Problem 35 - Biocalculus Calculus for the Life Sciences

Question

Fitzhugh-Nagumo equations Consider the following alternative form of the Fitzhugh-Nagumo equations: $$\frac{d v}{d t}=(v-a)(1-v) v-w \quad \frac{d w}{d t}=\varepsilon(v-w)$$ where \(\varepsilon>0\) and \(0<\)a\(<1\). $$\begin{array}{l}{\text { (a) Verify that the origin is an equilibrium. }} \\\ {\text { (b) Calculate the Jacobian matrix. }} \\ {\text { (c) Determine the local stability properties of the origin as }} \\ {\text { a function of the constants. }}\end{array}$$

Step-by-Step Solution

Answer

The origin \((0, 0)\) is an equilibrium, with Jacobian \(\begin{bmatrix} -a & -1 \\ \varepsilon & -\varepsilon \end{bmatrix}\). Stability depends on eigenvalues.

Step 1: Determine Equilibrium Points

To find equilibrium points, we need to set the derivatives \( \frac{dv}{dt} \) and \( \frac{dw}{dt} \) equal to zero. The system of equations becomes: 1. \((v-a)(1-v)v-w = 0\)2. \(\varepsilon(v-w) = 0\)From equation (2), we have \(v = w\). Substituting \(v = w\) into equation (1), we get:\((v-a)(1-v)v-v = 0\)Which simplifies to:\((v-a)(1-v)v = v\)Since \(v = 0\) is a solution, and plugging \(v = 0\) into equation (2) gives \(w = 0\), the origin \((v, w) = (0, 0)\) is an equilibrium point.

Step 2: Calculate the Jacobian Matrix

The Jacobian matrix of the system is the matrix of first derivatives with respect to the variables \(v\) and \(w\). For the given equations, compute: Let \(f(v, w) = (v-a)(1-v)v-w\) and \(g(v, w) = \varepsilon(v-w)\). The Jacobian matrix \(J\) is:\[J = \begin{bmatrix} \frac{\partial f}{\partial v} & \frac{\partial f}{\partial w} \ \frac{\partial g}{\partial v} & \frac{\partial g}{\partial w} \end{bmatrix}\]Calculating the partial derivatives at the point \((v, w) = (0, 0)\):- \(\frac{\partial f}{\partial v} = (1-v)(2v-a) - v^2\) evaluated at \(v=0\) gives \(1\cdot(-a) = -a\).- \(\frac{\partial f}{\partial w} = -1\) at \(w=0\).- \(\frac{\partial g}{\partial v} = \varepsilon\) as a constant, and \(\frac{\partial g}{\partial w} = -\varepsilon\).Therefore, the Jacobian matrix is:\[J(0,0) = \begin{bmatrix} -a & -1 \ \varepsilon & -\varepsilon \end{bmatrix}\]

Step 3: Analyze Local Stability via Eigenvalues

To determine the local stability, analyze the eigenvalues of the Jacobian matrix \(J(0,0)\). The eigenvalues \(\lambda\) are solutions of the characteristic equation:\[ \det(J - \lambda I) = 0 \]For our Jacobian:\[ \begin{vmatrix} -a-\lambda & -1 \ \varepsilon & -\varepsilon-\lambda \end{vmatrix} = 0 \]Expanding the determinant gives:\[ (-a-\lambda)(-\varepsilon-\lambda) - (-1)(\varepsilon) = \lambda^2 + (a+\varepsilon)\lambda + a\varepsilon - \varepsilon = 0 \]This is a quadratic equation with solutions:\[ \lambda = \frac{-(a+\varepsilon) \pm \sqrt{(a+\varepsilon)^2 - 4(a\varepsilon - \varepsilon)}}{2} \]The nature of the roots \(\lambda\) (real or complex, positive or negative) determines stability:- If both eigenvalues have negative real parts, the origin is locally stable.- If any eigenvalue has a positive real part, the origin is unstable.

Key Concepts

Equilibrium Points

In the context of the Fitzhugh-Nagumo equations, equilibrium points are states where the system remains constant over time. To find these points, we set the time derivatives \( \frac{dv}{dt} \) and \( \frac{dw}{dt} \) to zero. This essentially means that there is no change in the system's variables \(v\) and \(w\) over time.

For the given equations, setting these derivatives to zero gives us:

• \( (v-a)(1-v)v-w = 0 \)
• \( \varepsilon(v-w) = 0 \)

From the second equation, \(v = w\), meaning both variables need to be equal for the system to potentially be in equilibrium. Substituting \(v = w\) into the first equation leads to the solution \(v = 0\), \(w = 0\). This confirms that the origin \((0, 0)\) is an equilibrium point, where neither \(v\) nor \(w\) changes over time.

Jacobian Matrix

The Jacobian matrix is a crucial tool for analyzing the behavior of a system near equilibrium points. It consists of all first-order partial derivatives of the system's equations. For our Fitzhugh-Nagumo system, the Jacobian matrix helps determine how small changes in \(v\) and \(w\) affect the system's behavior.

Given the system:

• \(f(v, w) = (v-a)(1-v)v-w\)
• \(g(v, w) = \varepsilon(v-w)\)

The Jacobian matrix \(J\) at any point \((v, w)\) is:\[J = \begin{bmatrix} \frac{\partial f}{\partial v} & \frac{\partial f}{\partial w} \ \frac{\partial g}{\partial v} & \frac{\partial g}{\partial w} \end{bmatrix}\]Evaluated specifically at the equilibrium point \((0, 0)\), it becomes:\[J(0,0) = \begin{bmatrix} -a & -1 \ \varepsilon & -\varepsilon \end{bmatrix}\]This matrix summarizes how sensitive the system's behavior is to changes around the equilibrium.

Local Stability

Once the Jacobian matrix is determined, we use its eigenvalues to assess local stability. Local stability concerns how the system behaves in the immediate vicinity of an equilibrium point. If perturbations or small changes result in the system returning to equilibrium, the point is locally stable. Otherwise, it's unstable.

The stability depends on:- The eigenvalues of the Jacobian matrix
- Their associated real parts, derived from the characteristic equationAt the origin \((0,0)\), the local stability of the system depends on these eigenvalues. If all eigenvalues have negative real parts, any perturbation will eventually decay, meaning the origin is a stable point. Conversely, if any eigenvalue has a positive real part, any small deviation from equilibrium will grow, making the point unstable.

Eigenvalues

Eigenvalues are computed from the Jacobian matrix to determine how trajectories evolve near an equilibrium point. They are crucial for understanding the system's local properties, acting as coefficients in the characteristic equation derived from \( \det(J - \lambda I) = 0 \).

In our Fitzhugh-Nagumo system, the characteristic equation formed from the Jacobian matrix \(\begin{bmatrix} -a & -1 \ \varepsilon & -\varepsilon \end{bmatrix}\) is:\[\lambda^2 + (a + \varepsilon)\lambda + (a\varepsilon - \varepsilon) = 0\]The solutions to this quadratic equation, the eigenvalues \(\lambda\), give insight into system dynamics. If all \(\lambda\) have negative real parts, the trajectories around the equilibrium point decay to the origin, signifying stability. If any \(\lambda\) has a positive real part, it indicates potential instability, as perturbations grow exponentially rather than diminish.