Space–Time Methods Based on Isogeometric Analysis for Time-fractional Schrödinger Equation

Abstract

In this paper, we propose a time discontinuous Galerkin scheme for solving the nonlinear time-fractional Schrödinger equation using B-splines in time and Non-Uniform Rational B-splines in space. The technique of comparing real and imaginary parts is utilized to obtain optimal $L^2([0,T];L^2(\varOmega ))$ norm error estimate. Specifically, we have achieved $r+1$ accuracy in time and $p+1$ accuracy in space, where r and p represent the spline degrees in time and space, respectively. The convergence analysis is also provided on time graded mesh, taking into account solutions with initial singularity. Additionally, the space–time isogeometric analysis method is employed to solve the linear time-fractional Schrödinger equation. A new discrete norm is constructed, and the well-posedness analysis and error estimate are performed based on this norm. We can attain $\hat{p}$ accuracy concerning the new discrete norm error in space–time domain, where $\hat{p}$ denotes space–time spline degree. Theoretical results are validated through using numerical examples.

Appendix

1.1 Auxiliary Lemmas

In this part, we will introduce several useful Lemmas about fractional operator ${_{a}}I_t^{\alpha }$.

Lemma 7

[57, Lemma 2.1] If $\alpha < 1$ and $q\in L^2(I_n)$, then

$$\begin{aligned} \int _{I_n}\vert {_{t_{n-1}}}I_t^{\alpha }q\vert ^2dt\le \frac{\tau _n^{2\alpha }}{\varGamma ^2(1+\alpha )}\int _{I_n}\vert q(s)\vert ^2ds,\;\;{1\le n\le N.} \end{aligned}$$

Specially, if $q\in L^2([0,t_n])$, we have

$$\begin{aligned} \int _{0}^{t_n}\vert {_{0}}I_t^{\alpha }q\vert ^2dt\le \frac{t_n^{\alpha }}{\varGamma (1+\alpha )\varGamma (\alpha )}\int _{0}^{t_n}(t_n-t)^{\alpha -1}\int _{0}^{t}\vert q(s)\vert ^2dsdt,\;\;{1\le n\le N.} \end{aligned}$$

Lemma 8

[42, Lemma 3.1] Let real functions $q,w\in L^2([a,b])$, then we have

• (i) $\int _{a}^{b}q(t)\left( {_{a}}I_t^{\alpha } q\right) dt=\frac{1}{K}\Vert {_{a}}I_t^{\frac{\alpha }{2}} \tilde{q}\Vert ^2_{L^2([0,\infty ))}$;

• (ii) $\left[ \int _{a}^{b}q(t)\left( {_{a}}I_t^{\alpha } w\right) dt\right] ^2\le K^2\left[ \int _{a}^{b}q(t)\left( {_{a}}I_t^{\alpha }q\right) dt\right] \left[ \int _{a}^{b}w(t)\left( {_{a}}I_t^{\alpha } w\right) dt\right]$,

where $K=1/\cos (\alpha \pi /2)$ and $\tilde{q}$ are the extension of q outside of [a, b] by zero.

Lemma 9

If $u\in L^2(I_n)$ is a complex function, it holds that

$$\begin{aligned}&\textrm{Re}\left[ \int _{I_n}u\overline{\left( {_{t_{n-1}}}I_t^{\alpha } u\right) }dt\right] \geqslant 0\;\;\;\;\text {and}\;\;\;\; \left| \textrm{Im}\left[ \int _{I_n}u\overline{\left( {_{t_{n-1}}}I_t^{\alpha } u\right) }dt\right] \right| \\&\quad \le K\textrm{Re}\left[ \int _{I_n}u\overline{\left( {_{t_{n-1}}}I_t^{\alpha } u\right) }dt\right] , \quad {1\le n\le N.} \end{aligned}$$

Proof

We assume that $u=u_1+iu_2$, where $u_1$ and $u_2$ are real functions. Using the positive semi-definite property of the fractional integral ${_{a}}I_t^{\alpha }$ [36, 37], we can derive the following result:

$$\begin{aligned} \textrm{Re}\left[ \int _{I_n}u\overline{\left( {_{t_{n-1}}}I_t^{\alpha } u\right) }dt\right] = \int _{I_n}u_1\left( {_{t_{n-1}}}I_t^{\alpha } u_1\right) dt+\int _{I_n}u_2\left( {_{t_{n-1}}}I_t^{\alpha } u_2\right) dt\geqslant 0. \end{aligned}$$

Besides, according to Lemma 8 and Cauchy inequality, one can get

$$\begin{aligned} \left| \int _{I_n}u_1\left( {_{t_{n-1}}}I_t^{\alpha } u_2\right) dt\right| \le \frac{K}{2}\left[ \int _{I_n}u_1\left( {_{t_{n-1}}}I_t^{\alpha } u_1\right) dt+\int _{I_n}u_2\left( {_{t_{n-1}}}I_t^{\alpha } u_2\right) dt\right] . \end{aligned}$$

(45)

Then using triangle inequality and (45), we have

$$\begin{aligned} \left| \textrm{Im}\left[ \int _{I_n}u\overline{\left( {_{t_{n-1}}}I_t^{\alpha } u\right) }dt\right] \right|&\le \left| \int _{I_n}u_2\left( {_{t_{n-1}}}I_t^{\alpha } u_1\right) dt\right| +\left| \int _{I_n}u_1\left( {_{t_{n-1}}}I_t^{\alpha } u_2\right) dt\right| \\&\le K\textrm{Re}\left[ \int _{I_n}u\overline{\left( {_{t_{n-1}}}I_t^{\alpha } u\right) }dt\right] . \end{aligned}$$

The proof is completed. $\square$

The following result ensures the rationality of discrete norm defined in space–time IGA method.

Lemma 10

For complex functions $u,v\in L^2([0,T])$, we have

$$\begin{aligned} \left\{ \textrm{Re}\left[ \int _{0}^T(u+v)\overline{{_{0}}I_t^{\alpha } (u+v)}dt\right] \right\} ^{\frac{1}{2}} \le \left\{ \textrm{Re}\left[ \int _{0}^T u\overline{\left( {_{0}}I_t^{\alpha } u\right) }dt\right] \right\} ^{\frac{1}{2}}+\left\{ \textrm{Re}\left[ \int _{0}^T v\overline{\left( {_{0}}I_t^{\alpha } v\right) }dt\right] \right\} ^{\frac{1}{2}}. \end{aligned}$$

Proof

Let us first consider the case where u and v are real functions. Using Lemma 8, one has

$$\begin{aligned} \left[ \int _{0}^T(u+v){_{0}}I_t^{\alpha } (u+v)dt\right] ^{\frac{1}{2}}&=\frac{1}{\sqrt{K}}\Vert {_{0}}I_t^{\alpha } (\tilde{u}+\tilde{v})dt\Vert _{L^2([0,\infty ))}\nonumber \\&\le \frac{1}{\sqrt{K}}\left( \Vert {_{0}}I_t^{\alpha } \tilde{u}\Vert _{L^2([0,\infty ))}+\Vert {_{0}}I_t^{\alpha } \tilde{v}\Vert _{L^2([0,\infty ))}\right) \nonumber \\&=\left[ \int _{0}^Tu\left( {_{0}}I_t^{\alpha } u\right) dt\right] ^{\frac{1}{2}}+\left[ \int _{0}^T v\left( {_{0}}I_t^{\alpha } v\right) dt\right] ^{\frac{1}{2}}, \end{aligned}$$

(46)

where $\tilde{u}$ and $\tilde{v}$ are the extension of u and v outside of [0, T] by zero, respectively.

Next, let us consider the case where u and v are complex functions, and set $u=u_1+iu_2$ and $v=v_1+iv_2$, where $u_1$, $u_2$, $v_1$ and $v_2$ are real functions. Making use of the result (46) for real functions, we have

$$\begin{aligned}&\left\{ \textrm{Re}\left[ \int _{0}^T(u+v)\overline{{_{0}}I_t^{\alpha } (u+v)}dt\right] \right\} ^{\frac{1}{2}}\\&\quad =\left[ \int _{0}^T(u_1+v_1){_{0}}I_t^{\alpha } (u_1+v_1)dt+\int _{0}^T(u_2+v_2){_{0}}I_t^{\alpha } (u_2+v_2)dt\right] ^{\frac{1}{2}}\\&\quad \le \Bigg \{\bigg \{\left[ \int _{0}^Tu_1\left( {_{0}}I_t^{\alpha } u_1\right) dt\right] ^{\frac{1}{2}}+\left[ \int _{0}^Tv_1\left( {_{0}}I_t^{\alpha } v_1\right) dt\right] ^{\frac{1}{2}}\bigg \}^2\\&\qquad +\bigg \{\left[ \int _{0}^Tu_2\left( {_{0}}I_t^{\alpha } u_2\right) dt\right] ^{\frac{1}{2}}+\left[ \int _{0}^Tv_2\left( {_{0}}I_t^{\alpha } v_2\right) dt\right] ^{\frac{1}{2}}\bigg \}^2\Bigg \}^{\frac{1}{2}}\\&\quad \le \left[ \int _{0}^Tu_1\left( {_{0}}I_t^{\alpha } u_1\right) dt+\int _{0}^Tu_2\left( {_{0}}I_t^{\alpha } u_2\right) dt\right] ^{\frac{1}{2}}\\&\qquad +\left[ \int _{0}^Tv_1\left( {_{0}}I_t^{\alpha } v_1\right) dt+\int _{0}^Tv_2\left( {_{0}}I_t^{\alpha } v_2\right) dt\right] ^{\frac{1}{2}}\\&\quad =\left\{ \textrm{Re}\left[ \int _{0}^T u\overline{\left( {_{0}}I_t^{\alpha } u\right) }dt\right] \right\} ^{\frac{1}{2}}+\left\{ \textrm{Re}\left[ \int _{0}^T v\overline{\left( {_{0}}I_t^{\alpha } v\right) }dt\right] \right\} ^{\frac{1}{2}}, \end{aligned}$$

where the last inequality is obtained by using the result $\sqrt{(a+b)^2+(c+d)^2}\le \sqrt{a^2+c^2}+\sqrt{b^2+d^2}.$

This proof is completed. $\square$

The following is a fractional Grönwall inequality.

Lemma 11

[29, Lemma 6.4] Let $\left\{ a_n\right\} _{n=1}^{N}$ and $\left\{ b_n\right\} _{n=1}^{N}$ be nonnegative sequences and $\left\{ b_n\right\} _{n=1}^{N}$ is monotonically increasing. Assume that

$$\begin{aligned} a_n\le b_n +c\sum _{j=1}^{n} \omega _{n,j}(\alpha ) a_n\;\;\text {for}\;\;1\le n\le N, \end{aligned}$$

where $c\geqslant 0$ and $\omega _{n,j}(\alpha )=\int _{I_j}(t_n-t)^{\alpha -1}dt$. When $0< \alpha \le 1$ and $\delta =\frac{c \tau ^{\alpha }}{\alpha }<1$, it holds that

$$\begin{aligned} a_n\le Cb_n\;\;\text {for}\;\;1\le n\le N, \end{aligned}$$

where C is a constant related to $\delta$, $\alpha$, c and T.

1.2 Implementation of Space–Time Methods

Here we present algorithms for solving problems (8) and (32). For the former, our approach involves integrating first in time direction and deriving a nonlinear system of equations about spatial degrees of freedom on every time slice. We then use an Picard iteration to solve this nonlinear system, which is feasible due to the conclusion of Theorem 2. Specifically, we set the iteration termination error to be $10^{-12}$. For the latter, we solve a sparse linear system of equations using LU decomposition technology.

Firstly, we assume that $\left\{ B^n_{j,r_n}(t)\right\} _{j=1}^{r_n+1}$, $\left\{ S_j(\textbf{x})\right\} _{j=1}^{K}$, $\left\{ N_j(\textbf{x},t)\right\} _{j=1}^{W}$ form a set of basis for spaces $\mathcal {B}^{r_n}(I_n)$, $\mathcal {V}_{0h}(\varOmega )$ and $\mathcal {V}_{0h}(Q)$, respectively. Then the approximation for the solution of NTFS or LTFS takes the form:

$$\begin{aligned} U(\textbf{x},t)=\sum _{j=1}^{r_n+1}P_{j,n}(\textbf{x})B^n_{j,r_n}(t)\;\;\text {and}\;\;u_h(\textbf{x},t)=\sum _{j=1}^{W}c_jN_j(\textbf{x},t), \end{aligned}$$

where $P_{j,n}(\textbf{x})=\sum _{k=1}^{K}c^{n}_{jk}S_k(\textbf{x})$.

Furthermore, for the sake of convenience, we define

$$\begin{aligned} C_n&=\left( c^{n}_{jk}\right) _{r_n+1,K}, \;\;\textbf{P}_n=\left( P_{j,n}\right) _{r_n+1,1},\\ M_{n}&=\left( m_{jk}^{n}\right) _{r_n+1,r_n+1}\;\;\text {with}\;\;m_{jk}^{n}=\int _{I_n}B^n_{j,r_n}(t)B^n_{k,r_n}(t)dt,\\ D_{n}&=\left( d_{jk}^{n}\right) _{r_n+1,r_n+1}\;\;\text {with}\;\;d_{jk}^{n}=\frac{1}{\varGamma (\alpha )}\int _{I_n}B^n_{j,r_n}(t)\int _{t_{n-1}}^{t}(t-s)^{\alpha -1}B^n_{k,r_n}(s)dsdt,\\ H_{n}(\textbf{x},t)&=\frac{i\lambda }{\varGamma (\alpha )}\int _{t_{n-1}}^{t}(t-s)^{\alpha -1}f(U(\textbf{x},s))ds\\&\quad +\frac{i\lambda }{\varGamma (\alpha )}\sum _{m=1}^{n-1}\int _{I_m}(t-s)^{\alpha -1}f(U(\textbf{x},s))ds+u_0,\\ Y_{n}(\textbf{x},t)&=-\frac{i}{\varGamma (\alpha )}\sum _{m=1}^{n-1}\int _{I_m}(t-s)^{\alpha -1}U(\textbf{x},s)ds,\\ \textbf{L}_n&=(l_{j}^n)_{r_n+1,1},\quad \textbf{G}_n=(g_{j}^n)_{r_n+1,1},\quad \text {with}\quad l_{j}^n=\int _{I_n}B^n_{j,r_n}H_{n}dt\quad \text {and}\quad g_{j}^n\\&=\int _{I_n}B^n_{j,r_n}Y_{n}dt. \end{aligned}$$

Next, we will give the main steps to solve (8).

Algorithm 1

Solve problem (8).

When considering (32), formalism can be easily implemented as we only need to solve

$$\begin{aligned} A\textbf{c}=\textbf{f}, \end{aligned}$$

where $A=\left( a_{jk}\right) _{W,W}$ with $a_{jk}=a_h(N_k,N_j)$, $\textbf{c}=\left( c_1,c_2,\ldots ,c_W\right) ^T$ and $\textbf{f}=(f_{j})_{W,1}$ with $f_{j}=l_h(N_j)$. For larger matrices A, we can also consider using some precondition techniques or iterative methods in parallel.