Two-parameter Families of Linear Systems of Differential Equations
\(\newcommand{\trace}{\operatorname{tr}} \newcommand{\real}{\operatorname{Re}} \newcommand{\imaginary}{\operatorname{Im}} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} \)
Suppose that we have two tanks, Tank \(A\) and Tank \(B\text{,}\) that both have a book of \(Five\) 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 common salt mixture enters Tank \(A\) from Tank \(B\) at a rate of \(r_B\) liters per infinitesimal. 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 abiding (Figure 3.7.one).
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,
\brainstorm{align*} \frac{dx}{dt} \amp = \text{rate in} - \text{rate out} = - r_A \frac{10}{Five} + 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}. \cease{marshal*}
Furthermore, \(r_A = r_B + r_{\text{out}}\text{,}\) since the volume in Tank \(B\) is constant. Consequently, our system at present becomes
\begin{align*} \frac{dx}{dt} \amp = - r_A \frac{x}{V} + r_B \frac{y}{V}\\ \frac{dy}{dt} \amp = r_A \frac{ten}{V} - r_A \frac{y}{V}. \end{marshal*}
If we have initial conditions \(x(0) = x_0\) and \(y(0) = y_0\text{,}\) it is non 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 hard question. For example, is it ever possible that the equilibrium solution is a screw sink? I solution is provided by studying the trace-determinant plane.
Subsection 3.7.ane The Trace-Determinant Plane
¶The key to solving the organization
\brainstorm{equation*} \begin{pmatrix} 10' \\ y' \terminate{pmatrix} = \brainstorm{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \cease{pmatrix} = A \begin{pmatrix} 10 \\ y \finish{pmatrix} \end{equation*}
is determining the eigenvalues of \(A\text{.}\) To detect these eigenvalues, nosotros need to derive the characteristic polynomial of \(A\text{,}\)
\begin{equation*} \det(A - \lambda I) = \det \begin{pmatrix} a - \lambda & b \\ c & d - \lambda \stop{pmatrix} = \lambda^2 - (a + d) \lambda + (ad - bc). \finish{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 telephone call this quantity the trace of \(A\) and write \(\trace(A)\text{.}\) Thus, we tin rewrite the feature polynomial as
\begin{equation*} \det(A - \lambda I) = \lambda^two - T \lambda + D. \finish{equation*}
Nosotros can utilize the trace and determinant to constitute the nature of a solution to a linear system.
Theorem 3.7.2
If a \(2 \times two\) matrix \(A\) has eigenvalues \(\lambda_1\) and \(\lambda_2\text{,}\) so the trace of \(A\) is \(\lambda_1 + \lambda_2\) and \(\det(A) = \lambda_1 \lambda_2\text{.}\)
Proof
The proof follows from a direct computation. Indeed, nosotros tin rewrite the feature polynomial as
\begin{equation*} \det(A - \lambda I) = \lambda^2 - T \lambda + D. \stop{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 iii.7.2 tells u.s. that we can determine the determinant and trace of a \(two \times 2\) matrix from its eigenvalues. Thus, we should be able to determine the phase portrait of a system \({\mathbf ten}' = A {\mathbf x}\) by simply examining the trace and determinant of \(A\text{.}\) Since the eigenvalues of \(A\) are given past
\begin{equation*} \lambda = \frac{T \pm \sqrt{T^2 - 4D}}{2}, \end{equation*}
we can immediately see that the expression \(T^ii - 4D\) 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 ii circuitous eigenvalues, and these eigenvalues are complex conjugates.
- If \(T^two - 4D = 0\text{,}\) we have repeated eigenvalues.
If \(T^2 - 4D = 0\) or equivalently if \(D = T^2/4\text{,}\) we have repeated eigenvalues. In fact, we tin can represent those systems with repeated eigenvalues by graphing the parabola \(D= T^2/4\) on the \(TD\)-airplane or trace-determinant plane (Figure 3.vii.iii). Therefore, points on the parabola correspond to systems with repeated eigenvalues, points in a higher place the parabola (\(D \gt T^2/4\) or equivalently \(T^2 - 4D \lt 0\)) stand for to systems with circuitous eigenvalues, and points beneath the parabola (\(D \lt T^two/four\) or equivalently \(T^ii - 4D \gt 0\)) correspond to systems with existent eigenvalues.
Theorem 3.7.iv
The trace and determinant of a \(2 \times ii\) matrix are invariant under a alter of coordinates. That is, \(\det(T^{-ane} A T) = \det(A)\) and \(\trace(T^{-1} A T) = \trace(A)\) for whatsoever \(2 \times 2\) matrix \(A\) and whatsoever invertible \(2 \times 2\) matrix \(T\text{.}\)
Proof
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). \finish{equation*}
A straight 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 iii.vii.iv. That is, we only need to consider systems \({\mathbf ten}' = A {\mathbf x}\text{,}\) where \(A\) is one of the post-obit matrices:
\begin{equation*} \begin{pmatrix} \alpha & \beta \\ -\beta & \alpha \end{pmatrix}, \begin{pmatrix} \lambda & 0 \\ 0 & \mu \cease{pmatrix}, \brainstorm{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix}, \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}. \end{equation*}
The organization
\begin{equation*} {\mathbf x}' = \begin{pmatrix} \blastoff & \beta \\ - \beta & \alpha \end{pmatrix} {\mathbf 10} \terminate{equation*}
has eigenvalues \(\lambda = \blastoff \pm i \beta\text{.}\) The general solution to this system is
\begin{equation*} {\mathbf 10}(t) = c_1 due east^{\blastoff t} \brainstorm{pmatrix} \cos \beta t \\ - \sin \beta t \terminate{pmatrix} + c_2 e^{\blastoff t} \brainstorm{pmatrix} \sin \beta t \\ \cos \beta t \end{pmatrix}. \end{equation*}
The \(e^{\alpha t}\) factor tells usa that the solutions either screw into the origin if \(\blastoff \lt 0\text{,}\) spiral out to infinity if \(\alpha \gt 0\text{,}\) or stay in a closed orbit if \(\alpha = 0\text{.}\) The equilibrium points are screw sinks and screw sources, or centers, respectively.
The eigenvalues of \(A\) are given past
\begin{equation*} \lambda = \frac{T \pm \sqrt{T^2 - 4D}}{2}. \end{equation*}
If \(T^2 - 4D \lt 0\text{,}\) then we take a complex eigenvalues, and the type of equilibrium betoken depends on the real function of the eigenvalue. The sign of the real part is determined solely by \(T\text{.}\) If \(T \gt 0\) we have a source. If \(T \lt 0\text{,}\) nosotros accept a sink. If \(T = 0\text{,}\) we have a middle. See Figure 3.7.5.
The state of affairs for distinct existent eigenvalues is a flake more complicated. Suppose that we accept a system
\begin{equation*} {\mathbf ten}' = \begin{pmatrix} \lambda & 0 \\ 0 & \mu \end{pmatrix} {\mathbf 10} \end{equation*}
with distinct eigenvalues \(\lambda\) and \(\mu\text{.}\) Nosotros 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 past
\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 nosotros need only determine the sign of the second eigenvalue
\begin{equation*} \frac{T - \sqrt{T^2 - 4D}}{2} \end{equation*}
If \(D \lt 0\text{,}\) nosotros have one positive and i zilch 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^two - 4D \lt T^ii. \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 2d eigenvalue \((T - \sqrt{T^two - 4D}\,)/2\) is postive. Therefore, any signal in the offset quadrant below the parabola corresponds to a system with two positive eigenvalues and must represent to a nodal source.
I the other hand, suppose that \(T \lt 0\text{.}\) Then the eigenvalue \((T - \sqrt{T^ii - 4D}\,)/two\) is e'er negative, and we demand to determine if other eigenvalue is positive or negative. If \(D \lt 0\text{,}\) and so \(T^two - 4D \gt T^two\) and \(\sqrt{T^ii - 4D} \gt T\text{.}\) Therefore, the other eigenvalue \((T - \sqrt{T^two - 4D}\,)/two\) is positive, telling u.s.a. that any point in the 4th quadrant must correspond to a saddle. If \(D \gt 0\text{,}\) so \(\sqrt{T^2 - 4D} \lt T\) and the 2d eigenvalue is negative. In this case, we will take a nodal sink. Nosotros summarize our findings in Figure 3.seven.half-dozen.
For repeated eigenvalues, the analysis depends just on \(T\text{.}\) Since
\begin{equation*} T^two - 4D = 0, \stop{equation*}
the only eigenvalue is \(T/two\text{.}\) Thus, we have sources if \(T > 0\) and sinks if \(T \lt 0\) (Figure three.7.7).
Case 3.7.8
Let us return to the mixing trouble that we proposed at the beginning of this section. The trouble 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 = \brainstorm{pmatrix} -r_A/V \amp + r_B/V \\ r_A / V \amp - r_A / V \end{pmatrix}. \terminate{equation*}
Computing the trace and determinant of the matrix yields \(T = - 2 r_A/V\) and \(D = (r_A^ii - r_A r_B)/V^two\text{,}\) 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}}}{Five^ii} \gt 0. \cease{equation*}
Therefore, any solution must be stable. Finally, since
\brainstorm{equation*} 4D - T^ii = 4 \frac{r_A^2 - r_A r_B}{5^2} - \left( \frac{-ii r_A}{V} \right)^2 = -\frac{4r_A r_B}{V^two} \lt 0, \finish{equation*}
we are below the parabola in the trace-determinant plane and know that our solution must be a nodal sink.
Subsection 3.vii.ii Parameterized Families of Linear Systems
¶The trace-determinant aeroplane is an example of a parameter airplane. Nosotros can adapt the entries of a matrix \(A\) and, thus, change the value of the trace and the determinant.
Example 3.7.ix
Consider the arrangement
\brainstorm{equation*} \brainstorm{pmatrix} x' \\ y' \end{pmatrix} = A \mathbf x = \begin{pmatrix} -2 & a \\ -ii & 0 \stop{pmatrix} {\mathbf x}. \end{equation*}
The trace of \(A\) is always \(T = -ii\text{,}\) but \(D = \det(A) = 2a\text{.}\) Nosotros are on the parabola if
\brainstorm{equation*} T^2 - 4D = iv - 8a = 0 \qquad \text{or}\qquad a = \frac{1}{2}. \end{equation*}
Thus, a bifurcation occurs at \(a = 1/2\text{.}\) If \(a \gt 1/ii\text{,}\) we have a screw sink. If \(a \lt 1/2\text{,}\) we have a sink with real eigenvalues. Farther more, if \(a \lt 0\text{,}\) our sink becomes a saddle (Figure 3.vii.10).
Retrieve that a harmonic oscillator tin can be modeled by the second-society equation
\brainstorm{equation*} m \frac{d^ii x}{dt^ii} + b \frac{dx}{dt} + k x = 0, \cease{equation*}
where \(m > 0\) is the mass, \(b \geq 0\) is the damping coefficient, and \(k \gt 0\) is the leap constant. If nosotros rewrite this equation as a first-guild system, nosotros accept
\begin{equation*} {\mathbf 10}' = \begin{pmatrix} 0 & 1 \\ -k/chiliad & - b/m \end{pmatrix} {\mathbf x}. \cease{equation*}
Thus, for the harmonic oscillator \(T = -b/m\) and \(D= k/one thousand\text{.}\) If we utilise the trace-determinant plane to analyze the harmonic oscillator, nosotros need but concern ourselves with the second quadrant (Effigy Figure 3.7.eleven).
If \((T, D) = (-b/m, k/m)\) lies above the parabola, we have an underdamped oscillator. If \((T, D) = (-b/yard, k/grand)\) lies below the parabola, we have an overdamped oscillator. If \((T, D) = (-b/thousand, k/grand)\) lies on the parabola, nosotros have a critically damped oscillator. If \(b = 0\text{,}\) nosotros have an undamped oscillator.
Example 3.7.12
Now let usa see what happens to our harmonic oscillator when nosotros prepare \(m = i\) and \(1000 = iii\) and let the damping \(b\) vary betwixt zero and infinity. We tin rewrite our system every bit
\begin{align*} \frac{dx}{dt} & = y\\ \frac{dy}{dt} & = - 3x - by. \end{align*}
Thus, \(T = -b\) and \(D = 3\text{.}\) We can encounter how the phase portrait varies with the parameter \(b\) in Effigy Figure 3.seven.13.
The line \(D = 3\) in the trace-determinant plane crosses the repeated eigenvalue parabola, \(D = T^two/iv\) if \(b^2 = 12\) or when \(b = 2 \sqrt{3}\text{.}\) If \(b = 0\text{,}\) we accept purely imaginary eigenvalues. This is the undamped harmonic oscillator. If \(0 \lt b \lt 2 \sqrt{three}\text{,}\) 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, singled-out, and negative.
Instance 3.7.14
Although the trace-determinant aeroplane gives us a great deal of information well-nigh our organization, we can not determine everything from this parameter airplane. For case, the matrices
\begin{equation*} A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \qquad\text{and}\qquad B = \begin{pmatrix} 0 & -1 \\ ane & 0 \end{pmatrix} \end{equation*}
both have the same trace and determinant, but the solutions to \({\mathbf x}' = A {\mathbf 10}\) current of air around the origin in a clockwise direction while those of \({\mathbf x}' = B{\mathbf 10}\) wind around in a counterclockwise direction.
Subsection three.7.three Important Lessons
¶- The characteristic polynomial of a \(ii \times ii\) matrix can be written every bit
\brainstorm{equation*} \lambda^ii - 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 \(\lambda_2\text{,}\) and then \(\trace(A)\) is \(\lambda_1 + \lambda_2\) and \(\det(A) = \lambda_1 \lambda_2\text{.}\)
- The trace and determinant of a \(two \times 2\) matrix are invariant under a modify of coordinates.
- The trace-determinant airplane is determined by the graph of the parabola \(D= T^2/iv\) on the \(TD\)-plane. Points on the trace-determinant plane stand for to the trace and determinant of a linear organization \({\mathbf x}' = A {\mathbf 10}\text{.}\) Since the trace and the determinant of a matrix determine the eigenvalues of \(A\text{,}\) nosotros can utilize the trace-determinant airplane to parameterize the phase portraits of linear systems.
- The trace-determinant plane is useful for studying bifurcations.
Subsection Exercises
¶Classifiying Equilibrium Points
Classify the equilibrium points of the organization \(\mathbf ten' = A \mathbf x\) based on the position of \((T, D)\) in the trace-determinant plane in Exercise Grouping three.7.1–8. Sketch the phase portrait by hand and so use Sage to verify your result.
i
\(A = \begin{pmatrix} i \amp 2 \\ 3 \amp iv \finish{pmatrix}\)
Solution
\((T,D) = (5,-2)\) is a nodal saddle.
2
\(A = \begin{pmatrix} 4 \amp 2 \\ three \amp two \end{pmatrix}\)
Solution
\((T,D) = (six,2)\) is a nodal source.
3
\(A = \begin{pmatrix} -three \amp -8 \\ 4 \amp -6 \end{pmatrix}\)
Solution
\((T,D) = (-9,50)\) is a screw sink.
4
\(A = \begin{pmatrix} 4 \amp -5 \\ 3 \amp 2 \stop{pmatrix}\)
Solution
\((T,D) = (6,23)\) is a spiral source.
5
\(A = \brainstorm{pmatrix} -11 \amp 10 \\ iv \amp -5 \cease{pmatrix}\)
Solution
\((T,D) = (-16,xv)\) is a nodal sink.
vi
\(A = \begin{pmatrix} 5 \amp -3 \\ -8 \amp -6 \cease{pmatrix}\)
Solution
\((T,D) = (-ane,-54)\) is a nodal saddle.
seven
\(A = \begin{pmatrix} 4 \amp -fifteen \\ 3 \amp -8 \terminate{pmatrix}\)
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 screw source.
One-Parameter Families and Bifurcations
Each of the following matrices in Do Group 3.7.9–14 describes a family unit of differential equations \(\mathbf x' = A \mathbf x\) that depends on the parameter \(\blastoff\text{.}\) For each ane-parameter family sketch the curve in the trace-determinant plane determined by \(\alpha\text{.}\) Identify any values of \(\alpha\) where the blazon of system changes. These values are bifurcation values of \(\alpha\text{.}\)
9
\(A = \begin{pmatrix} \alpha \amp 3 \\ -1 \amp 0 \stop{pmatrix}\)
10
\(A = \brainstorm{pmatrix} \alpha \amp 3 \\ \alpha \amp 0 \terminate{pmatrix}\)
11
\(A = \begin{pmatrix} \blastoff \amp 2 \\ \alpha \amp \alpha \stop{pmatrix}\)
12
\(A = \begin{pmatrix} 1 \amp 2 \\ \alpha \amp 0 \end{pmatrix}\)
13
\(A = \brainstorm{pmatrix} \alpha \amp 1 \\ 1 \amp \blastoff - 1 \end{pmatrix}\)
fourteen
\(A = \brainstorm{pmatrix} 0 \amp 1 \\ \alpha \amp \sqrt{1 - \blastoff^2} \finish{pmatrix}\)
15
Consider the two-parameter family of linear systems
\begin{equation*} \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \alpha & \beta \\ 1 & 0 \stop{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}. \end{equation*}
Identify all of the regions in the \(\alpha\beta\)-plane where this organization possesses a saddle, a sink, a spiral sink, and so on.
xvi
Consider the ii-parameter family of linear systems
\begin{equation*} \brainstorm{pmatrix} ten' \\ y' \end{pmatrix} = \brainstorm{pmatrix} \alpha & \beta \\ \beta & \alpha \finish{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}. \end{equation*}
Place all of the regions in the \(\blastoff\beta\)-plane where this arrangement 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' \terminate{pmatrix} = \begin{pmatrix} \alpha & -\beta \\ \beta & \alpha \finish{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}. \end{equation*}
Place all of the regions in the \(\alpha\beta\)-airplane where this system possesses a saddle, a sink, a spiral sink, and and so on.
Subsection iii.7.4 Project—
¶Source: http://faculty.sfasu.edu/judsontw/ode/html-20180819/linear07.html
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