91Ó°ÊÓ

The system $$ x^{\prime}=3\left(x+y-\frac{1}{5} x^{3}-k\right), \quad y^{\prime}=-\frac{1}{3}(x+0.8 y-0.7) $$ is a special case of the Fitahugh-Nagumo equations, which model the transmission of neural impulses along an axon. The parameter \(k\) is the external stimulus. (a) For \(k=0\) show that there is one critical point. Find this point and show that it is an asymptotically stable spiral point. Repeat the analysis for \(k=0.5\) and show the critical point is now an unstable spiral point. Draw a phase portrait for the system in each case. (b) Find the value \(k_{0}\) where the critical point changes from asymptotically stable to unstable. Draw a phase portrait for the system for \(k=k_{0}\). (c) For \(k \geq k_{0}\) the system exhibits an asymptotically stable limit cycle. Plot \(x\) versus \(t\) for \(k=k_{0}\) for several periods and estimate the value of the period \(T\). (d) The limit cycle actually exists for a small range of \(k\) below \(k_{0}\). Let \(k_{1}\) be the smallest value of \(k\) for which there is a limit cycle. Find \(k_{1}\).

Step by step solution

01

Find the critical points for k=0 and k=0.5

Set the given functions equal to zero and solve for x and y. For k=0: $$ 0 = 3\left(x+y-\frac{1}{5} x^{3}\right), \quad 0 = -\frac{1}{3}(x+0.8 y-0.7) $$ Solve this system of equations to find the critical point. Repeat the process for k=0.5. Step 2 - Determine the Stability

02

Find the Jacobian matrix and its eigenvalues

Calculate the Jacobian matrix of the given functions, taking partial derivatives with respect to x and y: J(x,y) = $$\begin{pmatrix} \frac{\partial x^{\prime}}{\partial x} & \frac{\partial x^{\prime}}{\partial y} \\ \frac{\partial y^{\prime}}{\partial x} & \frac{\partial y^{\prime}}{\partial y} \end{pmatrix} = \begin{pmatrix} 3-\frac{3}{5}x^2 & 3 \\ -\frac{1}{3} & -\frac{8}{15} \end{pmatrix}$$ Evaluate the Jacobian matrix at both critical points previously found and find its eigenvalues in each case. Determine the stability based on the sign of the real part of the eigenvalues. Step 3 - Drawing Phase Portraits

03

Plot the phase portraits for both cases

Using a numerical solution method, such as the Euler method, solve the given system of equations for different initial conditions and plot the trajectories in the respective x-y plane. The nature of the solution (spiral, asymptotically stable, or unstable) should be consistent with the eigenvalues found in Step 2. Step 4 - Find \(k_{0}\) and Draw a Phase Portrait

04

Determine the value of \(k_0\) and draw a phase portrait

Equate the determinant of the Jacobian matrix to zero and solve for \(k\) to find \(k_{0}\). Then plot a phase portrait for the given system when \(k = k_{0}\). Step 5 - Finding Stability and Estimating Period T

05

Find stability and estimate period T for \(k=k_0\)

Using a numerical solution method, solve the given system with \(k=k_{0}\) for a range of initial conditions. Plot \(x\) vs. \(t\) and based on the behavior of the system, determine if it exhibits a stable limit cycle and estimate the period \(T\). Step 6 - Finding the Smallest Value of k with a Limit Cycle

06

Find \(k_{1}\), the smallest value of k with a limit cycle

Experiment with different initial conditions and values of \(k<k_{0}\) to find the smallest value of \(k\) for which the system exhibits a limit cycle. You can use various numerical methods to find the appropriate values. Record the smallest \(k\) value and report it as \(k_{1}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Asymptotically Stable Spiral Point

An asymptotically stable spiral point occurs in dynamic systems, where solutions spiral inwards, ultimately settling at the critical point over time. For a given system like the FitzHugh-Nagumo equations, identifying such a point means examining the system’s behavior as time progresses. When analyzing the system using the Jacobian matrix, you determine stability by looking at the eigenvalues at the critical point.

If both eigenvalues have negative real parts, the critical point is deemed stable and depicted as an inward spiral on a phase portrait. If only the real parts of these eigenvalues are negative, the system approaches the critical point in a spiral fashion.

In this problem, when the external stimulus parameter, \(k\), equals zero, the system shows an asymptotically stable spiral due to its eigenvalues.

Phase Portrait

A phase portrait is a visual representation of the trajectories of a dynamical system in the phase plane. It provides insights into the behavior of solutions over time. Each point in a phase portrait represents a unique state of the system, described by variables such as \(x\) and \(y\), and shows how these states evolve based on the system's equations.

To draw a phase portrait for a system like the FitzHugh-Nagumo equations, you'll plot trajectories over time for various initial conditions. For different values of parameter \(k\), the system can exhibit distinct patterns, such as asymptotically stable spirals, unstable spirals, or even complex limit cycles.

Phase portraits reveal whether solutions converge to critical points, diverge, or form cycles, allowing us to visualize the qualitative behavior of the system.

Limit Cycle

A limit cycle is a closed trajectory in a phase portrait signifying that the system's behavior is periodic. In the context of the FitzHugh-Nagumo equations, a limit cycle indicates that, after transients fade, the system will continue in a repetitive loop, regardless of small disturbances.

When parameter \(k\) is set to specific values at or beyond some critical threshold, \(k_0\), the system transitions from stable behavior to demonstrating a limit cycle. This means the system will oscillate with a fixed period and amplitude. You can identify this by observing periodic solutions that do not settle into a spiral or stability point.

Finding a limit cycle involves examining phase portraits for closed loops and using numerical simulations to estimate periods and validate ongoing cycles.

Jacobian Matrix

The Jacobian matrix is a derivative matrix that captures how a system's equations change with respect to each variable. It is essential for understanding the local behavior of dynamical systems like those modeled by the FitzHugh-Nagumo equations.

The matrix consists of all first-order partial derivatives of the system and provides critical insights into the system's stability around critical points. For a two-variable system, the Jacobian appears as:
\[ J(x,y) = \begin{pmatrix} \frac{\partial x^{\prime}}{\partial x} & \frac{\partial x^{\prime}}{\partial y} \ \frac{\partial y^{\prime}}{\partial x} & \frac{\partial y^{\prime}}{\partial y} \end{pmatrix} \]

By evaluating the Jacobian matrix at critical points, and calculating its eigenvalues, you can determine whether those points are stable, unstable or if they exhibit limit cycle behavior.

Critical Points

Critical points in a dynamical system are where the system can either remain in a steady state or change its behavior significantly. These are found by setting the system’s differential equations to zero and solving for the variables involved.

In the FitzHugh-Nagumo system, critical points correspond to the states of the system where both derivatives, \(x'\) and \(y'\), are zero. Thus, these points are essential in determining equilibrium states, stability, and potential oscillations.

Each identified critical point needs analysis through the Jacobian matrix to determine whether it's a point of stability, instability, or part of a more complex dynamic like a limit cycle. Depending on the parameter \(k\), which acts as an external stimulus, these points can suggest various behavioral shifts in the system's dynamics.