Problem 38 - Calculus for Biology and Medicine

Question

We consider differential equations of the form \(\frac{d \mathbf{x}}{d t}=A \mathbf{x}(t)\) where $$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] $$ The eigenvalues of A will be real, distinct, and nonzero. Analyze the stability of the equilibrium \((0,0)\), and classify the equilib\mathrm{\\{} r i u m ~ a c c o r d i n g ~ t o ~ w h e t h e r ~ i t ~ i s ~ a ~ s i n k , ~ a ~ s o u r c e , ~ o r ~ a ~ s a d d l e ~ point. \(A=\left[\begin{array}{rr}-3 & 1 \\ 1 & -2\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is stable and classified as a sink.

Step 1: Identify the Matrix

The given matrix \( A \) is \( \begin{bmatrix} -3 & 1 \ 1 & -2 \end{bmatrix} \). This matrix will be used to find the eigenvalues, which are crucial for determining stability and classifying the equilibrium.

Step 2: Compute Eigenvalues of A

We need to find the eigenvalues by solving the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. This becomes \(\det\left(\begin{bmatrix} -3 - \lambda & 1 \ 1 & -2 - \lambda \end{bmatrix}\right) = \lambda^2 + 5\lambda + 5 = 0.\)

Step 3: Solve the Characteristic Equation

The quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is used to solve the characteristic equation \( \lambda^2 + 5\lambda + 5 = 0 \). Set \( a = 1, b = 5, c = 5 \) to obtain the solutions: \(\lambda = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{-5 \pm \sqrt{5}}{2}.\)

Step 4: Analyze the Signs of the Eigenvalues

The eigenvalues are \( \lambda_1 = \frac{-5 + \sqrt{5}}{2} \) and \( \lambda_2 = \frac{-5 - \sqrt{5}}{2} \). Both are negative as \( \sqrt{5} < 5 \), indicating that they are real and distinct.

Step 5: Determine Stability and Classification

Since both eigenvalues are negative, the equilibrium at \((0,0)\) is stable and it is a sink. This is because a negative eigenvalue corresponds to solutions of the differential equation decreasing to zero as \( t \to \infty \).

Key Concepts

Understanding Differential Equations

Differential equations are mathematical tools used to describe how things change. They involve functions and their derivatives, expressing rates of change and often embody ideals from physics, economics, and other sciences. In the given exercise, we explore a differential equation associated with a matrix, specifically in the form \( \frac{d \mathbf{x}}{d t} = A \mathbf{x}(t) \), where \( A \) is a constant matrix and \( \mathbf{x}(t) \) is a vector of functions.

• This type of equation is a linear system, common in modeling ant illusions of interconnected systems like population growth, chemical reactions, or electrical-circuit behaviors.
• The key to analyzing these systems is often in the matrix \( A \), which governs the system's behavior.

When you solve such differential equations, you're effectively predicting how the system's state will evolve over time. The solution requires finding specific values, known as eigenvalues, which tell us how stable or unstable an equilibrium will be.

Stability Analysis in Differential Equations

Stability analysis examines whether a system will return to its equilibrium state after a disturbance. In our exercise, we analyze stability by looking at the eigenvalues of the matrix \( A \). This matrix determines how each variable affects each other in the system and how the system behaves as a whole.

• If all eigenvalues have negative real parts, the system will return to equilibrium over time, indicating stability.
• If any eigenvalue has a positive real part, the system will diverge from equilibrium, indicating instability.

By computing the eigenvalues in our example, we noted they were real, distinct, and negative, confirming that the equilibrium was stable. A stable system converges to an equilibrium point, much like a ball settling at the bottom of a bowl no matter how gently it is pushed.

The Role of Equilibrium Points

Equilibrium points are crucial concepts in the analysis of differential equations. An equilibrium point is where the system remains unchanged if it starts there; mathematically, it's where the derivatives or rates of change equal zero.

• Stable equilibrium: If small perturbations or deviations from the equilibrium cause the system to return, it is classified as stable.
• Unstable equilibrium: If those perturbations lead the system to move further away, it is considered unstable.
• Saddle point: A mix where movement might be stable in one direction but unstable in another.

In our matrix model, the equilibrium at \((0,0)\) was identified as stable because the eigenvalues were negative. Furthermore, since both eigenvalues were negative and distinct, our equilibrium point can be classified as a sink, meaning all trajectories in the phase plane move towards it, akin to how water flows towards and down a sink drain.