1

I came across a problem in the dynamic systems theory, which is very similar to the following simple example.

Consider the simple model equation $$ y''(x)=y(x)-y(x)^2. \tag{*}$$ Let us denote by $y_\infty(x)$ the solution to $(*)$ satisfying the initial conditions $$ y(0)=\frac{3}{2},\quad y'(0)=0.$$ Then by integration, it is not hard show that $$ y_\infty(x)=\frac{3}{2}\text{sech}^2(\frac{x}{2}),$$ where $\text{sech}(x)=2/(e^{x}+e^{-x})$ is the hyperbolic secant function.

On the other hand. Let $L>0$ be a real number, and denote by $y_L(x)$ be the solution to $(*)$ subject to the boundary conditions $$ y'(0)=y'(L)=0.$$ We also require that $y_L'(x)\neq 0$ for $0<x<L$.

Numeric calculations seem to suggest that when $L$ is large, say $L\geq 4$, we have the estimate $$ |y_L(0)-y_\infty(0)|\leq Ce^{-\alpha L}\tag{**}$$ for some positive constants $C$ and $\alpha$. For instance, the following table shows that some values of $|y_L(0)-y_\infty(0)|$ and $L$: $$ \begin{array}{l l} \hline |y_L(0)-y_\infty(0)| & L\\ \hline 0.1 & 3.56\\ \hline 0.01 & 4.60\\ \hline 0.001 & 5.74\\ \hline 10^{-4} & 6.89\\ \hline 10^{-5}& 8.03\\ \hline 10^{-6} & 9.12\\ \hline 10^{-7} & 9.89 \\ \hline \end{array} $$

I would like to ask, is estimate $(**)$ true? In the case that it holds, how to prove it?

Xiang Yu
  • 4,835
  • Idle thought: obviously your initial conditions provide $y'(x)\rightarrow0$ as $x\rightarrow\infty$. Can you integrate for any other value of $y(0)$ and possibly show exponential growth in $y'$? That might at least provide a starting point for 'running it backwards' as it were. – Steven Stadnicki Dec 11 '21 at 17:43
  • 1
    I tried in this direction. For other values of $y(0)$, the solutions involve elliptic integrals which I do not know how to simplify. – Xiang Yu Dec 11 '21 at 17:46
  • 1
    I think the estimate is true. We can consider this equation as a system $\dot{y} = v, ; \dot{v} = y - y^2$. It has a saddle equilibrium at $(0, 0)$ that has a homoclinic loop to it: basically, $y_{\infty}(x)$ is a time-parametrization for the solution corresponding to it. This loop encloses a region of closed trajectories and BVP $y'(0) = y'(L) = 0$ allows to select a specific trajectory. The bigger $L$ you get, the closer to homoclinic loop you become and the bigger period becomes. I guess you can adjust this estimate for your problem. – Evgeny Dec 11 '21 at 19:10

0 Answers0