Problem 14 - Calculus for Biology and Medicine

Question

Use a graphing calculator to study the following example of the Fitzhugh- Nagumo model: $$ \begin{array}{l} \frac{d V}{d t}=-V(V-0.6)(V-1)-w \\ \frac{d w}{d t}=V-c w \end{array} $$ Explain whether or not the model predicts multiple equilibria for the following values of \(c\) : (a) \(c=8\), (b) \(c=20\), (c) \(c=50\).

Step-by-Step Solution

Answer

(a) 3 equilibria, (b) 3 equilibria, (c) 1 or 2 equilibria.

Step 1: Understand Equilibrium Points

Equilibrium points occur where the derivatives \( \frac{dV}{dt} \) and \( \frac{dw}{dt} \) are both equal to zero. This means we need to solve the equations: 1. \(-V(V-0.6)(V-1) - w = 0\)2. \(V - cw = 0\).Solving these will give us the potential equilibria.

Step 2: Solve for \( w \) in terms of \( V \)

From equation (2), solve for \( w \): \[ w = \frac{V}{c} \] Substitute this expression for \( w \) into equation (1) to express it entirely in terms of \( V \).

Step 3: Substitute and Simplify Equation (1)

Substitute \( w = \frac{V}{c} \) into the first equation:\[-V(V-0.6)(V-1) - \frac{V}{c} = 0\]Simplify the equation:\[-V(V-0.6)(V-1) = \frac{V}{c}\]Factor out \( V \) from both sides (assuming \( V eq 0 \)), then rearrange the equation.

Step 4: Analyze the Polynomial for Roots

Continue simplifying:\[-c(V-0.6)(V-1) = 1\]This implies that we need to find the roots of \((V-0.6)(V-1) + \frac{1}{c} = 0\). The number of real solutions (roots) will give the number of equilibria.

Step 5: Apply Specific Values for \( c \)

For different values of \( c \), evaluate the number of solutions:- **(a) \(c = 8\):** Solve \( (V-0.6)(V-1) = -\frac{1}{8} \). This is a quadratic equation, determine the number of real roots.- **(b) \(c = 20\):** Solve \( (V-0.6)(V-1) = -\frac{1}{20} \). Check for real roots.- **(c) \(c = 50\):** Solve \( (V-0.6)(V-1) = -\frac{1}{50} \).Graph these equations if needed, to confirm the presence of solutions visually.

Step 6: Conclude on Number of Equilibria

Each \( c \) value yields a different equation; we observe the number of real solutions they produce:- For \( c = 8 \), there are typically 3 equilibria.- For \( c = 20 \), it still predicts 3 equilibria.- For \( c = 50 \), we might find only 1 or 2 equilibria due to reduced slope steepness, as larger \( c \) values might diminish oscillatory behavior according to the FitzHugh-Nagumo model.

Key Concepts

Differential Equations

Differential equations are mathematical expressions involving functions and their derivatives. In simple terms, they describe the rate of change of a quantity over time. In the context of the FitzHugh-Nagumo model, we deal with a system of two differential equations. These represent changes in variables related to nerve impulses: \

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• \( \frac{dV}{dt} = -V(V-0.6)(V-1) - w \) describes the rate of change of the voltage \( V \). It is a cubic polynomial, indicating it has several turning points.
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• \( \frac{dw}{dt} = V - cw \) describes the rate of change of the recovery variable \( w \).
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The FitzHugh-Nagumo model is important in neuroscience, specifically in understanding how nerve cells generate signals. It simplifies the Hodgkin-Huxley model and allows us to study excitability in a mathematically manageable way.

Equilibrium Points

Equilibrium points in a system of differential equations occur when variables no longer change with time. For a system to be at equilibrium, the derivatives of all the variables must be zero. In our FitzHugh-Nagumo system, this means both \( \frac{dV}{dt} \) and \( \frac{dw}{dt} \) must equal zero simultaneously. \
To find these points, we solve the equations: \

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• \(-V(V-0.6)(V-1) - w = 0 \)
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• \(V - cw = 0 \)
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By substituting \( w = \frac{V}{c} \) into our first equation, we reduce the problem to finding the roots of a single variable polynomial in \( V \). These roots tell us the possible values for \( V \) at equilibrium. Each root corresponds to a potential steady state for the system.

Polynomial Roots

Finding polynomial roots is crucial in determining solutions to the equilibrium condition in the FitzHugh-Nagumo model. The reduced equation, after substituting \( w \), is \(-c(V-0.6)(V-1) = 1\). By solving the resulting polynomial equation, we determine equilibrium points. \
But what are polynomial roots? They are the values of \( V \) that make the polynomial equal zero. In our adjusted equation, these values change with different constants, \( c \), affecting system behavior. The roots give information about the nature and number of equilibria. A root indicates a point where changes in \( V \) stop, influencing system dynamics profoundly. Understanding these roots is key in predicting multiple equilibria and their behavior.

Mathematical Modeling

Mathematical modeling is a method of representing real-world phenomena with mathematical formulas. The FitzHugh-Nagumo model is an example of using mathematical modeling to understand biological systems. \
This model simplifies complex biological processes into manageable equations that capture essential dynamics of neuron activity. By studying a pair of differential equations, the model allows us to predict how nerve cells will respond under different conditions, such as varying the parameter \( c \). \
Modeling helps in: \

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• Understanding fundamental mechanisms in nature.
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• Allowing for predictions of system behavior under varying parameters.
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• Providing insights that guide experimental and theoretical biology.
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\By using mathematical models like FitzHugh-Nagumo, researchers can explore hypotheses and refine theories about biological systems efficiently and effectively.