Problem 17 - Biocalculus Calculus for the Life Sciences

Question

Solve the initial value problem \(d \mathbf{x} / d t=A \mathbf{x}\) with \(\mathbf{x}(0)=\mathbf{x}_{0} .\) \(A=\left[ \begin{array}{rr}{-\frac{3}{2}} & {\frac{1}{2}} \\ {\frac{1}{2}} & {-\frac{3}{2}}\end{array}\right] \quad \mathbf{x}_{0}=\left[ \begin{array}{l}{1} \\ {2}\end{array}\right]\)

Step-by-Step Solution

Answer

\( \mathbf{x}(t) = \begin{bmatrix} \frac{3}{2} e^{-t} - \frac{1}{2} e^{-2t} \\ \frac{3}{2} e^{-t} + \frac{1}{2} e^{-2t} \end{bmatrix} \)

Step 1: Setup the Equation

The given problem is a system of linear differential equations. We need to solve \( \frac{d \mathbf{x}}{dt} = A \mathbf{x} \) where \( A = \begin{bmatrix} -\frac{3}{2} & \frac{1}{2} \ \frac{1}{2} & -\frac{3}{2} \end{bmatrix} \) and \( \mathbf{x}(0) = \mathbf{x}_0 = \begin{bmatrix} 1 \ 2 \end{bmatrix} \).

Step 2: Find the Eigenvalues of A

To solve the system, we first find the eigenvalues of matrix \( A \) by solving the characteristic equation \( \det(A - \lambda I) = 0 \). Compute the determinant: \[ \det \left( \begin{bmatrix} -\frac{3}{2}-\lambda & \frac{1}{2} \ \frac{1}{2} & -\frac{3}{2}-\lambda \end{bmatrix} \right) = \left(-\frac{3}{2}-\lambda\right)^2 - \left(\frac{1}{2}\right)^2 \]Simplify to get the quadratic equation: \[ \lambda^2 + 3\lambda + 1 = 0 \]Solving this gives eigenvalues \( \lambda_1 = -1 \) and \( \lambda_2 = -2 \).

Step 3: Find the Eigenvectors for Each Eigenvalue

For \( \lambda_1 = -1 \): Solve \( (A + I)\mathbf{v} = 0 \).Substitute in \( \lambda_1 \): \[ \begin{bmatrix} -\frac{3}{2}+1 & \frac{1}{2} \ \frac{1}{2} & -\frac{3}{2}+1 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \]Simplifies to: \[ \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & -\frac{1}{2} \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \]Eigenvector: \( v_1 = v_2 \). Choose eigenvector \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 1 \end{bmatrix} \).For \( \lambda_2 = -2 \): Solve similarly to find \( \mathbf{v}_2 = \begin{bmatrix} 1 \ -1 \end{bmatrix} \).

Step 4: Form the General Solution

The general solution of the differential equation system can be written as a linear combination of exponentials involving the eigenvectors, given by:\( \mathbf{x}(t) = c_1 e^{-t} \begin{bmatrix} 1 \ 1 \end{bmatrix} + c_2 e^{-2t} \begin{bmatrix} 1 \ -1 \end{bmatrix} \).

Step 5: Apply Initial Conditions

We apply the initial condition \( \mathbf{x}(0) = \begin{bmatrix} 1 \ 2 \end{bmatrix} \) to find constants \( c_1 \) and \( c_2 \).\( \begin{bmatrix} 1 \ 2 \end{bmatrix} = c_1 \begin{bmatrix} 1 \ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \ -1 \end{bmatrix} \)Solve the system: \( c_1 + c_2 = 1 \) and \( c_1 - c_2 = 2 \).Solving gives \( c_1 = \frac{3}{2} \) and \( c_2 = -\frac{1}{2} \).

Step 6: Write the Solution Satisfying Initial Conditions

Substitute the computed \( c_1 \) and \( c_2 \) into the general solution:\( \mathbf{x}(t) = \frac{3}{2} e^{-t} \begin{bmatrix} 1 \ 1 \end{bmatrix} - \frac{1}{2} e^{-2t} \begin{bmatrix} 1 \ -1 \end{bmatrix} \)Combine and simplify:\( \mathbf{x}(t) = \begin{bmatrix} \frac{3}{2} e^{-t} - \frac{1}{2} e^{-2t} \ \frac{3}{2} e^{-t} + \frac{1}{2} e^{-2t} \end{bmatrix} \)

Key Concepts

Initial Value Problem

An initial value problem in the context of differential equations refers to a differential equation along with a specific condition called an initial value. This initial value dictates the state of the dependent variable at a particular point, often serving as a starting point for finding a unique solution to the differential equation. In this exercise, the problem was to solve the system of linear differential equations
\[ \frac{d \mathbf{x}}{dt} = A \mathbf{x} \]with initial condition:
\[ \mathbf{x}(0) = \begin{bmatrix} 1 \ 2 \end{bmatrix} \]This defines the state of \( \mathbf{x} \) at time \( t = 0 \), ensuring a unique solution by restricting the infinite possibilities to one particular trajectory. Typically, solving an initial value problem requires integrating the differential equations, often through various techniques or tricks when the system or equations are too complex for direct integration.

• Ensures a unique solution by providing a specific condition.
• Often requires eigenvalue and eigenvector analysis in the case of systems of equations.
• Limits the general solution to satisfy the initial constraint at \( t = 0 \).

By solving the initial value problem, we found a specific expression for \( \mathbf{x}(t) \) that satisfies both the differential equations and the given starting condition.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra that help in analyzing linear transformations that frequently appear in solving systems of linear differential equations. In brief, an eigenvalue is a scalar, and an eigenvector is a nonzero vector that, when a matrix is multiplied by it, results in a vector that is a scaled version of the original vector.
For any square matrix \( A \), an eigenvalue \( \lambda \) and an eigenvector \( \mathbf{v} \) satisfying:
\[ A \mathbf{v} = \lambda \mathbf{v} \]
In this exercise, we found the eigenvalues by solving:
\[ \det(A - \lambda I) = 0 \]
Resulting in eigenvalues \( \lambda_1 = -1 \) and \( \lambda_2 = -2 \). To find corresponding eigenvectors, we solve:
\[ (A - \lambda I)\mathbf{v} = 0 \]
For each eigenvalue, giving us their associated vectors.

• Help simplify complex systems by turning them into separate, simpler problems.
• Determine important properties of matrices, like stability and system behavior.
• In diagonalizing matrices or systems, they make solving differential equations straightforward.

These calculations allowed us to express the solution in terms of exponential functions of these eigenvalues, facilitating the resolution of the initial value problem.

System of Linear Equations

A system of linear equations consists of multiple equations involving the same set of variables. In this exercise, the differential equation can be seen as a system of equations:\[ \frac{d \mathbf{x}}{dt} = A \mathbf{x} \]Here \( A \) is a matrix and \( \mathbf{x} \) is the variable vector. Solving systems of linear equations using matrices simplifies addressing multiple equations with multiple unknowns simultaneously.

• Matrices provide a compact representation of systems of equations.
• Eases manipulation compared to dealing with each equation separately.
• Beneficial when applying algebraic techniques like row reduction or finding inverse matrices.

In our problem, eigenvalue techniques broke down the matrix \( A \) into simpler systems for each eigenvalue, which we then represented using their respective eigenvectors. This greatly simplified the process of solving the initial value problem.