Breathers as metastable states for the discrete NLS equation

Figure 1.  Numerical illustration of the dynamics with (weak) dissipation. The parameters are $N = 4$, $\gamma = 0.2$, and $\epsilon = 0.01$. Shown is the ''energy'' of the degree of freedom $j$, $j = 1,\dots,4$. Note that the transients vanish after some time and then the energies settle at about $\epsilon ^{-2(j-1)}$. Here we define them as $p_j^2+q_j^2$. Note also that the dissipation is so slow that no decrease can be observed in the graph of $p_1^2+q_1^2$ over the time scale considered.

Figure 2.  The cylinder illustrates the set of periodic solutions of Eq.(9), with $\varphi$ changing (very little) from left to right and the circle illustrating the angle $\vartheta $. The spiral illustrates the way a time-dependent solution of Eq.(3) slides along the cylinder. It actually does not converge to it but will stay at some small, finite, distance from it. So the cylinder is Lyapunov stable in the sense of [1], at least as long as $\epsilon $ stays small.

Figure 3.  The figure shows the $\gamma$ dependence of the absolute value of the real parts of the eigenvalues, for $N = 3$ and $\epsilon = 0.01$. The three curves are linear with intercept 0 and slopes $3.2\cdot 10^{-10}$, $0.0027$, and $0.00727$. Note that the first eigenvalue has an extremely small positive real part, while the others are stable.

Figure 4.  This graph illustrates the behavior of $p_1(t)$, for $N = 3$ and several values of $\epsilon $ and $\gamma = 0.2\epsilon $. One measures the downcrossing times $T_k$ of the $k^{\rm th}$ downcrossing of $p_1$ through 0. The theory predicts that $X = \bigl({T_k/T_{k-1}}-1\bigr)/(\sqrt{k/(k-1)}-1) = 1$. As noted in the text, the transient behavior is not yet understood.

Figure 5.  This graph illustrates the decay properties of the $\ell_2$ norm as a function of time, for various values of $N$ and $\epsilon$. The horizontal axis is $\epsilon $ and the vertical axis is an estimate of the decay rate, obtained as follows: If $m_t$ is the $\ell_2$ norm at time $t$ and $m_{t'}$ that at time $t'$, then we compute $ k = k(\epsilon ) = \log\left( \frac{\log(m_t/m_{t'})}{\gamma\cdot (t'-t)}\right)/\log (\epsilon ). $ If $m_t$ decays like $\exp(-ct\gamma\epsilon^s)$, then the calculation will lead to $k = s$. Indeed, we see that the decay rate is $\epsilon ^{2N-1}$. The calculations shown were done for $\gamma = 0.2$.