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Chapter 9: Problem 20

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 螖. 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

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.

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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

This problem extends Problem 22 to a damped pendulum . The equations of motion are $$ d x / d t=y, \quad d y / d t=-4 \sin x-\gamma y . $$ where \(\gamma\) is the damping coefficient, with the initial conditions \(x(0)=0, y(0)=v\) (a) For \(\gamma=1 / 4\) plot \(x\) versus \(t\) for \(v=2\) and for \(v=5 .\) Explain these plots in terms of the motions of the pendulum that they represent. Also explain how they relate to the corresponding graphs in Problem 22 (a). (b) Estimate the critical value \(v_{c}\) of the initial velocity where the transition from one type of motion to the other occurs. (c) Repeat part (b) for other values of \(\gamma\) and determine how \(v_{c}\) depends on \(\gamma\).

(a) Find all the critical points (equilibrium solutions). (b) Use a computer to draw a direction field and portrait for the system. (c) From the plot(s) in part (b) determine whether each critical point is asymptotically stable, stable, or unstable, and classify it as to type. $$ \text { The van der Pol equation: } \quad d x / d t=y, \quad d y / d t=\left(1-x^{2}\right) y-x $$

We will prove part of Theorem 9.3 .2 : If the critical point \((0,0)\) of the almost linear system $$ d x / d t=a_{11} x+a_{12} y+F_{1}(x, y), \quad d y / d t=a_{21} x+a_{22} y+G_{1}(x, y) $$ is an asymptotically stable critical point of the corresponding linear system $$ d x / d t=a_{11} x+a_{12} y, \quad d y / d t=a_{21} x+a_{22} y $$ then it is an asymptotically stable critical point of the almost linear system (i). Problem 12 deals with the corresponding result for instability. In this problem we show that the Liapunov function constructed in the preceding problem is also a Liapunov function for the almost linear system (i). We must show that there is some region containing the origin for which \(\hat{V}\) is negative definite. (a) Show that $$ \hat{V}(x, y)=-\left(x^{2}+y^{2}\right)+(2 A x+B y) F_{1}(x, y)+(B x+2 C y) G_{1}(x, y) $$ (b) Recall that \(F_{1}(x, y) / r \rightarrow 0\) and \(G_{1}(x, y) / r \rightarrow 0\) as \(r=\left(x^{2}+y^{2}\right)^{1 / 2} \rightarrow 0 .\) This means that given any \(\epsilon>0\) there exists a circle \(r=R\) about the origin such that for \(0

(a) Show that the system $$ d x / d t=-y+x f(r) / r, \quad d y / d t=x+y f(r) / r $$ has periodic solutions corresponding to the zeros of \(f(r) .\) What is the direction of motion on the closed trajectories in the phase plane? (b) Let \(f(r)=r(r-2)^{2}\left(r^{2}-4 r+3\right)\). Determine all periodic solutions and determine their stability characteristics.

(a) Determine all critical points of the given system of equations. (b) Find the corresponding linear system near each critical point. (c) Find the eigenalues of each linear system. What conclusions can you then draw about the nonlinear system? (d) Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system. $$ d x / d t=1-x y, \quad d y / d t=x-y^{3} $$
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