Approximate Analytical Solution for the Temperature Field in Welding

Abstract

To determine the temperature fields associated with welding, significant efforts have been made to establish the relative merits of numerical approaches with variable material properties and the analytical approaches with constant material properties. Currently, analytical solutions are either based on the temperature field generated by a point source of heat or are developed for a finite domain derived approximately by using an infinite or semi-infinite heat kernel. Furthermore, the heat kernel applied in these solutions is derived from the Image method (for example, Nguyen’s book (Thermal Analysis of Welds, 2004)). The main problem with the heat kernels obtained from Image method is that they face the problem of singularity at and around the point where the heat source is located, and they do not satisfy the boundary condition accurately. That is why the Laplace transform method has been applied here instead of using the Image method to formulate a heat kernel that (1) converges rapidly, (2) avoids the problem of singularity, and (3) gives a good and robust approximation of the real analytic solution for the temperature field. The results obtained from the analytical solutions were compared with the results obtained from finite element method. The current work is believed to make a considerable contribution to the avoidance of previously mentioned problems by deriving a new approximate analytical solution for the temperature field on a three-dimensional finite body.

Appendix: Derivation of Finite Dimensional Heat Kernel by the Laplace Transform Method

The Laplace transform method has been used widely for the solution of time-dependent heat conduction problems because the partial differential operator with respect to time variable contained in the heat equation can be removed by the Laplace transformation.[13, p. 257] Because the temperature field in the welding process is a time-dependent problem and it is necessary to calculate the temperature field for a small period of time, it is useful to apply the techniques of the Laplace transform for constructing the heat kernel of the solution for the finite dimensional heat conduction problem. Considering the nonhomogeneous partial differential equation for the heat conduction problem in one-dimensional bar of length L, which can be written as

$$ \frac{{\partial {\rm T}\left( {x,t} \right)}}{\partial t} = K\frac{{\partial^{2} T\left( {x,t} \right)}}{{\partial x^{2} }} + {\mathbf{q}}\left( {x,t} \right),\quad \forall x \in [0,L] $$

(76)

where ${\mathbf{q}}\left( {x,t} \right)$ is a static continuous external heat source. The thermal diffusivity coefficient is defined as $K = \frac{k}{\rho c},$ where k is thermal conductivity and ρ is the mass density of the given body Ω. If the two ends of the bar have the boundary conditions of second kind (i.e., the Neumann boundary conditions), then the values of $\frac{{\partial T\left( {x,t} \right)}}{\partial x}$ at end points are known and can be written as

$$ \frac{\partial T}{\partial x}\left| {_{x = 0} } \right. = g_{1} \left( t \right),\quad \frac{\partial T}{\partial x}|_{x = L} = g_{2} \left( t \right) $$

(77)

where $g_{1} \left( t \right)\,{\text{and}}\,g_{2} (t)$ are two given time-dependent functions. The general integral representation of the solution for heat given by Eq. [76], which is associated with initial temperature $T\left( {x,0} \right)$ and the boundary conditions given by Eq. [77], can be written as

$$ \begin{aligned} T\left( {x,t} \right) = \int_{0}^{L} {T\left( {\xi ,0} \right)} {\mathbf{G}}\left( {x,\xi ,t} \right)d\xi & + K\int_{0}^{t} {\left[ {{\mathbf{G}}\left( {x,L,t - \tau } \right)g_{2} \left( t \right) - {\mathbf{G}}\left( {x,0,t - \tau } \right)g_{1} \left( t \right)} \right]} d\tau \\ & \quad + \int_{0}^{t} {\int_{0}^{L} {{\mathbf{q}}\left( {\xi ,\tau } \right){\mathbf{G}}\left( {x,\xi ,t - \tau } \right)d\xi \,d\tau } } \\ \end{aligned} $$

(78)

provided that $\tau \in \left[ {0,t} \right]$ and $\xi \in \left[ {0,L} \right]$. Note that a complete proof for the derivation of the solution in Eq. [78] can be found in Antimirov et al.[14, pp. 101–105]. For the heat conduction problem in welding phenomena, it can be assumed that heat Eq. [76] is associated with the adiabatic boundary conditions (i.e., the ends of the bar are insulated)

$$ \frac{\partial T}{\partial x}|_{x = 0} = 0,\quad \frac{\partial T}{\partial x}|_{x = L} = 0 $$

(79)

Therefore, the solution to heat Eq. [76] with boundary conditions given by Eq. [79] becomes

$$ T\left( {x,t} \right) = \int_{0}^{L} {T\left( {\xi ,0} \right){\mathbf{G}}\left( {x,\xi ,t} \right)d\xi + \int_{0}^{t} {\int_{0}^{L} {{\mathbf{q}}\left( {\xi ,\tau } \right){\mathbf{G}}\left( {x,\xi ,t - \tau } \right)d\xi \;d\tau } } } $$

(80)

Note that Eq. [80] is similar to the solution for 3-D heat conduction problem, which is described in Eq. [4] of the main article. The main problem now lies in determining the heat kernel ${\mathbf{G}}\left( {x,\xi ,t} \right)$ (For a one-dimensional case, this is normally called the Green’s function), which satisfies the associated initial and boundary conditions. In the particular case studied in this article, the Green’s function ${\mathbf{G}}\left( {x,\xi ,t} \right)$ can be chosen in such a way that it satisfies the following initial and boundary value problem:

$$ \frac{{\partial {\mathbf{G}}}}{\partial t} = - K\frac{{\partial^{2} G}}{{\partial x^{2} }} $$

(81)

$$ {\mathbf{G}}\left( {x,\xi ,0} \right) = \delta \left( {x - \xi } \right) $$

(82)

$$ \frac{{\partial {\mathbf{G}}}}{\partial x}\left| {_{x = 0} = 0,\quad \frac{{\partial {\mathbf{G}}}}{\partial x}} \right|_{x = L} = 0 $$

(83)

where $\delta \left( {x - \xi } \right)$ is the Dirac-delta function and defined as $\delta \left( {x - \xi } \right) = \mathop {\lim }\limits_{t \to 0} {\mathbf{G}}\left( {x,\xi ,t} \right)$. Before solving the initial boundary value problem (Eqs. [81] through [83]), the arbitrary function ${\mathbf{G}}\left( {x,\xi ,t} \right)$ and its Laplace transform properties defined for $x,\xi \in \left| {0,L} \right|\,{\text{and}}\,\forall t > 0,$ where $L$ is an arbitrary constant must be dealt with.

• (a) If the Laplace transform of ${\mathbf{G}}\left( {x,\xi ,t} \right)$ is denoted by $\widehat{{\mathbf{G}}}\left( {x,\xi ,p} \right)$, then by the definition of the Laplace transform

$$ \widehat{{\mathbf{G}}}\left( {x,\xi ,p} \right) = \mathcal{L}\left\{ {G\left( {x,\xi ,t} \right)} \right\} = \int_{0}^{\infty } {G\left( {x,\xi ,t} \right)e^{ - pt} dt} $$

(84)

• (b) The Laplace transform of $\frac{\partial {\mathbf{G}}}{\partial t}$ is given by

  $$ \mathcal{L}\left\{ {\frac{{\partial {\mathbf{G}}}}{\partial t}} \right\} = \int\limits_{0}^{\infty } {e^{ - pt} \frac{\partial G}{\partial t}dt} $$

  (85)

  Integrating the right side of Eq. [85] by parts and then using Eq. [84], gives

  $$ \mathcal{L}\left\{ {\frac{{\partial {\mathbf{G}}}}{\partial t}} \right\} = \left[ {{\mathbf{G}}e^{ - pt} } \right]_{0}^{\infty } + p\int_{0}^{\infty } {{\mathbf{G}}e^{ - pt} dt} = p\widehat{{\mathbf{G}}}\left( {x,\xi ,p} \right) - {\mathbf{G}}\left( {x,\xi ,0} \right) $$

  (86)

  If ${\mathbf{G}}\left( {x,\xi ,0} \right) = \delta \left( {x - \xi } \right)$ where $\delta \left( {x - \xi } \right)$ is the Dirac-delta distribution defined previously, then

  $$ \mathcal{L}\left\{ {\frac{{\partial {\mathbf{G}}}}{\partial t}} \right\} = p\widehat{{\mathbf{G}}}\left( {x,\xi ,p} \right) - \delta \left( {x - \xi } \right) $$

  (87)

• (c) Similarly, Laplace transform of $\frac{{\partial^{2} {\text{G}}}}{{\partial x^{2} }}$ is given by

$$ L\left\{ {\frac{{\partial^{2} {\mathbf{G}}}}{{\partial x^{2} }}} \right\} = \int_{0}^{\infty } {e^{ - pt} } \frac{{\partial^{2} {\mathbf{G}}\left( {x,\xi ,t} \right)}}{{\partial x^{2} }}dt = \frac{{\partial^{2} }}{{\partial x^{2} }}\left[ {\underbrace {{\int_{0}^{\infty } {e^{ - pt} {\mathbf{G}}\left( {x,\xi ,t} \right)dt} }}_{{ = \widehat{{\mathbf{G}}}\left( {x,\xi ,p} \right)}}} \right] = \frac{{d^{2} \widehat{{\mathbf{G}}}\left( {x,\xi ,p} \right)}}{{dx^{2} }} $$

(88)

Now the solution of PDE problem (Eqs. [81] through [83]) can be derived by the Laplace transform method in the following two steps.

First Step

With the application of Eqs. [84], [85], and [88], the Laplace transforms of Eqs. [81] and [83] become

$$ K\frac{{d^{2} \widehat{{\mathbf{G}}}}}{{dx^{2} }} = p\widehat{{\mathbf{G}}} - \delta \left( {x - \xi } \right) $$

(89)

$$ \left. {\frac{{\partial \widehat{{\mathbf{G}}}}}{\partial x}} \right|_{x = 0} = 0,\quad \left. {\frac{{\partial \widehat{{\mathbf{G}}}}}{\partial x}} \right|_{x = L} = 0 $$

(90)

To avoid the problem of singularity, which is generally at the point $x = \xi$ (i.e., the location of heat source), it is assumed that the Laplace transform of $\widehat{\bf{G}}$ of the Green’s function ${\mathbf{G}}$ is defined as

$$ \widehat{{\mathbf{G}}} = \left\{ {\begin{array}{*{20}c} {\widehat{{{\mathbf{G}}_{{\mathbf{1}}} }},} & {0 < x < \xi ,} \\ {\widehat{{{\mathbf{G}}_{{\mathbf{2}}} }},} & {\xi < x < L} \\ \end{array} } \right. $$

(91)

In such a case, problem (Eqs. [89] and [90]) can be divided into the following two subproblems:

$$ K\frac{{d^{2} \widehat{{{\mathbf{G}}_{{\mathbf{1}}} }}}}{{dx^{2} }} = p\widehat{{{\mathbf{G}}_{{\mathbf{1}}} }},\quad \left. {\frac{{\partial \widehat{{{\mathbf{G}}_{{\mathbf{1}}} }}}}{\partial x}} \right|_{x = 0} = 0 $$

(92)

$$ K\frac{{d^{2} \widehat{{{\mathbf{G}}_{{\mathbf{2}}} }}}}{{dx^{2} }} = p\widehat{{{\mathbf{G}}_{{\mathbf{2}}} }},\quad \left. {\frac{{\partial \widehat{{{\mathbf{G}}_{{\mathbf{2}}} }}}}{\partial x}} \right|_{x = L} = 0 $$

(93)

The function ${\widehat{\mathbf{G}}}$ is continuous, which is why the following may be written:

$$ \widehat{{{\mathbf{G}}_{{\mathbf{1}}} }}\left| {_{x = \xi } = \widehat{{{\mathbf{G}}_{{\mathbf{2}}} }}} \right|_{x = \xi } $$

(94)

Integrating Eq. [89] with respect to $x$ from $x = \xi - \epsilon\;{\text{to}}\;x = \xi + \epsilon$ gives

$$ K\left. {\frac{{d\widehat{{\mathbf{G}}}}}{dx}} \right|_{x = \xi - \epsilon }^{x = \xi + \epsilon } = p\int_{\xi - \epsilon }^{\xi + \epsilon } {\widehat{{\mathbf{G}}}dx - 1} $$

(95)

Because ${\widehat{\bf{G}}}$ is a continuous function at $x = \xi$, the limit of Eq. [95] may be taken as $\epsilon \to 0$ which will yield

$$ \begin{gathered} K\left[ {\left. {\frac{{d\widehat{{\mathbf{G}}}}}{dx}} \right|_{x = \xi + 0} - \left. {\frac{{d\widehat{{\mathbf{G}}}}}{dx}} \right|_{x = \xi - 0} } \right] = - 1, \hfill \\ \Rightarrow \left. {\frac{{d\widehat{{{\mathbf{G}}_{{\mathbf{2}}} }}}}{dx}} \right|_{x = \xi } - \left. {\frac{{d\widehat{{{\mathbf{G}}_{{\mathbf{1}}} }}}}{dx}} \right|_{x = \xi } = - \frac{1}{K} \hfill \\ \end{gathered} $$

(96)

Now the general solutions of Eqs. [92] and [93] can, respectively, be given as

$$ \widehat{{{\mathbf{G}}_{{\mathbf{1}}} }} = c_{1} \cosh \left( {x\sqrt \lambda } \right) + c_{2} \sinh \left( {x\sqrt \lambda } \right) $$

(97)

$$ \widehat{{{\mathbf{G}}_{{\mathbf{2}}} }} = c_{3} \cosh \left( {\left( {x - l} \right)\sqrt \lambda } \right) + c_{4} \sinh \left( {\left( {x - l} \right)\sqrt \lambda } \right) $$

(98)

where $\lambda = \frac{p}{K}$. By applying boundary conditions given in Eqs. [92] and [93], the coefficients $c_{2} \,{\text{and}}\,c_{4}$ can be determined as $c_{2} = 0\,{\text{and}}\,c_{4} = 0$. Therefore, Eqs. [97] and [98] are reduced to

$$ \widehat{{{\mathbf{G}}_{{\mathbf{1}}} }} = c_{1} \cosh \left( {x\sqrt \lambda } \right) $$

(99)

$$ \widehat{{{\mathbf{G}}_{{\mathbf{2}}} }} = c_{3} \cosh \left( {\left( {x - 1} \right)\sqrt \lambda } \right) $$

(100)

From Eqs. [94], [99], and [100], it follows that

$$ c_{1} \cosh \left( {\xi \sqrt \lambda } \right) - c_{3} \cosh \left( {\left( {\xi - L} \right)\sqrt \lambda } \right) = 0 $$

(101)

Similarly, from Eqs. [96], [99], and [100] it follows that

$$ c_{1} \sqrt \lambda \sinh \left( {\xi \sqrt \lambda } \right) - c_{3} \sqrt \lambda \sinh \left( {\left( {\xi - L} \right)\sqrt \lambda } \right) + \frac{1}{K} = 0 $$

(102)

Solving Eqs. [101] and [102], the values of $c_{1} \,{\text{and}}\,c_{3}$ are

$$ c_{1} = \frac{{\cosh \left( {\left( {L - \xi } \right)\sqrt \lambda } \right)}}{{K\sqrt \lambda \sinh \left( {\sqrt \lambda L} \right)}}\quad c_{3} = \frac{{\cosh \left( {\xi \sqrt \lambda } \right)}}{{K\sqrt \lambda \sinh \left( {\sqrt \lambda L} \right)}} $$

Finally, substituting the values of $c_{1} \,{\text{and}}\,c_{2}$, ${\widehat{\mathbf{G}}}_{1} \,{\text{and}}\,{\widehat{\mathbf{G}}}_{ 2}$ can be written in the form

$$ \widehat{{{\mathbf{G}}_{{\mathbf{1}}} }} = \frac{{\cosh \left( {\left( {L - \xi } \right)\sqrt \lambda } \right)\cosh \left( {x\sqrt \lambda } \right)}}{{K\sqrt \lambda \sinh \left( {\sqrt \lambda L} \right)}} $$

(103)

$$ \widehat{{{\mathbf{G}}_{{\mathbf{2}}} }} = \frac{{\cosh \left( {\xi \sqrt \lambda } \right)\cosh \left( {\left( {x - 1} \right)\sqrt \lambda } \right)}}{{K\sqrt \lambda \sinh \left( {\sqrt \lambda L} \right)}} $$

(104)

Second Step

The first step basically solved the ordinary differential equation problem (Eqs. [89] and [90]). In the second step, the inverse Laplace transform of ${\widehat{\text{G}}}$ is considered; this can be obtained by the formula[14, p. 108]

$$ L\left[ {\frac{{e^{\alpha \sqrt p } }}{\sqrt p }} \right] = \frac{1}{{\sqrt {\pi t} }}e^{{ - \frac{{\alpha^{2} }}{4t}}} $$

(105)

where $\alpha$ is any constant. Before using the preceding formula (Eq. [105]), the following transformation must be applied:

$$ \begin{aligned} \sinh \left( {\sqrt \lambda L} \right) = \frac{{e^{\sqrt \lambda L} - e^{{ - \sqrt \lambda_{{}} }} }}{2}, \hfill \\ \Rightarrow \frac{1}{{\sinh \left( {\sqrt \lambda L} \right)}} = \frac{{2e^{ - \sqrt \lambda L} }}{{1 - e^{ - 2\sqrt \lambda L} }} = 2e^{ - \sqrt \lambda L} \left[ {1 - e^{ - 2\sqrt \lambda - L} } \right]^{ - 1} \hfill \\ \end{aligned} $$

Expanding the term $\left[ {1 - e^{ - 2\sqrt \lambda L} } \right]^{ - 1}$ with the help of the binomial series,

$$ \frac{1}{{\sinh \left( {\sqrt \lambda L} \right)}} = 2e^{ - \sqrt \lambda L} \sum\limits_{n = 0}^{\infty } {e^{ - 2nL\sqrt \lambda } } $$

(106)

From Eqs. [103] and [106],

$$ \widehat{{{\mathbf{G}}_{{\mathbf{1}}} }} = \frac{{e^{ - L\sqrt \lambda } \sum\limits_{n = 0}^{\infty } {e^{ - 2nL\sqrt \lambda } } \left[ {2\cosh \left( {\left( {L - \xi } \right)\sqrt \lambda } \right)\cosh \left( {x\sqrt \lambda } \right)} \right]}}{K\sqrt \lambda } $$

(107)

Using the identity $\cosh \left( {A + B} \right) + \cosh \left( {A - B} \right) = 2\cosh \left( A \right)\cosh \left( B \right),$ Eq. [107] can be written as

$$ \widehat{{{\mathbf{G}}_{{\mathbf{1}}} }} = \frac{{e^{ - L\sqrt \lambda } \sum\limits_{n = 0}^{\infty } {e^{ - 2nL\sqrt \lambda } } \left[ {\cosh \left( {\sqrt \lambda \left( {L - \xi + x} \right)} \right) + \cosh \left( {\sqrt \lambda \left( {L - \xi - x} \right)} \right)} \right]}}{K\sqrt \lambda } $$

(108)

Again using the identity $\cosh \left( x \right) = \frac{{e^{x} + e^{ - x} }}{2}$ and substituting the value $\sqrt \lambda = \sqrt \frac{p}{K}$ into Eq. [108] gives

$$ \begin{gathered} \widehat{{{\mathbf{G}}_{{\mathbf{1}}} }} = \frac{1}{{2K\sqrt \frac{p}{K} }}e^{{ - L\sqrt \frac{p}{K} }} \sum\limits_{n = 0}^{\infty } {e^{{ - 2nL\sqrt \frac{p}{K} }} } \left[ {e^{{\sqrt \frac{p}{K} \left( {L - \xi + x} \right)}} + e^{{ - \sqrt \frac{p}{K} \left( {L - \xi + x} \right)}} + e^{{\sqrt \frac{p}{K} \left( {L - \xi - x} \right)}} + e^{{ - \sqrt \frac{p}{K} \left( {L - \xi - x} \right)}} } \right], \hfill \\ \Rightarrow \widehat{{{\mathbf{G}}_{{\mathbf{1}}} }} = \frac{1}{2\sqrt K \sqrt p }\sum\limits_{n = 0}^{\infty } {\left[ {e^{{-\sqrt \frac{p}{K} \left( {2nL} \right) - x + \xi }} + e^{{ - \sqrt \frac{p}{K} \left( {2\left( {n + 1} \right)L + x - \xi } \right)}} + e^{{ - \sqrt \frac{p}{K} \left( {2n L + x + \xi} \right)}} + e^{{ - \sqrt \frac{p}{K} \left( {2\left( {n + 1} \right)L - x - \xi} \right)}} } \right]} \hfill \\ \end{gathered} $$

(109)

From Eq. [105], we can find the inverse transform of $\widehat{{{\mathbf{G}}_{{\mathbf{1}}} }}$ for the interval $0 < x < \xi$ as

$$ \begin{gathered} {\mathbf{G}} = \frac{1}{{\sqrt {4\pi Kt} }}\sum\limits_{n = 0}^{\infty } {\left[ {e^{{ - \frac{{\left( {2nL - x + \xi } \right)^{2} }}{4Kt}}} + e^{{ - \frac{{\left( {2\left( {n + 1} \right)L + x - \xi } \right)^{2} }}{4Kt}}} + e^{{ - \frac{{\left( {2nL + x + \xi } \right)^{2} }}{4Kt}}} + e^{{ - \frac{{\left( {2\left( {n + 1} \right)L - x - \xi } \right)^{2} }}{4Kt}}} } \right]} ,\quad \forall 0 < x < \xi , \hfill \\ = \frac{1}{{\sqrt {4\pi Kt} }}\sum\limits_{n = 0}^{\infty } {\left[ {e^{{ - \frac{{\left( {x - \left( {2nL + \xi } \right)} \right)^{2} }}{4Kt}}} + e^{{ - \frac{{\left( {x + \left( {2nL + \xi } \right)} \right)^{2} }}{4Kt}}} + e^{{ - \frac{{\left( {x - \left( {2\left( {n + 1} \right)L - \xi } \right)} \right)^{2} }}{4Kt}}} + e^{{ - \frac{{\left( {x + \left( {2\left( {n + 1} \right)L - \xi } \right)} \right)^{2} }}{4Kt}}} } \right]} ,\quad \forall 0 < x < \xi \hfill \\ \end{gathered} $$

Proceeding in the same way as in the derivation of ${\mathbf{G}}\,{\text{for}}\,0 < x < \xi \,$, we can also find the inverse transform of $\widehat{{{\mathbf{G}}_{{\mathbf{2}}} }}$ for the interval $\xi < x < L$, which has the form

$$ {\mathbf{G}} = \frac{1}{{\sqrt {4\pi Kt} }}\sum\limits_{n = 0}^{\infty } {\left[ {e^{{ - \frac{{\left( {x + \left( {2nL - \xi } \right)} \right)^{2}}}{4Kt}}} + e^{{ - \frac{{\left( {x + \left( {2nL + \xi } \right)} \right)^{2} }}{4Kt}}} + e^{{ - \frac{{\left( {x - \left( {2\left( {n + 1} \right)L - \xi } \right)} \right)^{2} }}{4Kt}}} + e^{{ - \frac{{\left( {x - \left( {2\left( {n + 1} \right)L + \xi } \right)} \right)^{2} }}{4Kt}}} } \right]} ,\quad \forall \xi < x < L $$

Therefore, the Green’s function $G = \left( {x,\xi ,t} \right)$ can be given as

$$ {\mathbf{G}}\left( {x,\xi ,t} \right) = \left\{ {\begin{array}{*{20}c} {\frac{1}{{\sqrt {4\pi Kt} }}\sum\limits_{n = 0}^{\infty } {\left[ {e^{{ - \frac{{\left( {x - \left( {2nL + \xi } \right)} \right)^{2} }}{4Kt}}} + e^{{ - \frac{{\left( {x + \left( {2nL + \xi } \right)} \right)^{2} }}{4Kt}}} + e^{{ - \frac{{\left( {x - \left( {2\left( {n + 1} \right)L - \xi } \right)} \right)^{2} }}{4Kt}}} + e^{{ - \frac{{\left( {x + \left( {2\left( {n + 1} \right)L - \xi } \right)} \right)^{2} }}{4Kt}}} } \right]} ,} & {\forall 0 < x < \xi ,} \\ {\frac{1}{{\sqrt {4\pi Kt} }}\sum\limits_{n = 0}^{\infty } {\left[ {e^{{ - \frac{{\left( {x + \left( {2nL - \xi } \right)} \right)^{2} }}{4Kt}}} + e^{{ - \frac{{\left( {x + \left( {2nL + \xi } \right)} \right)^{2} }}{4Kt}}} + e^{{ - \frac{{\left( {x - \left( {2\left( {n + 1} \right)L - \xi } \right)} \right)^{2} }}{4Kt}}} + e^{{ - \frac{{\left( {x-\left( {2\left( {n + 1} \right)L + \xi } \right)} \right)^{2} }}{4Kt}}} } \right],} } & {\forall \xi < x < L} \\ \end{array} } \right. $$

(110)

Finally, the solution to Eq. [76] with zero initial temperature (i.e., $T\left( {x,0} \right) = 0$) and the boundary condition in Eq. [79] can be found from Eq. [80] in the form

$$ T\left( {x,t} \right) = \int_{0}^{t} {\int_{0}^{L} {{\mathbf{q}}\left( {\xi ,T} \right){\mathbf{G}}\left( {x,\xi ,t - \tau } \right)d\xi \,d\tau} } $$

where ${\mathbf{G}}\left( {x,\xi ,t} \right)$ is defined as in Eq. [110].