Here uy
¹
,t ¼ lim
x ↑ y
ux,t denotes the limit in y from the left and uy
þ
,t ¼ lim
x ↓ y
ux,t the limit from the right. But cstˆ,tˆ ¼ c
and thus by eqns 3 and 20 u
s ˆt
¹
, ˆt
u
S
c
p
26 a contradiction. In other words, eqn 24 implies over-
saturation for u. But this is not allowed under equilibrium conditions. We further note that if v is discontinuous at the
dissolution front, i.e. vst
þ
,t . 0 then s˙t ¼ q cannot occur. This is a direct consequence of eqn 22. This obser-
vation implies that stˆ ¼ qtˆ and s˙tˆ ¼ q for some tˆ . 0 can also not occur as vstˆ
þ
,tˆ ¼ v . 0. Hence
s t
, qt for all t . 0 27
The ordering of the fronts and eqns 20 and 21 imply u
x ,
t ¼
u
S
c
p
¹ ` , x , s
t u
S
c
p
s t
, x , qt u
S
c
p
qt , x , þ ` 8
: 28
Consequently by eqn 22 0 ˙s
t q for all t . 0
29 and s˙tˆ , q occurs at points tˆ where v is discontinuous. In
paxticular this shows that all dissolution fronts are mono- tone in time. Since v vanishes in the region x , st, we
have there
]u ]t
þ q
]u ]x
¼ 30
The initial condition on u for x , 0, the upperbound in eqns 29 and 30 imply that u ¼ u
for x , st. To determine v in the region x . st we use eqns 15 and 28. Combined,
they imply that ]vx,t]t ¼ 0 for x . st, x Þ qt. Then using the initial condition on v for x . 0 and the lower
bound on s˙, we find after integration v ¼ v , for x . st.
Thus we have constructed a piecewise constant solution of eqns 3, 15 and 20 which satisfies the initial distribution
eqn 7. The dissolution front follows from the Rankine– Hugoniot condition eqn 22: st ¼ at, with
a ¼ u
S
c
p
¹ u
p
u
S
c
p
¹ u
p
þ v
p
q , q
31 Across the other shock, x ¼ qt, v is constant. This is con-
sistent with eqn 22. In the chemical engineering literature this front is called the salinity front, see e.g. Bryant et al.
1
. In Fig. 2 we show the level set of the solution {u, v, w, c} at
equilibrium. The separating curves are shock curves at x ¼ at and x ¼ qt. Fig. 3 shows a sketch of the profiles of the vaxi-
ables for some t . 0. A qualitative comparison with the com- putations of Willis and Rubin
15
will be given in Section 5. One may raise the question if the solution as constructed
in this section is the unique solution of the initial value problem. For the following reasons we believe that it is.
In the construction, inequalities eqn 29 are crucial. They imply directly that the concentrations u and v are
constant to the left and the right of a dissolution front, lead- ing to the constant speed eqn 31. In eqn 29 the inequal-
ities are a consequence of the Rankine–Hugoniot condition and the fact that oversaturation is ruled out by requiring w , 1
in eqn 3. In other words, eqn 29 exhibits local properties of any dissolution front. As outlined above they lead to a
piecewise constant solution as presented in this section.
3 NON-EQUILIBRIUM
When precipitation–dissolution reactions cannot assumed to be at equilibrium, one needs to incorporate the first-
order reaction eqn 2 in the description. This leads to a much more involved analysis. In this section we construct
solutions of the Riemann problem eqns 2–4, 7 and 15 for two distinct cases. In Section 3.1 we assume eqn
10 to be satisfied, implying that crystalline solid is present only in part of the flow domain, and in Section 3.2 we
assume v
, v . 0. In this first case a dissolution front
exists for all t 0. In the second case it may appear in finite time.
3.1 Crystalline solid partly present
Inspired by the equilibrium results, i.e. ‘k ¼ `’, we start with the assumption that a dissolution front exists, as in eqn
Fig. 2. Level sets of concentrations at equilibrium.
Fig. 3. Sketch of profiles at some t . 0.
4 C.J. van Duijn et al.
23, which satisfies inequalities eqn 29. These inequal- ities are crucial for the construction of a solution. Unfortu-
nately there are no obvious physical or mathematical axguments to support these assumptions. In contrast, the
weaker statement st qt for all t 0, which is obviously physical, can be justified similarly as in Section 2. We return
to the possibility of existence of solutions not fulfilling these assumptions when discussing the uniqueness at the end of
Section 4. The main goal is to derive an equation for the location x ¼ st of the dissolution front.
As in
Section 2
we conclude,
because of
]v=]t ¼
0 for x , s t
, that u ¼ const ¼ u for x , st
and that w ¼ gu , c
K there. Similarly for x . qt we have u ¼ const ¼ u
¼ u
S
c and thus v ¼ const ¼ v
. With reference to Fig. 4, we are going to consider the fol-
lowing problem: Find u, v and s such that
]u ]t
þ q
]u ]x
¼ k{K ¹ g
u; c
p
} 32
]v ]t
¼ k{g
u; c
p
¹ K}
33 for st , x , qt and for t . 0, subject to
u s
t ,
t ¼
u
p
, v
s t
, t
¼ 34
and v
qt ,
t ¼
v
p
35 Note that in the composite solution the crystalline concen-
tration v is continuous across x ¼ qt, due to eqn 22 and then eqn 35 holds, while the fluid concentration u possibly
has a discontinuity there. Eqn 33 implies continuity of v across the dissolution front x ¼ st, which in turn, due to
eqn 22, implies continuity of u across x ¼ st.
We solve eqns 32 and 33 by the method of character- istics. Choose any point y,t in the domain {x,t:st , x ,
qt, t . 0}, see also Fig. 4. The characteristics of eqn 32 are straight lines in the x,t-plane, having slope q with respect
to the t-axis. The characteristic passing through the point y,t, i.e. the curve x ¼ y þ qt ¹ t, intersects the dissolu-
tion front in the point st
, t , which satisfies
s t
¼ y þ q
t ¹ t
36 Due to s˙t q this point is unique and thus is the same for
all starting points y,t satisfying eqn 36. For a given dissolution front st, this would determine t
as a function of y and t, i.e. t
¼ t
y,t. Integrating eqn 32 along the characteristic and using
eqn 36 yields Z
u y
, t
u
p
1 k{K ¹ g
z; c
p
} dz ¼ t ¹ t
y ,
t ¼
y ¹ s t
y ,
t q
37 The idea is now to use eqn 33 and the boundary condi-
tions on v to determine the location of the dissolution front, i.e. to find the function st. Before we proceed we first
introduce for the case u , u
S
c the function
ƒ :[0, þ `
→ [u ,u
S
c defined implicitly by the integral
Z
ƒ d
u
p
1 k{K ¹ g
z; c
p
} dz ¼
d q
for all d 0 38
and the function F: [0, þ ` → 0, þ `, defined by F
d ¼
k{K ¹ g ƒ
d ; c
p
} for all d 0 39
Here d denotes a distance y ¹ st y,t, measured along the
characteristic from the dissolution front x ¼ st to a point
y,t, while ƒ
d ¼ uy,t denotes the fluid concentration at that point. The fluid concentration
ƒ and reaction rate F at
any point of a characteristic depend only on this distance d, which is due to the fact that convection is the only transport
mechanism in this single reaction problem. Note that due to the differentiability of g.;c
at u ¼ u
S
c Z
u u
p
1 k{K ¹ g
z; c
p
} dz → ` for u → u
S
c p 40
and thus ƒ
and F are well defined.
Examples. Let the rate function g be given by the law of mass action eqn 11. We can explicitly compute the
cases:
n ¼ 1, m ¼ 0 The linear case, see also Ref.
7
, eqn 40. Then gu;c ¼ u and u
S
c ¼ K, independent of c. We find ƒ
d ¼
K ¹ K ¹ u
p
e
¹ k
q d
41 and
F d
¼ k
K ¹ u
p
e
¹ k
q d
42
n ¼ 1, m ¼ 1. See Fig. 5.
Then gu;c ¼ uu ¹ c and u
S
c ¼
c=2 þ 1=2
c
2
þ 4K
p We find that
ƒ satisfies
ƒ d
¹ u
u
S
¹ ƒ
d ¼
u
p
¹ u
u
S
¹ u
p
e
ad
43
Fig. 4. Dissolution front in the x,t-plane.
Analysis of crystal dissolution fronts 5
where u
S
¼ u
S
c ,
u ¼
u c
¼ c
2 ¹
1 2
c
2
þ 4K
p and a ¼
k q
u
S
¹ u
Consequently F
d ¼
k u
S
¹ u
2
u p ¹ u u
S
¹ up
e
¹ ad
e
¹ ad
þ u p ¹ u
u
S
¹ up
2
44
When u ¼
u
S
c , which we consider as a degenerate case,
we extend the definitions eqns 38 and 39 by setting f
d ¼
u
S
c
p
and F d
¼ 0 for all d 0
45 Unless stated otherwise we avoid this degeneracy by taking
u , u
S
c . This implies Fd . 0 and F9d , 0 for all
d 0. Next we continue the analysis of eqns 32 and 33, by
rewriting eqn 37. Take any t . 0 and let y ¼ st. Using eqns 38 and 39 we now write
k{K ¹ g u
s t
, t
; c
p
} ¼ F s
t ¹
s t
46 for any stq , t , t. Here we have used s˙t 0. In this
expression, t ¼
t st,t satisfies t
¼ t when t ¼ t. Sub-
stituting eqn 46 into eqn 33, integrating the result in time and applying the v-boundary condition in eqns 34
and 35, yields Z
t s
t q
F s
t ¹
s t
dt ¼ v
p
47 Note that in deriving this equation we in particular assumed
s to be monotone, but not to be strictly monotone. In the derivation of eqn 47 we can allow for constant parts of s,
i.e. for times 0 t , t
2
such that st ¼ st
1
for all t
1
t t
2
. In such a case the definition of t ¼
t st,t gives for t
such that t
1
t t
2
: t
s t
, t
¼ t for t [
[ t
1
, t
] 48
and the whole derivation of eqn 47 holds true with the following exception: if st ¼ 0 for 0 t t
2
, then the integration leading to eqn 47 cannot be performed for t [
[0,t
2
] as v0,0 is not defined. But on the other hand a dissolution front as sketched in Fig. 4, i.e. s0 ¼ 0 and
st . 0 for t . 0, would lead to a contradiction in eqn 47. Letting t ↓ 0 would make the left hand side zero while v
. 0 as given. Therefore, we have a waiting time t
. 0, see Fig. 6, such that
s t
¼ for
0 t t
p
, . 0
for t . t
p
49 From eqn 47 for t ↓ t
, we conclude t
p
F ¼
v
p
50 Using eqns 38 and 39 we find for the waiting time the
expression t
p
¼ v
p
k{K ¹ g u
p
, c
p
} 51
Note that in the degenerate case u ¼
u
S
c eqn 51
implies t ¼ `
. In other words, when u equals the solubi-
lity concentration then the dissolution front remains stag- nant. Furthermore, apart from the initial waiting time, no
further constant parts of s can occur: if this would be the case, say on the interval [t
1
,t
2
], then we conclude from eqn 47
v
p
¼ Z
t
2
s t
2
= q
F s
t
2
¹ s
t dt
52 ¼
Z
t
2
t
1
F s
t
1
¹ s
t dt þ
Z
t
1
s t
1
= q
F s
t
1
¹ s
t dt
¼ Z
t
2
t
1
F dt þ v
p
, which is a contradiction. We want to rewrite the integral in
eqn 47 in terms of t and we do this by using eqn 36.
From that equality, with y ¼ st, we obtain due to s˙t q a unique correspondence of the points t with the points t
, where
t : s
t q
→ t implies t : 0 → t
and ]t
]t ¼
1 ¹ 1
q ˙s
t 54
Fig. 5. Rate function g for n ¼ m ¼ 1. Note that u c. Fig. 6. Waiting time in dissolution front.
6 C.J. van Duijn et al.
Applying these observations to eqn 47 yields Z
t
F s
t ¹
s t
dt ¼
v
p
þ 1
q Z
t
F s
t ¹
s t
˙s t
dt 55
¼ v
p
þ 1
q Z
s t
F z
dz ,
where the left-hand side can be slightly rewritten by intro- ducing the waiting time:
Z
t
F s
t ¹
s t
dt ¼
t
p
F s
t þ
Z
t t
p
F s
t ¹
s t
dt 56
To summarize, we have obtained an integral equation from which the location of the dissolution front can be deter-
mined. The precise formulation is: Let t be given by eqn
51. Then find the function st, satisfying eqn 49 and the dissolution front equation DFE
DFE Z
t t
p
F s
t ¹
s t
dt ¼
B s
t for t t
p
where B
s t
: ¼ t
p
{F ¹
F s
t } þ
1 q
Z
s t
F z
dz: 8
: 57
The expression for B follows from eqns 50, 55 and 56. In general we have to rely on numerical methods to solve
DFE. One such method will be discussed in Section 5. Only very special cases can be solved analytically, for
instance the case n ¼ 1 and m ¼ 0 the linear case in the examples, where F is given by eqn 42. For that
form of F it is straightforward to solve DFE explicitly. The result is
s t
¼ q
1 þ kt
p
t ¹ t
p
for t t
p
58 where
t
p
¼ v
p
k K ¹ u
p
59 In Section 4 we show how to transform DFE into a
standard integral equation, from which some characteristic properties of the front can be derived. Having found an
expression or approximation for st, one has to go back to eqns 37 and 38 to determine u. The concentration of
the crystalline solid is obtained from integrating eqn 33.
3.2 Solid present everywhere
