Question

In each of Problems 15 through 18 solve the given initial value problem. Describe the behavior of the solution as \(t \rightarrow \infty\). $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{5} & {-1} \\ {3} & {1}\end{array}\right) \mathbf{x}, \quad \mathbf{x}(0)=\left(\begin{array}{r}{2} \\ {-1}\end{array}\right) $$

Short answer

As t approaches infinity, the solution to the given initial value problem will grow exponentially along the direction of the eigenvector $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

Step-by-step solution

Find the Eigenvalues

First, let's find the eigenvalues of the given matrix A: $$ A = \begin{pmatrix} 5 & -1 \\ 3 & 1 \end{pmatrix} $$ To find the eigenvalues, we need to solve the characteristic equation, which is given by: $$ \det(A - \lambda I) = 0 $$ where \(\lambda\) is the eigenvalue and \(I\) is the identity matrix. $$ \begin{vmatrix} 5 - \lambda & -1 \\ 3 & 1 - \lambda \end{vmatrix} = (5 - \lambda)(1 - \lambda) - (-1)(3) = \lambda^{2} - 6\lambda + 8 $$ Solving the quadratic equation \(\lambda^{2} - 6\lambda + 8 = 0\), we get two eigenvalues: $$ \lambda_1 = 2, \quad \lambda_2 = 4. $$

Find the Eigenvectors

Now, we will find the eigenvectors corresponding to the eigenvalues \(\lambda_1\) and \(\lambda_2\). For \(\lambda_1 = 2\): \((A - 2I)\mathbf{v}_1 = 0\) $$ \begin{pmatrix} 3 & -1 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} v_{11} \\ v_{12} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$ We get the following system of linear equations: $$ 3v_{11} - v_{12} = 0 $$ Taking \(v_{12} = 3\), we get \(v_{11}=1\). Thus, the eigenvector corresponding to the eigenvalue \(\lambda_1=2\) is: $$ \mathbf{v}_1 = \begin{pmatrix} 1 \\ 3 \end{pmatrix} $$ For \(\lambda_2 = 4\): \((A - 4I)\mathbf{v}_2 = 0\) $$ \begin{pmatrix} 1 & -1 \\ 3 & -3 \end{pmatrix} \begin{pmatrix} v_{21} \\ v_{22} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$ We get the following system of linear equations: $$ v_{21} - v_{22} = 0 $$ Taking \(v_{21} = 1\), we get \(v_{22}=1\). Therefore, the eigenvector associated with the eigenvalue \(\lambda_2 = 4\) is: $$ \mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} $$

Solve the System of ODEs

Now, we will construct the general solution using these eigenvalues and eigenvectors. The general solution to the system of ODEs is given by: $$ \mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 $$ Substituting the values of eigenvalues and eigenvectors, we get: $$ \mathbf{x}(t) = c_1 e^{2t} \begin{pmatrix} 1 \\ 3 \end{pmatrix} + c_2 e^{4t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} $$

Apply Initial Condition

To find the constants \(c_1\) and \(c_2\), we will use the given initial condition $\mathbf{x}(0) = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$. Substituting \(t = 0\) and the initial condition, we get: $$ \begin{pmatrix} 2 \\ -1 \end{pmatrix} = c_1 \begin{pmatrix} 1 \\ 3 \end{pmatrix} + c_2 \begin{pmatrix} 1 \\ 1 \end{pmatrix} $$ Solving for \(c_1\) and \(c_2\), we get: $$ c_1 = \frac{5}{2}, \quad c_2=-\frac{1}{2} $$ So, the solution to the initial value problem becomes: $$ \mathbf{x}(t) = \frac{5}{2} e^{2t} \begin{pmatrix} 1 \\ 3 \end{pmatrix} - \frac{1}{2} e^{4t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} $$

Analyze the Behavior as \(t \rightarrow \infty\)

As \(t\) approaches infinity, we need to find the dominant term in the solution. In this case, since the term with the eigenvalue \(\lambda_2 = 4\) has a larger exponent than the term with the eigenvalue \(\lambda_1 = 2\), it will dominate as \(t\) goes to infinity. Thus, as \(t \rightarrow \infty\), the solution \(\mathbf{x}(t)\) approaches: $$ \frac{-1}{2} e^{4t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} $$ Therefore, as \(t\) goes to infinity, the solution will grow exponentially along the direction of the eigenvector $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

Key concepts

Eigenvalues and Eigenvectors

In the study of differential equations, eigenvalues and eigenvectors play a crucial role in understanding the solutions' behavior. To put it simply, eigenvalues measure the factor by which a transformation scales a vector, while eigenvectors are vectors that do not change direction during that transformation.

An eigenvalue is a scalar \( \lambda \) that satisfies the equation \( A\textbf{v} = \lambda \textbf{v} \) for a given square matrix \( A \) and non-zero vector \( \textbf{v} \). For each eigenvalue, its corresponding eigenvector can be found by solving the system \( (A - \lambda I)\textbf{v} = 0 \), where \( I \) is the identity matrix. In the context of systems of linear ODEs, these concepts help construct solutions that describe the system's dynamics over time.

When the initial value problem was addressed, finding the eigenvalues and associated eigenvectors allowed the construction of a solution that illustrates how the system evolves in phase space, effectively showcasing the essence of the system's behavior.

System of ODEs

A system of Ordinary Differential Equations (ODEs) consists of several equations involving derivatives of dependent variables with respect to one independent variable, commonly time. For the system given by \( \textbf{x}' = A\textbf{x} \), the solution approach involved finding eigenvalues and eigenvectors of matrix \( A \) and using them to build a general solution that expresses the dependent variables in terms of the independent variable.

The general solution of a linear system with distinct eigenvalues is a linear combination of solutions, each involving an exponentiated term multiplied by the corresponding eigenvector, represented as \( \textbf{x}(t) = c_1 e^{\lambda_1 t} \textbf{v}_1 + c_2 e^{\lambda_2 t} \textbf{v}_2 \). This form allows us to plug in initial conditions to solve for the constants \( c_1 \) and \( c_2 \) that render the solution uniquely fitted to the problem's initial state.

Long-term Behavior of Solutions

The long-term behavior of solutions to a system of ODEs is greatly influenced by the eigenvalues of the matrix that defines the system. If all eigenvalues have negative real parts, the system is stable, and solutions tend towards zero as time approaches infinity. Conversely, if any eigenvalue has a positive real part, solutions can grow without bound, indicating instability.

In the given problem, the eigenvalues are positive, implying that the system solutions will grow exponentially. The term with the highest exponent in the solution—associated with the largest eigenvalue—will dominate in the long run, essentially determining the direction and rate of growth. Therefore, the behavior of the solution as \( t \rightarrow \infty \) clearly shows the exponential divergence along the eigenvector corresponding to the largest eigenvalue.

Characteristic Equation

The characteristic equation of a matrix is a fundamental component in finding the eigenvalues that shape the behavior of a system. It is derived from the determinant of \( A - \lambda I \), set to zero. Solving this equation, which involves a polynomial whose degree equals the size of the matrix, yields the eigenvalues.

In this scenario, the characteristic equation was formed by computing the determinant resulting in a quadratic equation of \( \lambda \). The solutions to this equation provided the values needed to progress further in finding the general solution to the system of ODEs. Conceptually, the characteristic equation is the bridging link between a matrix and the dynamics it governs within a system described by differential equations.