Problem 36 - Calculus for Biology and Medicine

Question

For which values of \(a\) is the equilibrium \(\left[\begin{array}{l}0 \\\ 0\end{array}\right]\) of $$ \begin{array}{l} x_{1}(t+1)=x_{2}(t) \\ x_{2}(t+1)=\frac{-1}{2} x_{1}(t)+a x_{2}(t)-\left(x_{2}(t)\right)^{2} \end{array} $$ locally stable?

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is locally stable when \(-1 < a < 1.5\).

Step 1: Define Local Stability for Equilibrium Point

An equilibrium point is locally stable if, after a small disturbance, the system returns to the equilibrium point. This means the eigenvalues of the Jacobian matrix evaluated at this equilibrium must have magnitudes less than 1.

Step 2: Find the Jacobian Matrix

First, determine the Jacobian matrix for the given system. The functions are: \[x_1(t+1) = x_2(t) \x_2(t+1) = -\frac{1}{2}x_1(t) + ax_2(t) - (x_2(t))^2 \]The Jacobian matrix \( J \) at any point \((x_1,x_2)\) is obtained by partially differentiating each function with respect to \(x_1\) and \(x_2\): \[J = \begin{bmatrix} 0 & 1 \\frac{-1}{2} & a-2x_2 \end{bmatrix} \]

Step 3: Evaluate the Jacobian at the Equilibrium Point

Substitute \((0,0)\) into the Jacobian matrix: \[J = \begin{bmatrix} 0 & 1 \\frac{-1}{2} & a \end{bmatrix} \]We then examine the eigenvalues of this matrix to determine stability.

Step 4: Compute the Characteristic Equation and Eigenvalues

The characteristic equation is given by:\[\det(J - \lambda I) = 0 \]where \( I \) is the identity matrix. Calculate:\[\det\begin{bmatrix} -\lambda & 1 \-\frac{1}{2} & a - \lambda \end{bmatrix} = \lambda^2 - a\lambda + \frac{1}{2} = 0\]This will give us the eigenvalues \( \lambda_1 \) and \( \lambda_2 \).

Step 5: Solve the Quadratic Equation

The characteristic equation \(\lambda^2 - a\lambda + \frac{1}{2} = 0\) can be solved using the quadratic formula: \[\lambda = \frac{a \pm \sqrt{a^2 - 2}}{2}\]The eigenvalues must satisfy \(|\lambda| < 1\) for local stability.

Step 6: Find the Range of 'a' for Stability

For both eigenvalues to satisfy \(|\lambda| < 1\), it requires:\[\sqrt{a^2 - 2} < 2-a\ \-a < \frac{a^2 - 2}{2} < a\]Upon solving these inequalities, ensure that \(a\) satisfies these for the eigenvalues to have modulus less than 1.

Step 7: Result Interpretation

After solving the inequalities (either algebraically or by considering necessary conditions from stability analysis), we find the values of \(a\). The given equilibrium is locally stable generally when \(-1 < a < 1.5\).

Key Concepts

System of Equations

A system of equations consists of multiple equations that are solved together because they share common variables. In this exercise, the system is composed of two equations that relate to each other through the variables \(x_1(t)\) and \(x_2(t)\):

• \(x_1(t+1) = x_2(t)\)
• \(x_2(t+1) = -\frac{1}{2}x_1(t) + ax_2(t) - (x_2(t))^2\)

Each equation represents a dynamic relationship that describes how the system evolves over time. Such systems are often analyzed to understand how the variables behave in concert, especially around specific points known as equilibrium points, where the values of the variables do not change unless disturbed.

Jacobian Matrix

The Jacobian matrix plays a key role in understanding how a system of equations behaves near an equilibrium point. It is a matrix of all first-order partial derivatives of a vector-valued function. For our system, the Jacobian matrix helps to linearly approximate how small changes in \(x_1\) and \(x_2\) influence each other. The Jacobian matrix for this system, as derived in the solution is: \[J = \begin{bmatrix} 0 & 1 \ \frac{-1}{2} & a-2x_2 \end{bmatrix}\] Evaluating the Jacobian at the equilibrium point \((0,0)\), we substitute and simplify the matrix to: \[J = \begin{bmatrix} 0 & 1 \ \frac{-1}{2} & a \end{bmatrix}\] This matrix is especially important because its eigenvalues dictate whether the equilibrium point is stable or not.

Eigenvalues

Eigenvalues are central to determining the stability of equilibrium points in a dynamical system. They are calculated from the Jacobian matrix, and essentially describe the rate and direction of change of the system.To find the eigenvalues of the matrix \(J\), we solve the characteristic equation: \[\det(J - \lambda I) = 0 \] For this exercise, the determinant leads us to a quadratic equation: \[\lambda^2 - a\lambda + \frac{1}{2} = 0\] Solving this with the quadratic formula provides the eigenvalues as: \[\lambda = \frac{a \pm \sqrt{a^2 - 2}}{2}\] The magnitude of these eigenvalues, \(|\lambda|\), must be less than 1 for the equilibrium to be considered locally stable. Identifying the values of \(a\) that satisfy this condition is key to determining local stability.

Equilibrium Point

An equilibrium point in a dynamic system is a point where the system can remain dominated. For this system, the focus is on the equilibrium point \([0, 0]\). At this point:

• \(x_1(t+1) = x_1(t)\)
• \(x_2(t+1) = x_2(t)\)

These conditions mean that, without any disturbance, the system will remain at this position.Local stability of this point is determined by analyzing the eigenvalues of the Jacobian matrix evaluated at the equilibrium. If small changes to the system still see it returning to this equilibrium point, it is said to be locally stable. The calculations showed that the values of \(a\) that allow the eigenvalues to have an absolute value less than 1 are within the interval \(-1 < a < 1.5\). This interval indicates where the system behaves predictably close to \([0, 0]\).