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Chapter 9: Problem 4

Verify that \((0,0)\) is a critical point, show that the system is almost linear, and discuss the type and stability of the critical point \((0,0)\) by examining the corresponding linear system. $$ d x / d t=x+y^{2}, \quad d y / d t=x+y $$

Short Answer

Expert verified
In summary, the critical point of the given system of differential equations is (0,0), and after analyzing the type and stability through linearization, we found that the system has a source-type critical point at (0,0) with both eigenvalues being real and positive. This results in an unstable system, as the solutions will move away from the critical point.

Step by step solution

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01

Find the critical points of the system

To find the critical points of the given system of differential equations, set both \(dx/dt\) and \(dy/dt\) to zero and solve for \(x\) and \(y\). $$ \frac{dx}{dt}=x+y^{2}=0, ~~~~~\frac{dy}{dt}=x+y=0 $$ Now, we just need to solve for \(x\) and \(y\): $$ x + y^{2} = 0, ~~~~x + y = 0 $$ Clearly, \((0,0)\) is the solution of the above system, which therefore is a critical point.
02

Linearize the system near \((0,0)\)

To linearize the system near the critical point \((0,0)\), we need to find the Jacobian matrix of the system. The Jacobian matrix, \(J(x, y)\), consists of the partial derivatives of the given functions with respect to \(x\) and \(y\). So, for our system, the Jacobian matrix is: $$ J(x, y)= \begin{bmatrix} \frac{\partial(x+y^2)}{\partial x} & \frac{\partial(x+y^2)}{\partial y} \\ \frac{\partial(x+y)}{\partial x} & \frac{\partial(x+y)}{\partial y} \end{bmatrix} = \begin{bmatrix} 1 & 2y \\ 1 & 1 \end{bmatrix} $$ Now, we find the Jacobian matrix at the critical point \((0,0)\): $$ J(0, 0) = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} $$ Our linearized system near the critical point \((0,0)\) is then given by: $$ \frac{d\begin{bmatrix} x \\ y \end{bmatrix}}{dt} = J(0,0) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} $$
03

Analyze the type and stability of the critical point \((0,0)\)

To analyze the type and stability of the critical point \((0,0)\), we need to find the eigenvalues of the linearized system's Jacobian matrix at \((0,0)\). Let's compute the eigenvalues of the Jacobian matrix \(J(0,0)\): $$ \det(J(0,0)-\lambda I) = \det\begin{bmatrix}1-\lambda & 0 \\ 1 & 1-\lambda\end{bmatrix} = (1-\lambda)^2 -(0)(1) = (1-\lambda)^2 $$ Setting the determinant to zero, we get: $$ (1-\lambda)^2 = 0 $$ The two eigenvalues are both equal to \(1\). Since both eigenvalues are real and positive, the linearized system, and hence the original nonlinear system, has a source-type critical point at \((0,0)\). This means the solutions will move away from the critical point, making the system unstable.

Key Concepts

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

Differential Equations
Understanding the behavior of dynamical systems often starts with the study of differential equations. These mathematical expressions relate the rate of change of a quantity to the quantity itself. In many scientific fields, from physics to biology, they are used to model real-world phenomena. For instance, the equations \(dx/dt = x + y^2,\quad dy/dt = x + y\) present a system where the rate of change in \(x\) and \(y\) is tied to their current values, capturing a dynamic interplay between them.

Critical points, or equilibrium points, are vital in such systems because they represent states where the system doesn't change - the dynamics are in balance. By setting the rate of change to zero and solving for \(x\) and \(y\), you can find these points of equilibrium. This process is a cornerstone in studying system behavior, helping predict long-term trends based on these fixed points.
Jacobian Matrix
To dive deeper into the local behavior of a system of differential equations near a critical point, we use the Jacobian matrix. This matrix is a grid of first-order partial derivatives. It serves as a tool to assess how small changes in the variables will impact the rates of change. It's like taking a magnifying glass and examining the system's behavior in the vicinity of a critical point.

By evaluating the Jacobian matrix at the critical point, we can infer the dynamics of the system around that point. The crucial aspect of this matrix is that it encapsulates the system's first-order behavior, which is often sufficient to determine the local dynamics, particularly near equilibrium points.
Eigenvalues
The concept of eigenvalues is central to understanding the stability of critical points. Eigenvalues are special numbers associated with a matrix that give insight into the system's response to perturbations. When you find the eigenvalues of the Jacobian matrix at a critical point, you're essentially looking for clues about how the system will behave over time if it starts near that point.

For a two-dimensional system, like our example, the sign of these eigenvalues is particularly revealing: positive eigenvalues indicate that the system tends to move away from the critical point over time, suggesting instability; while negative eigenvalues imply a return to equilibrium, indicating stability. Consequently, the eigenvalues tell us a story about the system's potential to resist or amplify disturbances.
Linearization
The principle of linearization is about simplifying the complex nature of nonlinear dynamical systems to make them more manageable. By approximating a nonlinear system with a linear one around a critical point, you get a clearer picture of the system's immediate response to small disturbances. This linear approximation is made concrete through the Jacobian matrix we have found for the fixed point.

Linearization provides us with an invaluable simplification of the system's behavior near the critical point, but it is important to remember that it's only an approximation. The farther you go from the critical point, the less reliable the linear model might become. Nonetheless, understanding this linear 'shadow' of the original system is usually the first step in a thorough analysis of system behavior.
System Stability
The stability of a system at a critical point essentially answers the question: 'If I nudge the system slightly, will it return to where it started, or will it drift away?' System stability is a crucial property that determines the robustness of physical, biological, or economic systems to external or internal disturbances.

In our example, determining system stability involves verifying the nature of the equilibrium point by analyzing the eigenvalues of the Jacobian matrix. In this case, since both eigenvalues are positive, any small move away from the critical point will only get magnified over time, indicating that the system is unstable around \( (0,0)\). This insight into stability not only forecasts the behavior around the critical point but also provides clues for controlling such systems or predicting their long-term behavior.

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Most popular questions from this chapter

an autonomous system is expressed in polar coordinates. Determine all periodic solutions, all limit cycles, and determine their stability characteristics. $$ d r / d t=r|r-2|(r-3), \quad d \theta / d t=-1 $$

Consider the system \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x},\) and suppose that \(\mathbf{A}\) has one zero eigenvalue. (a) Show that \(\mathbf{x}=\mathbf{0}\) is a critical point, and that, in addition, every point on a certain straight line through the origin is also a critical point. (b) Let \(r_{1}=0\) and \(r_{2} \neq 0,\) and let \(\boldsymbol{\xi}^{(1)}\) and \(\boldsymbol{\xi}^{(2)}\) be corresponding eigenvectors. Show that the trajectories are as indicated in Figure \(9.1 .8 .\) What is the direction of motion on the trajectories?

(a) By solving Eq. (9) numerically show that the real part of the complex roots changes sign when \(r \cong 24.737\). (b) Show that a cubic polynomial \(x^{3}+A x^{2}+B x+C\) has one real zero and two pure imaginary zeros only if \(A B=C\). (c) By applying the result of part (b) to Eq. (9) show that the real part of the complex roots changes sign when \(r=470 / 19\).

In this problem we indicate how to show that the trajectories are ellipses when the eigen- values are pure imaginary. Consider the system $$ \left(\begin{array}{l}{x} \\\ {y}\end{array}\right)^{\prime}=\left(\begin{array}{ll}{a_{11}} & {a_{12}} \\\ {a_{21}} & {a_{22}}\end{array}\right)\left(\begin{array}{l}{x} \\\ {y}\end{array}\right) $$ (a) Show that the eigenvalues of the coefficient matrix are pure imaginary if and only if $$ a_{11}+a_{22}=0, \quad a_{11} a_{22}-a_{12} a_{21}>0 $$ (b) The trajectories of the system (i) can be found by converting Eqs. (i) into the single equation $$ \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a_{21} x+a_{22} y}{a_{11} x+a_{12} y} $$ Use the first of Eqs. (ii) to show that Eq. (iii) is exact. (c) By integrating Eq. (iii) show that $$ a_{21} x^{2}+2 a_{22} x y-a_{12} y^{2}=k $$ where \(k\) is a constant. Use Eqs. (ii) to conclude that the graph of Eq. (iv) is always an ellipse. Hint: What is the discriminant of the quadratic form in Eq. (iv)?

(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. $$ d x / d t=(1+x) \sin y, \quad d y / d t=1-x-\cos y $$
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