Question

Determine the critical point \(\mathbf{x}=\mathbf{x}^{0},\) and then classify its type and examine its stability by making the transformation \(\mathbf{x}=\mathbf{x}^{0}+\mathbf{u} .\) \(\frac{d \mathbf{x}}{d t}=\left(\begin{array}{rr}{0} & {-\beta} \\ {\delta} & {0}\end{array}\right) \mathbf{x}+\left(\begin{array}{r}{\alpha} \\\ {-\gamma}\end{array}\right) ; \quad \alpha, \beta, \gamma, \delta>0\)

Short answer

Question: Determine the critical point of a linear system of first-order differential equations, classify its type, and examine its stability given the following differential equation: \( \frac{d\mathbf{x}}{dt} = \left(\begin{array}{rr}{0} & {-\beta} \\ {\delta} &{0}\end{array}\right) \mathbf{x}+\left(\begin{array}{r}{\alpha} \\ {-\gamma}\end{array}\right) \) Answer: The critical point is \(\mathbf{x}^0 = \left(\begin{array}{r}{\dfrac{\gamma}{\delta}} \\ {\dfrac{\alpha}{\beta}}\end{array}\right)\) and it is a center with neutral stability.

Step-by-step solution

Find the critical point \(\mathbf{x}^0\)

To determine the critical point \(\mathbf{x}^0\), we set the rate of change \(\frac{d \mathbf{x}}{d t}\) to zero and solve for \(\mathbf{x}\): $$\frac{d \mathbf{x}}{d t}=\left(\begin{array}{rr}{0} & {-\beta} \\\ {\delta} &{0}\end{array}\right) \mathbf{x}+\left(\begin{array}{r}{\alpha} \\\ {-\gamma}\end{array}\right) = \mathbf{0}$$ To find \(\mathbf{x}^0\), we can solve the linear system: $$\left(\begin{array}{rr}{0} & {-\beta} \\\ {\delta} &{0}\end{array}\right) \mathbf{x}^0 = -\left(\begin{array}{r}{\alpha} \\\ {-\gamma}\end{array}\right)$$ This yields the following system of linear equations: $$\begin{cases} -\beta x_2^0 = -\alpha \\ \delta x_1^0 = \gamma \end{cases}$$ From this system, we can solve for \(x_1^0\) and \(x_2^0\) as follows: $$x_1^0 = \frac{\gamma}{\delta}, \quad x_2^0 = \frac{\alpha}{\beta}$$ Therefore, the critical point is: $$\mathbf{x}^0 = \left(\begin{array}{r}{\dfrac{\gamma}{\delta}} \\\ {\dfrac{\alpha}{\beta}}\end{array}\right)$$

Perform the transformation \(\mathbf{x} = \mathbf{x}^0 + \mathbf{u}\)

Now, we will make the transformation \(\mathbf{x} = \mathbf{x}^0 + \mathbf{u}\) and find the resulting system in terms of \(\mathbf{u}\): $$\frac{d (\mathbf{x}^0 + \mathbf{u})}{d t} = \left(\begin{array}{rr}{0} & {-\beta} \\\ {\delta} &{0}\end{array}\right) (\mathbf{x}^0 + \mathbf{u})+\left(\begin{array}{r}{\alpha} \\\ {-\gamma}\end{array}\right)$$ Since \(\frac{d \mathbf{x}^0}{d t} = \mathbf{0}\), we have \(\frac{d \mathbf{u}}{d t} = \frac{d (\mathbf{x}^0 + \mathbf{u})}{d t}\) and rewrite the system in terms of \(\mathbf{u}\): $$\frac{d \mathbf{u}}{d t} = \left(\begin{array}{rr}{0} & {-\beta} \\\ {\delta} &{0}\end{array}\right) \mathbf{u}$$

Analyze eigenvalues and classify the critical point

To examine the stability and type of the critical point, we can analyze the eigenvalues of the system matrix. We first need to find the characteristic equation of the matrix: $$\begin{vmatrix}{-\lambda} & {-\beta} \\\ {\delta} & {-\lambda}\end{vmatrix} = \lambda^2 + \beta \delta = 0$$ The eigenvalues \(\lambda\) are given by the following equation: $$\lambda^2 = -\beta \delta$$ or $$\lambda = \pm \sqrt{-\beta \delta} i$$ Since the eigenvalues are purely imaginary and have zero real parts, the critical point \(\mathbf{x}^0\) is a center, meaning that trajectories near the critical point are periodic and closed. The system is neutrally stable, as perturbations neither grow nor decay in amplitude. In summary, the critical point \(\mathbf{x}^0 = \left(\begin{array}{r}{\dfrac{\gamma}{\delta}} \\\ {\dfrac{\alpha}{\beta}}\end{array}\right)\) is a center with neutral stability.

Key concepts

Stability Analysis

Stability analysis in differential equations involves assessing if a system returns to equilibrium after a small disturbance. This is crucial to determine whether solutions to the system are predictable in the long term. When we explore stability, we analyze the behavior of solutions in the vicinity of a critical point, which in our case is \(\mathbf{x}^0 = \left(\begin{array}{r}{\dfrac{\gamma}{\delta}} \ {\dfrac{\alpha}{\beta}}\end{array}\right)\).

• If perturbations decay over time, the system is stable.
• If they grow, it is unstable.
• If they neither grow nor decay, as in this case, the system is neutrally stable.

This means minor disturbances do not lead to changes in amplitude but are periodic and closed around the critical point. This neutral behavior is an important attribute in physical systems, indicating that like a pendulum, the system maintains its oscillatory nature over time without losing energy.

Linear Systems

Linear systems can be expressed in terms of matrices and vectors. In our exercise, the system is given by a matrix \[ \left(\begin{array}{rr}{0} & {-\beta} \ {\delta} & {0}\end{array}\right) \].

The governing differential equation, \( \frac{d \mathbf{x}}{dt} = \mathbf{A}\mathbf{x} + \mathbf{b} \), where \( \mathbf{A} \) is the coefficient matrix and \( \mathbf{b} \) is a constant vector, shows us how the variables change with time.

• For linear systems, solutions can often be found using straightforward algebraic techniques.
• This allows for a more predictable understanding of system behavior.

Understanding linear systems provides us with insight into more complex, nonlinear behaviors encountered in real-world phenomena.
The linear approximation often serves to simplify the stability analysis, as it highlights key behaviors without complicating calculations.

Eigenvalues

Eigenvalues are key components in understanding the behavior of a system. They tell us about the nature of equilibrium points and system stability. Here, we calculated the eigenvalues by solving the characteristic equation:
\[ \lambda^2 + \beta \delta = 0 \].
The resulting eigenvalues are purely imaginary, \( \lambda = \pm \sqrt{-\beta \delta} i \).
This finding is pivotal because

• it indicates oscillatory behavior around the critical point.
• The lack of a real part in the eigenvalues signifies that disturbances do not amplify or diminish over time, thereby asserting neutral stability.

Moreover, eigenvalues classified the critical point as a "center," a specific type of equilibrium that indicates perpetual cycles or orbits.
The system's behavior can hence be likened to circular motion, without any spiraling inward or outward movements.

Phase Space Analysis

Phase space is a crucial concept in analyzing dynamic systems by representing all possible states of the system. Each point in this space represents a unique state, defined here through the variables \( x_1 \) and \( x_2 \). In our example, the transformation \( \mathbf{x} = \mathbf{x}^0 + \mathbf{u} \) helps in simplifying our analysis to focus on deviations from the critical point.

Phase space analysis visualizes how systems evolve over time. It provides a comprehensive picture of the system's behavior and stability by showing trajectories of states.

Features of phase space include:

• Trajectories illustrate how starting at one point will take you to another, highlighting paths of future states.
• In our case, the closed, circular trajectories indicate a neutrally stable center.

Through phase space, complex movements in a system are represented graphically, which aids in qualitatively understanding the dynamics without solving complex equations.