Question

Suppose the eigenvalues of the \(2 \times 2\) matrix \(A\) are \(\lambda=0\) and \(\mu \neq 0\), with corresponding eigenvectors \(\mathbf{x}_{1}\) and \(\mathbf{x}_{2} .\) Let \(L_{1}\) be the line through the origin parallel to \(\mathbf{x}_{1}\). (a) Show that every point on \(L_{1}\) is the trajectory of a constant solution of \(\mathbf{y}^{\prime}=A \mathbf{y}\). (b) Show that the trajectories of nonconstant solutions of \(\mathbf{y}^{\prime}=A \mathbf{y}\) are half-lines parallel to \(\mathbf{x}_{2}\) and on either side of \(L_{1},\) and that the direction of motion along these trajectories is away from \(L_{1}\) if \(\mu>0,\) or toward \(L_{1}\) if \(\mu<0\)

Short answer

Question: Prove that (a) every point on \(L_{1}\) is the trajectory of a constant solution, and (b) the trajectories of nonconstant solutions are half-lines parallel to \(\mathbf{x}_{2}\), with the direction depending on the sign of \(\mu\) for the linear system of differential equations \(\mathbf{y}^{\prime}=A \mathbf{y}\), where \(A\) is a \(2\times 2\) matrix with eigenvalues \(\lambda = 0\) and \(\mu \neq 0\), and corresponding eigenvectors \(\mathbf{x}_1\) and \(\mathbf{x}_2\). Answer: We have shown that (a) by setting \(c_{1} = k\) and \(c_{2} = 0\), every point on \(L_{1}\) is the trajectory of a constant solution, and (b) the trajectories of nonconstant solutions are half-lines parallel to \(\mathbf{x}_{2}\), with the direction of motion depending on the sign of \(\mu\).

Step-by-step solution

Find the general solution of the system

First, let's write the general solution for the given system of differential equations \(\mathbf{y}^{\prime}=A \mathbf{y}\) in terms of the eigenvalues and eigenvectors of the matrix \(A\). The general solution is given by: \[\mathbf{y}(t)=c_{1} e^{\lambda t} \mathbf{x}_{1} + c_{2} e^{\mu t} \mathbf{x}_{2}\] where \(c_{1}\) and \(c_{2}\) are arbitrary constants.

Analyze the constant solution on L1

Consider a point \(P\) on the line \(L_{1}\). \(P\) can be expressed as \(P=k\mathbf{x}_{1}\), where \(k\) is a constant. Now, let's find the solution \(\mathbf{y}(t)\) for this point. Plugging \(P\) into the general solution, we get: \[\mathbf{y}(t)=c_{1} e^{0 t} \mathbf{x}_{1} + c_{2} e^{\mu t} \mathbf{x}_{2} = c_{1} \mathbf{x}_{1} + c_{2} e^{\mu t} \mathbf{x}_{2}\] Since \(P\) is on \(L_{1}\) and we want a constant solution, we have to find \(c_{1}\) and \(c_{2}\) such that this equation holds for all \(t\). By setting \(c_{1}=k\) and \(c_{2}=0\), we have: \[\mathbf{y}(t) = k \mathbf{x}_{1}\] This shows that every point on \(L_{1}\) is the trajectory of a constant solution.

Analyze nonconstant solutions

To study the trajectories of nonconstant solutions, consider a point not on \(L_{1}\). In the general solution equation, since the eigenvalue \(\lambda = 0\), the term \(c_{1} e^{\lambda t} \mathbf{x}_{1}\) remains constant, while the term \(c_{2} e^{\mu t} \mathbf{x}_{2}\) is variable. The trajectories of nonconstant solutions are determined by \(c_{2} e^{\mu t} \mathbf{x}_{2}\). These half-lines are parallel to the eigenvector \(\mathbf{x}_{2}\).

Determine the direction of motion

The direction of motion along the trajectories depends on the value of \(\mu\). If \(\mu > 0\), the term \(e^{\mu t}\) is an increasing function of \(t\), so the direction of motion is away from \(L_{1}\). If \(\mu < 0\), the term \(e^{\mu t}\) is a decreasing function of \(t\), so the direction of motion is towards \(L_{1}\). In summary, we have shown that every point on \(L_{1}\) is the trajectory of a constant solution, and the trajectories of nonconstant solutions are half-lines parallel to \(\mathbf{x}_{2}\), with the direction of motion depending on the sign of \(\mu\).

Key concepts

Eigenvectors and Eigenvalues

Understanding eigenvectors and eigenvalues is essential when dealing with differential equations and linear algebra. Eigenvectors of a matrix are non-zero vectors that only change by a scalar factor when that matrix is applied to them. This scalar is the eigenvalue associated with the eigenvector. To put it simply, if your matrix is a transformation, then its eigenvectors point in directions unaltered by the transformation, while the eigenvalues tell you how much the length of the vector is stretched or compressed.

For a given square matrix A, the eigenvector x and eigenvalue λ must satisfy the equation:
\[ A\textbf{x} = \textbf{λx} \]
Finding these eigenvectors and eigenvalues is crucial for solving many types of differential equations, as they help us understand the behavior of the system described by the matrix A.

Differential Equations Solutions

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are pivotal in modeling the dynamics of various systems in physics, engineering, and other disciplines. Solutions to differential equations can take many forms, ranging from simple functions to complex expressions involving exponentials, sines, and cosines, among others.

The general solution to a system of linear differential equations contains a combination of terms involving eigenvalues and eigenvectors. These terms present exponentially growing or decaying solutions depending on the eigenvalues. Particularly, eigenvalues serve as exponents in these terms and these exponential solutions describe how the system evolves over time. Having real and distinct eigenvalues greatly simplifies the general solution, leading to clear trajectories for the system's behavior.

Trajectory of a Differential Equation

The trajectory of a differential equation refers to the path traced out by the solution in the phase space, which is a graphical representation of all possible states of a system. Essentially, the phase space is a plot with axes corresponding to the variables of the system, and the trajectory is a curve depicting how these variables evolve together over time.

For systems described by linear differential equations, trajectories can often be characterized as straight lines or curves based on the associated eigenvectors and eigenvalues. When the eigenvalue is zero, the trajectory remains constant along the line parallel to the corresponding eigenvector. In contrast, if the eigenvalue is non-zero, trajectories change their position in the phase space, illustrating non-constant solutions. Analyzing these trajectories gives valuable insight into the stability and long-term behavior of the system in question.

Eigenvalue Problem

The eigenvalue problem is a fundamental issue in linear algebra which involves finding the eigenvalues and eigenvectors for a given matrix. The problem is to solve the characteristic equation derived from the matrix, which takes the form of a polynomial equation:
\[ \text{det}(A - \textbf{λI}) = 0 \]
Here I represents the identity matrix and det stands for determinant. This equation reveals the eigenvalues, and once these are known, one can solve for the corresponding eigenvectors, completing the eigenvalue problem. Solving this problem is not merely a mathematical exercise; it has real-world applications. For example, in our exercise scenario, the solution to the eigenvalue problem helps us articulate the stability and behavior of solutions to a system of differential equations, providing insight into physical systems' states or behaviors over time.