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Two occurrences of fractional actions in nonlinear dynamics

  • Rami Ahmad El-Nabulsi EMAIL logo
Veröffentlicht/Copyright: 12. August 2021
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Abstract

Fractional theories have gained recently an increasing interest in dynamical systems since they offer some solutions to a number of puzzles in particular nonconservative and dissipative issues. Most of fractional dynamical theories are formulated by means of one occurrence of action that group kinetic energy and potential energy in one single package. In this work, we introduce a modified fractional dynamics based on the occurrence of two independent actions where fractional and nonfractional Euler–Lagrange equations are mixed together. We show that their combination divulge some properties that offer new insights in nonlinear dynamics. In particular, it was observed that a large family of solutions that could be used to model dissipative systems may be derived from the action with two occurrences of integrals. Moreover, damping systems may be obtained by means of simple Lagrangian functionals.

MSC 2010: 49S05; 34C15

Corresponding author: Rami Ahmad El-Nabulsi, Faculty of Science, Research Center for Quantum Technology, Chiang Mai University, Chiang Mai 50200, Thailand; Department of Physics and Materials Science, Faculty of Science, Chiang Mai University, Chiang Mai 50200, Thailand; and Mathematics and Physics Divisions, Athens Institute for Education and Research, 8 Valaoritou Street, Kolonaki, 10671, Athens, Greece, E-mail: el-nabulsi@atiner.gr

Funding source: Chiang Mai University 10.13039/501100002842

Acknowledgments

The author is indebted to the group of anonymous referees for their useful comments and valuable suggestions.

  1. Author contribution: The author has accepted responsibility for the entire content of this submitted manuscript and approved submission.

  2. Research funding: The author would like to thank Chiang Mai University for funding this research.

  3. Conflict of interest statement: The author declares that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

  4. Data availability statement: The author confirms the absence of sharing data.

Appendix A: Proof of Theorem 1

We give x(τ) an increment h(x) with x ɛ (τ) = x(τ) + ɛh(τ), ɛ ≪ 1 and h(x) is a differentiable function satisfying boundary conditions h(a) = h(t) = 0.

We let S 1 , ε = a t L 1 , ε ( x ̇ ε ( τ ) , x ε ( τ ) , τ ) d τ and S 2 , ε = a t L 2 , ε ( x ̇ ε ( τ ) , x ε ( τ ) , τ ) d τ . Having S ɛ = [S 1,ɛ , S 2,ɛ ] we have:

d S ε d ε = S ε ε = 0 s i n c e S d o e s n o t d e p e n d e x p l i c i t l y o n ε + S ε S 1 , ε d S 1 , ε d ε + S ε S 2 , ε d S 2 , ε d ε .

Here we have:

d S 1 , ε d ε = a t h ( τ ) L 1 x ε + d h d τ L 1 x ̇ ε d τ ,

d S 2 , ε d ε = 1 Γ ( α ) a t h ( τ ) ( t τ ) α 1 L 2 x ε + d h d τ ( t τ ) α 1 L 2 x ̇ ε d τ .

Using the following integration by parts for second terms of both integrals yield:

a t d h d τ L 1 x ̇ ε d τ = h ( τ ) L 1 x ̇ ε a t = 0 h ( a ) = h ( τ ) = 0 a b h ( τ ) d d t L 1 x ̇ ε d τ ,

a t d h d τ ( t τ ) α 1 L 1 x ̇ ε d τ = h ( τ ) ( t τ ) α 1 L 1 x ̇ ε a t = 0 h ( a ) = h ( τ ) = 0 a b h ( τ ) d d t ( t τ ) α 1 L 1 x ̇ ε d τ ,

we find:

d S 1 , ε d ε = a t h ( τ ) L 1 x ε d d t L 1 x ̇ ε d τ ,

d S 2 , ε d ε = 1 Γ ( α ) a t h ( τ ) ( t τ ) α 1 L 2 x ε d d t ( t τ ) α 1 L 1 x ̇ ε d τ .

After substituting into d S ε d ε , we get:

d S ε d ε = a t h ( τ ) S ε S 1 , ε L 1 x ε d d t L 1 x ̇ ε + 1 Γ ( α ) S ε S 2 , ε ( t τ ) α 1 L 2 x ε d d t ( t τ ) α 1 L 1 x ̇ ε d τ .

A necessary condition for S = [S 1, S 2] to have an extremum is that d S ε d ε = 0 . For ɛ = 0, we get:

a t h ( τ ) S ε S 1 , ε L 1 x ε d d t L 1 x ̇ ε + 1 Γ ( α ) S ε S 2 , ε ( t τ ) α 1 L 2 x ε d d t ( t τ ) α 1 L 1 x ̇ ε d τ = 0 .

For any increment h, the previous integral must be equal to zero, and therefore, we get:

L 1 x ε d d t L 1 x ̇ ε L 2 x ε d d t L 2 x ̇ ε + α 1 t τ L 2 x ̇ ε = ( t τ ) α 1 Γ ( α ) S ε S 2 , ε S ε S 1 , ε .

Appendix B: Proof of Theorem 2

Let x ɛ (τ) = x(τ) + ɛQ(τ), ɛ ≪ 1 where Q ( τ ) C 1 ( [ a , b ] ; R ) satisfying the boundary conditions Q(a) = Q(b) = 0 and that its left and right fractional derivatives are continuous. Using the chain rule, we calculate at the beginning the derivative of S = [S α , S β ] with respect to ɛ as follows:

d S d ε = S ε = 0 s i n c e S d o e s n o t d e p e n d e x p l i c i t l y o n ε + S S 1 d S 1 d ε + S S 2 d S 2 d ε .

Here:

d S 1 d ε = 1 Γ ( α ) a b Q ( τ ) ( t τ ) β 1 L 1 x ε + D τ α a Q ( τ ) ( t τ ) β 1 L 1 D τ α a x ε d τ ,

d S 2 d ε = 1 Γ ( β ) a b Q ( τ ) ( t τ ) γ 1 L 2 x ε + D τ β a Q ( τ ) ( t τ ) γ 1 L 2 D τ β a x ε d τ .

Using the integration by parts rule for fractional derivatives:

a b D τ α a Q ( τ ) ( t τ ) β 1 L 1 D τ α a x ε d τ = Q ( τ ) I b 1 α τ ( t τ ) β 1 L 1 D τ α a x ε a b = 0 Q ( a ) = Q ( b ) = 0 + a b Q ( τ ) D b α τ ( t τ ) β 1 L 1 D τ α a x ε d τ ,

a b D τ β a Q ( τ ) ( t τ ) γ 1 L 2 D τ β a x ε d τ = Q ( τ ) I b 1 β τ ( t τ ) γ 1 L 2 D τ β a x ε a b = 0 Q ( a ) = Q ( b ) = 0 + a b Q ( τ ) D b β τ ( t τ ) γ 1 L 2 D τ β a x ε d τ .

I b α t q = t b ( τ t ) α 1 q ( τ ) Γ ( α ) d τ , t < b , τ [ a , b ] is the right Riemann–Liouville fractional integral of order α, we can write:

d S 1 d ε = a b Q ( τ ) ( t τ ) β 1 L 1 x ε + τ D b α ( t τ ) β 1 L 1 D τ α a x ε d τ ,

d S 2 d ε = a b Q ( τ ) ( t τ ) γ 1 L 2 x ε + τ D b β ( t τ ) γ 1 L 2 a D τ β x ε d τ .

After substituting into d S d ε , we get:

d S d ε = S S 1 a b Q ( τ ) ( t τ ) β 1 L 1 x ε + τ D b α ( t τ ) β 1 L 1 D τ α a x ε d τ + S S 2 a b Q ( τ ) ( t τ ) γ 1 L 2 x ε + τ D b β ( t τ ) γ 1 L 2 a D τ β x ε d τ , = S S 1 a b Q ( τ ) ( t τ ) β 1 L 1 x ε + ( t τ ) β 1 D b α τ L 1 D τ α a x ε + τ D b α ( t τ ) β 1 L 1 D τ α a x ε d τ + S S 2 a b Q ( τ ) ( t τ ) γ 1 L 2 x ε + ( t τ ) γ 1 D b β τ L 2 a D τ β x ε + τ D b β ( t τ ) γ 1 L 2 a D τ β x ε d τ .

Using the following derivatives:

D b α τ ( t τ ) β 1 = ( 1 ) β + 1 ( b t ) β α 1 Γ ( 1 α ) , β > α ,

D b β τ ( t τ ) γ 1 = ( 1 ) γ + 1 ( b t ) γ β 1 Γ ( 1 β ) , γ > β ,

we find:

d S d ε = S S 1 a b Q ( τ ) ( t τ ) β 1 L 1 x ε + ( t τ ) β 1 D b α τ L 1 D τ α a x ε + ( 1 ) β + 1 ( b t ) β α 1 Γ ( 1 α ) L 1 D τ α a x ε d τ + S S 2 a b Q ( τ ) ( t τ ) α 1 L 2 x ε + ( t τ ) α 1 D b β τ L 2 a D τ β x ε + ( 1 ) γ + 1 ( b t ) γ β 1 Γ ( 1 β ) L 2 a D τ β x ε d τ .

A necessary condition for S = [S 1, S 2] to have an extremum is that d S d ε = 0 . For any increment Q we get:

S S 1 ( t τ ) β 1 L 1 x ε + ( t τ ) β 1 D b α τ L 1 D τ α a x ε + ( 1 ) β + 1 ( b t ) β α 1 Γ ( 1 α ) L 1 D τ α a x ε

+ S S 2 ( t τ ) α 1 L 2 x ε + ( t τ ) α 1 D b β τ L 2 a D τ β x ε + ( 1 ) γ + 1 ( b t ) γ β 1 Γ ( 1 β ) L 2 a D τ β x ε = 0 ,

or

L 1 x ε + τ D b α L 1 D τ α a x ε + ( 1 ) β + 1 ( b t ) β α 1 Γ ( 1 α ) ( t τ ) β 1 L 1 D τ α a x ε L 2 x ε + τ D b β L 2 a D τ β x ε + ( 1 ) γ + 1 ( b t ) γ β 1 Γ ( 1 β ) ( t τ ) α 1 L 2 a D τ β x ε = ( t τ ) α β S S 2 S S 1 .

Appendix C: Proof of Theorem 3

Let again x ɛ (τ) = x(τ) + ɛY(τ), ɛ ≪ 1 where Y ( τ ) C 1 ( [ a , b ] ; R ) satisfying the boundary conditions Y(a) = Y(b) = 0 and that its left and right fractional derivatives are continuous. From S = [S α , S β ] we have

d S d ε = S ε = 0 s i n c e S d o e s n o t d e p e n d e x p l i c i t l y o n ε + S S 1 d S 1 d ε + S S 2 d S 2 d ε ,

where

d S 1 d ε = 1 Γ ( α ) a b Y ( τ ) ( t τ ) β 1 L 1 x ε + a D τ α Y ( τ ) ( t τ ) β 1 L 1 D τ α a x ε + τ D b χ Y ( τ ) ( t τ ) β 1 L 1 τ D b χ q ε d τ ,

d S 2 d ε = 1 Γ ( γ ) a b Y ( τ ) ( t τ ) γ 1 L 2 x ε + a D τ η Y ( τ ) ( t τ ) γ 1 L 2 a D τ η x ε + τ D b ζ Y ( τ ) ( t τ ) γ 1 L 2 τ D b ζ q ε d τ .

Using the integration by parts rule for fractional derivatives:

a b D τ α a Y ( τ ) ( t τ ) β 1 L 1 D τ α a x ε d τ = Y ( τ ) I b 1 α τ ( t τ ) β 1 L 1 D τ α a x ε a b = 0 Y ( a ) = Y ( b ) = 0 + a b Y ( τ ) D b α τ ( t τ ) β 1 L 1 D τ α a x ε d τ ,

a b D b ζ τ Y ( τ ) ( t τ ) γ 1 L 2 a D τ ζ x ε d τ = Y ( τ ) I τ 1 ζ a ( t τ ) γ 1 L 2 a D τ ζ x ε a b = 0 Y ( a ) = Y ( b ) = 0 + a b Y ( τ ) τ D b ζ ( t τ ) γ 1 L 2 a D τ ζ x ε d τ ,

we can write:

d S 1 d ε = 1 Γ ( α ) a b Y ( τ ) ( t τ ) β 1 L 1 x ε + Y ( τ ) D b α τ ( t τ ) β 1 L 1 D τ α a x ε + Y ( τ ) τ D b χ ( t τ ) β 1 L 2 a D τ χ x ε d τ ,

d S 2 d ε = 1 Γ ( β ) a b Y ( τ ) ( t τ ) γ 1 L 2 x ε + Y ( τ ) τ D b ζ ( t τ ) γ 1 L 1 a D τ ζ x ε + Y ( τ ) τ D b ζ ( t τ ) γ 1 L 2 a D τ ζ x ε d τ .

Substituting into d S d ε , we get:

d S d ε = S S 1 a b Y ( τ ) ( t τ ) β 1 L 1 x ε + τ D b α ( t τ ) β 1 L 1 D τ α a x ε + τ D b χ ( t τ ) β 1 L 2 a D τ χ x ε d τ + S S 2 a b Y ( τ ) ( t τ ) γ 1 L 2 x ε + D b η τ ( t τ ) γ 1 L 2 a D τ η x ε + τ D b ζ ( t τ ) γ 1 L 2 a D τ ζ x ε d τ .

Since we have

D b α τ ( t τ ) β 1 L 1 D τ α a x ε = ( t τ ) β 1 D b α τ L 1 D τ α a x ε + τ D b α ( t τ ) β 1 L 1 D τ α a x ε ,

D b χ τ ( t τ ) β 1 L 2 a D τ χ x ε = ( t τ ) β 1 D b χ τ L 2 a D τ χ x ε + τ D b χ ( t τ ) β 1 L 2 a D τ χ x ε ,

D b η τ ( t τ ) γ 1 L 2 a D τ η x ε = ( t τ ) γ 1 D b η τ L 2 a D τ η x ε + D b η τ ( t τ ) γ 1 L 2 a D τ η x ε ,

D b ζ τ ( t τ ) γ 1 L 2 a D τ ζ x ε = ( t τ ) γ 1 D b ζ τ L 2 a D τ ζ x ε + τ D b ζ ( t τ ) γ 1 L 2 a D τ ζ x ε ,

with

D b α τ ( t τ ) β 1 = ( 1 ) β + 1 ( b t ) β α 1 Γ ( 1 α ) , β > α ,

D b χ τ ( t τ ) β 1 = ( 1 ) β + 1 ( b t ) β χ 1 Γ ( 1 χ ) , β > χ ,

D b η τ ( t τ ) γ 1 = ( 1 ) γ + 1 ( b t ) γ η 1 Γ ( 1 η ) , γ > η ,

D b ζ τ ( t τ ) γ 1 = ( 1 ) γ + 1 ( b t ) γ ζ 1 Γ ( 1 ζ ) , γ > ζ ,

we find:

d S d ε = S S 1 a b Y ( τ ) ( t τ ) β 1 L 1 x ε + ( t τ ) β 1 D b α τ L 1 D τ α a x ε + ( 1 ) β + 1 ( b t ) β α 1 Γ ( 1 α ) L 1 D τ α a x ε + ( t τ ) β 1 D b χ τ L 1 a D τ χ x ε + ( 1 ) β + 1 ( b t ) β χ 1 Γ ( 1 χ ) L 1 a D τ β x ε d τ + S S 2 a b Y ( τ ) ( t τ ) γ 1 L 2 x ε + ( t τ ) γ 1 D b η τ L 2 a D τ η x ε + D b η τ ( t τ ) γ 1 L 2 a D τ η x ε + ( t τ ) γ 1 D b ζ τ L 2 a D τ ζ x ε + ( 1 ) γ + 1 ( b t ) γ ζ 1 Γ ( 1 ζ ) L 2 a D τ ζ x ε d τ .

Since d S d ε = 0 is the necessary condition for S = [S 1, S 2] to have an extremum, then for any increment Y we get

S S 1 ( t τ ) β 1 L 1 x ε + ( t τ ) β 1 D b α τ L 1 D τ α a x ε + ( 1 ) β + 1 ( b t ) β α 1 Γ ( 1 α ) L 1 D τ α a x ε

+ ( t τ ) β 1 D b χ τ L 1 a D τ χ x ε + ( 1 ) β + 1 ( b t ) β χ 1 Γ ( 1 χ ) L 1 a D τ β x ε d τ

+ S S 2 ( t τ ) γ 1 L 2 x ε + ( t τ ) γ 1 D b η τ L 2 a D τ η x ε + ( 1 ) γ + 1 ( b t ) γ η 1 Γ ( 1 η ) L 2 a D τ η x ε

+ ( t τ ) γ 1 D b ζ τ L 2 a D τ ζ x ε + ( 1 ) γ + 1 ( b t ) γ ζ 1 Γ ( 1 ζ ) L 2 a D τ ζ x ε d τ ,

or

L 1 x ε + τ D b α L 1 D τ α a x ε + D b χ τ L 1 a D τ χ x ε + ( 1 ) β + 1 ( b t ) β α 1 Γ ( 1 α ) ( t τ ) β 1 L 1 D τ α a x ε + Γ ( 1 α ) ( b t ) α χ Γ ( 1 χ ) L 1 a D τ χ x ε L 2 x ε + D b η τ L 2 a D τ η x ε + D b ζ τ L 2 a D τ ζ x ε + ( 1 ) γ + 1 ( b t ) γ η 1 Γ ( 1 η ) ( t τ ) γ 1 L 2 a D τ η x ε + Γ ( 1 η ) ( b t ) η ζ Γ ( 1 ζ ) L 2 a D τ ζ x ε = ( t τ ) γ β S S 2 S S 1

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Received: 2020-12-15
Revised: 2021-06-09
Accepted: 2021-07-20
Published Online: 2021-08-12

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