© meg/aol ‘02 solutions to the linear diffusion equation martin eden glicksman afina lupulescu...
DESCRIPTION
© meg/aol ‘02 Transform Methods where Kernel of the transform Laplace kernel Independent variable (time). Laplace transform converts an object function, F(t), to its image function, F(p). ~TRANSCRIPT
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Solutions To The Linear Diffusion Equation
Martin Eden GlicksmanAfina Lupulescu
Rensselaer Polytechnic InstituteTroy, NY, 12180
USA
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Outline• Transform methods
• Linear diffusion into semi-infinite medium– Boundary conditions
– Laplace transforms
• Behavior of the concentration field
• Instantaneous planar diffusion source in an infinite medium
• Conservation of mass for a planar source– Error function and its complement
– Estimation of erf(x) and erfc(x)
• Thin-film configuration
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Transform Methods
whereK t, p Kernel of the transform
e pt Laplace kernel
t Independent variable (time).
L F t F t K t, p d t
a
b
˜ F p
L F t Laplace transform converts an object function, F(t), to its image function, F(p). ~
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Linear diffusion into a semi-infinite
medium
time
Initial state
Final state
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
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Boundary Conditions
1) Initial state:C = 0, for x > 0, t = 0.
2) Left-hand boundary: At x = 0, C0 is maintained for all t > 0.
Ct
D 2Cx2
• The diffusion equation is a 2nd-order PDE and requires two boundary or initial conditions to obtain a unique solution.
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Laplace Transform of the Diffusion Equation
object functionC x, t
e ptLaplace kernel
e pt
0
2 Cx2 dt
2
x2 C e pt
0
dt d2 ˜ C d x2
˜ C x, p C x,t e pt d t0
2 Cx 2
Linear Diffusion Equation
˜ C image function
Laplace transformof C(x,t)
˜ C x, p
1D
Ct
0
e pt
0
2 Cx2 dt
Laplace transform ofthe spatial derivative.
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Laplace Transforms
The Laplace transform of Fick’s second law when C(x,0)=0
d2 ˜ C d x2
pD
˜ C 0
1D
e pt
0
Ct
dt
1D
Ce pt0
p Ce pt
0
dt
pD
˜ C
integration by parts
Ce pt0
0 C t 0 0 boundary
condition
1D
e pt
0
Ct
dt
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Laplace Transforms
˜ C 0 C0e pt
0
dt C0
pe pt
0
˜ C 0 C0
p
Transform of the boundary condition:
˜ C C0
pe
pD
xGeneral transform solution:
˜ C x C0
pe
pD
x
The particular transform solution for the image function arises from the negative root, because the positive root leads to non-physical behavior, . ˜ C (x)
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Laplace Transforms
C x, t C0 erfc x2 Dt
The concentration field associated with the image field is found by inverting the transform either by formal means, a look-up table,or using a computer-based mathematics package.
erfc z 1 erf z 1 2
e 2
d0
z
The error function, erf (z) , and its complement, erfc (z) , are defined
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Estimation of the Error Function
erf(z) erf( z) 2
z z3
31 !
z5
52 !
z7
73 ! ...
, ( z 1)
• For small arguments:
erf(z) 1 e z 2
z 1
12z2 ...
, (z )
• For very large arguments:
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Estimation of the Error Function
erf(z)
1.12838z, (0 z 0.15) 0.0198 z 1.2911 0.4262z , (0.15 z 1.5)0.8814 0.0584z, (1.5 z 2)1, (2 z)
• Piecewise approximations for restricted ranges of the argument:
erf z 1 10.278393z 0.230389z2
0.000972z3 0.078108z4
4
z 5 10 5
• Rational approximation for positive arguments, z > 0:
Useful for spreadsheet calculations.
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Error Function
-1
-0.5
0
0.5
1
-3 -2 -1 0 1 2 3
Err
or F
unct
ion,
erf
(z)
zz
Erf(
z)
Antisymmetric: erf(z)=-erf(-z)
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Complementary Error Function
0
0.5
1
1.5
2
-3 -2 -1 0 1 2 3
erfc
(z)
zz
Erfc
(z)
Non-antisymmetric: erfc(z) -erfc(-z)
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Linear diffusion into a semi-infinite
medium
time
Initial state
Final state
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
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Concentration versus the similarity variable
0
0.2
0.4
0.6
0.8
1
0 0.5 1 1.5 2 2.5 3
Rel
ativ
e co
ncen
tratio
n, C
/C0
Space-time similarity variable, x/2(Dt)1/2
C x,t C0 erfc x2 Dt
Rel
ativ
e C
onc.
C/C
0
Similarity Variable, x/2(Dt)1/2
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Concentration field versus distance
= 2(Dt)1/2 is the “time tag”
(Note: has the units of distance!)
0
0.2
0.4
0.6
0.8
1
0 2 4 6 8 10Rela
tive
Conc
entra
tion,
C(x
)/C0
Distance, x [ in units of =1]
0.20.5
1
23
510
20=100
Rel
ativ
e C
onc.
C/C
0
Distance, x [units of =1]
time
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Diffusion Penetration X*
0
0.2
0.4
0.6
0.8
1
0 2 4 6 8 10Rela
tive
Conc
entra
tion,
C(x
)/C0
Distance, x [ in units of =1]
0.20.5
1
23
510
20=100
X* = K t1/2
Rel
ativ
e C
onc.
C(x
)/C0
Rel
ativ
e C
onc.
C/C
0
Distance, x [units of =1]
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Penetration versus square-root of time
0
0.2
0.4
0.6
0.8
1
0 1 2 3 4 5
X*(0.8)X*(0.6)X*(0.4)X*(0.9)
Pen
etra
tion
Dis
tanc
e, X
* (C/C
0)
Time tag, 2(Dt)1/2
Pen
etra
tion
Dis
tanc
e, X
*(C
/C0)
0.40.6
0.8
0.9
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Instantaneous planar diffusion source in an
infinite medium
These diffusion problems concern placing a finite amount of diffusant that spreads into the adjacent semi-infinite solid.
Initial state
Final state
Tim
e
0 x
M
t = 0
-
t
t =1
t =10
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0 x
M
t = 0
-
t
t =1
t =10
0 x
M
t = 0
-
t
t =1
t =10
0 x
M
t = 0
-
t
t =1
t =10
0 x
M
t = 0
-
t
t =1
t =10
Instantaneous planar diffusion source in an
infinite medium
These diffusion problems concern placing a finite amount of diffusant that spreads into the adjacent semi-infinite solid. Ti
me
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Instantaneous planar diffusion source
pD
˜ C x, p d2 ˜ C x, p
d x2 C x, 0
D
C x,t
dx M
Application of the Laplace transform to Fick’s second law gives:
The diffusion process is subject to the mass constraint for a unit area:
t = 0, C (x, 0), for all x 0
Initial condition
C (∞, t) = 0
Boundary condition
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Instantaneous planar source
pD
˜ C x, p d2 ˜ C x, p
d x2 0
˜ C Ae p D x Be p D x
Reduction of the Laplace transform:
The general solution for which is:
˜ C Ae p Dx , x 0, B 0
˜ C Be p D x, x 0 , A 0
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Instantaneous planar source
or e pt
0
C x,t dx dt0
M20
e ptdt
or Bp
De
pD
x
0
M2 p
and so B M
2 pD
C x,t 0
dx M2
Mass constraint for the field:
L C x,t d x
0
L M
2
Laplace transform the mass constraint:
˜ C x, p d x0
M2p
The integral constraint for the image function is:
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Instantaneous planar source solution
˜ C x, p M
2 De
xD
p
p
Laplace transforms table shows, L-1 e a p
p
1 t
e a24t
The transform solution
where a = x / (D)1/2.
L-1 ˜ C C x,t = M2 D
L-1 p 12e a p
Inverting the transform solution
C x,t = M
2 Dte
x2
4 Dt .
Diffusion solution
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0
0.5
1
1.5
2
-3 -2 -1 0 1 2 3
C(x
)/M [
leng
th]-1
x [distance]
.075
.05
Dt=0
0.025
0.25
13
C (x
) M [l
engt
h] -1
x [distance]
Normalized plot of the planar source solution
time
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Conservation of mass for a planar source
x
dx 1C x,t M
(x), thus
C x,t M
d x
1
1
e u2
du
1Gauss’s integral
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Thin-film configuration
Thin-film diffusion configuration is used in many experimental studies for determining tracer diffusion coefficients. It is mathematically similar to the instantaneous planar source solution.
C x,t = M
2 Dte
x2
4 Dt .C x, t Mthin film
Dte x 2
4 Dt
where Mthin-film represents the instantaneous thin-film source “strength.”
2Mthin-film=M
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Procedure for Analysis of Thin-Film Data
C x,t Mthin film
Dte x 2
4 DtTake logs of both sides of
ln C x, t lnMthin film
Dt
x 2
4Dt
A plot of lnC versus x2 yields a slope=-1/4Dt
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Thin-Film Experiment
0
1 105
2 105
3 105
4 105
5 105
0 10 20 30 40 50
Counts 100s
Cou
nts
in 1
00 s
Distance, [microns]
Geiger counter data after microtoning 25 slices from the thin-film specimen.
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Log Concentration versus x2
0.3
0.4
0.5
0.6
0.7
0.8
0 0.5 1 1.5 2 2.5 3 3.5 4
log e R
adio
activ
ty, l
nA*
x2, [cm2 25104]
Slope=-1/4Dt
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Exercise1. Show by formal integration of the concentration distribution, C(x,t), given by eq.(3.31), that the initial surface mass, M, redistributed by the diffusive flow is conserved at all times, t>0.
C(x,t)d x
M
2 Dte
x 2
4 Dt d x
The mass conservation integral is given by:
C(x, t)d x
2 M2 Dt
e
x2
4Dt d x0
Symmetry of diffusion flow allows:
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Exercise
Introduce the variable substitution u= x /2 (Dt )1/2 and obtain:
C(x,t)d x
MDt
e u2
2 Dt d u0
Simplification yields:
C(x, t)d x
M 2
e u2
du0
The diffusant mass is conserved according to:
C(x,t)d x
M erf M
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Exercise2a) Two instantaneous planar diffusion sources, each of “strength” M, are symmetrically placed about the origin (x=0) at locations =1, respectively, and released at time t=0. Using the linearity of the diffusion solution develop an expression for the concentration, C(x,t), developed at any arbitrary point in the material at a fixed time t>0. Plot the concentration field as a function of x for several fixed values of the parameter Dt to expose its temporal behavior.
2b) Find the peak concentration at x=0 and determine the time, t*, at which it develops, if D=10-11 cm2/sec, M=25 g/cm2, and the sources are both located 1m to either side of the origin.
2c) Plot the concentration at the plane x=0 against time.
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Exercise
2a) C x, t M
2 Dte x 1 2 4 Dt e x1 2 4 Dt
0
0.5
1
1.5
2
-4 -3 -2 -1 0 1 2 3 4
C(x
,t)/M
x
0.5
0.05
2
0.025
Dt=0
0.1
0.25
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Exercise
C 0,t;ˆ x MDt
e ˆ x 2 4 Dt
2b)For two sources of strength M the concentration is:
C 0,t;ˆ x t
MDt
e ˆ x 24 Dt ˆ x 2
4Dt 2 12t
Differentiate with respect to t :
Dt ˆ x 2
2
The concentration reaches its maximum at t *:
Cmax 0,t M
ˆ x e / 2 0.484 M
ˆ x
The maximum concentration reached at x = 0
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Exercise
2c)
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0 1 2 3 4 5 6 7
C(0
,t) [g
/cm
3 ]
Time [103 sec]
D=10-11 cm/sM=5.10-4 g/cm2
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Key Points• Solutions to the linear diffusion equations require two initial or boundary conditions. Examples of
problems with constant composition and constant diffusing mass are demonstrated.
• Laplace transform methods were employed to obtain the desired solutions.
• Solutions are in the form of fields, C(r, t). Exposing the behavior of such fields requires careful parametric description and plotting.
• Similarity variables and time tags are used, because they capture special space-time relationships that hold in diffusion.
• Diffusion solutions in infinite, or semi-infinite, domains often contain error and complementary error functions. These functions can be “called” as built-in subroutines in standard math packages, like Maple® or Mathematica® or programmed for use in spreadsheets.
• The theory for the classical “thin-film” method of measuring diffusion coefficients was derived using the concept of an instantaneous planar source in linear flow.