Problem 1 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Find the local linear approximation to the system $$ \begin{array}{l} x^{\prime}=x-x^{2}-x y \\ y^{\prime}=y-0.5 x y-2 y^{2} \end{array} $$ a. At the equilibrium point (0,0) . b. At the equilibrium point (0,0.5) . c. At the equilibrium point (1,0) . d. At the equilibrium point \((2 / 3,1 / 3)\). For each of the local linear approximations, determine whether it is stable.

Step-by-Step Solution

Answer

Equilibrium points (0,0), (0,0.5), (1,0) are unstable; (2/3,1/3) is stable.

Step 1: Identify the Equilibrium Points

The equilibrium points occur when the derivatives are zero: \[ x' = x - x^2 - xy = 0 \]\[ y' = y - 0.5xy - 2y^2 = 0 \]Solve these equations for x and y to confirm the given equilibrium points: - (0,0)- (0,0.5)- (1,0)- \((2/3, 1/3)\)

Step 2: Jacobian Matrix Derivation

To find the local linear approximation near an equilibrium point, compute the Jacobian matrix \( J \) from the system of equations. The Jacobian \( J \) is given by:\[ J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} \]Calculate partial derivatives:- \( \frac{\partial x'}{\partial x} = 1 - 2x - y \)- \( \frac{\partial x'}{\partial y} = -x \)- \( \frac{\partial y'}{\partial x} = -0.5y \)- \( \frac{\partial y'}{\partial y} = 1 - 0.5x - 4y \)

Step 3: Evaluate Jacobian at the Equilibrium Points

Substitute each equilibrium point into the Jacobian matrix to obtain specific matrices for linear approximations:1. At \((0,0)\):\[ J = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \]2. At \((0,0.5)\):\[ J = \begin{bmatrix} 1 - 0.5 = 0.5 & 0 \ -0.5(0.5) = -0.25 & 1 \end{bmatrix} \]3. At \((1,0)\):\[ J = \begin{bmatrix} 1 - 2 = -1 & -1 \ 0 & 1 \end{bmatrix} \]4. At \((2/3, 1/3)\):\[ J = \begin{bmatrix} 1 - \frac{4}{3} = -\frac{1}{3} & -\frac{2}{3} \ -\frac{1}{6} & 1 - \frac{1}{3} - \frac{4}{3} = -\frac{2}{3} \end{bmatrix} \]

Step 4: Determine Stability from Eigenvalues

The system's stability at each point depends on the eigenvalues of the Jacobian matrix.1. For \((0,0)\), eigenvalues are \(1, 1\), hence the point is unstable.2. For \((0,0.5)\), eigenvalues include a positive \(0.5\), indicating instability.3. For \((1,0)\), eigenvalues \(-1, 1\) also show instability (saddle point).4. For \((2/3, 1/3)\), the eigenvalues have negative real parts, indicating stability.

Key Concepts

Equilibrium Points

Equilibrium points, sometimes known as fixed points, are critical values in a system of differential equations where the system is at rest. This means that at these points, the derivatives of the system are zero. In a two-variable system like the one given, we find equilibrium points by setting each equation for the derivatives to zero:

• The equation for \( x' \) is \( x - x^2 - xy = 0 \)
• The equation for \( y' \) is \( y - 0.5xy - 2y^2 = 0 \)

By solving these equations simultaneously, we can find the equilibrium points of the system.
These points are key in understanding the system's behavior because they indicate states where changes cease, and thus tell us much about the dynamics of the system.

Jacobian Matrix

The Jacobian matrix is a crucial tool in analyzing systems of equations, especially when examining their local behavior near equilibrium points. It is essentially a matrix of first-order partial derivatives and represents the tangent linear approximation of a nonlinear system.
For a system with variables \( x \) and \( y \), the Jacobian matrix \( J \) takes the form:\[J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix}\]To find \( J \), we must compute the partial derivatives of each differential equation with respect to each variable:

• \( \frac{\partial x'}{\partial x} = 1 - 2x - y \)
• \( \frac{\partial x'}{\partial y} = -x \)
• \( \frac{\partial y'}{\partial x} = -0.5y \)
• \( \frac{\partial y'}{\partial y} = 1 - 0.5x - 4y \)

By evaluating this Jacobian at each equilibrium point, we get a linear approximation of our system near those points, which helps in the analysis of stability.

Stability Analysis

Stability analysis is a method used to determine the behavior of a system when it is perturbed slightly from an equilibrium point. This is generally done by examining the eigenvalues of the Jacobian matrix at each equilibrium point.

• If all the eigenvalues of the Jacobian at an equilibrium point have negative real parts, the point is stable, meaning that the system will return to equilibrium after a small disturbance.
• If any eigenvalue has a positive real part, the equilibrium point is unstable, meaning the system will move away from equilibrium after a disturbance.

For the given problem:
- At \((0,0)\), the eigenvalues are \(1, 1\), indicating instability because neither is negative.- At \((0,0.5)\), one eigenvalue is \(0.5\), demonstrating instability.- At \((1,0)\), eigenvalues \(-1, 1\) suggest instability (saddle point).- At \((2/3, 1/3)\), both eigenvalues are negative, indicating the system is stable here.
Understanding stability through eigenvalues helps predict the system's behavior and informs decisions about controlling or altering the system.