UBC Mathematics 345 Section 201
Midterm Exam (March 8, 2016)
Duration: 75 min
I. Answer “True” or “False” to the statements below.
Put your answers in the boxes.
(2 mark each)
(a) For the initial value problem ˙ = x (1 − x )( x −
1
2
) and x (0) = 0 .
6, x ( t ) → 1 as t → ∞ .
True.
(b) For ˙ = r + x − ln(1 + x ), the normal form at ( x s
, r c
) = (0 , 0) is ˙ = r + 1
2 x 2 .
True.
(c) The equation describing the motion of a mass attached to a spring ¨ = f ( x ) is always a conservative system for any smooth function f ( x ).
True.
(d) For the system
˙ = sin y, y ˙ = sin x,
E ( x, y ) = cos x − cos y is a conserved quantity.
True
1

II. These questions do not require lengthy calculations.
Put your answers in the boxes.
(2 marks each)
(a) The equation dN dt dx x form dτ
= rx 1 − k terms of A, B, K, R.
= RN 1 −
− x
1 +
2 x 2
N
K
−
A 2
BN 2
+ N 2
, ( A, B, K, R > 0) can be nondimensionalized into the using the expression x = X/U and τ = t/T . Express r , k , U and T in
U = A , k = K/A , r = RA/B , T = A/B .
(b) For x ′ = x − rx 2
1 + x 2
, r c
= 2 is a bifurcation point. Determine its type.
r c
= 2 is a SN bifurcation point.
(c) For x ′ = x + rx
1 + x 2
, r c
= − 1 is a bifurcation point. Determine its type.
r c
= − 1 is a (subcritical) PF bifurcation point.
(d) For x ′ = 2 x − rx (1 − x ), find and determine the type of the bifurcation point.
r c
= 2 is a TC bifurcation.
(e) For the following system, there exists is a homoclinic orbit that connects the unstable invariant set to the stable invariant set of the saddle point at (0 , 0). Find the equation of this orbit.
˙ = y,
˙ = x − x 3 .
2 x 2
− x 4
− 2 y 2 = 0 or x 2
− x
4
2
− y 2 = 0.
2

III. Given the system
˙ = x [4 x (1 − x ) − y ] , y ˙ = y (4 x − 3) .
(18 in total)
(a) (3 marks) Find all fixed points.
(b) (4 marks) Sketch the nullclines. Put a circle at the location of each fixed point.
(c) (4 marks) Put a representative flow direction in each distinctive region and each distinctive segment of the nullclines.
(d) (1 mark) Based on the direction field of the nonlinear system obtained so far, can you determine the stability of the fixed point at (0 , 0)?
(e) (4 marks) Compute the Jacobian. Use it to help sketch the local phase portrait near each fixed point
(calculate the eigenvectors only when it is necessary for sketching the phase portrait). Classify the type and determine the stability of each fixed point.
(f) (2 marks) Explain any discrepancy between the phase portrait of the linearized system and that of the nonlinear system as obtaned in (c) and (d). Determine the long term behaviour of the system for x (0) = 1 = y (0).
Answer:
(a) x [4 x (1 − x ) − y ] = 0 and y (4 x − 3) = 0. Thus, (0 , 0) , (1 , 0) , ( 3
4
, 3
4
) are the fixed points.
(b) For ˙ y = 0, y=0, x=3/4.
(c) (4 marks) Put a representative flow direction in each distinctive region and each distinctive segment of the nullclines.
(d) We can tell that (0 , 0) is unstable since the phase trajectories flow away from it on the x -axis for
0 < x < 1.
(e) The Jacobian is: J ( x ) =
8 x − 12 x 2
4 y
− y − x
4 x − 3
.
Thus, at (0,0): J (0 , 0) = eigenvectors
0
0
0
− 3 which yields eigenvalues λ
1
= 0 , λ
2
= − 3 with corresponding v
1
=
1
0
, v
2
=
0
1
.
This is non-isolated fixed points on the line y = 0. The stability is neutral based on the linearized system.
At (1,0): J (1 , 0) =
− 4 − 1
0 1 which yields eigenvalues λ
1
= − 4 , λ
2
= 1 with corresponding eigenvectors v
1
=
1
0
, v
2
=
1
− 5
.
This is a saddle point and is unstable.
3

At ( 3
4
, 3
4
): J ( 3
4
, 3
4
) =
−
6
4
−
3 0
3
4 which yields eigenvalues λ
1
,
2
= −
3
4
[1 ± i
√
3] . This is a stable spiral that spirals in the counter-clockwise direction based on the direction field.
(f) The stability and local phase portrait near the fixed point (0 , 0) differ from the non-linear system.
It is an isolated fixed point in the nonlinear system but not in the linearized system. Also, the stability is not neutral but unstable in the nonlinear system. This is because (0 , 0) in the linearized system is not hyperbolic (there exists a zero eigenvalue) making the phase portrait unstable to small perturbations of the vector field.
The initial condition x (0) = 1 = y (0) is in t he basin of attraction of the stable fixed point at (
3
4
,
So, as time approaches infinity, the system should evolve toward this fixed point and stays there.
3
4
).
0.8
y
0.6
0.4
0.2
0
1.2
1
0 0.2
0.4
x
0.6
0.8
1 1.2
4