chapter 7 7.5 homogeneous system with constant coefficients

§7.5 Homogeneous System with Constant CoefficientsDirection Fields

Sample Problems

Chapter 7

§7.5 Homogeneous System with Constant

Coefficients

Satya Mandal, KU

Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients


Sample Problems

Homogeneous System with constant coefficients

◮ We consider homogeneous linear systems:

x′ = Ax A is an n × n matrix with constant entries

(1)

◮ As in §7.4, solutions of (1) will be denoted by

x(1)(t) =

x11(t)x21(t)· · ·

xn1(t)

, · · · , x(k)(t) =

x1k(t)x2k(t)· · ·

xnk(t)

.



Sample Problems

Principle of superposition

◮ Recall from §7.4 the Principle of superposition and theconverse: If x

(1), . . . , x(n) are solution of (1), then, anyconstant linear combination

x = c1x(1) + · · ·+ cnx

(n) (2)

is also a solution of the same system (1).

◮ The converse is also true, if Wronskian W 6= 0.



Sample Problems

Solutions of (1) and Eigenvectors

◮ Some solutions of (1) are given by

x = ξertwhere Aξ = rξ, (3)

which means that r is an eigenvalue of A and ξ is aneigenvector, corresponding to r .

◮ Proof. For x = ξert , we have

x′ = (rξ)ert = A (ξert) = Ax



Sample Problems

n eigenvalues and vectors

◮ Suppose A has n eigenvalues r1, r2, . . . , rn. Pick ξi , aneigenvector corresponding to ri . So, Aξi = riξi .

◮ A set of n solutions of (1) is given by

x(1) = ξ1e

r1t , . . . , x(n) = ξnernt (4)

So, the Wronskian

W := W (x(1), . . . x(n)) =∣

∣ x(1) · · · x

(n)∣

∣

= e(r1+···+rn)t

∣

∣ ξ1 · · · ξn

∣

∣



Sample Problems

Real and Distinct Eigenvalue

◮ Now assume r1, r2, . . . , rn are distinct.

◮ Then, ξ1, . . . , ξn are linearly independent. (It needs aproof). Hence the Wronskian W 6= 0.

◮ Now assume r1, r2, . . . , rn are real. By §7.4, x(1), . . . , x(n)

form a fundamental set of solutions of (1). In otherwords, any solution of (1) has the form

x = c1x(1) + · · ·+ cnx

(n)= c1ξ1er1t + · · ·+ cnξne

rnt (5)

where c1, . . . , cn are constants, to be determined by theinitial value conditions.



Sample Problems

Direction Fields

◮ When n = 2, system of homogeneous linear system, withconstant coefficients, looks like

(

x ′

1

x ′

2

)

=

(

a b

c d

)(

x1

x2

)

◮ Then,

dx2

dx1

=x ′

1

x ′

2

=ax1 + bx2

cx1 + dx2

is independent of t

.◮ In x1x2-plane, at each (x1, x2) draw a small line segment

with slope dx2

dx1

. This will be called a direction field.◮ In Matlab, use "pplane8" to draw direction fields.



Sample Problems

Continued

◮ This direction field provides different information thandirection field we saw in chapter 1, which was timedependent. For a direction field analogous to that inchapter 1, we would require 3-D diagrams.

◮ I will deemphasize direction field in this chapter.Motivated students should read everything given in thetextbook, on direction fields and/or discuss with to me.



Sample Problems

Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15

Computing eigenvalues and eigenvectors

◮ Matlab command [V, D]=eig(A) could be used to findeigenvalues and eigenvectors. Advantage with this is thatMatlab can handle complex numbers (see §7.6). However,after experimenting with it, I concluded that it does notwork very well. It uses the floating numbers, which makethings misleading.

◮ Throughout, the following would be my strategy:◮ Compute eigenvalues analytically.◮ If an eigenvalue is real, use use TI-84 (rref) to solve and

compute eigenvectors.◮ If an eigenvalue is complex (in §7.6), compute

eigenvectors analytically.



Sample Problems


Sample I Ex. 7

Find a general solution of

x′ =

(

4 −38 −6

)

x

◮ First, find the eigenvalues:

∣

∣

∣

∣

4 − r −38 −6 − r

∣

∣

∣

∣

= 0 =⇒ r2+2r = 0 =⇒ r = 0,−2



Sample Problems


Eigenvectors fo r = 0

◮ Eigenvetors for r = 0 is given by (use TI-84 "rref"):(

4 −38 −6

)(

ξ1

ξ2

)

=

(

00

)

=⇒

(

1 −.750 0

)(

ξ1

ξ2

)

=

(

00

)

◮ Taking ξ2 = 1, ξ =

(

.751

)

is an eigenvector for r = 0.

◮ So, a solution corresponding to r = 0 is:

x(1) = ξert =

(

.751

)



Sample Problems


Eigenvectors fo r = −2

◮ Eigenvetors for r = −2 is given by (use TI-84 "rref"):

(

4 − r −38 −6 − r

)(

ξ1

ξ2

)

=

(

00

)

=⇒

(

6 −38 −4

)(

ξ1

ξ2

)

=

(

00

)

=⇒

(

1 −.50 0

)(

ξ1

ξ2

)

=

(

00

)


(

.51

)

is an eigenvector for r = −2.



Sample Problems


Continued

◮ So, a solution corresponding to r = −2 is:

x(2) = ξert =

(

.51

)

e−2t

◮ So, the general solution is:

x = c1x(1) + c2x

(2) = c1

(

.751

)

+ c2

(

.51

)

e−2t



Sample Problems


Sample II Ex. 9


x′ =

(

1 i

−i 1

)

x


∣

∣

∣

∣

1 − r i

−i 1 − r

∣

∣

∣

∣

= 0 =⇒ r2 − 2r = 0 =⇒ r = 0, 2



Sample Problems



◮ Eigenvetors for r = 0 is given by :(

1 i

−i 1

)(

ξ1

ξ2

)

=

(

00

)

=⇒

{

ξ1 + iξ2 = 0−iξ1 + ξ2 = 0

=⇒

{

ξ1 + iξ2 = 00 = 0


(

−i

1

)



x(1) = ξert =

(

−i

1

)



Sample Problems


Eigenvectors for r = 2

◮ Eigenvetors for r = 2 is given by (use TI-84 "rref"):

(

1 − r i

−i 1 − r

)(

ξ1

ξ2

)

=

(

00

)

=⇒

(

−1 i

−i −1

)(

ξ1

ξ2

)

=

(

00

)

=⇒

{

−ξ1 + iξ2 = 0−iξ1 − ξ2 = 0

=⇒

{

−ξ1 + iξ2 = 00 = 0


(

i

1

)

is an eigenvector for r = −2.



Sample Problems


Continued

◮ So, a solution corresponding to λ = 2 is:

x(2) = ξert =

(

i

1

)

e2t


x = c1x(1) + c2x

(2) = c1

(

−i

1

)

+ c2

(

i

1

)

e2t



Sample Problems


Sample II Ex. 11


x′ =

1 1 21 2 12 1 1

x


∣

∣

∣

∣

∣

∣

1 − r 1 21 2 − r 12 1 1 − r

∣

∣

∣

∣

∣

∣

= 0



Sample Problems


Continued

(1 − r)

∣

∣

∣

∣

2 − r 11 1 − r

∣

∣

∣

∣

−

∣

∣

∣

∣

1 12 1 − r

∣

∣

∣

∣

+ 2

∣

∣

∣

∣

1 2 − r

2 1

∣

∣

∣

∣

= 0

−r3 + 4r

2 + r − 4 = 0 =⇒ r3 − 4r

2 − r + 4 = 0

r2(r − 1)− 3r(r − 1)− 4(r − 1) = 0

(r − 1)(r 2 − 3r − 4) = 0 =⇒ (r − 1)(r + 1)(r − 4) = 0

r = −1, 1, 4



Sample Problems


An eigenvector and solution for r = −1

◮ Eigenvetors for r = −1 is given by (use TI-84 "rref"):

1 − r 1 21 2 − r 12 1 1 − r

ξ1

ξ2

ξ3

=

000

=⇒

2 1 21 3 12 1 2

ξ1

ξ2

ξ3

=

000

=⇒

2ξ1 + ξ2 + 2ξ3 = 0ξ1 + 3ξ2 + ξ3 = 02ξ1 + ξ2 + 2ξ3 = 0

=⇒ (use rref)



Sample Problems


◮

ξ1 + ξ3 = 0ξ2 = 00 = 0


−101

is an eigenvector for

r = −1.

◮ So, a solution corresponding to r = −1 is:

x(1) = ξert =

−101

e−t



Sample Problems


An eigenvector and solution for r = 1

◮ Similarly, an eigenvector for r = 1 is ξ =

1−21


x(2) = ξert =

1−21

et



Sample Problems


An eigenvector and solution for r = 4

◮ Similarly, an eigenvector for r = 4 is ξ =

111


x(2) = ξert =

111

e4t



Sample Problems


General Solution


x = c1x(1) + c2x

(2) + c3x(3)

= c1

−101

e−t + c2

1−21

et + c3

111

e4t



Sample Problems


Sample IV Ex. 15

Solve the initial value problem:

x′ =

(

5 −13 1

)

x, x(0) =

(

2−1

)

.


∣

∣

∣

∣

5 − r −13 1 − r

∣

∣

∣

∣

= 0 =⇒ r2 − 6r + 8 =⇒ r = 2, 4



Sample Problems



◮ Eigenvetors for r = 2 is given by

(

5 − r −13 1 − r

)(

ξ1

ξ2

)

=

(

00

)

=⇒

(

3 −13 −1

)(

ξ1

ξ2

)

=

(

00

)

=⇒

{

3ξ1 − ξ2 = 03ξ1 − ξ2 = 0

=⇒

{

3ξ1 − ξ2 = 00 = 0


(

13

)




Sample Problems


A solution corresponding to r = 2


x(1) = ξert =

(

13

)

e2t



Sample Problems



◮ Eigenvetors for r = 4 is given by

(

5 − r −13 1 − r

)(

ξ1

ξ2

)

=

(

00

)

=⇒

(

1 −13 −3

)(

ξ1

ξ2

)

=

(

00

)

=⇒

{

ξ1 − ξ2 = 03ξ1 − 3ξ2 = 0

=⇒

{

ξ1 − ξ2 = 00 = 0


(

11

)




Sample Problems


A solution corresponding to r = 4


x(2) = ξert =

(

11

)

e4t



Sample Problems


General Solution


x = c1x(1) + c2x

(2) = c1

(

13

)

e2t + c2

(

11

)

e4t



Sample Problems


The Particular Solution

◮ To find the particular solution, use the initial condition:

x(0) = c1

(

13

)

+ c2

(

11

)

=

(

2−1

)

(

1 13 1

)(

c1

c2

)

=

(

2−1

)

=⇒ (use rref)

(

1 00 1

)(

c1

c2

)

=

(

−1.53.5

)

=⇒ c1 = −1.5, c2 = 3.5



Sample Problems


Continued

◮ So, the particular solution is:

x = c1x(1) + c2x

(2) = c1

(

13

)

e2t + c2

(

11

)

e4t

= −1.5

(

13

)

e2t + 3.5

(

11

)

e4t



Sample Problems


§7.5 Assignments and Homework

◮ Read Example 1, 2, 3, 4 (They are helpful).

◮ Homework: §7.5 (page 405) See the Homework Site!


chapter 7 7.5 homogeneous system with constant coefficients

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