mathematics of diffusion problems · r. abart and e. petrishcheva mathematics of diffusion...
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Mathematics of Diffusion Problems
R. Abart and E. Petrishcheva
Free University Berlin
19 April 2006
R. Abart and E. Petrishcheva Mathematics of Diffusion Problems
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Outline
1 Physical background;
2 General derivation of the diffusion equation;
3 One-dimensional problems;
4 Examples of applied problems;
5 Measurements of the diffusion coefficient.
R. Abart and E. Petrishcheva Mathematics of Diffusion Problems
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Particle density
Number of particles in a unit volume:
n(x , y , z , t)
Number of particles in an arbitrary volume Ω:
N(t) =
∫
Ω
n d3x
Change of N(t) in an arbitrary volume:
dN
dt= (income) − (outcome)
R. Abart and E. Petrishcheva Mathematics of Diffusion Problems
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Income
It is convenient to let “incomes” be positive and negative so thatwe may write:
dN
dt=
∑
(income)
There are two different reasons for changes in N(t)
Volume income (“birth” (positive) and “death” (negative))
Fluxes over the boundaries (“migration”)
R. Abart and E. Petrishcheva Mathematics of Diffusion Problems
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Volume sources
Number of particles that are created in a unit volume per unit time
q(x , y , z , t, n . . .)
The corresponding change of N(t) can be quantified
dN
dt=
∫
Ω
q d3x
Different source terms are possible
Predefined external source q(x , y , z , t)
Chemical reaction q = ±kn
Nonlinear source, e.g., q ∼ n2
R. Abart and E. Petrishcheva Mathematics of Diffusion Problems
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Flux
Number of particles that flow through a unit surface per unit time
j(x , y , z , t)
The corresponding change of N(t) can be quantified
dN
dt= −
∫
∂Ω
j dS
Divergence theorem
dN
dt= −
∫
Ω
(∇j) d3x
Where
∇j =∂jx
∂x+
∂jy
∂y+
∂jz
∂z
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Derivation
By the definition of N(t)
dN
dt=
d
dt
∫
Ω
n d3x =
∫
Ω
∂n
∂td3x
Fluxes and sources
dN
dt=
∫
Ω
(−∇j + q) d3x
Therefore∫
Ω
(∂tn + ∇j − q) d3x = 0
that is valid for an arbitrary volume.
R. Abart and E. Petrishcheva Mathematics of Diffusion Problems
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Continuity equation
We have just derived
One of the most important physical equations
∂n
∂t+ ∇j = q
However we have only one equation for both n and j.The continuity equation can not be used as is, a physical model,e.g.,
j = j(n)
is required.
R. Abart and E. Petrishcheva Mathematics of Diffusion Problems
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Flux models
Flux is caused by spatial changes in n(x , y , z , t).
It is natural to assume that j ∼ ∇n.
The proportionality coefficient is denoted −D.
Fick’s first law (for particles) j = −D∇n.
Nonuniform diffusion j = −D(x , y , z)∇n.
Non-isotropic diffusion jk = −∑
i=x ,y ,z
Dki∂in.
Non-linear diffusion j = −D(n)∇n.
Fourier’s law (for energy) and many others.
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Diffusion equation
Having a physical model (diffusion) we can proceed with thederivation of a self-consistent mathematical equation:
∂tn + ∇j = q
j = −D∇n
We have just obtained another fundamental equation
∂tn = ∇(D∇n) + q
Let us consider only a uniform isotropic medium
∂tn = D∇(∇n) + q
R. Abart and E. Petrishcheva Mathematics of Diffusion Problems
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Laplacian
It is convenient to introduce the following notation
n = ∇(∇n)
where
n =∂2n
∂x2+
∂2n
∂y2+
∂2n
∂z2
E.g., stationary distributions of particles are given by the famous
Laplace (Poisson) equation
D n = −q
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Resume
Linear diffusion equation in 1D
∂n
∂t= D
∂2n
∂x2+ q(x , t)
n = n(x , t) should be found;
D = const and q = q(x , t) are known;
the solution should exist for all t ≥ 0;
the initial distribution n0(x) = n(x , 0) is known;
a < x < b with the boundary conditions at x = a and x = b;
x > 0 and −∞ < x < ∞ are also possible.
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Boundary conditions
1 Dirichlet: edge concentrations are given, e.g.,
n(x = a, t) = 1
2 Neumann: edge fluxes are given, e.g.,
∂xn(x = b, t) = 0
3 More complicated: e.g., combined Dirichlet and Neumann
[
α(t)n(x , t) + β(t)∂xn(x , t)
]
x=a,b
= γ(t)
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Delta function
It was first introduced by Dirac to represent a unit point mass.
The physical definition of δ(x − a) (e.g., mass density) is
δ(x − a) = 0 if x 6= a
δ(x − a) = ∞ if x = a
and∫ ∞
−∞δ(x − a) dx = 1
so that we have a unit mass at x = a.
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Point source
Let N particles be initially placed at x = a. Their concentrationn(x , t) is subject to the diffusion equation
∂tn(x , t) = D∂2xn(x , t)
with
−∞ < x < ∞,
where one usually assumes that
n(x , t) → 0 at x → ±∞;
and the following initial condition
n(x , 0) = Nδ(x − a).
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Kernel solution
The “point source” problem can be solved explicitly
n(x , t) = NK (x , a, t),
where the kernel solution
K (x , a, t) =1√
4πDtexp
[
−(x − a)2
4Dt
]
.
Here
∂tK = D∂2xK ;
K (x , a, t) = K (a, x , t);
limx→±∞ K (x , a, t) = 0;
limt→+∞ K (x , a, t) = 0;
limt→0 K (x , a, t) = δ(x − a).
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Initial value problem
We can now give a formal solution of the initial value problem
∂tn(x , t) = D∂2xn(x , t),
where
−∞ < x < ∞;
lim|x |→∞ n(x , t) is finite;
n(x , 0) = n0(x).
One replaces N with n0(a)da for each x = a and uses linearity
n(x , t) =
∫ ∞
−∞K (x , a, t)n0(a)da.
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Diffusion problem with a source term
Problem with the trivial initial distribution n0(x) = 0 and a sourceterm
∂tn(x , t) = D∂2xn(x , t) + q(x , t)
where
−∞ < x < ∞;
lim|x |→∞ n(x , t) is finite;
n(x , 0) = 0.
The idea of solution is similar. The results reads
n(x , t) =
∫ ∞
−∞
∫ t
0
K (x , a, t − τ)q(a, τ)dτda.
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General diffusion problem
It is also easy to solve the general problem
∂tn(x , t) = D∂2xn(x , t) + q(x , t),
where
−∞ < x < ∞, n(x , 0) = n0(x);
lim|x |→∞ n(x , t) is finite.
The solution is a simple combination of the two previous results:
n(x , t) =
∫ ∞
−∞K (x , a, t)n0(a)da+
+
∫ ∞
−∞
∫ t
0
K (x , a, t − τ)q(a, τ)dτda.
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Application
This is an important standard situation for many applications:
∂tn(x , t) = D∂2xn(x , t),
where
n0(x) =
1 for x < 0,0 for x > 0.
−5 0 5−0.5
0
0.5
1
1.5 time: 200000 diffusion coefficient:1e−005
distance x
conc
entr
atio
n
The solution reads
n(x , t) =1
2Erf
(
x
2√
Dt
)
,
Erf(x) =2√π
∫ x
0
e−s2
ds.
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Another application
Another important stan-dard solution of
∂tn(x , t) = D∂2xn(x , t),
where
n0(x) =
1 if a < x < b,
0 otherwise.
−5 0 5−0.5
0
0.5
1
1.5 time: 100000 diffusion coefficient:1e−005
distance x
conc
entr
atio
n
The solution is found by a direct integration
n(x , t) =1
2
[
Erf
(
x − b
2√
Dt
)
− Erf
(
x − a
2√
Dt
)]
.
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How D can be measured
We start from the standard solution n(x , t) =1
2Erf
(
x
2√
Dt
)
,
and measure amount of particles m(T ) that have passed throughx = 0 into the region x > 0 as a function of time T
m(T ) = Area
∫ T
0
(−D∂xn)x=0 dt = Q
√
DT
π
The dependence m vs√T is linear and there-
fore
D =π tan2 α
Q2.
R. Abart and E. Petrishcheva Mathematics of Diffusion Problems