MATS3150 Data Cleaning and Visualization Assignment Sample

MATS3150 Data Cleaning and Visualization

Question 1: Consider The Following System of First Order Differential Equations

x′ = 10 − x2 − x2
1 1 2
′ = (x2 − 1)(x2 − 3x1)
Determine all of the equilibrium solutions of the above system. You must include your working.

b) Determine the Jacobian of the system.

c) Determine the type and stability of each equilibrium point. Justify your answers.

Question 2: Consider the predator-prey model
x′ = x 2 − x2 , x (0) = 200
1 1 200 1
x′ = x −1 + x1 , x (0) = 500
2 2 250 2
where x1 denotes the size of the prey population, and x2 denotes the size of the predator population. The time t is measured in years.

Without plotting the solution to the above system, determine whether the predator and prey populations are increasing or decreasing when t = 0.

b) Use the Runge-Kutta method to approximate the population sizes of the prey and predator populations over a period of 15 years. Plot the population sizes for MBA assignment expert on a single set of axes (use appropriate labeling).

Solution

Q1 a)
Equilibrium solutions occur when;
n11 =1 and n12 =0
from n11 = 10 – n21 – n21 =0;
n21 + n21 =10
from n12 =( n-12 ) (n1 -3n1) =0
either n2-1 =0 or n2-3n1 =10
n2-1 =0 given n2=1
n22 +1 =10 => n12 =g =>n1 = +-3
this gives the equilibrium points (3,1) and (-3,1)

n2 -3n1=0 gives n2=3n1
n12 + (3n1)2 =10 => n12 +gn12 =10 => 10n12 =10
=>n12 =1
=>n1 =+-1
This gives equilibrium points (1,3) and (-1,-3)

So the equilibrium solutions are (3,1), (-3,1). (1,3) (-1, -3)

Solution Q 2 a)
Evaluate the derivatives t=0 with n1(0) =200 and n2(0) =500
n1 (0) =100(2-500/200) =200( 2-2.5)
=200(0.5) = -100
n21(0) =500 (-7 +200/500) =500 (-1 +0.8)
=500(-0.2) =-100
Since n1(0) <0 and n21(0) <0, both the predation and prey population are decreasing when t=0

Solution Q 2b)

Using range-kutta method to approximate the populations, we can write the algorithm and code. However, due to space and complexity, we provide a step by step outline.
Initiate t=0, n1(0) =200, n2(0) =500, step size h= some small value
Iteration, for t from 0 to 15 with step size, compute intermediate values k1, k2, k3, k4 for both n1 and n2.
Plot: Create a plot with item on the x- axis and the population size on the y-axis

Solution Q1 b)

The Jacobian Matrix J of the system is given by
J = ∂f1/∂x1 ∂f1/∂x2
∂f2/∂x1 ∂f2/∂x2

Where f1 =10 –x12 =x2 2 and F2 =( x2-1)( x2-3x1 )
∂f1/∂x1 =2x1 , ∂f2/∂x2 =-2x2
∂f2/∂x1 = -3 (x2-1), ∂f2/∂x2 = 2x2 -3x1-1

So the Jacobian matrix is
J= -2x1 -2x2
-3(x2-1 ) 2x2 -3x1-1

Solution Q1 C)

To determine the type and stability of each equilibrium point and analyze the eigen values
Equilibrium (3,1);
T(3,1) = -6 -2
0 -4

The eigen values are -6 and -4. Both are negatives indicating that 3,1 is a stable node

Equilibrium (-3,1);
T(-3,1) = 6 -2
0 -4

The eigen values are 6 and -4. One is positive and another is negative which shows that the node is an instable node hence a saddle point
Equilibrium (1,3 );
T(3,1) = -2 -6
-6 1

The eigen values are found by solving |J-λT| =0
λ2 + λ - 34 =0
λ=(-1±√13 7)/2
Both eigenvalues have negative points including that 1,3 is a stable focus

Equilibrium (-1,-3);
J(-1,-3) = 2 6
12 -4

The eigen values are found by solving |J-λT| =0
λ 2 + 2 λ +40 =0
λ=(-2±√(4-160))/2=-1±ⅈ√39
Both eigenvalues have positive real points including that -1,-3 is unstable focus.