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Consider the linear system $$ d x / d t=a_{11} x+a_{12} y, \quad d y / d t=a_{21} x+a_{22} y $$ where \(a_{11}, \ldots, a_{22}\) are real constants. Let \(p=a_{11}+a_{22}, q=a_{11} a_{22}-a_{12} a_{21},\) and \(\Delta=\) \(p^{2}-4 q\). Show that the critical point \((0,0)\) is a (a) Node if \(q>0\) and \(\Delta \geq 0\) (b) Saddle point if \(q<0\); (c) Spiral point if \(p \neq 0\) and \(\Delta<0\); (d) Center if \(p=0\) and \(q>0\). Hint: These conclusions can be obtained by studying the eigenvalues \(r_{1}\) and \(r_{2}\). It may also be helpful to establish, and then to use, the relations \(r_{1} r_{2}=q\) and \(r_{1}+r_{2}=p\).

Short Answer

Expert verified
Answer: The behavior of the critical point (0,0) can be classified as a node, saddle point, spiral point, or center, depending on the values of p, q, and Δ. To determine the specific behavior, find the eigenvalues r1 and r2 of the system's matrix and analyze the critical point according to the given conditions.

Step by step solution

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01

Write down the linear system's matrix

First, let's write down the matrix of the linear system: $$ A=\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} $$
02

Solve the characteristic equation

Next, we need to find the eigenvalues \(r_1\) and \(r_2\) of the matrix \(A\). To do this, we solve the characteristic equation: $$ \det(A - rI) = 0 $$ where \(I\) is the identity matrix. In this case, the characteristic equation is $$ \begin{vmatrix} a_{11}-r & a_{12} \\ a_{21} & a_{22}-r \end{vmatrix} = (a_{11}-r)(a_{22}-r) - a_{12}a_{21} = r^2 - pr + q = 0 $$ Using the given relations \(r_1r_2 = q\) and \(r_1+r_2=p\), we will find the behavior of the critical point \((0,0)\).
03

Analyze the critical point using eigenvalues

Now, let's analyze the critical point according to the conditions given in the problem statement: (a) Node if \(q>0\) and \(\Delta \geq 0\) If \(q>0\), then \(r_1r_2 > 0\). This implies that either both eigenvalues are positive or both eigenvalues are negative. Moreover, if \(\Delta \geq 0\), the eigenvalues are real. Therefore, the critical point \((0,0)\) is a node because the solutions either move away from or towards the critical point in every direction. (b) Saddle point if \(q<0\) If \(q<0\), then \(r_1r_2 < 0\). This means that one eigenvalue is positive and one is negative, making the critical point a saddle point because the solutions move away from the critical point in one direction and towards it in another direction. (c) Spiral point if \(p \neq 0\) and \(\Delta<0\) If \(p\neq 0\) and \(\Delta < 0\), the eigenvalues are complex conjugates with non-zero real parts. This makes the critical point a spiral point because the solutions move away or towards the critical point in a spiral pattern. (d) Center if \(p=0\) and \(q>0\) If \(p=0\) and \(q>0\), the eigenvalues are purely imaginary (\(r_1 = ai\) and \(r_2 = -ai\) for some real number \(a\)). This results in the critical point being a center because the solutions move around it along closed curves.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Eigenvalues and Eigenvectors
In the study of differential equations, eigenvalues and eigenvectors play a crucial role in understanding the dynamics of linear systems. An eigenvalue, denoted as \(r\), is a scalar that results from the transformation of a vector by a square matrix, and the corresponding eigenvector is a non-zero vector that changes at most by its scalar factor during this transformation.

For the linear system given in the exercise, the eigenvalues are found by solving the characteristic equation \( r^2 - pr + q = 0 \) for the matrix that represents the system. The roots of this equation, \(r_1\) and \(r_2\), help determine the behavior of the system near the critical point \( (0,0) \). The relationships \(r_1r_2 = q\) and \(r_1+r_2=p\) further assist us in characterizing the stability and nature of the critical point, based on the aforementioned conditions.
Stability of Critical Points
The stability of a critical point refers to the system's behavior in the vicinity of that point. When analyzing a linear system of differential equations, the stability can be determined using the eigenvalues of the system's coefficient matrix.

If both eigenvalues are real and have the same sign, the critical point is a node; it's stable if both eigenvalues are negative (attracting node), and unstable if they are positive (repelling node). In the case where the eigenvalues are real but with opposite signs, the critical point becomes a saddle point, which is inherently unstable as trajectories approach the point along one eigendirection but repel along another. Conversely, if the eigenvalues are purely imaginary, we encounter a center, around which trajectories circulate in a neutral stability, neither approaching nor repelling from the critical point. Lastly, when the eigenvalues are complex with non-zero real parts, a spiral point emerges, indicating either a stable or unstable spiral, depending on the direction of the spiral, which is dictated by the sign of the real part of the eigenvalues.
Phase Plane Analysis
Phase plane analysis is a graphical method to study the behavior of solutions to a system of two first-order linear differential equations. This analysis involves plotting the trajectories of solutions in a plane, which is known as the phase plane, with one variable on the x-axis and the other on the y-axis.

In the context of the given exercise, the dynamics of the system near the critical point \( (0,0) \) can be visualized on the phase plane. The nature of the critical point, influenced by the eigenvalues and eigenvectors, defines the shape and stability of trajectories on this plane. For example, a node will appear as trajectories that stem from or converge to a single point, a saddle point will have trajectories that bend away in one direction while coming close in another, a center will show circular paths around the critical point, and a spiral point will have trajectories that wrap around the point in a spiral manner.

The visual examination of phase plane plots offers an intuitive understanding of system dynamics that can complement the algebraic insights provided by the eigenvalues. Moreover, phase plane analysis is not just restricted to linear systems and can also be applied to study the qualitative behavior of non-linear systems.

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Most popular questions from this chapter

(a) Find the eigenvalues and eigenvectors. (b) Classify the critical point \((0,0)\) as to type and determine whether it is stable, asymptotically stable, or unstable. (c) Sketch several trajectories in the phase plane and also sketch some typical graphs of \(x_{1}\) versus \(t .\) (d) Use a computer to plot accurately the curves requested in part (c). \(\frac{d \mathbf{x}}{d t}=\left(\begin{array}{ll}{3} & {-4} \\ {1} & {-1}\end{array}\right) \mathbf{x}\)

In this problem we show how small changes in the coefficients of a system of linear equations can affect a critical point that is a center. Consider the system $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{0} & {1} \\ {-1} & {0}\end{array}\right) \mathbf{x} $$ Show that the eigenvalues are \pmi so that \((0,0)\) is a center. Now consider the system $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{\epsilon} & {1} \\ {-1} & {\epsilon}\end{array}\right) \mathbf{x} $$ where \(|\epsilon|\) is arbitrarily small. Show that the eigenvalues are \(\epsilon \pm i .\) Thus no matter how small \(|\epsilon| \neq 0\) is, the center becomes a spiral point. If \(\epsilon<0,\) the spiral point is asymptotically stable; if \(\epsilon>0,\) the spiral point is unstable.

an autonomous system is expressed in polar coordinates. Determine all periodic solutions, all limit cycles, and determine their stability characteristics. $$ d r / d t=r^{2}\left(1-r^{2}\right), \quad d \theta / d t=1 $$

show that the given system has no periodic solutions other than constant solutions. $$ d x / d t=x+y+x^{3}-y^{2}, \quad d y / d t=-x+2 y+x^{2} y+y^{3} / 3 $$

Two species of fish that compete with each other for food, but do not prey on each other, are bluegill and redear. Suppose that a pond is stocked with bluegill and redear, and let \(x\) and \(y\) be the populations of bluegill and redear, respectively, at time \(t\). Suppose further that the competition is modeled by the equations $$\frac{d x}{d t}=x\left(\epsilon_{1}-\sigma_{1} x-\alpha_{1} y\right), \frac{d y}{d t}=y\left(\epsilon_{2}-\sigma_{2} y-\alpha_{2} x\right)$$ a. If \(\epsilon_{2} / \alpha_{2}>\epsilon_{1} / \sigma_{1}\) and \(\epsilon_{2} / \sigma_{2}>\epsilon_{1} / \alpha_{1},\) show that the only equilibrium populations in the pond are no fish, no redear, or no bluegill. What will happen for large \(t ?\) b. If \(\epsilon_{1} / \sigma_{1}>\epsilon_{2} / \alpha_{2}\) and \(\epsilon_{1} / \alpha_{1}>\epsilon_{2} / \sigma_{2}\), show that the only equilibrium populations in the pond are no fish, no redear, or no bluegill. What will happen for large \(t ?\)

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