Problem 15 - Calculus for Biology and Medicine

Question

Assume the following example of the Fitzhugh-Nagumo model: $$ \begin{array}{l} \frac{d V}{d t}=-V(V-1 / 2)(V-1)-w \\ \frac{d w}{d t}=V-c w \end{array} $$ Show that the model predicts multiple equilibria provided \(c \geq 16\). What happens if \(c<16 ?\)

Step-by-Step Solution

Answer

If \(c \geq 16\), there are multiple equilibria. With \(c < 16\), there is only one equilibrium at \(V = 0\).

Step 1: Define Equilibrium Conditions

To find the equilibria of the system, set the derivatives \(\frac{dV}{dt}\) and \(\frac{dw}{dt}\) to zero. This gives us the system of equations:\[-V(V - \frac{1}{2})(V - 1) - w = 0\] \[V - cw = 0\]

Step 2: Solve for Variables

From the second equation, solve for \(w\). This gives:\[w = \frac{V}{c}\] Substitute this expression of \(w\) into the first equation:\[-V(V-\frac{1}{2})(V-1) - \frac{V}{c} = 0\]

Step 3: Simplify and Factor

Simplify and factor the equation from Step 2 for \(V\):\[V^3 - \frac{3}{2}V^2 + V - \frac{V}{c} = 0\] This can be rewritten as:\[V^3 - \frac{3}{2}V^2 + (1 - \frac{1}{c})V = 0\]

Step 4: Analyze Equilibria for Different \(c\)

Factor out \(V\) from the equation in Step 3:\[V(V^2 - \frac{3}{2}V + (1 - \frac{1}{c})) = 0\]This gives equilibria at \(V = 0\) and the solutions of the quadratic:\[V^2 - \frac{3}{2}V + (1 - \frac{1}{c}) = 0\]

Step 5: Determine Nature of Roots

Using the discriminant \(\Delta = b^2 - 4ac\) of the quadratic equation, where \(a = 1\), \(b = -\frac{3}{2}\), and \(c = 1 - \frac{1}{c}\):\[\Delta = \left(-\frac{3}{2}\right)^2 - 4\times1\times\left(1 - \frac{1}{c}\right) = \frac{9}{4} - 4 + \frac{4}{c}\]When \(c \geq 16\), the discriminant \(\Delta \geq 0\), indicating two real roots for \(V\). If \(c < 16\), one root is complex since \(\Delta < 0\).

Step 6: Conclude on Equilibria Conditions

For \(c \geq 16\), there are three equilibria: \(V = 0\) and two additional roots. If \(c < 16\), only \(V = 0\) exists because the other roots are complex.

Key Concepts

Equilibrium Points

In the Fitzhugh-Nagumo model, equilibrium points are where the system is in a steady state, meaning the variables do not change over time. For this model, we set the time derivatives

• \(\frac{dV}{dt} = -V(V-\frac{1}{2})(V-1)-w = 0\)
• \(\frac{dw}{dt} = V - cw = 0\)

This system sets the stage to find the equilibrium conditions. From the second equation, we solve for \(w\) as \(w = \frac{V}{c}\) and substitute it into the first equation.
Now, the task simplifies to analyzing \(V\) when no change occurs over time. This results in the equation:

• \[V(V^2 - \frac{3}{2}V + (1 - \frac{1}{c})) = 0\]

This means one equilibrium is always at \(V = 0\). When the quadratic term equals zero, there are additional equilibria determined by solving \(V^2 - \frac{3}{2}V + (1 - \frac{1}{c}) = 0\). This exploration helps us understand where and how the system settles into equilibrium points.

Differential Equations

Differential equations are mathematical expressions defining how functions change. In the Fitzhugh-Nagumo model, we have two:

• \(\frac{dV}{dt} = -V(V-\frac{1}{2})(V-1)-w\)


• \(\frac{dw}{dt} = V - cw\)

These equations describe the dynamics of the voltage \(V\) and recovery variable \(w\) over time.
The first equation represents how \(V\) changes. It involves a cubic polynomial in \(V\) that captures the nonlinear behavior typical in neuron models. In simple terms, \(V\) evolves as a function of itself and \(w\).
The second equation indicates how \(w\) changes and adjusts based on \(V\) and itself via the parameter \(c\).
Together, they map out the trajectory or path the system will take, transitioning from one state to another according to these rates of change.

Bifurcation Analysis

Bifurcation analysis in this context examines how changing a parameter, like \(c\), affects the number of equilibrium points.
In the Fitzhugh-Nagumo model, the discriminant \(\Delta\) of the quadratic equation associated with \(V^2 - \frac{3}{2}V + (1 - \frac{1}{c}) = 0\) guides bifurcation. The discriminant is calculated as:

• \[\Delta = \left(-\frac{3}{2}\right)^2 - 4 \left(1 - \frac{1}{c}\right)\]

When \(c \geq 16\), \(\Delta \geq 0\). There are three real equilibria: \(V = 0\) and two more from the quadratic's roots.
However, if \(c < 16\), \(\Delta < 0\), leading to complex roots with only one real equilibrium at \(V = 0\).
This analysis tells us how the system's structure changes: when \(c\) varies, the nature of solutions changes from having multiple equilibrium points to only one, signaling a bifurcation.
Understanding this shift helps predict changes in behavior or stability in dynamic systems as parameters evolve.