Analysis of a Time-Stepping Discontinuous Galerkin Method for Fractional Diffusion-Wave Equations with Nonsmooth Data

Abstract

This paper analyzes a time-stepping discontinuous Galerkin method for fractional diffusion-wave problems. This method uses piecewise constant functions in the temporal discretization and continuous piecewise linear functions in the spatial discretization. Nearly optimal convergence with respect to the regularity of the solution is established when the source term is nonsmooth, and nearly optimal convergence rate $\scriptstyle \ln (1/\tau )(\sqrt{\ln (1/h)}h^2+\tau )$ is derived under appropriate regularity assumption on the source term. Convergence is also established without smoothness assumption on the initial value. Finally, numerical experiments are performed to verify the theoretical results.

1 Introduction

This paper considers the following time fractional diffusion-wave problem:

$$\begin{aligned} \left\{ \begin{aligned}&u' - \Delta {{\,\mathrm{D}\,}}_{0+}^{-\alpha } u = f&\text {in }~~ \Omega \times (0,T), \\&u = 0&\text {on } \partial \Omega \times (0,T) , \\&u(\cdot ,0) = u_0 &\text {in }~~ \Omega , \end{aligned} \right. \end{aligned}$$

(1)

where $0<\alpha <1$, $0< T < \infty$, $\Omega \subset {\mathbb {R}}^d$ ($d=1,2,3$) is a convex d-polytope, ${{\,\mathrm{D}\,}}_{0+}^{-\alpha }$ is a Riemann-Liouville fractional integral operator of order $\alpha$, and f and $u_0$ are two given functions. The above fractional diffusion-wave equation is an intermediate equation between diffusion and wave equations, and it also belongs to the class of evolution equations with a positive-type memory term (or integro-differential equations with a weakly singular convolution kernel), which has attracted a lot of works in the past thirty years.

Let us first briefly summarize some works devoted to the numerical treatments of problem (1). McLean and Thomée [9] proposed and analyzed two discretizations: the first uses the backward Euler method to approximate the first-order time derivative and a first-order integration rule to approximate the fractional integral; the second uses a second-order backward difference scheme to approximate the first-order time derivative and a second-order integration rule to approximate the fractional integral. Then McLean et al. [10] analyzed two discretizations with variable time steps: the first is a simple variant of the first one analyzed in [9]; the second combines the Crank–Nicolson scheme and two integral rules to approximate the fractional integral (but the temporal accuracy is not better than ${\mathcal {O}}(\tau ^{1+\alpha })$). By combining the first-order and second-order backward difference schemes and the convolution quadrature rules [4], Lubich et al. [5] proposed and analyzed two discretizations for problem (1), where optimal order error bounds were derived for positive time without spatial regularity assumption on the data. Cuesta et al. [1] proposed and studied a second-order discretization for problem (1) and its semilinear version.

By representing the solution as a contour integral by the Laplace transform technique and approximating this contour integral, McLean and Thomée [7, 8] developed and analyzed three numerical methods for problem (1). These methods use $2N+1$ quadrature points, and the first method possesses temporal accuracies ${\mathcal {O}}(e^{-cN})$ away from $t=0$, the second and third have temporal accuracy ${\mathcal {O}}(e^{-c\sqrt{N}})$.

McLean and Mustapha [11] studied a generalized Crank–Nicolson scheme for problem (1), and obtained accuracy ${\mathcal {O}}(h^2 + \tau ^2)$ on appropriately graded temporal grids under the condition that the solution and the forcing term satisfy some growth estimates. Mustapha and McLean [14] applied the famous time-stepping discontinuous Galerkin (DG) method [18, Chapter 12] to an evolution equation with a memory term of positive type. For the low-order DG method, they derived the accuracy order ${\mathcal {O}}(\ln (1/\tau )h^2 + \tau )$ on appropriately graded temporal grids under the condition that the time derivatives of the solution satisfy some growth estimates. We notice that this low-order DG method is identical to the first-order discretization analyzed in the aforementioned work [10]. They also analyzed an hp-version of the DG method in [13]. So far, to our best knowledge, the convergence of the low-order DG algorithm has not been established with nonsmooth data.

This paper is devoted to the convergence analysis of the aforementioned low-order DG method for problem (1) with nonsmooth data, which is a further development of the works in [10, 14]. For $f = 0$, we derive the error estimate

$$\begin{aligned} \Vert {u(\cdot ,t_j) - U_j} \Vert _{L^2(\Omega )} \leqslant C(h^2 t_j^{-\alpha -1} + \tau t_j^{-1}) \Vert {u_0} \Vert _{L^2(\Omega )}. \end{aligned}$$

For $u_0 = 0$, we obtain the following estimates:

$$\begin{aligned} \Vert {u-U} \Vert _{L^\infty (0,T;L^2(\Omega ))}&\leqslant C \big (h + \sqrt{\ln (1/h)}\,\tau ^{1/2}\big ) \Vert {f} \Vert _{L^2(0,T;\dot{H}^{\alpha /(\alpha +1)}(\Omega ))}, \\ \Vert {u-U} \Vert _{L^\infty (0,T;L^2(\Omega ))}&\leqslant C \ln (T/\tau ) \big (\sqrt{\ln (1/h)}h^2 + \tau \big ) \Vert {f} \Vert _{{}_0H^{\alpha +1/2}(0,T;L^2(\Omega ))}, \end{aligned}$$

where the first estimate is nearly optimal with respect to the regularity of the solution, and the second is nearly optimal. We note that $f(\cdot ,t)$ in the first estimate has no boundary condition on (0, T) due to $\alpha /(\alpha +1) < 1/2$. In addition, to investigate the effect of the nonvanishing $f(\cdot ,0)$ on the accuracy of the numerical solution, we establish the error estimate

$$\begin{aligned} \Vert {u(\cdot , t_j) - U_j} \Vert _{L^2(\Omega )} \leqslant C (t_j^{-\alpha } h^2 + \tau ) \Vert {{\tilde{f}}} \Vert _{L^2(\Omega )}, \end{aligned}$$

in the case that $u_0 = 0$ and $f(x,t) = {\tilde{f}}(x) \in L^2(\Omega )$, $0 \leqslant t \leqslant T$.

The rest of this paper is organized as follows. Section 2 introduces some Sobolev spaces, fractional calculus operators, the time-stepping discontinuous Galerkin method, the weak solution of problem (1) and its regularity results. Section 3 investigates two discretizations of two fractional ordinary equations, respectively. Section 4 establishes the convergence of the numerical method. Section 5 performs four numerical experiments to confirm the theoretical results. Finally, Sect. 6 provides some concluding remarks.

2 Preliminaries

Assume that $-\infty< a< b < \infty$. For each $m \in {\mathbb {N}}$, define

$$\begin{aligned} {}_0H^m(a,b)&:= \{v\in H^m(a,b): v^{(k)}(a)=0,\,\, 0\leqslant k<m\}, \\ {}^0H^m(a,b)&:= \{v\in H^m(a,b): v^{(k)}(b)=0,\,\, 0\leqslant k<m\}, \end{aligned}$$

where $H^m(a,b)$ is a usual Sobolev space [17] and $v^{(k)}$ is the k-th weak derivative of v. We equip the above two spaces with the norms

$$\begin{aligned} \Vert {v} \Vert _{{}^0H^m(a,b)}&:= \Vert {v^{(m)}} \Vert _{L^2(a,b)} \quad \forall v \in {}^0H^m(a,b), \\ \Vert {v} \Vert _{{}_0H^m(a,b)}&:= \Vert {v^{(m)}} \Vert _{L^2(a,b)} \quad \forall v \in {}_0H^m(a,b), \end{aligned}$$

respectively. For any $m \in {\mathbb {N}}_{>0}$ and $0< \theta < 1$, define

$$\begin{aligned} {}_0H^{m-\theta }(a,b)&:= [ {}_0H^{m-1}(a,b),\ {}_0H^m(a,b) ]_{1-\theta ,2}, \\ {}^0H^{m-\theta }(a,b)&:= [ {}^0H^{m-1}(a,b),\ {}^0H^m(a,b) ]_{1-\theta ,2}, \end{aligned}$$

where $[X,Y]_{1-\theta ,2}$ means the interpolation space of X and Y constructed by the famous K-method [17, Chapter 22]. For $0< \gamma < \infty$, we use ${}^0H^{-\gamma }(a,b)$ and ${}_0H^{-\gamma }(a,b)$ to denote the dual spaces of ${}_0H^\gamma (a,b)$ and ${}^0H^\gamma (a,b)$, respectively. Conversely, since ${}_0H^\gamma (a,b)$ and ${}^0H^\gamma (a,b)$ are reflexive, they are the dual spaces of ${}^0H^{-\gamma }(a,b)$ and ${}_0H^{-\gamma }(a,b)$, respectively. Moreover, for any $0< \gamma < 1/2$, ${}_0H^\gamma (a,b) = {}^0H^\gamma (a,b) = H^\gamma (a,b)$ with equivalent norms (cf. [3, Chapter 1]), and hence ${}_0H^{-\gamma }(a,b) = {}^0H^{-\gamma }(a,b)$ with equivalent norms.

It is well known that there exists an orthonormal basis $\{\phi _n: n \in {\mathbb {N}} \}$ of $L^2(\Omega )$ such that

$$\begin{aligned} \left\{ \begin{aligned}&-\Delta \phi _n = \lambda _n \phi _n&\, \mathrm{~in~}~\,\,\Omega ,\\&\phi _n=0&\mathrm{~on~}\partial \Omega , \end{aligned} \right. \end{aligned}$$

where $\{ \lambda _n: n \in {\mathbb {N}} \}$ is a positive non-decreasing sequence with $\lambda _n\rightarrow \infty$ as $n\rightarrow \infty$. For any $-\infty< \beta < \infty$, define

$$\begin{aligned} \dot{H}^\beta (\Omega ) := \left\{ \sum _{n=0}^\infty v_n \phi _n:\ \sum _{n=0}^\infty \lambda _n^\beta v_n^2 < \infty \right\} , \end{aligned}$$

and endow this space with the norm

$$\begin{aligned} \left\| \sum _{n=0}^\infty v_n \phi _n \right\| _{\dot{H}^\beta (\Omega )} := \left( \sum _{n=0}^\infty \lambda _n^\beta v_n^2 \right) ^{1/2} . \end{aligned}$$

For any $\beta ,\gamma \in {\mathbb {R}}$, define

$$\begin{aligned} {}^0H^\gamma (a,b;\dot{H}^\beta (\Omega )) := \left\{ \sum _{n=0}^\infty c_n \phi _n:\ \sum _{n=0}^\infty \lambda _n^\beta \Vert {c_n} \Vert _{{}^0H^\gamma (a,b)}^2 < \infty \right\} , \end{aligned}$$

and equip this space with the norm

$$\begin{aligned} \left\| \sum _{n=0}^\infty c_n \phi _n \right\| _{ {}^0H^\gamma (a,b;\dot{H}^\beta (\Omega )) } := \left( \sum _{n = 0}^\infty \lambda _n^\beta \Vert {c_n} \Vert _{{}^0H^\gamma (a,b)}^2 \right) ^{1/2}. \end{aligned}$$

The space ${}_0H^\gamma (a,b;\dot{H}^\beta (\Omega ))$ is analogously defined, and it is evident that ${}^0H^{-\gamma }(a,b;\dot{H}^{-\beta }(\Omega ))$ is the dual space of ${}_0H^\gamma (a,b;\dot{H}^\beta (\Omega ))$ in the sense that

$$\begin{aligned} \left\langle { \sum _{n=0}^\infty c_n \phi _n, \sum _{n=0}^\infty d_n \phi _n } \right\rangle _{{}_0H^\gamma (a,b;\dot{H}^\beta (\Omega ))} := \sum _{n=0}^\infty \langle {c_n, d_n} \rangle _{{}_0H^\gamma (a,b)} \end{aligned}$$

for all $\sum _{n=0}^\infty c_n \phi _n \in {}^0H^\gamma (a,b;\dot{H}^{-\beta }(\Omega ))$ and $\sum _{n=0}^\infty d_n \phi _n \in {}_0H^\gamma (a,b;\dot{H}^\beta (\Omega ))$. Since ${}_0H^\gamma (a,b;\dot{H}^\beta (\Omega ))$ is reflexive, it is the dual space of ${}^0H^{-\gamma }(a,b;\dot{H}^{-\beta }(\Omega ))$. Above and in what follows, for any Banach space W, the notation $\langle {\cdot ,\cdot } \rangle _W$ means the duality paring between $W^*$ (the dual space of W) and W.

2.1 Fractional Calculus Operators

This section introduces fractional calculus operators on a domain (a, b) , $-\infty< a< b < \infty$, and summarizes several properties of these operators to be used in this paper. Assume that X is a separable Hilbert space.

Definition 2.1

For $-1< \gamma < 0$, define

$$\begin{aligned} \left( {{\,\mathrm{D}\,}}_{a+}^\gamma v\right) (t)&:= \frac{1}{ \Gamma (-\gamma ) } \int _a^t (t-s)^{-\gamma -1} v(s) \, \mathrm {d}s, \quad t\in (a,b), \\ \left( {{\,\mathrm{D}\,}}_{b-}^\gamma v\right) (t)&:= \frac{1}{ \Gamma (-\gamma ) } \int _t^b (s-t)^{-\gamma -1} v(s) \, \mathrm {d}s, \quad t\in (a,b), \end{aligned}$$

for all $v \in L^1(a,b;X)$, where $\Gamma (\cdot )$ is the Gamma function. In addition, let ${{\,\mathrm{D}\,}}_{a+}^0$ and ${{\,\mathrm{D}\,}}_{b-}^0$ be the identity operator on $L^1(a,b;X)$. For $0 < \gamma \leqslant 1$, the left-sided and right-sided Riemann–Liouville fractional differential operators of order $\gamma$ are defined respectively as

$$\begin{aligned} {{\,\mathrm{D}\,}}_{a+}^\gamma v&:= {{\,\mathrm{D}\,}}{{\,\mathrm{D}\,}}_{a+}^{\gamma -1}v, \\ {{\,\mathrm{D}\,}}_{b-}^\gamma v&:= -{{\,\mathrm{D}\,}}{{\,\mathrm{D}\,}}_{b-}^{\gamma -1}v, \end{aligned}$$

for all $v \in L^1(a,b;X)$, where ${{\,\mathrm{D}\,}}$ is the first-order differential operator in the distribution sense.

Let $\{e_n:n \in {\mathbb {N}}\}$ be an orthonormal basis of X. For any $-\infty< \gamma < \infty$, define

$$\begin{aligned} {}^0H^\gamma (a,b;X) := \left\{ \sum _{n=0}^\infty c_n e_n:\ \sum _{n=0}^\infty \Vert {c_n} \Vert _{{}^0H^\gamma (a,b)} < \infty \right\} \end{aligned}$$

and endow this space with the norm

$$\begin{aligned} \left\| \sum _{n=0}^\infty c_n e_n\right\| _{{}^0H^\gamma (a,b;X)} := \left( \sum _{n=0}^\infty \Vert {c_n} \Vert _{{}^0H^\gamma (a,b)}^2 \right) ^{1/2}. \end{aligned}$$

The space ${}_0H^\gamma (a,b;X)$ is analogously defined. It is standard that ${}_0H^{-\gamma }(a,b;X)$ is the dual space of ${}^0H^\gamma (a,b;X)$ in the sense that

$$\begin{aligned} \left\langle { \sum _{n=0}^\infty c_n e_n, \sum _{n=0}^\infty d_n e_n } \right\rangle _{{}^0H^\gamma (a,b;X)} := \sum _{n=0}^\infty \langle {c_n, d_n} \rangle _{{}^0H^\gamma (a,b)} \end{aligned}$$

for all $\sum _{n=0}^\infty c_n e_n \in {}_0H^{-\gamma }(a,b;X)$ and $\sum _{n=0}^\infty d_n e_n \in {}^0H^\gamma (a,b;X)$.

Remark 2.1

For any $0< \gamma < 1$, a simple calculation gives that ${}_0H^\gamma (a,b;X)$ is identical to $[L^2(a,b;X), {}_0H^1(a,b;X)]_{\gamma ,2}$, and

$$\begin{aligned} \Vert {v} \Vert _{{}_0H^\gamma (a,b;X)} \leqslant \sqrt{2}\,\Vert {v} \Vert _{[L^2(a,b;X), {}_0H^1(a,b;X)]_{\gamma ,2}} \leqslant 2\Vert {v} \Vert _{{}_0H^\gamma (a,b;X)} \end{aligned}$$

for all $v \in {}_0H^\gamma (a,b;X)$.

Lemma 2.1

If $-1/2< \gamma < 1/2$, then

$$\begin{aligned} \cos (\gamma \pi ) \Vert {{{\,\mathrm{D}\,}}_{a+}^\gamma v} \Vert _{L^2(a,b;X)}^2 \leqslant ({{\,\mathrm{D}\,}}_{a+}^\gamma v, {{\,\mathrm{D}\,}}_{b-}^\gamma v)_{L^2(a,b;X)} \leqslant \sec (\gamma \pi ) \Vert {{{\,\mathrm{D}\,}}_{a+}^\gamma v} \Vert _{L^2(a,b;X)}^2, \\ \cos (\gamma \pi ) \Vert {{{\,\mathrm{D}\,}}_{b-}^\gamma v} \Vert _{L^2(a,b;X)}^2 \leqslant ({{\,\mathrm{D}\,}}_{a+}^\gamma v, {{\,\mathrm{D}\,}}_{b-}^\gamma v)_{L^2(a,b;X)} \leqslant \sec (\gamma \pi ) \Vert {{{\,\mathrm{D}\,}}_{b-}^\gamma v} \Vert _{L^2(a,b;X)}^2, \end{aligned}$$

for all $v \in {}_0H^\gamma (a,b;X)$ (equivalent to ${}^0H^\gamma (a,b;X)$), where $(\cdot ,\cdot )_{L^2(a,b;X)}$ is the usual inner product in $L^2(a,b;X)$.

Lemma 2.2

If $-1< \beta < 0$ and $-1 < \gamma \leqslant \beta$, then

$$\begin{aligned} C_1 \Vert {v} \Vert _{{}_0H^\beta (a,b;X)} \leqslant \Vert {{{\,\mathrm{D}\,}}_{a+}^\gamma v} \Vert _{{}_0H^{\beta -\gamma }(a,b;X )} \leqslant C_2 \Vert {v} \Vert _{{}_0H^\beta (a,b;X)} \,\forall v \in {}_0H^\beta (a,b;X), \\ C_1 \Vert {v} \Vert _{{}^0H^\beta (a,b;X)} \leqslant \Vert {{{\,\mathrm{D}\,}}_{b-}^\gamma v} \Vert _{{}^0H^{\beta -\gamma }(a,b;X )} \leqslant C_2 \Vert {v} \Vert _{{}^0H^\beta (a,b;X)} \,\forall v \in {}^0H^\beta (a,b;X), \end{aligned}$$

where $C_1$ and $C_2$ are two positive constants depending only on $\beta$ and $\gamma$.

Lemma 2.3

If $-1<\beta <1/2$ and $\beta< \gamma < \beta +1/2$, then

$$\begin{aligned} \langle {{{\,\mathrm{D}\,}}_{a+}^\gamma v, w} \rangle _{{}^0H^{\gamma -\beta }(a,b;X)} = \langle {{{\,\mathrm{D}\,}}_{a+}^\beta v, {{\,\mathrm{D}\,}}_{b-}^{\gamma -\beta } w} \rangle _{(a,b;X)} \end{aligned}$$

for all $v \in {}_0H^\beta (a,b;X)$ and $w \in {}^0H^{\gamma -\beta }(a,b;X)$.

Remark 2.2

For the proofs of the above lemmas, we refer the reader to [6, Section 3].

2.2 Algorithm Definition

Given $J \in {\mathbb {N}}_{>0}$, we set $\tau := T/J$ and $t_j := j\tau$, $0 \leqslant j \leqslant J$, and use $I_j$ to denote the interval $(t_{j-1},t_j)$ for each $1 \leqslant j \leqslant J$. Let ${\mathcal {K}}_h$ be a shape-regular triangulation of $\Omega$ consisting of d-simplexes, and we use h to denote the maximum diameter of the elements in ${\mathcal {K}}_h$. Define

$$\begin{aligned} S_h&:= \Big \{ v_h \in \dot{H}^1(\Omega ):\ v_h \text { is a linear polynomial on each } K \in {\mathcal {K}}_h \Big \},\\ W_{\tau ,h}&:= \Big \{ V \in L^2(0,T;S_h):\ V \text { is constant on } I_j,\, 1 \leqslant j \leqslant J \Big \}. \end{aligned}$$

For any $V \in W_{\tau ,h}$, we set

$$\begin{aligned} V_j&:= \lim \limits _{t\rightarrow t_{j}-}V(\cdot ,t), \quad 1 \leqslant j \leqslant J, \\ V^+_j&:= \lim \limits _{t\rightarrow t_{j}+}V(\cdot ,t), \quad 0 \leqslant j \leqslant J-1, \\ {[\![ {V_j} ]\!]}&:= V_j^{+} - V_j, \quad 0 \leqslant j \leqslant J, \end{aligned}$$

where the value of $V_0$ or $V_J^{+}$ will be explicitly specified whenever needed.

Assuming that $u_0 \in S_h^*$ and $f \in (W_{\tau ,h})^*$, we consider the following time-stepping discontinuous Galerkin scheme for problem (1): seek $U \in W_{\tau ,h}$ such that $U_0 = P_hu_0$ and

$$\begin{aligned} \sum _{j=0}^{J-1}\langle {{[\![ {U_j} ]\!]},V^+_j} \rangle _\Omega + \langle {\nabla {{\,\mathrm{D}\,}}_{0+}^{-\alpha } U,\nabla V} \rangle _{\Omega \times (0,T)} = \langle {f,V} \rangle _{W_{\tau ,h}} \end{aligned}$$

(2)

for all $V \in W_{\tau ,h}$, where $P_h$ is the $L^2$-orthogonal projection onto $S_h$. Above and afterwards, for a Lebesgue measurable set $\omega$ of $\mathbb R^l$ ($l= 1,2,3,4$), the symbol $\langle {p,q} \rangle _\omega$ means $\int _\omega pq$ whenever $pq \in L^1(\omega )$. In addition, the symbol $C_\times$ means a positive constant depending only on its subscript(s), and its value may differ at each occurrence.

Theorem 2.1

Assume that $u_0 \in L^2(\Omega )$. If $f \in L^1(0,T;L^2(\Omega ))$, then

$$\begin{aligned} \Vert {U} \Vert _{L^\infty (0,T;L^2(\Omega ))} \leqslant \sqrt{2}\ \Vert {u_0} \Vert _{L^2(\Omega )} + 2\Vert {f} \Vert _{L^1(0,T;L^2(\Omega ))}. \end{aligned}$$

(3)

If $f \in {}_0H^{\alpha /2}(0,T;\dot{H}^{-1}(\Omega ))$, then

$$\begin{aligned} \begin{aligned}&\Vert {U} \Vert _{L^\infty (0,T;L^2(\Omega ))} + \Vert {U} \Vert _{{}_0H^{-\alpha /2}(0,T;\dot{H}^1(\Omega ))} \\&\quad \leqslant {} C_\alpha \big ( \Vert {u_0} \Vert _{L^2(\Omega )} + \Vert {f} \Vert _{{}_0H^{\alpha /2}(0,T;\dot{H}^{-1}(\Omega ))} \big ). \end{aligned} \end{aligned}$$

(4)

For the proof of (3), we refer the reader to [14, Theorem 2.1]. By the techniques used in the proof of Theorem 4.3 (in Sect. 4), the proof of (4) is trivial and hence omitted.

2.3 Weak Solution and Regularity

Following [6], we introduce the weak solution to problem (1) as follows. Define

$$\begin{aligned} W&:= {}_0H^{(\alpha +1)/4}(0,T;L^2(\Omega )) \cap {}_0H^{-(\alpha +1)/4}(0,T;\dot{H}^1(\Omega )), \\ {\widehat{W}}&:= {}^0H^{(3-\alpha )/4}(0,T;L^2(\Omega )) \cap {}^0H^{(1-3\alpha )/4}(0,T;\dot{H}^1(\Omega )), \end{aligned}$$

and endow them with the norms

$$\begin{aligned} \Vert {\cdot } \Vert _W&:= \max \left\{ \Vert {\cdot } \Vert _{{}_0H^{(\alpha +1)/4}(0,T;L^2(\Omega )) },\ \Vert {\cdot } \Vert _{{}_0H^{-(\alpha +1)/4}(0,T;\dot{H}^1(\Omega ))} \right\} , \\ \Vert {\cdot } \Vert _{{\widehat{W}}}&:= \max \left\{ \Vert {\cdot } \Vert _{{}^0H^{(3-\alpha )/4}(0,T;L^2(\Omega )) },\ \Vert {\cdot } \Vert _{{}^0H^{(1-3\alpha )/4}(0,T;\dot{H}^1(\Omega ))} \right\} , \end{aligned}$$

respectively. Assuming that $u_0 t^{-(\alpha +1)/2} \in W^*$ and $f \in \widehat{W}^*$, we call $u \in W$ a weak solution to problem (1) if

$$\begin{aligned} \begin{aligned}&{} \left\langle { {{\,\mathrm{D}\,}}_{0+}^{(\alpha \!+\!1)/2} u, v } \right\rangle _{ {}^0\!H^{(\alpha \!+\!1)/4}(0,T;L^2(\Omega )\!) } \!+\! \left\langle { \nabla \! {{\,\mathrm{D}\,}}_{0+}^{\!-(\alpha \!+\!1)/4}u, \nabla \! {{\,\mathrm{D}\,}}_{T-}^{\!-(\alpha \!+\!1)/4} v } \right\rangle _{ \Omega \times (0,T)} \\&\quad ={} \left\langle { f,\ {{\,\mathrm{D}\,}}_{T-}^{(\alpha -1)/2} v } \right\rangle _{{\widehat{W}}} + \left\langle {\frac{t^{-(\alpha +1)/2}}{\Gamma ((1-\alpha )/2)} u_0,\ v} \right\rangle _W \end{aligned} \end{aligned}$$

(5)

for all $v \in W$.

By [6], we readily obtain the following regularity results.

Theorem 2.2

Assume that $u_0 t^{-(\alpha +1)/2} \in W^*$ and $f \in \widehat{W}^*$, then the weak solution u in (5) is well-defined and

$$\begin{aligned} \Vert {u} \Vert _W \leqslant C_\alpha \Big ( \Vert {f} \Vert _{{\widehat{W}}^*} + \Vert {t^{-(\alpha +1)/2} u_0} \Vert _{W^*} \Big ). \end{aligned}$$

Moreover, if $u_0 = 0$ and $f \in {}_0H^\gamma (0,T;\dot{H}^\beta (\Omega ))$ with $(\alpha -3)/4 \leqslant \gamma < \infty$ and $0 \leqslant \beta < \infty$, then the following conclusions hold:

• The solution u satisfies that

  $$\begin{aligned}&{{\,\mathrm{D}\,}}_{0+}^{\gamma +1} u - \Delta {{\,\mathrm{D}\,}}_{0+}^{\gamma -\alpha } u = {{\,\mathrm{D}\,}}_{0+}^\gamma f, \end{aligned}$$

  (6)
  $$\begin{aligned}&\Vert {u} \Vert _{{}_0H^{\gamma +1}(0,T;\dot{H}^\beta (\Omega ))} + \Vert {u} \Vert _{{}_0H^{\gamma -\alpha }(0,T;\dot{H}^{2+\beta }(\Omega ))} \leqslant C_{\alpha ,\gamma } \Vert {f} \Vert _{{}_0H^\gamma (0,T;\dot{H}^\beta (\Omega ))}; \end{aligned}$$

  (7)

• If $0 \leqslant \gamma < \alpha +1/2$, then

  $$\begin{aligned} \Vert {u} \Vert _{C([0,T];\dot{H}^{\beta +(2\gamma +1)/(\alpha +1)}(\Omega ))} \leqslant C_{\alpha ,\gamma } \Vert {f} \Vert _{{}_0H^\gamma (0,T;\dot{H}^\beta (\Omega ))}; \end{aligned}$$

  (8)

• If $\gamma = \alpha + 1/2$, then

  $$\begin{aligned} \Vert {u} \Vert _{ C([0,T];\dot{H}^{\beta +2(1-\epsilon )}(\Omega )) } \leqslant \frac{C_\alpha }{\sqrt{\epsilon }} \Vert {f} \Vert _{{}_0H^{\alpha +1/2}(0,T;\dot{H}^\beta (\Omega ))} \end{aligned}$$

  (9)

  for all $0< \epsilon < 1$.

Remark 2.3

For any $v \in W$, since [17, Lemma 33.2] implies

$$\begin{aligned} \sqrt{ \int _0^T t^{-(\alpha +1)/2} \Vert {v(t)} \Vert _{L^2(\Omega )}^2 \, \mathrm {d}t }\, \leqslant C_\alpha \Vert {v} \Vert _{{}_0H^{(\alpha +1)/4}(0,T;L^2(\Omega ))} \leqslant C_\alpha \Vert {v} \Vert _W, \end{aligned}$$

we have

$$\begin{aligned}&\left|{ \int _0^T t^{-(\alpha +1)/2} \langle {u_0, v(t)} \rangle _\Omega \, \mathrm {d}t } \right| \\&\quad \leqslant {} \Vert {u_0} \Vert _{L^2(\Omega )} \sqrt{ \int _0^T t^{-(\alpha +1)/2} \, \mathrm {d}t } \,\, \sqrt{ \int _0^T t^{-(\alpha +1)/2} \Vert {v(t)} \Vert _{L^2(\Omega )}^2 \, \mathrm {d}t } \\&\quad \leqslant {} C_\alpha \Vert {u_0} \Vert _{L^2(\Omega )} \Vert {v} \Vert _W. \end{aligned}$$

Therefore, $t^{-(\alpha +1)/2} u_0 \in W^*$ and hence the above weak solution is well-defined for the case $u_0 \in L^2(\Omega )$.

Next, we briefly summarize two other methods to define the weak solution to problem (1). The first method [11] uses the Mittag–Leffler function to define the weak solution of problem (1) with $f = 0$ and $u_0 \in \dot{H}^{-2}(\Omega )$, by that

$$\begin{aligned} u(t) = \sum _{n=0}^{\infty } \langle {u_0,\phi _n} \rangle _{\dot{H}^{2}(\Omega )} E_{\alpha +1,1}\big (-\lambda _n t^{\alpha +1}\big ) \phi _n, \quad 0 \leqslant t \leqslant T, \end{aligned}$$

where, for any $\beta , \gamma > 0$, the Mittag–Leffler function $E_{\beta ,\gamma }$ is defined by

$$\begin{aligned} E_{\beta ,\gamma }(z) := \sum _{n=0}^\infty \frac{z^n}{\Gamma (n\beta +\gamma )}, \quad z \in {\mathbb {C}}. \end{aligned}$$

Then we can investigate the regularity of this weak solution by a growth estimate [15]

$$\begin{aligned} |{E_{\beta ,\gamma }(-t)} | \leqslant {} \frac{C_{\beta ,\gamma }}{1+t},\quad t>0. \end{aligned}$$

The second method uses the well-known transposition technique to define the weak solution to problem (1) as follows. Define

$$\begin{aligned} G := {}^0H^1(0,T;L^2(\Omega )) \cap {}^0H^{-\alpha }(0,T;\dot{H}^2(\Omega )), \end{aligned}$$

and equip this space with the norm

$$\begin{aligned} \Vert {\cdot } \Vert _G := \max \left\{ \Vert {\cdot } \Vert _{{}^0H^1(0,T;L^2(\Omega ))},\ \Vert {\cdot } \Vert _{{}^0H^{-\alpha }(0,T;\dot{H}^2(\Omega ))} \right\} . \end{aligned}$$

Also, define

$$\begin{aligned} G_\mathrm {tr} := \big \{ v(0):\ v \in G \big \}, \end{aligned}$$

and endow this space with the norm

$$\begin{aligned} \Vert {v_0} \Vert _{G_\mathrm {tr}} := \inf _{v \in G,\ v(0) = v_0} \Vert {v} \Vert _G \quad \forall v_0 \in G_\mathrm {tr}. \end{aligned}$$

Assuming that $u_0 \in G_\mathrm {tr}^*$ and $f \in G^*$, we call u a weak solution to problem (1) if

$$\begin{aligned} \langle {u, -v' - \Delta {{\,\mathrm{D}\,}}_{T-}^{-\alpha } v} \rangle _{\Omega \times (0,T)} = \langle {f, v} \rangle _G + \langle {u_0, v(0)} \rangle _{G_\mathrm {tr}} \end{aligned}$$

for all $v \in G$. By the symmetric version of Theorem 2.2, applying the famous Babuška-Lax-Milgram theorem proves that the above weak solution is well-defined.

3 Discretizations of Two Fractional Ordinary Equations

3.1 An Auxiliary Function

For any $z \in \{x+iy:\ 0< x< \infty , -\infty< y < \infty \}$, define

$$\begin{aligned} \psi (z) := \frac{e^z-1}{\Gamma (2+\alpha )} \sum _{k=1}^\infty k^{1+\alpha } e^{-kz}. \end{aligned}$$

(10)

By the standard analytic continuation technique, $\psi$ has a Hankel integral representation (cf. [19, (12.1)] and [12, (21)])

$$\begin{aligned} \psi (z)&= \frac{e^z-1}{2\pi i} \int _{-\infty }^{({0+})} \frac{w^{-2-\alpha }}{e^{z-w}-1} \, \mathrm {d}w, \quad z \in {\mathbb {C}} \setminus (-\infty ,0], \end{aligned}$$

where $\int _{-\infty }^{({0+})}$ means an integral on a piecewise smooth and non-self-intersecting path enclosing the negative real axis and orienting counterclockwise, 0 and $\{z+2k\pi i \ne 0: k \in {\mathbb {Z}}\}$ lie on the different sides of this path, and $w^{-2-\alpha }$ is evaluated in the sense that

$$\begin{aligned} w^{-2-\alpha } = e^{-(2+\alpha ) {\text {Log}}w}. \end{aligned}$$

By Cauchy’s integral theorem and Cauchy’s integral formula, it is clear that (cf. [19, (13.1)])

$$\begin{aligned} \psi (z) = (e^z-1) \sum _{k \in {\mathbb {Z}}} (z+2k\pi i)^{-2-\alpha }, \end{aligned}$$

(11)

for all $z \in {\mathbb {C}} \setminus (-\infty , 0]$ satisfying $-2\pi< {\text {Im}} z < 2\pi$. From this series representation, it follows that

$$\begin{aligned} \psi (z) = \overline{\psi ({\overline{z}})} \quad \text {for all}\, z \in {\mathbb {C}} \setminus (-\infty ,0] \text { with } |{{\text {Im}}z} | < 2\pi . \end{aligned}$$

(12)

Moreover,

$$\begin{aligned} \psi (z) - (e^z-1)z^{-2-\alpha } \quad \text {is analytic on } \{w \in {\mathbb {C}}:\ |{{\text {Im}} w} | < 2\pi \}, \end{aligned}$$

(13)

and hence

$$\begin{aligned} \begin{aligned}&\lim _{r \rightarrow 0+} \frac{\psi (re^{i\theta })}{ r^{-1-\alpha }\big ( \cos ((1+\alpha )\theta ) - i \sin ((1+\alpha )\theta ) \big ) } = 1 \\&\text { uniformly for all } -\pi< \theta < \pi . \end{aligned} \end{aligned}$$

(14)

Lemma 3.1

For any $\mu >0$, there exist $\theta _\alpha$ and $\delta _{\alpha ,\mu }$, depending only on $\alpha$ and on $\alpha$ and $\mu$, respectively, with $\pi /2 < \theta _\alpha \leqslant (\alpha +3)/(4\alpha +4)\pi$ and $0< \delta _{\alpha ,\mu } < \infty$, such that

$$\begin{aligned} \begin{aligned}&1 + \mu \psi (z) \ne 0 \quad \text { for all } z \in \big \{ w \in {\mathbb {C}} \setminus \{0\}: -\pi \leqslant {\text {Im}} w \leqslant \pi \big \} \bigcap {} \\&\quad \big \{ w \in {\mathbb {C}}:\ 0 < {\text {Re}} w \leqslant \delta _{\alpha ,\mu } \text { or } \pi /2 \leqslant |{{\text {Arg}} w} | \leqslant \theta _\alpha \big \}. \end{aligned} \end{aligned}$$

(15)

Proof

By (14), there exists $0< \delta _\alpha < \pi$, depending only on $\alpha$, such that ${\text {Im}} \psi (z) < 0$ and hence

$$\begin{aligned} 1+\mu \psi (z) \ne 0 \text { for all } z \in \left\{ w \in {\mathbb {C}}:\ \pi /2 \leqslant {\text {Arg}} w \leqslant \frac{\alpha +3}{4(\alpha +1)}\pi ,\ 0 < {\text {Im}} w \leqslant \delta _\alpha \right\} . \end{aligned}$$

(16)

For $0 < y \leqslant \pi$, by (11) we have

$$\begin{aligned} \psi (iy)&= (e^{iy}-1) \sum _{k=-\infty }^\infty (iy+2k\pi i)^{-2-\alpha } \nonumber \\&= (e^{iy}-1) \left( \sum _{k=-\infty }^{-1} (-2k\pi - y)^{-2-\alpha } (-i)^{-2-\alpha } + \sum _{k=0}^\infty (2k\pi +y)^{-2-\alpha } i^{-2-\alpha } \right) \nonumber \\&= (1-e^{iy})\left( \sum _{k=1}^\infty (2k\pi - y)^{-2-\alpha } e^{i\alpha \pi /2} + \sum _{k=0}^\infty (2k\pi +y)^{-2-\alpha } e^{-i\alpha \pi /2} \right) \nonumber \\&= (1-e^{iy}) (A + iB), \end{aligned}$$

(17)

where

$$\begin{aligned} A&:= \cos (\alpha \pi /2) \sum _{k=0}^\infty \Big ( (2k\pi + 2\pi - y)^{-2-\alpha } + (2k\pi +y)^{-2-\alpha } \Big ), \\ B&:= \sin (\alpha \pi /2) \sum _{k=0}^\infty \Big ( (2k\pi +2\pi -y)^{-2-\alpha } - (2k\pi +y)^{-2-\alpha } \Big ). \end{aligned}$$

It follows that

$$\begin{aligned} {\text {Re}} \psi (iy) = A (1-\cos y) + B \sin y, \\ {\text {Im}} \psi (iy) = B(1-\cos y) - A \sin y. \end{aligned}$$

A straightforward calculation then gives

$$\begin{aligned}&{\text {Re}} \psi (i\pi ) = 4\pi ^{-2-\alpha } \cos (\alpha \pi /2) \sum _{k=1}^\infty (2k-1)^{-2-\alpha } > 0, \end{aligned}$$

(18)

$$\begin{aligned}&{\text {Im}} \psi (iy)< 0, \quad 0< y < \pi , \end{aligned}$$

(19)

and hence, by the continuity of $\psi$ in

$$\begin{aligned} \Phi := \left\{ z \in {\mathbb {C}} \setminus (-\infty ,0]: -2\pi< {\text {Im}} \psi (z) < 2\pi \right\} , \end{aligned}$$

a routine argument yields that there exists $0 < r_\alpha \leqslant \delta _\alpha \tan ((1-\alpha )/(4\alpha +4)\pi )$, depending only on $\alpha$, such that

$$\begin{aligned} 1+\mu \psi (z) \ne 0 \text { for all } z \in \left\{ w \in {\mathbb {C}}:\ -r_\alpha \leqslant {\text {Re}} w \leqslant 0,\ \delta _\alpha \leqslant {\text {Im}} w \leqslant \pi \right\} . \end{aligned}$$

(20)

By (16) and (20), letting $\theta _\alpha := \pi /2 + {\text {arctan}}(r_\alpha /\pi )$ yields

$$\begin{aligned} 1 + \mu \psi (z) \ne 0 \text { for all } z \in \{w \in {\mathbb {C}}:\ \pi /2 \leqslant {\text {Arg}}w \leqslant \theta _\alpha ,\ 0 < {\text {Im}} w \leqslant \pi \}. \end{aligned}$$

(21)

In addition, by (18), (19), (14) and the continuity of $\psi$ in $\Phi$, there exists $\delta _{\alpha ,\mu } > 0$ depending only on $\alpha$ and $\mu$ such that

$$\begin{aligned} 1+\mu \psi (z) \ne 0 \text { for all } z \in \{ w \in {\mathbb {C}} \setminus \{0\}:\ 0 \leqslant {\text {Re}} w \leqslant \delta _{\alpha ,\mu },\ 0 \leqslant {\text {Im}} w \leqslant \pi \}. \end{aligned}$$

(22)

Finally, combining (12), (21) and (22) proves (15) and hence this lemma. $\square$

Lemma 3.2

For any $\mu > 0$ and $0 < y \leqslant \pi$,

$$\begin{aligned} |{1+\mu \psi (iy)} | > C_\alpha (1+\mu y^{-1-\alpha }). \end{aligned}$$

(23)

Proof

By (14), (18) and (19), there exists $0< y_\alpha < \pi$, depending only on $\alpha$, such that

$$\begin{aligned} {\text {Re}} \psi (iy) > C_\alpha y^{-1-\alpha } \quad \forall \, y_\alpha \leqslant y \leqslant \pi , \\ {\text {Im}} \psi (iy)< -C_\alpha y^{-1-\alpha } \quad \forall \, 0 < y \leqslant y_\alpha . \end{aligned}$$

It follows that

$$\begin{aligned} |{1+\mu \psi (iy)} | > C_\alpha \mu y^{-1-\alpha } \quad \forall \, 0 < y \leqslant \pi , \end{aligned}$$

and hence

$$\begin{aligned} \inf _{ \begin{array}{c} 0< y \leqslant \pi \\ y^{1+\alpha } \leqslant \mu< \infty \end{array} } \frac{|{1+\mu \psi (iy)} |}{1+\mu y^{-1-\alpha }} \geqslant \inf _{ \begin{array}{c} 0< y \leqslant \pi \\ y^{1+\alpha } \leqslant \mu < \infty \end{array} } \frac{y^{1+\alpha }}{2\mu } |{1+\mu \psi (iy)} | > C_\alpha . \end{aligned}$$

It remains, therefore, to prove

$$\begin{aligned} \inf _{ \begin{array}{c} 0 < \mu \leqslant \pi ^{1+\alpha } \\ \mu ^{1/(1+\alpha )} \leqslant y \leqslant \pi \end{array} } \frac{|{1+\mu \psi (iy)} |}{1+\mu y^{-1-\alpha }} > C_\alpha . \end{aligned}$$

(24)

By (11), there exists a continuous function g on $[0,\pi ]$ such that $g(0) = 0$ and

$$\begin{aligned} \psi (iy) = (iy)^{-1-\alpha } + y^{-1-\alpha } g(y), \quad 0 < y \leqslant \pi . \end{aligned}$$

A straightforward calculation gives

$$\begin{aligned}&2|{1+\mu \psi (iy)} |^2 \\&\quad ={} 2|{ 1+\mu (iy)^{-1-\alpha } + \mu y^{-1-\alpha } g(y) } |^2 \\&\quad \geqslant {} |{1+\mu (iy)^{-1-\alpha }} |^2 - 2\mu ^2 y^{-2-2\alpha } |{g(y)} |^2 \\&\quad ={} 1 + \mu ^2 y^{-2-2\alpha } + 2\mu y^{-1-\alpha } \cos ((1+\alpha )\pi /2) - 2\mu ^2 y^{-2-2\alpha } |{g(y)} |^2 \\&\quad ={} \Big ( \mu y^{-1-\alpha } + \cos ^2\big ((1+\alpha )\pi /2\big ) \Big ) + \sin ^2\big ((1+\alpha )\pi /2\big ) - 2\mu ^2 y^{-2-2\alpha }|{g(y)} |^2 \\&\quad \geqslant {} \sin ^2\big ((1+\alpha )\pi /2\big ) - 2\mu ^2 y^{-2-2\alpha }|{g(y)} |^2, \end{aligned}$$

so that, by the fact $g(0) = 0$, there exists $0< y_\alpha < \pi$, depending only on $\alpha$, such that

$$\begin{aligned} \inf _{ \begin{array}{c} 0 < \mu \leqslant y_\alpha ^{1+\alpha } \\ \mu ^{1/(1+\alpha )} \leqslant y \leqslant y_\alpha \end{array} } |{1+\mu \psi (iy)} | > C_\alpha . \end{aligned}$$

In addition, applying the extreme value theorem implies, by (15), that

$$\begin{aligned} \inf _{ \begin{array}{c} 0 \leqslant \mu \leqslant \pi ^{1+\alpha } \\ y_\alpha \leqslant y \leqslant \pi \end{array} } |{1+\mu \psi (iy)} | > C_\alpha . \end{aligned}$$

Consequently, (24) follows from the above two estimates and the estimate

$$\begin{aligned} \inf _{ \begin{array}{c} 0< \mu \leqslant \pi ^{1+\alpha } \\ \mu ^{1/(1+\alpha )} \leqslant y \leqslant \pi \end{array} } \frac{|{1+\mu \psi (iy)} |}{1+\mu y^{-1-\alpha }} \geqslant \frac{1}{2} \inf _{ \begin{array}{c} 0 < \mu \leqslant \pi ^{1+\alpha } \\ \mu ^{1/(1+\alpha )} \leqslant y \leqslant \pi \end{array} } |{1+\mu \psi (iy)} |. \end{aligned}$$

This completes the proof. $\square$

Lemma 3.3

For any $\mu > 0$ and $0 < y \leqslant \pi$,

$$\begin{aligned} |{g'(y)} | < C_\alpha \frac{\mu y^{-2-\alpha }}{(1+\mu y^{-1-\alpha })^2}, \end{aligned}$$

(25)

where $g(y) := (1+\mu \psi (iy))^{-1}$.

Proof

By (17), $\psi (iy)$ can be expressed in the form

$$\begin{aligned} \psi (iy) = F(y) + G(y), \quad 0 < y \leqslant \pi , \end{aligned}$$

where F is analytic on $[0,\pi ]$ and

$$\begin{aligned} G(y) = (1-e^{iy}) y^{-2-\alpha } \big ( \cos (\alpha \pi /2) - i \sin (\alpha \pi /2) \big ). \end{aligned}$$

A direct calculation gives

$$\begin{aligned} |{G'(y)} |< C_\alpha y^{-2-\alpha }, \quad 0 < y \leqslant \pi , \end{aligned}$$

so that

$$\begin{aligned} \left|{i\psi '(iy)} \right| = |{F'(y) + G'(y)} |< C_\alpha y^{-2-\alpha }, \quad 0 < y \leqslant \pi . \end{aligned}$$

In addition, Lemma 3.2 implies

$$\begin{aligned} |{1+\mu \psi (iy)} |^{-2}< C_\alpha (1+\mu y^{-1-\alpha })^{-2}, \quad 0 < y \leqslant \pi . \end{aligned}$$

Therefore, (25) follows from the equality

$$\begin{aligned} g'(y) = \frac{i\mu \psi '(iy)}{(1+\mu \psi (iy))^2}. \end{aligned}$$

This completes the proof. $\square$

In the next two subsections, we use $\theta$ to abbreviate $\theta _\alpha$ given in Lemma 3.1, define

$$\begin{aligned} \Upsilon := (\infty ,0]e^{-i\theta } \cup [0,\infty )e^{i\theta }, \end{aligned}$$

and let $\Upsilon$ be oriented so that ${\text {Im}} z$ increases along $\Upsilon$. In addition, set

$$\begin{aligned} \Upsilon _1 := \{z \in \Upsilon :\ |{{\text {Im}} z} | \leqslant \pi \}, \end{aligned}$$

which inherits the orientation of $\Upsilon$.

3.2 The First Fractional Ordinary Equation

This subsection considers the fractional ordinary equation

$$\begin{aligned} \xi '(t) + \lambda {{\,\mathrm{D}\,}}_{0+}^{-\alpha } \xi (t) = 0, \quad t > 0, \end{aligned}$$

(26)

subject to the initial value condition $\xi (0) = \xi _0$, where $\lambda$ is a positive constant and $\xi _0 \in {\mathbb {R}}$. By [5, (2.1)], the solution $\xi$ of Eq. (26) can be expressed by a contour integral

$$\begin{aligned} \xi (t) = \frac{\xi _0}{2\pi i} \int _\Upsilon e^{tz} z^\alpha (z^{1+\alpha }+\lambda )^{-1} \, \mathrm {d}z, \quad t > 0. \end{aligned}$$

(27)

Applying the temporal discretization used in (2) to Eq. (26) yields the following discretization: let $Y_0 = \xi _0$; for $k \in {\mathbb {N}}$, the value of $Y_{k+1}$ is determined by that

$$\begin{aligned} \mu \left( \sum _{j=1}^k Y_j \big ( b_{k-j+2} - 2b_{k-j+1} + b_{k-j} \big ) + b_1 Y_{k+1} \right) + Y_{k+1} - Y_k = 0, \end{aligned}$$

where $\mu := \lambda \tau ^{1+\alpha }$ and $b_j := j^{1+\alpha }/\Gamma (2+\alpha )$, $j \in {\mathbb {N}}$.

Theorem 3.1

For any $k \in {\mathbb {N}}_{>0}$, it holds

$$\begin{aligned} |{Y_{k+1} - Y_k} | \leqslant C_\alpha k^{-1} |{\xi _0} |. \end{aligned}$$

(28)

Theorem 3.2

For any $k \in {\mathbb {N}}_{>0}$, it holds

$$\begin{aligned} |{\xi (t_k) - Y_k} |&\leqslant C_\alpha k^{-1} |{\xi _0} | . \end{aligned}$$

(29)

The main task of the rest of this subsection is to prove the above two theorems by the well-known Laplace transform method (the basic idea comes from [2, 5, 12]). To this end, we introduce the discrete Laplace transform of $(Y_k)_{k=0}^\infty$ by that

$$\begin{aligned} {\widehat{Y}}(z) := \sum _{k=0}^\infty Y_k e^{-kz} \quad \forall z \in H, \end{aligned}$$

(30)

where $H := \{x+iy: 0 < x \leqslant \delta _{\alpha ,\mu },\, -\pi \leqslant y \leqslant \pi \}$, with $\delta _{\alpha ,\mu }$ being defined in Lemma 3.1. By the definition of the sequence $(Y_k)_{k=0}^\infty$, a straightforward calculation gives

$$\begin{aligned} \mu ({\widehat{Y}}(z)-\xi _0) (e^z-1)^2\, {\widehat{b}}(z) + ({\widehat{Y}}(z) - \xi _0) e^z - {\widehat{Y}}(z) = 0, \quad z \in H, \end{aligned}$$

where ${\widehat{b}}$ is the discrete transform of the sequence $(b_k)_{k=0}^\infty$, namely

$$\begin{aligned} {\widehat{b}}(z) = \sum _{k=1}^\infty \frac{k^{1+\alpha }}{\Gamma (2+\alpha )} e^{-kz}. \end{aligned}$$

For any $z \in H$, combining like terms yields

$$\begin{aligned} (e^z-1 + \mu (e^z-1)^2 {\widehat{b}}(z)) {\widehat{Y}}(z) - \big ( e^z + \mu (e^z-1)^2\,{\widehat{b}}(z) \big ) \xi _0 = 0, \end{aligned}$$

which, together with (10) and Lemma 3.1, indicates

$$\begin{aligned} {\widehat{Y}}(z)&= \frac{ e^z + \mu (e^z-1)^2\, {\widehat{b}}(z) }{ e^z-1 + \mu (e^z-1)^2\,{\widehat{b}} } \, \xi _0 \\&= \Big ( 1 + \frac{1}{e^z-1 + \mu (e^z-1)^2\,{\widehat{b}}} \Big ) \xi _0 \\&= \Big ( 1 + \frac{(e^z-1)^{-1}}{1+\mu \psi (z)} \Big ) \xi _0. \end{aligned}$$

Therefore, a routine calculation (cf. [12, (28)]) yields that, for any $0 < a \leqslant \delta _{\alpha ,\mu }$ and $k \in {\mathbb {N}}_{>0}$,

$$\begin{aligned} Y_k&= \frac{\xi _0}{2\pi i} \int _{a-i\pi }^{a+i\pi } {\widehat{Y}}(z) e^{kz} \, \mathrm {d}z = \frac{\xi _0}{2\pi i} \int _{a-i\pi }^{a+i\pi } \frac{e^{kz}}{1+\mu \psi (z)} \frac{\mathrm {d}z}{e^z-1}. \end{aligned}$$

By (14) and (15), letting $a \rightarrow {0+}$ and applying Lebesgue’s dominated convergence theorem then lead to

$$\begin{aligned} Y_k = \frac{\xi _0}{2\pi i}\, \int _{-i\pi }^{i\pi } \frac{e^{kz}}{1+\mu \psi (z)} \, \frac{\mathrm {d}z}{e^z-1}. \end{aligned}$$

(31)

From (15) it follows that the integrand in (31) is analytic on

$$\begin{aligned} \omega := \left\{ z \in {\mathbb {C}}:\ 0< |{{\text {Im}} z} |< \pi ,\ \pi /2< |{{\text {Arg}} z} | < \theta \right\} \end{aligned}$$

and continuous on $\partial \omega \setminus \{0\}$, and that (14) implies that

$$\begin{aligned} \lim _{\omega \ni z \rightarrow 0} |{z} |^{-\alpha } |{e^{kz}(1+\mu \psi (z))^{-1}(e^z-1)^{-1}} | = \mu ^{-1}. \end{aligned}$$

Additionally,

$$\begin{aligned} \frac{e^{kz}}{(1+\mu \psi (z))(e^z-1)} = \frac{e^{k(z+2\pi i)}}{(1+\mu \psi (z+2\pi i))(e^{z+2\pi i}-1)} \end{aligned}$$

for all $z = x-i\pi$, $-\pi \tan \theta \leqslant x \leqslant 0$. Therefore, an elementary calculation yields

$$\begin{aligned} Y_k = \frac{\xi _0}{2\pi i} \int _{\Upsilon _1} \frac{e^{kz}}{1+\mu \psi (z)} \frac{\mathrm {d}z}{e^z-1}, \end{aligned}$$

(32)

by (31) and Cauchy’s integral theorem.

Remark 3.1

By the techniques used in the proof of Theorem 2.1, it is easy to obtain that $|{Y_k} | \leqslant |{\xi _0} |$ for all $k \in {\mathbb {N}}_{>0}$. Thus, the series in (30) converges absolutely for all $z \in H$.

Finally, we show the proofs of Theorems 3.1 and 3.2 as follows.

Proof of Theorem 3.1

Firstly, let us prove

$$\begin{aligned} \Big |\int _0^\pi \cos (ky) g(y) \, \mathrm {d}y \Big |\leqslant C_\alpha k^{-1}, \end{aligned}$$

(33)

where $g(y) := (1+\mu \psi (iy))^{-1}$, $0 < y \leqslant \pi$. A straightforward calculation gives

$$\begin{aligned} \int _0^\pi \cos (ky) g(y) \, \mathrm {d}y&= \sum _{j=1}^k \int _{(j-1)\pi /k}^{j\pi /k} \cos (ky) g(y) \, \mathrm {d}y \\&= \sum _{j=1}^k \int _{(j-1)\pi /k}^{j\pi /k} \cos (ky) \big ( g(y)-g((j-1)\pi /k) \big ) \, \mathrm {d}y \\&= \sum _{j=1}^k \int _{(j-1)\pi /k}^{j\pi /k} \cos (ky) \int _{(j-1)\pi /k}^y g'(s) \, \mathrm {d}s \, \mathrm {d}y. \end{aligned}$$

It follows that

$$\begin{aligned}&\left|{\int _0^\pi \cos (ky) g(y) \, \mathrm {d}y} \right| \leqslant \sum _{j=1}^k \int _{(j-1)\pi /k}^{j\pi /k} \int _{(j-1)\pi /k}^y |{g'(s)} | \, \mathrm {d}s \, \mathrm {d}y \\&\quad<{} \pi k^{-1} \int _0^\pi |{g'(y)} | \, \mathrm {d}y< C_\alpha k^{-1} \int _0^\pi \frac{\mu y^{-2-\alpha }}{(1+\mu y^{-1-\alpha })^2} \, \mathrm {d}y \quad (\text {by Lemma}~\text {3.3}) \\&\quad<{} C_\alpha k^{-1} \bigg ( \int _0^{\mu ^{1/(1+\alpha )}} \frac{\mu y^{-2-\alpha }}{(1+\mu y^{-1-\alpha })^2} \, \mathrm {d}y + \int _{\mu ^{1/(1+\alpha )}}^{\max \{\mu ^{1/(1+\alpha )}, \pi \}} \frac{\mu y^{-2-\alpha }}{(1+\mu y^{-1-\alpha })^2} \, \mathrm {d}y \bigg ) \\&\quad<{} C_\alpha k^{-1} \bigg ( \int _0^{\mu ^{1/(1+\alpha )}} \mu ^{-1} y^\alpha \, \mathrm {d}y + \int _{\mu ^{1/(1+\alpha )}}^{\max \{\mu ^{1/(1+\alpha )},\pi \}} \mu y^{-2-\alpha } \, \mathrm {d}y \bigg ) < C_\alpha k^{-1}, \end{aligned}$$

which proves (33).

Secondly, let us prove

$$\begin{aligned} \Big |\int _0^\pi \sin (ky) g(y) \, \mathrm {d}y \Big |< C_\alpha k^{-1}. \end{aligned}$$

(34)

If $k = 1 + 2m$, $m \in {\mathbb {N}}$, then a similar argument as that to derive (33) yields

$$\begin{aligned} \Big |\int _0^{2m\pi /(1+2m)} \sin (ky) g(y) \, \mathrm {d}y \Big |< C_\alpha k^{-1}, \end{aligned}$$

and hence (34) follows from the estimate

$$\begin{aligned} \Big |\int _{2m\pi /(1+2m)}^\pi \sin (ky) g(y) \, \mathrm {d}y \Big |< C_\alpha k^{-1}, \end{aligned}$$

which is evident by Lemma 3.2. If $k = 2m$, $m \in {\mathbb {N}}_{>0}$, then a simple modification of the above analysis proves that (34) still holds.

Finally, combining (33) and (34) yields

$$\begin{aligned} \Big |\int _0^\pi e^{iky} g(y) dy \Big |\leqslant C_\alpha k^{-1}, \end{aligned}$$

so that

$$\begin{aligned} \Bigl |{\text {Re}} \int _0^\pi e^{iky} g(y) \, \mathrm {d}y \Big |\leqslant C_\alpha k^{-1}. \end{aligned}$$

Therefore, (28) follows from

$$\begin{aligned} Y_{k+1} - Y_k = \frac{\xi _0}{\pi }\, {\text {Re}} \int _0^\pi e^{iky}g(y) \, \mathrm {d}y, \end{aligned}$$

which is evident by (12) and (31). This concludes the proof of Theorem 3.1. $\square$

Proof of Theorem 3.2

Substituting $\eta := \tau z$ into (27) yields

$$\begin{aligned} \xi (t_k) = \frac{\xi _0}{2\pi i} \int _\Upsilon e^{k\eta }(\eta + \mu \eta ^{-\alpha })^{-1} \, \mathrm {d}\eta , \end{aligned}$$

and then subtracting (32) from this equation gives

$$\begin{aligned} \xi (t_k) - Y_k = {\mathbb {I}}_1 + {\mathbb {I}}_2, \end{aligned}$$

(35)

where

$$\begin{aligned} {\mathbb {I}}_1&:= \frac{\xi _0}{2\pi i} \int _{\Upsilon \setminus \Upsilon _1} e^{kz} (z+\mu z^{-\alpha })^{-1} \, \mathrm {d}z, \\ {\mathbb {I}}_2&:= \frac{\xi _0}{2\pi i} \int _{\Upsilon _1} e^{kz} \Big ( (z + \mu z^{-\alpha })^{-1} - (1+\mu \psi (z))^{-1} (e^z-1)^{-1} \Big ) \, \mathrm {d}z. \end{aligned}$$

Since ${\mathbb {I}}_1$ is a real number, a simple calculation gives

$$\begin{aligned} {\mathbb {I}}_1&= \frac{\xi _0}{\pi }{\text {Im}} \int _{\pi /\sin \theta }^\infty e^{kre^{i\theta }} (re^{i\theta } + \mu (re^{i\theta })^{-\alpha })^{-1} e^{i\theta } \,\mathrm {d}r \\&= \frac{\xi _0}{\pi }{\text {Im}} \int _{\pi /\sin \theta }^\infty e^{kre^{i\theta }} \frac{(re^{i\theta })^\alpha }{ (re^{i\theta })^{1+\alpha } + \mu } e^{i\theta } \,\mathrm {d}r, \end{aligned}$$

and the fact $\pi /2< \theta < (\alpha +3)/(4\alpha +4)\pi$ implies

$$\begin{aligned} \left|{\frac{(re^{i\theta })^\alpha }{(re^{i\theta })^{1+\alpha }+\mu }} \right| = \frac{r^\alpha }{ |{ r^{1+\alpha }\cos ((1+\alpha )\theta ) + \mu + i r^{1+\alpha }\sin ((1+\alpha )\theta ) } | } < C_\alpha r^{-1}. \end{aligned}$$

Consequently,

$$\begin{aligned} |{{\mathbb {I}}_1} |&\leqslant C_\alpha |{\xi _0} | \int _{\pi /\sin \theta }^\infty e^{kr\cos \theta } r^{-1} \, \mathrm {d}r \leqslant C_\alpha |{\xi _0} | \int _{\pi /\sin \theta }^\infty e^{kr\cos \theta } \, \mathrm {d}r \\&\leqslant C_\alpha k^{-1} e^{k\pi \cot \theta } |{\xi _0} |. \end{aligned}$$

Then let us estimate ${\mathbb {I}}_2$. For any $z \in \Upsilon _1 \setminus \{0\}$, since

$$\begin{aligned}&z + \mu z^{-\alpha } = z^{-\alpha }(z^{1+\alpha } + \mu ) \\&\quad ={} |{z} |^{-\alpha } e^{-i\alpha \theta } \Big ( |{z} |^{1+\alpha } \cos \big ((1+\alpha )\theta \big ) + \mu + i |{z} |^{1+\alpha } \sin \big ((1+\alpha )\theta \big ) \Big ), \end{aligned}$$

from the fact $\pi /2< \theta < (\alpha +3)/(4\alpha +4)\pi$ it follows that

$$\begin{aligned} |{z+\mu z^{-\alpha }} | > C_\alpha |{z} |. \end{aligned}$$

By (11), a routine calculation gives

$$\begin{aligned} |{(1+\mu \psi (z))(e^z-1) - (z+\mu z^{-\alpha })} | \leqslant C_\alpha \big ( |{z} |^2 + \mu |{z} |^{1-\alpha } \big ), \end{aligned}$$

and, similar to (23), we have

$$\begin{aligned} |{1+\mu \psi (z)} | > C_\alpha (1+\mu |{z} |^{-1-\alpha }). \end{aligned}$$

In addition, it is clear that

$$\begin{aligned} |{e^z-1} | > C_\alpha |{z} |, \quad z \in \Upsilon _1 \setminus \{0\}. \end{aligned}$$

Using the above four estimates, we obtain

$$\begin{aligned}&|{ (z+\mu z^{-\alpha })^{-1} - (1+\mu \psi (z))^{-1}(e^z-1)^{-1} } | \\&\quad ={} \left|{ \frac{ (1+\mu \psi (z))(e^z-1) - (z+\mu z^{-\alpha }) }{ (z+\mu z^{-\alpha })(1+\mu \psi (z))(e^z-1) } } \right| \\&\quad <{} C_\alpha \frac{ |{z} |^2 + \mu |{z} |^{1-\alpha } }{ |{z} |^2(1+\mu |{z} |^{-1-\alpha }) } = C_\alpha \end{aligned}$$

for all $z \in \Upsilon _1 \setminus \{0\}$. Therefore,

$$\begin{aligned} |{{\mathbb {I}}_2} |&= \left|{ \frac{\xi _0}{\pi }{\text {Im}} \int _0^{\pi \!/\!\sin \theta } \! e^{kre^{i\theta }} \! \Big ( \big ( re^{i\theta } \!+\! \mu (re^{i\theta })^{-\alpha } \big )^{-1} \!-\! \big (1\!+\!\mu \psi (re^{i\theta })\big )^{-1} \big ( e^{re^{i\theta }} \!-\! 1 \big )^{-1} \Big ) e^{i\theta } \mathrm {d}r } \right| \\&\leqslant C_\alpha |{\xi _0} | \int _0^{\pi /\sin \theta } e^{kr\cos \theta } \, \mathrm {d}r \leqslant C_\alpha k^{-1} |{\xi _0} |. \end{aligned}$$

Finally, combing (35) and the above estimates for ${\mathbb {I}}_1$ and ${\mathbb {I}}_2$ proves (29) and thus concludes the proof. $\square$

3.3 The Second Fractional Ordinary Equation

This subsection considers the fractional ordinary equation

$$\begin{aligned} \xi '(t) + \lambda {{\,\mathrm{D}\,}}_{0+}^{-\alpha } \xi (t) = 1, \quad t > 0, \end{aligned}$$

(36)

subject to the initial value condition $\xi (0) = 0$. Applying the temporal discretization in (2) yields the following discretization: let $Y_0 = 0$; for $k \in {\mathbb {N}}$, the value of $Y_{k+1}$ is determined by that

$$\begin{aligned} \mu \left( \sum _{j=1}^k Y_j(b_{k-j+2} - 2b_{k-j+1} + b_{k-j}) + b_1 Y_{k+1} \right) + Y_{k+1} - Y_k = \tau . \end{aligned}$$

(37)

Similar to (27) and (32), we have

$$\begin{aligned} \xi (t)&= \frac{1}{2\pi i} \int _{\Upsilon } e^{tz} (z^2+\lambda z^{1-\alpha })^{-1} \, \mathrm {d}z, \quad t > 0, \end{aligned}$$

(38)

$$\begin{aligned} Y_k&= \frac{\tau }{2\pi i} \int _{\Upsilon _1} \frac{e^{kz+z}}{1+\mu \psi (z)} \frac{\mathrm {d}z}{(e^z-1)^2}, \quad k \in {\mathbb {N}}_{>0}. \end{aligned}$$

(39)

Theorem 3.3

For any $k \in {\mathbb {N}}_{>0}$,

$$\begin{aligned} |{\xi (t_k) - Y_k} | < C_\alpha \tau . \end{aligned}$$

(40)

Proof

Since the proof of this theorem is similar to that of Theorem 3.2, we only highlight the differences. Proceeding as in the proof of Theorem 3.2 yields

$$\begin{aligned} \xi (t_k) - Y_k = {\mathbb {I}}_1 + {\mathbb {I}}_2, \end{aligned}$$

where

$$\begin{aligned} {\mathbb {I}}_1&:= \frac{\tau }{2\pi i} \int _{\Upsilon \setminus \Upsilon _1} e^{kz} (z^2+\mu z^{1-\alpha })^{-1} \, \mathrm {d}z, \\ {\mathbb {I}}_2&:= \frac{\tau }{2\pi i} \int _{\Upsilon _1} e^{kz} \Big ( (z^2 + \mu z^{1-\alpha })^{-1} - (1+\mu \psi (z))^{-1} (e^z-1)^{-2} e^z \Big ) \, \mathrm {d}z. \end{aligned}$$

Moreover,

$$\begin{aligned} |{{\mathbb {I}}_1} |< C_\alpha \tau \int _{\pi /\sin \theta }^\infty e^{kr\cos \theta } r^{-2} \, \mathrm {d}r< C_\alpha \tau \int _{\pi /\sin \theta }^\infty e^{kr\cos \theta } \, \mathrm {d}r < C_\alpha \tau k^{-1} e^{k\pi \cot \theta }. \end{aligned}$$

For any $z \in \Upsilon _1 \setminus \{0\}$, since

$$\begin{aligned}&z^2 + \mu z^{1-\alpha } = z^{1-\alpha }(z^{1+\alpha } + \mu ) \\&\quad ={} |{z} |^{1-\alpha } e^{i(1-\alpha )\theta } \Big ( |{z} |^{1+\alpha } \cos \big ((1+\alpha )\theta \big ) + \mu + i |{z} |^{1+\alpha } \sin \big ((1+\alpha )\theta \big ) \Big ), \end{aligned}$$

from the fact $\pi /2< \theta < (\alpha +3)/(4\alpha +4)\pi$ it follows that there exists a positive constant c, depending only on $\alpha$, such that

$$\begin{aligned} |{z^2 + \mu z^{1-\alpha }} | > {\left\{ \begin{array}{ll} C_\alpha \mu |{z} |^{1-\alpha } &{} \text {if}\quad 0 < |{z} | \leqslant c\mu ^{1/(1+\alpha )} , \\ C_\alpha |{z} |^2 &{} \text {if}\quad c \mu ^{1/(1+\alpha )} \leqslant |{z} | \leqslant \pi /\sin \theta . \end{array}\right. } \end{aligned}$$

By (11), a routine calculation gives

$$\begin{aligned} |{(1+\mu \psi (z))(e^z-1)^2 - (z^2+\mu z^{1-\alpha }) e^z} | < C_\alpha \big ( |{z} |^4 + \mu |{z} |^{2-\alpha } \big ), \end{aligned}$$

and, similar to (23), it holds

$$\begin{aligned} |{1+\mu \psi (z)} | > C_\alpha (1+\mu |{z} |^{-1-\alpha }). \end{aligned}$$

Using the above three estimates, we obtain

$$\begin{aligned}&|{ (z^2+\mu z^{1-\alpha })^{-1} - (1+\mu \psi (z))^{-1}(e^z-1)^{-2}e^z } | \\&\quad ={} \left|{ \frac{ (1+\mu \psi (z))(e^z-1)^2 - (z^2+\mu z^{1-\alpha })e^z }{ (z^2+\mu z^{1-\alpha })(1+\mu \psi (z))(e^z-1)^2 } } \right| \\&\quad<{} \left\{ \begin{aligned}&C_\alpha \frac{|{z} |^4 + \mu |{z} |^{2-\alpha }}{\mu (|{z} |^{3-\alpha } + \mu |{z} |^{2-2\alpha })} \qquad \quad \text { if }\,\, 0< |{z} | \leqslant c\mu ^{1/(1+\alpha )}, \\&C_\alpha \frac{|{z} |^4 + \mu |{z} |^{2-\alpha }}{|{z} |^4 + \mu |{z} |^{3-\alpha }} \qquad \qquad \qquad \text { if}\,\, c\mu ^{1/(1+\alpha )}< |{z} | \leqslant \pi /\sin \theta , \end{aligned} \right. \\&\quad<{} \left\{ \begin{aligned}&C_\alpha \big ( 1 + \mu ^{-1} |{z} |^\alpha \big ) \qquad \qquad \qquad \,\,\text { if } \,\, 0< |{z} | \leqslant c\mu ^{1/(1+\alpha )}, \\&C_\alpha \big ( 1 + \mu |{z} |^{-2-\alpha } \big ) \qquad \qquad \qquad \text { if} \,\, c\mu ^{1/(1+\alpha )}< |{z} | < \pi /\sin \theta , \end{aligned} \right. \end{aligned}$$

for all $z \in \Upsilon _1 \setminus \{0\}$. Therefore, if $c\mu ^{1/(1+\alpha )} \leqslant \pi /\sin \theta$, then

$$\begin{aligned} |{{\mathbb {I}}_2} |&< C_\alpha \tau \bigg ( \int _0^{c\mu ^{1/(1+\alpha )}} e^{kr\cos \theta } \big ( 1 + \mu ^{-1} r^\alpha \big ) \, \mathrm {d}r\\&\quad {} + \int _{ c\mu ^{1/(1+\alpha )} }^{ \pi /\sin \theta } e^{kr\cos \theta }(1 + \mu r^{-2-\alpha }) \, \mathrm {d}r \bigg ) \\&< C_\alpha \tau \Big ( \int _0^{c\mu ^{1/(1+\alpha )}} (1 + \mu ^{-1} r^\alpha ) \, \mathrm {d} r + \int _{c\mu ^{1/(1+\alpha )}}^{\pi /\sin \theta } (1 + \mu r^{-2-\alpha }) \, \mathrm {d}r \Big ) \\&< C_\alpha \tau , \end{aligned}$$

and, if $c\mu ^{1/(1+\alpha )} > \pi /\sin \theta$, then

$$\begin{aligned} |{{\mathbb {I}}_2} |< C_\alpha \tau \int _0^{\pi \sin \theta } e^{kr\cos \theta } \big ( 1 + \mu ^{-1} r^\alpha \big ) \, \mathrm {d}r < C_\alpha \tau . \end{aligned}$$

Finally, combing the above estimates for ${\mathbb {I}}_1$ and ${\mathbb {I}}_2$ proves (40) and hence this theorem. $\square$

4 Main Results

In the rest of this paper, we assume that $h < e^{-2(1+\alpha )}$ and $\tau < T/e$. The symbol $a\lesssim b$ means $a \leqslant Cb$, where C is a generic positive constant depending only on $\alpha$, T, $\Omega$, the shape-regular parameter of ${\mathcal {K}}_h$, and the ratio of h to the minimum diameter of the elements in ${\mathcal {K}}_h$. Additionally, since the following properties are frequently used in the forthcoming analysis, we shall use them implicitly (cf. [16]): for $-\infty< a< b < \infty$, $-1< \beta , \gamma < 1$ and $-1< \beta +\gamma < 1$,

$$\begin{aligned}&{{\,\mathrm{D}\,}}_{a+}^\beta {{\,\mathrm{D}\,}}_{a+}^\gamma = {{\,\mathrm{D}\,}}_{a+}^{\beta +\gamma }, \quad {{\,\mathrm{D}\,}}_{b-}^\beta {{\,\mathrm{D}\,}}_{b-}^\gamma = {{\,\mathrm{D}\,}}_{b-}^{\beta +\gamma },\\&\langle {{{\,\mathrm{D}\,}}_{a+}^\beta v, w} \rangle _{(a,b)} = \langle {v, {{\,\mathrm{D}\,}}_{b-}^\beta w} \rangle _{(a,b)}, \quad v,w \in L^2(a,b). \end{aligned}$$

Let u be the weak solution to problem (1) and U be the numerical solution defined by (2).

Theorem 4.1

If $u_0 \in L^2(\Omega )$ and $f = 0$, then

$$\begin{aligned} \Vert {u(\cdot ,t_j) - U_j} \Vert _{L^2(\Omega )} \lesssim \big ( h^2 t_j^{-\alpha -1} + \tau t_j^{-1} \big ) \Vert {u_0} \Vert _{L^2(\Omega )} \end{aligned}$$

(41)

for all $1 \leqslant j \leqslant J$.

Proof

Let $u_h$ be the solution of the spatially discrete problem:

$$\begin{aligned} u_h'(\cdot ,t) - \Delta _h {{\,\mathrm{D}\,}}_{0+}^{-\alpha } u_h(\cdot ,t) = 0, \quad t > 0, \end{aligned}$$

subject to the initial value condition $u_h(\cdot ,0) = P_h u_0$, where the discrete Laplace operator $\Delta _h:S_h \rightarrow S_h$ is defined by that

$$\begin{aligned} \langle {-\Delta _h v_h,w_h} \rangle _\Omega := \langle {\nabla v_h, \nabla w_h} \rangle _\Omega \quad \text {for all}\, v_h, w_h \in S_h. \end{aligned}$$

By [5, Theorem 2.1] we have

$$\begin{aligned} \Vert {u(\cdot ,t) - u_h(\cdot ,t)} \Vert _{L^2(\Omega )} \lesssim h^2 t^{-\alpha -1} \Vert {u_0} \Vert _{L^2(\Omega )}, \quad t > 0, \end{aligned}$$

and by Theorem 3.2 we obtain

$$\begin{aligned} \Vert {U_j - u_h(\cdot ,t_j)} \Vert _{L^2(\Omega )} \lesssim \tau t_j^{-1} \Vert {u_0} \Vert _{L^2(\Omega )}. \end{aligned}$$

Combining the above two estimates proves (41). $\square$

Theorem 4.2

If $u_0 = 0$ and $f(x,t) = {\tilde{f}}(x) \in L^2(\Omega )$, $0< t < T$, then

$$\begin{aligned} \Vert {u(\cdot ,t_j) - U_j} \Vert _{L^2(\Omega )} \lesssim \big ( t_j^{-\alpha } h^2 + \tau \big ) \Vert {{\tilde{f}}} \Vert _{L^2(\Omega )} \end{aligned}$$

(42)

for all $1 \leqslant j \leqslant J$.

Proof

Let $u_h$ be the solution of the spatially discrete problem:

$$\begin{aligned} u_h'(\cdot ,t) - \Delta _h {{\,\mathrm{D}\,}}_{0+}^{-\alpha } u_h(\cdot ,t) = P_h {\tilde{f}}, \quad t > 0, \end{aligned}$$

subject to the initial value condition $u_h(\cdot ,0) = 0$. By [5, Theorem 2.2] it holds

$$\begin{aligned} \Vert {u(\cdot ,t) - u_h(\cdot ,t)} \Vert _{L^2(\Omega )} \lesssim t^{-\alpha } h^2 \Vert {{\tilde{f}}} \Vert _{L^2(\Omega )}, \quad t > 0, \end{aligned}$$

and Theorem 3.3 implies

$$\begin{aligned} \Vert {U_j - u_h(\cdot ,t_j)} \Vert _{L^2(\Omega )} \lesssim \tau \Vert {P_h{\tilde{f}}} \Vert _{L^2(\Omega )} \lesssim \tau \Vert {{\tilde{f}}} \Vert _{L^2(\Omega )}. \end{aligned}$$

Combining the above two estimates proves (42). $\square$

Theorem 4.3

If $u_0 = 0$ and $f \in L^2(0,T;\dot{H}^{\alpha /(\alpha +1)}(\Omega )\!)$, then

$$\begin{aligned} \Vert {u-U} \Vert _{L^\infty (0,T;L^2(\Omega ))} \lesssim \left( h + \sqrt{\ln (1/h)} \tau ^{1/2} \right) \Vert {f} \Vert _{L^2(0,T;\dot{H}^{\alpha /(\alpha +1)}(\Omega )\!)}. \end{aligned}$$

(43)

Remark 4.1

Since Theorem 2.2 implies

$$\begin{aligned} \Vert {u} \Vert _{{}_0H^1(0,T;\dot{H}^{\alpha /(\alpha +1)}(\Omega )\!)} + \Vert {u} \Vert _{C([0,T];\dot{H}^1(\Omega )\!)}&\leqslant C_\alpha \Vert {f} \Vert _{L^2(0,T;\dot{H}^{\alpha /(\alpha +1)}(\Omega )\!)}, \end{aligned}$$

the estimate (43) is nearly optimal with respect to the regularity of u.

Theorem 4.4

If $u_0 = 0$ and $f \in {}_0H^{\alpha +1/2}(0,T;L^2(\Omega ))$, then

$$\begin{aligned} \Vert {u\!-\!U} \Vert _{L^\infty (0,T;L^2(\Omega ))} \lesssim \ln (T/\tau ) \left( \sqrt{\ln (1/h)}\, h^2 \!+\! \tau \right) \Vert {f} \Vert _{{}_0H^{\alpha +1/2}(0,T;L^2(\Omega ))}. \end{aligned}$$

(44)

Remark 4.2

Assume that u satisfies the following regularity assumptions: for any $0 < t \leqslant T$,

$$\begin{aligned} \Vert {u(\cdot ,t)} \Vert _{\dot{H}^2(\Omega )} + t\Vert {u'(\cdot ,t)} \Vert _{\dot{H}^2(\Omega )}&\leqslant M, \\ \Vert {u'(\cdot ,t)} \Vert _{L^2(\Omega )} + t\Vert {u''(\cdot ,t)} \Vert _{L^2(\Omega )}&\leqslant M t^{\sigma -1}, \\ t\Vert {u'(\cdot ,t)} \Vert _{\dot{H}^2(\Omega )} + t^2 \Vert {u''(\cdot ,t)} \Vert _{\dot{H}^2(\Omega )}&\leqslant Mt^{\sigma -1}, \end{aligned}$$

where M and $\sigma$ are two positive constants. By taking

$$\begin{aligned} t_j = (j/J)^\gamma T \text { for all } 1 \leqslant j \leqslant J, \quad \gamma > 1/\sigma , \end{aligned}$$

Mustapha and McLean [14] obtained

$$\begin{aligned} \Vert {u(\cdot ,t_j)-U_j} \Vert _{L^2(\Omega )} \lesssim \Vert {u_0 - U_0} \Vert _{L^2(\Omega )} + M \big ( \ln (t_j/t_1)h^2 + T/J \big ). \end{aligned}$$

Hence, no convergence rate is available in the case that $u_0 \in L^2(\Omega )$. Besides, under the condition that $u_0 = 0$ and $f \in {}_0H^{\alpha +1/2}(0,T;L^2(\Omega ))$, Theorem 2.2 only yields

$$\begin{aligned} \Vert {u} \Vert _{{}_0H^{\alpha +3/2}(0,T;L^2(\Omega ))} + \Vert {u} \Vert _{{}_0H^{1/2}(0,T;\dot{H}^2(\Omega ))} \leqslant C_\alpha \Vert {f} \Vert _{{}_0H^{\alpha +1/2}(0,T;L^2(\Omega ))}, \end{aligned}$$

so that u does not satisfy the above three regularity assumptions necessarily.

Remark 4.3

Theorems 4.2 and 4.4 imply that if $f \in H^{\alpha +1/2}(0,T;L^2(\Omega ))$ and $f(\cdot ,0) \ne 0$, then

$$\begin{aligned} \Vert {u(\cdot ,t_j) - U_j} \Vert _{L^2(\Omega )} \lesssim \left( \left( \ln (T/\tau ) \sqrt{\ln (1/h)} + t_j^{-\alpha } \right) h^2 + \ln (T/\tau ) \tau \right) \Vert {f} \Vert _{H^{\alpha +1/2}(0,T;L^2(\Omega ))} \end{aligned}$$

for all $1 \leqslant j \leqslant J$, where $H^{\alpha +1/2}(0,T;L^2(\Omega ))$ is defined analogously to the space ${}_0H^{\alpha +1/2}(0,T;L^2(\Omega ))$. Furthermore, Theorems 4.1 and 4.2 imply that if the accuracy of U near $t = 0$ is unimportant, then it is unnecessary to use graded temporal grids to tackle the singularity caused by nonsmooth $u_0$ and f(0) .

The rest of this section is devoted to the proofs of Theorems 4.3 and 4.4. Let X be a separable Hilbert space. For any $w \in C((0,T];X)$ and $v \in L^1(0,T;X)$, we define, for $1 \leqslant j \leqslant J,$

$$\begin{aligned} (P_\tau w)|_{I_j}:= & {} w(\cdot ,t_j), \\ (Q_\tau v)|_{I_j}:= & {} \tau ^{-1} \int _{I_j} v. \end{aligned}$$

The operator $Q_\tau$ possesses the standard estimates

$$\begin{aligned} \Vert {(I-Q_\tau )v} \Vert _{L^2(0,T;X)}&\leqslant \Vert {v} \Vert _{L^2(0,T;X)} \quad \forall v \in L^2(0,T;X), \\ \Vert {(I-Q_\tau )v} \Vert _{L^2(0,T;X)}&\lesssim \tau \Vert {v} \Vert _{{}_0H^1(0,T;X)} \quad \forall v \in {}_0H^1(0,T;X). \end{aligned}$$

Hence, for any $v \in {}_0H^\beta (0,T;X)$ with $0< \beta < 1$, applying [17, Lemma 22.3] yields

$$\begin{aligned} \Vert {(I-Q_\tau )v} \Vert _{[L^2(0,T;X),\ L^2(0,T;X)]_{\beta ,2}} \lesssim \tau ^\beta \Vert {v} \Vert _{{}_0H^\beta (0,T;X)}, \end{aligned}$$

so that [17, (23.11)] implies

$$\begin{aligned} \Vert {(I-Q_\tau )v} \Vert _{L^2(0,T;X)} \lesssim \tau ^\beta \sqrt{\beta (1-\beta )} \, \Vert {v} \Vert _{{}_0H^\beta (0,T;X)}. \end{aligned}$$

(45)

Here we have used the fact that ${}_0H^\beta (0,T;X) = [L^2(0,T;X), {}_0H^1(0,T;X)]_{\beta ,2}$ with equivalent norms (cf. Remark 2.1). Similarly, for any $v \in {}^0H^\beta (0,T;X)$ with $0< \beta < 1$,

$$\begin{aligned} \Vert {(I-Q_\tau )v} \Vert _{L^2(0,T;X)} \lesssim \tau ^\beta \sqrt{\beta (1-\beta )} \, \Vert {v} \Vert _{{}^0H^\beta (0,T;X)}. \end{aligned}$$

(46)

Moreover, from [17, Lemmas 12.4, 16.3, 22.3, 23.1] it follows the following lemma.

Lemma 4.1

If $v \in {}_0H^\beta (0,1)$ with $0< \beta < 1$, then

$$\begin{aligned} \left( \int _0^1 \int _0^1 \frac{|{v(t)-v(s)} |^2}{|{t-s} |^{1+2\beta }} \, \mathrm {d}t \, \mathrm {d}s \right) ^{1/2} \leqslant C \Vert {v} \Vert _{{}_0H^\beta (0,1)}. \end{aligned}$$

(47)

Furthermore, if $1/2< \beta < 1$, then

$$\begin{aligned} \Vert {v} \Vert _{C[0,1]}\leqslant & {} C \sqrt{\frac{1-\beta }{2\beta -1}} \Vert {v} \Vert _{{}_0H^\beta (0,1)}, \end{aligned}$$

(48)

$$\begin{aligned} \Vert {v} \Vert _{C[0,1]}\leqslant & {} \frac{C}{\sqrt{2\beta \!-\!1}} \left( \Vert {v} \Vert _{L^2(0,1)} \!+\! \sqrt{1\!-\!\beta } \left( \int _0^1\!\!\int _0^1 \frac{|{v(t)\!-\!v(s)} |^2}{|{t\!-\!s} |^{1+2\beta }} \, \mathrm {d}s \, \mathrm {d}t \right) ^{1/2} \right) , \end{aligned}$$

(49)

where C is a positive constant independent of $\beta$ and v.

Lemma 4.2

If $v \in {}_0H^\beta (0,T)$ with $1/2< \beta < 1$, then

$$\begin{aligned} \Vert {(I-P_\tau )v} \Vert _{L^2(0,T)} \lesssim \tau ^\beta \sqrt{\frac{1-\beta }{2\beta -1}}\, \Vert {v} \Vert _{{}_0H^\beta (0,T)}. \end{aligned}$$

(50)

Proof

By the definition of $P_\tau$ and (49), a scaling argument yields

$$\begin{aligned}&\Vert {(I-P_\tau )v} \Vert _{L^2(I_j)}^2 \\&\quad \lesssim {} \frac{1}{2\beta -1} \left( \Vert {(I-Q_\tau )v} \Vert _{L^2(I_j)}^2 + (1-\beta )\tau ^{2\beta } \int _{I_j} \int _{I_j} \frac{|{v(t)-v(s)} |^2}{|{t-s} |^{1+2\beta }} \, \mathrm {d}s \, \mathrm {d}t \right) , \end{aligned}$$

so that from (45) it follows

$$\begin{aligned}&\sqrt{2\beta -1} \Vert {(I-P_\tau )v} \Vert _{L^2(0,T)} \\&\quad \lesssim {} \Vert {(I-Q_\tau )v} \Vert _{L^2(0,T)} + \sqrt{1-\beta }\, \tau ^\beta \left( \int _0^T \int _0^T \frac{|{v(t)-v(s)} |^2}{|{t-s} |^{1+2\beta }} \, \mathrm {d}s \, \mathrm {d}t \right) ^{1/2} \\&\quad \lesssim {} \tau ^\beta \sqrt{1-\beta }\left( \Vert {v} \Vert _{{}_0H^\beta (0,T)} + \left( \int _0^T \int _0^T \frac{|{v(t)-v(s)} |^2}{|{t-s} |^{1+2\beta }} \, \mathrm {d}s \, \mathrm {d}t \right) ^{1/2} \right) . \end{aligned}$$

Another scaling argument, together with (47), gives that

$$\begin{aligned} \left( \int _0^T \int _0^T \frac{|{v(t)-v(s)} |^2}{|{t-s} |^{1+2\beta }} \, \mathrm {d}s \, \mathrm {d}t \right) ^{1/2} \lesssim \Vert {v} \Vert _{{}_0H^\beta (0,T)}. \end{aligned}$$

Combining the above two estimates proves (50). $\square$

Lemma 4.3

[6]. Assume that $-\infty< \beta ,\gamma ,r,s < \infty$ and $0< \theta < 1$. If $v \in {}_0H^\beta (0,T;\dot{H}^r(\Omega )) \cap {}_0H^\gamma (0,T;\dot{H}^s(\Omega ))$, then

$$\begin{aligned} \begin{aligned}&\Vert {v} \Vert _{ {}_0H^{(1-\theta )\beta + \theta \gamma } (0,T;\dot{H}^{(1-\theta )r+\theta s}(\Omega )) } \\&\quad \leqslant {} C_{\beta ,\gamma ,\theta } \Vert {v} \Vert _{{}_0H^\beta (0,T;\dot{H}^r(\Omega )))}^{1-\theta } \Vert {v} \Vert _{{}_0H^\gamma (0,T;\dot{H}^s(\Omega ))}^\theta . \end{aligned} \end{aligned}$$

(51)

In particular, if $\beta = 0$ and $\gamma = 1$, then

$$\begin{aligned} \Vert {v} \Vert _{{}_0H^\theta (0,T;\dot{H}^{(1-\theta )r + \theta s}(\Omega )} \leqslant \frac{1}{\sqrt{2\theta (1-\theta )}} \Vert {v} \Vert _{L^2(0,T;\dot{H}^r(\Omega )}^{1-\theta } \Vert {v} \Vert _{{}_0H^1(0,T;\dot{H}^s(\Omega ))}^\theta \end{aligned}$$

(52)

for all $v \in L^2(0,T;\dot{H}^r(\Omega )) \cap {}_0H^1(0,T;\dot{H}^s(\Omega ))$.

Lemma 4.4

[18]. If $V \in W_{\tau ,h}$ and $0 \leqslant i < k \leqslant J$, then

$$\begin{aligned} \sum _{j=i}^k \langle {{[\![ {V_j} ]\!]}, V_j^{+}} \rangle _\Omega \geqslant \frac{1}{2} \big ( \Vert {V_k^{+}} \Vert _{L^2(\Omega )}^2 - \Vert {V_i} \Vert _{L^2(\Omega )}^2 \big ) \geqslant \sum _{j=i}^k \langle {V_j, {[\![ {V_j} ]\!]}} \rangle _\Omega . \end{aligned}$$

4.1 Proof of Theorem 4.3

Let us first prove

$$\begin{aligned} \begin{aligned}&\Vert {U-P_\tau P_hu} \Vert _{L^\infty (0,T;L^2(\Omega ))} \\&\quad \lesssim {} \Vert {(I-P_h){{\,\mathrm{D}\,}}_{0+}^{-\alpha /2} u} \Vert _{L^2(0,T;\dot{H}^{1}(\Omega )\!)} + \Vert {(I-P_\tau )P_hu} \Vert _{L^2(0,T;\dot{H}^1(\Omega )\!)}. \end{aligned} \end{aligned}$$

(53)

For any $1 \leqslant j \leqslant J$, by (2) and (6) we have

$$\begin{aligned} \sum _{i=1}^{j} \langle {u',\theta } \rangle _{\Omega \times I_i} + \langle { \nabla {{\,\mathrm{D}\,}}_{0+}^{-\alpha } (u-U),\nabla \theta } \rangle _{\Omega \times (0,t_j)} = \sum _{i=0}^{j-1}\langle {{[\![ {U_i} ]\!]},\theta ^+_i} \rangle _\Omega , \end{aligned}$$

where $\theta :=U-P_\tau P_hu$ and we set $(P_\tau P_h u)_0 = 0$. By the definitions of $P_h$ and $P_\tau$, a routine calculation (see [18, Chapter 12]) then yields

$$\begin{aligned} {}&\sum _{i=0}^{j-1}\langle {{[\![ {\theta _i} ]\!]},\theta ^+_i} \rangle _\Omega + \big \langle \nabla {{\,\mathrm{D}\,}}_{0+}^{-\alpha } \theta ,\nabla \theta \big \rangle _{\Omega \times (0,t_j)} \\&\quad ={} \langle { \nabla {{\,\mathrm{D}\,}}_{0+}^{-\alpha }(u-P_\tau P_hu), \nabla \theta } \rangle _{\Omega \times (0,t_j)} \\&\quad ={} \langle { \nabla {{\,\mathrm{D}\,}}_{0+}^{-\alpha }(I-P_h)u, \nabla \theta } \rangle _{\Omega \times (0,t_j)} + \langle { \nabla {{\,\mathrm{D}\,}}_{0+}^{-\alpha }(I-P_\tau )P_hu, \nabla \theta } \rangle _{\Omega \times (0,t_j)} \\&\quad ={} \langle { \nabla (I-P_h){{\,\mathrm{D}\,}}_{0+}^{-\alpha /2}u, \nabla {{\,\mathrm{D}\,}}_{t_j-}^{-\alpha /2}\theta } \rangle _{\Omega \times (0,t_j)} + \langle { \nabla (I-P_\tau )P_hu, {{\,\mathrm{D}\,}}_{t_j-}^{-\alpha } \nabla \theta } \rangle _{\Omega \times (0,t_j)}, \end{aligned}$$

so that using Lemma 4.4, Lemma 2.1, Sobolev inequality and Young’s inequality with $\epsilon$ gives

$$\begin{aligned} {}&\Vert {\theta _j} \Vert _{L^2(\Omega )}+ \Vert {\theta _1} \Vert _{L^2(\Omega )}+ \Vert { {{\,\mathrm{D}\,}}_{0+}^{-\alpha /2} \theta } \Vert _{L^{2}(0,t_j;\dot{H}^1(\Omega )\!)} \\&\quad \lesssim {} \Vert {(I-P_h){{\,\mathrm{D}\,}}_{0+}^{-\alpha /2} u} \Vert _{L^2(0,T;\dot{H}^{1}(\Omega )\!)} + \Vert {(I-P_\tau )P_hu} \Vert _{L^2(0,T;\dot{H}^1(\Omega )\!)}. \end{aligned}$$

Since $1 \leqslant j \leqslant J$ is arbitrary, this implies (53).

Next, let us prove

$$\begin{aligned} \Vert {U - P_\tau P_hu} \Vert _{L^\infty (0,T;L^2(\Omega ))} \lesssim \big ( h + \sqrt{\ln (1/h)} \, \tau ^{1/2} \big ) \Vert {f} \Vert _{ L^2(0,T;\dot{H}^{\alpha /(\alpha +1)}(\Omega )) }. \end{aligned}$$

(54)

By the inverse estimate and Lemma 4.2, a straightforward calculation gives that, for any $0< \epsilon < 1/(\alpha +1)$,

$$\begin{aligned}&\Vert {(I-P_\tau )P_hu} \Vert _{L^2(0,T;\dot{H}^1(\Omega )\!)} \lesssim h^{-\epsilon } \Vert {(I-P_\tau )P_hu} \Vert _{L^2(0,T;\dot{H}^{1-\epsilon }(\Omega )\!)} \\&\quad \lesssim {} h^{-\epsilon } \Vert {(I-P_\tau )u} \Vert _{L^2(0,T;\dot{H}^{1-\epsilon }(\Omega ))} \\&\quad \lesssim {} h^{-\epsilon } \tau ^{(1+\epsilon +\epsilon \alpha )/2} \sqrt{ \frac{1-(1+\alpha )\epsilon }{\epsilon }} \, \Vert {u} \Vert _{ {}_0H^{(1+\epsilon +\epsilon \alpha )/2} (0,T;\dot{H}^{1-\epsilon }(\Omega )) }, \end{aligned}$$

and hence letting $\epsilon = (2\ln (1/h)\!)^{-1}$ yields

$$\begin{aligned} \Vert {(I-P_\tau )P_hu} \Vert _{L^2(0,T;\dot{H}^1(\Omega )\!)} \lesssim \sqrt{\ln (1/h)} \, \tau ^{1/2} \Vert {u} \Vert _{ {}_0H^{(1+\epsilon +\epsilon \alpha )/2} (0,T;\dot{H}^{1-\epsilon }(\Omega )) }. \end{aligned}$$

Moreover, by Lemma 2.2 it holds

$$\begin{aligned} \Vert {(I-P_h){{\,\mathrm{D}\,}}_{0+}^{-\alpha /2} u} \Vert _{L^2(0,T;\dot{H}^1(\Omega )\!)}&\lesssim h \Vert {{{\,\mathrm{D}\,}}_{0+}^{-\alpha /2} u} \Vert _{L^2(0,T;\dot{H}^2(\Omega )\!)} \\&\lesssim h \Vert {u} \Vert _{{}_0H^{-\alpha /2}(0,T;\dot{H}^2(\Omega )\!)}. \end{aligned}$$

Therefore, by Theorem 2.2 and Lemma 4.3, combining (53) and the above two estimates yields (54).

Finally, a routine calculation gives

$$\begin{aligned}&\Vert {u-P_\tau P_hu} \Vert _{L^\infty (0,T;L^2(\Omega ))} \\&\quad \leqslant {} \Vert {(I-P_h)u} \Vert _{L^\infty (0,T;L^2(\Omega ))} + \Vert {P_h(I-P_\tau )u} \Vert _{L^\infty (0,T;L^2(\Omega ))} \\&\quad \leqslant {} \Vert {(I-P_h)u} \Vert _{L^\infty (0,T;L^2(\Omega ))} + \Vert {(I-P_\tau )u} \Vert _{L^\infty (0,T;L^2(\Omega ))} \\&\quad \lesssim {} h \Vert {u} \Vert _{C([0,T];\dot{H}^1(\Omega ))} + \tau ^{1/2} \Vert {u} \Vert _{H^1(0,T;L^2(\Omega ))} \\&\quad \lesssim {} \big ( h + \tau ^{1/2} \big ) \Vert {f} \Vert _{L^2(0,T;\dot{H}^{\alpha /(\alpha +1)}(\Omega ))} \quad (\text {by Theorem}~\text {2.2}), \end{aligned}$$

so that (43) follows from (54) and the triangle inequality

$$\begin{aligned} \Vert {u-U} \Vert _{L^\infty (0,T;L^2(\Omega ))} \leqslant \Vert {U-P_\tau P_hu} \Vert _{L^\infty (0,T;L^2(\Omega ))} + \Vert {u-P_\tau P_hu} \Vert _{L^\infty (0,T;L^2(\Omega ))}. \end{aligned}$$

This completes the proof of Theorem 4.3.

Remark 4.4

From the above proof, it is easy to see that Theorem 4.3 still holds for the case of variable time steps.

4.2 Proof of Theorem 4.4

Lemma 4.5

If $W \in W_{\tau ,h}$ satisfies that $W_0 := v_h \in S_h$ and

$$\begin{aligned} \sum _{j=0}^{J-1} \langle {{[\![ {W_j} ]\!]}, V_j^{+}} \rangle _\Omega + \langle {\nabla {{\,\mathrm{D}\,}}_{0+}^{-\alpha } W, \nabla V} \rangle _{\Omega \times (0,T)} = 0 \quad \forall V \in W_{\tau ,h}, \end{aligned}$$

(55)

then

$$\begin{aligned} \Vert {W} \Vert _{{}_0H^{-\alpha /2}(0,T;\dot{H}^1(\Omega ))}&\leqslant C_\alpha \Vert {v_h} \Vert _{L^2(\Omega )}, \end{aligned}$$

(56)

$$\begin{aligned} \Vert { Q_\tau {{\,\mathrm{D}\,}}_{0+}^{-\alpha }(-\Delta _h W) } \Vert _{L^1(0,T;L^2(\Omega ))}&\leqslant C_\alpha \ln (T/\tau ) \Vert {v_h} \Vert _{L^2(\Omega )}. \end{aligned}$$

(57)

Proof

Since Lemma 4.4 implies

$$\begin{aligned} \sum _{j=0}^{J-1} \langle {{[\![ {W_j} ]\!]}, W_j^{+}} \rangle _\Omega \geqslant \frac{1}{2} \big ( \Vert {W_J} \Vert _{L^2(\Omega )} - \Vert {v_h} \Vert _{L^2(\Omega )}^2 \big ), \end{aligned}$$

inserting $V = W$ into (55) yields

$$\begin{aligned} \frac{1}{2} \Vert {W_J} \Vert _{L^2(\Omega )}^2 + \langle {\nabla {{\,\mathrm{D}\,}}_{0+}^{-\alpha } W, \nabla W} \rangle _{\Omega \times (0,T)} \leqslant \frac{1}{2} \Vert {v_h} \Vert _{L^2(\Omega )}^2. \end{aligned}$$

Hence, using Lemmas 2.1 and 2.2 proves (56).

Now let us prove (57). Let $\{\phi _{n,h}: 1 \leqslant n \leqslant N\}$ be an orthonormal basis of $S_h$ endowed with the $L^2(\Omega )$ inner-product such that

$$\begin{aligned} -\Delta _h \phi _{n,h} = \lambda _{n,h} \phi _{n,h}, \end{aligned}$$

where $\{\lambda _{n,h}: 1 \leqslant n \leqslant N \}$ is the set of all eigenvalues of $-\Delta _h$. For each $1 \leqslant n \leqslant N$, define $(Y_k^n(t))_{k=0}^\infty$ by

$$\begin{aligned} (Y_k^n(t))' + \lambda _{n,h} {{\,\mathrm{D}\,}}_{0+}^{-\alpha } Y_k^n(t) = 0, \quad t > 0, \end{aligned}$$

(58)

with $Y_k^n(0)=\langle {v_h,\phi _{n,h}} \rangle _\Omega$. Set $W^n(t) := \langle {W(t), \phi _{n,h}} \rangle _\Omega$, $0< t < T$, and it is easy to verify that

$$\begin{aligned} W^n = Y_j^n \quad \text { on } I_j,\quad 1 \leqslant j \leqslant J. \end{aligned}$$

Hence, Theorem 3.1 implies

$$\begin{aligned} \Vert {{[\![ {W_j} ]\!]}} \Vert _{L^2(\Omega )} \leqslant C_\alpha j^{-1} \Vert {v_h} \Vert _{L^2(\Omega )},\quad 1 \leqslant j < J, \end{aligned}$$

and then it follows that

$$\begin{aligned} \sum _{j=1}^{J-1} \Vert {{[\![ {W_j} ]\!]}} \Vert _{L^2(\Omega )} \leqslant C_\alpha \Vert {v_h} \Vert _{L^2(\Omega )} \sum _{j=1}^{J-1} j^{-1} \leqslant C_\alpha \ln (T/\tau ) \Vert {v_h} \Vert _{L^2(\Omega )}. \end{aligned}$$

(59)

In addition, inserting $V = W \chi _{(0,t_1)}$ into (55) yields, by Lemma 2.1, that

$$\begin{aligned} \Vert {W_1} \Vert _{L^2(\Omega )} \leqslant \Vert {W_0} \Vert _{L^2(\Omega )}, \end{aligned}$$

which implies

$$\begin{aligned} \Vert {{[\![ {W_0} ]\!]}} \Vert _{L^2(\Omega )} \leqslant 2 \Vert {W_0} \Vert _{L^2(\Omega )} = 2\Vert {v_h} \Vert _{L^2(\Omega )}. \end{aligned}$$

(60)

Consequently, since (55) leads to

$$\begin{aligned} \tau Q_\tau {{\,\mathrm{D}\,}}_{0+}^{-\alpha }(-\Delta _h W) = {[\![ {W_{j-1}} ]\!]} \quad \text { on } I_j, \quad 1 \leqslant j \leqslant J, \end{aligned}$$

combining (59) and (60) proves (57) and hence this lemma. $\square$

Lemma 4.6

If $f \in {}_0H^{\alpha /2}(0,T;L^2(\Omega ))$, then

$$\begin{aligned} \begin{aligned} \Vert {(U-P_\tau P_hu)_j} \Vert _{L^2(\Omega )}&\lesssim \ln (T/\tau ) \Vert {R_hu - P_\tau P_hu} \Vert _{L^\infty (0,T;L^2(\Omega ))} \\&\qquad {} + \tau ^{\alpha /2} \Vert {(I-Q_\tau )u} \Vert _{L^2(0,T;\dot{H}^1(\Omega ))} \end{aligned} \end{aligned}$$

(61)

for each $1 \leqslant j \leqslant J$.

Proof

Let $\theta = U - P_\tau P_h u$ and set $(P_\tau P_hu)_0 = 0$. Define $W \in W_{\tau ,h}$ by that $W_J^{+} = \theta _J$ and

$$\begin{aligned} -\sum _{j=1}^J \langle {V_j, {[\![ {W_j} ]\!]}} \rangle _\Omega + \langle {\nabla V, \nabla {{\,\mathrm{D}\,}}_{T-}^{-\alpha } W} \rangle _{\Omega \times (0,T)} = 0 \quad \forall V \in W_{\tau ,h}. \end{aligned}$$

A simple calculation then yields

$$\begin{aligned}&\Vert {\theta _J} \Vert _{L^2(\Omega )}^2 = \langle {\theta _J, W_J^{+}} \rangle _\Omega = \sum _{j=0}^{J-1} \langle {{[\![ {\theta _j} ]\!]}, W_j^{+}} \rangle _\Omega + \sum _{j=1}^J \langle {\theta _j, {[\![ {W_j} ]\!]}} \rangle _\Omega \\&\quad ={} \sum _{j=0}^{J-1} \langle {{[\![ {\theta _j} ]\!]}, W_j^{+}} \rangle _\Omega + \langle {\nabla \theta , \nabla {{\,\mathrm{D}\,}}_{T-}^{-\alpha } W} \rangle _{\Omega \times (0,T)} \\&\quad ={} \sum _{j=0}^{J-1} \langle {{[\![ {\theta _j} ]\!]}, W_j^{+}} \rangle _\Omega + \langle {\nabla {{\,\mathrm{D}\,}}_{0+}^{-\alpha }\theta , \nabla W} \rangle _{\Omega \times (0,T)}, \end{aligned}$$

and proceeding as in the proof of Theorem 4.3 shows

$$\begin{aligned} \sum _{j=0}^{J-1} \langle {{[\![ {\theta _j} ]\!]}, W_j^{+}} \rangle _\Omega + \langle {\nabla {{\,\mathrm{D}\,}}_{0+}^{-\alpha } \theta , \nabla W} \rangle _{\Omega \times (0,T)} = \langle {\nabla {{\,\mathrm{D}\,}}_{0+}^{-\alpha } (u-P_\tau P_hu), \nabla W} \rangle _{\Omega \times (0,T)}. \end{aligned}$$

Consequently,

$$\begin{aligned} \Vert {\theta _J} \Vert _{L^2(\Omega )}^2&= \langle { \nabla (u-P_\tau P_hu), \nabla {{\,\mathrm{D}\,}}_{T-}^{-\alpha } W } \rangle _{\Omega \times (0,T)} \nonumber \\&= \langle { \nabla (R_hu - P_\tau P_hu),\ \nabla {{\,\mathrm{D}\,}}_{T-}^{-\alpha } W } \rangle _{\Omega \times (0,T)} \nonumber \\&= \langle { R_hu - P_\tau P_h u,\ {{\,\mathrm{D}\,}}_{T-}^{-\alpha }(-\Delta _h W } \rangle _{\Omega \times (0,T)} \nonumber \\&= {\mathbb {I}}_1 + {\mathbb {I}}_2, \end{aligned}$$

(62)

where

$$\begin{aligned} {\mathbb {I}}_1&:= \langle { R_hu - P_\tau P_hu,\ Q_\tau {{\,\mathrm{D}\,}}_{T-}^{-\alpha }(-\Delta _h W) } \rangle _{\Omega \times (0,T)}, \\ {\mathbb {I}}_2&:= \langle { R_hu-P_\tau P_hu,\ (I-Q_\tau ){{\,\mathrm{D}\,}}_{T-}^{-\alpha }(-\Delta _h W) } \rangle _{\Omega \times (0,T)}. \end{aligned}$$

Next, it is evident that

$$\begin{aligned} {\mathbb {I}}_1 \leqslant \Vert {R_hu - P_\tau P_hu} \Vert _{L^\infty (0,T;L^2(\Omega ))} \Vert {Q_\tau {{\,\mathrm{D}\,}}_{T-}^{-\alpha }(-\Delta _h W)} \Vert _{L^1(0,T;L^2(\Omega ))}. \end{aligned}$$

(63)

By the definitions of $Q_\tau$ and $R_h$,

$$\begin{aligned} {\mathbb {I}}_2&= \langle { R_hu, (I-Q_\tau ) {{\,\mathrm{D}\,}}_{T-}^{-\alpha }(-\Delta _h W) } \rangle _{\Omega \times (0,T)} \\&= \langle { \nabla R_hu, \nabla (I-Q_\tau ){{\,\mathrm{D}\,}}_{T-}^{-\alpha }W } \rangle _{\Omega \times (0,T)} \\&= \langle { \nabla u, \nabla (I-Q_\tau ){{\,\mathrm{D}\,}}_{T-}^{-\alpha }W } \rangle _{\Omega \times (0,T)} \\&= \langle { \nabla (I-Q_\tau )u, \nabla (I-Q_\tau ){{\,\mathrm{D}\,}}_{T-}^{-\alpha }W } \rangle _{\Omega \times (0,T)} \\&\leqslant \Vert {(I-Q_\tau )u} \Vert _{L^2(0,T;\dot{H}^1(\Omega ))} \Vert {(I-Q_\tau ){{\,\mathrm{D}\,}}_{T-}^{-\alpha }W} \Vert _{L^2(0,T;\dot{H}^1(\Omega ))}. \end{aligned}$$

In addition,

$$\begin{aligned}&\Vert {(I-Q_\tau ){{\,\mathrm{D}\,}}_{T-}^{-\alpha }W} \Vert _{L^2(0,T;\dot{H}^1(\Omega ))} \\&\quad \lesssim {} \tau ^{\alpha /2} \Vert {{{\,\mathrm{D}\,}}_{T-}^{-\alpha } W} \Vert _{{}^0H^{\alpha /2}(0,T;\dot{H}^1(\Omega ))} \quad (\text {by }(\text {46})) \\&\quad \lesssim {} \tau ^{\alpha /2} \Vert {W} \Vert _{{}^0H^{-\alpha /2}(0,T;\dot{H}^1(\Omega ))} \quad (\text {by Lemma}~\text {2.2}). \end{aligned}$$

Thus,

$$\begin{aligned} {\mathbb {I}}_2 \lesssim \tau ^{\alpha /2} \Vert {(I-Q_\tau )u} \Vert _{L^2(0,T;\dot{H}^1(\Omega ))} \Vert {W} \Vert _{{}^0H^{-\alpha /2}(0,T;\dot{H}^1(\Omega ))}. \end{aligned}$$

(64)

Finally, from the symmetric version of Theorem 4.5 it follows

$$\begin{aligned}&\Vert {W} \Vert _{{}^0H^{-\alpha /2}(0,T;\dot{H}^1(\Omega ))} \leqslant C_\alpha \Vert {\theta _J} \Vert _{L^2(\Omega )}, \\&\Vert {Q_\tau {{\,\mathrm{D}\,}}_{T-}^{-\alpha }(-\Delta _h W)} \Vert _{L^1(0,T;L^2(\Omega ))} \leqslant C_\alpha \ln (T/\tau ) \Vert {\theta _J} \Vert _{L^2(\Omega )}, \end{aligned}$$

and hence combining (62)–(64) yields that (61) holds for $j = J$. Since the case $1 \leqslant j < J$ can be proved analogously, this completes the proof. $\square$

Finally, we conclude the proof of Theorem 4.4 as follows. By Lemma 4.6, a straightforward calculation yields

$$\begin{aligned}&\Vert {u-U} \Vert _{L^\infty (0,T;L^2(\Omega ))} \nonumber \\&\quad \lesssim {} \tau ^{\alpha /2} \Vert {(I-Q_\tau )u} \Vert _{L^2(0,T;\dot{H}^1(\Omega ))} + \ln (T/\tau ) \Big ( \Vert {(I-R_h)u} \Vert _{L^\infty (0,T;L^2(\Omega ))} \nonumber \\&\qquad {} + \Vert {(I-P_h)u} \Vert _{L^\infty (0,T;L^2(\Omega ))} + \Vert {(I-P_\tau )u} \Vert _{L^\infty (0,T;L^2(\Omega ))} \Big ). \end{aligned}$$

(65)

By Theorem 2.2 it holds

$$\begin{aligned}&\Vert {(I-R_h)u} \Vert _{L^\infty (0,T;L^2(\Omega ))} + \Vert {(I-P_h)u} \Vert _{L^\infty (0,T;L^2(\Omega ))} \\&\quad \lesssim {} h^{2(1-\epsilon )} \Vert {u} \Vert _{C([0,T];\dot{H}^{2(1-\epsilon )}(\Omega ))} \\&\quad \lesssim {} \frac{h^{2(1-\epsilon )}}{\sqrt{\epsilon }} \Vert {f} \Vert _{{}_0H^{\alpha +1/2}(0,T;L^2(\Omega ))} \end{aligned}$$

for all $0< \epsilon < 1/2$, so that, by the assumption $h < e^{-2(1+\alpha )}$ (cf. the first paragraph of Sect. 4), letting $\epsilon := (\ln (1/h))^{-1}$ yields

$$\begin{aligned} \begin{aligned}&\Vert {(I-R_h)u} \Vert _{L^\infty (0,T;L^2(\Omega ))} + \Vert {(I-P_h)u} \Vert _{L^\infty (0,T;L^2(\Omega ))} \\&\quad \lesssim {} \sqrt{\ln (1/h)}\, h^2 \Vert {f} \Vert _{ {}_0H^{\alpha +1/2}(0,T;L^2(\Omega )) }. \end{aligned} \end{aligned}$$

(66)

In addition, by Theorem 2.2 and Lemma 4.3, it is standard that

$$\begin{aligned}&\Vert {(I-Q_\tau )u} \Vert _{L^2(0,T;\dot{H}^1(\Omega ))} + \Vert {(I-P_\tau )u} \Vert _{L^\infty (0,T;L^2(\Omega ))} \nonumber \\&\quad \lesssim {} \tau \Vert {f} \Vert _{{}_0H^{\alpha +1/2}(0,T;L^2(\Omega ))}. \end{aligned}$$

(67)

Combining (65)–(67) proves (44) and thus concludes the proof of Theorem 4.4.

5 Numerical Experiments

This section performs some numerical experiments in one dimensional space to verify the theoretical results. Throughout this section, $\Omega = (0,1)$, $T = 1$, the spatial grids are uniform, and $U^{m,n}$ is the numerical solution with $h = 2^{-m}$ and $\tau = 2^{-n}$. Additionally, $\Vert {\cdot } \Vert _{L^\infty (0,T;L^2(\Omega ))}$ is abbreviated to $\Vert {\cdot } \Vert$ for convenience, and, for any $\beta > 0$,

$$\begin{aligned} \Vert {v} \Vert _{\beta ,n} := \max _{1 \leqslant j \leqslant 2^n} (j/2^n)^\beta \Vert {v((j/2^n)-)} \Vert _{L^2(\Omega )}, \end{aligned}$$

where $v((j/2^n)-)$ means the left limit of v at $j/2^n$.

Table 1 Convergence behavior with respect to $\tau$

Table 2 Convergence behavior with respect to h

Experiment 1. This experiment verifies Theorem 4.1 by setting

$$\begin{aligned} u_0(x) = x^{-0.49}, \quad x \in \Omega , \end{aligned}$$

which is slightly smoother than the $L^2(\Omega )$-regularity. Table 1 validates the theoretical prediction that the convergence behavior of U is close to $\mathcal O(\tau )$ when h is fixed and sufficiently small. Table 2 confirms the theoretical prediction that the convergence behavior of U is close to ${\mathcal {O}}(h^2)$ when $\tau$ is fixed and sufficiently small.

Table 3 Convergence behavior with respect to $\tau$

Table 4 Convergence behavior with respect to h

Table 5 Convergence behavior with respect to h

Experiment 2. This experiment verifies Theorem 4.2 by setting

$$\begin{aligned} {\tilde{f}}(x) = x^{-0.49}, \quad x \in \Omega . \end{aligned}$$

Table 3 confirms the theoretical prediction that the convergence behavior of U is close to ${\mathcal {O}}(\tau )$ when h is fixed and sufficiently small. Table 4 confirms the theoretical prediction that the accuracy of $U(T-)$ (the left limit of U at T) in the norm $\Vert {\cdot } \Vert _{L^2(\Omega )}$ is close to ${\mathcal {O}}(h^2)$ when $\tau$ is fixed and sufficiently small.

Experiment 3. This experiment verifies Theorem 4.3 by setting

$$\begin{aligned} f(x,t) = x^{\alpha /(\alpha +1)-0.49} t^{-0.49}, \quad (x,t) \in \Omega \times (0,T), \end{aligned}$$

which has slightly higher regularity than the $L^2(0,T;\dot{H}^{\alpha /(\alpha +1)}(\Omega ))$-regularity. Theorem 4.3 predicts that the convergence behavior of U is close to ${\mathcal {O}}(h)$ when $\tau$ is fixed and sufficiently small, as is in good agreement with the numerical results in Table 5. Moreover, Theorem 4.3 predicts that the convergence behavior of U is close to ${\mathcal {O}}(\tau ^{1/2})$ when h is fixed and sufficiently small, which agrees well with the numerical results in Table 6.

Table 6 Convergence behavior with respect to $\tau$

Experiment 4. This experiment verifies Theorem 4.4 by setting

$$\begin{aligned} f(x,t) = x^{-0.49} t^{\alpha +0.01}, \quad (x,t) \in \Omega \times (0,T), \end{aligned}$$

which is slightly smoother than the ${}_0H^{\alpha +1/2}(0,T;L^2(\Omega ))$-regularity. Table 7 confirms the theoretical prediction that the convergence behavior of U is close to ${\mathcal {O}}(h^2)$ when $\tau$ is fixed and sufficiently small, and Table 8 confirms the theoretical prediction that the convergence behavior of U is close to ${\mathcal {O}}(\tau )$ when h is fixed and sufficiently small.

Table 7 Convergence behavior with respect to h

Table 8 Convergence behavior with respect to $\tau$

Table 9 Convergence behavior with respect to J

Experiment 5. This experiment verifies the effect of graded temporal grids by setting

$$\begin{aligned} u_0(x) = x^{-0.49}. \end{aligned}$$

Let $U^{m,J,\sigma }$ be the numerical solution with $h = 2^{-m}$ and temporal grids

$$\begin{aligned} t_j := \left( \frac{j}{2^J}\right) ^{\sigma }, \ j=0,1,\dots ,2^J. \end{aligned}$$

For simplicity, $\Vert {\cdot } \Vert _{L^2(0,T;L^2(\Omega ))}$ is abbreviated to $\Vert {\cdot } \Vert _2$. Table 9 gives the numerical results with different $\alpha$ and $\sigma$, which show that the graded temporal grids can improve the accuracy in the $L^2(0,T;L^2(\Omega ))$ norm significantly.

6 Conclusion

A time-stepping discontinuous Galerkin method has been analyzed in this paper. Nearly optimal error estimates with respect to the regularity of the solution have been derived with nonsmooth and smooth source terms. The error estimation with nonsmooth initial value has been carried out by the Laplace transform technique. In addition, the effect of the nonvanishing $f(\cdot ,0)$ on the accuracy of the numerical solution has been investigated. Finally, numerical results have confirmed the theoretical results.