88
Since η is not a function of time, the time derivative is:
∂
θ
‘
∂
τ
=∂
∂
τ
f
η
⎛
⎝
⎜⎞
⎠
⎟=1
η
∂f
∂
τ
(6.17.4)
6.18. (a) The conservation relation for unsteady, one-dimensional diffusion without chemical
reaction in a rectangular coordinate system is given by Equation (6.8.30). The boundary conditions
are:
boundary condition at η = 0 is not homogeneous. A homogeneous problem can be developed by
noting that at long times, a steady state concentration gradient exists across the membrane, i.e.,
θ(η,τ→∞)=ψ(η). In order for this to be the case, the time dependent solution must consist of two
parts, one ψ(η) which represents the steady state solution and a second term φ(η,τ) which represents
the transient change in concentration. Thus,
η = 0
τ ≥ 0
φ = 0
89
d2ψ
dη2=0
φ = Ansin(nπη)exp(−n2π2τ)
n=1
∞
∑
90
(c) The following MATLAB program was used to generate the concentration profile.
91
∂Cm
∂xx=0
=Co
L
∂θ
∂η η=0
=Co
L-1– 2 cos(nπη)exp(−n2π2τ)
n=1
∞
∑
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
η=0
=−Co
L1 + 2 (−1)nexp(−n2π2τ)
n=1
∞
∑
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
The change in C2 with time is simply:
dC2
dt
=CoADm
LV 1+2 (−1)nexp −n2π2tL2
Dm
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
n =1
∞
∑
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
Thus, after a dimensionless time interval of 0.5, the exponential terms decline to 1% or less of 1 and
the rate of loss from compartment 1 equals the rate of gain in compartment 2. Further, the change in
concentration in either reservoir is proportional to LV/ADm. Thus, the quasi-steady state
approximation is valid for times longer than 0.5L2/Dm. Note that a more accurate answer can be
determined by solving for the time-dependent boundary conditions.
6.19. For a cell of radius 15 µm, there are 3.54 x 109 receptors cm-2. The distance between receptors
is 1/(3.54 x 109 receptors cm-2)1/2 = 168 nm. From Equation (6.9.15), the fractional reduction in the
rate constant for a surface containing this density of receptors is:
6.20. For diffusion-limited dissociation of a ligand from a receptor, Equation (6.9.17) applies subject
to the boundary conditions (6.9.22a,b). Integrating Equation (6.9.17) yields:
C=A ln r +B
Co=A lns +B
92
Nrr=s=−DL
dC
dr r=s
=
DLCo−Cb
( )
sln b/s
( )
6.21. Begin with the definition of the error function:
erfc
η
( )
=1−erf
η
( )
=1−2
π
e−x2
dx
η
∫
(S6.21.1)
6.22. For a three-dimensional random walk, the mean square displacement is related to time and the
diffusion coefficient by Equation (6.5.10).
〈 〉 =r D t
ij
26 .
6.23. (a) Beginning with Equation (6.8.22) for the flux of solute at the surface with concentration C1
diffusing into a semi-infinite medium initially at a concentration C0.
Since, there is no drug initially in the tissue, C1 = 0. The total amount of drug entering the tissue by
time t, M, is the integral of the flux times surface area over time.
93
M=Nix (x=0)
t
∫Adt =Dij
π
tC0
t
∫Adt =2Dijt
π
C0A
Dij =
π
4t
M
C0A
⎛
⎝
⎜⎞
⎠
⎟
2
6.24. This is a case of unsteady diffusion in finite dimensions. C1 = 0, C0 = 0.1 mg/ml. We want to
find the time at which C/C0 = 0.05. For the results in Figure 6.20, this corresponds to (C- C0)/(C1 –
Programming this equation in MATLAB (or a spreadsheet) yields q = 0.9503 at t = 0.375. Revising
the calculation yields 347.2 days.
6.25. Taking the logarithm of both sides of Equation (6.8.69):
ln Ci/M0
( )
=−ln 2
π
Dijt
( )
1/ 2
−x2
4Dijt
6.26. (a) At steady state, the exponential term in the equation given in the problem statement is zero
and this expression reduces to: C = ΦC0(1-y/l)
The flux at y = L is
Ny=l=−DdC
dy y=l
=DΦC0
L
94
The release rate R is:
R=ANy=l=ADΦC0
L
in Section 6.7. Using Equation (6.7.17b) for the effective permeability:
6.28. (a)
D
r2
d
dr
r2dC
dr
⎛
⎝
⎜⎞
⎠
⎟=0
Solving for A:
A=C0R0Ri
Ri−R0
( )
6.29. (a) This problem is identical to the one discussed in Section 6.8.1, except that C0 = 0. Thus,
the flux is given by Equation (6.8.22)
96
Solution to Problems in Chapter 7, Section 7.11
7.1. The effective diffusion coefficient is given by Equation (7.5.23) and the diffusion potential is
given by Equation (7.5.27).
7.2. (a) Using the definition of the permeability and Equation (7.11.1)
(c) For the data given, PeL = 10. The permeability from part a for CL = 0 is: Pi = DijPeL/L= 1 x 10-6
cms-1. The diffusive permeability is Pi = DijPeL = 1 x 10-7 cm s-1. Thus, convection leads to a
7.3. Assuming steady state, no convection, no chemical reactions, and one-dimensional transport,
the material balance is:
97
98
Solving for A yields:
RT
⎝
⎠
RT
⎝
⎠
which agrees with Equation (6.7.9).
To obtain the flux, first calculate the derivative of Ci with respect to z:
RT
⎝
⎠
⎣
⎢
⎦
⎥
99
In the limit as ziFVm/RT approaches zero, the flux becomes
RT →0 −ΦiziDij FVm
RT −1
⎣
⎢
⎦
⎥
⎩
⎪
⎭
⎪
Typically, the derivation performed in the electrophysiology literature takes the origin of the
coordinate system at the intracellular surface and the flux and ion currents are have a sign that is
opposite to that noted above.
7.4. Based on concentration differences, potassium is transported from the cell and sodium and
chlorine are transport into the cell. If potassium transport is associated with the outward current,
then the outward current is FNK+z. In the limit as Vm goes to zero, the outward current becomes:
iK+z=FNK+z=FP
K+CK+o−CK+L
( )
Rearranging yields the following expression for the permeability:
PK+=
iK+z
F CK+o−CK+L
( )
7.5. To use the data provided, equation (7.4.32) is rearranged by dividing each permeability by PK+.
7.6. If we assume that sodium, potassium and chlorine are the major small molecule ions within the
cell, then proteins must balance the excess positive charge from cations. To satisfy electroneutrality,
Equation (7.4.5), the following relation must hold
CK+ + CNa+ – CCl– – |zp|Cp = 0