3.185 - Recitation Notes
September 11/12, 2003
Topics covered
• Transient state diffusion
⇒ Error function solution
⇒ Point source solution
________________________________________________________________________
Can you think of any real or hypothetical physical example?
Error function solution: Unlimited supply on one boundary in a semi-infinite domain
Point source solution: Depleting supply from a point source in an infinite domain
Fourier series solution: General solution for finite domain
One-dimensional step function initial condition (Error function solution)
Example
⎧c − ∞ < x < 0
c( x, t = 0) = ⎨ L
⎩c R 0 < x < ∞
c(−∞, t ) = c L
c ( ∞, t ) = c R
We want to know the concentration profile at any time t;
i.e., we have to solve the diffusion equation
∂c
∂ 2c
=D 2
∂t
∂x
x
(note: η is a dimensionless variable)
Change of variable: let η =
4 Dt
Applying the chain rule
∂c ∂η ∂c − η ∂
=
=
, and
∂t
∂t ∂η
2t ∂η
∂c ∂η ∂c
=
∂x ∂x ∂η
∂ 2 c ∂ ⎛ ∂c ⎞ ∂η ∂ ⎛ ∂η ∂c ⎞
⎜
⎟
= ⎜ ⎟=
∂x 2 ∂x ⎝ ∂x ⎠ ∂x ∂η ⎜⎝ ∂x ∂η ⎟⎠
∂ 2 c ∂η ∂ ⎛ ∂η ∂c ⎞ ⎛ ∂η ⎞ ∂ 2 c
⎜
⎟=⎜
=
⎟
∂x 2 ∂x ∂η ⎜⎝ ∂x ∂η ⎟⎠ ⎝ ∂x ⎠ ∂η 2
1 ∂ 2c
∂ 2c
=
∂x 2 4 Dt ∂η 2
2
Substituting back to the diffusion equation
dc d 2 c
− 2η
=
dη dη 2
Notice that c now depends only on η; therefore, the diffusion equation has
transformed from a partial differential equation into an ordinary differential equation,
which can be solved by another change of variable, for example
∂c
q=
∂η
After two integrations, we get the concentration profile in terms of η as
η
c(η ) − c(η = η o ) = a1 ∫ e −ζ dζ
2
ηo
Applying boundary conditions
c(η = −∞) = cL
η
c(η ) − c(−∞) = a1 ∫ e −ζ dζ
2
−∞
2 η −ζ 2 ⎞
⎛ 2 0 −ζ 2
c(η ) = cL + a2 ⎜
e dζ +
e dζ ⎟
∫
π ∫0
⎝ π −∞
⎠
c(η ) = cL + a2 (erf (∞) + erf (η ) )
Using the second boundary conditions
c(η = ∞) = cR
c(η = ∞) = cR = cL + a2 (1 + erf (∞) )
c −c
a2 = R L
2
The concentration profile for the 1-D step function initial condition is
c + cR cR − c L
⎛ x ⎞
c ( x, t ) = L
+
erf ⎜
⎟
2
2
⎝ 4 Dt ⎠
Note:
•
•
Since the concentration is a function of the dimensionless variable η =
x
4 Dt
concentration profile is identical if the parameters and the initial & boundary
conditions are scaled by
x′ = λx and t ′ = λ2t
The diffusion depth or penetration distance for the error function solution is
∆c
. This gives a
roughly estimated to be at a location where c( x, t ) − c± ∞ =
8
, the
penetration distance at any time t to be x ≈ 1.6 Dt , approximately x ≈ 2 Dt .
The time to reach steady state can be estimated by setting the penetration distance
to be larger than the largest characteristic length (L) over which the diffusion
L2
takes place. Hence, steady state might be achieved when t >> . Steady-state
D
conditions cannot be achieved when the boundary conditions are time-dependent
or the system is infinite or semi-infinite. However, under certain circumstances, a
transient system can be assumed to be at a quasi steady-state
Point source solution
Several error function solutions can be “superpositioned” together to obtain some
“funny-looking” concentration profiles (a.k.a. superposition method).
For example, if we merge the two following step functions
⎧0 −∞ < x < 0
c( x, t = 0) = ⎨
⎩co 0 < x < ∞
⎧ 0 − ∞ < x < ∆x
c( x, t = 0) = ⎨
⎩− co ∆x < x < ∞
we get a point source of width ∆x (somewhat similar to a delta function).
The solution for each step function is
c c
⎛ x ⎞
c( x, t ) = o + o erf ⎜
⎟
2 2
⎝ 4 Dt ⎠
c ( x, t ) = −
Combining the two solutions give
c c 2
c ( x, t ) = o + o
2 2 π ∫0
co co
⎛ x − ∆x ⎞
− erf ⎜
⎟
2 2
⎝ 4 Dt ⎠
x
4 Dt
c ( x, t ) =
e − ζ dζ −
2
co
π ∫
co co 2
−
2 2 π
x
4 Dt − ζ
x − ∆x
e
2
∫
x − ∆x
0
4 Dt
e − ζ dζ
2
dζ
4 Dt
For ∆x << x, the solution for the transient state concentration profile of a point source
is
x2
c ∆x − 4 Dt
e
c ( x, t ) = o
4πDt
Note:
• The penetration depth for diffusion from a point source is also roughly equals to
L2
x ≈ 2 Dt . Similarly, steady-state may be expected when t >> .
D
• The solution is symmetric about x = 0.
• The concentration gradient is zero at x = 0, i.e. no flux @ x = 0, hence the two
sides of the solution are independent
• In semi-infinite case, the solution has the same general form but concentration is
x2
•
N − 4 Dt
e
twice as large, c( x, t ) =
πDt
The solution is identical for point source diffusion in a line, line source diffusion
into a plane and planar source diffusion into a volume.
•
The solution is similar for 1-D, 2-D and 3-D diffusion of a point source
x2
−
N
e 4 Dt
4πDt
Point source in 1-D
c ( x, t ) =
Point source in 2-D
N − 4 Dt
e
c ( x, t ) =
4πDt
r2
r2
•
−
N
Point source in 3-D
e 4 Dt
c ( x, t ) =
3/ 2
8(πDt )
The point source solution can be used to find the concentration profile from
complicated initial conditions by the Green’s function method. The point source
solution is integrated over the domain of the initial condition:
c ( x, t ) = ∫
b
a
co dξ −
e
4πDt
(ξ − x ) 2
4 Dt