Reference no: EM131014456

Problem 1. Find and classify the fixed points of x? = x³ - 6x² - 13x + 42 as stable or unstable using the following 3 methods:

• a one-dimensional analysis. (Hint: Maple has a factor command which will help to find fixed points.);

• a two-dimensional Maple analysis. That is, plot the solutions on the plane with different initial values as on the one dimensional flow examples Maple worksheet;

• a linear stability analysis (i.e. find f′(x0) and explain);

• a Liapunov function as on the one dimensional flow examples Maple worksheet.

Problem 2. A particle travels on the half-line x ≥ 0 according to x? = -x^c, where c is a real number.

• Find all the c such that x = 0 is a stable fixed point.

• Given a c for which x = 0 is stable, is it possible that the flow reaches x = 0 in a finite amount of time? Specifically, calculate how long it takes for a particle at x = 1 to get to x = 0. (Hint: (1) you have to consider di?erent cases for c; (2) separate variables and integrate to get T = ₀∫¹ -? - dx.)

• For the two different cases of c which you determine above, choose representative values for c (e.g. c = 1/2, c = 2), solve the differential equation numerically as in class and plot the different solutions. Explain why these plots verify your conclusions above about how long it takes a particle to reach x = 0.

Problem 3. (Non-Uniform Oscillators)

• For the vector field on the circle θ? = µ sin(θ) - sin(2θ), find the fixed points and draw the flow on the circle assuming µ ≥ 0. Do the cases µ ≥ 2, µ = 2, 0 < µ < 2 and µ = 0 separately. (Hint: expand sin(2θ).)

• Show that θ? = µ + sin(θ), where µ is slightly less than 1, is an excitable system. That is, it satisfies two properties: it has a unique globally attracting rest state; and a large enough stimulus can send the system on a long excursion until it returns to the rest state. What is the rest state? And what threshold do you have to overcome with a stimulus to send the system on the excursion? Hint: draw the flow on the circle.

• For the excitable system θ?= 0.95 + sin(θ), modify the procedure nonunifosciani from the non-uniform oscillator worksheet to see what happens when θ(0) = -π/2. For this situation, how much of a initial impulse would you have to give to the system (i.e. the angle θ) to make it enter the long excursion  state?

Problem 6. The Picard-Lindel of Theorem says that the initial value problem

dx/dt = f (t, x(t)),          x(0) = a

has a unique solution. The proof is via the Banach Fixed Point Theorem applied to the operator T (taking continuous functions to other continuous functions) defined by

T (x)(t) = a + _{t_0}∫^t f (s, x(s)) ds.

The method of the proof actually allows an iterative scheme for solving initial value problems to be invented.  Here it is.

The following Maple commands implement the Picard iteration scheme for solving an initial value problem

dx/dt = f (t, x(t)),          x(0) = a.

The method is given by the following: start with any function x₀(t) and x(0) = a and let

?₁(t) = a + _{t_0}∫^t f (s, x₀(s)) ds.

Continue by defining

?_n(t) = a + _{t_0}∫^t f (s, ?_n-1(s)) ds.

Picard's Theorem says that the ?n converge to the solution x∗ of the differential equation (with initial condition). So if you take ?n for n big enough, you should have a good approximation to the true solution.

Fill in the blanks (denoted by a dash -) in the procedure and commands below and carry out the rest of the commands to solve the indicated initial value problems.

>     with(LinearAlgebra):with(plots):with(DEtools):

>     picard:=proc(g,x_start,x0,t0,N)
local i,x;x:=x_start;
for  i  from  1  to  -  do
x:=x0+int(g(t,-),t=t0..t);
od:
x;
end:

Solve dx/dt = t with x(0) = 1.

>     f:=(s,y)->s;

>     picard(f,1,1,0,2);

Solve dx/ dt = x with x(0) = 1.

>     f2:=(s,y)->y;

>     picard(-,1,-,0,10);

Use dsolve to find the true solution of this initial value problem. Does your solution make sense in this light? Explain. (Hint: apply the command "taylor" to dsolve's solution to compare.  Look up the help on "taylor" by typing ?taylor.)

Take a di?erent starting function, x_start = sin(t), and do Picard iteration for the same initial value problem.

>     picard(f2,-,1,0,11);

Show that the answers are the same by taking a Taylor series of cos(t) and then subtract- ing cos(t) and adding the Taylor series as follows. Explain what you are  doing.

>     taylor(cos(t),t=0,11);

>     sort(simplify(picard(f2,sin(t),1,0,11)-cos(t)+ convert(taylor(cos(t),t=0,11),polynom)),  t,ascending);

Now use the Picard iteration method on the following two initial value problems. Check your answers with dsolve.

dx/dt = x(1 - x),           x(0) = 2;

dx/dt = t -x²,              x(0) = 0.

For these problems, take N larger and larger (up to 10 say) and show that you get better and better approximations to the true solution (from dsolve). But be careful. The Picard iteration could give you a gigantic answer. You can truncate the answer that picard gives you by using something like

taylor(sort(picard(f,0,0,0,9),t,ascending),t=0,30);

Here we take the Taylor series of the picard answer polynomial up to t²⁹. (You may wind  up with strange functions for answers - such as Airy functions - but they still have Taylor polynomials that Maple will find. So compare your answers by using these.)