sec. 7.5: homogeneous linear systems with constant...

Sec. 7.5: Homogeneous Linear Systems with ConstantCoefficients

MATH 351

California State University, Northridge

April 20, 2014

MATH 351 (Differential Equations) Sec. 7.5 April 20, 2014 1 / 27

Homogeneous linear systems with constant coefficients

x′ = Ax (1)

where A is a constant n × n matrix.

We assume that all the elements of A are real.

If n = 1, then the system reduces to

dx

dt= ax , (2)

its solutions is

x = 0 is the only equilibrium solution if a 6= 0.

If a < 0, other solutions approach x = 0 as t increases, and in this case we saythat x = 0 is ;

If a > 0, other solutions depart from x = 0 as t increases, and in this case we saythat x = 0 is .


For systems of n equations, (n ≥ 2)

x′ = Ax (3)

If n ≥ 2,How to get equilibrium solutions?

Questions: Whether other solutions approach this equilibrium solution or departfrom it as t increases; in other words, is x = 0 asymptotically stable or unstable?Or are there still other possibilities?


2-dim homogeneous linear system with constant coefficients

If n = 2,

x′ = Ax, i.e.,

(x1x2

)′=

(a bc d

)(x1x2

)(4)

phase plane x1x2-plane includes a direction field of tangent vectors to solutions ofthe system of DEs.

phase portrait a plot includes a representative sample of trajectories for thesystem of DEs.


Example 1

Find the general solution of the system

x′ =

(3 00 −2

)x (5)

Solutions:


What we learn from Example 1?

? exponential solutions

To solve the general system ofx′ = Ax, (6)

let us try to seek solutions of the form

x = ξert (7)

where the exponent r and the vector ξ are to be determined.


Example 2

Consider the system

x′ =

(1 14 −2

)x (8)

Plot a direction field and determine the qualitative behavior of solutions. Then findthe general solution and draw a phase portrait showing several trajectories.

Solutions:


Direction Field for the System (8) in Example 2

−4 −3 −2 −1 0 1 2 3 4−4

−3

−2

−1

0

1

2

3

4

x1

x 2

Figure: Direction Field for the system (8) in Example 2


−4 −3 −2 −1 0 1 2 3 4−4

−3

−2

−1

0

1

2

3

4

x1

x 2

Figure: Direction Field for the system (8) in Example 2.


−4 −3 −2 −1 0 1 2 3 4−4

−3

−2

−1

0

1

2

3

4

x1

x 2

Figure: A phase portrait for the system (8) in Example 2


0 0.5 1 1.5 2 2.5 3−25

−20

−15

−10

−5

0

5

10

15

20

25

t

x 1

Figure: Typical solutions of x1 versus t for the system (8) in Example 2


Example 3

Consider the system

x′ =

(−2 11 −2

)x (9)

Draw a direction field for this system and find its general solution. Then plot a phaseportrait showing several typical trajectories in the phase plane.

Solutions:


Direction Field for the System (9) in Example 3

−4 −3 −2 −1 0 1 2 3 4−4

−3

−2

−1

0

1

2

3

4

x1

x 2

Figure: Direction Field for the system (9) in Example 3


−4 −3 −2 −1 0 1 2 3 4−4

−3

−2

−1

0

1

2

3

4

x1

x 2

Figure: Direction Field for the system (9) in Example 3.


−4 −3 −2 −1 0 1 2 3 4−4

−3

−2

−1

0

1

2

3

4

x1

x 2

Figure: A phase portrait for the system (9) in Example 3.


0 0.5 1 1.5 2 2.5 3−25

−20

−15

−10

−5

0

5

10

15

20

25

t

x 1

Figure: Typical solutions of x1 versus t for the system (9) in Example 3.


Example 4

Consider the system

x′ =

(5 −13 1

)x (10)


−4 −3 −2 −1 0 1 2 3 4−4

−3

−2

−1

0

1

2

3

4

x1

x 2

Figure: Direction field for the system (10) in Example 4.


−4 −3 −2 −1 0 1 2 3 4−4

−3

−2

−1

0

1

2

3

4

x1

x 2



Example 5

Consider the system

x′ =

(1 11 1

)x (11)


−4 −3 −2 −1 0 1 2 3 4−4

−3

−2

−1

0

1

2

3

4

x1

x 2

Figure: Direction field for the system (11) in Example 5.


−4 −3 −2 −1 0 1 2 3 4−4

−3

−2

−1

0

1

2

3

4

x1

x 2



For the general system x′ = Ax,

To solve it, we need find the eigenvalues and eigenvectors by solving the nth degreepolynomial equation

det(A− r I) = 0, (12)

If we assume that A is a real-valued matrix, then we have the following possibilities forthe eigenvalues of A:

All eigenvalues are real and different from each other;

Some eigenvalues occur in complex conjugate pairs;

Some eigenvalues, either real or complex, are repeated.

If the n eigenvalues are all real and different,

eigenvalue ri

eigenvector ξ(i) (the n eigenvectors ξ(1), . . . , ξ(n) are linearly independent)

The corresponding solutions of the system are


The general solutions for x′ = Ax


If A is real and symmetric (a special case of Hermitian matrices), then

all the eigenvalues r1, . . . , rn must be real;

even if some of the eigenvalues are repeated, there is always a full set of neigenvectors ξ(1), . . . , ξ(n) that are linearly independent (in fact, orthogonal)


Example 5

Find the general solution of

x′ =

3 2 42 0 24 2 3

x (13)


Summary: x′ = Ax

If A is real-valued

1 All eigenvalues are real and different from each other; (Sec. 7.5)

2 Some eigenvalues occur in complex conjugate pairs; (Sec. 7.6)

3 Some eigenvalues, either real or complex, are repeated. (Sec.7.8)

If A is complex-valued

complex eigenvalues need not occur in conjugate pairs

the eigenvectors are normally complex-valued even though the associatedeigenvalue may be real

the solutions of the system (in general complex-valued) are


sec. 7.5: homogeneous linear systems with constant...

Documents