/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Give a direct definition of an u... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 22

Give a direct definition of an unstable equilibrium of \(y^{\prime \prime}+p(y)=0\).

Short Answer

Expert verified
In the context of the given differential equation, \(y^{\prime \prime} + p(y) = 0\), an unstable equilibrium is an equilibrium point \(y_e\) where the function \(p(y)\) satisfies the conditions: \(y > y_e\) implies \(y^{\prime \prime} < 0\) (concave down), and \(y < y_e\) implies \(y^{\prime \prime} > 0\) (concave up). These conditions lead to nearby trajectories moving away from the equilibrium point over time, causing the equilibrium to be unstable.

Step by step solution

01

Understanding Equilibrium Points and Stability

An equilibrium point for a differential equation is a constant solution that makes it true. For the given second-order differential equation, let's find the equilibrium points by setting \(y^{\prime \prime}\) equal to zero: $$y^{\prime \prime} + p(y) = 0 \implies y^{\prime \prime} = -p(y).$$ In this case, at the equilibrium points, \(y^{\prime \prime} = 0\), so \(-p(y) = 0\). Hence, $$p(y) = 0.$$ An equilibrium point is said to be stable if nearby trajectories return toward it over time, unstable if nearby trajectories move away from it over time, or semi-stable if trajectories either converge to or diverge from the equilibrium point. To determine the stability, we'll need to analyze the behavior of solutions around the equilibrium point.
02

Finding the Equilibrium Points for the Given Equation

To find the equilibrium points for the given equation, we set $$p(y) = 0.$$ Since \(p(y)\) is an arbitrary function, we cannot solve this further. However, let's denote the equilibrium points as \(y_e\).
03

Checking the Stability of the Equilibrium Points

To check the stability of the equilibrium points, we will analyze the behavior of the second derivative, i.e., \(y^{\prime \prime} = -p(y)\). At an equilibrium point \(y_e\), the second derivative is zero (\(y^{\prime \prime} = 0\)). For the equilibrium points to be unstable, nearby solutions should have the second derivative with opposite signs as the function moves away from the equilibrium point. If the second derivative is positive, the function is concave up, and if it is negative, the function is concave down. So, to put it in simple words, an equilibrium point \(y_e\) is unstable if whenever \(y > y_e\), we have \(y^{\prime \prime} < 0\) (concave down), and whenever \(y < y_e\), we have \(y^{\prime \prime} > 0\) (concave up). This means that if a solution starts near the equilibrium point, it will move further away from it. Hence, the equilibrium point is unstable.
04

Providing the Direct Definition of an Unstable Equilibrium

So, a direct definition of an unstable equilibrium of the given equation \(y^{\prime \prime} + p(y) = 0\) is as follows: An equilibrium point \(y_e\) for which the function \(p(y)\) is such that $$y > y_e \implies y^{\prime \prime} < 0$$ and $$y < y_e \implies y^{\prime \prime} > 0,$$ which leads to the nearby trajectories moving away from the equilibrium point over time, making it unstable.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Equilibrium Points
Equilibrium points in the context of differential equations are specific solutions where the system remains constant, or in other words, still over time. For the differential equation \(y^{\prime \prime} + p(y) = 0\), these points are crucial in understanding dynamics.

We identify equilibrium points by setting the acceleration (or the second derivative) to zero, which means that the effect of the function \(p(y)\) is nullified:
  • Set \(y^{\prime \prime} = 0\), which implies that \(-p(y) = 0\).
  • This leads to \(p(y) = 0\).
Equilibrium points provide essential insights into the behavior of the system defined by the differential equation as they define where the system can potentially stabilize.
Stability Analysis
Stability analysis is a method used to determine if an equilibrium point will attract or repel nearby trajectories over time. It's vital for predicting long-term system behavior.

Understanding whether a system will return to stability or diverge further from it relies on stability analysis:
  • Stable equilibrium points attract nearby trajectories, meaning perturbations fade over time.
  • Unstable equilibrium points cause nearby trajectories to drift away.
  • Semi-stable points might see some trajectories converge while others diverge.
By analyzing the surrounding environment of equilibrium points through differential calculus, specifically using the second derivative, one can conclude the type of stability displayed by the system.
Second-Order Differential Equations
Second-order differential equations, like \(y^{\prime \prime} + p(y) = 0\), involve second derivatives and are essential for modeling acceleration-based systems. These equations often define how a system evolves over time across various fields, such as physics and engineering.

The structure of a second-order differential equation includes:
  • \(y^{\prime \prime}\): The acceleration or the second derivative of the function.
  • \(p(y)\): A function of \(y\) that influences the rate of change.
While solving these equations, finding equilibrium points and analyzing stability is a common approach, as they help elucidate possible system behaviors and responses to perturbations.
Concavity
In the context of differential equations, concavity of a function provides information on how the graph of the function curves, which is integral for determining stability of equilibrium points. Concavity is directly linked to the sign of the second derivative \(y^{\prime \prime}\):

  • A positive second derivative \(y^{\prime \prime} > 0\) implies the function is concave up, resembling a U-shape.
  • A negative second derivative \(y^{\prime \prime} < 0\) indicates the function is concave down, like an inverted U.
Concavity helps predict how a solution behaves around equilibrium points. For stability analysis, these signs determine whether a solution at equilibrium will revert to the point or diverge, thus classifying the point as stable or unstable.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the mixing problem of Example 4.2.4 in a tank with infinite capacity, but without the assumption that the mixture is stirred instantly so that the salt is always uniformly distributed throughout the mixture. Assume instead that the distribution approaches uniformity as \(t \rightarrow \infty .\) In this case the differential equation for \(Q\) is of the form $$ Q^{\prime}+\frac{a(t)}{t+100} Q=1 $$ where \(\lim _{t \rightarrow \infty} a(t)=1\) (a) Let \(K(t)\) be the concentration of salt at time \(t\). Assuming that \(Q(0)=Q_{0},\) can you guess the value of \(\lim _{t \rightarrow \infty} K(t) ?\) (b) Use numerical methods to confirm your guess in the these cases: (i) \(a(t)=t /(1+t)\) $$ \text { (ii) } a(t)=1-e^{-t^{2}} $$ (iii) \(a(t)=1+\sin \left(e^{-t}\right)\).

Tank \(T_{1}\) initially contain 50 gallons of pure water. Starting at \(t_{0}=0,\) water that contains 1 pound of salt per gallon is poured into \(T_{1}\) at the rate of \(2 \mathrm{gal} / \mathrm{min} .\) The mixture is drained from \(T_{1}\) at the same rate into a second tank \(T_{2}\), which initially contains 50 gallons of pure water. Also starting at \(t_{0}=0,\) a mixture from another source that contains 2 pounds of salt per gallon is poured into \(T_{2}\) at the rate of 2 gal/min. The mixture is drained from \(T_{2}\) at the rate of 4 gal/min. (a) Find a differential equation for the quantity \(Q(t)\) of salt in \(\operatorname{tank} T_{2}\) at time \(t>0\). (b) Solve the equation derived in (a) to determine \(Q(t)\). (c) Find \(\lim _{t \rightarrow \infty} Q(t)\).

A homebuyer borrows \(P_{0}\) dollars at an annual interest rate \(r,\) agreeing to repay the loan with equal monthly payments of \(M\) dollars per month over \(N\) years. (a) Derive a differential equation for the loan principal (amount that the homebuyer owes) \(P(t)\) at time \(t>0\), making the simplifying assumption that the homebuyer repays the loan continuously rather than in discrete steps. (See Example \(4.1 .6 .)\) (b) Solve the equation derived in (a). (c) Use the result of (b) to determine an approximate value for \(M\) assuming that each year has exactly 12 months of equal length. (d) It can be shown that the exact value of \(M\) is given by $$ M=\frac{r P_{0}}{12}\left(1-(1+r / 12)^{-12 N}\right)^{-1} $$ Compare the value of \(M\) obtained from the answer in (c) to the exact value if (i) \(P_{0}=\) (ii) $$P_{0}=\$ 150,000, r=9.0 \%, N=30$$ $$\$ 50,000, r=7 \frac{1}{2} \%, N=20$$

A bank pays interest continuously at the rate of \(6 \% .\) How long does it take for a deposit of \(Q_{0}\) to grow in value to \(2 Q_{0}\) ?

A tank initially contains 100 liters of a salt solution with a concentration of \(.1 \mathrm{~g} /\) liter. A solution with a salt concentration of \(.3 \mathrm{~g} /\) liter is added to the tank at 5 liters/min, and the resulting mixture is drained out at the same rate. Find the concentration \(K(t)\) of salt in the tank as a function of \(t\).
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.