1
Facilitated Diffusion - Corner Layer Analysis
The equations are
ǫ1 σxx = σ(1 − u) − u = −ǫ2 uxx
(1)
with boundary conditions
σ(0) = σ0 ,
u′ (0) = u′(1) = 0.
σ(1) = σ1 ,
(2)
There is a conservation law
ǫ1 σxx + ǫ2 uxx = 0,
(3)
ǫ1 σx + ǫ2 ux = ǫ1 J,
(4)
ǫ1 σ + ǫ2 u = ǫ1 Jx + ǫ1 σ0 + ǫ2 u(0).
(5)
so that
and
Thus we solve for σ and substitute into (1) to find
ǫ2 uxx = u − (Jx + σ0 + γu(0) − γu)(1 − u),
(6)
where γ = ǫǫ12 .
The outer solution is straightforward. We set ǫ1 = ǫ2 = 0 in (1) and observe that
σ(1 − u) − u = 0,
so that
(7)
σ=
u
,
1−u
(8)
u=
σ
,
1+σ
(9)
or
from which we determine
u(0) =
σ0
,
1 + σ0
u(1) =
σ1
1 + σ1
(10)
so that
ǫ1 J = ǫ1 σ1 + ǫ2 u(1) − ǫ1 σ0 − ǫ2 u(0) = ǫ1 (σ1 − σ0 ) + ǫ2 (
σ0
σ1
−
).
1 + σ1 1 + σ0
(11)
We can also determine the outer solution as a solution of the quadratic equation
u − (Jx + σ0 + γu(0) − γu)(1 − u) = 0.
(12)
It follows that the derivative of u satisfies
(1 + γ(1 − u) + (Jx + σ0 + γu(0) − γu))u′(x) − J(1 − u) = 0,
so that at x = 0
u′ (0) =
1
J(1 − u(0))
= 2 J(1 − u(0)),
(1 + γ(1 − u(0)) + σ0 )
α
1
(13)
(14)
where α2 = (1 + σ0 + γ(1 − U0 )), which is not zero.
We need a corner layer expansion in order to satisfy the boundary conditions for u. For
x
to find
the corner layer near x = 0, we set y = 1/2
ǫ2
1/2
uyy = u − (ǫ2 Jy + σ0 + γu(0) − γu)(1 − u).
(15)
We expand u = U(y) as
1/2
U(y) = U0 (y) + ǫ2 U1 (y) + · · · ,
(16)
and the resulting hierarchy of equations is
U0′′ = U0 − (σ0 + γU0 (0) − γU0 )(1 − U0 ),
(17)
U1′′ = U1 − (Jy + γU1 (0) − γU1 )(1 − U0 ) + (σ0 + γU0 (0) − γU0 )U1 .
(18)
The first of these we solve with U0′ (0) = 0, and the only solution is the constant solution
U0 = U(0) = u0 . The second order equation can be written as
U1′′ − α2 U1 = −(Jy + γU1 (0))(1 − U0 ).
The solution is
U1 (y) = A exp(−αy) +
For consistency,
A = U1 (0) −
1
(Jy + γU1 (0))(1 − U0 ).
α2
1
γU1 (0)(1 − U0 ),
α2
(19)
(20)
(21)
and U ′ (0) = 0 implies that
U1′ (0) = −αA +
so that
A=
1
J(1 − U0 ),
α3
1
J(1 − U0 ) = 0
α2
U1 (0) = α
(α2
1 − U0
.
− γ(1 − U0 )
(22)
(23)
Now let’s see if these match. (They had better since there are no free parameters to
1/4
x
z
adjust.) We introduce the intermediate variable z = 1/4
, so that x = ǫ2 z and y = 1/4
. We
ǫ2
ǫ2
must examine
1/4
lim
ǫ2 →0
u0(x) − U(y)
1/4
ǫ2
u0 (ǫ2 z) − U(
=
lim
z
1/4 )
ǫ2
1/4
ǫ2 →0
ǫ2
1/4
1/2
u′0 (0)(ǫ2 z) − ǫ2 A exp(−α
=
lim
ǫ2 →0
1/4
=
lim
1/4
1
zǫ2 J(1
α2
1/4
ǫ2
u′0 (0)(ǫ2 z) −
ǫ2 →0
= z(u′0 (0) −
1
J(1 − u0 )) = 0,
α2
2
1/2 1
z
z
(J 1/4
1/4 ) − ǫ2
α2
ǫ2
ǫ2
1/4
ǫ2
− u0 )
+ γU1 (0))(1 − U0 )
as hoped. The end result of this is that the composite solution is given by
x
1/2
u(x) = u0 (x) + ǫ2 A exp( ),
ǫ
(24)
which takes care of the corner layer at x = 0. A similar analysis implies a similar corner
layer at x = 1, yielding a solution of the form
u(x) = u0 (x) +
′
1/2 u0 (0)
ǫ2
α
exp(−α
x
1/2
ǫ2
3
)+
′
1/2 u0 (1)
ǫ2
β
exp(β
1−x
1/2
ǫ2
) + ···,
(25)