Problem 2 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Suppose \(y(t)\) solves $$ y^{\prime}=k y(t)(1-y(t)) \quad \text { and } \quad z(t)=M y(k t) $$ Show that \(z(t)\) solves $$ z^{\prime}(t)=k z(t)\left(1-\frac{z(t)}{M}\right) $$

Step-by-Step Solution

Answer

Yes, \(z(t)\) satisfies \(z'(t) = k z(t)(1 - \frac{z(t)}{M})\).

Step 1: Understand the Function and Transformation

We have the differential equation \( y'(t) = k y(t)(1-y(t)) \) and a transformation given by \( z(t) = M y(kt) \). Our goal is to show that \( z(t) \) satisfies \( z'(t) = k z(t) \left(1 - \frac{z(t)}{M}\right) \).

Step 2: Differentiate z(t) with Respect to t

We need to compute \( z'(t) \). Start by applying the chain rule to \( z(t) = M y(kt) \). The differentiation is \( z'(t) = M \cdot \frac{d}{dt}[y(kt)] = M \cdot y'(kt) \cdot \frac{d}{dt}(kt) = M \cdot y'(kt) \cdot k \).

Step 3: Substitute y'(kt) using the Original Differential Equation

From the differential equation, we have \( y'(t) = k y(t)(1-y(t)) \). Substitute \( t \) with \( kt \): \( y'(kt) = k y(kt)(1-y(kt)) \). Substitute this into the expression for \( z'(t) \): \( z'(t) = M \cdot (k y(kt)(1-y(kt))) \cdot k = k M y(kt)(1-y(kt)) \).

Step 4: Express z(t) in Terms of y(kt)

From the transformation \( z(t) = M y(kt) \), solve for \( y(kt) \): \( y(kt) = \frac{z(t)}{M} \). Substitute it back: \( z'(t) = k M \left( \frac{z(t)}{M} \right) \left(1 - \frac{z(t)}{M}\right) \).

Step 5: Simplify the Expression for z'(t)

Simplify the expression for \( z'(t) \): \( z'(t) = k \cdot z(t) \cdot \left(1 - \frac{z(t)}{M}\right) \). This matches the desired form \( z'(t) = k z(t) \left(1 - \frac{z(t)}{M}\right) \). The equality shows that \( z(t) \) is a solution to the given differential equation.

Key Concepts

chain rule

The chain rule is a fundamental concept in calculus that facilitates the differentiation of composite functions. The process involves differentiating the outer function while multiplying it by the derivative of the inner function.
For instance, when given a composite function like \( z(t) = M y(kt) \), to find \( z'(t) \), the chain rule is applied:

• First, differentiate \( y(kt) \) as the inner function concerning \( kt \).
• Then, multiply the result by the derivative of \( kt \), which is simply \( k \).

This sequence reveals:
\[ z'(t) = M rac{d}{dt} y(kt) = M imes y'(kt) imes k \]
Overall, the chain rule efficiently resolves the differentiation of nested or multi-layer functions.

solution transformation

Solution transformation involves converting a differential equation to another form that can be more easily analyzed or solved using different conditions or variables.
In our example, the transformation involves taking the solution \( y(t) \) of a logistic equation and converting it to \( z(t) = M y(kt) \).

• This transformation uses a scaling factor \( M \) applied to \( y(kt) \), which changes the function's amplitude.
• The transformation is typically motivated by a need to match specific initial or boundary conditions in problems.

The elegance of the transformation lies in maintaining the equation's core behavior while placing it in a more convenient form for our analysis or application, such as substituting \( kt \) in place of \( t \).

logistic growth model

The logistic growth model is a powerful tool used extensively in modeling population dynamics and other scenarios where growth is naturally limited by resources.
Consider the equation \( y'(t) = k y(t)(1 - y(t)) \):

• It describes growth where an initial exponential phase slows down as it approaches a carrying capacity, here implicitly considered 1.
• The term \( k y(t)(1 - y(t)) \) embodies this behavior, causing the growth rate to decrease as \( y(t) \) nears its limit.

This type of growth ensures an S-shaped curve (sigmoid) over time, reflecting an initial rapid increase, a decrease in growth, and eventually stabilization. The model is a cornerstone in ecological and biomedical studies, providing insights into stable populations.

substitution method

Substitution is a valuable technique in solving differential equations, involving replacing variables or expressions to simplify the equation.
In our exercise, the function \( y(t) \) is represented by \( z(t) = M y(kt) \), as shown:

• First, express \( y(kt) \) in terms of \( z(t) \): \( y(kt) = \frac{z(t)}{M} \).
• Next, substitute this expression into the differential equation derived from the original form.

This results in:
\[ z'(t) = kM \left( \frac{z(t)}{M} \right) \left( 1 - \frac{z(t)}{M} \right) = k z(t) \left( 1 - \frac{z(t)}{M} \right) \]
The substitution method streamlines solving complex equations by reducing them to simpler forms that can be directly interpreted or solved. By converting variables, the method often reveals solutions or properties that may otherwise remain obscured.