Problem 10 - Calculus: Early Transcendentals

Question

The Fitzhugh-Nagumo model for the electrical impulse in a neuron states that, in the absence of relaxation effects, the electrical potential in a neuron \( v(t) \) obeys the differential equation \( \frac {dv}{dt} = - v [v^2 - (1 + a) v + a] \) where \( a \) is a positive constant such that \( 0 < a < 1\. \) (a) For what values of \( v \) is \( v \) unchanging (that is, \( dv/dt = 0)?. \) (b) For what values of \( v \) is \( v \) increasing? (c) For what values of \( v \) is \( v \) decreasing?

Step-by-Step Solution

Answer

(a) \( v = 0, a, 1 \); (b) \( 0 < v < a \); (c) \( v < 0 \) or \( v > 1 \).

Step 1: Set Up Equation for Equilibrium

To determine where the electrical potential \( v(t) \) is unchanging, we need to set \( \frac{dv}{dt} \) equal to zero. This gives us the equation: \\[-v \left(v^2 - (1 + a)v + a \right) = 0.\] \This expression factors into two parts: a \( v \) term and a quadratic polynomial.

Step 2: Solve for Equilibrium Points

Since \( v \times (v^2 - (1+a)v + a) = 0 \), there are two cases to consider: \1. \( v = 0 \) \2. \( v^2 - (1+a)v + a = 0 \), which is a quadratic equation.

Step 3: Solve the Quadratic Equation

The quadratic equation \( v^2 - (1+a)v + a = 0 \) is solved using the quadratic formula: \[ v = \frac{-(b) \pm \sqrt{b^2 - 4ac}}{2a}, \] where \( a=1, b=-(1+a), c=a \). Substituting these values, \[ v = \frac{(1+a) \pm \sqrt{(1+a)^2 - 4 \cdot 1 \cdot a}}{2} \] \[ = \frac{(1+a) \pm \sqrt{1 + 2a + a^2 - 4a}}{2} \] \[ = \frac{(1+a) \pm \sqrt{1 - 2a + a^2}}{2}. \] This simplifies to \[ = \frac{(1+a) \pm (1-a)}{2}. \] This results in two solutions: \[ v = a \text{ and } v = 1.\]

Step 4: Determine Values Where \( v \) is Increasing

The potential \( v(t) \) is increasing where \( \frac{dv}{dt} > 0 \). Since \( \frac{dv}{dt} = -v(v^2 - (1+a)v + a) \), we examine the sign of this expression. \- \( v > 0 \) and \( v^2 - (1+a)v + a < 0 \): This occurs between the roots of the quadratic equation, hence for \( 0 < v < a \). - \( v < 0 \) creates a positive sign, but the quadratic should also be positive for this combination to make a positive derivative.

Step 5: Determine Values Where \( v \) is Decreasing

The potential \( v(t) \) is decreasing where \( \frac{dv}{dt} < 0 \). The function \( -v(v^2 - (1 + a)v + a) < 0 \) can have: \- \( v > 0 \) with \( v^2 - (1+a)v + a > 0 \) which occurs for \( v > 1 \). - \( v < 0 \) with \( v^2 - (1+a) v + a < 0 \), but negative quadratic roots are already considered and result in an overall negative outcome.

Key Concepts

Fitzhugh-Nagumo model

The Fitzhugh-Nagumo model is a mathematical model that simplifies how electrical impulses propagate through neurons. It is a simplification of the more complex Hodgkin-Huxley model, which describes how action potentials in neurons are initiated and propagated. By capturing only the essential dynamics, the Fitzhugh-Nagumo model allows for easier analysis and understanding of neuronal behavior.
In this model, the electrical potential, denoted as \( v(t) \), changes over time based on a differential equation that includes both linear and nonlinear components. The core idea is to capture the oscillatory behavior of the neuron as it transmits signals, balancing between excitation and recovery processes.

• Excitation describes the rapid increase in potential when a neuron sends a signal.
• Recovery corresponds to the reset period, allowing the neuron to be ready for the next signal.

The equation considered here simplifies to focus on the electrical signal without accounting for relaxation effects, giving insights into how neurons behave under certain conditions.

Quadratic equation

In many mathematical models, including the Fitzhugh-Nagumo model, we encounter situations where we need to solve a quadratic equation. A quadratic equation takes the general form \( ax^2 + bx + c = 0 \). The solutions to this equation can be found using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]
This formula provides the conditions under which a quadratic equation will have two, one, or no real solutions, which is determined by the discriminant \( b^2 - 4ac \):

• If the discriminant is positive, two distinct real solutions exist.
• If it is zero, there is exactly one real solution.
• If negative, the solutions are complex or imaginary.

In our current problem, solving the quadratic equation helps us find critical values of \( v \) where the rate of change of electric potential is zero, thereby identifying the equilibrium points.

Electrical potential in neurons

The electrical potential in neurons, represented by \( v(t) \), is vital for understanding how signals travel within the nervous system. This potential represents the difference in electrical charge between the inside and outside of a neuron, creating a voltage across the neuronal membrane.
When a neuron is stimulated sufficiently, it generates an action potential—a rapid spike in electrical potential that travels along the axon, transmitting information throughout the nervous system.

• The model considers the differential equation \( \frac{dv}{dt} = -v[v^2 - (1 + a)v + a] \), capturing the dynamics of this potential over time.
• The result provides insight into when the potential is constant, increasing, or decreasing, which is essential for comprehending neuronal communication and response to stimuli.

This equation helps in predicting how the electrical potential will behave under different values of \( v \), indicating whether a neuron is likely to be sending a signal or resting.

Equilibrium points

Equilibrium points in a dynamic system are values where the system remains constant over time—in this case, where the electrical potential \( v(t) \) does not change. Finding equilibrium points involves setting the rate of change to zero, thus solving \( \frac{dv}{dt} = 0 \).
Active equilibrium points indicate where the system can temporarily remain without outside forces altering it.

• When studied in the context of our differential equation, setting \( -v[v^2 - (1+a)v + a] = 0 \) leads to equilibrium conditions.
• These correspond to the solutions of quadratic equations, specifically where \( v = 0 \), \( v = a \), and \( v = 1 \).

Understanding these equilibrium points allows researchers and students alike to predict neuron behavior during normal functions or under specific conditions that may lead to constant potential scenarios.