Problem 10 - Calculus for Biology and Medicine

Question

Find all equilibria of each system of differential equations and determine the stability of each equilibrium. \(\begin{aligned} & \frac{d x_{1}}{d t}=2 x_{1}\left(5-x_{1}-x_{2}\right) \\ & \frac{d x_{2}}{d t}=3 x_{2}\left(7-3 x_{1}-x_{2}\right) \end{aligned}\)

Step-by-Step Solution

Answer

Equilibria: (0,0) - unstable, (0,7) - saddle point, (5,0) - unstable, (1,4) - stable.

Step 1: Find Equilibrium Points

Equilibrium points occur where the rate of change of both variables is zero. This means solving the equations: 1. \(2x_1(5 - x_1 - x_2) = 0\) 2. \(3x_2(7 - 3x_1 - x_2) = 0\).For equation 1, either \(x_1 = 0\) or \(5 - x_1 - x_2 = 0\). For equation 2, either \(x_2 = 0\) or \(7 - 3x_1 - x_2 = 0\). Solve these simultaneously to find equilibrium points.

Step 2: Solve for Equilibrium Cases

Consider the cases:- \((x_1 = 0, x_2 = 0)\)- \((x_1 = 0, 7 - 3x_1 - x_2 = 0)\) simplifies to \(x_2 = 7\)- \((x_2 = 0, 5 - x_1 - x_2 = 0)\) simplifies to \(x_1 = 5\)- \((5 - x_1 - x_2 = 0, 7 - 3x_1 - x_2 = 0)\).For the simultaneous equations, substitute \(x_2 = 5 - x_1\) into \(7 - 3x_1 - x_2 = 0\) resulting in \(7 - 3x_1 - (5 - x_1) = 0\), thus \(2x_1 = 2\) giving \(x_1 = 1\), and consequently \(x_2 = 4\). Thus, the equilibria are \((0,0)\), \((0,7)\), \((5,0)\), and \((1,4)\).

Step 3: Determine Stability of Equilibria

Calculate the Jacobian matrix:\[J = \begin{bmatrix}\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \\end{bmatrix}\]Where \(f_1 = 2x_1(5 - x_1 - x_2)\) and \(f_2 = 3x_2(7 - 3x_1 - x_2)\).Calculate partial derivatives:- \(\frac{\partial f_1}{\partial x_1} = 2(5 - 2x_1 - x_2)\)- \(\frac{\partial f_1}{\partial x_2} = -2x_1\)- \(\frac{\partial f_2}{\partial x_1} = -9x_2\)- \(\frac{\partial f_2}{\partial x_2} = 3(7 - 3x_1 - 2x_2)\).Evaluate at each equilibrium and determine the eigenvalues of \(J\). An equilibrium is stable if all eigenvalues have negative real parts.

Step 4: Evaluate Jacobian and Eigenvalues at Equilibria

For each equilibrium point:1. \((0,0)\): \[J = \begin{bmatrix} 10 & 0 \ 0 & 21 \end{bmatrix} \rightarrow \text{eigenvalues: } (10, 21)\]2. \((0,7)\): \[J = \begin{bmatrix} -24 & 0 \ 0 & 0 \end{bmatrix} \rightarrow \text{eigenvalues: } (-24, 0)\]3. \((5,0)\): \[J = \begin{bmatrix} -10 & -10 \ 0 & 21 \end{bmatrix} \rightarrow \text{eigenvalues: } (-10, 21)\]4. \((1,4)\): \[J = \begin{bmatrix} -2 & -2 \ -12 & -4 \end{bmatrix} \rightarrow \text{eigenvalues: } (-5.46, -0.54)\].Analyzing the eigenvalues results in: \((0,0)\) and \((5,0)\) are unstable, \((0,7)\) is a saddle point, and \((1,4)\) is stable as both eigenvalues are negative.

Key Concepts

Equilibrium Points

In the realm of differential equations, equilibrium points are where the system's state does not change over time. These are points where the rate of change of all variables in the system is zero. To find equilibrium points, we set the differential equations to zero and solve for the variables.
When looking at the system given in the exercise, we need to solve for:

• The equation \( 2x_1(5 - x_1 - x_2) = 0 \) written for the first variable \( x_1 \)
• The equation \( 3x_2(7 - 3x_1 - x_2) = 0 \) written for the second variable \( x_2 \)

These equations give us several cases that include setting each individual term to zero, allowing us to determine potential equilibrium points like \((0,0), (0,7), (5,0),\) and \((1,4)\). Each of these equilibrium points is found by making one or more terms of the equations zero and checking where these intersect.

Stability Analysis

Stability analysis tells us whether small perturbations to an equilibrium point will decay over time, meaning the system returns to equilibrium, or grow over time, meaning the system is unstable. After identifying the equilibrium points, stability is determined by examining the Jacobian matrix at those points.
Once you have the Jacobian evaluated at the equilibrium points, the stability is checked by finding the eigenvalues of that matrix.
To determine stability, recall the following:

• An equilibrium point is stable if all eigenvalues of the Jacobian have negative real parts.
• It is unstable if any eigenvalue has a positive real part.
• A saddle point is a mix of some stable and some unstable eigenvalues, common in two-dimensional systems.

For instance, in our exercise, the point \((1,4)\) is stable because both eigenvalues are negative, suggesting that any trajectory that starts near this point will remain close or move towards it.

Jacobian Matrix

The Jacobian matrix is a pivotal tool in stability analysis as it linearly approximates a system near an equilibrium point. For a system of differential equations, the Jacobian contains all first-order partial derivatives of the system. This matrix captures how each variable influences the others and itself.
In our problem, we calculate partial derivatives for each function:

• \( rac{\partial f_1}{\partial x_1} = 2(5 - 2x_1 - x_2) \) and \( \frac{\partial f_1}{\partial x_2} = -2x_1 \)
• \( \frac{\partial f_2}{\partial x_1} = -9x_2 \) and \( \frac{\partial f_2}{\partial x_2} = 3(7 - 3x_1 - 2x_2) \)

Evaluating this Jacobian at each equilibrium point gives a matrix whose eigenvalues we check to determine stability. The simplicity or complexity of evaluating depends heavily on how these partial derivatives interact, showing us whether any small change in one variable leads to small or large changes in others.