Analyses of the Contour Integral Method for Time Fractional Normal-Subdiffusion Transport Equation

Abstract

In this work, we theoretically and numerically discuss a class of time fractional normal-subdiffusion transport equation, which depicts a crossover from normal diffusion (as $t\rightarrow 0$) to sub-diffusion (as $t\rightarrow \infty$). Firstly, the well-posedness and regularities of the model are studied by using the bivariate Mittag-Leffler function. Theoretical results show that after introducing the first-order derivative operator, the regularity of the solution can be improved in substance. Then, a numerical scheme with high-precision is developed no matter the initial value is smooth or non-smooth. More specifically, we use the contour integral method (CIM) with parameterized hyperbolic contour to approximate the temporal local and non-local operators, and employ the standard Galerkin finite element method for spatial discretization. Rigorous error estimates show that the proposed numerical scheme has spectral accuracy in time and optimal convergence order in space. Besides, we further improve the algorithm and reduce the computational cost by using the barycentric Lagrange interpolation. Finally, the obtained theoretical results as well as the acceleration algorithm are verified by several 1-D and 2-D numerical experiments, which also show that the numerical scheme developed in this paper is effective and robust.

Appendices

Appendix A: Limiting Asymptotic Behavior of the MSD of Subordinator $\textbf{B}[\mathbb {E}_t]$

Let $\textbf{B}(t)$ denote the d-dimensional Brownian motion and $\mathbb {E}_t$ be the inverse process or hitting time process of the $\beta$-stable subordinator $\mathbb {S}_t= Kt +\overline{S}_t$ with drift, where $0<\beta <1$ and $K>0$ ($K=0$ implies drift-less). Then, by the celebrated Lévy-Khintchine formula [1], we have $\mathbb {E}[e^{i(\lambda ,\mathbb {S}_t)}]=e^{-t(K\lambda +\lambda ^{\beta })}$.

Denote G and H as the PDFs of $\textbf{B}(t)$ and $\mathbb {E}_t$, respectively. Then, there is $u(t,x)=\int _{0}^{\infty }G(x,s)H(s,t)ds$. Since $\partial G(x,t)=\varDelta G(x,t)$ and $H(u,t)=-\partial _u\int _{0}^{t}g(y,u)dy$ with g being the PDF of $\mathbb {S}_t$, then by Laplace transformation, we obtain $\widehat{H}(u,z)=-\partial _u\widehat{g}(u,z)z^{-1}=\frac{Kz+z^{\beta }}{z}e^{-u(Kz+z^{\beta })}$. Based on these facts, the mean-squared displacement (MSD) of $\textbf{B}[\mathbb {E}_t]$ is

$$\begin{aligned} \begin{aligned} D(t):=\left\langle (\textbf{B}[\mathbb {E}_t])^2\right\rangle =\int _{\mathbb {R}^d}|x|^{2}u(t,x)dx =\int _{0}^{\infty }H(u,t)\int _{\mathbb {R}^d}|x|^{2}G(x,u)dxdu=2d\langle \mathbb {E}_t\rangle , \end{aligned}\end{aligned}$$

where d is the dimension of space. Transforming $\langle \mathbb {E}_t\rangle$ by Laplace w.r.t. t, we have

$$\begin{aligned} \mathscr {L}_{t\rightarrow z}\{\langle \mathbb {E}_t\rangle \}(z):=\int _{0}^{\infty }u\widehat{H}(u,z)du=\frac{1}{z(Kz+z^{\beta })}\simeq \left\{ \begin{aligned}&K^{-1}z^{-2},\quad{} & {} z\rightarrow \infty ,\\&z^{-(1+\beta )},\quad{} & {} z\rightarrow 0. \end{aligned}\right. \end{aligned}$$

Further, by Tauberian’s theorem (cf. [1, Theorem 1.5.7]), the limiting asymptotic behaviors of MSD are

$$\begin{aligned} D(t)\simeq \left\{ \begin{aligned}&2dK^{-1}t,\quad{} & {} t\rightarrow 0,\\&\frac{2d}{\varGamma (1+\beta )}t^{\beta },{} & {} t\rightarrow \infty . \end{aligned}\right. \end{aligned}$$

Now, it can be clearly seen that the model we discuss in this paper depicts an crossover from normal diffusion to sub-diffusion. That is, when t is small enough, it portrays normal diffusion; as t grows large enough, it reflects sub-diffusion.

Appendix B

1.1 B.1 Definition of the Multivariate Mittag-Leffler Function

Let $\alpha ,\beta ,\gamma \in \mathbb {C}$, and $\textrm{Re}(\alpha )>0,\textrm{Re}(\beta )>0$. The bivariate Mittag-Leffler function is defined as

$$\begin{aligned} E_{(\alpha ,\beta ),\gamma }^{\delta }(z_1, z_2):=\sum \limits _{k=0}^{\infty }\sum \limits _{l=0}^{\infty }\frac{(\delta )_{k+l}}{\varGamma (\alpha k+\beta l+\gamma )}\cdot \frac{z_1^k z_2^l}{k!l!}, \end{aligned}$$

(59)

where the numerator $(\delta )_{k+l}$ is the Pochhammer symbol, i.e.,

$$\begin{aligned} (a)_n=\frac{\varGamma (a+n)}{\varGamma (a)}=a(a+1)(a+2)...(a+n-1). \end{aligned}$$

As mentioned in [3], the multiple power series (59), which converges absolutely and locally uniformly, defines an entire function in $z_1$ and $z_2$.

1.2 B.2 Proof of Lemma 1

Proof

Let $\alpha$, $\beta$, $\gamma$, $\omega _1$, $\omega _2\in \mathbb {R}$ with $0<\alpha <\beta \le 1$, $\gamma \ge 0$ and $\omega _1$, $\omega _2<0$. Firstly, for $t=0$, according to the series representation of the bivariate Mittag-Leffler function in Eq. (59), there exist a positive constant C such that

$$\begin{aligned} \left| E^1_{(\alpha ,~\beta ),~\gamma }\left( \omega _1 t^{\alpha }, \omega _2 t^{\beta }\right) \right| =\frac{1}{\varGamma (\gamma )}\le C. \end{aligned}$$

(60)

For $t>0$, by Corollary 2 in [13], the contour integral representation of the bivariate Mittag-Leffler function $E^1_{(\alpha ,~\beta ),~\gamma }(\omega _1 t^{\alpha }, \omega _2 t^{\beta })$ is

$$\begin{aligned} \begin{aligned} E^1_{(\alpha ,~\beta ),~\gamma }\left( \omega _1 t^{\alpha }, \omega _2 t^{\beta }\right) =\frac{t^{1-\gamma }}{2\pi i}\int _{\varGamma _{\theta ,\delta }}\frac{e^{zt}z^{-\gamma }}{1+|\omega _1|z^{-\alpha }+|\omega _2|z^{-\beta }}dz. \end{aligned} \end{aligned}$$

(61)

Case I:

for given $\beta \in (0,1)$ and any $\alpha \in (0,1)$ with $0<\alpha<\beta <1$. When $z\in \varGamma _{\theta ,\delta }$, we choose $\theta \in (\frac{\pi }{2},\pi )$, which only depends on $\beta$ and closes to $\frac{\pi }{2}$ enough such that $\arg (|\omega _2|z^{-\beta })>-\frac{\pi }{2}$. Then the angle between $|\omega _2|z^{-\beta }$ and $1+|\omega _1|z^{-\alpha }$ is less then $\pi /2$ and the denominator in (61) satisfies

$$\begin{aligned} \left| 1+|\omega _1|z^{-\alpha }+|\omega _2|z^{-\beta }\right| \ge |\omega _2||z|^{-\beta }\cos \beta \theta . \end{aligned}$$

(62)

Based on these, choosing $\delta =1/t>0$ large enough, then we get

$$\begin{aligned}\begin{aligned} \left| E^1_{(\alpha ,~\beta ),~\gamma }\left( \omega _1 t^{\alpha }, \omega _2 t^{\beta }\right) \right|&\le \frac{t^{1-\gamma }}{ 2\pi }\int _{\varGamma _{\theta ,\delta }}\frac{e^{|z|t}|z|^{-\gamma }}{\left| 1+|\omega _1|z^{-\alpha }+|\omega _2|z^{-\beta }\right| }|dz|\\&\le \frac{t^{1-\gamma }}{2\pi \left| \omega _2\right| \cos (\beta \theta )} \int _{\varGamma _{\theta ,\delta }}e^{|z|t}|z|^{\beta -\gamma }|dz| \\&\le \frac{t^{1-\gamma }}{2\pi \left| \omega _2\right| \cos (\beta \theta )}Ct^{\gamma -1-\beta } \le \frac{C}{|\omega _1t^{\beta }|}. \end{aligned} \end{aligned}$$

Case II:

for given $\alpha \in (0,1)$, $\beta =1$. Similar to (i), when $z\in \varGamma _{\theta ,\delta }$, by choosing $\theta \in (\frac{\pi }{2},\pi )$, which depends on $\alpha$ and closes to $\frac{\pi }{2}$ enough such that $\alpha \theta +\arg (1+|\omega _2|z^{-1})\ge -\frac{\pi }{2}$. In this case, the denominator in (61) satisfies

$$\begin{aligned} \left| 1+|\omega _2|z^{-1}+|\omega _1|z^{-\alpha }\right| >\left| 1+|\omega _2|z^{-1}\right| \ge |\omega _2||z|^{-1}\cos \left( \theta -\frac{\pi }{2}\right) . \end{aligned}$$

(63)

Similarly, there holds

$$\begin{aligned}\begin{aligned} \left| E^1_{(\alpha ,~\beta ),~\gamma }\left( \omega _1 t^{\alpha }, \omega _2 t^{\beta }\right) \right|&\le \frac{t^{1-\gamma }}{ 2\pi }\int _{\varGamma _{\theta ,\delta }}\frac{e^{|z|t}|z|^{-\gamma }}{\left| 1+|\omega _1|z^{-\alpha }+|\omega _2|z^{-\beta }\right| }|dz|\\&\le \frac{t^{1-\gamma }}{2\pi \left| \omega _2\right| \sin (\theta )} \int _{\varGamma _{\theta ,\delta }}e^{|z|t}|z|^{1-\gamma }|dz|\\&\le \frac{t^{1-\gamma }}{2\pi \left| \omega _2\right| \sin (\theta )}Ct^{\gamma -1-1} \le \frac{C}{|\omega _2t|}. \end{aligned}\end{aligned}$$

Thus the proof of this lemma is completed. $\square$

Appendix C: Proof of Corollary 1

Proof

Firstly, we let $t_0<t\le \varLambda t_0$ and take the solution to Problem (1) as $u(x,t)=\delta (t-t_0)\cdot \textbf{1}_{(\varOmega )}$, where $\delta (t)$ is the Dirac delta function. Then $u_0\equiv 0$, $\widehat{u}(x,z)=e^{-zt_0}\cdot \textbf{1}_{(\varOmega )}$, $\widehat{f}(x,z)=(Kz+z^{\beta })e^{-zt_0}\cdot \textbf{1}_{(\varOmega )}$, and

$$\begin{aligned} u^{N}(t)=\frac{\tau }{2\pi i}\sum _{k=-(N-1)}^{N-1}e^{z(\phi _k)t}e^{-z(\phi _k)t_0}z'(\phi _k). \end{aligned}$$

From (36)) and (14), there is a constant $C_2$, which depends on $\varOmega ,K,\beta ,\alpha ,\mu$, such that

$$\begin{aligned} \left\| \widehat{f}(z)\right\| _{Nr} \le \sup \limits _{z\in D\subset \Sigma _{\theta }}|\varOmega |\left| Kz+z^{\beta }\right| \left| e^{-zt_0}\right| \le C|z|\cdot \left| e^{-zt_0}\right| \le C_2. \end{aligned}$$

Thus, for $t_0< t\le \varLambda t_0$, by Theorem 3, we can obtain

$$\begin{aligned}&\left\| \frac{\tau }{2\pi i}\sum _{k=-(N-1)}^{N-1}e^{z(\phi _k)(t-t_0)}z'(\phi _k)\right\| _{L^2(\varOmega )}\\&\quad \le C_2 L\left( \mu t_0\sin (\alpha -\tilde{d})\right) \varphi (\alpha ,\tilde{d})\cdot \left( \varepsilon \cdot (\varepsilon _N(\varrho ))^{\varrho -1}+\frac{(\varepsilon _N(\varrho ))^{\varrho }}{1-\varepsilon _N(\varrho )}\right) , \end{aligned}$$

i.e.,

$$\begin{aligned}&\left\| \sum _{k=-(N-1)}^{N-1}e^{z(\phi _k)(t-t_0)}z'(\phi _k)\right\| _{L^2(\varOmega )}\\&\quad \le C_1 L\left( \mu t_0\sin (\alpha -\tilde{d})\right) \varphi (\alpha ,\tilde{d}) \cdot \left( \varepsilon \cdot (\varepsilon _N(\rho ))^{\rho -1}+\frac{(\varepsilon _N(\varrho ))^{\rho }}{1-\varepsilon _N(\rho )}\right) . \end{aligned}$$

Actually, the above estimation is obtained by using the fact that $\int _{0}^{\infty }e^{-\mu t\sin (\alpha -\tilde{d})\cosh (x)}\le \int _{0}^{\infty }e^{-\mu t_0\sin (\alpha -\tilde{d})\cosh (x)}\le L(\mu t_0\sin (\alpha -\tilde{d}))$ (cf. Lemma 3). In other words,

$$\begin{aligned}&\left\| \sum _{k=-(N-1)}^{N-1}e^{z(\phi _k)(t-t_0)}z'(\phi _k)\right\| _{L^2(\varOmega )}\\&\quad \le C_1\int _{0}^{\infty }e^{-\mu t\sin (\alpha -\tilde{d})\cosh (x)}\varphi (\alpha ,\tilde{d})\cdot \left( \varepsilon \cdot (\varepsilon _N(\rho ))^{\rho -1} +\frac{(\varepsilon _N(\rho ))^{\rho }}{1-\varepsilon _N(\rho )}\right) , \end{aligned}$$

still holds.

So, for $0< t\le (\varLambda -1) t_0$, there is

$$\begin{aligned}&\left\| \sum _{k=-(N-1)}^{N-1}e^{z(\phi _k)t}z'(\phi _k)\right\| _{L^2(\varOmega )}\\&\quad \le C_1\int _{0}^{\infty }e^{-\mu (t+t_0)\sin (\alpha -\tilde{d})\cosh (x)}\varphi (\alpha ,\tilde{d})\cdot \left( \varepsilon \cdot (\varepsilon _N(\varrho ))^{\varrho -1}+\frac{(\varepsilon _N(\varrho ))^{\varrho }}{1-\varepsilon _N(\varrho )}\right) \\&\quad \le C_1L\left( \mu t_0\sin (\alpha -\tilde{d})\right) \varphi (\alpha ,\tilde{d})\cdot \left( \varepsilon \cdot (\varepsilon _N(\varrho ))^{\varrho -1} +\frac{(\varepsilon _N(\rho ))^{\rho }}{1-\varepsilon _N(\rho )}\right) . \end{aligned}$$