Math 345 Final Exam (April 2010)
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Instructions:
1. No notes or books except those mentioned below are to be used in this exam.
2. You are allowed to bring a letter-sized formula sheet and a small-screen, non-graphic, nonprogrammable calculator.
3. Justify every answer whenever is necessary, and show your work. Unsupported answers will
receive no credit.
4. You will be given 2.5 hrs to write this exam. Read over the exam before you begin. You are
asked to stay in your seat during the last 5 minutes of the exam, until all exams are collected.
5. At the end of the hour you will be given the instruction “Put away all writing implements and
remain seated.” Continuing to write after this instruction will be considered as cheating.
6. Academic dishonesty: Exposing your paper to another student, copying material from another
student, or representing your work as that of another student constitutes academic dishonesty.
Cases of academic dishonesty may lead to a zero grade in the exam, a zero grade in the course,
and other measures, such as suspension from this university.
Question grade value
1
24
2
16
3
16
4
20
5
24
Total
100
Math 345 Final
Name:
Question 1:
[24 marks]
Consider the one-dimensional differential equation
ẋ = f (x, r) = x(r + 2x − x2 ),
where the variable x(t) and the parameter r both belong to the real line R.
(a) Note that xs = 0 is a fixed point for all r. (i) Use linear stability analysis to determine the value
rc at which its stability changes. (ii) Determine the intervals on the r-axis on which xs = 0 is either
stable or unstable. (iii) Determine the type of this bifurcation point (0, rc ) (SN, TC, or PF). (iv) Find
its normal form.
(b) (i) Find all fixed points for all possible values of r (clearly state the conditions under which they
exist). (ii) Plot the set Z = {(x, r) : f (x, r) = 0} (horizontal axis r, vertical axis x). (iii) Sketch all
the possible qualitatively different one-dimensional phase portraits for different values of r. (No need
to sketch the graph of ẋ versus x).
(c) Based on results obtained in (b)(iii) or any other method(s), determine the stability of each fixed
point in the plot obtained in (b)(ii). Sketch a bifurcation diagram (horizontal axis r, vertical axis x),
showing all fixed points. Draw the stable branch(s) of fixed points with solid curves and the unstable
branch(s) with dashed curves. (This diagram is identical in shape as that in (b)(ii) but is different only
in separating the stable from the unstable fixed points).
(d) There is another bifurcation point in addition to that in (a). (i) Determine the values of xs and rc
for this fixed point. (ii) Determine its type. (iii) Find its normal form.
Solution:
(a) (i) f ′ (x) = r + 4x − 3x2 . f ′ (0) = r. Thus, for r < 0 it is stable and for r > 0 it is unstable. rc = 0.
(ii) As found previously, for r < 0 it is stable and for r > 0 it is unstable. (iii) Note that f (0, 0) = 0,
fx (0, 0) = 0, fxr (0, 0) = 1 and fxx (0, 0) = 4, it is a TC bifurcation point. (iv) The normal form is
ẋ = ax(r − rc ) + bx2 , where a = fxr (0, 0) = 1 and b = fxx (0, 0)/2 = 2. Thus, ẋ = rx + 2x2 is the normal
form.
(b) (i) f (x) = x(r + 2x − x2 ) = 0 yields xs = 0 and xs = 1 ±
r > −1. (ii) See Fig. 1(b)(ii). (iii) See Fig. 1(b)(iii).
√
1 + r, the latter two exist only when
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Math 345 Final
Name:
x
0
x
(r<−1)
0
x
(r=−1)
−1
0
0
r
−1
x
(−1<r<0)
0
x
(r=0)
−2
0
(Fig. 1b(ii))
(Fig. 1b(iii))
x
(r>0)
√
(c) It is clear from the results in (b) the branch xs = 1 − 1 + r is unstable on the interval r ∈ (−1, 0).
See Fig. 1(c).
x
−1
0
r
−1
−2
(Fig. 1c)
(d) Yes, there is another bifurcation point located at (xs , rc ) = (1, −1) as we can clearly see in
Fig. 1c. It is a SN point as can be verified by: (SN1) f (xs , rc ) = 1(−1 + 2 − 1) = 0; (SN2):
fx (xs , rc ) = −1 + 4 − 3 = 0; (SN3): fr (xs , rc ) = 1 6= 0; (SN4): fxx = 4 − 6 = −2 6= 0. The normal form
is given by ẋ = a(r − rc ) + b(x − xs )2 , where a = fr (xs , rc ) = 1 and b = fxx (xs , rc )/2 = −2/2 = −1.
Thus, ẋ = (r + 1) − (x − 1)2 .
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Math 345 Final
Name:
Question 2:
[16 marks]
Answer “True” or “False” to the statements below. Put your answers in the boxes.(4 mark each)
(a) (xs , rc ) = (0, 1) is a transcritical bifurcation point for the equation ẋ = r − x − e−x .
False. (It is a SN point with normal form: ẋ = r − 1 − x2 /2.)
(b) The system ẋ = −x + 4y and ẏ = −x − y 3 has no closed orbits.
True. (V (x, y) = x2 + 4y 2 ⇒ V̇ = −2x2 − 8y 2 < 0 for all (x, y) 6= (0, 0).)
(c) Consider a map xn+1 = f (xn ). A period-2 cycle (characterized by f (p) = q and f (q) = p) is stable
if the multiplier λ = f ′ (p)f ′ (q) < 1.
False. (|λ| = |f ′ (p)f ′(q)| < 1.)
(d) For the system ẋ = x(3 − x − 2y) and ẏ = y(2 − x − y), the fixed point (0, 0) is a hyperbolic
unstable node.
True. (λ1 = 3, λ2 = 2.)
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Math 345 Final
Name:
Question 3:
[16 marks]
These questions do not require lengthy calculations. Put your answers in the boxes.
each)
(4 marks
dz
(a) The first-order differential equation
= λ[N − z − x0 exp(−kz/λ)] can be non-dimensionalized to
dt
dx
= a − bx − e−x by introducing the non-dimensional variables x = z/Z and τ = t/T . Determine the
dτ
expressions for Z, T, a, b in terms of λ, N, x0 , k.
Z = λ/k, T = 1/(kx0 ), a = N/x0 , b = λ/(kx0 ).
2x,
0 ≤ x < 1/2
. Find out its orbit if
2x − 1, 1/2 ≤ x < 1
x0 = 57/60. On which periodic orbit does it evolve to? Determine the stability of this periodic orbit.
(b) For the doubling map xn+1 = f (xn ) where f (x) =
57
27
4
3
1
2
4
→
→ → → → → = 4-cycle { 51 , 52 , 54 , 53 }. |µ| = |f ′ ( 15 )f ′ ( 52 )f ′ ( 45 )f ′( 53 )| = 24 > 1 ⇒ unstabl
60
30
5
5
5
5
5
(c) Suppose ẋ = f (x, y), ẏ = g(x, y). Then, ẋ = −▽V (where x = (x, y)T and ▽V = (∂V /∂x, ∂V /∂y)T )
implies f (x, y) = −∂V /∂x and g(x, y) = −∂V /∂y. These can be “partially integrated” to find V for
gradient systems. Find V for the system ẋ = y + 2xy, ẏ = x + x2 − y 2 .
V (x, y) = −xy − x2 y + y 3/3.
(d) For the system ẋ = −xy and ẏ = xy − y (x, y > 0). Find a quantity V (x, y) that is conserved
(Hint: Form a differential equation for dy/dx. Solve it using separation of variables).
V (x, y) = x + y − ln x.
Question 4:
[20 marks]
Consider the quadratic map xn+1 = f (xn ) = x2n + r, with x, r ∈ R, r is the parameter.
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Math 345 Final
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(a) Find all the fixed points and determine for which values of r do they exist.
(b) Determine the stability of the fixed points found in (a) as a function of r. Find the value of r at
which the fixed point is superstable (i.e. λ = f ′ (x∗ ) = 0.)
(c) Find all the period-2 cycles. (Hint: Suppose f (p) = q and f (q) = p. Show that p, q are roots of
the equation (x2 − x + r)[x2 + x + r + 1] = 0.)
(d) Determine the stability of the period-2 cycles as a function of r. Find the value of r at which the
2-cycle solution is superstable (i.e. λ = f ′ (p)f ′ (q) = 0.)
(e) Plot a bifurcation diagram based on the information obtained above for −2 ≤ r ≤ 1.
Solution:
(a) The fixed points are obtained by solvingpx = f (x) = x2 +r ⇒ x2 −x+r = 0. We find immediately
that the fixed points are x∗± = 1/2 ± 1/4 − r. They only exist for r ≤ 1/4. For r > 1/4, no
fixed point exists.
(b) The stability
is determined by the magnitude
of the multiplier λ = f ′ (xs ) = 2x∗ . For x∗ =
p
p
1/2 + 1/4 − r, λ = 2x∗ = 1 + 2 1/4 − r > 1 for r < 1/4. Thus, it is untable.
For
p
∗
r = 1/4, λ = 1 ⇒
p it is a non-hyperbolic bifurcation point. For x = 1/2 − 1/4 − r,
∗
|λ| = |2x | = |1 − 2 1/4 − r| < 1 for −3/4 < r < 1/4; |λ| > 1 for r < −3/4. λ = 1 for
r = 1/4, −3/4 p
⇒ it is a non-hyperbolic bifurcation
point.
p
∗
∗
For x = 1/2 − 1/4 − r, |λ| = |2x | = |1 − 2 1/4 − r| = 0 if r = 0. Thus, x∗ = 0 at r = 0 is a
superstable fixed point.
(c) The period-2 cycles are roots of x = f 2 (x) = (x2 + r)2 + r] ⇒ (x2 + r)2 − x + r = 0 ⇒
(x2 − x + r + x)2 − x + r = 0 ⇒ (x2 − x + r)2 + 2x(x2 − x + r) + x2 − x + r = 0 ⇒
(x2 − x + r)[(x2 − x + r) + 2x + 1] = 0 ⇒ (x2 − x + r)[x2 + x + r + 1] = 0 .
x2 − x + p
r = 0 gives the fixed points
obtained in (a). Thus, the 2-cycle solutions are {p, q} =
p
{−1/2 + −(r + 3/4), −1/2 − −(r + 3/4)} and exist for r < −3/4.
(d) The stability of the period-2 cycles are determined by the magnitude of the multiplier λ =
f ′ (p)f ′ (q).
The the period cycle (i), λ = f ′ (p)f ′ (q) = (2p)(2q) = 4pq = 4(1/4 + r + 3/4) = 4(r + 1) ⇒ |λ| < 1
for −5/4 < r < −3/4 which indicates that the 2-cycle solution is stable on this interval of r
values. At rc = −5/4, it is a non-hyperbolic bifurcation point. It becomes unstable for r < −5/4.
At r = −1, λ = 0. Thus, the 2-cycle {p, q} = {0, −1} at r = −1 is a superstable 2-cycle solution.
(e) See Fig. 4e.
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Math 345 Final
Name:
x
1
1/2
−3/4
−3/2
−1
−1/2
−1/2
−1
(Fig. 4e)
−5/4
1/4
r
Black=fixed points
Blue=2−cycles
Solid=stable
Dashed=unstable
= superstable soln
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Math 345 Final
Name:
Question 5:
[24 marks]
The Brusselator is a simple model of hypothetical chemical oscillator, named after the home of scientists
who proposed it.
ẋ = 1 − (b + 1)x + ax2 y,
ẏ = bx − ax2 y,
where x, y ≥ 0 are the dimensionless concentrations of two chemicals and a, b > 0 are parameters.
(a) Find all fixed point(s). Use the Jacobian to classify each one and determine for which values of
a, b it is hyperbolic and for which value(s) it is not.
(b) Sketch the nullclines in the first quadrant x, y ≥ 0. Determine the direction of the flow in each
separate region and on each section of the nullclines.
(c) Construct a trapping region for the flow in the first quadrant x, y ≥ 0.
(d) Show that a Hopf bifurcation occurs at some parameter value b = bc and express bc in terms of
parameter a.
(e) Does the limit cycle exist for b > bc or b < bc ? Explain, using the Poincaré-Bendixson theorem.
(f) Find the approximate period of the limit cycle for b ≈ bc using the imaginary part of the eigenvalues at b = bc .
Solution:
(a) The x-nullcline is: x = 0 which is the curve 1 − x − bx + ax2 y = 0 ⇒ y = [(1 + b)x − 1]/(ax2 ).
The y-nullcline is: y = 0 which is the curve bx − ax2 y = 0 ⇒ y = b/(ax) = ys /x.
There is only one fixed point (xs , ys ) = (1, b/a).
Let’s begin first by finding the Jacobian matrix:
b−1 a
−(1 + b) + 2axy ax2
⇒ J(1, b/a) =
J(x, y) =
−b −a
b − 2axy
−ax2
Thus, T r = b − (1 + a) and Det = a > 0. It is non-hyperbolic only when T r = 0 (i.e. b = 1 + a)
were it is a center. It is hyperbolic otherwise.
(b) See Fig. 5b.
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Math 345 Final
Name:
y
b(1+b)/a
dy/dt=0
. .
x − |y| = 1−x < 0 for x>1.
(1,b/a)
dx/dt=0
1/(1+b)
x
1
(Fig. 5b)
(c) See the area enclosed in the dashed lines. To show that the flow direction is indeed pointed to
the interior of this area along the -45 degree line, we need to show that ẋ < |ẏ| = −ẏ along this
line. This is obvious if we check and verify the following inequality
ẋ − |ẏ| = ẋ + ẏ = 1 − x − bx + ax2 y + (bx − ax2 y) = 1 − x < 0.
On the other boundaries of this area, it is easy to see that the flow is directed into this area. This
is based on the direction field established in (b).
√
(d) At b = bc = 1 +pa, the eigenvalues are λ± = ±i a, (a 6= 0). For b = b − ǫ < bc (0 < ǫ << 1),
λ± = −ǫ/2 ± i a − ǫ2 /4 with negative real p
part, (xs , ys ) = (1, b/a) is a stable spiral. For
b = b + ǫ > bc (0 < ǫ << 1), λ± = ǫ/2 ± i a − ǫ2 /4 with positive real part, (1, b/a) is an
unstable spiral. Thus, b = bc is a Hopf bifurcation point.
(e) Based on result in (d), (xs , ys ) = (1, b/a) is repeller when b > bc . Combining results in (c) and
Poincaré-Bendixsen theorem, a limit cycle exist for b > bc .
(f) At the Hopf point b = bc , the Hopf frequency
is ω =
√
limit cycle for b ≈ bc is T ≈ 2π/ω = 2π/ a.
√
a. Thus, the approximate period of the
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