Question

In each of Problems 9 and 10 find the solution of the given initial value problem. Describe the behavior of the solution as \(t \rightarrow \infty\). $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{1} & {-5} \\ {1} & {-3}\end{array}\right) \mathbf{x}, \quad \mathbf{x}(0)=\left(\begin{array}{l}{1} \\ {1}\end{array}\right) $$

Short answer

In summary, given the matrix differential equation with initial value \(\mathbf{x}(0) = \left(\begin{array}{l}{1} \\\ {1}\end{array}\right)\), we found the eigenvalues (\(\lambda_1 = 2\) and \(\lambda_2 = -4\)) and the corresponding eigenvectors (\(\mathbf{x}_1 = \begin{bmatrix}5\\1\end{bmatrix}\) and \(\mathbf{x}_2 = \begin{bmatrix}1\\1\end{bmatrix}\)). We then formed the general solution and applied the initial condition to find constants \(c_1 = -1\) and \(c_2 = 2\). Analyzing the behavior of the solution as \(t \rightarrow \infty\), we found that the \(\begin{bmatrix}5\\1\end{bmatrix}\) component dominates the behavior and the solution grows without bound along this direction.

Step-by-step solution

Find the eigenvalues and eigenvectors of the matrix

To find the eigenvalues of the given matrix, \(A=\left(\begin{array}{ll}{1} & {-5} \\\\ {1} & {-3}\end{array}\right)\), we need to solve the characteristic equation: det\((A-\lambda I)=0\), where \(\lambda\) are the eigenvalues and \(I\) is the identity matrix. The characteristic equation is: $$\textrm{det}\left(\begin{array}{ll}{1-\lambda} & {-5} \\\\ {1} & {-3-\lambda}\end{array}\right) = (1-\lambda)(-3-\lambda) - (-5)(1)=0$$ Solving the equation, we get: $$(\lambda-1)(\lambda+3) - 5=0$$ $$\lambda^2 + 2\lambda - 3 - 5 = 0$$ $$\lambda^2 + 2\lambda - 8 = 0$$ The solutions are \(\lambda_1 = 2\) and \(\lambda_2 = -4\). Now, we will find the corresponding eigenvectors. For \(\lambda_1 = 2\): $$A - 2I = \left(\begin{array}{ll}{-1} & {-5} \\\\ {1} & {-5}\end{array}\right)$$ Row reducing the matrix, we get: $$\begin{array}{ll}{1} & {5}\\ {0} & {0}\end{array} \Rightarrow \mathbf{x}_1 = \begin{bmatrix}5\\1\end{bmatrix}$$ For \(\lambda_2 = -4\): $$A + 4I = \left(\begin{array}{cc}{5} & {-5} \\\\ {1} & {1}\end{array}\right)$$ Row reducing the matrix, we get: $$\begin{array}{ll}{1} & {-1}\\ {0} & {0}\end{array} \Rightarrow \mathbf{x}_2 = \begin{bmatrix}1\\1\end{bmatrix}$$

Form the general solution of the matrix differential equation

With the eigenvalues and eigenvectors, we can write the general solution to the matrix differential equation as: $$\mathbf{x}(t) = c_1e^{2t}\begin{bmatrix}5\\1\end{bmatrix} + c_2e^{-4t}\begin{bmatrix}1\\1\end{bmatrix}$$

Apply the initial condition to determine the constants

We are given that \(\mathbf{x}(0)=\left(\begin{array}{l}{1} \\\ {1}\end{array}\right)\). Plugging this into our general solution, we get: $$\begin{bmatrix}1\\1\end{bmatrix} = c_1e^{2(0)}\begin{bmatrix}5\\1\end{bmatrix} + c_2e^{-4(0)}\begin{bmatrix}1\\1\end{bmatrix}$$ $$\begin{bmatrix}1\\1\end{bmatrix} = c_1\begin{bmatrix}5\\1\end{bmatrix} + c_2\begin{bmatrix}1\\1\end{bmatrix}$$ Solving this system of equations, we find \(c_1=-1\) and \(c_2=2\).

Analyze the behavior of the solution as \(t \rightarrow \infty\)

Since we have found the constants \(c_1\) and \(c_2\), we can write the complete solution as: $$\mathbf{x}(t) = -e^{2t}\begin{bmatrix}5\\1\end{bmatrix} + 2e^{-4t}\begin{bmatrix}1\\1\end{bmatrix}$$ As \(t \rightarrow \infty\), the term \(e^{2t}\) grows without bound, while the term \(e^{-4t}\) goes to zero. Thus, the behavior of the solution as \(t \rightarrow \infty\) is primarily determined by the first term, which corresponds to the eigenvalue \(\lambda_1=2\): $$\mathbf{x}(t) \approx -e^{2t}\begin{bmatrix}5\\1\end{bmatrix}$$ As \(t\) goes to infinity, the solution vector \(\mathbf{x}(t)\) grows without bound along the direction of the eigenvector \(\begin{bmatrix}5\\1\end{bmatrix}\), such that the overall behavior becomes unbounded and tends to dominate.

Key concepts

Eigenvalues and Eigenvectors

At the heart of understanding matrix differential equations lies the concept of eigenvalues and eigenvectors. These are critical in determining the behavior of systems described by matrices. An eigenvalue, denoted by \(\lambda\), is a scalar that indicates how much an eigenvector is scaled during a linear transformation. An eigenvector is a non-zero vector that changes at most by its scalar factor when that linear transformation is applied.

In the context of the provided problem, finding the eigenvalues and corresponding eigenvectors of the coefficient matrix \(A\) is the first step. We do this by setting up and solving the characteristic equation \(\text{det}(A - \lambda I) = 0\). The eigenvalues found (\(\lambda_1 = 2\) and \(\lambda_2 = -4\)) provide us with critical information about how solutions to the system evolve over time—whether they grow, decay, or oscillate, for instance. The associated eigenvectors give us the direction in which these changes occur.

Matrix Differential Equations

Matrix differential equations are systems of ordinary differential equations that can be compactly written in a matrix form. An essential advantage of using matrices is that they allow us to solve a system of equations simultaneously. The general solution to a matrix differential equation of the form \( \mathbf{x}' = A\mathbf{x} \) involves expressing \( \mathbf{x}(t) \) as a linear combination of eigenvectors, each multiplied by the exponential of the product of time \( t \) and the respective eigenvalue.

In our exercise, the general solution is shown as \( \mathbf{x}(t) = c_1e^{\lambda_1 t}\mathbf{x}_1 + c_2e^{\lambda_2 t}\mathbf{x}_2 \) with eigenvectors \( \mathbf{x}_1 \) and \( \mathbf{x}_2 \) and their corresponding eigenvalues \( \lambda_1 \) and \( \lambda_2 \) respectively. This form lets us capture both the magnitude and direction of the system's state at any time \( t \) and is pivotal to understanding the dynamic behavior of systems modeled by differential equations.

Behavior of Solutions as t Approaches Infinity

The long-term behavior of solutions to matrix differential equations is essential for understanding stability and predicting system behavior. As \( t \) approaches infinity, we evaluate how each component of the solution to the initial value problem behaves. The term associated with a positive eigenvalue \( (\lambda_1 = 2) \) grows exponentially, while the one with the negative eigenvalue \( (\lambda_2 = -4) \) decays to zero.

Specifically for the given problem, as \( t \rightarrow \infty \), the term \( -e^{2t}\begin{bmatrix}5\1\end{bmatrix} \) will dominate. This means that the solution grows unbounded in the direction of the corresponding eigenvector \( \begin{bmatrix}5\1\end{bmatrix} \). The negative sign indicates that the growth is in the opposite direction to the eigenvector. Understanding this helps ascertain stability and the long-term tendency of systems like populations, chemical reactions, or even financial models.