Problem 28 - Biocalculus Calculus for the Life Sciences

Question

Solve the initial value problem \(d \mathbf{x} / d t=A \mathbf{x}\) with \(\mathbf{x}(0)=\mathbf{x}_{0} .\) \(A=\left[ \begin{array}{ll}{3} & {-4} \\ {1} & {-3}\end{array}\right] \quad \mathbf{x}_{0}=\left[ \begin{array}{l}{2} \\ {3}\end{array}\right]\)

Step-by-Step Solution

Answer

\(\mathbf{x}(t) = \frac{5}{2}\begin{bmatrix} 4\\ 1\end{bmatrix} e^{2t} + \frac{1}{2}\begin{bmatrix} 2\\ 1\end{bmatrix} e^{-3t}\) is the solution.

Step 1: Identify the System of Differential Equations

The problem is given in matrix form as \(\frac{d \mathbf{x}}{dt} = A \mathbf{x}\). Here, \(A\) is a constant matrix, \(\mathbf{x}\) is a vector function, and \(\mathbf{x}_0\) is the initial condition vector.

Step 2: Compute Eigenvalues of Matrix A

To solve the system, first find the eigenvalues of the matrix \(A\). The characteristic equation is found from \(\det(A - \lambda I) = 0\), where \(I\) is the identity matrix. \(\begin{align*}&\det \left( \begin{bmatrix} 3 & -4 \ 1 & -3 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \right) = 0 \&\det \left( \begin{bmatrix} 3 - \lambda & -4 \ 1 & -3 - \lambda \end{bmatrix} \right) = (3-\lambda)(-3-\lambda) - (-4)(1)\& = \lambda^2 - 0\lambda -3 + 12 = \lambda^2 + \lambda - 9 = 0\end{align*}\)Solving this quadratic equation gives the eigenvalues: \(\lambda_1 = 2\), \(\lambda_2 = -3\).

Step 3: Compute Eigenvectors of Matrix A

To find the eigenvectors, solve \((A - \lambda_i I) \mathbf{v}_i = \mathbf{0}\) for each eigenvalue \(\lambda_i\).1. For \(\lambda_1 = 2\):\[\begin{bmatrix} 1 & -4 \ 1 & -5 \end{bmatrix} \begin{bmatrix} v_{11} \ v_{12} \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\]One solution is \(\mathbf{v}_1 = \begin{bmatrix} 4 \ 1 \end{bmatrix}\).2. For \(\lambda_2 = -3\):\[\begin{bmatrix} 6 & -4 \ 1 & 0 \end{bmatrix} \begin{bmatrix} v_{21} \ v_{22} \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\]One solution is \(\mathbf{v}_2 = \begin{bmatrix} 2 \ 1 \end{bmatrix}\).

Step 4: Form the General Solution

The general solution to the differential equation is a linear combination of solutions derived from eigenvalues and eigenvectors. \(\mathbf{x}(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t}\).Plugging the eigenvalues and eigenvectors, we have:\[\mathbf{x}(t) = c_1 \begin{bmatrix} 4 \ 1 \end{bmatrix} e^{2t} + c_2 \begin{bmatrix} 2 \ 1 \end{bmatrix} e^{-3t}\]

Step 5: Apply Initial Condition

Apply the initial condition \(\mathbf{x}(0) = \mathbf{x}_0 = \begin{bmatrix} 2 \ 3 \end{bmatrix}\) to determine \(c_1\) and \(c_2\):\[c_1 \begin{bmatrix} 4 \ 1 \end{bmatrix} + c_2 \begin{bmatrix} 2 \ 1 \end{bmatrix} = \begin{bmatrix} 2 \ 3 \end{bmatrix}\]This leads to the following system of equations:1. \(4c_1 + 2c_2 = 2\)2. \(c_1 + c_2 = 3\).Solving these equations, we find \(c_1 = \frac{5}{2}\) and \(c_2 = \frac{1}{2}\).

Step 6: Write the Particular Solution

With the constants \(c_1\) and \(c_2\) determined, substitute into the general solution:\[\mathbf{x}(t) = \frac{5}{2} \begin{bmatrix} 4 \ 1 \end{bmatrix} e^{2t} + \frac{1}{2} \begin{bmatrix} 2 \ 1 \end{bmatrix} e^{-3t}\].This represents the particular solution to the initial value problem.

Key Concepts

Eigenvalues

When you encounter the term eigenvalues, think of them as special scalars that offer valuable insight into a matrix's transformation behavior. They're pivotal in analyzing systems of differential equations, like in our exercise. Eigenvalues solve the so-called characteristic equation, formed by:

• Computing the determinant of \(A - \lambda I\), where \(A\) denotes our matrix, \(\lambda\) is the eigenvalue, and \(I\) represents the identity matrix.
• Setting this determinant to zero, which gives a polynomial. Solving this polynomial results in the eigenvalues.

Remember, each eigenvalue reflects how a matrix scales its eigenvectors along particular directions. In the given example, solving \(\det(A - \lambda I) = 0\) yielded \(\lambda_1 = 2\) and \(\lambda_2 = -3\). These reveal that two distinct scaling actions occur under transformation by matrix \(A\).

Eigenvectors

Just as eigenvalues are crucial in matrix analysis, eigenvectors, in tandem with them, provide profound insights into how matrices behave. An eigenvector is essentially a non-zero vector that, when multiplied by the corresponding matrix, results in a vector scaled by its eigenvalue. This means:

• When you find \((A - \lambda_i I) \mathbf{v}_i = \mathbf{0}\), \(\mathbf{v}_i\) represents the eigenvector associated with the eigenvalue \(\lambda_i\).

For our given matrix \(A\), we calculated two eigenvectors: \(\mathbf{v}_1 = \begin{bmatrix} 4 \ 1 \end{bmatrix}\) for the eigenvalue \(2\) and \(\mathbf{v}_2 = \begin{bmatrix} 2 \ 1 \end{bmatrix}\) for the eigenvalue \(-3\). These vectors indicate directions along which the initial vector \(\mathbf{x}_0\) will stretch or compress when transformed. If you visualize transformations, eigenvectors stay in their own line, pointing to the transformational power the matrix holds.

Differential Equations

Differential equations are at the heart of understanding systems that change continuously over time. They are crucial tools for pinpointing the dynamics of such systems, as they model how a vector function changes according to some rules. In our exercise, we look at a first-order system represented as \(\frac{d\mathbf{x}}{dt} = A\mathbf{x}\). Here:- The equation describes how our vector function \(\mathbf{x}\) evolves over time.- The matrix \(A\) determines the nature of this transformation.
Solving these equations often involves identifying the relationship between the rate of change and the current state. For our exercise, the key to solving is leveraging eigenvalues and eigenvectors to express the solution as a function of time, capturing the interaction between the dynamics expressed by \(A\) and the initial condition \(\mathbf{x}_0\). Thus, differential equations serve as a powerful framework to model and predict complex phenomena's behavior over time.

Matrix Exponentiation

Matrix exponentiation might sound intimidating, but it's simply about applying the concept of exponents to matrices, especially in the context of solving linear differential equations. In problems like ours, calculating matrix exponentials aids in deriving a time-dependent solution that fits the given initial condition. Imagine:

• The general solution of the differential equation \(\mathbf{x}(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t}\) involves exponentiating the eigenpairs.
• Each eigenpair's contribution to the solution is scaled by an exponential factor. This reflects how quickly or slowly each mode grows or decays over time.

In the given exercise, matrix exponentiation incorporates both our computed eigenvalues and eigenvectors, creating expressions like \(e^{\lambda_1 t}\) and \(e^{\lambda_2 t}\) which, when multiplied, tailor the behavior of the solution over time. Understanding matrix exponentiation provides insight into the longevity and stability of solutions in evolving systems, making it fundamental for solving differential equations with matrices.