Galerkin Finite Element Method for Caputo–Hadamard Time-Space Fractional Diffusion Equation

Abstract

In this paper, we study the Caputo–Hadamard time-space fractional diffusion equation, where the Caputo derivative is defined in the temporal direction and the Hadamard derivative is defined in the spatial direction separately. We first use the Laplace transform and the modified Fourier transform to study the analytical solution of the Cauchy problem. Then, using the Galerkin finite element method in space, we generate a semi-discrete scheme and study the convergence analysis. Furthermore, using the L1 scheme of the Caputo derivative in time, we construct a fully discrete scheme and then discuss the stability and error estimation in detail. Finally, the numerical experiments are displaced to verify the theoretical results.

1. Introduction

Fractional calculus has been applied in many fields to describe various complex phenomena [1,2,3]. Frequently used fractional derivatives include the Riemann–Liouville derivative, Caputo derivative, Gr$\ddot{\mathrm{u}}$nwald–Letnikov derivative, Hadamard derivative, and so on. Hadamard fractional calculus has attracted many scholars, which can describe the logarithmic creep law and ultra-slow dynamics of the complex substances, such as solid-like viscoelastic materials [4]. In 2012, Mainardi and Spada [5] provided the rheological properties related to this general model, which includes the original Lomnitz creep law. In 2017, Garra et al. [6] proposed a generalized Lomnitz logarithmic creep law, which is molded by the tool of the Hadamard fractional calculus.

However, the non-locality and weak singularity of the fractional derivative brings many difficulties and challenges to the numerical calculation of fractional differential equations. In this regard, many scholars have conducted extensive and in-depth research. In [7], the authors use a robust higher-order finite difference technique to study a Caputo time-fractional singularly perturbed problem, in which the stability analysis and the convergence analysis are discussed for the numerical scheme. In [8], the authors use a finite difference scheme for a novel reaction-diffusion equation with an anisotropic behavior and a time-fractional derivative, to study the problem of textured image recovery. In [9], the authors construct a new approach to study the image multi-frame superresolution field based on a fractional variable order for total variation regularization. In [10], the authors propose the variational formulation for the stationary fractional advection–dispersion equation in one dimension. In [11], the authors study the Galerkin finite element method for a time dependent nonlinear, space-fractional diffusion equation. In [12], the author presents the finite element method for the space and time fractional Fokker–Planck equation. In [13], the authors propose a finite difference/element approach for solving the time-fractional subdiffusion equation, where the fractional linear multistep method is applied. In [14], the authors study the two-dimensional multi-term time-space fractional Bloch–Torrey equation, where the fast and efficient finite difference/finite element method is used. In [15], the authors discuss a class of time-space fractional diffusion equations with nonsmooth data using the Galerkin finite element method. In [16], the authors present a $\theta$–scheme/finite element method for the space-time fractional diffusion equations. In [17], the authors develop the finite element method in a new fractional derivative space to solve inhomogeneous boundary value problems of space fractional differential equations. In [18], the authors study a finite element/finite difference technique for a 2D fractional integro-partial differential equation.

In published literature, the Riemann–Liouville derivative is mainly used as one part of the Riesz derivative with a spatial variable in the fractional partial differential equation, while the Caputo derivative is mainly used in the temporal direction because of the generalized Fick law during the modeling process. Similar to the Caputo derivative, the Hadamard derivative is also used in the temporal direction, defined as the Caputo–Hadamard derivative [19]. Nevertheless, many different numerical algorithms have been studied for time Hadamard fractional problems, such as the local discontinuous Galerkin method [20], finite difference method [21], logarithmic Jacobi collocation method [22], Haar wavelets method [23], Krawtchouk wavelets method [24], generalized Legendre method [25], and so on. Furthermore, there are a few literature reports that consider the Hadamard derivative in the spatial direction except [26]. In [26], the authors study the Hadamard fractional calculus and solve the space Hadamard fractional differential equation numerically by using the Galerkin finite element method, where a novel piecewise log-polynomial is defined as the shape function. Naturally, to study the time-space fractional differential equation with the Caputo and Hadamard derivative numerically is an interesting work.

In this paper, we shall consider a classic initial-boundary value problem with time-space fractional diffusion operators,

$${}_{C}D_{0,t}^{\alpha }u(x,t)-\lambda {}_{H}D_{a,x}^{2\beta }u(x,t)=f(x,t),$$

(1)

where ${}_{C}D_{0,t}^{\alpha }$ belongs to the time Caputo derivative, and ${}_{H}D_{a,x}^{2\beta }$ belongs to the space Hadamard derivative. Together with the homogeneous boundary condition, the initial condition $u(x,0)=u_0\left(x\right)$ and the source term $f(x,t)$ are given functions, $\mathsf{\Omega }=(a,b),0<a<b$. Throughout this paper, the order in the spatial direction is $1/2<\beta <1$ and the order in the temporal direction is $0<\alpha <1$.

The rest of this paper is constructed as follows: In Section 2, we study the modified Fourier transform and give some related conclusions of fractional calculus. In Section 3, we study the analytic solution for the Caputo–Hadamard time-space fractional diffusion equation. In Section 4, we construct the Hadamard fractional derivative space and fractional Sobolev space and present the semi-discrete scheme to solve the Hadamard fractional diffusion equation. Then, we study the convergence analysis of the semi-discrete scheme. In Section 5, we present the fully discrete scheme and then discuss its stability analysis and convergence analysis, which is the combination of the $L1$ scheme and the Galerkin finite element method. A numerical example is provided in the last section to validate the efficiency of the theoretical analyses.

2. Preliminaries

In the following, we give some definitions of fractional derivatives, which will be used in the subsequent sections. Some classic works are described in [27,28,29].

Definition 1. 

The α-th $(n-1<\alpha <n\in {\mathbb{Z}}^{+})$ order Caputo derivative of function $f\left(t\right)$ is defined as follows:

$$\begin{array}{c}\hfill {}_{C}D_{0,t}^{\alpha }f\left(t\right)={\displaystyle \frac{1}{\mathsf{\Gamma }(n-\alpha )}}{\int }_0^t{(t-\tau )}^{n-\alpha -1}{\displaystyle \frac{d^{n}f\left(\tau \right)}{d{\tau }^n}}d\tau .\end{array}$$

Definition 2. 

The β-th order left and right space Riemann–Liouville fractional derivatives of function $f\left(x\right)$ are separately defined as the following:

$${}_{RL}D_{a,x}^{\beta }f\left(x\right)={\displaystyle \frac{1}{\mathsf{\Gamma }(m-\beta )}}{\displaystyle \frac{d^m}{d{x}^m}}{\int }_a^x{(x-s)}^{m-\beta -1}f\left(s\right)ds,$$

$${}_{RL}D_{x,b}^{\beta }f\left(x\right)={\displaystyle \frac{{(-1)}^m}{\mathsf{\Gamma }(m-\beta )}}{\displaystyle \frac{d^m}{d{x}^m}}{\int }_x^b{(s-x)}^{m-\beta -1}f\left(s\right)ds.$$

Definition 3. 

The β-th order left and right space Hadamard fractional derivatives of function $f\left(x\right)$ are separately defined as the following:

$${}_{H}D_{a,x}^{\beta }f\left(x\right)={\displaystyle \frac{1}{\mathsf{\Gamma }(m-\beta )}}{\delta }^m{\int }_a^x{\left(log{\displaystyle \frac{x}{s}}\right)}^{m-\beta -1}f\left(s\right){\displaystyle \frac{ds}{s}},$$

$${}_{H}D_{x,b}^{\beta }f\left(x\right)={\displaystyle \frac{1}{\mathsf{\Gamma }(m-\beta )}}{(-\delta )}^m{\int }_x^b{\left(log{\displaystyle \frac{s}{x}}\right)}^{m-\beta -1}f\left(s\right){\displaystyle \frac{ds}{s}},$$

where $\delta =x{\displaystyle \frac{d}{dx}},0<a<x<b$.

3. The Modified Fourier Transform

It is well known that the classic Fourier transform plays an important role not only for the classic Sobolev space but also for the fractional Sobolev space with Riemann–Liouville derivatives in the finite element analysis. However, the Fourier transform may be modified correspondingly for the Hadamard fractional derivative, which has the logarithmic singular kernel. Therefore, we define the modified Fourier transform in the following way.

Definition 4 

([26]). For a absolutely integrable function $f\left(x\right)$ on $(-\infty ,\infty )$, we define the modified Fourier transform as follows:

$${\mathcal{F}}_m\{f\left(x\right);\omega \}=\widehat{f}\left(\omega \right)={\displaystyle \frac{1}{\sqrt{2\pi }}}{\int }_{-\infty }^{\infty }{e}^{-ilog(\mid \omega \mid )log(\mid x\mid )}f\left(x\right){\displaystyle \frac{dx}{\mid x\mid }}.$$

(2)

Theorem 1. 

If $f\left(x\right)$ is absolutely integrable on $(-\infty ,\infty )$, then the following modified Fourier integral converges to the function $2\left[f\right(x)+f(-x\left)\right]$, i.e.,

$$\begin{array}{c}2[f\left(x\right)+f(-x)]={\displaystyle \frac{1}{2\pi }}{\int }_{-\infty }^{\infty }{e}^{ilog(\mid \omega \mid )log(\mid x\mid )}\left[{\int }_{-\infty }^{\infty }{e}^{-ilog(\mid \omega \mid )log(\mid \xi \mid )}f\left(\xi \right){\displaystyle \frac{d\xi }{\mid \xi \mid }}\right]{\displaystyle \frac{d\omega }{\mid \omega \mid }}.\hfill \end{array}$$

(3)

Proof. 

For the right hand of Equation (3), we have the following expression:

$$\begin{array}{c}{\displaystyle \frac{1}{2\pi }}{\int }_{-\infty }^{\infty }{e}^{ilog(\mid \omega \mid )log(\mid x\mid )}\left[{\int }_{-\infty }^{\infty }{e}^{-ilog(\mid \omega \mid )log(\mid \xi \mid )}f\left(\xi \right){\displaystyle \frac{d\xi }{\mid \xi \mid }}\right]{\displaystyle \frac{d\omega }{\mid \omega \mid }}\hfill \\ ={\displaystyle \frac{1}{2\pi }}{\int }_0^{\infty }{e}^{ilog(\mid \omega \mid )log(\mid x\mid )}\left[{\int }_0^{\infty }{e}^{-ilog(\mid \omega \mid )log(\mid \xi \mid )}f\left(\xi \right){\displaystyle \frac{d\xi }{\xi }}\right]{\displaystyle \frac{d\omega }{\omega }}\hfill \\ +{\displaystyle \frac{1}{2\pi }}{\int }_{-\infty }^0{e}^{ilog(\mid \omega \mid )log(\mid x\mid )}\left[{\int }_0^{\infty }{e}^{-ilog(\mid \omega \mid )log(\mid \xi \mid )}f\left(\xi \right){\displaystyle \frac{d\xi }{\xi }}\right]{\displaystyle \frac{d\omega }{-\omega }}\hfill \\ +{\displaystyle \frac{1}{2\pi }}{\int }_0^{\infty }{e}^{ilog(\mid \omega \mid )log(\mid x\mid )}\left[{\int }_{-\infty }^0{e}^{-ilog(\mid \omega \mid )log(\mid \xi \mid )}f\left(\xi \right){\displaystyle \frac{d\xi }{-\xi }}\right]{\displaystyle \frac{d\omega }{\omega }}\hfill \\ +{\displaystyle \frac{1}{2\pi }}{\int }_{-\infty }^0{e}^{ilog(\mid \omega \mid )log(\mid x\mid )}\left[{\int }_{-\infty }^0{e}^{-ilog(\mid \omega \mid )log(\mid \xi \mid )}f\left(\xi \right){\displaystyle \frac{d\xi }{-\xi }}\right]{\displaystyle \frac{d\omega }{-\omega }}\hfill \\ :=F_1+F_2+F_3+F_4.\hfill \end{array}$$

(4)

For the right hand of (4), we have the following expression:

$$\begin{array}{ccc}\hfill F_1& =\hfill & {\displaystyle \frac{1}{2\pi }}{\int }_0^{\infty }{e}^{ilog\left(\omega \right)log\left(\right|x\left|\right)}\left[{\int }_0^{\infty }{e}^{-ilog\left(\omega \right)log\left(\xi \right)}f\left(\xi \right){\displaystyle \frac{d\xi }{\xi }}\right]{\displaystyle \frac{d\omega }{\omega }}\hfill \\ \hfill & =\hfill & {\displaystyle \frac{1}{2\pi }}{\int }_{-\infty }^{\infty }{e}^{i\mu z}{\int }_{-\infty }^{\infty }{e}^{-i\mu \eta }f\left(e^{\eta }\right)d\eta d\mu \hfill \\ \hfill & =\hfill & {\displaystyle \frac{1}{2}}[\underset{\eta \to z+0}{lim}f\left(e^{\eta }\right)+\underset{\eta \to z-0}{lim}f\left(e^{\eta }\right)]\hfill \\ \hfill & =\hfill & f\left(\right|x\left|\right),\hfill \end{array}$$

where $\mu =log\left(\omega \right),\eta =log\left(\xi \right),z=log\left(\right|x\left|\right)$ are denoted, and the conclusion of the classic Fourier transform is as follows:

$$\begin{array}{c}{\displaystyle \frac{1}{2\pi }}{\int }_{-\infty }^{\infty }{e}^{ikx}\left[{\int }_{-\infty }^{\infty }{e}^{-ik\xi }f\left(\xi \right)d\xi \right]dk={\displaystyle \frac{1}{2}}[f(x+0)+f(x-0)]\hfill \end{array}$$

And $F_2=f\left(\right|x\left|\right)$ can be proved similarly.

$$\begin{array}{ccc}\hfill F_3& =\hfill & {\displaystyle \frac{1}{2\pi }}{\int }_0^{\infty }{e}^{ilog\left(\omega \right)log\left(\right|x\left|\right)}\left[{\int }_{-\infty }^0{e}^{-ilog\left(\omega \right)log(-\xi )}f\left(\xi \right){\displaystyle \frac{d\xi }{-\xi }}\right]{\displaystyle \frac{d\omega }{\omega }}\hfill \\ \hfill & =\hfill & {\displaystyle \frac{1}{2\pi }}{\int }_{-\infty }^{\infty }{e}^{i\mu z}{\int }_{-\infty }^{\infty }{e}^{-i\mu \eta }f(-e^{\eta })d\eta d\mu \hfill \\ \hfill & =\hfill & {\displaystyle \frac{1}{2}}[\underset{\eta \to z+0}{lim}f(-e^{\eta })+\underset{\eta \to z-0}{lim}f(-e^{\eta })]\hfill \\ \hfill & =\hfill & f(-|x\left|\right),\hfill \end{array}$$

where $\mu =log\left(\omega \right),\eta =log(-\xi ),z=log\left(\right|x\left|\right)$ are denoted. $F_4=f(-|x\left|\right)$ can be proved similarly. □

Conclusion 1. 

If $f\left(x\right)$ is absolutely integrable on $(-\infty ,\infty )$, and satisfies $f(-x)=0$, when $x>0$, then the following conclusion is expressed:

$$\begin{array}{c}f\left(x\right)={\displaystyle \frac{1}{2\sqrt{2\pi }}}{\int }_{-\infty }^{\infty }{e}^{ilog(\mid \omega \mid )log\left(x\right)}\left[{\displaystyle \frac{1}{\sqrt{2\pi }}}{\int }_{-\infty }^{\infty }{e}^{-ilog(\mid \omega \mid )log(\mid \xi \mid )}f\left(\xi \right){\displaystyle \frac{d\xi }{\mid \xi \mid }}\right]{\displaystyle \frac{d\omega }{\mid \omega \mid }}.\hfill \end{array}$$

(5)

Remark 1. 

It can be concluded that under the conditions of Conclusion 1, the formula for the inverse modified Fourier transform of $\widehat{f}\left(\omega \right)$ should be

$${\mathcal{F}}_m^{-1}\left\{\widehat{f}\left(\omega \right)\right\}=f\left(x\right)={\displaystyle \frac{1}{2\sqrt{2\pi }}}{\int }_{-\infty }^{\infty }{e}^{ilog(\mid \omega \mid )log\left(x\right)}\widehat{f}\left(\omega \right){\displaystyle \frac{d\omega }{\mid \omega \mid }}.$$

(6)

4. The Analytic Solution for the Cauchy Problem of the Caputo–Hadamard Time-Space Fractional Diffusion Equation

In this section, by using the modified Fourier transform and the Laplace transform, the analytic solution for the Cauchy problem of the Caputo–Hadamard fractional diffusion equation are discussed as follows:

$${}_{C}D_{0,t}^{\alpha }u(x,t)-\lambda {}_{H}D_{a,x}^{2\beta }u(x,t)=f(x,t),a<x<\infty ,t>0,$$

(7)

with the initial condition

$$u(x,0)=u_0\left(x\right),\mathrm{as}a<x<\infty .$$

(8)

In order to satisfy the condition of the modified Fourier transform, we always assume that

$$u(x,t)=0,\mathrm{as}-\infty <x\le a,t\ge 0,$$

(9)

which can be considered as zero extension of $u(x,t)$, defined on $(a,\infty ),t>0.$

The Mittag–Leffler function is an important tool in this section, defined as

$$E_{\alpha ,\beta }\left(z\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{z^n}{\mathsf{\Gamma }(\alpha n+\beta )}},z\in \mathbb{C}.$$

$E_{\alpha ,1}\left(z\right)$ is often written as $E_{\alpha }\left(z\right)$. $E_1\left(z\right)$ is just the exponential function of $e^z$.

Lemma 1 

([30]). The Laplace transform formula of the Mittag–Leffler function can be represented as follows

$${\int }_0^{\infty }{e}^{-st}{t}^{\alpha k+\beta -1}{E}_{\alpha ,\beta }^{\left(k\right)}(\pm \lambda t^{\alpha })dt={\displaystyle \frac{k!s^{\alpha -\beta }}{{(s^{\alpha }\mp \lambda )}^{k+1}}},Re\left(s\right)>\mid \lambda {\mid }^{\frac{1}{\alpha }}.$$

Lemma 2. 

Denote $\mathcal{L}\left\{f\left(t\right)\right\}=\tilde{f}\left(s\right)$ and $\mathcal{L}\left\{g\left(t\right)\right\}=\tilde{g}\left(s\right)$, then

$$\mathcal{L}\{f\left(t\right)\ast g\left(t\right)\}=\tilde{f}\left(s\right)\tilde{g}\left(s\right),$$

(10)

where $f\left(t\right)\ast g\left(t\right)$ is defined by the integral

$$f\left(t\right)\ast g\left(t\right)={\int }_0^{t}f(t-\tau )g\left(\tau \right)d\tau ,$$

which is called the convolution of $f\left(t\right)$ and $g\left(t\right)$.

Lemma 3 

([30]). The Laplace transform formula of the Caputo derivative can be represented as follows:

$$\mathcal{L}{\{}_C{D}_{0,t}^{\alpha }u\left(t\right)\}=s^{\alpha }\mathcal{L}\left\{u\left(t\right)\right\}-\sum _{k=0}^{n-1}{s}^{p-k-1}{u}^{\left(k\right)}\left(0\right)$$

(11)

Lemma 4 

([26]). Let $m-1<\beta <m$. Assume ${\delta }^{k}u\left(x\right)\to 0$ as $\left|x\right|\to \infty ,k=0,\cdots ,m-1,$ and ${\delta }^{m}u\left(x\right)$ is continuously differential, the modified Fourier transform of the Hadamard derivatives

$${\mathcal{F}}_m\{{}_{H}D_{a,x}^{\beta }u\left(x\right);\omega \}={(ilog(\mid \omega \mid ))}^{\beta }\widehat{u}\left(\omega \right),$$

(12)

holds.

Theorem 2. 

Assume ${\delta }^{2}u(x,t)$ is continuously differential and ${\delta }^{k}u(x,t)\to 0$ as $\left|x\right|\to \infty ,k=0,1$, then Equations (7)–(9) have the analytic solution in the form of

$$\begin{array}{ccc}u(x,t)\hfill & =\hfill & {\displaystyle \frac{1}{2\sqrt{2\pi }}}{\int }_{-\infty }^{\infty }{e}^{ilog\left(x\right)log(\mid \omega \mid )}{E}_{\alpha ,1}\left({(ilog(\mid \omega \mid ))}^{2\beta }{t}^{\alpha }\right)\widehat{u_0}\left(\omega \right){\displaystyle \frac{d\omega }{\mid \omega \mid }}\hfill \\ \hfill & \hfill & +{\displaystyle \frac{1}{2\sqrt{2\pi }}}{\int }_0^t{\tau }^{\alpha -1}d\tau {\int }_{-\infty }^{\infty }{e}^{ilog\left(x\right)log(\mid \omega \mid )}{E}_{\alpha ,\alpha }\left({(ilog(\mid \omega \mid ))}^{2\beta }{\tau }^{\alpha }\right)\hfill \\ \hfill & \hfill & \widehat{f}(\omega ,t-\tau ){\displaystyle \frac{d\omega }{\mid \omega \mid }},\hfill \end{array}$$

(13)

where ${\mathcal{F}}_m{u}_0\left(x\right)=\widehat{u_0}\left(\omega \right)$ and ${\mathcal{F}}_{m}f(x,t)=\widehat{f}(\omega ,t),0<a<x<\infty .$

Proof. 

We first apply the Laplace transform on both sides of Equation (7) with respect to t and then obtain the following result:

$$s^{\alpha }\tilde{u}(x,s)-s^{\alpha -1}{u}_0\left(x\right)-\lambda {}_{H}D_{a,x}^{2\beta }\tilde{u}(x,s)=\tilde{f}(x,s).$$

(14)

Next, applying the modified Fourier transform with respect to x in the above equation yields

$$s^{\alpha }\widehat{\tilde{u}}(\omega ,s)-s^{\alpha -1}\widehat{u_0}\left(\omega \right)-\lambda {(ilog(\mid \omega \mid ))}^{2\beta }\widehat{\tilde{u}}(\omega ,s)=\widehat{\tilde{f}}(\omega ,s).$$

So we have the following result

$$\widehat{\tilde{u}}(\omega ,s)=\widehat{\widetilde{G_0}}(\omega ,s)\widehat{u_0}\left(\omega \right)+\widehat{\widetilde{G_f}}(\omega ,s)\widehat{\tilde{f}}(\omega ,s),$$

where

$$\widehat{\widetilde{G_0}}(\omega ,s)={\displaystyle \frac{s^{\alpha -1}}{s^{\alpha }-\lambda {(ilog(\mid \omega \mid ))}^{2\beta }}},$$

$$\widehat{\widetilde{G_f}}(\omega ,s)={\displaystyle \frac{1}{s^{\alpha }-\lambda {(ilog(\mid \omega \mid ))}^{2\beta }}}.$$

By using the inverse Laplace transform, we obtain

$$\widehat{u}(\omega ,t)=\widehat{G_0}(\omega ,t)\widehat{u_0}\left(\omega \right)+{\int }_0^t\widehat{G_f}(\omega ,t-\tau )\widehat{f}(\omega ,\tau )d\tau ,$$

(15)

where

$$\widehat{G_0}(\omega ,t)=E_{\alpha ,1}\left(\lambda {(ilog(\mid \omega \mid ))}^{2\beta }{t}^{\alpha }\right),$$

$$\widehat{G_f}(\omega ,t)=t^{\alpha -1}{E}_{\alpha ,\alpha }\left(\lambda {(ilog(\mid \omega \mid ))}^{2\beta }{t}^{\alpha }\right).$$

Because $u(x,t)$ is absolutely integrable on $(-\infty ,\infty )\times (0,\infty )$, and $\mathrm{supp}\left(u_0\left(x\right)\right)=(a,\infty )$, $\mathrm{supp}\left(u(x,t)\right)=\mathrm{supp}\left({}_{H}D_{a,x}^{\beta }u(x,t)\right)=(a,\infty )\times (0,\infty )$, we obtain $\mathrm{supp}\left(f\right(x,t\left)\right)=(a,\infty )\times (0,\infty )$, which (15) satisfies the condition of the inverse modified Fourier transform (6). Taking the inverse modified Fourier transform to (15), we obtain the analytic solution in the following representation:

$$\begin{array}{ccc}u(x,t)\hfill & =\hfill & {\displaystyle \frac{1}{2\sqrt{2\pi }}}{\int }_{-\infty }^{\infty }{e}^{ilog\left(x\right)log(\mid \omega \mid )}\widehat{G_0}(\omega ,t)\widehat{u_0}\left(\omega \right){\displaystyle \frac{d\omega }{\mid \omega \mid }}\hfill \\ \hfill & \hfill & +{\displaystyle \frac{1}{2\sqrt{2\pi }}}{\int }_0^t{\tau }^{\alpha -1}d\tau {\int }_{-\infty }^{\infty }{e}^{ilog\left(x\right)log(\mid \omega \mid )}\widehat{G_f}(\omega ,\tau )\widehat{f}(\omega ,t-\tau ){\displaystyle \frac{d\omega }{\mid \omega \mid }}.\hfill \end{array}$$

□

Remark 2. 

From the analytic solution (13), we can see that it belongs to the improper integral, which is inconvenient in the practical application. Therefore, it is necessary to study its numerical solution.

5. The Hadamard Fractional Derivative Spaces and the Semi-Discrete Scheme

To further discuss the finite element analysis of the Hadamard fractional derivative, we define modified the inner product ${(·,·)}_{X^2\left(\mathbb{R}\right)}$ by

$${(u,v)}_{X^2\left(\mathbb{R}\right)}={\int }_{-\infty }^{\infty }u\left(x\right)\overline{v\left(x\right)}{\displaystyle \frac{dx}{\mid x\mid }},$$

and define the modified norm ${\parallel ·\parallel }_{X^2\left(\mathbb{R}\right)}$ by

$${\parallel u\parallel }_{X^2\left(\mathbb{R}\right)}^2={(u,u)}_{X^2\left(\mathbb{R}\right)}={\int }_{-\infty }^{\infty }u\left(x\right)\overline{u\left(x\right)}{\displaystyle \frac{dx}{\mid x\mid }}$$

Lemma 5 

([26]). Denote ${\mathcal{F}}_m\{u\left(x\right);\omega \}=\widehat{u}\left(\omega \right)$ and ${\mathcal{F}}_m\{v\left(x\right);\omega \}=\widehat{v}\left(\omega \right)$, then

$${(\widehat{u}\left(\omega \right),\overline{\widehat{v}\left(\omega \right)})}_{X^2\left(\mathbb{R}\right)}=2{(u\left(x\right)+u(-x),\overline{v\left(x\right)})}_{X^2\left(\mathbb{R}\right)}$$

(16)

and

$$\parallel \widehat{u}{\parallel }_{X^2\left(\mathbb{R}\right)}={\parallel u\left(x\right)+u(-x)\parallel }_{X^2\left(\mathbb{R}\right)}.$$

(17)

Lemma 6 

([26]). Denote $u\in H^{\alpha }(0,T)\bigcap H_0^{\frac{\alpha }{2}}(0,T),v\in H_0^{\frac{\alpha }{2}}(0,T)$, we have

$${({}_{Rl}D_{0,t}^{\alpha }u\left(t\right),v\left(t\right))}_{L^2(0,T)}={({}_{RL}D_{0,t}^{\frac{\alpha }{2}}u\left(t\right),{}_{RL}D_{x,T}^{\frac{\alpha }{2}}v\left(t\right))}_{L^2(0,T)}.$$

(18)

According to assumption (9), we obtain ${}_{H}D_{a,x}^{\beta }u(-x)=0$ when $x\in \mathsf{\Omega }$. Together with Equation (17) in Lemma 5, we obtain

$$\begin{array}{ccc}\parallel {}_{H}D_{a,x}^{\beta }{u\left(x\right)\parallel }_{X^2\left(\mathsf{\Omega }\right)}\hfill & =& \parallel {}_{H}D_{a,x}^{\beta }{u\left(x\right)\parallel }_{X^2\left(\mathbb{R}\right)}\hfill \\ \hfill & =\hfill & \parallel {\mathcal{F}}_m\left({}_{H}D_{a,x}^{\beta }u\left(x\right)\right){\parallel }_{X^2\left(\mathbb{R}\right)}\hfill \\ \hfill & =\hfill & \parallel \mid log(\mid \omega \mid ){\mid }^{\beta }{u\parallel }_{X^2\left(\mathbb{R}\right)}.\hfill \end{array}$$

(19)

In the next part, we will introduce the Hadamard fractional derivative space and Hadamard fractional Sobolev space.

Definition 5 

([26]). Let $\beta >0$. Equip the semi-norm and and the norm of the Hadamard fractional derivative space as follows

$${\left|u\right|}_{J_L^{\beta }\left(\mathbb{R}\right)}:={\parallel {}_{H}D_{a,x}^{\beta }u\parallel }_{X^2\left(\mathbb{R}\right)},$$

$${\parallel u\parallel }_{J_L^{\beta }\left(\mathbb{R}\right)}:={(\parallel u\parallel }_{X^2\left(\mathbb{R}\right)}^2+{\left|u\right|}_{J_L^{\beta }\left(\mathbb{R}\right)}^2{)}^{1/2},$$

respectively, where $J_L^{\beta }\left(\mathbb{R}\right)$ denotes the closure of $C_0^{\infty }\left(\mathbb{R}\right)$ with respect to ${\parallel ·\parallel }_{J_L^{\beta }}\left(\mathbb{R}\right)$.

Definition 6 

([26]). Let $\beta >0$. Equip the semi-norm and the norm of the Hadamard fractional Sobolev space as follows

$${\mid u\mid }_{H^{\beta }\left(\mathbb{R}\right)}:={\parallel \mid log(\mid \omega \mid ){\mid }^{\beta }\widehat{u}\parallel }_{X^2\left(\mathbb{R}\right)},$$

$${\parallel u\parallel }_{H^{\beta }\left(\mathbb{R}\right)}:={(\parallel u\parallel }_{X^2\left(\mathbb{R}\right)}^2+\mid u{\mid }_{H^{\beta }\left(\mathbb{R}\right)}^2{)}^{1/2},$$

respectively, where $H_0^{\beta }\left(\mathbb{R}\right)$ denote the closure of $C_0^{\infty }\left(\mathbb{R}\right)$ under the norm ${\parallel ·\parallel }_{H^{\beta }\left(\mathbb{R}\right)}$.

According to Equation (19), we can obtain the following conclusion.

Theorem 3 

([26]). Let $\beta >0$. Then $J_{L,0}^{\beta }\left(\mathbb{R}\right)$ and $H_0^{\beta }\left(\mathbb{R}\right)$, $J_{L,0}^{\beta }\left(\mathsf{\Omega }\right)$ and $H_0^{\beta }\left(\mathsf{\Omega }\right)$ are equivalent with each other separately. □

Then multiplying by $v\in H_0^{\beta }\left(\mathsf{\Omega }\right)$ on both sides of Equation (1), we obtain the variational form

$${({}_{C}D_{0,t}^{\alpha }u(x,t),v)}_{X^2}-\lambda {({}_{H}D_{a,x}^{\beta }u(x,t),{}_{H}D_{x,b}^{\beta }v)}_{X^2}={(f,v)}_{X^2}.$$

(20)

Denoted by

$$A(u,v)=-\lambda {({}_{H}D_{a,x}^{\beta }u(x,t),{}_{H}D_{x,b}^{\beta }v)}_{X^2},$$

(21)

then,

$${({}_{C}D_{0,t}^{\alpha }u(x,t),v)}_{X^2}+A(u,v)={(f,v)}_{X^2}.$$

(22)

In the following, we present the semi-discrete approximation for Equation (1) as $u_h\left(t\right)\in X_{h0}^r\left(\mathsf{\Omega }\right)$ such that

$$\left\{\begin{array}{c}{({}_{C}D_{0,t}^{\alpha }{u}_h\left(t\right),v)}_{X^2}+A(u_h\left(t\right),v)={({\mathcal{I}}_{h}f,v)}_{X^2},v\in S_r^h,\hfill \\ u_h\left(0\right)=R_h{u}_0,\hfill \end{array}\right.$$

(23)

where $S_r^h$ is the finite element subspace, which is generated by the basis of the $r-1$ order piecewise log-Lagrangian polynomial, and ${\mathcal{I}}_{h}f$ is the piecewise log-Lagrangian polynomial interpolation of f in $S_r^h$ satisfies the property ([26])

$$\parallel {\mathcal{I}}_h{f-f\parallel }_{X^2\left(\mathsf{\Omega }\right)}\le C{h}^r{\parallel f\parallel }_{H^r\left(\mathsf{\Omega }\right)}.$$

(24)

Meanwhile, $R_h:H_0^{\beta }\left(\mathsf{\Omega }\right)\to S_r^h$ is the orthogonal projection operation, defined as

$${({}_{H}D_{a,x}^{\beta }(u-R_{h}u),{}_{H}D_{x,b}^{\beta }v)}_{X^2}=0,u\in H_0^{\beta }\left(\mathsf{\Omega }\right),\forall v\in S_r^h.$$

(25)

Lemma 7 

([31]). Assume $s,r$ are nonnegative real numbers, and $u\in H_0^{\beta }\cap H^r\left(\mathsf{\Omega }\right)$. If $\beta \le s\le r$, then we have the following inequality

$$\parallel R_h{u-u\parallel }_{H^{\beta }\left(\mathsf{\Omega }\right)}\le C{h}^{s-\beta }{\parallel u\parallel }_{H^s\left(\mathsf{\Omega }\right)},$$

where C is a positive constant.

Theorem 4. 

Let $u\in H_0^{\beta }\left(\mathsf{\Omega }\right)\cap H^r\left(\mathsf{\Omega }\right)$ be the exact solution of Equation (1), and $u_h$ be the numerical solution of semi-discrete scheme (23). Then we have the following inequality

$${\int }_0^t{\parallel {}_{C}D_{0,s}^{\frac{\alpha }{2}}(u\left(s\right)-u_h\left(s\right))\parallel }_{X^2}^{2}ds\le C{h}^{2r},$$

(26)

where C is a positive constant.

Proof. 

Denote by $\theta =R_{h}u-u_h\left(t\right)$ and $\rho =u-R_{h}u$. Letting $v=\theta$ in Equation (23) satisfies

$$\begin{array}{ccc}\hfill {({}_{C}D_{0,t}^{\alpha }\theta ,\theta )}_{X^2}+A(\theta ,\theta )& =& -{({}_{C}D_{0,t}^{\alpha }\rho ,\theta )}_{X^2}+{(f-{\mathcal{I}}_{h}f,\theta )}_{X^2}\hfill \\ \hfill & \le & {ϵ\parallel \theta \parallel }_{X^2}^2+{\displaystyle \frac{1}{4ϵ}}(\parallel {}_{C}D_{0,t}^{\alpha }{\rho \parallel }_{X^2}^2+\parallel f-{\mathcal{I}}_{h}f{\parallel }_{X^2}^2),\hfill \end{array}$$

where $ϵ$ is a suitable positive constant satisfying ${ϵ\parallel \theta \parallel }^2\le A(\theta ,\theta )$, and $A(\rho ,\theta )=0$ (because of Equation (25)). Therefore, we obtain

$${({}_{C}D_{0,t}^{\alpha }\theta ,\theta )}_{X^2}\le C\parallel {}_{C}D_{0,t}^{\alpha }{\rho \parallel }_{X^2}^2+{\parallel f-{\mathcal{I}}_{h}f\parallel }_{X^2}^2.$$

(27)

Integrating in time both sides of (27) obtains

$${\int }_0^t{({}_{C}D_{0,t}^{\alpha }\theta \left(s\right),\theta \left(s\right))}_{X^2}ds\le C{\int }_0^t(\parallel {}_{C}D_{0,t}^{\alpha }{\rho \left(s\right)\parallel }_{X^2}^2+\parallel f\left(s\right)-{\mathcal{I}}_{h}f\left(s\right){\parallel }_{X^2}^2)ds.$$

Because of $\theta \left(0\right)=0$, and ${}_{C}D_{0,t}^{\alpha }\theta \left(s\right)={}_{RL}D_{0,t}^{\alpha }\theta \left(s\right)$, we have

$$\begin{array}{c}{\int }_0^t{({}_{C}D_{0,t}^{\alpha }\theta \left(s\right),\theta \left(s\right))}_{X^2}ds={\int }_0^t{({}_{RL}D_{0,t}^{\alpha }\theta \left(s\right),\theta \left(s\right))}_{X^2}ds\hfill \\ ={\int }_0^t{({}_{RL}D_{0,s}^{\frac{\alpha }{2}}\theta \left(s\right),{}_{Rl}D_{s,t}^{\frac{\alpha }{2}}\theta \left(s\right))}_{X^2}ds.\hfill \end{array}$$

By using Lemma 6, we obtain

$$\begin{array}{ccc}{\int }_0^t{\parallel {}_{C}D_{0,s}^{\frac{\alpha }{2}}\theta \left(s\right)\parallel }_{X^2}^{2}ds\hfill & \le \hfill & C{\int }_0^t(\parallel {}_{C}D_{0,t}^{\alpha }{\rho \left(s\right)\parallel }_{X^2}^2+\parallel f\left(s\right)-{\mathcal{I}}_{h}f\left(s\right){\parallel }_{X^2}^2)ds\hfill \\ \hfill & \le \hfill & C{h}^{2r}{\int }_0^t(\parallel {}_{C}D_{0,s}^{\alpha }{u\left(s\right)\parallel }_{r}ds+{\parallel f\left(s\right)\parallel }_r^2)ds.\hfill \end{array}$$

Applying

$$\parallel {}_{C}D_{0,s}^{\alpha }(u\left(s\right)-u_h\left(s\right)){\parallel }_{X^2}^2\le \parallel {}_{C}D_{0,s}^{\frac{\alpha }{2}}{\theta \left(s\right)\parallel }_{X^2}^2+{\parallel {}_{C}D_{0,s}^{\frac{\alpha }{2}}\rho \left(s\right)\parallel }_{X^2}^2,$$

we obtain the desired results. The proof of Theorem 4 is completed. □

6. Fully Discrete Schemes

In this section, we first give the fully discrete scheme, then analyze the stability analysis and the convergence analysis.

It is well known that the Caputo derivative can be directly discretized by the $L1$ scheme or the Gr$\ddot{\mathrm{u}}$nwald–Letnikov formula and so on [32]. Here we selected the $L1$ scheme.

The fully discrete finite element method for Equation (1) with the time direction approximated by the $L1$ method is present as follows: Find $u_h^{n+1},n=0,1,\cdots ,N_T-1$ such that

$$\left\{\begin{array}{c}{({\mathsf{\Lambda }}^{\alpha }{u}_h^{n+1},v)}_{X^2}+A(u_h^{n+1},v)={({\mathcal{I}}_h{f}^{n+1},v)}_{X^2},\hfill \\ u_h\left(0\right)=R_h{u}_0,\hfill \end{array}\right.$$

(28)

where ${\mathsf{\Lambda }}^{\alpha }$ is defined by

$${\mathsf{\Lambda }}^{\alpha }{u}^{n+1}={\displaystyle \frac{{(\Delta t)}^{-\alpha }}{\mathsf{\Gamma }(2-\alpha )}}\sum _{j=0}^n{\nu }_{n-j}(u_h\left(t_{j+1}\right)-u_h\left(t_j\right)),{\nu }_j={(j+1)}^{1-\alpha }-j^{1-\alpha }.$$

After some adjustment, a fully discrete scheme (28) can be rewritten as

$${(u_h^{n+1},v)}_{X^2}+\lambda {\alpha }_0{({}_{H}D_{a,x}^{\beta }{u}_h^{n+1},{}_{H}D_{x,b}^{\beta }v)}_{X^2}=\sum _{j=0}^n{a}_j^n{(u_h^j,v)}_{X^2}+{\alpha }_0{({\mathcal{I}}_h{f}^{n+1},v)}_{X^2},$$

(29)

where ${\alpha }_0=\Delta t\mathsf{\Gamma }(1-\alpha ),$ and $a_j^n$ is the j-th element of $a^n$ satisfying

$$\begin{array}{ccc}{a}_j^n\hfill & =\hfill & \left\{\begin{array}{cc}{\nu }_n,\hfill & j=0,\hfill \\ {\nu }_{n-j}-{\nu }_{n-j+1},\hfill & 1\le j\le n,\hfill \end{array}\right.\hfill \end{array}$$

(30)

$j=0,1,\cdots ,n.$

Lemma 8. 

For sufficiently small step sizes $\Delta t$ and h, the fully discrete form (28) is unconditionally stable such that

$$\parallel u_h^{n+1}{\parallel }_{X^2}^2\le 2\parallel u_h^0{\parallel }_{X^2}^2+C\underset{0\le k\le N_T}{max}{\parallel f^k\parallel }_{X^2}^2,$$

(31)

holds.

Proof. 

Taking $v=u_h^{n+1}$ in Equation (29) gives

$$\begin{array}{c}{(u_h^{n+1},u_h^{n+1})}_{X^2}+\lambda {\alpha }_0{({}_{H}D_{a,x}^{\beta }{u}_h^{n+1},{}_{H}D_{x,b}^{\beta }{u}_h^{n+1})}_{X^2}\hfill \\ =\sum _{j=0}^n{a}_j^n{(u_h^j,u_h^{n+1})}_{X^2}+{\alpha }_0{({\mathcal{I}}_h{f}^{n+1},u_h^{n+1})}_{X^2}.\hfill \end{array}$$

By using the Cauchy–Schwarz inequality, one has

$$\begin{array}{c}\parallel u_h^{n+1}{\parallel }_{X^2}^2+\lambda {\alpha }_0{\parallel {}_{H}D_{a,x}^{\beta }{u}_h^{n+1}\parallel }_{X^2}^2\hfill \\ \le {\displaystyle \frac{1}{2}}\sum _{j=0}^n{a}_j^n(\parallel u_h^j{\parallel }_{X^2}^2+\parallel u_h^{n+1}{\parallel }_{X^2}^2)+{\alpha }_0(\parallel {\mathcal{I}}_h{f}^{n+1}{\parallel }_{X^2}^2+\parallel u_h^{n+1}{\parallel }_{X^2}^2).\hfill \end{array}$$

Then,

$$\parallel u_h^{n+1}{\parallel }_{X^2}^2\le \sum _{j=0}^n{a}_j^n\parallel u_h^j{\parallel }_{X^2}^2+C{\parallel {\mathcal{I}}_h{f}^{n+1}\parallel }_{X^2}^2.$$

(32)

Now The mathematical induction method is used to prove

$$\parallel u_h^{n+1}{\parallel }_{X^2}^2\le 2\parallel u_h^0{\parallel }_{X^2}^2+C{\parallel {\mathcal{I}}_h{f}^{n+1}\parallel }_{X^2}^2$$

(33)

holds. For $n=0$, we have

$${(u_h^1,v)}_{X^2}+\lambda {\alpha }_0({}_{H}D_{a,x}^{\beta }{u}_h^1,{}_{H}D_{x,b}^{\beta }v)={(u_h^0,v)}_{X^2}+{\alpha }_0{({\mathcal{I}}_h{f}^1,v)}_{X^2}.$$

Taking $v=u_h^1$, one immediately has

$$\parallel u_h^1{\parallel }_{X^2}^2\le 2\parallel u_h^0{\parallel }_{X^2}^2+C{\parallel {\mathcal{I}}_h{f}^1\parallel }_{X^2}^2.$$

Suppose that

$${\parallel u_h^j\parallel }_{X^2}^2\le 2\parallel u_h^0{\parallel }_{X^2}^2+C{\parallel {\mathcal{I}}_h{f}^1\parallel }_{X^2}^2,j=1,2,\cdots ,n$$

(34)

holds, next we will show that

$${\parallel u_h^{n+1}\parallel }_{X^2}^2\le 2\parallel u_h^0{\parallel }_{X^2}^2+C{\parallel {\mathcal{I}}_h{f}^1\parallel }_{X^2}^2.$$

By using Equations (32) and (34), one obtains

$$\begin{array}{ccc}\parallel u_h^{n+1}{\parallel }_{X^2}^2\hfill & \le \hfill & \sum _{j=0}^m{a}_j^m\parallel u_h^j{\parallel }_{X^2}^2+C{\parallel {\mathcal{I}}_h{f}^{n+1}\parallel }_{X^2}^2\hfill \\ \hfill & \le \hfill & \sum _{j=0}^m{a}_j^m(2\parallel u_h^0{\parallel }_{X^2}^2+C\parallel {\mathcal{I}}_h{f}^{n+1}{\parallel }_{X^2}^2)+C\parallel {\mathcal{I}}_h{f}^{n+1}{\parallel }_{X^2}^2\hfill \\ \hfill & \le \hfill & 2\parallel u_h^0{\parallel }_{X^2}^2+C{\parallel {\mathcal{I}}_h{f}^{n+1}\parallel }_{X^2}^2.\hfill \end{array}$$

Therefore,

$$\begin{array}{c}\parallel u_h^{n+1}{\parallel }_{X^2}^2\le 2\parallel u_h^0{\parallel }_{X^2}^2+C{\parallel {\mathcal{I}}_h{f}^{n+1}\parallel }_{X^2}^2\hfill \end{array}$$

holds for all n. The proof is completed. □

Theorem 5. 

Assume that $u\left(t_{n+1}\right)$ is the exact solution of Equation (1), and ${\left\{u_h^{n+1}\right\}}_{n=0}^{N_T-1}$ is the numerical solution of the fully discrete scheme (28) with the following initial boundary conditions

$$u^0={\phi }_0\left(x\right),x\in \mathsf{\Omega },u^{n+1}=0,x\in \partial \mathsf{\Omega },t\in [0,T];$$

Then, $u_h^{n+1}$ satisfies the following error estimate:

$$\Vert u\left(t_{n+1}\right)-u_h^{n+1}{\Vert }_{X^2}\le C({(\Delta t)}^{2-\alpha }+h^r).$$

(35)

Proof. 

Denote ${\eta }^n=u^{n+1}-R_h^{n+1}$ and ${\rho }^{n+1}=R_h^{n+1}-u_h^{n+1}$. According to the $L1$ scheme [33], we have

$${\mathsf{\Lambda }}_t^{\alpha }{u}^{n+1}=\lambda D_{a,x}^{\alpha }{u}^{n+1}+f^{n+1}+O(\Delta t^{2-\alpha }).$$

Taking $u_h^{n+1}$ in Equation (28) yields

$${({\mathsf{\Lambda }}_t^{\alpha }{u}_h^{n+1},v)}_{X^2}+\lambda {({}_{H}D_{a,x}^{\beta }{u}_h^{n+1},{}_{H}D_{x,b}^{\beta }v)}_{X^2}={({\mathcal{I}}_h{f}_h^{n+1},v)}_{X^2}+r^{n+1}.$$

Removing $r^{n+1}$ from Equation (28) obtains

$${({\mathsf{\Lambda }}_t^{\alpha }{u}_h^{n+1},v)}_{X^2}+\lambda {({}_{H}D_{a,x}^{\beta }{u}_h^{n+1},{}_{H}D_{x,b}^{\beta }v)}_{X^2}={(E^{n+1},v)}_{X^2},$$

where ${({}_{H}D_{a,x}^{\alpha }(u^{n+1}-R_{h}u),{}_{H}D_{x,b}^{\alpha }v)}_{X^2}=0$ for $v\in X_{h0}^r,$ and $E^{n+1}=E_1^{n+1}+E_2^{n+1}+E_3^{n+1}$ satisfies

$$E_1^{n+1}=O(\Delta t^{2-\alpha }),E_2^{n+1}=f^{n+1}-{\mathcal{I}}_h{f}^{n+1},E_3^{n+1}=-{\mathsf{\Lambda }}^{\alpha }{\eta }^{n+1}.$$

Similar to the above Lemma 8, we need only estimate

$$\parallel {\rho }^0{\parallel }_{X^2}^2+C\underset{0\le k\le N_T}{max}{\parallel E^k\parallel }_{X^2}^2.$$

Obviously, ${\rho }^0=0$, and

$$\parallel E^{n+1}{\parallel }_{X^2}\le \parallel E_1^{n+1}{\parallel }_{X^2}+\parallel E_2^{n+1}{\parallel }_{X^2}+{\parallel E_3^{n+1}\parallel }_{X^2}\le C(\Delta t^{2-\alpha }+h^r).$$

Hence,

$$\parallel {\rho }^{n+1}{\parallel }_{X^2}\le C(\Delta t^{2-\alpha }+h^r).$$

Using

$$\parallel u_h^{n+1}-u^{n+1}{\parallel }_{X^2}\le \parallel {\rho }^{n+1}{\parallel }_{X^2}+{\parallel {\eta }^{n+1}\parallel }_{X^2}$$

yields the results. The proof is completed. □

7. Numerical Experiments

In this section, the numerical experiments are presented to test the theoretical analysis, where the Galerkin finite element method is used in the spatial direction, in which the piecewise linear log-polynomials [26] are adapted as the shape functions. The $L1$ scheme is applied in the temporal direction, with the uniform time stepping. According to the theoretical analysis, the expected convergence order with $X^2$ norm should be $\mathcal{O}(\Delta t^{2-\alpha }+h^2).$

Example 1. 

We consider the following fractional diffusion equation containing the Caputo derivative and Hadamard derivative as follows:

$${}_{C}D_{0,t}^{\alpha }u(x,t)=\lambda {}_{H}D_{1,x}^{2\beta }u(x,t)+f(x,t).$$

The space region is $\mathsf{\Omega }=(1,e)$ and the time region is $T=[0,1]$. The source term

$$f(x,t)={\displaystyle \frac{2t^{2-\alpha }}{\mathsf{\Gamma }(3-\alpha )}}{log}^{2}x(1-logx)-\lambda t^2[{\displaystyle \frac{2{log}^{2-2\beta }x}{\mathsf{\Gamma }(3-2\beta )}}-{\displaystyle \frac{6{log}^{3-2\beta }x}{\mathsf{\Gamma }(4-2\beta )}}],$$

and $\lambda =0.01$ are chosen. Then $u(x,t)=t^2{log}^{2}x(1-logx)$, which generates the homogenous initial-boundary values.

Table 1. $X^2$-norm errors and spatial convergence orders with $\alpha =0.9,\Delta t=ch$.

h	$\mathit{\beta }=0.9$	cv.rate	$\mathit{\beta }=0.7$	cv.rate	$\mathit{\beta }=0.5$	cv.rate
${\displaystyle \frac{1}{4}}$	$5.4725\times {10}^{-3}$	-	$5.4654\times {10}^{-3}$	-	$5.4565\times {10}^{-3}$	-
${\displaystyle \frac{1}{8}}$	$1.5161\times {10}^{-3}$	1.6238	$1.5025\times {10}^{-3}$	1.8630	$1.4881\times {10}^{-3}$	1.8745
${\displaystyle \frac{1}{16}}$	$4.6714\times {10}^{-4}$	1.6984	$4.3959\times {10}^{-4}$	1.7731	$4.1780\times {10}^{-4}$	1.8326
${\displaystyle \frac{1}{32}}$	$2.1521\times {10}^{-4}$	1.1181	$1.9019\times {10}^{-4}$	1.2087	$1.6285\times {10}^{-4}$	1.3593
${\displaystyle \frac{1}{64}}$	$1.0198\times {10}^{-4}$	1.0775	$8.9032\times {10}^{-5}$	1.0950	$7.5982\times {10}^{-5}$	1.0998

shows the $X^2$-norm errors and spatial convergence rates with $\alpha =0.9$ and $\beta =0.9,0.7,0.5$, separately.

Table 2. $X^2$-norm errors and temporal convergence orders with $\beta =0.9,\Delta t=c{h}^2$.

$\mathbf{\Delta }\mathit{t}$	$\mathit{\alpha }=0.8$	cv.rate	$\mathit{\alpha }=0.6$	cv.rate	$\mathit{\alpha }=0.4$	cv.rate
${\displaystyle \frac{1}{4}}$	$1.0007\times {10}^{-2}$	-	$1.0097·{10}^{-2}$	-	$1.0157\times {10}^{-2}$	-
${\displaystyle \frac{1}{8}}$	$1.7722\times {10}^{-3}$	2.4974	$1.7763\times {10}^{-4}$	2.5070	$1.7992\times {10}^{-3}$	2.4970
${\displaystyle \frac{1}{16}}$	$3.3374\times {10}^{-4}$	2.4087	$3.1284\times {10}^{-4}$	2.5054	$3.0068\times {10}^{-4}$	2.5811
${\displaystyle \frac{1}{32}}$	$8.1798\times {10}^{-5}$	2.0286	$8.1182\times {10}^{-5}$	1.9462	$8.5607\times {10}^{-5}$	1.8124
${\displaystyle \frac{1}{64}}$	$2.2641\times {10}^{-5}$	1.8531	$2.7893\times {10}^{-5}$	1.5413	$3.6984\times {10}^{-5}$	1.2108

shows $X^2$-norm errors and temporal convergence rates with $\beta =0.9$ and $\alpha =0.8,0.6,0.4$, separately. From the above tables, we can see that the experimental results support the theoretical analysis.

Figure 1 and Figure 2 show the numerical simulation between the exact solution and the numerical solution with $\alpha =0.9$ and $\beta =0.9$, separately. Figure 3 shows the experimental error results under the same condition. From the above figures, we can see that the numerical solution is in good agreement with the exact solution.

8. Conclusions

In this research, we studied the theoretical analysis and numerical approximation of the Caputo–Hadamard time-space fractional diffusion equation in one dimension. By using the novel modified Fourier transform, we studied the analytic solution of the Cauchy problem of the Caputo–Hadamard time-space fractional diffusion equation. By using the novel piecewise linear log-polynomials for the space Hadamard fractional derivative and the L1 scheme for the Caputo derivative, we studied the finite difference/finite element algorithm of the Caputo-Hadamard time-space fractional diffusion equation. The experimental results provided well validate the theoretical analysis. Furthermore, the theoretical analysis and numerical approximation of the multidimensional problem will be studied in the future.