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Research ArticleFinite DifferenceFourier Spectral for a Time FractionalBlackndashScholes Model with Option Pricing

Juan He12 and Aiqing Zhang 1

1Business School Central University of Finance and Economics Beijing 100081 China2Accounting School Guizhou University of Finance and Economics Guiyang 550025 Guizhou China

Correspondence should be addressed to Aiqing Zhang zhangaiqingcufeeducn

Received 4 May 2020 Accepted 27 July 2020 Published 4 September 2020

Academic Editor Hamdy Nabih Agiza

Copyright copy 2020 Juan He and Aiqing Zhang)is is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

We study the fractional BlackndashScholes model (FBSM) of option pricing in the fractal transmission system In this work wedevelop a full-discrete numerical scheme to investigate the dynamic behavior of FBSM )e proposed scheme implements aknown L1 formula for the α-order fractional derivative and Fourier-spectral method for the discretization of spatial directionEnergy analysis indicates that the constructed discrete method is unconditionally stable Error estimate indicates that the2 minus α-order formula in time and the spectral approximation in space is convergent with order O(Δt2minus α + N1minus m) where m is theregularity of u and Δt and N are step size of time and degree respectively Several numerical results are proposed to confirm theaccuracy and stability of the numerical scheme At last the present method is used to investigate the dynamic behavior of FBSM aswell as the impact of different parameters

1 Introduction

)e classical option pricing model is proposed by Black andScholes [1] which is based on the assumption that stocks andoptions are in an ldquoideal staterdquo in the market and Samuelsonrsquosmodel [2]

dS μSdt + σSdB(t) (1)

where S be the the stock value B(t) is the Brownian motionwith the unit variance and μ and σ are two constants But inmany cases fractional Brownian motion is more accuratethan integer order [3] On the other hand more and morediffusion processes were found to be non-Fickian [4 5] andthe fractional order stochastic differential equation is con-sidered as an extension of the stochastic differential equa-tion One view is that fractional order option tradingequation is regarded as nonrandom growth process causedby Brownian motion )erefore Jumarie [6] and Liang et al[7] considered fractional Brownian motion in Samuelsonrsquosmodel equation

dS μSdt + σSdB(t α)

dS μSdt + σSw(t)(dt)α

(2)

Combining Ito lemma and fractional Taylor expansionof the option price V Jumarie [6] obtained the followingFBSM

zαV(S t)

ztα

rV(S t)

(1 minus α)minus rS

αzαV(S t)

zSα1113888 1113889t

1minus α

minus(α)

3[(1 minus α)]

2

(2α)σ2S2

z2α

V(S t)

zS2α 0

zαV(S t)

ztα rV(S t) minus rS

zV

zS1113888 1113889

t1minus α

(1 minus α)

minusα

2σ2S2

z2V(S t)

zS2 0

(3)

Jumarie [8] derived new families of the exact solution ofthe above equations Moreover traditional pricing models

HindawiMathematical Problems in EngineeringVolume 2020 Article ID 1393456 9 pageshttpsdoiorg10115520201393456

for double barrier options are often biased when pricechanges are considered as fractal transmission systemsChen et al [9] revealed that it would be better to usefractional order BlackndashScholes equation to explain thepricing in fractal transmission systems

)ere is a lot of work in the modeling and calculation offractional equations Yang et al [10 11] developed a newdefinition of fractional derivative )e advantage of thisdefinition is that it does not contain singular kernel Inc et al[12 13] studied the isolated solutions of a class of fractionalequations with Kerr law nonlinearity by RiccatindashBernoullimethod )e exact dark optical and periodic singular solitonsolution is obtained Singhet al [14ndash16] solved a series offractional equations by using homotopy analysis techniqueand Laplace transform algorithm

As we all know as an effective method L1 formula [17]has been widely used in the calculation of fractional dif-ferential equations Langlands and Henry [18] Sun and Wu[19] and Lin and Xu [20] discussed the error estimate for L1scheme De Staelena and Hendybc [21] constructed a nu-merical method of fourth-order finite difference in space and2 minus α in time Stability uniqueness and error estimates areanalyzed Zhang et al [22] presented a discrete implicit finitescheme to solve time fractional BlackndashScholes model )eydiscussed the stability and error estimation of numericalschemes by Fourier analysis Unfortunately their analysismethods are local approach Due to the importance ofFBSM it is necessary to reconstruct an efficient numericalmethod and analyze global stability and error estimates

In this work we will develop an efficient full-discretescheme to approximate the BlackndashScholes model withα-order fractional derivative We apply L1 method to dis-cretize the direction of time and Fourier-spectral method todiscretize the direction of space Using the energy analysismethod we discuss the stability and error estimate of thefully discrete numerical method)e detailed analysis showsthat the scheme is unconditionally energy stable and theerror estimates indicate that our full-discrete scheme canachieve 2 minus α-order accuracy in time and exponential ac-curacy in space direction Finally some numerical examplesare conducted to support the theoretical claims At the sametime the dynamic behavior of FBSM is studied by theproposed method

We organize the rest of the paper as follows Section 2will briefly introduce the FBSM In Section 3 we develop atime-discrete method for FBSM and then present its discreteenergy law In Section 4 we will study the error estimate ofthe full-discrete scheme In Section 5 we present accuracystability tests and numerous numerical examples to dem-onstrate the validity of the full-discrete method In additionwe will discuss the properties of the solution of the FBSMSome concluding remarks are given in Section 6

2 BlackndashScholes Model

In this work we will consider the following time fractionalBlackndashScholes model

zαV(S t)

ztα +

12σ2S2

z2V(S t)

zS2

+ rSzV(S t)

zSminus rV(S t) 0 (S t) isin (0 +infin) times(0 T)

(4)

V(0 t) p(t)

V(+infin t) q(t)(5)

V(S T) w(S) (6)

where V is the price of the option S the price of the un-derlying asset r the interest rate and σ the volatility of thestock price 0lt αle 1 the time fractional derivativezαV(S t)ztα is defined by

zαV(middot t)

ztα z

αt V(middot t)

1Γ(1 minus α)

1113946t

0(t minus μ)

minus αzV(middot μ)

zμdμ

(7)

)is is a linear parabolic partial differential equationwhich has been studied extensively

We transform the problem to an initial value problem byusing the time to mature t T minus τ and we then setx ln S V(S τ) u(ex T minus τ) we can rewrite (4) as

zαt u(x τ) minus ξz

2xu(x τ) minus ωzxu(x τ)

+ ru(x τ) 0 (x τ) isin (0 +infin) times(0 T)(8)

where ξ 12σ2ω r minus ξ and with the following boundary(barrier) and initial conditions

u(minus infin τ) p(τ)

u(minus infin τ) q(τ)

u(x 0) u0(x) altxlt b

(9)

In order to solve the above model by numerical methodit is necessary to truncate the original unbounded regioninto a finite interval )erefore we will consider problem (8)in bounded interval (0 2π) )en we will study the fol-lowing problem

zατu(x τ) minus ξz


+ ru(x τ) 0 (x τ) isin (0 2π) times(0 T)(10)

u(0 τ) u(2π τ) (11)

u(x 0) u0(x) (12)

Remark 1 In fact one can choose homogeneous or inho-mogeneous boundary conditions It all depends on theactual option price We have tested it and it does not makeany difference in actual numerical examples

2 Mathematical Problems in Engineering

3 2minus αOrder Numerical Method

Here we will develop the time-discrete method for equation(10) First given a positive integer K set Δt TK be thetime step size and denote τn nΔt (0le nleK) as the meshpoint )en we introduce an L1 method to discrete theCaputo fractional derivative of order α

zαt u(x τ)

1Γ(2 minus α)

1113944

n

j0bj

u middot τn+1minus j1113872 1113873 minus u middot τnminus j1113872 1113873

Δtα+ O Δt2minus α

1113872 1113873

(13)

where bj (j + 1)1minus α minus j1minus α

Lemma 1 (see [20 23 24]) -e coefficients of formula (13)satisfy the following properties

1 b0 gt b1 gt b2 gt middot middot middot bj⟶ 0 asj⟶ +infin (14)

)en we can obtain the following time-discrete scheme

1a0

un+1minus 1113944

nminus 1

j0bj minus bj+11113872 1113873unminus j

minus bnu0⎛⎝ ⎞⎠ minus ξz

2xu

n+1

minus ωzxun+1

+ run+1 0 nge 0

(15)

where a0 ΔtαΓ(2 minus α) It should be noted that if n 0 wecan rewrite the above equation as

1a0

u1 minus u01113872 1113873 minus ξz2xu

1minus ωzxu

1+ ru1 0 (16)

First of all we have the following energy stability resultsfor time-discrete (15)

Theorem 1 -e time-discrete scheme (15) is unconditionallystable It satisfies the following energy dissipation law

uk+1

le u0

k 0 1 K (17)

Proof When n 0 computing the L2 inner product of (16)with 2u1 we obtain

2 u1 minus u0 u11113872 1113873 minus 2a0ξ z2xu

1 u11113872 1113873

minus a0ω zxu1u11113872 1113873 + a0r u1 u11113872 1113873 0

(18)

It is easy to verify that the following formula is correct

2(A minus B A) A2

minus B2

+(A minus B)2 (19)

)us

u1

2

minus u0

2

+ u1 minus u0

2

+ 2a0a zxu12

+ 2a0r u12

0

(20)

Giving up some positive terms we have

u1

2le u0

2 (21)

Assume the following inequality holds

uj

2le u0

2 j 2 3 n (22)

Next we will show un+12 le u02 is still valid Ifj n + 1 taking the L2 inner product of (15) with 2un+1 wederive

2 un+12

+ 2a0ξ zxun+1

2

+ 2a0r un+1

2

2 1113944nminus 1


+ bnu0un+1⎛⎝ ⎞⎠

le 1113944nminus 1

j0bj minus bj+11113872 1113873 unminus j

2

+ un+12

1113874 1113875

+ bn u0

2

+ un+12

1113874 1113875

(23)

Note the fact that

1113944

nminus 1

j0bj minus bj+11113872 1113873 + bn 1 (24)

)us we get

un+12le 1113944

nminus 1


2

+ bn u0

2

le 1113944nminus 1

j0bj minus bj+11113872 1113873 + bn

⎛⎝ ⎞⎠ u0

2

u0

2

(25)

)is yields (17)

4 Error Estimate for Full Discretization

In this part we will study the Fourier-spectral method forthe time-discrete method (15) First we define SN as thepolynomial space Define πN L2(Ω)⟶ SN be theL2-projection operator which satisfies

πNϕ minus ϕψ( 1113857 0 forallψ isin SN (26)

We have the following estimate [25]

ϕ minus πNϕ

lleCNlminus m

ϕm forallϕ isin Hm

(Ω) mgt lge 0

(27)

)en we can develop the following full-discrete scheme

1a0

un+1N minus 1113944

nminus 1


N minus bnu0NϕN

⎛⎝ ⎞⎠

minus ξ z2xu

n+1N ϕN1113872 1113873 minus ω zxu

n+1ϕN1113872 1113873 + r un+1

ϕN1113872 1113873 0 ϕN isin SN

(28)

We now present the stability results of the fully discretescheme (28)

Theorem 2 Let unN1113864 1113865

Mminus 1n1 be the solution of (28) then we

derive

Mathematical Problems in Engineering 3

un+1N

2le u0N

2 (29)

Next we begin to analyze the error estimates of the full-discrete scheme (28) Define the following error function

Rn+1

1Γ(2 minus α)

1113944

n

j0bj

u middot tn+1minus j1113872 1113873 minus u middot tnminus j1113872 1113873

Δtα

minus1Γ(1 minus α)

1113946tn+1

0

zu(middot μ)

zμdμ

tn+1 minus μ( 1113857α

(30)

From [20 19 24 23] we know that Rn+1 satisfies

Rn+1

leCΔt2minus α (31)

We also define the following error functions

1113957enu πNu tn( 1113857 minus un

N

1113954enu u tn( 1113857 minus πNu tn( 1113857

enu 1113954e

nu + 1113957e

nu u tn( 1113857 minus un

N

(32)

Lemma 2 For a0 and bn we have the following results

a0 le 2bnΓ(1 minus α)Tα (33)

Proof

bn (n + 1)1minus α

minus (n)1minus α

1113872 1113873

n1minus α 1 +

1n

1113874 11138751minus α

minus 11113888 1113889

n1minus α (1 minus α)

n+

(1 minus α)(minus α)

2

1n2 + middot middot middot1113888 1113889

ge n1minus α (1 minus α)

n+


2

1n21113888 1113889


2n+

(1 minus α)

2n+


2

1n21113888 1113889

(34)

Note that

(1 minus α)

2n+


2

1n2 ge 0 (35)

)erefore we obtain

a0

bn

le2n

αΔtαΓ(2 minus α)

(1 minus α)le 2Γ(1 minus α)T

α (36)

Remark 2 In [23 24] readers will also find similar pa-rameter estimation)is result is very useful for us to analyzeerror estimate

Theorem 3 For the constructed numerical scheme (28) wehave the following error estimate

u τk1113872 1113873 minus uk

leC Δt2minus α+ N

1minus m1113872 1113873 k 0 1 K

T

Δt

(37)

Proof For n 0 we can write equation (28) as

1a0

u1N minus u0NϕN1113872 1113873 minus ξ z2xu

1NϕN1113872 1113873 minus ω zxu

1N ϕN1113872 1113873 + r u1N ϕN1113872 1113873 0

(38)

Subtracting (38) from (10) at τ1 we note that

zατu middot τ1( 1113857 minus

u1N minus u0Na0


u middot τ1( 1113857 minus u middot τ0( 1113857

a0

+u middot τ1( 1113857 minus u middot τ0( 1113857

a0minusπNu middot τ1( 1113857 minus πNu middot τ0( 1113857

a0

+πNu middot τ1( 1113857 minus πNu middot τ0( 1113857

a0minusu1N minus u0N

a0

R1

+1a0

I minus πN( 1113857 u middot τ1( 1113857 minus u middot τ0( 1113857( 1113857 +1a0

1113957e1u minus 1113957e

0u1113872 1113873

z2xu middot τ1( 1113857 minus z

2xu

1N z

2xu middot τ1( 1113857 minus πNz

2xu middot τ1( 1113857

+ πNz2xu middot τ1( 1113857 minus z

2xu

1N

z2x I minus πN( 1113857u middot τ1( 11138571113858 1113859 + z

2x1113957e

1u

zxu middot τ1( 1113857 minus zxu1N zxu middot τ1( 1113857 minus πNzxu middot τ1( 1113857

+ πNzxu middot τ1( 1113857 minus zxu1N

zx I minus πN( 1113857u middot τ1( 11138571113858 1113859 + zx1113957e1u

u middot τ1( 1113857 minus u1N u middot τ1( 1113857 minus πNu middot τ1( 1113857 + πNu middot τ1( 1113857 minus u1N

I minus πN( 1113857u middot τ1( 1113857 + 1113957e1u

(39)

)en we have

1113957e1u minus 1113957e

0u ϕN1113872 1113873 + a0ξ zx1113957e

1u zxϕN1113872 1113873 minus a0ω zx1113957e

1u ϕN1113872 1113873 + a0r 1113957e

1uϕN1113872 1113873

minus a0 R1 ϕN1113872 1113873 + πN minus I( 1113857 u middot τ1( 1113857 minus u middot τ0( 1113857( 1113857ϕN( 1113857

+ a0ξ πN minus I( 1113857zxu middot τ1( 1113857 zxϕN( 1113857

minus a0ω I minus πN( 1113857u middot τ1( 1113857 zxϕN( 1113857

+ a0r πN minus I( 1113857u middot τ1( 1113857 ϕN( 1113857 ϕN isin SN

(40)

Set ϕN 21113957e1u we have


2 1113957e1u

2

+ 2a0ξ zx1113957e1u

2

+ 2a0r 1113957e1u

2

le a01

rb0R12

+ rb0 1113957e1u

2

1113888 1113889

+ a0ξ1b0

πN minus I( 1113857zxu middot τ1( 1113857

2

+ b0 zx1113957e1u

2

1113888 1113889

+ a0ωωξb0

I minus πN( 1113857u middot τ1( 1113857

2

+ξb0

ωzx1113957e

1u

2

1113888 1113889

(41)

Dropping some positive terms we find

1113957e1u

2leC1

a0

b0Δt4minus 2α

+ C2a0

b0N

1minus m+ C3

a0

b0N

minus m (42)

Assume

1113957eku

2leC1

a0

bkminus 1Δt4 + C2

a0

bkminus 1N

1minus m+ C3

a0

bjminus 1N

minus m k 2 3 n

(43)

Next we will prove that it holds also for k n + 1Subtracting (28) from a reformulation of (10) at tn+1 we find

1113957en+1u minus 1113944

nminus 1

j0bj minus bj+11113872 11138731113957e

nminus ju minus bn1113957e

0u ϕN

⎛⎝ ⎞⎠ + a0ξ zx1113957en+1u zxϕN1113872 1113873 minus a0ω zx1113957e

n+1u ϕN1113872 1113873 + a0r 1113957e

n+1u ϕN1113872 1113873

minus a0 Rn+1

ϕN1113872 1113873 + πN minus I( 1113857 u middot τn+1( 1113857 minus 1113944nminus 1

j0bj minus bj+11113872 1113873u middot τnminus j1113872 1113873 minus u middot τ0( 1113857⎛⎝ ⎞⎠ ϕN

⎛⎝ ⎞⎠

+ a0ξ πN minus I( 1113857zxu middot τn+1( 1113857 zxϕN( 1113857 minus a0ω I minus πN( 1113857u middot τn+1( 1113857 zxϕN( 1113857 + a0r πN minus I( 1113857u middot τn+1( 1113857 ϕN( 1113857

(44)

Let ϕN 21113957en+1u we have

2 1113957en+1u

2

+ 2a0ξ zx1113957en+1u

2

+ 2a0r 1113957en+1u

2

le 2 1113944nminus 1

j0bj minus bj+11113872 11138731113957e


0u 1113957e

n+1u

⎛⎝ ⎞⎠ minus 2a0 Rn+1

1113957en+1u1113872 1113873

+ 2a0ξ πN minus I( 1113857zxu middot τn+1( 1113857 zx1113957en+1u1113872 1113873 minus 2a0ω I minus πN( 1113857u middot τn+1( 1113857 zx1113957e

n+1u1113872 1113873

le 1113944nminus 1

j0bj minus bj+11113872 1113873 1113957e

nminus ju

2

+ 1113957en+1u

2

1113874 1113875 + a01r

Rn+1

2

+ r 1113957en+1u

2

1113874 1113875

+ a0ξ πN minus I( 1113857zxu middot τn+1( 1113857

2

+ zx1113957en+1u

2

1113874 1113875 + a0ωωξ

I minus πN( 1113857u middot τn+1( 1113857

2

+ξω

zx1113957en+1u

2

1113888 1113889

(45)

)us we have

1113957en+1u

2le 1113944

nminus 1

j0bj minus bj+11113872 1113873 1113957e

nminus ju

2

+a0

rR

n+12

+ a0a πN minus I( 1113857zxu middot τn+1( 1113857

2

+a0b

2

aI minus πN( 1113857u middot τn+1( 1113857

2

le 1113944nminus 1

j0bj minus bj+11113872 1113873 C1

a0

bjminus 1Δt4 + C2

a0

bjminus 1N

1minus m+ C3

a0

bjminus 1N

minus m1113888 1113889

+ bn C1a0

bn

Δt4 + C2a0

bn

N1minus m

+ C3a0

bn

Nminus m

1113888 1113889

(46)

Note that bnminus j ge bn thus

1113957en+1u

2le C1

a0

bn

Δt4 + C2a0

bn

N1minus m

+ C3a0

bn

Nminus m

1113888 1113889

1113944

nminus 1

j0bj minus bj+11113872 1113873 + bn

⎛⎝ ⎞⎠

C1a0

bn

Δt4 + C2a0

bn

N1minus m

+ C3a0

bn

Nminus m

(47)

Note that

u τk1113872 1113873 minus uk

le 1113954eku

+ 1113957eku

(48)

)is ends the proof


5 Numerical Examples

In this section several numerical examples will be present toconfirm the accuracy and applicability of the full-discretescheme (28) We consider a rectangular computed domainof [0 2π] In order to better simulate the periodic boundaryconditions the Fourier-spectral method will be used todiscretize space direction

51 Verification of Convergence of Numerical MethodFirst in order to conduct a time accuracy test an exactsolution will be constructed to evaluate the convergence ofthe full-discrete scheme (28)

Example 1 We consider the following FBSM with α-orderCaputo derivative


2xu(x τ) minus ωzxu(x τ) + ru(x τ) f(x τ)

(49)

where

f(x τ) 2T

2minus α

Γ(3 minus α)sinx + ξτ2 sinx minus ωτ2 cosx + rτ2 sinx

(50)

It is easy to verify that the exact solution will beu τ2 sinx

We set N 128 )e default values for the parametersare set as T 1 ξ 1ω 0 and r 1 In Table 1 we showthe temporal convergence orders of various time steps Ascan be seen from Table 1 our full-discrete scheme is close to2 minus α-order accuracy in time which is confirm with theresult in )eorem 3

Fix N 128 in Figure 1 we give L2 error for different αIt is obvious that our numerical scheme has good conver-gence in time direction Let T 01Δt 10minus 6 ξ r 05

ω 0 and u0 cos 8x Figure 2 shows that the full-discrete

Table 1 Temporal convergence orders of various time steps forExample 1

αΔt Δt

1200Δt

1400Δt

1800Δt

11600Δt

13200α 01 18166 18263 18345 18415 18475α 03 16680 16746 16797 16837 16869α 05 14892 14925 14947 14963 14974α 06 13939 13960 13974 13983 13989α 07 12964 12979 12987 12992 12995α 09 10980 10990 10995 10997 10999

10ndash4 10ndash3 10ndash210ndash8

10ndash6

10ndash4

10ndash2

Time step

Erro

rs in

tim

e

L2 error at α = 01L2 error at α = 03L2 error at α = 05


Figure 1 Error in time direction for different α

Erro

rs in

spac

e

10ndash15

10ndash10

10ndash5

100

N

14 16 18 20 22 24 26

L2 error

Linfin error

Figure 2 )e L2 and Linfin errors in space direction with α 01

1

05

06

4

2x0 0

0204

0608

1

τ

u (x

τ)

Figure 3 )e dynamic behavior of solution to FBSM equation atα 02


1

05

06

4

2x0 0

0204

0608

1

τ

u (x

τ)

Figure 4 )e dynamic behavior of solution to FBSM equation at α 05

1

05

06

42x

0 002

0406

081

τ

u (x

τ)


01

02

03

04

05

06

0 1 2 3 4 5 6 7

ξ = 01

ξ = 02

ξ = 03

Figure 6 )e influence of various ξ(ξ 01 02 03) on the option price


scheme (28) has excellent convergence behavior in spacedirection

52 Effect of Various Parameters )is section is devoted toinvestigate the dynamic behavior of FBSM equation withdifferent α In the following numerical experiments we fixξ 05 w 01 r 06 N 64Δt 001 and u0 |sinx|From Figures 3ndash5 we know that α has certain influence onthe solution as α increases and the solution becomessmoother In order to test the influence of ξ on the optionprice we set r 06 u0 sin(x2) and let ξ change at thesame time Figure 6 shows that the parameter ξ has asignificant effect on the price of options and there will bean inflection point around x 28 Finally we investigatedthe influence of r on the option price and the results areshown in Figure 7 it can be seen that when r increases theoption price also increases

6 Conclusion

In this paper a new full-discrete numerical method is de-veloped to solve the FBSM An efficient 2 minus α-order andunconditionally energy stable method is constructed bycombining the L1 approach in time and Fourier method inspace direction It is proved that the full-discrete convergesto the order O(Δt2minus α + Nminus s + Nminus m) globally Numericalexamples demonstrate the robustness and accuracy of thedeveloped full-discrete method numerically Finally we alsostudy the properties of the solution of the FBSM

Data Availability

)e data used to support the findings of this study areavailable from the corresponding author upon request

Conflicts of Interest

)e authors declare that they have no conflicts of interest

Authorsrsquo Contributions

Juan He carried out an efficient numerical approach to timefractional BlackndashScholes model Aiqing Zhang helped todraft the manuscript All authors read and approved the finalmanuscript

Acknowledgments

)e work of Juan He was supported by the China Schol-arship Council (no 202008520027) )e work of AiqingZhang was supported by the Cultivation Project of MajorScientic Research Projects of Central University of Financeand Economics (no 14ZZD007)

References

[1] F Black andM Scholes ldquo)e pricing of options and corporateliabilitiesrdquo Journal of Political Economy vol 81 no 3pp 637ndash654 1973

[2] R C Merton ldquo)eory of rational option pricingrdquo -e BellJournal of Economics and Management Science vol 4 no 1pp 141ndash183 1973

[3] R Cioczek-Georges and B B Mandelbrot ldquoAlternativemicropulses and fractional brownian motionrdquo StochasticProcesses and -eir Applications vol 64 no 2 pp 143ndash1521996

[4] W Min B P English G Luo B J Cherayil S C Kou andX S Xie ldquoFluctuating enzymes lessons from single-moleculestudiesrdquo Accounts of Chemical Research vol 38 no 12pp 923ndash931 2005

[5] T A M Langlands B I Henry and S L Wearne ldquoFractionalcable equation models for anomalous electrodiffusion in

01

02

03

05

04

06

07

08

0 1 2 3 4 5 6 7

r = 03

r = 04

r = 05

Figure 7 )e influence of various r(ξ 03 04 05) on the option price


nerve cells infinite domain solutionsrdquo Journal of Mathe-matical Biology vol 59 no 6 pp 761ndash808 2009

[6] G Jumarie ldquoStock exchange fractional dynamics defined asfractional exponential growth driven by (usual) Gaussianwhite noise application to fractional black-scholes equa-tionsrdquo Insurance Mathematics and Economics vol 42 no 1pp 271ndash287 2007

[7] J R Liang J Wang W J Zhang W Y Qiu and F Y Renldquo)e solution to a bifractional black-scholes-merton differ-ential equationrdquo International Journal of Pure and AppliedMathematics vol 58 no 1 pp 99ndash112 2010

[8] G Jumarie ldquoDerivation and solutions of some fractionalBlack-Scholes equations in coarse-grained space and timeApplication to Mertonrsquos optimal portfoliordquo Computers ampMathematics with Applications vol 59 no 3 pp 1142ndash11642010

[9] W Chen X Xu and S-P Zhu ldquoAnalytically pricing doublebarrier options based on a time-fractional black-scholesequationrdquo Computers amp Mathematics with Applicationsvol 69 no 12 pp 1407ndash1419 2015

[10] X J Yang F Gao J A Machado and D Baleanu ldquoA newfractional derivative involving the normalized sinc functionwithout singular kernelrdquo European Physical Journal SpecialTopics vol 226 no 16-18 pp 3567ndash3575 2017

[11] X-J Yang F Gao Y Ju and H-W Zhou ldquoFundamentalsolutions of the general fractional-order diffusion equationsrdquoMathematical Methods in the Applied Sciences vol 41 no 18pp 9312ndash9320 2018

[12] M Inc A Yusuf A I Aliyu and D Baleanu ldquoDark andsingular optical solitons for the conformable space-timenonlinear Schrodinger equation with Kerr and power lawnonlinearityrdquo Optik vol 162 pp 65ndash75 2018

[13] M Inc A I Aliyu and A Yusuf ldquoDark optical singularsolitons and conservation laws to the nonlinear Schrodingerrsquosequation with spatio-temporal dispersionrdquo Modern PhysicsLetters B vol 31 no 14 p 1750163 2017

[14] J Singh D Kumar and D Baleanu ldquoOn the analysis offractional diabetes model with exponential lawrdquo Advances inDifference Equations vol 231 no 1 2018

[15] J Singh D Kumar and D Baleanu ldquoNew aspects of fractionalbiswas-milovic model with mittag-leffler lawrdquo MathematicalModelling of Natural Phenomena vol 14 no 3 p 303 2019

[16] J Singh D Kumar D Baleanu and S Rathore ldquoOn the localfractional wave equation in fractal stringsrdquo MathematicalMethods in the Applied Sciences vol 42 no 5 pp 1588ndash15952019

[17] K B Oldham and J Spanier ldquo)e fractional calculusrdquoMathematical Gazette vol 56 no 247 pp 396ndash400 1974

[18] T A M Langlands and B I Henry ldquo)e accuracy andstability of an implicit solution method for the fractionaldiffusion equationrdquo Journal of Computational Physicsvol 205 no 2 pp 719ndash736 2005

[19] Z-z Sun and X Wu ldquoA fully discrete difference scheme for adiffusion-wave systemrdquo Applied Numerical Mathematicsvol 56 no 2 pp 193ndash209 2006

[20] Y Lin and C Xu ldquoFinite differencespectral approximationsfor the time-fractional diffusion equationrdquo Journal of Com-putational Physics vol 225 no 2 pp 1533ndash1552 2007

[21] R H De Staelena and A S Hendybc ldquoFractional cableequationmodels for anomalous electrodiffusion in nerve cellsinfinite domain solutionsrdquo Computers and Mathematics withApplications vol 74 no 6 pp 1166ndash1175 2017

[22] H Zhang F Liu I Turner and Q Yang ldquoNumerical solutionof the time fractional black-scholes model governing

european optionsrdquo Computers amp Mathematics with Appli-cations vol 71 no 9 pp 1772ndash1783 2016

[23] C Li T Zhao W Deng and Y Wu ldquoOrthogonal splinecollocation methods for the subdiffusion equationrdquo Journal ofComputational and Applied Mathematics vol 255 pp 517ndash528 2014

[24] C-M Chen F Liu I Turner and V Anh ldquoA fourier methodfor the fractional diffusion equation describing sub-diffusionrdquoJournal of Computational Physics vol 227 no 2 pp 886ndash8972007

[25] A Quarteroni and A Valli Numerical Approximation ofPartial Differential Equations Springer Berlin Germany1994


for double barrier options are often biased when pricechanges are considered as fractal transmission systemsChen et al [9] revealed that it would be better to usefractional order BlackndashScholes equation to explain thepricing in fractal transmission systems

)ere is a lot of work in the modeling and calculation offractional equations Yang et al [10 11] developed a newdefinition of fractional derivative )e advantage of thisdefinition is that it does not contain singular kernel Inc et al[12 13] studied the isolated solutions of a class of fractionalequations with Kerr law nonlinearity by RiccatindashBernoullimethod )e exact dark optical and periodic singular solitonsolution is obtained Singhet al [14ndash16] solved a series offractional equations by using homotopy analysis techniqueand Laplace transform algorithm

As we all know as an effective method L1 formula [17]has been widely used in the calculation of fractional dif-ferential equations Langlands and Henry [18] Sun and Wu[19] and Lin and Xu [20] discussed the error estimate for L1scheme De Staelena and Hendybc [21] constructed a nu-merical method of fourth-order finite difference in space and2 minus α in time Stability uniqueness and error estimates areanalyzed Zhang et al [22] presented a discrete implicit finitescheme to solve time fractional BlackndashScholes model )eydiscussed the stability and error estimation of numericalschemes by Fourier analysis Unfortunately their analysismethods are local approach Due to the importance ofFBSM it is necessary to reconstruct an efficient numericalmethod and analyze global stability and error estimates

In this work we will develop an efficient full-discretescheme to approximate the BlackndashScholes model withα-order fractional derivative We apply L1 method to dis-cretize the direction of time and Fourier-spectral method todiscretize the direction of space Using the energy analysismethod we discuss the stability and error estimate of thefully discrete numerical method)e detailed analysis showsthat the scheme is unconditionally energy stable and theerror estimates indicate that our full-discrete scheme canachieve 2 minus α-order accuracy in time and exponential ac-curacy in space direction Finally some numerical examplesare conducted to support the theoretical claims At the sametime the dynamic behavior of FBSM is studied by theproposed method

We organize the rest of the paper as follows Section 2will briefly introduce the FBSM In Section 3 we develop atime-discrete method for FBSM and then present its discreteenergy law In Section 4 we will study the error estimate ofthe full-discrete scheme In Section 5 we present accuracystability tests and numerous numerical examples to dem-onstrate the validity of the full-discrete method In additionwe will discuss the properties of the solution of the FBSMSome concluding remarks are given in Section 6

2 BlackndashScholes Model

In this work we will consider the following time fractionalBlackndashScholes model

zαV(S t)

ztα +

12σ2S2

z2V(S t)

zS2

+ rSzV(S t)

zSminus rV(S t) 0 (S t) isin (0 +infin) times(0 T)

(4)

V(0 t) p(t)

V(+infin t) q(t)(5)

V(S T) w(S) (6)

where V is the price of the option S the price of the un-derlying asset r the interest rate and σ the volatility of thestock price 0lt αle 1 the time fractional derivativezαV(S t)ztα is defined by

zαV(middot t)

ztα z

αt V(middot t)

1Γ(1 minus α)

1113946t

0(t minus μ)

minus αzV(middot μ)

zμdμ

(7)

)is is a linear parabolic partial differential equationwhich has been studied extensively

We transform the problem to an initial value problem byusing the time to mature t T minus τ and we then setx ln S V(S τ) u(ex T minus τ) we can rewrite (4) as

zαt u(x τ) minus ξz


+ ru(x τ) 0 (x τ) isin (0 +infin) times(0 T)(8)

where ξ 12σ2ω r minus ξ and with the following boundary(barrier) and initial conditions

u(minus infin τ) p(τ)

u(minus infin τ) q(τ)

u(x 0) u0(x) altxlt b

(9)

In order to solve the above model by numerical methodit is necessary to truncate the original unbounded regioninto a finite interval )erefore we will consider problem (8)in bounded interval (0 2π) )en we will study the fol-lowing problem



+ ru(x τ) 0 (x τ) isin (0 2π) times(0 T)(10)

u(0 τ) u(2π τ) (11)

u(x 0) u0(x) (12)

Remark 1 In fact one can choose homogeneous or inho-mogeneous boundary conditions It all depends on theactual option price We have tested it and it does not makeany difference in actual numerical examples




zαt u(x τ)

1Γ(2 minus α)

1113944

n

j0bj



1113872 1113873

(13)





1a0

un+1minus 1113944

nminus 1



2xu

n+1

minus ωzxun+1

+ run+1 0 nge 0

(15)


1a0


1minus ωzxu

1+ ru1 0 (16)



uk+1

le u0

k 0 1 K (17)



1 u11113872 1113873


(18)


2(A minus B A) A2

minus B2

+(A minus B)2 (19)

)us

u1

2

minus u0

2

+ u1 minus u0

2

+ 2a0a zxu12

+ 2a0r u12

0

(20)


u1

2le u0

2 (21)


uj

2le u0

2 j 2 3 n (22)


2 un+12

+ 2a0ξ zxun+1

2

+ 2a0r un+1

2

2 1113944nminus 1


+ bnu0un+1⎛⎝ ⎞⎠

le 1113944nminus 1


2

+ un+12

1113874 1113875

+ bn u0

2

+ un+12

1113874 1113875

(23)

Note the fact that

1113944

nminus 1

j0bj minus bj+11113872 1113873 + bn 1 (24)

)us we get

un+12le 1113944

nminus 1


2

+ bn u0

2

le 1113944nminus 1

j0bj minus bj+11113872 1113873 + bn

⎛⎝ ⎞⎠ u0

2

u0

2

(25)

)is yields (17)





ϕ minus πNϕ

lleCNlminus m


(Ω) mgt lge 0

(27)


1a0

un+1N minus 1113944

nminus 1


N minus bnu0NϕN

⎛⎝ ⎞⎠

minus ξ z2xu

n+1N ϕN1113872 1113873 minus ω zxu

n+1ϕN1113872 1113873 + r un+1

ϕN1113872 1113873 0 ϕN isin SN

(28)




derive


un+1N

2le u0N

2 (29)


Rn+1

1Γ(2 minus α)

1113944

n

j0bj


Δtα


1113946tn+1

0

zu(middot μ)

zμdμ


(30)


Rn+1




N


enu 1113954e

nu + 1113957e


N

(32)



Proof

bn (n + 1)1minus α

minus (n)1minus α

1113872 1113873

n1minus α 1 +

1n

1113874 11138751minus α

minus 11113888 1113889


n+


2



n+


2

1n21113888 1113889


2n+

(1 minus α)

2n+


2

1n21113888 1113889

(34)

Note that

(1 minus α)

2n+


2

1n2 ge 0 (35)

)erefore we obtain

a0

bn

le2n



α (36)



u τk1113872 1113873 minus uk

leC Δt2minus α+ N

1minus m1113872 1113873 k 0 1 K

T

Δt

(37)


1a0


1NϕN1113872 1113873 minus ω zxu

1N ϕN1113872 1113873 + r u1N ϕN1113872 1113873 0

(38)



u1N minus u0Na0



a0



a0



a0

R1

+1a0


1113957e1u minus 1113957e

0u1113872 1113873


2xu

1N z




2xu

1N


2x1113957e

1u






(39)

)en we have

1113957e1u minus 1113957e

0u ϕN1113872 1113873 + a0ξ zx1113957e


1u ϕN1113872 1113873 + a0r 1113957e

1uϕN1113872 1113873





(40)



2 1113957e1u

2

+ 2a0ξ zx1113957e1u

2

+ 2a0r 1113957e1u

2

le a01

rb0R12

+ rb0 1113957e1u

2

1113888 1113889

+ a0ξ1b0


2

+ b0 zx1113957e1u

2

1113888 1113889

+ a0ωωξb0


2

+ξb0

ωzx1113957e

1u

2

1113888 1113889

(41)


1113957e1u

2leC1

a0

b0Δt4minus 2α

+ C2a0

b0N

1minus m+ C3

a0

b0N

minus m (42)

Assume

1113957eku

2leC1

a0

bkminus 1Δt4 + C2

a0

bkminus 1N

1minus m+ C3

a0

bjminus 1N

minus m k 2 3 n

(43)


1113957en+1u minus 1113944

nminus 1

j0bj minus bj+11113872 11138731113957e


0u ϕN


n+1u ϕN1113872 1113873 + a0r 1113957e

n+1u ϕN1113872 1113873

minus a0 Rn+1



⎛⎝ ⎞⎠


(44)


2 1113957en+1u

2

+ 2a0ξ zx1113957en+1u

2

+ 2a0r 1113957en+1u

2


j0bj minus bj+11113872 11138731113957e


0u 1113957e

n+1u


1113957en+1u1113872 1113873


n+1u1113872 1113873

le 1113944nminus 1

j0bj minus bj+11113872 1113873 1113957e

nminus ju

2

+ 1113957en+1u

2

1113874 1113875 + a01r

Rn+1

2

+ r 1113957en+1u

2

1113874 1113875


2

+ zx1113957en+1u

2

1113874 1113875 + a0ωωξ


2

+ξω

zx1113957en+1u

2

1113888 1113889

(45)

)us we have

1113957en+1u

2le 1113944

nminus 1

j0bj minus bj+11113872 1113873 1113957e

nminus ju

2

+a0

rR

n+12


2

+a0b

2


2

le 1113944nminus 1

j0bj minus bj+11113872 1113873 C1

a0

bjminus 1Δt4 + C2

a0

bjminus 1N

1minus m+ C3

a0

bjminus 1N

minus m1113888 1113889

+ bn C1a0

bn

Δt4 + C2a0

bn

N1minus m

+ C3a0

bn

Nminus m

1113888 1113889

(46)


1113957en+1u

2le C1

a0

bn

Δt4 + C2a0

bn

N1minus m

+ C3a0

bn

Nminus m

1113888 1113889

1113944

nminus 1

j0bj minus bj+11113872 1113873 + bn

⎛⎝ ⎞⎠

C1a0

bn

Δt4 + C2a0

bn

N1minus m

+ C3a0

bn

Nminus m

(47)

Note that

u τk1113872 1113873 minus uk

le 1113954eku

+ 1113957eku

(48)

)is ends the proof








(49)

where

f(x τ) 2T

2minus α


(50)






αΔt Δt

1200Δt

1400Δt

1800Δt

11600Δt

13200α 01 18166 18263 18345 18415 18475α 03 16680 16746 16797 16837 16869α 05 14892 14925 14947 14963 14974α 06 13939 13960 13974 13983 13989α 07 12964 12979 12987 12992 12995α 09 10980 10990 10995 10997 10999


10ndash6

10ndash4

10ndash2

Time step

Erro

rs in

tim

e




Erro

rs in

spac

e

10ndash15

10ndash10

10ndash5

100

N

14 16 18 20 22 24 26

L2 error

Linfin error


1

05

06

4

2x0 0

0204

0608

1

τ

u (x

τ)



1

05

06

4

2x0 0

0204

0608

1

τ

u (x

τ)


1

05

06

42x

0 002

0406

081

τ

u (x

τ)


01

02

03

04

05

06

0 1 2 3 4 5 6 7

ξ = 01

ξ = 02

ξ = 03





6 Conclusion


Data Availability






Acknowledgments


References






01

02

03

05

04

06

07

08

0 1 2 3 4 5 6 7

r = 03

r = 04

r = 05




























zαt u(x τ)

1Γ(2 minus α)

1113944

n

j0bj



1113872 1113873

(13)





1a0

un+1minus 1113944

nminus 1



2xu

n+1

minus ωzxun+1

+ run+1 0 nge 0

(15)


1a0


1minus ωzxu

1+ ru1 0 (16)



uk+1

le u0

k 0 1 K (17)



1 u11113872 1113873


(18)


2(A minus B A) A2

minus B2

+(A minus B)2 (19)

)us

u1

2

minus u0

2

+ u1 minus u0

2

+ 2a0a zxu12

+ 2a0r u12

0

(20)


u1

2le u0

2 (21)


uj

2le u0

2 j 2 3 n (22)


2 un+12

+ 2a0ξ zxun+1

2

+ 2a0r un+1

2

2 1113944nminus 1


+ bnu0un+1⎛⎝ ⎞⎠

le 1113944nminus 1


2

+ un+12

1113874 1113875

+ bn u0

2

+ un+12

1113874 1113875

(23)

Note the fact that

1113944

nminus 1

j0bj minus bj+11113872 1113873 + bn 1 (24)

)us we get

un+12le 1113944

nminus 1


2

+ bn u0

2

le 1113944nminus 1

j0bj minus bj+11113872 1113873 + bn

⎛⎝ ⎞⎠ u0

2

u0

2

(25)

)is yields (17)





ϕ minus πNϕ

lleCNlminus m


(Ω) mgt lge 0

(27)


1a0

un+1N minus 1113944

nminus 1


N minus bnu0NϕN

⎛⎝ ⎞⎠

minus ξ z2xu

n+1N ϕN1113872 1113873 minus ω zxu

n+1ϕN1113872 1113873 + r un+1

ϕN1113872 1113873 0 ϕN isin SN

(28)




derive


un+1N

2le u0N

2 (29)


Rn+1

1Γ(2 minus α)

1113944

n

j0bj


Δtα


1113946tn+1

0

zu(middot μ)

zμdμ


(30)


Rn+1




N


enu 1113954e

nu + 1113957e


N

(32)



Proof

bn (n + 1)1minus α

minus (n)1minus α

1113872 1113873

n1minus α 1 +

1n

1113874 11138751minus α

minus 11113888 1113889


n+


2



n+


2

1n21113888 1113889


2n+

(1 minus α)

2n+


2

1n21113888 1113889

(34)

Note that

(1 minus α)

2n+


2

1n2 ge 0 (35)

)erefore we obtain

a0

bn

le2n



α (36)



u τk1113872 1113873 minus uk

leC Δt2minus α+ N

1minus m1113872 1113873 k 0 1 K

T

Δt

(37)


1a0


1NϕN1113872 1113873 minus ω zxu

1N ϕN1113872 1113873 + r u1N ϕN1113872 1113873 0

(38)



u1N minus u0Na0



a0



a0



a0

R1

+1a0


1113957e1u minus 1113957e

0u1113872 1113873


2xu

1N z




2xu

1N


2x1113957e

1u






(39)

)en we have

1113957e1u minus 1113957e

0u ϕN1113872 1113873 + a0ξ zx1113957e


1u ϕN1113872 1113873 + a0r 1113957e

1uϕN1113872 1113873





(40)



2 1113957e1u

2

+ 2a0ξ zx1113957e1u

2

+ 2a0r 1113957e1u

2

le a01

rb0R12

+ rb0 1113957e1u

2

1113888 1113889

+ a0ξ1b0


2

+ b0 zx1113957e1u

2

1113888 1113889

+ a0ωωξb0


2

+ξb0

ωzx1113957e

1u

2

1113888 1113889

(41)


1113957e1u

2leC1

a0

b0Δt4minus 2α

+ C2a0

b0N

1minus m+ C3

a0

b0N

minus m (42)

Assume

1113957eku

2leC1

a0

bkminus 1Δt4 + C2

a0

bkminus 1N

1minus m+ C3

a0

bjminus 1N

minus m k 2 3 n

(43)


1113957en+1u minus 1113944

nminus 1

j0bj minus bj+11113872 11138731113957e


0u ϕN


n+1u ϕN1113872 1113873 + a0r 1113957e

n+1u ϕN1113872 1113873

minus a0 Rn+1



⎛⎝ ⎞⎠


(44)


2 1113957en+1u

2

+ 2a0ξ zx1113957en+1u

2

+ 2a0r 1113957en+1u

2


j0bj minus bj+11113872 11138731113957e


0u 1113957e

n+1u


1113957en+1u1113872 1113873


n+1u1113872 1113873

le 1113944nminus 1

j0bj minus bj+11113872 1113873 1113957e

nminus ju

2

+ 1113957en+1u

2

1113874 1113875 + a01r

Rn+1

2

+ r 1113957en+1u

2

1113874 1113875


2

+ zx1113957en+1u

2

1113874 1113875 + a0ωωξ


2

+ξω

zx1113957en+1u

2

1113888 1113889

(45)

)us we have

1113957en+1u

2le 1113944

nminus 1

j0bj minus bj+11113872 1113873 1113957e

nminus ju

2

+a0

rR

n+12


2

+a0b

2


2

le 1113944nminus 1

j0bj minus bj+11113872 1113873 C1

a0

bjminus 1Δt4 + C2

a0

bjminus 1N

1minus m+ C3

a0

bjminus 1N

minus m1113888 1113889

+ bn C1a0

bn

Δt4 + C2a0

bn

N1minus m

+ C3a0

bn

Nminus m

1113888 1113889

(46)


1113957en+1u

2le C1

a0

bn

Δt4 + C2a0

bn

N1minus m

+ C3a0

bn

Nminus m

1113888 1113889

1113944

nminus 1

j0bj minus bj+11113872 1113873 + bn

⎛⎝ ⎞⎠

C1a0

bn

Δt4 + C2a0

bn

N1minus m

+ C3a0

bn

Nminus m

(47)

Note that

u τk1113872 1113873 minus uk

le 1113954eku

+ 1113957eku

(48)

)is ends the proof








(49)

where

f(x τ) 2T

2minus α


(50)






αΔt Δt

1200Δt

1400Δt

1800Δt

11600Δt

13200α 01 18166 18263 18345 18415 18475α 03 16680 16746 16797 16837 16869α 05 14892 14925 14947 14963 14974α 06 13939 13960 13974 13983 13989α 07 12964 12979 12987 12992 12995α 09 10980 10990 10995 10997 10999


10ndash6

10ndash4

10ndash2

Time step

Erro

rs in

tim

e




Erro

rs in

spac

e

10ndash15

10ndash10

10ndash5

100

N

14 16 18 20 22 24 26

L2 error

Linfin error


1

05

06

4

2x0 0

0204

0608

1

τ

u (x

τ)



1

05

06

4

2x0 0

0204

0608

1

τ

u (x

τ)


1

05

06

42x

0 002

0406

081

τ

u (x

τ)


01

02

03

04

05

06

0 1 2 3 4 5 6 7

ξ = 01

ξ = 02

ξ = 03





6 Conclusion


Data Availability






Acknowledgments


References






01

02

03

05

04

06

07

08

0 1 2 3 4 5 6 7

r = 03

r = 04

r = 05


























un+1N

2le u0N

2 (29)


Rn+1

1Γ(2 minus α)

1113944

n

j0bj


Δtα


1113946tn+1

0

zu(middot μ)

zμdμ


(30)


Rn+1




N


enu 1113954e

nu + 1113957e


N

(32)



Proof

bn (n + 1)1minus α

minus (n)1minus α

1113872 1113873

n1minus α 1 +

1n

1113874 11138751minus α

minus 11113888 1113889


n+


2



n+


2

1n21113888 1113889


2n+

(1 minus α)

2n+


2

1n21113888 1113889

(34)

Note that

(1 minus α)

2n+


2

1n2 ge 0 (35)

)erefore we obtain

a0

bn

le2n



α (36)



u τk1113872 1113873 minus uk

leC Δt2minus α+ N

1minus m1113872 1113873 k 0 1 K

T

Δt

(37)


1a0


1NϕN1113872 1113873 minus ω zxu

1N ϕN1113872 1113873 + r u1N ϕN1113872 1113873 0

(38)



u1N minus u0Na0



a0



a0



a0

R1

+1a0


1113957e1u minus 1113957e

0u1113872 1113873


2xu

1N z




2xu

1N


2x1113957e

1u






(39)

)en we have

1113957e1u minus 1113957e

0u ϕN1113872 1113873 + a0ξ zx1113957e


1u ϕN1113872 1113873 + a0r 1113957e

1uϕN1113872 1113873





(40)



2 1113957e1u

2

+ 2a0ξ zx1113957e1u

2

+ 2a0r 1113957e1u

2

le a01

rb0R12

+ rb0 1113957e1u

2

1113888 1113889

+ a0ξ1b0


2

+ b0 zx1113957e1u

2

1113888 1113889

+ a0ωωξb0


2

+ξb0

ωzx1113957e

1u

2

1113888 1113889

(41)


1113957e1u

2leC1

a0

b0Δt4minus 2α

+ C2a0

b0N

1minus m+ C3

a0

b0N

minus m (42)

Assume

1113957eku

2leC1

a0

bkminus 1Δt4 + C2

a0

bkminus 1N

1minus m+ C3

a0

bjminus 1N

minus m k 2 3 n

(43)


1113957en+1u minus 1113944

nminus 1

j0bj minus bj+11113872 11138731113957e


0u ϕN


n+1u ϕN1113872 1113873 + a0r 1113957e

n+1u ϕN1113872 1113873

minus a0 Rn+1



⎛⎝ ⎞⎠


(44)


2 1113957en+1u

2

+ 2a0ξ zx1113957en+1u

2

+ 2a0r 1113957en+1u

2


j0bj minus bj+11113872 11138731113957e


0u 1113957e

n+1u


1113957en+1u1113872 1113873


n+1u1113872 1113873

le 1113944nminus 1

j0bj minus bj+11113872 1113873 1113957e

nminus ju

2

+ 1113957en+1u

2

1113874 1113875 + a01r

Rn+1

2

+ r 1113957en+1u

2

1113874 1113875


2

+ zx1113957en+1u

2

1113874 1113875 + a0ωωξ


2

+ξω

zx1113957en+1u

2

1113888 1113889

(45)

)us we have

1113957en+1u

2le 1113944

nminus 1

j0bj minus bj+11113872 1113873 1113957e

nminus ju

2

+a0

rR

n+12


2

+a0b

2


2

le 1113944nminus 1

j0bj minus bj+11113872 1113873 C1

a0

bjminus 1Δt4 + C2

a0

bjminus 1N

1minus m+ C3

a0

bjminus 1N

minus m1113888 1113889

+ bn C1a0

bn

Δt4 + C2a0

bn

N1minus m

+ C3a0

bn

Nminus m

1113888 1113889

(46)


1113957en+1u

2le C1

a0

bn

Δt4 + C2a0

bn

N1minus m

+ C3a0

bn

Nminus m

1113888 1113889

1113944

nminus 1

j0bj minus bj+11113872 1113873 + bn

⎛⎝ ⎞⎠

C1a0

bn

Δt4 + C2a0

bn

N1minus m

+ C3a0

bn

Nminus m

(47)

Note that

u τk1113872 1113873 minus uk

le 1113954eku

+ 1113957eku

(48)

)is ends the proof








(49)

where

f(x τ) 2T

2minus α


(50)






αΔt Δt

1200Δt

1400Δt

1800Δt

11600Δt

13200α 01 18166 18263 18345 18415 18475α 03 16680 16746 16797 16837 16869α 05 14892 14925 14947 14963 14974α 06 13939 13960 13974 13983 13989α 07 12964 12979 12987 12992 12995α 09 10980 10990 10995 10997 10999


10ndash6

10ndash4

10ndash2

Time step

Erro

rs in

tim

e




Erro

rs in

spac

e

10ndash15

10ndash10

10ndash5

100

N

14 16 18 20 22 24 26

L2 error

Linfin error


1

05

06

4

2x0 0

0204

0608

1

τ

u (x

τ)



1

05

06

4

2x0 0

0204

0608

1

τ

u (x

τ)


1

05

06

42x

0 002

0406

081

τ

u (x

τ)


01

02

03

04

05

06

0 1 2 3 4 5 6 7

ξ = 01

ξ = 02

ξ = 03





6 Conclusion


Data Availability






Acknowledgments


References






01

02

03

05

04

06

07

08

0 1 2 3 4 5 6 7

r = 03

r = 04

r = 05


























2 1113957e1u

2

+ 2a0ξ zx1113957e1u

2

+ 2a0r 1113957e1u

2

le a01

rb0R12

+ rb0 1113957e1u

2

1113888 1113889

+ a0ξ1b0


2

+ b0 zx1113957e1u

2

1113888 1113889

+ a0ωωξb0


2

+ξb0

ωzx1113957e

1u

2

1113888 1113889

(41)


1113957e1u

2leC1

a0

b0Δt4minus 2α

+ C2a0

b0N

1minus m+ C3

a0

b0N

minus m (42)

Assume

1113957eku

2leC1

a0

bkminus 1Δt4 + C2

a0

bkminus 1N

1minus m+ C3

a0

bjminus 1N

minus m k 2 3 n

(43)


1113957en+1u minus 1113944

nminus 1

j0bj minus bj+11113872 11138731113957e


0u ϕN


n+1u ϕN1113872 1113873 + a0r 1113957e

n+1u ϕN1113872 1113873

minus a0 Rn+1



⎛⎝ ⎞⎠


(44)


2 1113957en+1u

2

+ 2a0ξ zx1113957en+1u

2

+ 2a0r 1113957en+1u

2


j0bj minus bj+11113872 11138731113957e


0u 1113957e

n+1u


1113957en+1u1113872 1113873


n+1u1113872 1113873

le 1113944nminus 1

j0bj minus bj+11113872 1113873 1113957e

nminus ju

2

+ 1113957en+1u

2

1113874 1113875 + a01r

Rn+1

2

+ r 1113957en+1u

2

1113874 1113875


2

+ zx1113957en+1u

2

1113874 1113875 + a0ωωξ


2

+ξω

zx1113957en+1u

2

1113888 1113889

(45)

)us we have

1113957en+1u

2le 1113944

nminus 1

j0bj minus bj+11113872 1113873 1113957e

nminus ju

2

+a0

rR

n+12


2

+a0b

2


2

le 1113944nminus 1

j0bj minus bj+11113872 1113873 C1

a0

bjminus 1Δt4 + C2

a0

bjminus 1N

1minus m+ C3

a0

bjminus 1N

minus m1113888 1113889

+ bn C1a0

bn

Δt4 + C2a0

bn

N1minus m

+ C3a0

bn

Nminus m

1113888 1113889

(46)


1113957en+1u

2le C1

a0

bn

Δt4 + C2a0

bn

N1minus m

+ C3a0

bn

Nminus m

1113888 1113889

1113944

nminus 1

j0bj minus bj+11113872 1113873 + bn

⎛⎝ ⎞⎠

C1a0

bn

Δt4 + C2a0

bn

N1minus m

+ C3a0

bn

Nminus m

(47)

Note that

u τk1113872 1113873 minus uk

le 1113954eku

+ 1113957eku

(48)

)is ends the proof








(49)

where

f(x τ) 2T

2minus α


(50)






αΔt Δt

1200Δt

1400Δt

1800Δt

11600Δt

13200α 01 18166 18263 18345 18415 18475α 03 16680 16746 16797 16837 16869α 05 14892 14925 14947 14963 14974α 06 13939 13960 13974 13983 13989α 07 12964 12979 12987 12992 12995α 09 10980 10990 10995 10997 10999


10ndash6

10ndash4

10ndash2

Time step

Erro

rs in

tim

e




Erro

rs in

spac

e

10ndash15

10ndash10

10ndash5

100

N

14 16 18 20 22 24 26

L2 error

Linfin error


1

05

06

4

2x0 0

0204

0608

1

τ

u (x

τ)



1

05

06

4

2x0 0

0204

0608

1

τ

u (x

τ)


1

05

06

42x

0 002

0406

081

τ

u (x

τ)


01

02

03

04

05

06

0 1 2 3 4 5 6 7

ξ = 01

ξ = 02

ξ = 03





6 Conclusion


Data Availability






Acknowledgments


References






01

02

03

05

04

06

07

08

0 1 2 3 4 5 6 7

r = 03

r = 04

r = 05
































(49)

where

f(x τ) 2T

2minus α


(50)






αΔt Δt

1200Δt

1400Δt

1800Δt

11600Δt

13200α 01 18166 18263 18345 18415 18475α 03 16680 16746 16797 16837 16869α 05 14892 14925 14947 14963 14974α 06 13939 13960 13974 13983 13989α 07 12964 12979 12987 12992 12995α 09 10980 10990 10995 10997 10999


10ndash6

10ndash4

10ndash2

Time step

Erro

rs in

tim

e




Erro

rs in

spac

e

10ndash15

10ndash10

10ndash5

100

N

14 16 18 20 22 24 26

L2 error

Linfin error


1

05

06

4

2x0 0

0204

0608

1

τ

u (x

τ)



1

05

06

4

2x0 0

0204

0608

1

τ

u (x

τ)


1

05

06

42x

0 002

0406

081

τ

u (x

τ)


01

02

03

04

05

06

0 1 2 3 4 5 6 7

ξ = 01

ξ = 02

ξ = 03





6 Conclusion


Data Availability






Acknowledgments


References






01

02

03

05

04

06

07

08

0 1 2 3 4 5 6 7

r = 03

r = 04

r = 05


























1

05

06

4

2x0 0

0204

0608

1

τ

u (x

τ)


1

05

06

42x

0 002

0406

081

τ

u (x

τ)


01

02

03

04

05

06

0 1 2 3 4 5 6 7

ξ = 01

ξ = 02

ξ = 03





6 Conclusion


Data Availability






Acknowledgments


References






01

02

03

05

04

06

07

08

0 1 2 3 4 5 6 7

r = 03

r = 04

r = 05




























6 Conclusion


Data Availability






Acknowledgments


References






01

02

03

05

04

06

07

08

0 1 2 3 4 5 6 7

r = 03

r = 04

r = 05

















































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