Question

The matrices of the systems in exercises are singular. Describe and graph the trajectories of nonconstant solutions of the given systems. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 1 & -3 \\ -1 & 3 \end{array}\right] \mathbf{y} $$

Short answer

Based on the given singular matrix and the solution steps, describe the trajectories of nonconstant solutions of the system and explain the behavior of the solutions as time varies. The nonconstant solution trajectories can be described using the general solution: $$ \mathbf{y}(t)=\begin{bmatrix}3c_1 + c_2 \text{e}^{2t}\\c_1 - c_2 \text{e}^{2t}\end{bmatrix} $$ As time progresses, the first component of the general solution has a term that grows exponentially, while the second component has a term that decays exponentially. This implies that the nonconstant solution trajectories generally move away from the origin. By choosing different values of \(c_1\) and \(c_2\), various trajectories can be plotted, giving a visual representation of the behavior of nonconstant solutions for this system of ODEs.

Step-by-step solution

Find the eigenvalues of the matrix

To find the eigenvalues of the matrix, we need to solve the characteristic equation. The given matrix is $$ A = \left[\begin{array}{rr} 1 & -3 \\ -1 & 3 \end{array}\right] $$ To solve the characteristic equation, compute \(\text{det}(A - \lambda I)\), where \(I\) is the identity matrix and \(\lambda\) are the eigenvalues we are trying to find: $$ \text{det}(A - \lambda I) = \text{det} \left(\begin{array}{cc} 1-\lambda & -3 \\ -1 & 3-\lambda \end{array}\right)=(1-\lambda)(3-\lambda)+3=(-\lambda)(2-\lambda) $$ Setting the determinant equal to zero and solving for \(\lambda\) gives us the eigenvalues: $$ -\lambda(2-\lambda) = 0 $$ Thus, the eigenvalues are \(\lambda_1 = 0\) and \(\lambda_2 = 2\).

Calculate the eigenvectors corresponding to each eigenvalue

For the eigenvalue \(\lambda_1 = 0\), we need to find the eigenvectors that satisfy \((A - \lambda_1 I)\mathbf{v_1}=0\). This gives us: $$ \left[\begin{array}{cc} 1 & -3 \\ -1 & 3 \end{array}\right]\mathbf{v_1}=0 $$ Solving the above system of equations yields the eigenvector: $$ \mathbf{v_1}=\begin{bmatrix}3\\1\end{bmatrix} $$ For the eigenvalue \(\lambda_2 = 2\), we need to find the eigenvectors that satisfy \((A - \lambda_2 I)\mathbf{v_2}=0\). This gives us: $$ \left[\begin{array}{cc} -1 & -3 \\ -1 & 1 \end{array}\right]\mathbf{v_2}=0 $$ Solving the above system of equations yields the eigenvector: $$ \mathbf{v_2}=\begin{bmatrix}1\\-1\end{bmatrix} $$

Construct the general solution to the system of ODEs

The general solution to the system of ODEs can be expressed as $$ \mathbf{y}(t)=c_1 \text{e}^{\lambda_1t}\mathbf{v_1} + c_2 \text{e}^{\lambda_2t}\mathbf{v_2} $$ Where \(\mathbf{y}(t)\) represents the solution trajectories. Substituting the eigenvalues and eigenvectors, we get: $$ \mathbf{y}(t)=c_1\begin{bmatrix}3\\1\end{bmatrix} + c_2 \text{e}^{2t}\begin{bmatrix}1\\-1\end{bmatrix} $$

Describe the trajectories

The general solution can be rewritten as $$ \mathbf{y}(t)=\begin{bmatrix}3c_1 + c_2 \text{e}^{2t}\\c_1 - c_2 \text{e}^{2t}\end{bmatrix} $$ Since the first component of the general solution has a term that grows exponentially and the second component has a term that decays exponentially as time progresses, the trajectories of the nonconstant solutions generally move away from the origin.

Graph some trajectories of the nonconstant solutions

Choose different values of \(c_1\) and \(c_2\) to generate some trajectories. For instance, with \(c_1=1\) and \(c_2=1\), the trajectory will be: $$ \mathbf{y}(t)=\begin{bmatrix}3 + \text{e}^{2t}\\1 - \text{e}^{2t}\end{bmatrix} $$ For \(c_1=-1\) and \(c_2=1\), $$ \mathbf{y}(t)=\begin{bmatrix}-3 + \text{e}^{2t}\\-1 - \text{e}^{2t}\end{bmatrix} $$ Plot different trajectories by varying \(c_1\) and \(c_2\) on a graph to visualize the behavior of nonconstant solutions for this system of ODEs.

Key concepts

Eigenvalues and Eigenvectors

The study of eigenvalues and eigenvectors is essential in understanding the behavior of a system of differential equations. In such systems, a square matrix represents the linear transformation of the variable vector over time. An eigenvalue of a matrix is a scalar that indicates the factor by which an eigenvector, which is a non-zero vector, is scaled during the transformation. To find the eigenvalues of the matrix from our exercise, one must solve the characteristic equation, which is derived from the matrix's determinant.The importance of these values and vectors cannot be overstated: they provide insight into the system's stability and the nature of the solution trajectories. As observed in the provided exercise, every eigenvalue has a corresponding eigenvector, which points in a direction that is invariant to the linear transformation. Hence, solving for eigenvectors allows us to reconstruct the general form of the system's solutions. Moreover, the behavior of the system can be predicted by the sign and magnitude of the eigenvalues: a positive eigenvalue will generally lead to solutions that grow exponentially, while a negative value indicates exponential decay.

Characteristic Equation

The characteristic equation is fundamental in finding the eigenvalues of a matrix. This equation is formed by setting the determinant of the matrix subtracted by a scalar multiple of the identity matrix to zero, as shown in \( \text{det}(A - \lambda I) = 0 \). This equation is a polynomial whose degree corresponds to the size of the matrix, and its roots are the eigenvalues. In our exercise, the characteristic equation was a quadratic polynomial, and solving it resulted in two distinct eigenvalues.Understanding this part of the process is crucial because the characteristic equation bridges the matrix's properties and the eigenvalues, which are central to analyzing the dynamics of the system. When the term coefficients relate to the transition dynamics, the equation's solutions directly inform how solution trajectories behave. In many real-world scenarios, like in physics or engineering, interpreting the characteristic equation's roots allows us to predict system behavior over time.

General Solution of ODEs

A system of ordinary differential equations (ODEs) often seeks to describe the evolution of certain variables over time. To construct the general solution of ODEs, particularly the linear homogenous systems, one combines the eigenvectors multiplied by the exponential of the product of the corresponding eigenvalue and time. This form reflects how each eigenvalue-eigenvector pair contributes to the system's overall behavior. In the provided solution, each term of the general solution corresponds to a specific mode of the system, one for each eigenvalue.The general solution's format is typically as follows: \( \mathbf{y}(t) = c_1 \text{e}^{\lambda_1 t}\mathbf{v_1} + c_2 \text{e}^{\lambda_2 t}\mathbf{v_2} + \ldots + c_n \text{e}^{\lambda_n t}\mathbf{v_n} \), where the constants \( c_i \) are determined by initial conditions. This solution provides a complete description of the system's evolution from any given starting point, thereby allowing the prediction of future states from past behavior. It's the versatility and comprehensiveness of the general solution that makes it a powerful tool in various scientific and engineering disciplines.

Trajectory in Phase Space

When studying the behavior of dynamical systems, the trajectory in phase space is a concept used to represent the state of the system at any given time. Phase space is a multidimensional space in which each axis corresponds to one of the system's variables; the system's state can be represented as a point in this space. A trajectory then is the path that this point follows as time evolves.In the context of our exercise, each nonconstant solution of the system corresponds to a unique trajectory in the phase space. When the system is linear, as it is in our case, these trajectories are typically straight lines or curves that exhibit exponential growth or decay, depending on the system's eigenvalues. The trajectories may approach or recede from fixed points, which are the system's equilibria. Plotting these trajectories for varying initial conditions reveals the system's dynamic nature and aids in visualizing the stability or instability of the equilibria. By understanding the configuration of these trajectories, one can gain valuable insights into the behavior of more complex nonlinear systems as well.