Chapter 7.4, Problem 1E

Interpretation Introduction

Interpretation:

To show equation ${\ddot{\text{x}}\text{ + μ(x}}^{\text{2}}\text{-1)}\dot{\text{x}}\text{ + tanhx =0}$ has exactly one solution for $\mu >0$ and classify its stability.

Concept Introduction:

Equation of the form $\ddot{\text{x}}\text{ +f(x)}\dot{\text{x}}\text{+g(x)=0}$ is known as Lienard’s equation. This equation is equivalent to the system

$\dot{\text{x}}\text{=y}$

$\dot{\text{y}}\text{ = -g(x) - f(x)y}$

Lienard’s Theorem: If $\text{f(x)}$ and $\text{g(x)}$ satisfy the following conditions

$\text{f(x)}$ and $\text{g(x)}$ are continuously differentiable for all x;

$\text{g(- x)= - g(x)}$ for all x; i.e., $\text{g(x)}$ is an odd function;

$\text{g(x)>0 for x>0;}$

$\text{f(-x)= f(x)}$ for all x; i.e., $\text{f(x)}$ is an even function;

The odd function $\text{F(x)=}{\displaystyle {\int }_{\text{0}}^{\text{x}}\text{f(u)du}}$ has exactly one positive zero at $\text{x=0}$, is negative for $\text{0 < x < a}$, is positive and nondecreasing for $\text{x >a}$, and $\text{F(x)}\to \infty \text{ as x}\to \infty$

Then the system

$\dot{\text{x}}\text{=y}$

$\dot{\text{y}}\text{ = -g(x) - f(x)y}$ has a unique, stable limit cycle surrounding the origin in the phase plane.