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Chapter 9: Problem 6

A generalization of the undamped pendulum equation is $$ d^{2} u / d t^{2}+g(u)=0 $$ where \(g(0)=0, g(u)>0\) for \(00\) for \(u \neq 0,-k

Short Answer

Expert verified
Question: Rewrite the given second-order differential equation as a system of two first-order equations in terms of x and y. Determine the critical point (x=0, y=0). Show that the given function V(x, y) is positive definite and use this to prove the stability of the critical point (0,0). Answer: The second-order differential equation can be rewritten as a system of two first-order equations as follows: 1. \( \frac{d x}{d t} = y \) 2. \( \frac{d y}{d t} = - g(u)\) The critical point is (0, 0) as it satisfies the conditions \(\frac{d x}{d t} = 0\) and \(\frac{d y}{d t} = 0\). The function V(x, y) is positive definite as it satisfies the criteria \(V(x, y) > 0\) for all \(x \neq 0\) and \(y \neq 0\), and \(V(0, 0) = 0\). The time derivative of the Liapunov function is zero, thereby proving that the critical point (0, 0) is stable.

Step by step solution

01

Part (a) - Rewrite the equation as a system of two first-order equations

Determine the derivatives x and y given the equations \(x = u\) and \( y = \frac{d u}{d t}\). According to this, we have: \( \frac{d x}{d t} = \frac{d u}{d t} = y \) and \( \frac{d y}{d t} = \frac{d^2 u}{d t^2} = - g(u).\) Now our system of two first-order equations is: 1. \( \frac{d x}{d t} = y \) 2. \( \frac{d y}{d t} = - g(u)\)
02

Part (a) - Determine the critical point

A critical point occurs when the derivatives are zero, i.e., \(\frac{d x}{d t} = 0\) and \(\frac{d y}{d t} = 0\). From the first equation, we have \(y = 0\). Plugging this into the second equation gives us: \(0 = - g(u)\) Since \(g(0) = 0\), we find that \(u = x = 0\). Therefore, the critical point is \((0, 0)\).
03

Part (b) - Show that V(x, y) is positive definite

We are given the function: $$ V(x, y) = \frac{1}{2} y^2 + \int_0^x g(s) ds, \quad -k < x < k$$ For a function to be positive definite, it needs to satisfy two criteria: 1. \(V(x, y) > 0\) for all \(x \neq 0\) and \(y \neq 0\). 2. \(V(0, 0) = 0\). Let's examine these criteria. 1. The integral \(\int_0^x g(s) ds\) is positive for \(0 < x < k\) and negative for \(-k < x < 0\). Since we are considering the range \(-k < x < k\), this integral is non-negative for this range. The term \(\frac{1}{2}y^2\) is always non-negative. Therefore, the sum of these two terms is always non-negative, and it is positive for any \(x \neq 0\) and \(y \neq 0\). Hence, the first criterion is satisfied. 2. Plugging the critical point \((0, 0)\) into the function V(x, y) yields: $$ V(0, 0) = \frac{1}{2} 0^2 + \int_0^0 g(s) ds = 0$$ Hence, the second criterion is also satisfied. Since both criteria are fulfilled, the function V(x, y) is positive definite.
04

Part (b) - Show the stability of the critical point (0,0)

To show that the critical point is stable, we need to find the time derivative of the Liapunov function \(V(x, y)\): $$ \frac{d V(x, y)}{d t} = \frac{d}{d t}\left( \frac{1}{2} y^2 + \int_0^x g(s) ds \right)$$ Using the chain rule and the fundamental theorem of calculus, we have: $$ \frac{d V(x, y)}{d t} = y \frac{d y}{d t} + g(x) \frac{d x}{d t}$$ From our first-order equations system, we know that: 1. \( \frac{d x}{d t} = y\) 2. \(\frac{d y}{d t} = - g(u)\) Substitute these into the above equation: $$ \frac{d V(x, y)}{d t} = y(- g(u)) + g(x) y = - y g(u) + y g(x) = 0$$ Since the time derivative of the Liapunov function is zero, the critical point \((0, 0)\) is stable.

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Key Concepts

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

Liapunov Function
In the realm of differential equations, the concept of a Liapunov function is an essential tool for determining the stability of critical points. It is a scalar function that serves as a measure of the system's energy; the function generally increases or decreases along trajectories of the system and thus provides insight into the behavior of the system over time.

A Liapunov function, often denoted as V(x,y), must satisfy certain conditions to conclude stability. First, it must be positive definite, which means it is positive for all non-zero values in its domain and equals zero at the critical point. Moreover, its time derivative, taken along trajectories of the system, should be negative semi-definite, suggesting that the function does not increase with time.

In the context of the exercise provided, the Liapunov function is constructed by summing the kinetic and potential energy of the system, which involves the integral of the restoring force function, g(u), over the displacement, and the quadratic term involving velocity, y. This function's positive definiteness speaks volumes about the system's inherent tendency to resist divergence from the critical point, indicating potential stability.
Critical Point
A critical point (also known as an equilibrium point) of a system of differential equations is a point where all derivatives of the system vanish鈥攎eaning, it's a point where the system doesn't change over time. In simpler terms, it's like a ball at the bottom of a bowl; it is stable and doesn't move unless perturbed.

Identifying a critical point involves setting the system's derivatives equal to zero and solving for the variable(s) in question. In the exercise example, the condition of \(g(0)=0\) helps us pinpoint the critical point at (0,0). This implies, intuitively, that when the system (the pendulum in this case) is at rest with no displacement and no velocity, it's at the critical point. Understanding where these points lie and their stability is fundamental in predicting the long-term behavior of dynamical systems.
Positive Definite
Describing a function as positive definite relates directly to the concept of stability in different contexts, including optimization and system dynamics. A positive definite function represents a scenario where the output of the function is always greater than zero for any input other than the function's minimum - often at the origin, in the case of dynamical systems.

For the function to be positive definite, it must meet two critical requirements: it should be zero at the origin (or critical point) and positive elsewhere in its domain. A clear understanding of these conditions allows us to assert that a Liapunov function exhibiting such properties, as shown in the given exercise, aligns with a system that favors stability. When a Liapunov function is established to be positive definite, it effectively encapsulates the idea that the system鈥檚 鈥榚nergy鈥� doesn鈥檛 dissipate or escalate arbitrarily, thereby indicating a stable configuration around the critical point.

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Most popular questions from this chapter

By introducing suitable dimensionless variables, the system of nonlinear equations for the damped pendulum [Frqs. (8) of Section 9.3] can be written as $$ d x / d t=y, \quad d y / d t=-y-\sin x \text { . } $$ (a) Show that the origin is a critical point. (b) Show that while \(V(x, y)=x^{2}+y^{2}\) is positive definite, \(f(x, y)\) takes on both positive and negative values in any domain containing the origin, so that \(V\) is not a Liapunov function. Hint: \(x-\sin x>0\) for \(x>0\) and \(x-\sin x<0\) for \(x<0 .\) Consider these cases with \(y\) positive but \(y\) so small that \(y^{2}\) can be ignored compared to \(y .\) (c) Using the energy function \(V(x, y)=\frac{1}{2} y^{2}+(1-\cos x)\) mentioned in Problem \(6(b),\) show that the origin is a stable critical point. Note, however, that even though there is damping and we can epect that the origin is asymptotically stable, it is not possible to draw this conclusion using this Liapunov function. (d) To show asymptotic stability it is necessary to construct a better Liapunov function than the one used in part (c). Show that \(V(x, y)=\frac{1}{2}(x+y)^{2}+x^{2}+\frac{1}{2} y^{2}\) is such a Liapunov function, and conclude that the origin is an asymptotically stable critical point. Hint: From Taylor's formula with a remainder it follows that \(\sin x=x-\alpha x^{3} / 3 !,\) where \(\alpha\) depends on \(x\) but \(0<\alpha<1\) for \(-\pi / 2

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. $$x^{\prime}=-4 x+y+x^{2}, \quad y^{\prime}=-\alpha-x+y$$

(a) Find all the critical points (equilibrium solutions). (b) Use a computer to draw a direction field and portrait for the system. (c) From the plot(s) in part (b) determine whether each critical point is asymptotically stable, stable, or unstable, and classify it as to type. $$ d x / d t=(2+x)(y-x), \quad d y / d t=y\left(2+x-x^{2}\right) $$

(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. $$ d x / d t=y, \quad d y / d t=2 x+y $$

Verify that \((0,0)\) is a critical point, show that the system is almost linear, and discuss the type and stability of the critical point \((0,0)\) by examining the corresponding linear system. $$ d x / d t=(1+x) \sin y, \quad d y / d t=1-x-\cos y $$
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