Question

show that the given system has no periodic solutions other than constant solutions. $$ d x / d t=x+y+x^{3}-y^{2}, \quad d y / d t=-x+2 y+x^{2} y+y^{3} / 3 $$

Short answer

Question: Show that the given system has no periodic solutions other than constant solutions: $$ \begin{cases} \frac{dx}{dt} = x + y + x^3 - y^2 \\ \frac{dy}{dt} = -x + 2y + x^2 y + \frac{y^3}{3} \end{cases} $$ Answer: The given system has no periodic solutions other than constant solutions because the only equilibrium point \((0, 0)\) is an unstable node, as shown by analyzing the eigenvalues of the Jacobian matrix at this point. Since there are no stable equilibrium points, there cannot be any periodic solutions around them.

Step-by-step solution

Find equilibrium points

The equilibrium points (or constant solutions) of the system are the points \((x, y)\) where the derivatives are zero: $$\begin{cases} \frac{dx}{dt} = x + y + x^3 - y^2 = 0 \\ \frac{dy}{dt} = -x + 2y + x^2 y + \frac{y^3}{3} = 0 \end{cases}$$ It's clear that \((0,0)\) is an equilibrium point.

Jacobian matrix

Compute the Jacobian matrix of the system, which contains the partial derivatives of the functions with respect to x and y: $$ J = \begin{bmatrix} \frac{\partial(dx/dt)}{\partial x} & \frac{\partial(dx/dt)}{\partial y} \\ \frac{\partial(dy/dt)}{\partial x} & \frac{\partial(dy/dt)}{\partial y} \end{bmatrix}. $$ $$ J(x, y) = \begin{bmatrix} 1 + 3x^2 & 1 - 2y \\ -1 + 2xy & 2 + x^2 + y^2 \end{bmatrix}. $$

Find the Jacobian matrix at the equilibrium point (0,0)

Evaluate the Jacobian matrix at the equilibrium point \((0, 0)\): $$ J(0, 0) = \begin{bmatrix} 1 & 1 \\ -1 & 2 \end{bmatrix}. $$

Compute eigenvalues

To analyze the stability of the equilibrium point, compute the eigenvalues of the Jacobian matrix at the point \((0, 0)\). We have to solve the characteristic equation $$\text{det}(J - \lambda I) = 0$$ where \(I\) is the identity matrix. Using the Jacobian matrix at \((0,0)\): $$\text{det}(J - \lambda I) = \text{det}\left(\begin{bmatrix} 1 - \lambda & 1 \\ -1 & 2 - \lambda \end{bmatrix}\right) = (1-\lambda)(2-\lambda) - (-1)(1) = \lambda^2 - 3\lambda + 1 = 0$$.

Analyze stability

Solve the quadratic equation \(\lambda^2 - 3\lambda + 1 = 0\). By applying the quadratic formula, we get: $$\lambda_{1,2} = \frac{3 \pm \sqrt{3^2 - 4 \times 1 \times 1}}{2 \times 1} = \frac{3 \pm \sqrt{5}}{2}.$$ The eigenvalues are real and positive, which means that the equilibrium point \((0, 0)\) is an unstable node. Since the only equilibrium point is unstable, the conclusion is that the given system doesn't have any periodic solutions other than constant solutions.

Key concepts

Equilibrium Points

Equilibrium points in differential equations are like the 'resting positions' in a dynamic system where all changes cease. To find these points, we set the derivatives to zero because these are the spots where the system doesn't change over time.

In our case, the system of equations given by \(\frac{dx}{dt} = x + y + x^3 - y^2\) and \(\frac{dy}{dt} = -x + 2y + x^2y + \frac{y^3}{3}\) provides the equilibrium point \( (0,0) \), by solving for when both derivatives equate to zero. Recognizing these points is an essential first step in determining the behavior of a system over time and plays a crucial role in analyzing the system's stability.

Jacobian Matrix

The Jacobian matrix is a powerful tool representing how a dynamical system changes around an equilibrium point. It's constructed from the partial derivatives of the system's equations.

For our differential equation system, this involves calculating \(\frac{\partial(dx/dt)}{\partial x}\), \(\frac{\partial(dx/dt)}{\partial y}\), \(\frac{\partial(dy/dt)}{\partial x}\), and \(\frac{\partial(dy/dt)}{\partial y}\) to assemble the Jacobian matrix. Simply put, it captures the essence of the system's dynamics in a matrix form, which we can analyze further to understand the behavior of the system near the equilibrium points.

Eigenvalues

Eigenvalues are a set of scalars associated with a system of linear equations or a matrix. They play a central role in understanding the behavior of differential equations. Once we compute the Jacobian, we can find its eigenvalues by solving the characteristic equation \(\text{det}(J - \lambda I) = 0\), where \(I\) denotes the identity matrix.

In our exercise, we derived the eigenvalues \(\lambda_{1,2} = \frac{3 \pm \sqrt{5}}{2} \). These values give us critical information about the system's tendencies near the equilibrium, such as whether solutions will grow, decay, or oscillate.

Stability Analysis

Stability analysis in the context of differential equations is concerned with understanding how a system behaves after small perturbations around an equilibrium point. The nature and sign of the eigenvalues of the Jacobian at an equilibrium point determine stability.

For the equilibrium point \( (0,0) \) in our exercise, the eigenvalues are real and positive, which indicates that any small perturbation will compound over time, making \( (0,0) \) an unstable node. Stability analysis helps us conclude that our system won't settle into a periodic orbit – instead, it will depart from the equilibrium, which confirms the absence of non-constant periodic solutions.