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Find the equilibrium points and analyze their stability. $$ \begin{aligned} &x^{\prime}=x(2-y) \\ &y^{\prime}=y(-3+x) \end{aligned} $$

Short Answer

Expert verified
Equilibrium points: (0,0), (0,2), (3,0), (3,2); stability: (0,0) unstable, (0,2) stable (non-hyperbolic), (3,0) unstable, (3,2) center.

Step by step solution

01

Finding the Equilibrium Points

To find the equilibrium points, we set the derivatives equal to zero: \ 1. \( x' = x(2-y) = 0 \) \ 2. \( y' = y(-3+x) = 0 \) \ From the first equation, either \( x = 0 \) or \( y = 2 \). From the second equation, either \( y = 0 \) or \( x = 3 \). \ Combining these, we find the equilibrium points: \((0, 0)\), \((0, 2)\), \((3, 0)\), and \((3, 2)\).
02

Analyzing Stability at Equilibrium Points

To analyze stability, we find the Jacobian matrix of the system of equations. The Jacobian is: \ \[ J = \begin{pmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{pmatrix} = \begin{pmatrix} 2 - y & -x \ y & -3 + x \end{pmatrix} \] \ We will evaluate this Jacobian at each equilibrium point to determine stability.
03

Evaluating Jacobian at (0, 0)

At \((0, 0)\), the Jacobian is: \ \[ J(0, 0) = \begin{pmatrix} 2 & 0 \ 0 & -3 \end{pmatrix} \] \ The eigenvalues are 2 and -3. Since one eigenvalue is positive, \((0, 0)\) is an unstable saddle point.
04

Evaluating Jacobian at (0, 2)

At \((0, 2)\), the Jacobian is: \ \[ J(0, 2) = \begin{pmatrix} 0 & 0 \ 2 & -3 \end{pmatrix} \] \ The eigenvalues are 0 and -3, indicating that \((0, 2)\) is stable but non-hyperbolic (more analysis is needed to be conclusive).
05

Evaluating Jacobian at (3, 0)

At \((3, 0)\), the Jacobian is: \ \[ J(3, 0) = \begin{pmatrix} 2 & -3 \ 0 & 0 \end{pmatrix} \] \ The eigenvalues are 2 and 0, indicating that \((3, 0)\) is unstable.
06

Evaluating Jacobian at (3, 2)

At \((3, 2)\), the Jacobian is: \ \[ J(3, 2) = \begin{pmatrix} 0 & -3 \ 2 & 0 \end{pmatrix} \] \ The characteristic polynomial is \( \lambda^2 + 6 = 0 \), giving eigenvalues \( \lambda = \pm i\sqrt{6} \). This indicates \((3, 2)\) is a center, suggesting neutral (Lyapunov) stability.

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Key Concepts

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

Equilibrium Points
Equilibrium points are where the system remains in a steady state. For a system of differential equations, these points occur when the rate of change of all variables is zero. In simple terms, the system stops evolving at these points.
To find equilibrium points, one must set the derivatives
  • \( x' = x(2-y) \)
  • \( y' = y(-3+x) \)
to zero and solve the resulting equations. This helps identify combinations of \( x \) and \( y \) where neither changes over time.
For example, if we solve these conditions together, we find the equilibrium points for our specific system:
  • \((0, 0)\)
  • \((0, 2)\)
  • \((3, 0)\)
  • \((3, 2)\)
Each point is an equilibrium state—meaning at these coordinates, the system won't change unless perturbed.
Stability Analysis
Once equilibrium points are found, understanding their stability gives insight into how the system behaves near these points. Stability analysis helps determine if a small disturbance will cause the system to return to equilibrium or move away.
Stability uses eigenvalues derived from the Jacobian matrix (which we will cover in the next section). The eigenvalues indicate how solutions evolve around equilibrium points.
Analyzing the stability involves:
  • If all eigenvalues are negative, the point is stable—systems return to equilibrium after a disturbance.
  • If at least one eigenvalue is positive, the point is unstable—systems move away from equilibrium.
  • If eigenvalues are imaginary without a real part, the equilibrium point might be a center, leading to periodic motion (neutral stability).
This method of analysis provides qualitative understanding of the behaviors around equilibrium points.
Jacobian Matrix
The Jacobian matrix is a fundamental tool in analyzing systems of differential equations, especially for stability. The matrix is essentially a snapshot of how variables in a system influence each other at an equilibrium.
For a two-variable system like \( x \) and \( y \), the Jacobian matrix is calculated by taking partial derivatives:
  • The element \( J_{11} \) is the partial derivative of \( x' \) with respect to \( x \).
  • The element \( J_{12} \) is the partial derivative of \( x' \) with respect to \( y \).
  • The element \( J_{21} \) is the partial derivative of \( y' \) with respect to \( x \).
  • The element \( J_{22} \) is the partial derivative of \( y' \) with respect to \( y \).
For our exercise, we obtained:\[J = \begin{pmatrix} 2-y & -x \ y & -3+x \end{pmatrix}\]This matrix is evaluated at each equilibrium point to reveal the system's local behavior and allows us to determine the nature (stable, unstable, or neutral) based on the eigenvalues. Understanding this process is key to interpreting how dynamic systems behave around stable and unstable points.

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Most popular questions from this chapter

Prove that the quadratic polynomial \(V(x, y)=a x^{2}+2 b x y+\) \(c y^{2}\) is positive definite with a minimum at \((x, y)=(0,0)\) if and only if \(a>0\) and \(a c-b^{2}>0\).

Perform each of the following tasks. (i) Sketch the nullclines for each equation. Use a distinctive marking for each nullcline so they can be distinguished. (ii) Use analysis to find the equilibrium points for the system. Label each equilibrium point on your sketch with its coordinates. (iii) Use the Jacobian to classify each equilibrium point (spiral source, nodal sink, etc.). \(x^{\prime}=2 x-2 x^{2}-x y\) \(y^{\prime}=2 y-x y-2 y^{2}\)

Determine if the given system is Hamiltonian. If the system is Hamiltonian, find its Hamiltonian function. $$ \begin{aligned} &x^{\prime}=\cos x \\ &y^{\prime}=-y \sin x+2 x \end{aligned} $$

Consider the system $$ \begin{aligned} &y^{\prime}=v, \\ &v^{\prime}=f(y) . \end{aligned} $$ A potential function \(U(y)=-\int f(y) d y\) is drawn. Make a copy of the potential function; then align the \(y v\)-phase plane below the potential axes, as shown in Figure \(1 .\) Use the potential function \(U\) to create the corresponding phase portrait in the \(y v\)-phase plane. $$ f(y)=y $$

Consider the second-order equation $$ x^{\prime \prime}+x^{\prime}-\frac{1}{3}\left(x^{\prime}\right)^{3}+x=0 . $$ Use the standard substitutions \(x_{1}=x\) and \(x_{2}=x^{\prime}\) to write equation (7.14) as a system of first-order equations having an equilibrium point at the origin. (a) Using the positive definite function \(V\left(x_{1}, x_{2}\right)=x_{1}^{2}+\) \(x_{2}^{2}\), show that $$ \dot{V}\left(x_{1}, x_{2}\right)=-\frac{2}{3} x_{2}^{2}\left(3-x_{2}^{2}\right) $$ on solution trajectories of the system. Argue that \(\dot{V}\) is negative semidefinite on \(\left\\{\left(x_{1}, x_{2}\right):\left|x_{2}\right|<\sqrt{3}\right\\}\) and show that the equilibrium point at the origin is stable. (b) Given that \(\dot{V}\) is negative semidefinite on \(\left\\{\left(x_{1}, x_{2}\right)\right.\) : \(\left.\left|x_{2}\right|<\sqrt{3}\right\\}\), use Theorem \(7.10\) to argue that the equilibrium point at the origin is asymptotically stable. Provide a phase portrait of the system highlighting the asymptotic stability of the equilibrium point at the origin.

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