Problem 25 - Biocalculus Calculus for the Life Sciences

Question

$$x^{\prime}=a x^{2}+a y-x, \quad y^{\prime}=x-y, \quad a \neq 0$$

Step-by-Step Solution

Answer

Equilibrium at (0, 0) is stable for \(0 < a < 1\); unstable if \(a > 1\).

Step 1: Identify the system of differential equations

We are given two differential equations: \(x' = ax^2 + ay - x\) and \(y' = x - y\). Here, \(x'\) and \(y'\) denote the derivatives of \(x\) and \(y\) with respect to some independent variable, usually time (\(t\)). \(a\) is a constant parameter, and we are analyzing the system when \(a eq 0\).

Step 2: Determine the equilibrium points

To find equilibrium points, set \(x' = 0\) and \(y' = 0\). So, we solve the system:1. \(ax^2 + ay - x = 0\)2. \(x - y = 0\)From the second equation, we find \(x = y\). Substitute \(y = x\) in the first equation to get \(ax^2 + ax - x = 0\), which simplifies to \(x(ax + a - 1) = 0\). This yields the solutions \(x = 0\) or \(ax + a - 1 = 0\). Solve the second part to find \(x = \frac{1-a}{a}\), given \(a eq 0\). Thus, the equilibrium points are \((0, 0)\) and \(\left(\frac{1-a}{a}, \frac{1-a}{a}\right)\).

Step 3: Analyze the stability of equilibrium points

The stability can be determined by constructing the Jacobian matrix of the system at the equilibrium points. The Jacobian matrix, \(J\), of the given system is:\[ \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \end{bmatrix} = \begin{bmatrix} 2ax - 1 & a \ 1 & -1 \end{bmatrix} \]Analyze \(J\) at each equilibrium:1. At \((0,0)\), \(J = \begin{bmatrix} -1 & a \ 1 & -1 \end{bmatrix}\)2. At \(\left(\frac{1-a}{a}, \frac{1-a}{a}\right)\), calculate the Jacobian values using the equilibrium coordinates to determine stability.

Step 4: Determine eigenvalues for stability

For stability analysis at an equilibrium, find the eigenvalues of the Jacobian:- For \((0,0)\): \[ \text{det}(J - \lambda I) = \begin{vmatrix} -1 - \lambda & a \ 1 & -1 - \lambda \end{vmatrix} = (-1-\lambda)^2 - a = 0 \] Solving gives the eigenvalues: \(\lambda_{1,2} = -1 \pm \sqrt{a}\). - If \(a > 1\), one eigenvalue is positive (unstable saddle point when \(a > 0\)). - If \(0 < a < 1\), both eigenvalues are negative (stable node when \(a > 0\)).

Step 5: Short Answer for stability

For the point \((0, 0)\):- If \(0 < a < 1\), it is a stable node.- If \(a > 1\), it is an unstable saddle point.

Key Concepts

Eigenvalues in Differential Equations

Eigenvalues are crucial in understanding how solutions to a differential equation behave over time. When analyzing stability, eigenvalues give insight into how small perturbations around an equilibrium point react. In a nutshell, eigenvalues depict the tendencies of solutions to either converge or diverge. You might wonder why we calculate eigenvalues in the first place. When we linearize our system of equations using a Jacobian matrix, the eigenvalues of this matrix help us determine the behavior around equilibrium points. If all eigenvalues have negative real parts, any disturbance from equilibrium will fade away, suggesting stability. Conversely, a positive real part indicates an increase in the perturbation, implying instability. Here's a quick checklist:

• Negative Real Parts: Leading towards stability (solutions converge).
• Positive Real Parts: Indicating instability (solutions diverge).
• Zero Eigenvalue: Can signify marginal stability, where further investigation is needed.

Remember, eigenvalues are a tool for analyzing how linear approximations of our system behave in the vicinity of equilibrium points. Keep an eye on those real parts!

Finding Equilibrium Points

Equilibrium points in a system of differential equations hold a special significance, as they represent the states where the system doesn’t change over time. These are the solutions to the system obtained by setting the derivatives to zero.For our exercise, we found equilibrium points by solving:

• Setting derivatives equal to zero: This simplifies the dynamic equations into algebraic ones.
• Identifying relations among variables: Such as, in our case, where we identified that \(x = y\).
• Solving the resulting equations: As seen in finding solutions like \((0, 0)\) and \(\left(\frac{1-a}{a}, \frac{1-a}{a}\right)\).

Equilibrium points are essentially the 'resting states' of the system, where the system remains constant unless disturbed. They form the basis for understanding the local behavior of the system, helping us understand how solutions behave from various initial conditions.

Stability Analysis of Equilibrium Points

Stability analysis involves determining how a system behaves near its equilibrium points. It's like checking if a calm lake will remain calm or become turbulent when we toss in a stone.For our system, we use the Jacobian matrix to inspect local stability:The process includes:

• Constructing the Jacobian: Derived from partial derivatives of the system’s equations.
• Evaluating the Jacobian at equilibrium points: For example, \((0, 0)\) in our case.
• Determining Eigenvalues: As seen in our exercise, negative values indicate stability.

Stability is classified into several types:

• Stable Node: Where solutions tend to return to equilibrium.
• Unstable Saddle Point: Where solutions tend to diverge from equilibrium.
• Marginal Stability: Sometimes needing deeper analysis if eigenvalues are zero.

These analyses are foundational for predicting the behavior of dynamic systems over time, offering insights into whether solutions return to equilibrium or diverge.