We present a new discretization for the Hadamard fractional derivative, that simplifies
the computations. We then apply the method to solve a fractional differential equation
and a fractional variational problem with dependence on the Hadamard fractional
derivative

Almeida, Ricardo

Bastos, Nuno

Mathematical Sciences and Applications E-Notes

arXiv

Repositório Institucional da Universidade de Aveiro

MATHEMATICAL SCIENCES AND APPLICATIONS E-NOTES4 (1) 31-39 (2016) c©MSAENA Discretization of the Hadamard fractionalderivativeRicardo Almeida* and Nuno R. O. Bastos(Communicated by Murat TOSUN)AbstractWe present a new discretization for the Hadamard fractional derivative, that simplifiesthe computations. We then apply the method to solve a fractional differential equationand a fractional variational problem with dependence on the Hadamard fractionalderivative.Keywords: fractional calculus, discretization methods.AMS Subject Classification (2010): Primary: 26A33; Secondary: 49M25, 49M25.*Corresponding author1. IntroductionSimilarly to ordinary calculus, we can find in the literature distinct definitions for fractionalderivatives and for fractional integrals, which are generalizations of the integer-order derivativesand multiple integrals, respectively. The most common ones and consequently more studiedare the Riemann–Liouville, Caputo and Grünwald–Letnikov definitions. We deal in this paperwith the Hadamard fractional derivative, introduced in [4]. Recently, it has call the attention ofresearchers and numerous results have appeared, with an extensive study of properties of suchkind of operators [1, 2, 8]. For recent results we suggest [5, 10, 11].Due the complexity of solving equations involving fractional operators, in most cases isimpossible to determine the exact solution and so numerical methods are used to determine anapproximated solution of the problem. This is an emerging field, and we can find already in theliterature several methods to deal with these problems, at least for the most common fractionalderivative types. For the Hadamard fractional derivative, we mention the recent paper [9], wherethe fractional operator is replaced by a finite sum involving only integer-order derivatives of thefunction. Replacing the fractional derivative by this sum, we rewrite the initial problem in termsof integer-order derivatives and thus we are able to apply classical known methods. In [3] anotherapproximation formula is obtained, using also integer-order derivatives only. The disadvantage isReceived : 25–September–2015, Accepted : 14–January–2016This article is the written version of author’s plenary talk delivered on September 03- August 31, 2015 at 4th International EurasianConference on Mathematical Sciences and Applications (IECMSA-2015) in Athens, Greece.32 R. Almeida & N. R. O. Bastosthat in order to have a good approximation, we need to use higher-order derivatives, which maynot be adequate for fractional problems. In this paper we follow a different path, by discretizingthe fractional derivative, and then convert continuous problems into discrete ones.To start, let us recall the definition of the Hadamard fractional derivative.Definition 1.1. Let a, b be two reals with 0 < a < b, and x : [a, b]→ R be a function. For α ∈ (0, 1),the left Hadamard fractional derivative of order α is defined byaDαt x(t) =tΓ(1− α)ddt∫ ta(lntτ)−αx(τ)τdτ,while the right Hadamard fractional derivative of order α bytDαb x(t) =−tΓ(1− α)ddt∫ bt(lnτt)−α x(τ)τdτ,where Γ denotes the Gamma function.When x is an absolutely continuous function, there exists an equivalent definition (cf. [6])aDαt x(t) =x(a)Γ(1− α)(lnta)−α+1Γ(1− α)∫ ta(lntτ)−αx˙(τ) dτ,andtDαb x(t) =x(b)Γ(1− α)(lnbt)−α− 1Γ(1− α)∫ bt(lnτt)−αx˙(τ)dτ.More properties can be found in references at the end. The paper is organized in the followingway. In Section 2 we present the main result of the paper. Starting with the definition, and withan appropriate grid on time, we present a new discrete version for the left Hadamard fractionalderivative. To show the efficiency of the method, in Section 3 we compare the exact expression ofa fractional derivative with some numerical experiments, for different values of α and differentstep sizes n. In Section 4 we appply the technique to solve a fractional differential equation and afractional calculus of variation problem.2. The discretization methodThe discretization method is described in the following way. Given a function x : [a, b]→ R ,fix a positive integer n, and define the time step4T = lnban.Given N ∈ {0, 1, . . . , n}, denote the time and space grid bytN = a exp(N4T ) = a n√(ba)Nand xN = x(tN ).Theorem 2.1. Let x : [a, b]→ R be a function of class C2 and n ∈ N. Denoteψ =(4T )1−αa(1− exp(−4T ))Γ(2− α) and (ωαk ) = k1−α − (k − 1)1−α.Then, for all N ∈ {1, . . . , n},aDαtNxN = ˜aDαtNxN +O(4T ),where˜aDαtNxN =x(a)Γ(1− α)(lntNa)−α+ ψN∑k=1(ωαN−k+1) xk − xk−1exp(k4T ) · tk, (2.1)A Discretization of the Hadamard fractional derivative 33andlim4T→0O(4T ) = 0.Proof.aDαtNxN =x(a)Γ(1− α)(lntNa)−α+1Γ(1− α)∫ tNa(lntNτ)−αx˙(τ) dτ=x(a)Γ(1− α)(lntNa)−α+1Γ(1− α)N∑k=1∫ tktk−1(lntNτ)−α1τ(x˙(τ)τ) dτ≈ x(a)Γ(1− α)(lntNa)−α+1Γ(1− α)N∑k=1∫ tktk−1(lntNτ)−α1τ(xk − xk−1a(1− exp(−4T )) exp(k4T ) · tk)dτ=x(a)Γ(1− α)(lntNa)−α+1a(1− exp(−4T ))Γ(1− α)N∑k=1xk − xk−1exp(k4T ) · tk∫ tktk−1(lntNτ)−α1τdτ=x(a)Γ(1− α)(lntNa)−α+(4T )1−αa(1− exp(−4T ))Γ(2− α)N∑k=1xk − xk−1exp(k4T ) · tk[(N − k + 1)1−α − (N − k)1−α] .Thus, we get the desired approximation formula. Now, let us determine an upper bound for theerror when we use formula (2.1). The error is given byE =1Γ(1− α)N∑k=1∫ tktk−1(lntNτ)−α1τ∣∣∣∣x˙(τ)τ − xk − xk−1tk − tk−1 · tk∣∣∣∣ dτ.LetMi = maxτ∈[a,b]∣∣∣x(i)(τ)∣∣∣ , i = 1, 2.Then, using Taylor’s Theorem, we get that, for all k ∈ {1, . . . , N} and for all τ ∈ [tk−1, tk],∣∣∣∣x˙(τ)τ − xk − xk−1tk − tk−1 · tk∣∣∣∣=∣∣∣∣x˙(τ)τ − (x˙(tk−1) + x¨(ξ1) tk − tk−12)· tk∣∣∣∣≤ |x˙(τ)τ − x˙(tk−1)tk|+M2 tk − tk−12· tk= |(x˙(tk−1) + x¨(ξ2)(τ − tk−1)) τ − x˙(tk−1)tk|+M2 tk − tk−12· tk≤M1(tk − τ) +M2(τ − tk−1)τ +M2 tk − tk−12· tk≤ (tk − tk−1)[M1 +32M2b].Therefore, the error is bounded byE ≤ 1Γ(1− α)N∑k=1∫ tktk−1(lntNτ)−α1τ(tk − tk−1)[M1 +32M2b]dτ=M1 +32M2bΓ(1− α)N∑k=1(tk − tk−1)∫ tktk−1(lntNτ)−α1τdτ.34 R. Almeida & N. R. O. BastosHaving into consideration that, for all k ∈ {1, . . . , N} and for all τ ∈ [tk−1, tk),0 ≤(lntNτ)−α≤(lntkτ)−α,we have that0 ≤∫ tktk−1(lntNτ)−α1τdτ ≤ (4T )1−α1− α .Also, sincetk − tk−1 = a(1− exp(−4T )) exp(k4T ),thenE ≤ M1 +32M2bΓ(2− α) a(1− exp(−4T ))(4T )1−αN∑k=1exp(k4T )=M1 +32M2bΓ(2− α) a(4T )1−α(exp(N4T )− 1)≤ M1 +32M2bΓ(2− α) a(4T )1−α(ba− 1).In conclusion, we obtain the upper bound formula for our approximation (2.1):E ≤ M1 +32M2bΓ(2− α) (b− a)(4T )1−α (2.2)which converges to zero as4T → 0.In opposite to the classical case, where the concept of derivative is local, a fractional derivativecontains memory, and thus to compute the approximation obtained in Eq. (2.1) at a point tN , weneed to know the values of x(tn) from the beginning until de end-point, i.e., from n = 0 to n = N.For the right Hadamard fractional derivative, we have in a similar way the following approxi-mation formula:tNDαb xN ≈x(b)Γ(1− α)(lnbtN)−α− ψn∑k=N+1(ωαk−N) xk − xk−1exp(k4T ) · tk.3. ExampleLet x(t) = ln t, for t ∈ [1, 2]. Then (see [7])1Dαt x(t) =(ln t)1−αΓ(2− α) .In Figure 1 we show the accuracy of the procedure, for different values of α ∈ {0.2, 0.5, 0.7, 0.9}and for different values of n ∈ {10, 30, 50}. The error of the numerical experiments is measuredusing the normd(x, y) =∑nk=1 |xk − yk|n. (3.1)We can see that, for a greater value of n, the error decreases as expected.A Discretization of the Hadamard fractional derivative 351 1.2 1.4 1.6 1.8 200.10.20.30.40.50.60.70.80.9t1D0.2tln(t)  AnalyticApproximate, n=10, E=0.016965Approximate, n=30, E=0.0053124Approximate, n=50, E=0.00314611.445 1.45 1.4550.480.50.52(a) α = 0.21 1.2 1.4 1.6 1.8 200.10.20.30.40.50.60.70.80.91t1D0.5tln(t)  AnalyticApproximate, n=10, E=0.0234Approximate, n=30, E=0.0074314Approximate, n=50, E=0.00441271.48 1.49 1.50.680.70.720.740.76(b) α = 0.51 1.2 1.4 1.6 1.8 200.20.40.60.811.21.4t1D0.7tln(t)  AnalyticApproximate, n=10, E=0.028163Approximate, n=30, E=0.0090573Approximate, n=50, E=0.00539151.475 1.48 1.4850.840.860.88(c) α = 0.71 1.2 1.4 1.6 1.8 200.20.40.60.811.21.4t1D0.9tln(t)  AnalyticApproximate, n=10, E=0.032897Approximate, n=30, E=0.010763Approximate, n=50, E=0.00643071.43 1.44 1.450.951(d) α = 0.9Figure 1. Analytic vs. numerical approximation.4. ApplicationsExample 4.1. Consider a fractional differential equation with dependence on the left Hadamardfractional derivative: {f (t, x(t), aDαt x(t)) = 0, t ∈ [a, b]x(a) = xa.The procedure to solve numerically the system is described next. Fix n ∈ N and for N ∈ {1, . . . , n},definetN = a exp(N4T ), xN = x(tN ) with 4T =ln ban.Replacing the fractional operator by the approximation given in Eq. (2.1), we obtain a classicaldifference equation with n unknown points x1, . . . , xn,{f(tN , xN , ˜aDαtNxN)= 0, N ∈ {1, . . . , n}x0 = xa.For example, consider the system 1Dαt x(t) + x(t) =(ln t)1−αΓ(2− α) + ln t, t ∈ [1, 2]x(1) = 0.36 R. Almeida & N. R. O. BastosThe obvious solution is x(t) = ln t. Applying the discussed method, we obtain ψN∑k=1(ωαN−k+1) xk − xk−1exp(k4T ) · tk + xN =(ln tN )1−αΓ(2− α) + ln tN , N ∈ {1, . . . , n}x0 = 0,with4T = ln 2n, tN = exp(N4T ), ψ = (4T )1−α(1− exp(−4T ))Γ(2− α) and ωk = k1−α − (k − 1)1−α.In Figure 2 we show the numerical results, for different values of α ∈ {0.2, 0.5, 0.7, 0.9} and fordifferent values of n ∈ {5, 15, 30}.1 1.2 1.4 1.6 1.8 200.10.20.30.40.50.60.7tx(t)α = 0.2  AnalyticApproximate, n=5, E=0.013201Approximate, n=15, E=0.0044804Approximate, n=30, E=0.00225011.44 1.46 1.48 1.5 1.520.340.360.380.40.42(a) α = 0.21 1.2 1.4 1.6 1.8 200.10.20.30.40.50.60.7tx(t)α = 0.5  AnalyticApproximate, n=5, E=0.015074Approximate, n=15, E=0.005199Approximate, n=30, E=0.0026211.44 1.46 1.48 1.5 1.520.340.360.380.40.42(b) α = 0.51 1.2 1.4 1.6 1.8 200.10.20.30.40.50.60.7tx(t)α = 0.7  AnalyticApproximate, n=5, E=0.016174Approximate, n=15, E=0.0056458Approximate, n=30, E=0.00285531.44 1.46 1.48 1.5 1.520.340.360.380.40.42(c) α = 0.71 1.2 1.4 1.6 1.8 200.10.20.30.40.50.60.7tx(t)α = 0.9  AnalyticApproximate, n=5, E=0.017093Approximate, n=15, E=0.0060434Approximate, n=30, E=0.0030681.44 1.46 1.48 1.5 1.520.340.360.380.40.42(d) α = 0.9Figure 2. Analytic and approximated solutions for problem of Example 4.1.Example 4.2. For our next application, we show how to solve fractional variational problems withthe Lagrangian depending on the Hadamard fractional derivative. Consider the functionalJ(x) =∫ baL (t, x(t), aDαt x(t)) dt,on the set on functions that satisfy the boundary conditionsx(a) = xa and x(b) = xb,A Discretization of the Hadamard fractional derivative 37where xa, xb are two fixed reals. The procedure how to find a numerical approximation is explainednext. First, divide the interval [a, b] into n subintervals [tN−1, tN ] for N ∈ {1, . . . , n}, wheretN = a exp(N4T ) and 4T =ln ban.Denoting xN = x(tN ), applying the trapezoidal rule and taking into consideration Eq. (2.1), thevariational integral is approximated asJ(x) =n∑N=1∫ tNtN−1f (t, x(t), aDαt x(t)) dt≈ f(t0, x0, aDαt0x0)(t1 − t0) + f(tn, xn, aDαtnxn)(tn − tn−1)2+n−1∑N=1f(tN , xN , aDαtNxN)2(tN+1 − tN−1)≈f(a, xa, ˜aDαt1x1)(t1 − t0) + f(tn, xn, ˜aDαtnxn)(tn − tn−1)2+n−1∑N=1f(tN , xN , ˜aDαtNxN)2(tN+1 − tN−1).Observe that we used here the approximationaDαaxa ≈ ˜aDαt1x1.We can regard this sum as a function of n − 1 unknown variables Ψ(x1, . . . , xn−1), and then tofind the optimal solution one needs to solve the system∂Ψ∂xN= 0, for N ∈ {1, . . . , n− 1},and with this we track the desired values (x1, . . . , xn−1). Observe that, in opposite to the classicalcase, ∂Ψ/∂xN depends on the points xN , xN+1, . . . , xn−1.For example, we want the global minimizer forJ(x) =∫ 21(1Dαt x(t)−(ln t)1−αΓ(2− α))2dt,with the restrictionsx(1) = 0 and x(2) = ln 2.The optimal solution is x(t) = ln t since the functional takes only non-negative values and vanisheswhen evaluated at x.In Figure 3 we show the solution of the problem, for different values of α ∈ {0.2, 0.5, 0.7, 0.9}and for different values of n ∈ {5, 15, 30}.5. ConclusionsFor all numerical experiments presented above we used MatLab to obtain the results. InExamples 4.1 and 4.2, when we take a small number of mesh points (n = 5) we get a not so goodsolution. However, increasing the value of n, the error decreases and the numerical solutionapproaches the analytic solution, converging to it. From the numerical results we also notice that,for the same values of n, as α increases the error also increases, which makes sense taking intoaccount formula (2.2). We fix all parameters except α, it is easy to check that the maximum forerror value increases as we increase the value of α.38 R. Almeida & N. R. O. Bastos1 1.2 1.4 1.6 1.8 200.10.20.30.40.50.60.7tx(t)α = 0.2  AnalyticApproximate, n=5, E=0.018783Approximate, n=15, E=0.0081261Approximate, n=30, E=0.00416411.44 1.46 1.48 1.5 1.520.340.360.380.40.42(a) α = 0.21 1.2 1.4 1.6 1.8 200.10.20.30.40.50.60.7tx(t)α = 0.5  AnalyticApproximate, n=5, E=0.026079Approximate, n=15, E=0.013255Approximate, n=30, E=0.00663031.44 1.46 1.48 1.5 1.520.340.360.380.40.42(b) α = 0.51 1.2 1.4 1.6 1.8 200.10.20.30.40.50.60.7tx(t)α = 0.7  AnalyticApproximate, n=5, E=0.037751Approximate, n=15, E=0.021007Approximate, n=30, E=0.010661.44 1.46 1.48 1.5 1.520.320.340.360.380.4(c) α = 0.71 1.2 1.4 1.6 1.8 200.10.20.30.40.50.60.7tx(t)α = 0.9  AnalyticApproximate, n=5, E=0.058731Approximate, n=15, E=0.031629Approximate, n=30, E=0.0166841.46 1.48 1.5 1.52 1.540.320.340.360.380.4(d) α = 0.9Figure 3. Analytic and approximated solutions for problem of Example 4.2.α n E n E n EExample 4.10.2 5 0.013201 15 0.0044804 30 0.00225010.5 5 0.015074 15 0.005199 30 0.0026210.7 5 0.016174 15 0.0056458 30 0.00285530.9 5 0.017093 15 0.0060434 30 0.003068Example 4.20.2 5 0.018783 15 0.0081261 30 0.00416410.5 5 0.026079 15 0.013255 30 0.00663030.7 5 0.037751 15 0.021007 30 0.010660.9 5 0.058731 15 0.031629 30 0.016684Table 1. Number of mesh points, n, with corresponding error, E from formula (3.1).AcknowledgmentsWork supported by Portuguese funds through the CIDMA - Center for Research and De-velopment in Mathematics and Applications, and the Portuguese Foundation for Science andTechnology (FCT-Fundação para a Ciência e a Tecnologia), within project UID/MAT/04106/2013.A Discretization of the Hadamard fractional derivative 39References[1] Butzer, P.L., Kilbas, A. A. and Trujillo, J. J., Mellin transform analysis and integration by partsfor Hadamard-type fractional integrals, J. Math. Anal. Appl. 270 (2002), no. 1, 1-15.[2] Butzer, P.L., Kilbas, A. A. and Trujillo, J. J., Fractional calculus in the Mellin setting andHadamard-type fractional integrals, J. Math. Anal. Appl. 269 (2002), no. 1, 1-27.[3] Butzer, P.L., Kilbas, A. A. and Trujillo, J. J., Stirling functions of the second kind in the settingof difference and fractional calculus, Numer. Funct. Anal. Optim. 24 (2003), no. 7-8, 673-711.[4] Hadamard, J., Essai sur l’etude des fonctions donnees par leur developpment de Taylor, J.Pure Appl. Math. 4 (1892), no. 8, 101-186.[5] Jarad, F., Abdeljawad, T. and Baleanu, D., Caputo-type modification of the Hadamard frac-tional derivatives, Advances in Difference Equations August 2012 (2012), 2012–142.[6] Kilbas, A.A., Hadamard-type fractional calculus, J. Korean Math. Soc. 38 (2001), no. 6, 1191-1204.[7] Kilbas, A.A., Srivastava, H.M. and Trujillo, J.J., Theory and applications of fractional differen-tial equations. North-Holland Mathematics Studies, 204. Elsevier Science B.V., Amsterdam,2006.[8] Kilbas, A.A. and Titioura, A.A., Nonlinear differential equations with Marchaud-Hadamard-type fractional derivative in the weighted space of summable functions, Math. Model. Anal. 12(2007), no. 3, 343-356.[9] Pooseh, S., Almeida, R. and Torres, D. F. M., Expansion formulas in terms of integer-orderderivatives for the Hadamard fractional integral and derivative, Numer. Funct. Anal. Optim.33 (2012), no. 3, 301–319.[10] Qassim, M. D., Furati, K. M. and Tatar, N.E., On a Differential Equation Involving Hilfer-Hadamard Fractional Derivative, Abstract and Applied Analysis vol. 2012 (2012), Article ID391062, 17 pages, doi:10.1155/2012/391062.[11] Qian, D., Gong, Z. and Li, C., A generalized Gronwall inequality and its application tofractional differential equations with Hadamard derivatives, 3rd Conference on NonlinearScience and Complexity (NSC10), Cankaya University, Ankara, Turkey, 28–31 July, 2010.AffiliationsRICARDO ALMEIDAADDRESS: Center for Research and Development in Mathematics and Applications (CIDMA)Department of Mathematics, University of Aveiro, 3810–193 Aveiro, Portugal.E-MAIL: ricardo.almeida@ua.ptNUNO R. O. BASTOSADDRESS: Centre for the Study of Education, Technologies and Health (CI&DETS), Departmentof Mathematics, School of Technology and Management of ViseuPolytechnic Institute of Viseu, 3504–510 Viseu, Portugal.E-MAIL: nbastos@estv.ipv.pt

A Discretization of the Hadamard fractional derivative

Abstract

Similar works

Full text

Available Versions

Repositório Institucional da Universidade de Aveiro