## 6 Numerical study on the error and applications to planar-symmetric problems
### One-phase freezing problem  
A planar-symmetric freezing problem in a semi-infinite domain is considered. Due to symmetry and material isotropy, the problem can be reduced to one dimension and the coordinate \(x\) in the direction orthogonal to the constant-temperature planes suffices to the problem description. The one-dimensional domain \(\Omega=\left[0,\infty\right[\) is initially at the freezing/melting temperature, with \(\Omega_{\text{sol}}=\left[0,x_{\text{s}}\right[\), \(\Omega_{\text{big}}=\left]x_{\text{s}},\infty\right[\), and \(\Gamma_{\text{tan}}=\left\{x_{\text{s}}\right\}\). The boundary condition is \(T\left(0,t\right)=T_{0}<T_{\text{tan}}\), whereas the initial position of the interface is \(x_{\text{s}}\left(0\right)=a\in\Omega\). The values used in the numerical examples are reported in Table 1. Stefan number is computed with the temperature difference between phase-transition and coolant thermostat,  
\[\alpha=\frac{L}{c_{\text{sol}}\left(T_{\text{tan}}-T_{0}\right)}=15. \tag{48}\]  
With these initial and boundary conditions, no heat flow occurs in the liquid domain, which remains at a constant temperature, and the interface is driven only by the temperature gradient in the solid domain, so that the problem can be considered as a one-phase problem with moving boundary. [PERSON]'s exact solution is available for this case. The temperature field in the solid region is given by  
\[T_{\text{exact}}\left(x,t\right)=\frac{\text{erf}\left(\frac{x}{x_{\text{s}}^ {\text{exact}}\left(t\right)2}\right)}{\text{erf}\left(\frac{\gamma}{2} \right)}\big{(}T_{\text{tan}}-T_{0}\big{)}+T_{0},\quad x\in\Omega_{\text{sol}}\, \tag{49}\]  
where the non-dimensional coefficient \(\gamma\) is the positive real root of  
\[\exp\left(\frac{\gamma}{2}\right)\sqrt{\frac{\pi\gamma}{2}}\ \text{erf}\left(\sqrt{\frac{ \gamma}{2}}\right)=\frac{1}{\alpha}\, \tag{50}\] \begin{table}
\begin{tabular}{l l l l} \hline One-phase & Two-phase & Polar-symmetric & Water-pipe \\ example data & example data & example data & example data \\ \hline \(k_{\text{sol}}=2.18\) W/m/K & \(k_{\text{sol}}=250\) W/m/K & \(k_{\text{sol}}=2.18\) W/m/K & \(k_{\text{sol}}=18\) W/m/K \\ \(k_{\text{big}}=0.6\) W/m/K & \(k_{\text{big}}=190\) W/m/K & \(k_{\text{big}}=0.6\) W/m/K & \(c_{\text{sol}}=2260\) J/kg/K \\ \(c_{\text{sol}}=2260\) J/kg/K & \(c_{\text{sol}}=c_{\text{big}}=880\) J/kg/K & \(c_{\text{sol}}=2260\) J/kg/K & \(\rho=1000\) kg/m\({}^{3}\) \\ \(c_{\text{big}}=4186\) J/kg/K & \(\rho=2700\) kg/m\({}^{3}\) & \(c_{\text{big}}=4186\) J/kg/K & \(L=335000\) J/kg \\ \(\rho=1000\) kg/m\({}^{3}\) & \(L=267000\) J/kg & \(\rho=1000\) kg/m\({}^{3}\) & \(k_{\text{ins}}=0.05\) W/m/K \\ \(L=335000\) J/kg & \(T_{\text{tan}}=660\)\({}^{\circ}\)C & \(L=335000\) J/kg & \(c_{\text{ins}}=600\) J/kg/K \\ \(T_{\text{tan}}=0\)\({}^{\circ}\)C & \(T_{0}=652.5\)\({}^{\circ}\)C & \(T_{\text{tan}}=0\)\({}^{\circ}\)C & \(\rho_{\text{ins}}=100\) kg/m\({}^{3}\) \\ \(T_{0}=-100\)\({}^{\circ}\)C & \(T_{1}=670\)\({}^{\circ}\)C & \(T_{0}=1\)\({}^{\circ}\)C & \(T_{\text{tan}}=0\)\({}^{\circ}\)C \\ \(a=1\) m & \(\ell=1\) m & \(R=0.01\) m & \(T_{0}=-10\)\({}^{\circ}\)C \\  & \(a=0.02\) m & \(a=0.9R\) & \\ \hline \end{tabular}
\end{table} 
Table 1: Data used in the numerical examples. and the position of the freezing front is given by  
\[x_{*}^{\text{exact}}\left(t\right)=a\sqrt{1+2\gamma\frac{k_{\text{sol}}}{\rho c_ {\text{sol}}a^{2}}}t. \tag{51}\]  
As further comparison, we also consider the PSS solution, although the Stefan number of the specific example, see Eq. (48), is not large enough for the PSS solution to constitute a good approximation [19]. Under the PSS assumption, the one-dimensional governing equation becomes  
\[\frac{\partial^{2}T_{\text{PSS}}}{\partial x^{2}}=0\, \tag{52}\]  
with the boundary conditions  
\[T_{\text{PSS}}\left(0,t\right)=T_{0}\,\qquad T_{\text{PSS}}\left(x_{*}^{ \text{PSS}},t\right)=T_{\text{van}}. \tag{53}\]  
Integration of Eq.