Question

Consider the autonomous system $$ d x / d t=x, \quad d y / d t=-2 y+x^{3} $$ (a) Show that the critical point \((0,0)\) is a saddle point. (b) Sketch the trajectories for the corresponding linear system and show that the trajectory for which \(x \rightarrow 0, y \rightarrow 0\) as \(t \rightarrow \infty\) is given by \(x=0\). (c) Determine the trajectories for the nonlinear system for \(x \neq 0\) by integrating the equation for \(d y / d x\). Show that the trajectory corresponding to \(x=0\) for the linear system is unaltered, but that the one corresponding to \(y=0\) is \(y=x^{3} / 5 .\) Sketch several of the trajectories for the nonlinear system.

Short answer

Answer: The critical point (0,0) is a saddle point.

Step-by-step solution

Determine the critical points

Set the system of differential equations equal to zero: $$ x = 0 \\ -2y + x^3 = 0. $$ Notice that \((0,0)\) is the only critical point.

Determine the stability at the critical point using Jacobian matrix

Calculate the Jacobian matrix for the system: $$ J(x,y) = \begin{pmatrix} \frac{\partial( x)}{\partial x} & \frac{\partial( x)}{\partial y} \\ \frac{\partial(-2y + x^3)}{\partial x} & \frac{\partial(-2y + x^3)}{\partial y} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 3x^2 & -2 \end{pmatrix}. $$ Evaluate the Jacobian matrix at the critical point \((0,0)\): $$ J(0,0) = \begin{pmatrix} 1 & 0 \\ 0 & -2 \end{pmatrix}. $$ Find the eigenvalues of \(J(0,0)\): $$ \det(J(0,0) - \lambda I) = \det\begin{pmatrix} 1-\lambda & 0 \\ 0 & - 2 - \lambda \end{pmatrix} = (1-\lambda)(-2-\lambda) = 0. $$ We get two eigenvalues: \(\lambda_1 = 1\) and \(\lambda_2 = -2\). Since we have one positive and one negative eigenvalue, the critical point \((0,0)\) is a saddle point.

Linear system's trajectories

The Jacobian matrix for the linearized system is $$ J_L(x,y) = \begin{pmatrix} 1 & 0 \\ 0 & -2 \end{pmatrix}. $$ The system of equations for the linear system is $$ \frac{dx}{dt} = x \\ \frac{dy}{dt} = -2y. $$ Let's sketch the trajectories in the phase plane. The \(x\)-axis (\(y=0\)) has positive growth (since \(x' = x > 0\) for \(x>0\)) and negative growth for \(x<0\). The \(y\)-axis (\(x=0\)) has negative growth (since \(y' = -2y < 0\) for \(y>0\)) and eventually goes to the origin, so the trajectory for which \(x \to 0\) and \(y \to 0\) as \(t \to \infty\) is given by \(x=0\).

Trajectories for nonlinear system

Now we will determine the trajectories for the nonlinear system, which are given by the equation $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2y + x^3}{x}. $$ Separate variables and integrate: $$ \int \frac{dy}{-2y + x^3} = \int \frac{dx}{x}. $$ Substitute \(u = -2y + x^3\), then \(du = -2dy\). We have, $$ -\frac{1}{2} \int \frac{du}{u} = \int \frac{dx}{x}. $$ Integrate both sides: $$ -\frac{1}{2} \ln |u| = \ln |x| + C. $$ Solve for \(y\): $$ y = -\frac{1}{2}(x^3-2Ce^{\ln(x)}) = -\frac{1}{2}(x^3 - 2Cx). $$ At \(x=0\), we get $$ y = -\frac{1}{2} (0-2Cx) = -Cx, $$ which indicates that the trajectory given as \(x=0\) remains unaltered. For \(y=0\), we have $$ 0 = -\frac{1}{2}(x^3 - 2Cx) \Rightarrow x^3=2Cx \Rightarrow y = \frac{x^3}{2C} = \frac{x^3}{5}. $$ Now we can sketch the trajectories for the nonlinear system considering \(x=0\) and \(y = \frac{x^3}{5}\). The overall process involved finding critical points, evaluating the stability using eigenvalues, sketching linear system's trajectories, and finally finding and sketching trajectories for the nonlinear system.

Key concepts

Critical Point

A critical point in the context of a differential equation is a point where the system does not change, that is, its derivative is zero. For the autonomous system
\[ \frac{dx}{dt} = x \quad \frac{dy}{dt} = -2y + x^3 \]
we find that setting the derivatives equal to zero yields the point \((0,0)\). This critical point is where all trajectories in the phase plane either originate, terminate, or change direction. Understanding the behavior of the system near this point is crucial in analyzing the system's overall behavior.

Jacobian Matrix

The Jacobian matrix is a powerful tool in analyzing nonlinear systems. It contains all first-order partial derivatives of a function with respect to its variables and helps in linearizing the system near the critical points. For our system, the Jacobian matrix is calculated as
\[ J(x,y) = \begin{pmatrix} 1 & 0 \ 3x^2 & -2 \end{pmatrix} \]
At the critical point \((0,0)\), the Jacobian simplifies to a diagonal matrix. The eigenvalues of this matrix help determine the stability and type of the critical point.

Saddle Point

A saddle point is a type of critical point where the behavior of trajectories is such that they are attracted towards the point along one axis and repelled along another. With eigenvalues \(\lambda_1 = 1\) and \(\lambda_2 = -2\), we find that our critical point \((0,0)\) is indeed a saddle point since one eigenvalue is positive (indicative of a source along that direction) and the other is negative (indicative of a sink along that direction).

Trajectory Sketching

Trajectory sketching involves drawing the paths in the phase plane that solutions to the differential equations follow. For the linear system, the equations
\[ \frac{dx}{dt} = x \quad \frac{dy}{dt} = -2y \]
tell us that the trajectory along the x-axis (where \(y=0\)) moves away from the origin, and along the y-axis (where \(x=0\)), it moves towards the origin. Sketching trajectories helps visualize the stability and the behavior of the system around critical points.

Nonlinear System Analysis

Nonlinear system analysis is more complex than linear system analysis due to the nonlinearity present in equations. It typically involves examining how solutions behave using tools such as the phase plane, Jacobian matrix, and trajectory sketching. Through these methods, we can gain insight into the unique behavior of nonlinear systems, such as the existence of limit cycles, bifurcations, and complex equilibrium structures.

Eigenvalues

Eigenvalues play a pivotal role in determining the stability and nature of critical points in a differential system. By finding the eigenvalues of the Jacobian matrix at the critical point, we can infer whether the point is a node, spiral, or saddle, and whether it is stable or unstable. In our example, the differing signs of the eigenvalues reveal the saddle nature of the critical point \((0,0)\).

Phase Plane

The phase plane is a graphical representation of the possible trajectories of a dynamical system. It is a two-dimensional plot with axes representing the system's variables. The plane's points correspond to the system's states, and the trajectories represent the path that the state of the system will follow over time. Analyzing the phase plane allows us to visualize the behavior of the system without solving the equations explicitly.

Separation of Variables

Separation of variables is a technique used to solve differential equations by separating the variables involved into two sides of an equation. For our system, the process involves separating the variables \(x\) and \(y\) so that each is only on one side of the equation. As seen in the solution, integrating these separated variables helps us find expressions for the trajectories of the nonlinear system, such as \(y = \frac{x^3}{5}\).