Krogh Cylinder
Steven A. Jones
BIEN 501
Wednesday, May 7, 2008
Louisiana Tech University
Ruston, LA 71272
Slide 1
Announcements
1. All homeworks have been assigned.
2. Final exam will be taken from parts of the
homework.
3. No homework on Krogh cylinder.
4. Krogh cylinder will not be on the exam.
5. Friday – finish Krogh and do Comparmental
Models
6. Monday – Review and Course Evaluations
7. Wednesday - Exam
8. Friday – No class, will be available for
questions.
9. Today – Office Hours will start at 10:30.
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Slide 2
Energy Balance
Major Learning Objectives:
1. Learn a simple model of capillary
transport.
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Slide 3
The Krogh Cylinder
Capillary
Tissue
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Slide 4
Assumptions
• The geometry follows the Krogh cylinder
configuration
– Radial symmetry
– Transport from capillary
– Capillary influences a region of radius Rk.
• Reactions are continuously distributed
• There is a radial location at which there is
no flux
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Slide 5
Capillary Transport
Consider the following simple model for capillary transport:
Capillary
Interior
Reactive Tissue
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Matrix
Slide 6
Capillary Transport
What are appropriate reaction rates and boundary
conditions?
cO2  0 or
J O2  0
Constant rate of
consumption (determined
by tissue metabolism, not
O2 concentration)
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cO2  C0 or J O2  J 0
cO2  continuous
Not metabolic (no
consumption)
Slide 7
Diffusion Equation
c
 D 2 c  rx
t
c
1   c 
D
 r   rx
t
r r  r 
For steady state:
D
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1   cr  
r
  rx r , cr 
r r  r 
Slide 8
Constant Rate of Reaction
Assume the rate of reaction, rx, is constant:
1   c 
D
r   M
r r  r 
(M will be numerially negative
since the substance is being
consumed).
And that the concentration is constant at the capillary
wall and zero at the edge of the Krogh cylinder
cRc   c0
cRk 
J Rk    D
0
r
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Slide 9
Constant Reaction: Roadmap
The roadmap for solving the steady state problem is
as follows:
• Integrate the differential equation once to obtain a
solution for flux as a function of r.
• Use the zero flux boundary condition at Rk to determine
the first constant of integration.
• Substitute the constant back into the previouslyintegrated differential equation.
• Integrate again to obtain the form for concentration.
• Use the constant concentration boundary condition to
determine the second constant of integration.
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Slide 10
First Integration
Because there is only one independent variable,   d :
1   c 
1 d  dc 
D
 r   M  D
r   M
r r  r 
r dr  dr 
Integrate
once:
d  dc 
Mr
dc
Mr 2
 r

a
r   
dr  dr 
D
dr
2D
dc Mr aD
Write in terms of flux: J   D


dr
2
r
We will use this form to satisfy the no-flux boundary condition
at Rk.
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Slide 11
Flux Boundary Condition at Rk
Since flux is 0 at the edge of the cylinder (Rk),
MRk aD
MRk2
dc
D


0  a
dr r  Rk
2
Rk
2D
Substitute back into the (once-integrated) differential
equation (boxed equation in slide 10):
2
MR
dc
Mr
dc
Mr
k
r

a  r


dr
2D
dr
2D
2D
dc M 2 2
r

Rk  r
dr 2 D
2
2

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
Slide 12
Second Integration for Concentration
With the previous differential equation:

Divide by r:
Integrate:
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dc M 2 2
r

Rk  r
dr 2 D

dc M  Rk2


 r 
dr 2 D  r


M  2
r2 
 Rk ln r    b
c
2D 
2
Slide 13
Boundary Condition at Capillary Wall
From the problem statement (Slide 9) c(Rc) = c0.
Evaluate the previous solution for c(r) at r=Rc.
Rc2 
M  2
 Rk ln Rc 
  b  c0
cRc  
2D 
2 
Solve for b.
Rc2 
M  2
 Rk ln Rc  
b  c0 
2D 
2 
Rc2 
M  2
r2 
M  2
 Rk ln Rc 

 Rk ln r    c0 
 cr  
2D 
2
2D 
2 
b
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Slide 14
Simplify
Combine like terms, recalling that ln r  ln Rc  ln r Rc :

2
2

r

R
M
r
2
c

cr   c0 
R
ln

k
2 D 
Rc
2



Or, in terms of partial pressures:

2
2

r  Rc
M
r
2
 Rk ln
 Pr   Pc 

2 D 
Rc
2
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


Slide 15
Concentration
(nmoles/L)
Efffect of Different M Values
60
50
40
30
20
10
0
-10
-20
-30
Rc
Krogh Cylinder Solution
Critical Supply
Starvation
Ample Supply
Rk
0
0.02
0.01
0.03
r (cm)
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Slide 16
Plot of the Solution
Note that the solution is not valid beyond Rk.
Concentration
(nmoles/L)
60
Krogh Cylinder Solution
50
40
30
20
10
0
0
0.005
0.01
0.015
0.02
0.025
0.03
r (cm)
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Slide 17
Finding Rk
The steady state equation is a function of Rk,
but an important question is, “What is Rk,
given a certain metabolic rate?”
Non-starvation: Halfway between capillaries.
Starvation: Is the solution still valid?
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Slide 18
Non Steady State
Diffusion equation:
Initial Condition:
Boundary Conditions:
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c
 D 2 c  M
t
c
1   c 
D
r   M
t
r r  r 
c  0 at t  0 for all r
c(rc , t )  c0u t 
crk , t 
D
0
r
Slide 19
Homogeneous Boundary Conditions
The problem will be easier to solve if we can make the
boundary conditions homogeneous, i.e. of the form:
f r , t   0
f r , t 
and/or
0
dr
Our boundary condition at r = rc is not homogeneous
because it is in the form:
f r , t   c(rc , t )  c0ut   0
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Slide 20
Homogeneous Boundary Conditions
However, if we define the following new variable:
c  C0
  r, t  
C0
The boundary condition at rc becomes:
  rc , t   0 for t  0
And the boundary condition at rk is still homogeneous:
rk , t 
0
r
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Slide 21
Non-Dimensionalization
The new concentration variable also has the
advantage of being dimensionless. We can nondimensionalize the rest of the problem as follows:
r
 ,
rc
Let:
Dt
 2
rc
Why are these forms
obvious?
The boundary conditions become:
1,   0
 k , 
0
r
The initial condition becomes:
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  0 at   0 for all 
Slide 22
Non-Dimensionalization
The diffusion equation can now be non-dimensionalized:
c
1   c 
D
r   M
t
r r  r 
Use:
So that:
c  r, t   C0  r, t   C0
 C0   C0 
1    C0   C0  
D
r
M
t
r r 
r


1    
C0
 DC 0
r
M
t
r r  r 
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Slide 23
Non-Dimensionalization (Continued)
r

rc
Use:
tD
 2
rc
To determine that:
      1 

 

r    r  rc 
      D 
     2
t    t  rc 
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Slide 24
Non-Dimensionalization (Continued)
Now apply:
r  rc ,
t
 rc2
D
 1 
 D 

,
 2
r rc  t rc 
To:
To get:

1    
C0
 DC 0
r
M
t
r r  r 
DC0 
1 1  
1  
 rc
  M
 DC0
2
rc 
rc rc   rc  
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Slide 25
Non-Dimensionalization (Cont)
Simplify
Multiply by
DC0 
1 1  
1  
 rc
  M
 DC0
2
rc 
rc rc   rc  
rc2
DC0
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DC0 
1    

  M
 DC0 2
2
rc 
rc     
:
 1     Mrc2

 

      DC 0
Slide 26
Non-Dimensionalization (Cont)
Examine
 1     Mrc2

 

      DC 0
The term on the right hand side must be non-dimensional
because the left hand side of the equation is nondimensional. Thus, we have found the correct nondimensionalization for the reaction rate.
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Slide 27
The Mathematical Problem
The problem reduces mathematically to:
Differential Equation
 1     Mrc2

 

      DC 0
Boundary Conditions
1, t   0
Initial Condition
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 k , t 
0
r
 ,0  0
Slide 28
Change to Homogeneous
Follow the approach of section 3.4.1 (rectangular channel) to
change the non-homogeneous equation to a homogeneous
equation and a simpler non-homogeneous equation.
Diffusion equation:
 1     Mrc2

 

      DC 0
Let:  ,   f    g  , 
Then:
g  ,  1   g  ,   1   f    Mrc2

 




  
r       DC 0
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Slide 29
Divide the Equation
The equation is solved if we solve both of the following
equations:
g  ,  1   g  ,  


0

  
r 
Mrc2
1   f   

  
    
DC 0
In other words, we look for a time-dependent part and a
time independent (steady state) solution. Note that since
the reaction term does not depend on time, it can be
satisfied completely by the time-independent term.
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Slide 30
Solution to the Spatial Part
We already know that the solution to:
1  f r  
 D r
M
r  r 
Is:


M  r 2  rc2
r
2

 cr   cc 
 rk ln 
2D 
2
rc 
And that this form satisfies the boundary conditions at rc
and rk. It will do so for all time (because it does not
depend on time).
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Slide 31
Non-Dimensionalize the Spatial Part
In terms of the non-dimensional variables:



Mrc2   2 c2
2

f   
k ln  
2 DC0 
2

And this form also becomes zero at the two boundaries.
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Slide 32
Transient Part
We therefore require that:
g  ,  1   g  ,  


0

  
r 
With Boundary Conditions
g c , t   0
And the Initial Condition
g  k , t 
0
r
,0  f    g ,0  0



Mrc2   2 c2
2

 g  ,0  
k ln  
2DC0 
2

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Slide 33
Separation of Variables
Assume that g ,   R T 
Homogeneous
diffusion equation:
T 
1   R  

  0
R 
 T 

    
T'  
1
  R  

  0

T  R     
Function of  only
Function of  only
T'  
1
  R  

  2  constant

T  R     
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Slide 34
The Two ODEs
Solutions:
T'  
2
 2
   T    Ae
T 
1
d  dR  

  2 
R  d  d 
d  dR   2

   R   0 
dr  d 
2
d
R 
dR  2 2
2


   R   0
2
d
d
What does this equation remind you of?
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Slide 35
Radial Dependence
Could it perhaps be a zero-order Bessel Function?
 y
y
2
2


z

z

z

p
y0
2
z
z
2
2
2
d
R 
dR  2 2
2


   R   0
2
d
d
Let z  
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Slide 36
Radial Dependence
let z  
d
dz d
d


d d dz
dz
2
d2
d  d 
d  d 
d
2

       

2
d
d  d 
dz  dz 
dz 2
Differential equation for radial dependence becomes:
 z 2  2 d 2Rz    z  dRz   2
 2  
    
  z Rz   0
2
dz    
dz 
  
So the solution is: R   AJ 0    BY0  
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Slide 37
Radial Dependence
 z 2  2 d 2Rz    z  dRz   2
 2  
    
  z Rz   0
2
dz    
dz 
  
2
d
Rz 
dRz  2
2
z
z
 z Rz   0
2
dz
dz
This is Bessel’s equation, so the solution is:
Rz   AJ 0 z   BY0 z  with z  
or R   AJ 0    BY0  
Louisiana Tech University
Ruston, LA 71272
Slide 38
Bessel Functions
J0
Y0
J1
Y1
Bessel Function
1
0
-1
0
2
4
6
8
z
Louisiana Tech University
Ruston, LA 71272
Slide 39
Derivatives of Bessel Functions
The derivatives of Bessel functions can be
obtained from the general relations:
dJ n  x 

dx
dYn  x 

dx
n
J n  x   J n 1  x 
x
n
Yn  x   Yn 1  x 
x
Specifically:
dJ 0 x 
 J1 x ,
dx
Louisiana Tech University
Ruston, LA 71272
dY0 x 
 Y1 x 
dx
Slide 40
Flux Boundary Condition
In the solution we will have terms like:
 A J 0    B Y0   e
 2
We will be requiring the gradient of these terms to go to
zero at k. I.e.
 A J1 k   B Y1 k  e
 2
0
The only way these terms can go to zero for all  is if:
A J1 k   B Y1 k   0
For every value of .
Louisiana Tech University
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Slide 41
Relationship between A and B
In other words:
A J1 k   B Y1 k   0
 B  Y1  k 
A  
J1  k 
And for the boundary condition at the capillary wall:
c c ,   0 (recall  c  1)
 AJ 0  1  BY0  1  0
  Y1  k 

 B
J 0    Y0    0
 J1  k 

Louisiana Tech University
Ruston, LA 71272
Slide 42
Characteristic Values
  Y1 k 

From B
J 0    Y0    0
 J1 k 

We conclude that the allowable values of  are
those which satisfy:
Y1 k J 0    J1 k Y0    0
This is not as simple as previous cases, where the Y0
term became zero, but it is possible to find these
values from MatLab or Excel, given a value for k.
Louisiana Tech University
Ruston, LA 71272
Slide 43
The Characteristic Function (k=10)
0.8
f    Y1 k J 0    J1 k Y0  
0.6
f ( )
0.4
0.2
0
-0.2
-0.4
0
1
2
3
4
5

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Ruston, LA 71272
Slide 44
To Calculate (and Plot) in Excel
0.1
=BESSELY(A1*etak,1)*BESSELJ(A1,0)
- BESSELJ(A1*etak,1)*BESSELY(A1,0)
0.15 =BESSELY(A2*etak,1)*BESSELJ(A2,0)
- BESSELJ(A2*etak,1)*BESSELY(A2,0)
0.2
=BESSELY(A3*etak,1)*BESSELJ(A3,0)
- BESSELJ(A3*etak,1)*BESSELY(A3,0)
0.25 =BESSELY(A4*etak,1)*BESSELJ(A4,0)
- BESSELJ(A4*etak,1)*BESSELY(A4,0)
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Slide 45
Finding the Roots
Use “Tools | Goal Seek” to find the roots of the equation.
For example, the plot indicates that one root is near =1.2.
With the value 1.2 in cell A1 and the formula in cell A2:
Goal Seek
Set cell:
A2
To value:
0
By changing cell:
A1
OK
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Cancel
“OK” will
change
the value
of cell A1
to the 4th
root,
0.110266.
Slide 46
First Six Roots
This process gives the following value for the first
6 roots:
n
1
2
3
4
5
6
Louisiana Tech University
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n
0.110266
0.497895
0.855445
1.208701
1.560322
1.8
Slide 47
Complete Solution
The complete solution has the form:
  Y1 n k 
 n2
c ,    Bn 
J 0 n   Y0 n e
n 1
 J1 nk 


Where the values of n are obtained as described
above.
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Slide 48
A Prettier Form
If we define:
n  
Bn
J1 n k 
We obtain the somewhat more aesthetically
pleasing form:

c ,     n Y1 n k J 0 n   J1 n k Y0 n e
 n2
n 1
Louisiana Tech University
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Slide 49
Initial Condition
Now apply the initial condition:



Mrc2   2  c2
2

g  ,0  
  k ln  
2 DC0 
2


To:
c ,0    n Y1 n k J 0 n   J1 n k Y0 n 
n 1

So:
  Y   J     J   Y   
n 1
n
1
n
k
0
n
1

n
k
0
n

2
2


Mr
 c
2


  k ln  
2 DC 0 
2

2
c
Louisiana Tech University
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Slide 50
Orthogonality
For simplicity, define:
C0  , n   Y1  nk  J 0  n   J1  nk  Y0  n 
These are called “Cylindrical Functions.” They satisfy
the radial Laplacian operator, are zero for =1, and
have zero derivative for =. According to the
Sturm-Liouville Theorem, the following orthogonality
relationship holds for these functions:

k
1
C0  , n  C0  , m  d  0 if m  n
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Slide 51
Sturm-Liouville

Compare:
To:
p  x
d 2   , x 
dx
2
2
d 2 R  
d 2
 p  x 

dR  
d
d   , x 
dx
  2 2 R    0
  q  x    w  x     , x   0,
a xb
Note that  is not the first derivative of 2, but if
we divide by :

d 2 R  
d
2

dR  
d
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  2 R    0
and 1 is the
1st derivative of 
Slide 52
Sturm-Liouville
Compare:

d 2 R  
To:
p  
d 2   , 
dx
2
d 2
 p  

dR  
d
  2 R    0
d   , 
dx
  q     w      ,   0,
a   b
p     , p    1, q    0, w    
The w() is particularly important because it tells
us the weighting factor.
Louisiana Tech University
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Slide 53
Initial Condition
With the initial condition:


2
2





Mr
c
2

 n C 0  , n   
 k ln  


2 DC0 
2
n 1



2
c
Multiply both sides by C0  , m  and integrate:

 
n 1
k
n 1
 C 0  , n  C 0  , m  d
2
2


Mr k   c 
2





ln   C 0  , m  d
k


2 DC0 1 
2


2
c
Louisiana Tech University
Ruston, LA 71272
Slide 54
Initial Condition
Now use the orthogonality condition:
k
 n   C 02  , n  d
1


2
2





Mr k
c
2



 k ln   C 0  , n  d


2 DC0 1 
2


2
c
These integrals can be found in tables and worked
out to determine the values for the n’s, from which
the final answer is obtained.
Louisiana Tech University
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Slide 55
Fourier Bessel Series
For Example:
b
2
n a


 C 2  n  d 
2
2
b2 2
a2 2
b
a
C   nb   C   n a   n2 C 2  nb   n2 C 2  n a 
2
2
2
2
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Slide 56
Integral of Cylindrical Functions
To obtain the previous result, take Bessel’s
Equation, substituting in the cylindrical function, and
d C  n  / d
multiply by
and then integrate.

2
d 2C  n 
d 2
b

a
2

d C  n 
d
d C  n  d 2C  n 
d
d
2
 n2 2 C  n   0 
b
d   
d
a
b
2
n a

Louisiana Tech University
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d C  n  d C  n 

2
d C  n 
d
d
d
C  n d  0
Slide 57
Norm of Cylindrical Functions
For the first term:
2


b
b


d
C


d
C


d
C









d
n
n
n
2

 d

d




2
a

a 2  d
d
d
d  



2
2
Integrate by parts with:
u
2
2
du   d
 d C  n  
d  d C  n  
dv 

 d v  

d  d 
 d 
2
Louisiana Tech University
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2
Slide 58
Integration by Parts, First Term
2

b
d  d C  n   

 d 
I1  


a 2  d
d  



2
  d C  n  


2  d 
2
2 b
 d C  n  
 
 d
a
d 

a
2
b
b  d C  n  
b 2
a 2
 C   nb   C   n a     
 d
a
2
2
 d 
2
Louisiana Tech University
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2
2
Slide 59
Integration by Parts, Second Term
The second integral of Slide 54 is:
 d C  n  
I2    
 d
a
 d 
2
b
But:
I1 

2
C    b  C    a 

2
2
2
n
n
 d C  n  
 
 d
a
 d 
2
2
b
So the second integral of Slide 54 is cancels with
the last term of I1, leaving:
b2 2
a2 2
I1  I 2  C   nb   C   n a 
2
2
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Slide 60
Last Term
For the last term:
b
d C  n 
a
d
I 3  n2   2
u
b
2
n a


2
2
2
d
C
n 


2
C  n  d  n 
d
a 2
d
, du   d ; dv 
b
d C 2  n 
d
2
d , v  C 2  n 
b
 d C  n 
2
2
2
d  n
C  n   n   C 2  n  d
a
2
d
2
a
2
2
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2
b
Slide 61
Last Term
2
2
b
b
a
2
2
2
2
2
I 3  n
C  nb   n
C  n a   n   C 2  n  d
a
2
2
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Slide 62
Combine the Three Integrals
b2 2
a2 2
I1  I 2  I 3  C   nb   C   n a  
2
2
2
2
b
b
a
2
2
2
2
2
n C  nb   n
C  n a   n   C 2  n  d  0
a
2
2
b
2
n a


 C 2  n  d 
2
2
b 2 2
a 2 2
b
a
C  nb   C  n a   n2 C 2  nb   n2 C 2  n a 
2
2
2
2
The expression will simplify further, given that we will
be working with problems for which either the
function or its derivative is zero at each boundary.
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Slide 63
What about the similarity solution
As it turns out, we did not need to abandon
the similarity solution. We could have done
the same thing we did with the separation of
variables solution. I.e. we could have said
that the complete solution is the sum of the
particular solution and a sum of similarity
solutions.
Louisiana Tech University
Ruston, LA 71272
Slide 64
Louisiana Tech University
Ruston, LA 71272
Slide 65
Similarity Solution?
We could attempt a
similarity solution:
r

1 r

1


,

t
2 4 Dt 3 r
4 Dt
4 Dt
  
1 r



t t  2 4 Dt 3 
  
1



r r 
4 Dt 

Which transforms
c
1   c 
the equation to:
D
r   M 
t
r r  r 
 1 r  c
1  1


D 
 2 4 Dt 3  
r  4 Dt


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    1
 

   r  4 Dt

 
 c 


    M


Slide 66
Transform Variables
From Previous:
1 r

 2 4 Dt 3

 c
1  1

D 
 
r  4 Dt

    1
 r

    4 Dt

 
 c 
M

  


 1  c
1      c 
 

  M
 D  
r  r     
 2t  
Multiply by r2/D
 r 2  c
 r2 
  c 



  M  

   
 2 Dt  
D
2

  c 
r 
3 c

  M  



   
D
 
Louisiana Tech University
Ruston, LA 71272
Slide 67
The Problem
From Previous:
2




c
r 
3 c

  M  



   
D
 
If this had been a “no reaction” problem, the
method would work. Unfortunately, the reaction
term prevents the similarity solution from working,
so we need to take another approach.
Louisiana Tech University
Ruston, LA 71272
Slide 68