Question

Theorem 9.3 .2 provides no information about the stability of a critical point of an almost linear system if that point is a center of the corresponding linear system. That this must be the case is illustrated by the systems $$ \begin{aligned} d x / d t &=y+x\left(x^{2}+y^{2}\right) \\ d y / d t &=-x+y\left(x^{2}+y^{2}\right) \end{aligned} $$ and $$ \begin{aligned} d x / d t &=y-x\left(x^{2}+y^{2}\right) \\ d y / d t &=-x-y\left(x^{2}+y^{2}\right) \end{aligned} $$ (a) Show that \((0,0)\) is a critical point of each system and, furthermore, is a center of the corresponding linear system. (b) Show that each system is almost linear. (c) Let \(r^{2}=x^{2}+y^{2},\) and note that \(x d x / d t+y d y / d t=r d r / d t\). For system (ii) show that \(d r / d t<0\) and that \(r \rightarrow 0\) as \(t \rightarrow \infty\); hence the critical point is asymptotically stable. For system (i) show that the solution of the initial value problem for \(r\) with \(r=r_{0}\) at \(t=0\) becomes unbounded as \(t \rightarrow 1 / 2 r_{0}^{2},\) and hence the critical point is unstable.

Short answer

Question: Analyze the stability of the critical point (0,0) for the following systems of differential equations: (i) $$ \begin{aligned} d x / d t &=y+x\left(x^{2}+y^{2}\right) \\\ d y / d t &=-x+y\left(x^{2}+y^{2}\right) \end{aligned} $$ (ii) $$ \begin{aligned} d x / d t &=y-x\left(x^{2}+y^{2}\right) \\\ d y / d t &=-x-y\left(x^{2}+y^{2}\right) \end{aligned} $$ Answer: For system (i) the critical point (0,0) is unstable, while for system (ii) the critical point (0,0) is asymptotically stable.

Step-by-step solution

Find the critical points of the first system

Let \(y = 0\) and solve for \(x\): $$x\left(x^{2} + y^{2}\right) = 0 \Rightarrow x = 0$$ Now let \(x = 0\) and solve for \(y\): $$-x + y\left(x^{2} + y^{2}\right) = 0 \Rightarrow y = 0$$ Thus, for system (i), the critical point is \((0, 0)\). For system (ii):

Find the critical points of the second system

Let \(y = 0\) and solve for \(x\): $$-x\left(x^{2} + y^{2}\right) = 0 \Rightarrow x = 0$$ Now let \(x = 0\) and solve for \(y\): $$-x - y\left(x^{2} + y^{2}\right) = 0 \Rightarrow y = 0$$ Thus, for system (ii), the critical point is also \((0, 0)\). Now let's consider the corresponding linear systems: For system (i): $$ \begin{aligned} d x / d t &=y \\\ d y / d t &=-x \end{aligned} $$ For system (ii): $$ \begin{aligned} d x / d t &=y \\\ d y / d t &=-x \end{aligned} $$ Both of these linear systems have the same critical point \((0,0)\) and have eigenvalues \(\lambda = \pm i\) (imaginary), which means they are both centers. (b) Show that each system is almost linear. A system is "almost linear" if it can be expressed as the sum of the linear part and a nonlinear part that goes to zero as the point \((x, y)\) goes to the critical point \((0, 0)\). Since both systems are of the form: $$ \begin{aligned} d x / d t &=y \pm x\left(x^{2}+y^{2}\right) \\\ d y / d t &=-x \pm y\left(x^{2}+y^{2}\right) \end{aligned} $$ They are indeed almost linear because the nonlinear part \(x(x^2+y^2)\) and \(y(x^2+y^2)\) goes to zero as \((x, y)\) goes to \((0, 0)\). (c) Let \(r^{2}=x^{2}+y^{2}\), and note that \(x d x / d t+y d y / d t=r d r / d t\). For system (ii):

Compute dr/dt for the second system

Let's compute \(d r/d t\): $$\begin{aligned} x ( y - x \left( x^2 + y^2 \right) ) + y ( - x - y\left( x^2 + y^2 \right) ) &= r dr/dt \end{aligned}$$ From \(r^2 = x^2 + y^2\), we get \(2r dr/dt = 2x dx/dt + 2y dy/dt\). Simplify this to get the equation for \(dr/dt\): $$\begin{aligned} 2r dr/dt &= - 2x^2\left( x^2 + y^2 \right) - 2y^2\left( x^2 + y^2 \right) \\ dr/dt &= - r\left( x^2 + y^2 \right) \end{aligned}$$ Since all terms are negative, we can conclude that \(dr/dt < 0\) for system (ii), which means \(r \rightarrow 0\) as \(t \rightarrow \infty\), and therefore, the critical point is asymptotically stable. For system (i):

Find the solution of the initial value problem for r with r=r0 at t=0

In this case, the equation for \(dr/dt\) is: $$\begin{aligned} dr/dt &= r\left( x^2 + y^2 \right) \end{aligned}$$ Now, we'll solve the initial value problem for \(r\) with \(r = r_0\) at \(t = 0\): $$\begin{aligned} \frac{dr}{dt} &= r\left( x^2 + y^2 \right) \\ \frac{dr}{r^3} &= dt \\ \int_{r_0}^r \frac{1}{\tau^3} d\tau &= \int_0^t d\tau \end{aligned}$$ Solve the integrals: $$\begin{aligned} \left. -\frac{1}{2\tau^2}\right|_{r_0}^r &= t \\ -\frac{1}{2r^2} + \frac{1}{2r_0^2} &= t \end{aligned}$$ Since \(r\) is unbounded when \(t \rightarrow \frac{1}{2r_0^2}\), we can conclude that the critical point is unstable for system (i).

Key concepts

Almost Linear System

An almost linear system is characterized by a blend of linear and nonlinear components, where the nonlinear part vanishes as the solution approaches a critical point. Typically, such systems can be represented as a linear system plus a nonlinear perturbation that becomes insignificant near the equilibrium.

In the given exercise, both differential equation systems meet this criterion because they contain linear terms with the variables 'x' and 'y', plus nonlinear terms involving higher powers of 'x' and 'y' that diminish as \( (x, y) \) approaches the critical point \( (0, 0) \). Thus, at the critical point, the behavior of the systems resembles that of their linear counterparts, making them easier to analyze around that point.

Eigenvalues

In the context of linear systems of differential equations, eigenvalues determine the stability and behavior of solutions near critical points. They are derived from the system's matrix by solving the characteristic equation. The sign and nature (real or complex) of the eigenvalues offer insights into how a system behaves.

For instance, purely imaginary eigenvalues, as found in the linear parts of our systems (\( \lambda = \pm i \) ), indicate that the critical point is a center. This means solutions around this critical point move in closed orbits, neither gravitating towards nor away from it, in the absence of the nonlinear parts.

Asymptotically Stable

A critical point is considered asymptotically stable if, apart from the solutions staying close when perturbed slightly, they also tend to return to the equilibrium point over time. A hallmark of asymptotic stability is when all the eigenvalues of the corresponding linear system have negative real parts.

In the exercise's second system, we derive that \( dr/dt < 0 \) using polar coordinates, signaling that the radius \( r \) decreases over time, meaning solutions spiral towards the origin as time advances toward infinity. This behavior indicates that the critical point at the origin is asymptotically stable – a significant insight for predicting the system's long-term behavior.

Initial Value Problem

An initial value problem (IVP) in the realm of differential equations consists of a differential equation accompanied by specific conditions at the outset, known as initial conditions. The purpose is to find a function or solution curve that not only satisfies the differential equation but also aligns with the given initial values.

The exercise exhibits such a problem where \( r \) is set to \( r_0 \) at \( t = 0 \) for both system (i) and (ii). By solving the IVP, we understand how the solution evolves over time from a specific starting point. In system (i), the solution to the IVP reveals the unpredictable nature of the solution, affirming the instability of the critical point.

Differential Equations

At their core, differential equations are mathematical equations that relate functions with their derivatives. In the realm of dynamics, they are indispensable tools for modeling the rate of change of physical quantities.

The systems provided in the exercise represent two-dimensional differential equations, capturing the interplay between two variables, 'x' and 'y', over time. The systems' solutions, especially when graphically represented, elucidate the trajectories of the system in the phase plane. Such equations are foundational in the fields of physics, engineering, economics, biology, and beyond, providing a mathematical framework for understanding and predicting complex systems behavior.