QUARTERLY OF APPLIED MATHEMATICS 341
OCTOBER, 1972
ON THE SOLUTE DISTRIBUTION AT A MOVING PHASE BOUNDARY*
By K. A. HEIMES (Iowa Stale University)
1. Introduction. Consider an infinite rod of homogeneous binary alloy with a planar
solid-liquid interface advancing at constant velocity R. Let D, , v'(x, t) denote respec-
tively the diffusion coefficient and solute concentration in the solid (i = 1) and liquid
(i = 2) regions. For t = 0 we locate the interface at x = 0 and describe the initial solute
compositions by /,(x) for (— l)'x > 0. In order to obtain equilibrium at the interface,
we require
V\Rt, t) = kV\Rt, t),
D1V1x(Rt, t) - D.VliRt, t) = R[V\Rt, t) - V\Rt, t)],
where k is a constant equilibrium distribution coefficient. Assuming no convection in
the liquid, the diffusion equations are DXV'IX = V\ in their respective regions. By putting
C\z, t) = V'(z + Rt, t) we fix the interface at z = 0 and move the rod into the solid
region (z < 0). Consequently, this one-dimensional liquid-solid transformation (solidi-
fication) can be described as
Problem S: For i — 1, 2 let , k, R be positive constants and let /t(z) be continuous
real functions defined for (— l)'z > 0 with /,(0) = kj2(0). Find functions C'(z, t) for
t > 0, (— l)'z > 0 satisfying
(51) DtC*.. + RCl = C\ ;
(52) C\z, 0) = /.(2);
(SO C\0, t) = kC\0, <), ~R( 1 - k)C\0, t) = D2C](0, t) - AC!(0, t).
The corresponding solid-liqiud transformation (Problem M for melting) is the same
as Problem S except that R is replaced by —R in (SO and (S3). Problem S has been
solved for cases Di = D2 and I), = 0 with particular initial conditions in [1] and [2].
In this paper we give sufficient conditions that both the above problems have unique
solutions and explicit solutions are obtained by Laplace transforms methods. Both
problems reduce to solving an integral-differential for the function g(t) = C'(0, t) =
kC2(0, t) which describes the time behavior of solutions at the interface.
2. The reduced problem. We first show that problems M and S are equivalent to
Problem A: Given positive constants K, X and real functions g^x) for x > 0 with
ffi(0) = ?2(0), find u'(x, y) for x > 0, y > 0 so that
(Ai) uxx Up ,
* Received December 30, 1970; revised version received April 12, 1971. Work performed in part
at Ames Laboratory, U. S. Atomic Energy Commission. Contribution No. 2911.
342 K. A. HEIMES
(A2) u\x, 0) = g,(x)]
(A,) e~'u\0, y) = e"V(0, Xy), (1 - K)e"V(0, y) = e^O, y) + Ke~Xvu\{0, Xj/).
Functions u'(x, y) will be called solutions to problem A when (i) for x > 0, y > 0
they are continuous, satisfy (A2) and estimates of the form \u(x, y)\ < Me"*' on compact
y intervals; (ii) for x > 0, y > 0 u\x and u\ are continuous and satisfy (A,); (iii) for
x > 0, y > 0 u'x is continuous and (A^) holds.
Straightforward calculations verify
Lemma 1. Let u1, u2 solve problem A with K = /c, X = D2/Dx and </i(Xx) = ex7i( — z)>
g2(x) = Ke~*f2(z) where x = Rz/2D2 , y = R?t/AD2 . Then
C\z, t) = exu~vV(—Xx, Xy)
1 (1)
C\z, t) = y)
solve problem M. Similarly, if u , u solve problem A for K — l/k, X = Di/D2 and
<7i(Xx) = eXxf2(—z), g2{x) = Ke~*fi(z) where x = —Rz/2DX , y = R2t/4D1 , then
C'(z, t) = ±e'-u\x,y) (2)
C2(z, 0 = eX(*-"V(-Xx, Xy)
solve problem S.
Lemma 2. For p, 5 £ C[0, 00), solutions to = wv for x > 0, y > 0 with u(x, 0) =
p(x), w(0, y) = <7(2/) are unique.
Proof: This is a standard application of the maximum principle for the heat equa-
tion. Details are similar to those given on p. 48 of [3].
Define the functions
S(x, y) = (47r7/)~1/2 exp {—x/Ay)
U'(x, y) = ^.(x), y = 0,
= f [S(x - r, y) + S(x + r, y)]g<(r) dr, y > 0. (3)
Jo
Then the U' solve (^1,), {A2) with U'x(0, y) = 0 and
U'(0, y) = 2 f S(r, y)gi(r) dr = -y- [ e~"'gi(2uVy) du (4)
Jo VT Jo
(see p. 53 of [3]).
Suppose that u1, u solve Problem A with gx , g2 , g £ C[0, ») n C'(0, <») where
?(y) = e~"u2(0, y) = e~*vu(0, \y) is the interface function. It then follows from Lemma 2
that
u\x, y) = U\x, 1i) + ^ (erfc * ^1/2) Jr [e'gW - U\0, r)] dr ^
m2(x, ?/) = U\x, y) + J (erfc f ^1/2) £ [e'ff(r) - U\0, r)] dr
NOTES 343
where n = 1/X. That is, the right-hand sides in (5) solve (A,) and agree with u'(0, y)
and u\x, 0). Substituting (5) in the last equation of condition (A3), we see that g(y)
solves
(1 - K)g(y) = e~" f" W(y - r)]"1/21 [tf2(0, r) - e'g(r)} dr
+ Ke~* J* - r)]-/21 [^(0, r) - eg^r)] dr. (6)
These steps are reversible, giving
Lemma 3. Let W = C[0, ») f~\ C"(0, <»). For initial conditions g< £ W, Problem A
has solutions u' with u'(0, y) E.W if and only if there exists a function g £ W solving (6)
with 0(0) = ff,(0).
For the trivial case K = X = 1 we see that g(y) = (e~v/2)[[/1(0, j/) + £/2(0, ?/)].
Denote the transform of a function by a superscript *. All the formulas used below
may be found in [4].
From (4) we have U'*(0, s) = g*i(\/s)/Vs. Transforming (6) and solving for g*,
we obtain
g*(s) = D^iKvg^s + 1)1/2) + g%((s + l)1/2)j, (7)
where D*(s) = [((s + 1)I/2 + 1) + K((ns + 1)1/2 — 1)]_1 and p = 1/X. Except for the
easy case K = X = 1 we can write
n*M _ «' + D1/2 + 1) ~ Mgg + 1)'/2 + 1) {Ki
U [S) ~ ((« + 1)1/2 + l)(A(s + 1)1/2 + 5) ' W
where A = 1 — KV and B = 1 — 2K + K2^. Given specific initial conditions gt , g% ,
one would now simplify (7) using (8) prior to inversion. In general we get the convolution
g(y) = ^ Jo dt Jo rVte~T'' D(y - t)[e"g2(2rt) + e~ugx{2^\ rt)] dr, (9)
where
D(s)" TTKvi + " G<»>1 + 2C^ri) [ IW - o« ]«» - » a,
m = [e"x« - e'"]/(W)U2, F(y) = (e"7W"!) - erfc Vj/, (10)
G(t/) = 0 K2 = \
= (e"V(x2/)1/2) - Q exp ((Q2 - l)y) erfc K2 * Q = B/A.
When if/x = 1 we have Q = 1 and F = G so the correct formula for Z) is obtained by
computing lim^i [F(y) - G(y)]/(K» - 1).
From Eq. (9) it follows that if the initial conditions g! , g2 belong to the class of
functions W defined in Lemma 3 and have Laplace transforms, the same is true for g.
The function g(y) in (9) then solves Eq. (6) and provides solutions, via (5), to Problem A.
Uniqueness of such solutions is clear from Eq. (7). Thus we have
Theorem: For initial conditions £,(x) in class W with \gi(x)\ < M, exp (rriiX) for
large x, Problem A has a unique solution with interface function satisfying the same
conditions. The solution is given by (5) for g(y) in (9).
344 K. A. HEIMES
The formulas in (5) and (9) are less than elegant. However, for any particular problem
one can usually perform some additional simplifications, transform (5) back to problem
M or S using (1) or (2) and study the solutions numerically. Approximations are also
easily accessible. For example, one can choose a simple approximation for g(y) in (9),
select one of the initial conditions g! or g2 and solve for the remaining function (g2 or gt)
using (7). Then Eq. (5) gives solutions to problem A except that one of the initial condi-
tions differs from the original. If the difference is small, an approximate solution is
obtained.
A study of problem M for specific initial conditions using the above procedure is
given in [5].
References
[1] W. A. Tiller and R. F. Sekerka, Redistribution of solute during phase transformation, J. Appl. Phys.
33, 2726 (1964)
[2] V. G. Smith, W. A. Tiller and J. W. Rutter, A mathematical analysis of solute redistribution during
solidification, Canad. J. Phys. 33, 723 (1955)
[3] G. Hellwig, Partial differential equations, Blaisdell, 1964
[4] G. E. Roberts and H. Kaufman, Table of Laplace transforms, W. B. Saunders, 1966
[5] J. D. Verhoeven and K. A. Heimes, The initial transient in melting and solidification experiments.
Int. J. Crystal Growth (to appear)