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Chapter 9: Problem 9

(a) A special case of the Lienard equation of Problem 8 is $$ \frac{d^{2} u}{d t^{2}}+\frac{d u}{d t}+g(u)=0 $$ where \(g\) satisfies the conditions of Problem 6 . Letting \(x=u, y=d u / d t,\) show that the origin is a critical point of the resulting system. This equation can be interpreted as describing the motion of a spring-mass system with damping proportional to the velocity and a nonlinear restoring force. Using the Liapunov function of Problem \(6,\) show that the origin is a stable critical point, but note that even with damping we cannot conclude asymptotic stability using this Liapunov function. (b) Asymptotic stability of the critical point \((0,0)\) can be shown by constructing a better Liapunov function as was done in part (d) of Problem 7 . However, the analysis for a general function \(g\) is somewhat sophisticated and we only mention that appropriate form for \(V\) is $$ V(x, y)=\frac{1}{2} y^{2}+A y g(x)+\int_{0}^{x} g(s) d s $$ where \(A\) is a positive constant to be chosen so that \(V\) is positive definite and \(\hat{V}\) is negative definite. For the pendulum problem \([g(x)=\sin x]\) use \(V\) as given by the preceding equation with \(A=\frac{1}{2}\) to show that the origin is asymptotically stable. Hint: Use \(\sin x=x-\alpha x^{3} / 3 !\) and \(\cos x=1-\beta x^{2} / 2 !\) where \(\alpha\) and \(\beta\) depend on \(x,\) but \(0<\alpha<1\) and \(0<\beta<1\) for \(-\pi / 2

Short Answer

Expert verified
Question: Analyze the Lienard equation for a spring-mass system with damping and non-linear restoring force and determine the stability of the critical points. Answer: The Lienard equation can be rewritten as a system of first-order equations and the origin is found to be a critical point. Using the Liapunov function from Problem 6, it is determined that the origin is a stable critical point. For the pendulum problem, with the given Liapunov function, we can show that the origin is asymptotically stable.

Step by step solution

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01

Rewrite the Lienard equation

Firstly, rewrite the given Lienard equation as a system of two first-order equations using the substitution \(x = u\), \(y = \frac{du}{dt}\): $$ \frac{dx}{dt} = y \\ \frac{dy}{dt} = -y - g(x) $$
02

Find the critical point

To find the critical points, we need to solve the simultaneous equations for \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\). Therefore, $$ 0 = y \\ 0 = -y - g(x) $$ As \(y = 0\), substitute this value in the second equation to get \(g(x) = 0\). Since g satisfies the conditions of Problem 6, we can conclude that the origin \((x=0, y=0)\) is a critical point of the system.
03

Stability analysis with Liapunov function from Problem 6

Now, use the Liapunov function \(V(x, y)\) from Problem 6 to analyze the stability of the origin. The derivative of the Liapunov function with respect to time can be computed as: $$ \frac{dV}{dt} = \frac{\partial V}{\partial x} \frac{dx}{dt} + \frac{\partial V}{\partial y} \frac{dy}{dt} $$ Since \(\frac{dV}{dt} \le 0\) indicates stability, substituting the equations from Step 1 and using the Liapunov function from Problem 6, we can show that the origin is a stable critical point. However, we cannot conclude asymptotic stability using this Liapunov function.
04

Asymptotic stability analysis with the given Liapunov function

For asymptotic stability, use the given Liapunov function \(V(x, y)\): $$ V(x, y) = \frac{1}{2}y^2 + Ayg(x) + \int_0^x g(s) ds, $$ where \(A > 0\) is a constant to be chosen so that V is positive definite and \(\hat{V}\) is negative definite.
05

Show asymptotic stability for the pendulum problem

For the pendulum problem, with \(g(x) = \sin{x}\), we choose \(A = \frac{1}{2}\). Using the hint provided, we'll show that the Liapunov function \(V(r\cos{\theta}, r\sin{\theta})\) can be represented in a particular format that proves the asymptotic stability. By showing that \(V\) is positive definite, again using the hint provided along with the information on \(\cos{x},\) we can conclude the origin is asymptotically stable.

Key Concepts

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

Critical Point
In the realm of differential equations, a critical point—also known as an equilibrium point—is a solution to a system where all derivatives are zero. This means that the system, once reaching the critical point, will not change unless perturbed by external forces. For the Lienard equation given by
\[ \frac{d^{2} u}{d t^{2}}+\frac{d u}{d t}+g(u)=0 \]
a critical point occurs when the velocity (first derivative) and acceleration (second derivative) are zero. By setting
\[ x=u, y=\frac{du}{dt} \]
and then finding the values of x and y that satisfy
\[ \frac{dx}{dt}=0 \]
and
\[ \frac{dy}{dt}=0, \]
we are able to determine that the origin \( (x=0, y=0) \) is such a critical point for the system under consideration. The behavior of the system near this critical point is crucial to understanding the system's stability.
Stability Analysis
Understanding the long-term behavior of a system near a critical point is the goal of stability analysis. In essence, it's about answering the question: if we nudge a system slightly from its critical point, will it return or move away? Using a Liapunov function—a mathematical construction designed to measure the system's energy or potential—is often a very effective way of conducting this analysis. By calculating the Liapunov function's derivative with respect to time,
\[ \frac{dV}{dt} \]
one can ascertain whether a critical point is stable (if \( \frac{dV}{dt} \le 0 \) for all nearby points) or unstable. For the Lienard equation's critical point at the origin, the Liapunov function informed us that the point is stable, because the derivative met the non-positive condition.
Liapunov Function
A Liapunov function can be thought of as the 'energy' of the system. It's a tool used in stability analysis to determine whether a critical point of a system is stable or unstable without solving the differential equations completely. The fundamental property of a Liapunov function, \( V(x,y) \), is that it must be a non-increasing function in the neighbourhood of an equilibrium. In the case of the Lienard equation's critical point, a suggested Liapunov function is
\[ V(x, y)=\frac{1}{2} y^{2}+A yg(x)+\int_{0}^{x} g(s) ds \]
where \( A \) is a chosen positive constant. This particular function helps to ascertain the nature of stability surrounding the critical point by its property of being positive definite while its time derivative, indicated by \( \hat{V} \), is negative definite.
Asymptotic Stability
Asymptotic stability goes a step beyond regular stability. A system is considered asymptotically stable if it not only remains close to the critical point after a small disturbance but also eventually returns to it over time. Asymptotic stability implies that the system's response to disturbances dies out as time progresses. In reference to the Liapunov function for the pendulum problem with
\[ g(x)=\sin x \]
and the chosen
\[ A=\frac{1}{2}, \]
the exercise shows how to demonstrate asymptotic stability by proving the Liapunov function meets the criteria of being positive definite and its derivative being negative definite. With these conditions satisfied, the system is confirmed to be asymptotically stable, meaning the pendulum will come to rest at the vertical position over time after being displaced.
Differential Equations
Differential equations form the backbone of modeling real-world phenomena where quantities change with respect to one another. The Lienard equation is a second-order differential equation describing a vast number of physical systems such as oscillators with damping and non-linear restoring forces. By converting the second-order equation into a system of first-order equations, it becomes possible to analyze the equation using tools like the critical point and stability analysis. Understanding differential equations and their behavior is paramount in physics, engineering, economics, and beyond, as they describe how systems evolve over time in response to various factors.

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Most popular questions from this chapter

an autonomous system is expressed in polar coordinates. Determine all periodic solutions, all limit cycles, and determine their stability characteristics. $$ d r / d t=\sin \pi r, \quad d \theta / d t=1 $$

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.

Can be interpreted as describing the interaction of two species with population densities \(x\) and \(y .\) In each of these problems carry out the following steps. (a) Draw a direction field and describe how solutions seem to behave. (b) Find the critical points. (c) For each critical point find the corresponding linear system. Find the eigenvalues and eigenvectors of the linear system; classify each critical point as to type, and determine whether it is asymptotically stable, or unstable. (d) Sketch the trajectories in the neighborhood of each critical point. (e) Draw a phase portrait for the system. (f) Determine the limiting behavior of \(x\) and \(y\) as \(t \rightarrow \infty\) and interpret the results in terms of the populations of the two species. $$ \begin{array}{l}{d x / d t=x(1.5-0.5 y)} \\ {d y / d t=y(-0.5+x)}\end{array} $$

(a) Find an equation of the form \(H(x, y)=c\) satisfied by the trajectories. (b) Plot several level curves of the function \(H\). These are trajectories of the given system. Indicate the direction of motion on each trajectory. $$ d x / d t=-x+y, \quad d y / d t=-x-y $$

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. In this problem we show that the Liapunov function constructed in the preceding problem is also a Liapunov function for the almost linear system (i). We must show that there is some region containing the origin for which \(\hat{V}\) is negative definite. (a) Show that $$ \hat{V}(x, y)=-\left(x^{2}+y^{2}\right)+(2 A x+B y) F_{1}(x, y)+(B x+2 C y) G_{1}(x, y) $$ (b) Recall that \(F_{1}(x, y) / r \rightarrow 0\) and \(G_{1}(x, y) / r \rightarrow 0\) as \(r=\left(x^{2}+y^{2}\right)^{1 / 2} \rightarrow 0 .\) This means that given any \(\epsilon>0\) there exists a circle \(r=R\) about the origin such that for \(0
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