A Comparative Study of Responses of Fractional Oscillator to Sinusoidal Excitation in the Weyl and Caputo Senses

Abstract

The fractional oscillator equation with the sinusoidal excitation $m{x}^{″}\left(t\right)+b{D}_t^{\alpha }x\left(t\right)+kx\left(t\right)=Fsin\left(\omega t\right)$, $m,b,k,\omega >0$ and $0<\alpha <2$ is comparatively considered for the Weyl fractional derivative and the Caputo fractional derivative. In the sense of Weyl, the fractional oscillator equation is solved to be a steady periodic oscillation $x_W\left(t\right)$. In the sense of Caputo, the fractional oscillator equation is solved and subjected to initial conditions. For the fractional case $\alpha \in (0,1)\cup (1,2)$, the response to excitation, $S\left(t\right)$, is a superposition of three parts: the steady periodic oscillation $x_W\left(t\right)$, an exponentially decaying oscillation and a monotone recovery term in negative power law. For the two responses to initial values, $S_0\left(t\right)$ and $S_1\left(t\right)$, either of them is a superposition of an exponentially decaying oscillation and a monotone recovery term in negative power law. The monotone recovery terms come from the Hankel integrals which make the fractional case different from the integer-order case. The asymptotic behaviors of the solutions removing the steady periodic response are given for the four cases of the initial values. The Weyl fractional derivative is suitable for a describing steady-state problem, and can directly lead to a steady periodic solution. The Caputo fractional derivative is applied to an initial value problem and the steady component of the solution is just the solution in the corresponding Weyl sense.

1. Introduction

Fractional calculus has been attracting significant attention in recent decades since it can be used to describe the memory and hereditary properties of various materials and processes [1,2,3,4,5,6,7,8,9]. Its application fields are extensive, including viscoelastic a constitutive relationship [4,7,10], anomalous diffusion [2,4], vibration and relaxation [2,4,7,8], control theory [2,5], stochastic process [11,12], etc.

Let $f\left(t\right)$ be integrable on any finite subinterval of $(a,+\infty )$. Then, the Riemann–Liouville fractional integral of $f\left(t\right)$ of order $\nu$ is defined as

$$\begin{array}{c}\hfill J_{a,t}^{\nu }f\left(t\right)={\int }_a^t\frac{{(t-\tau )}^{\nu -1}}{\mathsf{\Gamma }\left(\nu \right)}f\left(\tau \right)d\tau ,t>a,\end{array}$$

(1)

for $\nu >0$, and $J_{a,t}^{0}f\left(t\right)=f\left(t\right)$ for $\nu =0$, where $\mathsf{\Gamma }(·)$ is the gamma function.

Let $\alpha$ be a real positive number, $m-1<\alpha \le m$, $m\in \mathbb{N}$, and the function $f\left(t\right)$ has an absolutely continuous $(m-1)$st derivative on any finite subinterval of $(a,+\infty )$. Then, the Riemann–Liouville fractional derivative of $f\left(t\right)$ of order $\alpha$ is defined as

$$\begin{array}{c}\hfill {}^{RL}{D}_{a,t}^{\alpha }f\left(t\right)=\frac{d^m}{d{t}^m}\left(J_{a,t}^{m-\alpha }f\left(t\right)\right),t>a,\end{array}$$

(2)

while the Caputo fractional derivative of $f\left(t\right)$ of order $\alpha$ is defined as

$$\begin{array}{c}\hfill {}^C{D}_{a,t}^{\alpha }f\left(t\right)=J_{a,t}^{m-\alpha }\left(f^{\left(m\right)}\left(t\right)\right),t>a.\end{array}$$

(3)

Usually, a is a finite number and without loss of generality, a is taken as 0. If we choose $a=-\infty$, then the operator in (1) is known as the Weyl fractional integral, while the operator in (2) or (3), which are consistent in this case, is known as the Weyl fractional derivative [13]. Theoretical analyses and experimental simulations have indicated that the stress–strain relationship of a viscoelastic body can be better described by introducing the fractional calculus, such as the Scott–Blair model [4], the Kelvin–Voigt, Maxwell and Zener models [4,10,14], among others [7,15,16]. Thus, vibration problems involving a viscoelastic medium with a fractional stress–strain relationship lead to fractional oscillator equations [7,8,17,18,19,20,21,22,23,24,25].

In [22], fractional oscillator equations in the sense of the Weyl fractional calculus were clarified and analyzed. In [11], the fractional oscillator was described by a stochastic differential equation with the Weyl fractional calculus. In [26,27], the Weyl fractional derivative was used to derive the steady response of a fractional oscillator system. Comparatively, the Caputo fractional derivative is more commonly used in modeling physical or mechanical problems partly due to the feasible classical initial conditions [4,7,9,19,20,21,24,28]. The objective of this paper was to understand the differences and relationship between the Weyl fractional derivative and the Caputo fractional derivative in modeling the fractional oscillator equations. By imposing the sinusoidal excitation, we derive the steady periodic response for the Weyl case, while an initial value problem needs to be solved for the Caputo case. We will show that the steady component of the response for the Caputo case is just the response for the corresponding Weyl case.

For short, we used ${}^W{D}_t^{\alpha }f\left(t\right)$ to denote the Weyl fractional derivative ${}^{RL}{D}_{-\infty ,t}^{\alpha }f\left(t\right)$ or ${}^C{D}_{-\infty ,t}^{\alpha }f\left(t\right)$, and ${}^C{D}_t^{\alpha }f\left(t\right)$ the Caputo fractional derivative ${}^C{D}_{0,t}^{\alpha }f\left(t\right)$. For the model based on the Caputo fractional derivative, we will use the Laplace transform

$$\tilde{f}\left(s\right)=\mathcal{L}\left[f\left(t\right)\right]={\int }_0^{+\infty }f\left(t\right)e^{-st}dt$$

(4)

and its complex inverse integral formula

$$f\left(t\right)={\mathcal{L}}^{-1}\left[\tilde{f}\left(s\right)\right]=\frac{1}{2\pi \mathrm{i}}{\int }_{\mathrm{Br}}\tilde{f}\left(s\right)e^{st}ds,$$

(5)

and the Laplace transform formula of the Caputo fractional derivative

$$\mathcal{L}\left[{}^C{D}_t^{\alpha }f\left(t\right)\right]=s^{\alpha }\tilde{f}\left(s\right)-\sum _{k=0}^{m-1}{f}^{\left(k\right)}\left(0\right)s^{\alpha -1-k},m-1<\alpha \le m.$$

(6)

In this paper, we consider the response of the fractional oscillator to sinusoidal excitation. In the next Section 2, the fractional derivative is in the sense of Weyl and a steady periodic response is directly obtained by the method of undetermined coefficients. In Section 3, the fractional derivative is in the sense of Caputo and an initial value problem needs to be solved. The response is much more complicated than the Weyl case, and is solved in detail using the Laplace transform and its complex inverse integral formula. A comparative study of the responses of fractional oscillator in the Weyl and Caputo senses is conducted. Section 4 summarizes our conclusions.

2. Oscillator Equation with the Weyl Fractional Derivative

Consider the fractional oscillator equation with the Weyl fractional derivative and the sinusoidal excitation

$$\begin{array}{ccc}& & m{x}^{″}\left(t\right)+b\left({}^W{D}_t^{\alpha }x\left(t\right)\right)+kx\left(t\right)=Fsin\left(\omega t\right),m,b,k,\omega >0,\hfill \end{array}$$

(7)

where F is a constant and the order $\alpha$ is limited in the interval $0<\alpha <2$.

Suppose the response has the amplitude X and phase lag $\psi$,

$$\begin{array}{c}\hfill x\left(t\right)=Xsin(\omega t-\psi ),\end{array}$$

(8)

then the Weyl fractional derivative is [13]

$$\begin{array}{c}\hfill {}^W{D}_t^{\alpha }x\left(t\right)=X{\omega }^{\alpha }sin\left(\omega t-\psi +\frac{\alpha \pi }{2}\right).\end{array}$$

(9)

Substituting into (7) and equating the coefficients of $sin\left(\omega t\right)$ and $cos\left(\omega t\right)$, respectively, on both sides, we obtain

$$\begin{array}{ccc}& & X(k-m{\omega }^2)cos\left(\psi \right)+bX{\omega }^{\alpha }cos\left(\frac{\alpha \pi }{2}-\psi \right)=F,\hfill \\ & & (k-m{\omega }^2)sin\left(\psi \right)-b{\omega }^{\alpha }sin\left(\frac{\alpha \pi }{2}-\psi \right)=0.\hfill \end{array}$$

Furthermore, the amplitude and phase lag are solved as

$$\begin{array}{ccc}\hfill X& =& \frac{F}{\sqrt{{\left(k-m{\omega }^2+b{\omega }^{\alpha }cos\left(\frac{\alpha \pi }{2}\right)\right)}^2+{\left(b{\omega }^{\alpha }sin\left(\frac{\alpha \pi }{2}\right)\right)}^2}},\hfill \end{array}$$

(10)

$$\begin{array}{ccc}\hfill tan\left(\psi \right)& =& \frac{b{\omega }^{\alpha }sin\left(\frac{\alpha \pi }{2}\right)}{k-m{\omega }^2+b{\omega }^{\alpha }cos\left(\frac{\alpha \pi }{2}\right)},\hfill \end{array}$$

(11)

where the phase lag is limited to the interval $0\le \psi <\pi .$ Thus, the steady periodic response is derived as

$$\begin{array}{c}\hfill x_W\left(t\right)=\frac{Fsin(\omega t-\psi )}{\sqrt{{\left(k-m{\omega }^2+b{\omega }^{\alpha }cos\left(\frac{\alpha \pi }{2}\right)\right)}^2+{\left(b{\omega }^{\alpha }sin\left(\frac{\alpha \pi }{2}\right)\right)}^2}},\end{array}$$

(12)

where the subscript W denotes the solution for Equation (7) with the Weyl fractional derivative. The response is a sinusoidal oscillation with same frequency as the excitation with frequency-dependent amplitude and phase lag.

We look into the amplitude ratio

$$\beta \left(\omega \right)=\frac{X}{F/k}=\frac{k}{\sqrt{{\left(k-m{\omega }^2+b{\omega }^{\alpha }cos\left(\frac{\alpha \pi }{2}\right)\right)}^2+{\left(b{\omega }^{\alpha }sin\left(\frac{\alpha \pi }{2}\right)\right)}^2}},$$

(13)

and the phase lag $\psi \left(\omega \right)$ as functions of the excitation frequency $\omega$. It is obvious that $\beta \left(\omega \right)\to 1$ as $\omega \to 0^{+}.$ The derivative of amplitude ratio $\beta \left(\omega \right)$ satisfies

$$\begin{array}{c}\hfill {\beta }^{\prime }\left(\omega \right)\sim -\alpha b{k}^{-1}cos\left(\frac{\alpha \pi }{2}\right){\omega }^{\alpha -1},\omega \to 0^{+}.\end{array}$$

Thus, the limit as $\omega \to 0^{+}$ is

$$\begin{array}{c}\hfill {\beta }^{\prime }\left(0^{+}\right)=\left\{\begin{array}{cc}0,& 1\le \alpha <2,\hfill \\ -\infty ,& 0<\alpha <1.\hfill \end{array}\right.\end{array}$$

(14)

The amplitude–frequency curves $\beta \left(\omega \right)$ vs. $\omega$ are plotted in Figure 1 for $m=1,$$b=2,$$k=3$ and $\alpha =0.3,0.65,1,1.35,1.7$. The plot of the phase lag $\psi \left(\omega \right)$ vs. $\omega$ is shown in Figure 2 for the same parameter values. The amplitude–frequency curve in Figure 1 displays a decreasing initial passage for the case of $0<\alpha <1$. For the two limiting cases $\alpha =0$ and $\alpha =2$, the undamped natural frequencies are $\sqrt{(k+b)/m}=\sqrt{5}$ and $\sqrt{k/(m+b)}=1$. The resonant frequency in Figure 1 shifts to them nearby as $\alpha$ is close to the two endpoints 0 and 2. As $\alpha$ approaches to 1, the format becomes low and a large damping is implied. The fractional case exhibits a stronger resonance phenomenon than the case of $\alpha =1$. Figure 2 displays the transition process of the phase lag $\psi \left(\omega \right)$ with $\omega$. The transition happens around the resonant frequency. As $\alpha$ approaches towards 1, the transition becomes gentle.

3. Oscillator Equation with the Caputo Fractional Derivative

Consider the fractional oscillator equation with the Caputo fractional derivative and the sinusoidal excitation

$$\begin{array}{ccc}& & m{x}^{″}\left(t\right)+b\left({}^C{D}_t^{\alpha }x\left(t\right)\right)+kx\left(t\right)=Fsin\left(\omega t\right),m,b,k,\omega >0,\hfill \end{array}$$

(15)

$$\begin{array}{ccc}& & x\left(0\right)=x_0,x^{\prime }\left(0\right)=x_1,\hfill \end{array}$$

(16)

where F is a constant, $x_0$ and $x_1$ are specified initial values and the order $\alpha$ satisfies $0<\alpha <2$.

Operating the Laplace transform to Equation (15) with the initial condition (16) gives

$$\begin{array}{c}\hfill m(s^2\tilde{x}\left(s\right)-s{x}_0-x_1)+b(s^{\alpha }\tilde{x}\left(s\right)-s^{\alpha -1}{x}_0-\lceil \alpha -1\rceil s^{\alpha -2}{x}_1)+k\tilde{x}\left(s\right)=\frac{F\omega }{s^2+{\omega }^2}.\end{array}$$

where $\tilde{x}\left(s\right)=\mathcal{L}\left[x\left(t\right)\right]$ is the Laplace transform of $x\left(t\right)$, for $\alpha \ne 1$, the one-valued branch $s^{\alpha }=e^{\alpha (ln|s|+\mathrm{i} arg s)}$, $-\pi <args<\pi$, is taken, and $\lceil \alpha -1\rceil$ denotes the least integer no less than $\alpha -1$. Thus, if $0<\alpha \le 1$, then $\lceil \alpha -1\rceil =0$, and if $1<\alpha <2$, then $\lceil \alpha -1\rceil =1$. Like an integer-order equation, refer to the trinomial

$$\begin{array}{c}\hfill C\left(s\right)=m{s}^2+b{s}^{\alpha }+k,\end{array}$$

(17)

as the characteristic polynomial with a fractional power. As such, the Laplace image function has the expression

$$\tilde{x}\left(s\right)=\frac{F\omega }{(s^2+{\omega }^2)C\left(s\right)}+x_0\frac{ms+b{s}^{\alpha -1}}{C\left(s\right)}+x_1\frac{m+\lceil \alpha -1\rceil b{s}^{\alpha -2}}{C\left(s\right)}.$$

We introduce three special responses, i.e., the response to excitation and the two responses to initial values,

$$\begin{array}{ccc}\hfill S\left(t\right)& =& F\omega {\mathcal{L}}^{-1}\left[\frac{1}{(s^2+{\omega }^2)C\left(s\right)}\right],\hfill \end{array}$$

(18)

$$\begin{array}{ccc}\hfill S_0\left(t\right)& =& x_0{\mathcal{L}}^{-1}\left[\frac{ms+b{s}^{\alpha -1}}{C\left(s\right)}\right],\hfill \end{array}$$

(19)

$$\begin{array}{ccc}\hfill S_1\left(t\right)& =& x_1{\mathcal{L}}^{-1}\left[\frac{m+\lceil \alpha -1\rceil b{s}^{\alpha -2}}{C\left(s\right)}\right].\hfill \end{array}$$

(20)

Thus, the solution is a superposition of them

$$x\left(t\right)=S\left(t\right)+S_0\left(t\right)+S_1\left(t\right).$$

(21)

In order to obtain results that are convenient to compare with those in last section, we consider the inverses in (18)–(20) by using the complex inverse integral formula, e.g., for (18), the formula reads

$$\begin{array}{ccc}\hfill S\left(t\right)& =& \frac{F\omega }{2\pi \mathrm{i}}{\int }_{\mathrm{Br}}\frac{e^{st}}{(s^2+{\omega }^2)C\left(s\right)}ds,\hfill \end{array}$$

(22)

where Br denotes the Bromwich path, i.e., a straight-line $\mathrm{Re}\left(s\right)=\gamma >0$. The poles of the integrand from the sinusoidal excitation are two purely imaginary numbers and denoted by

$$\begin{array}{c}\hfill s_{1,2}=\pm \mathrm{i}\omega .\end{array}$$

If $\alpha =1$, then the power function $s^{\alpha }$ is a single-valued holomorphic function on the whole complex plane, while if $\alpha \in (0,1)\cup (1,2)$, the function $s^{\alpha }$ has a branch cut, which is taken as the nonpositive real axis here. Moreover, for a distribution of zeros of $C\left(s\right)$, the fractional case is different from the integer-order case. Accordingly, we solve the problem separately in the two following subsections in a contrasting way.

3.1. The Integer-Order Case $\alpha =1$

The integral in (22) can be reduced from the loop integral on the contour shown in Figure 3. By the residue theorem and Jordan’s lemma, Equation (22) is expressed as a sum of residues of the integrand. The singularities comprise the poles from the sinusoidal excitation, $s_{1,2}=\pm \mathrm{i}\omega$ and the poles from the characteristic polynomial

$$C\left(s\right)=m{s}^2+bs+k.$$

We denote the residue contributions from the two types of poles by $S_{RE}\left(t\right)$ and $S_{RC}\left(t\right)$, respectively. Thus, we have

$$\begin{array}{ccc}\hfill S\left(t\right)& =& S_{RE}\left(t\right)+S_{RC}\left(t\right)\hfill \\ & =& F\omega \sum _{i=1}^2\mathrm{Res}\left(\frac{e^{st}}{(s^2+{\omega }^2)C\left(s\right)},s_i\right)+F\omega \sum _i\mathrm{Res}\left(\frac{e^{st}}{(s^2+{\omega }^2)C\left(s\right)},s_i\right),\hfill \end{array}$$

(23)

where the $s_i$ in the second sum are poles from the characteristic polynomial $C\left(s\right)$ and have negative real parts. By calculating the residues at the two simple poles $s_{1,2}$ we obtain $S_{RE}\left(t\right)$ as follows and it is just the special case of $x_W\left(t\right)$ as $\alpha =1$ in Section 2,

$$\begin{array}{ccc}\hfill S_{RE}\left(t\right)& =& F\omega \left(\frac{e^{s_{1}t}}{2s_{1}C\left(s_1\right)}+\frac{e^{s_{2}t}}{2s_{2}C\left(s_2\right)}\right)\hfill \\ & =& \frac{F\left(\left(k-m{\omega }^2\right)sin\left(\omega t\right)-b\omega cos\left(\omega t\right)\right)}{{(k-m{\omega }^2)}^2+{\left(b\omega \right)}^2}=x_W(t;\alpha =1).\hfill \end{array}$$

(24)

For $S_0\left(t\right)$ and $S_1\left(t\right)$ in (19) and (20), we derive in a similar manner that

$$\begin{array}{ccc}\hfill S_0\left(t\right)& =& x_0{\mathcal{L}}^{-1}\left[\frac{ms+b}{C\left(s\right)}\right]=x_0\sum _i\mathrm{Res}\left(\frac{(ms+b)e^{st}}{C\left(s\right)},s_i\right),\hfill \end{array}$$

(25)

$$\begin{array}{ccc}\hfill S_1\left(t\right)& =& x_1{\mathcal{L}}^{-1}\left[\frac{m}{C\left(s\right)}\right]=x_{1}m\sum _i\mathrm{Res}\left(\frac{e^{st}}{C\left(s\right)},s_i\right),\hfill \end{array}$$

(26)

where, like $S_{RC}\left(t\right)$, only the poles from the characteristic polynomial $C\left(s\right)$ are involved. Such poles are clarified to the three cases as shown in Figure 3. Thus, we calculate $S_{RC}\left(t\right)$, $S_0\left(t\right)$ and $S_1\left(t\right)$ according to the following three cases.

Case i.$b^2-4mk<0$

The poles are a pair of conjugated complex numbers $s_{3,4}=-\lambda \pm \mathrm{i}\mu$, where $\lambda =\frac{b}{2m}$ and $\mu =\frac{\sqrt{4mk-b^2}}{2m}.$ By calculating the residues at the two simple poles $s_{3,4}$, we obtain

$$\begin{array}{ccc}\hfill S_{RC}\left(t\right)& =& \frac{F\omega \left(b\sqrt{4mk-b^2}cos\left(\mu t\right)+\left(b^2-2mk+2m^2{\omega }^2\right)sin\left(\mu t\right)\right)}{m^2\sqrt{4mk-b^2}\left({\lambda }^4+2{\lambda }^2\left({\mu }^2+{\omega }^2\right)+{\left({\mu }^2-{\omega }^2\right)}^2\right)}{e}^{-\lambda t},\hfill \end{array}$$

(27)

$$\begin{array}{ccc}\hfill S_0\left(t\right)& =& x_0\left(cos\left(\mu t\right)+\frac{b}{\sqrt{4mk-b^2}}sin\left(\mu t\right)\right)e^{-\lambda t},\hfill \end{array}$$

(28)

$$\begin{array}{ccc}\hfill S_1\left(t\right)& =& \frac{2x_{1}msin\left(\mu t\right)}{\sqrt{4mk-b^2}}{e}^{-\lambda t}.\hfill \end{array}$$

(29)

Case ii.$b^2-4mk>0$

The poles are two negative real numbers $s_{3,4}=-\lambda \pm \nu$, where $\lambda =\frac{b}{2m}$ and $\nu =\frac{\sqrt{b^2-4mk}}{2m}.$ By calculating the residues at the two simple poles $s_{3,4}$, we obtain

$$\begin{array}{ccc}\hfill S_{RC}\left(t\right)& =& \frac{F\omega }{\sqrt{b^2-4mk}}\left(\frac{e^{-(\lambda -\nu )t}}{{(\lambda -\nu )}^2+{\omega }^2}-\frac{e^{-(\lambda +\nu )t}}{{(\lambda +\nu )}^2+{\omega }^2}\right),\hfill \end{array}$$

(30)

$$\begin{array}{ccc}\hfill S_0\left(t\right)& =& \frac{x_0}{2}\left(\left(1+\frac{b}{\sqrt{b^2-4mk}}\right)e^{-(\lambda -\nu )t}+\left(1-\frac{b}{\sqrt{b^2-4mk}}\right)e^{-(\lambda +\nu )t}\right),\hfill \end{array}$$

(31)

$$\begin{array}{ccc}\hfill S_1\left(t\right)& =& \frac{x_{1}m}{\sqrt{b^2-4mk}}\left(e^{-(\lambda -\nu )t}-e^{-(\lambda +\nu )t}\right).\hfill \end{array}$$

(32)

Case iii.$b^2-4mk=0$

There is one second-ordered pole $s_3=-\lambda =-\frac{b}{2m}$. By calculating the residues at the second-ordered pole $s_3$, we obtain

$$\begin{array}{ccc}\hfill S_{RC}\left(t\right)& =& \frac{4Fm\omega \left(4bm+b^{2}t+4m^2{\omega }^{2}t\right)}{{\left(b^2+4m^2{\omega }^2\right)}^2}{e}^{-\lambda t},\hfill \end{array}$$

(33)

$$\begin{array}{ccc}\hfill S_0\left(t\right)& =& x_0(1+\lambda t)e^{-\lambda t},\hfill \end{array}$$

(34)

$$\begin{array}{ccc}\hfill S_1\left(t\right)& =& x_{1}t{e}^{-\lambda t}.\hfill \end{array}$$

(35)

In summary, the residue contribution from the sinusoidal excitation is a periodic oscillation in (24), whereas other solution components in Equations (27)–(35) all decay in negative exponential rates. We notice that the response to excitation, $S\left(t\right)$, is divided into two parts, $S_{RE}\left(t\right)$ and $S_{RC}\left(t\right)$. The former represents a steady periodic oscillation; the latter exhibits a negative exponential decay, respectively, in three patterns (27), (30), and (33).

3.2. The Fractional Case $\alpha \in (0,1)\cup (1,2)$

In this case, the integral in (22) needs to be derived from a loop integral on the contour shown in Figure 4. Thus, in addition to the residue contribution, the Hankel integral contribution is added to the fractional case. For $S\left(t\right)$ in (22), the poles from the sinusoidal excitation are still $s_{1,2}=\pm \mathrm{i}\omega .$ For the fractional case, the poles from the characteristic polynomial $C\left(s\right)=m{s}^2+b{s}^{\alpha }+k$ only have one case compared with the integer-order case $\alpha =1$, that is, a pair of conjugated complex numbers with a negative real part on the principal Riemann sheet $-\pi <args<\pi$ as shown in Figure 4 [19,29]. Denote the pair of simple poles by

$$s_{3,4}=-\lambda \pm \mathrm{i}\mu =\eta e^{\pm \mathrm{i}\varphi },$$

(36)

where $\lambda ,\mu >0$, $\eta =\sqrt{{\lambda }^2+{\mu }^2}$ and $\varphi =\pi -arctan\frac{\mu }{\lambda }$.

By the residue theorem and Jordan’s lemma, $S\left(t\right)$ in (22) is deformed to a sum of the residue contribution and the Hankel integral,

$$\begin{array}{c}\hfill S\left(t\right)=S_R\left(t\right)+S_H\left(t\right),\end{array}$$

(37)

where the residue contribution is further classified according to the origins of poles,

$$\begin{array}{ccc}\hfill S_R\left(t\right)& =& S_{RE}\left(t\right)+S_{RC}\left(t\right)\hfill \\ & =& F\omega \sum _{i=1}^2\mathrm{Res}\left[\frac{e^{st}}{(s^2+{\omega }^2)C\left(s\right)},s_i\right]+F\omega \sum _{i=3}^4\mathrm{Res}\left[\frac{e^{st}}{(s^2+{\omega }^2)C\left(s\right)},s_i\right],\hfill \end{array}$$

(38)

and the Hankel integral is

$$\begin{array}{ccc}\hfill S_H\left(t\right)& =& \frac{F\omega }{2\pi \mathrm{i}}{\int }_{\mathrm{Ha}}\frac{e^{st}}{(s^2+{\omega }^2)C\left(s\right)}ds,\hfill \end{array}$$

(39)

where Ha denotes the Hankel loop encompassing the nonpositive real axis.

For (38), the residue contribution from the sinusoidal excitation is computed to be exactly the solution in Section 2 by using the Weyl fractional derivative

$$\begin{array}{c}\hfill S_{RE}\left(t\right)=x_W\left(t\right),\end{array}$$

(40)

while the residue contribution from the characteristic polynomial is computed as

$$\begin{array}{ccc}\hfill S_{RC}\left(t\right)& =& F\omega \left(\frac{e^{s_{3}t}}{(s_3^2+{\omega }^2)C^{\prime }\left(s_3\right)}+\frac{e^{s_{4}t}}{(s_4^2+{\omega }^2)C^{\prime }\left(s_4\right)}\right)\hfill \\ & =& 2F\omega \mathrm{Re}\left(\frac{e^{(-\lambda +\mathrm{i}\mu )t}}{\left({(-\lambda +\mathrm{i}\mu )}^2+{\omega }^2\right)\left(2m(-\lambda +\mathrm{i}\mu )+b\alpha {\eta }^{\alpha -1}{e}^{\mathrm{i}(\alpha -1)\varphi }\right)}\right)\hfill \\ & =& \frac{2F\omega P{e}^{-\lambda t}}{\left({\lambda }^4+2{\lambda }^2\left({\mu }^2+{\omega }^2\right)+{\left({\mu }^2-{\omega }^2\right)}^2\right)Q},\hfill \end{array}$$

(41)

where Re(·) denotes the real part and

$$\begin{array}{ccc}\hfill P& =& 2{\eta }^2\mu m\left(3{\lambda }^2-{\mu }^2+{\omega }^2\right)sin\left(\mu t\right)+2{\eta }^2\lambda m\left(-{\lambda }^2+3{\mu }^2-{\omega }^2\right)cos\left(\mu t\right)\hfill \\ & & -2\alpha b\lambda \mu {\eta }^{\alpha +1}sin(\mu t-\alpha \varphi +\varphi )+\alpha b{\eta }^{\alpha +1}\left({\lambda }^2-{\mu }^2+{\omega }^2\right)cos(\mu t-\alpha \varphi +\varphi ),\hfill \\ \hfill Q& =& {\alpha }^2{b}^2{\eta }^{2\alpha }-4\alpha b\lambda m{\eta }^{\alpha +1}cos(\alpha \varphi -\varphi )+4\alpha b\mu m{\eta }^{\alpha +1}sin(\alpha \varphi -\varphi )\hfill \\ & & +4{\eta }^2{\lambda }^2{m}^2+4{\eta }^2{\mu }^2{m}^2.\hfill \end{array}$$

(42)

The Hankel integral in (39) is derived from the limit of an integral on the path $CD+C_{\epsilon }+BA$ in Figure 4. The integral on the small circle $C_{\epsilon }$ approaches zero as follows

$$\begin{array}{c}\hfill \underset{\epsilon \to 0^{+}}{lim}\frac{F\omega }{2\pi \mathrm{i}}{\int }_{C_{\epsilon }}\frac{e^{st}}{(s^2+{\omega }^2)C\left(s\right)}ds=\underset{\epsilon \to 0^{+}}{lim}\frac{F\omega }{2\pi }{\int }_{-\pi }^{\pi }\frac{e^{\epsilon t(cos\theta +\mathrm{i}sin\theta )}\epsilon e^{\mathrm{i}\theta }}{({\epsilon }^2{e}^{\mathrm{i}2\theta }+{\omega }^2)(m{\epsilon }^2{e}^{\mathrm{i}2\theta }+b{\epsilon }^{\alpha }{e}^{\mathrm{i}\alpha \theta }+k)}d\theta =0.\end{array}$$

Thus, the Hankel integral in (39) becomes

$$\begin{array}{ccc}\hfill S_H\left(t\right)& =& -\frac{F\omega }{2\pi \mathrm{i}}{\int }_0^{+\infty }\frac{e^{-rt}}{r^2+{\omega }^2}\left(\frac{1}{C\left(r{e}^{\mathrm{i}\pi }\right)}-\frac{1}{C\left(r{e}^{-\mathrm{i}\pi }\right)}\right)dr\hfill \\ & =& -\frac{F\omega }{\pi }{\int }_0^{+\infty }\frac{e^{-rt}}{r^2+{\omega }^2}\mathrm{Im}\left(\frac{1}{m{r}^2+b{r}^{\alpha }{e}^{\mathrm{i}\alpha \pi }+k}\right)dr\hfill \\ & =& \frac{F\omega bsin\left(\pi \alpha \right)}{\pi }{\int }_0^{\infty }\frac{r^{\alpha }{e}^{-rt}}{(r^2+{\omega }^2)V\left(r\right)}dr,\hfill \end{array}$$

(43)

where Im(·) denotes the imaginary part and

$$\begin{array}{c}\hfill V\left(r\right)=b^2{r}^{2\alpha }+2bcos\left(\pi \alpha \right)\left(k+m{r}^2\right)r^{\alpha }+k^2+2km{r}^2+m^2{r}^4.\end{array}$$

(44)

We checked whether $\alpha$ is taken as 1, the Hankel integral in (43) vanishes and $S\left(t\right)$ of the fractional case is simplified to that in Case i of Section 3.1. In Figure 5, the response $S\left(t\right)$ is shown for $m=1$, $b=2$, $k=3$, $F=1$, $\omega =2$ and for different values of $\alpha$. After excluding the steady periodic component $x_W\left(t\right)$, the curves of $S\left(t\right)-x_W\left(t\right)$ are shown in Figure 6. Under the zero initial value conditions, the transitions of responses to the steady state are displayed in Figure 5. From Figure 6, it can be observed that the decay in the case of $\alpha =1$ is more rapid than the fractional cases.

The two responses to initial values in (19) and (20) are computed in a similar manner. We list the results as follows

$$\begin{array}{ccc}\hfill S_0\left(t\right)& =& S_{0R}\left(t\right)+S_{0H}\left(t\right),\hfill \end{array}$$

(45)

$$\begin{array}{ccc}\hfill S_{0R}\left(t\right)& =& x_0\frac{2P_0{e}^{-\lambda t}}{Q},\hfill \end{array}$$

(46)

$$\begin{array}{ccc}\hfill S_{0H}\left(t\right)& =& \frac{x_{0}bksin\left(\pi \alpha \right)}{\pi }{\int }_0^{\infty }\frac{r^{\alpha -1}{e}^{-rt}}{V\left(r\right)}dr,\hfill \end{array}$$

(47)

$$\begin{array}{ccc}\hfill S_1\left(t\right)& =& S_{1R}\left(t\right)+S_{1H}\left(t\right),\hfill \end{array}$$

(48)

$$\begin{array}{ccc}\hfill S_{1R}\left(t\right)& =& x_1\frac{2P_1{e}^{-\lambda t}}{Q},\hfill \end{array}$$

(49)

$$\begin{array}{ccc}\hfill S_{1H}\left(t\right)& =& \frac{x_{1}bsin\left(\pi \alpha \right)}{\pi }{\int }_0^{\infty }\frac{r^{\alpha -2}\left(m{r}^2-\lceil \alpha -1\rceil (m{r}^2+k)\right)}{V\left(r\right)}{e}^{-rt}dr,\hfill \end{array}$$

(50)

where

$$\begin{array}{ccc}\hfill P_0& =& \left(\alpha b^2{\eta }^{2\alpha }+2{\eta }^2{m}^2{\lambda }^2+2{\eta }^2{m}^2{\mu }^2\right)cos\left(\mu t\right)-2bm{\eta }^{\alpha +1}\lambda cos(\mu t+\alpha \varphi -\varphi )\hfill \\ & & -\alpha bm{\eta }^{\alpha +1}\lambda cos(\mu t-\alpha \varphi +\varphi )+2bm{\eta }^{\alpha +1}\mu sin(\mu t+\alpha \varphi -\varphi )-\alpha bm{\eta }^{\alpha +1}\mu sin(\mu t-\alpha \varphi +\varphi ),\hfill \\ \hfill P_1& =& 2{\eta }^2\mu m^{2}sin\left(\mu t\right)-2{\eta }^2\lambda m^{2}cos\left(\mu t\right)+\lceil \alpha -1\rceil \alpha b^2{\eta }^{2\alpha -1}cos(\mu t-\varphi )+\alpha bm{\eta }^{\alpha +1}cos(\mu t-\alpha \varphi +\varphi )\hfill \\ & & -2\lceil \alpha -1\rceil b\lambda m{\eta }^{\alpha }cos(\mu t+(\alpha -2)\varphi )+2\lceil \alpha -1\rceil b\mu m{\eta }^{\alpha }sin(\mu t+(\alpha -2)\varphi ),\hfill \end{array}$$

Q and $V\left(r\right)$ are the same as in (42) and (44).

We checked that if $\alpha$ is taken as 1, the Hankel integrals in (47) and (50) vanish and the results of the fractional case for $S_0\left(t\right)$ and $S_1\left(t\right)$ are simplified to that in Case i of Section 3.1. Taking $m=1$, $b=2$ and $k=3$, the responses $S_0\left(t\right)$ in Figure 7 and the responses $S_1\left(t\right)$ in Figure 8 are shown for $x_0=1$ and $x_1=1$, respectively, and for different values of $\alpha$. From the two figures, more rapid decays in the case of $\alpha =1$ are observed than that in the fractional cases.

In Figure 8, there is an abnormal jump between the curves of $\alpha =1$ and $\alpha =1.35$. In fact, $S_1(t;\alpha )$ is not right continuous with respect to the order $\alpha$ at the integer $\alpha =1$. Specifically, $S_1(t;\alpha )$ at $\alpha =1$ is

$$\begin{array}{ccc}\hfill S_1(t;\alpha =1)& =& x_1{\mathcal{L}}^{-1}\left[\frac{m}{m{s}^2+bs+k}\right],\hfill \end{array}$$

however, the right limit of $S_1(t;\alpha )$ as $\alpha \to 1^{+}$ is

$$\begin{array}{ccc}\hfill S_1(t;1^{+})& =& x_1{\mathcal{L}}^{-1}\left[\frac{m+b{s}^{-1}}{m{s}^2+bs+k}\right],\hfill \end{array}$$

from (20). Compared with (25), the right limit can be calculated as

$$\begin{array}{ccc}\hfill S_1(t;1^{+})& =& x_1{\int }_0^t\frac{S_0(t;\alpha =1)}{x_0}dt,\hfill \end{array}$$

where $S_0(t;\alpha =1)$ is given in Equations (28), (31) and (34) in Section 3.1. In Figure 9, a more detailed demonstration of $S_1\left(t\right)$ for $1<\alpha <2$ is shown, where $\alpha$ is taken as 1.2, 1.4, 1.6, and 1.8, and the right limit $S_1(t;1^{+})$ is plotted by dot–dot–dash line.

The curves in Figure 6, Figure 7 and Figure 8 exhibit declines over time t. This can be made clear by looking into the asymptotic behaviors. First, we consider the Hankel integral contribution $S_H\left(t\right)$ in (43). Let $r=u/t$, and the integral in (43) is substituted as

$$\begin{array}{c}\hfill S_H\left(t\right)=\frac{F\omega bsin\left(\pi \alpha \right)}{\pi t}{\int }_0^{\infty }\frac{{(u/t)}^{\alpha }{e}^{-u}}{({(u/t)}^2+{\omega }^2)V(u/t)}du.\end{array}$$

(51)

As $t\to +\infty$, we have $V(u/t)\to k^2$, and hence the denominator of the integrand in (51) approaches to $k^2{\omega }^2$. Thus, we obtain the asymptotic representation for $S_H\left(t\right)$,

$$\begin{array}{c}\hfill S_H\left(t\right)\sim \frac{Fbsin\left(\pi \alpha \right)}{\pi k^2\omega }{t}^{-\alpha -1}{\int }_0^{\infty }{u}^{\alpha }{e}^{-u}du=\frac{Fbsin\left(\pi \alpha \right)\mathsf{\Gamma }(\alpha +1)}{\pi k^2\omega }{t}^{-\alpha -1},t\to +\infty .\end{array}$$

Considering the relationship $S\left(t\right)=x_W\left(t\right)+S_{RC}\left(t\right)+S_H\left(t\right)$, where $S_{RC}\left(t\right)$ decays in an oscillatory manner at a negative exponential rate, as a consequence, we obtain the asymptotic representation for the response to excitation

$$\begin{array}{c}\hfill S\left(t\right)-x_W\left(t\right)\sim \frac{Fbsin\left(\pi \alpha \right)\mathsf{\Gamma }(\alpha +1)}{\pi k^2\omega }{t}^{-\alpha -1},t\to +\infty .\end{array}$$

(52)

By a similar method, the two responses to the initial values have the asymptotic representations

$$\begin{array}{ccc}\hfill S_0\left(t\right)& \sim & \frac{x_{0}bsin\left(\pi \alpha \right)\mathsf{\Gamma }\left(\alpha \right)}{\pi k}{t}^{-\alpha },t\to +\infty ,\hfill \end{array}$$

(53)

$$\begin{array}{ccc}\hfill S_1\left(t\right)& \sim & \left\{\begin{array}{c}\frac{x_{1}bm sin\left(\pi \alpha \right)\mathsf{\Gamma }(\alpha +1)}{\pi k^2}{t}^{-\alpha -1},t\to +\infty ,0<\alpha <1,\hfill \\ -\frac{x_{1}b sin\left(\pi \alpha \right)\mathsf{\Gamma }(\alpha -1)}{\pi k}{t}^{-(\alpha -1)},t\to +\infty ,1<\alpha <2.\hfill \end{array}\right.\hfill \end{array}$$

(54)

By combining the above results in Equations (52)–(54), the asymptotic behavior of the solution of the initial value problem, (15) and (16), depends on the four cases of the initial values as follows.

Case A.$x_0=0$, $x_1=0$

$$\begin{array}{c}\hfill x\left(t\right)-x_W\left(t\right)\sim \frac{Fbsin\left(\pi \alpha \right)\mathsf{\Gamma }(\alpha +1)}{\pi k^2\omega }{t}^{-\alpha -1},t\to +\infty ,\end{array}$$

Case B.$x_0\ne 0$, $x_1=0$

$$\begin{array}{c}\hfill x\left(t\right)-x_W\left(t\right)\sim \frac{x_{0}bsin\left(\pi \alpha \right)\mathsf{\Gamma }\left(\alpha \right)}{\pi k}{t}^{-\alpha },t\to +\infty ,\end{array}$$

Case C.$x_0=0$, $x_1\ne 0$

$$\begin{array}{c}\hfill x\left(t\right)-x_W\left(t\right)\sim \left\{\begin{array}{c}\frac{b sin\left(\pi \alpha \right)\mathsf{\Gamma }(\alpha +1)}{\pi k^2}(\frac{F}{\omega }+x_{1}m)t^{-\alpha -1},t\to +\infty ,0<\alpha <1,\hfill \\ -\frac{x_{1}b sin\left(\pi \alpha \right)\mathsf{\Gamma }(\alpha -1)}{\pi k}{t}^{-(\alpha -1)},t\to +\infty ,1<\alpha <2,\hfill \end{array}\right.\end{array}$$

Case D.$x_0\ne 0$, $x_1\ne 0$

$$\begin{array}{c}\hfill x\left(t\right)-x_W\left(t\right)\sim \left\{\begin{array}{c}\frac{x_{0}b sin\left(\pi \alpha \right)\mathsf{\Gamma }\left(\alpha \right)}{\pi k}{t}^{-\alpha },t\to +\infty ,0<\alpha <1,\hfill \\ -\frac{x_{1}b sin\left(\pi \alpha \right)\mathsf{\Gamma }(\alpha -1)}{\pi k}{t}^{-(\alpha -1)},t\to +\infty ,1<\alpha <2.\hfill \end{array}\right.\end{array}$$

In summary, for the initial problem, (15) and (16), in the fractional case, we derive the response to excitation, $S\left(t\right)$, and the two responses to initial values, $S_0\left(t\right)$ and $S_1\left(t\right)$, and find that $S\left(t\right)$ is a superposition of a periodic oscillation $x_W\left(t\right)$, an exponentially decaying oscillation and a monotone recovery term in negative power law, and either $S_0\left(t\right)$ or $S_1\left(t\right)$ is a superposition of an exponentially decaying oscillation and a monotone recovery term in negative power law. The monotone recovery terms come from the Hankel integrals, and their attenuations in negative power laws dominate the exponential decays, which makes the fractional case different from the integer-order case. The steady periodic component $x_W\left(t\right)$ is just the response of the Weyl fractional oscillator in Section 2.

4. Conclusions

We consider the fractional oscillator equation with the fractional order $0<\alpha <2$ and the sinusoidal excitation, and conduct a comparative study of responses for the Weyl fractional derivative and the Caputo fractional derivative.

In Section 2, the fractional oscillator equation in the sense of Weyl is solved to be a steady periodic oscillation. The amplitude–frequency and the phase–frequency relations are analyzed with the variation of the order $\alpha$. In Section 3, the fractional oscillator equation in the sense of Caputo subject to initial conditions is solved. For comparison, we first consider the integer-order case of $\alpha =1$, and except for the steady periodic response, other solution components all decay in negative exponential rates. For the fractional case $\alpha \in (0,1)\cup (1,2)$, the response to excitation, $S\left(t\right)$, is a superposition of three parts: a steady periodic response in the sense of Weyl, an exponentially decaying oscillation, and a monotone recovery term in negative power law. For the two responses to initial values, $S_0\left(t\right)$ and $S_1\left(t\right)$, either of them is a superposition of an exponentially decaying oscillation and a monotone recovery term in negative power law. The monotone recovery terms come from Hankel integrals, and their attenuations in negative power laws dominate over the exponential decays, which makes the fractional case different from the integer-order case. Finally, the asymptotic behaviors of the solutions removing the steady periodic response are given for the four cases of the initial values.

The Weyl fractional derivative is suitable for describing the steady-state problem, and can directly lead to a steady periodic solution. The Caputo fractional derivative is applied to an initial value problem, the method of the Laplace transform yields results and the steady component of the solution is just the solution of the corresponding Weyl fractional differential equation.