# The Solid-Liquid Phase Interface Dynamics in an Undercooled Melt with a Solid Wall
## 2 2D Boundary Integral Equation
Next, multiplying Equation (1) taken at point \(\mathbf{p}_{1}\) by \(G(\mathbf{p}|\mathbf{p}_{1})\) and then subtracting Equation (5) multiplied by \(T(p_{1})\), we obtain  
\[ \begin{split} D_{T}\Bigl{(}G(\mathbf{p}|\mathbf{p}_{1})\
abla_{1 }^{2}T-T(\mathbf{p}_{1})\
abla_{1}^{2}G(\mathbf{p}|\mathbf{p}_{1})\Bigr{)}- G(\mathbf{p}|\mathbf{p}_{1})\frac{\partial T}{\partial t_{1}}-T(\mathbf{p}_{1}) \frac{\partial G(\mathbf{p}|\mathbf{p}_{1})}{\partial t_{1}}\\ +V\biggl{(}G(\mathbf{p}|\mathbf{p}_{1})\frac{\partial T}{\partial z _{1}}+T(\mathbf{p}_{1})\frac{\partial G}{\partial z_{1}}\biggr{)}=\delta( \mathbf{p}-\mathbf{p}_{1})T(\mathbf{p}_{1}).\end{split}  \tag{7}\]  
Now, we integrate Equation (7) over time and volume  
\[ \begin{split} D_{T}\int\limits_{-\infty}^{t+\varepsilon}\int \limits_{\Lambda_{1}}\Bigl{(}G(\mathbf{p}|\mathbf{p}_{1})\
abla_{1}^{2}T-T( \mathbf{p}_{1})\
abla_{1}^{2}G(\mathbf{p}|\mathbf{p}_{1})\Bigr{)}dt_{1}dx_{1} dx_{1}\\ -\int\limits_{\Lambda_{1}}G(\mathbf{p}|\mathbf{p}_{1})T(\mathbf{ p}_{1})|_{t_{1}\rightarrow-\infty}^{t_{1}\to t+\varepsilon}dx_{1}dz_{1}+V\int \limits_{-\infty}^{t+\varepsilon}\int\limits_{-\infty}^{\infty}G(\mathbf{p}| \mathbf{p}_{1})T(\mathbf{p}_{1})|_{z_{1}\rightarrow-\infty}^{z_{1}\to z _{0}}dx_{1}dt_{1}=T(\mathbf{p}),\end{split}  \tag{8}\]  
where \(\varepsilon>0\) is an infinitely small parameter and \(\Lambda_{1}\) is an integration domain without the interface \(\zeta(x,t)\) (\(-\infty<x_{1}<\infty,-\infty<z_{1}<z_{0}\)). The second integral in Equation (8) vanishes since Equation (6) gives zero in the limit \(t_{1}\rightarrow-\infty\), and \(G(\mathbf{p}|\mathbf{p}_{1})=0\) when \(t_{1}>t\) (causality condition). Considering the third  
Figure 1: A scheme of solid/liquid interface propagation into an undercooled liquid in the presence of a solid wall \(z=z_{0}\). integral in (8) with allowance for \(T(z_{1}\to z_{0})=T_{1}=\text{const. }\) and \(T(z_{1}\to-\infty)=\text{const. }\), we are able to integrate [PERSON]'s function as follows  
\[ \begin{split} V\int\limits_{-\infty}^{t+\varepsilon}\int\limits_ {-\infty}^{\infty}G(\mathbf{p}|\mathbf{p}_{1})dx_{1}dt_{1}=\left\{\begin{array} []{c}\exp\biggl{(}(z_{1}-z)\frac{V}{D_{T}}\biggr{)},\quad z>z_{1},\\ 1,\quad z<z_{1}.\end{array}\right.\\ \left.-\left\{\begin{array}{c}\exp\biggl{(}(z_{1}+z-2z_{0}) \frac{V}{D_{T}}\biggr{)},\quad z-2z_{0}+z_{1}<0,\\ 1,\quad z-2z_{0}+z_{1}>0.\end{array}\right.\end{split}  \tag{9}\]  
Here, the coordinate \(z_{1}\in(z_{0},-\infty)\). At \(z_{1}\to-\infty\), both terms on the right-hand side of expression (9) are vanishing. Considering the boundary \(z_{1}=z_{0}\) and taking into account that the total domain of the problem under question is bounded by the solid wall, we arrive at  
\[V\int\limits_{-\infty}^{t+\varepsilon}\int\limits_{-\infty}^{\infty}G( \mathbf{p}|\mathbf{p}_{1})T(\mathbf{p}_{1})|_{z_{1}\to-\infty}^{z_{1}\to z_{0 }}dx_{1}dt_{1}=T_{1}\biggl{(}1-\exp\biggl{(}(z-z_{0})\frac{V}{D_{T}}\biggr{)} \biggr{)}. \tag{10}\]  
Now, we apply [PERSON]'s second identity to the first term in (8)  
\[ \begin{split}&\int\limits_{-\infty}^{t+\varepsilon}\int\limits _{\Lambda_{1}}\Bigl{(}G(\mathbf{p}|\mathbf{p}_{1})\
abla_{1}^{2}T-T(\mathbf{p }_{1})\
abla_{1}^{2}G(\mathbf{p}|\mathbf{p}_{1})\Bigr{)}dt_{1}dx_{1}dz_{1}\\ &=\int\limits_{-\infty}^{t+\varepsilon}\int\limits_{S_{1}}(G( \mathbf{p}|\mathbf{p}_{1})\
abla_{1}T-T(\mathbf{p}_{1})\
abla_{1}G(\mathbf{p }|\mathbf{p}_{1}))\cdot dS_{1}dt_{1},\end{split}  \tag{11}\]  
where \(S_{1}\) is a boundary encompassing the volume \(\Lambda_{1}\) in 3D. By substituting (10) and (11) into (8), we have  
\[ \begin{split} D_{T}\int\limits_{-\infty}^{t+\varepsilon}\int \limits_{S_{1}}(G(\mathbf{p}|\mathbf{p}_{1})\
abla_{1}T-T(\mathbf{p}_{1}) \
abla_{1}G(\mathbf{p}|\mathbf{p}_{1}))\cdot dS_{1}dt_{1}\\ +T_{1}\biggl{(}1-\exp\biggl{(}(z-z_{0})\frac{V}{D_{T}}\biggr{)} \biggr{)}=T(\mathbf{p}).\end{split}  \tag{12}\]  
The integral containing \(\
abla_{1}G(\mathbf{p}|\mathbf{p}_{1})\) vanishes due to the continuity of temperature \(T\)[28; 30].