Podlubny I., Skovranek T, Verbickij V., Vinagre B., Chen YQ., Petras I.

Transcription

1 DISCRETE FRACTIONAL CALCULUS: NON-EQUIDISTANT GRIDS AND VARIABLE STEP LENGTH Podlubny I., Skovranek T, Verbickij V., Vinagre B., Chen YQ., Petras I. Technical University of Kosice, Slovakia Odessa National University, Odessa, Ukraine University of Extremadura, Badajoz, Spain Utah State University, Logan, USA

2 History: OFDEs: 2000, PFDEs: 2008, SOFDEs: 2009 MATRIX APPROACH TO DISCRETE FRACTIONAL CALCULUS Igor Podlubny 2

3 Implemented in MATLAB (2008, updated 2009)

4 Triangular strip matrices (TSMs)

5 Generating functions for TSMs

6 Operations with TSMs

7 Left-sided fractional derivatives

8 Discrete analog of left-sided fractional derivatives

9 Solving ordinary FDEs Bagley-Torvik equation: ay (t)+b 0 D 3/2 t y(t)+cy(t) =f(t) add zero initial conditions: ab (2) + bb n n (3/2) + cb n (0) Yn = F n, 9

10 Solving partial FDEs β U x β β U x β 10

11 Left-sided fractional integrals: inverse of fractional derivatives

12 Non-equidistant grid: discretization is cumbersome even in simplest cases... w (x) : 1 ĥ i ( wi+1 w i h i w i w i 1 h i 1 ) h i = x i+1 x i, ĥ i =ˆx i ˆx i 1, e difference proble

13 Recall that I α N = BN α 1 Change the viewpoint: Left-sided fractional derivatives: inverse of fractional integrals; then B α N =(I α N ) 1 Any approximation of fractional integration after inversion gives an approximation for fractional differentiation on the same grid!

14 The simplest approach: approximation of a function under integration by a piecewise constant function B α N =(I α N ) 1 Coefficients of I α N as I k,j = (t k t j 1 ) α (t k t j ) α Γ(α + 1) j =1,...,k; k =1,...,N., For non-equidistant grids, the matrix is not a TSM.

15 Example 1: fractional integrals of sin(x) y = 0.1 = 0.3 = 0.5 = 0.7 = [ ] with non equidistant step t

16 Example 2: fractional derivatives of sin(x) y = 0.1 = 0.3 = 0.5 = 0.7 = [0.1, 0.3, 0.5, 0.7], with non equidistant step t

17 Example 3: two-term ordinary FDE y (α) (t)+y(t) =1, Exact solution: y(0) = 0, y (0) = 0. y(t) =t α E α,α+1 ( t α ) y analytical solution 0.2 numerical solution (non equidistant step) t α = 1.8, number of discretization nodes N = 500

18 Example 4(a): Bagley-Torvik equation Ay (t)+by 3/2 (t)+cy(t) =F (t), F (t) = 8, (0 t 1) 0, (t >1), y(0) = y (0) = equidistant step non equidistant step 3 y t A = 1, B = 1, C = 1.

19 Example 4(b): Bagley-Torvik equation Ay (t)+by 3/2 (t)+cy(t) =F (t), F (t) = 8, (0 t 1) 0, (t >1), y(0) = y (0) = equidistant step non equidistant step 4 y t A = 1, B = 0.5, C = 0.5.

20 Can we have variable step length? As seen in MATLAB: ode23.m and ode45.m solvers optimized for a variable step using a variable step ensures that a large step size is used for low frequencies and a small step size is used for high frequencies make a step, estimate the error at this step, check if the value is greater than or less than the tolerance, and adapt the step size accordingly There was nothing like this available for FDEs so far...

21 Method of large steps 0Dt α y(t) =f(y(t),t), (t >0), y(0) = 0, Suppose we obtained its solution in the interval (0,a) (and the final value y a at t =a), then we can use this for transforming the above problem to adt α y(t) =f(y(t),t) 0 Ra α y(t), (t >a), y(a) =y a, where 0R α a y(t) = 1 Γ(1 α) 0R α a y(t) = 0 D α t a 0 (t τ) α 1 y(τ)dτ, (t >a). (1 H(t a))y(t)

22 First large step in [0, a] 0D α t y(t) =f(y(t),t), (t >0), y(0) = 0, ad α t y(t) =f(y(t),t) 0 R α a y(t), (t >a), Auxiliary function: y(a) =y a, Second large step in [a, b] y(t) = u(t) +y a, the function u(t), w ad α t u(t) =f(u(t)+y a,t) 0 R α a y(t) y a, u(a) =0. (t >a),

23 Method of large steps : example (1) 0D 1/2 t y(t)+y(t) = t1.5 Γ(1.5) y(0) = 0. + t, (t >0), Exact solution: y(t) =t. First large step : interval [0, 1]: clear all h = 0.01; Using the matrix approach t = 0:h:1; N = 1/h + 1; M = zeros(n,n); M = ban(0.5, N, h) + eye(n,n); F = (t.^(0.5)/gamma(1.5) + t) ; M = eliminator(n,[1])*m*eliminator(n, [1]) ; F = eliminator(n,[1])*f; Y = M\F; Y0 = [0; Y]; plot (t,y0, b ) set(gca, xlim, [0 2], ylim, [0 2] ) grid on, hold on

24 Method of large steps : example (2) Second large step : interval [1, 2] 0D 1/2 t y(t) = 1 D 1/2 t y(t)+ 1 Γ(0.5) 1 0 y (τ)dτ, (t >1) (t τ) 1/2 1D 1/2 t y(t)+y(t) = t1.5 Γ(1.5) + t 1 Γ(0.5) 1 0 dτ (t τ) 1/2 (t>1). 1D α t y(t)+y(t) = t1.5 Γ(1.5) + t 2t0.5 Γ(0.5) + 2(t 1)0.5 Γ(0.5) ; (t>1) y(1) = 1. y(t) =u(t)+1,

25 Method of large steps : example (3) The problem to solve in [1, 2]: 1D α t u(t)+u(t) = t1.5 Γ(1.5) + t 2t0.5 Γ(0.5) + 2(t 1)0.5 Γ(0.5) 1; (t >1) u(1) = 0. Using the matrix approach clear all h = 0.01; t = 1:h:2; N = 1/h + 1; M = zeros(n,n); M = ban(0.5, N, h) + eye(n,n); F = (t.^(0.5)/gamma(1.5) + t - 2*t.^(0.5)/gamma(0.5) *(t-1).^(0.5)/gamma(0.5) - 1) ; M = eliminator(n,[1])*m*eliminator(n, [1]) ; F = eliminator(n,[1])*f; U = M \F; U0 = [0; U]; Y0 = U0 + 1; plot(t, Y0, g )

26 Conclusions Easy, transparent, uniform approach to fractional integration and differentiation on non-equidistant grids No need for cumbersome formulas for finite differences on nonequidistant grids Any numerical approximation of a fractional-order integral on a non-equidistant mesh gives a matrix for fractional-order differentiation on the same grid Works for left-sided, right-sided, and symmetric (Riesz) fractional integrals and derivatives Can be used for solving PDEs on non-uniform spatial-temporal meshes Moving towards variable-length and adaptive-length step methods in numerical solution of ordinary and partial FDEs.

27 Thank you!