservative averaging method, see [9]-[11] and references there) 0 0 1 ( , ) ( , , , ) Y Z def u x t dy U x y z t dz Y Z ≡ ⋅ ⋅ ∫ ∫ , and we will integrate the basis equation (1), the boundary conditions (3)-(6) and also the initial condition (2) on variables y and z . Then we will use the conditions (7), (8) in obtained expression. For example, after abovementioned procedure from (1) we get 0 0 0 0 1 ( , ) ( , ) ( , ), y Y z Z Z Y t j xx j xx z y U U u x t k u x t dz dy k u x t Y Z y z = = = = ⎤ ⎧⎪ ⎥ ⎨ ⎥ ⎪⎩ ⎦ ∂ ∂ = ⋅ + + = ⋅ ⋅ ∂ ∂ ∫ ∫ from (2) we have ( ) 0 0 0 0 0 0 0 0 0 1 , , , 0 ( , ) ( , , , ) 1 ( , , , ) 0, ( , ) 0. Y Z