ODE-Project Repeated Eigenvalues

Section 3.5 Repeated Eigenvalues

Objectives

To understand and be able to solve systems $x^{'} = A x,$ where $A$ is a $2 \times 2$ matrix with a single eigenvalue $λ .$

Consider the following system

\begin{matrix} (3.5.1) & (\begin{matrix} d x / d t \\ d y / d t \end{matrix}) = (\begin{matrix} 2 & 1 \\ - 1 & 4 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) . \end{matrix}

The characteristic polynomial of the system (3.5.1) is

λ^{2} - 6 λ + 9

and

λ^{2} - 6 λ + 9 = (λ - 3)^{2} .

This polynomial has a single root

λ = 3

with eigenvector

v = (1, 1) .

There is a single straight-line solution for this system (Figure 3.5.1). The strategy that we used to find the general solution to a system with distinct real eigenvalues will clearly have to be modified if we are to find a general solution to a system with a single eigenvalue.

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Figure 3.5.1. A system with one straight-line solution
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Subsection 3.5.1 Repeated Eigenvalues

The remaining case that we must consider is when the characteristic equation of a matrix

A

has repeated roots. The simplest such case is

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(\begin{matrix} d x / d t \\ d y / d t \end{matrix}) = (\begin{matrix} λ & 0 \\ 0 & λ \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = A (\begin{matrix} x \\ y \end{matrix}) .

The eigenvalues of

A

are both

λ .

Since

A v = λ v,

any nonzero vector in

R^{2}

is an eigenvector for

λ .

Thus, solutions to this system are of the form

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x (t) = α e^{λ t} v .

Each solution to our system lies on a straight line through the origin and either tends to the origin if

λ < 0

or away from zero if

λ > 0 .

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A more interesting case occurs if

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

Again, both eigenvalues are

λ;

however, there is only one linearly independent eigenvector, which we can take to be

(1, 0) .

Therefore, we have a single straight-line solution

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x_{1} (t) = α e^{λ t} (\begin{matrix} 1 \\ 0 \end{matrix}) .

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To find other solutions, we will rewrite the system as

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\begin{aligned} x^{'} & = λ x + y \\ y^{'} & = λ y . \end{aligned}

This is a partially coupled system (Subsection 2.4.1). If

y \neq 0,

the solution of the second equation is

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y (t) = β e^{λ t} .

Therefore, the first equation becomes

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x^{'} = λ x + β e^{λ t},

which is a first-order linear differential equation with solution

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x (t) = α e^{λ t} + β t e^{λ t} .

Consequently, a solution to our system is

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α e^{λ t} (\begin{matrix} 1 \\ 0 \end{matrix}) + β e^{λ t} (\begin{matrix} t \\ 1 \end{matrix}) .

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Example 3.5.2.

Consider the linear system

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\begin{aligned} x^{'} & = - x + y \\ y^{'} & = - y \\ x (0) & = 1 \\ y (0) & = 3. \end{aligned}

The matrix that corresponds to this system is

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A = (\begin{matrix} - 1 & 1 \\ 0 & - 1 \end{matrix})

has a single eigenvalue,

λ = - 1 .

An eigenvector for

λ

v = (1, 0) .

The general solution to our system is

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

Applying the initial conditions

x (0) = 1

and

y (0) = 3,

the solution to our initial value problem is

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\begin{aligned} x (t) & = e^{- t} + 3 t e^{- t} \\ y (t) & = 3 e^{- t} . \end{aligned}

Notice that we have only one straight-line solution (Figure 3.5.3).

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Figure 3.5.3. Phase portrait for repeated eigenvalues

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Activity 3.5.1. Systems with Repeated Eigenvalues.

Consider the system

d x / d t = A x,

where

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

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(a)

Find the eigenvalues of

A .

There should be a single real eigenvalue

λ .

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(b)

Find the eigenvectors

v

for the eigenvalues

λ .

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(c)

Find the straight-line solution of

d x / d t = A x .

Plot the solution in the

x y

-plane.

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(d)

Find the general solution of

d x / d t = A x .

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(e)

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 solution?

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Subsection 3.5.2 Solving Systems with Repeated Eigenvalues

If the characteristic equation has only a single repeated root, there is a single eigenvalue. If this is the situation, then we actually have two separate cases to examine, depending on whether or not we can find two linearly independent eigenvectors.

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Example 3.5.4.

Suppose we have the system

x^{'} = A x,

where

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

The single eigenvalue is

λ = 2,

but there are two linearly independent eigenvectors,

v_{1} = (1, 0)

and

v_{2} = (0, 1) .

In this case our solution is

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

This is not too surprising since the system

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\begin{aligned} x^{'} & = 2 x \\ y^{'} & = 2 y \end{aligned}

is uncoupled and each equation can be solved separately.

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Example 3.5.5.

Now let us consider the example

x^{'} = A x,

where

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A = (\begin{matrix} 5 & 1 \\ - 4 & 1 \end{matrix}) .

Since the characteristic polynomial of

A

λ^{2} - 6 λ + 9 = (λ - 3)^{2},

we have only a single eigenvalue

λ = 3

with eigenvector

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

This gives us one solution to our system,

x_{1} (t) = e^{3 t} v_{1};

however, we still need a second solution.

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Since all other eigenvectors of

A

are a multiple of

v,

we cannot find a second linearly independent eigenvector, and we need to obtain the second solution in a different manner. Furthermore, since this system is not partially coupled, we will need a more general strategy.

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First, we must find a vector

v_{2}

such that

(A - λ I) v_{2} = v_{1} .

To do this we can start with any nonzero vector

w

that is not a multiple of

v_{1},

say

w = (1, 0) .

We then compute

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(A - λ I) w = (A - 3 I) w = (\begin{matrix} 2 & 1 \\ - 4 & - 2 \end{matrix}) (\begin{matrix} 1 \\ 0 \end{matrix}) = (\begin{matrix} 2 \\ - 4 \end{matrix}) = 2 v_{1} .

Thus, we can take

v_{2} = (1 / 2) w = (1 / 2, 0),

and our second solution is

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x_{2} = e^{λ t} (v_{2} + t v_{1}) = e^{3 t} (\begin{matrix} 1 / 2 + t \\ - 2 t \end{matrix})

Thus, our general solution is

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

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If the eigenvalue is positive, the origin is a source. If it is negative, we will have a sink. Notice that we have only given a recipe for finding a solution to

x^{'} = A x,

where

A

has a repeated eigenvalue and any two eigenvectors are linearly dependent. We will justify our procedure in the next section (Subsection 3.6.6).

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Activity 3.5.2. Systems with Repeated Eigenvalues—Finding a Second Solution.

Consider the system

d x / d t = A x,

where

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A = (\begin{matrix} 4 & 3 \\ - 3 & - 2 \end{matrix})

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(a)

Find the eigenvalues of

A .

There should be a single real eigenvalue

λ .

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(b)

Find the eigenvectors

v_{1}

for the eigenvalues

λ .

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(c)

Find the straight-line solution of

d x / d t = A x .

Plot the solution in the

x y

-plane.

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(d)

Find one solution,

x_{1},

d x / d t = A x .

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(e)

To find a second solution of

d x / d t = A x,

choose a vector

w

that is not a multiple of

v_{1}

and compute

(A - λ I) w .

This should give you a vector of the form

α v_{1} .

Let

v_{2} = (1 / α) w .

The second solution is

x_{2} = e^{λ t} (v_{2} + t v_{1}) .

What is the general solution?

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(f)

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 solution?

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Subsection 3.5.3 Important Lessons

If

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

then $A$ has one repeated real eigenvalue. The general solution to the system $x^{'} = A x$ is

$x (t) = α e^{λ t} (\begin{matrix} 1 \\ 0 \end{matrix}) + β e^{λ t} (\begin{matrix} t \\ 1 \end{matrix}) .$

If $λ < 0,$ then the solutions tend towards the origin as $t \to \infty .$ For $λ > 0,$ the solutions tend away from the origin.
Suppose that a system $d x / d t = A x$ has a single eigenvalue with an $v_{1}$ and that all other eigenvectors are multiples of $v_{1} .$ Then one solution is $x_{1} = e^{λ t} v_{1} .$ To find a second linearly independent solution of $d x / d t = A x,$ choose a vector $w$ that is not a multiple of $v_{1}$ and compute $(A - λ I) w .$ This should give you a vector of the form $α v_{1} .$ Let $v_{2} = (1 / α) w .$ The second solution is $x_{2} = e^{λ t} (v_{2} + t v_{1}) .$ The general solution of $d x / d t = A x$ will be

$x (t) = c_{1} e^{λ t} v_{1} + c_{2} e^{λ t} (v_{2} + t v_{1}) .$

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Reading Questions 3.5.4 Reading Questions

1.

Given a

2 \times 2

system with repeated eigenvalues, how many straight-line solutions are there?

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2.

Given a

2 \times 2

system with repeated eigenvalues, explain why it is necessary to find a second linearly independent solution.

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Exercises 3.5.5 Exercises

Solving Linear Systems with Repeated Eigenvalues.

Find the general solution of each of the linear systems in Exercise Group 3.5.5.1–8.

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1.

\begin{aligned} x^{'} & = 9 x + 4 y \\ y^{'} & = - 9 x - 3 y \end{aligned}

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2.

\begin{aligned} x^{'} & = 5 x + 4 y \\ y^{'} & = - 9 x - 7 y \end{aligned}

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3.

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

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4.

\begin{aligned} x^{'} & = 2 x + y \\ y^{'} & = - x \end{aligned}

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5.

\begin{aligned} x^{'} & = 8 x + 4 y \\ y^{'} & = - 9 x - 4 y \end{aligned}

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6.

\begin{aligned} x^{'} & = 3 x + 4 y \\ y^{'} & = - 9 x - 9 y \end{aligned}

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7.

\begin{aligned} x^{'} & = 11 x + 4 y \\ y^{'} & = - 9 x - y \end{aligned}

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8.

\begin{aligned} x^{'} & = 13 x / 2 + 4 y \\ y^{'} & = - 9 x - 11 y / 2 \end{aligned}

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Solving Initial Value Problems.

Solve each of the following linear systems for the given initial values in Exercise Group 3.5.5.9–16.

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9.

\begin{aligned} x^{'} & = 9 x + 4 y \\ y^{'} & = - 9 x - 3 y \\ x (0) & = 2 \\ y (0) & = - 3 \end{aligned}

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10.

\begin{aligned} x^{'} & = 5 x + 4 y \\ y^{'} & = - 9 x - 7 y \\ x (0) & = 2 \\ y (0) & = 1 \end{aligned}

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11.

\begin{aligned} x^{'} & = - x + y \\ y^{'} & = - x - 3 y \\ x (0) & = 2 \\ y (0) & = 2 \end{aligned}

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12.

\begin{aligned} x^{'} & = 2 x + y \\ y^{'} & = - x \\ x (0) & = 0 \\ y (0) & = - 5 \end{aligned}

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13.

\begin{aligned} x^{'} & = 8 x + 4 y \\ y^{'} & = - 9 x - 4 y \\ x (0) & = 1 \\ y (0) & = - 1 \end{aligned}

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14.

\begin{aligned} x^{'} & = 3 x + 4 y \\ y^{'} & = - 9 x - 9 y \\ x (0) & = - 1 \\ y (0) & = 0 \end{aligned}

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15.

\begin{aligned} x^{'} & = 11 x + 4 y \\ y^{'} & = - 9 x - y \\ x (0) & = - 3 \\ y (0) & = 2 \end{aligned}

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16.

\begin{aligned} x^{'} & = 13 x / 2 + 4 y \\ y^{'} & = - 9 x - 11 y / 2 \\ x (0) & = - 2 \\ y (0) & = - 2 \end{aligned}

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17.

Consider the linear system

d x / d t = A x,

where

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A = (\begin{matrix} 3 & 1 \\ - 4 & - 1 \end{matrix}) .

Suppose the initial conditions for the solution curve are

x (0) = - 2

and

y (0) = 5 .

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

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x, y, t = var('x y t') #declare the variables
F = [3*x + y, -4*x - 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, -20, 20), (y, -20, 20), aspect_ratio = 1)
# solve the system for the initial condition t = 0, x = -2, y = 5
P1 = desolve_system_rk4(F, [x, y], ics=[0, -2, 5], 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 = -20, xmax = 20, ymin = -20, ymax = 20)
# plot the straight-line solutions
p += line([(-10, 20), (10, -20)], 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 3.5.5.9–16. Plot the straight-line solutions and the solution curve for the given initial condition.

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