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## Section3.7The Trace-Determinant Plane

Suppose that we have two tanks, Tank $A$ and Tank $B\text{,}$ 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 $r_{\text{in}}$ 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 $r_{\text{out}}$ so that the volume in each tank is constant (Figure 3.7.1).

If $x(t)$ and $y(t)$ are the amounts of salt in Tank $A$ and Tank $B\text{,}$ respectively, then our problem can be modeled with a linear system of two equations,

\begin{align*} \frac{dx}{dt} \amp = \text{rate in} - \text{rate out} = - r_A \frac{x}{V} + r_B \frac{y}{V}\\ \frac{dy}{dt} \amp = \text{rate in} - \text{rate out} = r_A \frac{x}{V} - r_B \frac{y}{V} - r_{\text{out}} \frac{y}{V}. \end{align*}

Furthermore, $r_A = r_B + r_{\text{out}}\text{,}$ since the volume in Tank $B$ is constant. Consequently, our system now becomes

\begin{align*} \frac{dx}{dt} \amp = - r_A \frac{x}{V} + r_B \frac{y}{V}\\ \frac{dy}{dt} \amp = r_A \frac{x}{V} - r_A \frac{y}{V}. \end{align*}

If we have initial conditions $x(0) = x_0$ and $y(0) = y_0\text{,}$ it is not too difficult to deduce that the amount of salt in each tank will approach zero as $t \to \infty\text{,}$ and we will have a stable equilibrium solution at $(0, 0)\text{.}$ 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.

### Subsection3.7.1The Trace-Determinant Plane

The key to solving the system

\begin{equation*} \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = A \begin{pmatrix} x \\ y \end{pmatrix} \end{equation*}

is determining the eigenvalues of $A\text{.}$ To find these eigenvalues, we need to derive the characteristic polynomial of $A\text{,}$

\begin{equation*} \det(A - \lambda I) = \det \begin{pmatrix} a - \lambda & b \\ c & d - \lambda \end{pmatrix} = \lambda^2 - (a + d) \lambda + (ad - bc). \end{equation*}

Of course, $D = \det(A) = ad -bc$ is the determinant of $A\text{.}$ The quantity $T = a + d$ is the sum of the diagonal elements of the matrix $A\text{.}$ We call this quantity the trace of $A$ and write $\trace(A)\text{.}$ Thus, we can rewrite the characteristic polynomial as

\begin{equation*} \det(A - \lambda I) = \lambda^2 - T \lambda + D. \end{equation*}

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

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

\begin{equation*} \det(A - \lambda I) = \lambda^2 - T \lambda + D. \end{equation*}

The eigenvalues of $A$ are now given by

\begin{equation*} \lambda_1 = \frac{T + \sqrt{T^2 - 4D}}{2} \quad \text{and} \quad \lambda_2 = \frac{T - \sqrt{T^2 - 4D}}{2}. \end{equation*}

Hence, $T = \lambda_1 + \lambda_2$ and $D = \lambda_1 \lambda_2\text{.}$

Theorem 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 ${\mathbf x}' = A {\mathbf x}$ by simply examining the trace and determinant of $A\text{.}$ Since the eigenvalues of $A$ are given by

\begin{equation*} \lambda = \frac{T \pm \sqrt{T^2 - 4D}}{2}, \end{equation*}

we can immediately see that the expression $T^2 - 4D$ determines the nature of the eigenvalues of $A\text{.}$

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

If $T^2 - 4D = 0$ or equivalently if $D = T^2/4\text{,}$ we have repeated eigenvalues. In fact, we can represent those systems with repeated eigenvalues by graphing the parabola $D= T^2/4$ on the $TD$-plane or trace-determinant plane (Figure 3.7.3). Therefore, points on the parabola correspond to systems with repeated eigenvalues, points above the parabola ($D \gt T^2/4$ or equivalently $T^2 - 4D \lt 0$) correspond to systems with complex eigenvalues, and points below the parabola ($D \lt T^2/4$ or equivalently $T^2 - 4D \gt 0$) correspond to systems with real eigenvalues.

It is straightforward to verify that $\det(AB) = \det(A) \det(B)$ and $\det(T^{-1}) = 1/\det(T)$ for $2 \times 2$ matrices $A$ and $B\text{.}$ Therefore,

\begin{equation*} \det(T^{-1} A T) = \det(T^{-1}) \det(A) \det(T) = \frac{1}{\det(T)} \det(A) \det(T) = \det(A). \end{equation*}

A direct computation shows that $\trace(AB) = \trace(BA)\text{.}$ Thus,

\begin{equation*} \trace(T^{-1} A T) = \trace (T^{-1} T A ) = \trace(A). \end{equation*}

Furthermore, each of the expression $T^2 - 4D$ is not affected by a change of coordinates by Theorem Theorem 3.7.4. That is, we only need to consider systems ${\mathbf x}' = A {\mathbf x}\text{,}$ where $A$ is one of the following matrices:

\begin{equation*} \begin{pmatrix} \alpha & \beta \\ -\beta & \alpha \end{pmatrix}, \begin{pmatrix} \lambda & 0 \\ 0 & \mu \end{pmatrix}, \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix}, \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}. \end{equation*}

The system

\begin{equation*} {\mathbf x}' = \begin{pmatrix} \alpha & \beta \\ - \beta & \alpha \end{pmatrix} {\mathbf x} \end{equation*}

has eigenvalues $\lambda = \alpha \pm i \beta\text{.}$ The general solution to this system is

\begin{equation*} {\mathbf x}(t) = c_1 e^{\alpha t} \begin{pmatrix} \cos \beta t \\ - \sin \beta t \end{pmatrix} + c_2 e^{\alpha t} \begin{pmatrix} \sin \beta t \\ \cos \beta t \end{pmatrix}. \end{equation*}

The $e^{\alpha t}$ factor tells us that the solutions either spiral into the origin if $\alpha \lt 0\text{,}$ spiral out to infinity if $\alpha \gt 0\text{,}$ or stay in a closed orbit if $\alpha = 0\text{.}$ The equilibrium points are spiral sinks and spiral sources, or centers, respectively.

The eigenvalues of $A$ are given by

\begin{equation*} \lambda = \frac{T \pm \sqrt{T^2 - 4D}}{2}. \end{equation*}

If $T^2 - 4D \lt 0\text{,}$ 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 $T\text{.}$ If $T \gt 0$ we have a source. If $T \lt 0\text{,}$ we have a sink. If $T = 0\text{,}$ we have a center. See Figure 3.7.5. Figure 3.7.5 $D \gt T^2/4$

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

\begin{equation*} {\mathbf x}' = \begin{pmatrix} \lambda & 0 \\ 0 & \mu \end{pmatrix} {\mathbf x} \end{equation*}

with distinct eigenvalues $\lambda$ and $\mu\text{.}$ We will have three cases to consider if none of our eigenvalues are zero:

• Both eigenvalues are positive (source).
• Both eigenvalues are negative (sink).
• One eigenvalue is negative and the other is positive (saddle).

Our two eigenvalues are given by

\begin{equation*} \lambda = \frac{T \pm \sqrt{T^2 - 4D}}{2}. \end{equation*}

If $T \gt 0\text{,}$ then the eigenvalue

\begin{equation*} \frac{T + \sqrt{T^2 - 4D}}{2} \end{equation*}

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

\begin{equation*} \frac{T - \sqrt{T^2 - 4D}}{2} \end{equation*}

If $D \lt 0\text{,}$ we have one positive and one zero eigenvalue. That is, we have a saddle if $T \gt 0$ and $D \lt 0\text{.}$

If $D \gt 0\text{,}$ then

\begin{equation*} T^2 - 4D \lt T^2. \end{equation*}

Since we are considering the case $T \gt 0\text{,}$ we have

\begin{equation*} \sqrt{T^2 - 4D} \lt T \end{equation*}

and the value of the second eigenvalue $(T - \sqrt{T^2 - 4D}\,)/2$ 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.

One the other hand, suppose that $T \lt 0\text{.}$ Then the eigenvalue $(T - \sqrt{T^2 - 4D}\,)/2$ is always negative, and we need to determine if other eigenvalue is positive or negative. If $D \lt 0\text{,}$ then $T^2 - 4D \gt T^2$ and $\sqrt{T^2 - 4D} \gt T\text{.}$ Therefore, the other eigenvalue $(T - \sqrt{T^2 - 4D}\,)/2$ is positive, telling us that any point in the fourth quadrant must correspond to a saddle. If $D \gt 0\text{,}$ then $\sqrt{T^2 - 4D} \lt T$ and the second eigenvalue is negative. In this case, we will have a nodal sink. We summarize our findings in Figure 3.7.6.

For repeated eigenvalues, the analysis depends only on $T\text{.}$ Since

\begin{equation*} T^2 - 4D = 0, \end{equation*}

the only eigenvalue is $T/2\text{.}$ Thus, we have sources if $T > 0$ and sinks if $T \lt 0$ (Figure 3.7.7). Figure 3.7.7 $D = T^2/4$
###### Example3.7.8

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{align*} \frac{dx}{dt} \amp = - r_A \frac{x}{V} + r_B \frac{y}{V}\\ \frac{dy}{dt} \amp = r_A \frac{x}{V} - r_A \frac{y}{V}\\ x(0) \amp = x_0\\ y(0) \amp = y_0. \end{align*}

The matrix corresponding to this system is

\begin{equation*} A = \begin{pmatrix} -r_A/V \amp + r_B/V \\ r_A / V \amp - r_A / V \end{pmatrix}. \end{equation*}

Computing the trace and determinant of the matrix yields $T = - 2 r_A/V$ and $D = (r_A^2 - r_A r_B)/V^2\text{,}$ where $r_A$ and $r_B$ are both positive. Certainly, $T \lt 0$ and

\begin{equation*} D = \frac{r_A^2 - r_A r_B}{V^2} = \frac{r_A(r_A - r_B)}{V^2} = \frac{r_A r_{\text{out}}}{V^2} \gt 0. \end{equation*}

Therefore, any solution must be stable. Finally, since

\begin{equation*} 4D - T^2 = 4 \frac{r_A^2 - r_A r_B}{V^2} - \left( \frac{-2 r_A}{V} \right)^2 = -\frac{4r_A r_B}{V^2} \lt 0, \end{equation*}

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

### Subsection3.7.2Parameterized 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.

###### Example3.7.9

Consider the system

\begin{equation*} \begin{pmatrix} x' \\ y' \end{pmatrix} = A \mathbf x = \begin{pmatrix} -2 & a \\ -2 & 0 \end{pmatrix} {\mathbf x}. \end{equation*}

The trace of $A$ is always $T = -2\text{,}$ but $D = \det(A) = 2a\text{.}$ We are on the parabola if

\begin{equation*} T^2 - 4D = 4 - 8a = 0 \qquad \text{or}\qquad a = \frac{1}{2}. \end{equation*}

Thus, a bifurcation occurs at $a = 1/2\text{.}$ If $a \gt 1/2\text{,}$ we have a spiral sink. If $a \lt 1/2\text{,}$ we have a sink with real eigenvalues. Further more, if $a \lt 0\text{,}$ our sink becomes a saddle (Figure 3.7.10).

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

\begin{equation*} m \frac{d^2 x}{dt^2} + b \frac{dx}{dt} + k x = 0, \end{equation*}

where $m > 0$ is the mass, $b \geq 0$ is the damping coefficient, and $k \gt 0$ is the spring constant. If we rewrite this equation as a first-order system, we have

\begin{equation*} {\mathbf x}' = \begin{pmatrix} 0 & 1 \\ -k/m & - b/m \end{pmatrix} {\mathbf x}. \end{equation*}

Thus, for the harmonic oscillator $T = -b/m$ and $D= k/m\text{.}$ If we use the trace-determinant plane to analyze the harmonic oscillator, we need only concern ourselves with the second quadrant (Figure Figure 3.7.11).

If $(T, D) = (-b/m, k/m)$ 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 $b = 0\text{,}$ we have an undamped oscillator.

###### Example3.7.12

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{align*} \frac{dx}{dt} & = y\\ \frac{dy}{dt} & = - 3x - by. \end{align*}

Thus, $T = -b$ and $D = 3\text{.}$ We can see how the phase portrait varies with the parameter $b$ in Figure Figure 3.7.13.

The line $D = 3$ in the trace-determinant plane crosses the repeated eigenvalue parabola, $D = T^2/4$ if $b^2 = 12$ or when $b = 2 \sqrt{3}\text{.}$ If $b = 0\text{,}$ we have purely imaginary eigenvalues. This is the undamped harmonic oscillator. If $0 \lt b \lt 2 \sqrt{3}\text{,}$ the eigenvalues are complex with a nonzero real part—the underdamped case. If $b = 2 \sqrt{3}\text{,}$ the eigenvalues are negative and repeated—the critically damped case. Finally, if $b \gt 2 \sqrt{3}\text{,}$ we have the overdamped case. In this case, the eigenvalues are real, distinct, and negative.

###### Example3.7.14

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

\begin{equation*} A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \qquad\text{and}\qquad B = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \end{equation*}

both have the same trace and determinant, but the solutions to ${\mathbf x}' = A {\mathbf x}$ wind around the origin in a clockwise direction while those of ${\mathbf x}' = B{\mathbf x}$ wind around in a counterclockwise direction.

### Subsection3.7.3Important Lessons

• The characteristic polynomial of a $2 \times 2$ matrix can be written as
\begin{equation*} \lambda^2 - T \lambda + D, \end{equation*}
where $T = \trace(A)$ and $D = \det(A)\text{.}$
• If a $2 \times 2$ matrix $A$ has eigenvalues $\lambda_1$ and $\lambda_2\text{,}$ then $\trace(A)$ is $\lambda_1 + \lambda_2$ and $\det(A) = \lambda_1 \lambda_2\text{.}$
• The trace and determinant of a $2 \times 2$ matrix are invariant under a change of coordinates.
• The trace-determinant plane is determined by the graph of the parabola $D= T^2/4$ on the $TD$-plane. Points on the trace-determinant plane correspond to the trace and determinant of a linear system ${\mathbf x}' = A {\mathbf x}\text{.}$ Since the trace and the determinant of a matrix determine the eigenvalues of $A\text{,}$ we can use the trace-determinant plane to parameterize the phase portraits of linear systems.
• The trace-determinant plane is useful for studying bifurcations.

### SubsectionExercises

###### Classifiying Equilibrium Points

Classify the equilibrium points of the system $\mathbf x' = A \mathbf x$ based on the position of $(T, D)$ in the trace-determinant plane in Exercise Group 3.7.1–8. Sketch the phase portrait by hand and then use Sage to verify your result.

###### 1

$A = \begin{pmatrix} 1 \amp 2 \\ 3 \amp 4 \end{pmatrix}$

Solution

$(T,D) = (5,-2)$ is a nodal saddle.

###### 2

$A = \begin{pmatrix} 4 \amp 2 \\ 3 \amp 2 \end{pmatrix}$

Solution

$(T,D) = (6,2)$ is a nodal source.

###### 3

$A = \begin{pmatrix} -3 \amp -8 \\ 4 \amp -6 \end{pmatrix}$

Solution

$(T,D) = (-9,50)$ is a spiral sink.

###### 4

$A = \begin{pmatrix} 4 \amp -5 \\ 3 \amp 2 \end{pmatrix}$

Solution

$(T,D) = (6,23)$ is a spiral source.

###### 5

$A = \begin{pmatrix} -11 \amp 10 \\ 4 \amp -5 \end{pmatrix}$

Solution

$(T,D) = (-16,15)$ is a nodal sink.

###### 6

$A = \begin{pmatrix} 5 \amp -3 \\ -8 \amp -6 \end{pmatrix}$

Solution

$(T,D) = (-1,-54)$ is a nodal saddle.

###### 7

$A = \begin{pmatrix} 4 \amp -15 \\ 3 \amp -8 \end{pmatrix}$

Solution

$(T,D) = (-4,13)$ is a spiral sink.

###### 8

$A = \begin{pmatrix} 4 \amp 11 \\ -8 \amp -3 \end{pmatrix}$

Solution

$(T,D) = (1,76)$ is a spiral source.

###### One-Parameter Families and Bifurcations

Each of the following matrices in Exercise Group 3.7.9–14 describes a family of differential equations $\mathbf x' = A \mathbf x$ that depends on the parameter $\alpha\text{.}$ For each one-parameter family sketch the curve in the trace-determinant plane determined by $\alpha\text{.}$ Identify any values of $\alpha$ where the type of system changes. These values are bifurcation values of $\alpha\text{.}$

###### 9

$A = \begin{pmatrix} \alpha \amp 3 \\ -1 \amp 0 \end{pmatrix}$

###### 10

$A = \begin{pmatrix} \alpha \amp 3 \\ \alpha \amp 0 \end{pmatrix}$

###### 11

$A = \begin{pmatrix} \alpha \amp 2 \\ \alpha \amp \alpha \end{pmatrix}$

###### 12

$A = \begin{pmatrix} 1 \amp 2 \\ \alpha \amp 0 \end{pmatrix}$

###### 13

$A = \begin{pmatrix} \alpha \amp 1 \\ 1 \amp \alpha - 1 \end{pmatrix}$

###### 14

$A = \begin{pmatrix} 0 \amp 1 \\ \alpha \amp \sqrt{1 - \alpha^2} \end{pmatrix}$

###### 15

Consider the two-parameter family of linear systems

\begin{equation*} \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \alpha & \beta \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}. \end{equation*}

Identify all of the regions in the $\alpha\beta$-plane where this system possesses a saddle, a sink, a spiral sink, and so on.

###### 16

Consider the two-parameter family of linear systems

\begin{equation*} \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \alpha & \beta \\ \beta & \alpha \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}. \end{equation*}

Identify all of the regions in the $\alpha\beta$-plane where this system possesses a saddle, a sink, a spiral sink, and so on.

###### 17

Consider the two-parameter family of linear systems

\begin{equation*} \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \alpha & -\beta \\ \beta & \alpha \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}. \end{equation*}

Identify all of the regions in the $\alpha\beta$-plane where this system possesses a saddle, a sink, a spiral sink, and so on.