ODE-ProjectODE-Project The Trace-Determinant Plane

Section 3.7 The Trace-Determinant Plane

Objectives

To understand that the characteristic polynomial of a $2 \times 2$ matrix can be written as

$λ^{2} - T λ + D,$

where $T = tr (A)$ and $D = det (A) .$ Furthermore, if a $2 \times 2$ matrix $A$ has eigenvalues $λ_{1}$ and $λ_{2},$ then $tr (A)$ is $λ_{1} + λ_{2}$ and $det (A) = λ_{1} λ_{2},$ and the trace and determinant of a $2 \times 2$ matrix are invariant under a change of coordinates.

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To understand that the trace-determinant plane is determined by the graph of the parabola $D = T^{2} / 4$ on the $T D$ -plane and that the trace-determinant plane can be used to determine the phase portrait of a linear system.
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To understand that the trace-determinant plane is useful for studying bifurcations.
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Suppose that we have two tanks, Tank

A

and Tank

B,

that both have a volume of

V

liters and are both filled with a brine solution. Suppose that pure water enters Tank

A

at a rate of

r_{in}

liters per minute, and a salt mixture enters Tank

A

from Tank

B

at a rate of

r_{B}

liters per minute. Brine also enters Tank

B

from Tank

A

at a rate of

r_{A}

liters per minute. Finally, brine is drained from Tank

B

at a rate of

r_{out}

so that the volume in each tank is constant (Figure 3.7.1).

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Figure 3.7.1. Mixing example with two tanks
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x (t)

and

y (t)

are the amounts of salt in Tank

A

and Tank

B,

respectively, then our problem can be modeled with a linear system of two equations,

\begin{aligned} \frac{d x}{d t} & = rate in - rate out = - r_{A} \frac{x}{V} + r_{B} \frac{y}{V} \\ \frac{d y}{d t} & = rate in - rate out = r_{A} \frac{x}{V} - r_{B} \frac{y}{V} - r_{out} \frac{y}{V} . \end{aligned}

Furthermore,

r_{A} = r_{B} + r_{out},

since the volume in Tank

B

is constant. Consequently, our system now becomes

\begin{aligned} \frac{d x}{d t} & = - r_{A} \frac{x}{V} + r_{B} \frac{y}{V} \\ \frac{d y}{d t} & = r_{A} \frac{x}{V} - r_{A} \frac{y}{V} . \end{aligned}

If we have initial conditions

x (0) = x_{0}

and

y (0) = y_{0},

it is not too difficult to deduce that the amount of salt in each tank will approach zero as

t \to \infty,

and we will have a stable equilibrium solution at

(0, 0) .

Determining the nature of the equilibrium solution is a more difficult question. For example, is it ever possible that the equilibrium solution is a spiral sink? One solution is provided by studying the trace-determinant plane.

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Subsection 3.7.1 The Trace-Determinant Plane

The key to solving the system

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

is determining the eigenvalues of

A .

To find these eigenvalues, we need to derive the characteristic polynomial of

A,

det (A - λ I) = det (\begin{matrix} a - λ & b \\ c & d - λ \end{matrix}) = λ^{2} - (a + d) λ + (a d - b c) .

Of course,

D = det (A) = a d - b c

is the determinant of

A .

The quantity

T = a + d

is the sum of the diagonal elements of the matrix

A .

We call this quantity the trace of

A

and write

tr (A) .

Thus, we can rewrite the characteristic polynomial as

det (A - λ I) = λ^{2} - T λ + D .

We can use the trace and determinant to establish the nature of a solution to a linear system.

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Theorem 3.7.2.

If a

2 \times 2

matrix

A

has eigenvalues

λ_{1}

and

λ_{2},

then the trace of

A

λ_{1} + λ_{2}

and

det (A) = λ_{1} λ_{2} .

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Proof.

The proof follows from a direct computation. Indeed, we can rewrite the characteristic polynomial as

det (A - λ I) = λ^{2} - T λ + D .

The eigenvalues of

A

are now given by

λ_{1} = \frac{T + \sqrt{T^{2} - 4 D}}{2} and λ_{2} = \frac{T - \sqrt{T^{2} - 4 D}}{2} .

Hence,

T = λ_{1} + λ_{2}

and

D = λ_{1} λ_{2} .

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Theorem 3.7.2 tells us that we can determine the determinant and trace of a

2 \times 2

matrix from its eigenvalues. Thus, we should be able to determine the phase portrait of a system

x^{'} = A x

by simply examining the trace and determinant of

A .

Since the eigenvalues of

A

are given by

λ = \frac{T \pm \sqrt{T^{2} - 4 D}}{2},

we can immediately see that the expression

T^{2} - 4 D

determines the nature of the eigenvalues of

A .

If $T^{2} - 4 D > 0,$ we have two distinct real eigenvalues.
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If $T^{2} - 4 D < 0,$ we have two complex eigenvalues, and these eigenvalues are complex conjugates.
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If $T^{2} - 4 D = 0,$ we have repeated eigenvalues.
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T^{2} - 4 D = 0

or equivalently if

D = T^{2} / 4,

we have repeated eigenvalues. In fact, we can represent those systems with repeated eigenvalues by graphing the parabola

D = T^{2} / 4

on the

T D

-plane or trace-determinant plane (Figure 3.7.3). Therefore, points on the parabola correspond to systems with repeated eigenvalues, points above the parabola (

D > T^{2} / 4

or equivalently

T^{2} - 4 D < 0

) correspond to systems with complex eigenvalues, and points below the parabola (

D < T^{2} / 4

or equivalently

T^{2} - 4 D > 0

) correspond to systems with real eigenvalues.

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Figure 3.7.3. The trace-determinant plane
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Theorem 3.7.4.

The trace and determinant of a

2 \times 2

matrix are invariant under a change of coordinates. That is,

det (T^{- 1} A T) = det (A)

and

tr (T^{- 1} A T) = tr (A)

for any

2 \times 2

matrix

A

and any invertible

2 \times 2

matrix

T .

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Proof.

It is straightforward to verify that

det (A B) = det (A) det (B)

and

det (T^{- 1}) = 1 / det (T)

for

2 \times 2

matrices

A

and

B .

Therefore,

det (T^{- 1} A T) = det (T^{- 1}) det (A) det (T) = \frac{1}{det (T)} det (A) det (T) = det (A) .

A direct computation shows that

tr (A B) = tr (B A) .

Thus,

tr (T^{- 1} A T) = tr (T^{- 1} T A) = tr (A) .

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Furthermore, the expression

T^{2} - 4 D

is not affected by a change of coordinates by Theorem 3.7.4. That is, we only need to consider systems

x^{'} = A x,

where

A

is one of the following matrices:

(\begin{matrix} α & β \\ - β & α \end{matrix}), (\begin{matrix} λ & 0 \\ 0 & μ \end{matrix}), (\begin{matrix} λ & 0 \\ 0 & λ \end{matrix}), (\begin{matrix} λ & 1 \\ 0 & λ \end{matrix}) .

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The system

x^{'} = (\begin{matrix} α & β \\ - β & α \end{matrix}) x

has eigenvalues

λ = α \pm i β .

The general solution to this system is

x (t) = c_{1} e^{α t} (\begin{matrix} \cos β t \\ - \sin β t \end{matrix}) + c_{2} e^{α t} (\begin{matrix} \sin β t \\ \cos β t \end{matrix}) .

The

e^{α t}

factor tells us that the solutions either spiral into the origin if

α < 0,

spiral out to infinity if

α > 0,

or stay in a closed orbit if

α = 0 .

The equilibrium points are spiral sinks and spiral sources, or centers, respectively.

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The eigenvalues of

A

are given by

λ = \frac{T \pm \sqrt{T^{2} - 4 D}}{2} .

T^{2} - 4 D < 0,

then we have a complex eigenvalues, and the type of equilibrium point depends on the real part of the eigenvalue. The sign of the real part is determined solely by

T .

T > 0

we have a source. If

T < 0,

we have a sink. If

T = 0,

we have a center. See Figure 3.7.5.

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The situation for distinct real eigenvalues is a bit more complicated. Suppose that we have a system

x^{'} = (\begin{matrix} λ & 0 \\ 0 & μ \end{matrix}) x

with distinct eigenvalues

λ

and

μ .

We will have three cases to consider if none of our eigenvalues are zero:

Both eigenvalues are positive (source).

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Both eigenvalues are negative (sink).

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One eigenvalue is negative and the other is positive (saddle).

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Our two eigenvalues are given by

λ = \frac{T \pm \sqrt{T^{2} - 4 D}}{2} .

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T > 0,

then the eigenvalue

\frac{T + \sqrt{T^{2} - 4 D}}{2}

is positive and we need only determine the sign of the second eigenvalue

\frac{T - \sqrt{T^{2} - 4 D}}{2}

D < 0,

we have one positive and one zero eigenvalue. That is, we have a saddle if

T > 0

and

D < 0 .

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D > 0,

then

T^{2} - 4 D < T^{2} .

Since we are considering the case

T > 0,

we have

\sqrt{T^{2} - 4 D} < T

and the value of the second eigenvalue

(T - \sqrt{T^{2} - 4 D}) / 2

is postive. Therefore, any point in the first quadrant below the parabola corresponds to a system with two positive eigenvalues and must correspond to a nodal source.

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One the other hand, suppose that

T < 0 .

Then the eigenvalue

(T - \sqrt{T^{2} - 4 D}) / 2

is always negative, and we need to determine if other eigenvalue is positive or negative. If

D < 0,

then

T^{2} - 4 D > T^{2}

and

\sqrt{T^{2} - 4 D} > T .

Therefore, the other eigenvalue

(T - \sqrt{T^{2} - 4 D}) / 2

is positive, telling us that any point in the fourth quadrant must correspond to a saddle. If

D > 0,

then

\sqrt{T^{2} - 4 D} < T

and the second eigenvalue is negative. In this case, we will have a nodal sink. We summarize our findings in Figure 3.7.6.

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Figure 3.7.6. The trace-determinant plane for real and complex eigenvalues
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For repeated eigenvalues, the analysis depends only on

T .

Since

T^{2} - 4 D = 0,

the only eigenvalue is

T / 2 .

Thus, we have sources if

T > 0

and sinks if

T < 0

(Figure 3.7.7).

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Example 3.7.8.

Let us return to the mixing problem that we proposed at the beginning of this section. The problem could be modeled by the system of equations

\begin{aligned} \frac{d x}{d t} & = - r_{A} \frac{x}{V} + r_{B} \frac{y}{V} \\ \frac{d y}{d t} & = r_{A} \frac{x}{V} - r_{A} \frac{y}{V} \\ x (0) & = x_{0} \\ y (0) & = y_{0} . \end{aligned}

The matrix corresponding to this system is

A = (\begin{matrix} - r_{A} / V & + r_{B} / V \\ r_{A} / V & - r_{A} / V \end{matrix}) .

Computing the trace and determinant of the matrix yields

T = - 2 r_{A} / V

and

D = (r_{A}^{2} - r_{A} r_{B}) / V^{2},

where

r_{A}

and

r_{B}

are both positive. Certainly,

T < 0

and

D = \frac{r_{A}^{2} - r_{A} r_{B}}{V^{2}} = \frac{r_{A} (r_{A} - r_{B})}{V^{2}} = \frac{r_{A} r_{out}}{V^{2}} > 0.

Therefore, any solution must be stable. Finally, since

4 D - T^{2} = 4 \frac{r_{A}^{2} - r_{A} r_{B}}{V^{2}} - {(\frac{- 2 r_{A}}{V})}^{2} = - \frac{4 r_{A} r_{B}}{V^{2}} < 0,

we are below the parabola in the trace-determinant plane and know that our solution must be a nodal sink.

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Subsection 3.7.2 Parameterized Families of Linear Systems

The trace-determinant plane is an example of a parameter plane. We can adjust the entries of a matrix

A

and, thus, change the value of the trace and the determinant.

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Recall that a harmonic oscillator can be modeled by the second-order equation

m \frac{d^{2} x}{d t^{2}} + b \frac{d x}{d t} + k x = 0,

where

m > 0

is the mass,

b \geq 0

is the damping coefficient, and

k > 0

is the spring constant. If we rewrite this equation as a first-order system, we have

x^{'} = (\begin{matrix} 0 & 1 \\ - k / m & - b / m \end{matrix}) x .

Thus, for the harmonic oscillator

T = - b / m

and

D = k / m .

If we use the trace-determinant plane to analyze the harmonic oscillator, we need only concern ourselves with the second quadrant (Figure Figure 3.7.9).

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Figure 3.7.9. A one-parameter family for a harmonic oscillator
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(T, D) = (- b / m, k / m)

lies above the parabola, we have an underdamped oscillator. If

(T, D) = (- b / m, k / m)

lies below the parabola, we have an overdamped oscillator. If

(T, D) = (- b / m, k / m)

lies on the parabola, we have a critically damped oscillator. If

b = 0,

we have an undamped oscillator.

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Example 3.7.10.

Now let us see what happens to our harmonic oscillator when we fix

m = 1

and

k = 3

and let the damping

b

vary between zero and infinity. We can rewrite our system as

\begin{aligned} \frac{d x}{d t} & = y \\ \frac{d y}{d t} & = - 3 x - b y . \end{aligned}

Thus,

T = - b

and

D = 3 .

We can see how the phase portrait varies with the parameter

b

in Figure Figure 3.7.11.

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Figure 3.7.11. The trace-determinant plane for varying damping
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The line

D = 3

in the trace-determinant plane crosses the repeated eigenvalue parabola,

D = T^{2} / 4

b^{2} = 12

or when

b = 2 \sqrt{3} .

b = 0,

we have purely imaginary eigenvalues. This is the undamped harmonic oscillator. If

0 < b < 2 \sqrt{3},

the eigenvalues are complex with a nonzero real part—the underdamped case. If

b = 2 \sqrt{3},

the eigenvalues are negative and repeated—the critically damped case. Finally, if

b > 2 \sqrt{3},

we have the overdamped case. In this case, the eigenvalues are real, distinct, and negative. A bifurcation occurs at

b = 2 \sqrt{3} .

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Activity 3.7.1. Harmonic Oscillator with a Varying Spring Constant.

Consider a harmonic oscillator modeled by the second-order equation

\begin{matrix} (3.7.1) & m \frac{d^{2} x}{d t^{2}} + b \frac{d x}{d t} + k x = 0, \end{matrix}

where

m = 2

is the mass,

b = 2

is the damping coefficient, and

k > 0

is the spring constant.

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(a)

Rewrite (3.7.1) as a system of first-order differential equations,

d x / d t = A x .

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(b)

Calculate the trace and determinant of

A .

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(c)

Sketch a line in the trace-determinant plane that parameterizes the family of equations

d x / d t = A x .

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(d)

For what values of

k

is the harmonic oscillator underdamped? Overdamped? For what value of

k

do we have a bifurcation?

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Example 3.7.12.

Consider the system

(\begin{matrix} x^{'} \\ y^{'} \end{matrix}) = A x = (\begin{matrix} - 2 & a \\ - 2 & 0 \end{matrix}) x .

The trace of

A

is always

T = - 2,

but

D = det (A) = 2 a .

We are on the parabola if

T^{2} - 4 D = 4 - 8 a = 0 or a = \frac{1}{2} .

Thus, a bifurcation occurs at

a = 1 / 2 .

a > 1 / 2,

we have a spiral sink. If

a < 1 / 2,

we have a sink with real eigenvalues. Further more, if

a < 0,

our sink becomes a saddle (Figure 3.7.13).

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Figure 3.7.13. A one-parameter family of linear systems
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Activity 3.7.2. Parameterized Families of Linear Systems.

Consider the parameterized system of linear differential equations

d x / d t = A x,

where

A = (\begin{matrix} α & β \\ 1 & α \end{matrix}) .

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(a)

Find the trace,

T,

and determinant,

D,

A .

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(b)

Calculate

T^{2} = 4 D .

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(c)

For what values of

α

and

β

is the origin a spiral sink of

d x / d t = A x ?

A spiral source? A center?

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(d)

For what values of

α

and

β

is the origin a nodal sink of

d x / d t = A x ?

A nodal source? A saddle?

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(e)

Identify all of the regions in the

α β

-plane where the system

d x / d t = A x

possesses a saddle, a sink, a spiral sink, and so on. Plot your results on the

α β

-plane.

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Example 3.7.14.

Although the trace-determinant plane gives us a great deal of information about our system, we can not determine everything from this parameter plane. For example, the matrices

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

both have the same trace and determinant, but the solutions to

x^{'} = A x

wind around the origin in a clockwise direction while those of

x^{'} = B x

wind around in a counterclockwise direction.

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Subsection 3.7.3 Important Lessons

The characteristic polynomial of a $2 \times 2$ matrix can be written as

$λ^{2} - T λ + D,$

where $T = tr (A)$ and $D = det (A) .$

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If a $2 \times 2$ matrix $A$ has eigenvalues $λ_{1}$ and $λ_{2},$ then $tr (A)$ is $λ_{1} + λ_{2}$ and $det (A) = λ_{1} λ_{2} .$
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The trace and determinant of a $2 \times 2$ matrix are invariant under a change of coordinates.
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The trace-determinant plane is separated by the graph of the parabola $D = T^{2} / 4$ on the $T D$ -plane. Points on the trace-determinant plane correspond to the trace and determinant of a linear system $x^{'} = A x .$ Since the trace and the determinant of a matrix determine the eigenvalues of $A,$ we can use the trace-determinant plane to parameterize the phase portraits of linear systems.
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The trace-determinant plane is useful for studying bifurcations.
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