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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).

Figure3.7.1Mixing example with two tanks

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.

Figure3.7.3The trace-determinant plane

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.

Figure3.7.5\(D \gt T^2/4\)

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.

Figure3.7.6The trace-determinant plane for real and complex eigenvalues

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).

Figure3.7.7\(D = T^2/4\)

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.


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).

Figure3.7.10A one-parameter family of linear systems

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).

Figure3.7.11A one-parameter family for a harmonic oscillator

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.

Figure3.7.13The trace-determinant plane for varying damping

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.

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.



Consider the one-parameter family of linear systems given by

\begin{equation*} \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} a & \sqrt{2} + a/2 \\ \sqrt{2} - a/2 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}. \end{equation*}
  1. Sketch the path traced out by this family of linear systems in the trace-determinant plane as \(a\) varies.
  2. Discuss any bifurcations that occur along this path and compute the corresponding values of \(a\text{.}\)

Consider the two-parameter family of linear systems

\begin{equation*} \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} a & b \\ b & a \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}. \end{equation*}

Identify all of the regions in the \(ab\)-plane where this system possesses a saddle, a sink, a spiral sink, and so on.