Boundary Element Method in Inverse Heat Conduction Problem

Synonyms

BEM; Boundary element method

Overview

The essence of the boundary element method consists in transformation of a partial differential equation with given conditions into the form of an integral equation. This transformation may be achieved, for example, by using the weighted residual method followed by integration by parts method or by using the Green’s second identity [1–3]. A result of the transformation mentioned above is an integral connected with differential operator which by using a fundamental solution is reduced to an unknown function. If the differential equation has non-zero right hand side then as a result of the transformation of the differential equation into the integral equation appears a domain integral which may be transformed into a boundary integral [4, 5]. In order to determine the value of the boundary integrals a boundary discretization and an interpolation of integrand are made which in the end leads to the problem of solving a system of linear algebraic equations.

The first advantage of the boundary element method is carrying out the boundary discretization which simplifies considerably a generation of the grid on the boundary and setting boundary conditions. The dimension of a matrix connected with a vector of unknowns is significantly smaller in the boundary element method when compared with the finite element method (in the first case this matrix is a dense one, and in the second – a sparse one).

The second advantage of the boundary element method is connected with the use of the fundamental solution which improves the accuracy of the solution and enables considering the boundary conditions in infinity. To the contrary, in the finite element method, considering the boundary conditions in infinity requires a huge number of elements.

The third advantage results from the appearance of the unknown function and its derivative in boundary integrals of the domain. Both the functions are interpolated with the same accuracy which allows to obtain better approximation of temperature and its gradient.

The fourth advantage of the boundary element method is the possibility of using discontinuous elements (taking into account discontinuity of heat flux while passing from one element to another).

The boundary element method is more difficult to implement than the finite element method.

Boundary Element Method for Helmholtz’s Equation

Since many heat conduction problems may be reduced to the problem of solving the Helmholtz’s equation, thus we will discuss the boundary element method on the example of this typical equation.

Let us consider a heat conduction equation in the form of

$$\begin{array}{ll} \rho c\frac{{\partial T\left( {x,t} \right)}}{{\partial t}} & =\mathrm{ div}\left( {k\nabla T\left( {x,t} \right)} \right)+p\left( {x,t} \right) \hfill \\ \qquad \quad x & =\left( {{x_1},{x_2},{x_3}} \right),\kern0.75em t>0,\kern0.75em x\in \Omega \hfill \cr \end{array}$$

(1)

Since the domain Ω is very often irregular in order to find a solution under given conditions we use the numerical methods to both the spatial variables and time variable. Applying backward difference quotient formula to approximate partial time derivative we have

$$\frac{{\partial T\left( {x,t} \right)}}{{\partial t}}\approx \frac{{T\left( {x,t} \right)-T\left( {x,t-\Delta t} \right)}}{{\Delta t}}$$

and the (1) will take the approximate form

$$\begin{gathered} \mathrm{ div}\left( {k\nabla T\left( {x,t} \right)} \right)-\frac{{\rho \cdot c}}{{\Delta t}}\cdot T\left( {x,t} \right)\hfill \cr = -\frac{{\rho \cdot c}}{{\Delta t}}\cdot T\left( {x,t-\Delta t} \right)-p\left( {x,t} \right) \hfill \cr \end{gathered}$$

(2)

Now, assuming that the parameter k, ρ, c are constant, the (2) can be written as follows

$$\Delta T\left( {x,t} \right)-{\beta^2}T\left( {x,t} \right)=-{\beta^2}Q\left( {x,t} \right)$$

(3)

$$\begin{array}{ll} {\beta^2}=-\frac{{\rho \cdot c}}{{k \cdot \Delta t}},\kern0.75em Q=T\left( {x,t-\Delta t}\right)+\frac{{\Delta t}}{{\rho \cdot c}}\cdot p\left( {x,t}\right)\end{array}$$

Here, in the (3) time t plays the role of parameter as well. Let us notice that for β = 0 the (3) describes a stationary distribution of temperature modelled by the equation

$$\Delta T(x)=-\frac{1}{k}p(x),\kern0.75em x=\left( {{x_1},{x_2},{x_3}} \right)\in \Omega$$

(4)

We can transform the (3) to the boundary integral equation with the boundary conditions of the three possible forms, Fig. 1

Fig. 1

Calculation domain

• 1. Dirichlet’s type boundary condition

  $$T(x)=f(x),\;x\in {\Gamma_A}$$

  (5)

• 2. Neumann’s type boundary condition

  $$q(x)=-k\frac{{\partial T}}{{\partial n}}=g(x),\;x\in {\Gamma_B}$$

  (6)

• 3. Mixed type boundary condition

  $$-k\frac{{\partial T}}{{\partial n}}=h\left( {T(x)-{T_{\mathrm{ fluid}}}} \right),\;x\in {\Gamma_C},\,h>0$$

  (7)

The first step to formulate the boundary element method is the transformation of the (3) to an integral equation. Multiplying the (3) by the test function T ^* and integrating over the domain Ω we have

$$\int\limits_{\Omega } {\left( {\Delta T-{\beta^2}T+{\beta^2}Q} \right)} {\,}{T^{*}}d\Omega =0$$

(8)

Since

$$\begin{gathered} {\rm div}\left( {{T^{*}}\nabla T} \right) =\nabla {T^{*}}\nabla T+{T^{*}}\cdot \Delta T \hfill \cr {\rm div}\left( {T\cdot \nabla {T^{*}}} \right) =\nabla T\cdot \nabla {T^{*}}+T\cdot \Delta {T^{*}} \hfill \cr \end{gathered}$$

(9)

thus for the first term in (8) on the basis of (9) and Gauss’ theorem we have

$$\begin{array}{ll} {\int\limits_{\Omega } {\Delta T\cdot{T^{*}}d\Omega =\int\limits_{\Omega } {\mathrm{ div}\left({{T^{*}}\cdot \nabla T} \right)d\Omega }}} \cr\quad-\int\limits_{\Omega } {\nabla {T^{*}}\nabla Td\Omega=\int\limits_{\Omega } {\mathrm{ div}\left( {{T^{*}}\cdot \nabla T}\right)d\Omega } } \hfill \cr \quad-\int\limits_{\Omega } {\mathrm{div}\left( {T\cdot \nabla {T^{*}}} \right)d\Omega+\int\limits_{\Omega } {T\cdot {T^{*}}d\Omega } } \hfill \cr=\int\limits_{\Gamma } {{T^{*}}\cdot \nabla T} \cdot \vec{n}d\Omega-\int\limits_{\Gamma } {T\cdot \nabla {T^{*}}} \cdot \vec{n}d\Omega\hfill \cr \quad +\int\limits_{\Omega } {T\cdot \Delta {T^{*}}\cdot d\Omega } \hfill \cr \end{array}$$

(10)

Therefore the (8) may be transformed into the form

$$\begin{array}{ll} \int\limits_{\Omega } {T\left( {\Delta{T^{*}}-{\beta^2}{T^{*}}} \right)d\Omega} = \int\limits_{\Gamma }{\left( {T\frac{{\partial {T^{*}}}}{{\partial n}}-{u^{*}}\frac{{\partial T}}{{\partial n}}} \right)d\Gamma }\hfill \cr \quad -{\beta^2}\int\limits_{\Omega } Q\cdot{T^{*}}\cdot d\Omega\end{array}$$

(11)

The integral on the left hand side (11) is reduced to the function T if the test function T ^* satisfies the fundamental equation

$$\Delta {T^{*}}-{\beta^2}{T^{*}}=-\delta \left( {x,\xi } \right)$$

(12)

If it is so, we have

$$\int\limits_{\Omega } {T(x)\cdot \delta \left( {x,\xi } \right)d\Omega =T(x)}, \kern0.5em x,\xi \in \Omega$$

(13)

The possible explicite forms of the function T ^* and its derivative ${q^{*}}=T_i^{*}\cdot {n_i}$ are as follows [3, 5], with the distance $r=\sqrt{{\left( {{\xi_i}-{x_i}} \right)\left( {{\xi_i}-{x_i}} \right)}}$

• For Laplace’s equation for the 2D case

  $${T^{*}}\left( {x,\xi } \right)=-\frac{1}{2}\cdot \ln r,\;{q^{*}}\left( {x,\xi } \right)=-\frac{{\left( {{x_i}-{\xi_i}} \right)\cdot {n_i}}}{{2\pi r}}$$

• For Laplace’s equation for the 3D case

  $${T^{*}}\left( {x,\xi } \right)=\frac{1}{{4\pi r}},\;{q^{*}}\left( {x,\xi } \right)=-\frac{{\left( {{x_i}-{\xi_i}} \right)\cdot {n_i}}}{{4\pi \cdot {r^{3/2 }}}}$$

• For Helmholtz’s equation for the 2D case

  $$\begin{gathered} {T^{*}}\left( {r,\beta } \right) =\frac{1}{{2\pi }}\cdot {K_0}\left( {\beta r} \right), \hfill \cr {q^{*}}\left( {r,\beta } \right) =-\frac{\beta }{{2\pi }}\cdot {K_1}\left( {\beta r} \right) \hfill \cr \end{gathered}$$

• For Helmholtz’s equation for the 3D case

  $$\begin{gathered} {T^{*}}\left( {r,\beta } \right) =\frac{{\exp \left( {\beta r} \right)}}{{4\pi r}}, \hfill \cr {q^{*}}\left( {r,\beta } \right) =-\left( {1-\beta \cdot r} \right)\cdot \frac{{\exp \left( {\beta r} \right)}}{{4\pi {r^2}}}\cdot \frac{{\partial r}}{{\partial n}} \hfill \cr \end{gathered}$$

Taking into account the property (13) in (11) we obtain

$$\begin{array}{lll} T\left( \xi \right)= & \int\limits_{\Gamma } {\left( {{T^{*}}\cdot q-T\cdot {q^{*}}} \right)d\Gamma } \hfill \\ & +{\beta^2}\int\limits_{\Omega } {Q\cdot } {T^{*}}d\Omega,\;q=\frac{{\partial T}}{{\partial n}},\kern0.5em \xi \in \Omega \hfill \cr \end{array}$$

(14)

If temperature T distribution and heat flux q on the boundary Γ are known then the temperature distribution within the domain Ω may be determined from the (14). To obtain an equation including only data on the boundary Γ in (14) one should approach with the coordinate ξ to the boundary Γ. The equation obtained in such a way is called a boundary integral equation (BIE). Because of the singularity of the fundamental solution T ^* and the derivate q ^*, if $\xi \to x\in \Gamma$ one should make a boundary transition. In order to do that we divide the integration boundary into two $\Gamma -{\Gamma_{\epsilon }}$ and ${\Gamma_{\epsilon }}$, Fig. 2, and split the integral disjoint over Γ into two boundary integrals

Fig. 2

Boundary Γ augmented by Γ _ε

$$\int\limits_{\Gamma } {{q^{*}}Td\Gamma =\int\limits_{{\Gamma -{\Gamma_{\epsilon }}}} {{q^{*}}Td\Gamma +} } \int\limits_{{{\Gamma_{\epsilon }}}} {{q^{*}}Td\Gamma ={I_1}+{I_2}}$$

(15)

Introducing the polar coordinates in the domain bounded by the arc Γ _ε we have

$${q^{*}}=-\frac{1}{{2\pi r}}\cdot \frac{{\partial r}}{{\partial n}}=-\frac{1}{{2\pi \epsilon }},\kern0.5em x\in {\Gamma_e};$$

therefore

$$\begin{array}{ll} {I_2} =\int\limits_{{{\Gamma_{\epsilon }}}} {{q^{*}}u\cdot d\Gamma =-\frac{{u\left( \xi \right)}}{{2\pi }}\int\limits_{{{\Gamma_{\epsilon }}}} {\frac{1}{\epsilon}\epsilon \cdot d\Theta } } \cr = -\frac{{u\left( \xi \right)}}{{2\pi }}\int\limits_{{{\Theta_1}}}^{{{\Theta_2}}} {d\Theta =-} \frac{{u\left( \xi \right)}}{{2\pi }}\left( {{\Theta_2}-{\Theta_1}} \right) \end{array}$$

(16)

The integral $\int\limits_{\Gamma } {{T^{*}}\cdot q\cdot d\Gamma }$ is a weakly singular one and is equal to zero over the arc Γ _ε for ε → 0.

Thus, for ξ → x ∈ Γ the boundary integral equation has the form

$$\begin{array}{ll} c\left( \xi \right)\cdot T\left( \xi \right) =\int\limits_{\Gamma } {\left( {{T^{*}}q-T\cdot {q^{*}}} \right)d\Gamma } \cr \quad+{\beta^2}\int\limits_{\Omega } {Q\cdot {T^{*}}d\Omega, \kern0.5em \xi \in \Gamma } \end{array}$$

(17)

$$c=\left\{\begin{matrix} 1 & {x\in \Omega } \cr 0 & {x\ \not in \Omega } \cr {1+\mathop{\lim}\limits_{{\epsilon \to0}}\int\limits_{{{\Gamma_{\epsilon }}}} {{q^{*}}d\Gamma, } } & {\xi\in \Gamma } \end{matrix} \right.$$

(18)

and for the 2D case

$$\mathop{\lim}\limits_{{\epsilon \to 0}}\int\limits_{{{\Gamma_{\epsilon }}}} {{q^{*}}d\Gamma =1-\frac{{{\Theta_2}-{\Theta_1}}}{{2\pi }}}$$

(19)

For the boundary Γ ∈ C¹ , in the 2D case ${\Theta_2}-{\Theta_1}=\pi$, then c =1/2, and by analogy for the 3D case, c = ¼.

Discretization of Boundary

Approximation of the boundary Γ is done by dividing it to elements Γ ⁽¹⁾,Γ ⁽²⁾,…, Γ ^(NE) with a disjoint interiors, NE is the number of boundary elements. Each of elements may possesses one or more nodes, Fig. 3.

Fig. 3

Division of the boundary into elements: (a) collocation points coincide with nodes, (b) the collocation point is removed from the node (a single corner), (c) collocation points are inside the element

The boundary Γ can be approximated by straight segments, Figs. 4 and 5 or curved ones for the 2D case and plain or curved surfaces for the 3D case. Figure 4 shows the subdivision of the boundary into the straight segments which imposes the interpolation of temperature T and heat flux q by the piecewise linear function. Namely

Fig. 4

Linear approximation of temperature and heat flux in element Γ ⁽ⁱ⁾

Fig. 5

Linear continuous approximation of temperature

$$\left.\ \begin{array}{ll}{T\left( \eta \right)={T_i}\cdot {\varphi_1}\left( \eta \right)+{T_{i+1 }}\cdot{\varphi_2}\left( \eta \right)={{{\left\{ \varphi\right\}}}^T}\left\{ T \right\}} \\ {q\left( \eta \right)={q_i}\cdot {\varphi_1}\left( \eta \right)+{q_{i+1 }}\cdot{\varphi_2}\left( \eta \right)={{{\left\{ \varphi\right\}}}^T}\left\{ q \right\}} \\ {\varphi_1}\left( \eta \right)=1-\eta, \kern0.75em {\varphi_2}\left( \eta \right)=\eta,\kern0.75em \eta \in \left.\langle {0,1} \right.\rangle \end{array}\right\}$$

(20)

$$\begin{array}{ll} d\Gamma =\sqrt{{{{{\left( {{x_{1i}}-{x_{1,i+1 }}} \right)}}^2}+{{{\left( {{x_{2i }}-{x_{2,i+1 }}}\right)}}^2}}}\cdot d\eta =\Delta {s_i}\cdot d\eta . \end{array}$$

We can use the discontinuous approximation of temperature and heat flux to the boundaries with corners or in polygonal domains.

Introducing a discrete form of the solution (20) to the boundary (18) we obtain

$$\begin{array}{ll} c\left( \xi \right)\cdot T\left( \xi\right) & + \sum\limits_{i=1}^{\mathrm{ NE}} {\Delta{s_i}\int\limits_{{{\Gamma^{(i) }}}} {{q^{*}}\left( {\xi, \eta }\right)\cdot {{{\left\{ {\varphi \left( \xi \right)}\right\}}}^T}\cdot d\eta \cdot \left\{ T \right\}} } \hfill \\ & \quad =\sum\limits_{{\mathrm{ i}=1}}^{\mathrm{ NE}} {\Delta {{\mathrm{s}}_{\mathrm{ i}}}\cdot \int\limits_{{{\Gamma^{(i) }}}}{{T^{*}}\left( {\xi, \eta } \right)} } \cdot {{\left\{ {\varphi\left( \xi \right)} \right\}}^T}d\eta \cdot \left\{ q \right\} \\ \quad & \qquad + {\beta^2}\int\limits_{\Omega } {{T^{*}}\left( {\xi, x}\right)\cdot Q(x)\cdot d\Omega } \hfill \cr \end{array}$$

(21)

For subsequent collocation points ξ _k , k = 1,…, N equal to the number of nodes (piecewise continuous linear approximation) after discretization the solution (21) takes the matrix form

$$\left[ H \right]\left\{ T \right\}=\left[ G \right]\left\{ q \right\}+\left[ P \right]\left\{ Q \right\}$$

(22)

$$\dim \left[ H \right]=\dim \left[ G \right]=N\times N,\;\dim \left[ P \right]=N\times NQ$$

In case of discontinuous approximation (Fig. 6) the number of unknowns and collocation points is equal to 2 N.

Fig. 6

Segmental discontinuous approximation of temperature

Boundary Conditions

Separating from the dependence (22) matrices connected with the boundaries Γ _A , Γ _B and Γ _C , we may write this dependence in the matrix form

$$\left[ {{H_A},{H_B},{H_C}} \right]\left\{ {\begin{matrix}{\left\{ {{T_A}} \right\}} \cr {\left\{ {{T_B}} \right\}} \cr{\left\{ {{T_C}} \right\}} \cr \end{matrix}} \right\}=\left[{{G_A},{G_B},{G_C}} \right]\left\{ {\begin{matrix} {{q_A}} \cr{{q_B}} \cr {{q_C}} \cr \end{matrix}} \right\}+\left[ P\right]\left\{ Q \right\}$$

or equivalently

$$\begin{array}{ll} \left[ {{H_A}} \right]\left\{{{T_A}} \right\} +\left[ {{H_B}} \right]\left\{ {{T_B}}\right\}+\left[ {{H_C}} \right]\left\{ {{T_C}} \right\}=\left[{{G_A}} \right]\left\{ {{q_A}} \right\} \cr +\left[ {{G_B}}\right]\left\{ {{q_B}} \right\}+\left[ {{G_C}} \right]\left\{{{q_C}} \right\}+\left[ {P\left\{ Q \right\}} \right]\end{array}$$

(23)

Now taking into account the conditions (5)–(7) we obtain

$$\begin{array}{ll} \left[ {-{G_A},{H_B},{H_C}+\frac{h}{k}{G_C}}\right]\left\{ \begin{matrix}{\left\{ {{q_A}} \right\}} \\{\left\{ {{T_B}} \right\}} \cr {\left\{ {{T_C}} \right\}} \cr\end{matrix} \right\}\hfill \cr = \left[ {-{H_A},-\frac{1}{\lambda}{G_B},\frac{h}{k}{G_C}} \right]\left\{ \begin{matrix} \left\{f \right\} \\ {\left\{ g \right\}} \cr {{T_{\mathrm{ fluid}}}} \cr\end{matrix} \right\}+\left[ P \right]\left\{ Q \right\} \hfill \cr\end{array}$$

(24)

or, in a compact form

$$\left[ {HS} \right]\left\{ \begin{matrix} {\left\{ {{q_A}}\right\}} \\ \left\{ {{T_B}} \right\} \\ {\left\{ {{T_C}}\right\}} \cr \end{matrix} \right\}=\left[ {GS} \right]\left\{\begin{matrix} {\left\{ f \right\}} \\ {\left\{ g \right\}}\cr \left\{ {{T_{\mathrm{ fluid}}}} \right\} \\ \end{matrix}\right\}+\left[ P \right]\left\{ Q \right\}$$

Premultiplying both sides by the matrix [HS]^-1

$$\begin{array}{ll} \left\{ \begin{matrix} {\left\{ {{q_A}}\right\}} \\ {\left\{ {{T_B}} \right\}} \cr {\left\{ {{T_C}}\right\}} \\ \end{matrix} \right\}=\left[ {HG} \right]\left\{\begin{matrix} {\left\{ f \right\}} \cr {\left\{ g\right\}} \cr {\left\{ {{T_{\mathrm{ fluid}}}} \right\}} \cr\end{matrix} \right\}+\left[ {HP} \right]\left\{ Q\right\}\end{array}$$

(25)

For determined temperature distribution {T _C } (25) we have ${q_C}=-\frac{{\partial T}}{{\partial n}}=h{{{\left( {{T_C}-{T_{\mathrm{ fluid}}}} \right)}} \left/ {k} \right.}$, thus the vector of temperature {T} and heat flux {q} on the entire boundary Γ is known. Therefore the temperature distribution at a given point ξ ∈ Ω is represented by the dependence (17), which may be written in the following form c = 1 (Γ ∈ C¹)

$$\begin{gathered} T\left( \xi \right)= {{\left\{ {HU\left( \xi \right)} \right\}}^T}\left\{ q \right\}+{{\left\{ {GU\left( \xi \right)} \right\}}^T}\left\{ T \right\} \hfill \cr +{{\left\{ {PU\left( \xi \right)} \right\}}^T}\left\{ Q \right\},\;\xi \in \Omega \hfill \cr \end{gathered}$$

(26)

Perturbed temperature T(ξ) for perturbed data T + δT, q + δq, Q + δQ on the basis of the dependence (26) is represented as follows

$$\begin{array}{ll} T\left( \xi \right)+\delta T\left( \xi\right)={{\left\{ {HU\left( \xi \right)} \right\}}^T}\left\{{q+\delta q} \right\} \hfill \cr \qquad +{{\left\{ {GU\left( \xi\right)} \right\}}^T}\left\{ {T+\delta T} \right\}+{{\left\{{PU\left( \xi \right)} \right\}}^T}\left\{ {Q+\delta Q} \right\}\hfill \cr \end{array}$$

(27)

Subtracting the dependence (26) from the (27) one, we obtain a temperature error at the point ξ ∈ Ω as the function of a perturbed value (data error).

$$\begin{array}{ll} \delta T\left( \xi \right)= {{\left\{{HU\left( \xi \right)} \right\}}^T}\left\{ {\delta q}\right\}+{{\left\{ {GU\left( \xi \right)} \right\}}^T}\left\{{\delta T} \right\} \cr + {{\left\{ {PU\left( \xi \right)}\right\}}^T}\left\{ {\delta Q} \right\} \end{array}$$

(28)

Having considered the boundary conditions represented by the vector (25) we may write the dependence (26) in the form

$$\begin{array}{ll} T\left( \xi \right)= \left\{ {{{{\left\{{H{U_A}} \right\}}}^T},{{{\left\{ {H{U_B}} \right\}}}^T},{{{\left\{{H{U_C}} \right\}}}^T}} \right\}\left\{ \begin{matrix} {\left\{{{q_A}} \right\}} \cr {\left\{ {{q_B}} \right\}} \cr {\left\{{{q_C}} \right\}} \cr \end{matrix} \right\} \hfill \cr \qquad \quad +\left\{{{{{\left\{ {G{U_A}} \right\}}}^T},{{{\left\{ {G{U_B}}\right\}}}^T},{{{\left\{ {G{U_C}} \right\}}}^T}} \right\}\left\{\begin{matrix} {\left\{ {{T_A}} \right\}} \cr {\left\{ {{T_B}}\right\}} \cr {\left\{ {{T_C}} \right\}} \cr \end{matrix} \right\}\hfill \cr \quad \quad = \ {{\left\{ {SH} \right\}}^T}\left\{\begin{matrix} {\left\{ {{q_A}} \right\}} \cr {\left\{ {{q_B}}\right\}} \cr {\left\{ {{q_C}} \right\}} \cr \end{matrix}\right\}+{{\left\{ {SG} \right\}}^T}\left\{ \begin{matrix}{\left\{ f \right\}} \cr {\left\{ g \right\}} \cr {\left\{{{T_{\mathrm{ fluid}}}} \right\}} \cr \end{matrix} \right\} \hfill\cr \qquad = {{\left\{ {SH} \right\}}^T}\left[ {HG} \right]\left\{\begin{matrix} {\left\{ f \right\}} \cr {\left\{ g \right\}}\cr {\left\{ {{T_{\mathrm{ fluid}}}} \right\}} \cr \end{matrix}\right\}+ {{\left\{ {SG} \right\}}^T}\left\{ \begin{matrix}{\left\{ f \right\}} \cr {\left\{ g \right\}} \cr {\left\{{{T_{\mathrm{ fluid}}}} \right\}} \cr \end{matrix} \right\} \hfill\cr \qquad +{{\left\{ {SH} \right\}}^T}\left[ {HP} \right]\left\{ Q\right\}={{\left\{ {{S_A}} \right\}}^T}\left\{ f \right\}+{{\left\{{{S_B}} \right\}}^T}\left\{ g \right\} \hfill \cr \qquad +{{\left\{ {{S_C}}\right\}}^T}\left\{ {{T_{\mathrm{ fluid}}}} \right\}+{{\left\{ {SP}\right\}}^T}\left\{ Q \right\} \hfill \cr \end{array}$$

(29)

For each ξ value the vectors {HU}, {GU} and {PU} have to be determined from scratch.

Calculation of Domain Integral

An appearance of the integral ${I_{\Omega }}=\int\limits_{\Omega } {Q\cdot {T^{*}}d\Omega }$ is a certain inconvenience in the boundary element method. A domain integral can be determined by using the finite element method or by reducing it to a sequence of boundary integrals (it is particularly useful if the domain Ω is unbounded) [5].

Figure 7 shows the discretization of the domain Ω with the use of the simplest possible finite elements [3]. For better approximation of the boundary (with the same number of boundary nodes) we can use curvilinear elements.

Fig. 7

Division of the domain Ω into finite elements

The second approach consists on transformation of the integral I _Ω into the boundary integrals [6].

To determine the integral I _Ω we will use the fundamental solution u ^∗ of n-th order for which the following equality occurs

$$\begin{gathered} {\Delta^n}{u^{*}}={T^{*}}\quad \mathrm{ hence}\quad {\Delta^{n+1 }}{u^{*}}=\Delta {T^{*}}=\delta \hfill \cr \mathrm{ and}\quad {\Delta^{n-j }}{u^{*}}={\Delta^{-(\,j+1) }}\delta,\;\ n>1 \hfill \cr \end{gathered}$$

(30)

Therefore n + 1 times using the integration by parts we have

$$\begin{array}{ll} {I_{\Omega }} & = \int\limits_{\Omega } Q\cdot {T^{*}}d\Omega =\int\limits_{\Omega } {\Delta^{n+1 }}Q\cdot {u^{*}}d\Omega \\ & +\sum\limits_{j=0}^n \left[ {\frac{\partial }{{\partial n}}\left( {{\Delta^j}Q} \right)\cdot {\Delta^{n-j }}{u^{*}}-{\Delta^j}Q } \right. \\ & \left. \times\cdot \frac{\partial }{{\partial n}}\left( {{\Delta^{n-j }}{u^{*}}} \right) \right] \cdot d\Gamma \end{array}$$

where ${\Delta^{n-j }}{u^{*}}={\Delta^{-(j+1) }}\delta$ and

$${\Delta^{-(j+1) }}\delta =\left\{ \begin{matrix}\displaystyle{\frac{1}{2}\cdot \frac{{{r^{2j+1 }}}}{{2\left( {2j+1}\right)!}},\ r=\left| {x-\xi } \right|, \ 1\mathrm{ D}\ \mathrm{case}} \hfill \cr \displaystyle{\frac{1}{{2\pi }}\cdot \frac{{{r^{2j}}}}{{{{{\left( {{2^{\,j}}\cdot j!} \right)}}^2}}}\cdot \left({\sum\limits_{k=1}^j {} \frac{1}{k}-\ln r} \right), \ 2\mathrm{ D}\ \mathrm{ case}} \hfill \cr \displaystyle{\frac{1}{{4\pi }}\cdot\frac{{{r^{2j+1 }}}}{{\left( {2\left( {{\,j}+1} \right)}\right)!}},\ 3\mathrm{ D}\ \mathrm{ case}} \hfill \cr \end{matrix}\right.$$

(31)

Similarity, for the 2D case [4]

$$\begin{array}{ll} \left| {\int\limits_{\Omega } {{\Delta^{n+1 }}Q\cdot {u^{*}}\cdot d\Omega } } \right|\leq \frac{{\max \left( {\left| {{\Delta^{n+1 }}Q} \right|} \right)}}{{\pi \cdot {2^{2(n+1) }}\left( {n+1} \right)\cdot {{{\left( {n!} \right)}}^2}}}\hfill \cr\cdot \left( {\frac{1}{{2\left( {n+1} \right)}}+\sum\limits_{k=`}^n {\frac{1}{k}} } \right)\end{array}$$

(32)

Inverse Problem

The well posed problems for the heat conduction (1) require: knowledge of temperature coefficients λ, ρ, c and the source function, the determination of the domain Ω, the initial condition and the boundary conditions. If one of these data is missing then from the physical point of view it can be completed by the temperature measurement in the interior points of the domain Ω [7]. The problem stated in such a way and called an inverse problem is ill posed in Hadamard’s sense. It means that a small perturbation of data may lead to huge perturbation in solving the (1). To minimize this negative phenomenon a regularization of the inverse problem is used.

The inverse problems mentioned above may be one of the following types: initial (unknown initial temperature distribution), boundary (unknown boundary condition on a part of the boundary), coefficient (unknown coefficient λ, ρ and c), geometric (the shape of a part of the domain’s boundary is unknown).

One of the problems which most often appears in practice is an inverse boundary problem for which a distribution of temperature T = f on the boundary ${\Gamma_A}$, Fig. 8, is sought after. Let the number of measuring points on the interior boundary Γ _w be denoted by ${N_w}$ while the number ${N_T}$ of nodes on the boundary Γ _T be ${N_T}\leq {N_w}$.

Fig. 8

Calculation domain with interior temperature measurement points

In order to find the unknown distribution of the function f in the nodes of the boundary Γ _A we use the dependence (29), that is

$$\begin{array}{ll} T\left( {{\xi_{w,k }}} \right)= \ {{\left\{ {{S_A}\left( {{\xi_{w,k }}} \right)} \right\}}^T}\left\{ f \right\}+{{\left\{ {{S_B}\left( {{\xi_{w,k }}} \right)} \right\}}^T}\left\{ g \right\} \hfill \cr \qquad \qquad \quad+{{\left\{ {{S_C}\left( {{\xi_{w,k }}} \right)} \right\}}^T}\left\{ {{T_{\mathrm{ fluid}}}} \right\}+{{\left\{ {SP\left( {{\xi_{w,k }}} \right)} \right\}}^T}\left\{ Q \right\} \hfill \cr \qquad \quad = \ {{\left\{ {{S_A}\left( {{\xi_{w,k }}} \right)} \right\}}^T}\left\{ f \right\}+{R_k}\hbox{,}\kern0.5em k=1,2,\ldots,{N_w} \hfill \cr \end{array}$$

(33)

We obtain the discrete distribution of the function f from the minimization of the distance in the least squares sense between the measured temperature T _w,k , k = 1,2,…, N _w and the one determined from the solution (33). Namely the error-functional, is represented as follows:

$$\begin{array}{ll} \qquad I\left( {\left\{ f \right\}} \right) & =\sum\limits_{k=1}^{{{N_w}}} {{{{\left( {T\left( {{\xi_{w,k }}} \right)-{T_{w,k }}} \right)}}^2}} \hfill \cr & =\sum\limits_{k=1}^{{{N_w}}} {{{{\left( {{{{\left\{ {{S_A}\left( {{\xi_{w,k }}} \right)} \right\}}}^T}\left\{ f \right\}+{R_k}-{T_{w,k }}} \right)}}^2}} \hfill \cr & ={{\left\| {\left[ {AA} \right]\left\{ f \right\}-\left\{ {TR} \right\}} \right\|}^2},\;{{\left\{ {A{A_i}} \right\}}^T} ={{\left\{ {{S_A}\left( {{\xi_{w,i }}} \right)} \right\}}^T} \hfill \\\qquad T{R_i} & ={T_{w,i }}-{R_i} \\ \end{array}$$

(34)

The minimization of the functional (34) is equivalent to solving the system of linear equation

$$\begin{array}{ll}\left[ {AA} \right]\left\{ f \right\}=\left\{ {TR} \right\}\ \mathrm{ hence}\ \left\{ f \right\}={{\left[ {AA} \right]}^{+}}\left\{ {TR} \right\}\end{array}$$

(35)

where [AA]⁺ is a pseudoinverse matrix.

The influence of the data error δTR on the solution f results from the dependence (35), that is

$$\left\{ {\delta f} \right\}={{\left[ {AA} \right]}^{+}}\left\{ {\delta TR} \right\}$$

(36)

If small changes in $\left\| {\delta TR} \right\|$ produce huge changes in $\left\| {\delta f} \right\|$ then the inverse problem requires regularization [7]. After making regularization in the Tichonov’s sense the functional (34) takes the form

$$\begin{gathered} {I_{\alpha }}\left( {\left\{ f \right\}} \right)= {{\left\| {\left[ {AA} \right]\left\{ f \right\}-\left\{ {TR} \right\}} \right\|}^2} \hfill \cr + {\alpha^2}{{\left\| {\left[ B \right]\left\{ {f-{f_0}} \right\}} \right\|}^2},\kern0.75em \alpha>0 \hfill \cr \end{gathered}$$

(37)

where B is called the regularization matrix (in the special case B = I) while {f ₀} is the approximation of {f} (in many cases {f ₀} = {0}).

The minimization of the functional (37) is equivalent to solving the overdetermined system of linear algebraic equations in the least squares sense

$$\begin{array}{ll}\left[ {\begin{matrix} {\left[ {AA} \right]} \cr {\alpha \left[ B \right]} \cr \end{matrix}} \right]=\left[ \begin{matrix} {\left\{ {TR} \right\}} \cr {\alpha \left[ {{f_0}} \right]} \cr \end{matrix} \right]\ \mathrm{ or}\ \left[ {A{A_{\alpha }}} \right]\left\{ f \right\}=\left\{ {T{R_{\alpha }}} \right\}\end{array}$$

(38)

Thus

$$\left\{ f \right\}={{\left[ {A{A_{\alpha }}} \right]}^{+}}\left\{ {T{R_{\alpha }}} \right\}$$

(39)

A number of methods to determine the optimum value of the parameter $\alpha$ is shown in the paper [8]. A solution of a non-stationary inverse problem with the use of the fundamental solution dependent on space and time variables is presented in the paper [9].