ODE-ProjectODE-Project Phase Plane Analysis of Linear Systems

Section 3.3 Phase Plane Analysis of Linear Systems

Objectives

To understand that given a system of linear differential equations

$(\begin{matrix} d x / d t \\ d y / d t \end{matrix}) = (\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = A (\begin{matrix} x \\ y \end{matrix}),$

with distinct real eigenvalues, we can classify the origin as sink, saddle, or source depending on the signs of the eigenvalues.

In Section 3.2, we learned how to solve the system

(\begin{matrix} d x / d t \\ d y / d t \end{matrix}) = (\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = A (\begin{matrix} x \\ y \end{matrix})

provided the system has distinct real eigenvalues. If

A

has distinct real eigenvalues

λ

and

μ

with eigenvectors

u

and

v,

respectively, then the general solution of the system is

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x (t) = c_{1} e^{λ t} u + c_{2} e^{μ t} v .

Furthermore, we can use the general solution of such a system to find the straight-line solutions to the system. If

c_{2} = 0,

then all solutions will lie along the line in the

x y

-plane that contains the vector

u .

Similarly, if

c_{1} = 0,

then all solutions will lie along the line in the

x y

-plane that contains the vector

v .

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Subsection 3.3.1 The Case $λ_{1} < 0 < λ_{2}$

Example 3.3.1.

The system

\begin{aligned} x^{'} & = x + 3 y \\ y^{'} & = x - y \end{aligned}

can be written in matrix form

x^{'} = A x,

where

A = (\begin{matrix} 1 & 3 \\ 1 & - 1 \end{matrix}) .

The eigenvalues of

A

are

λ = - 2

λ = 2

with eigenvectors

u = (1, - 1)

and

v = (3, 1),

respectively. Therefore, the straight-line solutions must be lines containing

u

and

v

(Figure 3.3.2).

Let us consider the special case of the system

x^{'} = A x,

where

λ_{1} < 0 < λ_{2}

and

🔗

A = (\begin{matrix} λ_{1} & 0 \\ 0 & λ_{2} \end{matrix}) .

Since this is a decoupled system,

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\begin{aligned} \frac{d x}{d t} & = λ_{1} x \\ \frac{d y}{d t} & = λ_{2} y, \end{aligned}

we already know how to find the solutions. However, in keeping with the spirit of our investigation, we will find the eigenvalues of

A .

The characteristic equation of

A

🔗

(λ - λ_{1}) (λ - λ_{2}) = 0,

and our eigenvalues are

λ_{1}

and

λ_{2} .

It is easy to see that we can associate eigenvectors

(1, 0)

and

(0, 1)

λ_{1}

and

λ_{2},

respectively. Thus, our general solution is

🔗

x (t) = c_{1} e^{λ_{1} t} (\begin{matrix} 1 \\ 0 \end{matrix}) + c_{2} e^{λ_{2} t} (\begin{matrix} 0 \\ 1 \end{matrix}) .

🔗

Since

λ_{1} < 0,

the straight-line solutions of the form

c_{1} e^{λ_{1} t} (1, 0)

lie on the

x

-axis. These solutions approach zero as

t \to \infty .

On the other hand, the solutions

c_{2} e^{λ_{2} t} (0, 1)

lie on the

y

-axis and approach infinity as

t \to \infty .

The

x

-axis is a stable line of solutions, while the

y

-axis is an unstable line of solutions. All other solutions

🔗

x (t) = c_{1} e^{λ_{1} t} (\begin{matrix} 1 \\ 0 \end{matrix}) + c_{2} e^{λ_{2} t} (\begin{matrix} 0 \\ 1 \end{matrix})

(with

c_{1}, c_{2} \neq 0

) tend to infinity in the direction of the unstable line, since

x (t)

approaches

(0, c_{2} e^{λ_{2} t})

t \to \infty .

The phase portrait for the system

🔗

\begin{aligned} x^{'} & = - x \\ y^{'} & = y \end{aligned}

is given in Figure 3.3.3. The equilibrium point of such systems is called a saddle.

🔗

In general, a straight-line solution is called a stable line of solutions if all solutions approach

(0, 0) .

A straight-line solution is called an unstable line if all of the non-zero solutions approach infinity.

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Example 3.3.4.

For the system in Example 3.3.1, the unstable line of solutions is

x_{1} (t) = c_{1} e^{2 t} (\begin{matrix} 3 \\ 1 \end{matrix}) .

Each solution tends away from the origin as

t \to \infty .

The stable line of solutions is given by

x_{2} (t) = c_{2} e^{- 2 t} (\begin{matrix} 1 \\ - 1 \end{matrix}),

and each solution on this line approaches the origin as

t \to \infty .

By the Principle of Superposition, the general solution to the system is

x (t) = c_{1} e^{2 t} (\begin{matrix} 3 \\ 1 \end{matrix}) + c_{2} e^{- 2 t} (\begin{matrix} 1 \\ - 1 \end{matrix}) .

c_{1} \neq 0,

we have

x (t) \to x_{1} (t)

t \to \infty .

c_{2} \neq 0,

we have

x (t) \to x_{2} (t)

t \to - \infty .

Thus, we have the phase portrait in Figure 3.3.5.

For the general case, where

A

has eigenvalues

λ_{1} < 0 < λ_{2},

we always have a stable line of solutions and an unstable line of solutions. All other solutions approach the unstable line as

t \to \infty

and the stable line as

t \to - \infty .

🔗

Activity 3.3.1. Planar Systems with Eigenvalues of Different Signs.

Consider the system

d x / d t = A x,

where

🔗

A = (\begin{matrix} 8 & - 3 \\ 18 & - 7 \end{matrix})

🔗

(a)

Find the eigenvalues of

A .

You should find distinct real eigenvalues

λ

and

μ .

🔗

(b)

Find eigenvectors

v_{1}

and

v_{2}

for the eigenvalues

λ

and

μ,

respectively.

🔗

(c)

Find the straight-line solutions of

d x / d t = A x .

Plot the solutions in the

x y

-plane.

🔗

(d)

Sketch several solution curves for the system

d x / d t = A x .

What do you notice about the solution curves, especially with respect to the straight-line solutions?

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Subsection 3.3.2 The Case $λ_{1} < λ_{2} < 0$

Suppose

λ_{1} < λ_{2} < 0

and consider the diagonal system

🔗

(\begin{matrix} x^{'} (t) \\ y^{'} (t) \end{matrix}) = (\begin{matrix} λ_{1} & 0 \\ 0 & λ_{2} \end{matrix}) (\begin{matrix} x (t) \\ y (t) \end{matrix}) .

The general solution of this system is

🔗

x (t) = c_{1} e^{λ_{1} t} (\begin{matrix} 1 \\ 0 \end{matrix}) + c_{2} e^{λ_{2} t} (\begin{matrix} 0 \\ 1 \end{matrix}),

but unlike the case of the saddle, all solutions tend towards the origin as

t \to \infty .

To see how the solutions approach the origin, we will compute

d y / d x

for

c_{2} \neq 0 .

🔗

\begin{aligned} x (t) & = c_{1} e^{λ_{1} t} \\ y (t) & = c_{2} e^{λ_{2} t}, \end{aligned}

then

🔗

\frac{d y}{d x} = \frac{y^{'} (t)}{x^{'} (t)} = \frac{λ_{2} c_{2} e^{λ_{2} t}}{λ_{1} c_{1} e^{λ_{1} t}} = \frac{λ_{2} c_{2}}{λ_{1} c_{1}} e^{(λ_{2} - λ_{1}) t} .

Since

λ_{2} - λ_{1} > 0,

the derivative,

d y / d x,

must approach

\pm \infty,

provided

c_{2} \neq 0 .

Therefore, the solutions tend towards the origin tangentially to the

y

-axis (Figure 3.3.6). We say that the equilibrium point for this system is a sink.

🔗

Since

λ_{1} < λ_{2} < 0,

we say that

λ_{1}

is the dominant eigenvalue. The

x

-coordinates of the solutions approach the origin much faster than the

y

-coordinates.

🔗

To see what happens in the general case, suppose that

λ_{1} < λ_{2} < 0,

the eigenvectors associated with

λ_{1}

and

λ_{2}

are

(u_{1}, u_{2})

and

(v_{1}, v_{2}),

respectively. The general solution of our system is

🔗

x (t) = c_{1} e^{λ_{1} t} (\begin{matrix} u_{1} \\ u_{2} \end{matrix}) + c_{2} e^{λ_{2} t} (\begin{matrix} v_{1} \\ v_{2} \end{matrix}) .

The slope of a solution curve at

(x, y)

is given by

🔗

\begin{aligned} \frac{d y}{d x} & = \frac{λ_{1} c_{1} e^{λ_{1} t} u_{2} + λ_{2} c_{2} e^{λ_{2} t} v_{2}}{λ_{1} c_{1} e^{λ_{1} t} u_{1} + λ_{2} c_{2} e^{λ_{2} t} v_{1}} \\ = (\frac{λ_{1} c_{1} e^{λ_{1} t} u_{2} + λ_{2} c_{2} e^{λ_{2} t} v_{2}}{λ_{1} c_{1} e^{λ_{1} t} u_{1} + λ_{2} c_{2} e^{λ_{2} t} v_{1}}) \frac{e^{- λ_{2} t}}{e^{- λ_{2} t}} \\ = \frac{λ_{1} c_{1} e^{(λ_{1} - λ_{2}) t} u_{2} + λ_{2} c_{2} v_{2}}{λ_{1} c_{1} e^{(λ_{1} - λ_{2}) t} u_{1} + λ_{2} c_{2} v_{1}} . \end{aligned}

This last expression tends toward the slope

v_{2} / v_{1}

of the eigenvector of

λ_{2}

(unless

c_{2} = 0

). If

c_{2} = 0,

then we have the straight-line solution corresponding to the eigenvalue

λ_{1} .

Hence, all the solutions for this case (except those on the straight-line belonging to the dominant eigenvalue) tend toward the origin tangentially to the straight-line solution corresponding to the weaker eigenvalue,

λ_{2} .

🔗

Example 3.3.7.

Consider the system

(\begin{matrix} x^{'} (t) \\ y^{'} (t) \end{matrix}) = (\begin{matrix} - 5 & - 2 \\ - 1 & - 4 \end{matrix}) (\begin{matrix} x (t) \\ y (t) \end{matrix}) .

The eigenvalues of this system are

λ_{1} = - 6

and

λ_{2} = - 3

with eigenvectors

v_{1} = (2, 1)

and

v_{2} = (1, - 1),

respectively. Since the dominant eigenvalue is

λ_{1} = - 6,

solutions tend towards the straight-line solution containing the vector

v_{1} = (2, 1)

more quickly (Figure 3.3.8).

Activity 3.3.2. Planar Systems with Two Negative Eigenvalues.

Consider the system

d x / d t = A x,

where

🔗

A = (\begin{matrix} 6 & 14 \\ - 4 & - 9 \end{matrix})

🔗

(a)

Find the eigenvalues of

A .

You should find distinct real eigenvalues

λ

and

μ .

🔗

(b)

Find eigenvectors

v_{1}

and

v_{2}

for the eigenvalues

λ

and

μ,

respectively.

🔗

(c)

Find the straight-line solutions of

d x / d t = A x .

Plot the solutions in the

x y

-plane.

🔗

(d)

Sketch several solution curves for the system

d x / d t = A x .

What do you notice about the solution curves, especially with respect to the straight-line solutions?

🔗

(e)

Which of the two eigenvalues is the dominant eigenvalue? Why?

🔗

Subsection 3.3.3 The Case $λ_{1} > λ_{2} > 0$

λ_{1} > λ_{2} > 0,

we can regard our direction field as the negative of the direction field of the previous case. The general solution and the direction field are the same, but the arrows are reversed (Figure 3.3.9). In this case, we say that the equilibrium point is a source.

🔗

Example 3.3.10.

Consider the system

(\begin{matrix} x^{'} (t) \\ y^{'} (t) \end{matrix}) = (\begin{matrix} 4 & - 3 \\ - 1 & 2 \end{matrix}) (\begin{matrix} x (t) \\ y (t) \end{matrix}) .

The eigenvalues of this system are

λ_{1} = 5

and

λ_{2} = 1

with eigenvectors

v_{1} = (3, - 1)

and

v_{2} = (1, 1),

respectively. Since the dominant eigenvalue is

λ_{1} = 5,

solutions are closer to the straight-line solution containing the vector

v_{2} = (3, - 1)

more as

t \to \infty

(Figure 3.3.11).

🔗

Subsection 3.3.4 Important Lessons

Given a system of linear differential equations

$(\begin{matrix} d x / d t \\ d y / d t \end{matrix}) = (\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = A (\begin{matrix} x \\ y \end{matrix}),$

we can use the eigenvalues of $A$ to find and classify the solutions of the system.
If

$A = (\begin{matrix} λ_{1} & 0 \\ 0 & λ_{2} \end{matrix}),$

then $A$ has two distinct real eigenvalues. The general solution to the system $x^{'} = A x$ is

$x (t) = α e^{λ_{1} t} (\begin{matrix} 1 \\ 0 \end{matrix}) + β e^{λ_{2} t} (\begin{matrix} 0 \\ 1 \end{matrix}) .$
- For the case $λ_{1} < 0 < λ_{2},$ the equilibrium point of the system $x^{'} = A x$ is a saddle.
- For the case $λ_{1} < λ_{2} < 0,$ the equilibrium point of the system $x^{'} = A x$ is a sink.
- For the case $0 < λ_{1} < λ_{2},$ the equilibrium point of the system $x^{'} = A x$ is a source.

🔗

Reading Questions 3.3.5 Reading Questions

1.

What is a stable line of solutions?

🔗

2.

For a

2 \times 2

linear system with distinct real eigenvalues, what are the three different possibilities for the phase plane of the system?

🔗

Exercises 3.3.6 Exercises

Phase Plane Analysis of Linear Systems with Distinct Real Eigenvalues.

For each of the linear systems

d x / d t = A x

in Exercise Group 3.3.6.1–8

🔗

Find the eigenvalues of $A .$
What is the dominant eigenvalue?
Find the eigenvectors for each eigenvalue of $A .$
What are the straight-line solutions of $d x / d t = A x ?$
Describe the nature of the equilibrium solution at $0 .$
Sketch the phase plane and several solution curves.

🔗

1.

A = (\begin{matrix} - 1 & 2 \\ - 6 & 6 \end{matrix})

🔗

2.

A = (\begin{matrix} - 12 & 30 \\ - 5 & 13 \end{matrix})

🔗

3.

A = (\begin{matrix} - 9 & - 2 \\ 10 & 0 \end{matrix})

🔗

4.

A = (\begin{matrix} 11 & 8 \\ - 12 & - 9 \end{matrix})

🔗

5.

A = (\begin{matrix} 7 & 12 \\ - 4 & - 7 \end{matrix})

🔗

6.

A = (\begin{matrix} 10 & 12 \\ - 4 & - 4 \end{matrix})

🔗

7.

A = (\begin{matrix} - 2 & - 6 \\ 2 & 5 \end{matrix})

🔗

8.

A = (\begin{matrix} - 18 & - 30 \\ 10 & 17 \end{matrix})

🔗

9.

Solve each linear systems

d x / d t = A x

in Exercise Group 3.3.6.1–8 for the initial condition

x (0) = (2, 2) .

🔗

10.

Consider the linear system

d x / d t = A x,

where

🔗

A = (\begin{matrix} 3 & 2 \\ 3 & - 2 \end{matrix}) .

Suppose the initial conditions for the solution curve are

x (0) = 1

and

y (0) = 1 .

We can use the following Sage code to plot the phase portrait of this system, including the straight-line solutions and a solution curve.

🔗

xxxxxxxxxx
 
x, y, t = var('x y t') #declare the variables
F = [3*x + 2*y, 3*x - 2*y] #declare the system
# normalize the vector fields so that all of the arrows are the same length
n = sqrt(F[0]^2 + F[1]^2)
# plot the vector field
p = plot_vector_field((F[0]/n, F[1]/n), (x, -4, 4), (y, -4, 4), aspect_ratio = 1)
# solve the system for the initial condition t = 0, x = 1, y = 1
P1 = desolve_system_rk4(F, [x, y], ics=[0, 1, 1], ivar = t, end_points = 5, step = 0.01)
# grab the x and y values
S1 = [ [j, k] for i, j, k in P1]
# plot the solution
# Setting xmin, xmax, ymin, ymax will clip the window
# Try plotting without doing this to see what happens
p += line(S1, thickness = 2, axes_labels=['$x(t)$','$y(t)$'], xmin = -4, xmax = 4, ymin = -4, ymax = 4) 
# plot the straight-line solutions
p += line([(-4, -2), (4, 2)], thickness = 2, color = "red") 
p += line([(-4/3, 4), (4/3, -4)], thickness = 2, color = "red")
p

Use Sage to graph the direction field for the system linear systems

d x / d t = A x

in Exercise Group 3.3.6.1–8. Plot a solution curve for the initial condition

x (0) = (2, 2) .

Be sure to show the corresponding straight-line solutions on your graph.

🔗

xxxxxxxxxx
 
​

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You have attempted 1 of 3 activities on this page.

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Section 3.3 Phase Plane Analysis of Linear Systems

Objectives

Subsection 3.3.1 The Case λ1<0<λ2

Example 3.3.1.

Example 3.3.4.

Activity 3.3.1. Planar Systems with Eigenvalues of Different Signs.

(a)

(b)

(c)

(d)

Subsection 3.3.2 The Case λ1<λ2<0

Example 3.3.7.

Activity 3.3.2. Planar Systems with Two Negative Eigenvalues.

(a)

(b)

(c)

(d)

(e)

Subsection 3.3.3 The Case λ1>λ2>0

Example 3.3.10.

Subsection 3.3.4 Important Lessons

Reading Questions 3.3.5 Reading Questions

1.

2.

Exercises 3.3.6 Exercises

Phase Plane Analysis of Linear Systems with Distinct Real Eigenvalues.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Subsection 3.3.1 The Case $λ_{1} < 0 < λ_{2}$

Subsection 3.3.2 The Case $λ_{1} < λ_{2} < 0$

Subsection 3.3.3 The Case $λ_{1} > λ_{2} > 0$