# The Solid-Liquid Phase Interface Dynamics in an Undercooled Melt with a Solid Wall
## 2 2D Boundary Integral Equation  
Let us consider the evolution of a curved solid/liquid phase transition boundary in an undercooled one-component liquid in the presence of a solid wall (Figure 1). Below, we assume that the one-component melt is motionless, and the BIE is written out in a reference frame that is moving with a constant (\(V\)) toward the liquid phase. Also, we assume that the temperature conductivity coefficient \(D_{T}\) is the same constant in both phases. In addition, the solid wall is smooth and planar; no particle nucleation occurs in the bulk melt or at the solid wall. The temperature field in the solid and liquid phases satisfies the temperature conductivity equation in a reference frame that is moving with a constant velocity (\(V\)) [28]  
\[D_{T}\
abla^{2}T-\frac{\partial T}{\partial t}+V\frac{\partial T}{\partial z} =0, \tag{1}\]  
where \(T\) is the temperature, \(D_{T}\) is the thermal conductivity, \(t\) is the time and \(V\) is the constant velocity of moving Cartesian coordinate systems \(x,y\), and \(z\) fixed on the growing crystal (steady-state growth rate). The boundary conditions at the solid/liquid interface take the form [17; 20; 22; 25]  
\[ \begin{split} T=T_{i}=T_{0}-d_{c}\mathcal{K}\frac{Q}{c_{p}}- \tilde{\beta}\Big{(}V+\frac{\partial\zeta}{\partial t}\Big{)},\\ D_{T}\Big{(}\
abla T_{solid}-\
abla T_{liquid}\Big{)}\cdot d \mathbf{s}=\frac{Q}{c_{p}}\Big{(}V+\frac{\partial\zeta}{\partial t}\Big{)}d^ {2}x,\end{split}  \tag{2}\]  
where \(d_{c}\) is the anisotropy capillary length, \(\mathcal{K}\) is the average interface curvature, \(Q\) is the latent heat of crystallization, \(c_{p}\) is the heat capacity, \(\tilde{\beta}\) is the kinetic coefficient, \(T_{i}\) is the interfacial temperature, \(T_{0}\) is the phase transition temperature for a flat crystallization front, \(d\mathbf{s}\) is the surface area vector element directed towards the liquid phase, and \(\zeta(x,t)\) is the interface function. Let us especially emphasize that \(\zeta(x,t)\) represents a curved line in 2D whereas \(\zeta(x,y,t)\) is a crystal surface in 3D. Note that the boundary conditions (2) represent the Gibbs-Thomson and heat balance conditions, respectively. The average interface curvature \(\mathcal{K}\) in a two-dimensional case is determined as  
\[\mathcal{K}(x,t)=-\frac{\partial^{2}\zeta/\partial x^{2}}{\Big{[}1+\left( \partial\zeta/\partial x\right)^{2}\Big{]}^{3/2}}. \tag{3}\]  
Here, we consider how a solid wall \(z=z_{0}\) kept at a fixed temperature  
\[T(z_{0})=T_{1}=\text{const}. \tag{4}\]  
influences the boundary integral equation. To do this, we use [PERSON]'s function \(G(\mathbf{p}|\mathbf{p}_{1})\) to derive an integro-differential equation for the interface function \(\zeta(x,t)\). [PERSON]'s function satisfies the following equation and boundary condition [28; 29]  
\[ \begin{split}\frac{\partial G(\mathbf{p}|\mathbf{p}_{1})}{ \partial t_{1}}+D_{T}\
abla_{1}^{2}G(\mathbf{p}|\mathbf{p}_{1})-V\frac{ \partial G(\mathbf{p}|\mathbf{p}_{1})}{\partial z_{1}}=-\delta(\mathbf{p}- \mathbf{p}_{1}),\\ G(x,z_{0},t|x_{1},z_{1},t_{1})=0,\end{split}  \tag{5}\]where \(\mathbf{p}=(x,z,t)\) and \(\mathbf{p}_{1}=(x_{1},z_{1},t_{1})\). Applying the method of images to [PERSON]'s function for an infinite domain [29], we construct [PERSON]'s function for the problem under question with a solid wall as follows  
\[ \begin{split} G(\mathbf{p}|\mathbf{p}_{1})=\frac{1}{4\pi D_{T}(t -t_{1})}\left[\exp\biggl{(}-\frac{(x-x_{1})^{2}+(z-z_{1}+V(t-t_{1}))^{2}}{4D_{ T}(t-t_{1})}\biggr{)}\right.\\ \left.-\exp\biggl{(}-\frac{(x-x_{1})^{2}+(z-2z_{0}+z_{1}-V(t-t_{1} ))^{2}}{4D_{T}(t-t_{1})}\biggr{)}\right].\end{split}  \tag{6}\]  
It is easily seen that [PERSON]'s function (6) satisfies the boundary value problem (5). Note that the first term in expression (6) coincides with [PERSON]'s function for the unbounded problem [30] (when a solid wall is absent).