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Next: Computer simulations Up: No Title Previous: Introduction

Steady Progressing Wave Solutions

We consider here only the case where $s\in E^1$ and $\theta\in E^m$ where $m=1 \mbox{ or } 2$to illustrate several interesting features of such systems. The general case can be treated in similar ways.

The observables of this system are usually not $\theta$, but some periodic wave form or function of $\theta$ such as $\cos\theta$ or $\cos_+\theta$. This suggests that we seek a steady progressing wave solution in the form

\begin{displaymath}\theta=\alpha s + c t

where $\alpha$ is called the wave number and c is a vector of wave speeds.

Direct substitution of the steady progressing wave solution into the equation gives

\int_{-L}^L { {K}}(s-s')

There is a unique constant solution of this equation for cprovided the components of $\alpha$ have the form $k\pi/L$ for some integer $k=0,\pm 1,\pm 2,\cdots$.

Let us fix $\alpha = N\pi/L$for some integer N. According to the calculation above, there is a steady progressing wave solution of the system having the form

\begin{displaymath}\theta=N\pi s/L + c_N t

where cN is given by the formula

\int_{-L}^L K(s-s')

To test the stability of this steady progressing wave we define

\begin{displaymath}x_n(s,t)\delta =\theta(s,t)-N\pi s/L-c_Nt

The result is

\begin{displaymath}\delta \dot x(s,t)=\int_{-L}^LK(s-s')

Expanding the right hand side in powers of $\delta$ and ignoring higher order terms gives

\begin{displaymath}\dot x(s,t)=\int_{-L}^LK(s-s')

Expanding x in its Fourier series

\begin{displaymath}x(s,t)=\sum_{n=-\infty}^{\infty} x_n(t)e^{in\pi s/L}

and substituting this into the equation gives

\begin{displaymath}\dot x_n(t)=(A_0-A_n)x_n


\begin{displaymath}K(s)f'(N\pi s/L+\psi(s))=
\sum_{-\infty}^{\infty} A_ne^{in\pi s/L}\equiv A(s)

Thus, depending on the properties of the coefficients in A, we get the following stability results for the deviation x: If $K(s)=f(s)=\cos s$ and $\psi(s)\equiv 0$, then the leading components of x are harmonic and the remaining modes are damped. If $\psi\equiv\pi/2$, then all modes (except for n=0) are damped.

next up previous
Next: Computer simulations Up: No Title Previous: Introduction
Hans Mittelmann