Problem 64 - Calculus for Biology and Medicine

Question

We consider differential equations of the form \(\frac{d x}{d t}=A x(t)\) where \(A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]\) Analyze the stability of the equilibrium \(\mathbf{( 0 , 0 ) , \text { and classify the }}\) equilibrium. \(A=\left[\begin{array}{rr}-2 & -7 \\ 1 & 2\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium point is a stable spiral node.

Step 1: Determine the Equilibrium Point

For a linear system of differential equations \( \frac{d\mathbf{x}}{dt} = A \mathbf{x} \), the equilibrium point is where \( \frac{d\mathbf{x}}{dt} = 0 \). As given, the system is already analyzed at the equilibrium point \((0, 0)\), so no further action is required in this step.

Step 2: Find the Eigenvalues of Matrix A

The matrix \( A = \begin{bmatrix} -2 & -7 \ 1 & 2 \end{bmatrix} \). To analyze stability, calculate the eigenvalues by solving \( \text{det}(A - \lambda I) = 0 \), where \( I \) is the identity matrix. We have:\[ \det \left(\begin{bmatrix} -2 & -7 \ 1 & 2 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \ 0 & \lambda \end{bmatrix}\right) = 0 \] This simplifies to:\[ \det \begin{bmatrix} -2 - \lambda & -7 \ 1 & 2 - \lambda \end{bmatrix} = (-2 - \lambda)(2 - \lambda) - (-7)(1) = 0 \] Expand and simplify:\[ \lambda^2 + 4\lambda + 11 = 0 \]

Step 3: Solve the Characteristic Equation

The characteristic equation \( \lambda^2 + 4\lambda + 11 = 0 \) is quadratic. Solve for \( \lambda \) using the quadratic formula: \[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = 4, \text{ and } c = 11 \). Calculate the discriminant: \[ b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot 11 = 16 - 44 = -28 \] Since the discriminant is negative, the eigenvalues are complex: \[ \lambda = \frac{-4 \pm \sqrt{-28}}{2} = -2 \pm i \sqrt{7} \]

Step 4: Analyze Stability and Classify Equilibrium

The eigenvalues \( \lambda = -2 \pm i \sqrt{7} \) have negative real parts. This indicates that the equilibrium point is stable (as solutions decay over time), and since they are complex, the equilibrium is a spiral node, specifically a stable spiral node or focus.

Key Concepts

Eigenvalues

An essential task in analyzing differential equations is determining the eigenvalues of a matrix, which provide insight into the system's behavior. Consider a system represented by the matrix:\[A = \begin{bmatrix}-2 & -7 \1 & 2\end{bmatrix}\]Eigenvalues are crucial to understanding how solutions to differential equations evolve over time. They are found by setting up the characteristic equation derived from the matrix \(A\): - Substitute into the equation \( \text{det}(A - \lambda I) = 0 \) to form the characteristic equation.Given the matrix \(A\):- \[\det \begin{bmatrix}-2 - \lambda & -7 \1 & 2 - \lambda\end{bmatrix} = 0\]- This results in the quadratic equation \( \lambda^2 + 4\lambda + 11 = 0 \).Use the quadratic formula to solve it: \[\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For this system, we find complex eigenvalues: \( -2 \pm i\sqrt{7} \).

• The presence of complex eigenvalues indicates oscillatory behavior in the system.
• The real part of these eigenvalues \(-2\) is negative, which gives more information in the next concept.

Stability Analysis

Stability analysis involves determining whether solutions to the system's differential equation converge to an equilibrium point as time progresses. When dealing with eigenvalues, the sign of the real part is crucial:- Real part < 0: The solution tends to decay over time, indicating stability.- Real part > 0: The solution grows over time, suggesting instability.For our matrix, the eigenvalues are \(-2 \pm i \sqrt{7}\) with a negative real part of \(-2\). This indicates stability because:

• Solutions to the differential equation will eventually settle down to a stable point.
• But due to their oscillatory nature, they will spiral towards this point.

Stability is especially significant in understanding the long-term behavior of dynamic systems in real-world applications such as control systems or population models.

Equilibrium Classification

Equilibrium classification helps in understanding not only if an equilibrium point is stable or unstable but also how the system behaves as it approaches equilibrium.Once eigenvalues are determined, we classify the equilibrium:- If the eigenvalues are complex conjugates with a negative real part, the equilibrium behaves as a stable spiral node or focus.- This means the system exhibits spiraling behavior as it approaches the equilibrium point.In our example, since eigenvalues are \(-2 \pm i \sqrt{7}\):

• The real part of \(-2\) ensures that the equilibrium in question is stable.
• The imaginary component indicates this stability takes the form of a spiral.

This classification is essential for anticipating how a system returns to equilibrium after a disturbance or how it behaves naturally over time.