Fractional Hamilton’s Canonical Equations and Poisson Theorem of Mechanical Systems with Fractional Factor

Abstract

Because of the nonlocal and nonsingular properties of fractional derivatives, they are more suitable for modelling complex processes than integer derivatives. In this paper, we use a fractional factor to investigate the fractional Hamilton’s canonical equations and fractional Poisson theorem of mechanical systems. Firstly, a fractional derivative and fractional integral with a fractional factor are presented, and a multivariable differential calculus with fractional factor is given. Secondly, the Hamilton’s canonical equations with fractional derivative are obtained under this new definition. Furthermore, the fractional Poisson theorem with fractional factor is presented based on the Hamilton’s canonical equations. Finally, two examples are given to show the application of the results.

1. Introduction

Fractional calculus, as a branch of mathematics, is one of the modern mathematical tools to solve complex problems in science and engineering technology [1,2,3,4,5,6,7,8,9]. The significance of fractional calculus is to extend the integer order calculus in the general sense to any fractional order, which is an extension of integer order calculus operation. As we all know, integer derivatives are local in nature, so these derivatives cannot accurately describe the problem, especially for processes with historical memory. Fractional equations can better model real-world problems than the integer order equations, and reveal the general, universal and deep physical laws and mathematical principles. For example, in viscoelastic theory, the stress-strain constitutive relations of various materials can be accurately described with few parameters in a wide range by using a fractional derivative [10]. In the control problem, a fractional controller not only expands the range of controller selection, but also has better robustness [11]. In the field of dissipative processes, it has been shown that the anomalous dissipative processes can be described by fractional differential equations, and fractional calculus has become a powerful tool to model anomalous dissipative phenomena [12]. In the study of dynamics, Lagrange mechanics and Hamiltonian mechanics can be established by using fractional derivatives, which write non-conservative forces directly into the Lagrange and Hamiltonian functions in terms of fractional derivatives [13,14,15,16]. In addition to the application of the above problems, fractional calculus has also been applied to Maxwell fluid [17,18], unsmooth boundary [19,20,21,22], porous medium [23,24], numerical calculations [25,26], etc. This is an active field, and various fractional models have been proposed by many researchers studying in it, such as fractional Biswas–Milovic model [27], fractional (1+1)-dimensional SRLW equation [28], fractional Riccati differential equation [29], time-fractional K(m, n) equation [30], fractional optical fiber Schrodinger models [31], etc. With the development of science and technology, the application of fractional calculus in various fields will become more and more important.

There are various definitions of fractional derivative and fractional integral, such as Riemann–Liouville (RL), Caputo, Riesz, Grunwald–Letnikov, He’s fractional derivative, etc. These definitions are popular among mathematicians and physicists [32,33,34]. However, previous studies are based on fractional derivatives defined in integral form, so it is very difficult to use these definitions. Recently, Fu et al. [35] used a new method of fractional factor to study the fractional order calculation of functions, and gave the equations of motion and cyclic integral equations of fractional holonomic mechanical systems. Then, Xiang et al. [36] further studied time and space fractional Schrodinger equation with fractional factor. Compared with other types of fractional derivatives, fractional derivatives with fractional factors are defined in limiting form rather than integral form, and have the same form and operational properties as integer derivatives. It becomes ordinary derivative when $\alpha =1$, and it is more convenient to use fractional factors to deal with fractional problems.

For fractional mechanics problems, the fractional differential equation obtained by modeling is generally complex and difficult to solve. The commonly used methods are numerical simulation, Fourier transform, Laplace transform, Adomian decomposition and path integral method. However, the methods of these fractional problems are far from satisfying the practical needs of engineering. Seeking the methods of fractional problems is a major scientific problem of mathematics, mechanics, physics and other basic disciplines. In this paper, we investigate fractional Hamilton systems using fractional factors. The Hamilton’s canonical equations with fractional factor are obtained according to the Hamilton principle. Further, we also study the Poisson theorem with fractional factor based on the Hamilton’s canonical equations. The first integral of the system is found by Poisson theorem and the solution of the fractional motion equation is obtained based on the first integral. This study shows that the fractional differential equation given by the fractional factor is a general differential equation containing fractional factors, so the fractional differential equation can be solved by the method of general differential equation.

2. Definitions and Properties of Fractional Derivative and Integral

In this section, a fractional factor is introduced to present the definitions of fractional derivative and integral. The properties of fractional derivatives and integrals used in the following sections are given. Moreover, we also study the multivariable differential calculus with fractional factor.

2.1. Definition and Properties of Fractional Derivative [35]

At first, let us introduce the fractional factor and fractional increment, and the definition of fractional derivative with fractional factor is as follows.

Suppose y = f (x) is defined in an open interval (a, b), then, for two distinct points x and x₀ in (a, b), we introduce the fractional increment ${\Delta }_{\alpha }x$ and fractional factor $e^{-(1-\alpha )x}$ which satisfy $\Delta x=x-x_0={\Delta }_{\alpha }x·e^{-(1-\alpha )x}$ (where $0<\alpha <1$). Hence, the increment of the function is $\Delta y=f(x_0+{\Delta }_{\alpha }x·e^{-(1-\alpha )x_0})-f(x_0)$. The difference quotient is

$$\frac{f\left(x_0+{\Delta }_{\alpha }x{e}^{-(1-\alpha )x_0}\right)-f\left(x_0\right)}{{\Delta }_{\alpha }x}.$$

Then, we keep x₀ fixed and study the behavior of this difference quotient as Δ_α x → 0.

Definition 1.

Let y = f(x) be defined in an open interval (a, b) and assume that x₀ ∈ (a, b). Then f is said to be fractional derivable at x₀, if and only if

$$\underset{{\Delta }_{\alpha }x\to 0}{\lim }\frac{f\left(x_0+{\Delta }_{\alpha }x{e}^{-(1-\alpha )x_0}\right)-f\left(x_0\right)}{{\Delta }_{\alpha }x}$$

(1)

exists. The limit, denoted by $f^{(\alpha )}(x)$, is called the fractional derivative of f at x₀. The definition of the fractional derivative can also be expressed as

$$f^{(\alpha )}(x_0)=\underset{{\Delta }_{\alpha }x\to 0}{\lim }\frac{f\left(x_0+{\Delta }_{\alpha }x{e}^{-(1-\alpha )x_0}\right)-f\left(x_0\right)}{{\Delta }_{\alpha }x}=\underset{{\Delta }_{\alpha }x\to 0}{\lim }\frac{\Delta y}{{\Delta }_{\alpha }x}=\frac{dy}{d_{\alpha }x}|{}_{x=x_0},$$

(2)

where $d_{\alpha }x=e^{(1-\alpha )x}dx$ and ${\Delta }_{\alpha }x=e^{(1-\alpha )x}\Delta x$. Hence, we can obtain

$$f^{(\alpha )}(x)=D_{\alpha }\left(f\right)=\frac{dy}{d_{\alpha }x}=e^{-(1-\alpha )x}\frac{dy}{dx}=e^{-(1-\alpha )x}{f}^{\prime }(x).$$

(3)

Let α∈(0, 1] and f, g be differentiable at a point x > 0. Then the following properties hold:

$$\left(1\right)D_{\alpha }\left(af+bg\right)=a{D}_{\alpha }f+b{D}_{\alpha }g,\mathrm{for}\mathrm{all}a,b\in \mathrm{R};$$

(4)

$$\left(2\right)D_{\alpha }(fg)=f{D}_{\alpha }(g)+g{D}_{\alpha }(f);$$

(5)

$$(3)D_{\alpha }\left(\frac{f}{g}\right)=\frac{g{D}_{\alpha }(f)-f{D}_{\alpha }(g)}{g^2}.$$

(6)

2.2. Definition and Properties of Fractional Integral [35]

Let the function f(x) be continuous in a closed interval [a, b]. We divide the interval by (n − 1) points x₁, x₂, …, x_n−1 into n equal or unequal subintervals with fractional lengths

$${\Delta }_{\alpha }{x}_i=(x_i-x_{i-1})e^{(1-\alpha )x_i}=\Delta x_i{e}^{(1-\alpha )x_i},\left(i=1,2,\cdots ,n\right),$$

(7)

where in addition, we put x₀ = a, x_n = b. For all ${\xi }_i\in [x_{i-1},x_i]$, we form the following sum:

$$I_{\alpha }=f\left({\xi }_1\right){\Delta }_{\alpha }{x}_1+f\left({\xi }_2\right){\Delta }_{\alpha }{x}_2+\cdots +f\left({\xi }_n\right){\Delta }_{\alpha }{x}_n={\displaystyle \sum _{i=1}^{n}f\left({\xi }_i\right){\Delta }_{\alpha }{x}_i}.$$

(8)

Further, taking limit for Δ_αx_i → 0 on both sides of Equation (8), we present the following:

Definition 2.

For any continuous function f(x) in a closed interval [a, b], the fractional integral is written by

$$I_{\alpha }^{ab}\left[f\left(x\right)\right]=\underset{\max \{\Delta x_i\}\to 0}{\lim }{\displaystyle \sum _{i=1}^{n}f({\xi }_i){\Delta }_{\alpha }{x}_i}={\displaystyle {\int }_{\alpha }^{ab}f\left(x\right)d_{\alpha }x},$$

(9)

$$I_{\alpha }\left[f\left(x\right)\right]={\displaystyle {\int }_{\alpha }f\left(x\right)d_{\alpha }x}.$$

(10)

Fractional integrals have the following properties:

(1)

Fractional integral can be expressed as usual form of integral. Since Δx = e^−(1−α)xΔ_α x, dx = e^−(1−α)x d_αx, we have

$${\displaystyle {\int }_{\alpha }f(x)}{d}_{\alpha }x={\displaystyle {\int }_{}^{}{e}^{(1-\alpha )x}f\left(x\right)dx}.$$

(11)

(2)

Constant λ can be settled outside of the fraction integral symbol

$${\displaystyle {\int }_{\alpha }\lambda f(x)}{d}_{\alpha }x=\lambda {\displaystyle {\int }_{\alpha }f(x)}{d}_{\alpha }x.$$

(12)

(3)

The integral of the algebraic sum of the functions is equal to the algebraic sum of the functions’ integral.

$${\displaystyle {\int }_{\alpha }\left[f(x)+\phi \left(x\right)\right]}{d}_{\alpha }x={\displaystyle {\int }_{\alpha }f(x)}{d}_{\alpha }x+{\displaystyle {\int }_{\alpha }\phi \left(x\right)}{d}_{\alpha }x.$$

(13)

(4)

Let f, g ∈ [a, b] → R be two functions such that f, g are differentiable. Then

$${\displaystyle {\int }_{\alpha }^{ab}f{D}_{\alpha }(g)}{d}_{\alpha }x=fg|{}_a^b-{\displaystyle {\int }_{\alpha }^{ab}g{D}_{\alpha }(f)}{d}_{\alpha }x.$$

(14)

(5)

If f is a continuous function in the domain on I_α or D_α, for x ≥ a, then we have the following formulae:

$$\frac{d}{d_{\alpha }x}{\displaystyle {\int }_{\alpha }f(x)d_{\alpha }x}=f\left(x\right),$$

(15)

$${\displaystyle {\int }_{\alpha }\frac{d}{d_{\alpha }x}}f\left(x\right)d_{\alpha }x=f\left(x\right)-f\left(a\right).$$

(16)

2.3. Multivariable Differential Calculus with Fractional Factor

Definition 3.

Suppose that f(x, y) is a function of two variables x and y on a region D. If y is held constant, say y = y₀, then f(x, y₀) is a function of the single variable x. Its fractional derivative of α order at x = x₀ is called the fractional partial derivative of f(x, y) with respect to x at (x₀, y₀) and is denoted by $\frac{\partial f}{{\partial }_{\alpha }x}{|}_{(x_0,y_0)}$ or f_αx(x₀, y₀). Thus,

$$\frac{\partial f}{{\partial }_{\alpha }x}{|}_{(x_0,y_0)}=\underset{{\Delta }_{\alpha }x\to 0}{\lim }\frac{{\Delta }_{x}f(x_0,y_0)}{{\Delta }_{\alpha }x}=\underset{{\Delta }_{\alpha }x\to 0}{\lim }\frac{f(x_0+e^{-(1-\alpha )x_0}{\Delta }_{\alpha }x,y_0)-f(x_0,y_0)}{{\Delta }_{\alpha }x}.$$

(17)

Similarly, the fractional partial derivative of f(x, y) with respect to y at (x₀, y₀) is denoted by $\frac{\partial f}{{\partial }_{\alpha }y}{|}_{(x_0,y_0)}$ or f_αy(x₀, y₀) and is given by

$$\frac{\partial f}{{\partial }_{\alpha }y}{|}_{(x_0,y_0)}=\underset{{\Delta }_{\alpha }y\to 0}{\lim }\frac{{\Delta }_{\alpha y}f(x_0,y_0)}{{\Delta }_{\alpha }y}=\underset{{\Delta }_{\alpha }y\to 0}{\lim }\frac{f(x_0,y_0+e^{-(1-\alpha )y_0}{\Delta }_{\alpha }y)-f(x_0,y_0)}{{\Delta }_{\alpha }y}.$$

(18)

Substituting the relation $\Delta x={\Delta }_{\alpha }x·e^{-(1-\alpha )x}$ into Equation (17), we have

$$\begin{gathered} \frac{\partial f}{{\partial }_{\alpha }x}{|}_{(x_0,y_0)}=\underset{{\Delta }_{\alpha }x\to 0}{\lim }\frac{f(x_0+e^{-(1-\alpha )x_0}{\Delta }_{\alpha }x,y_0)-f(x_0,y_0)}{{\Delta }_{\alpha }x} \\ =\underset{{\Delta }_{\alpha }x\to 0}{\lim }\frac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x}·e^{-(1-\alpha )x_0} \\ =\frac{\partial f}{\partial x}·e^{-(1-\alpha )x}{|}_{(x_0,y_0)}. \end{gathered}$$

(19)

If the fractional partial derivative exists for any (x, y) ∈ D, from Equation (19), we obtain

$$\frac{\partial f}{{\partial }_{\alpha }x}=\frac{\partial f}{\partial x}{e}^{-(1-\alpha )x},$$

(20)

$$\frac{\partial f}{{\partial }_{\alpha }y}=\frac{\partial f}{\partial y}{e}^{-(1-\alpha )y}.$$

(21)

In the integer order, the total differential of z = f(x, y) at (x, y) is given by

$$dz=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy.$$

(22)

Considering dx = e^−(1−α)xd_αx, we obtain

$$dz=\frac{\partial f}{\partial x}{e}^{-(1-\alpha )x}{d}_{\alpha }x+\frac{\partial f}{\partial y}{e}^{-(1-\alpha )y}{d}_{\alpha }y$$

(23)

Inserting Equations (20) and (21) into Equation (23), we have the fractional total differential

$$dz=\frac{\partial f}{{\partial }_{\alpha }x}{d}_{\alpha }x+\frac{\partial f}{{\partial }_{\alpha }y}{d}_{\alpha }y.$$

(24)

Theorem 1.

Let x = x(t) and y = y(t) be differentiable at t, and let z = f(x, y) be differentiable at (x(t), y(t)). Then z = f(x(t), y(t)) is differentiable at t, and

$$\frac{dz}{d_{\alpha }t}=\frac{\partial f}{\partial x}\frac{dx}{d_{\alpha }t}+\frac{\partial f}{\partial y}\frac{dy}{d_{\alpha }t}.$$

(25)

Proof. 

$$\frac{dz}{d_{\alpha }t}=\underset{{\Delta }_{\alpha }t\to 0}{\lim }\frac{f(x(t+e^{-(1-\alpha )t}{\Delta }_{\alpha }t),y)-f(x,y)}{x(t+e^{-(1-\alpha )t}{\Delta }_{\alpha }t)-x(t)}·\frac{x(t+e^{-(1-\alpha )t}{\Delta }_{\alpha }t)-x(t)}{{\Delta }_{\alpha }t}$$

$$+\underset{{\Delta }_{\alpha }t\to 0}{\lim }\frac{f(x,y(t+e^{-(1-\alpha )t}{\Delta }_{\alpha }t))-f(x,y)}{y(t+e^{-(1-\alpha )t}{\Delta }_{\alpha }t)-y(t)}·\frac{y(t+e^{-(1-\alpha )t}{\Delta }_{\alpha }t)-y(t)}{{\Delta }_{\alpha }t}$$

$$=\underset{\Delta t\to 0}{\lim }\frac{f(x+\Delta x,y)-f(x,y)}{x(t+\Delta t)-x(t)}·\underset{{\Delta }_{\alpha }t\to 0}{\lim }\frac{x(t+e^{-(1-\alpha )t}{\Delta }_{\alpha }t)-x(t)}{{\Delta }_{\alpha }t}$$

$$+\underset{\Delta t\to 0}{\lim }\frac{f(x,y+\Delta y)-f(x,y)}{y(t+\Delta t)-y(t)}·\underset{{\Delta }_{\alpha }t\to 0}{\lim }\frac{y(t+e^{-(1-\alpha )t}{\Delta }_{\alpha }t)-y(t)}{{\Delta }_{\alpha }t}$$

$$=\frac{\partial f}{\partial x}\frac{dx}{d_{\alpha }t}+\frac{\partial f}{\partial y}\frac{dy}{d_{\alpha }t}.$$

□

3. The Hamilton’s Canonical Equations with Fractional Derivative

In this section, we study the fractional Hamilton’s canonical equations based on the Hamilton principle. In [35], the exchanging relationship with respect to the isochronous variation and the fractional derivatives is the following:

$$D_{\alpha }\delta q=\delta D_{\alpha }q.$$

(26)

The fractional Hamilton principle can be written as

$$\delta S=\delta {\displaystyle {\int }_{\alpha }^{ab}L}(t,q_s,D_{\alpha }{q}_s)d_{\alpha }t=0,$$

(27)

where the terminal conditions are given below:

$$\delta q_s{|}_{t=a}=\delta q_s{|}_{t=b}=0$$

(28)

At first, let us introduce the generalized momenta as

$$p_{\alpha s}=\frac{\partial L}{\partial D_{\alpha }{q}_s}0<\alpha <1,s=(1,2,\dots ,n),$$

(29)

where the Lagrangian depends on the fractional time derivatives of coordinates in the form $L(t,q_s,D_{\alpha }{q}_s)$. Then, the Hamiltonian function depending on the fractional time derivatives can be expressed as

$$H=-L+{\displaystyle \sum _{s=1}^n{p}_{\alpha s}{D}_{\alpha }{q}_s}.$$

(30)

Substituting Equation (30) into Equation (27), we have

$${\displaystyle {\int }_{\alpha }^{ab}\delta ({\displaystyle \sum _{s=1}^n{p}_{\alpha s}{D}_{\alpha }{q}_s}-H)}{d}_{\alpha }t=0.$$

(31)

Then, we obtain

$${\displaystyle {\int }_{\alpha }^{ab}{\displaystyle \sum _{s=1}^n(D_{\alpha }{q}_s\delta p_{\alpha s}+p_{\alpha s}\delta D_{\alpha }{q}_s-\frac{\partial H}{\partial q_s}\delta q_s-\frac{\partial H}{\partial p_{\alpha s}}\delta p_{\alpha s})}}{d}_{\alpha }t=0.$$

(32)

Taking the commutable relation (26), we have

$${\displaystyle {\int }_{\alpha }^{ab}{\displaystyle \sum _{s=1}^n(D_{\alpha }{q}_s\delta p_{\alpha s}+p_{\alpha s}{D}_{\alpha }\delta q_s-\frac{\partial H}{\partial q_s}\delta q_s-\frac{\partial H}{\partial p_{\alpha s}}\delta p_{\alpha s})}}{d}_{\alpha }t=0.$$

(33)

According to Equation (5), we obtain

$$D_{\alpha }(p_{\alpha s}\delta q_s)=p_{\alpha s}{D}_{\alpha }\delta q_s+\delta q_s{D}_{\alpha }{p}_{\alpha s},$$

(34)

Substituting Equation (34) into Equation (33), we derive

$${\displaystyle {\int }_{\alpha }^{ab}{\displaystyle \sum _{s=1}^n[D_{\alpha }{q}_s\delta p_{\alpha s}+D_{\alpha }(p_{\alpha s}\delta q_s)-\delta q_s{D}_{\alpha }{p}_{\alpha s}-\frac{\partial H}{\partial q_s}\delta q_s-\frac{\partial H}{\partial p_{\alpha s}}\delta p_{\alpha s}]}}{d}_{\alpha }t=0.$$

(35)

Taking notice of Equation (16), we have

$${\displaystyle \sum _{s=1}^n{p}_{\alpha s}\delta q_s{|}_a^b}+{\displaystyle {\int }_{\alpha }^{ab}{\displaystyle \sum _{s=1}^n[(D_{\alpha }{q}_s-\frac{\partial H}{\partial p_{\alpha s}})\delta p_{\alpha s}-(D_{\alpha }{p}_{\alpha s}+\frac{\partial H}{\partial q_s})\delta q_s]}}{d}_{\alpha }t=0.$$

(36)

Using the terminal condition (28), we obtain

$$p_{\alpha s}\delta q_s{|}_a^b=0.$$

(37)

Then, we have

$${\displaystyle {\int }_{\alpha }^{ab}{\displaystyle \sum _{s=1}^n[(D_{\alpha }{q}_s-\frac{\partial H}{\partial p_{\alpha s}})\delta p_{\alpha s}-(D_{\alpha }{p}_{\alpha s}+\frac{\partial H}{\partial q_s})\delta q_s]}}{d}_{\alpha }t=0.$$

(38)

Calculating the derivative of Equation (30) with respect to $p_{\alpha }$, we acquire

$$\frac{\partial H}{\partial p_{\alpha s}}=D_{\alpha }{q}_s.$$

(39)

Substituting Equation (39) into Equation (38) and considering that $\delta q_s$ (s = 1, 2, …, n) are independent, we obtain

$$\frac{\partial H}{\partial q_s}=-D_{\alpha }{p}_{\alpha s}.$$

(40)

According to Equation (3), we can change Equations (39) and (40) into the following form:

$$\frac{\partial H}{\partial p_{\alpha s}}=e^{-(1-\alpha )t}{\dot{q}}_s$$

(41)

and

$$\frac{\partial H}{\partial q_s}=-e^{-(1-\alpha )t}{\dot{p}}_{\alpha s}.$$

(42)

Equations (39) and (40), as well as Equations (41) and (42), are all called Hamilton’s canonical equations with fractional factor. Obviously, they are consistent with normal equations when $\alpha =1$.

4. The Fractional Poisson Theorem with Fractional Factor of Hamilton Systems

In this section, we study the fractional Poisson theorem with fractional factor of Hamilton systems. Firstly, let us introduce the Poisson conditions and Poisson brackets. Suppose that the first integral of the Hamilton equations exists in the following form:

$$f(q_s,p_{\alpha s},t)=C(s=1,2,\dots ,n).$$

(43)

Then, calculating the fractional total derivative of Equation (43) with respect to t, we have

$$\frac{\partial f}{{\partial }_{\alpha }t}+{\displaystyle \sum _{s=1}^n(\frac{\partial f}{\partial q_s}{D}_{\alpha }{q}_s+\frac{\partial f}{\partial p_{\alpha s}}{D}_{\alpha }{p}_{\alpha s})=0}.$$

(44)

According to Equation (39) and (40), we obtain the Poisson condition of the first integral

$$\frac{\partial f}{{\partial }_{\alpha }t}+{\displaystyle \sum _{s=1}^n(\frac{\partial f}{\partial q_s}\frac{\partial H}{\partial p_{\alpha s}}-\frac{\partial f}{\partial p_{\alpha s}}\frac{\partial H}{\partial q_s})=0}.$$

(45)

From Equations (41) and (42), it can also be written as

$$\frac{\partial f}{\partial t}{e}^{-(1-\alpha )t}+{\displaystyle \sum _{s=1}^n(\frac{\partial f}{\partial q_s}{e}^{-(1-\alpha )t}\stackrel{}{{\dot{q}}_s}-\frac{\partial f}{\partial p_{\alpha s}}{e}^{-(1-\alpha )t}\stackrel{}{{\dot{p}}_{\alpha s}})=0}.$$

(46)

Obviously, it is the necessary and sufficient condition.

Then, we define the Poisson bracket of two given functions φ and ψ in the following form:

$$(\phi ,\psi )={\displaystyle \sum _{s=1}^n(\frac{\partial \phi }{\partial q_s}\frac{\partial \psi }{\partial p_{\alpha s}}-\frac{\partial \phi }{\partial p_{\alpha s}}\frac{\partial \psi }{\partial q_s})}.$$

(47)

Thus, the Poisson condition becomes

$$\frac{\partial f}{{\partial }_{\alpha }t}+(f,H)=0$$

(48)

or

$$\frac{\partial f}{\partial t}{e}^{-(1-\alpha )t}+(f,H)=0.$$

(49)

We can easily prove that the following properties hold:

$$\left(\mathrm{i}\right)(\phi ,\phi )=0;$$

(50)

$$\left(\mathrm{ii}\right)(\phi ,\psi )=-(\psi ,\phi );$$

(51)

$$\left(\mathrm{iii}\right)(\phi ,-\psi )=-(\phi ,\psi ),(-\phi ,\psi )=-(\phi ,\psi );$$

(52)

$$(\mathrm{iv})\frac{\partial }{{\partial }_{\alpha }t}(\phi ,\psi )=(\frac{\partial \phi }{{\partial }_{\alpha }t},\psi )+(\phi ,\frac{\partial \psi }{{\partial }_{\alpha }t}).$$

(53)

Secondly, we study the Jacobi’s identity with fractional factor. The fractional composite Poisson bracket can be written as

$$(f,(\phi ,\psi ))={\displaystyle \sum _{s=1}^n(\frac{\partial f}{\partial q_s}\frac{\partial (\phi ,\psi )}{\partial p_{\alpha s}}-\frac{\partial f}{\partial p_{\alpha s}}\frac{\partial (\phi ,\psi )}{\partial q_s})}.$$

(54)

If f₁(t, q, p_α) and f₂(t, q, p_α) satisfy the condition

$$(f_1,f_2)\equiv 0,$$

(55)

we call that f₁ and f₂ are internal rotation. Then, the Jacobi’s identity is given below:

$$(f,(\phi ,\psi ))+(\phi ,(\psi ,f))+(\psi ,(f,\phi ))=0.$$

(56)

Now, let us prove the identity. At first, we discuss the system with one degree q, and then we have

$$(f,(\phi ,\psi ))=\frac{\partial f}{\partial q}·\frac{\partial (\phi ,\psi )}{\partial p_{\alpha }}-\frac{\partial f}{\partial p_{\alpha }}·\frac{\partial (\phi ,\psi )}{\partial q}.$$

(57)

Since

$$\frac{\partial (\phi ,\psi )}{\partial p_{\alpha }}=\frac{{\partial }^2\phi }{\partial q\partial p_{\alpha }}\frac{\partial \psi }{\partial p_{\alpha }}+\frac{{\partial }^2\psi }{\partial p_{\alpha }{}^2}\frac{\partial \phi }{\partial q}-\frac{{\partial }^2\phi }{\partial p_{\alpha }{}^2}\frac{\partial \psi }{\partial q}-\frac{{\partial }^2\psi }{\partial p_{\alpha }\partial q}\frac{\partial \phi }{\partial p_{\alpha }}$$

(58)

and

$$\frac{\partial (\phi ,\psi )}{\partial q}=\frac{{\partial }^2\phi }{\partial q^2}\frac{\partial \psi }{\partial p_{\alpha }}+\frac{{\partial }^2\psi }{\partial p_{\alpha }\partial q}\frac{\partial \phi }{\partial q}-\frac{{\partial }^2\phi }{\partial p_{\alpha }\partial q}\frac{\partial \psi }{\partial q}-\frac{{\partial }^2\psi }{\partial q^2}\frac{\partial \phi }{\partial p_{\alpha }},$$

(59)

we obtain

$$\begin{gathered} (f,(\phi ,\psi ))=\frac{\partial f}{\partial q}\left[\frac{{\partial }^2\phi }{\partial q\partial p_{\alpha }}\frac{\partial \psi }{\partial p_{\alpha }}+\frac{{\partial }^2\psi }{\partial p_{\alpha }{}^2}\frac{\partial \phi }{\partial q}-\frac{{\partial }^2\phi }{\partial p_{\alpha }{}^2}\frac{\partial \psi }{\partial q}-\frac{{\partial }^2\psi }{\partial p_{\alpha }\partial q}\frac{\partial \phi }{\partial p_{\alpha }}\right] \\ -\frac{\partial f}{\partial p_{\alpha }}\left[\frac{{\partial }^2\phi }{\partial q^2}\frac{\partial \psi }{\partial p_{\alpha }}+\frac{{\partial }^2\psi }{\partial p_{\alpha }\partial q}\frac{\partial \phi }{\partial q}-\frac{{\partial }^2\phi }{\partial p_{\alpha }\partial q}\frac{\partial \psi }{\partial q}-\frac{{\partial }^2\psi }{\partial q^2}\frac{\partial \phi }{\partial p_{\alpha }}\right]. \end{gathered}$$

(60)

Similarly, we can calculate that

$$\begin{gathered} (\phi ,(\psi ,f))=\frac{\partial \phi }{\partial q}\left[\frac{{\partial }^2\psi }{\partial q\partial p_{\alpha }}\frac{\partial f}{\partial p_{\alpha }}+\frac{{\partial }^{2}f}{\partial p_{\alpha }{}^2}\frac{\partial \psi }{\partial q}-\frac{{\partial }^2\psi }{\partial p_{\alpha }{}^2}\frac{\partial f}{\partial q}-\frac{{\partial }^{2}f}{\partial p_{\alpha }\partial q}\frac{\partial \psi }{\partial p_{\alpha }}\right] \\ -\frac{\partial \phi }{\partial p_{\alpha }}\left[\frac{{\partial }^2\psi }{\partial q^2}\frac{\partial f}{\partial p_{\alpha }}+\frac{{\partial }^{2}f}{\partial p_{\alpha }\partial q}\frac{\partial \psi }{\partial q}-\frac{{\partial }^2\psi }{\partial p_{\alpha }\partial q}\frac{\partial f}{\partial q}-\frac{{\partial }^{2}f}{\partial q^2}\frac{\partial \psi }{\partial p_{\alpha }}\right] \end{gathered}$$

(61)

and

$$\begin{gathered} (\psi ,(f,\phi ))=\frac{\partial \psi }{\partial q}\left[\frac{{\partial }^{2}f}{\partial q\partial p_{\alpha }}\frac{\partial \phi }{\partial p_{\alpha }}+\frac{{\partial }^2\phi }{\partial p_{\alpha }{}^2}\frac{\partial f}{\partial q}-\frac{{\partial }^{2}f}{\partial p_{\alpha }{}^2}\frac{\partial \phi }{\partial q}-\frac{{\partial }^2\phi }{\partial p_{\alpha }\partial q}\frac{\partial f}{\partial p_{\alpha }}\right] \\ -\frac{\partial \psi }{\partial p_{\alpha }}\left[\frac{{\partial }^{2}f}{\partial q^2}\frac{\partial \phi }{\partial p_{\alpha }}+\frac{{\partial }^2\phi }{\partial p_{\alpha }\partial q}\frac{\partial f}{\partial q}-\frac{{\partial }^{2}f}{\partial p_{\alpha }\partial q}\frac{\partial \phi }{\partial q}-\frac{{\partial }^2\phi }{\partial q^2}\frac{\partial f}{\partial p_{\alpha }}\right]. \end{gathered}$$

(62)

When we add Equations (60)–(62) together, the sum is zero. That is, for one degree, the Jacobi’s identity is true. From the above calculation, we can observe that the composite Poisson brackets contain a second partial derivative of a function and the coefficients of the partial derivative are just opposite, so the sum of the three composite Poisson brackets is zero. The Jacobi’s identity holds for the general conditions.

Finally, let us discuss the Poisson theorem with fractional factor using the previous conclusions. The Poisson theorem can be expressed as follows:

Theorem 2.

If f₁ and f₂ are two first integrals of the Hamilton equations, that is,

$$f_1(t,q,p_{\alpha })=C_1,f_2(t,q,p_{\alpha })=C_2.$$

(63)

In addition, f₁ and f₂ are not internal rotation. Then, (f₁, f₂) is also the first integral of the Hamilton equations.

Now let us prove the theorem. Since f₁ and f₂ are the first integrals, we have

$$\frac{\partial f_1}{{\partial }_{\alpha }t}+(f_1,H)=0,\frac{\partial f_2}{{\partial }_{\alpha }t}+(f_2,H)=0.$$

(64)

Then, we study the Jacobi’s identity of the functions H, f₁ and f₂, i.e.,

$$(H,(f_1,f_2))+(f_1,(f_2,H))+(f_2,(H,f_1))=0.$$

(65)

Substituting Equation (64) into Equation (65) and taking Equation (51) into consideration, we obtain

$$(H,(f_1,f_2))+(f_1,-\frac{\partial f_2}{{\partial }_{\alpha }t})+(f_2,\frac{\partial f_1}{{\partial }_{\alpha }t})=0.$$

(66)

Then, from the properties of (51)–(53), we have

$$\frac{\partial (f_1,f_2)}{{\partial }_{\alpha }t}+((f_1,f_2),H)=0.$$

(67)

As f₁ and f₂ are not internal rotation, we derive $(f_1,f_2)\ne 0$, and then $(f_1,f_2)$ is another function $(f_1,f_2)=f_3(t,q,p_{\alpha })$. Therefore, it meets the Poisson condition

$$\frac{\partial f_3}{{\partial }_{\alpha }t}+(f_3,H)=0,$$

(68)

which indicates that $f_3=C_3$ is also the first integral.

According to the previous results, we can easily prove the following theorems.

Theorem 3.

If the Hamiltonian H does not contain the time variable t, then H is the first integral of the system.

Theorem 4.

If f (t, q, p_α) is the first integral of the Hamilton equations containing the time variable t, and the Hamiltonian H does not contain the time variable t, then $\frac{\partial f}{{\partial }_{\alpha }t},\frac{{\partial }^{2}f}{{\partial }_{\alpha }{t}^2},\cdots$ are all the first integrals of the systems.

Theorem 5.

If f (t, q, p_α) is the first integral of the Hamilton equations containing the variable q (or p_α), and the Hamiltonian H does not contain the corresponding variable q (or p_α), then $\frac{\partial f}{{\partial }_{\alpha }q},\frac{{\partial }^{2}f}{{\partial }_{\alpha }{q}^2},$⋯$(\mathrm{or}\frac{\partial f}{{\partial }_{\alpha }{p}_{\alpha }},\frac{{\partial }^{2}f}{{\partial }_{\alpha }{p}_{\alpha }{}^2},\cdots )$ are all the first integrals of the systems.

5. Examples

Example 1.

Considering a system with two free degrees, the Lagrangian is given by

$$L=\frac{1}{2}{D}_{\alpha }{}^2{q}_1+\frac{1}{2}{D}_{\alpha }{}^2{q}_2{\sin }^2{q}_1+a\cos q_1.$$

(69)

Study its Hamiltonian function and establish the canonical equations.

In terms of Equation (29), we have the generalized momenta

$$p_{\alpha 1}=\frac{\partial L}{\partial D_{\alpha }{q}_1}=D_{\alpha }{q}_1,$$

(70)

$$p_{\alpha 2}=\frac{\partial L}{\partial D_{\alpha }{q}_2}={\sin }^2{q}_1{D}_{\alpha }{q}_2.$$

(71)

Then, we construct the Hamiltonian function in the following form:

$$H=-L+p_{\alpha 1}{D}_{\alpha }{q}_1+p_{\alpha 2}{D}_{\alpha }{q}_2.$$

(72)

Substituting Equations (69)–(71) into Equation (72), we obtain the Hamiltonian function

$$H=\frac{1}{2}(p_{\alpha 1}{}^2+\frac{p_{\alpha 2}{}^2}{{\sin }^2{q}_1})-a\cos q_1.$$

(73)

According to Equations (39) and (40), we have the canonical equations

$$D_{\alpha }{q}_1=p_{\alpha 1},$$

(74)

$$D_{\alpha }{q}_2=\frac{p_{\alpha 2}}{{\sin }^2{q}_1},$$

(75)

$$-D_{\alpha }{p}_{\alpha 1}=-\frac{p_{\alpha 2}{}^2\cos q_1}{{\sin }^3{q}_1}+a\sin q_1,$$

(76)

$$-D_{\alpha }{p}_{\alpha 2}=0.$$

(77)

They can also be written in the form with fractional factor as

$$e^{-(1-\alpha )t}\stackrel{}{{\dot{q}}_1}=p_{\alpha 1},$$

(78)

$$e^{-(1-\alpha )t}\stackrel{}{{\dot{q}}_2}=\frac{p_{\alpha 2}}{{\sin }^2{q}_1},$$

(79)

$$e^{-(1-\alpha )t}\stackrel{}{{\dot{p}}_{\alpha 1}}=\frac{p_{\alpha 2}{}^2\cos q_1}{{\sin }^3{q}_1}-a\sin q_1,$$

(80)

$$-e^{-(1-\alpha )t}\stackrel{}{{\dot{p}}_{\alpha 2}}=0.$$

(81)

Example 2.

The Hamiltonian function of a mechanical system with fractional derivative is given by

$$H=p_{\alpha 1}{p}_{\alpha 2}+q_1{q}_2.$$

(82)

Give the proof that the functions ${\phi }_1=p_{\alpha 1}{}^2+q_2{}^2,{\phi }_2=p_{\alpha 2}{}^2+q_1{}^2,{\phi }_3=({\phi }_1,{\phi }_2)$ are the first integrals of the system and try to study the motion equations of the mechanical system.

Proof. 

At first, let us calculate the Poisson brackets

$$\begin{gathered} ({\phi }_1,H)=\frac{\partial {\phi }_1}{\partial q_1}\frac{\partial H}{\partial p_{\alpha 1}}-\frac{\partial {\phi }_1}{\partial p_{\alpha 1}}\frac{\partial H}{\partial q_1}+\frac{\partial {\phi }_1}{\partial q_2}\frac{\partial H}{\partial p_{\alpha 2}}-\frac{\partial {\phi }_1}{\partial p_{\alpha 2}}\frac{\partial H}{\partial q_2} \\ =-2p_{\alpha 1}{q}_2+2p_{\alpha 1}{q}_2 \\ =0, \end{gathered}$$

(83)

$$\begin{gathered} ({\phi }_2,H)=\frac{\partial {\phi }_2}{\partial q_1}\frac{\partial H}{\partial p_{\alpha 1}}-\frac{\partial {\phi }_2}{\partial p_{\alpha 1}}\frac{\partial H}{\partial q_1}+\frac{\partial {\phi }_2}{\partial q_2}\frac{\partial H}{\partial p_{\alpha 2}}-\frac{\partial {\phi }_2}{\partial p_{\alpha 2}}\frac{\partial H}{\partial q_2} \\ =2p_{\alpha 2}{q}_1-2p_{\alpha 2}{q}_1 \\ =0. \end{gathered}$$

(84)

Since $\frac{\partial {\phi }_1}{{\partial }_{\alpha }t}=0,\frac{\partial {\phi }_2}{{\partial }_{\alpha }t}=0$, we derive that ${\phi }_1$ and ${\phi }_2$ meet the Poisson conditions, which indicates that ${\phi }_1$ and ${\phi }_2$ are the first integrals. Then, according to the Poisson theorem, we acquire that ${\phi }_3=({\phi }_1,{\phi }_2)$ is also the first integral.

Now, let us study the motion equations of the system. Since φ₁ and φ₂ are the first integrals, we assume that

$$p_{\alpha 1}{}^2+q_2{}^2={\beta }_1,$$

(85)

$$p_{\alpha 2}{}^2+q_1{}^2={\beta }_2,$$

(86)

where β₁ and β₂ are constant. According to Equations (41) and (42), we acquire the Hamilton’s canonical equations

$$p_{\alpha 2}=e^{-(1-\alpha )t}{\dot{q}}_1,$$

(87)

$$p_{\alpha 1}=e^{-(1-\alpha )t}{\dot{q}}_2,$$

(88)

$$q_2=-e^{-(1-\alpha )t}{\dot{p}}_{\alpha 1},$$

(89)

$$q_1=-e^{-(1-\alpha )t}{\dot{p}}_{\alpha 2}.$$

(90)

Inserting Equation (87) into Equation (86) and Equation (88) into Equation (85), we have

$$e^{-2(1-\alpha )t}{\dot{q}}_1{}^2+q_1{}^2={\beta }_1,$$

(91)

$$e^{-2(1-\alpha )t}{\dot{q}}_2{}^2+q_2{}^2={\beta }_2.$$

(92)

Calculating the Equations (91) and (92), we obtain the equations of motion of the system as follows:

$$q_1=\sqrt{{\beta }_1}\sin (\frac{1}{1-\alpha }{e}^{(1-\alpha )t}+C_1),$$

(93)

$$q_2=\sqrt{{\beta }_2}\sin (\frac{1}{1-\alpha }{e}^{(1-\alpha )t}+C_2),$$

(94)

where C₁ and C₂ are constants. □

6. Conclusions

Fractional calculus is widely used to describe the dynamics of complex systems. This paper introduces fractional calculus with a new method of fractional factor. Using this fractional factor, we further study the fractional mechanical systems and some significant results are given in the following:

(1)

The form and calculation rules of the fractional derivative with fractional factor are similar to those of the integer derivative, and it becomes an ordinary derivative when $\alpha =1$.

(2)

Fractional Hamilton’s canonical equation is a common Hamilton’s canonical equation with fractional factor.

(3)

Fractional Poisson theorem is acquired by the fractional factor.

(4)

All fractional derivatives can be expressed as ordinary derivatives with fractional factor, and all the fractional differential equations can be transformed into general differential equations with fractional factor. Therefore, we can solve the fractional differential equations with usual solutions.

(5)

A new method for solving the motion equation of fractional Hamilton system is presented. We can easily establish the fractional Hamilton’s equation by using our results, the first integral of the fractional Hamilton system can be obtained by using the fractional Poisson theorem, and then the solution of the fractional motion equation can be given according to the first integral.

In this paper, the fundamental equations of fractional mechanical systems are studied by using fractional factors, and a series of meaningful results are obtained. However, only fractional holonomic systems are considered in this paper. Using this fractional factor, we can further study the fractional nonholonomic systems, the fractional electromechanical coupling systems, fractional anomalous diffusion problems and engineering problems with fractional derivatives.