Problem 88 - Calculus for Biology and Medicine

Question

Find all equilibria, and, by calculating the eigenvalue of the differential equation, determine which equilibria are stable and which are unstable. \(d x / d t=h x-x^{3}\), where \(h\) is a constant and (a) \(h>0\), (b) \(h<0\)

Step-by-Step Solution

Answer

For (a) \( h > 0 \), equilibria are \( x = 0 \) (unstable) and \( x = \pm \sqrt{h} \) (stable). For (b) \( h < 0 \), equilibrium is \( x = 0 \) (stable).

Step 1: Finding Equilibria

To find the equilibria, set the right-hand side of the differential equation to zero: \[ hx - x^3 = 0 \] This can be factored as: \[ x(h - x^2) = 0 \] Therefore, the equilibria are \( x = 0 \) and \( x = \pm \sqrt{h} \) if \( h > 0 \). If \( h < 0 \), then \( x = 0 \) is the only real equilibrium.

Step 2: Calculating the Jacobian

The Jacobian of \( dx/dt = hx - x^3 \) is found by taking the derivative with respect to \( x \): \[ J = h - 3x^2 \]

Step 3: Evaluating Eigenvalues for Equilibrium Points (a)

For \( h > 0 \): 1. At \( x = 0 \), \( J = h \). Thus, \( J > 0 \), making \( x = 0 \) unstable. 2. At \( x = \pm \sqrt{h} \), \( J = h - 3h = -2h \), so \( J < 0 \). Therefore, \( x = \pm \sqrt{h} \) are stable.

Step 4: Evaluating Eigenvalues for Equilibrium Points (b)

For \( h < 0 \): At \( x = 0 \), \( J = h \). Since \( J < 0 \) for \( h < 0 \), \( x = 0 \) is stable.

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In this context, we are looking at a simple first-order differential equation given by: \( \frac{d x}{d t} = hx - x^3 \). This equation examines the rate of change of a variable \( x \) with respect to time \( t \). Here, \( h \) is a constant. The structure means that the rate of change of \( x \) depends both linearly and non-linearly on \( x \) itself.

These equations are powerful tools in modeling various dynamic systems. They help us understand how a system evolves over time.

• Linear terms like \( hx \) provide proportional growth or decay.
• Non-linear terms like \( x^3 \) introduce more complex behavior such as saturation or feedback.

Solutions to these equations are functions or sets of functions describing the system's behavior over time. Identifying equilibrium in such systems involves solving the differential equation by setting its derivative to zero.

Stability Analysis

Stability analysis involves understanding how a system responds to small disturbances around its equilibrium points. By analyzing stability, we determine whether a slight deviation from equilibrium will return to equilibrium or diverge further away.

To assess stability, we consider the Jacobian matrix of the system, which gives us insight into how changes in state affect changes in time. The eigenvalues of the Jacobian play an important role.

• If an eigenvalue has a positive real part, the equilibrium is unstable, meaning disturbances grow over time.
• If all eigenvalues have negative real parts, the system returns to equilibrium and is stable.

This analysis helps predict long-term system behavior and is crucial in fields like engineering and ecology, where stability signifies system reliability.

Eigenvalues

Eigenvalues are a key concept in linear algebra, crucial for stability analysis in differential equations. When dealing with the Jacobian matrix of a system, eigenvalues indicate the nature of equilibrium behavior.

For the differential equation \( \frac{d x}{d t} = hx - x^3 \), the eigenvalues stem from the Jacobian, \( J = h - 3x^2 \). By evaluating these eigenvalues at different equilibrium points, we can infer whether the system is stable or not.

Specifically:

• If an eigenvalue is positive, small perturbations will grow, indicating instability.
• If an eigenvalue is negative, disturbances will decay, showing stability.

Hence, computing eigenvalues is essential for determining equilibrium stability.

Equilibrium Points

Equilibrium points are where the system’s state does not change over time. These occur where the derivative of the state is zero. For the equation \( \frac{d x}{d t} = hx - x^3 \):

• Set \( hx - x^3 = 0 \) to find equilibria.
• This gives \( x(h - x^2) = 0 \), leading to equilibria at \( x = 0 \), and for \( h > 0 \), \( x = \pm \sqrt{h} \).

In the case where \( h < 0 \), \( x = 0 \) is the only real equilibrium. By examining these points, we understand where the system naturally tends to settle. Analyzing such points helps predict how real-world systems might behave over time, ensuring stability or signaling potential changes that can be vital for strategic planning in dynamic systems.