Problem 47 - 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. \(A=\left[\begin{array}{ll}0 & -1 \\ 1 & -1\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium (0,0) is a stable spiral.

Step 1: Setup the Characteristic Equation

To analyze the stability of the equilibrium point (0,0), we first need to find the eigenvalues of matrix \(A\). The eigenvalues are found using the determinant of \( (A - \lambda I) \), where \( I \) is the identity matrix. \[\det \left( A - \lambda I \right) = \det \left( \begin{bmatrix} -\lambda & -1 \ 1 & -1 - \lambda \end{bmatrix} \right) = 0\]

Step 2: Calculate the Determinant

The determinant of the matrix \(\begin{bmatrix} -\lambda & -1 \ 1 & -1 - \lambda \end{bmatrix}\) is given by the formula \((-\lambda)(-1-\lambda) - (-1)(1) = \lambda^2 + \lambda + 1\). Set this equal to zero to find the eigenvalues: \[\lambda^2 + \lambda + 1 = 0.\]

Step 3: Solve the Quadratic Equation

To find the eigenvalues, we need to solve the quadratic equation \( \lambda^2 + \lambda + 1 = 0 \). Using the quadratic formula: \[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = 1, c = 1 \). This gives: \[ \lambda = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}. \]

Step 4: Interpret the Eigenvalues

The eigenvalues \( \lambda = \frac{-1}{2} \pm \frac{i\sqrt{3}}{2} \) are complex with a negative real part. This indicates the system is a stable spiral. Complex eigenvalues with negative real parts imply the trajectories spiral towards the equilibrium point over time.

Step 5: Classify the Equilibrium

Since the eigenvalues have negative real parts and are complex, the equilibrium point \((0,0)\) is classified as a stable spiral. This means trajectories near \((0,0)\) will spiral inward as time progresses towards the equilibrium.

Key Concepts

Eigenvalues

Understanding eigenvalues is key when working with differential equations. They offer insight into the behavior of systems described by such equations, especially regarding stability.

• An eigenvalue is a scalar that determines whether or not a certain transformation, represented by a matrix, has any effect on a vector, known as an eigenvector, aside from possible scaling.
• The eigenvalues are calculated from the matrix associated with the system, in this case, the matrix A. The characteristic equation, typically of the form \(\det(A - \lambda I) = 0\), is solved to find these values.
• For our system, we found that the characteristic equation is \( \lambda^2 + \lambda + 1 = 0 \), which led us to calculate the eigenvalues \( \lambda = \frac{-1}{2} \pm \frac{i\sqrt{3}}{2} \).

This mathematical process involves using the quadratic formula, particularly when complex conjugates appear, as they do in our case. It's crucial to recognize how these conjugate pairs relate to stability and the nature of the system.

Stability Analysis

Stability analysis in differential equations refers to understanding how solutions behave as they approach an equilibrium point over time.

• Specifically, we are looking at the stability of the equilibrium point \((0,0)\).
• The real part of the eigenvalues plays a critical role in determining stability. If their sum (or individual parts in the case of complex numbers) is negative, it implies exponential decay towards the equilibrium.
• For the system described by matrix A, the eigenvalues, \( \frac{-1}{2} \pm \frac{i\sqrt{3}}{2} \), have negative real parts.
• This negative real component indicates that the system is stable, with solutions eventually converging towards equilibrium.

In essence, the behavior of solutions as time tends to infinity tells us about the long-term trends of the system. For negative real parts, those trends involve moving ever closer to stability and equilibrium, without oscillation.

Equilibrium Classification

Classifying equilibria helps in understanding the qualitative behavior of dynamical systems near an equilibrium point.

• An equilibrium point could be categorized as a nodal sink, nodal source, saddle point, stable spiral, unstable spiral, or a center, based on the nature of the eigenvalues.
• In our scenario, the equilibrium at \((0,0)\) is evaluated using the nature of eigenvalues, which are complex conjugates with negative real parts.
• The classification as a stable spiral arises because the negative real parts imply solutions spiral inward, while the imaginary components result in oscillatory motion.
• This means as time progresses, solutions not only spiral inwards but do so in a manner that exhibits rotation around the equilibrium point, eventually settling at the point itself.

Recognizing this behavior is critical for predicting system responses and for developing intuitive understanding of physical phenomena modeled by these equations.