Question

Verify that \((0,0)\) is a critical point, show that the system is almost linear, and discuss the type and stability of the critical point \((0,0)\) by examining the corresponding linear system. $$ d x / d t=-x+y+2 x y, \quad d y / d t=-4 x-y+x^{2}-y^{2} $$

Short answer

Based on the given system of equations, we analyzed the critical point (0,0) and found that the system is almost linear. After finding the Jacobian matrix and its eigenvalues, we concluded that the critical point (0,0) corresponds to a stable spiral.

Step-by-step solution

Verify \((0,0)\) as a critical point

To check if \((0,0)\) is a critical point, we need to evaluate the given system of equations at \((x, y) = (0, 0)\): $$\begin{aligned} \frac{dx}{dt} &= -x + y + 2xy \\ \frac{dy}{dt} &= -4x - y + x^2 - y^2 \end{aligned}$$ At \((x, y) = (0, 0)\), we have: $$\begin{aligned} \frac{dx}{dt} &= -0 + 0 + 2(0)(0) = 0 \\ \frac{dy}{dt} &= -4(0) - 0 + (0)^2 - (0)^2 = 0 \end{aligned}$$ Since both equations equal zero at this point, \((0,0)\) is a critical point.

Determine if the system is almost linear

A system is almost linear if it can be written as: $$\begin{aligned} \frac{dx}{dt} &= f(x,y) = a(x,y)x + b(x,y)y \\ \frac{dy}{dt} &= g(x,y) = c(x,y)x + d(x,y)y \end{aligned}$$ Here, \(a(x,y), b(x,y), c(x,y),\) and \(d(x,y)\) are continuous functions in a neighborhood of the critical point \((0,0)\). Comparing this with the given system of equations, we see that it can be written in the form of an almost linear system.

Analyze the corresponding linear system for type and stability

Let's find the Jacobian matrix of the given system of equations at the critical point \((0,0)\): $$J(x,y) = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \end{bmatrix} = \begin{bmatrix} -1+2y & 1+2x \\ -4+2x & -1-2y \end{bmatrix}$$ Evaluating the Jacobian at the critical point \((0,0)\), we get: $$J(0,0) = \begin{bmatrix} -1 & 1 \\ -4 & -1 \end{bmatrix}$$ Now, let's find the eigenvalues of the Jacobian matrix \(J(0,0)\): $$\text{det}(J(0,0) - \lambda I) = \begin{vmatrix} -1-\lambda & 1 \\ -4 & -1-\lambda \end{vmatrix} = \lambda^2 + 2\lambda +3$$ The characteristic equation is: $$\lambda^2 + 2\lambda + 3 = 0$$ The eigenvalues are complex and can be found as \(\lambda = -1 \pm i\sqrt{2}\). Since the real part of both eigenvalues is negative, the critical point \((0,0)\) corresponds to a stable spiral.

Key concepts

Differential Equations

Differential equations are a foundational concept in mathematics and physics that describe how a quantity changes over time or space. The equation presented in the exercise translates real-world phenomena into a language that can be analyzed mathematically: the rate of change over time. In our example, the system of differential equations

\[\[\begin{align*}\frac{dx}{dt} &= -x + y + 2xy, \quad \frac{dy}{dt} &= -4x - y + x^2 - y^2 \end{align*}\]\]
defines how the variables x and y evolve over time, t. By solving these equations, or examining their critical points, we gain insight into the behavior of the system such as equilibrium and stability.

Jacobian Matrix

The Jacobian matrix is a powerful tool in understanding multivariable systems. It consists of the partial derivatives of each equation with respect to each variable, capturing the system's local behavior at a specific point. When evaluating the Jacobian matrix at the critical point \((0,0)\), as our exercise dictates, it simplifies to a constant matrix

\[J(0,0) = \begin{bmatrix} -1 & 1 \ -4 & -1\end{bmatrix}\]
which represents the system's linear approximation around that point. This matrix will be essential in further analysis to determine the stability of the critical point.

Eigenvalues

Eigenvalues are scalar values associated with a linear system of equations that provide significant insights into the system's behavior. In the context of our problem, we calculate the eigenvalues by solving the characteristic equation \(\lambda^2 + 2\lambda + 3 = 0\) of the Jacobian matrix at the critical point, which yields complex solutions \(\lambda = -1 \pm i\sqrt{2}\). Here, the real parts of the eigenvalues indicate the rate of approach or divergence at the critical point: negative for approaching (stable), positive for divergence (unstable), and zero for a marginally stable position.

Linear Stability Analysis

Linear stability analysis is a method to predict the behavior of a nonlinear system close to its critical points using linear approximations. This analysis relies on the sign of the real part of the eigenvalues calculated from the Jacobian matrix. For our problem, since the real parts of the eigenvalues \(-1 \pm i\sqrt{2}\) are negative, it signals that perturbations from the critical point \((0,0)\) will decay over time, hence indicating that this critical point is stable. Furthermore, because the eigenvalues are complex with non-zero imaginary parts, they suggest oscillatory behavior – in this case, leading to a stable spiral pattern as the system evolves.