Ch 7.5: Homogeneous Linear Systems with Constant Coefficients
Chapter 7 7.5 Homogeneous System with Constant Coefficients
Transcript of 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
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
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)
.
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
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.
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
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
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
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
∣
∣
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
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.
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
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.
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection 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.
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
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.
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
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
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
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
)
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
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
)
◮ Taking ξ2 = 1, ξ =
(
.51
)
is an eigenvector for r = −2.
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
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
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
Sample II Ex. 9
Find a general solution of
x′ =
(
1 i
−i 1
)
x
◮ First, find the eigenvalues:
∣
∣
∣
∣
1 − r i
−i 1 − r
∣
∣
∣
∣
= 0 =⇒ r2 − 2r = 0 =⇒ r = 0, 2
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
Eigenvectors fo r = 0
◮ Eigenvetors for r = 0 is given by :(
1 i
−i 1
)(
ξ1
ξ2
)
=
(
00
)
=⇒
{
ξ1 + iξ2 = 0−iξ1 + ξ2 = 0
=⇒
{
ξ1 + iξ2 = 00 = 0
◮ Taking ξ2 = 1, ξ =
(
−i
1
)
is an eigenvector for r = 0.
◮ So, a solution corresponding to r = 0 is:
x(1) = ξert =
(
−i
1
)
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
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
◮ Taking ξ2 = 1, ξ =
(
i
1
)
is an eigenvector for r = −2.
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
Continued
◮ So, a solution corresponding to λ = 2 is:
x(2) = ξert =
(
i
1
)
e2t
◮ So, the general solution is:
x = c1x(1) + c2x
(2) = c1
(
−i
1
)
+ c2
(
i
1
)
e2t
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
Sample II Ex. 11
Find a general solution of
x′ =
1 1 21 2 12 1 1
x
◮ First, find the eigenvalues:
∣
∣
∣
∣
∣
∣
1 − r 1 21 2 − r 12 1 1 − r
∣
∣
∣
∣
∣
∣
= 0
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
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
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
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)
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
◮
ξ1 + ξ3 = 0ξ2 = 00 = 0
◮ Taking ξ3 = 1, ξ =
−101
is an eigenvector for
r = −1.
◮ So, a solution corresponding to r = −1 is:
x(1) = ξert =
−101
e−t
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
An eigenvector and solution for r = 1
◮ Similarly, an eigenvector for r = 1 is ξ =
1−21
◮ So, a solution corresponding to r = 1 is:
x(2) = ξert =
1−21
et
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
An eigenvector and solution for r = 4
◮ Similarly, an eigenvector for r = 4 is ξ =
111
◮ So, a solution corresponding to r = 4 is:
x(2) = ξert =
111
e4t
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
General Solution
◮ So, the general solution is:
x = c1x(1) + c2x
(2) + c3x(3)
= c1
−101
e−t + c2
1−21
et + c3
111
e4t
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
Sample IV Ex. 15
Solve the initial value problem:
x′ =
(
5 −13 1
)
x, x(0) =
(
2−1
)
.
◮ First, find the eigenvalues:
∣
∣
∣
∣
5 − r −13 1 − r
∣
∣
∣
∣
= 0 =⇒ r2 − 6r + 8 =⇒ r = 2, 4
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
Eigenvectors fo r = 2
◮ 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
◮ Taking ξ1 = 1, ξ =
(
13
)
is an eigenvector for r = 2.
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
A solution corresponding to r = 2
◮ So, a solution corresponding to r = 2 is:
x(1) = ξert =
(
13
)
e2t
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
Eigenvectors fo r = 4
◮ 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
◮ Taking ξ1 = 1, ξ =
(
11
)
is an eigenvector for r = 2.
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
A solution corresponding to r = 4
◮ So, a solution corresponding to r = 4 is:
x(2) = ξert =
(
11
)
e4t
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
General Solution
◮ So, the general solution is:
x = c1x(1) + c2x
(2) = c1
(
13
)
e2t + c2
(
11
)
e4t
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
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
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
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
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients
§7.5 Homogeneous System with Constant CoefficientsDirection Fields
Sample Problems
Strategies to compute eigenvalues and eigenvectorsSample I Ex. 7Sample II Ex. 9Sample III Ex. 11Sample IV Ex. 15
§7.5 Assignments and Homework
◮ Read Example 1, 2, 3, 4 (They are helpful).
◮ Homework: §7.5 (page 405) See the Homework Site!
Satya Mandal, KU Chapter 7 §7.5 Homogeneous System with Constant Coefficients