Problem 6
Question
Write the given nonlinear second-order differential equation as a plane autonomous system. Find all critical points of the resulting system. \(x^{\prime \prime}+x-\epsilon x|x|=0\) for \(\epsilon>0\)
Step-by-Step Solution
Verified Answer
Critical point: (0, 0).
1Step 1: Convert to first-order system
The given equation is a second-order differential equation: \(x'' + x - \epsilon x|x| = 0\). To convert it into a system of first-order ODEs, let's introduce a new variable \(y = x'\). Hence, we have \(x'' = y'\). This gives us the system \(y' = -x + \epsilon x|x|\) and \(x' = y\). The system can be written as:\( \begin{cases} x' = y \ y' = -x + \epsilon x|x| \end{cases} \)
2Step 2: Find critical points
Critical points of the system are found by setting \(x' = 0\) and \(y' = 0\). From \(x' = y\), we set \(y = 0\). From \(y' = -x + \epsilon x|x|\), set \(-x + \epsilon x|x| = 0\). Solving \(-x + \epsilon x|x| = 0\) gives us \(x = 0\). Substituting \(x = 0\) into \(y = 0\) gives the critical point \((0, 0)\).
Key Concepts
Autonomous SystemsCritical PointsFirst-Order System
Autonomous Systems
An autonomous system in mathematics, especially in the context of differential equations, refers to a system where the variables evolve over time based on the current state, not explicitly dependent on time itself. This is crucial when dealing with equations, as it simplifies the analysis. You can think of it as a self-contained system, where everything you need to predict future behavior is determined by the current state of the system.
In the context of our exercise, converting the second-order differential equation into an autonomous system involves rewriting the equation in terms of first-order equations. The new system of equations derived does not explicitly involve the independent variable, usually time, which helps in simplifying the problem.
In the context of our exercise, converting the second-order differential equation into an autonomous system involves rewriting the equation in terms of first-order equations. The new system of equations derived does not explicitly involve the independent variable, usually time, which helps in simplifying the problem.
- Autonomous systems are time-invariant; the rules that govern the behavior do not change over time.
- Such systems allow for more streamlined theoretical analysis, like identifying critical points and analyzing stability.
Critical Points
The critical points of a differential equation system are values where the system's behavior changes or stabilizes. These points usually occur where the derivatives (the rates of change) are zero. It's like finding equilibrium in the system, where movement ceases or transitions.
In our exercise, to find these critical points of the autonomous system, we set both equations in the system to zero. That means finding the values of variables where neither changes - a moment of rest.
In our exercise, to find these critical points of the autonomous system, we set both equations in the system to zero. That means finding the values of variables where neither changes - a moment of rest.
- For the equation set in the problem, the equations derived are: \( x' = y \) and \( y' = -x + \epsilon x|x| \).
- Setting these to zero helps identify where the system reaches a steady state.
First-Order System
A first-order system is a type of differential equation where the highest derivative is a first derivative. Converting higher-order systems, like second-order equations, into first-order systems is common practice as these are easier to analyze and interpret.
In the exercise, we took a second-order differential equation and introduced a new variable to convert it into a system of first-order differential equations. The steps involved are systematic and ensure that the system is easier to handle mathematically.
In the exercise, we took a second-order differential equation and introduced a new variable to convert it into a system of first-order differential equations. The steps involved are systematic and ensure that the system is easier to handle mathematically.
- For the given exercise, by setting \( y = x' \), the equation becomes a set of first-order equations: \( y' = -x + \epsilon x|x| \) and \( x' = y \).
- Such systems are often easier to visualize and solve numerically, as you can plot them on a plane with two variables interacting.
Other exercises in this chapter
Problem 6
Show that the given plane autonomous system (or second-order differential equation) has no periodic solutions. $$ \begin{aligned} &x^{r}=2 x+y^{2} \\ &y^{r}=x y
View solution Problem 6
Without solving explicitly, classify the critical points of the given first- order autonomous differential equation as either asymptotically stable or unstable.
View solution Problem 7
In Problems, show that the given plane autonomous system (or second-order differential equation) has no periodic solutions. $$ x^{\prime \prime}-2 x+\left(x^{\p
View solution Problem 7
The general solution of the linear system \(\mathbf{X}^{\prime}=\mathbf{A} \mathbf{X}\) is given. (a) In each case discuss the nature of the solution in a neigh
View solution