I plotted the vector field of this system of linear equations using MATLAB. These are the sequence of commands I’ve ran in order to produce this image:

[x,y] = meshgrid(-1:1/10:1,-1:1/10:1);

u = x + y;

v = x;

quiver(x,y,u,v,1)

(b) Find any invariant lines, and write down the differential equations corresponding to a flow for which the vectors in the vector field are tangent to the flow.

I followed a sequence of steps to find these invariant lines for this linear system of equations. First, I produced the coefficient matrix:

With this matrix I could find the eigen values in MATLAB and use those eigen values to find the function of the invariant lines. I explored finding eigen values on paper as well; however, a powerful calculating tool such as MATLAB allows me to work faster. Regardless, I came up with these equations for some value, and Then, using algebra, the roots of the equation can be found. In this case the eigen values were -0.6180 and 1.6180. With these eigen values I can come up with the equations of the invariant lines. They are x = y/-0.618 and y = x/1.618. Now that I have the functions of the invariant lines, I may plot them. I did this using MATLAB. One note about plotting these in MATLAB is that the normal plot function won’t achieve the  exact result that I want. I would have to come up with an array of values. I used the fplot function instead. This command plots a function. It was a little tricky getting things to work out but using a function handle and the equation I was able to get the invariant lines to appear in the picture along with the quiver plot. The hold on command should be used so that these three different plots can all appear in the same figure.

f = @(x)  x/(-0.6180);

g = @(x)  x/1.6180;

fplot(g,[-1.5,1.5])

fplot(f,[-1.5,1.5])

The two images above and the one below are all of the same linear system with the invariant lines drawn. The only difference between them is the zoom. Above and below are zoomed in views of this system.

(c) Try to describe the flow geometrically, in words.

Generally, the vector fields seem to be flowing away from the origin. Some vectors are aimed towards it then the swerve away from it. Of course the closer the vector is to an invariant line the more it behaves like it. The invariant line go straight through the origin.

2. Repeat this for the linear functions for several choices of integers chosen randomly in the range

Now, right away I need to do the same thing but experiment with the values of four coefficients. This can be automated in MATLAB making more work in less time: Efficiency. Understanding the sequence of steps I performed for the first exercise, I can formulate it into an algorithm which I can write as a MATLAB function. Below is the MATLAB function that I wrote to automate this algorithm and make MATLAB do the work for me. The file is called hw4scr.m

function hw4scr(a,b,c,d)

% Homework 4 - Systems of linear equations and systems of differential

% linear equations.

% Playing with the form T[(x,y)] = (ax + by, cx + dy). a,b,c,d are the

% choices for integers.

clf;

[x,y] = meshgrid(-1:1/10:1, -1:1/10:1);

u = a*x + b*y;

v = c*x + d*y;

Matrix = [a b; c d];

eigen = eig(Matrix);

f = @(x) x/eigen(1);

g = @(x) x/eigen(2);

% Plotting the system. . .

quiver(x,y,u,v,1,'k')

hold on

fplot(g,[-1.5,1.5]);

fplot(f,[-1.5,1.5]);

This MATLAB function takes four arguments: a,b,c,d. These are the integers that I am supposed to play with. Pass the arguments to the function and run it and MATLAB does the rest. Below is the output from my MATLAB function after experimenting with random values. The invariant lines are blue and the vector field is obvoiusly the black arrows that populate the figure.

1.

with initial conditions

Here is an approximation of this system with t ranging from 0 to 10.

Problem 1 Figure A

This is what happens when the the range of t changes to 0 to 20.

Problem 1 Figure B

The range of t then extends to 30 (0-30).

Problem 1 Figure C

When the the step size is increased, the Euler solution becomes more precise.

Problem 1 Figure D

Basically the lines grow closer together when there are more steps while the lines will be more spaced out with lower steps.

2.

with initial conditions
When I began this problem I ran into an unconvincing answer.

Here you see an undefined slope

Problem 2 Figure A

The range of t seems to be too large. In Figure A it is {0-10} the number of steps is 2,000.

Below in figure B the range of t is shortened to {0-4}.

Problem 2 Figure B

As seen in figure B the slope is irrational. It seems that it is getting better, that meaning the Euler solution is becoming more precise and convincing. The number of steps used to plot Figure B was 1000.

Problem 3 Figure C

Finally in Figure C, I’ve arrived at this curve. T ranges from 0 to 0.4 and the number of steps here is 500.

This rigid curve obtained from using Euler’s method for a system of differential equations demonstrates that this

solution curve is only approximate in relation to the accurate solution.

Problem 3 Figure D

Here, in figure D I increase the number of steps to the plot so that a more smoother solution curve.

The number of steps is 1,000.

Problem 3 Figure E

In figure E, I increased the number of steps to 2,000. still while the range of t is {0-0.4}.

The difference between figure E and D is nearly unnoticeable. Now I’ll shorten the range of t.

Problem 3 Figure F

In Figure F, I lowered the range of t to {0-0.2}. The Euler solution curve gets more smooth and precise. The step size is 1,000.

Problem 3 Figure G

In figure G, the range of t is 0 to 0.1 and a 1,000 steps.

The solution curve is most convincing approximate Euler solution to this system of differential equations.

Problem 3 Figure H

In the above figure H, I plot both the Euler solution with a t range of 0 to 0.1 and 0 to 0.2 with 2,000 steps. The figure allows the viewer to compare the two curves. The blue line with dots indicates the 0 to 0.1 range Euler solution. The green line with the x’s indicates the 0 to 0.2 range Euler solution.

MATLAB: I used the given euler_system.m and larenzsystem.m files from Gary Davis’s blog at mth21202f09.wordpress.com

For problem 2, euler_system.m needed to be modified so that it would acommodate for the 2 differential equations from the system specified in Problem 2.

Another interesting note about problem 2 is that the Euler solutions seemed to blow up when t’s range was greater than .5

I believe it may have been technical problem. As the curve has a discontinuity of some sort around t = 0.5 and onward towards .

1. [Exact solution: ]

The euler approximation is seen how inaccurate it can be. It roughly takes the form of the solution on the graph.

The closer to , the closer that these two lines are. As x gets bigger and more positive, the lines diverge. At very large positive numbers, the euler approximation will be too inaccurate.

Euler (x,y)           Solution (x,y)

(0,5)                                (0,5)

(1,1.5)                              (1,3.694528)

(2,4.5)                              (2,27.29908)

2. [Exact solution: ]

The Euler approximation is more accurate this time. With a larger differences when moving closer to

Euler (x,y)           Solution (x,y)

(0,1)                                (0,1)

(0.5,0.5)                        (.5,0.713)

(1,0.5)                             (1,0.7357589)

(1.5,.75)                         (1.5,0.94626)

(2,1.125)                         (2,1.27)

3.   [Exact solution: ]

Both plots converge as we move towards

Euler (x,y)           Solution (x,y)

(0,2)                            (0,2)

(0.5,2)                        (.5,1.5576)

(1,1)                            (1,0.7357589)

(1.5,0)                        (1.5,0.21)

(2,0)                           (2,0.366313)

4.

Euler (x,y)

(0,1)

(0,1.5)

(1,2.375)

(1.5,4.6953)

(2,14.968)

(i)

C is some constant

3 K = 4 

(ii) 

C is some constant

(iii)

so

Q2. (i) Find all functions satisfying .

Two equations come from this

(ii) Find all functions   satisfying and .

(iii) Find all functions   satisfying and .

y(-1) = 2 so