A Time-Spectral Algorithm for Fractional Wave Problems

Abstract

This paper develops a high-accuracy algorithm for time fractional wave problems, which employs a spectral method in the temporal discretization and a finite element method in the spatial discretization. Moreover, stability and convergence of this algorithm are derived, and numerical experiments are performed, demonstrating the exponential decay in the temporal discretization error provided the solution is sufficiently smooth.

Appendix A: Weak Solution

We call

$$\begin{aligned} u \in H^{(\gamma +1)/2}(0,T; L^2(\Omega )) \cap L^2(0,T; H_0^1(\Omega )) \end{aligned}$$

a weak solution to problem (1) if $u(0) = u_0$ and

(18)

for all $v \in H^{(\gamma -1)/2}(0,T; L^2(\Omega )) \cap L^2(0,T; H_0^1(\Omega ))$.

To prove that problem (1) admits a unique weak solution, we first consider the following problem: given $c_0$, $c_1 \in \mathbb R$ and $g \in L^2(0,T)$, seek $y \in H^\gamma (0,T)$ such that

$$\begin{aligned} D_{0+}^\gamma (y - c_0 - c_1t) + \lambda y = g, \end{aligned}$$

(19)

where $\lambda$ is a positive constant such that $\lambda \geqslant 1$.

Lemma A.1

Suppose that $v\in H^{(\gamma +1)/2}(0,T)$ and $D_{0+}^{\gamma }v\in L^2(0,T)$, then

$$\begin{aligned} \left\| v \right\| _{H^\gamma (0,T)} \lesssim \left\| D_{0+}^\gamma v \right\| _{L^2(0,T)}. \end{aligned}$$

(20)

Proof

Since $D_{0+}^\gamma v\in L^2(0,T)$, by [9, Lemmas A.4] we conclude that $I_{0+}^\gamma D_{0+}^\gamma v\in H^{\gamma }(0,T)$ with

$$\begin{aligned} \left\| I_{0+}^\gamma D_{0+}^\gamma v \right\| _{H^\gamma (0,T)} \lesssim \left\| D_{0+}^\gamma v \right\| _{L^2(0,T)}. \end{aligned}$$

(21)

A simple calculation yields

$$\begin{aligned} v = c_0t^{\gamma -2} + c_1t^{\gamma -1} +I_{0+}^\gamma D_{0+}^\gamma v, \end{aligned}$$

which indicates that $c_0=c_1=0$ by the fact $v\in H^{(\gamma +1)/2}(0,T)$. Then (20) follows from (21). This completes the proof. $\square$

Lemma A.2

Suppose that $v\in H^{(\gamma +1)/2}(0,T)$ with $v(0)=0$, then we have the following properties.

• (a) It holds that

  $$\begin{aligned} \left( D_{0+}^\frac{\gamma +1}{2} v, D_{T-}^\frac{\gamma -1}{2} v' \right) _{L^2(0,T)} \sim \left\| v \right\| _{H^{(\gamma +1)/2}(0,T)}^2. \end{aligned}$$

  (22)

• (b) For any $w\in H^{(\gamma -1)/2}(0,T)$, it holds that

  $$\begin{aligned} \left( D_{0+}^\frac{\gamma +1}{2} v, D_{T-}^\frac{\gamma -1}{2} w \right) _{L^2(0,T)}\lesssim \left\| v \right\| _{H^{(\gamma +1)/2}(0,T)} \left\| w \right\| _{H^{(\gamma -1)/2}(0,T)}. \end{aligned}$$

  (23)

• (c) For any $\varphi \in C_0^{\infty }(0,T)$, it holds that

  $$\begin{aligned} \left\langle {D_{0+}^\gamma v, \varphi } \right\rangle = \left( D_{0+}^\frac{\gamma +1}{2} v, D_{T-}^\frac{\gamma -1}{2} \varphi \right) _{L^2(0,T)}. \end{aligned}$$

  (24)

Proof

Let us first prove (a). Since $v\in H^{(\gamma +1)/2}(0,T)$ and $v(0)=0$, we have

$$\begin{aligned} \left\| v' \right\| _{H^{(\gamma -1)/2}(0,T)}\sim \left\| v \right\| _{H^{(\gamma +1)/2}(0,T)}. \end{aligned}$$

(25)

In addition, a straightforward calculation gives

$$\begin{aligned} D_{0+}^\frac{\gamma +1}{2} v = D^{2}I_{0+}^{\frac{3-\gamma }{2}}I_{0+}v' = D^{2}I_{0+}^{\frac{5-\gamma }{2}}v' = D_{0+}^{\frac{\gamma -1}{2}}v'. \end{aligned}$$

(26)

So (22) follows from (25), (26) and Lemma 5.4.

Then let us prove (b). In view of (25), (26), using Lemma 5.4 yields (23).

Finally we prove (c). Observe that (26) implies $I_{0+}^{\frac{3-\gamma }{2}}v'\in H^1(0,T)$, and a simple computation implies

$$\begin{aligned} \left( I_{0+}^{\frac{3-\gamma }{2}}v'\right) (t)\leqslant \frac{t^{\frac{1-\gamma }{2}}\left\| v' \right\| _{L^2(0,t)}}{\Gamma (\frac{3-\gamma }{2})\sqrt{2-\gamma }},\quad 0<t\leqslant T. \end{aligned}$$

Thus,

$$\begin{aligned} \left( I_{0+}^{\frac{3-\gamma }{2}}v'\right) (0) = 0. \end{aligned}$$

Using integration by parts gives

$$\begin{aligned} \begin{aligned} {}&\left\langle {D_{0+}^\gamma v, \varphi } \right\rangle = \left\langle {D^2I_{0+}^{2-\gamma } v, \varphi } \right\rangle \\&\quad = (I_{0+}^{2-\gamma } v, \varphi '')_{L^2(0,T)} = (I_{0+}^{3-\gamma } v', \varphi '')_{L^2(0,T)} \\&\quad = (I_{0+}^{\frac{3-\gamma }{2}}v' , I_{T-}^{\frac{3-\gamma }{2}}\varphi '')_{L^2(0,T)} =-(DI_{0+}^{\frac{3-\gamma }{2}}v' , I_{T-}^{\frac{3-\gamma }{2}}\varphi ')_{L^2(0,T)} \\&\quad = (D_{0+}^{\frac{\gamma -1}{2}}v' , D_{T-}^{\frac{\gamma -1}{2}}\varphi )_{L^2(0,T)} =(D_{0+}^{\frac{\gamma +1}{2}}v , D_{T-}^{\frac{\gamma -1}{2}}\varphi )_{L^2(0,T)} \end{aligned} \end{aligned}$$

for all $\varphi \in C_0^\infty (0,T)$. This shows (24) and completes the proof of this lemma. $\square$

Lemma A.3

Problem (19) has a unique solution $y \in H^\gamma (0,T)$, and y satisfies that $y(0) = c_0$ and

$$\begin{aligned} \left( D_{0+}^\frac{\gamma +1}{2} ( y - c_0 - c_1t ), \ D_{T-}^\frac{\gamma -1}{2} z \right) _{L^2(0,T)} + \lambda (y,z)_{L^2(0,T)} = (g,z)_{L^2(0,T)} \end{aligned}$$

(27)

for all $z \in H^\frac{\gamma -1}{2} (0,T)$. Moreover,

$$\begin{aligned} \left\| y \right\| _{ H^\frac{\gamma +1}{2}(0,T) } + \lambda ^\frac{1}{2} \left\| y \right\| _{L^2(0,T)} \lesssim \left\| g \right\| _{L^2(0,T)} + \lambda ^\frac{1}{2} \left| c_0 \right| + \left| c_1 \right| . \end{aligned}$$

(28)

Proof

Set

$$\begin{aligned} b(z) := \left( g,z \right) _{L^2(0,T)} + \left( D_{0+}^\frac{\gamma +1}{2} (c_1t), D_{T-}^\frac{\gamma -1}{2} z \right) _{L^2(0,T)} - \lambda \left( c_0, z \right) _{L^2(0,T)} \end{aligned}$$

for all $z \in H^\frac{\gamma -1}{2}(0,T)$. Since Lemma 5.4 implies $b \in H^\frac{1-\gamma }{2}(0,T)$, Lemma A.2 and the well-known Lax-Milgram Theorem guarantee the unique existence of $w \in H^\frac{\gamma +1}{2}(0,T)$ with $w(0) = 0$ such that

$$\begin{aligned} \left( D_{0+}^\frac{\gamma +1}{2} w, D_{T-}^\frac{\gamma -1}{2} z \right) _{L^2(0,T)} + \lambda (w, z)_{L^2(0,T)} = b(z) \end{aligned}$$

(29)

for all $z \in H^\frac{\gamma -1}{2}(0,T)$. Using Lemma A.2 gives

$$\begin{aligned} \left\langle {D_{0+}^\gamma w, \varphi } \right\rangle&= \left( D_{0+}^\frac{\gamma +1}{2} w, D_{T-}^\frac{\gamma -1}{2} \varphi \right) _{L^2(0,T)}, \\ \left\langle {D_{0+}^\gamma (c_1t), \varphi } \right\rangle&= \left( D_{0+}^\frac{\gamma +1}{2} (c_1t), D_{T-}^\frac{\gamma -1}{2} \varphi \right) _{L^2(0,T)} \end{aligned}$$

for all $\varphi \in C_0^\infty (0,T)$, so that from (29) it follows that

$$\begin{aligned} D_{0+}^\gamma ( w - c_1t ) = g - \lambda ( w+ c_0 ). \end{aligned}$$

Putting $y := w + c_0$ gives

$$\begin{aligned} D_{0+}^\gamma ( y - c_0 - c_1t ) + \lambda y = g, \end{aligned}$$

and then by Lemma A.1 and A.2 it is evident that y is the unique $H^\gamma (0,T)$-solution to problem (19). Also, $y(0) = c_0$ is obvious, and (27) follows directly from (29).

Now let us prove (28). Firstly, substituting $z := y'$ into (27) and using integration by parts yield

$$\begin{aligned} \left( D_{0+}^\frac{\gamma +1}{2} ( y-c_0-c_1t ), \ D_{T-}^\frac{\gamma -1}{2} y' \right) _{L^2(0,T)} + \frac{\lambda }{2} y^2(T) = \frac{\lambda }{2} c_0^2+ (g,y')_{L^2(0,T)}. \end{aligned}$$

Therefore, Lemma A.2, the Cauchy–Schwarz inequality and the Young’s inequality with $\epsilon$ imply

$$\begin{aligned} \left\| y - c_0 \right\| _{ H^\frac{\gamma +1}{2} (0,T) }^2 + \lambda y^2(T) \lesssim \left\| g \right\| _{L^2(0,T)}^2 + \lambda c_0^2 + c_1^2, \end{aligned}$$

and so

$$\begin{aligned} \left\| y \right\| _{ H^\frac{\gamma +1}{2}(0,T) } \lesssim \left\| g \right\| _{L^2(0,T)} + \lambda ^\frac{1}{2} \left| c_0 \right| + \left| c_1 \right| . \end{aligned}$$

(30)

Secondly, substituting $z := y$ into (27) yields

$$\begin{aligned} \lambda \left\| y \right\| _{L^2(0,T)}^2 = \left( g,y \right) _{L^2(0,T)} - \left( D_{0+}^\frac{\gamma +1}{2} (y - c_0 - c_1t), \ D_{T-}^\frac{\gamma -1}{2} y \right) _{L^2(0,T)}, \end{aligned}$$

so that using Lemmas 5.4 and A.2, the Cauchy–Schwarz inequality and the Young’s inequality with $\epsilon$ gives

$$\begin{aligned} \lambda \left\| y \right\| _{L^2(0,T)}^2 \lesssim \left\| y - c_0 - c_1t \right\| _{ H^\frac{\gamma +1}{2}(0,T) } \left\| y \right\| _{H^\frac{\gamma -1}{2}(0,T)} + \lambda ^{-1} \left\| g \right\| _{L^2(0,T)}^2, \end{aligned}$$

which, together with (30), yields

$$\begin{aligned} \lambda ^\frac{1}{2} \left\| y \right\| _{L^2(0,T)} \lesssim \left\| g \right\| _{L^2(0,T)} + \lambda ^\frac{1}{2} \left| c_0 \right| + \left| c_1 \right| . \end{aligned}$$

(31)

Finally, collecting (30), (31) proves (28), and thus proves this lemma. $\square$

Finally, by the above lemma and the Galerkin method, we readily conclude that problem (1) admits a unique weak solution indeed. We summarize the result as follows.

Theorem A.1

The weak solution u of problem (1) satisfies that $u(0) = u_0$ and that

$$\begin{aligned} \left( D_{0+}^\frac{\gamma +1}{2} ( u - u_0 - tu_1 ), \ D_{T-}^\frac{\gamma -1}{2} v \right) _{L^2(\Omega _T)} + (\nabla u,\nabla v)_{L^2(\Omega _T)} = (f,v)_{L^2(\Omega _T)} \end{aligned}$$

(32)

for all $v \in H^\frac{\gamma -1}{2} (0,T;H_0^1(\Omega ))$. Furthermore, we have