Question

Construct a suitable Liapunov function of the form \(a x^{2}+c y^{2}\) where \(a\) and \(c\) are to be determined. Then show that the critical point at the origin is of the indicated type. $$ d x / d t=x^{3}-y^{3}, \quad d y / d t=2 x y^{2}+4 x^{2} y+2 y^{3} ; \quad \text { unstable } $$

Short answer

The Liapunov function proving that the origin is an unstable critical point for the given system of differential equations is \(V(x, y) = ax^2 + cy^2\), where \(a\) and \(c\) are positive constants.

Step-by-step solution

Find the critical point of the given system of ODEs

To find the critical point of the given system of ODEs, set both \(dx/dt\) and \(dy/dt\) to zero and solve the following equations. $$ x^3 - y^3 = 0 \quad (1) \\ 2 xy^2 + 4 x^2 y + 2 y^3 = 0 \quad (2) $$ From equation (1), we have \(x^3 = y^3,\) which implies that either \(x=y\) or \(x=-y.\) For the case \(x=y\), equation (2) becomes \(2y^3 + 4y^2 y + 2y^3 = 0 \Rightarrow y^3=0 \Rightarrow y=0.\) So, \(x=y=0\) is a critical point.

Choose a Liapunov function

Choose a Liapunov function of the form \(V(x, y) = ax^2 + cy^2\), where \(a\) and \(c\) are positive constants to be determined.

Calculate the time derivative of the Liapunov function

We will now find \(\frac{dV}{dt}\). Using the original ODEs: $$ \frac{dV}{dt} = \frac{\partial V}{\partial x} \frac{dx}{dt} + \frac{\partial V}{\partial y} \frac{dy}{dt} $$ Differentiate the Liapunov function \(V(x, y)\) with respect to \(x\) and \(y\): $$ \frac{\partial V}{\partial x} = 2ax, \quad \frac{\partial V}{\partial y} = 2cy $$ Substitute the derivatives and the given ODEs into the equation for \(\frac{dV}{dt}\): $$ \frac{dV}{dt} = 2ax(x^3 - y^3) + 2cy(2xy^2 + 4x^2y + 2y^3) $$

Determine the signs of \(a\) and \(c\)

In order for the Liapunov function to be valid as per the given conditions (unstable system), the following must hold: 1. \(V(0, 0) = 0\) 2. \(V(x, y) > 0\) for all \((x, y) \neq (0, 0)\) 3. \(\frac{dV}{dt} > 0\) for all \((x, y) \neq (0, 0)\) Since the Liapunov function is \(V(x, y) = ax^2 + cy^2\), the first two conditions are automatically satisfied for positive values of \(a\) and \(c\). Now we need to ensure that \(\frac{dV}{dt} > 0\) for all \((x, y) \neq (0, 0)\) to show the system is unstable. From the equation for \(\frac{dV}{dt}\): $$ \frac{dV}{dt} = 2ax(x^3 - y^3) + 2cy(2xy^2 + 4x^2y + 2y^3) $$ Both terms in \(\frac{dV}{dt}\) need to have the same sign and be non-zero to satisfy the condition. If we take \(a\) and \(c\) positive, this condition will hold true for any \(x, y \neq 0,\) hence making the origin an unstable critical point.

Show that the origin is an unstable critical point

We have determined the Liapunov function \(V(x, y) = ax^2 + cy^2\), where \(a\) and \(c\) are positive constants, ensuring that \(V(0, 0) = 0\) and \(V(x, y) > 0\) for all \((x, y) \neq (0, 0)\). We have also calculated the time derivative of the Liapunov function and found that \(\frac{dV}{dt} > 0\) for all \((x, y) \neq (0, 0)\) when \(a\) and \(c\) are positive. Thus, by the Liapunov criteria, the critical point at the origin is unstable for the given system of ODEs.

Key concepts

Differential Equations

Differential equations play a pivotal role in sciences and engineering to model various phenomena. A differential equation expresses the rate of change of a quantity as a function of the quantity itself and possibly other variables. For instance, when we describe the growth of a population over time, we might use a differential equation to represent the rate at which the population increases.

In the given exercise, the system of ordinary differential equations (ODEs) comprises two equations that define how two variables, x and y, change with respect to time. Solving these equations often involves finding functions of x and y that satisfy the given relationships. A unique aspect of differential equations is their ability to describe dynamical systems, which are systems that change over time. Here, the given equations suggest a complex interaction between x and y, with their rates of change depending on cubic powers and products of both variables.

Critical Point

In the landscape of a differential equation, a critical point (or equilibrium point) is where the system doesn't change — it's in equilibrium. Identifying critical points is crucial because they highlight where something interesting happens; for example, a predator-prey model might have an equilibrium where both species coexist in balance.

Locating these critical points typically involves setting the rates of change to zero and solving the resulting algebraic equations. In our exercise, when we solve the given system of ODEs by setting the derivatives equal to zero, we find that the origin \(x=y=0\) is a critical point. This step is often the starting point for analyzing the behavior of the system near these points. Establishing the nature of a critical point as stable or unstable gives us powerful insights into how small changes can dissipate over time or escalate into larger fluctuations.

Stability Analysis

Stability analysis involves determining whether small perturbations or changes to a system will grow over time, leading to instability, or if the system will return to its original state, reflecting stability. This concept is essential across various fields, for instance, in mechanical engineering, the stability of a structure is critical to its safety and performance.

One method used in stability analysis for ODEs is the Liapunov function, a clever mathematical tool that helps us understand the behavior around critical points without solving the entire differential equation. In our textbook problem, we construct a Liapunov function with a specific form and determine conditions that show the origin is an unstable critical point. By testing if the Liapunov function grows over time (positive derivative), we can conclude that any slight disturbance will not settle down, thereby characterizing the critical point as unstable, which was the goal of our exercise.