Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

For Duffing's equation given in Problem 13, the behaviour of the solutions changes as r changes sign. When $r > 0$ , the restoring force $k y + r y^{3}$ becomes stronger than for the linear spring $(r = 0)$ . Such a spring is called hard. When $r < 0$ , the restoring force becomes weaker than the linear spring and the spring is called soft. Pendulums act like soft springs.

(a) Redo Problem 13 with $r = - 1$ . Notice that for the initial conditions , $y (0) = 0, y^{'} (0) = 1$ the soft and hard springs appear to respond in the same way for small.

(b) Keeping $k = A = 1$ and , $ω = 10$ change the initial conditions to $y (0) = 1$ and $y^{'} (0) = 0$ . Now redo Problem 13 with $r = \pm 1$ .

(c) Based on the results of part (b), is there a difference between the behavior of soft and hard springs for small? Describe.

(a) The required polynomial is, $p_{3} (t) = t + \frac{t^{2}}{2} - \frac{t^{3}}{6}$

(b) For $r = 1, p_{3} (t) = 1 - \frac{t^{2}}{2} - \frac{33 t^{4}}{8}$ and for $r = - 1, p_{3} (t) = 1 + \frac{t^{2}}{2} - \frac{101 t^{3}}{24}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1:To Find the Taylor polynomial of degree

The formula for the Taylor polynomial of degree n centered at $x_{0}$ , approximating a function $f (x)$ possessing n derivatives at , $x_{0}$ is given by

$p_{n} (x) = f (x_{0}) + f^{'} (x_{0}) \times (x - x_{0}) + f^{''} (x_{0}) \times \frac{{(x - x_{0})}^{2}}{2!} + \dots + f^{n} (x_{0}) \times \frac{{(x - x_{0})}^{n}}{n!}$

The differential equation is given as

$y^{''} + k y + r y^{3} = A \cos ω t$

By substituting given values $k = A = 1, r = - 1$ and $ω = 10$ , differential equation becomes

$y^{''} + y - y^{3} = \cos 10 t$

It is given that for the function $y (x)$ ,

$y (0) = 0 and y^{'} (0) = 1$

The Taylor's polynomial cantered around $x_{0} = 0$ is given by

$p_{n} (x) = y (0) + y^{'} (0) \times (x - 0) + y^{''} (0) \times \frac{{(x - 0)}^{2}}{2!} + \dots + y^{n} (0) \times \frac{{(x - 0)}^{n}}{n!}$

We need the value of $y (0), y^{'} (0), y^{''} (0)$ and $y^{'''} (0)$ etc for finding the behaviour of spring. The first two are provided by the initial conditions.

The value of $y^{''} (0)$ can be deduced from the differential equation itself and the values of the lower derivatives.

$\begin{array}{l} y^{''} (0) = \cos 10 t - y (0) + y^{3} (0) \\ y^{''} (0) = 1 \end{array}$

Now since $y^{''} = \cos 10 t - y + y^{3}$ holds for some interval around $x_{0} = 0$ , we can differentiate both sides to derive,

$\begin{matrix} y^{'''} = - 10 \sin 10 t - y^{'} + 3 y^{2} y^{'} \\ y^{'''} (0) = - 10 \sin 10 t - y^{'} (0) (1 - 3 y^{2} (0)) \\ = - 1 \end{matrix}$

The Taylor polynomial for the soft spring when $r < 0$ in the form of equation is given by. $p_{3} (t) = t + \frac{t^{2}}{2} - \frac{t^{3}}{6} \dots \dots$

Step 2:To Find the oscillatory equation

$x_{0} = 0$ First, we will find the oscillatory equation in case where $r > 0$ ,the restoring force is stronger and the spring is hard. The formula for the Taylor polynomial of degree centered at $x_{0}$ , approximating a function $f (x)$ possessing n derivatives at , $x_{0}$ is given by,

$p_{n} (x) = f (x_{0}) + f^{'} (x_{0}) \times (x - x_{0}) + f^{''} (x_{0}) \times \frac{{(x - x_{0})}^{2}}{2!} + \dots + f^{n} (x_{0}) \times \frac{{(x - x_{0})}^{n}}{n!}$

The differential equation is given as

$y^{''} + k y + r y^{3} = A \cos ω t$

By substituting given values $k = A = 1, r = 1$ and $ω = 10$ , differential equation becomes

$y^{''} + y + y^{3} = \cos 10 t$

It is given that for the function $y (x)$ ,

$y (0) = 1 and y^{'} (0) = 0$

The Taylor's polynomial cantered around $x_{0} = 0$ is given by

$p_{n} (x) = y (0) + y^{'} (0) \times (x - 0) + y^{''} (0) \times \frac{{(x - 0)}^{2}}{2!} + \dots + y^{n} (0) \times \frac{{(x - 0)}^{n}}{n!}$

We need the value of $y (0), y^{'} (0), y^{''} (0)$ and $y^{'''} (0)$ etc for finding the behaviour of spring. The first two are provided by the initial conditions.

The value of $y^{''} (0)$ can be deduced from the differential equation itself and the values of the lower derivatives.

$\begin{array}{l} y^{''} (0) = \cos 10 t - y (0) - y^{3} (0) \\ y^{''} (0) = - 1 \end{array}$

Now since $y^{''} = \cos 10 t - y - y^{3}$ holds for some interval around $x_{0} = 0$ , we can differentiate both sides to derive,

$\begin{matrix} y^{'''} = - 10 \sin 10 t - y^{'} - 3 y^{2} y^{'} \\ y^{'''} (0) = - 10 \sin 10 t - y^{'} (0) (1 + 3 y^{2} (0)) \\ = 0 \end{matrix}$

Similarly,

$\begin{matrix} y^{4} = - 100 \cos 10 t - y^{''} - 6 y {(y^{'})}^{2} \\ y^{4} (0) = - 100 \cos 10 t - y^{''} (0) - 6 y (0) {(y^{'} (0))}^{2} \\ y^{4} (0) = - 99 \end{matrix}$

The Taylor polynomial for the hard spring when $r > 0$ in the form of equation is given by,

$p_{3} (t) = 1 - \frac{t^{2}}{2} - \frac{33 t^{4}}{8} \dots \dots$

Now, we will find the oscillatory equation in case where ,the restoring force is weaker and the spring is soft. The formula for the Taylor polynomial of degree n

centered at $x_{0}$ , approximating a function $f (x)$ possessing n derivatives at , $x_{0}$ is given by

$p_{n} (x) = f (x_{0}) + f^{'} (x_{0}) \times (x - x_{0}) + f^{''} (x_{0}) \times \frac{{(x - x_{0})}^{2}}{2!} + \dots + f^{n} (x_{0}) \times \frac{{(x - x_{0})}^{n}}{n!}$

The differential equation is given as

$y^{''} + k y + r y^{3} = A \cos ω t$

By substituting given values $k = A = 1, r = - 1$ and $ω = 10$ , differential equation becomes

$y^{''} + y - y^{3} = \cos 10 t$

It is given that for the function , $y (x)$ . $y (0) = 1 and y^{'} (0) = 0$

The Taylor's polynomial centered around $x_{0} = 0$ is given by

$p_{n} (x) = y (0) + y^{'} (0) \times (x - 0) + y^{''} (0) \times \frac{{(x - 0)}^{2}}{2!} + \dots + y^{n} (0) \times \frac{{(x - 0)}^{n}}{n!}$

We need the value of $y (0), y^{'} (0), y^{''} (0)$ and $y^{'''} (0)$ etc. for finding the behaviour of spring. The first two are provided by the initial conditions.

The value of $y^{''} (0)$ can be deduced from the differential equation itself and the values of the lower derivatives.

$\begin{array}{l} y^{''} (0) = \cos 10 t - y (0) + y^{3} (0) \\ y^{''} (0) = 1 \end{array}$

Now since $y^{''} = \cos 10 t - y + y^{3}$ holds for some interval around $x_{0} = 0$ , we can differentiate both sides to derive,

$\begin{matrix} y^{'''} = - 10 \sin 10 t - y^{'} + 3 y^{2} (y^{'}) \\ y^{'''} (0) = - 10 \sin 10 t - y^{'} (0) (1 - 3 y^{2} (0)) \\ = 0 \end{matrix}$

Similarly,

$\begin{matrix} y^{4} = - 100 \cos 10 t - y^{''} + 6 y {(y^{'})}^{2} \\ y^{4} (0) = - 100 \cos 10 t - y^{''} (0) + 6 y (0) (y^{'} {(0)}^{2}) \\ = - 101 \end{matrix}$

The Taylor polynomial for the hard spring when $r > 0$ in the form of equation is given by

$p_{3} (t) = 1 + \frac{t^{2}}{2} - \frac{101 t^{4}}{24} \dots \dots$

Step 3:The initial condition

From problem 13 and $14 (a)$ , the solution describing the behaviour of both soft and hard spring, for the initial conditions $y (0) = 0$ and $y^{'} (0) = 1$ , is approximated by the polynomial

$p_{3} (t) = t + \frac{t^{2}}{2} - \frac{t^{3}}{6} + \dots$

For small , the higher order terms will be negligible and the polynomial can be written as

$p_{3} (t) \approx t$

Similarly, from problem $14 (b)$ , the solution describing the behaviour of soft and hard spring, for the initial conditions $y (0) = 1$ and $y^{'} (0) = 0$ , are approximated by the polynomial,

$\begin{array}{l} p_{3} (t) = 1 - \frac{t^{2}}{2} - \frac{33 t^{4}}{8} + \dots & [ForHardSpring] \\ p_{3} (t) = 1 + \frac{t^{2}}{2} - \frac{101 t^{4}}{24} + \dots & [ForSoftSpring] \end{array}$

For small t , the higher order terms will be negligible and both the polynomial can be written as

$p_{3} (t) \approx 1$

We can see from the above analysis that, in both the cases the soft and hard spring show same behaviour for small , the difference however is that for the initial conditions $y (0) = 0$ and $y^{'} (0) = 1$ , the springs behave in direct proportion to the small , whereas for the initial conditions $y (0) =$ 1 and , $y^{'} (0) = 0$ the springs behave as constant with value 1 , to the small t

Step 4:Final proof

(a) the required Taylor’s polynomial is, $p_{3} (t) = t + \frac{t^{2}}{2} - \frac{t^{3}}{6}$

(b) For $r = 1, p_{3} (t) = 1 - \frac{t^{2}}{2} - \frac{33 t^{4}}{8}$ and for $r = - 1, p_{3} (t) = 1 + \frac{t^{2}}{2} - \frac{101 t^{3}}{24}$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Step by Step Solution

Step 1:To Find the Taylor polynomial of degree

Step 2:To Find the oscillatory equation

Step 3:The initial condition

Step 4:Final proof

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Step by Step Solution

Step 1:To Find the Taylor polynomial of degree

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Step 3:The initial condition

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