2D Homogeneous Linear Systems withConstant Coefficients

— repeated eigenvalues

Xu-Yan Chen

Systems of Diff Eqs:d~x

dt= A~x

where ~x(t) =

[x1(t)x2(t)

], A is a 2× 2 real constant matrix

Things to explore:

I General solutions

I Initial value problems

I Geometric figures

I Solutions graphs x1 vs t & x2 vs tI Direction fields in the (x1, x2) planeI Phase portraits in the (x1, x2) plane

I Stability/instability of equilibrium (x1, x2) = (0, 0)

2D Systems:d~x

dt= A~x

What if A has repeated eigenvalues?Assume that the eigenvalues of A are: λ1 = λ2.

• Easy Cases: A =

[λ1 00 λ1

];

• Hard Cases: A 6=[λ1 00 λ1

], but λ1 = λ2.

Find Solutions in the Easy Cases: A = λ1I

I All vector ~x ∈ R2 satisfy (A− λ1I)~x = 0.The eigenspace of λ1 is the entire plane.

We can pick ~u1 =

[10

], ~u2 =

[01

]as linearly indep eigenvectors.

I General solutions are ~x(t) = C1eλ1t~u1 + C2e

λ1t~u2

⇒ ~x(t) = C1eλ1t

[10

]+ C2e

λ1t

[01

]⇒ ~x(t) = eλ1t

[C1

C2

]

2D Systems:d~x

dt= A~x

What if A has repeated eigenvalues?Assume that the eigenvalues of A are: λ1 = λ2.

• Easy Cases: A =

[λ1 00 λ1

];

• Hard Cases: A 6=[λ1 00 λ1

], but λ1 = λ2.

Find Solutions in the Easy Cases: A = λ1I

I All vector ~x ∈ R2 satisfy (A− λ1I)~x = 0.The eigenspace of λ1 is the entire plane.

We can pick ~u1 =

[10

], ~u2 =

[01

]as linearly indep eigenvectors.

I General solutions are ~x(t) = C1eλ1t~u1 + C2e

λ1t~u2

⇒ ~x(t) = C1eλ1t

[10

]+ C2e

λ1t

[01

]⇒ ~x(t) = eλ1t

[C1

C2

]

2D Systems:d~x

dt= A~x

What if A has repeated eigenvalues?Assume that the eigenvalues of A are: λ1 = λ2.

• Easy Cases: A =

[λ1 00 λ1

];

• Hard Cases: A 6=[λ1 00 λ1

], but λ1 = λ2.

Find Solutions in the Easy Cases: A = λ1I

I All vector ~x ∈ R2 satisfy (A− λ1I)~x = 0.The eigenspace of λ1 is the entire plane.

We can pick ~u1 =

[10

], ~u2 =

[01

]as linearly indep eigenvectors.

I General solutions are ~x(t) = C1eλ1t~u1 + C2e

λ1t~u2

⇒ ~x(t) = C1eλ1t

[10

]+ C2e

λ1t

[01

]⇒ ~x(t) = eλ1t

[C1

C2

]

2D Systems:d~x

dt= A~x

What if A has repeated eigenvalues?Assume that the eigenvalues of A are: λ1 = λ2.

• Easy Cases: A =

[λ1 00 λ1

];

• Hard Cases: A 6=[λ1 00 λ1

], but λ1 = λ2.

Find Solutions in the Easy Cases: A = λ1I

I All vector ~x ∈ R2 satisfy (A− λ1I)~x = 0.The eigenspace of λ1 is the entire plane.

We can pick ~u1 =

[10

], ~u2 =

[01

]as linearly indep eigenvectors.

I General solutions are ~x(t) = C1eλ1t~u1 + C2e

λ1t~u2

⇒ ~x(t) = C1eλ1t

[10

]+ C2e

λ1t

[01

]⇒ ~x(t) = eλ1t

[C1

C2

]

2D Systems:d~x

dt= A~x

What if A has repeated eigenvalues?Assume that the eigenvalues of A are: λ1 = λ2.

• Easy Cases: A =

[λ1 00 λ1

];

• Hard Cases: A 6=[λ1 00 λ1

], but λ1 = λ2.

Find Solutions in the Easy Cases: A = λ1I

I All vector ~x ∈ R2 satisfy (A− λ1I)~x = 0.The eigenspace of λ1 is the entire plane.

We can pick ~u1 =

[10

], ~u2 =

[01

]as linearly indep eigenvectors.

I General solutions are ~x(t) = C1eλ1t~u1 + C2e

λ1t~u2

⇒ ~x(t) = C1eλ1t

[10

]+ C2e

λ1t

[01

]⇒ ~x(t) = eλ1t

[C1

C2

]

2D Systems:d~x

dt= A~x

What if A has repeated eigenvalues?Assume that the eigenvalues of A are: λ1 = λ2.

• Easy Cases: A =

[λ1 00 λ1

];

• Hard Cases: A 6=[λ1 00 λ1

], but λ1 = λ2.

Find Solutions in the Easy Cases: A = λ1I

I All vector ~x ∈ R2 satisfy (A− λ1I)~x = 0.The eigenspace of λ1 is the entire plane.

We can pick ~u1 =

[10

], ~u2 =

[01

]as linearly indep eigenvectors.

I General solutions are ~x(t) = C1eλ1t~u1 + C2e

λ1t~u2

⇒ ~x(t) = C1eλ1t

[10

]+ C2e

λ1t

[01

]

⇒ ~x(t) = eλ1t

[C1

C2

]

2D Systems:d~x

dt= A~x

What if A has repeated eigenvalues?Assume that the eigenvalues of A are: λ1 = λ2.

• Easy Cases: A =

[λ1 00 λ1

];

• Hard Cases: A 6=[λ1 00 λ1

], but λ1 = λ2.

Find Solutions in the Easy Cases: A = λ1I

I All vector ~x ∈ R2 satisfy (A− λ1I)~x = 0.The eigenspace of λ1 is the entire plane.

We can pick ~u1 =

[10

], ~u2 =

[01

]as linearly indep eigenvectors.

I General solutions are ~x(t) = C1eλ1t~u1 + C2e

λ1t~u2

⇒ ~x(t) = C1eλ1t

[10

]+ C2e

λ1t

[01

]⇒ ~x(t) = eλ1t

[C1

C2

]

2D Systems:d~x

dt= A~x

Find Solutions in the Hard Cases: λ1 = λ2, but A 6= λ1I

I Find an eigenvector ~u for λ1, by solving

(A− λ1I)~x = 0.

Since A 6= λ1I, we can only pick one linearly indep eigenvector ~u.This gives partial solutions: ~x(t) = Ceλ1t~u.

Need more to get complete solution formula.

I Find a generalized eigenvector ~v by solving

(A− λ1I)~x = ~u.

Solutions of (A− λ1I)~x = ~u satisfy (A− λ1I)2~x = (A− λ1I)~u = 0.

Nonzero solutions of (A− λ1I)2~x = 0 are called generalized eigenvectors.

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u)

2D Systems:d~x

dt= A~x

Find Solutions in the Hard Cases: λ1 = λ2, but A 6= λ1I

I Find an eigenvector ~u for λ1, by solving

(A− λ1I)~x = 0.

Since A 6= λ1I, we can only pick one linearly indep eigenvector ~u.This gives partial solutions: ~x(t) = Ceλ1t~u.

Need more to get complete solution formula.

I Find a generalized eigenvector ~v by solving

(A− λ1I)~x = ~u.

Solutions of (A− λ1I)~x = ~u satisfy (A− λ1I)2~x = (A− λ1I)~u = 0.

Nonzero solutions of (A− λ1I)2~x = 0 are called generalized eigenvectors.

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u)

2D Systems:d~x

dt= A~x

Find Solutions in the Hard Cases: λ1 = λ2, but A 6= λ1I

I Find an eigenvector ~u for λ1, by solving

(A− λ1I)~x = 0.Since A 6= λ1I, we can only pick one linearly indep eigenvector ~u.This gives partial solutions: ~x(t) = Ceλ1t~u.

Need more to get complete solution formula.

I Find a generalized eigenvector ~v by solving

(A− λ1I)~x = ~u.

Solutions of (A− λ1I)~x = ~u satisfy (A− λ1I)2~x = (A− λ1I)~u = 0.

Nonzero solutions of (A− λ1I)2~x = 0 are called generalized eigenvectors.

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u)

2D Systems:d~x

dt= A~x

Find Solutions in the Hard Cases: λ1 = λ2, but A 6= λ1I

I Find an eigenvector ~u for λ1, by solving

(A− λ1I)~x = 0.Since A 6= λ1I, we can only pick one linearly indep eigenvector ~u.This gives partial solutions: ~x(t) = Ceλ1t~u.

Need more to get complete solution formula.

I Find a generalized eigenvector ~v by solving

(A− λ1I)~x = ~u.

Solutions of (A− λ1I)~x = ~u satisfy (A− λ1I)2~x = (A− λ1I)~u = 0.

Nonzero solutions of (A− λ1I)2~x = 0 are called generalized eigenvectors.

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u)

2D Systems:d~x

dt= A~x

Find Solutions in the Hard Cases: λ1 = λ2, but A 6= λ1I

I Find an eigenvector ~u for λ1, by solving

(A− λ1I)~x = 0.Since A 6= λ1I, we can only pick one linearly indep eigenvector ~u.This gives partial solutions: ~x(t) = Ceλ1t~u.

Need more to get complete solution formula.

I Find a generalized eigenvector ~v by solving

(A− λ1I)~x = ~u.

Solutions of (A− λ1I)~x = ~u satisfy (A− λ1I)2~x = (A− λ1I)~u = 0.

Nonzero solutions of (A− λ1I)2~x = 0 are called generalized eigenvectors.

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u)

2D Systems:d~x

dt= A~x

Find Solutions in the Hard Cases: λ1 = λ2, but A 6= λ1I

I Find an eigenvector ~u for λ1, by solving

(A− λ1I)~x = 0.Since A 6= λ1I, we can only pick one linearly indep eigenvector ~u.This gives partial solutions: ~x(t) = Ceλ1t~u.

Need more to get complete solution formula.

I Find a generalized eigenvector ~v by solving

(A− λ1I)~x = ~u.

Solutions of (A− λ1I)~x = ~u satisfy (A− λ1I)2~x = (A− λ1I)~u = 0.

Nonzero solutions of (A− λ1I)2~x = 0 are called generalized eigenvectors.

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u)

2D Systems ~x′ = A~x: phase portraits & stability

λ1 = λ2 < 0, A = λ1I :Attractive proper node,asymtotically stable

λ1 = λ2 < 0, A 6= λ1I :Attractive degenerate node,asymtotically stable

λ1 = λ2 > 0, A = λ1I :Repulsive proper node,unstable

λ1 = λ2 > 0, A 6= λ1I :Repulsive degenerate node,unstable

λ1 = λ2 = 0, A = 0 :Every point is a stable equilib-rium, but not asymp stable

λ1 = λ2 = 0, A 6= 0 :Laminated flow,unstable

Example 1. (Attractive Proper node)

Consider ~x′ = A~x, where A =

[−2 00 −2

].

(a) Find general solutions of ~x′ =

[−2 00 −2

]~x.

(b) Solve the initial value problem ~x′ =

[−2 00 −2

]~x, ~x(0) =

[23

].

(c) Sketch the phase portrait.

(d) Is the equilibrium (0, 0) stable, asymptotically stable, or unstable?

Example 1.d~x

dt=

[−2 00 −2

]~x

(a) Find general solutions.

I Eigenvalues of A are λ1 = λ2 = −2.

I (A− λ1I)~x = 0⇔ 0~x = 0: All ~x ∈ R2 are eigenvectors.

I General solutions are

~x(t) = C1e−2t

[10

]+ C2e

−2t[01

]⇔ ~x(t) = e−2t

[C1

C2

]

(b) Solved~x

dt=

[−2 00 −2

]~x, ~x(0) =

[23

].

I Use the initial condition:

~x(0) =

[23

]⇒

[C1

C2

]=

[23

]I The solution to the initial value problem:

~x(t) = e−2t[23

]

Example 1.d~x

dt=

[−2 00 −2

]~x

(a) Find general solutions.

I Eigenvalues of A are λ1 = λ2 = −2.

I (A− λ1I)~x = 0⇔ 0~x = 0: All ~x ∈ R2 are eigenvectors.

I General solutions are

~x(t) = C1e−2t

[10

]+ C2e

−2t[01

]⇔ ~x(t) = e−2t

[C1

C2

]

(b) Solved~x

dt=

[−2 00 −2

]~x, ~x(0) =

[23

].

I Use the initial condition:

~x(0) =

[23

]⇒

[C1

C2

]=

[23

]I The solution to the initial value problem:

~x(t) = e−2t[23

]

Example 1.d~x

dt=

[−2 00 −2

]~x

(a) Find general solutions.

I Eigenvalues of A are λ1 = λ2 = −2.

I (A− λ1I)~x = 0⇔ 0~x = 0: All ~x ∈ R2 are eigenvectors.

I General solutions are

~x(t) = C1e−2t

[10

]+ C2e

−2t[01

]⇔ ~x(t) = e−2t

[C1

C2

]

(b) Solved~x

dt=

[−2 00 −2

]~x, ~x(0) =

[23

].

I Use the initial condition:

~x(0) =

[23

]⇒

[C1

C2

]=

[23

]I The solution to the initial value problem:

~x(t) = e−2t[23

]

Example 1.d~x

dt=

[−2 00 −2

]~x

(a) Find general solutions.

I Eigenvalues of A are λ1 = λ2 = −2.

I (A− λ1I)~x = 0⇔ 0~x = 0: All ~x ∈ R2 are eigenvectors.

I General solutions are

~x(t) = C1e−2t

[10

]+ C2e

−2t[01

]⇔ ~x(t) = e−2t

[C1

C2

]

(b) Solved~x

dt=

[−2 00 −2

]~x, ~x(0) =

[23

].

I Use the initial condition:

~x(0) =

[23

]⇒

[C1

C2

]=

[23

]I The solution to the initial value problem:

~x(t) = e−2t[23

]

Example 1.d~x

dt=

[−2 00 −2

]~x

(a) Find general solutions.

I Eigenvalues of A are λ1 = λ2 = −2.

I (A− λ1I)~x = 0⇔ 0~x = 0: All ~x ∈ R2 are eigenvectors.

I General solutions are

~x(t) = C1e−2t

[10

]+ C2e

−2t[01

]⇔ ~x(t) = e−2t

[C1

C2

]

(b) Solved~x

dt=

[−2 00 −2

]~x, ~x(0) =

[23

].

I Use the initial condition:

~x(0) =

[23

]⇒

[C1

C2

]=

[23

]

I The solution to the initial value problem:

~x(t) = e−2t[23

]

Example 1.d~x

dt=

[−2 00 −2

]~x

(a) Find general solutions.

I Eigenvalues of A are λ1 = λ2 = −2.

I (A− λ1I)~x = 0⇔ 0~x = 0: All ~x ∈ R2 are eigenvectors.

I General solutions are

~x(t) = C1e−2t

[10

]+ C2e

−2t[01

]⇔ ~x(t) = e−2t

[C1

C2

]

(b) Solved~x

dt=

[−2 00 −2

]~x, ~x(0) =

[23

].

I Use the initial condition:

~x(0) =

[23

]⇒

[C1

C2

]=

[23

]I The solution to the initial value problem:

~x(t) = e−2t[23

]

Example 1 (c) Phase portrait ofd~x

dt=

[−2 00 −2

]~x

General solutions: ~x(t) = e−2t[C1

C2

]

All solutions decay to 0 in the same exponential rate λ1 = −2.The trajectories are lines converging to the origin.

(d) Stability?

The equilibrium (0, 0) isasymptotically stable.

We have anattractive proper node,when A = λ1I and λ1 < 0.

Example 1 (c) Phase portrait ofd~x

dt=

[−2 00 −2

]~x

General solutions: ~x(t) = e−2t[C1

C2

]All solutions decay to 0 in the same exponential rate λ1 = −2.The trajectories are lines converging to the origin.

(d) Stability?

The equilibrium (0, 0) isasymptotically stable.

We have anattractive proper node,when A = λ1I and λ1 < 0.

Example 1 (c) Phase portrait ofd~x

dt=

[−2 00 −2

]~x

General solutions: ~x(t) = e−2t[C1

C2

]All solutions decay to 0 in the same exponential rate λ1 = −2.The trajectories are lines converging to the origin.

(d) Stability?

The equilibrium (0, 0) isasymptotically stable.

We have anattractive proper node,when A = λ1I and λ1 < 0.

Example 1 (c) Phase portrait ofd~x

dt=

[−2 00 −2

]~x

General solutions: ~x(t) = e−2t[C1

C2

]All solutions decay to 0 in the same exponential rate λ1 = −2.The trajectories are lines converging to the origin.

(d) Stability?

The equilibrium (0, 0) isasymptotically stable.

We have anattractive proper node,when A = λ1I and λ1 < 0.

Example 1 (c) Phase portrait ofd~x

dt=

[−2 00 −2

]~x

General solutions: ~x(t) = e−2t[C1

C2

]All solutions decay to 0 in the same exponential rate λ1 = −2.The trajectories are lines converging to the origin.

(d) Stability?

The equilibrium (0, 0) isasymptotically stable.

We have anattractive proper node,when A = λ1I and λ1 < 0.

Example 2. (Repulsive proper node)

~x′ = A~x, where A =

[3 00 3

].

(a) Find general solutions of ~x′ =

[3 00 3

]~x.

(b) Solve the initial value problem ~x′ =

[3 00 3

]~x, ~x(0) =

[23

].

(c) Sketch the phase portrait.

(d) Is the equilibrium (0, 0) stable, asymptotically stable, or unstable?

Example 2.d~x

dt=

[3 00 3

]~x

(a) Find general solutions.

General solutions are ~x(t) = e3t[C1

C2

]

(b) Solved~x

dt=

[3 00 3

]~x, ~x(0) =

[23

].

The solution to the initial value problem: ~x(t) = e3t[23

]

Example 2.d~x

dt=

[3 00 3

]~x

(a) Find general solutions.

General solutions are ~x(t) = e3t[C1

C2

]

(b) Solved~x

dt=

[3 00 3

]~x, ~x(0) =

[23

].

The solution to the initial value problem: ~x(t) = e3t[23

]

Example 2.d~x

dt=

[3 00 3

]~x

(a) Find general solutions.

General solutions are ~x(t) = e3t[C1

C2

]

(b) Solved~x

dt=

[3 00 3

]~x, ~x(0) =

[23

].

The solution to the initial value problem: ~x(t) = e3t[23

]

Example 2.d~x

dt=

[3 00 3

]~x

(a) Find general solutions.

General solutions are ~x(t) = e3t[C1

C2

]

(b) Solved~x

dt=

[3 00 3

]~x, ~x(0) =

[23

].

The solution to the initial value problem: ~x(t) = e3t[23

]

Example 2 (c) Phase portrait ofd~x

dt=

[3 00 3

]~x

General solutions: ~x(t) = e3t[C1

C2

]

All solutions grow in the same exponential rate λ1 = 3.The trajectories are lines emanating from the origin.

(d) Stability?

The equilibrium (0, 0) isunstable.

We have arepulsive proper node,when A = λ1I and λ1 > 0.

Example 2 (c) Phase portrait ofd~x

dt=

[3 00 3

]~x

General solutions: ~x(t) = e3t[C1

C2

]All solutions grow in the same exponential rate λ1 = 3.The trajectories are lines emanating from the origin.

(d) Stability?

The equilibrium (0, 0) isunstable.

We have arepulsive proper node,when A = λ1I and λ1 > 0.

Example 2 (c) Phase portrait ofd~x

dt=

[3 00 3

]~x

General solutions: ~x(t) = e3t[C1

C2

]All solutions grow in the same exponential rate λ1 = 3.The trajectories are lines emanating from the origin.

(d) Stability?

The equilibrium (0, 0) isunstable.

We have arepulsive proper node,when A = λ1I and λ1 > 0.

Example 2 (c) Phase portrait ofd~x

dt=

[3 00 3

]~x

General solutions: ~x(t) = e3t[C1

C2

]All solutions grow in the same exponential rate λ1 = 3.The trajectories are lines emanating from the origin.

(d) Stability?

The equilibrium (0, 0) isunstable.

We have arepulsive proper node,when A = λ1I and λ1 > 0.

Example 2 (c) Phase portrait ofd~x

dt=

[3 00 3

]~x

General solutions: ~x(t) = e3t[C1

C2

]All solutions grow in the same exponential rate λ1 = 3.The trajectories are lines emanating from the origin.

(d) Stability?

The equilibrium (0, 0) isunstable.

We have arepulsive proper node,when A = λ1I and λ1 > 0.

Example 3. (attractive degenerate node)

Consider ~x′ = A~x, where A =

[−7 8−2 1

].

(a) Find general solutions of ~x′ =

[−7 8−2 1

]~x.

(b) Solve the initial value problem ~x′ =

[−7 8−2 1

]~x, ~x(0) =

[23

].

(c) Sketch the phase portrait.

(d) Is the equilibrium (0, 0) stable, asymptotically stable, or unstable?

Example 3 (a) ~x′ =

[−7 8−2 1

]~x

I Eigenvalues of A, by solving det(A− λI) = 0:

det

[−7− λ 8−2 1− λ

]= λ2 + 6λ+ 9 = 0 ⇒ λ1 = λ2 = −3

I Eigenvectors of A for λ1 = λ2 = −3, by solving (A− λ1I)~x = 0:

(A+ 3I)~x = 0⇔[−4 8−2 4

] [x1x2

]=

[00

]⇔

[x1x2

]= x2

[21

]

• Can only pick one linear indep eigenvector ~u =

[21

].

• Partial solutions: ~x(t) = Ce−3t[21

].

• Need more to get complete solution formula.

Example 3 (a) ~x′ =

[−7 8−2 1

]~x

I Eigenvalues of A, by solving det(A− λI) = 0:

det

[−7− λ 8−2 1− λ

]= λ2 + 6λ+ 9 = 0 ⇒ λ1 = λ2 = −3

I Eigenvectors of A for λ1 = λ2 = −3, by solving (A− λ1I)~x = 0:

(A+ 3I)~x = 0⇔[−4 8−2 4

] [x1x2

]=

[00

]⇔

[x1x2

]= x2

[21

]

• Can only pick one linear indep eigenvector ~u =

[21

].

• Partial solutions: ~x(t) = Ce−3t[21

].

• Need more to get complete solution formula.

Example 3 (a) ~x′ =

[−7 8−2 1

]~x

I Eigenvalues of A, by solving det(A− λI) = 0:

det

[−7− λ 8−2 1− λ

]= λ2 + 6λ+ 9 = 0 ⇒ λ1 = λ2 = −3

I Eigenvectors of A for λ1 = λ2 = −3, by solving (A− λ1I)~x = 0:

(A+ 3I)~x = 0⇔[−4 8−2 4

] [x1x2

]=

[00

]⇔

[x1x2

]= x2

[21

]

• Can only pick one linear indep eigenvector ~u =

[21

].

• Partial solutions: ~x(t) = Ce−3t[21

].

• Need more to get complete solution formula.

Example 3 (a) ~x′ =

[−7 8−2 1

]~x

I Eigenvalues of A, by solving det(A− λI) = 0:

det

[−7− λ 8−2 1− λ

]= λ2 + 6λ+ 9 = 0 ⇒ λ1 = λ2 = −3

I Eigenvectors of A for λ1 = λ2 = −3, by solving (A− λ1I)~x = 0:

(A+ 3I)~x = 0⇔[−4 8−2 4

] [x1x2

]=

[00

]⇔

[x1x2

]= x2

[21

]• Can only pick one linear indep eigenvector ~u =

[21

].

• Partial solutions: ~x(t) = Ce−3t[21

].

• Need more to get complete solution formula.

Example 3 (a)d~x

dt=

[−7 8−2 1

]~x

I Eigenvalues of A: λ1 = λ2 = −3

I An eigenvector for λ1 = λ2 = −3: ~u =

[21

]

I Find a generalized eigenvector, by solving (A− λ1I)~x = ~u:

(A+ 3I)~x = ~u⇔[−4 8−2 4

] [x1x2

]=

[21

]⇔− 2x1 + 4x4 = 1⇔x1 = − 1

2 + 2x2⇔[x1x2

]=

[− 1

2 + 2x2x2

]

⇒ A generalized eigenvector ~v =

[− 1

20

].

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u),

~x(t) = C1e−3t

[21

]+ C2e

−3t([− 1

20

]+ t

[21

])

Example 3 (a)d~x

dt=

[−7 8−2 1

]~x

I Eigenvalues of A: λ1 = λ2 = −3

I An eigenvector for λ1 = λ2 = −3: ~u =

[21

]I Find a generalized eigenvector, by solving (A− λ1I)~x = ~u:

(A+ 3I)~x = ~u⇔[−4 8−2 4

] [x1x2

]=

[21

]⇔− 2x1 + 4x4 = 1⇔x1 = − 1

2 + 2x2⇔[x1x2

]=

[− 1

2 + 2x2x2

]

⇒ A generalized eigenvector ~v =

[− 1

20

].

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u),

~x(t) = C1e−3t

[21

]+ C2e

−3t([− 1

20

]+ t

[21

])

Example 3 (a)d~x

dt=

[−7 8−2 1

]~x

I Eigenvalues of A: λ1 = λ2 = −3

I An eigenvector for λ1 = λ2 = −3: ~u =

[21

]I Find a generalized eigenvector, by solving (A− λ1I)~x = ~u:

(A+ 3I)~x = ~u⇔[−4 8−2 4

] [x1x2

]=

[21

]⇔− 2x1 + 4x4 = 1⇔x1 = − 1

2 + 2x2⇔[x1x2

]=

[− 1

2 + 2x2x2

]⇒ A generalized eigenvector ~v =

[− 1

20

].

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u),

~x(t) = C1e−3t

[21

]+ C2e

−3t([− 1

20

]+ t

[21

])

Example 3 (a)d~x

dt=

[−7 8−2 1

]~x

I Eigenvalues of A: λ1 = λ2 = −3

I An eigenvector for λ1 = λ2 = −3: ~u =

[21

]I Find a generalized eigenvector, by solving (A− λ1I)~x = ~u:

(A+ 3I)~x = ~u⇔[−4 8−2 4

] [x1x2

]=

[21

]⇔− 2x1 + 4x4 = 1⇔x1 = − 1

2 + 2x2⇔[x1x2

]=

[− 1

2 + 2x2x2

]⇒ A generalized eigenvector ~v =

[− 1

20

].

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u),

~x(t) = C1e−3t

[21

]+ C2e

−3t([− 1

20

]+ t

[21

])

Example 3 (a)d~x

dt=

[−7 8−2 1

]~x

I Eigenvalues of A: λ1 = λ2 = −3

I An eigenvector for λ1 = λ2 = −3: ~u =

[21

]I Find a generalized eigenvector, by solving (A− λ1I)~x = ~u:

(A+ 3I)~x = ~u⇔[−4 8−2 4

] [x1x2

]=

[21

]⇔− 2x1 + 4x4 = 1⇔x1 = − 1

2 + 2x2⇔[x1x2

]=

[− 1

2 + 2x2x2

]⇒ A generalized eigenvector ~v =

[− 1

20

].

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u),

~x(t) = C1e−3t

[21

]+ C2e

−3t([− 1

20

]+ t

[21

])

Example 3 (b) Solved~x

dt=

[−7 8−2 1

]~x, ~x(0) =

[23

]

I General solutions:

~x(t) = C1e−3t

[21

]+ C2e

−3t([− 1

20

]+ t

[21

])I Use the initial condition:

~x(0) =

[23

]⇒ C1

[21

]+ C2

[− 1

20

]=

[23

]⇒

[2 − 1

21 0

] [C1

C2

]=

[23

]⇒

[C1

C2

]=

[38

]I The solution to the initial value problem:

~x(t) = 3e−3t[21

]+ 8e−3t

([− 1

20

]+ t

[21

])= e−3t

[2 + 16t3 + 8t

]

Example 3 (b) Solved~x

dt=

[−7 8−2 1

]~x, ~x(0) =

[23

]

I General solutions:

~x(t) = C1e−3t

[21

]+ C2e

−3t([− 1

20

]+ t

[21

])

I Use the initial condition:

~x(0) =

[23

]⇒ C1

[21

]+ C2

[− 1

20

]=

[23

]⇒

[2 − 1

21 0

] [C1

C2

]=

[23

]⇒

[C1

C2

]=

[38

]I The solution to the initial value problem:

~x(t) = 3e−3t[21

]+ 8e−3t

([− 1

20

]+ t

[21

])= e−3t

[2 + 16t3 + 8t

]

Example 3 (b) Solved~x

dt=

[−7 8−2 1

]~x, ~x(0) =

[23

]

I General solutions:

~x(t) = C1e−3t

[21

]+ C2e

−3t([− 1

20

]+ t

[21

])I Use the initial condition:

~x(0) =

[23

]⇒ C1

[21

]+ C2

[− 1

20

]=

[23

]⇒

[2 − 1

21 0

] [C1

C2

]=

[23

]⇒

[C1

C2

]=

[38

]

I The solution to the initial value problem:

~x(t) = 3e−3t[21

]+ 8e−3t

([− 1

20

]+ t

[21

])= e−3t

[2 + 16t3 + 8t

]

Example 3 (b) Solved~x

dt=

[−7 8−2 1

]~x, ~x(0) =

[23

]

I General solutions:

~x(t) = C1e−3t

[21

]+ C2e

−3t([− 1

20

]+ t

[21

])I Use the initial condition:

~x(0) =

[23

]⇒ C1

[21

]+ C2

[− 1

20

]=

[23

]⇒

[2 − 1

21 0

] [C1

C2

]=

[23

]⇒

[C1

C2

]=

[38

]I The solution to the initial value problem:

~x(t) = 3e−3t[21

]+ 8e−3t

([− 1

20

]+ t

[21

])= e−3t

[2 + 16t3 + 8t

]

Example 3 (c) Phase portrait ofd~x

dt=

[−7 8−2 1

]~x

General solutions: ~x(t) = C1e−3t

[21

]+ C2e

−3t([− 1

20

]+ t

[21

])

• When C2 = 0, ~x(t) = C1e−3t

[21

]decays to the origin,

along the eigenspace of λ1 = λ2 = −3.

Example 3 (c) Phase portrait ofd~x

dt=

[−7 8−2 1

]~x

General solutions: ~x(t) = C1e−3t

[21

]+ C2e

−3t([− 1

20

]+ t

[21

])• When C2 = 0, ~x(t) = C1e

−3t[21

]decays to the origin,

along the eigenspace of λ1 = λ2 = −3.

Example 3 (c) Phase portrait ofd~x

dt=

[−7 8−2 1

]~x

General solutions: ~x(t) = C1e−3t

[21

]+ C2e

−3t([− 1

20

]+ t

[21

])• When C2 = 0, ~x(t) = C1e

−3t[21

]decays to the origin,

along the eigenspace of λ1 = λ2 = −3.

Example 3 (c) Phase portrait ofd~x

dt=

[−7 8−2 1

]~x

General solutions: ~x(t) = C1e−3t

[21

]+ C2e

−3t([− 1

20

]+ t

[21

])• When C2 6= 0, ~x(t) ≈ C2te

−3t[21

]for t ≈ ∞,

decaying to the origin along the eigenspace of λ1 = λ2 = −3.

(d) Stability or instability?

The equilibrium (0, 0) isasymptotically stable.

We have anattractive degenerate node,when λ1 = λ2 < 0, but A 6= λ1I.

Example 3 (c) Phase portrait ofd~x

dt=

[−7 8−2 1

]~x

General solutions: ~x(t) = C1e−3t

[21

]+ C2e

−3t([− 1

20

]+ t

[21

])• When C2 6= 0, ~x(t) ≈ C2te

−3t[21

]for t ≈ ∞,

decaying to the origin along the eigenspace of λ1 = λ2 = −3.

(d) Stability or instability?

The equilibrium (0, 0) isasymptotically stable.

We have anattractive degenerate node,when λ1 = λ2 < 0, but A 6= λ1I.

Example 3 (c) Phase portrait ofd~x

dt=

[−7 8−2 1

]~x

General solutions: ~x(t) = C1e−3t

[21

]+ C2e

−3t([− 1

20

]+ t

[21

])• When C2 6= 0, ~x(t) ≈ C2te

−3t[21

]for t ≈ ∞,

decaying to the origin along the eigenspace of λ1 = λ2 = −3.

(d) Stability or instability?

The equilibrium (0, 0) isasymptotically stable.

We have anattractive degenerate node,when λ1 = λ2 < 0, but A 6= λ1I.

Example 3 (c) Phase portrait ofd~x

dt=

[−7 8−2 1

]~x

General solutions: ~x(t) = C1e−3t

[21

]+ C2e

−3t([− 1

20

]+ t

[21

])• When C2 6= 0, ~x(t) ≈ C2te

−3t[21

]for t ≈ ∞,

decaying to the origin along the eigenspace of λ1 = λ2 = −3.

(d) Stability or instability?

The equilibrium (0, 0) isasymptotically stable.

We have anattractive degenerate node,when λ1 = λ2 < 0, but A 6= λ1I.

Example 3 (c) Phase portrait ofd~x

dt=

[−7 8−2 1

]~x

General solutions: ~x(t) = C1e−3t

[21

]+ C2e

−3t([− 1

20

]+ t

[21

])• When C2 6= 0, ~x(t) ≈ C2te

−3t[21

]for t ≈ ∞,

decaying to the origin along the eigenspace of λ1 = λ2 = −3.

(d) Stability or instability?

The equilibrium (0, 0) isasymptotically stable.

We have anattractive degenerate node,when λ1 = λ2 < 0, but A 6= λ1I.

Example 4. (repulsive degenerate node)

Consider ~x′ = A~x, where A =

[1 −14 5

].

(a) Find general solutions of ~x′ =

[1 −14 5

]~x.

(b) Solve the initial value problem ~x′ =

[1 −14 5

]~x, ~x(0) =

[23

].

(c) Sketch the phase portrait.

(d) Is the equilibrium (0, 0) stable, asymptotically stable, or unstable?

Example 4 (a) ~x′ =

[1 −14 5

]~x

I Eigenvalues of A, by solving det(A− λI) = 0:

det

[1− λ −1

4 5− λ

]= λ2 − 6λ+ 9 = 0 ⇒ λ1 = λ2 = 3

I Eigenvectors of A for λ1 = λ2 = 3, by solving (A− λ1I)~x = 0:

(A− 3I)~x = 0⇔[−2 −14 2

] [x1x2

]=

[00

]⇔

[x1x2

]= x2

[− 1

21

]

• Can only pick one linear indep eigenvector ~u =

[− 1

21

].

• Partial solutions: ~x(t) = Ce3t[− 1

21

].

• Need more to get complete solution formula.

Example 4 (a) ~x′ =

[1 −14 5

]~x

I Eigenvalues of A, by solving det(A− λI) = 0:

det

[1− λ −1

4 5− λ

]= λ2 − 6λ+ 9 = 0 ⇒ λ1 = λ2 = 3

I Eigenvectors of A for λ1 = λ2 = 3, by solving (A− λ1I)~x = 0:

(A− 3I)~x = 0⇔[−2 −14 2

] [x1x2

]=

[00

]⇔

[x1x2

]= x2

[− 1

21

]

• Can only pick one linear indep eigenvector ~u =

[− 1

21

].

• Partial solutions: ~x(t) = Ce3t[− 1

21

].

• Need more to get complete solution formula.

Example 4 (a) ~x′ =

[1 −14 5

]~x

I Eigenvalues of A, by solving det(A− λI) = 0:

det

[1− λ −1

4 5− λ

]= λ2 − 6λ+ 9 = 0 ⇒ λ1 = λ2 = 3

I Eigenvectors of A for λ1 = λ2 = 3, by solving (A− λ1I)~x = 0:

(A− 3I)~x = 0⇔[−2 −14 2

] [x1x2

]=

[00

]⇔

[x1x2

]= x2

[− 1

21

]

• Can only pick one linear indep eigenvector ~u =

[− 1

21

].

• Partial solutions: ~x(t) = Ce3t[− 1

21

].

• Need more to get complete solution formula.

Example 4 (a) ~x′ =

[1 −14 5

]~x

I Eigenvalues of A, by solving det(A− λI) = 0:

det

[1− λ −1

4 5− λ

]= λ2 − 6λ+ 9 = 0 ⇒ λ1 = λ2 = 3

I Eigenvectors of A for λ1 = λ2 = 3, by solving (A− λ1I)~x = 0:

(A− 3I)~x = 0⇔[−2 −14 2

] [x1x2

]=

[00

]⇔

[x1x2

]= x2

[− 1

21

]• Can only pick one linear indep eigenvector ~u =

[− 1

21

].

• Partial solutions: ~x(t) = Ce3t[− 1

21

].

• Need more to get complete solution formula.

Example 4 (a)d~x

dt=

[1 −14 5

]~x

I Eigenvalues of A: λ1 = λ2 = 3

I An eigenvector for λ1 = λ2 = 3: ~u =

[− 1

21

]

I Find a generalized eigenvector, by solving (A− λ1I)~x = ~u:

(A− 3I)~x = ~u⇔[−2 −14 2

] [x1x2

]=

[− 1

21

]⇔− 2x1 − x2 = − 1

2⇔x1 = 14 −

12x2⇔

[x1x2

]=

[14 −

12x2

x2

]

⇒ A generalized eigenvector ~v =

[140

].

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u),

~x(t) = C1e3t

[− 1

21

]+ C2e

3t

([140

]+ t

[− 1

21

])

Example 4 (a)d~x

dt=

[1 −14 5

]~x

I Eigenvalues of A: λ1 = λ2 = 3

I An eigenvector for λ1 = λ2 = 3: ~u =

[− 1

21

]I Find a generalized eigenvector, by solving (A− λ1I)~x = ~u:

(A− 3I)~x = ~u⇔[−2 −14 2

] [x1x2

]=

[− 1

21

]⇔− 2x1 − x2 = − 1

2⇔x1 = 14 −

12x2⇔

[x1x2

]=

[14 −

12x2

x2

]

⇒ A generalized eigenvector ~v =

[140

].

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u),

~x(t) = C1e3t

[− 1

21

]+ C2e

3t

([140

]+ t

[− 1

21

])

Example 4 (a)d~x

dt=

[1 −14 5

]~x

I Eigenvalues of A: λ1 = λ2 = 3

I An eigenvector for λ1 = λ2 = 3: ~u =

[− 1

21

]I Find a generalized eigenvector, by solving (A− λ1I)~x = ~u:

(A− 3I)~x = ~u⇔[−2 −14 2

] [x1x2

]=

[− 1

21

]⇔− 2x1 − x2 = − 1

2⇔x1 = 14 −

12x2⇔

[x1x2

]=

[14 −

12x2

x2

]⇒ A generalized eigenvector ~v =

[140

].

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u),

~x(t) = C1e3t

[− 1

21

]+ C2e

3t

([140

]+ t

[− 1

21

])

Example 4 (a)d~x

dt=

[1 −14 5

]~x

I Eigenvalues of A: λ1 = λ2 = 3

I An eigenvector for λ1 = λ2 = 3: ~u =

[− 1

21

]I Find a generalized eigenvector, by solving (A− λ1I)~x = ~u:

(A− 3I)~x = ~u⇔[−2 −14 2

] [x1x2

]=

[− 1

21

]⇔− 2x1 − x2 = − 1

2⇔x1 = 14 −

12x2⇔

[x1x2

]=

[14 −

12x2

x2

]⇒ A generalized eigenvector ~v =

[140

].

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u),

~x(t) = C1e3t

[− 1

21

]+ C2e

3t

([140

]+ t

[− 1

21

])

Example 4 (a)d~x

dt=

[1 −14 5

]~x

I Eigenvalues of A: λ1 = λ2 = 3

I An eigenvector for λ1 = λ2 = 3: ~u =

[− 1

21

]I Find a generalized eigenvector, by solving (A− λ1I)~x = ~u:

(A− 3I)~x = ~u⇔[−2 −14 2

] [x1x2

]=

[− 1

21

]⇔− 2x1 − x2 = − 1

2⇔x1 = 14 −

12x2⇔

[x1x2

]=

[14 −

12x2

x2

]⇒ A generalized eigenvector ~v =

[140

].

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u),

~x(t) = C1e3t

[− 1

21

]+ C2e

3t

([140

]+ t

[− 1

21

])

Example 4 (b) Solved~x

dt=

[1 −14 5

]~x, ~x(0) =

[23

]

I General solutions:

~x(t) = C1e3t

[− 1

21

]+ C2e

3t

([140

]+ t

[− 1

21

])I Use the initial condition:

~x(0) =

[23

]⇒ C1

[− 1

21

]+ C2

[140

]=

[23

]⇒

[− 1

214

1 0

] [C1

C2

]=

[23

]⇒

[C1

C2

]=

[314

]I The solution to the initial value problem:

~x(t) = 3e3t[− 1

21

]+ 14e3t

([140

]+ t

[− 1

21

])= e3t

[2− 7t3 + 14t

]

Example 4 (b) Solved~x

dt=

[1 −14 5

]~x, ~x(0) =

[23

]

I General solutions:

~x(t) = C1e3t

[− 1

21

]+ C2e

3t

([140

]+ t

[− 1

21

])

I Use the initial condition:

~x(0) =

[23

]⇒ C1

[− 1

21

]+ C2

[140

]=

[23

]⇒

[− 1

214

1 0

] [C1

C2

]=

[23

]⇒

[C1

C2

]=

[314

]I The solution to the initial value problem:

~x(t) = 3e3t[− 1

21

]+ 14e3t

([140

]+ t

[− 1

21

])= e3t

[2− 7t3 + 14t

]

Example 4 (b) Solved~x

dt=

[1 −14 5

]~x, ~x(0) =

[23

]

I General solutions:

~x(t) = C1e3t

[− 1

21

]+ C2e

3t

([140

]+ t

[− 1

21

])I Use the initial condition:

~x(0) =

[23

]⇒ C1

[− 1

21

]+ C2

[140

]=

[23

]⇒

[− 1

214

1 0

] [C1

C2

]=

[23

]⇒

[C1

C2

]=

[314

]

I The solution to the initial value problem:

~x(t) = 3e3t[− 1

21

]+ 14e3t

([140

]+ t

[− 1

21

])= e3t

[2− 7t3 + 14t

]

Example 4 (b) Solved~x

dt=

[1 −14 5

]~x, ~x(0) =

[23

]

I General solutions:

~x(t) = C1e3t

[− 1

21

]+ C2e

3t

([140

]+ t

[− 1

21

])I Use the initial condition:

~x(0) =

[23

]⇒ C1

[− 1

21

]+ C2

[140

]=

[23

]⇒

[− 1

214

1 0

] [C1

C2

]=

[23

]⇒

[C1

C2

]=

[314

]I The solution to the initial value problem:

~x(t) = 3e3t[− 1

21

]+ 14e3t

([140

]+ t

[− 1

21

])= e3t

[2− 7t3 + 14t

]

Example 4 (c) Phase portrait ofd~x

dt=

[1 −14 5

]~x

General solutions: ~x(t) = C1e3t

[− 1

21

]+ C2e

3t

([140

]+ t

[− 1

21

])

• When C2 = 0, ~x(t) = C1e3t

[− 1

21

]leaves the origin,

along the eigenspace of λ1 = λ2 = 3.

Example 4 (c) Phase portrait ofd~x

dt=

[1 −14 5

]~x

General solutions: ~x(t) = C1e3t

[− 1

21

]+ C2e

3t

([140

]+ t

[− 1

21

])• When C2 = 0, ~x(t) = C1e

3t

[− 1

21

]leaves the origin,

along the eigenspace of λ1 = λ2 = 3.

Example 4 (c) Phase portrait ofd~x

dt=

[1 −14 5

]~x

General solutions: ~x(t) = C1e3t

[− 1

21

]+ C2e

3t

([140

]+ t

[− 1

21

])• When C2 = 0, ~x(t) = C1e

3t

[− 1

21

]leaves the origin,

along the eigenspace of λ1 = λ2 = 3.

Example 4 (c) Phase portrait ofd~x

dt=

[1 −14 5

]~x

General solutions: ~x(t) = C1e3t

[− 1

21

]+ C2e

3t

([140

]+ t

[− 1

21

])• When C2 6= 0, ~x(t) ≈ C2te

3t

[− 1

21

]is very large, for t ≈ ∞;

~x(t) ≈ C2te3t

[− 1

21

]is very small, for t ≈ −∞.

(d) Stability or instability?

The equilibrium (0, 0) isunstable.

We have a repulsive degenerate node,when λ1 = λ2 > 0, but A 6= λ1I.

Example 4 (c) Phase portrait ofd~x

dt=

[1 −14 5

]~x

General solutions: ~x(t) = C1e3t

[− 1

21

]+ C2e

3t

([140

]+ t

[− 1

21

])• When C2 6= 0, ~x(t) ≈ C2te

3t

[− 1

21

]is very large, for t ≈ ∞;

~x(t) ≈ C2te3t

[− 1

21

]is very small, for t ≈ −∞.

(d) Stability or instability?

The equilibrium (0, 0) isunstable.

We have a repulsive degenerate node,when λ1 = λ2 > 0, but A 6= λ1I.

Example 4 (c) Phase portrait ofd~x

dt=

[1 −14 5

]~x

General solutions: ~x(t) = C1e3t

[− 1

21

]+ C2e

3t

([140

]+ t

[− 1

21

])• When C2 6= 0, ~x(t) ≈ C2te

3t

[− 1

21

]is very large, for t ≈ ∞;

~x(t) ≈ C2te3t

[− 1

21

]is very small, for t ≈ −∞.

(d) Stability or instability?

The equilibrium (0, 0) isunstable.

We have a repulsive degenerate node,when λ1 = λ2 > 0, but A 6= λ1I.

Example 4 (c) Phase portrait ofd~x

dt=

[1 −14 5

]~x

General solutions: ~x(t) = C1e3t

[− 1

21

]+ C2e

3t

([140

]+ t

[− 1

21

])• When C2 6= 0, ~x(t) ≈ C2te

3t

[− 1

21

]is very large, for t ≈ ∞;

~x(t) ≈ C2te3t

[− 1

21

]is very small, for t ≈ −∞.

(d) Stability or instability?

The equilibrium (0, 0) isunstable.

We have a repulsive degenerate node,when λ1 = λ2 > 0, but A 6= λ1I.

Example 4 (c) Phase portrait ofd~x

dt=

[1 −14 5

]~x

General solutions: ~x(t) = C1e3t

[− 1

21

]+ C2e

3t

([140

]+ t

[− 1

21

])• When C2 6= 0, ~x(t) ≈ C2te

3t

[− 1

21

]is very large, for t ≈ ∞;

~x(t) ≈ C2te3t

[− 1

21

]is very small, for t ≈ −∞.

(d) Stability or instability?

The equilibrium (0, 0) isunstable.

We have a repulsive degenerate node,when λ1 = λ2 > 0, but A 6= λ1I.

Example 5. (laminated flow)

Consider ~x′ = A~x, where A =

[−6 4−9 6

].

(a) Find general solutions of ~x′ =

[−6 4−9 6

]~x.

(b) Solve the initial value problem ~x′ =

[−6 4−9 6

]~x, ~x(0) =

[23

].

(c) Sketch the phase portrait.

(d) Is the equilibrium (0, 0) stable, asymptotically stable, or unstable?

Example 5 (a) ~x′ =

[−6 4−9 6

]~x

I Eigenvalues of A, by solving det(A− λI) = 0:

det

[−6− λ 4−9 6− λ

]= λ2 = 0 ⇒ λ1 = λ2 = 0

I Eigenvectors of A for λ1 = λ2 = 0, by solving (A− λ1I)~x = 0:

A~x = 0⇔[−6 4−9 6

] [x1x2

]=

[00

]⇔

[x1x2

]= x2

[231

]

• Can only pick one linear indep eigenvector ~u =

[231

].

• Equilibrium solutions: ~x(t) = C

[231

].

• Need more to get complete solution formula.

Example 5 (a) ~x′ =

[−6 4−9 6

]~x

I Eigenvalues of A, by solving det(A− λI) = 0:

det

[−6− λ 4−9 6− λ

]= λ2 = 0 ⇒ λ1 = λ2 = 0

I Eigenvectors of A for λ1 = λ2 = 0, by solving (A− λ1I)~x = 0:

A~x = 0⇔[−6 4−9 6

] [x1x2

]=

[00

]⇔

[x1x2

]= x2

[231

]

• Can only pick one linear indep eigenvector ~u =

[231

].

• Equilibrium solutions: ~x(t) = C

[231

].

• Need more to get complete solution formula.

Example 5 (a) ~x′ =

[−6 4−9 6

]~x

I Eigenvalues of A, by solving det(A− λI) = 0:

det

[−6− λ 4−9 6− λ

]= λ2 = 0 ⇒ λ1 = λ2 = 0

I Eigenvectors of A for λ1 = λ2 = 0, by solving (A− λ1I)~x = 0:

A~x = 0⇔[−6 4−9 6

] [x1x2

]=

[00

]⇔

[x1x2

]= x2

[231

]

• Can only pick one linear indep eigenvector ~u =

[231

].

• Equilibrium solutions: ~x(t) = C

[231

].

• Need more to get complete solution formula.

Example 5 (a) ~x′ =

[−6 4−9 6

]~x

I Eigenvalues of A, by solving det(A− λI) = 0:

det

[−6− λ 4−9 6− λ

]= λ2 = 0 ⇒ λ1 = λ2 = 0

I Eigenvectors of A for λ1 = λ2 = 0, by solving (A− λ1I)~x = 0:

A~x = 0⇔[−6 4−9 6

] [x1x2

]=

[00

]⇔

[x1x2

]= x2

[231

]• Can only pick one linear indep eigenvector ~u =

[231

].

• Equilibrium solutions: ~x(t) = C

[231

].

• Need more to get complete solution formula.

Example 5 (a)d~x

dt=

[−6 4−9 6

]~x

I Eigenvalues of A: λ1 = λ2 = 0

I An eigenvector for λ1 = λ2 = 0: ~u =

[231

]

I Find a generalized eigenvector, by solving (A− λ1I)~x = ~u:

A~x = ~u⇔[−6 4−9 6

] [x1x2

]=

[231

]⇔− 6x1 + 4x2 = 2

3⇔x1 = − 19 + 2

3x2⇔[x1x2

]=

[− 1

9 + 23x2

x2

]

⇒ A generalized eigenvector ~v =

[− 1

90

].

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u),

~x(t) = C1

[231

]+ C2

([− 1

90

]+ t

[231

])

Example 5 (a)d~x

dt=

[−6 4−9 6

]~x

I Eigenvalues of A: λ1 = λ2 = 0

I An eigenvector for λ1 = λ2 = 0: ~u =

[231

]I Find a generalized eigenvector, by solving (A− λ1I)~x = ~u:

A~x = ~u⇔[−6 4−9 6

] [x1x2

]=

[231

]⇔− 6x1 + 4x2 = 2

3⇔x1 = − 19 + 2

3x2⇔[x1x2

]=

[− 1

9 + 23x2

x2

]

⇒ A generalized eigenvector ~v =

[− 1

90

].

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u),

~x(t) = C1

[231

]+ C2

([− 1

90

]+ t

[231

])

Example 5 (a)d~x

dt=

[−6 4−9 6

]~x

I Eigenvalues of A: λ1 = λ2 = 0

I An eigenvector for λ1 = λ2 = 0: ~u =

[231

]I Find a generalized eigenvector, by solving (A− λ1I)~x = ~u:

A~x = ~u⇔[−6 4−9 6

] [x1x2

]=

[231

]⇔− 6x1 + 4x2 = 2

3⇔x1 = − 19 + 2

3x2⇔[x1x2

]=

[− 1

9 + 23x2

x2

]⇒ A generalized eigenvector ~v =

[− 1

90

].

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u),

~x(t) = C1

[231

]+ C2

([− 1

90

]+ t

[231

])

Example 5 (a)d~x

dt=

[−6 4−9 6

]~x

I Eigenvalues of A: λ1 = λ2 = 0

I An eigenvector for λ1 = λ2 = 0: ~u =

[231

]I Find a generalized eigenvector, by solving (A− λ1I)~x = ~u:

A~x = ~u⇔[−6 4−9 6

] [x1x2

]=

[231

]⇔− 6x1 + 4x2 = 2

3⇔x1 = − 19 + 2

3x2⇔[x1x2

]=

[− 1

9 + 23x2

x2

]⇒ A generalized eigenvector ~v =

[− 1

90

].

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u),

~x(t) = C1

[231

]+ C2

([− 1

90

]+ t

[231

])

Example 5 (a)d~x

dt=

[−6 4−9 6

]~x

I Eigenvalues of A: λ1 = λ2 = 0

I An eigenvector for λ1 = λ2 = 0: ~u =

[231

]I Find a generalized eigenvector, by solving (A− λ1I)~x = ~u:

A~x = ~u⇔[−6 4−9 6

] [x1x2

]=

[231

]⇔− 6x1 + 4x2 = 2

3⇔x1 = − 19 + 2

3x2⇔[x1x2

]=

[− 1

9 + 23x2

x2

]⇒ A generalized eigenvector ~v =

[− 1

90

].

I General solutions are ~x(t) = C1eλ1t~u + C2e

λ1t(~v + t~u),

~x(t) = C1

[231

]+ C2

([− 1

90

]+ t

[231

])

Example 5 (b) Solved~x

dt=

[−6 4−9 6

]~x, ~x(0) =

[23

]

I General solutions:

~x(t) = C1

[231

]+ C2

([− 1

90

]+ t

[231

])I Use the initial condition:

~x(0) =

[23

]⇒ C1

[231

]+ C2

[− 1

90

]=

[23

]⇒

[23 − 1

91 0

] [C1

C2

]=

[23

]⇒

[C1

C2

]=

[30

]I The solution to the initial value problem:

~x(t) = 3

[231

]=

[23

](an equilibrium)

Example 5 (b) Solved~x

dt=

[−6 4−9 6

]~x, ~x(0) =

[23

]

I General solutions:

~x(t) = C1

[231

]+ C2

([− 1

90

]+ t

[231

])

I Use the initial condition:

~x(0) =

[23

]⇒ C1

[231

]+ C2

[− 1

90

]=

[23

]⇒

[23 − 1

91 0

] [C1

C2

]=

[23

]⇒

[C1

C2

]=

[30

]I The solution to the initial value problem:

~x(t) = 3

[231

]=

[23

](an equilibrium)

Example 5 (b) Solved~x

dt=

[−6 4−9 6

]~x, ~x(0) =

[23

]

I General solutions:

~x(t) = C1

[231

]+ C2

([− 1

90

]+ t

[231

])I Use the initial condition:

~x(0) =

[23

]⇒ C1

[231

]+ C2

[− 1

90

]=

[23

]⇒

[23 − 1

91 0

] [C1

C2

]=

[23

]⇒

[C1

C2

]=

[30

]

I The solution to the initial value problem:

~x(t) = 3

[231

]=

[23

](an equilibrium)

Example 5 (b) Solved~x

dt=

[−6 4−9 6

]~x, ~x(0) =

[23

]

I General solutions:

~x(t) = C1

[231

]+ C2

([− 1

90

]+ t

[231

])I Use the initial condition:

~x(0) =

[23

]⇒ C1

[231

]+ C2

[− 1

90

]=

[23

]⇒

[23 − 1

91 0

] [C1

C2

]=

[23

]⇒

[C1

C2

]=

[30

]I The solution to the initial value problem:

~x(t) = 3

[231

]=

[23

](an equilibrium)

Example 5 (c) Phase portrait ofd~x

dt=

[−6 4−9 6

]~x

General solutions:

~x(t) = C1

[231

]+ C2

([− 1

90

]+ t

[231

])

• When C2 = 0,

~x(t) = C1

[231

]are equilibria,

lining along the eigenspace.

Example 5 (c) Phase portrait ofd~x

dt=

[−6 4−9 6

]~x

General solutions:

~x(t) = C1

[231

]+ C2

([− 1

90

]+ t

[231

])

• When C2 = 0,

~x(t) = C1

[231

]are equilibria,

lining along the eigenspace.

Example 5 (c) Phase portrait ofd~x

dt=

[−6 4−9 6

]~x

General solutions:

~x(t) = C1

[231

]+ C2

([− 1

90

]+ t

[231

])

• When C2 = 0,

~x(t) = C1

[231

]are equilibria,

lining along the eigenspace.

Example 5 (c) Phase portrait ofd~x

dt=

[−6 4−9 6

]~x

General solutions:

~x(t) = C1

[231

]+C2

([− 1

90

]+ t

[231

])=

(C1

[231

]+ C2

[− 1

90

])+tC2

[231

]

• When C2 6= 0,

~x(t) are linear functions;

~x(0) = C1

[231

]+ C2

[− 1

90

];

velocity = C2

[231

]parallel to

the eigenspace.

Example 5 (c) Phase portrait ofd~x

dt=

[−6 4−9 6

]~x

General solutions:

~x(t) = C1

[231

]+C2

([− 1

90

]+ t

[231

])=

(C1

[231

]+ C2

[− 1

90

])+tC2

[231

]

• When C2 6= 0,

~x(t) are linear functions;

~x(0) = C1

[231

]+ C2

[− 1

90

];

velocity = C2

[231

]parallel to

the eigenspace.

Example 5 (d) Is the equilibrium (0, 0) stable,asymptotically stable, or unstable?

General solutions: ~x(t) = C1

[231

]+ C2

([− 1

90

]+ t

[231

])

The equilibrium (0, 0) isunstable.

We have a laminar flow,when λ1 = λ2 = 0, but A 6= 0.

Example 5 (d) Is the equilibrium (0, 0) stable,asymptotically stable, or unstable?

General solutions: ~x(t) = C1

[231

]+ C2

([− 1

90

]+ t

[231

])

The equilibrium (0, 0) isunstable.

We have a laminar flow,when λ1 = λ2 = 0, but A 6= 0.

Example 5 (d) Is the equilibrium (0, 0) stable,asymptotically stable, or unstable?

General solutions: ~x(t) = C1

[231

]+ C2

([− 1

90

]+ t

[231

])

The equilibrium (0, 0) isunstable.

We have a laminar flow,when λ1 = λ2 = 0, but A 6= 0.