A solution to the Smoluchowski-problem
May 10, 2019
The concentration n(r, t) of particles is given by the solution of the dif- fusion equation
∂
∂tn(r, t) =D∆n(r, t). (1)
with the boundary conditions n(a, t) = 0, n(r → ∞, u) =c, where c is the initial concentration outside of the sink. The initial condition is n(r,0) = cθ(r−a), which, in the case of sperical symmetry reads
∂
∂tn(r, t) = D1 r2
∂
∂r
"
r2∂n(r, t)
∂r
#
(2) Applying the Laplace-transform in time, ψ(r, u) = R0∞ψ(r, t)e−utdt, we get:
uψ(r, u)−cθ(r−a) = D1 r2
∂
∂r r2∂n(r, u)
∂r
!
. (3)
The initial condition is absorbed into the equation under the Laplace trans- form. The boundary conditions to this equation are given by the the Laplace- transform of the ones for Eq.(2):
n(a, u) = 0, n(r → ∞, u) = c u.
To solve this equation we change the dependent variablen→ψwithn(r, u) = ψ(r, u)/r:
1
ruψ−cθ(r−a) = D1 r2
∂
∂r r∂ψ
∂r −ψ
!
= D1 r2
∂ψ
∂r +r∂2ψ
∂r2 − ∂ψ
∂r
!
= 1
rD∂2ψ
∂r2 1
which is then an inhomogeneous linear equation with constant coefficients:
D∂2ψ
∂r2 −uψ =−crθ(r−a).
The general solution of the homogeneous equation is ψ1(r, u) =Ae
√
u/Dr+Be−
√
u/Dr
and aparticular solution of the inhomogeneous equation can be taken very simple, the one with ψ(r, u)∝r: it is easy to see that
ψ2(r, u) = c ur
solves the equation for all r > asince its second derivative vanishes. There- fore
ψ(r, u) =Ae
√
u/Dr+Be−
√
u/Dr+ c ur.
To fit the boundary condition at infinity we have to takeA= 0. Then, to fit the boudary condition at a we have to take B such that
Be−
√
u/Da+ c ua= 0, i.e.
B =−c uae
√
u/Da.
Therefore the full solution in the Laplace domain is ψ(r, u) = c
ur− c uae−
√u/D(r−a)
= c ur
1−a re−
√u/D(r−a)
.
Now we return back to
n(r, u) = c
1 u − 1
u a re−
√u/D(r−a)
and perform the inverse Laplace transform. The inverse transforms of both terms are known; you can look them up in tables:
L−1
1 u
= 1, L−1
1 ue−C
√u
= erfc C 2√
t
!
(C > 0), and get
2
n(r, t) =c
"
1− a
rerfc r−a 2√
Dt
!#
.
Note that for t→ ∞ the solution tends to a stationary one, n(r, u) =c
1−a r
,
i.e. to the solution of the three-dimensional Laplace equation fulfilling the boundary conditions.
The equation in the box gives us the flux onto the reacting surface J =−4πa2 ∂
∂rn(r, t)
r=a
= 4πDca 1 + a 2√
Dt
!
which after a transient tends to a constant, giving the famous result kdif f = J
c = 4πDa
for the reaction rate constant (Smoluchowski, 1917), albeit very slow.
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