Question

Each exercise lists a nonlinear system \(\mathbf{z}^{\prime}=A \mathbf{z}+\mathbf{g}(\mathbf{z})\), where \(A\) is a constant ( \(2 \times 2\) ) invertible matrix and \(\mathbf{g}(\mathbf{z})\) is a \((2 \times 1)\) vector function. In each of the exercises, \(\mathbf{z}=\mathbf{0}\) is an equilibrium point of the nonlinear system. (a) Identify \(A\) and \(\mathbf{g}(\mathbf{z})\). (b) Calculate \(\|\mathbf{g}(\mathbf{z})\|\). (c) Is \(\lim _{\mid \mathbf{z} \| \rightarrow 0}\|\mathbf{g}(\mathbf{z})\| /\|\mathbf{z}\|=0\) ? Is \(\mathbf{z}^{\prime}=A \mathbf{z}+\mathbf{g}(\mathbf{z})\) an almost linear system at \(\mathbf{z}=\mathbf{0}\) ? (d) If the system is almost linear, use Theorem \(6.4\) to choose one of the three statements: (i) \(\mathbf{z}=\mathbf{0}\) is an asymptotically stable equilibrium point. (ii) \(\mathbf{z}=\mathbf{0}\) is an unstable equilibrium point. (iii) No conclusion can be drawn by using Theorem \(6.4\). $$ \begin{aligned} &z_{1}^{\prime}=2 z_{2}+z_{2}^{2} \\ &z_{2}^{\prime}=-2 z_{1}+z_{1} z_{2} \end{aligned} $$

Short answer

$$ \begin{aligned} z_1^{\prime}=2z_2+z_2^2 \\ z_2^{\prime}=-2z_1+z_1z_2 \end{aligned} $$ Answer: No conclusion can be drawn by using Theorem 6.4.

Step-by-step solution

Identify A and \(\mathbf{g}(\mathbf{z})\)

First, we need to identify the matrix A and the vector function \(\mathbf{g}(\mathbf{z})\). Comparing the given equations to the general form: \(\mathbf{z}^{\prime}=A\mathbf{z}+\mathbf{g}(\mathbf{z})\), we can identify the following: $$ \begin{aligned} &A= \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix} \\\ \\ &\mathbf{g}(\mathbf{z}) = \begin{pmatrix} z_2^2 \\ z_1z_2 \end{pmatrix} \end{aligned} $$

Calculate \(\|\mathbf{g}(\mathbf{z})\|\)

To calculate the norm of the vector function \(\mathbf{g}(\mathbf{z})\), we can use the Euclidean norm formula: $$ \|\mathbf{g}(\mathbf{z})\| = \sqrt{(z_2^2)^2 + (z_1z_2)^2} = \sqrt{z_2^4 + z_1^2z_2^2} $$

Determine if the system is almost linear

To check if the system is almost linear, we need to find the limit \(\lim_{\|\mathbf{z}\|\rightarrow 0}\frac{\|\mathbf{g}(\mathbf{z})\|}{\|\mathbf{z}\|}\). If this limit is equal to 0, then the system is almost linear at \(\mathbf{z}=\mathbf{0}\). Using the result from Step 2, we obtain: $$ \lim_{\|\mathbf{z}\|\rightarrow 0}\frac{\sqrt{z_2^4 + z_1^2z_2^2}}{\sqrt{z_1^2 + z_2^2}} = 0 $$ Since the limit is 0, the system is almost linear at \(\mathbf{z}=\mathbf{0}\).

Apply Theorem 6.4 to determine the nature of the equilibrium point

Now that the system is almost linear, we can apply Theorem 6.4 to conclude the stability of the equilibrium point \(\mathbf{z}=\mathbf{0}\). For this, we need to find the eigenvalues of matrix A. Matrix A is given by: $$ A= \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix} $$ The characteristic equation is: $$ \det(A-\lambda I) = \det\begin{pmatrix}-\lambda & 2 \\ -2 & -\lambda\end{pmatrix} = \lambda^2 + 4 $$ Solving for the eigenvalues: $$ \lambda^2 + 4 = 0 \\ \Rightarrow \lambda = \pm 2i $$ Both eigenvalues have zero real parts. Since we don't have any eigenvalue with a positive real part, we can't conclude about the stability or instability of the equilibrium point \(\mathbf{z}=\mathbf{0}\) using Theorem 6.4. Therefore, the correct statement is: (iii) No conclusion can be drawn by using Theorem 6.4.

Key concepts

Equilibrium Point

An equilibrium point in a system of differential equations is a point where the system is at rest. This means no change occurs at this point; the derivatives of the system variables are zero. In the context of nonlinear systems, particularly in the exercise described, the equilibrium point is \( \mathbf{z} = \mathbf{0} \). This signifies that when the values of all variables in the system are zero, the system remains unchanged over time.

To find equilibrium points in general, you set the system of equations equal to zero and solve for the variables. For the given nonlinear system with \( z_1' = 2z_2 + z_2^2 \) and \( z_2' = -2z_1 + z_1z_2 \), setting both equations to zero leads us to confirm that \( \mathbf{z} = \mathbf{0} \) is indeed an equilibrium point. Finding equilibrium points helps understand the system's behavior at specific states.

• Equilibrium points are essential in studying dynamical systems because they indicate potential system stability.
• Nonlinear systems can have multiple equilibrium points.
• The stability of these points determines how the system responds to disturbances.

Almost Linear System

An almost linear system is an approximation where the behavior of a nonlinear system near an equilibrium point is similar to a linear system. This concept is instrumental in simplifying the analysis of nonlinear systems.

In the exercise, we need to determine if the given system \( \mathbf{z}' = A\mathbf{z} + \mathbf{g}(\mathbf{z}) \) is almost linear around the equilibrium point \( \mathbf{z} = \mathbf{0} \). To do this, we evaluate the limit \( \lim_{\|\mathbf{z}\| \rightarrow 0}\frac{\|\mathbf{g}(\mathbf{z})\|}{\|\mathbf{z}\|} \). If the limit equals zero, the system behaves like a linear system near this point.

This is seen when comparing the non-linear parts of a system to a linear one by evaluating \( \|\mathbf{g}(\mathbf{z})\| \) to assess its effects as \( \|\mathbf{z}\| \) tends towards zero. When successful, this indicates that the system's nonlinear components become negligible, affirming the system's almost linear nature.

• Determining almost linearity aids in simplifying complex systems into more manageable linear forms.
• Linear approximations are powerful in predicting system behavior near equilibrium.
• Even though a system is nonlinear, its resemblance to linear systems can help apply linear theory for analysis.

Matrix Eigenvalues

Matrix eigenvalues are pivotal in understanding the stability of equilibrium points in systems of differential equations. These values are derived from the matrix formed by the linear component of the system.

In the exercise, given the matrix \( A = \begin{pmatrix} 0 & 2 \ -2 & 0 \end{pmatrix} \), eigenvalues are found by solving the characteristic equation:
\[ \det(A-\lambda I) = \lambda^2 + 4 = 0 \]

which gives the eigenvalues \( \lambda = \pm 2i \). These eigenvalues are purely imaginary, meaning they have no real part.

• Eigenvalues indicate stability: if all eigenvalues have negative real parts, the equilibrium is asymptotically stable.
• If any have positive real parts, the equilibrium is unstable.
• Purely imaginary or zero real part eigenvalues, as in this case, make it impossible to deduce the stability with linear terms alone.

In this scenario, since all eigenvalues are purely imaginary, Theorem 6.4 indicates that no direct conclusion about the stability can be made, demonstrating the central role eigenvalues play in system stability analysis.