Problem 13 - Calculus for Biology and Medicine

Question

Use a graphing calculator to study the following example of the Fitzhugh- Nagumo model: $$ \begin{array}{l} \frac{d V}{d t}=-V(V-0.3)(V-1)-w \\ \frac{d w}{d t}=V-c w \end{array} $$ Explain whether or not the model predicts multiple equilibria for the following values of \(c\) : (a) \(c=2\), (b) \(c=16\).

Step-by-Step Solution

Answer

The Fitzhugh-Nagumo model predicts multiple equilibria for both \(c = 2\) and \(c = 16\).

Step 1: Identifying Equilibrium Points

An equilibrium point is where both derivatives \( \frac{dV}{dt} \) and \( \frac{dw}{dt} \) equal zero. This provides the system of equations:1. \(-V(V-0.3)(V-1) - w = 0\)2. \(V - cw = 0\)We'll use these to find solutions \((V, w)\).

Step 2: Solving for Equilibrium Points in terms of c

From the second equation \(V = cw\). Substitute \(w = \frac{V}{c}\) into the first equation:\[-V(V-0.3)(V-1) - \frac{V}{c} = 0\]Factor out \(V\):\[V[-(V-0.3)(V-1) - \frac{1}{c}] = 0\]Thus, \(V = 0\) or solve \(-(V-0.3)(V-1) = \frac{1}{c}\).

Step 3: Case Analysis for Different Values of c

For \(c = 2\):- Substituting \(c = 2\) yields:\(-(V-0.3)(V-1) = \frac{1}{2}\).- This results in the quadratic equation in \(V\).For \(c = 16\):- Substituting \(c = 16\) results in another quadratic equation:\(-(V-0.3)(V-1) = \frac{1}{16}\).

Step 4: Solving the Quadratic Equation for Both Cases

For \(c = 2\), solve:\((V-0.3)(V-1) = -0.5\).- Upon expanding and solving, you'll find distinct \(V\) solutions.For \(c=16\), solve:\((V-0.3)(V-1) = -0.0625\).- Expanding and solving this, you'll also get distinct \(V\) values.

Step 5: Determine the Number of Equilibrium Points

Upon solving:- For \(c = 2\), you typically get 3 equilibria points as \(V\) values satisfy the quadratic.- For \(c = 16\), you also find distinct \(V\) values suggesting multiple equilibria.

Key Concepts

Equilibrium Points

Equilibrium points are critical in understanding dynamic systems like the FitzHugh-Nagumo model. These points occur where the system's variables do not change over time. This means both derivatives, \( \frac{dV}{dt} \) and \( \frac{dw}{dt} \), must equal zero.

To find these points in our model, we set up the system of equations:

• \(-V(V-0.3)(V-1) - w = 0\)
• \(V - cw = 0\)

The approach is to solve these equations simultaneously. From the second equation, we have \( V = cw \). By substituting \( w = \frac{V}{c} \) into the first equation, we simplify the equations to find potential equilibrium values for \( V \).

By factoring and rearranging, we find that \( V \) must either be zero, or a solution to the equation \(-(V-0.3)(V-1) = \frac{1}{c}\). Each value of \( c \) provides different scenarios for possible configurations and numbers of equilibrium points.

Differential Equations

Differential equations are equations that involve the rates of change of quantities. In the FitzHugh-Nagumo model, these equations describe how the variables \( V \) and \( w \) evolve over time.

Specifically:

• \( \frac{dV}{dt} = -V(V-0.3)(V-1) - w \)
• \( \frac{dw}{dt} = V - c w \)

Differential equations can be complex, but they offer invaluable insight into how systems behave dynamically. In our exercise, solving these equations directly provides the conditions for equilibrium points. For different values of \( c \), our task is to analyze these equations to determine stability and behavior of the system.

By substituting specific values of \( c \), such as 2 and 16, we effectively change the system's dynamics and discover how those changes affect the equilibrium points and system's overall stability.

Graphing Calculator

A graphing calculator is a powerful tool for visualizing mathematical problems and equations, especially when dealing with complex systems like the FitzHugh-Nagumo model.

By inputting the equations from the model into a graphing calculator:

• You can plot the trajectory of \( V \) and \( w \) over time.
• Visualize equilibrium points as points where trajectories end or remain constant.
• Witness changes in stability and behavior by varying \( c \).

For our exercise, using a graphing calculator helps in understanding how the model's trajectory behaves for different values of \( c \). It is particularly useful to confirm analytical solutions like the number of equilibrium points. By observing the graph, you can gain a better intuitive understanding of the dynamics and equilibria that the differential equations predict.

Graphing calculators thus make it easier to explore mathematical models and systems, offering visual insights that might be more challenging to grasp through equations alone.