Problem 27
Question
Show that the equilibrium \(\left[\begin{array}{l}0 \\ 0\end{array}\right]\) of $$ \begin{array}{l} x_{1}(t+1)=\frac{x_{2}(t)}{4\left(1+\left(x_{1}(t)\right)^{2}\right)} \\ x_{2}(t+1)=\frac{3 x_{1}(t)}{1+\left(x_{2}(t)\right)^{2}} \end{array} $$ is locally stable.
Step-by-Step Solution
Verified Answer
The equilibrium is locally stable because its Jacobian has purely imaginary eigenvalues, indicating a center-type stability.
1Step 1: Identify the System of Equations
We are given the system of equations: \[ x_{1}(t+1) = \frac{x_{2}(t)}{4\left(1+\left(x_{1}(t)\right)^{2}\right)} \] \[ x_{2}(t+1) = \frac{3 x_{1}(t)}{1+\left(x_{2}(t)\right)^{2}} \] Our goal is to show that the equilibrium \( \left[\begin{array}{l}0 \ 0\end{array}\right] \) is locally stable.
2Step 2: Calculate the Jacobian Matrix
The Jacobian matrix \( J \) of the system is obtained by calculating the partial derivatives of the functions with respect to \( x_1 \) and \( x_2 \):\[J = \begin{bmatrix} \frac{\partial x_{1}(t+1)}{\partial x_{1}(t)} & \frac{\partial x_{1}(t+1)}{\partial x_{2}(t)} \ \frac{\partial x_{2}(t+1)}{\partial x_{1}(t)} & \frac{\partial x_{2}(t+1)}{\partial x_{2}(t)} \end{bmatrix}.\]
3Step 3: Compute the Partial Derivatives
Compute the necessary partial derivatives to construct the Jacobian matrix:- \( \frac{\partial x_{1}(t+1)}{\partial x_{1}(t)} = -\frac{2x_{1}(t)x_{2}(t)}{4(1 + (x_{1}(t))^2)^2}\).- \( \frac{\partial x_{1}(t+1)}{\partial x_{2}(t)} = \frac{1}{4(1 + (x_{1}(t))^2)}\).- \( \frac{\partial x_{2}(t+1)}{\partial x_{1}(t)} = \frac{3}{1 + (x_{2}(t))^2}\).- \( \frac{\partial x_{2}(t+1)}{\partial x_{2}(t)} = -\frac{6x_{1}(t)x_{2}(t)}{(1 + (x_{2}(t))^2)^2}\).
4Step 4: Evaluate the Jacobian at the Equilibrium
Evaluate each partial derivative at the equilibrium \( \left[\begin{array}{l}0 \ 0\end{array}\right] \):- \( \frac{\partial x_{1}(t+1)}{\partial x_{1}(t)} \bigg|_{(0,0)} = 0 \).- \( \frac{\partial x_{1}(t+1)}{\partial x_{2}(t)} \bigg|_{(0,0)} = \frac{1}{4} \).- \( \frac{\partial x_{2}(t+1)}{\partial x_{1}(t)} \bigg|_{(0,0)} = 3 \).- \( \frac{\partial x_{2}(t+1)}{\partial x_{2}(t)} \bigg|_{(0,0)} = 0 \).Thus, \( J(0,0) = \begin{bmatrix} 0 & \frac{1}{4} \ 3 & 0 \end{bmatrix} \).
5Step 5: Determine the Eigenvalues of the Jacobian
To find the eigenvalues, solve the characteristic equation:\[\det(\lambda I - J) = \left| \begin{bmatrix} \lambda & -\frac{1}{4} \ -3 & \lambda \end{bmatrix} \right| = \lambda^2 - \frac{3}{4} = 0.\]Solve for \( \lambda \) to get \( \lambda = \pm \frac{\sqrt{3}}{2} i \).
6Step 6: Conclude Local Stability
Since the eigenvalues \( \lambda = \pm \frac{\sqrt{3}}{2} i \) are purely imaginary, the real parts are zero. This indicates that the equilibrium is a center and, by linearization, locally stable (Lyapunov's indirect method confirms near-linear stability).
Key Concepts
system of equationsJacobian matrixeigenvalues
system of equations
A system of equations is a set of two or more equations that have common variables and are solved together. In the context of dynamic systems, these equations often represent changes in variables over time. In our exercise, we have a system describing the evolution of two variables, \(x_1(t+1)\) and \(x_2(t+1)\), in terms of previous states \(x_1(t)\) and \(x_2(t)\). Each equation works like an instruction manual, defining how one step leads into the next.The primary objective when dealing with such systems is to find solutions that satisfy all equations simultaneously at each point in time. Specifically, we often seek equilibrium solutions, where the system doesn't change as time progresses. For the system given, solving for equilibrium means finding conditions under which both equations lead to no change over time, resulting in the fixed point \([0, 0]\). Understanding these systems is crucial for analyzing how changes in one variable affect others and for predicting future states of a system.
Jacobian matrix
The Jacobian matrix is a powerful tool in analyzing dynamic systems of equations. Specifically, it contains the first-order partial derivatives of a vector function and helps in understanding how the function behaves in the vicinity of a point, such as equilibrium.In our system, the Jacobian matrix, denoted as \( J \), is constructed by evaluating the partial derivatives of each component of the system concerning each variable:
- \( \frac{\partial x_{1}(t+1)}{\partial x_{1}(t)} \) and \( \frac{\partial x_{1}(t+1)}{\partial x_{2}(t)} \) for the first equation.
- \( \frac{\partial x_{2}(t+1)}{\partial x_{1}(t)} \) and \( \frac{\partial x_{2}(t+1)}{\partial x_{2}(t)} \) for the second equation.
eigenvalues
Eigenvalues are central to understanding the stability of equilibrium points in linear and nonlinear systems. After the Jacobian matrix is determined, the next step is finding its eigenvalues, which are solutions to the characteristic equation \( \det(\lambda I - J) = 0 \).For our exercise:
- The Jacobian matrix at equilibrium is \( \begin{bmatrix} 0 & \frac{1}{4} \ 3 & 0 \end{bmatrix} \).
- The characteristic equation becomes \( \lambda^2 - \frac{3}{4} = 0 \), leading to the eigenvalues \( \lambda = \pm \frac{\sqrt{3}}{2} i \).
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