Problem 7 - Biocalculus Calculus for the Life Sciences

Question

Find all equilibria. Then find the linearization near each equilibrium. $$\begin{array}{l}{\frac{d x_{1}}{d t}=-5 x_{1}+x_{1} x_{2}} \\ {\frac{d x_{2}}{d t}=x_{2}-5 x_{1} x_{2}}\end{array}$$

Step-by-Step Solution

Answer

Equilibria are (0,0) and (1/5,5). (0,0) is unstable; (1/5,5) is a center.

Step 1: Equilibrium Conditions

Equilibria occur when the derivatives are zero: \( \frac{d x_{1}}{d t} = 0 \) and \( \frac{d x_{2}}{d t} = 0 \). Substitute into the system: \(-5x_1 + x_1 x_2 = 0\) and \(x_2 - 5x_1 x_2 = 0\).

Step 2: Solve for Equilibria

From the first equation, factor out \(x_1\) to get \(x_{1}(-5 + x_{2}) = 0\), yielding solutions \(x_1 = 0\) or \(x_2 = 5\). From the second equation, factor out \(x_2\) to get \(x_{2}(1 - 5x_{1}) = 0\), which gives \(x_2 = 0\) or \(x_1 = \frac{1}{5}\). Combine these for equilibria: \((0, 0)\), \((\frac{1}{5}, 5)\).

Step 3: Find Jacobi Matrix

The linearization involves the Jacobian matrix of the system. Compute partial derivatives: \(\frac{\partial}{\partial x_1}(-5x_1 + x_1x_2) = -5 + x_2\), \(\frac{\partial}{\partial x_2}(-5x_1 + x_1x_2) = x_1\), \(\frac{\partial}{\partial x_1}(x_2 - 5x_1x_2) = -5x_2\), and \(\frac{\partial}{\partial x_2}(x_2 - 5x_1x_2) = 1 - 5x_1\). The Jacobian is \[ J(x_1, x_2) = \begin{bmatrix} -5 + x_2 & x_1 \ -5x_2 & 1 - 5x_1 \end{bmatrix} \].

Step 4: Evaluate Jacobian at Equilibria

Calculate the Jacobian matrix at each equilibrium point. At \((0, 0)\): \[ J(0, 0) = \begin{bmatrix} -5 & 0 \ 0 & 1 \end{bmatrix} \]. At \((\frac{1}{5}, 5)\): \[ J(\frac{1}{5}, 5) = \begin{bmatrix} 0 & \frac{1}{5} \ -25 & 0 \end{bmatrix} \].

Step 5: Analyze Stability of Each Equilibrium

For \((0, 0)\), the eigenvalues of \[ J(0, 0) = \begin{bmatrix} -5 & 0 \ 0 & 1 \end{bmatrix} \] are \(-5\), and \(1\). Since one eigenvalue is positive, \((0, 0)\) is unstable. For \((\frac{1}{5}, 5)\), find the eigenvalues of \[ J(\frac{1}{5}, 5) = \begin{bmatrix} 0 & \frac{1}{5} \ -25 & 0 \end{bmatrix} \] by solving \(\lambda^2 + 5 = 0\), giving eigenvalues \(\pm i\sqrt{5}\), indicating a center at \((\frac{1}{5}, 5)\).

Key Concepts

Equilibrium Points

Equilibrium points or equilibria in differential equations are vital concepts that represent steady states where the system doesn't change over time. These are the specific values of variables where the system's rate of change is zero. In mathematical terms for our example, equilibrium is found when both the derivatives in the system equal zero:

• \( \frac{d x_{1}}{d t} = 0 \)
• \( \frac{d x_{2}}{d t} = 0 \)

To find these points, you substitute these conditions back into the original system of equations. For our exercise, the solutions to these equations are

• \((0, 0)\)
• \((\frac{1}{5}, 5)\)

These are equilibrium points because, at these pairs, the functions defining the system's behavior are zero, meaning the state remains constant when analyzed over time.

Jacobian Matrix

The Jacobian matrix is a crucial concept in understanding how systems of equations behave near equilibrium points. Simply put, it is a matrix composed of first-order partial derivatives of a function. In the context of non-linear systems, like our provided system of differential equations, the Jacobian matrix helps in examining and approximating how small changes in variables near an equilibrium point influence the system.For our system, the Jacobian matrix is constructed by calculating the partial derivatives of the system's equations with respect to each variable. The computed Jacobian matrix for this exercise is:\[ J(x_1, x_2) = \begin{bmatrix} -5 + x_2 & x_1 \ -5x_2 & 1 - 5x_1 \end{bmatrix} \]Evaluating the Jacobian matrix at each equilibrium point allows us to study their local stability attributes and system dynamics.

Eigenvalues

Eigenvalues are intrinsic to analyzing the behavior of differential equations when considering their stability. They represent characteristic values that tell us how a transformation represented by the Jacobian changes space. Essentially, eigenvalues are derived from the Jacobian matrix and are crucial in determining the stability properties of equilibrium points.For each equilibrium point, you evaluate the Jacobian matrix and solve the characteristic equation: \[ \det(J - \lambda I) = 0 \]Here, \(\lambda\) represents the eigenvalues, and \(I\) is the identity matrix. In our system, we found:

• For \((0, 0)\), eigenvalues are \(-5\) and \(1\).
• For \((\frac{1}{5}, 5)\), eigenvalues are \(\pm i\sqrt{5}\).

Analyzing eigenvalues reveals how the system responds to small disturbances from equilibrium, indicating whether it returns to equilibrium (stability) or diverges away (instability).

Stability Analysis

Stability analysis is essential for determining how systems react to minor changes. In essence, it tells us if an equilibrium point will return to stability after a slight perturbation or diverge further away.In the context of our exercise, stability is assessed by examining the eigenvalues of the Jacobian matrix at each equilibrium point:

• For the point \((0, 0)\), the eigenvalues \(-5\) and \(1\) indicate instability because having at least one positive eigenvalue suggests that disturbances can grow over time.
• At \((\frac{1}{5}, 5)\), the purely imaginary eigenvalues \(\pm i\sqrt{5}\) correspond to a center, meaning it remains neutrally stable. Here, while perturbations neither grow nor shrink, they could cause oscillating motion around the equilibrium.

Understanding stability helps predict a system's behavior in real-world scenarios, where returning to or deviating from equilibrium can significantly impact strategic decision-making or system control.