Appendix: Derivation of the half-order time derivative method
The algorithm and derivation of the half-order time derivative method was first proposed by Wang and Bras (1999). Here, details of the derivation are provided.
First of all, define two new variables: \(\tilde z = {z \mathord{\left/ {\vphantom {z {\sqrt {D_0 } }}} \right. \kern-\nulldelimiterspace} {\sqrt {D_0 } }}\) and \(\Theta = T - T_0 \), and then transfer Eqs. 7, 9, and 10 to:
$$\frac{{\partial \Theta }}{{\partial t}} = \frac{{\partial ^2 \Theta }}{{\partial \tilde z^2 }}$$
(A1)
$$\Theta = 0,\,{\text{for t = 0,}}\,\tilde z <0$$
(A2)
$$\Theta = 0,\,{\text{for t}}\,{\text{ $>$ }}\,{\text{0,}}\,\tilde z \to - \infty $$
(A3)
Now, take the Laplace transform of both sides of Eq. A1. According to the definition of Laplace transform, \(L\left\{ {\frac{\partial }{{\partial t}}f\left( t \right)} \right\}\left( s \right) = sF\left( s \right) - f\left( 0 \right)\), the left hand side of A1 then becomes \(s\tilde \Theta \left( {\tilde z,s} \right) - \Theta \left( {\tilde z,0} \right) = s\tilde \Theta \left( {\tilde z,s} \right)\) since A2 requires \(\Theta \left( {\tilde z,0} \right) = 0\). The Laplace transform of the right hand side remains the same since the Laplace transform is based on the variable t. Hence, Eq. A1 is transformed to
$$s\tilde \Theta \left( {\tilde z,s} \right) = \frac{{\partial ^2 \tilde \Theta }}{{\partial \tilde z^2 }}$$
(A4)
where \(\tilde \Theta \) is the Laplace transform of \(\Theta \) and defined as:
$$\tilde \Theta \left( {\tilde z,s} \right) = \int\limits_0^\infty {\exp \left( { - st} \right)\,} \Theta \left( {\tilde z,t} \right)dt$$
(A5)
The general solution of Eq. A4 is
$$\tilde \Theta \left( {\tilde z,s} \right) = A\left( s \right)\exp \left( {\tilde z\sqrt s } \right) + B\left( s \right)\exp \left( { - \tilde z\sqrt s } \right)$$
(A6)
where A(s) and B(s) are arbitrary functions of s and can be determined by the boundary conditions. By Eq. A3, B(s) is found to be zero and Eq. A6 then becomes:
$$\tilde \Theta \left( {\tilde z,s} \right) = A\left( s \right)\exp \left( {\tilde z\sqrt s } \right)$$
(A7)
Differentiating both sides of Eq. A7 with respect to \(\tilde z\) gives:
$$\frac{\partial }{{\partial \tilde z}}\tilde \Theta \left( {\tilde z,s} \right) = \sqrt s A\left( s \right)\exp \left( {\tilde z\sqrt s } \right)$$
(A8)
Now, substituting A7 into A8 yields
$$\frac{\partial }{{\partial \tilde z}}\tilde \Theta \left( {\tilde z,s} \right) = \sqrt s \tilde \Theta \left( {\tilde z,s} \right)$$
(A9)
Based on the equation derived by the fractional calculus (Miller and Ross 1993),
$$L\left\{ {D^\nu f\left( t \right)} \right\} = s^\nu F\left( s \right) - \sum\limits_{k = 0}^{m - 1} {s^{m - k - 1} D^{k - m + \nu } f\left( 0 \right)} $$
(A10)
the right hand side of Eq. A9 becomes
$$\sqrt s \tilde \Theta \left( {\tilde z,s} \right) = L\left\{ {\frac{{\partial ^{\frac{1}{2}} }}{{\partial t^{\frac{1}{2}} }}\tilde \Theta \left( {\tilde z,t} \right)} \right\}$$
(A11)
By the initial condition Eq. A2, the last term of Eq. A10, f(0), is found to be zero. Replacing the right hand side of Eq. A9 by Eq. A11, we have
$$\frac{\partial }{{\partial \tilde z}}\tilde \Theta \left( {\tilde z,s} \right) = L\left\{ {\frac{{\partial ^{\frac{1}{2}} }}{{\partial t^{\frac{1}{2}} }}\tilde \Theta \left( {\tilde z,t} \right)} \right\}$$
(A12)
Inverting the Laplace transform of Eq. A12 leads to
$$\frac{\partial }{{\partial \tilde z}}\Theta \left( {\tilde z,t} \right) = \frac{{\partial ^{\frac{1}{2}} }}{{\partial t^{\frac{1}{2}} }}\Theta \left( {\tilde z,t} \right)$$
(A13)
With the definition of Θ, Eq. A13 becomes
$$\frac{\partial }{{\partial z}}T\left( {z,t} \right) = \frac{1}{{\sqrt {D_o } }}\frac{{\partial ^{\frac{1}{2}} }}{{\partial t^{\frac{1}{2}} }}\left[ {T\left( {z,t} \right) - T\left( 0 \right)} \right].$$
(A14)
Now, applying the fractional calculus (Miller and Ross 1993)
$$\frac{{d^\alpha f\left( t \right)}}{{dt^\alpha }} = \frac{1}{{\Gamma \left( {1 - \alpha } \right)}}\frac{d}{{dt}}\int\limits_0^t {\frac{{f\left( s \right)}}{{\left( {t - s} \right)^\alpha }}ds} $$
(A15)
to the right hand side of Eq. A14, we have
$$
\begin{array}{*{20}c}
{\frac{\partial }
{{\partial z}}T{\left( {z,t} \right)} = \frac{1}
{{{\sqrt {\pi D_{0} } }}}{\int\limits_0^t {\frac{{\partial T{\left( {z,s} \right)}}}
{{\partial s}}\frac{{ds}}
{{{\sqrt {t - s} }}}} }} \\
{ = \frac{1}
{{{\sqrt {\pi D_{0} } }}}{\int\limits_0^t {\frac{{dT{\left( {z,s} \right)}}}
{{{\sqrt {t - s} }}}} }} \\
\end{array}
$$
(A16)
Finally, the prognostic result can be evaluated by applying Eq. A16 to Fourier’s law, and then we have
$$G\left( {z,t} \right) = \sqrt {\frac{{k\rho _s c_s }}{\pi }} \int\limits_0^t {\frac{{dT\left( {z,s} \right)}}{{\sqrt {t - s} }}} $$
(A17)
Equation A17 can also be written as:
$$G\left( {z,t} \right) = \sqrt {\frac{{k\rho _s c_s }}{\pi }} \int\limits_0^t {\frac{{\partial T\left( {z,s} \right)}}{{\partial s}}\frac{{ds}}{{\sqrt {t - s} }}} = 2\sqrt {\frac{{k\rho _s c_s }}{\pi }} \int\limits_0^t {\frac{{\partial T\left( {z,s} \right)}}{{\partial s}}} d\sqrt {t - s} $$
(A18)
And its discrete form is
$$G = 2\sqrt {\frac{{k\rho _s c_s }}{\pi }} \sum\limits_{i = 0}^N {\frac{{T_{i + 1} - T_i }}{{t_{i + 1} - t_i }}\left[ {\sqrt {t_N - t_{i + 1} } - \sqrt {t_N - t_i } } \right]} $$
(A19)
where N is the number of intervals.