Problem 33 - 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 equilibrium according to whether it is a sink, a source, or a saddle point. $$ A=\left[\begin{array}{ll} -4 & 2 \\ -5 & 3 \end{array}\right] $$

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is a saddle point, indicating instability.

Step 1: Find the Characteristic Equation

To analyze the stability of the equilibrium, we first need to find the eigenvalues of matrix \(A\). We do this by solving the characteristic equation \( \det(A - \lambda I) = 0 \). With the given matrix \(A\), we have:\[A = \begin{bmatrix} -4 & 2 \ -5 & 3 \end{bmatrix}\]The characteristic equation is found by:\[\det\left(\begin{bmatrix} -4 & 2 \ -5 & 3 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\right) = 0\]\[\det\left(\begin{bmatrix} -4 - \lambda & 2 \ -5 & 3 - \lambda \end{bmatrix}\right) = 0\]Calculate the determinant:\[(-4 - \lambda)(3 - \lambda) - (2)(-5) = 0\]\[\lambda^2 + \lambda - 2 = 0\]

Step 2: Solve for Eigenvalues

We solve the quadratic equation \( \lambda^2 + \lambda - 2 = 0 \) using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a=1, b=1, c=-2 \):\[\lambda = \frac{-1 \pm \sqrt{1^2 - 4(1)(-2)}}{2(1)}\]\[\lambda = \frac{-1 \pm \sqrt{1 + 8}}{2}\]\[\lambda = \frac{-1 \pm 3}{2}\]This gives us \( \lambda_1 = 1 \) and \( \lambda_2 = -2 \).

Step 3: Determine Stability

Since the eigenvalues \( \lambda_1 = 1 \) and \( \lambda_2 = -2 \) are real, distinct, and of opposite signs, the equilibrium point \((0,0)\) is a saddle point. A saddle point means that the equilibrium is unstable because small perturbations will grow in some directions.

Step 4: Classify the Equilibrium

With one eigenvalue positive and the other negative, the equilibrium can be classified as a saddle point. Saddle points lead to instability as trajectories move away from the equilibrium in one direction.

Key Concepts

Eigenvalues

In the realm of differential equations, eigenvalues help us understand how solutions behave over time around an equilibrium point. They originate from the concept of linear transformations and matrices. When you have a matrix, say \( A \), and you multiply it by a vector \( x \), sometimes the vector only stretches or shrinks but does not change direction. This interaction is fascinating, and these particular vectors are called eigenvectors, while the factors by which the vectors stretch or shrink, are the eigenvalues.
To find the eigenvalues, you subtract \( \lambda I \) from the matrix \( A \) and set the determinant of the result equal to zero. For our example matrix:

• \( A = \begin{bmatrix} -4 & 2 \ -5 & 3 \end{bmatrix} \)
• Solve the characteristic equation \( \det(A - \lambda I) = 0 \)
• This gives the quadratic equation \( \lambda^2 + \lambda - 2 = 0 \)

Solving this, you discover the eigenvalues \( \lambda_1 = 1 \) and \( \lambda_2 = -2 \). These numeric eigenvalues become crucial in predicting the system's dynamics.

Stability Analysis

Stability analysis is an essential part of studying differential equations because it reveals the behavior of solutions near equilibrium points. An equilibrium point is stable if small disturbances to the system result in behavior that eventually returns to equilibrium.
For a matrix \( A \) in the form \( \frac{d \mathbf{x}}{dt} = A \mathbf{x}(t) \), the eigenvalues are the key to determining stability:

• If all eigenvalues have negative real parts, the equilibrium point is stable (often called a sink).
• If any eigenvalue has a positive real part, the equilibrium point is unstable (often termed a source or a saddle point).

In our scenario with eigenvalues \( \lambda_1 = 1 \) and \( \lambda_2 = -2 \), one eigenvalue is positive and the other negative, directly indicating instability as the solution runs away in at least one direction.

Saddle Point

A saddle point occurs in systems described by differential equations when you have eigenvalues of opposite signs, as in our case with \( \lambda_1 = 1 \) and \( \lambda_2 = -2 \). This contrast in signs means that while one direction seems to be pulled back towards the equilibrium, the other direction pushes away.
You can think about it like rolling a marble on a saddle-shaped surface: in some directions, the marble will roll away from the center, while in others, it will roll back towards it. This analogy helps visualize why a saddle point is inherently unstable.

• The positive eigenvalue suggests divergence from equilibrium along certain vectors.
• The negative eigenvalue suggests convergence along others.

As a result of this mixed behavior, any small change in the system leads it to stray away from being balanced at the saddle point, thus illustrating why systems modeled with such conditions tend to instability.