Problem 3 - Calculus for Biology and Medicine

Question

The point \((0,0)\) is always an equilibrium. Use the analytical approach to investigate its stability \(\frac{d x_{1}}{d t}=x_{1}+x_{1}^{2}-2 x_{1} x_{2}+x_{2}\) \(\frac{d x_{2}}{d t}=x_{1}\)

Step-by-Step Solution

Answer

The equilibrium point \((0,0)\) is unstable.

Step 1: Identify Equilibrium Points

Equilibrium points occur where the derivatives are zero. We start by setting the right-hand side of the differential equations to zero. For \( \frac{dx_1}{dt} \), we have \( x_1 + x_1^2 - 2x_1x_2 + x_2 = 0 \). For \( \frac{dx_2}{dt} \), we have \( x_1 = 0 \). The point \((0,0)\) satisfies both equations, confirming it is an equilibrium point.

Step 2: Linearize the System

To analyze stability, we linearize the system at the equilibrium point \((0,0)\). Compute the Jacobian matrix \( J \) by calculating the partial derivatives with respect to \( x_1 \) and \( x_2 \). The Jacobian matrix is \[ J = \begin{bmatrix} \frac{\partial}{\partial x_1}(x_1 + x_1^2 - 2x_1x_2 + x_2) & \frac{\partial}{\partial x_2}(x_1 + x_1^2 - 2x_1x_2 + x_2) \ \frac{\partial}{\partial x_1}(x_1) & \frac{\partial}{\partial x_2}(x_1) \end{bmatrix} \] Evaluating at \((0,0)\), we find \[ J = \begin{bmatrix} 1 & 1 \ 1 & 0 \end{bmatrix} \].

Step 3: Calculate the Eigenvalues of the Jacobian

The stability of the equilibrium point is determined by the eigenvalues of the Jacobian matrix. The characteristic equation of \( J \) is given by \( \det(J - \lambda I) = 0 \), where \( I \) is the identity matrix. For this system, \[ \det \begin{bmatrix} 1-\lambda & 1 \ 1 & -\lambda \end{bmatrix} = (1-\lambda)(-\lambda) - 1 \cdot 1 = \lambda^2 - \lambda - 1 = 0 \]. Solve for \( \lambda \).

Step 4: Solve the Characteristic Equation

Solve the quadratic equation \( \lambda^2 - \lambda - 1 = 0 \) using the quadratic formula: \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a=1, b=-1, \) and \( c=-1 \). \[ \lambda = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]. The eigenvalues are \( \lambda_1 = \frac{1 + \sqrt{5}}{2} \) and \( \lambda_2 = \frac{1 - \sqrt{5}}{2} \).

Step 5: Determine Stability from Eigenvalues

For stability analysis, we examine the real parts of the eigenvalues. Since \( \lambda_1 = \frac{1 + \sqrt{5}}{2} \) is positive, the equilibrium point \((0,0)\) is unstable as trajectories will move away from \((0,0)\) over time due to the exponential growth.

Key Concepts

Jacobian Matrix

When analyzing the stability of equilibrium points in a system of differential equations, the Jacobian matrix plays a crucial role. It is constructed by taking partial derivatives of each equation with respect to each variable in the system. For our system:

• The Jacobian helps transform the original nonlinear system around its equilibrium into a linear form, making it easier to analyze potential stability.
• The matrix is evaluated at the equilibrium point, giving us insight into how small changes around this point evolve over time.

To derive the Jacobian matrix, we calculate the partial derivatives of the right-hand side of the equations with respect to each variable, resulting in:\[J = \begin{bmatrix} \frac{\partial}{\partial x_1}(x_1 + x_1^2 - 2x_1x_2 + x_2) & \frac{\partial}{\partial x_2}(x_1 + x_1^2 - 2x_1x_2 + x_2) \\frac{\partial}{\partial x_1}(x_1) & \frac{\partial}{\partial x_2}(x_1) \end{bmatrix}\] Substituting at the equilibrium point \((0,0)\), this matrix becomes:\[J = \begin{bmatrix} 1 & 1 \1 & 0 \end{bmatrix}\]This is the foundation for further stability analysis through calculation of eigenvalues.

Eigenvalues

Eigenvalues represent critical insights into the behavior of a system near equilibrium points. They are derived from the Jacobian matrix, and their real parts indicate whether trajectories will converge to or diverge from equilibrium points.

• For a stable equilibrium, all eigenvalues should have negative real parts.
• Conversely, if any eigenvalue has a positive real part, even just one, the equilibrium is classified as unstable.

The eigenvalues are solutions to the characteristic equation:\[\det(J - \lambda I) = 0\]For our system, this is:\[\det \begin{bmatrix} 1-\lambda & 1 \1 & -\lambda \end{bmatrix} = \lambda^2 - \lambda - 1 = 0\]Solving this quadratic equation using the formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we find:\[\lambda_1 = \frac{1 + \sqrt{5}}{2}, \lambda_2 = \frac{1 - \sqrt{5}}{2}\]Observing these eigenvalues, we note that \(\lambda_1\) is positive, confirming the equilibrium is unstable.

Linearization

Linearization is an ingenious technique used to approximate nonlinear systems near equilibrium points. It simplifies the analysis of complex systems through a process that makes them easier to understand and study.

• By considering the Jacobian matrix at the equilibrium, we obtain a linear approximation of the system. This serves to predict the system's behavior in the vicinity of this point.
• Linearization focuses on small perturbations around equilibrium, allowing a breakdown of the system into manageable linear equations. This is where the Jacobian's composition of partial derivatives is particularly helpful.

In our exercise, linearization around \((0,0)\), provides a roadmap to explore the vicinity of equilibrium:

• The eigenvalues of the linearized system dictate whether it is stable or not.
• By analyzing these, we understand that for any initial state near \((0,0)\), the system is prone to diverge due to one of the eigenvalues having a positive real part.

Thus, linearization not only simplifies but also speeds up the stability examination process by reducing complex systems to their core linear components.