From the above considerations and insights from both 
experiments and the cellular automata model of chapter 
Y, it can be seen that the two 
pertinent quantities of interest are the concentration 
of impurity in the liquid $C$, and how this impurity 
diffuses, which will be characterized by the diffusivity matrix
${\tilde{D}}$ = 
$\bigl(
\begin{smallmatrix}
D_{xx}&D_{xy}\\
D_{yx}&D_{yy}
\end{smallmatrix}
\bigr)$.  
In principle, it is necessary to solve the 
exact problem resolving each and every dendrite, but 
if only information about the envelope is desired, it 
seems reasonable to consider the more coarse-grained 
problem of one phase of a given diffusive nature ${\tilde{D}}_1$ 
(the dendritic array) propagating into another phase 
of a different diffusive nature ${\tilde{D}}_2$ (the pristine liquid).  
Now this ``diffusive nature'' or diffusivity matrix ${\tilde{D}}$ 
is a reasonably straightforward quantity and is dictated 
by the solid-liquid microstructure.  In the pure liquid 
phase, diffusion is isotropic 
${\tilde{D}}$ = $D_o$
$\bigl(
\begin{smallmatrix}
1&0\\
0&1
\end{smallmatrix}
\bigr)$, 
where $D_o$ is the diffusivity of the 
solute B through the liquid and is taken to be independent 
of the concentration of B in the liquid.  Now, imagine 
solid barriers going through the liquid, forming channels 
in the x-direction as in Fig. \ref{fig:patt_to_matrx_horz}.  
\begin{figure}[thp]
        \includegraphics[angle=0,scale=.80]{mickey.eps}
        \caption[A schematic diagram indicating how the pattern of a 
	growing dendritic array might dictate the diffusional 
	matrix.]{A schematic diagram indicating how the pattern of a 
	growing dendritic array might dictate the diffusional matrix.  
	$\vec{J} = \tilde{D}(-\overrightarrow{grad}C)$.  The direction
	of $(-\overrightarrow{grad}C)$ is the direction in which 
	there is a driving force for diffusion and is indicated by 
	the arrows.  The magnitude of the concentration gradient 
	in this direction is $(-gradC)$.}
        \label{fig:patt_to_matrx_horz}
\end{figure}
In this case a 
concentration gradient purely in the x-direction 
(parallel to the channels), would 
result in diffusion of solute in the x-direction, just as in 
the isotropic case.  However, a concentration gradient in 
the y-direction (perpendicular to the channels) 
would result in no flux (diffusion) since 
the solute B can not penetrate/cross the solid A phase.  
This would therefore result in 
${\tilde{D}}$ = $D_o$
$\bigl(
\begin{smallmatrix}
1&0\\
0&0
\end{smallmatrix}
\bigr)$.  
By considering 
simple rotations of this same structure, as observed in 
other dendritic patterns Fig. \ref{fig:patt_to_matrx_diag},  
\begin{figure}[thp]
        \includegraphics[angle=0,scale=.75]{mickey.eps}
        \caption[Another schematic diagram indicating how the pattern of a 
	growing dendritic array might dictate the diffusional 
	matrix.]{Another schematic diagram indicating how the pattern of a 
	growing dendritic array might dictate the diffusional matrix.  
	The purpose of the figure is to show that the 
	diffusion matrix is simply 
	comprised of the coefficients of the component diffusion 
	equations, which can be discussed intuitively when 
	considering a given array geometry.  
	$\vec{J} = \tilde{D}(-\overrightarrow{grad}C)$.}
        \label{fig:patt_to_matrx_diag}
\end{figure}
the resulting diffusion matrix can be 
$D_o\bigl(
\begin{smallmatrix}
0.5&0.5\\
0.5&0.5
\end{smallmatrix}
\bigr)$ 
(the bottom pattern) or 
$D_o\bigl(
\begin{smallmatrix}
0.5&-0.5\\
-0.5&0.5
\end{smallmatrix}
\bigr)$ 
(the top pattern), as well 
as any other rotation of 
$D_o\bigl(
\begin{smallmatrix}
1&0\\
0&0
\end{smallmatrix}
\bigr)$, 
depending on the 
orientation of the dendrites.  Another simple case to consider is a dendritic 
structure which is completely closed off inside.
In this case, diffusion will not be possible over very long lengthscales, 
and the tighter the mesh, the closer ${\tilde{D}}$ will approach 
$D_o\bigl(
\begin{smallmatrix}
0&0\\
0&0
\end{smallmatrix}
\bigr)$.  
To study a propagating envelope, it will be assumed 
that the ${\tilde{D}}$'s of the growing needle networks 
are given.  In other words, the problem to solve is: 

Given two 
microstructural phases with known macroscopic 
${\tilde{D}}$'s (when averaged over a suitably large area), 
find how the separating envelope evolves as one phase consumes the other.  
It will always be considered that one phase of 
${\tilde{D}}_1 \not= D_o\bigl(
\begin{smallmatrix}
1&0\\
0&1
\end{smallmatrix}
\bigr)$ 
will be growing into the liquid phase with 
${\tilde{D}}_2 = D_o\bigl(
\begin{smallmatrix}
1&0\\
0&1
\end{smallmatrix}
\bigr)$.  

The idealized underlying assumption, allowing the use of 
an averaged diffusivity $\tilde{D}_1$ for the array of 
needles is that the array is periodic over some lenghtscale 
$l_{nat}$.  For figs. \ref{fig:patt_to_matrx_horz} and 
\ref{fig:patt_to_matrx_diag}, $l_{nat}$ is the spacing 
of the dendrites.  The averaging should then be taken over 
a region spanning several times $l_{nat}$ (see the dotted 
regions in figs. \ref{fig:patt_to_matrx_horz} and 
\ref{fig:patt_to_matrx_diag}).  This averaging can 
be approached most rigorously using homogenization 
techniques \cite{Sanchez-Palencia:80}.  
To carry through the derivation of the PDE's, one must first consider 
the three quantities: $V_{A_{sol}}$ the volume fraction of A which has 
solidified; $V_{A_{liq}}$ the volume fraction of A which is in the liquid; 
and $V_B$ the volume fraction of B (which is always in liquid since 
it is rejected from the solid).  Any total volume change 
upon solidification is being ignored.  There are three 
conservation statements available.  Two are global, stating that 
the total amounts of A and B in the entire system do not change
\begin{align}
	\frac{\partial}{{\partial}t}\int\int\limits_{system}
                 (V_{A_{sol}} + V_{A_{liq}})dxdy = 0        \label{eq:cons1a}\\
	\frac{\partial}{{\partial}t}\int\int\limits_{system} V_B dxdy = 0
                                                            \label{eq:cons1b}
\end{align}
\noindent and one is local, i.e. the total local volume must be 
filled by these three volume fractions
\medskip
\begin{equation} \label{eq:cons2a}
	V_{A_{sol}} + V_{A_{liq}} + V_B = 1
\end{equation}
\medskip
\noindent yielding
\medskip
\begin{equation} \label{eq:cons2b}
	\frac{{\partial}V_{A_{sol}}}{{\partial}t} + 
	\frac{{\partial}V_{A_{liq}}}{{\partial}t} +
	\frac{{\partial}V_B}{{\partial}t} = 0
\end{equation}
\medskip

\noindent This local conservation statement 
(Eq. \ref{eq:cons2b}), allows one to 
reduce the required number of fields from three 
($V_{A_{sol}}$, $V_{A_{liq}}$, $V_B$) to two.  
These can be chosen to correspond 
to the physically significant quantities of the problem
\begin{align}
	\Phi &= \frac{V_{A_{sol}}}{f_s}			\label{eq:phi_def}\\
	C &= \frac{V_{B}}{V_{liq}} = \frac{V_B}{(V_B + V_{A_{liq}})} = 
	\frac{V_B}{(1 - V_{A_{sol}})} = \frac{V_B}{(1 - f_s\Phi)}
							\label{eq:C_def}
\end{align}

Here $f_s$ is the volume fraction of solid A present in the 
fully developed dendritic array and is equivalent to 
$\pi$ (from Eq. A) for the system 
before any growth begins.  $\Phi$ will be 
appropriately referred to as the phase field, but the reader 
should not look for any similarity between this derivation and 
the traditional phase-field models based on Ginsburg-Landau 
energy functional formulations.   
In the fully developed 
pattern, $\Phi$ = 1, corresponding to $V_{A_{sol}} = f_s$ and 
$C = \frac{V_B}{V_{liq}} = \frac{V_B}{(1-f_s)}$.  
In the liquid with no solid present at all, $\Phi = 0$, corresponding
to $V_{A_{sol}} = 0$ and $C = V_B$.  
Intermediate values represent the 
interface between the two phases.
$C$ can be seen to be the concentration 
(in volume fraction) of B in the liquid 
portion of the system regardless of whether it is interdendritic liquid, 
or liquid far from any solid.  This is the physically 
pertinent variable, since it is this concentration which 
drives both solidification and the diffusion of B 
throughout the liquid.  The equation governing this 
concentration field is

\begin{equation}  \label{eq:Ceq1}
	\frac{{\partial}C}{{\partial}t}=
	\vec{\nabli}{\tilde{D}(\Phi)}\vec{\nabli}C + 
	\frac{{\partial}\Phi}{{\partial}t}  \frac{f_s C}{(1 - f_s\Phi)}
\end{equation}
The source term 
$\frac{{\partial}\Phi}{{\partial}t}  \frac{f_s C}{(1 - f_s\Phi)}$ 
comes from straightforward differentiation of Eq. \ref{eq:C_def}, 
and couples the equation for $\frac{{\partial}C}{{\partial}t}$  
to the equation for $\frac{{\partial}\Phi}{{\partial}t}$, 
(Eq. \ref{eq:phi_eq}).  In the diffusive term 
$\vec{\nabli}{\tilde{D}(\Phi)}\vec{\nabli}C$, 
the only discussion required is about the form of $\tilde{D}(\Phi)$.  
It has already been illustrated that $\tilde{D}$ takes on the value 
of $\tilde{D}(\Phi = 0) = \tilde{D_1} = D_o\bigl(
\begin{smallmatrix}
1&0\\
0&1
\end{smallmatrix}
\bigr)$ 
in the liquid phase ahead of 
the interface, and in the fully evolved dendritic pattern 
$\tilde{D}(\Phi = 1) = 
\tilde{D_2} = D_o\bigl( 
\begin{smallmatrix}
d_{xx}&d_{xy}\\
d_{yx}&d_{yy}
\end{smallmatrix}
\bigr)$
$(0 \le d_{\alpha\alpha} \le 1; -0.5 \le d_{\alpha\beta} 
\le 0.5; \alpha \ne \beta)$ 
where the $d_{\alpha\alpha}$ and $d_{\alpha\beta}$ 
are dictated by
the pattern as illustrated in figures \ref{fig:patt_to_matrx_horz} and 
\ref{fig:patt_to_matrx_diag}.  
When $d_{xy}$ is treated as differing from $d_{yx}$, one sees that 
these quantities only appear as ($d_{xy} + d_{yx}$).  This implies 
that there is no need to treat them as two independent 
quantities and the assertion that $d_{xy} = d_{yx}$ can be made 
with confidence.  To connect these known values of $\tilde{D}(\Phi = 0)$ 
and $\tilde{D}(\Phi = 1)$ across the interface where $0<\Phi<1$, any number 
of relations can be invented.  Several such relations have 
been used in studying these equations numerically.  A 
linear relation will be presented here because the 
calculations are so straightforward.

The linear relation resulting in a $\tilde{D}(\Phi)$ which connects 
$\tilde{D}(\Phi = 0) = D_o\bigl(
\begin{smallmatrix}
1&0\\
0&1
\end{smallmatrix}
\bigr)$ 
and 
$\tilde{D}(\Phi = 1) = D_o\bigl(
\begin{smallmatrix}
d_{xx}&d_{xy}\\
d_{yx}&d_{yy}
\end{smallmatrix}
\bigr)$ 
is 

\begin{equation}  \label{eq:connect}
	\tilde{D}(\Phi) = D_o\biggl(
	\begin{matrix}
	(1 - (1 - d_{xx})\Phi) & d_{xy}\Phi\\
	d_{yx}\Phi & (1 - (1 - d_{yy})\Phi)
	\end{matrix}	
	\biggr)
\end{equation}
With this, Eq. \ref{eq:Ceq1} can be written in a more expanded form.
\begin{multline*}
\frac{{\partial}C}{{\partial}t} = 
D_o\frac{{\partial}}{{\partial}x}\biggl\{
[1 - (1 - d_{xx})\Phi]\frac{{\partial}C}{{\partial}x} + 
d_{xy}\Phi \frac{{\partial}C}{{\partial}y}
\biggr\} \\
+
D_o\frac{{\partial}}{{\partial}y}\biggl\{
d_{yx}\Phi \frac{{\partial}C}{{\partial}x} + 
[1 - (1 - d_{yy})\Phi]\frac{{\partial}C}{{\partial}y}
\biggr\} \\
+ 
\frac{{\partial}\Phi}{{\partial}t}  
\frac{f_s C}{(1 - f_s\Phi)} 
\end{multline*}

or equivalently

\begin{multline*} %\label{eq:Ceq2}
\frac{{\partial}C}{{\partial}t} = D_o\biggl\{
[1 - (1 - d_{xx})\Phi]\frac{{\partial}^2C}{{\partial}x^2} +
d_{xx}\frac{{\partial}\Phi}{{\partial}x}\frac{{\partial}C}{{\partial}x} +
d_{xy}\Phi\frac{{\partial}^2C}{{\partial}y{\partial}x} +
d_{xy}\frac{{\partial}\Phi}{{\partial}x}\frac{{\partial}C}{{\partial}y} \\
+
[1 - (1 - d_{yy})\Phi]\frac{{\partial}^2C}{{\partial}y^2} +
d_{yy}\frac{{\partial}\Phi}{{\partial}y}\frac{{\partial}C}{{\partial}y} +
d_{yx}\Phi\frac{{\partial}^2C}{{\partial}x{\partial}y} +
d_{yx}\frac{{\partial}\Phi}{{\partial}y}\frac{{\partial}C}{{\partial}x} 
\biggr\} \\
+ 
\frac{{\partial}\Phi}{{\partial}t}  
\frac{f_s C}{(1 - f_s\Phi)}
\end{multline*}
To describe the evolution of $\Phi$ is more difficult.  
The problem is that of nucleated growth where $\Phi$ 
can not grow at a certain position in space unless 
growth has already begun either at that position or 
at a neighboring position.  The driving force at a 
given position for this growth is also proportional 
to how far the concentration of impurity is below the 
equilibrium concentration $C_{eq}(T_o)$, below which growth 
is thermodynamically favorable.  This results in 

\begin{equation}\label{eq:phi_eq}
\frac{{\partial}\Phi}{{\partial}t}=\beta(C_{eq} - C)
       [1 - \Theta(\int\limits_\epsilon dx \int\limits_\epsilon dy 
                                                       \Phi/\epsilon^2)]
\end{equation}
$\beta$ is a constant of proportionality, 
$
  \Theta(x) = 
  \begin{cases}
	 1, &\text{if $(x \le 0 )$}\\
	 0, &\text{if $(x > 0)$}
  \end{cases}
$ 
is the Heaviside step 
function, and the integration is taken over a small 
region of area $\epsilon^2$ surrounding the point in question.  
Again, Eq. \ref{eq:phi_eq} is simply a statement of the growth of the
solid being proportional to the driving force, and occuring only 
when nucleated by nearby solid.  There should be no confusion
with any type of Ginsburg-Landau based phase-field model.  
The integral in Eq. \ref{eq:phi_eq} is very easy to work with numerically, 
but relatively intractable analytically.  Its purpose is to  
allow growth of new solid to occur in 
supersaturated liquid only if there is old solid 
already present, off of which the new solid can grow.
Equations \ref{eq:Ceq1} and \ref{eq:phi_eq} then 
describe the coupled behaviour of 
both the concentration and phase fields.  
Solving these two equations in tandem leads to 
solving first for the diffusion of $C(x,\ y,\ t + \Delta t)$ given 
the fixed array $\Phi(x,\ y,\ t)$ and then solving for 
the growth of $\Phi(x,\ y,\ t + \Delta t)$ given the fixed array 
$C(x,\ y,\ t + \Delta t)$.  In other words, the first step allows 
the impurity B to diffuse through a fixed network 
of solid and liquid, and the second step calculates 
the evolution of the solid/liquid network based 
on the local concentration of impurity.  This system
behaves quite nicely.  

It is presently being investigated if there is any validity
in collapsing these two equations into a single equation
of one variable in the limit of fast solidification.
Simply plugging Eq. \ref{eq:phi_eq} into the expanded form of 
Eq. \ref{eq:Ceq1}, then allowing diffusion
to take care of nucleation (this is explained 
after Fig. \ref{fig:potwell}), and then using Eq. \ref{eq:C_def} 
to rewrite everything in terms of either $C$ or $\Phi$
results in the equation


\begin{multline}\label{eq:combined}
\frac{{\partial}C}{{\partial}t} = D_o\Biggl\{
\biggl[1 - (1 - d_{xx})\frac{(1 - C_o/C)}{f_s}\biggr]
\frac{{\partial}^2C}{{\partial}x^2} +
d_{xx}\frac{C_o}{f_sC^2}(\frac{{\partial}C}{{\partial}x})^2 \\
+
2d_{xy}\biggl[\frac{(1 - C_o/C)}{f_s}
\frac{{\partial}^2C}{{{\partial}x}{\partial}y} +
\frac{C_o}{f_s C^2}
(\frac{{\partial}C}{{\partial}x}\frac{{\partial}C}{{\partial}y})\biggr] \\
+
\biggl[1 - (1 - d_{yy})\frac{(1 - C_o/C)}{f_s}\biggr]
\frac{{\partial}^2C}{{\partial}y^2} +
d_{yy}\frac{C_o}{f_sC^2}(\frac{{\partial}C}{{\partial}y})^2 +
\Biggr\} \\
+ 
\beta \frac{f_s}{C_o} C^2(C_{eq} - C) 
\bigl(1 - \Theta [C - C_{nuc}]\bigr)
\end{multline}
or equivalently 

\begin{multline*}
\frac{{\partial}\Phi}{{\partial}t}=
D_o\Biggl\{
(1 - d_{xx}\Phi)\frac{{\partial}^2\Phi}{{\partial}x^2} -
d_{xx}(\frac{{\partial}\Phi}{{\partial}x})^2 \\
+
2d_{xy}(\Phi\frac{{\partial}^2\Phi}{{{\partial}x}{\partial}y} +
\frac{{\partial}\Phi}{{\partial}x}\frac{{\partial}\Phi}{{\partial}y}) +
(1 - d_{yy}\Phi)\frac{{\partial}^2\Phi}{{\partial}y^2} -
d_{yy}(\frac{{\partial}\Phi}{{\partial}y})^2 \\
+
\frac{2f_s}{(1 - f_s\Phi)}\biggl[
(1 - d_{xx}\Phi)(\frac{{\partial}\Phi}{{\partial}x})^2 +
2d_{xy}\Phi
\frac{{\partial}\Phi}{{\partial}x}\frac{{\partial}\Phi}{{\partial}y} +
(1 - d_{yy}\Phi)(\frac{{\partial}\Phi}{{\partial}y})^2
\biggr]
\Biggr\} \\
+ 
\beta \frac{(C_{eq} - C_o)(1 - \Phi)}{(1 - f_s\Phi)} 
\bigl(1 - \Theta [\Phi - \Phi_{nuc}]\bigr)
\end{multline*}
Here, $C_{nuc}$ and $\phi_{nuc}$ are simply 
values, above which the field value has to increase before 
growth can take place.  
In this equation, the phase and concentration fields are 
exactly coupled.  Since the role of solute diffusion plays 
a less significant role at greater growth rates, 
it was initially thought that this equation might 
describe the dynamics of the envelope
during quite rapid solidification.  
However, in the equation, diffusion plays a critical 
role in the nucleation process.  
To see this more clearly, it is helpful to consider 
these equations in their 1-dimensional forms, 
for the simplest case of the 
$\tilde{D}(\Phi = 1) = D_o\bigl(
\begin{smallmatrix}
0&0\\
0&0
\end{smallmatrix}
\bigr)$ 
phase
propagating into the 
$\tilde{D}(\Phi = 0) = D_o\bigl(
\begin{smallmatrix}
1&0\\
0&1
\end{smallmatrix}
\bigr)$ 
phase yields

$$
\frac{{\partial}C}{{\partial}t}=
D_o\biggl\{
\bigl[1 - \frac{1}{f_s}(1 - \frac{C_o}{C})\bigr]
\frac{{\partial}^2C}{{\partial}x^2} + 
\frac{C_o}{f_s}\frac{1}{C^2}
(\frac{{\partial}C}{{\partial}x})^2
\biggr\} +
\beta \frac{f_s}{C_o}C^2(C_{eq} - C)
\bigl(1 - \Theta [C - C_{nuc}]\bigr)
$$
and

$$\frac{{\partial}\Phi}{{\partial}t}=
D_o\bigl[
\frac{{\partial}^2\Phi}{{\partial}x^2} + 
\frac{2f_s}{(1 - f_s\Phi)}
(\frac{{\partial}\Phi}{{\partial}x})^2
\bigr] +
\beta (C_{eq} - C_o)\frac{(1 - \Phi)}{(1 - f_s\Phi)} 
\bigl(1 - \Theta [\Phi - \Phi_{nuc}]\bigr)
$$
In a gradient flow context, if looking at the 
source term in Eq. \ref{eq:combined}, and then considering the
``potential function'' $V(C)$ from which it can be derived, 
the influence of the Heaviside function can be seen from a 
different point of view (Fig. \ref{fig:potwell}).  
Without the step function, any
value of $C \ne 0$ will proceed readily to the equilibrium 
value of $C_{eq}$.  The step function actually creates an
extended region of ``neutrality'', and it then becomes
the role of diffusion to drive the state past the nucleation 
concentration $C_{nuc}$, thus allowing growth to proceed. 

A numerical study and an investigation of additional applications 
of the equations presented here are clearly called for.

\begin{figure}[thp]
        \includegraphics[angle=0,scale=1.0]{mickey.eps}
        \caption[The source term of the combined equation 
	(Eq. \ref{eq:combined}) can be seen to come 
	from a potential function $V(C)$ of the form 
	$\frac{1}{4}C^4 - \frac{C_{eq}}{3} C^3$ (the dotted line), 
	modified by the Heaviside step function as 
	shown in the diagram.]{The source term of the combined equation 
	(Eq. \ref{eq:combined}) can be seen to come 
	from a potential function $V(C)$ of the form 
	$\frac{1}{4}C^4 - \frac{C_{eq}}{3} C^3$ (the dotted line), 
	modified by the Heaviside step function as 
	shown in the diagram.  There is still a minimum 
	at $C = C_{eq}$, but the system only flows 
	towards this minimum once the local concentration $C$
	has been pushed to a value $C_o \le C_{nuc} \le C \le C_{eq}$ 
	by diffusion.  The step function therefore plays 
	the role of a type of nucleation barrier, which 
	must be surmounted by diffusion.  NOTE:  There should be no 
	confusion of this with the mention of surface diffusion and 
	nucleation barriers in the Chapter X discussion of the 
	temperature dependence of $\pi_{th}$.}
        \label{fig:potwell}
\end{figure}