MATH 345: Assignment 2 Solutions
1
˙ = f ( x, r ) = x [ r − ( x − 1) 2 ] , 0 ≤ x < ∞ (1)
(a) Saddle-node bifurcation at x ∗ = 1 , r c
1
= 0
From HW1, Question 4: f r
(1 , 0) = 1 , f xx
(1 , 0) = − 2. The normal form is:
˙ = r − ( x − 1) 2
Below are some local phase portraits (light grey, dark grey, black points indicate unstable, half-stable, stable equilibria resp.), then the bifurcation diagram.
r < 0 r = 0 r > 0
1
0
1
0
1
0
−1
−2
−3
−0.5
0 0.5
1 x
1.5
2 2.5
−1
−2
−3
−0.5
0 0.5
1 x
1.5
2 2.5
−1
−2
−3
−0.5
0 0.5
1 x
1.5
2 2.5
SN bifurcation at (x = 1, r = 0)
1.5
1
0.5
−0.1
0 0.1
0.2
0.3
0.4
r
(b) Transcritical bifurcation at x ∗ = 0 , r c
2
= 1
From HW1, Question 5: f xr
(0 , 1) = 1 , f xx
(0 , 1) = 4. The normal form is:
˙ = x ( r − 1) + 2 x 2 = x ( r − 1 + 2 x )
1

r < 1
0.4
0.3
0.2
0.1
0
−0.1
−0.2
−1 −0.5
0 x
0.5
0.2
0.1
0
−0.1
−0.2
0.5
r = 1
1
0.4
0.3
0.2
0.1
0
−0.1
−0.2
−1 −0.5
0 x
0.5
1
TC bifurcation at (x = 0, r = 1)
0.3
0.4
0.3
0.2
0.1
0
−0.1
−0.2
−1 −0.5
r > 1
0 x
0.5
1.5
r
1
1
Note that in the above bifurcation diagram the stable equilibrium is coloured grey. This is to indicate that this equilibrium exists, but we will ignore it because we are only concerned with x ≥ 0.
2

(c) r < 0
0
−0.2
−0.4
−0.6
−0.8
−1
−1.2
0 0.5
x r = 1
1
1
0.5
0
−0.5
0 0.5
1 x
1.5
2
1.5
r = 0
0.1
0
−0.1
−0.2
−0.3
−0.4
0 0.5
x r > 1
1
1.5
1
0.5
0
−0.5
0 0.5
1 x
1.5
2
1.5
0
−0.1
−0.2
0.3
0.2
0.1
0
0 < r < 1
0.5
x
1 1.5
(d) Z = { ( x, r ) | f ( x, r ) = 0 , 0 ≤ x < ∞} x [ r − ( x − 1) 2 ] = 0 ⇔ x = 0 or r − ( x − 1) 2 = 0 ⇔ x = 0 or x = 1 ±
√ r, r ≥ 0
2.5
2 x = 1 + r
1.5
1
0.5
0
−0.5
x = 0
0
1/2
0.5
r x = 1 − r
1/2
1 1.5
(e) Z and phase portraits
Near the bifurcation points, local and global phase portraits and bifurcation diagrams agree qualitatively (near x = 1 , r = 0 for a), near x = 0 , r = 1 for b))
3

2.5
2
1.5
1
0.5
0
−0.5
0 0.5
r
1 1.5
(f)
Overall shape of branches and stability agree. Numerical values agree exactly or very closely. The author of these solutions received the following output from XPPAUT:
BR PT TY LAB
1 1 EP
PAR(1)
1 -1.000000E+00
L2-NORM
0.000000E+00
U(1)
0.000000E+00
1 50 2 -1.999990E-02 0.000000E+00 0.000000E+00
1 100 3 9.800002E-01 0.000000E+00 0.000000E+00
1 101 BP 4 1.000000E+00 0.000000E+00 0.000000E+00
1 150 5 1.980000E+00 0.000000E+00 0.000000E+00
1 151 EP 6 2.000000E+00 0.000000E+00 0.000000E+00
BR PT TY LAB PAR(1) L2-NORM U(1)
2 50 7 2.103928E-01 5.413141E-01 5.413141E-01
2 76 LP 8 4.587019E-11 9.999978E-01 9.999978E-01
2 100 9 1.860324E-01 1.431315E+00 1.431315E+00
2 150
2 200
10
11
1.000978E+00
1.922805E+00
2.000489E+00
2.386652E+00
2.000489E+00
2.386652E+00
2 205 EP 12 2.017005E+00 2.420213E+00 2.420213E+00
BR PT TY LAB
2 50 13
PAR(1)
1.886473E+00
L2-NORM U(1)
3.734892E-01 -3.734892E-01
2 57 EP 14 2.018285E+00 4.206635E-01 -4.206635E-01
Bifurcation i) Transcritical ( ii) Saddle-node (
BP
LP
)
)
Bifurcation point (exact) r = 1 , x = 0 r = 0 , x = 1
Bifurcation point (AUTO) r = 1 .
000000 , x = 0 .
000000 r = 4.587019E-11 , x = 9.999978E-01
Note: The points with type EP represent endpoints in the bifurcation plot calculation. In this case, we calculated the bifurcation diagram from r = − 1 to r = 2.
The bifurcation diagram is on the next page. Note that the thicker lines indicate a stable fixed point and the thin lines indicate an unstable fixed point.
4

1
0.5
2
1.5
X
3
2.5
0
-0.5
-1
-0.4
-0.2
0 0.2
0.4
r
0.6
0.8
1 1.2
1.4

2
φ
˙
= f ( φ, γ ) = γ sin φ cos φ − sin φ, φ ∈ S
1
(a) Fixed points
Solve f ( φ, γ ) = sin φ ( γ cos φ − 1) = 0 ⇔ sin φ = 0 or γ cos φ − 1 = 0
⇔ φ = 0 mod 2 π, π mod 2 π or γ cos φ − 1 = 0
(2) i) 0 < γ < 1
γ cos φ = 1 has no solutions. (cos φ = 1 > 1) r
The fixed points are φ ∗ = 0 mod 2 π, φ ∗ = π mod 2 π ii) γ = 1
γ cos φ = 1 has one solution: φ ∗ = 0 mod 2 π , which we already have.
The fixed points are φ ∗ = 0 mod 2 π, φ ∗ = π mod 2 π iii) γ > 1
γ cos φ = 1 has two solutions: φ ∗ = ± Arccos (1 /γ ).
The fixed points are φ ∗ = 0 mod 2 π, φ ∗ = π mod 2 π, φ ∗ = + Arccos (1 /γ ) mod 2 π, and φ ∗ = − Arccos (1 /γ ) mod 2 π
(b)
See XPP plots attached on the next few pages. The hand drawn plots should look something like this:
γ
> 1
γ
< 1
γ
= 1
φ
= Arccos(
γ −1
)
φ
=
π
φ
=
π φ
=
π 2
2 2 identify identify identify
0
0 0
−2
−2
φ
= −
π
−2
φ
= −
π
φ
= −Arccos(
γ −1
) φ
= −
π
0 2 t
4 6 8 0 2 4 t
6 8
0 2 t
4 6 8
6

10
8
6 gamma = 0.5
4
2
0
0 1 2 3 4 5 6 7 8

10
8
6 gamma = 1
4
2
0
0 1 2 3 4 5 6 7 8

10
8
6 gamma = 2
4
2
0
0 1 2 3 4 5 6 7 8

(c) i)
γ
< 1
1
0.5
0
−0.5
−1
−2 0
φ
2
γ
< 1
φ
=
π
mod(2
π
)
φ
= 0 mod (2
π
) ii)
γ
= 1
1
0.5
0
−0.5
−1
1.5
1
0.5
0
−0.5
−1
−1.5
−2
−2 0
φ
γ
> 1
2
0
φ
2 iii)
γ
φ
=
π
mod(2
π
)
φ
= 0 mod (2
π
)
γ
= 1
> 1
φ
=
π
mod(2
π
)
φ
= −Arccos(
γ −1
)
φ
= 0 mod (2
π
)
φ
= Arccos(
γ −1
)
10

(d)
Z = { ( φ, γ ) | f ( φ, γ ) = sin φ ( γ cos φ − 1) = 0 , φ ∈ S
1 } sin φ = 0 or γ cos φ − 1 = 0
φ = 0 or π, or γ = sec φ
Z
3
0
−1
2
1
−2
−3
0 0.5
1
φ
1.5
2
(e)
−1
−2
1
0
3
2
−3
0
Phase portraits
0.5
1
γ
1.5
identify
2
0
−1
−2
3
Global Bifurcation diagram
φ
=
π
2 identify
1
φ
= 0,
γ
= 1
φ
= −
π
−3
0 0.5
1
γ
1.5
2
Near φ = 0 , γ = 1, the bifurcation diagram and phase portraits agree qualitatively with global versions. There are some differences:
• Global bifurcation diagram has asymptotes φ = ± π/ 2. Local bifurcation does not.
11

• Global bifurcation diagram shows unstable solution φ not.
∗ = π mod 2 π . Local bifurcation does
• Global bifurcation diagram is on cylinder. Local bifurcation is on a plane.
• Global phase portraits are on a circle, local phase portraits are on a line.
12