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Chapter 8: Problem 29

Let \(y(t)\) be the proportion of crystallizable fat in a sample after \(t\) hours. Then \(y\) satisfies the differential equation $$y^{\prime}=k\left(y^{n}-y\right).$$ where \(k\) is a constant and \(n\) is the Avrami exponent for decrystallization reactions. In practice, \(n\) is an integer greater than 1 that is computed from the time dependence of nucleation and the number of dimensions in which crystal growth occurs. \({ }^{15}\) Use this differential equation to solve Exercises \(29-32\). Show that \(y=0\) is an asymptotically stable equilibrium point.

Short Answer

Expert verified
The equilibrium point \( y = 0 \) is asymptotically stable.

Step by step solution

01

Identify the differential equation

The given differential equation is \[ y^{\text{'} }= k(y^{n}- y) \] where \( y(t) \) is the proportion of crystallizable fat and \( k \) is a constant, and \( n \) is the Avrami exponent.
02

Find the equilibrium points

Equilibrium points occur where \[ y' = 0 \] For the given differential equation: \[ k(y^n - y) = 0 \] or \[ y^n - y = 0 \] Factoring out \( y \): \[ y(y^{n-1} - 1) = 0 \] This gives the equilibrium points: \[ y = 0 \] or \[ y^{n-1} = 1 \] Since \( n > 1 \), the second equilibrium point is: \[ y = 1 \]
03

Analyze the stability of the equilibrium points

To analyze stability, examine the sign of \( y' \) around the equilibrium points. \[ y' = k(y^n - y) \] For small \( y > 0 \): \[ y' = k(y^n - y) \] Since \( y < 1 \) and \( n > 1 \), \[ y^n < y \] so \[ y^n - y < 0 \] Hence, \[ y' < 0 \] indicating that \( y = 0 \) is asymptotically stable.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
Differential equations are mathematical equations that involve functions and their derivatives. They describe how a particular quantity changes over time or space. In this problem, the differential equation represents how the proportion of crystallizable fat, denoted as \(y(t)\), changes over time in a sample.
The given differential equation is \( y' = k(y^n - y) \), where:
  • \( y(t) \) is the proportion of crystallizable fat after time \( t \).
  • \( k \) is a constant that affects how fast the reaction occurs.
  • \( n \) is the Avrami exponent, related to the kinetics of decrystallization.
Understanding differential equations helps us model real-world phenomena, from physics and engineering to biology and economics.
Equilibrium Points
Equilibrium points in differential equations are values where the function remains constant over time. For a differential equation \( y'=f(y) \), an equilibrium point is a value \( y_e \) such that \( f(y_e)=0 \).
In our problem, to find equilibrium points, we set \( y' = k(y^n - y) = 0 \).
This simplifies to finding the roots of the equation \( y^n - y = 0 \). Factoring gives us:
  • \( y(y^{n-1} - 1) = 0 \)
  • The solutions to this are \( y = 0 \) and \( y = 1 \) since \( n > 1 \).
These equilibrium points tell us where the proportion of crystallizable fat does not change over time.
Stability Analysis
Stability analysis determines whether equilibrium points are stable or unstable. An equilibrium point is asymptotically stable if, when displaced slightly, the system returns to equilibrium.
For the differential equation \( y' = k(y^n - y) \), we analyzed the stability of the equilibrium points by looking at the behavior of \( y' \) near \( y = 0 \) and \( y = 1 \).
For small \( y > 0 \):
  • When \( y < 1 \) and \( n > 1 \), we have \( y^n < y \), so \( y^n - y < 0 \).
  • This means \( y' < 0 \), indicating that the proportion \( y \) decreases when slightly positive.
  • Thus, \( y = 0 \) is asymptotically stable because the system returns to \( y = 0 \) if displaced slightly.
Understanding stability helps predict the long-term behavior of systems modeled by differential equations.
Avrami Exponent
The Avrami exponent, denoted as \( n \), is a parameter in the context of phase transformation kinetics, such as crystallization and decrystallization. It is significant in determining the nature of these processes.
\( n \) is computed based on:
  • The time dependence of nucleation.
  • The number of dimensions in which crystal growth occurs.
In the given differential equation, \( y' = k(y^n - y) \), the Avrami exponent \( n \) shapes the function's behavior by affecting how the proportion of crystallizable fat changes over time.
A higher value of \( n \) typically indicates more complex behaviors in the crystallization process, highlighting the critical role of the exponent in understanding phase transformations in material science.

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