A space-time finite element method for fractional wave problems

Abstract

In this paper, we analyze a space-time finite element method for fractional wave problems involving the time fractional derivative of order γ (1 < γ < 2). We first establish the stability of the proposed method and then derive the optimal convergence rate in H¹(0,T;L²(Ω))-norm and suboptimal rate in discrete $L^{\infty }(0,T;{H_{0}^{1}}({\Omega }))$-norm. Furthermore, we discuss the performance of this method when the true solution has singularity at t = 0 and show that optimal convergence rate with respect to H¹(0,T;L²(Ω))-norm can still be achieved by using graded temporal grids. Finally, numerical experiments are performed to verify the theoretical results.

Appendices

Appendix A: Properties of fractional calculus operators

Lemma A.1

[4, 29, 31] Let $-\infty < a < b < \infty$. If $0 < \alpha , \beta < \infty$, then

$$ \text{I}_{a+}^{\alpha} \text{I}_{a+}^{\beta} = \text{I}_{a+}^{\alpha+\beta}, \quad \text{I}_{b-}^{\alpha} \text{I}_{b-}^{\beta}= \text{I}_{b-}^{\alpha+\beta}. $$

If $0 < \alpha < \beta < \infty$, then

$$ \text{D}_{a+}^{\beta} \text{I}_{a+}^{\alpha} = \text{D}_{a+}^{\beta-\alpha}, \quad \text{D}_{b-}^{\beta} \text{I}_{b-}^{\alpha} = \text{D}_{b-}^{\beta-\alpha}. $$

Lemma A.2

[5] Assume that $-\infty < a < b < \infty$ and 0 < α < 1/2. If v ∈H^α(a,b), then

$$ \begin{array}{@{}rcl@{}} \left\lVert {\text{D}_{a+}^{\alpha} v} \right\rVert_{L^{2}(a,b)} &\leqslant& \left\lvert {v} \right\rvert_{H^{\alpha}(a,b)}, \\ \left\lVert {\text{D}_{b-}^{\alpha} v} \right\rVert_{L^{2}(a,b)} &\leqslant& \left\lvert {v} \right\rvert_{H^{\alpha}(a,b)}, \\ \left\langle {\text{D}_{a+}^{\alpha} v, \text{D}_{b-}^{\alpha} v} \right\rangle_{(a,b)} &=& \cos(\alpha\pi) \left\lvert {v} \right\rvert_{H^{\alpha}(a,b)}^{2}, \\ \left\langle {\text{D}_{a+}^{\alpha} v, \text{D}_{b-}^{\alpha} w} \right\rangle_{(a,b)}&\leqslant& \left\lvert {v} \right\rvert_{H^{\alpha}(a,b)} \left\lvert {w} \right\rvert_{H^{\alpha}(a,b)},\quad\forall w \in H^{\alpha}(a,b), \\ \left\langle {\text{D}_{a+}^{2\alpha} v, w} \right\rangle_{H^{\alpha}(a,b)} &=& \left\langle {\text{D}_{a+}^{\alpha} v, \text{D}_{b-}^{\alpha} w} \right\rangle_{(a,b)} = \left\langle {\text{D}_{b-}^{2\alpha} w, v} \right\rangle_{H^{\alpha}(a,b)}, \quad\forall w \in H^{\alpha}(a,b). \end{array} $$

Lemma A.3

Suppose that $-\infty < a < b < \infty$ and 0 < α < 1/2. If v,w ∈H^α(a,b), then

$$ \left\langle {\text{D}_{a+}^{\alpha} v, \text{D}_{b-}^{\alpha} w} \right\rangle_{(a,b)} \leqslant \left\lvert {v} \right\rvert_{H^{\alpha}(a,b)} \left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert\kern-0.25ex{w}\right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert\kern-0.25ex_{H^{\alpha}(a,b)}. $$

Proof

By the definition of $\left \vert \kern -0.25ex\left \vert \kern -0.25ex\left \vert \kern -0.25ex{\cdot }\right \vert \kern -0.25ex\right \vert \kern -0.25ex\right \vert \kern -0.25ex_{H^{\alpha }(a,b)}$, this lemma is a direct consequence of Lemma A.2. □

Lemma A.4

If α ∈ [0, 1) ∖{0.5} and $0 < \beta < \infty$, then

$$ \left\lVert {\text{I}_{1-}^{\beta} v} \right\rVert_{H^{\alpha+\beta}(0,1)} \leqslant C_{\alpha,\beta} \left\lVert {v} \right\rVert_{H^{\alpha}(0,1)} $$

(26)

for all $v \in H_{0}^{\alpha }(0,1)$.

Proof

The proof is a simple modification of that of [16, Lemma 5.7]. Let us first prove that

$$ \left\lVert {\text{I}_{1-}^{\beta} w} \right\rVert_{H^{\beta}(0,1)} \leqslant C_{\beta} \left\lVert {w} \right\rVert_{L^{2}(0,1)} $$

(27)

for all w ∈L²(0, 1) and 0 < β < 1. Extending w to $\mathbb{R} \backslash (0,1)$ by zero, we define

$$ G(t) := \frac1{\Gamma(\beta/2)} {\int}_{t}^{\infty} (s-t)^{\beta/2-1} w(s) \mathrm{d}s, \quad -\infty < t < \infty. $$

Since 0 < β/2 < 1/2, a routine calculation yields $G \in L^{2}(\mathbb{R})$, and then [31, Theorem 7.1] implies

$$ \mathcal{F}G(\xi) = (-\mathrm{i}\xi)^{-\beta/2} \mathcal{F}w(\xi), \quad -\infty < \xi < \infty, $$

where $\mathcal{F}: L^{2}(\mathbb{R}) \to L^{2}(\mathbb{R})$ is the Fourier transform operator, and i is the imaginary unit. From the well-known Plancherel Theorem, it follows

$$ \left\lVert {G} \right\rVert_{H^{\beta/2}(\mathbb{R})} \leqslant C_{\beta} \left\lVert {w} \right\rVert_{L^{2}(0,1)}, $$

and hence

$$ \left\lVert {\text{I}_{1-}^{\beta/2} w} \right\rVert_{H^{\beta/2}(0,1)} \leqslant C_{\beta} \left\lVert {w} \right\rVert_{L^{2}(0,1)}. $$

(28)

In addition, if $w \in {H_{0}^{1}}(0,1),$ then, since

$$ \text{D} \text{I}_{1-}^{\beta/2} w = -\text{D} \text{I}_{1-}^{\beta/2} \text{I}_{1-} w' = \text{I}_{1-}^{\beta/2} w^{\prime}, $$

the estimate (28) implies

$$ \left\lVert { \text{I}_{1-}^{\beta/2}w } \right\rVert_{H^{1+\beta/2}(0, 1) } \leqslant C_{\beta} \left\lVert {w} \right\rVert_{{H_{0}^{1}}(0,1)}. $$

Consequently, [35, Lemma 22.3] yields

$$ \left\lVert { \text{I}_{1-}^{\beta/2} w } \right\rVert_{H^{\beta}(0,1) } \leqslant C_{\beta} \left\lVert { w } \right\rVert_{H_{0}^{\beta/2}(0,1) } \quad \text{ for all } w \in H_{0}^{\beta/2}(0,1). $$

(29)

Therefore, since $\text {I}_{1-}^{\beta } w = \text {I}_{1-}^{\beta /2} \text {I}_{1-}^{\beta /2} w$, combining (28) and (29) indicates that (27) holds for all w ∈L²(0, 1) and 0 < β < 1.

Next, let us proceed to prove (26) . Since the case of $\beta \in \mathbb{N}$ is trivial, we assume that k < β < k + 1 with $k \in \mathbb{N}$, and so it suffices to prove

$$ \left\lVert {\text{I}_{1-}^{\beta-k} v} \right\rVert_{H^{\alpha+\beta-k}(0,1)} \leqslant C_{\alpha,\beta} \left\lVert {v} \right\rVert_{H^{\alpha}(0,1)}. $$

(30)

Since we have already prove that (27) holds for all w ∈L²(0, 1) and 0 < β < 1, we obtain

$$ \begin{array}{@{}rcl@{}} \left\lVert {\text{I}_{1-}^{\beta-k} w} \right\rVert_{H^{\beta-k}(0,1)} \leqslant C_{\beta} \left\lVert {w} \right\rVert_{L^{2}(0,1)} \quad \text{ for all } w \in L^{2}(0,1), \\ \left\lVert {\text{I}_{1-}^{\beta-k} w} \right\rVert_{H^{1+\beta-k}(0,1)} \leqslant C_{\beta} \left\lVert {w} \right\rVert_{{H_{0}^{1}}(0,1)} \quad \text{ for all } w \in {H_{0}^{1}}(0,1). \end{array} $$

Therefore, using [35, Lemma 22.3] again proves (30) and thus concludes the proof of this lemma. □

Appendix B: Three inequalities

Lemma A.5

Let $0\leqslant a < b < \infty$ and $0 \leqslant \alpha < 1$. If $v^{\prime } \in L^{2}_{\delta }(a,b)$ with $0\leqslant \delta < 1$, then

$$ {{\int}_{a}^{b}} \mathrm{d}t {{\int}_{a}^{b}} \left\lvert {v(s)-v(t)} \right\rvert^{2} \left\lvert {s-t} \right\rvert^{-1-\alpha} \mathrm{d}s \leqslant \frac{8b^{-\delta}}{1-\delta}(b-a)^{2-\alpha} \left\lVert {v'} \right\rVert_{L_{\delta}^{2}(a,b)}^{2}. $$

(31)

Moreover, if v(b) = 0, then

$$ \begin{array}{@{}rcl@{}} {{\int}_{a}^{b}} v^{2}(t) (t-a)^{-\alpha} \mathrm{d}t &\leqslant& \frac{b^{-\delta}(b-a)^{2-\alpha}}{(1-\delta)(1-\alpha)} \left\lVert {v'} \right\rVert_{L_{\delta}^{2}(a,b)}^{2}, \end{array} $$

(32)

$$ \begin{array}{@{}rcl@{}} {{\int}_{a}^{b}} v^{2}(t) (b-t)^{-\alpha} \mathrm{d}t &\leqslant& \frac{b^{-\delta}(b-a)^{2-\alpha}}{(1-\delta)(1-\alpha)} \left\lVert {v'} \right\rVert_{L_{\delta}^{2}(a,b)}^{2}. \end{array} $$

(33)

Proof

Let us first establish (32). For a < t < b, a simple computing gives

$$ \begin{array}{@{}rcl@{}} \left\lvert {v(t)} \right\rvert &=& \left\lvert {{{\int}_{t}^{b}} v'(s) \mathrm{d}s } \right\rvert \leqslant {{\int}_{t}^{b}} \left\lvert {v^{\prime}(s)} \right\rvert \mathrm{d}s\\ &\leqslant{}& \left( {{\int}_{t}^{b}} s^{-\delta} \mathrm{d}s \right)^{\frac{1}{2}} \left( {{\int}_{t}^{b}} s^{\delta} \left\lvert {v^{\prime}(s)} \right\rvert^{2} \mathrm{d}s \right)^{\frac{1}{2}} \\ &=&\sqrt{ \frac{b^{1-\delta} - t^{1-\delta}}{1-\delta} } \left\lVert {v^{\prime}} \right\rVert_{L_{\delta}^{2}(a,t)} \leqslant \sqrt{ \frac{b^{-\delta}(b-a)}{1-\delta} } \left\lVert {v^{\prime}} \right\rVert_{L_{\delta}^{2}(a,b)}, \end{array} $$

so that we obtain

$$ \begin{array}{@{}rcl@{}} {{\int}_{a}^{b}} v^{2}(t) (t-a)^{-\alpha} \mathrm{d}t \leqslant {} & \frac{b^{-\delta}(b-a)}{1-\delta} {{\int}_{a}^{b}} (t-a)^{-\alpha} \mathrm{d}t \left\lVert {v'} \right\rVert_{L_{\delta}^{2}(a,b)}^{2} \\ = {} & \frac{b^{-\delta}(b- a)^{2-\alpha}}{(1-\delta)(1-\alpha)} \left\lVert {v'} \right\rVert_{L_{\delta}^{2}(a,b)}^{2}, \end{array} $$

namely, estimate (32). Similarly, we can derive (33) by that

$$ \begin{array}{@{}rcl@{}} {{\int}_{a}^{b}} v^{2}(t) (b-t)^{-\alpha} \mathrm{d}t \leqslant {} &\frac{b^{-\delta}(b-a)}{1-\delta} {{\int}_{a}^{b}} (b-t)^{-\alpha} \mathrm{d}t \left\lVert {v'} \right\rVert_{L_{\delta}^{2}(a,b)}^{2} \\ = {} & \frac{b^{-\delta}(b- a)^{2-\alpha}}{(1-\delta)(1-\alpha)} \left\lVert {v'} \right\rVert_{L_{\delta}^{2}(a,b)}^{2}. \end{array} $$

Then, let us prove (31). Since

$$ \begin{array}{@{}rcl@{}} && {{\int}_{a}^{b}} \mathrm{d}t {{\int}_{a}^{b}} \left\lvert {v(s)-v(t)} \right\rvert^{2} \left\lvert {s-t} \right\rvert^{-1-\alpha} \mathrm{d}s \\ &={} & 2{{\int}_{a}^{b}} \mathrm{d}t {{\int}_{t}^{b}} \left\lvert {{{\int}_{t}^{s}} v'(\tau) \mathrm{d}\tau} \right\rvert^{2} (s-t)^{-1-\alpha} \mathrm{d}s \\ &={} & 2{{\int}_{a}^{b}} \mathrm{d}t {{\int}_{t}^{b}} \left( {{\int}_{0}^{1}} v'(t+\theta(s-t)) \mathrm{d}\theta\right)^{2} (s-t)^{1-\alpha} \mathrm{d}s\\ &\leqslant{} & 2(b-a)^{1-\alpha} {{\int}_{a}^{b}} \mathrm{d}t {{\int}_{t}^{b}} \left( {{\int}_{0}^{1}} v'(t+\theta(s-t)) \mathrm{d}\theta\right)^{2} \mathrm{d}s, \end{array} $$

applying Minkowski’s integral inequality (cf. [8, Eq. 6.13.9]) yields that

$$ \begin{array}{@{}rcl@{}} && {{\int}_{a}^{b}} \mathrm{d}t {{\int}_{a}^{b}} \left\lvert {v(s)-v(t)} \right\rvert^{2} \left\lvert {s-t} \right\rvert^{-1-\alpha} \mathrm{d}s \\ &\leqslant{} & 2(b-a)^{1-\alpha} {{\int}_{a}^{b}} \left( {{\int}_{0}^{1}} \sqrt{ {{\int}_{t}^{b}} \left\lvert {v'(t+\theta(s-t))} \right\rvert^{2} \mathrm{d}s } \mathrm{d}\theta \right)^{2} \mathrm{d} t \\ &={} & 2(b-a)^{1-\alpha} {{\int}_{a}^{b}} \left( {{\int}_{0}^{1}} \theta^{-1/2}\sqrt{ {\int}_{t}^{t+\theta(b-t)} \left\lvert {v'(\eta)} \right\rvert^{2} \mathrm{d}\eta } \mathrm{d}\theta \right)^{2} \mathrm{d}t. \end{array} $$

Finally, the inequality (31) is a direct consequence of

$$ \begin{array}{@{}rcl@{}} && {{\int}_{a}^{b}} \left( {{\int}_{0}^{1}} \theta^{-1/2}\sqrt{ {\int}_{t}^{t+\theta(b-t)} \left\lvert {v'(\eta)} \right\rvert^{2} \mathrm{d}\eta } \mathrm{d}\theta \right)^{2} \mathrm{d}t\\ &\leqslant{} & {{\int}_{a}^{b}} \left( {{\int}_{0}^{1}} \theta^{-1/2}\sqrt{ {\int}_{t}^{t+\theta(b-t)} (\eta/t)^{\delta} \left\lvert {v'(\eta)} \right\rvert^{2} \mathrm{d}\eta } \mathrm{d}\theta \right)^{2} \mathrm{d}t \\ &={} & {{\int}_{a}^{b}} t^{-\delta} \left( {{\int}_{0}^{1}} \theta^{-1/2} \sqrt{ {\int}_{t}^{t+\theta(b-t)} \eta^{\delta} \left\lvert {v'(\eta)} \right\rvert^{2} \mathrm{d}\eta } \mathrm{d}\theta \right)^{2} \mathrm{d}t \\ &\leqslant{} & {{\int}_{a}^{b}} t^{-\delta} \left( {{\int}_{0}^{1}} \theta^{-1/2} \mathrm{d}\theta \right)^{2} \mathrm{d}t \left\lVert {v'} \right\rVert_{L_{\delta}^{2}(a,b)}^{2} \\ &\leqslant{} & \frac{4b^{-\delta}(b-a)}{1-\delta} \left\lVert {v'} \right\rVert_{L_{\delta}^{2}(a,b)}^{2}. \end{array} $$

This lemma is thus proved. □