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Chapter 11: Problem 2

The corresponding plane autonomous system is \\[ x^{\prime}=y, \quad y^{\prime}=-2 x-y^{2} \\] If \((x, y)\) is a critical point, then \(y=0\) and so \(-2 x-y^{2}=-2 x=0 .\) Therefore (0,0) is the sole critical point.

Short Answer

Expert verified
The critical point is (0, 0).

Step by step solution

01

Understand Critical Points

A critical point of a system of differential equations occurs when the derivatives are both zero. For the system given by \( x' = y \) and \( y' = -2x - y^2 \), we are looking for the values of \( x \) and \( y \) such that both equations are satisfied.
02

Set First Equation to Zero

For the first equation \( x' = y \), set \( y = 0 \). This is because critical points require that the derivative is zero.
03

Solve Second Equation at y = 0

Substitute \( y = 0 \) into the second equation \( y' = -2x - y^2 \) to get \( y' = -2x \). For a critical point, \( y' \) must also be zero, hence, solve \(-2x = 0\).
04

Solve for x

From \(-2x = 0\), we find \( x = 0 \). This gives us the critical point \((x, y) = (0, 0)\).

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Key Concepts

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

Differential Equations
Differential equations are mathematical expressions that involve derivatives of a function. Simply put, they describe how a particular quantity changes over time or space. In our exercise, we have two differential equations:
  • \(x' = y\)
  • \(y' = -2x - y^2\)
The prime symbol (') denotes a derivative, often representing a rate of change with respect to time.
Differential equations are powerful tools used to model a wide range of phenomena in physics, engineering, and other sciences.
They help in understanding how complex systems behave by breaking them down into related rates of changes.
In this problem, our task was to find the critical points or equilibrium states of a system, where all derivative values are zero. This means that at these points, the variables do not change, providing us a stable system state.
Critical Points
Critical points play a crucial role in understanding the behavior of dynamic systems described by differential equations. At these points, the derivatives of the system are equal to zero. In other words, there is no change happening at these points.
To find a critical point, you analyze the system by setting each derivative equation to zero. For the equations given:
  • \(x' = y\) is set to zero, leading to \(y = 0\)
  • Substitute \(y = 0\) into \(y' = -2x - y^2\), resulting in \(y' = -2x = 0\)
  • Solve \(-2x = 0\) to find \(x = 0\)
Therefore, the only critical point in this system is \((0, 0)\).
This single point is where both equations show no change over time, making it a point of equilibrium. Understanding these critical points helps classify the stability and behavior patterns that can occur around them.
Plane Systems
Plane systems refer to systems of differential equations that are set up in two dimensions, such as given in this exercise with variables \(x\) and \(y\). Each point in the Cartesian plane (the grid defined by the \(x\) and \(y\) axes) can represent a state of the system.
For a system like ours, analysis involves finding and studying these points where the system doesn't change; these are known as critical points.
The goal is to visualize the behavior of solutions across the plane, often using phase portraits鈥攁 graphical tool to represent trajectories of systems in a phase space.
  • The horizontal axis represents changes in \(x\) (the state variable)
  • The vertical axis represents changes in \(y\) (the rate of change of state variable)
By exploring these systems, one can determine paths (trajectories) that show how the system evolves over time starting from different initial points. Overall, plane systems provide a clear visual understanding of how variables in the system interact dynamically.

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Most popular questions from this chapter

We have \(d y / d x=y^{\prime} / x^{\prime}=-f(x) / y\) and so, using separation of variables, $$\frac{y^{2}}{2}=-\int_{0}^{x} f(\mu) d \mu+c \quad \text { or } \quad y^{2}+2 F(x)=c$$ . We can conclude that for a given value of \(x\) there are at most two corresponding values of \(y .\) If (0,0) were a stable spiral point there would exist an \(x\) with more than two corresponding values of \(y .\) Note that the condition \(f(0)=0\) is required for (0,0) to be a critical point of the corresponding plane autonomous system \(x^{\prime}=y, y^{\prime}=-f(x)\).

Critical points are (1,0) and \((-1,0),\) and $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{rr} 2 x & -2 y \\ 0 & 2 \end{array}\right)$$ At \(\mathbf{X}=(1,0), \tau=4, \Delta=4,\) and so \(\tau^{2}-4 \Delta=0 .\) We can conclude that (1,0) is unstable but we are unable to classify this critical point any further. At \(\mathbf{X}=(-1,0), \Delta=-4<0\) and so (-1,0) is a saddle point.

The corresponding plane autonomous system is \\[ x^{\prime}=y, \quad y^{\prime}=-x+\epsilon x|x| \\] If \((x, y)\) is a critical point, \(y=0\) and \(-x+\epsilon x|x|=x(-1+\epsilon|x|)=0 .\) Hence \(x=0,1 / \epsilon,-1 / \epsilon .\) The critical points \(\operatorname{are}(0,0),(1 / \epsilon, 0)\) and \((-1 / \epsilon, 0)\)

since \(x^{2}-y^{2}=0, y^{2}=x^{2}\) and so \(x^{2}-3 x+2=(x-1)(x-2)=0 .\) It follows that the critical points are (1,1) \((1,-1),(2,2),\) and \((2,-2) .\) We next use the Jacobian $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{rr} -3 & 2 y \\ 2 x & -2 y \end{array}\right)$$ to classify these four critical points. For \(\mathbf{X}=(1,1), \tau=-5, \Delta=2,\) and so \(\tau^{2}-4 \Delta=17>0 .\) Therefore (1,1) is a stable node. For \(\mathbf{X}=(1,-1), \Delta=-2<0\) and so (1,-1) is a saddle point. For \(\mathbf{X}=(2,2), \Delta=-4<0\) and so we have another saddle point. Finally, if \(\mathbf{X}=(2,-2), \tau=1, \Delta=4,\) and so \(\tau^{2}-4 \Delta=-15<0 .\) Therefore (2,-2) is an unstable spiral point.

The corresponding plane autonomous system is $$x^{\prime}=y, \quad y^{\prime}=-x+\left(\frac{1}{2}+3 y^{2}\right) y-x^{2}$$ and so \(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}=0+\frac{1}{2}+9 y^{2} > 0 .\) Therefore there are no periodic solutions by Theorem 11.5.
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