Application of variable- and distributed-order fractional operators to the dynamic analysis of nonlinear oscillators

Abstract

Many physical processes in nature exhibit complex dynamics that result from a combination of multiscale, nonlinear, non-local, and memory effects. Recent experimental measurements conducted in a variety of physical domains have shown that, at the macroscale level, these effects typically result in significant deviations from the behavior predicted by classical models. Notably, the underlying dynamics was often shown to be of non-integer order and possibly better captured by fractional-order models. Fractional operators are intrinsically multiscale; thus, they provide a natural approach to account for non-local and memory effects. In this study, we present the possible application of variable-order (VO) and distributed-order (DO) fractional operators to a few classes of nonlinear lumped parameter models that have great practical relevance in mechanics and dynamics. More specifically, we present a methodology to define VO and DO fractional operators that are capable of capturing various physical transitions characteristic of contact dynamics, nonlinear reversible systems, hysteretic systems, and nonlinear damped oscillator systems. Despite using simplified lumped parameters models to illustrate the application of VO and DO operators to mechanics, we show numerical evidence of their unique modeling capabilities as well as their connection to more complex systems at the continuum scale. Further, for a selected problem involving distributed nonlinear damping, we provide approximate analytical solutions that are helpful to better understand the underlying dynamics and to quantify the accuracy of our numerical models.

Appendix

Consider the averaging integral given in Eq. (30):

$$\begin{aligned} I=\lim \limits _{T\rightarrow \infty }\frac{1}{T}\int _0^T\frac{\cos (\omega _0t+\phi )}{\varGamma [1-\alpha (x)]}\bigg \{\int _{0}^{t}\frac{\cos (\omega _0\tau +\phi )}{(t-\tau )^{\alpha (x)}}\mathrm {d}\tau \bigg \}\mathrm {d}t. \end{aligned}$$

(48)

Using the coordinate transformation $\tau =t-s$, and trigonometric relations for double angles, we obtain:

$$\begin{aligned} I=&\lim \limits _{T\rightarrow \infty }\frac{1}{2T}\int _0^T\frac{1+\cos (2\omega _0t+2\phi )}{\varGamma [1-\alpha (x)]}\bigg \{\int _{0}^{t}\frac{\cos (\omega _0s)}{s^{\alpha (x)}}\mathrm {d}s\bigg \}\mathrm {d}t\nonumber \\&+ \lim \limits _{T\rightarrow \infty }\frac{1}{2T}\int _0^T\frac{\sin (2\omega _0t+2\phi )}{\varGamma [1-\alpha (x)]}\bigg \{\int _{0}^{t}\frac{\sin (\omega _0s)}{s^{\alpha (x)}}\mathrm {d}s\bigg \}\mathrm {d}t. \end{aligned}$$

(49)

The right-hand side of the above equation consists of two integrals which we denote as $I_1$ (containing $\cos (\cdot )$ terms) and $I_2$ (containing $\sin (\cdot )$ terms). Note that $\alpha (x)$ is an implicit function of time such that $\alpha (x)=\alpha (x(t))$. We first evaluate $I_1$ by using integration by parts in the following manner:

$$\begin{aligned}&I_1\nonumber \\&\quad =\lim \limits _{T\rightarrow \infty }\frac{1}{4T}\bigg |\frac{2\omega _0 t+\sin (2\omega _0 t+ 2\phi )}{\omega _0\varGamma [1-\alpha (x)]}\bigg \{\int _{0}^{t}\frac{\cos (\omega _0s)}{s^{\alpha (x)}}\mathrm {d}s\bigg \}\bigg |^{T}_{0}\nonumber \\&\qquad -\lim \limits _{T\rightarrow \infty }\frac{1}{4T}\int _0^T \frac{2\omega _0 t+\sin (2\omega _0 t+ 2\phi )}{\omega _0}\frac{\mathrm {d}}{\mathrm {d}t}\nonumber \\&\quad \qquad \bigg [\frac{1}{{\varGamma [1-\alpha (x)]}} \times \int _{0}^{t}\frac{\cos (\omega _0s)}{s^{\alpha (x)}}\mathrm {d}s\bigg ]\mathrm {d}t. \end{aligned}$$

(50)

We use Leibniz’s rule to perform the differentiation in the second integral of $I_1$. Further, under the limiting condition $T\rightarrow \infty$, the above integral simplifies to the following:

$$\begin{aligned} I_1=&\lim \limits _{T\rightarrow \infty }\frac{1}{2\varGamma [1-\alpha (x(T))]}\int _{0}^{T}\frac{\cos (\omega _0s)}{s^{\alpha (x(T))}}\mathrm {d}s\nonumber \\&-\lim \limits _{T\rightarrow \infty }\frac{1}{2T}\Bigg [\int _0^T \bigg \{\frac{t^{1-\alpha (x)}\cos (\omega _0t)}{\varGamma [1-\alpha (x)]}+\dot{x}tg(x,t)\bigg \}\mathrm {d}t\Bigg ] \end{aligned}$$

(51)

where

$$\begin{aligned} g(x,t)= & {} \frac{\alpha '(x)}{\varGamma [1-\alpha (x)]}\int _{0}^{t}\big [\psi (1-\alpha (x))-\log s)\big ]\nonumber \\&\quad \frac{\cos (\omega _0s)}{s^{\alpha (x)}}\mathrm {d}s. \end{aligned}$$

(52)

In the above equation, $\psi (x)$ is the digamma function and is defined as the logarithmic derivative of the gamma function, i.e., $\psi (x)=\varGamma '(x)/\varGamma (x)$. Now consider the following part of the integration in Eq. (51) denoted as $I_3$:

$$\begin{aligned} I_3=\lim \limits _{T\rightarrow \infty }\frac{1}{2T}\int _0^T\dot{x}tg(x,t)\mathrm {d}t. \end{aligned}$$

(53)

For $\alpha (x)\in (0,1)$ and a smooth variation of $\alpha (x)$ (as already assumed in Sect. 5), it can be shown that g(x, t) is bounded. Note that g(x, t) is an implicit function of time. Assuming $M=\sup |g(x,t)|~\forall ~t\in (0,T)$ and using the slow flow $\dot{x}$ from Eq. (25), we obtain the inequality:

$$\begin{aligned} |I_3|<\lim \limits _{T\rightarrow \infty }\frac{MR\omega _0}{2T}\bigg |\int _0^Tt\cos (\omega _0 t+\phi )\mathrm {d}t\bigg |. \end{aligned}$$

(54)

which under the slow flow constraint simplifies to:

$$\begin{aligned} |I_3|<\lim \limits _{T\rightarrow \infty }\omega _0Mx(T). \end{aligned}$$

(55)

Thus, $|I_3|\rightarrow 0$ as $\lim \limits _{T\rightarrow \infty }x(T)\rightarrow 0$. This simplifies Eq. (51) to:

$$\begin{aligned} I_1= & {} \lim \limits _{T\rightarrow \infty }\frac{1}{2\varGamma [1-\alpha (x(T))]}\int _{0}^{T}\frac{\cos (\omega _0s)}{s^{\alpha (x(T))}}\mathrm {d}s\nonumber \\&-\lim \limits _{T\rightarrow \infty }\frac{1}{2T}\int _0^T \frac{t^{1-\alpha (x)}\cos (\omega _0t)}{\varGamma [1-\alpha (x)]}\mathrm {d}t. \end{aligned}$$

(56)

Applying integration by parts to the second integral above and by repeating the arguments used to obtain the Eqs. (53–55), it can be shown that the second integral similarly simplifies to zero. Now, using the standard integration result:

$$\begin{aligned} \int _{0}^{\infty }\frac{\cos (\omega _0s)}{s^{m}}\mathrm {d}s=\omega _0^{m-1}\varGamma (1-m)\sin \Big (\frac{m\pi }{2}\Big ) \end{aligned}$$

(57)

we obtain:

$$\begin{aligned} I_1=\frac{\omega _0^{\alpha _{\infty }-1}}{2}\sin \Big (\frac{\alpha _{\infty }\pi }{2}\Big ) \end{aligned}$$

(58)

where $\alpha _{\infty }=\lim \limits _{T\rightarrow \infty }\alpha (x(T))$. Similarly, it can be shown that the integral $I_2$ in Eq. (49) simplifies to zero. Now collecting all the above terms the integral I in Eq. (48) is obtained as:

$$\begin{aligned} I=\frac{\omega _0^{\alpha _{\infty }-1}}{2}\sin \Big (\frac{\alpha _{\infty }\pi }{2}\Big ). \end{aligned}$$

(59)

The above integration procedure becomes fairly straightforward when the order $\alpha$ of the Caputo operator is a constant and can be found in [23, 24].