section 3 - university of michigangmarple/oct2.pdf · 2017-10-02 · 5 3 = 0: (1 )(3 + ) ( 1)5 = 0...

Agenda

Section 3.4

Reminders

Written HW 2 due 10/3 or 10/5

WebHW due 10/6

Office hours Tues, Thurs1-2 pm (5852 East Hall)

MathLab office hourSun 7-8 pm (MathLab)

(Gary Marple) October 2nd, 2017 Math 216: Introduction to Differential Equations

§3.4 Complex Eigenvalues

Objectives

Be able to find real solutions to x′ = Ax when A hascomplex eigenvalues

Be able to properly label the critical point wheneigenvalues are complex

Be able to sketch a phase portrait for systems withcomplex eigenvalues

Example

Find the general solution to

x′ = Ax,

[1 −15 −3

]and plot the phase portrait.

Recall that we’re looking of solutions of the form

x = veλt ,

where λ and v are the eigenvalues and eigenvectors of A,respectively. We begin by finding the eigenvalues.

det(A− λI ) = 0

Example

x′ = Ax,

[1 −15 −3

x = veλt ,

det(A− λI ) = 0

Example

x′ = Ax,

[1 −15 −3

x = veλt ,

det(A− λI ) = 0

det(A− λI ) = 0∣∣∣∣ 1− λ −15 −3− λ

∣∣∣∣ = 0.

−(1− λ)(3 + λ)− (−1)5 = 0

λ2 + 2λ + 2 = 0

λ =−2±

√4− 4 · 22

λ = −1± i

Recall that complex roots of polynomials with real coefficientscome in conjugate pairs. In addition, we showed that thecorresponding eigenvectors also come in conjugate pairs.Therefore, we just need to find one eigenvector and then weshould be able to write down the second one. Let’s start byfinding the eigenvector with λ = −1 + i .

det(A− λI ) = 0∣∣∣∣ 1− λ −15 −3− λ

∣∣∣∣ = 0.

−(1− λ)(3 + λ)− (−1)5 = 0

λ2 + 2λ + 2 = 0

λ =−2±

√4− 4 · 22

λ = −1± i

det(A− λI ) = 0∣∣∣∣ 1− λ −15 −3− λ

∣∣∣∣ = 0.

−(1− λ)(3 + λ)− (−1)5 = 0

λ2 + 2λ + 2 = 0

λ =−2±

√4− 4 · 22

λ = −1± i

det(A− λI ) = 0∣∣∣∣ 1− λ −15 −3− λ

∣∣∣∣ = 0.

−(1− λ)(3 + λ)− (−1)5 = 0

λ2 + 2λ + 2 = 0

λ =−2±

√4− 4 · 22

λ = −1± i

det(A− λI ) = 0∣∣∣∣ 1− λ −15 −3− λ

∣∣∣∣ = 0.

−(1− λ)(3 + λ)− (−1)5 = 0

λ2 + 2λ + 2 = 0

λ =−2±

√4− 4 · 22

λ = −1± i

det(A− λI ) = 0∣∣∣∣ 1− λ −15 −3− λ

∣∣∣∣ = 0.

−(1− λ)(3 + λ)− (−1)5 = 0

λ2 + 2λ + 2 = 0

λ =−2±

√4− 4 · 22

λ = −1± i

det(A− λI ) = 0∣∣∣∣ 1− λ −15 −3− λ

∣∣∣∣ = 0.

−(1− λ)(3 + λ)− (−1)5 = 0

λ2 + 2λ + 2 = 0

λ =−2±

√4− 4 · 22

λ = −1± i

To find the eigenvector with eigenvalue λ = −1 + i , we needto solve

(A− (−1 + i)I )v−1+i = 0([1 −15 −3

]− (−1 + i)

[1 00 1

])v−1+i = 0[

1 + 1− i −15 −3 + 1− i

]v−1+i = 0[

2− i −15 −2− i

]v−1+i = 0

We can express the system as an augmented matrix[2− i −1 0

5 −2− i 0

(A− (−1 + i)I )v−1+i = 0([1 −15 −3

]− (−1 + i)

[1 00 1

])v−1+i = 0[

1 + 1− i −15 −3 + 1− i

]v−1+i = 0[

2− i −15 −2− i

]v−1+i = 0

5 −2− i 0

(A− (−1 + i)I )v−1+i = 0([1 −15 −3

]− (−1 + i)

[1 00 1

])v−1+i = 0[

1 + 1− i −15 −3 + 1− i

]v−1+i = 0[

2− i −15 −2− i

]v−1+i = 0

5 −2− i 0

(A− (−1 + i)I )v−1+i = 0([1 −15 −3

]− (−1 + i)

[1 00 1

])v−1+i = 0[

1 + 1− i −15 −3 + 1− i

]v−1+i = 0[

2− i −15 −2− i

]v−1+i = 0

5 −2− i 0

(A− (−1 + i)I )v−1+i = 0([1 −15 −3

]− (−1 + i)

[1 00 1

])v−1+i = 0[

1 + 1− i −15 −3 + 1− i

]v−1+i = 0[

2− i −15 −2− i

]v−1+i = 0

5 −2− i 0

[2− i −1 0

5 −2− i 0

2− iR1 to R2[

2− i −1 00 0 0

]The matrix is now in row echelon form. In this case, the firstcolumn is a pivot column, so x1 is a basic variable and x2 is afree variable. Therefore, we should set x2 equal to a parameterand try to solve for x1.

x2 = c

We can use the first equation to solve for x1. That is,

(2− i)x1 − 1c = 0 =⇒ x1 =c

2− i.

[2− i −1 0

5 −2− i 0

2− iR1 to R2[

2− i −1 00 0 0

x2 = c

(2− i)x1 − 1c = 0 =⇒ x1 =c

2− i.

[2− i −1 0

5 −2− i 0

2− iR1 to R2[

2− i −1 00 0 0

x2 = c

(2− i)x1 − 1c = 0 =⇒ x1 =c

2− i.

[2− i −1 0

5 −2− i 0

2− iR1 to R2[

2− i −1 00 0 0

x2 = c

(2− i)x1 − 1c = 0 =⇒ x1 =c

2− i.

[2− i −1 0

5 −2− i 0

2− iR1 to R2[

2− i −1 00 0 0

x2 = c

(2− i)x1 − 1c = 0 =⇒ x1 =c

2− i.

[2− i −1 0

5 −2− i 0

2− iR1 to R2[

2− i −1 00 0 0

x2 = c

(2− i)x1 − 1c = 0 =⇒ x1 =c

2− i.

[2− i −1 0

5 −2− i 0

2− iR1 to R2[

2− i −1 00 0 0

x2 = c

(2− i)x1 − 1c = 0 =⇒ x1 =c

2− i.

We can now write out the eigenvector.

v−1+i =

[c/(2− i)

[1/(2− i)

]If we drop the parameter and multiply by 2− i to remove thefraction, we get

v−1+i =

2− i

Since eigenvectors come in conjugate pairs when A is real, wehave

v−1−i =

]as the eigenvector when λ = −1− i .

v−1+i =

[c/(2− i)

[1/(2− i)

v−1+i =

2− i

v−1−i =

v−1+i =

[c/(2− i)

[1/(2− i)

v−1+i =

2− i

v−1−i =

v−1+i =

2− i

]and v−1−i =

]We can now write down two solutions

x−1+i =

2− i

]e(−1+i)t and x−1−i =

]e(−1−i)t .

While we could multiply each solution by a constant and addthem together to construct the general solution, the main issueis that our solutions have complex terms which are hard tomake sense of, especially if we’re trying to model a real worldproblem. However, if we could find two linearly independent,real solutions from these complex solutions, then we could usethem to write down the general solution with real terms.

v−1+i =

2− i

]and v−1−i =

]We can now write down two solutions

x−1+i =

2− i

]e(−1+i)t and x−1−i =

]e(−1−i)t .

While we could multiply each solution by a constant and addthem together to construct the general solution, the main issueis that our solutions have complex terms which are hard tomake sense of, especially if we’re trying to model a real worldproblem. However, if we could find two linearly independent,real solutions from these complex solutions, then we could usethem to write down the general solution with real terms.

Constructing Real Solutions from Complex Solutions

Suppose x′ = Ax has a complex eigenvalue λ = α + iβ witheigenvector

vα+iβ = a + ib,

where α, β, a, and b are real. Since complex eigenvalues andeigenvectors come in conjugate pairs, we also know thatλ = α− iβ and

vα−iβ = a− ib

are eigenvalues and eigenvectors of A, respectively. Therefore,two solutions of x′ = Ax are

xα+iβ = (a + ib)e(α+iβ)t

andxα−iβ = (a− ib)e(α−iβ)t .

vα+iβ = a + ib,

vα−iβ = a− ib

vα+iβ = a + ib,

vα−iβ = a− ib

Recall Euler’s formula:

e iθ = cos θ + i sin θ.

(This identity can be derived by adding the Taylor series for cos θand i sin θ.) We can now use this to rewrite xα+iβ and xα−iβ.That is,

= (a + ib)eαte iβt

= (a + ib)eαt(cosβt + i sinβt)

andxα−iβ = (a− ib)e(α−iβ)t

= (a− ib)eαte−iβt

= (a− ib)eαt(cos (−βt) + i sin (−βt))

= (a− ib)eαt(cosβt − i sinβt).

In the last step, we used the fact that cosine is even and sine isodd.

xα+iβ = (a + ib)eαt(cos βt + i sin βt)

xα−iβ = (a− ib)eαt(cos βt − i sin βt)

If we separate xα+iβ and xα−iβ into their real and imagineparts, we get

xα+iβ = eαt (a cos βt − b sin βt) + ieαt (a sin βt + b cos βt)

xα−iβ = eαt (a cos βt − b sin βt)− ieαt (a sin βt + b cos βt) .

By the principle of superposition, we know that xα+iβ + xα−iβis a solution to x′ = Ax. Computing this gives us

xα+iβ + xα−iβ = 2eαt (a cos βt − b sin βt) .

This is a real solution! For simplicity, let’s divide by 2. Ourfirst real solution is then

x1 = eαt (a cos βt − b sin βt) .

To find our second real solution, let’s compute xα+iβ − xα−iβ.That is,

xα+iβ − xα−iβ = 2ieαt (a sin βt + b cos βt) .

Recall that if we divide a solution by a nonzero scalar, it’s stilla solution, even if the scalar is complex. Therefore, if wedivide xα+iβ − xα−iβ by 2i , we’ll have our second real solution.That is,

x2 = eαt (a sin βt + b cos βt) .

To find our second real solution, let’s compute xα+iβ − xα−iβ.That is,

xα+iβ − xα−iβ = 2ieαt (a sin βt + b cos βt) .

Recall that if we divide a solution by a nonzero scalar, it’s stilla solution, even if the scalar is complex. Therefore, if wedivide xα+iβ − xα−iβ by 2i , we’ll have our second real solution.That is,

x2 = eαt (a sin βt + b cos βt) .

To summarize, if you happen to get a complex eigenvalue, start bywriting the solution in the form veλt , like we normally would:

xα+iβ = (a + ib)e(α+iβ)t .

Next, use Euler’s formula, e iθ = cos θ + i sin θ, to rewrite theexponential in terms of sine and cosine. That is,

xα+iβ = (a + ib)eαt(cosβt + i sinβt).

After that, you should separate the real and imaginary terms to get

xα+iβ = eαt (a cosβt − b sinβt) + ieαt (a sinβt + b cosβt) .

We’ve shown that Re(xα+iβ) and Im(xα+iβ) both satisfy x′ = Ax.Therefore, we can use the principle of superposition to write oursolution as

x = c1eαt (a cosβt − b sinβt) + c2e

αt (a sinβt + b cosβt) .

Let’s use our solution from the example to find two real solutions to x′ = Ax. Recallthat our first solution was

x−1+i =

2− i

]e(−1+i)t .

We need to first rewrite the exponential term using Euler’s formula.

x−1+i =

2− i

]e(−1+i)t

2− i

]e−te it

2− i

]e−t (cos t + i sin t)

Next, we need to separate real and imaginary terms.

x−1+i =

2− i

[0−1

])e−t (cos t + i sin t)

= e−t

]cos t −

[0−1

]sin t

)+ ie−t

]sin t +

[0−1

]cos t

x−1+i =

2− i

]e(−1+i)t .

x−1+i =

2− i

]e(−1+i)t

2− i

]e−te it

2− i

x−1+i =

2− i

[0−1

= e−t

]cos t −

[0−1

]sin t

)+ ie−t

]sin t +

[0−1

]cos t

x−1+i =

2− i

]e(−1+i)t .

x−1+i =

2− i

]e(−1+i)t

2− i

]e−te it

2− i

x−1+i =

2− i

[0−1

= e−t

]cos t −

[0−1

]sin t

)+ ie−t

]sin t +

[0−1

]cos t

x−1+i =

2− i

]e(−1+i)t .

x−1+i =

2− i

]e(−1+i)t

2− i

]e−te it

2− i

x−1+i =

2− i

[0−1

= e−t

]cos t −

[0−1

]sin t

)+ ie−t

]sin t +

[0−1

]cos t

x−1+i =

2− i

]e(−1+i)t .

x−1+i =

2− i

]e(−1+i)t

2− i

]e−te it

2− i

x−1+i =

2− i

[0−1

= e−t

]cos t −

[0−1

]sin t

)+ ie−t

]sin t +

[0−1

]cos t

x−1+i =

2− i

]e(−1+i)t .

x−1+i =

2− i

]e(−1+i)t

2− i

]e−te it

2− i

x−1+i =

2− i

[0−1

= e−t

]cos t −

[0−1

]sin t

)+ ie−t

]sin t +

[0−1

]cos t

x−1+i =

2− i

]e(−1+i)t .

x−1+i =

2− i

]e(−1+i)t

2− i

]e−te it

2− i

x−1+i =

2− i

[0−1

= e−t

]cos t −

[0−1

]sin t

)+ ie−t

]sin t +

[0−1

]cos t

x−1+i = e−t

]cos t −

[0−1

]sin t

)+ ie−t

]sin t +

[0−1

]cos t

Since we know that Re(x−1+i ) and Im(x−1+i ) are solutions to x′ = Ax we can use theprinciple of superposition to write the solution as

x = c1e−t

]cos t −

[0−1

]sin t

)+ c2e

]sin t +

[0−1

]cos t

or more compactly as

x = c1e−t

[cos t

2 cos t + sin t

]+ c2e

[sin t

2 sin t − cos t

To check that our solutions form a fundamental set, we can compute the Wronskianand show it’s nonzero.

W [x1, x2](t) =

∣∣∣∣ e−t cos t e−t sin te−t(2 cos t + sin t) e−t(2 sin t − cos t)

∣∣∣∣(Gary Marple) October 2nd, 2017 Math 216: Introduction to Differential Equations

x−1+i = e−t

]cos t −

[0−1

]sin t

)+ ie−t

]sin t +

[0−1

]cos t

x = c1e−t

]cos t −

[0−1

]sin t

)+ c2e

]sin t +

[0−1

]cos t

x = c1e−t

[cos t

2 cos t + sin t

]+ c2e

[sin t

2 sin t − cos t

W [x1, x2](t) =

x−1+i = e−t

]cos t −

[0−1

]sin t

)+ ie−t

]sin t +

[0−1

]cos t

x = c1e−t

]cos t −

[0−1

]sin t

)+ c2e

]sin t +

[0−1

]cos t

x = c1e−t

[cos t

2 cos t + sin t

]+ c2e

[sin t

2 sin t − cos t

W [x1, x2](t) =

x−1+i = e−t

]cos t −

[0−1

]sin t

)+ ie−t

]sin t +

[0−1

]cos t

x = c1e−t

]cos t −

[0−1

]sin t

)+ c2e

]sin t +

[0−1

]cos t

x = c1e−t

[cos t

2 cos t + sin t

]+ c2e

[sin t

2 sin t − cos t

W [x1, x2](t) =

∣∣∣∣= e−2t(cos t)(2 sin t − cos t)− e−2t(sin t)(2 cos t + sin t)

= e−2t(2 sin t cos t − cos2 t)− e−2t(2 sin t cos t + sin2 t)

= e−2t(2 sin t cos t − cos2 t − 2 sin t cos t − sin2 t)

= −e−2t(cos2 t + sin2 t)

= −e−2t

Since −e−2t 6= 0 for all t, we can conclude that our solution is thegeneral solution.

W [x1, x2](t) =

= −e−2t

W [x1, x2](t) =

= −e−2t

W [x1, x2](t) =

= −e−2t

W [x1, x2](t) =

= −e−2t

W [x1, x2](t) =

= −e−2t

W [x1, x2](t) =

= −e−2t

x = c1e−t

[cos t

2 cos t + sin t

]+ c2e

sin t2 sin t − cos t

]Recall that our eigenvalues were λ = −1± i . The imaginary part of theeigenvalue leads to rotations in our phase portrait due to the sines andcosines in our solution. In addition, the negative real part causestrajectories to tend to the critical point at the origin. As a result we endup getting a spiral sink which is asymptotically stable. To determinewhich direction the spiral is spinning, just evaluate x′ = Ax at a point.

Theorem

Let A have real or complex eigenvalues λ1 and λ2 such thatλ1 6= λ2, and let the corresponding eigenvectors be

]and v2 =

([x11 x12

x21 x22

])6= 0.

Suppose that

([x11 x12

x21 x22

])= 0. (∗)

We will show that this leads to a contradiction. If (∗) weretrue, then the system[

x11 x12

x21 x22

]= 0 (?)

would have a nonzero solution. However, (?) can be rewrittenas

or simplyc1v1 + c2v2 = 0.

c1v1 + c2v2 = 0 (#)

Since there exists a nonzero solution to (#), v1 and v2 must belinearly dependent. Therefore, there exists a constant k such that

v1 = kv2.

Furthermore, k 6= 0 since eigenvectors by definition are nonzero. Ifwe multiply both sides by A, we get that

Av1 = kAv2

λ1v1 = kλ2v2.

If we plug in v1 = kv2 for v1, we see that

λ1kv2 = kλ2v2

k(λ1 − λ2)v2 = 0.

Since k 6= 0 and v2 6= 0 by definition, we must have that λ1 = λ2.However, this contradicts that assumption that λ1 6= λ2.

Example

Suppose that a matrix A has complex eigenvalues λ = α± iβ,where β 6= 0, and eigenvectors v = a± ib. Show that

x1 = eαt (a cosβt − b sinβt)

andx2 = eαt (a sinβt + b cosβt)

form a fundamental set of solutions to x′ = Ax.

We can show that x1 and x2 form a fundamental set of solutions ifwe can show that

W [x1, x2](t) 6= 0, for all t.

We start by computing the Wronskian.

W [x1, x2](t) =

∣∣∣∣ eαt(a1 cosβt − b1 sinβt) eαt(a1 sinβt + b1 cosβt)eαt(a2 cosβt − b2 sinβt) eαt(a2 sinβt + b2 cosβt)

Example

W [x1, x2](t) 6= 0, for all t.

W [x1, x2](t) =

Example

W [x1, x2](t) 6= 0, for all t.

W [x1, x2](t) =

∣∣∣∣=e2αt(a1 cosβt − b1 sinβt)(a2 sinβt + b2 cosβt)

−e2αt(a1 sinβt + b1 cosβt)(a2 cosβt − b2 sinβt)

=e2αt[a1b2(cos2 βt + sin2 βt)− a2b1(sin2 βt + cos2 βt)

]=e2αt(a1b2 − a2b1)

We know that e2αt 6= 0 for all t. Now we just need to show thata1b2 − a2b1 6= 0. Recall that

vα+iβ =

]and vα−iβ =

]− i

Since ∣∣∣∣ a1 + ib1 a1 − ib1

a2 + ib2 a2 − ib2

∣∣∣∣ = −2i(a1b2 − a2b1) 6= 0

by the previous theorem, we can conclude that a1b2 − a2b1 6= 0 andhence W [x1, x2](t) 6= 0. Therefore, x1 and x2 form a fundamental set ofsolutions.

W [x1, x2](t) =

]=e2αt(a1b2 − a2b1)

vα+iβ =

]and vα−iβ =

]− i

Since ∣∣∣∣ a1 + ib1 a1 − ib1

a2 + ib2 a2 − ib2

∣∣∣∣ = −2i(a1b2 − a2b1) 6= 0

W [x1, x2](t) =

]=e2αt(a1b2 − a2b1)

vα+iβ =

]and vα−iβ =

]− i

Since ∣∣∣∣ a1 + ib1 a1 − ib1

a2 + ib2 a2 − ib2

∣∣∣∣ = −2i(a1b2 − a2b1) 6= 0

W [x1, x2](t) =

]=e2αt(a1b2 − a2b1)

vα+iβ =

]and vα−iβ =

]− i

Since ∣∣∣∣ a1 + ib1 a1 − ib1

a2 + ib2 a2 − ib2

∣∣∣∣ = −2i(a1b2 − a2b1) 6= 0

W [x1, x2](t) =

]=e2αt(a1b2 − a2b1)

vα+iβ =

]and vα−iβ =

]− i

Since ∣∣∣∣ a1 + ib1 a1 − ib1

a2 + ib2 a2 − ib2

∣∣∣∣ = −2i(a1b2 − a2b1) 6= 0

W [x1, x2](t) =

]=e2αt(a1b2 − a2b1)

vα+iβ =

]and vα−iβ =

]− i

Since ∣∣∣∣ a1 + ib1 a1 − ib1

a2 + ib2 a2 − ib2

∣∣∣∣ = −2i(a1b2 − a2b1) 6= 0

W [x1, x2](t) =

]=e2αt(a1b2 − a2b1)

vα+iβ =

]and vα−iβ =

]− i

Since ∣∣∣∣ a1 + ib1 a1 − ib1

a2 + ib2 a2 − ib2

∣∣∣∣ = −2i(a1b2 − a2b1) 6= 0

Example

x′ = Ax,

where A has eigenvalues λ = ±i and correspondingeigenvectors

[2± i

Plot the phase portrait.

We can start by writing down one of our solutions in the formveλt .

[2 + i

Example

x′ = Ax,

where A has eigenvalues λ = ±i and correspondingeigenvectors

[2± i

Plot the phase portrait.

We can start by writing down one of our solutions in the formveλt .

[2 + i

We can use Euler’s formula to rewrite the e it term. That is,

[2 + i

](cos t + i sin t).

Separating real and imaginary terms gives us

[2 + i

](cos t + i sin t)

])(cos t + i sin t)

]cos t −

]sin t

]sin t +

]cos t

[2 + i

](cos t + i sin t).

[2 + i

](cos t + i sin t)

])(cos t + i sin t)

]cos t −

]sin t

]sin t +

]cos t

[2 + i

](cos t + i sin t).

[2 + i

](cos t + i sin t)

])(cos t + i sin t)

]cos t −

]sin t

]sin t +

]cos t

[2 + i

](cos t + i sin t).

[2 + i

](cos t + i sin t)

])(cos t + i sin t)

]cos t −

]sin t

]sin t +

]cos t

[2 + i

](cos t + i sin t).

[2 + i

](cos t + i sin t)

])(cos t + i sin t)

]cos t −

]sin t

]sin t +

]cos t

]cos t −

]sin t

]sin t +

]cos t

)It has been shown that Re(xi) and Im(xi) form a fundamentalset of solutions to x′ = Ax. Therefore, the general solution is

x = c1

]cos t −

]sin t

]sin t +

]cos t

)or more compactly

x = c1

[2 cos t − sin t

[2 sin t + cos t

]cos t −

]sin t

]sin t +

]cos t

x = c1

]cos t −

]sin t

]sin t +

]cos t

)or more compactly

x = c1

[2 cos t − sin t

[2 sin t + cos t

]cos t −

]sin t

]sin t +

]cos t

x = c1

]cos t −

]sin t

]sin t +

]cos t

)or more compactly

x = c1

[2 cos t − sin t

[2 sin t + cos t

x = c1

[2 cos t − sin t

[2 sin t + cos t

]Recall that our eigenvalues were λ = ±i . The imaginary part of the eigenvalue leadsto rotations in our phase portrait due to the sines and cosines in our solution. Inaddition, since the real part of our eigenvalues is zero, the trajectories will not tend tothe origin or infinity as t →∞. Instead, they form closed orbits around the origin.This kind of critical point is known as a center. Since the trajectories don’t tend tothe origin as t →∞, we can’t call it asymptotically stable. Instead, we just call itstable since a particle that starts near the origin will remain close to the origin. Todetermine which direction the spiral is spinning, just evaluate x′ = Ax at a point.

λ1 6= λ2 λ1 6= λ2 λ1 6= λ2

Both positive Both negative Opposite signs

Node Node SaddleUnstable Asymptotically stable Unstable

λ1 = 0 λ1 = 0λ2 > 0 λ2 < 0

Line of Eq. Pts. Line of Eq. Pts.

λ = α± iβ λ = α± iβ λ = α± iβα < 0 α = 0 α > 0

Spiral sink Center Spiral sourceAsymptotically stable Stable Unstable

Example

Consider the system of DE’s

x′ =

[2 −5α −2

(a) Determine the eigenvalues in terms of α.

(b) Find the critical value or values of α where the qualitativenature of the phase portrait for the system changes.

(c) Draw a phase portrait for a value of α slightly below, andanother value slightly above, each critical value.

section 3 - university of michigangmarple/oct2.pdf · 2017-10-02 · 5 3 = 0: (1 )(3 + ) ( 1)5 = 0...

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