Non-Filippov dynamics arising from the smoothing of nonsmooth systems, and its robustness to noise

Abstract

Switch-like behaviour in dynamical systems may be modelled by highly nonlinear functions, such as Hill functions or sigmoid functions, or alternatively by piecewise-smooth functions, such as step functions. Consistent modelling requires that piecewise-smooth and smooth dynamical systems have similar dynamics, but the conditions for such similarity are not well understood. Here we show that by smoothing out a piecewise-smooth system one may obtain dynamics that is inconsistent with the accepted wisdom — so-called Filippov dynamics — at a discontinuity, even in the piecewise-smooth limit. By subjecting the system to white noise, we show that these discrepancies can be understood in terms of potential wells that allow solutions to dwell at the discontinuity for long times. Moreover we show that spurious dynamics will revert to Filippov dynamics, with a small degree of stochasticity, when the noise magnitude is sufficiently large compared to the order of smoothing. We apply the results to a model of a dry-friction oscillator, where spurious dynamics (inconsistent with Filippov’s convention or with Coulomb’s model of friction) can account for different coefficients of static and kinetic friction, but under sufficient noise the system reverts to dynamics consistent with Filippov’s convention (and with Coulomb-like friction).

Appendix: Supplementary calculations

1.1 Derivation of (5.9)

The probability that a solution to (4.6), starting from $\tilde{y} = 0$, eventually escapes $[-\tilde{r},\tilde{r}]$, is $1$.

Consequently integration of (5.1) with respect to $\tilde{t}$ over $[0,\infty )$ yields

$$\begin{aligned} -\delta (\tilde{{y}}) = \frac{d}{d\tilde{{y}}} \left( -\phi R + \frac{\tilde{\kappa }^2}{2} \frac{dR}{d\tilde{{y}}} \right) , \end{aligned}$$

(9.1)

where we have let $R(\tilde{{y}}) = \int \nolimits _0^\infty p(\tilde{{y}},\tilde{t}) \,d\tilde{t}$. To solve (9.1) subject to the boundary conditions $R(\pm \tilde{r}) = 0$, we first integrate (9.1) to obtain

$$\begin{aligned} -\phi R + \frac{\tilde{\kappa }^2}{2} \frac{dR}{d\tilde{{y}}} = C - H(\tilde{{y}}), \end{aligned}$$

(9.2)

where $H$ is the Heaviside function and $C$ is a constant. Through the use of an integrating factor, further integration produces

$$\begin{aligned} R(\tilde{{y}}) = \frac{2}{\tilde{\kappa }^2} \,\mathrm{e}^{\frac{-2 V(\tilde{{y}})}{\tilde{\kappa }^2}} \int \limits _{\tilde{{y}}}^{\tilde{r}} \left( H(v) - C \right) \mathrm{e}^{\frac{2 V(v)}{\tilde{\kappa }^2}}\,dv, \end{aligned}$$

(9.3)

where we have chosen the limits of integration to automatically satisfy $R(\tilde{r}) = 0$. The requirement $R(-\tilde{r}) = 0$ implies that $C$ is given by (5.10). By (5.8) we have $\tilde{{T}} = \int \nolimits _{-\tilde{r}}^{\tilde{r}} R(u) \,du$, and the expression (5.9) follows from reversing the order of integration.

1.2 Bounding $C$ for (6.7)

Here we derive an upper bound for $C$ (5.10). Given any $\Delta > 0$, we have

$$\begin{aligned} \int \limits _{-\tilde{r}}^0 \mathrm{e}^{\frac{2 V(w)}{\tilde{\kappa }^2}} \,dw&\ge \int \limits _{-\tilde{r}}^{-1} \mathrm{e}^{\frac{2 V(w)}{\tilde{\kappa }^2}} \,dw = \frac{\tilde{\kappa }^2}{2 a^-} \left( \mathrm{e}^{\frac{2 a^- (\tilde{r} - 1)}{\tilde{\kappa }^2}} - 1 \right) \nonumber \\&\ge \frac{\tilde{\kappa }^2 (1-\Delta )}{2 a^-} \,\mathrm{e}^{\frac{2 a^- \tilde{r} (1-\Delta )}{\tilde{\kappa }^2}}, \end{aligned}$$

(9.4)

for sufficiently small $\varepsilon , \kappa$ and $r$. We let $V_\mathrm{max} = \mathrm{max}_{w \in [0,1]} V(w)$. Then

$$\begin{aligned} \int \limits _0^{\tilde{r}} \mathrm{e}^{\frac{2 V(w)}{\tilde{\kappa }^2}} \,dw&\le \mathrm{e}^{\frac{2 V_\mathrm{max}}{\tilde{\kappa }^2}} \left( 1 + \int \limits _1^{\tilde{r}} \mathrm{e}^{\frac{-2 a^+ (w-1)}{\tilde{\kappa }^2}} \,dw \right) \nonumber \\&\le \left( 1 + \frac{\tilde{\kappa }^2}{2 a^+} \right) \mathrm{e}^{\frac{2 V_\mathrm{max}}{\tilde{\kappa }^2}}. \end{aligned}$$

(9.5)

From (5.10), we have

$$\begin{aligned} C&= \frac{1}{1 + \frac{\int _{-\tilde{r}}^0 \mathrm{e}^{\frac{2 V(w)}{\tilde{\kappa }^2}} \,dw}{\int _0^{\tilde{r}} \mathrm{e}^{\frac{2 V(w)}{\tilde{\kappa }^2}} \,dw}} \le \frac{1}{1 + \frac{\frac{\tilde{\kappa }^2 (1-\Delta )}{2 a^-} \,\mathrm{e}^{\frac{2 a^- \tilde{r} (1-\Delta )}{\tilde{\kappa }^2}}}{\left( 1 + \frac{\tilde{\kappa }^2}{2 a^+} \right) \mathrm{e}^{\frac{2 V_\mathrm{max}}{\tilde{\kappa }^2}}}} \nonumber \\&\le \frac{(1+\Delta ) a^-}{1-\Delta } \left( \frac{1}{a^+} + \frac{2}{\tilde{\kappa }^2} \right) \mathrm{e}^{\frac{-2 a^- \tilde{r} \left( 1-\Delta -\frac{V_\mathrm{max}}{\tilde{r}} \right) }{\tilde{\kappa }^2}} \nonumber \\&\le \left( 1 + 4 \Delta \right) a^- \left( \frac{1}{a^+} + \frac{2}{\tilde{\kappa }^2} \right) \mathrm{e}^{\frac{-2 a^- \tilde{r} \left( 1-2\Delta \right) }{\tilde{\kappa }^2}} \;, \end{aligned}$$

(9.6)

where, in the last inequality we have assumed $\tilde{r}$ is sufficiently large that $\frac{V_\mathrm{max}}{\tilde{r}} \le \Delta$. Substituting $\Delta = \frac{1}{4}$ gives the result in the text.

1.3 The double integral (6.11)

The double integral is independent of $\frac{1}{\tilde{r}}$, so we simply evaluate it asymptotically in $\tilde{\kappa }$. Since the maximum of the exponent is attained at $(u,v) = (\tilde{{y}}_1,\tilde{{y}}_2)$, we employ the integral substitution:

$$\begin{aligned} \hat{u} = \frac{u - \tilde{{y}}_1}{\tilde{\kappa }} \hat{v} = \frac{v - \tilde{{y}}_2}{\tilde{\kappa }}. \end{aligned}$$

(9.7)

This yields

$$\begin{aligned}&\frac{2}{\tilde{\kappa }^2} \int \limits _0^1 \mathrm{e}^{\frac{2 V(v)}{\tilde{\kappa }^2}} \int \limits _{-1}^v \mathrm{e}^{\frac{-2 V(u)}{\tilde{\kappa }^2}} \,du \,dv = 2 \int \limits _{\frac{-\tilde{{y}}_2}{\tilde{\kappa }}}^{\frac{1 - \tilde{{y}}_2}{\tilde{\kappa }}} \int \limits _{\frac{-1 - \tilde{{y}}_1}{\tilde{\kappa }}}^{\hat{v} + \frac{\tilde{{y}}_2 - \tilde{{y}}_1}{\tilde{\kappa }}}\nonumber \\&\quad \times \,\mathrm{e}^{\frac{-2}{\tilde{\kappa }^2} \left( V(\tilde{{y}}_1) + \frac{1}{2} V''(\tilde{{y}}_1) \tilde{\kappa }^2 \hat{u}^2 - V(\tilde{{y}}_2) - \frac{1}{2} V''(\tilde{{y}}_2) \tilde{\kappa }^2 \hat{v}^2 + O(\tilde{\kappa }^3) \right) } \,d\hat{u} \,d\hat{v}.\nonumber \\ \end{aligned}$$

(9.8)

Laplace’s method justifies taking the limits of integration to $\pm \infty$, with which explicit integration produces

$$\begin{aligned}&\frac{2}{\tilde{\kappa }^2} \int \limits _0^1 \mathrm{e}^{\frac{2 V(v)}{\tilde{\kappa }^2}} \int \limits _{-1}^v \mathrm{e}^{\frac{-2 V(u)}{\tilde{\kappa }^2}} \,du \,dv\nonumber \\&\quad = 2 \mathrm{e}^{\frac{2}{\tilde{\kappa }^2} \left( V(\tilde{{y}}_2) - V(\tilde{{y}}_1) \right) } \left( \frac{\sqrt{\pi }}{\sqrt{V''(\tilde{{y}}_1)}} \frac{\sqrt{\pi }}{\sqrt{-V''(\tilde{{y}}_2)}} + O(\tilde{\kappa }) \right) .\nonumber \\ \end{aligned}$$

(9.9)

Note $V''(\tilde{{y}}) = -A'(\tilde{{y}})$.