Problem 11 - Biocalculus Calculus for the Life Sciences

Question

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

Step-by-Step Solution

Answer

Equilibria are at (0, 0) and (-1, -1). Linearizations are given by the Jacobian matrices at these points.

Step 1: Find Equilibrium Points

To find the equilibrium points, set the derivatives equal to zero: \( \frac{dx_1}{dt} = 0 \) and \( \frac{dx_2}{dt} = 0 \). From the first equation, solving \( e^{-x_1}(x_1 - x_2) = 0 \) gives us two cases: \( x_1 = x_2 \) or \( e^{-x_1} = 0 \), but the latter is impossible. Thus, \( x_1 = x_2 \).For the second equation, substitute \( x_1 = x_2 \) into \( x_1 - x_2^2 + 2x_1x_2 = 0 \). This simplifies to \( x_1 - x_1^2 + 2x_1^2 = 0 \) yielding \( x_1(1 + x_1) = 0 \) or \( x_1 = 0 \) or \( x_1 = -1 \). Thus, equilibrium points are \((0, 0)\) and \((-1, -1)\).

Step 2: Calculate Jacobian Matrix

To find the linearization, compute the Jacobian of the system:\[J = \begin{bmatrix}\frac{\partial}{\partial x_1} \left(e^{-x_1}(x_1 - x_2)\right) & \frac{\partial}{\partial x_2} \left(e^{-x_1}(x_1 - x_2)\right) \\frac{\partial}{\partial x_1} \left(x_1 - x_2^2 + 2x_1x_2\right) & \frac{\partial}{\partial x_2} \left(x_1 - x_2^2 + 2x_1x_2\right)\end{bmatrix}\]Calculate partial derivatives:

Step 3: Evaluate Jacobian at Equilibrium (0, 0)

Substitute \((x_1, x_2) = (0, 0)\) into the Jacobian:\[J = \begin{bmatrix}-1 & 1 \1 & 0\end{bmatrix}\]This matrix represents the linearization around the equilibrium point \((0,0)\).

Step 4: Evaluate Jacobian at Equilibrium (-1, -1)

Substitute \((x_1, x_2) = (-1, -1)\) into the Jacobian:\[J = \begin{bmatrix}0 & 1 \-1 & 4\end{bmatrix}\]This matrix represents the linearization around the equilibrium point \((-1,-1)\).

Step 5: Conclude with Linearizations

The linearizations of the system near the equilibrium points \((0, 0)\) and \((-1, -1)\) are given by the Jacobian matrices evaluated in Steps 3 and 4, respectively. These matrices describe the local behavior of the system near each equilibrium.

Key Concepts

Equilibrium Points

The concept of equilibrium points is fundamental in understanding dynamic systems. These are the points where the system does not change over time. To find equilibrium points, we set the derivative functions equal to zero since this condition represents a state of no change. In our exercise, we set \( \frac{dx_1}{dt} = 0 \) and \( \frac{dx_2}{dt} = 0 \). This nullifies the changes in the system, allowing us to solve for \( x_1 \) and \( x_2 \).

In this specific problem, simplifying the first equation \( e^{-x_1}(x_1 - x_2) = 0 \) gives us \( x_1 = x_2 \) because \( e^{-x_1} \) cannot actually be zero. Subsequently, substituting this into the second equation yields equilibrium points at \((0, 0)\) and \((-1, -1)\). These points indicate the states of the system where the behavior will remain constant without any change over time.

Linearization

Linearization is a crucial method used to approximate the behavior of a nonlinear system near its equilibrium points by employing linear equations. It involves finding a linear approximation using the system's Jacobian matrix evaluated at those points.

In our example, once the equilibrium points were identified, linearization was conducted using the Jacobian Matrix. This matrix consists of the first-order partial derivatives of the system equations. Evaluating these derivatives at each equilibrium point gives a linear representation of the system's dynamics at those critical points. It helps to understand how small perturbations around the equilibrium will evolve.

For instance, the linearization at \((0,0)\) results in a Jacobian matrix of \[\begin{bmatrix}-1 & 1 \ 1 & 0\end{bmatrix}\] whereas at \((-1, -1)\) it becomes \[\begin{bmatrix}0 & 1 \ -1 & 4\end{bmatrix}\]. These matrices offer insights into the system's stability near these equilibrium points.

Partial Derivatives

Partial derivatives play a significant role in understanding how each variable in a multivariable function affects the outcome, holding other variables constant. In the context of our problem, they're essential for constructing the Jacobian matrix used in the linearization process.

When calculating partial derivatives for our equations, we are essentially looking for how changes in one variable individually influence the rate of change of the system, while keeping the other variable fixed. This method isolates the effects of each variable, which is important in analyzing complex systems.

• The partial derivative of \( e^{-x_1}(x_1-x_2) \) with respect to \(x_1\) shows how changes in \(x_1\) while \(x_2\) is constant affect the system.
• Similarly, calculating for \(x_2\) demonstrates its distinct impact.

Each of these derivatives contributes to forming the rows and columns of our Jacobian matrix. Understanding these effects through partial derivatives helps diagnose each variable’s impact on the system's dynamics around equilibrium points.