# The Ordinary Differential Equations Project

## 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 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, the expression $$T^2 - 4D$$ is not affected by a change of coordinates by 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. $$D = T^2/4$$
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.
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 3.7.9).
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 3.7.11.
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. A bifurcation occurs at $$b = 2 \sqrt{3}\text{.}$$

#### Activity3.7.1.Harmonic Oscillator with a Varying Spring Constant.

Consider a harmonic oscillator modeled by the second-order equation
\begin{equation} m \frac{d^2 x}{dt^2} + b \frac{dx}{dt} + k x = 0,\tag{3.7.1} \end{equation}
where $$m = 2$$ is the mass, $$b = 2$$ is the damping coefficient, and $$k \gt 0$$ is the spring constant.
##### (a)
Rewrite (3.7.1) as a system of first-order differential equations, $$d\mathbf x/dt = A \mathbf x\text{.}$$
##### (b)
Calculate the trace and determinant of $$A\text{.}$$
##### (c)
Sketch a line in the trace-determinant plane that parameterizes the family of equations $$d\mathbf x/dt = A \mathbf x\text{.}$$
##### (d)
For what values of $$k$$ is the harmonic oscillator underdamped? Overdamped? For what value of $$k$$ do we have a bifurcation?
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 3.7.13).

#### Activity3.7.2.Parameterized Families of Linear Systems.

Consider the parameterized system of linear differential equations $$d\mathbf x/dt = A \mathbf x\text{,}$$ where
\begin{equation*} A = \begin{pmatrix} \alpha \amp \beta \\ 1 \amp \alpha \end{pmatrix}. \end{equation*}
##### (a)
Find the trace, $$T\text{,}$$ and determinant, $$D\text{,}$$ of $$A\text{.}$$
##### (b)
Calculate $$T^2 = 4D\text{.}$$
##### (c)
For what values of $$\alpha$$ and $$\beta$$ is the origin a spiral sink of $$d\mathbf x/dt = A \mathbf x\text{?}$$ A spiral source? A center?
##### (d)
For what values of $$\alpha$$ and $$\beta$$ is the origin a nodal sink of $$d\mathbf x/dt = A \mathbf x\text{?}$$ A nodal source? A saddle?
##### (e)
Identify all of the regions in the $$\alpha\beta$$-plane where the system $$d\mathbf x/dt = A \mathbf x$$ possesses a saddle, a sink, a spiral sink, and so on. Plot your results on the $$\alpha\beta$$-plane.
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.

#### 1.

What is the trace of a matrix?

#### 2.

Explain what information the trace-determinant plane provides about a $$2 \times 2$$ linear system.

### Exercises3.7.5Exercises

#### 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 3.7.5.1–8. Sketch the phase portrait by hand and then use Sage to verify your result.
##### 1.
$$A = \begin{pmatrix} 1 \amp 2 \\ 3 \amp 4 \end{pmatrix}$$
##### 2.
$$A = \begin{pmatrix} 4 \amp 2 \\ 3 \amp 2 \end{pmatrix}$$
##### 3.
$$A = \begin{pmatrix} -3 \amp -8 \\ 4 \amp -6 \end{pmatrix}$$
##### 4.
$$A = \begin{pmatrix} 4 \amp -5 \\ 3 \amp 2 \end{pmatrix}$$
##### 5.
$$A = \begin{pmatrix} -11 \amp 10 \\ 4 \amp -5 \end{pmatrix}$$
##### 6.
$$A = \begin{pmatrix} 5 \amp -3 \\ -8 \amp -6 \end{pmatrix}$$
##### 7.
$$A = \begin{pmatrix} 4 \amp -15 \\ 3 \amp -8 \end{pmatrix}$$
##### 8.
$$A = \begin{pmatrix} 4 \amp 11 \\ -8 \amp -3 \end{pmatrix}$$

#### One-Parameter Families and Bifurcations.

Each of the following matrices in Exercise Group 3.7.5.9–14 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.