Download Diffusion theory

The Diffusion Equation
Derivation of the Diffusion Equation
from the Transport Equation
Tissue Optics
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Approximations
The transport equation is difficult to solve
analytically. In order to find an analytical
solution we need to simplify the problem:
äDiscretisation methods N (r , s, t ) = ∑ N (r , si , t )
i
ä discrete
ordinates method
ä 2-flux method or Kubelka-Munk theory
ä Adding-double method
äExpansion
methods
ä Diffusion theory
äProbabilistic
methods
ä Monte Carlo simulations
© Stefan Andersson-Engels
1
The P N - approximation
A formal method of solving a differential equation
is to find a solution to its homogenous part and
expand the general solution in terms of the
homogenous solution obtained.
Another approach would be to expand the
radiance or photon distribution function in wellknown appropriate function series. This is
performed in the diffusion approximation.
N ( r , s , t ) = N (r , t ) ∑ ai ( r , t )ψi (s )
i
© Stefan Andersson-Engels
Diffusion approximation
One of the most useful approaches for tissuelight interaction is the diffusion approximation,
as it can be solved analytically for simple
geometries. The diffusion approximation has
been used extensively in photon propagation
modelling in tissue in the time domain, as well as
in the frequency domain and in the steady state.
© Stefan Andersson-Engels
2
Diffusion approximation
The basic idea behind the diffusion approximation
is to expand the functions in the transport
equation into spherical harmonics, and to use only
the lowest order in the expansion. Thus the
position and directional variables can be
separated into new functions. Expanding the
distribution function N(r,s,t) yields:
∞
N (r , s , t ) = ∑
+l
∑
l = 0 m =− l
2l + 1
N lm ( r , t )Ylm ( s )
4π
© Stefan Andersson-Engels
Spherical harmonics
Spherical harmonics do form a complete set
of expansion of wave functions. They are
frequently used in quantum mechanics since
they are eigenfunctions to the angular
momentum operators in central force
problems.
© Stefan Andersson-Engels
3
Spherical harmonics
The spherical harmonics form an orthogornal
set , since
2π π
∫ ∫ (Y
l
m
(θ , ϕ )) Yl m' ' (θ , ϕ )sin θdθ dϕ = δl ,l 'δm ,m '
∗
0 0
© Stefan Andersson-Engels
Spherical harmonics
Definition:
Ylm (θ , ϕ ) =
2 l + 1 ( l − m)!
( − 1) m eimϕ Pl m (cosθ )
4π ( l + m)!
where
2 m/ 2
1
−
ζ
(
)
P m (ζ ) =
l
2l ⋅ l !
d l+ m
2
l
(
ζ
−
1
)
; m≤ l
l+ m
dζ
are the associated Legendre polynomials
© Stefan Andersson-Engels
4
Spherical harmonics
The first few spherical harmonics are:
Y00 =
1
4π
3
cos θ
4π
3 ±iϕ
Y1±1 = m
e sin θ
8π
Y10 =
© Stefan Andersson-Engels
The P 1-approximation
Truncating the expansion after the terms would
yield
N (r , s , t ) =
1
L 00 (r , t ) Y00 (s ) +
4π
3
4π
1
∑L
m =− 1
1m
(r , t ) Y1m ( s)
This expression could now be inserted into the
transport equation, which after integration over s
would give four equations and four unknowns.
© Stefan Andersson-Engels
5
The vector approach
Another approach is to observe that Y 00 is a scalar
±
and that Y1m (m=0, ±1) represent
the three
components of a vector. The expression for the
photon distribution function N thus can be
substituted by
N ( r , s, t ) = A( r , t ) + B (r , t ) ⋅ s
where A is the isotropic distribution and B the
linear gradient of the photon distribution function,
respectively, and s the vector of the gradient.
© Stefan Andersson-Engels
Arbitrary point in space
ä An
arbitrary point in space can be expressed
as coordinate values for a defined set of
$ + yy$ + zz$
base vectors. r = xx
ä To have a unique set of coordinates for each
point the base vectors need to be
orthogonal, i.e. x$ ⋅ y$ = 0 etc.
ä For a multidimensional function the base
vectors need to be function, but the
principle of expansion is otherwise identical
© Stefan Andersson-Engels
6
Point versus function
Describing a directional
function is the same as
having an intensity plot
over a spherical surface
Describing a point
in space can be done
with 3 co-ordinates
© Stefan Andersson-Engels
Diffusion: schematically
1
N =a Y +
0
0 0
∑a
m= −1
Y =A + B ⋅s
m
1m 1
Assume for the illustration that A = 1, Bx= 0.1, By = Bz = 0
a0
+ a10
Y00
=
Y10
© Stefan Andersson-Engels
7
Photon density function ρ(r,t)
To determine A, this expression is integrated
over ω:
ρ( r, t ) ≡ ∫ N ( r, s, t )dω = ∫ ( A + s ⋅ B )dω
where ρ(r,t) is the photon density. The last
term on the right-hand-side is equal to zero.
This yields the expression for A:
1
A=
ρ(r , t )
4π
© Stefan Andersson-Engels
Photon current density
To determine B, the expression is first multiplied
by s and then integrated over ω:
∫ sN (r , s, t )dω = ∫ s( A + s ⋅ B)dω
The left-hand-side is the definition of the photon
current density divided by the velocity of light.
The first term on the right-hand-side is equal to
zero. This yields the expression for B:
B=
3
J (r , t )
4πc
© Stefan Andersson-Engels
8
Diffusion approximation
Putting the expressions for A and B together
yields the angular expansion of N.
1 
3

N ( r , s, t ) =
ρ
(
r
,
t
)
+
J
(
r
,
t
)
⋅
s

4π 
c
In a similar way the source function can be
expanded and written as.
q( r , s, t ) =
[
1
q ( r , t ) + 3q1 (r , t ) ⋅ s
4π 0
]
© Stefan Andersson-Engels
Insertion in transport equation
This gives:
∂
∂t
3
3
3






ρ
+
J
⋅
s
=
−
c
s
⋅∇
ρ
+
J
⋅
s
−
c
µ
(
r
)
ρ
+
J ⋅ s +
t





c
c
c





3


cµs ( r) ∫ p( s'⋅s)  ρ + J ⋅ s ' dω ' + q 0 + 3q1 ⋅ s
c


4π
The integral can now be solved using
∫4 π p(s, s′ )( s ⋅ s' )dω '
p( s, s′ )dω ' = 1 and g =
∫ p( s, s' )dω '
4π
∫
4π
© Stefan Andersson-Engels
9
Rearrangements yields
The resulting expression is:
∂ρ 3 ∂ ( J ⋅ s)
+
= c( − s ⋅∇ − µt + µs ) ρ +
∂t c ∂t
3( − s ⋅∇ − µt + gµs ) J ⋅ s + q 0 + 3q1 ⋅ s
Rearranging and simplifying this expression, using
µtr = (1 − g ) µs + µa = µs ′+ µa
yields:
1  ∂ρ 3 ∂ ( J ⋅ s ) 
3
+
=
(
−
s
⋅∇
−
µ
)
ρ
+
( − s ⋅ ∇ − µtr ) J ⋅ s +
a
c  ∂t c ∂ t 
c
1
3
q 0 + q1 ⋅ s
c
c
© Stefan Andersson-Engels
The reduced scattering coefficient
The reduced scattering coefficient is defined as:
µs ' = (1 − g ) µs
© Stefan Andersson-Engels
10
Mathematical simplification I
Now, if this expression is integrated over ω and use
∫ (s ⋅ A )dω = 0
4π
(A ⋅ B)
3
∫ ( s ⋅ A )( s ⋅ B ) dω =
we get:
1 ∂ρ
1
1
= − µa ρ − ∇ ⋅ J + q 0
c ∂t
c
c
(1)
If the equation instead first is multiplied by s and then
integrated over ω using
© Stefan Andersson-Engels
Mathematical simplification II
∫ s(s ⋅ A )dω =
4π
A
3
∫ Csdω = 0
we get
1 ∂J
1
1
µtr
= − ∇ρ −
J + q1
c 2 ∂t
3
c
c
Here we make some approximations. Assuming that we
have an isotropic source, i.e. q1=0, at steady-state, we
obtain Fick’s law from this equation:
J = −c D ∇ρ
© Stefan Andersson-Engels
11
The diffusion coefficient
Here we introduce an important definition, the
diffusion coefficient:
1
1
D=
=
3µtr 3( µa + (1 − g ) µs )
Putting Fick’s law into Eq. (1) we arrive at the
final expression, the diffusion equation:
1 ∂ρ
1
− ∇ D ∇ρ + µa ρ = q 0
c ∂t
c
© Stefan Andersson-Engels
The Diffusion Equation!
By assuming a homogenous medium, that means
that the diffusion coefficient D is constant and
using the relation that the fluence rate is related
to the photon density by:
φ ( r , t ) = chνρ (r , t )
one arrives in the diffusion equation as it is usually
presented in the field of biomedical optics:
1∂
φ (r , t ) − D∇ 2φ (r , t ) + µa φ (r , t ) = S (r , t )
c ∂t
© Stefan Andersson-Engels
12
Diffusion approximation validity
The diffusion equation is valid only if the
propagating light is diffuse and this
implies that the reduced scattering
coefficient is much larger than the
absorption, i.e. (1-g)µs>>µa. The source
and detector must also be separated in
space and time to allow that the light is
diffuse when it reaches the detector.
© Stefan Andersson-Engels
Point source solution
The solution to the diffusion equation for an
infinite homogenous slab with a short pulse
isotropic point source S ( r , t ) = δ ( 0,0) is
φ( r , t ) = c( 4πDct )
− 3/ 2
r2
exp( −
− µa ct )
4 Dct
This Green’s function can be used to solve
problems for a slightly more complex geometry.
© Stefan Andersson-Engels
13