Problem 18 - Biocalculus Calculus for the Life Sciences

Question

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

Step-by-Step Solution

Answer

\( \mathbf{x}(t) = \frac{3}{2} e^{1.5t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} - \frac{1}{2} e^{-t} \begin{bmatrix} 1 \\ -1 \end{bmatrix}. \)

Step 1: Write the System of Differential Equations

Given the matrix \(A\) and the vector \(\mathbf{x}_0\), the system of differential equations is: \[ \frac{d \mathbf{x}}{dt} = A \mathbf{x} \rightarrow \frac{d}{dt}\begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & -\frac{3}{2} \ -\frac{3}{2} & \frac{1}{2} \end{bmatrix}\begin{bmatrix} x_1 \ x_2 \end{bmatrix}. \]

Step 2: Find Eigenvalues of Matrix A

To solve the system, we first find the eigenvalues of \(A\). Solve the characteristic equation \(\det(A - \lambda I) = 0\). For \(A\), this gives: \[ \det\begin{pmatrix} \frac{1}{2} - \lambda & -\frac{3}{2} \ -\frac{3}{2} & \frac{1}{2} - \lambda \end{pmatrix} = \left(\frac{1}{2} - \lambda\right)^2 - \left(-\frac{3}{2}\right)^2 = 0. \] Simplifying, \[ \lambda^2 - \lambda + 2.25 = 0. \] Solving this quadratic equation gives eigenvalues \(\lambda_1 = 1.5\) and \(\lambda_2 = -1.0\).

Step 3: Find Eigenvectors Corresponding to Eigenvalues

For \(\lambda_1 = 1.5\), solve \((A - 1.5I)\mathbf{v}_1 = 0\):\[ A - 1.5I = \begin{bmatrix} -1 & -\frac{3}{2} \ -\frac{3}{2} & -1 \end{bmatrix}. \] Solving \(A\mathbf{v}_1 = 0\), we find \(\mathbf{v}_1 = \begin{bmatrix} 1 \ 1 \end{bmatrix}\). For \(\lambda_2 = -1.0\), solve \( (A + I)\mathbf{v}_2 = 0\):\[ A + I = \begin{bmatrix} 1.5 & -\frac{3}{2} \ -\frac{3}{2} & 1.5 \end{bmatrix}. \] Solving \(A\mathbf{v}_2 = 0\), we find \(\mathbf{v}_2 = \begin{bmatrix} 1 \ -1 \end{bmatrix}\).

Step 4: Construct General Solution Using Eigenvalues and Eigenvectors

The general solution involves linear combinations of the eigenvectors scaled by their respective exponential terms: \[ \mathbf{x}(t) = c_1 e^{1.5t} \begin{bmatrix} 1 \ 1 \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} 1 \ -1 \end{bmatrix}. \]

Step 5: Solve for Constants Using Initial Conditions

Apply the initial condition \(\mathbf{x}(0) = \mathbf{x}_0\):\[ \mathbf{x}(0) = c_1 \begin{bmatrix} 1 \ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \ -1 \end{bmatrix} = \begin{bmatrix} 1 \ 2 \end{bmatrix}. \] This results in the system of equations:\[ c_1 + c_2 = 1 \quad \text{and} \quad c_1 - c_2 = 2. \] Solving these gives \(c_1 = \frac{3}{2}, c_2 = -\frac{1}{2}\).

Step 6: Write Final Solution with Constants

Substitute the constants back into the general solution to find the specific solution:\[ \mathbf{x}(t) = \frac{3}{2} e^{1.5t} \begin{bmatrix} 1 \ 1 \end{bmatrix} - \frac{1}{2} e^{-t} \begin{bmatrix} 1 \ -1 \end{bmatrix}. \]

Key Concepts

Eigenvalues

Eigenvalues are critical values in the context of systems of differential equations, especially when we deal with matrix equations. When determining eigenvalues for a matrix, you're essentially finding values that scale a vector in the linear transformation described by the matrix. Each eigenvalue corresponds to a factor by which an eigenvector is stretched or shrunk during transformation. To find eigenvalues, we use the characteristic equation, which involves the determinant of \( A - \lambda I \), where \( A \) is your matrix and \( I \) is the identity matrix.

In our case, we found that the eigenvalues of the matrix \( A \) were \( \lambda_1 = 1.5 \) and \( \lambda_2 = -1.0 \). These values are solutions to the equation that arises from the characteristic polynomial, which for our matrix \( A \) is \( \lambda^2 - \lambda + 2.25 = 0 \). These eigenvalues tell us how the dynamic system expands or contracts along certain directions in the solution space.

Eigenvectors

Once eigenvalues are determined, we shift focus to eigenvectors. An eigenvector is a non-zero vector that changes only by a scalar factor when a linear transformation is applied. In other words, if \( \mathbf{v} \) is an eigenvector of \( A \), applying \( A \) to \( \mathbf{v} \) simply scales \( \mathbf{v} \) by its corresponding eigenvalue \( \lambda \), i.e., \( A\mathbf{v} = \lambda \mathbf{v} \).

For our matrix \( A \), the eigenvectors we found were:

• \( \mathbf{v}_1 = \begin{bmatrix} 1 & 1 \end{bmatrix} \) for the eigenvalue \( \lambda_1 = 1.5 \)
• \( \mathbf{v}_2 = \begin{bmatrix} 1 & -1 \end{bmatrix} \) for the eigenvalue \( \lambda_2 = -1.0 \)

These eigenvectors point in the directions in which the scaling (represented by the eigenvalues) happens. The result is a clear picture of the system's dynamics and how each part contributes uniquely to the behavior of the solution.

Initial Value Problem

An initial value problem (IVP) is a differential equation along with specified values at a starting point, known as initial conditions. In our case, we are given \( \frac{d \mathbf{x}}{dt} = A \mathbf{x} \) and \( \mathbf{x}(0) = \mathbf{x}_0 \). The initial value \( \mathbf{x}_0 = \begin{bmatrix} 1 & 2 \end{bmatrix} \) tells us the starting state of the system.

Solving an IVP involves not only finding the general solution to the differential equation but also determining specific constants based on the initial conditions. This leads to a particular solution that satisfies both the differential equation and the initial condition \( \mathbf{x}(0) = \mathbf{x}_0 \). By applying these conditions, constants in the general solution are determined, helping accurately describe the system's behavior from the starting point.

Matrix Equations

Matrix equations are essential in representing systems of linear differential equations. Here, matrices simplify the representation and manipulation of linear systems. Our primary matrix equation is \( \frac{d \mathbf{x}}{dt} = A \mathbf{x} \), which is expressive enough to encompass the behavior of multiple linear equations simultaneously.

In our example, the matrix \( A \) is \( \begin{bmatrix} \frac{1}{2} & -\frac{3}{2} \ -\frac{3}{2} & \frac{1}{2} \end{bmatrix} \). This matrix encompasses all the coefficients of the system and dictates how each variable interacts with each other. Through matrix theory, we use techniques like determining eigenvalues and eigenvectors to reformulate and solve differential equations more straightforwardly, ultimately leading to precise solutions that describe the entire system comprehensively.