Problem 40 - Calculus for Biology and Medicine

Question

You should treat \(h\) as a constant. For what values of \(h\) (if any) does each equation have equilibria? Use a graphical argument to show which of the equilibria (if any) are stable. $$ \frac{d y}{d t}=y-h $$

Step-by-Step Solution

Answer

The equation has an equilibrium at \(y = h\), but it is unstable for any value of \(h\).

Step 1: Understand the differential equation

The differential equation given is \(\frac{d y}{d t} = y - h\). Here, \(y\) is a function of \(t\), and \(h\) is a constant. We are tasked with finding values of \(h\) for which the equation has equilibria. Equilibria occur where the rate of change \(\frac{d y}{d t}\) is zero.

Step 2: Set the equation to zero to find equilibria

To find equilibria, set \(\frac{d y}{d t} = y - h = 0\). Solving for \(y\), we get \(y = h\). Thus, the equilibrium value occurs where \(y = h\).

Step 3: Determine the stability of the equilibrium

To determine the stability, consider small perturbations around the equilibrium point \(y = h\). Suppose \(y = h + \epsilon\), where \(\epsilon\) is a small disturbance. The differential equation becomes \(\frac{d(h + \epsilon)}{d t} = (h + \epsilon) - h = \epsilon\).

Step 4: Analyze \(\epsilon\), the perturbation

\(\frac{d \epsilon}{d t} = \epsilon\) implies that \(\epsilon\) grows or shrinks exponentially with time, depending on its initial sign. If \(\epsilon\) is positive, \(\epsilon\) increases with time, leading \(y\) away from the equilibrium \(h\). Similarly, if \(\epsilon\) is negative, \(\epsilon\) decreases (more negative), also moving away from equilibrium.

Step 5: Conclusion on stability

Since perturbations \(\epsilon\) grow over time regardless of their initial sign, the equilibrium \(y = h\) is unstable. Any initial perturbation will cause \(y\) to diverge from the equilibrium rather than return to it.

Key Concepts

Equilibrium in Differential Equations

Equilibrium points in differential equations are values where the rate of change of the system is zero. This means that the system doesn't evolve over time when it is at these points. In our given differential equation \( \frac{dy}{dt} = y - h \), equilibrium occurs when \( y - h = 0 \).
Thus, the equilibrium for this equation is when \( y = h \).

• Equilibrium conditions ensure that the system maintains a constant state over time.
• No change occurs in the system at equilibrium — it is at rest.

In the context of this problem, finding equilibrium is about determining when the solution of the differential equation remains constant.

Stability Analysis of Equilibrium

Stability analysis helps us understand how a system behaves when it is disturbed slightly. For the equilibrium point \( y = h \) in our equation, we need to assess this. We introduce a small perturbation \( \epsilon \), so \( y = h + \epsilon \).
This yields \( \frac{d(h + \epsilon)}{dt} = \epsilon \), explaining how the disturbance \( \epsilon \) evolves.

• If \( \epsilon > 0 \), it indicates the system is away from \( h \) and continuing further away.
• If \( \epsilon < 0 \), similarly, the system moves away from \( h \) consistently.

Thus, regardless of the sign of \( \epsilon \), it expands over time, demonstrating that the equilibrium is not stable.

Graphical Argument for Stability

By visualizing the differential equation \( \frac{dy}{dt} = y - h \), we can gain insights into stability.
Graphically, plot \( \frac{dy}{dt} \) versus \( y \). The line \( y - h = 0 \) intersects the horizontal axis at \( y = h \).

• Above this line, \( \frac{dy}{dt} > 0 \), indicating \( y \) increases.
• Below this line, \( \frac{dy}{dt} < 0 \), suggesting \( y \) decreases.

This visual approach confirms our earlier analysis: any deviation from \( y = h \) results in \( y \) moving further away, confirming the equilibrium is unstable.

Perturbation Analysis

Perturbation analysis is a technique used to examine the behavior of a system near its equilibrium by introducing small changes. Here, we added a small disturbance \( \epsilon \) to \( y = h \).
This gave us \( \frac{dy}{dt} = \epsilon \), illustrating the response of the system to perturbations.

• This differential equation \( \frac{d\epsilon}{dt} = \epsilon \) means the disturbance grows exponentially, whether positive or negative.
• This growth reveals that any disturbance will cause the system to leave the equilibrium.

From a perturbation perspective, the behavior supports that the equilibrium \( y = h \) of the system is unstable.