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Approximate calculation of the Caputo-type fractional derivative from inaccurate data. Dynamical approach

  • Platon G. Surkov

Abstract

A specific formulation of the “classical” problem of mathematical analysis is considered. This is the problem of calculating the derivative of a function. The purpose of this work is to construct an algorithm for the approximate calculation of the Caputo-type fractional derivative based on the methods of control theory. The input data of the algorithm is represented by inaccurate measured function values at discrete, frequently enough, times. The proposed algorithm is based on two aspects: a local modification of the Tikhonov regularization method from the theory of ill-posed problems and the Krasovskii extremal shift method from the guaranteed control theory, both of which ensure the stability to informational noises and computational errors. Numerical experiments were carried out to illustrate the operation of the algorithm.

Appendix A

Appendix A Appendices

A.1. Monotonicity of functions fi(t). Let us investigate the following functions for monotonicity:

fk(t)=(tτkh)γ(τihτkh)γ(tτk+1h)γ+(τihτk+1h)γ,tδih,k=0,,i1,iNh.

At the left point of the interval δih the functions have the values fk(τih)=0 . The derivatives of these functions satisfy the inequalities

fk(t)=γ(tτk+1)1γ(tτk)1γ(tτk)1γ(tτk+1)1γ<0,tδih,k=0,,i1,iNh.

Therefore, the functions fk(t) are strictly monotone and reach the smallest values, which are the largest in absolute values, as tτi+1h . Let us calculate these values

|fk(τi+1h)|=|(τi+1hτkh)γ(τihτkh)γ(τi+1hτk+1h)γ+(τihτk+1h)γ|=|(i+1)δkδ)γ(iδkδ)γ((i+1)δ(k+1)δ)γ+(iδ(k+1)δ)γ|=δγ|(ik1)γ2(ik)γ+(ik+1)γ|=δγcik,

where cik are defined by formulas (4.12).

A.2. Estimate of sum k=0i1cik. . The considered sum consists of positive terms (see (4.12)), therefore, the following relations are valid

k=0i1cik=k=0i1|(ik1)γ2(ik)γ+(ik+1)γ|=m=1i|(m1)γ2mγ+(m+1)γ|m=1|(m1)γ2mγ+(m+1)γ|=:Σ.

Since the common term of the series Σ can be represented as |(m+1)γmγ–(mγ–(m –1)γ)|, we consider the function

g(x)=(x+1)γxγ,x>0,

Examining the monotonicity of g(x) we have

g(x)=γ((x+1)γx+1xγx)=γxγx+1((1+1x)γ(1+1x)).

Since 1 + 1/x >1 for x > 0, then for γ ϵ (0,1) the inequality

(1+1x)γ<(1+1x)

holds. Consequently, g'(x) < 0, which means g(x) decreases monotonically, and g(m + 1) – g(m) < 0. Then the expression for the series Σ can be rewritten without the modulus sign in the form

Σ=m=1(2mγ(m+1)γ(m1)γ).

Now, we write the elements of this series

a1=22γ,a2=22γ3γ1,a3=23γ4γ2γ,a4=24γ5γ3γ,a5=25γ6γ4γ,

then we obtain a formula for partial sums:

Sm=1+mγ(m+1)γ.

Transforming the expression for Sm and using the equivalence of infinitesimal functions, we calculate the limit

Σ=limmSm=1+limm(1+1m+1)γ1(m+1)γ=1+limmγ(m+1)1γ=1,

this implies

k=0i1cikΣ=1.

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Received: 2020-09-30
Revised: 2021-05-14
Published Online: 2021-06-23
Published in Print: 2021-06-25

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