For this block, use Laplace transforms to discuss the general solutions of the following differential equations. In each case plot solution curves for a variety of parameters and given initial conditions:

Specific first and second-order linear differential equations:

(a) The first two first-order equations on p. 466

(b) Any two second-order equations on p. 466.

(c) Two examples of linear systems from pp. 470-471.

Find the general solution using Laplace transforms (not by any other method) and plot solutions over a range of t values, for a given initial condition.

First-Order Equations

Problem 1:

,

The MATLAB code I used was:

>> syms s t;

>> ilaplace(3/(2*(s-1))+ 1/(2*(s-3)),s,t)

ans =

3/2*exp(t)+1/2*exp(3*t)

To plot I used:

>> ezplot ‘3/2*exp(t)+1/2*exp(3*t)’

The solution comes from -∞ to +∞ along the y-axis at 0 in the direction of the x-axis. When the graph approaches about x = 3, the slope of the graph increases towards +∞ in the y- direction. After x =4, the slope of the graph increases rapidly between x =4 & x =5 towards+∞ in the y- direction. (Following the exponential trend.)

Problem 2:

$latexs \mathscr{L}[y] – y(0) + \mathscr{L}[y] = \frac{3s}{s^2+1}$

The MATLAB code I used was:

>> syms s t;

>> ilaplace(3*s/(2*(s^2+1))+ 3/(2*(s^2+1)) – 7/(2*(s+1)),s,t)

ans =

3/2*cos(t)+3/2*sin(t)-7/2*exp(-t)

To plot I used:

>> ezplot ‘3/2*cos(t)+3/2*sin(t)-7/2*exp(-t)’

The function goes from -∞ to +∞ in the y direction. The slope is at first very steep then around x= -1.5 it looks as if it goes level; however, it doesn’t. It does show the exponential features, but the function also has ‘cos and sin’ in it. The graph below shows that at approximately x = .5, the function starts to follow the sin wave towards +∞ in the x direction. Don’t just trust your first graph.

Second-Order Equations

Problem 3: Number 3 on p. 466:

The MATLAB code I used was:

>> syms s t;

>> ilaplace((2*s – 3)/(s^2 + 4))

ans =

2*cos(2*t)-3/2*sin(2*t)

To plot I used:

>> ezplot ‘2*cos(2*t)-3/2*sin(2*t)’

The graph displays a function like a basic sin and cos wave with a magnitude of about 2.5.

Problem 4: Number 4 on p. 466:

The MATLAB code I used was:

>> syms s t;

>> ilaplace(7*s/(s^2+16))

ans =

7*cos(4*t)

To plot I used:

>> ezplot ‘7*cos(4*t)’

So you can see this is a basic cos function with the magnitude of 7 & the ’stretch’ of 4t from -∞ to +∞ in the x direction.

Linear Systems of Equations

Problem 5: Problem 1, p 470

dx/dt:

dy/dt:

Solving System for x(s):

x(t):

Solving System for y(s):

y(t):

The MATLAB code I used was:

For \mathscr{L}[y]:

>> syms s t;

>> ilaplace(-1/(s^2 – 3*s + 2))

ans =

-exp(2*t)+exp(t)

For \mathscr{L}[x]:

>> syms s t;

>> ilaplace(-1/(s+2) + 2/(s+1) + 2/s)

ans =

-exp(-2*t)+2*exp(-t)+2

To plot I used:

>> x= @(t) -exp(-2*t)+2*exp(-t)+2;

>> fplot (x, [-4,4,-4,4])

>> hold on

>>y =@(t) -exp(2*t)+exp(t);

>>fplot (y, [-4,4,-4,4], ‘r’)

>> grid on

The blue line displays the X and red line displays the Y. The X comes with a steep slope from -∞, hits x= 0 @ y =3 then drops slowly and levels off at y = 2 and continues to +∞ The Y: It is coming from -∞ with a slight slope to about y=.25 till about x=-.5 and then the function just drops rapidly between x=0 and x = 1.

Problem 6: Problem 2, p.460

dy/dt:

$ latex \mathscr{L} [x] = -s \mathscr{L} [y] + 0 $

Solving System for x(s):

x(t):

Solving System for y(s):

y(t):

The MATLAB code I used was:

>> syms s t;

>> ilaplace(-s*(3/((1-s^2)*(s-2)) + 2/(1 – s^2)))

ans =

-sinh(t)+2*exp(2*t)

>> syms s t;

>> ilaplace(3/((1-s^2)*(s-2)) + 2/(1 – s^2))

ans =

cosh(t)-exp(2*t)

To plot I used:

>> x= @(t) -sinh(t)+2*exp(2*t);

>> fplot (x, [-1,1,-10,10])

>> hold on

>>y =@(t) cosh(t)-exp(2*t);

>>fplot (y, [-1,1,-10,10], ‘r’)

>> grid on

Once again, the blue line displays the X while the red line displays the Y. The X shows a function coming from – ∞ in the x direction and its slope becomes increasingly greater. The Y shows a function coming from about the same location as the X but it slops drps but not as much as the X.

1.

(a) Plot the vector field for the linear function T[(x,y)] = (x + y, x).

(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.

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

2. Repeat this for the linear functions T[(x,y)] = (ax + by, cx + dy) for several choices of integers a, b, c, d chosen randomly in the range -5 ≤ a, b, c, d ≤ 5.

Introduction

For the fifth block, systems of linear differential equations was discussed–which is provided on pages 287-300, 325-337, 353-377 of the text.

An example of systems of linear differential equations is two differential equations such as:

where are constants

(*There is no higher order term in x and y since we are dealing with linear differential equations.)

Two linear differential equations: and give you the formula of

The unit square in the first graph transforms into the second graph displaying the parallelogram.

Part 1

Part A: Plot the vector field for the linear function T[(x,y)] = (x + y, x).

To plot this vector field I used the MATLAB code:

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

>> u=x+y;

>> v=x;

>> DW=sqrt(u.^2+v.^2);

>> quiver(x,y,u./DW,v./DW,.5)
A better (zoomed) version of the vector field is shown below:

The vector field has the slope of about 1 and the field lines are going from left to right in a diagonal. When it hits approximately 0, the vector lines switch direction. It seems as though there is a ‘X’ made and in between each line, the vector fields are in the shape of parabolas.

Part 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.

Problems:

1. $latex x + y = \lambda x$ and 2.

1. $latex ( \lambda – 1) x – y = 0$ (*Bringing all variables to one side.)

2. (*Bringing all variables to one side.)

1. (*Substitution)

2. (*Substitution)

1. (*Plugging it back into the equation.)

2. (*Plugging it back into the equation.)

(*X & Y are non-zeros.)

(*Quadratic for λ.)

1. (*Line solution.)

2. (*Line solution.)

The MATLAB code I used was:

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

>> u=x+y;

>> v=x;

>> DW=sqrt(u.^2+v.^2);

>> quiver(x,y,u./DW,v./DW,.5)

>> hold on

>> x = -1:.1:1;

>> y1 = -1/2*(1+sqrt(5))*x;

>> y2 = 1/2*(sqrt(5)-1)*x;

>> plot(x,y1)

>> plot(x,y2)

A better (zoomed) version of the vector field is shown below:

As you can see, the vector lines are split into four quadrants (on a 45 degree turn of the normal quadrants).

Part C: Try to describe the flow geometrically, in words.

The vector fields follow one of the tangent lines, and then curves and follows the others direction. Approximately a 90 degree direction change closer to the tangent lines and progressively less as it moves away from the lines.

Part 2

Part A: Plot the vector field for the linear function T[(x,y)] = (2x + y, x + y).

The MATLAB code I used was:

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

>> u=2*x+y;

>> v=x+y;

>> DW=sqrt(u.^2+v.^2);

>> quiver(x,y,u./DW,v./DW,.5)

A better (zoomed) version of the vector field is shown below:

From looking at the graphs, it seems as though there is a tangent line that has a slope of 1 going from left to right at a 45 degree angle. All of the vector segments seem to protrude around the origin (0,0) then splits to either side of the tangent line.

Part 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.

Problems: 1. and 2.

1. (*Bringing all variables to one side.)

2. (*Bringing all variables to one side.)

1. (*Using substitution.)

2. (*Using substitution.)

1. (*Plugging in.)

2. (*Plugging in.)

(*X & Y are non-zeros.)

(*Quadratic for λ.)

1. (*Line solution.)

2. (*Line solution.)

The MATLAB code I used was:

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

>> u=2*x+y;

>> v=x+y;

>> DW=sqrt(u.^2+v.^2);

>> quiver(x,y,u./DW,v./DW,.5)

>> hold on

>> x = -.5:.1:.5;

>> y1 = 1/2*(1+sqrt(5))*x;

>> plot(x,y1)

>> x = -1:.1:1;

>> y2 =1/2*(sqrt(5)-1)*x;

>> plot(x,y2)
A better (zoomed) version of the vector field is shown below:

You can see that the tangent line is where all of the vector lines originate from.

Part C: Try to describe the flow geometrically, in words.

The vector lines seem to spill outward as if the origin (0,0). If the field was 3D, it would be in the shape of a vortex, but its not a particular singular point, it would be along the tangent line in which the vector field would be come out from. A good visual example would be to take a piece of paper and crease it. now bend the two ends slightly, that would be the depiction.

1. Use Matlab, go through all the cases:

* Case 1: Two real roots

* Case 2: One real root

* Case 3: Two complex roots which are complex conjugates

for the roots of the characteristic equation, describing the geometry of the solutions of the differential equations.

for the linear function corresponding to the matrix

Case 1: Two real roots

a = 4, b = 1, c = 3, d = 2

Solve for roots:

(\lambda ^2 – 6 \lambda + 5 = 0)

(( \lambda – 1)( \lambda – 5) = 0)

Roots: λ = 1, 5

Invariant Lines:

((4 – \lambda) x + y = 0)

( y = – (4 – \lambda) x)

( y = – (4 – 1)x)

(y = – (4 – 5)x)

y = -3x , x

The MATLAB code I used was:

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

>> u=4*x+y;

>> v=3*x+2*y;

>>DW=sqrt(u.^2+v.^2);

>> quiver(x,y,u,v,1.25)

>>hold on

>> ezplot (’-3*x’, [-1,1,-1,1])

>> ezplot (’x’, [-1,1,-1,1])

The vector field is coming out from the origin of [0,0] and goes outward (almost like an inverted cone). However, at the 3x invariant line, the vector fields seem to curve to the left or the right. The farther the way from the line, the vector lines are getting longer which means they are increasing in magnitude, so they are protruding from the inverted cone and continuing in the Z direction at a rapid rate.

Case 2: One real root

a = -4, b = 7, c = 0, d = -4

(y’=-4y)

Solve for roots:

(-4x+7t=lambda x)

(-4y=lambda y)

(-4-lambda)x +7y=0)

(-4-lambda)y=0)

( (-4 – \lambda)(-4 – \lambda) + 7(0) = 0)

((-4 – \lambda)^2 = 0)

Roots:

Invariant Lines:

((-4 – \lambda)x + 7y = 0)

(y = \frac{-1}{7}(-4 – \lambda)x)

(y = \frac{-1}{7}(-4 – (-4)))

y = 0

The MATLAB code I used was:

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

>> u=-4*x+7*y;

>> v=-4*y;

>> quiver(x,y,u,v,1.25)

>>hold on

>>ezplot (’0′, [-1,1,-1,1])

There is downward spiral that goes from the origin of [0,0] and approaches the single invariant line. The vector lines get shorter and shorter with the approach of the invariant line from -∞ to +∞. (Almost looking like a whirlpool.)

Case 3: Two complex roots which are complex conjugates

a = 2, b = 1, c = 2, d = 1

(y’ = 2x + y)

Solve for roots:

(2x + y = \lambda x)

(2x + y = \lambda y)

Roots: ( \lambda = \frac {1}{2}y – sqrt{17})

Invariant Lines:

y = x

y = -2x

The MATLAB code I used was:

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

>> u=2*x+y;

>> v=2*x+y;

>>DW=sqrt(u.^2+v.^2);

>> quiver(x,y,u,v,1.25)

>>hold on

>>ezplot (’x’, [-1,1,-1,1])

>>ezplot(’-2*x’, [-1,1,-1,1])

1. Choose 6 examples from pp. 24, 25, 26, problems 12-16 pp. 76-77 of the text and carry out the integrations in the variables separable method outlined above, to obtain exact solutions for these differential equations.

2. In each case, compare your exact solution with that obtained using dsolve in MATLAB

Introduction

For my fourth post we learned about variable separable equations–which are differential equations that produce an exact solution.

A differential equation is expressed in the differential form of:

(Newton used whereas Leibniz used )

An example we learned in class was –which is a variable seperable equation unlike . A variable separable equation is when a differential equation, , can be written as the function for some functions and .

The example , , can be written as where and . We then rewrite the equation, as (not involving any values of for which ). In this form, we can then carry out an integration with respect to the variable to get (not involving any values of for which ).

Also the equation, , can be rewritten as when using substitution.

Example One

(provided on page 24 of the text, problem #1)

Using the Variables Seperable Method:

(*It is easier to bring the to the other side of the equal sign.)

(*My first step on trying to get the variables, x and y, on either side.)

(*The next step is to integrate.)

(*Whereas C is some constant.)

Therefore:

Using MATLAB’s DSOLVE:

>>dsolve(‘Dy = -4*x*y/(x^2+1)’, ‘x’)

ans =
C1/(x^2+1)^2

>>ezplot ‘-2/(x^2+1)^2′

Example Two

$latex tan x dy+2y dx$

(provided on page 24 of the text, problem #2)

Using the Variables Seperable Method:

(*We know this using Trigonemtric Identities.)

Therefore:

Using MATLAB’s DSOLVE:

>>dsolve(‘-2*y/tanx’, ‘x’)

ans =
-2*C1/(-1+cos(2*x))

>>ezplot ‘-2/(-1+cos(2*x)’

Example Three

(provided on page 25 of the text, problem #5)

Using the Variables Seperable Method:

Therefore:

Using MATLAB’s DSOLVE:

>>dsolve(‘Dy=x*y/(x+1)’, ‘x’)

ans =
C1*exp(-x)*(x+1)

>>ezplot ‘exp(-x)*(x+1)’

Example Four

(provided on page 25 of the text, problem #10)

Using the Variables Seperable Method:

(*Seperating the variables.)

(*Taking the integration.)

(*I multiply by 2 so x isn’t being divided by 2.)

(*This is the answer after taking the square root of both sides.)

Therefore:

Using MATLAB’s DSOLVE:

>>dsolve(‘Dx=(t-1)/(x)’, ‘t’)

ans =
-(-2*t+t^2+C1)^(1/2)

(-2*t+t^2+C1)^(1/2)

>>ezplot ‘-(-2*t+t^2+C1)^(1/2)’

Example Five

(provided on page 77 of the text, problem #16)

Using the Variables Seperable Method:

Using MATLAB’s DSOLVE:

>>dsolve(‘Dy=(y^2)/(x)’, ‘x’)

ans =
1/(1+C1*x)

>>ezplot ’1/(1+x)’

Example Six

(provided on page 26 of the text, problem #24)

Using the Variables Seperable Method:

Using MATLAB’s DSOLVE:

>>dsolve(‘(y^2+2*x*y)/(x^2)’, ‘x’)

ans =
-x^2/(x-C1)

>>ezplot ‘-x^2/(x-1)’

1. Explore how exact solutions to first order differential equations differ from Euler approximations.
2. Experiment with Euler approximations to the Lorenz equations for various parameter values and initial conditions.
3. Choose one of the following:

a. The Lotka-Volterra predator-prey equations are where y is the number of predators, x is the number of prey, and are parameters that tune the interaction of the predators and prey. Explore solutions of the predator-prey equations for varying parameter values and initial conditions.

b. Explore the behavior of the Rossler system of differential equations for various parameter values and initial conditions.

Introduction

For the third block we learned about Edward Lorenz (1917-2008) who was not only a mathematician but also a meteorologist from West Hartford, Connecticut. Lorenz studied at both Dartmouth College and Harvard University. He earned his two degrees at the Massachusetts Institute of Technology–where he also taught for many years. Edward Lorenz is also considered a “pioneer” of chaos theory (sensitivity to initial conditions). He is also known for the Lorenz attractor and the Butterfly effect.

A simplified model was set up by him of convection rolls in 1963. Strangely, Lorenz was not fascinated by biological or physical phenomena, but rather used his model to try and describe the weather. The simplified model consists of three linked differential equations using three different variables:

(where and are positive numbers)

The parameters and to physical properties of a fluid in motion: kinematic viscosity and heat flow (the Prandtl number and Rayleigh number, respectively). In these equations it is common to take , .

The method for setting up an Euler approximation to a system of first-order differential equations is more difficult than for a single first-order equation (while keeping in mind that the Euler approximation formula the variables are the linked equations above). Luckily, MATLAB has a 3D plot capability. The two options that MATLAB provides us with to execute the comparison of the Euler approximation and the Lorenz Attractor, is the“inline” following the MATLAB prompt, or by defining an M-file to run.

After observing the error associated with the relationship between the error and the step size of the Euler approximations, it is noticeable that the smaller the step size, the the more accurate the approximation usually is. Also, after observing the Euler approximation from the previous block, we know that there is some error for every approximation of a non-linear solution because Euler’s method approximates by using a straight line in very small increments. There is still error even when every approximated points are on a curve corresponds to the exact solution, because those points are connected with line segments instead of curves like I previously mention.

The Euler M-file is:

% Euler’s method for a single differential equation.
% The differential equation is dy/dt = f(t,y).
function [ t, y ] = euler ( f, t_range, y_initial, nstep )
% We set a range of time values, from t(1) to t(2).
t(1) = t_range(1);
% We define dt by dividing the time range into an equal number of specified steps.
dt = ( t_range(2) – t_range(1) ) / nstep;
% We set the initial value of y at the beginning time t(1).
y(1) = y_initial;
% We use Euler’s method to update the value of y at new time steps.
% “feval” is used instead of “eval” because we are passing the name of f to the program.
for i = 1 : nstep
t(i+1) = t(i) + dt;
y(i+1) = y(i) + dt * feval ( f, t(i), y(i) );
end
% The following command plots y as a function of t.
plot(t,y)

The two data vectors the M-file returns are t–the value of the independent variable at nstep+1 equally spaced points–and y–the value of the dependent variable at each t. It also plots y as a function of t over the time limit specified.

The next step is to write a function specifying:

• the name of the function appears in single quotes
• the name of the function test_example appears in single quotes
• the initial value of y
• the number of steps

The necessary function M-File is:

function yprime = test_example ( t, y )
yprime = y*(2./t-1);

For the example, for 0.1 ≤ t ≤ & y(0.1) = 0.1, the plot will look like:

Euler’s Approximation for Lorenz Equations

Now I will be exploring Euler Approximations for Lorenz Equations (previously mentioned in the beginning of this block). In order to find a first-order equation, we must modify euler.m file and call it euler_system.m

euler_system.m:

% Euler’s method for a system of differential equations.

% The differential equations are dy/dt = f(t,y) where y is a vector of unknown functions. function [ t, y ] = euler_system( f, t_range, y_initial, nstep )

% We set a range of time values, from t(1) to t(2). t(1) = t_range(1);

% We define dt by dividing the time range into an equal number of specified steps. dt = ( t_range(2) – t_range(1) ) / nstep;

% We set the initial value of the vector y at the beginning time t(1). y(:,1) = y_initial;

% We use Euler’s method to update the value of y at new time steps.

% “feval” is used instead of “eval” because we are passing the name of f to the program.

for i = 1 : nstep
t(i+1) = t(i) + dt;
y(:,i+1) = y(:,i) + dt * feval ( f, t(i), y(:,i) );

end

% The following command plots the components of y as functions of t. plot(t,y)

% We also want a 3D plot of the vector components as they vary with t.
plot3(y(1, : ),y(2, : ),y(3, : ))

The colon is used because y is a column vector (that is what the ’1′ is used for) which as as many components (= rows) as there are unknown functions. The graph will become more stable as it approaches 28 in which it peaks, but after 28 it becomes increasingly chaotic.

The following inputs are what I started with:
ρ =5, t=[0,5] , y initial=(0,5,0)

lorenz_system.m:

function yprime = lorenz_system ( t, y )
yprime = [ 10.0* (y(2)-y(1)); y(1)*(5.0-y(3))-y(2);y(1)*y(2)-8*y(3)/3 ];

>> y_init = [0; 5; 0];
>> [ t, y ] = euler_system (’lorenz_system’, [ 0.0, 5 ], y_init, 1000 );

Lorenz at p=5:

Lorenz at p=10, t= [0,5], y initial=(0,5,0):

Lorenz at p=15, t= [0,5], y initial=(0,5,0)

Lorenz at p=20, t= [0,10], y initial=(0,5,0)

Lorenz at p=25, t= [0,12], y initial=(0,5,0)

Lorenz at p=28, t= [0,15], y initial=(0,5,0)

Lorenz at p=50, y initial=(0,5,0)

at t=(0,5)

at t=(0,10)

at t=(0,11)

at t=(0,12)

at t=(0,13)

The graph has a lot less chaotic nature as the t value is smaller; it becomes progressively unrecognizable as it the value of t increases.

Runge-Kutta for Lorenz Equations

As was discovered in the previous block, the Runge-Kutta (Ode 45) Method usually provides a better approximation.

g.m

function xdot = g(t,x)
xdot = zeros(3,1);
sig = 10.0;
rho = 28.0; % change for desired rho
bet = 8.0/3.0;
xdot(1) = sig*(x(2)-x(1));
xdot(2) = rho*x(1)-x(2)-x(1)*x(3);
xdot(3) = x(1)*x(2)-bet*x(3);

lorenz_demo.m

function lorenz_demo(time)

% Usage: lorenz_demo(time)
% time=end point of time interval
% This function integrates the lorenz attractor

% from t=0 to t=time
[t,x] = ode45(’g’,[0 time],[1;2;3]);
disp(’press any key to continue …’)
pause

plot3(x(:,1),x(:,2),x(:,3))
print -deps lorenz.eps

(*The p value will be changing in the g.m like the euler.m file. The command line will call the “lorenz_demo(200)”–where the 200 specifies the time interval.)

at p=5:

at p=10:

at p=15:

at p=20:

at p=25:

at p=28:

The Runge-Kutta Method provides a much better approximation of the Lorentz attractor–compared to using the Euler’s Method–since the line is much smoother. It is a better representation of how the approximation is creating the object in 3D space. The Euler’s method was much more jagged, and wasn’t as defined where the Runge-Kutta Method produced a better graph using a much smaller time step.

Lotka-Volterra predator-prey equations

Proposed independently by Alfred J. Lotka in 1925 and Vito Volterra in 1926 are the Lotka-Volterra equations. Also known as the predator-prey equations, they are a pair of first order, non-linear, differential equations frequently used to describe the dynamics of biological systems in which two species interact (one a predator and one its prey).

The prey equation:

(*The change in the prey’s numbers is given by its own growth minus the rate at which it is preyed upon.)

The prey are assumed to have an unlimited food supply, and to reproduce exponentially unless subject to predation. The exponential growth is represented in equation above by the term αx. The rate of predation upon the prey is assumed to be proportional to the rate at which the predators and the prey meet; this is represented above by βxy. (There is no prediction if either x or y is zero.)

The predator equation:

In this equation, δxy represents the growth of the predator population. (Note the similarity to the predation rate; however, a different constant is used as the rate at which the predator population grows is not necessarily equal to the rate at which it consumes the prey). γy represents the natural death of the predators; it is an exponential decay. Hence the equation represents the change in the predator population as the growth of the predator population, minus natural death.

Both equations use:

• y is the number of some predator (for example, wolves)
• x is the number of its prey (for example, rabbits)
• and represent the growth of the two populations against time
• time is represented by the variable t
  
• α, β, γ and δ are parameters representing the interaction of the two species

The equations have periodic solutions which do not have a simple expression in terms of the usual trigonometric functions. However, an approximate linearised solution yields a simple harmonic motion with the population of predators following that of prey by 90°.

The equations would look like the following:

Another graph comparison:

The following are actual examples with varying parameter values and initial conditions using the m files shown below:

lotka.m

function yp = lotka(t,y)

yp = diag([1 - .01*y(2), -1 + .02*y(1)])*y;

LV.m

t0 = 0; % this is the start time of the experiment

tfinal = 20; % this is the end time of the experiment

y0 = [5 3]‘; % these are the initial numbers of predator and prey

[t,y] = ode45(’lotka’,[t0 tfinal],y0);

figure;

% Plot the result of the simulation two different ways.

subplot(1,2,1)

plot(t,y)

title(’Time history’)

legend(’Predators’,’Prey’)

subplot(1,2,2)

plot(y(:,1),y(:,2))

title(’Phase plane plot’)

An example of start time to=0, tfinal = 100 and y0= [300 300]:

The first plot is the time (time being dimensionless) in which it shows the progression of the two species over time. One can also plot a solution which corresponds to the oscillatory nature of the population of the two species. At any given time, the solution is somewhere on the inside of these elliptical solutions. As you can see, as the number of predators decreases, the number of prey increases drastically, but right when the number of predators to approximately \frac {1}{4}th to \frac {1}{2} of the maximum of prey, the prey drops drastically, close to zero. And the cycle repeats.

The second plot shows the model system in which the predators thrive when there are plentiful prey but, ultimately, outstrip their food supply and decline. As the predator population is low the prey population will increase again. These dynamics continue in a cycle of growth and decline. A better representation would be the one created from wikipedia.

This depicts that there is a problem with the ’survival of the fittest.’ In each cycle, the baboon population is reduced to extremely low numbers yet recovers (while the cheetah population remains sizeable at the lowest baboon density). Given chance fluctuations, discrete numbers of individuals, and the family structure and life-cycle of baboons, the baboons actually go extinct and by consequence the cheetahs as well. This modeling problem has been called the “atto-fox problem“ an atto-fox being an imaginary 10-18 of a fox, in relation to rabies modeling in the UK.

Here are a few more parameter values and initial conditions. (to=0, tfinal = 30 and y0= [2 5] so that we can see a better representation.)

]]> http://jcloutier1.wordpress.com/2009/03/04/block-3-assignment/feed/ 0 jcloutier1 Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket Photobucket http://jcloutier1.wordpress.com/2009/03/04/block-2-assignment/ http://jcloutier1.wordpress.com/2009/03/04/block-2-assignment/#comments Wed, 04 Mar 2009 13:15:03 +0000 jcloutier1 http://jcloutier1.wordpress.com/?p=25 ]]> Objective

Using:

1. Euler’s method
2. ode45
   
3. dsolve

examine a variety of first-order differential equations and compare and contrast Euler’s method, MATLAB’s ode45 numerical solver, and an exact solution when there is one. Choose two examples from each of the following problem sets (6 examples in all) for a variety of inital conditions (not just those given in the text):

• problems 2-5 on pp. 138-139 of the text, ;
• problems 6-11 on p. 139 of the text;
• problems 12-16 on p. 139 of the text

Introduction

In the previous block assignment I learned about Euler’s method. However, in this block assignment, it was proven that there are some disadvantages of this method. Euler’s method is generally used for approximating numerical solutions to differential equations, yet, this method is subject to errors and can be very inaccurate. Another method, such as the Runge-Kutta method, is more accurate. This new method we learned about uses parabolas (degree 2 polynomials) and quartics (degree 4 polynomials) in order to approximate a differential equation, whereas Euler’s method approximates by using a straight line in very small increments. To show the comparisons of these two methods we used MATLAB–which has a built in solver for the Runge-Kutta method called ode45.

Example One: Problem Two

Shown below is the differential equation using Euler’s method proving that the method is jagged and not very accurate.

[x,y]=ode45(’example1’,[-3,1],.1);

plot(x,y)

In the graph below that is comparing both methods, the methods appear to be pretty accurate to one another when x is negative. However, as x approaches 1.5 Euler’s method becomes less accurate.

[x,y]=ode45(‘example1′,[-3,1],.1);
plot(x,y)
hold on
[x,y]=euler(‘example1′,[-3,1],.1,.2);
plot(x,y,’r')
hold off

However, as you can see, by increasing the step size of the graph the methods differ greatly and proves that the Euler method isn’t the best option.

Example Two: Problem Three

[x,y]=ode45(‘example2′,[-3,1],.1);
plot(x,y)
hold on
[x,y]=euler(‘example2′,[-3,1],.1,.2);
plot(x,y,’r')
hold off

The graph above displays an example of using both Euler’s Method (in red) and MATLAB’s ode45 numerical solver (in blue).

Example Three: Problem Nine

[x,y]=ode45(‘example3′,[0,1.5],-3);
plot(x,y)
hold on
[x,y]=euler(‘example3′,[0,1.5],-3,.15);
plot(x,y,’r')
hold off

Example Four: Problem Ten

[x,y]=ode45(‘example4′,[0,2.5],.1);
plot(x,y)
hold on
[x,y]=euler(‘example4′,[0,2.5],.1,.1);
plot(x,y,’r')
hold off

Example Five: Problem Twelve

[x,y]=ode45(‘example5′,[-3,3],.1);
plot(x,y)
hold on
[x,y]=euler(‘example5′,[-3,3],.1,.4);
plot(x,y,’r')
hold off

Example Six: Problem Thirteen

[x,y]=ode45(‘example6′,[-1.5,1.5],.1);
plot(x,y)
hold on
[x,y]=euler(‘example6′,[-1.5,1.5],.1,.01);
plot(x,y,’r')
hold off

From the graph above you can determine that Euler’s Method (the line in red) for this step size exhibits an inaccurate graph. When reading from left to right, one can see that the line segments aren’t accurate. However, on the graph below, both Euler’s Method and MATLAB’s ode45 numerical solver look so similar they are almost exact. This graph is displayed when decreasing the step size.

For this block assignment I worked with Eric Donahoe.

To solve the following problems graphically and numerically.

Problem 1:
Assigned problem:

Problem 2:
Number 18 in the text:

Introduction

For my first post, I will describe two differential equations using both graphically and numerically methods.A differential equation is a mathematical equation for an unknown function involving one or more variables relating the function itself and its values and of its derivatives of different orders. Differential Equations occur whenever a deterministic relationship involving endless changing quantities (represented by functions) and their rates of change (expressed as derivatives) is known or suggested. The equations I will be explaining are —which was a problem given— and —which is problem 18 in the text, A Course in Ordinary Differential Equations by Randall J. Swift and Stephen A. Wirkus.

For the problem given: ( )

Using MATLAB, I will begin by plotting a direction field for this differential equation. To create the plot, the code for MATLAB I used was (using page 85 in the text as a guide):

> [t,y]=Euler1(‘Example1′,[0,2],1,.01);
> [T,Y]=meshgrid(-1:.5:3,-1:.5:2);
> DY=Y.^2+T;
> DT=ones(size(DY));
> DW=sqrt(DT.^2+DY.^2);
> quiver(T,Y,DT./DW,DY./DW,.2,’.');
> xlabel (‘t’)
> ylabel (‘y’)

As a function:
> function f=Example2(tn,yn)
> f=(yn.^2+tn)

*The meshgrid command used above generates X and Y arrays for 3-D plots. Each line segment has an associated length by default that is scaled so all the vectors fit in the window. Tangent lines of varying size would make the direction field very confusing, so each line is scaled by its magnitude to have the same length of 1.

*The quiver command used above creates a “quiver” or velocity plot–which is a plot that displays velocity vectors as arrows by default. The ‘.’ is used as the last entry of the quiver command to tell MATLAB not to put an arrow to produce clearer graphs to see possible solution curves.

The MATLAB code I used for the Euler’s Method is:

> function [xout,yout]=Euler1(fname,xvals,y0,h)
> x0=xvals(1);xf=xvals(2);
> xn=x0;
> yn=y0;
> xout=xn;
> yout=yn;
> steps=(xf-x0)/h;
> for j=1:steps
> fn=feval(fname,xn,yn);
> xn=xn+h;
> yn=yn+h*fn;
> xout=[xout;xn];
> yout=[yout;yn];
> end

By adjusting the code to :

> [T,Y]=meshgrid(-1:.5:3,-1:.5:2);

You will receive the following plot:

The first graph displays a large view of the direction field while the second graph displays a better view of what is going on. It is easier to see that the tangent lines have a “swooping” effect from left to right. The more the graph goes towards the right the direction field seems to become more vertical.

The Euler approximation yielded an appropriate solution to this equation.

Displayed below is the direction field lines that are tangent to the solution curve indicating it is a successful solution.

For problem 18 in the text: ( )

> [x,y]=Euler1(‘Example2′,[0,2],1,.01);

> [X,Y]=meshgrid(-1:.5:3,-1:.5:2);

> DY=X+Y;

> DX=ones(size(DY));

> DW=sqrt(DX.^2+DY.^2);

> quiver(X,Y,DX./DW,DY./DW,.2,’.');

> hold on

> plot (x,y)

> hold off

As a function:
> function f=Example2(x,y)
> f= x+y

Like the previous, assigned problem, my second example shows tangent lines have a “swooping” effect from left to right. Also, the more the graph goes towards the right the direction field seems to become more vertical.

The Euler approximation yielded an appropriate solution to this equation.

Displayed below is the direction field lines that are tangent to the solution curve indicating it is a successful solution.