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Chapter 9: Problem 5

Consider the system of equations $$ d x / d t=y-x f(x, y), \quad d y / d t=-x-y f(x, y) $$ where \(f\) is continuous and has continuous first partial derivatives. Show that if \(f(x, y)>0\) in some neighborhood of the origin, then the origin is an asymptotically stable critical point, and if \(f(x, y)<0\) in some neighborhood of the origin, then the origin is an unstable critical point. Hint: Construct a Liapunov function of the form \(c\left(x^{2}+y^{2}\right) .\)

Short Answer

Expert verified
Question: Determine the stability of the origin for the given system of differential equations depending on the sign of the unknown function \(f(x, y)\). Answer: If \(f(x, y) > 0\) in some neighborhood of the origin, the origin is an asymptotically stable critical point. If \(f(x, y) < 0\) in some neighborhood of the origin, the origin is an unstable critical point.

Step by step solution

01

Identifying Given Information and Liapunov Function

We have a system of differential equations $$ \begin{aligned} \frac{dx}{dt} &= y - xf(x, y) \\ \frac{dy}{dt} &= -x - yf(x, y), \end{aligned} $$ where \(f(x, y)\) is continuous and has continuous first partial derivatives. We are asked to determine the stability of the origin \((0, 0)\). For this purpose, we will consider a Liapunov function of the form \(V(x, y) = c(x^2 + y^2)\), where \(c\) is a constant.
02

Calculate the Time Derivative of the Liapunov Function

To study the stability of the origin, we need to find the time derivative of the Liapunov function. We have $$ \begin{aligned} \frac{dV}{dt} &= \frac{\partial V}{\partial x} \frac{dx}{dt} + \frac{\partial V}{\partial y}\frac{dy}{dt} \\ &= 2cx \left(y - xf(x, y)\right) + 2cy\left(-x - yf(x, y)\right). \end{aligned} $$
03

Analyze the Time Derivative of the Liapunov Function

Now, we need to analyze the sign of \(\frac{dV}{dt}\) to determine the stability of the origin. Let's consider the two cases: Case 1: \(f(x, y) > 0\) in some neighborhood of the origin. In this case, we have $$ \begin{aligned} \frac{dV}{dt} &= 2cx(y - x f(x, y)) - 2cy(x + y f(x, y))\\ &= -2cxyf(x, y) < 0. \end{aligned} $$ Hence, the origin is an asymptotically stable critical point in this case. Case 2: \(f(x, y) < 0\) in some neighborhood of the origin. In this case, we have $$ \begin{aligned} \frac{dV}{dt} &= 2cx(y - x f(x, y)) - 2cy(x + y f(x, y))\\ &= -2cxyf(x, y) > 0. \end{aligned} $$ Hence, the origin is an unstable critical point in this case.
04

Conclusion

We have analyzed the system of equations given, and shown that by constructing a Liapunov function of the form \(c(x^2 + y^2)\), we can conclude: if \(f(x, y) > 0\) in some neighborhood of the origin, the origin is an asymptotically stable critical point; whereas if \(f(x, y) < 0\) in some neighborhood of the origin, the origin is an unstable critical point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Liapunov Function
In the context of differential equations, particularly when analyzing the stability of critical points, a Liapunov function is a powerful tool. Created by the Russian mathematician Aleksandr Lyapunov, it serves as a mathematical 'energy-like' measure for systems described by differential equations.

Think of a Liapunov function as a sort of 'gauge' that can measure how the system's state changes over time. If we find a suitable Liapunov function, we can tell whether a critical point, such as the equilibrium of a system, is stable or not. The function is typically a scalar, and for a two-dimensional system, a common choice is a quadratic form like V(x, y) = c(x^2 + y^2), which represents a paraboloid in a 3D space.

The clever trick here is to choose a Liapunov function that, when we do the math, will have a time derivative that reveals something about the stability. If this derivative is negative in the vicinity of a critical point, we're holding a sign that points to stability, hinting that the system wants to stay put at equilibrium.
Asymptotically Stable Critical Point
Now, what exactly do we mean when we say a critical point is asymptotically stable? Imagine a ball settling at the bottom of a bowl. No matter how you nudge it, it always finds its way back to the center. An asymptotically stable critical point behaves similarly in the realm of differential equations.

For a critical point to be asymptotically stable, two criteria must be met: first, the system must return to the critical point if it's given a slight nudge (we call this stability), and second, the system's state must approach the critical point as time goes on (this is the 'asymptotic' part).

Back to our exercise, if the function f(x, y) is positive near the origin, any deviations will dwindle away over time, and the state will gravitate back to the origin, satisfying both criteria for asymptotic stability.
Time Derivative
The time derivative is essentially the rate at which a function changes with respect to time. When analyzing the system's stability with a Liapunov function, what matters is how the 'energy' (or the Liapunov function's value) changes as the system evolves.

For instance, if the system's energy decreases over time (meaning the time derivative of the Liapunov function is negative), then the system is losing its 'oomph' to stray far from the critical point - a hint that the point is stable. Conversely, if that energy tends to increase (positive time derivative), we're looking at a system that's eager to deviate from the critical point, spelling instability.

Mathematically, the time derivative of our chosen Liapunov function V(x, y) = c(x^2 + y^2) in our system's context was found by applying the chain rule, summing the partial derivatives, and setting the stage for assessing stability.
System of Differential Equations
A system of differential equations like the one we're dealing with can be seen as a set of equations that describes the behavior of a multi-dimensional system. Each equation represents the rate of change of one aspect of the system, typically with respect to time.

These systems can model a vast array of phenomena, from the orbits of planets to the way populations of animals change through time. The key to understanding them lies in the interplay between their individual components - in our case, how the variable x and y influence each other.

Any equilibrium points found within these systems, where the rates of change are zero, become critical to understanding the overall system behavior. Stability analysis using Liapunov functions is just one of the many ways we can peer into the system's future behavior without solving the equations explicitly.

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Most popular questions from this chapter

(a) Determine all critical points of the given system of equations. (b) Find the corresponding linear system near each critical point. (c) Find the eigenalues of each linear system. What conclusions can you then draw about the nonlinear system? (d) Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system. $$ d x / d t=x-x^{2}-x y, \quad d y / d t=\frac{1}{2} y-\frac{1}{4} y^{2}-\frac{3}{4} x y $$

We will prove part of Theorem 9.3 .2 : If the critical point \((0,0)\) of the almost linear system $$ d x / d t=a_{11} x+a_{12} y+F_{1}(x, y), \quad d y / d t=a_{21} x+a_{22} y+G_{1}(x, y) $$ is an asymptotically stable critical point of the corresponding linear system $$ d x / d t=a_{11} x+a_{12} y, \quad d y / d t=a_{21} x+a_{22} y $$ then it is an asymptotically stable critical point of the almost linear system (i). Problem 12 deals with the corresponding result for instability. Consider the linear system (ii). (a) Since \((0,0)\) is an asymptotically stable critical point, show that \(a_{11}+a_{22}<0\) and \(\left.a_{11} a_{22}-a_{12} a_{21}>0 . \text { (See Problem } 21 \text { of Section } 9.1 .\right)\) (b) Construct a Liapunov function \(V(x, y)=A x^{2}+B x y+C y^{2}\) such that \(V\) is positive definite and \(\hat{V}\) is negative definite. One way to ensure that \(\hat{V}\) is negative definite is to choose \(A, B,\) and \(C\) so that \(\hat{V}(x, y)=-x^{2}-y^{2} .\) Show that this leads to the result $$ \begin{array}{l}{A=-\frac{a_{21}^{2}+a_{22}^{2}+\left(a_{11} a_{22}-a_{12} a_{21}\right)}{2 \Delta}, \quad B=\frac{a_{12} a_{22}+a_{11} a_{21}}{\Delta}} \\\ {C=-\frac{a_{11}^{2}+a_{12}^{2}+\left(a_{11} a_{22}-a_{12} a_{21}\right)}{2 \Delta}}\end{array} $$ where \(\Delta=\left(a_{11}+a_{22}\right)\left(a_{11} a_{22}-a_{12} a_{21}\right)\) (c) Using the result of part (a) show that \(A>0\) and then show (several steps of algebra are required) that $$ 4 A C-B^{2}=\frac{\left(a_{11}^{2}+a_{12}^{2}+a_{21}^{2}+a_{22}^{2}\right)\left(a_{11} a_{22}-a_{12} a_{21}\right)+2\left(a_{11} a_{22}-a_{12} a_{21}\right)^{2}}{\Delta^{2}}>0 $$ Thus by Theorem 9.6.4, \(V\) is positive definite.

Consider the system (3) in Example 1 of the text. Recall that this system has an asymptotically stable critical point at (0.5,0.5) , corresponding to the stable coexistence of the two population species. Now suppose that immigration or emigration occurs at the constant rates of \(\delta a\) and \(\delta b\) for the species \(x\) and \(y,\) respectively. In this case equations ( 3 ) are replaced by $$\frac{d x}{d t}=x(1-x-y)+\delta a, \quad \frac{d y}{d t}=\frac{y}{4}(3-4 y-2 x)+\delta b$$ The question is what effect this has on the location of the stable equilibrium point. a. To find the new critical point, we must solve the equations $$\begin{aligned} x(1-x-y)+\delta a &=0 \\ \frac{y}{4}(3-4 y-2 x)+\delta b &=0 \end{aligned}$$ One way to proceed is to assume that \(x\) and \(y\) are given by power series in the parameter \(\delta ;\) thus $$x=x_{0}+x_{1} \delta+\cdots, \quad y=y_{0}+y_{1} \delta+\cdots$$ Substitute equations (44) into equations (43) and collect terms according to powers of \(\delta\). b. From the constant terms (the terms not involving \(\delta\) ), show that \(x_{0}=0.5\) and \(y_{0}=0.5,\) thus confirming that in the absence of immigration or emigration, the critical point is (0.5,0.5) . c. From the terms that are linear in \(\delta,\) show that \\[ x_{1}=4 a-4 b, \quad y_{1}=-2 a+4 b \\] d. Suppose that \(a>0\) and \(b>0\) so that immigration occurs for both species. Show that the resulting equilibrium solution may represent an increase in both populations, or an increase in one but a decrease in the other. Explain intuitively why this is a reasonable result.

The equation of motion of a spring-mass system with damping (see Section 3.8) is $$ m \frac{d^{2} u}{d t^{2}}+c \frac{d u}{d t}+k u=0 $$ where \(m, c,\) and \(k\) are positive. Write this second order equation as a system of two first order equations for \(x=u, y=d u / d t .\) Show that \(x=0, y=0\) is a critical point, and analyze the nature and stability of the critical point as a function of the parameters \(m, c,\) and \(k .\) A similar analysis can be applied to the electric circuit equation (see Section 3.8) $$L \frac{d^{2} I}{d t^{2}}+R \frac{d I}{d t}+\frac{1}{C} I=0.$$

a. Sketch the nullclines and describe how the critical points move as \(\alpha\) increases. b. Find the critical points. c. Let \(\alpha=2\). Classify each critical point by investigating the corresponding approximate linear system. Draw a phase portrait in a rectangle containing the critical points. d. Find the bifurcation point \(\alpha_{0}\) at which the critical points coincide. Locate this critical point, and find the eigenvalues of the approximate linear system. Draw a phase portrait. e. For \(\alpha>\alpha_{0},\) there are no critical points. Choose such a value of \(\alpha\) and draw a phase portrait. $$x^{\prime}=\frac{3}{2} \alpha-y, \quad y^{\prime}=-4 x+y+x^{2}$$
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