Lecture 4.6: Phase portraits with complex eigenvalues Matthew - - PowerPoint PPT Presentation

lecture 4 6 phase portraits with complex eigenvalues
SMART_READER_LITE
LIVE PREVIEW

Lecture 4.6: Phase portraits with complex eigenvalues Matthew - - PowerPoint PPT Presentation

Lecture 4.6: Phase portraits with complex eigenvalues Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 2080, Differential Equations M. Macauley (Clemson) Lecture 4.6: Phase


slide-1
SLIDE 1

Lecture 4.6: Phase portraits with complex eigenvalues

Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 2080, Differential Equations

  • M. Macauley (Clemson)

Lecture 4.6: Phase portraits, complex eigenvalues Differential Equations 1 / 6

slide-2
SLIDE 2

Complex-valued solutions

Lemma

Suppose x1(t) = u(t) + iw(t) solves x′ = Ax. Then so do u(t) and w(t).

  • M. Macauley (Clemson)

Lecture 4.6: Phase portraits, complex eigenvalues Differential Equations 2 / 6

slide-3
SLIDE 3

Re-writing the general solution

Example 2a

Consider the following homogeneous system x′

1

x′

2

  • =

− 1

2

1 −1 − 1

2

x1 x2

  • .

It is easily verified that the eigenvalues and eigenvectors of A are λ1 = −1 2 + i, v1 = 1 i

  • ;

λ2 = −1 2 − i, v2 = 1 −i

  • .

Thus, the general solution is x(t) = C1e(− 1

2 +i)t

1 i

  • + C2e(− 1

2 −i)t

1 −i

  • .
  • M. Macauley (Clemson)

Lecture 4.6: Phase portraits, complex eigenvalues Differential Equations 3 / 6

slide-4
SLIDE 4

Phase portraits

Example 2a (cont.)

Let’s plot the general solution x(t) = C1e−t/2 cos t − sin t

  • + C2e−t/2

sin t cos t

  • .
  • M. Macauley (Clemson)

Lecture 4.6: Phase portraits, complex eigenvalues Differential Equations 4 / 6

slide-5
SLIDE 5

Purely imaginary eigenvalues

Example 2a

Consider the following initial value problem x′

1

x′

2

  • =

1

2

− 5

4

2 − 1

2

x1 x2

  • ,

x(0) = −1 −2

  • .

It is easily verified that the eigenvalues and eigenvectors of A are λ1 = 3 2i, v1 =

  • 5

2 − 6i

  • ;

λ2 = −3 2i, v2 =

  • 5

2 + 6i

  • .

Thus, the general solution is x(t) = C1e

3 2 it

  • 5

2 − 6i

  • + C2e− 3

2 it

  • 5

2 + 6i

  • .
  • M. Macauley (Clemson)

Lecture 4.6: Phase portraits, complex eigenvalues Differential Equations 5 / 6

slide-6
SLIDE 6

Concluding remarks

Summary

Suppose we wish to solve x′ = Ax, where λ1,2 = a ± bi, b = 0. We have two solutions: x1(t) = e(a+bi)tv1 and x2(t) = e(a−bi)tv2. Take one of these (say x1(t)), and use Euler’s formula to write it as x1(t) = u(t) + iw(t). The general solution is x(t) = C1u(t) + C2w(t). The phase portrait will have ellipses, that are

spiraling inward if a < 0; spiraling outward if a > 0; stable if a = 0.

  • M. Macauley (Clemson)

Lecture 4.6: Phase portraits, complex eigenvalues Differential Equations 6 / 6