MATH 345: Assignment 3 Solutions

= , = + sin cos sin, (, ) S1 R (1)

mrd2

dt2+ b

d

dtmr2 sin cos+mg sin = 0, S1 (2)

Q1

To nondimensionalize eqn. (2), we let = tB

, where B > 0 has units of [time]. We will choose thevalue for B later. Applying the chain rule:

d

dt=d

d

d

dt=

1

B

d

d, d

2

dt2=

1

B2d2

d 2

Making the appropriate substitutions into equation (2) gives us:

mr

B2d2

d 2+b

B

d

dmr2 sin cos+mg sin = 0, S1

To get the equation into the appropriate form we divide it by mg, thus giving us the coefficient1 in front of the sin term.

r

gB2d2

d 2+

b

mgB

d

d r

2

gsin cos+ sin = 0, S1

We now choose B such that we get coefficient 1 in front of the d2d2

term. (i.e: We solve rgB2

= 1

for B. This gives us B =rg1. This is a sensible substitution as r has units of [length], g has

units [length][time]2 which gives B units of [time]. Making this substitution for B gives us:

d2

d 2+

b

mrg

d

d r

2

gsin cos+ sin = 0, S1

Therefore, we have our required form:

d2

d 2+

d

d sin cos+ sin = 0, = b

mrg, =

r2

g, =

g

rt, S1 (3)

(a) Fixed points

To get the fixed points for eqn. (1), we set = = 0 and solve the resulting equations for , .

0 = , 0 = + sin cos sin, (, ) S1 R = 0, sin cos sin = 0 sin = 0, or cos 1 = 0

sin = 0 has solutions = 0, mod 2. Notice that the cos 1 = 0 only has solutions for 1. These solutions are = Arccos (1). When = 1, these solutions coincide with the = 0 solution. Therefore, we have three cases.

(i) 0 < 1: There are two fixed points: (, ) = (0 mod 2, 0) and (, ) = ( mod 2, 0)

1

(ii) = 1: There are still two fixed points: (, ) = (0 mod 2, 0) and (, ) = ( mod 2, 0)

(iii) > 1: There are four fixed points: (, ) = (0 mod 2, 0), (, ) = ( mod 2, 0) and(, ) = (Arccos (1) mod 2, 0)

(b) Linear stability analysis

Calculating the linear part of the evolution function near a fixed point (, ) gives us:

A = Df(, ) =

[0 1

(cos2 sin2 ) cos

]=

[0 1

cos 2 cos

]Calculating the determinant, DetA, and trace, TrA, gives us:

DetA = cos cos 2, TrA =

(i) 0 < 1:

At (, ) = (0 mod 2, 0), we have DetA = 1 > 0, TrA = .When = 0, TrA = 0 and this fixed point is a linear centre.When > 0, this point is a hyperbolic attractor

At (, ) = ( mod 2, 0), we have DetA = 1 < 0, TrA = .For all 0, this fixed point is a hyperbolic saddle point.

(ii) = 1:

At (, ) = (0 mod 2, 0), we have DetA = 0, TrA = .When = 0, this point is a non-hyperbolic fixed point with a double zero eigenvalue.When > 0, this point is a non-hyperbolic fixed point with a single zero eigenvalue.

At (, ) = ( mod 2, 0), we have DetA = 1 = 2, TrA = .For all 0, this fixed point is a hyperbolic saddle point.

(iii) > 1:

At (, ) = (0 mod 2, 0), we have DetA = 1 < 0, TrA = .For all 0, this fixed point is a hyperbolic saddle point. At (, ) = ( mod 2, 0), we have DetA = 1 < 0, TrA = .

For all 0, this fixed point is a hyperbolic saddle point. At (, ) = (Arccos (1) mod 2, 0) we have DetA = 1 > 0, TrA = .

(Useful: cos = 1, sin =

21

)

When = 0, this point is a linear centre.When > 0, this point is a hyperbolic attractor

At (, ) = (Arccos (1) mod 2, 0) we have DetA = 1 > 0, TrA = .(Useful: cos = 1, sin =

21

)

When = 0, this point is a linear centre.When > 0, this point is a hyperbolic attractor

2

(d) Conserved quantity for the undamped system

The system (1) is undamped if = 0. To show the system is conservative we need to find afunction E(, ) such that the following hold:

dE

d= = ,

dE

d= = sin cos+ sin

Solving the first differential equation gives us:

E(, ) =1

22 + g()

Plugging this solution into the second equation gives us:

dg

d= sin cos+ sin g() =

2cos2 cos+ C

(Set C = 0). So we find that the conserved quantity is:

E(, ) =1

22 +

2cos2 cos

(You can check that this solution is correct by evaluating E = dEdt

and confirming that it isequal to 0, indicating that it is a conserved quantity that does not change in time. E is sometimesrefered to as the Hamiltonian)

(e) Trajectories of the damped system

Let V (, ) = E(, ) from part (d). Calculate V = dVdt

.

V =V

d

dt+V

d

dt= ( sin cos+ sin) + ( + sin cos sin) = 2

Since we have > 0, we can see that V 0 for all (, ) S1 R with equality when = 0.When 6= 0, we have V < 0, so V is decreasing along ((t), (t)), except where (t) = 0.

When = 0, but is not a fixed point, then 6= 0, so the trajectory moves away from = 0 and Vis again decreasing. (t) = 0 can only occur at some isolated time t = t0. For t < t0, near t0, wehave V ((t), (t)) > V ((t0), (t0)) and for t > t0, near t0, we have V ((t), (t)) < V ((t0), (t0)).So we know that for an open interval containing t0, V is decreasing.

Thus V ((t), (t)) is decreasing in time provided that ((t), (t)) is not a fixed point.

4

Q2

d2

d 2+d

d sin cos+ sin = 0, S1 (4)

To give the first order system that is equivalent to eqn. (4), we let = dd R and make the

appropriate changes to get the following:

d

d= ,

d

d=

1

( + sin cos sin) , (, ) S1 R (5)

(a) A different nondimensionalization

We could perform an analysis where we find the fixed points and look at the determinant andtrace of the linear part of the evolution equation near those fixed points. However as per therecommendation we will instead opt to determine the relationships between the parameters of thesystems (3) and (4).

To do this, we can assume a new rescaling of dimensionless time = oldA

, where A is dimen-sionless, and old is the dimensionless time from Question 1. Applying the chain rule to (3), weget:

1

A2d2

d 2+

A

d

d sin cos+ sin = 0

By inspection we can achieve the dimensionless form of (5) if we choose A = . This makes = 1

2, > 0. We find that the correspondence with (1) is = 1/2, > 0. (Our rescaled

dimensionless time will be s =

= mgbt. does not change.)

(, ) = (0 mod 2, 0): This fixed point exists for all 0, > 0 and is classified as:

a hyperbolic attractor if 0 < 1, > 0; a non-hyperbolic fixed point with single zero eigenvalue if = 1, > 0 a hyperbolic saddle point if > 1, > 0

(, ) = ( mod 2, 0): This fixed point exists for all 0, > 0 and is classified as ahyperbolic saddle point.

(, ) = (Arccos(1) mod 2, 0): These fixed points exist for all 0, > 0 and both areclassified as a hyperbolic attractors. (They are stable spirals if 0 < < 1

2, and stable nodes

if > 12.)

(b)

See the attached XPP plots. As gets close to zero we can see that the results using the RK4 andthe default step size are obviously not correct. To rectify this, try using a smaller step size (i.emake Dt smaller) or try using the (S)tiff method. To check that your plots are indeed correct,make a qualitative comparison of plots generated using these methods.

7

-4

-3

-2

-1

0

1

2

3

4

-4 -3 -2 -1 0 1 2 3 4

nu

phi

epsilon = 1

-4

-3

-2

-1

0

1

2

3

4

-4 -3 -2 -1 0 1 2 3 4

nu

phi

epsilon = 0.5

-4

-3

-2

-1

0

1

2

3

4

-4 -3 -2 -1 0 1 2 3 4

nu

phi

epsilon = 0.1

-4

-3

-2

-1

0

1

2

3

4

-4 -3 -2 -1 0 1 2 3 4

nu

phi

epsilon = 0.01

(c) Discussion of the overdamped system

In general, the initial condition, ((0), (0)) = (0, 0) will be away from the nullclines of thesystem. In other words ( + sin cos sin) is not small, so 1

( + sin cos sin)

will be very large.

> sin cos sin < 0, When is above the nullcline, is decreasing < sin cos sin > 0, When is below the nullcline, is increasing

So we can see that solutions to the system are attracted to the state = 0 and that they areattracted to this state very quickly because of the 1

factor. So the solution will move very quickly

in the -coordinate towards the nullcline.When the solution is near the nullcline, we have that sin cos sin, and thus is

not large. Since = , we can see that (6) is now a good approximation for (4).

d

d= sin cos sin, S1 (6)

So after a fast initial transcient where the system moves rapidly towards = sin cossin,(6) becomes a good approximation to the describe the evolution of ().

(For this assignment solution, the figures for 1c), 1f), and 2b) were generated by Prof. WayneNagata)

Q3

Solution:

(a) In order to find the critical points, we shall find the x- and y-nullclines.The x-nullclines are: x = 0 (line) and y = 2 x (line).The y-nullclines are: y = 0 (line) and x = 1/2 (line).Thus the critical points in this case are: (0, 0), (2, 0) and (1/2, 3/2).See the sketch for plotting these points in Fig. 2.

(b) Lets begin first by finding the Jacobian matrix:

J =

(1 x 0.5y 0.5x

0.5y 0.25 + 0.5x

)(i) At (0, 0):

J(