Optimal Error Estimates of a Time-Spectral Method for Fractional Diffusion Problems with Low Regularity Data

Abstract

This paper is devoted to the error analysis of a time-spectral algorithm for fractional diffusion problems of order $\alpha$ ($0< \alpha < 1$). The solution regularity in the Sobolev space is revisited and new regularity results in the Besov space are established. A time-spectral algorithm is developed which adopts a standard spectral method and a conforming linear finite element method for temporal and spatial discretizations, respectively. Optimal error estimates are derived with nonsmooth data. Particularly, a sharp temporal convergence rate $1+2\alpha$ is shown theoretically and numerically.

Appendices

A The Shifted Jacobi Polynomial

Given $a,b>-1$, the family of shifted Jacobi polynomial $\{S_k^{a,b}\}_{k=0}^\infty$ on (0, T) are defined as follows:

$$\begin{aligned} \mu ^{a,b}(t)S_{k}^{a,b}(t)=\frac{(-1)^{k}}{T^{k} k !} \frac{\,\mathrm{d}^{k}}{\,\mathrm{d}t^{k}}\mu ^{k+a,k+b}(t),\quad 0<t<T, \end{aligned}$$

(79)

where $\mu ^{\nu ,\theta }(t)=(T-t)^{\nu }t^{\theta }$ for all $-1<\nu ,\theta <\infty$. Note that (79) is also called Rodrigues’ formula [56], which implies $\{S_k^{a,b}\}_{k=0}^\infty$ is orthogonal with respect to the weight $\mu ^{a,b}$ on (0, T), i.e.,

$$\begin{aligned} \left\langle {S_{k}^{a, b},S_{l}^{a, b}} \right\rangle _{\mu ^{a,b}} =\xi _k^{a, b} \delta _{kl}, \end{aligned}$$

where $\delta _{kl}$ denotes the Kronecker product and

$$\begin{aligned} \xi _k^{a, b}:=\frac{T^{a+b+1} \Gamma (k+a+1) \Gamma (k+b+1)}{(2k+a+b+1) k! \Gamma (k+a+b+1)}. \end{aligned}$$

(80)

As $\{S_k^{a,b}\}_{k=0}^\infty$ forms a complete orthogonal basis of $L^2_{\mu ^{a,b}}(0,T)$, any $v\in L^2_{\mu ^{a,b}}(0,T)$ admits a unique decomposition

$$\begin{aligned} v=\sum _{k=0}^{\infty } v_kS_k^{a,b} \quad \text {with } v_k = \frac{1}{\xi _k^{a,b}} \left\langle {v,S_{k}^{a, b}} \right\rangle _{\mu ^{a,b}}, \end{aligned}$$

(81)

and the $L^2_{\mu ^{a,b}}$-orthogonal projection of v onto $P_M(0,T)$ is defined as $\Phi _{M}^{a,b}v:= \sum _{k=0}^{M} v_kS_k^{a,b}$. For ease of notation, we shall set $S_k^a = S_k^{a,a},\,\mu ^{a} = \mu ^{a,a},\,\Phi _M^a = \Phi _M^{a,a}$, and all the superscripts are omitted when $a = 0$.

Thanks to [7, Lemma 2.5], a standard calculation gives

$$\begin{aligned} {{\,\mathrm{D}\,}}_{0+}^{\theta } S_k^{\beta -\theta ,0} ={}&\frac{ \Gamma (k+1)}{\Gamma (k+1-\theta )}t^{-\theta } S_k^{\beta ,-\theta }(t), \end{aligned}$$

(82)

$$\begin{aligned} {{\,\mathrm{D}\,}}_{T-}^{\theta } S_k^{0,\beta -\theta } ={}&\frac{ \Gamma (k+1)}{\Gamma (k+1-\theta )} (T-t)^{-\theta } S_k^{-\theta ,\beta }(t), \end{aligned}$$

(83)

where $0<\theta <1$ and $-1<\beta <\infty$.

Lemma A.1

For any $v\in H^{\alpha /2}(0,T)$, it holds that

$$\begin{aligned} \left\langle {{{\,\mathrm{D}\,}}_{0+}^{\alpha }(I-\Phi _{M}^{-\alpha ,0})v,q} \right\rangle _{H^{\alpha /2}(0,T)} = 0 \quad \forall \,q\in P_M(0,T). \end{aligned}$$

(84)

Consequently, we have the stability:

$$\begin{aligned} \left|{\Phi _{M}^{-\alpha ,0}v} \right|_{H^{\alpha /2}(0,T)} \leqslant {}C_\alpha \left|{v} \right|_{H^{\alpha /2}(0,T)}, \end{aligned}$$

(85)

and the convergence: $\lim \limits _{M\rightarrow \infty }|(I-\Phi _{M}^{-\alpha ,0})v|_{H^{\alpha /2}(0,T)} = 0$.

Proof

Given any fixed $v\in H^{\alpha /2}(0,T)$, by [17, Theorem 1.4.4.3], we know that $v\in L_{\mu ^{-\alpha ,0}}^2(0,T)$. To prove (84), it is enough to consider $q = S_{k}^{0,-\alpha }$ for any $0\leqslant k\leqslant M$. Thanks to (83), we have

$$\begin{aligned} {{\,\mathrm{D}\,}}_{T-}^{\alpha }q = \frac{\Gamma (k+1)}{\Gamma (k+1-\alpha )}(T-t)^{-\alpha }S_k^{-\alpha ,0}. \end{aligned}$$

Again, it follows from [17, Theorem 1.4.4.3] that ${{\,\mathrm{D}\,}}_{T-}^{\alpha }q\in (H^{\alpha /2}(0,T))^*$. Thus using the definition of $\Phi _{M}^{-\alpha ,0}$ and Lemma 2.1 gives

$$\begin{aligned} \begin{aligned} \left\langle {{{\,\mathrm{D}\,}}_{0+}^{\alpha }(I-\Phi _{M}^{-\alpha ,0})v,q} \right\rangle _{H^{\alpha /2}(0,T)}&=\left\langle {(I-\Phi _{M}^{-\alpha ,0})v,{{\,\mathrm{D}\,}}_{T-}^{\alpha }q} \right\rangle _{(H^{\alpha /2}(0,T),(H^{\alpha /2}(0,T))^*) }\\&=\frac{\Gamma (k+1)}{\Gamma (k+1-\alpha )}\left\langle {(I-\Phi _{M}^{-\alpha ,0})v,(T-t)^{-\alpha }S_k^{-\alpha ,0}} \right\rangle _{(0,T)}\\&= 0. \end{aligned} \end{aligned}$$

This establishes (84) and by Lemma 2.1, we have

$$\begin{aligned} \begin{aligned} \cos (\alpha \pi /2)\left|{\Phi _{M}^{-\alpha ,0}v} \right|_{H^{\alpha /2}(0,T)}^2&= \left\langle {{{\,\mathrm{D}\,}}_{0+}^{\alpha }\Phi _{M}^{-\alpha ,0}v,\Phi _{M}^{-\alpha ,0}v} \right\rangle _{H^{\alpha /2}(0,T)}\\&=\left\langle {{{\,\mathrm{D}\,}}_{0+}^{\alpha }v,\Phi _{M}^{-\alpha ,0}v} \right\rangle _{H^{\alpha /2}(0,T)}\\&\leqslant \left|{\Phi _{M}^{-\alpha ,0}v} \right|_{H^{\alpha /2}(0,T)} \left|{v} \right|_{H^{\alpha /2}(0,T)}, \end{aligned} \end{aligned}$$

which implies (85).

By (84) and the proof of (85), we find that

$$\begin{aligned} |(I-\Phi _{M}^{-\alpha ,0})v|_{H^{\alpha /2}(0,T)} \leqslant \sec (\alpha \pi /2) |v-q|_{H^{\alpha /2}(0,T)} \quad \forall \,q\in P_M(0,T). \end{aligned}$$

Therefore, a standard density argument leads to

$$\begin{aligned} \lim \limits _{M\rightarrow \infty }|(I-\Phi _{M}^{-\alpha ,0})v|_{H^{\alpha /2}(0,T)} = 0. \end{aligned}$$

This finishes the proof of this lemma. $\square$

B Proof of the Commutativity (57)

Let $\{x_i\}_{i\in {\mathbb {N}}}$ and $\{y_j\}_{j\in {\mathbb {N}}}$ be the orthonormal basis of X and Y, respectively. Assume $Ax_i = \sum _{j=0}^{\infty }a_{ij}y_j$ with $a_{ij}\in \,{{\mathbb {R}}}$ for all $i\in {\mathbb {N}}$. It is clear that

$$\begin{aligned} \sum _{j=0}^{\infty }\left|{a_{ij}} \right|^2 = \left\Vert {Ax_i} \right\Vert _Y^2\leqslant \left\Vert {A} \right\Vert _{X\rightarrow Y}^2\left\Vert {x_i} \right\Vert _X^2\quad \text { for all }i\in {\mathbb {N}}, \end{aligned}$$

(86)

where $\left\Vert {A} \right\Vert _{X\rightarrow Y}$ denotes the operator norm of A.

We claim first that by definition, $\Phi _{M,Z}^{a,b}v_n\overset{n\rightarrow \infty }{\rightarrow } \Phi _{M,Z}^{a,b}v$ in $L^2_{\mu ^{a,b}}(0,T;Z)$ whenever $v_n\overset{n\rightarrow \infty }{\rightarrow } v$ in $L^2_{\mu ^{a,b}}(0,T;Z)$ for $Z = X$ or Y. Besides, we have the identity

$$\begin{aligned} \left\Vert {w} \right\Vert _{L^2_{\mu ^{a,b}}(0,T;Z)}^2= & {} \sum _{i=0}^{\infty }\left\Vert {w_i} \right\Vert _{L^2_{\mu ^{a,b}}(0,T)}^2 =\int _{0}^{T} \sum _{i=0}^{\infty }\left|{w_i(t)} \right|^2\mu ^{a,b}(t)\,\mathrm{d}t\\= & {} \int _{0}^{T} \left\Vert {w(t)} \right\Vert _{Z}^2\mu ^{a,b}(t)\,\mathrm{d}t, \end{aligned}$$

for all $w=\sum _{i=0}^{\infty }w_iz_i\in L^2_{\mu ^{a,b}}(0,T;Z)$, where $z_i=x_i$ or $y_i$ and we used the monotone convergence theorem (see [6, Theorem 4.1, pp.90]). Based on this, let us verify $Av_n\rightarrow Av$ in $L^2_{\mu ^{a,b}}(0,T;Y)$ provided that $v_n\overset{n\rightarrow \infty }{\rightarrow } v$ in $L^2_{\mu ^{a,b}}(0,T;X)$. Indeed,

$$\begin{aligned} \begin{aligned} \left\Vert {Av_n-Av} \right\Vert _{L^2_{\mu ^{a,b}}(0,T;Y)}^2&=\int _{0}^{T} \left\Vert {(Av_n-Av)(t)} \right\Vert _{Y}^2\mu ^{a,b}(t)\,\mathrm{d}t\\&\leqslant \left\Vert {A} \right\Vert ^2_{X\rightarrow Y}\int _{0}^{T} \left\Vert {(v_n-v)(t)} \right\Vert _{X}^2\mu ^{a,b}(t)\,\mathrm{d}t \\&=\left\Vert {A} \right\Vert ^2_{X\rightarrow Y}\left\Vert {v-v_n} \right\Vert _{L^2_{\mu ^{a,b}}(0,T;X)}^2. \end{aligned} \end{aligned}$$

Now, we take $v_n = \sum _{i=0}^{n}v_ix_i$, which converges to v in $L^2_{\mu ^{a,b}}(0,T;X)$. According to the above discussions, $\Phi _{M,X}^{a,b}v_n\overset{n\rightarrow \infty }{\rightarrow } \Phi _{M,X}^{a,b}v$ in $L^2_{\mu ^{a,b}}(0,T;X)$ and $Av_n\overset{n\rightarrow \infty }{\rightarrow } Av$ in $L^2_{\mu ^{a,b}}(0,T;Y)$. To prove (57), it is sufficient to establish

$$\begin{aligned} A \Phi _{M,X}^{a,b}v_n = \Phi _{M,Y}^{a,b}Av_n. \end{aligned}$$

(87)

Consider $\xi _i^m = \sum _{j=0}^{m}a_{ij}y_j$, which converges to $Ax_i$ by (86) and further implies that

$$\begin{aligned} \sum _{i=0}^{n}\xi _i^m \Phi _{M}^{a,b}v_i \overset{m\rightarrow \infty }{\rightarrow } \sum _{i=0}^{n}Ax_i \Phi _{M}^{a,b}v_i= A \Phi _{M,X}^{a,b}v_n\quad \text {in}\quad L^2_{\mu ^{a,b}}(0,T;Y). \end{aligned}$$

(88)

On the other hand, we find

$$\begin{aligned} \sum _{i=0}^{n}\xi _i^m \Phi _{M}^{a,b}v_i =\sum _{j=0}^{m}\left( \sum _{i=0}^{n}a_{i j} \Phi _{M}^{a,b}v_{i} \right) y_j=\Phi _{M,Y}^{a,b}\sum _{j=0}^{m}\left( \sum _{i=0}^{n}a_{i j} v_{i} \right) y_{j}=\Phi _{M,Y}^{a,b}\sum _{i=0}^{n}v_i\xi _i^m. \end{aligned}$$

Since

$$\begin{aligned} \sum _{i=0}^{n}v_i\xi _i^m\overset{m\rightarrow \infty }{\rightarrow }\sum _{i=0}^{n}v_iAx_i = Av_n\quad \text {in}\quad L^2_{\mu ^{a,b}}(0,T;Y), \end{aligned}$$

we conclude that

$$\begin{aligned} \sum _{i=0}^{n}\xi _i^m \Phi _{M}^{a,b}v_i \overset{m\rightarrow \infty }{\rightarrow } \Phi _{M,Y}^{a,b}Av_n\quad \text {in}\quad L^2_{\mu ^{a,b}}(0,T;Y). \end{aligned}$$

This together with (88) proves (87) and completes the proof.