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 \(6.4\). $$ \begin{aligned} &z_{1}^{\prime}=-z_{1}+3 z_{2}+z_{2} \cos \sqrt{z_{1}^{2}+z_{2}^{2}} \\ &z_{2}^{\prime}=-z_{1}-5 z_{2}+z_{1} \cos \sqrt{z_{1}^{2}+z_{2}^{2}} \end{aligned} $$

Short answer

To summarize: (a) The constant matrix \(A\) and the vector function \(\mathbf{g}(\mathbf{z})\) are identified as: $$ A = \begin{pmatrix} -1 & 3 \\ -1 & -5 \end{pmatrix}, \qquad \mathbf{g}(\mathbf{z}) = \begin{pmatrix} z_{2} \cos \sqrt{z_{1}^{2}+z_{2}^{2}} \\ z_{1} \cos \sqrt{z_{1}^{2}+z_{2}^{2}} \end{pmatrix}. $$ (b) The magnitude of the vector function is given by: $$ \|\mathbf{g}(\mathbf{z})\|= \sqrt{(z_{2} \cos \sqrt{z_{1}^{2}+z_{2}^{2}})^{2}+(z_{1} \cos \sqrt{z_{1}^{2}+z_{2}^{2}})^{2}}. $$ (c) The given system is not almost linear at \(\mathbf{z}=\mathbf{0}\), as the limit is equal to 1, not 0: $$ \lim_{\|\mathbf{z}\| \to 0} \frac{\|\mathbf{g}(\mathbf{z})\|}{\|\mathbf{z}\|} = 1. $$ (d) Since the system is not almost linear, we cannot use Theorem 6.4 to analyze the stability of the equilibrium point \(\mathbf{z}=\mathbf{0}\).

Step-by-step solution

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

From the given system, we can identify the constant matrix \(A\) and the vector function \(\mathbf{g}(\mathbf{z})\) as: $$ A = \begin{pmatrix} -1 & 3 \\ -1 & -5 \end{pmatrix}, \qquad \mathbf{g}(\mathbf{z}) = \begin{pmatrix} z_{2} \cos \sqrt{z_{1}^{2}+z_{2}^{2}} \\ z_{1} \cos \sqrt{z_{1}^{2}+z_{2}^{2}} \end{pmatrix}. $$

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

To find the magnitude of the vector function, we can use the standard formula for the Euclidean norm of a vector of length 2: $$ \|\mathbf{g}(\mathbf{z})\|= \sqrt{(z_{2} \cos \sqrt{z_{1}^{2}+z_{2}^{2}})^{2}+(z_{1} \cos \sqrt{z_{1}^{2}+z_{2}^{2}})^{2}}. $$

Check for almost linearity

To determine if the given system is almost linear at the equilibrium point \(\mathbf{z}=\mathbf{0}\), we need to check if $$ \lim_{\|\mathbf{z}\| \to 0} \frac{\|\mathbf{g}(\mathbf{z})\|}{\|\mathbf{z}\|} = 0. $$ Taking the limit as \(\mathbf{z}\) approaches the origin, we notice that both the numerator and denominator go to zero, making it an indeterminate form. To deal with this, we use L'Hôpital's rule: $$ \lim_{\|\mathbf{z}\| \to 0} \frac{\|\mathbf{g}(\mathbf{z})\|}{\|\mathbf{z}\|} = \lim_{\|\mathbf{z}\| \to 0} \frac{z_{1}^{2}+z_{2}^{2}}{z_{1}^{2}+z_{2}^{2}} = 1. $$ Since the limit is not equal to zero, the given system is not almost linear at \(\mathbf{z}=\mathbf{0}\).

Stability analysis (not applicable)

Since the system is not almost linear, we cannot use Theorem 6.4 to analyze the stability of the equilibrium point \(\mathbf{z}=\mathbf{0}\). Therefore, we cannot draw any conclusions regarding the stability of this equilibrium point.

Key concepts

Equilibrium Points

Equilibrium points in a differential equation system are locations where the system is in a state of balance. That means the derivatives of the variables, like \(z_1'\) and \(z_2'\), become zero at these points.
In our example, \(\mathbf{z} = \mathbf{0}\) is given as an equilibrium point.

• If you start at the equilibrium point, the system doesn't move, it's in equilibrium.
  
• To determine equilibrium points more generally, set the derivatives to zero and solve the resulting equations.
  
• In this exercise, equilibrium is already given, simplifying our task.

At \(\mathbf{z} = \mathbf{0}\), the forces from the system's dynamics, described by matrix \(A\) and vector function \(\mathbf{g}(\mathbf{z})\), balance out entirely, so there's no net change.
Understanding equilibrium points helps predict how systems behave, and they are crucial when analyzing the stability of these systems.

Almost Linear Systems

Almost linear systems are nonlinear systems that behave similarly to linear systems near an equilibrium point. Linear systems are simpler to analyze than nonlinear ones.
To check if a system is almost linear, examine what happens when the inputs are very small, approaching the equilibrium point.
Here, the condition\(\lim_{\|\mathbf{z}\| \to 0} \frac{\|\mathbf{g}(\mathbf{z})\|}{\|\mathbf{z}\|} = 0\)must be satisfied.

• This expression means the extra nonlinearity disappears as you get closer to the equilibrium point.
• If the limit evaluates to zero, the nonlinearity becomes insignificant, making the system almost linear near \(\mathbf{z} = \mathbf{0}\).

In this exercise, evaluating the limit showed it is not zero, indicating the system isn't almost linear at the given equilibrium point.
It's important to establish this because it determines whether linear approximation methods can be used to study the system's behavior.

Stability Analysis

Stability analysis in differential equations assesses whether an equilibrium point is stable, unstable, or neutrally stable. It involves studying how perturbations (small changes) affect the equilibrium.
For systems thought to be almost linear, Theorem 6.4 can be applied to determine the stability at the equilibrium point, but only if almost linearity is confirmed.
The importance of stability analysis is in predicting system behavior over time:

• **Stable Equilibrium:** If a system returns to equilibrium after slight disturbances, it is stable.
  
• **Unstable Equilibrium:** If it moves away after disturbances, it's unstable.
  
• **Neutral Stability:** If it neither converges back nor diverges away, it has neutral stability.

In the original exercise, since the system wasn’t almost linear, we couldn’t apply Theorem 6.4 for stability analysis.
Stability analysis gives insight into whether systems tend to return to a point of equilibrium or evolve to different states, which is crucial in many scientific and engineering contexts.