Question

Consider the linear system $$ \begin{aligned} &x^{\prime}=a_{1} x+b_{1} y \\ &y^{\prime}=a_{2} x+b_{2} y \end{aligned} $$ where \(a_{1}, b_{1}, a_{2}\), and \(b_{2}\) are real constants. Show that the condition \(a_{2} b_{1}>0\) is sufficient, but not necessary, for the system to have two real linearly independent solutions of the form $$ x=A e^{\lambda t}, \quad y=B e^{\lambda t} . $$

Short answer

The condition \(a_2 b_1 > 0\) is sufficient, but not necessary, for the linear system to have two real linearly independent solutions of the form \(x=A e^{\lambda t}\) and \(y=B e^{\lambda t}\). When the condition holds, there are two distinct real eigenvalues \(\lambda_1\) and \(\lambda_2\), yielding two linearly independent solutions. However, a counterexample shows that the condition is not necessary, as a linear system with \(a_2b_1 \leq 0\) can also have two real linearly independent solutions.

Step-by-step solution

Rewrite the linear system#

Rewrite the linear system in matrix form as: \[ \begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}. \] Notice that the given solutions \(x=Ae^{\lambda t}\) and \(y=Be^{\lambda t}\) are characterized by eigenvectors and eigenvalues of the coefficient matrix.

Calculate eigenvalues and eigenvectors#

To obtain the eigenvalues, we need to solve the following characteristic equation, which comes from the determinant of A-\(\lambda\)I, where A is the coefficient matrix and I is the identity matrix: \[ \begin{vmatrix} a_1-\lambda & b_1 \\ a_2 & b_2-\lambda \end{vmatrix} = 0, \] which simplifies to: \[ (a_1-\lambda)(b_2-\lambda) - a_2 b_1 = 0. \] Now, calculate the discriminant, which will give us information about the number of real solutions and their dependence: \[ \Delta = ((b_2 - a_1)^2 - 4(-a_2b_1)) \]

Evaluate the condition \(a_2 b_1 > 0\)#

If \(a_2 b_1 > 0\), then we get \(\Delta = (b_2 - a_1)^2 + 4(a_2b_1) > 0\). Thus, the discriminant is always greater than zero, and thus there are two distinct real eigenvalues \(\lambda_1\) and \(\lambda_2\). For each eigenvalue, there exists one eigenvector, which corresponds to a linearly independent solution of the form \(x=A e^{\lambda t}\) and \(y=B e^{\lambda t}\). Because there are two distinct real eigenvalues, we have two real linearly independent solutions of the given form. This shows that the condition \(a_2 b_1 > 0\) is sufficient but does not guarantee that it is necessary.

Finding a counterexample for necessity#

Let us find a counterexample to show that this condition is not necessary: Consider the linear system \[ \begin{aligned} &x^{\prime}=-x \\\ &y^{\prime}=-y \end{aligned}, \] where \(a_1 = -1\), \(b_1 = 0\), \(a_2 = 0\), and \(b_2 = -1\). Here, we have \(a_2 b_1 = (-1)(0) = 0 \leq 0\). The characteristic equation is: \[ (-1-\lambda)(-1-\lambda) - 0 = 0, \] which yields two distinct real eigenvalues \(\lambda_1 = -1\) and \(\lambda_2 = -1\). Therefore, the system has two linearly independent solutions \(x=A e^{-t}\) and \(y=B e^{-t}\). This counterexample shows that the condition \(a_2 b_1 > 0\) is not necessary to have two real linearly independent solutions of the given form. Thus, we have shown that the condition \(a_2 b_1 > 0\) is sufficient, but not necessary, for the system to have two real linearly independent solutions.

Key concepts

Eigenvalues

Eigenvalues are crucial in understanding the behavior of linear systems, such as \[\begin{aligned} &x^{\prime}=a_{1} x+b_{1} y \&y^{\prime}=a_{2} x+b_{2} y\end{aligned}\]They determine the magnitude and direction of the solutions.When solving for eigenvalues, you typically use the characteristic equation. For our system, the eigenvalues \(\lambda\) are found by solving \[\begin{vmatrix} a_1-\lambda & b_1 \ a_2 & b_2-\lambda \end{vmatrix} = 0\]Eigenvalues can reveal whether solutions decay or grow over time and help identify the nature of equilibrium points.
If a matrix has distinct eigenvalues, it often leads to distinct and independent solutions, aiding greatly in solving differential equations.

Eigenvectors

Eigenvectors, paired with eigenvalues, form the foundation of solutions for linear systems.They provide direction to the solution trajectories in the system. If you determine the eigenvalues \(\lambda\), you can then find eigenvectors by solving the equation \[(A-\lambda I) \mathbf{v} = 0,\]where \(A\) is the coefficient matrix, \(I\) is the identity matrix, and \(\mathbf{v}\) represents the eigenvectors. Each eigenvalue has corresponding eigenvectors, dictating the direction of the system's solutions.
These vectors will point along lines where the matrix transforms vectors simply by stretching or compressing, without changing direction.

Characteristic Equation

The characteristic equation is a central tool when analyzing linear systems as it helps find eigenvalues.In our linear system, the characteristic equation is derived from setting the determinant of \(A - \lambda I\) to zero: \[(a_1-\lambda)(b_2-\lambda) - a_2 b_1 = 0.\]Solving this equation gives us the eigenvalues \(\lambda\) of the matrix. The discriminant of the characteristic equation informs us about the number and nature of these eigenvalues, which are essential to determine if solutions are real or complex.
In cases where the discriminant is positive, you obtain distinct real eigenvalues, meaning the system has a better chance of producing distinct linearly independent solutions.

Matrix Form

Rewriting a linear system in matrix form simplifies analysis and solution finding.The system \[\begin{aligned} &x^{\prime}=a_{1} x+b_{1} y \&y^{\prime}=a_{2} x+b_{2} y\end{aligned}\]transforms into matrix form as: \[\begin{pmatrix} \dot{x} \ \dot{y} \end{pmatrix} = \begin{pmatrix} a_1 & b_1 \ a_2 & b_2 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix}.\]This transformation allows you to utilize matrix algebra, making it easier to apply various mathematical methods for finding solutions.
It turns the differential equations into an elegant framework that is recognized the same as a linear map, making computation and analysis more systematic.

Real Linearly Independent Solutions

Real linearly independent solutions are vital to fully resolving systems of linear differential equations.When system parameters meet certain conditions, like having two distinct eigenvalues, the solutions will be linearly independent.
This means one solution cannot be expressed as a constant multiple of the other. In particular, the system \[\begin{aligned} &x=A e^{\lambda_1 t} \&y=B e^{\lambda_2 t}\end{aligned}\]generates two solutions that do not overlap completely in direction.
They span the entire solution space for the system, enabling a complete description of the dynamics. Understanding linearly independent solutions is crucial for forming the general solution to any linear system.