Chapter 7.6, Problem 13E

Interpretation Introduction

Interpretation:

Consider the Duffing oscillator, withsystem equation ${\ddot{\text{x}}\text{ +x +εx}}^3=0$ where $\text{0 < ε << 1, x(0) = a and }\dot{\text{x}}\text{(0) = 0}$. Use the conservation of energy express oscillation period $\text{T}\left(\text{ε}\right)$ as certain integral. Expand the integral as power series in, and integrate term by term to obtain formula $\text{T}\left(\text{ε}\right)={\text{c}}_{\text{0}}{\text{+ c}}_{\text{1}}{\text{ε + c}}_{\text{2}}{\text{ε}}^{\text{2}}+{\rm O}\left({\text{ε}}^{\text{3}}\right)$, determine ${\text{c}}_{\text{0}}{\text{, c}}_{\text{1}}{\text{, c}}_{\text{2}}$ and check that ${\text{c}}_{\text{0}}{\text{, c}}_{\text{1}}$ are consistent with $\text{7}\text{.6}\text{.57}$.

Concept Introduction:

• Use energy conservation of system to determine time period of oscillation.

• Determine the coefficients.

• Use power series for the expansion and compare the time period equation of system and for $\text{7}\text{.6}\text{.57}$.

Answer to Problem 13E

Solution:

1. The oscillation period $\text{T}\left(\text{ε}\right)$ as certain integral is

   $\text{T = }\frac{\text{1}}{\sqrt{\left(\text{1 + ε}\frac{{\text{a}}^{\text{2}}}{\text{2}}\right)}}\text{K}\left(\text{k}\right)$

2. The time period of the given system equation and of $\text{7}\text{.6}\text{.57}$ is same for at least first two terms. ${\text{c}}_{\text{0}}\text{= 2π}$, ${\text{c}}_{\text{1}}\text{= - }\frac{{\text{3πa}}^{\text{2}}}{4}$ and ${\text{c}}_{\text{2}}=\frac{{\text{57a}}^{\text{4}}\text{π}}{\text{128}}$ and ${\text{c}}_{\text{0}}{\text{, c}}_{\text{1}}$ are consistent with $\text{7}\text{.6}\text{.57}$.

Explanation of Solution

Conservation of energy: This theorem states that the energy of an isolated system is constant over certain time.

The non-linear system equation is,

$\ddot{\text{x}}\text{ + x + εh}\left(\text{x, }\dot{\text{x}}\right)\text{ = 0}$

Here, x is the position.

The expression of $\ddot{\text{x}}$ and $\dot{\text{x}}$ is,

$\ddot{\text{x}}\text{ = }\frac{{\text{d}}^{\text{2}}\text{x}}{{\text{dt}}^{\text{2}}}$

$\dot{\text{x}}\text{ = }\frac{\text{dx}}{\text{dt}}$

The conserved energy expression for the non-linear system is,

$\text{E = }\frac{\text{1}}{\text{2}}{\dot{\text{x}}}^{\text{2}}\text{+ }{\displaystyle \int \left(\text{x + εh}\left(\text{x, }\dot{\text{x}}\right)\right)}\text{dx}$

Here E is the energy.

The expression of the time period for the system is,

$\text{T = }{\displaystyle \underset{\text{0}}{\overset{\text{T}}{\int }}\text{dt}}$

Here T is the time period of the system.

The expanded Taylor series is,

$\text{y}\left({\text{x}}_{\text{0}}\text{+ h}\right)\text{ = y}\left({\text{x}}_{\text{0}}\right)\text{ + hy'}\left({\text{x}}_{\text{0}}\right)\text{ + }\frac{{\text{h}}^{\text{2}}}{\text{2!}}{\text{y}}^{\text{"}}\left({\text{x}}_{\text{0}}\right)\text{ + }\mathrm{...}$

Here h is the constant parameter.

1. The non-linear system equation as per given is,

   ${\ddot{\text{x}}\text{ + x + εx}}^{\text{3}}\text{ = 0 , x}\left(\text{0}\right)\text{ = a, }\dot{\text{x}}\left(\text{0}\right)\text{ = 0}$

   By comparing with the nonlinear system equation,

   $\ddot{\text{x}}+\text{x}+\text{εh}\left(\text{x},\dot{\text{ x }}\right)=0$

   $\text{h}\left(\text{x, }\dot{\text{x}}\right)={\text{x}}^{\text{3}}$

   From the equation of energy, the system energy is,

   $\text{E = }\frac{\text{1}}{\text{2}}{\dot{\text{x}}}^{\text{2}}\text{+ }{\displaystyle \int \left(\text{x + εh}\left(\text{x, }\dot{\text{x}}\right)\right)}\text{dx}$

   $\text{E = }\frac{\text{1}}{\text{2}}{\dot{\text{x}}}^{\text{2}}\text{+ }{\displaystyle \int \left({\text{x + εx}}^{\text{3}}\right)}\text{dx}$

   $\text{E = }\frac{\text{1}}{\text{2}}{\dot{\text{x}}}^{\text{2}}\text{+ }\frac{{\text{x}}^{\text{2}}}{\text{2}}\text{ + ε}\frac{{\text{x}}^{\text{4}}}{\text{4}}$

   From the given initial position of the system $\text{t = 0}$, the maximum amplitude of the oscillation of the system is $\text{a}$.

   Substitute initial condition $\text{x}\left(\text{0}\right)=\text{a}$ and $\dot{\text{x}}\left(\text{0}\right)=0$ in the energy expression,

   $\text{E = }\frac{\text{1}}{\text{2}}{\left(\dot{\text{x}}\left(\text{0}\right)\right)}^{\text{2}}\text{+ }\frac{{\left(\text{x}\left(\text{0}\right)\right)}^{\text{2}}}{\text{2}}\text{+ ε}\frac{{\left(\text{x}\left(0\right)\right)}^{\text{4}}}{\text{4}}$

   $\text{E = 0 + }\frac{{\text{a}}^{\text{2}}}{\text{2}}\text{ + ε}\frac{{\text{a}}^{\text{4}}}{\text{4}}$

   $\text{E = }\frac{{\text{a}}^{\text{2}}}{\text{2}}\text{ + ε}\frac{{\text{a}}^{\text{4}}}{\text{4}}$

   Comparing the above initial energy condition equation the energy expression related with time is,

   $\frac{\text{1}}{\text{2}}{\dot{\text{x}}}^{\text{2}}\text{ + }\frac{{\text{x}}^{\text{2}}}{\text{2}}\text{ + ε}\frac{{\text{x}}^{\text{4}}}{\text{4}}\text{ = }\frac{{\text{a}}^{\text{2}}}{\text{2}}\text{ + ε}\frac{{\text{a}}^{\text{4}}}{\text{4}}$

   ${\dot{\text{x}}}^{\text{2}}{\text{ + x}}^{\text{2}}\text{ + ε}\frac{{\text{x}}^{\text{4}}}{\text{2}}{\text{ = a}}^{\text{2}}\text{ + ε}\frac{{\text{a}}^{\text{4}}}{\text{2}}$

   ${\dot{\text{x}}}^{\text{2}}{\text{ = a}}^{\text{2}}\text{ + ε}\frac{{\text{a}}^{\text{4}}}{\text{2}}{\text{ - x}}^{\text{2}}\text{ - ε}\frac{{\text{x}}^{\text{4}}}{\text{2}}$

   Square root of both sides,

   $\dot{\text{x}}\text{ = }\sqrt{{\text{a}}^{\text{2}}\text{ + ε}\frac{{\text{a}}^{\text{4}}}{\text{2}}{\text{ - x}}^{\text{2}}\text{ - ε}\frac{{\text{x}}^{\text{4}}}{\text{2}}}$

   The expression of $\dot{\text{x}}$ is,

   $\dot{\text{x}}\text{ = }\frac{\text{dx}}{\text{dt}}$

   Rearrange the above expression in terms of dt is,

   $\text{dt = }\frac{\text{dx}}{\dot{\text{x}}}$

   On integration,

   ${\displaystyle \int \text{dt}}\text{ = }{\displaystyle \int \frac{\text{dx}}{\dot{\text{x}}}}$

   The expression of the time period is,

   $\text{T = }{\displaystyle \underset{\text{0}}{\overset{\text{T}}{\int }}\text{dt}}$

   For the quarter cycle, the value of x decrease from $\text{a}$ to $\text{0}$. Hence the period of the one cycle is $\text{T = 4}{\displaystyle \underset{\text{0}}{\overset{\frac{\text{T}}{\text{4}}}{\int }}\text{dt}}$.

   Substitute $\text{dt}=\frac{\text{dx}}{\dot{\text{x}}}$

   $\text{T = 4}{\displaystyle \underset{\text{0}}{\overset{\frac{\text{T}}{\text{4}}}{\int }}\frac{\text{dx}}{\dot{\text{x}}}}$

   Substitute $\text{0}$ for the lower limit of the integration and a for the upper limit of the integration

   $\text{T = 4}{\displaystyle \underset{\text{0}}{\overset{\text{a}}{\int }}\frac{\text{dx}}{\dot{\text{x}}}}$

   Substitute $\dot{\text{x}}\text{ = }\sqrt{{\text{a}}^{\text{2}}\text{ + ε}\frac{{\text{a}}^{\text{4}}}{\text{2}}{\text{ - x}}^{\text{2}}\text{ - ε}\frac{{\text{x}}^{\text{4}}}{\text{2}}}$ in the above expression of the time period

   $\text{T = 4}{\displaystyle \underset{\text{0}}{\overset{\text{a}}{\int }}\frac{\text{dx}}{\sqrt{{\text{a}}^{\text{2}}\text{ + ε}\frac{{\text{a}}^{\text{4}}}{\text{2}}{\text{ - x}}^{\text{2}}\text{ - ε}\frac{{\text{x}}^{\text{4}}}{\text{2}}}}}$

   For easy calculation of the time period consider a function $\text{x = ay}$, $\text{dx = a dy}$.

   For the calculation of the integration limit cycle,

   by changing the limit,

   The lower limit, $\text{x= 0 , 0 = ay, hence y = 0}$

   Similarly, The higher limit, $\text{x= a , a = ay, hence y = 1}$.

   By substituting all the values in time period equation,

   $\text{T = 4}{\displaystyle \underset{\text{0}}{\overset{1}{\int }}\frac{\text{ady}}{\sqrt{{\text{a}}^{\text{2}}\text{ + ε}\frac{{\text{a}}^{\text{4}}}{\text{2}}\text{ - }{\left(\text{ay}\right)}^{\text{2}}\text{ - ε}\frac{{\left(\text{ay}\right)}^{\text{4}}}{\text{2}}}}}$

   $\text{T = 4}{\displaystyle \underset{\text{0}}{\overset{\text{1}}{\int }}\frac{\text{ady}}{\sqrt{{\text{a}}^{\text{2}}\text{ + ε}\frac{{\text{a}}^{\text{4}}}{\text{2}}{\text{ - a}}^{\text{2}}{\text{y}}^{\text{2}}\text{ - ε}\frac{{\text{a}}^{\text{4}}{\text{y}}^{\text{4}}}{\text{2}}}}}$

   $\text{T = 4}{\displaystyle \underset{\text{0}}{\overset{\text{1}}{\int }}\frac{\text{ady}}{\sqrt{{\text{a}}^{\text{2}}\left({\text{1 - y}}^{\text{2}}\right)\text{+ ε}\frac{{\text{a}}^{\text{4}}}{\text{2}}\left({\text{1 - y}}^{\text{4}}\right)}}}$

   $\text{T = 4}{\displaystyle \underset{\text{0}}{\overset{\text{1}}{\int }}\frac{\text{ady}}{\text{ a}\sqrt{\left({\text{1 - y}}^{\text{2}}\right)\text{+ ε}\frac{{\text{a}}^{\text{2}}}{\text{2}}\left({\text{1 - y}}^{\text{4}}\right)}}}$

   From trigonometric identity,

   $\left({\text{a}}^{\text{2}}{\text{ - b}}^{\text{2}}\right)\text{ = }\left(\text{a + b}\right)\left(\text{a - b}\right)$

   $\text{T = 4}{\displaystyle \underset{\text{0}}{\overset{\text{1}}{\int }}\frac{\text{dy}}{\sqrt{\left({\text{1 - y}}^{\text{2}}\right)\text{+ ε}\frac{{\text{a}}^{\text{2}}}{\text{2}}\left({\text{1 - y}}^{\text{2}}\right)\left({\text{1 + y}}^{\text{2}}\right)}}}$

   $\text{T = 4}{\displaystyle \underset{\text{0}}{\overset{\text{1}}{\int }}\frac{\text{dy}}{\sqrt{\left({\text{1 - y}}^{\text{2}}\right)\left(\text{1 + ε}\frac{{\text{a}}^{\text{2}}}{\text{2}}\left({\text{1 + y}}^{\text{2}}\right)\right)}}}$

   $\text{T = 4}{\displaystyle \underset{\text{0}}{\overset{\text{1}}{\int }}\frac{\text{dy}}{\sqrt{\left({\text{1 - y}}^{\text{2}}\right)\left(\text{1 + ε}\frac{{\text{a}}^{\text{2}}}{\text{2}}\right)\left(\text{1 + }\frac{\text{ε}\frac{{\text{a}}^{\text{2}}{\text{y}}^{\text{2}}}{2}}{\left(\text{1 + ε}\frac{{\text{a}}^{\text{2}}}{\text{2}}\right)}\right)}}}$

   Further solve,

   $\text{T = }\frac{\text{1}}{\sqrt{\left(\text{1 + ε}\frac{{\text{a}}^{\text{2}}}{\text{2}}\right)}}\text{4}{\displaystyle \underset{\text{0}}{\overset{\text{1}}{\int }}\frac{\text{dy}}{\sqrt{\left({\text{1 - y}}^{\text{2}}\right)\left(\text{1 + }\frac{\text{ε}\frac{{\text{a}}^{\text{2}}{\text{y}}^{\text{2}}}{\text{2}}}{\left(\text{1 + ε}\frac{{\text{a}}^{\text{2}}}{\text{2}}\right)}\right)}}}$

   Consider a constant term ${\text{k}}^{\text{2}}\text{ = }\frac{\text{- ε}\frac{{\text{a}}^{\text{2}}}{\text{2}}}{\left(\text{1 + ε}\frac{{\text{a}}^{\text{2}}}{\text{2}}\right)}$

   Hence the time period is

   $\text{T = }\frac{\text{1}}{\sqrt{\left(\text{1 + ε}\frac{{\text{a}}^{\text{2}}}{\text{2}}\right)}}\text{4}{\displaystyle \underset{\text{0}}{\overset{\text{1}}{\int }}\frac{\text{dy}}{\sqrt{\left({\text{1 - y}}^{\text{2}}\right)\left({\text{1 - k}}^{\text{2}}{\text{y}}^{\text{2}}\right)}}}$

   Consider a function K(k) This is the complete elliptic integral of the first kind.

   Therefore the time constant of the given system is, $\text{T = }\frac{\text{1}}{\sqrt{\left(\text{1 + ε}\frac{{\text{a}}^{\text{2}}}{\text{2}}\right)}}\text{K}\left(\text{k}\right)$.

2. The power series of the time period function is calculated as,

   The expression of the time period of the system is,

   $\text{T = 4}{\displaystyle \underset{\text{0}}{\overset{\text{1}}{\int }}\frac{\text{ady}}{\sqrt{{\text{a}}^{\text{2}}\text{ + ε}\frac{{\text{a}}^{\text{4}}}{\text{2}}{\text{ - a}}^{\text{2}}{\text{y}}^{\text{2}}\text{ - ε}\frac{{\text{a}}^{\text{4}}{\text{y}}^{\text{4}}}{\text{2}}}}}$

   $\text{T }\text{}\text{= 4}{\displaystyle \underset{\text{0}}{\overset{\text{1}}{\int }}\frac{\text{ady}}{\text{ a}\sqrt{\text{1 + ε}\frac{{\text{a}}^{\text{2}}}{\text{2}}{\text{ - y}}^{\text{2}}\text{ - ε}\frac{{\text{a}}^{\text{2}}{\text{y}}^{\text{4}}}{\text{2}}}}}$

   $\text{T = 4}{\displaystyle \underset{\text{0}}{\overset{\text{1}}{\int }}\frac{\text{dy}}{\sqrt{\text{1 + ε}\frac{{\text{a}}^{\text{2}}}{\text{2}}{\text{ - y}}^{\text{2}}\text{ - ε}\frac{{\text{a}}^{\text{2}}{\text{y}}^{\text{4}}}{\text{2}}}}}$

   $\text{T = 4}{\displaystyle \underset{\text{0}}{\overset{\text{1}}{\int }}\frac{\text{dy}}{\sqrt{{\text{1 - y}}^{\text{2}}\text{ + ε}\left(\frac{{\text{a}}^{\text{2}}}{\text{2}}\text{ - }\frac{{\text{a}}^{\text{2}}{\text{y}}^{\text{4}}}{\text{2}}\right)}}}$

   Further solve,

   $\text{T = 4}{\displaystyle \underset{\text{0}}{\overset{\text{1}}{\int }}\frac{\text{dy}}{\sqrt{{\text{1 - y}}^{\text{2}}\text{ + ε}\left(\frac{{\text{a}}^{\text{2}}}{\text{2}}\text{ - }\frac{{\text{a}}^{\text{2}}{\text{y}}^{\text{4}}}{\text{2}}\right)}}}$

   $\text{= 4}{\displaystyle \underset{\text{0}}{\overset{\text{1}}{\int }}\frac{\text{dy}}{\sqrt{{\text{1 - y}}^{\text{2}}\text{ + ε}\left(\frac{{\text{a}}^{\text{2}}}{\text{2}}\text{ - }\frac{{\text{a}}^{\text{2}}{\text{y}}^{\text{4}}}{\text{2}}\right)}}}$

   Apply Taylor series in the above expression of the time period

   $\text{T = 4}{\displaystyle \underset{\text{0}}{\overset{\text{1}}{\int }}\frac{\text{dy}}{\sqrt{{\text{1 - y}}^{\text{2}}\text{ + ε}\left(\frac{{\text{a}}^{\text{2}}}{\text{2}}\text{ - }\frac{{\text{a}}^{\text{2}}{\text{y}}^{\text{4}}}{\text{2}}\right)}}}$

   $\text{T}\text{}\text{ = }{\displaystyle \underset{\text{0}}{\overset{\text{1}}{\int }}\left(\frac{\text{4}}{\sqrt{{\text{1-y}}^{\text{2}}}}\text{ - ε}\frac{\left(\frac{{\text{a}}^{\text{2}}}{\text{2}}\text{ - }\frac{{\text{a}}^{\text{2}}{\text{y}}^{\text{4}}}{\text{2}}\right)}{\text{2}{\left({\text{1 - y}}^{\text{2}}\right)}^{\frac{\text{3}}{\text{2}}}}{\text{ + ε}}^{\text{2}}\frac{{\text{3a}}^{\text{4}}{\left(\frac{{\text{a}}^{\text{2}}}{\text{2}}\text{ - }\frac{{\text{a}}^{\text{2}}{\text{y}}^{\text{4}}}{\text{2}}\right)}^{\text{2}}}{\text{8}{\left({\text{1 - y}}^{\text{2}}\right)}^{\frac{\text{5}}{\text{2}}}}\text{ + O}\left({\text{ε}}^{\text{3}}\right)\right)}\text{dy}$

   $\text{T = 2π - }\frac{{\text{3πa}}^{\text{2}}}{4}\text{ε + }\frac{{\text{57a}}^{\text{4}}\text{π}}{\text{128}}{\text{ε}}^{\text{2}}\text{ + O}\left({\text{ε}}^{\text{3}}\right)$

   By comparing the $\text{T}\left(\text{ε}\right)={\text{c}}_{\text{0}}{\text{+ c}}_{\text{1}}{\text{ε + c}}_{\text{2}}{\text{ε}}^{\text{2}}+{\rm O}\left({\text{ε}}^{\text{3}}\right)$ to obtain ${\text{c}}_{\text{0}}{\text{, c}}_{\text{1}}{\text{, c}}_{\text{2}}$

   $\text{T = 2π - }\frac{{\text{3πa}}^{\text{2}}}{4}\text{ε + }\frac{{\text{57a}}^{\text{4}}\text{π}}{\text{128}}{\text{ε}}^{\text{2}}\text{ + O}\left({\text{ε}}^{\text{3}}\right)$

   Hence ${\text{c}}_{\text{0}}\text{= 2π}$, ${\text{c}}_{\text{1}}\text{= - }\frac{{\text{3πa}}^{\text{2}}}{4}$ and ${\text{c}}_{\text{2}}=\frac{{\text{57a}}^{\text{4}}\text{π}}{\text{128}}$

   $\text{T}\text{}\text{ = 2π }\left(\text{1 - }\frac{{\text{3a}}^{\text{2}}}{8}\text{ε + }\frac{{\text{57a}}^{\text{4}}}{\text{256}}{\text{ε}}^{\text{2}}\text{ + O}\left({\text{ε}}^{\text{3}}\right)\right)$

   The expression of the frequency in equation of $\text{7}\text{.6}\text{.57}$ is as mention in the book is,

   $\text{ω = 1 + }\frac{\text{3}}{\text{8}}{\text{εa}}^{\text{2}}\text{ + O}\left({\text{ε}}^{\text{2}}\right)$

   The time period of the equation $\text{7}\text{.6}\text{.57}$ is calculated as,

   The expression of the time period in terms of the angular frequency is,

   $\text{T = }\frac{\text{2π}}{\text{ω}}$

   Substituting $\text{ω}$ in the above expression,

   $\text{T = }\frac{\text{2π}}{\left(\text{1 + }\frac{\text{3}}{\text{8}}{\text{εa}}^{\text{2}}\text{ + O}\left({\text{ε}}^{\text{2}}\right)\right)}$

   $\text{T =2π}{\left(\text{1 + }\frac{\text{3}}{\text{8}}{\text{εa}}^{\text{2}}+{\rm O}\left({\text{ε}}^2\right)\right)}^{-1}$

   Apply Taylor series expansion in the above equation of the time period,

   $\text{T = 2π}\left(\text{1 - }\frac{\text{3}}{\text{8}}{\text{εa}}^{\text{2}}\text{ + O}\left({\text{ε}}^{\text{2}}\right)\right)$

   So two methods have agreement in at least the first two terms.

   Therefore the time period of the above system expression and time period for the equation $\text{7}\text{.6}\text{.57}$ is same as at least two terms in the system equation.