1. Trang chủ >
2. Công Nghệ Thông Tin >
3. Kỹ thuật lập trình >
Tải bản đầy đủ - 0 (trang)
6 Nonlinear Systems, Linearization, and Classification of Equilibrium Points

# 6 Nonlinear Systems, Linearization, and Classification of Equilibrium Points

Tải bản đầy đủ - 0trang

536

Chapter 6 Systems of Ordinary Differential Equations

2. Suppose that both eigenvalues are positive. If 0 < Λ2 < Λ1 , then eΛ1 t and

0, then X becomes

eΛ2 t both become unbounded as t increases. If c1

unbounded in either the direction of v1 or v1 . If c1 0, then X becomes

unbounded in the directions given by v2 or v2 . In this case, 0, 0 is an

unstable node.

3. Suppose that the eigenvalues have opposite sign. Then, if Λ2 < 0 < Λ1 and

0, X becomes unbounded in either the direction of v1 or v1 as it

c1

did in (2). However, if c1 0, then due to the fact that Λ2 < 0, limt X

0 along the line determined by v2 . If the initial point X 0 is not on the

line determined by v2 , then the line given by v1 is an asymptote for the

solution. We say that 0, 0 is a saddle point in this case.

EXAMPLE 6.6.1: Classify the equilibrium point 0, 0 of the systems:

(a)

x

y

5x

4x

3y

3y

; (b)

x

x

y

3x

2y

4y

; (c)

x

x

2y

y

3x

4y

.

SOLUTION: (a) We ﬁnd the eigenvalues and corresponding eigenvec5 3

tors of A

with Eigensystem.

4 3

In:= Clear a, x, y

a

Out=

5 3

4 3

Eigensystem a

1, 3 ,

1, 2 ,

3, 2

Because these eigenvalues have opposite sign, 0, 0 is a saddle point.

1

Eigenvectors corresponding to Λ1

1 and Λ2

3 are v1

and

2

3

, respectively. Hence the solution becomes unbounded in the

v2

2

3

and v2

directions associated with the positive eigenvalue, v2

2

3

1

. Along the line through 0, 0 determined by v1

, the solu2

2

tion approaches 0, 0 . We see this when we graph various solutions and

display the results together with the direction ﬁeld associated with the

system. First, we use DSolve to ﬁnd a general solution of the system.

6.6 Nonlinear Systems, Linearization, and Classiﬁcation of Equilibrium Points

In:= gensol

y t

Out=

x t

y t

DSolve x t

3y t ,

4x t

3y t

1 t

1 3 4t

2

3 t

4t

1

4

t

4t

1

C

1 t

4t

3

2

5x t

, x t ,y t

,t

C 1

C 2 ,

1

C 2

Then, we use Table and Flatten to create a list of ordered pairs

x t , y t , corresponding to the solution for various values of the

arbitrary constants. These functions are then graphed with

ParametricPlot.

In:= toplot

Flatten

Table

e

t

C 1

3e3t C 2 ,

t

2e C 1

2e3t C 2 /.

C 1 > i, C 2 > j ,

i, 0.5, 0.5, 0.25 ,

j, 0.5, 0.5, 0.25 , 1

In:= somegraphs ParametricPlot

Evaluate toplot , t, 3, 3 ,

PlotRange

1, 1 , 1, 1 ,

AspectRatio 1, PlotStyle GrayLevel 0 ,

DisplayFunction Identity

In:= p4

Plot

2 x,

2x

, x, 1, 1 ,

3

PlotStyle

GrayLevel 0 , Dashing 0.02 ,

Thickness 0.01 ,

GrayLevel 0.2 , Dashing 0.02 ,

Thickness 0.01

,

DisplayFunction Identity

We graph the direction ﬁeld associated with the system with PlotVectorField. Last, all graphs are displayed together with Show in

Figure 6-45.

537

538

Chapter 6 Systems of Ordinary Differential Equations

1

0.75

0.5

0.25

-1 -0.75-0.5-0.25

0.25 0.5 0.75

1

-0.25

-0.5

-0.75

-1

Figure 6-45

The origin is a saddle point

In:= << Graphics‘PlotField‘

In:= pvf1

PlotVectorField

5x 3y, 4 x 3y , x, 1, 1 ,

y, 1, 1 ,

DefaultColor GrayLevel 0.5 ,

PlotPoints 20,

ScaleFunction

0.05& ,

DisplayFunction Identity

In:= Show pvf1, somegraphs, p4,

DisplayFunction \$DisplayFunction,

PlotRange

1, 1 , 1, 1 ,

AspectRatio 1,

Axes Automatic, AxesOrigin

0, 0 ,

DisplayFunction \$DisplayFunction

(b) In this case, the eigenvalues Λ1

1 and Λ2

1 2

3 4

Eigensystem a

Out=

2, 1 , 2, 3 , 1, 1

In:= a

2 are both negative.

6.6 Nonlinear Systems, Linearization, and Classiﬁcation of Equilibrium Points

Hence, 0, 0 is a stable node. Corresponding eigenvectors are v1

1

1

2

. Therefore, the solutions approach 0, 0 along the lines

3

through the origin determined by these vectors, y x and y 32 x. We see

this in the graph of the direction ﬁeld and graphs of several solutions

to the system. First, we graph the direction ﬁeld associated with the

system.

and v2

In:= << Graphics‘PlotField‘

In:= Clear x, y, x0, y0, sol

pvf1

Then we use

x x 2y

y

x0

3x

PlotVectorField x 2y, 3x 4y ,

x, 1 , 1 , y, 1 , 1 ,

DefaultColor GrayLevel 0.5 ,

PlotPoints 20,

ScaleFunction

0.5& ,

DisplayFunction Identity

DSolve

to

solve

the

initial-value

problem

4y

x0 , y 0

y0 .

In:= gensol

DSolve

y t

y 0

Out=

x t

y t

x t

x t

2y t ,

3x t

4 y t ,x 0

x0,

y0 , x t , y t , t //Simplify

2t

2t

2

3

3

1

Given an ordered pair x0 , y0 , sol x0 , y0

y0 .

satisﬁes x 0

x0 and y 0

In:= sol x0 , y0

e t 3x0 2y0 2e t

e t 3x0 2y0 3e t

t

t

x0

x0

2

3

1

2

t

t

y0 ,

y0

returns the solution that

x0 y0

x0 y0

,

539

540

Chapter 6 Systems of Ordinary Differential Equations

We then generate several lists of ordered pairs with Table

In:= initconds1

Table

1 , i , i , 1, 1, 2/ 9

initconds2

Table 1, i , i, 1, 1, 2/9

initconds3

Table i, 1 , i, 1, 1, 2/9

initconds4

Table i, 1 , i, 1, 1, 2/9

and use Union to join them together.

In:= initconds initconds1 initconds2

initconds3 initconds4

Map is used to apply sol to the list initconds. The resulting list of

parametric functions is graphed with ParametricPlot.

In:= toplot

Map sol, initconds

In:= somegraphs

In:= p4

ParametricPlot Evaluate toplot ,

t, 3, 3 , PlotRange

1, 1 ,

1, 1 , AspectRatio 1,

PlotStyle GrayLevel 0 ,

DisplayFunction Identity

3x

, x, 1, 1 ,

2

PlotStyle

GrayLevel 0 , Dashing 0.02 ,

Thickness 0.01 ,

GrayLevel 0.2 , Dashing 0.02 ,

Thickness 0.01

,

DisplayFunction Identity

Plot

x,

Finally, all the graphics are displayed together with Show in Figure 6-46.

In:= Show pvf1, somegraphs, p4,

DisplayFunction \$DisplayFunction,

PlotRange

1, 1 , 1, 1 ,

AspectRatio 1,

Axes Automatic, AxesOrigin

0, 0

6.6 Nonlinear Systems, Linearization, and Classiﬁcation of Equilibrium Points

1

0.75

0.5

0.25

-1 -0.75-0.5-0.25

0.25 0.5 0.75

1

-0.25

-0.5

-0.75

-1

Figure 6-46 The origin is a stable node

(c) Because the eigenvalues Λ1

an unstable node.

2 and Λ1

1 are both positive, 0, 0 is

1 2

3 4

Eigensystem a

In:= a

Out=

1, 2 ,

1, 1 ,

2, 3

2

1

and v2

,

3

1

respectively. Hence, the solutions become unbounded along the lines

3

passing through the origin determined by these vectors, y

2 x and

y

x. As before, we see this in the graph of the direction ﬁeld and

various solutions of the system. See Figure 6-47.

Note that the corresponding eigenvectors are v1

In:= pvf1

PlotVectorField

x 2y, 3x 4y ,

x, 1 , 1 , y, 1 , 1 ,

DefaultColor GrayLevel 0.5 ,

PlotPoints 20,

ScaleFunction

0.05& ,

DisplayFunction Identity

541

542

Chapter 6 Systems of Ordinary Differential Equations

1

0.75

0.5

0.25

-1 -0.75-0.5-0.25

0.25 0.5 0.75

1

-0.25

-0.5

-0.75

-1

Figure 6-47 The origin is an unstable node

In:= gensol

DSolve x t

x t

2y t ,

y t

3x t

4y t ,

x 0

x0, y 0

y0 ,

x t ,y t ,t

t

Out=

x t

3 x0 2 t x0 2 y0 2 t y0 ,

t

y t

3 x0 3 t x0 2 y0 3 t y0

In:= sol

x0 , y0

et 3 x0 2y0 2et x0 y0

et 3 x0 2y0 3et x0 y0

,

In:= initconds

Table 0.5t Cos 2Πt , 0.5t Sin 2Πt

1

t, 0, 1,

24

In:= toplot

,

Map sol, initconds

In:= somegraphs

ParametricPlot Evaluate toplot ,

t, 3, 3 , PlotRange

1 , 1 , 1, 1 ,

AspectRatio 1, PlotStyle GrayLevel 0 ,

DisplayFunction Identity

6.6 Nonlinear Systems, Linearization, and Classiﬁcation of Equilibrium Points

In:= p4

3x

, x, 1, 1 ,

2

PlotStyle

GrayLevel 0 , Dashing 0.02 ,

Thickness 0.01 ,

GrayLevel 0.2 , Dashing 0.02 ,

Thickness 0.01

,

DisplayFunction Identity

x,

Plot

In:= Show pvf1, somegraphs, p4,

DisplayFunction \$DisplayFunction,

PlotRange

1, 1 , 1, 1 ,

AspectRatio 1,

Axes Automatic, AxesOrigin

0, 0

6.6.2 Repeated Eigenvalues

We recall from our previous experience with repeated eigenvalues of a 2 2 system that the eigenvalue can have two linearly independent eigenvectors associated with it or only one eigenvector associated with it. Hence, we investigate the

behavior of solutions in this case by considering both of these possibilities.

1. Suppose that the eigenvalue Λ Λ1 Λ2 has two corresponding linearly

independent eigenvectors v1 and v2 . Then, a general solution is

X

c1 v1 eΛt

c2 v2 eΛt .

Hence, if Λ > 0, then X becomes unbounded along the line through the origin determined by the vector c1 v1 c2 v2 where c1 and c2 are arbitrary constants. In this case, we call the equilibrium point a degenerate unstable

node (or an unstable star). On the other hand, if Λ < 0, then X approaches

0, 0 along these lines, and we call 0, 0 a degenerate stable node (or

stable star). Note that the name “star” was selected due to the shape

of the solutions.

Λ2 has only one corresponding eigenvector v1 .

2. Suppose that Λ

Λ1

Hence, a general solution is

X

c1 v1 eΛt

c2 v1t

w2 eΛt

c1 v1

c2 w2 eΛt

c2 v1teΛt ,

where A ΛI w2 v1 . We can more easily investigate the behavior of this

solution if we write this solution as

1

c1 v1 c2 w2 c2 v1 .

X teΛt

t

543

544

Chapter 6 Systems of Ordinary Differential Equations

1

c1 v1 c2 w2 c2 v1

c2 v1 . Hence,

t

the solutions approach 0, 0 along the line determined by v1 , and we call

0, 0 a degenerate stable node. If Λ > 0, the solutions become unbounded

along this line, and we say that 0, 0 is a degenerate unstable node.

If Λ < 0, limt

teΛt

0 and limt

EXAMPLE 6.6.2: Classify the equilibrium point 0, 0 in the systems:

(a)

x

y

x

9y

x

5y

; (b)

x

2x

y

2y

.

SOLUTION: (a) Using Eigensystem,

1 9

1 5

Eigensystem a

Out=

2, 2 ,

3, 1 , 0, 0

In:= a

we see that Λ1 Λ2

2 and that there is only one corresponding eigenvector. Therefore, because Λ

2 < 0, 0, 0 is a degenerate stable node.

Notice that in the graph of several members of the family of solutions of

this system along with the direction ﬁeld shown in Figure 6-48, which

1

0.75

0.5

0.25

-1 -0.75-0.5-0.25

0.25 0.5 0.75

1

-0.25

-0.5

-0.75

-1

Figure 6-48 The origin is a degenerate stable node

6.6 Nonlinear Systems, Linearization, and Classiﬁcation of Equilibrium Points

we generate using the same technique as in part (b) of the previous

example, the solutions approach 0, 0 along the line in the direction of

3

1

,y

v1

3 x.

1

In:= << Graphics‘PlotField‘

In:= Clear x, y

pvf1

PlotVectorField x 9y, x 5y ,

x, 1 , 1 , y, 1 , 1 ,

DefaultColor GrayLevel 0.5 ,

PlotPoints 20,

ScaleFunction

0.5& ,

DisplayFunction Identity

In:= Simplify

x t

9y t ,

DSolve x t

y t

x t

5 y t ,x 0

x0,

y 0

y0 , x t , y t , t

2t

Out=

x t

x0 3 t x0 9 t y0 ,

2t

y t

y0 t x0 3 y0

In:= sol x0 , y0

x0 3tx0 9ty0

,

e2t

tx0

y0

e2t

3ty0

In:= initconds1

Table

1, i , i , 1, 1, 2/ 9

initconds2

Table 1, i , i, 1, 1, 2/9

initconds3

Table i, 1 , i, 1, 1, 2/9

initconds4

Table i, 1 , i, 1, 1, 2/9

In:= initconds initconds1 initconds2

initconds3 initconds4

In:= toplot

Map sol, initconds

In:= somegraphs

ParametricPlot Evaluate toplot ,

t, 3, 3 , PlotRange

1 , 1 , 1, 1 ,

AspectRatio 1, PlotStyle GrayLevel 0 ,

DisplayFunction Identity

545

546

Chapter 6 Systems of Ordinary Differential Equations

In:= p4

x

, x, 1, 1 ,

3

PlotStyle

GrayLevel 0 , Dashing 0.02

Thickness 0.01

,

DisplayFunction Identity

Plot

,

In:= Show pvf1, somegraphs, p4,

DisplayFunction \$DisplayFunction,

PlotRange

1, 1 , 1, 1 ,

AspectRatio 1,

Axes Automatic, AxesOrigin

0, 0

(b) We have Λ1

Λ2

2 and two linearly independent vectors, v1

1

0

0

. (Note: The choice of these two vectors does not change

1

the value of the solution, because of the form of the general solution in

this case.)

and v2

2 0

0 2

Eigensystem a

Out=

2 , 2 , 0, 1 , 1 , 0

In:= a

Because Λ 2 > 0, we classify (0, 0) as a degenerate unstable node (or

star). Some of these solutions along with the direction ﬁeld are graphed

in Figure 6-49 in the same manner as in part (c) of the previous example.

Notice that they become unbounded in the direction of any vector in the

1

0

and v2

.

xy-plane because v1

0

1

In:= Clear x, y

pvf1

PlotVectorField 2x, 2y , x, 1, 1 ,

y, 1, 1 , DefaultColor

GrayLevel 0.5 , PlotPoints 20,

ScaleFunction

0.5& ,

DisplayFunction Identity

In:= Simplify DSolve x t

2x t ,

y t

2y t ,

x 0

x0, y 0

y0 , x t , y t

Out=

x t

2t

x0, y t

2t

y0

,t ### Tài liệu bạn tìm kiếm đã sẵn sàng tải về

6 Nonlinear Systems, Linearization, and Classification of Equilibrium Points

Tải bản đầy đủ ngay(0 tr)

×