Theory and Modern Applications

# A collocation method based on cubic trigonometric B-splines for the numerical simulation of the time-fractional diffusion equation

## Abstract

Fractional differential equations sufficiently depict the nature in view of the symmetry properties, which portray physical and biological models. In this paper, we present a proficient collocation method based on cubic trigonometric B-Splines (CuTBSs) for time-fractional diffusion equations (TFDEs). The methodology involves discretization of the Caputo time-fractional derivatives using the typical finite difference scheme with space derivatives approximated using CuTBSs. A stability analysis is performed to establish that the errors do not magnify. A convergence analysis is also performed The numerical solution is obtained as a piecewise sufficiently smooth continuous curve, so that the solution can be approximated at any point in the given domain. Numerical tests are efficiently performed to ensure the correctness and viability of the scheme, and the results contrast with those of some current numerical procedures. The comparison uncovers that the proposed scheme is very precise and successful.

## 1 Introduction

The time-fractional diffusion equation (TFDE) is given as

$${}^{C}_{a}{D}_{t}^{\gamma }u(s,t)- \frac{\partial ^{2}}{\partial s^{2}}u(s,t)=f(s,t), \quad 0< \gamma < 1,$$
(1)

subject to the following initial condition (IC) and boundary conditions (BCs):

\begin{aligned}& u(s,0)=\varphi (s),\quad a\leq s \leq b, \end{aligned}
(2)
\begin{aligned}& u(a,t)=\psi _{1}(t), \qquad u(b,t)=\psi _{2}(t),\quad t \geq 0, \end{aligned}
(3)

where the diffusion exponent is denoted by γ, and $${}^{C}_{a}{D}_{t}^{\gamma }u(s,t)$$ is the Caputo fractional derivative (CFD) of order γ given by 

$${}^{C}_{a}{D}_{s}^{\gamma } f(s)=J^{n-\gamma } D^{n} f(s)= \textstyle\begin{cases} \frac{1}{\Gamma (n-\gamma )} \int _{a}^{s} \frac{f^{(n)}(\xi )}{(s-\xi )^{\gamma +1-n}}\,d\xi , & n-1< \gamma < n \in \mathbb{N}, \\ \frac{d^{n}}{ds^{n}}f(s), & \gamma =n \in \mathbb{N}. \end{cases}$$
(4)

Note that $$\gamma =0$$ and $$\gamma =1$$ correspond to the classical Helmholtz and standard diffusion equations, respectively.

The fractional calculus  has gained keenness in numerous fields such as chemistry, plasma physics, material science, biology, fluid mechanics, and so on. The fractional-order differential and integral equations are reliable tools to describe physical models of interest more exactly than their integer-order counterparts. A variety of applications of fractional calculus and TFDEs can be found in . The numerical and approximate solutions play an important role in exploring applications of fractional partial differential equations. It is emphasized in many research papers that the fractional derivatives and integrals are more efficient tools for modeling the hereditary and memory effects of different processes and materials, in contrast with integer-order models, in which such effects are ignored.

Numerous analytical schemes are available for TFDEs [1, 1113]. Numerical techniques are developed continuously because exact solutions are available in very few cases. Numerous numerical procedures for solving TFDEs have been developed recently. Esmaeili and Garrappa  obtained numerical solutions of TFDEs by a pseudospectral scheme. Mustapha et al.  presented a discontinuous Petrov–Galerkin method for TFDEs. Zhuang and Liu  obtained implicit difference approximations for TFDEs. Karatay et al.  used the Crank–Nicholson approach to construct a scheme for TFDEs. A weighted average and explicit finite difference schemes were developed in [18, 19] for TFDEs. Murio  presented an unconditionally stable implicit scheme for TFDEs on a finite slab. Tasbozan et al.  introduced a numerical scheme using B-spline basis functions for space fractional subdiffusion equations. Huang et al.  presented a fully discrete discontinuous Galerkin method for TFDEs. Chen et al.  used the Fourier method to find approximate solutions of the fractional diffusion equation describing subdiffusion. Gao and Sun  presented a compact finite difference scheme for fractional subdiffusion equations using a compact finite difference scheme.

In this paper, we present a cubic trigonometric B-spline collocation method to obtain numerical solutions of TFDEs. The main motivation behind using B-splines is that the solutions are obtained in the form of piecewise continuous sufficiently smooth functions, enabling us to approximate the solution at any desired location in the domain. The stability and convergence analysis are also discussed to establish that the scheme does not propagate errors. Numerical tests are performed to affirm the feasibility and applicability of the method. The results are compared with those presented in [22, 23].

The rest of the paper is organized as follows. In Sect. 2, we derive a numerical procedure. The stability and convergence analysis of the scheme are presented in Sects. 3 and 4, respectively. In Sect. 5, we show a contrast of our numerical results with those of [22, 23]. Section 6 contains the outcomes of this study.

## 2 Materials and methods

### 2.1 Space discretization

Let the solution domain be $$[a,b]\times [0,T]$$. For given positive integers M and N, let $$\tau =\frac{T}{N}$$ be the temporal and $$h=\frac{b-a}{M}$$ the spatial step sizes, respectively. The space interval $$[a,b]$$ is uniformly partitioned as $$a=s_{0}< s_{1}<\cdots <s_{M}=b$$, where $$s_{i}=a+i h$$, $$i=0,\dots ,M$$. In this partition, the CuTBS function $$TB^{4}_{i} (x)$$  is defined as

$$TB^{4}_{i} (s)=\frac{1}{\beta } \textstyle\begin{cases} \sigma ^{3}(s_{i}), \\ \quad s\in [s_{i},s_{i+1}], \\ \sigma (s_{i})(\sigma (s_{i})\varphi (s_{i+2})+\varphi (s_{i+3}) \sigma (s_{i+1}))+\varphi (s_{i+4})\sigma ^{2}(s_{i+1}), \\ \quad s \in [s_{i+1},s_{i+2}], \\ \varphi (s_{i+4})(\sigma (s_{i+1})\varphi (s_{i+3})+\varphi (s_{i+4}) \sigma (s_{i+2}))+\sigma (s_{i})\varphi ^{2}(s_{i+3}), \\ \quad s \in [s_{i+2},s_{i+3}], \\ \varphi ^{3}(s_{i+4}), \\ \quad s \in [s_{i+3},s_{i+4}], \end{cases}$$
(5)

where

$$\sigma (s_{i} )=\sin \biggl( \frac{s-s_{i}}{2} \biggr), \qquad \varphi (s_{i} )=\sin \biggl( \frac{s_{i}-s}{2} \biggr),\qquad \beta =\sin \biggl(\frac{h}{2} \biggr) \sin (h ) \sin \biggl( \frac{3 h}{2} \biggr).$$

The support of the B-spline function $$TB^{4}_{i} (s)$$ is assumed to be $$[s_{i},s_{i+4}]$$. Note that each $$TB^{4}_{i}$$ is piecewise cubic and nonzero over four consecutive subintervals and vanishes otherwise. Consequently, each subinterval $$[s_{i},s_{i+1}]$$ contains three segments of $$TB^{4}_{i} (s)$$. Suppose that $$u(s,t)$$ and $$U(s,t)$$ are the analytic and numerical solutions of the given differential equation. We seek the approximation $$U(s,t)$$ to the solution $$u(s,t)$$ in terms of $$TB^{4}_{i}$$ as [26, 27]

$$u(s,t)\simeq U(s,t)=\sum_{i=-1}^{M+1} c_{i}(t) TB^{4}_{i} (s),$$
(6)

where $$c_{i}(t)$$ are unknowns, which are to be determined using the collocation method by utilizing the initial and boundary conditions. Using (5) and (6), the values of $$U(x,t)$$ and its necessary derivatives at the nodal points are determined in terms of the parameters $$c_{i}$$ as follows:

$$\textstyle\begin{cases} { U(s_{i},t)=\varrho _{1} c_{i-1}(t)+\varrho _{2} c_{i}(t)+ \varrho _{1} c_{i+1}(t)}, \\ { U_{s}(s_{i},t)=-\varrho _{3} c_{i-1}(t)+\varrho _{3} c_{i+1}(t)}, \\ { U_{ss}(s_{i},t)=\varrho _{4} c_{i-1}(t)+ \varrho _{5} c_{i}(t) +\varrho _{4} c_{i+1}(t),} \end{cases}$$
(7)

where

\begin{aligned}& {\varrho _{1}=-\frac{1}{2\sin (h)\sin (\frac{3h}{2})}\bigl( \cos (h)-1 \bigr)},\\& {\varrho _{2}=-\frac{2}{4\sin ^{2}(\frac{h}{2})-3}},\\& {\varrho _{3}=\frac{3}{4\sin (\frac{3h}{2})}},\\& {\varrho _{4}=\frac{3}{4} \biggl( \frac{3\sin ^{2}(\frac{h}{2})-2}{\sin ^{2}(\frac{h}{4})-\sin ^{2}(\frac{5h}{4})} \biggr)},\\& {\varrho _{5}=- \frac{3}{(4 \cos (h)+2)(\tan ^{2}(\frac{h}{2}))}}. \end{aligned}

### 2.2 Temporal discretization

To discretize the problem in time scale, we take the uniform partition on $$[0,T]$$ as $$0=t_{0}< t_{1}< t_{2}<\cdots <t_{N}=T$$ with $$\tau =t_{n+1}-t_{n}$$ for $$n=0,1,\ldots , N-1$$. Following , the CFD $${}^{C}_{a}{D}_{t}^{\gamma }u(s,t)$$ is discretized as

\begin{aligned} {}^{C}_{a}{D}_{t}^{\gamma }u(s,t_{n+1})&= \sum_{l=0}^{n} \kappa _{l} \frac{u(s,t_{n+1-l})-u(s,t_{n-l})}{\Gamma (2-\gamma )\tau ^{\gamma }}+r_{\tau }^{n+1} \\ &=\frac{1}{\alpha _{0}} \Biggl(u(s,t_{n+1})-\sum _{l=0}^{n-1} ( \kappa _{l}-\kappa _{l+1})u(s,t_{n-l})-\kappa _{n} u(s,t_{0}) \Biggr) +r_{\tau }^{n+1} \\ &= \frac{1}{\alpha _{0}} \Biggl(u^{n+1}-\sum _{l=0}^{n-1} ( \kappa _{l}-\kappa _{l+1})u^{n-l}-\kappa _{n} u^{0} \Biggr)+r_{\tau }^{n+1}, \end{aligned}
(8)

where $$u^{n}=u(s,t_{n})$$, $$\alpha _{0}=\tau ^{\gamma }\Gamma (2-\gamma )$$, $$\kappa _{l}=(l+1)^{1-\gamma }-l^{1-\gamma }$$, and $$r_{\tau }^{n+1}$$ is the truncation error. Note that

$$\textstyle\begin{cases} \kappa _{l} >0,\quad l=0,1,2, \ldots , N, \\ 1=\kappa _{0}>\kappa _{1}>\kappa _{2}>\cdots >\kappa _{l} \quad \text{and}\quad \kappa _{l}\rightarrow 0 \quad \text{as } l \rightarrow \infty , \\ \sum_{l=0}^{N-1} (\kappa _{l}-\kappa _{l+1})+\kappa _{N}=1. \end{cases}$$
(9)

It is shown in  that $$r_{\tau }^{n+1}$$ satisfies

$$r_{\tau }^{n+1}\leq C_{u} \tau ^{2-\gamma },$$

where the constant $$C_{u}$$ depends on u. Inserting (8) into (1) gives

$$u^{n+1}-\sum_{l=0}^{n-1} (\kappa _{l}-\kappa _{l+1})u^{n-l}- \kappa _{n} u^{0}-\alpha _{0} \bigl(u^{n+1} \bigr)_{ss}=\alpha _{0}f^{n+1}.$$
(10)

To obtain a full discretization, let $$c_{i}^{n}=c_{i}(t_{n})$$ and $$U_{i}^{n}=U(s_{i},t_{n})$$ for $$i=0,1,\ldots , M$$, $$n=0,1,\ldots , N$$. Now substituting (6) and (7) into (10), we get

\begin{aligned} &(\varrho _{1}-\alpha _{0} \varrho _{4})c_{i-1}^{n+1} +( \varrho _{2}-\alpha _{0} \varrho _{5})c_{i}^{n+1}+( \varrho _{1}- \alpha _{0} \varrho _{4})c_{i+1}^{n+1} \\ &\quad =\sum_{l=0}^{n-1}(\kappa _{l}-\kappa _{l+1}) \bigl(\varrho _{1} c_{i-1}^{n-l}+ \varrho _{2} c_{i}^{n-l}+\varrho _{1} c_{i+1}^{n-l} \bigr) +\kappa _{n} \bigl( \varrho _{1} c_{i-1}^{0}+\varrho _{2} c_{i}^{0}+\varrho _{1} c_{i+1}^{0} \bigr) \\ &\qquad {}+\alpha _{0} f \bigl(s_{i},t^{n+1} \bigr). \end{aligned}
(11)

Framework (11) is a system of $$(M+1)$$ equations in $$(M+3)$$ unknowns. Then we use given boundary conditions to obtain two additional equations. Consequently, we obtain a consistent diagonal system, which can be solved using any suitable algorithm based on Gaussian elimination.

### 2.3 Initial vector

The initial vector $$\mathbf{d}^{0}=[d_{-1}^{0},d_{0}^{0}, \ldots , d_{M+1}^{0}]^{T}$$ is required to commence the iterative process, which can be obtained using the IC and the derivatives of IC at the two boundaries as follows :

1. 1.

$$(u_{i}^{0})_{s}=\frac{d}{ds}\varphi (s_{i})$$, $$i=0$$,

2. 2.

$$u_{i}^{0}=\varphi (s_{i})$$, $$i=0,1,\dots ,M$$,

3. 3.

$$(u_{i}^{0})_{s}=\frac{d}{ds}\varphi (s_{i})$$, $$i=M$$,

which becomes the matrix equation

$$Ad^{0}=b,$$
(12)

where

$A=\left[\begin{array}{cccccccc}-{\varrho }_{3}& 0& {\varrho }_{3}& \cdots & \cdots & \cdots & \cdots & 0\\ {\varrho }_{1}& {\varrho }_{2}& {\varrho }_{1}& \ddots & & & & ⋮\\ 0& {\varrho }_{1}& {\varrho }_{2}& {\varrho }_{1}& \ddots & & & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & & ⋮\\ ⋮& & \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& & & & \ddots & {\varrho }_{1}& {\varrho }_{2}& {\varrho }_{1}\\ 0& \cdots & \cdots & \cdots & \cdots & -{\varrho }_{3}& 0& {\varrho }_{3}\end{array}\right]$
(13)

and $$b=[\varphi '(s_{0}), \varphi (s_{0}), \ldots , \varphi (s_{M}), \varphi '(s_{M})]^{T}$$.

## 3 Stability analysis

Here we test scheme (11) for the stability analysis. The Duhamels principle  states that for an inhomogeneous case, the stability estimates are the same as those of the corresponding homogeneous case. So we present the stability analysis only for the case $$f=0$$. Let $$\omega _{i}^{n}$$ and $$\tilde{\omega }_{i}^{n}$$ be the growth factor and its approximation, respectively, of a Fourier mode. Defining $$\Omega _{i}^{n}=\omega _{i}^{n}-\tilde{\omega }_{i}^{n}$$, from (11) we get

\begin{aligned} &(\varrho _{1}-\alpha _{0} \varrho _{4})\Omega _{i-1}^{n+1}+( \varrho _{2}-\alpha _{0} \varrho _{5})\Omega _{i}^{n+1}+(\varrho _{1}- \alpha _{0} \varrho _{4})\Omega _{i+1}^{n+1} \\ & \quad =\sum_{l=0}^{n-1}(\kappa _{l}-\kappa _{l+1}) \bigl(\varrho _{1} \Omega _{i-1}^{n-l}+\varrho _{2} \Omega _{i}^{n-l}+\varrho _{1} \Omega _{i+1}^{n-l} \bigr) \\ &\qquad {}+\kappa _{n} \bigl(\varrho _{1} \Omega _{i-1}^{0} + \varrho _{2} \Omega _{i}^{0}+\varrho _{1} \Omega _{i+1}^{0} \bigr). \end{aligned}
(14)

The initial and boundary conditions are satisfied as

$$\Omega _{i}^{0}=0, \quad i=1,2,\ldots , M-1,$$
(15)

and

$$\Omega _{0}^{n}=\psi _{1}(t_{n}),\qquad \Omega _{M}^{n}= \psi _{2}(t_{n}), \quad n=0,1, \ldots , N.$$
(16)

The grid function is defined as

$$\Omega ^{n}(s)= \textstyle\begin{cases} 0, & 0< s\leq \frac{h}{2} \text{ or } (b-a)-\frac{h}{2}< s\leq (b-a), \\ \Omega _{i}^{n}, & s_{i}-\frac{h}{2} < s \leq s_{i}+\frac{h}{2}, i=1, \ldots ,M-1. \end{cases}$$

The function $$\Omega ^{n}(s)$$ has the Fourier expansion

$$\Omega ^{n}(s)=\sum_{m=-\infty }^{\infty } \eta _{n}(m)e^{ \frac{i 2 \pi m s}{b-a}},\quad n=0,1,\ldots , N,$$

where $$\eta _{n}(m)=\frac{1}{b-a}\int _{a}^{b} \Omega ^{n}(s)e^{ \frac{-i 2 \pi m s}{b-a}}\,ds$$. Let $$\Omega ^{n}=[\Omega _{1}^{n},\Omega _{2}^{n},\ldots ,\Omega _{M-1}^{n}]^{T}$$ and

$$\bigl\Vert \Omega ^{n} \bigr\Vert _{2}= \Biggl(\sum _{i=1}^{M-1} h \bigl\vert \Omega _{i}^{n} \bigr\vert ^{2} \Biggr)^{\frac{1}{2}}= \biggl[ \int _{a}^{b} \bigl\vert \Omega ^{n}(s) \bigr\vert ^{2}\,ds \biggr]^{\frac{1}{2}}.$$

Using the Parseval equality, we see that

$$\int _{a}^{b} \bigl\vert \Omega ^{n}(s) \bigr\vert ^{2}\,ds=\sum _{m=- \infty }^{\infty } \bigl\vert \eta _{n}(m) \bigr\vert ^{2},$$

so that

$$\bigl\Vert \Omega ^{n} \bigr\Vert _{2}^{2}=\sum_{m=-\infty }^{\infty } \bigl\vert \eta _{n}(m) \bigr\vert ^{2}.$$
(17)

Suppose that $$\Omega _{i}^{n}=\eta _{n} e^{I \theta i s}$$ is the solution to system (14)–(15), where $$I=\sqrt{-1}$$ and $$\theta \in [-\pi ,\pi ]$$, so that equation (14) reduces to

\begin{aligned} &(\varrho _{1}-\alpha _{0} \varrho _{4})\eta _{n+1} e^{I \theta (i-1) s}+(\varrho _{2}-\alpha _{0} \varrho _{5})\eta _{n+1} e^{I \theta (i) s}+ (\varrho _{1}-\alpha _{0} \varrho _{4})\eta _{n+1} e^{I \theta (i+1) s} \\ &\quad =\sum_{l=0}^{n-1}(\kappa _{l}-\kappa _{l+1}) \bigl(\varrho _{1} \eta _{n-l} e^{I \theta (i-1) s}+ \varrho _{2} \eta _{n-l} e^{I \theta (i) s}+\varrho _{1} \eta _{n-l}{} e^{I \theta (i+1) s} \bigr) \\ &\qquad {} + \kappa _{n} \bigl(\varrho _{1} \eta _{0} e^{I \theta (i-1) s}+ \varrho _{2} \eta _{0} e^{I \theta i s}+\varrho _{1} \eta _{0} e^{I \theta (i+1) s} \bigr). \end{aligned}
(18)

Dividing (18) by $$e^{I \theta i s}$$, using $$e^{-I\theta s}+e^{I\theta s}=2\cos (\theta s)$$, and gathering like terms, we get

$$\biggl(1- \frac{\alpha _{0}(2\cos (\theta s)\varrho _{4}+\varrho _{5})}{2\varrho _{1}\cos (\theta s)+\varrho _{2}} \biggr)\eta _{n+1}= \sum_{l=0}^{n-1}(\kappa _{l}- \kappa _{l+1}) \eta _{n-l}+\kappa _{n} \eta _{0}.$$
(19)

Without loss of generality, let $$\theta =0$$, so that the last equation reduces to

$$\biggl(1- \frac{\alpha _{0}(2\varrho _{4}+\varrho _{5})}{2\varrho _{1}+\varrho _{2}} \biggr)\eta _{n+1}= \sum_{l=0}^{n-1}(\kappa _{l}- \kappa _{l+1}) \eta _{n-l}+\kappa _{n} \eta _{0}.$$
(20)

Then

$$\eta _{n+1}=\frac{1}{\zeta }\sum _{l=0}^{n-1}(\kappa _{l}- \kappa _{l+1})\eta _{n-l}+\frac{1}{\zeta } \kappa _{n} \eta _{0},$$
(21)

where $$\zeta =1- \frac{\alpha _{0}(2\varrho _{4}+\varrho _{5})}{2\varrho _{1}+\varrho _{2}}$$. Note that $$\frac{\alpha _{0}(2\varrho _{4}+\varrho _{5})}{2\varrho _{1}+\varrho _{2}}=- \frac{ 3\alpha _{0}}{4}\tan (\frac{h}{4})^{2}\leq 0$$, so that $$\zeta \geq 1$$.

### Proposition 1

If $$\eta _{k}$$ ($$k=0,1,\ldots , N$$) is the solution of equation (21), then $$|\eta _{k}|\leq |\eta _{0}|$$.

### Proof

We use induction on k. For $$k=0$$, equation (21) gives $$\eta _{1}=\frac{1}{\zeta } \eta _{0}$$, so that $$|\eta _{1}|=\frac{1}{\zeta } |\eta _{0}|\leq |\eta _{0}|$$ because $$\zeta \geq 1$$. Supposing $$|\eta _{i}| \leq |\eta _{0}|$$, $$i=1, 2,\dots , k$$, from (21) we get

\begin{aligned} \vert \eta _{k+1} \vert & \leq \frac{1}{\zeta }\sum _{l=0}^{k-1}(\kappa _{l}- \kappa _{l+1}) \vert \eta _{n-l} \vert +\frac{1}{\zeta } \kappa _{k} \vert \eta _{0} \vert \\ &\leq \sum_{l=0}^{k-1}(\kappa _{l}-\kappa _{l+1}) \vert \eta _{0} \vert + \kappa _{k} \vert \eta _{0} \vert \\ &= \vert \eta _{0} \vert \Biggl(\sum _{l=0}^{k-1}(\kappa _{l}-\kappa _{l+1})+ \kappa _{k} \Biggr) \\ &= \vert \eta _{0} \vert . \end{aligned}
(22)

□

### Theorem 1

Scheme (11) is unconditionally stable.

### Proof

By Proposition 1 and relation (17) we have

$$\bigl\Vert \Omega ^{k} \bigr\Vert _{2}\leq \bigl\Vert \Omega ^{0} \bigr\Vert _{2}, \quad k=0,1,\ldots , N,$$

which establishes the unconditional stability. □

## 4 Convergence analysis

Here we give convergence estimates for the discrete-time problem (10). As in the case of stability analysis, we present the convergence analysis for the homogeneous problem only.

### Theorem 2

Let $$\{u(s,t^{n})\}_{n=0}^{N-1}$$ be the exact solution of (1), and let $$\{u^{n}\}_{n=0}^{N-1}$$ be the discrete-time solution of (10). Then

$$\bigl\Vert e^{n+1} \bigr\Vert \leq D+C_{u} \tau ^{2-\gamma },$$

where $$e^{n+1}=u(s,t^{n+1})-u^{n+1}$$, and D is a constant.

### Proof

As before, we give a proof for $$f=0$$ only. Note that the exact solution u also satisfies the semidiscrete scheme (10), so that we have

$$u \bigl(s,t^{n+1} \bigr)=\sum _{l=0}^{n-1} (\kappa _{l}-\kappa _{l+1})u \bigl(s,t^{n-l} \bigr)+ \kappa _{n} u \bigl(s,t^{0} \bigr)+\alpha _{0} \bigl(u \bigl(s,t^{n+1} \bigr) \bigr)_{ss}$$
(23)

and

$$u^{n+1}=\sum_{l=0}^{n-1} (\kappa _{l}-\kappa _{l+1})u^{n-l}+ \kappa _{n} u^{0}+\alpha _{0} \bigl(u^{n+1} \bigr)_{ss}.$$
(24)

Subtracting (24) from (23), we obtain

\begin{aligned} e^{n+1} &= \sum_{l=0}^{n-1} (\kappa _{l}-\kappa _{l+1})e^{n-l}+ \kappa _{n} e^{0}+\alpha _{0} \bigl(e^{n+1} \bigr)_{ss}+r_{\tau }^{n+1} \\ &= \sum_{l=0}^{n-1} (\kappa _{l}-\kappa _{l+1})e^{n-l}+ \alpha _{0} \bigl(e^{n+1} \bigr)_{ss}+r_{\tau }^{n+1}, \end{aligned}
(25)

where we have used the fact that $$e^{0}=0$$. Now taking the inner product of both sides of (25) with $$e^{n+1}$$ and using $$\langle x,x\rangle =\|x\|^{2}\geq 0$$, we get

\begin{aligned} \bigl\langle e^{n+1},e^{n+1} \bigr\rangle &= \sum_{l=0}^{n-1} (\kappa _{l}-\kappa _{l+1}) \bigl\langle e^{n-l},e^{n+1} \bigr\rangle +\alpha _{0} \bigl\langle \bigl(e^{n+1} \bigr)_{ss},e^{n+1} \bigr\rangle + \bigl\langle r_{\tau }^{n+1},e^{n+1} \bigr\rangle \\ &= \sum_{l=0}^{n-1} (\kappa _{l}-\kappa _{l+1}) \bigl\langle e^{n-l},e^{n+1} \bigr\rangle -\alpha _{0} \bigl\langle \bigl(e^{n+1} \bigr)_{s}, \bigl(e^{n+1} \bigr)_{s} \bigr\rangle + \bigl\langle r_{\tau }^{n+1},e^{n+1} \bigr\rangle \\ &=\sum_{l=0}^{n-1} (\kappa _{l}-\kappa _{l+1}) \bigl\langle e^{n-l},e^{n+1} \bigr\rangle -\alpha _{0} \bigl\Vert \bigl(e^{n+1} \bigr)_{s} \bigr\Vert ^{2}+ \bigl\langle r_{\tau }^{n+1},e^{n+1} \bigr\rangle \\ &\leq \sum_{l=0}^{n-1} (\kappa _{l}-\kappa _{l+1}) \bigl\langle e^{n-l},e^{n+1} \bigr\rangle + \bigl\langle r_{\tau }^{n+1},e^{n+1} \bigr\rangle , \end{aligned}
(26)

where we have used the relations $$\langle u_{ss},u\rangle =-\langle u_{s},u_{s}\rangle$$ and $$\langle x,x\rangle =\|x\|^{2}$$. Applying the Cauchy–Schwarz inequality $$\langle x,y\rangle \leq \|x\|\|y\|$$ in (26), we obtain

\begin{aligned} \bigl\Vert e^{n+1} \bigr\Vert ^{2} & \leq \sum_{l=0}^{n-1} (\kappa _{l}-\kappa _{l+1}) \bigl\Vert e^{n-l} \bigr\Vert \bigl\Vert e^{n+1} \bigr\Vert + \bigl\Vert r_{\tau }^{n+1} \bigr\Vert \bigl\Vert e^{n+1} \bigr\Vert . \end{aligned}
(27)

Dividing (27) throughout by $$\|e^{n+1}\|$$, we obtain

\begin{aligned} \bigl\Vert e^{n+1} \bigr\Vert &\leq \sum _{l=0}^{n-1} (\kappa _{l}- \kappa _{l+1}) \bigl\Vert e^{n-l} \bigr\Vert + \bigl\Vert r_{\tau }^{n+1} \bigr\Vert \\ &\leq D_{n} \sum_{l=0}^{n-1} (\kappa _{l}-\kappa _{l+1})+ \bigl\Vert r_{\tau }^{n+1} \bigr\Vert \\ &\leq D_{n} (1-\kappa _{n})+ \bigl\Vert r_{\tau }^{n+1} \bigr\Vert \\ &\leq D+C_{u}\tau ^{2-\gamma }, \end{aligned}
(28)

where $$D_{n}=\max_{0\leq l\leq n-1}\|e^{n-l}\|$$ and $$D=\max_{0\leq n\leq N} D_{n}$$. We have also used the relation $$(1-\kappa _{n})<1$$. □

## 5 Numerical results and discussions

In this section, we present the results of the numerical tests for the TFDE (1) with initial (2) and boundary conditions (3). We use the following error norms to measure the accuracy of the method:

$$L_{2}= \bigl\Vert U^{\text{exact}}-U^{N} \bigr\Vert _{2}\simeq \sqrt{h\sum_{j=-1}^{M+1} \bigl( \bigl\vert U_{j}^{ \text{exact}}-U_{j}^{N} \bigr\vert \bigr)^{2}}$$

and

$$L_{\infty }= \bigl\Vert U^{\text{exact}}-U^{N} \bigr\Vert _{\infty }\simeq \max_{-1 \leq j\leq M+1} \bigl\vert U_{j}^{\text{exact}}-U_{j}^{N} \bigr\vert |,$$

where $$U^{\text{exact}}$$ is the exact solutions, and $$(U_{N})_{j}$$ is the approximate one. The order of convergence is given by

$$OC= \frac{\log (\operatorname{Error}(M)/\operatorname{Error}(2M) )}{\log (2M/M )},$$

where $$\operatorname{Error}(M)$$ and $$\operatorname{Error}(2M)$$ are the $$L_{\infty }$$ norms at M and 2M, respectively.

### Example 1

Consider the TFDE (1)  with initial condition $$u(s,0)=\sin s$$ and boundary conditions $$u(0,t) =u(\pi ,t)=0$$. This problem has the exact solution $$u(s,t)=E_{\gamma }(-t^{\gamma })\sin (s)$$, where $$E_{\gamma }(z)=\sum_{m=0}^{\infty } \frac{z^{m}}{\Gamma (\gamma m+1)}$$ is the ML function. The corresponding source term is $$f=0$$.

We apply the proposed algorithm (11) to the problem. The approximate solutions when $$\tau =0.01$$, $$h=\frac{\pi }{20}$$, and $$\gamma =0.5$$ at $$t=0.5$$ and $$t=1$$ are given by

\begin{aligned} U(s,0.5)= \textstyle\begin{cases} 0.4361 \cos (\frac{s}{2})-0.1707 \cos ^{3}(\frac{s}{2})-0.0618\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(0.0762 + 0.2560 \sin (s))\\ \quad {}+0.0464 \csc ( \frac{s}{2})\sin ^{2}(s),& s\in [0,\frac{\pi }{20}], \\ 0.0009 \cos (\frac{s}{2})-0.0009\cos ^{3}(\frac{s}{2})-0.1355\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(0.6433 + 0.0014 \sin (s)) \\ \quad {}+0.1016 \csc ( \frac{s}{2})\sin ^{2}(s),& s\in [\frac{\pi }{20},\frac{\pi }{10}], \\ 0.0047 \cos (\frac{s}{2})-0.0046\cos ^{3}(\frac{s}{2})-0.1426\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(0.6197 + 0.0068 \sin (s)) \\ \quad {}+0.1070\csc (\frac{s}{2}) \sin ^{2}(s),& s\in [\frac{\pi }{10},\frac{3\pi }{20}], \\ \vdots \\ 0.6197 \cos (\frac{s}{2})-0.1426\cos ^{3}(\frac{s}{2})-0.0046\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(0.0047 +0.2140 \sin (s)) \\ \quad {}+0.0034\csc (\frac{s}{2}) \sin ^{2}(s),& s\in [\frac{17\pi }{20},\frac{9\pi }{10}], \\ 0.6433 \cos (\frac{s}{2})-0.1355\cos ^{3}(\frac{s}{2})-0.0009\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(0.0009 +0.2033 \sin (s))+0.0007\csc (\frac{s}{2}) \sin ^{2}(s),& s\in [\frac{9\pi }{10},\frac{19\pi }{20}], \\ 0.6554 \cos (\frac{s}{2})-0.1316\cos ^{3}(\frac{s}{2})+1.9984\times 10^{-15} \sin ^{3}(\frac{s}{2}) \\ \quad {}+0.1974\sin (\frac{s}{2})\sin (s)-1.3323 \times 10^{-15}\csc ( \frac{s}{2})\sin ^{2}(s),& s\in [\frac{19\pi }{20},\pi ], \end{cases}\displaystyle \end{aligned}

and

\begin{aligned} U(s,1)= \textstyle\begin{cases} 4.4409 \times 10^{-16} \cos (\frac{s}{2})-1.6653 \times 10^{-16}\cos ^{3}( \frac{s}{2}) \\ \quad {}-0.1075\sin ^{3}(\frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(0.5352 + 1.1102\times 10^{-16} \sin (s)) \\ \quad {}+0.0806 \csc (\frac{s}{2})\sin ^{2}(s),& s\in [0,\frac{\pi }{20}], \\ 0.0008 \cos (\frac{s}{2})-0.0008\cos ^{3}(\frac{s}{2})-0.1107\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(0.5254 + 0.0012 \sin (s)) \\ \quad {}+0.0830 \csc ( \frac{s}{2})\sin ^{2}(s),& s\in [\frac{\pi }{20},\frac{\pi }{10}], \\ 0.0038 \cos (\frac{s}{2})-0.0037\cos ^{3}(\frac{s}{2})-0.1165\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(0.5060 + 0.0056 \sin (s)) \\ \quad {}+0.0874\csc (\frac{s}{2}) \sin ^{2}(s),& s\in [\frac{\pi }{10},\frac{3\pi }{20}], \\ \vdots \\ 0.5061 \cos (\frac{s}{2})-0.1165\cos ^{3}(\frac{s}{2})-0.0037\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(0.0038 +0.1747 \sin (s)) \\ \quad {}+0.0028\csc (\frac{s}{2}) \sin ^{2}(s),& s\in [\frac{17\pi }{20},\frac{9\pi }{10}], \\ 0.5254 \cos (\frac{s}{2})-0.1107\cos ^{3}(\frac{s}{2})-0.0008\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(0.0008 +0.1660 \sin (s)) \\ \quad {}+0.0006\csc (\frac{s}{2}) \sin ^{2}(s),& s\in [\frac{9\pi }{10},\frac{19\pi }{20}], \\ 0.5352 \cos (\frac{s}{2})-0.1075\cos ^{3}(\frac{s}{2})+1.3322\times 10^{-15} \sin ^{3}(\frac{s}{2}) \\ \quad {}+0.1612\sin (\frac{s}{2})\sin (s)-8.8818 \times 10^{-16}\csc ( \frac{s}{2})\sin ^{2}(s),& s\in [\frac{19\pi }{20},\pi ], \end{cases}\displaystyle \end{aligned}

respectively. Figure 1 displays the behavior of the numerical and exact solutions at different times. The graphs are in excellent affirmation. In Fig. 2 the absolute errors are presented in 2D and 3D at $$t=0.5$$. Figure 3 demonstrates an excellent 3D contrast between the exact and numerical solutions at time step $$t=1$$. In Table 1, a comparison of the error norms with those obtained in  is tabulated. Our methodology gives better precision for bigger τ over that obtained in . The order of convergence is tabulated for the $$L_{\infty }$$ norm in Table 2.

### Example 2

Consider the nonhomogenous TFDE 

\begin{aligned} &{}^{C}_{a}{D}_{t}^{0.9}u(s,t)- \frac{\partial ^{2}}{\partial s^{2}} u(s,t) \\ &\quad = \frac{2}{\Gamma (2.1)}t^{1.1} \sin (2 \pi s) \\ &\qquad {}+ 4\pi ^{2} t^{2} \sin (2\pi s),\quad s\in [0,1], t\in [0,T], \end{aligned}
(29)

with zero initial and boundary conditions. This problem has the exact solution $${u(s,t)=t^{2} \sin (2\pi s)}$$.

We solve (29) by using the proposed scheme (11). The approximate solutions when $$\tau =0.01$$ and $$h=\frac{1}{20}$$ at $$t=0.5$$ and $$t=1$$ are given by

\begin{aligned} U(s,0.5)= \textstyle\begin{cases} 3.5527\times 10^{-15}\cos (\frac{s}{2})-20.2482\sin ^{3}(\frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(-57.6234 -7.1054\times 10^{-15} \sin (s)) \\ \quad {}+15.1861 \csc (\frac{s}{2})\sin ^{2}(s),& s\in [0,\frac{1}{20}], \\ -0.1401 \cos (\frac{s}{2})+0.1310\cos ^{3}(\frac{s}{2}) \\ \quad {}-18.3852 \sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(-52.0206 -0.20997 \sin (s)) \\ \quad {}+13.7889 \csc ( \frac{s}{2})\sin ^{2}(s),& s\in [\frac{1}{20},\frac{1}{10}], \\ -0.6729 \cos (\frac{s}{2})+0.67101\cos ^{3}(\frac{s}{2}) \\ \quad {}-14.8716\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(-41.3734-1.0065 \sin (s)) \\ \quad {}+11.1537\csc ( \frac{s}{2})\sin ^{2}(s),& s\in [\frac{1}{10},\frac{3}{20}], \\ \vdots \\ 1.0159 \cos (\frac{s}{2})+0.6104\cos ^{3}(\frac{s}{2})+5.7332\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(10.1455-0.9156 \sin (s)) \\ \quad {}-4.2910\csc (\frac{s}{2}) \sin ^{2}(s),& s\in [\frac{17}{20},\frac{9}{10}], \\ 25.0629 \cos (\frac{s}{2})-18.3491\cos ^{3}-1.1608\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(-45.5852 +27.5236 \sin (s)) \\ \quad {}+0.8707\csc ( \frac{s}{2})\sin ^{2}(s),& s\in [\frac{9}{10},\frac{19}{20}], \\ 27.6261 \cos (\frac{s}{2})-20.1974\cos ^{3}(\frac{s}{2})-1.4323\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(-50.5693 + 30.2961 \sin (s)) \\ \quad {}+1.0742\csc ( \frac{s}{2})\sin ^{2}(s),& s\in [\frac{19}{20},1], \end{cases}\displaystyle \end{aligned}

and

\begin{aligned} U(s,1)= \textstyle\begin{cases} -1.4211\times 10^{-14}\cos (\frac{s}{2})+3.5527\times 10^{-14}\cos ^{3}( \frac{s}{2}) \\ \quad {}-80.9352 \sin ^{3}(\frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(-230.33 -2.8422\times 10^{-14} \sin (s)) \\ \quad {}+60.7014 \csc (\frac{s}{2})\sin ^{2}(s),& s\in [0,\frac{1}{20}], \\ -0.5600 \cos (\frac{s}{2})+0.5595\cos ^{3}(\frac{s}{2})-73.4887 \sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(-207.935 -0.8393 \sin (s)) \\ \quad {}+55.1165 \csc ( \frac{s}{2})\sin ^{2}(s),& s\in [\frac{1}{20},\frac{1}{10}], \\ -2.6897 \cos (\frac{s}{2})+2.6821\cos ^{3}(\frac{s}{2})-59.4443\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(-165.376-4.0232 \sin (s)) \\ \quad {}+44.5832\csc ( \frac{s}{2})\sin ^{2}(s),& s\in [\frac{1}{10},\frac{3}{20}], \\ \vdots \\ 81.6459 \cos (\frac{s}{2})-59.4851\cos ^{3}(\frac{s}{2})-1.5295\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(-143.842+89.2276 \sin (s)) \\ \quad {}+1.1471\csc ( \frac{s}{2})\sin ^{2}(s),& s\in [\frac{17}{20},\frac{9}{10}], \\ 100.181 \cos (\frac{s}{2})-73.3442\cos ^{3}-4.6402\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(-182.211 +110.016 \sin (s)) \\ \quad {}+3.4802\csc ( \frac{s}{2})\sin ^{2}(s),& s\in [\frac{9}{10},\frac{19}{20}], \\ 110.426 \cos (\frac{s}{2})-80.7324\cos ^{3}(\frac{s}{2})-5.7251\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(-202.134 + 121.099 \sin (s)) \\ \quad {}+4.2938\csc ( \frac{s}{2})\sin ^{2}(s),& s\in [\frac{19}{20},1], \end{cases}\displaystyle \end{aligned}

respectively. We get the numerical results by utilizing the proposed scheme. A close comparison between the exact and numerical solutions at different times is shown in Fig. 4. In Fig. 5 the 2D and 3D error profiles are displayed at $$t=0.5$$. Figure 6 deals with 3D comparison between the exact and approximate solutions. Table 3 reports a comparison of the error norms with those obtained in . Although we have chosen a larger time step than that of , we still obtained a better accuracy. Table 4 records the convergence orders for the $$L_{\infty }$$ norm.

### Example 3

Consider the TFDE describing subdiffusion 

\begin{aligned} \frac{\partial u(s,t)}{\partial t}={}&{}_{0}{D_{t}^{1-\gamma }} \biggl[ \frac{\partial ^{2} u(s,t)}{\partial s^{2}} \biggr] \\ &{}+ e^{s} \biggl[(1+\gamma )t^{ \gamma }- \frac{\Gamma (2+\gamma )}{\Gamma (1+2\gamma )}t^{2\gamma } \biggr],\quad 0< t\leq 1, 0< x< 1, \end{aligned}
(30)

with initial condition

$$u(s,0)=0,\quad 0\leq s\leq 1,$$

and boundary conditions

$$u(0,t)=t^{1+\gamma },\qquad u(1,t)=et^{1+\gamma },\quad 0\leq t\leq 1.$$

This problem has the exact solution $$u(s,t)=e^{s} t^{1+\gamma }$$.

Following , equation (30) can be equivalently written as

\begin{aligned} {}^{C}_{0}{D_{t}^{\gamma }}u(s,t)= \frac{\partial ^{2} u(s,t)}{\partial s^{2}}+ e^{s} \bigl[\Gamma (2+ \gamma )t-t^{1+\gamma } \bigr], \quad 0< t\leq 1, 0< s< 1, \end{aligned}
(31)

with same initial and boundary conditions. We apply scheme (11) to (30). The approximate solutions when $$\tau =0.01$$, $$h=\frac{1}{20}$$, and $$\gamma =0.5$$ when $$t=0.5$$ and $$t=1$$ are given by

\begin{aligned} U(s,0.5)= \textstyle\begin{cases} 0.5744 \cos (\frac{s}{2})-0.2208\cos ^{3}(\frac{s}{2})+0.1593\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(1.1850 +0.3313 \sin (s)) \\ \quad {}-0.1195 \csc (\frac{s}{2}) \sin ^{2}(s),& s\in [0,\frac{1}{20}], \\ 0.5725 \cos (\frac{s}{2})-0.2190\cos ^{3}(\frac{s}{2})+0.1844\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(1.2605 + 0.3284 \sin (s)) \\ \quad {}-0.1383 \csc ( \frac{s}{2})\sin ^{2}(s),& s\in [\frac{1}{20},\frac{1}{10}], \\ 0.5685 \cos (\frac{s}{2})-0.2150\cos ^{3}(\frac{s}{2})+0.2106\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(1.3398 + 0.3225 \sin (s)) \\ \quad {}-0.1579\csc (\frac{s}{2}) \sin ^{2}(s),& s\in [\frac{1}{10},\frac{3}{20}], \\ \vdots \\ 0.0807 \cos (\frac{s}{2})+0.2062\cos ^{3}(\frac{s}{2})+0.6030\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(3.0809 - 0.3093 \sin (s)) \\ \quad {}-0.4522\csc (\frac{s}{2}) \sin ^{2}(s),& s\in [\frac{17}{20},\frac{9}{10}], \\ 0.0039 \cos (\frac{s}{2})+0.2637\cos ^{3}(\frac{s}{2})+0.6159\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(3.2400 - 0.3956 \sin (s)) \\ \quad {}-0.4619\csc (\frac{s}{2}) \sin ^{2}(s),& s\in [\frac{9}{10},\frac{19}{20}], \\ -0.0811 \cos (\frac{s}{2})+0.3250\cos ^{3}(\frac{s}{2})+0.6249\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(3.4052 - 0.4874 \sin (s)) \\ \quad {}-0.4686\csc (\frac{s}{2}) \sin ^{2}(s),& s\in [\frac{19}{20},1], \end{cases}\displaystyle \end{aligned}

and

\begin{aligned} U(s,1)= \textstyle\begin{cases} 1.6249 \cos (\frac{s}{2})-0.6249\cos ^{3}(\frac{s}{2})+0.4508\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(3.3515 + 0.9374 \sin (s)) \\ \quad {}-0.3381 \csc ( \frac{s}{2})\sin ^{2}(s),& s\in [0,\frac{1}{20}], \\ 1.6196 \cos (\frac{s}{2})-0.6196\cos ^{3}(\frac{s}{2})+0.5218\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(3.5652 + 0.9294 \sin (s)) \\ \quad {}-0.3914 \csc ( \frac{s}{2})\sin ^{2}(s),& s\in [\frac{1}{20},\frac{1}{10}], \\ 1.6083 \cos (\frac{s}{2})-0.6084\cos ^{3}(\frac{s}{2})+0.5959\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(3.7896 + 0.9126 \sin (s)) \\ \quad {}-0.4469\csc (\frac{s}{2}) \sin ^{2}(s),& s\in [\frac{1}{10},\frac{3}{20}], \\ \vdots \\ 0.2280 \cos (\frac{s}{2})+0.5835\cos ^{3}(\frac{s}{2})+1.7062\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(8.7160 - 0.8753 \sin (s)) \\ \quad {}-1.2796\csc (\frac{s}{2}) \sin ^{2}(s),& s\in [\frac{17}{20},\frac{9}{10}], \\ 0.01057 \cos (\frac{s}{2})+0.7461\cos ^{3}(\frac{s}{2})+1.7427\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(9.1663 - 1.1192 \sin (s)) \\ \quad {}-1.3070\csc (\frac{s}{2}) \sin ^{2}(s),& s\in [\frac{9}{10},\frac{19}{20}], \\ -0.2298 \cos (\frac{s}{2})+0.9195\cos ^{3}(\frac{s}{2})+1.7682\sin ^{3}( \frac{s}{2}) \\ \quad {}+\sin (\frac{s}{2})(9.6338 - 1.3793 \sin (s)) \\ \quad {}-1.3261\csc (\frac{s}{2}) \sin ^{2}(s),& s\in [\frac{19}{20},1], \end{cases}\displaystyle \end{aligned}

respectively. Figure 7 analyzes the graphs of the exact and approximate solutions when $$\tau =0.01$$ $$h=\frac{1}{80}$$, and $$\gamma =0.5$$. Figure 8 depicts the 2D and 3D error profiles, which exhibit exactness of the method. Figure 9 shows exceptionally close comparison of 3D graphs of approximate and exact solutions using $$\tau =0.01$$, $$h=\frac{1}{60}$$, and $$\gamma =0.5$$. In Tables 58 the maximum errors contrast with those presented in  for various values of τ and h to show that the present scheme is increasingly precise and gives better precision.

## 6 Conclusions

In this study, we developed a cubic trigonometric B-spline collocation method for numerical approximation of time-fractional diffusion equations. The time discretization is done using the typical finite difference method, whereas the derivatives in space are approximated by utilizing the trigonometric B-splines. The approximate solution is obtained as a piecewise continuous function, so that the solution can be approximated at any desired location in the domain of interest. We also presented a stability and convergence analysis of the scheme to affirm that the errors do not propagate. The obtained numerical results contrast with those of some current numerical procedures. We infer that the present scheme is more precise and provides better accuracy.

Not applicable.

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### Acknowledgements

We thank Dr. Muhammad Amin for his assistance in proofreading of the manuscript. The authors are also grateful to anonymous referees for their valuable suggestions, which significantly improved this manuscript.

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