The concentration in the membrane satisfies the equation
with initial conditions
C(x,0) = F(x)
and boundary conditions
C(0,t) = C0
C(L,t) = C1
The solution can be written as its Steady-State, Css(x), plus a Transient C'(x,t) where the steady-state is
Css(x) = C0 + (C1 - C0)x / L
and the transient satisfies
with initial condition
(NOTE: the prime does NOT denote derivative, instead, it denotes the perturbation from the steady state) and boundary conditions
Solution: The solution is given by
where
Approach to Steady State: As t increases, each mode decays proportional to . For larger n, the decay is faster. Thus, the n = 1 mode decays the slowest. Even so, afer t = L2 / 2D, this mode will be 1/e of its original value. (Jon, can you supply values (L,D) here for an example?)