Molecular viewpoint on the crystal growth dynamics driven by solution ow
D. Maes
†and James F. Lutsko
∗,‡† Structural Biology Brussels, Vrije Universiteit Brussel, Pleinlaan 2, 1050, Brussels, Belgium
‡ Center for Nonlinear Phenomena and Complex Systems, Code Postal 231, Université Libre de Bruxelles, Boulevard du Triomphe, 1050 Brussels, Belgium
E-mail: jlutsko@ulb.ac.be
Multiscale analysis of growth rate
Model
A simple analysis for the interaction of diusion and growth in our model is possible under the assumption of planar symmetry. Specically, we assume that diusion is fast compared to step growth so that inhomogeneities in the density in the planes parallel to the crystal surface are very small or are short-lived compared to the time-scale of step growth. In this case, the concentration depends only on the z−coordinate, c
t(z) , and satises a diusion equation
∂c
t(z)
∂t = D ∂
2c
t(z)
∂z
2(1)
Boundary conditions are needed at the top and bottom of the simulation cell. Consider one
cell in the computational lattice in the top-most layer of the uid. The average number
of molecules expected in such a cell is c
t(L
z) a
3. Molecules can enter and leave via any of
the six faces of the cell but movement via the lateral faces is of no importance since we are assuming homogeneity within the plane: this means that the rates of molecules entering and leaving via the lateral faces exactly balance. The rate at which molecules leave via the top and bottom faces is simply ν
jumpc
t(L
z) a
3in each case. The rate at which molecules enter at the bottom is correspondingly ν
jumpc
t(L
z− a) a
3while the rate at which they enter via the top is the product of the rate at which insertions are attempted and the probability that the cell is empty, r
add(1 − c
t(L
z) a
3) . Putting this together,
∂a
3c
t(L
z)
∂t = r
add1 − c
t(L
z) a
3− ν
jumpc
t(L
z) a
3(2) + ν
jumpc
t(L
z− a) a
3− ν
jumpc
t(L
z) a
3or in the continuum limit,
∂c
t(L
z)
∂t ' a
−3r
add− (r
add+ ν
jump) c
t(L
z) − a
−1D ∂c
t(z)
∂z
z=L
(3)
At the interface between the crystal and the uid, taken to be at position z = H
t, there must be a balance between the rate at which material arrives at the surface, via diusion, and the rate at which the crystal grows giving
−D ∂c
t(z)
∂z
Ht
L
xL
y= a
−2v (c
t(H
t)) L
y(4)
where it is assumed that the steps are parallel to the y-direction and grow in the x−direction.
The crystal increases height by one lattice cell, so H
t→ H
t+ a , when a total of N
xN
ymolecules have been added so, in the continuum limit, we have that dH =
NxaNydM where dM is the change in mass of the crystal. As a step grows a distance dL
x= vdt , the crystal adds L
ydL
x/a
2molecules so dM/dt = L
yv/a
2and the balance requires
dH
tdt = a N N
L
yv (c
t(H
t))
a
2= a
L v (c
t(H
t)) (5)
So, the model for the rate of step growth can be written as:
∂c
t(z)
∂t = D ∂
2c
t(z)
∂z
2, c
t(L
z) = A
tand c
t(H
t) = B
t(6) dH
tdt = a
L
xv (B
t) dA
tdt = a
−3r
add− (r
add+ ν
jump) A
t− a
−1D ∂c
t(z)
∂z
z=L
v (B
t) = −a
2L
xD ∂c
t(z)
∂z
z=Ht
Quasi-steady state
The problem admits of a natural small parameter, ≡
Lax. This arises because the rate of growth in the z−direction is necessarily small compared to the rate of step advancement.
We can therefore use this to construct a simple multiscale analysis. We slightly rewrite the system of equations as
∂c
t(z)
∂t = D ∂
2c
t(z)
∂z
2, c
t(L
z; H
t) = A
t(H
t) and c
t(H
t; H
t) = B
t(H
t) (7) dH
tdt = v (B
t) dA
tdt = a
−3r
add− (r
add+ ν
jump) A
t− a
−2D L
xL
zL
z∂c
t(z)
∂z
z=L
v (B
t) = −a
2L
xL
zD
L
z∂c
t(z)
∂z
z=Ht
to emphasize the fact that we expect the concentration near the crystal to be close to its equilibrium value so that
∂c∂zt(z)∼
c(LzL)−ceqz
. Hence, we wish to treat L
z∂ct(z)
∂z
as being of
order
0. With this in mind, it is clear that the only consequence of being small is that
the rate of growth in the z-direction is small. We could use this as the basis for a general
multiscale analysis but here we restrict attention to the quasi-steady state in which the only
time-dependence of the concentration occurs via its dependence on H
t. In this case, we have
that c
t(z) → c (z; H
t) and
∂c
t(z)
∂t = ∂c (z; H
t)
∂H
tdH
tdt = v (B (H
t)) ∂c (z; H
t)
∂H
t(8)
so
v (B (H
t)) ∂c (z; H
t)
∂H
t= D ∂
2c (z; H
t)
∂z
2, c (L
z; H
t) = A (H
t) and c (H
t; H
t) = B (H
t) (9)
v (B (H
t)) dA (H
t)
dH
t= a
−3r
add− (r
add+ ν
jump) A (H
t) − a
−2D L
xL
zL
z∂c (z; H
t)
∂z
z=L
v (B (H
t)) = −a
2L
xL
zD
L
z∂c (z; H
t)
∂z
z=Ht
with the height of the crystal being determined by
dHdtt= v (B
t) . These equations can be solved perturbatively by expanding c (z; H
t) = c
(0)(z; H
t) + c
(1)(z; H
t) + ... and solving order by order in . At lowest order, this gives
0 = D ∂
2c
(0)(z; H
t)
∂z
2(10)
0 = a
−3r
add− (r
add+ ν
jump) A
(0)(H
t) v B
(0)(H
t) = −a
2L
xL
zD
L
z∂c
(0)(z; H
t)
∂z
z=Ht