Mathematics 562
Homework Assignment 3
Due Thursday Mar 3, 2016
Question 1 :
Show that the following system has a periodic solution using Poincar´e-Bendixson Theorem.
y ˙
˙ = x − y − x
= x + y − y
3
3 .
,
(Hint: find a square area − b ≤ x ≤ b, − b ≤ y ≤ b ( b > 0) in the phase space that is a trapping region). Determine the smallest possible number b m
> 0 such that for all b ≥ b m this square region is trapping but not if b < b m
.
Answer:
J ( x, y ) =
1 − 3 x 2
1
− 1
1 − 3 y 2
⇒ J (0 , 0) =
1 − 1
1 1
.
T r = 2 , Det = 2, λ
1
,
2
= 1 ±
√
1 − spiral. Now, if we find a number b ≥
√
± i . Thus, the fixed point at (0 , 0) is an unstable
2, we can show that the region defined by
− b ≤ x, y ≤ b is a trapping region. Thus,
This is because b =
√
2 is the smallest possible value of
2 is a root of the equation x − x 3 = − x b for it to be trapping.
. For all values of b ≥ 2, we guarantee that the x -nullcline only crosses y = ± b and y -nullcline only crosses x = ± b . This guarantees that the region is trapping at all four boundaries.

Question 2 :
Consider the following version of the BZ-reaction model modified by YXL based on the model developed by Lengyel-Rabai-Epstein (1990) (Experimental and modeling study of oscillations in the chlorine-iodine-malonic acid reaction.
J. Am. Chem. Soc.
112, 9104).
˙ = c − x − y ˙ = bx 1 − k 2
4 xy
+ x 2
, k y − a
2 + x 2
, where a, b, c are all positive parameters and that a = 0 .
06 , b = 2 , c = 3 , k = 0 .
08 for the following questions unless otherwise specified. Type the following lines in your ode file immediately before the last line containing “done”:
# Change some XPP internal parameters from their default values
{ @ TOTAL=50,DT=0.001,xlo=0,xhi=1,ylo=0,yhi=0.8,Bounds=100000,MAXSTOR=50000000
{ @ xplot=x,yplot=y
(Careful: the curled bracket in front of the last two lines above should not be there!)
(a) Use XPPAUT to find all steady states of the system for a = 0 .
113, print the eigenvalues for all of them, and classify them.
(b) Use XPPAUT to find the invariant sets of the saddle point found in (a), hand-in either a hand drawn copy of it or do a screen dump print of the graph produced by XPPAUT. Because the stable manifolds of the saddle serve as the threshold for generating an “excitation” in this system, describe what is special about the threshold for this particular system.
(c) Use XPP to produce the bifurcation diagram against the parameter a on the interval
[0 .
06 , 0 .
14]. Starting at a = 0 .
06, one can use “Sing pts” to find the steady-state values of ( x, y ), copy these values to the initial conditions. Then, goto AUTO to run the steady state bifurcation diagram. If there is any HB point, grab it and run “Periodic” starting from it. If you want to generate a copy of the diagram, you can either copy the data from the “xpp” window and use a spread sheet to plot it. Or simply do a screen dump to save a copy of the graph.
Suggested parameters in AUTO “Axes” are:
“hI-lo”, “Y-axis: x”, “Main param:a”, “Second param:b”, “Xmin:0.06”, “Ymin:-0.1”,
“Xmax:0.14”, “Ymax:0.8”.
Suggested parameters in AUTO “Numerics” are:
N tst = 50 , N max = 300 , N P r = 5 , Ds = 0 .
001 , Dsmin = 0 .
0001 , Dsmax = 0 .
0025 ,
P ar M in = 0 .
06 , P ar M ax = 0 .
14 , N col = 4 , EP SL = EP SU = EP SS = 0 .
000001.
2

(d) Following the work in (c), without exiting the program, now select “Period” in “Axes”, use the choose “Ymin” and “Ymax” values based on the period range shown on the
“xpp” window. Save a copy of the period as a function of parameter a .
(e) Following the work in (b) and (c), without exiting the program, now grab again the
HB point, select now select “Two par” in “Axes”, use “Ymin:1.5” and “Ymax:2.5”, and then run the “Two par” in the “Run” list. You typically only get half of the curve.
Now, go to “Numerics”, reverse the sign of “Ds”, run the “Two par” in the “Run” list again. Now, you get the whole curve.
Furthermore, now grab the 1st “LP” (SN) point, do the same “Two par” run twice with opposite signs of “Ds”. Then, grab the 2nd “LP” point, do the same as before.
After all this, you get three lines in the two parameter space, the sloped line for all HB bifurcation points and the two vertical lines for the LP or SN points on the steady state bifurcation branch. Save a copy of this digram. Remember after running “Two par”, values for both parameters a and b will be wildly different from the default values. To restart the whole process after a mistake, you have to remember to reset the parameters to their original values.
(f) Now, change the parameter b to b = 1 .
5, repeat question (c). You should find that there exists no HB point in this bifurcation diagram. In particular, at a = 0 .
06, there is only one steady state that is unstable. If you use “Initialconds” to integrate, you will find that there is a stable limit cycle solution. Now, integrate the system for long enough time by using “Initialconds” followed by choosing “Last” (or hit “I” followed by “L” several times). Then, open the “Data” window from there find what is the period of the limit cycle (I found T=1.854, please verify by yourself!)
Then select “nUmerics” and change from Total=50 to TOTAL=the period you found.
Hit “esc” to get out of it. Now, hit “I” followed by “L” for the last time. You’ve just saved an exact period of data of this periodic solution. Now, move into AUTO. Select
“Run” and choose “Periodic”, you are supposed to get the periodic solution branch of the system starting from the one you have just numerically calculated. Save a copy of this bifurcation diagram.
(g) Repeat the work in (d) for the case in (f), generate a plot of the period as a function of the parameter a for the periodic branch obtained in (f).
Solution:
(a) There exist 3 steady states:
(0.059331, 0.12292) with eigenvalues: − 13 .
661348 ± 5 .
576224 i . It is a stable spiral.
(0.17891, 0.15141) with eigenvalues: 5 .
243939 , − 5 .
051168. It is a saddle point.
(0.36176, 0.25027) with eigenvalues: 0 .
168165 ± 3 .
256079 i . It is an unstable spiral.
3

(b) Because the unstable spiral has complex eigenvalues with a small positive real part, the trajectories spiral strongly near this steady state. This makes the stable manifold of the saddle point spirals strongly around this steady state. See the figure below.
y
0.3
0.25
0.2
0.15
0.1
0.05
0.5
0.45
0.4
0.35
0
0 0.1
0.2
0.3
x
(c) See the left figure below.
0.4
0.5
0.6
0.6
0.5
0.4
0.3
X
0.8
0.7
0.2
0.1
0
-0.1
0.06
0.07
0.08
0.09
0.11
0.1
a
(d) See the right figure above.
(e) See the figure below.
0.12
0.13
0.14
b
2.4
2.2
2
1.8
1.6
0.06
0.07
0.08
0.09
0.1
a
(f) See the left figure below.
0.11
0.12
0.13
0.14
Period
14
12
10
8
6
4
2
0
0.06
0.07
0.08
0.09
0.1
a
0.11
0.12
0.13
0.14
4

X
1
0.8
0.6
0.4
0.2
0
0.06
0.07
0.08
0.09
0.11
0.1
a
(g) See the right figure above.
0.12
0.13
0.14
Period
25
20
15
10
5
0
0.06
0.07
0.08
0.09
0.1
a
0.11
0.12
0.13
0.14
5