Problem 67 - Calculus for Biology and Medicine

Question

The following system has two distinct real eigenvalues, but one eigenvalue is equal to 0 : $$ \frac{d \mathbf{x}}{d t}=\left[\begin{array}{ll} 4 & 2 \\ 2 & 1 \end{array}\right] \mathbf{x}(t), x(t)=\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ (a) Find both eigenvalues and the associated eigenvectors. (b) From the general solution of (11.27) find \(x_{1}(t)\) and \(x_{2}(t)\). (c) The vector field is shown in Figure \(11.38\). Sketch the lines corresponding to the eigenvectors. Compute \(d x_{2} / d x_{1}\), and conclude that all direction vectors are parallel to the line of eigenvectors corresponding to the nonzero eigenvalue. Describe in words how solutions starting at different points behave.

Step-by-Step Solution

Answer

Eigenvalues: 0, 5; Eigenvectors: \( \begin{bmatrix} 1 \\ -2 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \end{bmatrix} \). Solutions grow along \( \mathbf{v}_2 \).

Step 1: Define the System Matrix

The system matrix is given as \( A = \begin{bmatrix} 4 & 2 \ 2 & 1 \end{bmatrix} \). Our task is to find its eigenvalues and eigenvectors.

Step 2: Find the Characteristic Polynomial

To find the eigenvalues, we first determine the characteristic polynomial, which is given by \( \det(A - \lambda I) = 0 \). Let \( I \) be the identity matrix, \( \lambda \) be the eigenvalue we want to calculate:\[A - \lambda I = \begin{bmatrix} 4 - \lambda & 2 \ 2 & 1 - \lambda \end{bmatrix}\]Calculating the determinant results in:\[\det(A - \lambda I) = (4-\lambda)(1-\lambda) - (2 \times 2) = \lambda^2 - 5\lambda + 0\]

Step 3: Find Eigenvalues

The eigenvalues are the solutions to the characteristic polynomial \( \lambda^2 - 5\lambda = 0 \). Factorizing gives us:\[\lambda(\lambda - 5) = 0\]The eigenvalues are \( \lambda_1 = 0 \) and \( \lambda_2 = 5 \).

Step 4: Find Eigenvectors for Eigenvalue λ=0

Set \( \lambda = 0 \) in \( A - \lambda I \):\[\begin{bmatrix} 4 & 2 \ 2 & 1 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\]Solving this system, we get the equation system:\[4x_1 + 2x_2 = 0\]Simplifying, \( x_2 = -2x_1 \). Arbitrarily choosing \( x_1 = 1 \), we find the eigenvector \( \mathbf{v}_1 = \begin{bmatrix} 1 \ -2 \end{bmatrix} \).

Step 5: Find Eigenvectors for Eigenvalue λ=5

Set \( \lambda = 5 \) in \( A - \lambda I \):\[\begin{bmatrix} -1 & 2 \ 2 & -4 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\]Solving, we get:\[-x_1 + 2x_2 = 0 \Rightarrow x_1 = 2x_2\]Choosing \( x_2 = 1 \), the eigenvector is \( \mathbf{v}_2 = \begin{bmatrix} 2 \ 1 \end{bmatrix} \).

Step 6: Derive the General Solution

Given eigenvalues and eigenvectors, the general solution is:\[\mathbf{x}(t) = c_1 \left( \begin{bmatrix} 1 \ -2 \end{bmatrix} \right) + c_2 e^{5t} \left( \begin{bmatrix} 2 \ 1 \end{bmatrix} \right)\]

Step 7: Analyze the Solution and Vector Field

The direction vector \( \mathbf{v}_2 = \begin{bmatrix} 2 \ 1 \end{bmatrix} \) corresponding to \( \lambda = 5 \) dominates the behavior since it is non-zero. All lines in the phase space are parallel to this eigenvector, \( dx_2/dx_1 = 1/2 \), matching the slope of \( \mathbf{v}_2 \). Solutions grow explosively along this direction.

Key Concepts

Characteristic Polynomial

The characteristic polynomial is a foundational concept in linear algebra that allows us to find the eigenvalues of a matrix. Specifically, it originates from the determinant of the matrix equation \(A - \lambda I\), where \(A\) is your system matrix, \(I\) is the identity matrix, and \(\lambda\) represents the eigenvalues we are solving for. For our matrix:

• Given the matrix \(A = \begin{bmatrix} 4 & 2 \ 2 & 1 \end{bmatrix}\), the characteristic equation is determined by \(\det(A - \lambda I) = 0\).
• Substituting, we have: \(\begin{bmatrix} 4 - \lambda & 2 \ 2 & 1 - \lambda \end{bmatrix}\).
• Solving for the determinant: \((4-\lambda)(1-\lambda) - 4 = \lambda^2 - 5\lambda = 0\).
• The polynomial we derived is \(\lambda(\lambda - 5) = 0\), revealing two eigenvalues: \(\lambda_1 = 0\) and \(\lambda_2 = 5\).

Understanding and forming the characteristic polynomial is crucial because it unlocks the key to identifying behaviors within a dynamical system, like how certain solutions evolve over time.

General Solution

The general solution of a differential equation, especially when dealing with systems of equations, describes all possible solutions that satisfy the given conditions. Here, we use eigenvalues and eigenvectors to form the general solution for the system matrix.To form the general solution:

• Start by identifying the eigenvectors associated with each eigenvalue.
• For \(\lambda_1 = 0\), the eigenvector is \(\begin{bmatrix} 1 \ -2 \end{bmatrix}\).
• For \(\lambda_2 = 5\), the eigenvector is \(\begin{bmatrix} 2 \ 1 \end{bmatrix}\).

Combining these, the general solution is:\[\mathbf{x}(t) = c_1 \begin{bmatrix} 1 \ -2 \end{bmatrix} + c_2 e^{5t} \begin{bmatrix} 2 \ 1 \end{bmatrix}\]Here,- \(c_1\) and \(c_2\) are constants determined by initial conditions.- The term \(e^{5t}\) indicates exponential growth due to the positive eigenvalue \(\lambda_2 = 5\), causing the solution to grow exponentially over time along the direction of its corresponding eigenvector.

Direction Vectors

Direction vectors are critical for understanding the geometry of a solution in phase space, illustrating how solutions behave over time. They directly relate to eigenvectors, which denote directions in which the transformation (represented by the matrix) scales the vector by the eigenvalue.For this exercise:

• The direction vector associated with \(\lambda_2 = 5\) is \(\begin{bmatrix} 2 \ 1 \end{bmatrix}\).
• This vector significantly affects the system's long-term behavior, as all trajectories are drawn towards the direction it suggests due to the exponential growth term \(e^{5t}\).
• The slope \(\frac{dx_2}{dx_1} = \frac{1}{2}\) of this vector matches the direction, indicating that the trajectory will be parallel to this line.

This means that for different initial conditions, provided \(c_2\) is not zero, solutions will align along this slope, leading to outcomes where all but special cases (when \(c_2 = 0\)) are influenced heavily by the eigenvector corresponding to the non-zero eigenvalue.