###### Theorem 3.7.2

If a \(2 \times 2\) matrix \(A\) has eigenvalues \(\lambda_1\) and \(\lambda_2\text{,}\) then the trace of \(A\) is \(\lambda_1 + \lambda_2\) and \(\det(A) = \lambda_1 \lambda_2\text{.}\)

\(\newcommand{\trace}{\operatorname{tr}}
\newcommand{\real}{\operatorname{Re}}
\newcommand{\imaginary}{\operatorname{Im}}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\)

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.

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.

If a \(2 \times 2\) matrix \(A\) has eigenvalues \(\lambda_1\) and \(\lambda_2\text{,}\) then the trace of \(A\) is \(\lambda_1 + \lambda_2\) and \(\det(A) = \lambda_1 \lambda_2\text{.}\)

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.

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 \(\trace(T^{-1} A T) = \trace(A)\) for any \(2 \times 2\) matrix \(A\) and any invertible \(2 \times 2\) matrix \(T\text{.}\)

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.

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).

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.

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.

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.

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.

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.

- 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.

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.

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

Solution

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

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

Solution

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

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

Solution

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

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

Solution

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

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

Solution

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

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

Solution

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

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

Solution

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

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

Solution

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

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 \(\alpha\text{.}\)

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

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

\(A = \begin{pmatrix} \alpha \amp 2 \\ \alpha \amp \alpha \end{pmatrix}\)

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

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

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

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.

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.

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.