Question

Analyze the local stability of each of the following nonlinear systems: (a) \(x^{\prime}=e^{x}-1\) \(y^{\prime}=y e^{x}\) (b) \(x^{\prime}=x+2 y\) \(y^{\prime}=x^{2}+y\) (c) \(x^{\prime}=1-e^{y}\) \(y^{\prime}=5 x-y\) (d) \(x^{\prime}=x^{3}+3 x^{2} y+y\) \(y^{\prime}=x\left(1+y^{2}\right)\)

Short answer

(a) Center at (0, y). (b) Saddle point at (1/2, -1/4). (c) Spiral point at (0, 0), stable. (d) Saddle point at (0, 0).

Step-by-step solution

Find Equilibrium Points for System (a)

For system (a), solve for equilibrium points: \(x' = e^x - 1 = 0\) gives \(x = 0\), and substituting into \(y = ye^x = y\), any value of \(y\) is an equilibrium. Therefore, equilibrium points are \((0, y)\), \(y \in \mathbb{R}\).

Determine Stability of System (a)

Calculate the Jacobian matrix: \[ J(x, y) = \begin{bmatrix} e^x & 0 \ y e^x & e^x \end{bmatrix} \]. Evaluate at \((0, y)\), giving \[ J = \begin{bmatrix} 1 & 0 \ y & 1 \end{bmatrix} \]. Since both eigenvalues are \(1\), the equilibrium \((0, y)\) is a center.

Find Equilibrium Points for System (b)

Solve \(x' = x + 2y = 0\) and \(y' = x^2 + y = 0\). From \(y' = 0\), \(y = -x^2\). Substituting into \(x' = 0\), \(x - 2x^2 = 0\), gives solutions \(x = 0, x = 1/2\). Thus, equilibrium points are \((0, 0)\) and \((1/2, -1/4)\).

Determine Stability of System (b)

Calculate the Jacobian: \[ J(x, y) = \begin{bmatrix} 1 & 2 \ 2x & 1 \end{bmatrix} \]. At \((0, 0)\), \(J = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix}\), with eigenvalues \(\lambda = 1\). At \((1/2, -1/4)\), \(J = \begin{bmatrix} 1 & 2 \ 1 & 1 \end{bmatrix}\). The determinant \((-3) < 0\) indicates a saddle point.

Find Equilibrium Points for System (c)

Solve \(x' = 1 - e^y = 0\) gives \(y = 0\), substituting into \(y' = 5x - y = 0\) with \(y = 0\) gives \(x = 0\). Hence, the equilibrium point is \((0, 0)\).

Determine Stability of System (c)

Find Jacobian: \[ J(x, y) = \begin{bmatrix} 0 & -e^y \ 5 & -1 \end{bmatrix} \]. At \((0, 0)\), \(J = \begin{bmatrix} 0 & -1 \ 5 & -1 \end{bmatrix}\). Eigenvalues solve \(\lambda^2 + \lambda + 5 = 0\), with complex roots \(-1/2 \pm i\surd{19}/2\). The equilibrium is a spiral point, confirming it is locally stable.

Find Equilibrium Points for System (d)

Solve \(x' = x^3 + 3x^2y + y = 0\) and \(y' = x(1 + y^2) = 0\), leading to learning \(x = 0\) or \(y^2 = -1\). Thus, \(x=0, y=0\) as viable, or solutions aren't realistic.

Determine Stability of System (d)

Jacobian matrix is \[ J(x, y) = \begin{bmatrix} 3x^2 + 6xy & 3x^2 + 1 \ 1 + y^2 & 2xy \end{bmatrix} \]. At this, evaluate at \((0,0)\), \(J = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}\). Eigenvalues solve \(\lambda^2 - 1 = 0\), revealing roots \(\pm 1\), indicating a saddle point.

Key concepts

Local Stability

Local stability in the context of nonlinear dynamic systems refers to how solutions or trajectories of systems behave near equilibrium points. An equilibrium point is stable if nearby trajectories stay close to it when slightly disturbed. For complete understanding, it’s crucial to distinguish between different types of stability:

• Stable: Perturbations remain close to the equilibrium.
• Asymptotically Stable: Trajectories not only remain close but also tend to return to the equilibrium over time.
• Unstable: Perturbations cause trajectories to diverge away from the equilibrium.

Understanding local stability informs us about the behavior of dynamic systems around specified points, determining the feasibility of predictability and control in applications like engineering and natural systems.

Equilibrium Points

Equilibrium points in a nonlinear system are where the system doesn't change over time, meaning the derivatives of all variables are zero. At these points, the forces or influences in the system are balanced. Finding equilibrium points involves solving a set of equations derived from setting the system's derivatives to zero. For example, in system (a) with equations \(x' = e^x - 1 = 0\) and \(y' = ye^x = y\), substituting gives equilibrium points at \((0, y)\), where \(y\) can be any real number. Equilibrium points are critical for stability analysis since they are the points we assess using the Jacobian matrix and eigenvalues to determine the system's local stability behavior.

Jacobian Matrix

The Jacobian matrix is a crucial tool in analyzing the local behavior of nonlinear dynamic systems near equilibrium points. It consists of first-order partial derivatives of the system functions with respect to the state variables. For a system characterized by equations like \(x' = f(x, y)\) and \(y' = g(x, y)\), the Jacobian is:\[J(x, y) = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\end{bmatrix}.\]At equilibrium, the Jacobian provides a linear approximation of the system. By evaluating it at these points, we understand the system dynamics better and determine local stability by calculating its eigenvalues.

Eigenvalues

Eigenvalues are fundamental in determining the local stability of equilibrium points in nonlinear dynamic systems. They are derived from the Jacobian matrix evaluated at the equilibrium points. To find them, we solve the characteristic equation of the Jacobian: \(\det(J - \lambda I) = 0\), where \(I\) is the identity matrix. The properties of the eigenvalues inform us about stability:

• All eigenvalues with negative real parts: The equilibrium is asymptotically stable.
• At least one eigenvalue with a positive real part: The equilibrium is unstable.
• Purely imaginary or zero eigenvalues: The system might exhibit stable oscillations or other behaviors, requiring further analysis.

In the discussed systems, when eigenvalues include complex numbers with negative real parts like in system (c), the equilibrium behaves as a spiral point, indicating local stability.