Problem 20 - Advanced Engineering Mathematics

Question

Investigate global stability for the system $$ \begin{aligned} &x^{\prime}=y-x \\ &y^{\prime}=-x-y^{3} \end{aligned} $$

Step-by-Step Solution

Answer

The system is globally stable at the equilibrium point \((0,0)\).

Step 1: Identify Equilibrium Points

To identify the equilibrium points of the system, we set the derivatives to zero: \( x' = y-x = 0 \) and \( y' = -x-y^3 = 0 \). Solving these equations, \( y = x \) and \( -x = -x^3 \), we obtain \( x^3 = y^3 \), which implies \( y = -x \). Thus, the equilibrium point is at \((0,0)\).

Step 2: Linearize the System

We linearize the system around the equilibrium point \((0,0)\) using the Jacobian matrix: \( J = \begin{pmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \end{pmatrix} = \begin{pmatrix} -1 & 1 \ -1 & 0 \end{pmatrix} \).

Step 3: Determine Eigenvalues of the Jacobian

To find the stability, we calculate the eigenvalues of the Jacobian matrix \( J = \begin{pmatrix} -1 & 1 \ -1 & 0 \end{pmatrix} \) by solving the characteristic equation \( \det(J - \lambda I) = 0 \). The characteristic equation is \( \lambda^2 + \lambda + 1 = 0 \). Solving for \( \lambda \), we get the roots \( \lambda_{1,2} = -\frac{1}{2} \pm i \frac{\sqrt{3}}{2} \).

Step 4: Analyze Stability Using Eigenvalues

The eigenvalues \( \lambda_{1,2} = -\frac{1}{2} \pm i \frac{\sqrt{3}}{2} \) have negative real parts, which indicate that the equilibrium point at \((0,0)\) is locally asymptotically stable. This implies trajectories spiral towards the origin.

Step 5: Conclude Global Stability

Since the system is a planar (2-dimensional) system, the local asymptotic stability of a single equilibrium point often implies global stability if there are no other equilibrium points or limit cycles in the plane. The nature of the nonlinear terms should further suggest the absence of such components, confirming global stability of the equilibrium point \((0,0)\).

Key Concepts

Equilibrium Points

Equilibrium points are crucial spots in a dynamic system where all variables remain constant over time. To find these points, we set the time derivatives of the system to zero. In our exercise, this means solving the equations:

• \( x' = y-x = 0 \)
• \( y' = -x-y^3 = 0 \)

By setting these equal to zero, we find that \( y = x \) and consequently, \( x^3 = y^3 \). Thus, one key equilibrium point is (0,0), where neither \( x \) nor \( y \) changes with time. Identifying these points helps us understand how the system behaves at rest and sets the stage for further analysis like stability checks.

Jacobian Matrix

The Jacobian matrix is an essential tool that helps us determine the behavior of dynamic systems near equilibrium points. It is a matrix of first-order partial derivatives of the system and provides a linear approximation of the system near the equilibrium.
To assemble the Jacobian, we derive partial derivatives of each equation with respect to both variables. For our system:

• The partial derivative of \( x'=y-x \) with respect to \( x \) is \(-1\); with respect to \( y \) is \(1\).
• The partial derivative of \( y'=-x-y^3 \) with respect to \( x \) is \(-1\); with respect to \( y \) is \(0\).

Thus, the Jacobian matrix \( J \) for our system at the equilibrium point \((0,0)\) is:\[J = \begin{pmatrix} -1 & 1 \ -1 & 0 \end{pmatrix}\]The Jacobian simplifies the study of the system's behavior around the equilibrium point.

Eigenvalues

Eigenvalues are key to understanding the stability of a system at equilibrium points. They are solutions to the characteristic equation derived from a matrix, like the Jacobian, minus a scalar times the identity matrix.
We find eigenvalues by solving the equation \( \det(J - \lambda I) = 0 \). For our Jacobian matrix, this turns into:\[\lambda^2 + \lambda + 1 = 0\]Solving it yields eigenvalues \( \lambda_{1,2} = -\frac{1}{2} \pm i \frac{\sqrt{3}}{2} \), which have negative real parts. This is indicative of local asymptotic stability, showing us that small disturbances around the equilibrium point will decay over time, leading the system's trajectory to spiral towards the equilibrium point.

Linearization

Linearization is the process of approximating a nonlinear system around an equilibrium point with a linear one, using the Jacobian matrix. This technique simplifies analysis, especially when the system's behavior is complex.
By focusing on small deviations from an equilibrium, we effectively "zoom in" on the system's behavior, making it easier to predict how it might react to small changes.
For our given system, we've used the Jacobian to linearize around the equilibrium point \((0,0)\). This results in a system represented by:

• \( x' = -x + y \)
• \( y' = -x \)

The linearized system assists in evaluating stability and other dynamical properties by providing insight into whether disturbances grow or decay over time.