Math 1280
Sample Midterm Exam
Spring 2016
Solutions
1. For the equation
ẋ = 1 − 2 cos x
find all fixed points, classify their stability, and sketch the vector field.
x∗ = − π3 + 2πn or x∗ =
1 − 2 cos x < 0 when x ∈ − π3 + 2πn, π3 + 2πn
1 − 2 cos x > 0 when x ∈ π3 + 2πn, 5π
+ 2πn
3
Solution:
FPs: cos x = 12 ,
π
3
+ 2πn.
ẋ
x
FPs x∗ = − π3 + 2πn are stable and x∗ =
π
3
+ 2πn are unstable.
2. For the equation
ẋ = x(1 − x)(2 − x)
use linear stability analysis to classify the fixed points and their stability.
Solution:
f (x) = x(1 − x)(2 − x) = x3 − 3x2 + 2x
FPs: x∗1 = 0, x∗2 = 1, x∗3 = 2.
f 0 (x) = 3x2 − 6x + 2
f 0 (0) = 2 > 0 and the FP x∗1 = 0 is unstable
f 0 (1) = −1 < 0 and the FP x∗2 = 1 is stable.
f 0 (2) = 2 > 0 and the FP x∗3 = 2 is unstable.
3. For the equation
ẋ = − sinh x
find the potential function V (x) and classify all the equilibrium points and their stability.
Solution:
Equilibrium (fixed) points: f (x) = − sinh x = 0,
x∗ = 0.
dV
= −f (x) = sinh x, V (x) = cosh x. The curve y = cosh x attains its minimum
dx
at the vertex when x = 0. Hence the equilibrium point x∗ = 0 is stable.
4.
For the given equation sketch all the qualitatively different vector fields that occur as r
varies. Find the critical value of r∗ and determine the type of bifurcation at that value.
Sketch the bifurcation diagram of fixed points x∗ vs. r.
(a) ẋ = r + 12 x −
x
1+x
(b) ẋ = rx + 4x3
(c) ẋ = x − rx(1 − x)
Solution:
x
x
= 0, r + 12 x =
. Below are plots of the family of
1+x
1+x
x
lines y = 21 x + r and the curve y =
1+x
(a) FPs: f (x) = r + 12 x −
y
r > r∗
x
r = r∗
r < r∗
So, there is no FPs when r > r∗ , one FP when r = r∗ , and two FPs when r < r∗ .
Let’s find the critical value r∗ . For that we solve the system:
x
d d
x
1
1
r + 2x =
,
r + 2x =
(tangentiality condition).
1+x
dx
dx 1 + x
√
√
Its solution is x∗ = 2 − 1, r∗ = 1.5 − 2
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This is a saddle-node bifurcation.
ẋ
ẋ
ẋ
x
x
r < r∗
x
r > r∗
r = r∗
The bifurcation diagram is
x
r
(b) FPs: f (x) = rx + 4x3 = x(r + 4x2 ) = 0,
√
x∗ = 0, x∗ = ± −r/2.
This is a subcritical pitchfork bifurcation.
There is one FP when r ≥ 0 and three FPs when r < 0.
ẋ
ẋ
ẋ
x
r<0
x
r=0
The bifurcation diagram is
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x
r>0
x
r
(c) FPs: f (x) = x − rx(1 − x) = rx
1−r
r
+ x = 0,
x∗ = 0, x∗ = µ where µ =
This is a transcritical bifurcation.
There is always two FPs. Their stability depends on the sign of µ.
ẋ
ẋ
x
x
µ<0
(0 < r < 1)
µ>0
(r < 0 or r > 1)
The bifurcation diagram is
x
µ
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r−1
.
r
5.
For the given vector field find and classify all fixed points and sketch the phase portrait
on the circle.
(a) θ̇ = sin3 (θ)
(b) θ̇ = cos θ + sin θ
Solution:
FPs: f (θ) = sin3 (θ) = 0,
(a)
θ1∗ = 0, θ2∗ = π.
sin3 (θ) > 0 when θ ∈ (0, π); sin3 (θ) < 0 when θ ∈ (π, 2π).
Hence the FP θ1∗ = 0 is unstable and the FP θ2∗ = π is stable.
Below is the phase portrait.
π
0
(b)
FPs: f (θ) = cos θ + sin θ = 0,
θ1∗
= 3π/4,
θ2∗
sin θ = − cos θ,
tan θ = −1 and cos θ 6= 0,
= 7π/4.
cos θ + sin θ > 0 when θ ∈ (0, 3π/4) ∪ (7π/4, 2π)
cos θ + sin θ < 0 when θ ∈ (3π/4, 7π/4).
Hence the FP θ1∗ = 3π/4 is unstable and the FP θ2∗ = 7π/4 is stable.
Below is the phase portrait.
3π
4
7π
4
6. For the given linear system find eigenvalues, eigenvectors, and the general solution of the
system. Classify the fixed point and determine its stability. Sketch the phase portrait.
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(a) ẋ = 2x − 6y,
ẏ = −y
(b) ẋ = −x − 2y,
ẏ = 4x + 3y
(c) ẋ = 3x − y,
ẏ = x + y
Solution:
(a)
This is linear homogeneous system and its the only FP is the origin (0, 0).
2 −6
A=
, T = 1/2, ∆ = −2, λ1 = −1, λ2 = 2.
0 −1
u1
λ1 = −1: let v̄1 =
be an eigenvector associated with λ1 = −1. Then
u2
0
2
3u1 − 6u2
3 −6
u1
=
, u1 = 2u2 , v̄1 =
.
=
(A − λ1 I)v̄ =
0
0
1
u2
0
0
u1
be an eigenvector associated with λ2 = −1. Then
λ2 = 2: let v̄2 =
u2
0 −6
u1
−6u2
0
(A − λ2 I)v̄ =
=
=
, u2 = 0 and u1 is any number,
0 −3
u2
−3u2
0
1
say 1. Then v̄1 =
.
0
The general solution is x̄(t) = c1 e−t v̄1 + c2 e2t v̄2
x(t)
where x̄(t) =
y(t)
The FP is a saddle since ∆ = −2 < 0. It is unstable. Use PPLANE to see the phase
portrait.
(b)
This is linear homogeneous system and its the only FP is the origin (0, 0).
−1 −2
A=
, T = 1, ∆ = 5, λ = 1 + 2i, λ̄ = 1 − 2i.
4
3
w1
Let w̄ =
be an eigenvector.
w2
−2 − 2i
−2
w1
0
Then (A − λI)w̄ =
=
.
4 2 − 2i
w2
0
From the first line of the result we get −2(1 + i)w1 − 2w2 = 0 or (1 + i)w1 + w2 = 0.
Take w1 = 1 and w2 = −(1 + i) = −1 − i
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Then w̄ =
1
−1 − i
=
1
−1
+i
and hence the eigenvectors are v̄1 =
0
−1
1
0
, v̄2 =
−1
−1
The general solution is x̄(t) = et (c1 (cos 2t v̄1 − sin 2t v̄2 ) + c2 (sin 2t v̄1 + cos 2t v̄2 ))
x(t)
where x̄(t) =
y(t)
The phase portrait is a spiral which at the point (0, 1) is directed downward since
c = 4 > 0. Hence its rotation is counterclockwise. Real part of λ is positive and the
FP is unstable. Use PPLANE to see the phase portrait.
(c)
This is linear homogeneous system and its the only FP is the origin (0, 0).
3 −1
A=
, T = 2, ∆ = 4, λ = 2 is a repeated root.
1
1
u1
be an eigenvector.
Let v̄1 =
u2
0
1 −1
u1
=
.
Then (A − λI) v̄1 =
u2
0
1 −1
1
We get u1 − u2 = 0. Take u1 = 1 and u2 = 1. Then v̄1 =
1
To find v̄2 we use the equality (A − λI) v̄2 = v̄1
u1
1 −1
u1
1
Let v̄2 =
. Then (A − λI) v̄2 =
=
.
u2
1 −1
u2
1
2
We get u1 − u2 = 1. Take u1 = 2 and u2 = 1. Then v̄2 =
1
The general solution is x̄(t) = e2t (c1 v̄1 + c2 (v̄2 + t v̄1 ))
x(t)
where x̄(t) =
y(t)
The phase portrait is a degenerate node. λ is positive and the FP is unstable. Use
PPLANE to see the phase portrait.
7.
For the given system draw nullclines and find fixed points. Classify the fixed points and
determine their stability, when possible.
(a) ẋ = x(6 − 2x − 3y),
(b) ẋ = y,
ẏ = y(1 − x − y)
ẏ = − sin x − y
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Solution:
(a)
x-nullcline: x = 0, y = − 32 x + 2 (solid lines on the plot below).
y-nullcline: y = 0, y = −x + 1 (dashed lines).
(−3, 4)
(0, 1)
(3, 0)
(0, 0)
FPs are (0, 0), (0, 1), (3, 0), and (−3, 4).
6 − 4x − 3y
−3x
Jacobian: A =
−y 1 − x − 2y
6 0
At (0, 0): A =
, T = 7/2, ∆ = 6, ∆ < T 2 .
0 1
(0, 0) is a nodal source (unstable node). It is unstable.
3
0
At (0, 1): A =
, T = 1, ∆ = −3 < 0.
−1 −1
It is a saddle, unstable.
−6 −9
At (3, 0): A =
,
0 −2
T = −4, ∆ = 12, ∆ < T 2 .
(0, 0) is a nodal sink (stable node). It is stable.
6
9
At (−3, 4): A =
, T = 1, ∆ = 12, ∆ > T 2 .
−4 −9
It is a spiral source. It is unstable.
(b)
x-nullcline: y = 0 (solid line on the plot below).
y-nullcline: y = − sin x (dashed curve).
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(−2π, 0)
(−π, 0)
(0, 0)
(π, 0)
FPs are (πn, 0).
0
1
Jacobian: A =
− cos x −1
0
1
n is even: A =
, T = −1/2, ∆ = 1, ∆ > T 2 .
−1 −1
In this case the FP is a spiral sink. It is stable.
0
1
n is odd: A =
, T = −1/2, ∆ = −1 < 0.
1 −1
In this case the FP is a saddle. It is unstable.
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(2π, 0)