Biophysics and Physiological Modeling: Finite Difference Method

This assignment presents a comprehensive solution to a biophysics homework problem, specifically addressing the Finite Difference Method and its application to O2 diffusion modeling. The solution includes detailed responses to various questions, utilizing equations and algorithms to simulate and analyze the behavior of oxygen molecules within a biological system. The student's work encompasses the development and analysis of finite difference algorithms, the comparison of finite difference curves with kinetic Monte Carlo curves, and the application of mathematical models to understand the movement and distribution of O2. The assignment covers topics such as the total number of marbles, the change in the number of marbles over time, and the impact of various parameters on the system. The assignment provides a detailed, step-by-step approach to solving the problems, incorporating equations and calculations, and demonstrating a clear understanding of the principles of biophysical modeling.

Running Head: FINITE DIFFERENCE METHOD AND O2
FINITE DIFFERENCE METHOD AND O2
Name of the Student:
Name of the University:
Author’s Note:

1FINITE DIFFERENCE METHOD AND O2
Response to Q. 3.1 (a) and (b)
Given, N=10
k =0.05 s−1
δ t=0.5 s
x0=0.3
Using equation (3.2) and (3.3), the complete finite difference algorithm can be written as,
Total number of marbles can be given by, N=N 1+N 2
Instruction
N × x0=N1
N2=N −N1
δ N i=(N2 −N1)kδt
x1= N 1
N
Step 1 of Finite Difference Algorithm
The steps are as follows,
10 ×0.3=3 ; N1=3
Or, N2=10−3
Or, N2=7
Or, δ N 0= ( 7−3 )∗0.05∗0.5

2FINITE DIFFERENCE METHOD AND O2
Or, δ N 0=0.1
N1=3+ 0.1
N1=3.1
N2=6.9
Or, x1= 3.1
10
Or, x1=0.31
δ N 0= ( 6.9−3.1 )∗0.05∗0.5
δ N 1=0.095
Output Table
Variable Values
N1 3.1
N2 6.9
δN1 0.095
x1 0.3
Response to Q. 3.2 (a)
N k dt N1 N2 dN1 new X
Step 0 10 0.05 0.5 3 7 0.1 0.3
Step 1 10 0.05 0.5 3.1 6.9 0.095 0.31
Step 2 10 0.05 0.5 3.195 6.805 0.09025 0.3195

3FINITE DIFFERENCE METHOD AND O2
Step 3 10 0.05 0.5 3.28525 6.71475 0.085738 0.328525
The equation N2
new=N −N1
new signifies that, since N1 changes after a time interval of δt
that indicates the new average number of marblesN1
new, therefore the total number of marbles in
each box also changes with time and therefore the entire equation can reconstructed with the
above form. Here, N2
new is the new average number of marbles after a small interval of time with
respect to δt.
Response to Q. 3.2 (b)
Equation (3.2) has been drawn from equation (3.1), which is the finite difference equation
that signifies N1 changes (on average) during a short but finite amount of time δt. The
combination of both the equation signifies a great deal about the entire system. Equation (3.3)
deals with the system where average number of marbles is the sum of old value (N1
old) and a
small change in the value of N1 to δ N 1
new. Combination of both the equations depend on an
adjustable parameter and can be applicable to almost all the psychological systems.
Response to Q. 3.3
N k dt N1 N2 dN1 new X1
Step 0 10 0.05 0.5 3 7 0.1 0.3
Step 1 10 0.05 0.5 3.1 6.9 0.095 0.31
Step 2 10 0.05 0.5 3.195 6.805 0.09025 0.3195
Step 3 10 0.05 0.5 3.28525 6.71475 0.085738 0.328525

4FINITE DIFFERENCE METHOD AND O2
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10
0.600000000000001
0.650000000000001
0.700000000000001
0.750000000000001
0.800000000000001
0.850000000000001
0.900000000000001
0.950000000000001
1
1.05
X1
X
Time in t
value of X
Response to Q. 3.4
a) FD curves does not completely mach with the kMC curve, however most of the points have
intersected for t=10s
b) For t=20 the points of FD curve are deviating from kMC along with the progression towards
t=20
c) Both in t=40s and t=80s the deviations of FD curve from kMC.
Response to Q. 3.4
a) FD curve and kMC curve was similar in N=1000, however, with different values of N the
curve difference increases
b) YES
c) With the decreasing value of N, the difference of FD curve and kMC curve increased.
d)

5FINITE DIFFERENCE METHOD AND O2
0 10 20 30 40 50 60 70 80
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Approach to equilibrium
kMC sim
equilibrium
FD method
time t (s)
fraction
in box 1
x1
Response to Q. 3.6
Using equation 3.2
δ N 1= ( N2−N1 ) kδt
Using equation 2.6
N2=N −N1
Substituting the value of Equation 2.6 in equation 3.2
δ N 1= [ ( N−N 1 ) −N1 ] kδt
δ N 1= ( N−2 N1 ) kδt
Response to Q. 3.7
We know, δ x1=x1
new−x1
old

6FINITE DIFFERENCE METHOD AND O2
Also, x1
new= N1
new
N
And δ N 1=N1
new−N1
old
We can write, δ x1= N1
new
N − N 1
old
N
δ x1= N1
new−N 2
old
N
δ x1= δ N1
N
Response to Q. 3.8
Using equation 3.7 , 3.8 and 2.5, we get
δ N 1= ( N−2 N1 ) kδt
δ x1= δ N1
N
x1= N 1
N
Therefore, δ x1= ( N −2 N1 ) kδt
N
δ x1= (1−2 N 1
N )kδt
δ x1= ( 1−2 x1 ) kδt
Response to Q. 3.9

7FINITE DIFFERENCE METHOD AND O2
Using equation 3.9
δ x1= ( 1−2 x1 ) kδt
Since, the system is approaching towards equilibrium, there we can consider δ x1 to be 0(zero)
We get, x1= 1
2 kδt
Response to Q. 3.10
N k dt t N1 N2 dN1
new
X1 dx1
Step 0 10 0.05 2 0 3 7 0.4 0.3 0
Step 1 10 0.05 2 2 3.4 6.6 0.32 0.34 0.032
Step 2 10 0.05 2 4 3.72 6.28 0.256 0.372 0.0256
Step 3 10 0.05 2 6 3.976 6.024 0.2048 0.3976 0.02048
Response to Q. 3.12
u=2 x1−1
Or, u=2 N1
N −1
Or, u=2 N1−N
N
Or, , u=2 N1−N1−N 2
N
Or, , u= N1−N 2
N

8FINITE DIFFERENCE METHOD AND O2
Response to Q. 3.14
Using Equation 3.12
u=2 x1−1
Given by the definition, δu=unew−uold
However, the definition can be altered by, δ x1=x1
new−x1
old
Therefore, it can be written as, unew =2 x1
new−1
uold =2 x1
old−1
Therefore, equations can be written as,
δu= ( 2 x1
new−1 ) −(2 x1
old−1)
Or, δu=2 ( x1
new−x1
old )
δu=2 δ x1, Hence proved
Response to Q. 3.15
Given equation 3.9,
δ x1= ( 1−2 x1 ) kδt
Equation 3.14, δu=2 δ x1
Substituting equation 3.9 into 3.14, we get
δu=2 ( 1−2 x1 ) kδt

9FINITE DIFFERENCE METHOD AND O2
Or, δu=−2k ( 2 x1−1 ) kδt
Response to Q. 3.16
Given equation 3.15
δu=−2k ( 2 x1−1 ) δt
Equation 3.12
u=2 x1−1
Substituting equation 3.12 in 3.15, we get
δu=−2kuδt
Response to Q. 3.17
N k dt t N1 N2 dN1
new
u
Step 0 10 0.05 5 0 10 0 -2.5
Step 1 10 0.05 5 5 7.5 2.5 -1.25 0.5
Step 2 10 0.05 5 10 6.25 3.75 -0.625 0.25
Step 3 10 0.05 5 15 5.625 4.375 -0.3125 0.125
Response to Q. 3.19

10FINITE DIFFERENCE METHOD AND O2
0 10 20 30 40 50 60
-1
-0.75
-0.5
-0.25
0
0.25
0.5
0.75
1
1.25
Series2
Theory
Response to Q. 3.20
a) 3.8s
b) 3.88s
c) 3.885
Response to Q. 3.22
0 0.05 0.1 0.15 0.2 0.25 0.3
-0.505
-0.5
-0.495
-0.49
-0.485
-0.48
u
u

11FINITE DIFFERENCE METHOD AND O2