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Chapter 9: Problem 26

In this problem we show how small changes in the coefficients of a system of linear equations can affect the nature of a critical point when the eigenvalues are equal. Consider the system $$ \mathbf{x}^{\prime}=\left(\begin{array}{cc}{-1} & {1} \\ {0} & {-1}\end{array}\right) \mathbf{x} $$ Show that the eigenvalues are \(r_{1}=-1, r_{2}=-1\) so that the critical point \((0,0)\) is an asymptotically stable node. Now consider the system $$ \mathbf{x}^{\prime}=\left(\begin{array}{cc}{-1} & {1} \\ {-\epsilon} & {-1}\end{array}\right) \mathbf{x} $$ where \(|\epsilon|\) is arbitrararily small. Show that if \(\epsilon>0,\) then the eigenvalues are \(-1 \pm i \sqrt{\epsilon}\), so that the asymptotically stable node becomes an asymptotically stable spiral point. If \(\epsilon<0,\) then the roots are \(-1 \pm \sqrt{|\epsilon|},\) and the critical point remains an asymptotically stable node.

Short Answer

Expert verified
Question: Given two systems of linear equations, prove that the critical point (0,0) is an asymptotically stable node for the first system with eigenvalues r1 = -1, r2 = -1. Moreover, analyze the critical point's nature for a modified system introducing a term -ε, for ε > 0 and ε < 0. Answer: For the first system, the critical point (0,0) is an asymptotically stable node since both eigenvalues are negative. For the modified system, when ε > 0, the critical point (0,0) becomes an asymptotically stable spiral point because the real parts of both eigenvalues are negative, and the imaginary parts are non-zero. When ε < 0, the critical point (0,0) remains an asymptotically stable node, as both eigenvalues have negative real parts.

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01

System 1: Eigenvalue Calculation

To show that the eigenvalues of the first system are \(r_1=-1, r_2=-1\), we will compute the determinant of the matrix: \((A - rI)\), where A is the matrix of the system, r is the eigenvalue, and I is the identity matrix. $$ \mathrm{det}(A-rI)=\left|\begin{array}{cc}-1-r & 1 \\ 0 & -1-r\end{array}\right|= (-1-r)^2 $$ Setting the determinant equal to \(0\) to solve for the eigenvalues: \((-1-r)^2=0 \implies r_1=-1, r_2=-1\).
02

System 1: Stability Analysis

Since both eigenvalues are negative, the critical point \((0,0)\) is an asymptotically stable node.
03

System 2: Eigenvalue Calculation for \(\epsilon > 0\)

For the modified system, the determinant of \((A-rI)\) becomes: $$ \mathrm{det}(A-rI)=\left|\begin{array}{cc}-1-r & 1 \\ -\epsilon & -1-r\end{array}\right|= (1+r)^2+\epsilon $$ For \(\epsilon>0\), we must find the eigenvalues of the system. Setting the determinant equal to \(0\): \((1+r)^2+\epsilon=0\) This is a quadratic equation in r. As it is not easily factorized, we use the quadratic formula to find the roots: $$ r = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} $$ Where A=1, B=2, C=\(1+\epsilon\). The discriminant is given by: $$ B^2 - 4AC = 4 - 4(1+\epsilon) = 4(1-\epsilon) $$ For \(\epsilon > 0\), we have \(1-\epsilon<1\). Hence the discriminant is less than 4. Let the square root of the discriminant be \(i\sqrt{4(1-\epsilon)}\). Then, the roots are: $$ r=-1\pm i\sqrt{\epsilon} $$
04

System 2: Stability Analysis for \(\epsilon > 0\)

Since the real parts of both roots are negative and the imaginary parts are non-zero, the critical point \((0,0)\) becomes an asymptotically stable spiral point for \(\epsilon > 0\).
05

System 2: Eigenvalue Calculation for \(\epsilon < 0\)

For \(\epsilon<0\), the discriminant is: $$ B^2 - 4AC = 4 - 4(1+\epsilon) = 4(1-\epsilon) $$ For \(\epsilon < 0\), we have \(1-\epsilon>1\). Hence the discriminant is greater than 4. The roots can now be written as: $$ r=-1\pm\sqrt{4(1-\epsilon)-4} = -1 \pm \sqrt{|\epsilon|} $$
06

System 2: Stability Analysis for \(\epsilon < 0\)

As both eigenvalues have negative real parts, the critical point \((0,0)\) remains an asymptotically stable node for \(\epsilon<0\).

Key Concepts

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

Eigenvalues and Eigenvectors
When solving linear differential equations, eigenvalues and eigenvectors play a crucial role in determining the behavior of the solutions. An eigenvalue is a scalar that describes the factor by which a corresponding eigenvector is scaled during a linear transformation. To find the eigenvalues, particularly of a square matrix \( A \), we solve the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalues.

In the first system of the given exercise, we have a matrix \( \begin{pmatrix} -1 & 1 \ 0 & -1 \end{pmatrix} \). By computing \( \det(A - \lambda I) \), we find the eigenvalues are both \(-1\). This indicates a repeated eigenvalue, which is important for understanding the system's stability.

Eigenvectors, which are solutions to the equation \( (A - \lambda I)\mathbf{x} = 0 \), help us find directions in the solution space that undergo only scaling during the transformation. In this matrix, for \( \lambda = -1 \), any vector of the form \( \begin{pmatrix} a \ 0 \end{pmatrix} \) is an eigenvector, where \( a \) is any scalar, indicating that the direction along the x-axis remains unchanged under transformations.
Stability Analysis
Stability analysis involves examining the nature of critical points to determine if solutions approach (attract) or retreat from (repel) these points over time. This can be understood by analyzing the eigenvalues of the system's coefficient matrix. If all eigenvalues have negative real parts, the corresponding critical point is asymptotically stable, meaning solutions will converge towards it.

For the first system with eigenvalues \(-1\), both are negative, leading to a stable node. The trajectories of solutions will approach the critical point \((0,0)\) as time progresses. However, if the system undergoes a small change, as with the introduction of \( \epsilon \), the nature of these eigenvalues may alter the stability type.

When \( \epsilon > 0 \), the eigenvalues become \(-1 \pm i\sqrt{\epsilon}\), introducing imaginary parts, which suggests a spiral behavior around the critical point; but with negative real parts, stability is maintained albeit as a spiral point. For \( \epsilon < 0 \), the eigenvalues are real and distinct \(-1 \pm \sqrt{|\epsilon|}\), and the system remains a stable node, although possibly faster in decay as \(|\epsilon|\) increases.
Critical Points
Critical points in a differential equation system occur where all derivatives are zero. They are important as they represent equilibrium states that the system may settle into or repel away from depending on the stability. For the systems in the exercise, \((0,0)\) is a critical point.

Analyzing critical points, especially when eigenvalues change, helps us understand potential system behavior changes. With a small perturbation \(\epsilon\), if \(\epsilon > 0\), the critical point turns into an asymptotically stable spiral point. This reflects a system where trajectories spiral into the critical point, rather than directly converging, suggesting a potential oscillatory decay.

Conversely, when \(\epsilon < 0\), although still stable, these nodes act differently as decay can be more direct and varied due to the change in eigenvalues. The critical point remains attractive, ensuring solutions trend towards it over time, showing the system's ability to adapt to slight parameter changes while maintaining stability.

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Most popular questions from this chapter

Consider the autonomous system $$ d x / d t=x, \quad d y / d t=-2 y+x^{3} $$ (a) Show that the critical point \((0,0)\) is a saddle point. (b) Sketch the trajectories for the corresponding linear system and show that the trajectory for which \(x \rightarrow 0, y \rightarrow 0\) as \(t \rightarrow \infty\) is given by \(x=0\). (c) Determine the trajectories for the nonlinear system for \(x \neq 0\) by integrating the equation for \(d y / d x\). Show that the trajectory corresponding to \(x=0\) for the linear system is unaltered, but that the one corresponding to \(y=0\) is \(y=x^{3} / 5 .\) Sketch several of the trajectories for the nonlinear system.

(a) By solving the equation for \(d y / d x,\) show that the equation of the trajectories of the undamped pendulum of Problem 19 can be written as $$ \frac{1}{2} y^{2}+\omega^{2}(1-\cos x)=c $$ where \(c\) is a constant of integration. (b) Multiply Eq. (i) by \(m L L^{2} .\) Then express the result in terms of \(\theta\) to obtain $$ \frac{1}{2} m L^{2}\left(\frac{d \theta}{d t}\right)^{2}+m g L(1-\cos \theta)=E $$ where \(E=m L^{2} c .\) (c) Show that the first term in Eq. (ii) is the kinetic energy of the pendulum and that the second term is the potential energy due to gravity. Thus the total energy \(E\) of the pendulum is constant along any trajectory; its value is determined by the initial conditions.

(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. $$ d x / d t=y, \quad d y / d t=x-\frac{1}{6} x^{3}-\frac{1}{5} y $$

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. Consider the linear system (ii). (a) Since \((0,0)\) is an asymptotically stable critical point, show that \(a_{11}+a_{22}<0\) and \(\left.a_{11} a_{22}-a_{12} a_{21}>0 . \text { (See Problem } 21 \text { of Section } 9.1 .\right)\) (b) Construct a Liapunov function \(V(x, y)=A x^{2}+B x y+C y^{2}\) such that \(V\) is positive definite and \(\hat{V}\) is negative definite. One way to ensure that \(\hat{V}\) is negative definite is to choose \(A, B,\) and \(C\) so that \(\hat{V}(x, y)=-x^{2}-y^{2} .\) Show that this leads to the result $$ \begin{array}{l}{A=-\frac{a_{21}^{2}+a_{22}^{2}+\left(a_{11} a_{22}-a_{12} a_{21}\right)}{2 \Delta}, \quad B=\frac{a_{12} a_{22}+a_{11} a_{21}}{\Delta}} \\\ {C=-\frac{a_{11}^{2}+a_{12}^{2}+\left(a_{11} a_{22}-a_{12} a_{21}\right)}{2 \Delta}}\end{array} $$ where \(\Delta=\left(a_{11}+a_{22}\right)\left(a_{11} a_{22}-a_{12} a_{21}\right)\) (c) Using the result of part (a) show that \(A>0\) and then show (several steps of algebra are required) that $$ 4 A C-B^{2}=\frac{\left(a_{11}^{2}+a_{12}^{2}+a_{21}^{2}+a_{22}^{2}\right)\left(a_{11} a_{22}-a_{12} a_{21}\right)+2\left(a_{11} a_{22}-a_{12} a_{21}\right)^{2}}{\Delta^{2}}>0 $$ Thus by Theorem 9.6.4, \(V\) is positive definite.

an autonomous system is expressed in polar coordinates. Determine all periodic solutions, all limit cycles, and determine their stability characteristics. $$ d r / d t=\sin \pi r, \quad d \theta / d t=1 $$
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