Problem 27 - Calculus for Biology and Medicine

Question

We discuss the case of repeated eigenvalues. $$ \left[\begin{array}{l} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ (a) Show that $$ A=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] $$ has the repeated eigenvalues \(\lambda_{1}=\lambda_{2}=1\). (b) Show that \(\left[\begin{array}{l}1 \\ 0\end{array}\right]\) and \(\left[\begin{array}{l}0 \\ 1\end{array}\right]\) are eigenvectors of \(A\) and that any vector \(\left[\begin{array}{l}c_{1} \\ c_{2}\end{array}\right]\) can be written as $$ \left[\begin{array}{l} c_{1} \\ c_{2} \end{array}\right]=c_{1}\left[\begin{array}{l} 1 \\ 0 \end{array}\right]+c_{2}\left[\begin{array}{l} 0 \\ 1 \end{array}\right] $$ (c) Show that $$ \mathbf{x}(t)=c_{1} e^{t}\left[\begin{array}{l} 1 \\ 0 \end{array}\right]+c_{2} e^{t}\left[\begin{array}{l} 0 \\ 1 \end{array}\right] $$ is a solution of \((11.34)\) that satisfies the initial condition \(x_{1}(0)=c_{1}\) and \(x_{2}(0)=c_{2}\).

Step-by-Step Solution

Answer

Matrix A has repeated eigenvalues 1; all vectors in \( \mathbb{R}^2 \) are linear combinations of its eigenvectors. \( \mathbf{x}(t) = c_1 e^t \mathbf{v}_1 + c_2 e^t \mathbf{v}_2 \) satisfies the differential equation and initial conditions.

Step 1: Calculate the Eigenvalues of A

The matrix \( A = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \) is a diagonal matrix, so its eigenvalues are directly the entries on its diagonal. Therefore, the eigenvalues of \( A \) are \( \lambda_1 = 1 \) and \( \lambda_2 = 1 \). These are repeated eigenvalues since both are equal to 1.

Step 2: Verify the Eigenvectors

An eigenvector \( \mathbf{v} \) corresponding to an eigenvalue \( \lambda \) satisfies the equation \( A\mathbf{v} = \lambda\mathbf{v} \). For \( \lambda = 1 \), we check \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \) and \( \mathbf{v}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix} \). Calculating: \( A\mathbf{v}_1 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \ 0 \end{bmatrix} = \begin{bmatrix} 1 \ 0 \end{bmatrix} \), which equals \( 1 \cdot \mathbf{v}_1 \), and \( A\mathbf{v}_2 = \begin{bmatrix} 0 \ 1 \end{bmatrix} = 1 \cdot \mathbf{v}_2 \). This confirms both are eigenvectors corresponding to \( \lambda = 1 \).

Step 3: Express Any Vector as a Linear Combination

Any vector \( \mathbf{v} = \begin{bmatrix} c_1 \ c_2 \end{bmatrix} \) can indeed be expressed as \( c_1 \begin{bmatrix} 1 \ 0 \end{bmatrix} + c_2 \begin{bmatrix} 0 \ 1 \end{bmatrix} \). This is a standard representation for vectors in terms of the basis formed by the eigenvectors.

Step 4: Construct the General Solution

For the matrix differential equation \( \frac{d\mathbf{x}}{dt} = A\mathbf{x} \), the general solution is \( \mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 \). Substituting \( \lambda_1 = \lambda_2 = 1 \), \( \mathbf{x}(t) = c_1 e^t \begin{bmatrix} 1 \ 0 \end{bmatrix} + c_2 e^t \begin{bmatrix} 0 \ 1 \end{bmatrix} \).

Step 5: Verify Initial Condition

At \( t = 0 \), the solution \( \mathbf{x}(0) = c_1 e^0 \begin{bmatrix} 1 \ 0 \end{bmatrix} + c_2 e^0 \begin{bmatrix} 0 \ 1 \end{bmatrix} = \begin{bmatrix} c_1 \ c_2 \end{bmatrix} \). This satisfies the initial conditions \( x_1(0) = c_1 \) and \( x_2(0) = c_2 \), confirming the solution.

Key Concepts

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, vector spaces, and linear transformations. It is foundational for understanding complex mathematical concepts and is applied in various scientific and engineering disciplines.

In linear algebra, matrices play a central role. A matrix is essentially a rectangular array of numbers or functions that can represent linear transformations or systems of linear equations. In our exercise, the matrix \[A = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]is a simple diagonal matrix, which is particularly interesting because its eigenvalues are directly the elements on the diagonal.

Understanding these properties allows us to analyze and solve systems of equations efficiently. The initial system of differential equations \[\begin{bmatrix} \frac{dx_1}{dt} \ \frac{dx_2}{dt} \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix}\]utilizes concepts of linear algebra to describe the behavior of the system over time.

Eigenvectors

Eigenvectors are special vectors associated with a linear transformation represented by a matrix. When a matrix transforms a vector and the resulting vector is a scalar multiple of the original, that vector is called an eigenvector. The scalar is known as an eigenvalue.

In our example, the eigenvectors of matrix \[A = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]corresponding to eigenvalue \( \lambda = 1 \) are:

• \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \)
• \( \begin{bmatrix} 0 \ 1 \end{bmatrix} \)

These vectors span the space and serve as a basis for expressing any vector in the space as a linear combination of the eigenvectors. This is verified by showing that for each eigenvector \( \mathbf{v} \), \[A\mathbf{v} = \lambda \mathbf{v}\]both results in the original vector scaled by the eigenvalue.

Such properties are leveraged in solving systems of linear equations, simplifying calculations, and in various fields like quantum mechanics and stability analysis.

Differential Equations

Differential equations involve functions and their derivatives, describing how quantities change over time. They are essential in modeling real-world phenomena in physics, biology, economics, and engineering.

The system of differential equations given in the exercise is:\[\begin{bmatrix} \frac{dx_1}{dt} \ \frac{dx_2}{dt} \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix}\]It is a linear system with constant coefficients, making it straightforward to solve using methods from linear algebra. We find that the solution is of the form:\[\mathbf{x}(t) = c_1 e^{t} \begin{bmatrix} 1 \ 0 \end{bmatrix} + c_2 e^{t} \begin{bmatrix} 0 \ 1 \end{bmatrix}\]where \(c_1\) and \(c_2\) are constants determined by initial conditions.

This solution illustrates the direct application of eigenvalues and eigenvectors in constructing solutions to differential equations. Each term in the solution corresponds to an eigenvalue-eigenvector pair of the matrix \(A\), underscoring the intimate connection between linear algebra and differential equations.