Question

Find the solution of the given initial value problem. Draw the trajectory of the solution in the \(x_{1} x_{2}-\) plane and also the graph of \(x_{1}\) versus \(t .\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{3} & {9} \\ {-1} & {-3}\end{array}\right) \mathbf{x}, \quad \mathbf{x}(0)=\left(\begin{array}{l}{2} \\ {4}\end{array}\right) $$

Short answer

Based on the given initial value problem, we found a solution of: $$ \mathbf{x}(t)=-e^{-t} \left(\begin{array}{c}{1} \\\ {1}\end{array}\right)+2e^{-3t} \left(\begin{array}{c}{1} \\\ {-\frac{1}{2}}\end{array}\right) $$ When analyzing the behavior of the solution as \(t \rightarrow \infty\), we observe that \(\mathbf{x}(t)\) approaches the zero vector: $$ \lim_{t \to \infty} \mathbf{x}(t) = \mathbf{0} $$

Step-by-step solution

Finding Eigenvalues

We start by finding the eigenvalues \(\lambda\) of the matrix: $$ A = \left(\begin{array}{rr}{-3} & {2} \\\ {-1} & {-1}\end{array}\right) $$ We find the eigenvalues by solving the characteristic equation \(\text{det}(A-\lambda I) =0\). In our case, for the given matrix: $$ \text{det}\left(\begin{array}{cc}{(-3-\lambda)} & {2} \\\ {-1} & {(-1-\lambda)}\end{array}\right) = (-3-\lambda)((-1)-\lambda) - 2(-1) = 0 $$

Solving the characteristic equation

Now we solve the characteristic equation: $$ (-3-\lambda)((-1)-\lambda) +2= 0 \Rightarrow \lambda^2 +4\lambda +3 = 0 $$ Applying the quadratic equation formula to find the eigenvalues, we get: $$ \lambda_1 = \frac{-4+\sqrt{(-4)^2-4(3)}}{2} = -1 \\ \\ \lambda_2 = \frac{-4-\sqrt{(-4)^2-4(3)}}{2} = -3 $$

Finding eigenvectors

Next, we find the eigenvectors corresponding to each eigenvalue by solving the system \((A-\lambda_iI)\mathbf{v}_i = \mathbf{0}\). For \(\lambda_1 = -1\): $$ \left(\begin{array}{cc}{-2} & {2} \\\ {-1} & {0}\end{array}\right) \left(\begin{array}{c}{v_{11}} \\\ {v_{12}}\end{array}\right) = \left(\begin{array}{c}{0} \\\ {0}\end{array}\right) $$ The corresponding eigenvector is: $$ \mathbf{v}_{1}=\left(\begin{array}{c}{1} \\\ {1}\end{array}\right) $$ For \(\lambda_2 = -3\): $$ \left(\begin{array}{cc}{0} & {2} \\\ {-1} & {2}\end{array}\right) \left(\begin{array}{c}{v_{21}} \\\ {v_{22}}\end{array}\right) = \left(\begin{array}{c}{0} \\\ {0}\end{array}\right) $$ The corresponding eigenvector is: $$ \mathbf{v}_{2}=\left(\begin{array}{c}{1} \\\ {-\frac{1}{2}}\end{array}\right) $$

General solution of the system

With the eigenvalues and eigenvectors, we write down the general solution of the system: $$ \mathbf{x}(t)=c_{1} e^{-t} \left(\begin{array}{c}{1} \\\ {1}\end{array}\right)+c_{2} e^{-3t} \left(\begin{array}{c}{1} \\\ {-\frac{1}{2}}\end{array}\right) $$

Apply the initial condition

Next, we use the initial condition \(\mathbf{x}(0)=\left(\begin{array}{r}{1} \\\ {-2}\end{array}\right)\) to find the coefficients \(c_1\) and \(c_2\). Plugging in \(t=0\): $$ \left(\begin{array}{c}{1} \\\ {-2}\end{array}\right) = c_{1} \left(\begin{array}{c}{1} \\\ {1}\end{array}\right) + c_{2} \left(\begin{array}{c}{1} \\\ {-\frac{1}{2}}\end{array}\right) $$ Solving the system of linear equations, we get \(c_1=-1\) and \(c_2=2\). Thus, the solution to the initial value problem is: $$ \mathbf{x}(t)=-e^{-t} \left(\begin{array}{c}{1} \\\ {1}\end{array}\right)+2e^{-3t} \left(\begin{array}{c}{1} \\\ {-\frac{1}{2}}\end{array}\right) $$

Behavior as \(t\rightarrow \infty\)

As \(t \rightarrow \infty\), both exponentials, \(e^{-t}\) and \(e^{-3t}\), tend to zero. Therefore, the solution \(\mathbf{x}(t)\) approaches the zero vector: $$ \lim_{t \to \infty} \mathbf{x}(t) = \lim_{t \to \infty} \left( -e^{-t} \left(\begin{array}{c}{1} \\\ {1}\end{array}\right)+2e^{-3t} \left(\begin{array}{c}{1} \\\ {-\frac{1}{2}}\end{array}\right) \right) = \mathbf{0} $$

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation that comes with specific starting values, known as initial conditions. This initial condition helps us find a precise solution to the equation. The exercise we are looking at involves finding such a solution. It provides a matrix differential equation along with an initial condition for the vector \( \mathbf{x}(0) = \left( \begin{array}{c} 2 \ 4 \end{array} \right) \).
This condition tells us the state of the system at time \( t = 0 \). The initial value problem involves determining a function \( \mathbf{x}(t) \) that satisfies both the differential equation and the initial condition.
Solving an initial value problem often involves identifying constants that fit into the solution, allowing it to satisfy the given initial conditions. This specificity makes the initial value problem crucial in finding unique solutions to systems of differential equations like the one in the exercise.

Eigenvectors

In the context of solving differential equations, eigenvectors play a crucial role. When we look into a matrix, eigenvectors help us comprehend the structure and behavior of the solutions.
To find eigenvectors, we first need to determine the eigenvalues of a matrix \( A \). Once we ascertain the eigenvalues, each one provides us a corresponding eigenvector. These vectors show directions in which linear transformations represented by \( A \) do not change direction, only magnitude.

• The exercise involves finding eigenvectors for the matrix by solving \((A-\lambda I)\mathbf{v} = \mathbf{0}\). This system yields vectors that satisfy the matrix equation.
• Eigenspaces, or sets of all eigenvectors corresponding to an eigenvalue, help us construct the system's general solution, leading to a more straightforward resolution of the differential equation.

Understanding eigenvectors brings us closer to the complete picture of the system's dynamics.

System of Linear Equations

A system of linear equations is a set of equations where each is linear. Solving such systems is fundamental for many areas of applied mathematics, especially when dealing with matrices.
Often, as in our exercise, we come across these equations while solving initial value problems or determining eigenvectors.
Our solution process involves setting up and solving a system of linear equations derived from expressing \( (A-\lambda_iI)\mathbf{v}_i = \mathbf{0} \).
This is essential when finding eigenvectors. Similarly, solving for constants \( c_1 \) and \( c_2 \) requires solving a linear system based on the initial condition:

• \( \begin{pmatrix} 2 \ 4 \end{pmatrix} = c_1\begin{pmatrix} 1 \ 1 \end{pmatrix} + c_2\begin{pmatrix} 1 \ -\frac{1}{2} \end{pmatrix} \)

Gaining proficiency in resolving such systems helps understand underlying patterns in more complex problems and supports the crafting of the final solution.

Asymptotic Behavior

The asymptotic behavior of a function describes what happens to its values as time goes to infinity. It's crucial for understanding the long-term trends of solutions in differential equations.
In our exercise, after finding the solution, we examine what happens as \( t \to \infty \). The influence of each term in the solution diminishes as the exponential functions \( e^{-t} \) and \( e^{-3t} \) approach zero.
This results in the solution \( \mathbf{x}(t) \) converging to the zero vector:

• This convergence signifies the system stabilizing over time, displaying how initial effects vanish, and the system's trajectory calms to a steady state.

Understanding asymptotic behavior is vital for many applications as it reveals how systems behave in the long-run, providing insights into the stability and sustainability of solutions.