Problem 22 - Calculus for Biology and Medicine

Question

Show that \(\left[\begin{array}{l}0 \\ 0\end{array}\right]\) is an equilibrium of $$ \left[\begin{array}{l} x_{1}(t+1) \\ x_{2}(t+1) \end{array}\right]=\left[\begin{array}{ll} 1.5 & 0.2 \\ 0.08 & 0 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ and determine its stability.

Step-by-Step Solution

Answer

The point is an equilibrium, but it is unstable.

Step 1: Understanding Equilibrium

An equilibrium point \(\begin{array}{l} x_1^* \ x_2^* \end{array}\) occurs when the state does not change over time, meaning \(\begin{array}{l} x_1(t+1) \ x_2(t+1) \end{array}\) remains constant. From the given equations, this implies \[ \begin{array}{l} x_1^* \ x_2^* \end{array} = \begin{array}{ll} 1.5 & 0.2 \ 0.08 & 0 \end{array} \begin{array}{l} x_1^* \ x_2^* \end{array}. \]

Step 2: Solving for Equilibrium

We substitute \(\begin{array}{l} 0 \ 0 \end{array}\) into the matrix equation: \[ \begin{array}{l} x_1(t+1) \ x_2(t+1) \end{array} = \begin{array}{ll} 1.5 & 0.2 \ 0.08 & 0 \end{array} \begin{array}{l} 0 \ 0 \end{array} = \begin{array}{l} 0 \ 0 \end{array}. \] The resulting vector is \(\begin{array}{l} 0 \ 0 \end{array}\), confirming that \(\begin{array}{l} 0 \ 0 \end{array}\) is indeed an equilibrium.

Step 3: Determining Stability

The stability of the equilibrium point is determined by the eigenvalues of the matrix \(\begin{array}{ll} 1.5 & 0.2 \ 0.08 & 0 \end{array}\). Calculate the determinant of \[ \begin{array}{ll} \lambda - 1.5 & -0.2 \ -0.08 & \lambda \end{array}. \] This results in the characteristic equation: \[(\lambda - 1.5)(\lambda) - (0.08)(0.2) = 0,\] which simplifies to \( \lambda^2 - 1.5\lambda - 0.016 = 0 \).

Step 4: Solving the Characteristic Equation

Solve the quadratic equation \(\lambda^2 - 1.5\lambda - 0.016 = 0 \) using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a=1, b=1.5, c=-0.016\): \\lambda = \frac{-1.5 \pm \sqrt{1.5^2 + 4 \times 0.016}}{2}.\Calculating, we find the eigenvalues \(\lambda_1 \approx 1.516\) and \(\lambda_2 \approx -0.016\).

Step 5: Interpreting the Stability Result

An equilibrium point is stable if all eigenvalues have absolute values less than 1. Here, \(\lambda_1 \approx 1.516\) has an absolute value greater than 1, implying the system is unstable at \(\begin{array}{l} 0 \ 0 \end{array}\).

Key Concepts

Matrix Stability

Matrix stability helps us determine the behavior of a system over time. In the context of linear algebra, we often have dynamic systems represented by matrices, and we want to analyze if their states remain bounded or diverge.
Stability can be understood by looking at the characteristic properties of a matrix and its eigenvalues.
If all the eigenvalues of a matrix have absolute values less than 1, the system is stable; otherwise, it is unstable.
This provides insight into how the system will evolve: whether it will settle into equilibrium or diverge away.

Eigenvalues

Eigenvalues play a crucial role in understanding matrix behavior. They are special numbers that give us essential information about a matrix's properties.

• An eigenvalue tells us how a specific vector, connected to this eigenvalue, is "stretched" or "compressed" when passed through a matrix.
• In our exercise, finding eigenvalues helps with determining stability, as they indicate whether a system will remain in equilibrium.

When the absolute value of any eigenvalue is greater than 1, it suggests potential instability. This is because the related vector will grow instead of settling down over time.

Characteristic Equation

The characteristic equation is essential to finding eigenvalues. It is derived from the matrix by setting its determinant to zero.
For a 2x2 matrix, the characteristic equation is found using \[ ext{det}(A - \lambda I) = 0, \]where \(A\) is the original matrix and \(I\) is the identity matrix.
Solving the characteristic equation, which is often a quadratic, provides the eigenvalues of the matrix.
In our example, this process produced the equation \( \lambda^2 - 1.5\lambda - 0.016 = 0 \).
The solutions to this quadratic equation reveal the eigenvalues crucial for assessing stability.

Linear Algebra

Linear algebra is the mathematical foundation underlying the concepts used in this exercise. It provides the framework to work with vectors and matrices, essential tools for understanding dynamic systems.

• Linear algebra involves operations such as matrix multiplication and vector transformation, enabling us to set equations and solve for unknowns.
• It allows us to perform stability analysis, find equilibrium points, and understand how systems change over time.

Concepts like eigenvectors and eigenvalues stem from linear algebra, aiding in the analysis of complex systems via simple mathematical structures. This branch of math makes it possible to simplify many real-world problems into solvable components.