Problem 45 - 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 complex conjugates. Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium according to whether it is a stable spiral, an unstable spiral, or a center. \(A=\left[\begin{array}{rr}-2 & 4 \\ -3 & -2\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium is a stable spiral.

Step 1: Find the Eigenvalues

To determine the stability, we first need to find the eigenvalues of the matrix \(A\). The characteristic equation is given by \(\det(A - \lambda I) = 0\), where \(I\) is the identity matrix. For \(A\), this becomes \[\begin{vmatrix}a_{11} - \lambda & a_{12} \a_{21} & a_{22} - \lambda\end{vmatrix}= \lambda^2 - (a_{11} + a_{22})\lambda + (a_{11}a_{22} - a_{12}a_{21}) = 0.\]Substitute the specific values from matrix \(A\) to find the eigenvalues.

Step 2: Calculate the Characteristic Polynomial

With \(A = \begin{bmatrix} -2 & 4 \ -3 & -2 \end{bmatrix}\), the characteristic equation is:\[\begin{vmatrix}-2 - \lambda & 4 \-3 & -2 - \lambda\end{vmatrix}= (-2 - \lambda)^2 - (4)(-3) = 0.\]This simplifies to \(\lambda^2 + 4\lambda + 16 = 0\).

Step 3: Solve for Eigenvalues

The eigenvalues can be found by solving the quadratic equation \(\lambda^2 + 4\lambda + 16 = 0\). Use the quadratic formula: \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 4\), and \(c = 16\). This gives us \[\lambda = \frac{-4 \pm \sqrt{4^2 - 4 \times 1 \times 16}}{2} = \frac{-4 \pm \sqrt{-48}}{2}.\]Simplifying gives \(\lambda = -2 \pm 2i\sqrt{3}\).

Step 4: Determine Stability and Classification

Since the eigenvalues are complex conjugates with negative real parts (\(-2 \pm 2i\sqrt{3}\)), the equilibrium at \((0,0)\) is a stable spiral. If the real parts were positive, it would be an unstable spiral. If the real parts were zero, the equilibrium would be a center.

Key Concepts

Differential Equations

Differential equations are a tool used in mathematics to describe how things change over time. They show relationships that involve functions and their derivatives. In simple terms, they tell us how a quantity changes with respect to another, often time.In our problem, the differential equation is given by \[\frac{d \mathbf{x}}{d t} = A \mathbf{x}(t)\]which shows that the rate of change of \(\mathbf{x}\) over time \(t\) is determined by multiplying a matrix \(A\) with the vector \(\mathbf{x}(t)\). Differential equations are crucial in modeling real-world phenomena:

• Motions of planets in physics
• Population growth in biology
• Economics forecasting
• Engineering systems such as circuits or mechanical vibrations

By solving these equations, we can predict and understand the behavior of systems.

Stability

Stability in the context of differential equations is about the behavior of solutions as time progresses. We mainly ask, "Will a system settle down to a particular behavior over time, or will it become unpredictable?"For linear systems, like the one in the problem, stability is determined by the eigenvalues of the matrix \(A\). If the eigenvalues have negative real parts, the system will settle down and is called stable. Conversely, if the eigenvalues have positive real parts, the system diverges and becomes unstable.Understanding stability is crucial for:

• Predicting whether a mechanical system will stop oscillating
• Determining if a population will reach a steady state
• Ensuring electronics circuits remain operational without blowing up due to oscillations

In our exercise, the eigenvalues are \(-2 \pm 2i\sqrt{3}\), both having a negative real part, indicating a stable spiral.

Equilibrium

Equilibrium refers to a state where the system does not change as time moves forward, meaning it has reached a steady condition. In simpler terms, it's a "rest" state of a system.In differential equations, equilibrium occurs when the derivative is zero, meaning the system doesn’t change:\[\frac{d \mathbf{x}}{d t} = 0\]For our problem, \(\mathbf{x} = 0\) is the equilibrium point.Classifying equilibrium depends on eigenvalues:

• Stable spiral: Both eigenvalues have negative real parts.
• Unstable spiral: Both eigenvalues have positive real parts.
• Center: Eigenvalues have zero real parts.

This understanding helps:

• Predict if a pendulum will return to vertical or spin endlessly
• Decide if resting states in economies will persist or disappear
• Ensure infrastructures like bridges maintain their states without collapsing

Understanding equilibrium can help with designing systems that either maintain stability or provocatively change states.

Complex Conjugates

Complex conjugates are pairs of complex numbers that have the same real component but opposite imaginary components. If an eigenvalue is complex, it usually appears with its conjugate.In matrix problems, particularly like our exercise, when eigenvalues are complex this generally implies oscillatory behavior. The imaginary parts are responsible for the oscillations. The real parts of these eigenvalues will determine convergence or divergence over time.Examples of complex oscillatory systems include:

• Electrical circuits that oscillate currents and voltages
• Mechanical systems like springs that vibrate
• Economic cycles reflecting boom and bust sequences

In our case, the eigenvalues are \(-2 \pm 2i\sqrt{3}\). Their form shows that the system exhibits oscillations due to the imaginary part \(2i\sqrt{3}\) while eventually settling due to the negative real part \(-2\). This reinforces the stable spiral classification for the equilibrium.