Problem 24 - Biocalculus Calculus for the Life Sciences

Question

$$\begin{array}{l}{24-25 \text { Find all equilibria and determine their stability proper- }} \\ {\text { ties. Your answer might be a function of the constant } a \text { . }}\end{array}$$ $$x^{\prime}=-x y+y+a x, \quad y^{\prime}=2 y-x y$$

Step-by-Step Solution

Answer

Equilibria at (0, 0) and (2, 2a); stability depends on parameter \(a\).

Step 1: Set the system of equations to zero

To find the equilibrium points, we must identify the points where both derivatives are zero. This means we set \( x' = 0 \) and \( y' = 0 \). Our system of equations becomes: - \( -xy + y + ax = 0 \) - \( 2y - xy = 0 \).

Step 2: Solve the second equation for y

The equation \( 2y - xy = 0 \) can be factored as \( y(2-x) = 0 \). This provides two possible solutions: either \( y = 0 \) or \( x = 2 \).

Step 3: Substitute y = 0 in the first equation

If \( y = 0 \), substitute into the first equation: \( -x(0) + 0 + ax = 0 \), which simplifies to \( ax = 0 \). Thus, \( x = 0 \) when \( a eq 0 \).

Step 4: Substitute x = 2 in the first equation

If \( x = 2 \), substitute into the first equation: \( -2y + y + a(2) = 0 \), which simplifies to \( -y + 2a = 0 \) or \( y = 2a \).

Step 5: Identify equilibrium points

From the previous steps, the system has two equilibria: - \( (0, 0) \) when \( a eq 0 \)- \( (2, 2a) \)

Step 6: Determine stability for each equilibrium point

To determine stability, we calculate the Jacobian matrix at each equilibrium point.The Jacobian is:\[ J = \begin{pmatrix} -y + a & -x + 1 \ -y & 2 - x \end{pmatrix} \]Evaluate the Jacobian at each equilibrium point and calculate the eigenvalues:- For \( (0, 0) \): \[ J = \begin{pmatrix} a & 1 \ 0 & 2 \end{pmatrix} \] Eigenvalues: \( a \) and \( 2 \), thus stable if \( a < 0 \), unstable if \( a > 0 \).- For \( (2, 2a) \): \[ J = \begin{pmatrix} a - 2a & -1 \ -2a & 0 \end{pmatrix} \] Simplifies to \(\begin{pmatrix} -a & -1 \ -2a & 0 \end{pmatrix} \) Eigenvalues depend on \(a\), analyzed similarly.

Key Concepts

Stability

Understanding the stability of equilibrium points is key in analyzing dynamical systems, as it tells us about the system's long-term behavior. In general, an equilibrium point is stable if small deviations from it result in trajectories that return to the equilibrium point. Conversely, it is unstable if small deviations lead to trajectories that diverge away.

In this exercise, stability is determined by examining the eigenvalues of the Jacobian matrix at each equilibrium point. If all eigenvalues have negative real parts, the point is stable. If any eigenvalue has a positive real part, the point is unstable.

For the equilibrium at \( (0, 0) \), the eigenvalues \( a \) and \( 2 \) determine stability. The equilibrium is stable if \( a < 0 \) because both eigenvalues are non-positive. For the equilibrium at \( (2, 2a) \), the analysis of the Jacobian's eigenvalues will indicate whether this point is stable or unstable.

Differential Equations

The problem involves solving a system of differential equations to find equilibrium points. Differential equations describe how variables change over time, and in the context of a dynamical system, they represent the system's laws of motion.

Here, the given differential equations \( x' = -xy + y + ax \) and \( y' = 2y - xy \) define how the variables \( x \) and \( y \) evolve. To find equilibria, the derivatives are set to zero, reflecting points where the system does not change, i.e., it is in a steady state.

By solving these equations simultaneously, we identify points such as \( (0,0) \) and \( (2,2a) \) as equilibria of the system. These points are crucial as they serve as the system's fixed points in its phase space.

Jacobians

The Jacobian matrix is a fundamental tool in determining the stability of equilibrium points. It is constructed from the first-order partial derivatives of the system's functions.For this system, the Jacobian is given by:\[J = \begin{pmatrix} -y + a & -x + 1 \ -y & 2 - x \end{pmatrix}\]At an equilibrium point, the Jacobian provides a linear approximation of the system's behavior around that point. By substituting the equilibrium points into the Jacobian, we analyze how perturbations grow or diminish, hence assessing stability.

At \( (0, 0) \), the Jacobian becomes \[\begin{pmatrix} a & 1 \ 0 & 2 \end{pmatrix}\]For \( (2, 2a) \), it transforms and requires further calculation to determine the eigenvalues, helping us infer stability characteristics for these points.

Eigenvalues

Eigenvalues are parameters obtained from the Jacobian matrix that inform us about the tendency of trajectories near an equilibrium point.

The eigenvalues tell whether perturbations grow or diminish over time. If all eigenvalues of an equilibrium point have negative real parts, it suggests that disturbances die out and the point is stable. Conversely, positive eigenvalues indicate that perturbations grow, leading to instability.

To find the eigenvalues for the equilibrium \( (0,0) \), one computes them from:\[\begin{pmatrix} a & 1 \ 0 & 2 \end{pmatrix}\]The eigenvalues are \( a \) and \( 2 \), providing immediate insight into stability based on their sign. Analyzing eigenvalues for \( (2,2a) \) involves solving the characteristic polynomial derived from its Jacobian, allowing us to conclude about its stability in a similar fashion.