Problem 52 - 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. \(A=\left[\begin{array}{rr}3 & -2 \\ 1 & 3\end{array}\right]\)

Step-by-Step Solution

Answer

Unstable spiral, since the real part of eigenvalues is positive.

Step 1: Find the Eigenvalues of A

To find the eigenvalues of matrix \(A\), we solve the characteristic equation: \[ \det(A - \lambda I) = 0 \] where \(I\) is the identity matrix. For \(A = \left[\begin{array}{rr} 3 & -2 \ 1 & 3 \end{array}\right]\), the equation becomes: \[ \det \left( \begin{array}{cc} 3-\lambda & -2 \ 1 & 3-\lambda \end{array} \right) = 0 \] Calculate the determinant:\[(3-\lambda)^2 + 2 = 0\]\[\lambda^2 - 6\lambda + 13 = 0\] We now solve the quadratic equation to find eigenvalues.

Step 2: Solve the Quadratic Equation for Eigenvalues

Use the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = -6\), and \(c = 13\): \[\lambda = \frac{6 \pm \sqrt{(-6)^2 - 4\cdot1\cdot13}}{2\cdot1}\] \[\lambda = \frac{6 \pm \sqrt{36 - 52}}{2}\] \[\lambda = \frac{6 \pm \sqrt{-16}}{2}\] \[\lambda = \frac{6 \pm 4i}{2}\] \[\lambda = 3 \pm 2i\] The eigenvalues are \(3 + 2i\) and \(3 - 2i\).

Step 3: Analyze the Real Part of the Eigenvalues

The eigenvalues are complex numbers, \(3 \pm 2i\). The real part of these eigenvalues is \(3\). Since the real part is positive, the equilibrium at \((0,0)\) is classified as an unstable spiral.

Key Concepts

Eigenvalues

When dealing with differential equations, especially in the context of matrix equations, eigenvalues play a crucial role. They help in understanding how systems evolve over time.

In simple terms, eigenvalues are special numbers associated with a matrix. They give us insight into the behavior of systems described by linear equations.

• To find the eigenvalues of a matrix, we use the characteristic equation: \( \det(A - \lambda I) = 0 \). Here, \(\lambda\) is the eigenvalue, \(A\) is our matrix, and \(I\) is the identity matrix.

Solving this equation involves finding values of \(\lambda\) where the determinant equals zero.
For example, with the matrix \(A = \begin{bmatrix}3 & -2 \ 1 & 3 \end{bmatrix}\), we aim to determine values of \(\lambda\) that satisfy the equation. By doing so, we find that the eigenvalues are \(3 + 2i\) and \(3 - 2i\).

Understanding eigenvalues helps us analyze the long-term behavior of the system, shedding light on aspects such as stability and oscillation.

Complex Conjugates

In many differential equations, especially those that involve real-valued matrices, eigenvalues can be complex numbers. These numbers often appear as pairs of complex conjugates.

So, what are complex conjugates? Quite simply, if one eigenvalue of a matrix is complex, such as \(a + bi\), its complex conjugate will be \(a - bi\). Together, they form a pair.

• In our example, the eigenvalues \(3 + 2i\) and \(3 - 2i\) are complex conjugates.

Complex conjugates are crucial in understanding the rotational dynamics of a system.
They indicate that the system might exhibit spiral behavior, either moving towards or away from a central equilibrium point. This property is essential for classifying the nature of equilibrium, which directly connects to the system's stability.

Stability Analysis

Stability analysis is a fundamental concept in differential equations, helping us determine how systems respond over time. By analyzing the eigenvalues of a matrix, we can predict whether systems tend towards equilibrium or diverge away.

If we have complex eigenvalues, the real part of these values tells us about the system's stability.

• In our case, the eigenvalues \(3 \pm 2i\) have a real part of 3.
• Because this real part is positive, it signifies that the system's equilibrium at \((0,0)\) is an unstable spiral.

An unstable spiral means that, over time, the system's state will spiral outwards, moving further away from the equilibrium point.
This classification helps predict system behavior in various fields, from engineering to natural sciences, ensuring that we understand how systems react to perturbations. This knowledge is vital for designing safe, robust systems across different applications.