A fractional-order dependent collocation method with graded mesh for impulsive fractional-order system

Abstract

The impulsive differential equations are regarded as an optimal method to describe solute concentration fluctuation transport in unsteady flow field which are influenced by natural factors or human activities. The key difficulty of impulsive fractional-order system (IFS) in numerical discretization is that fractional-orders are different in different impulsive period. This paper proposes a double-scale-dependent mesh method considering the period memory, and makes a comparison with four collocation modes for the implict difference method. Furthermore, the stability and truncation error for graded meshes are estimated and analyzed. The analysis result reveals that the convergence rate mainly depends on the largest fractional order on the IFS. Numerical results show all graded meshes (producing the dense mesh at the early stage) provide better performance than uniform mesh. Meanwhile, the PDE cases show double-scale-dependent mesh is the most efficient numerical approximation method for the pulsation diffusion of contaminant in porous medium.

Appendices

Appendix A

Lemma 2.1.

For 0 < $\alpha$ _k+1 < 1 and $f_{k + 1} (s_{n} ) \in C^{2}$, $s_{n} \in [t_{k} ,t_{k + 1} ]\backslash \left\{ {t_{{1}} , \ldots ,t_{m} } \right\}$, k = 0, 1, … m (when k = m and 0, $t_{m + 1}^{ - } = t_{m + 1}^{ + }$ and $t_{0}^{ - } = t_{0}^{ + }$. j is the j-th node of s_n in [t_k,t_k+1]. when j = 0, s_n = t_k), it holds that

$$ \frac{{1}}{{\Gamma \left( {{1} - \alpha_{k + 1} } \right)}}\int\limits_{{t_{k} }}^{{s_{n} }} {u^{^{\prime}} (\xi )(s_{n} - \xi )^{{ - \alpha_{k + 1} }} d\xi } = \frac{{1}}{{\Gamma \left( {{1} - \alpha_{k + 1} } \right)}}\sum\limits_{p = 0}^{j - 1} {\frac{{u(s_{n - p} ) - u(s_{n - p - 1} )}}{{\tau_{n - p} }}} \int\limits_{{s_{n - p - 1} }}^{{s_{n - p} }} {(s_{n} - \xi )^{{ - \alpha_{k + 1} }} d\xi } + R_{k + 1}^{n} $$

$$ \begin{aligned} \left| {R^{n} } \right| & \le \frac{{max\left( {\tau_{{^{n - p} }}^{2} } \right)}}{8}\mathop {max}\limits_{{t_{k} \le s_{n - p} \le s_{n - 1} }} \left| {u^{\prime\prime}(s_{n - p} )} \right|(\tau_{n}^{{ - \alpha_{k + 1} }} - \left( {s_{n} - t_{k} } \right)^{{ - \alpha_{k + 1} }} )\\& \quad + \frac{{\tau_{n}^{{2 - \alpha_{k + 1} }} }}{{2\left( {1 - \alpha_{k + 1} } \right)}}\mathop {max}\limits_{{s_{n - 1} \le s \le s_{n} }} \left| {u^{^{\prime\prime}} (s)} \right| \hfill \\ & \le \mathop {max}\limits_{{t_{k} \le s \le s_{n} }} \left| {u^{^{\prime\prime}} (s)} \right|\left( {\frac{{max\left( {\tau_{{^{n - p} }}^{2} } \right)\tau_{n}^{{ - \alpha_{k + 1} }} }}{8} + \frac{{\tau_{n}^{{2 - \alpha_{k + 1} }} }}{{2\left( {1 - \alpha_{k + 1} } \right)}}} \right) \hfill \\ \end{aligned} $$

(18)

Proof

The discretization scheme of the integral shown in Eq. (18) can also be written as:

$$ \int\limits_{{t_{k} }}^{{s_{n} }} {u^{^{\prime}} (\xi )(s_{n} - \xi )^{{ - \alpha_{k + 1} }} d\xi } = \int\limits_{{t_{k} }}^{{s_{n - 1} }} {u^{^{\prime}} (\xi )(s_{n} - \xi )^{{ - \alpha_{k + 1} }} d\xi } + \int\limits_{{s_{n - 1} }}^{{s_{n} }} {u^{^{\prime}} (\xi )(s_{n} - \xi )^{{ - \alpha_{k + 1} }} d\xi } $$

(19)

Then, we divide the scheme into two parts. First, utilizing integration by parts, it is easy to obtain:

$$ \begin{aligned}& \int\limits_{{t_{k} }}^{{s_{n - 1} }} {u^{\prime} (\xi )(s_{n} - \xi )^{{ - \alpha_{k + 1} }} d\xi } \hfill \\ &= \left[ {(s_{n} - \xi )^{{ - \alpha_{k + 1} }} u(\xi )} \right]_{{t_{k} }}^{{s_{n - 1} }} - \alpha_{k + 1} \int\limits_{{t_{k} }}^{{s_{n - 1} }} {u(\xi )(s_{n} - \xi )^{{ - \alpha_{k + 1} - 1}} d\xi } \hfill \\ &= \left( \begin{aligned} & \tau_{n}^{{ - \alpha_{k + 1} }} u(s_{n - 1} ) - (s_{n} - t_{k} )^{{ - \alpha_{k + 1} }} u(t_{k} ) - \hfill \\ & \alpha_{k + 1} \sum\limits_{p = 1}^{j} {\int\limits_{{s_{n - 1 - p} }}^{{s_{n - p} }} {\frac{{\left( {s_{n - p} - \xi } \right)u(s_{n - 1 - p} ) + \left( {\xi - s_{n - 1 - p} } \right)u(s_{n - p} )}}{{\tau_{n - p} }}(s_{n} - \xi )^{{ - \alpha_{k + 1} - 1}} } } d\xi - \left( {R_{1}^{k} } \right)^{n} \hfill \\ \end{aligned} \right) \hfill \\ \end{aligned} $$

(20)

Note that:

$$ \begin{gathered} \int\limits_{{s_{n - 1 - p} }}^{{s_{n - p} }} {\left( {s_{n - p} - \xi } \right)(s_{n} - \xi )^{{ - \alpha_{k + 1} - 1}} } d\xi = - \frac{{\tau_{n - p} }}{{\alpha_{k + 1} }}\left( {s_{n} - s_{n - 1 - p} } \right)^{{ - \alpha_{k + 1} }} + \frac{1}{{\alpha_{k + 1} }}\int\limits_{{s_{n - 1 - p} }}^{{s_{n - p} }} {(s_{n} - \xi )^{{ - \alpha_{k + 1} }} } d\xi \hfill \\ \int\limits_{{s_{n - 1 - p} }}^{{s_{n - p} }} {\left( {\xi - s_{n - 1 - p} } \right)(s_{n} - \xi )^{{ - \alpha_{k + 1} - 1}} } d\xi = \frac{{\tau_{n - p} }}{{\alpha_{k + 1} }}\left( {s_{n} - s_{n - p} } \right)^{{ - \alpha_{k + 1} }} - \frac{1}{{\alpha_{k + 1} }}\int\limits_{{s_{n - 1 - p} }}^{{s_{n - p} }} {(s_{n} - \xi )^{{ - \alpha_{k + 1} }} } d\xi \hfill \\ \end{gathered} $$

(21)

By combining Eqs. (20) and (21), we obtains:

$$ \begin{aligned} &\int\limits_{{t_{k} }}^{{s_{n - 1} }} {u^{^{\prime}} (\xi )(s_{n} - \xi )^{{ - \alpha_{k + 1} }} d\xi } \hfill \\ &= \left( \begin{aligned} & \tau_{n}^{{ - \alpha_{k + 1} }} u(s_{n - 1} ) - (s_{n} - t_{k} )^{{ - \alpha_{k + 1} }} u(t_{k} ) + \sum\limits_{p = 1}^{j} {u(s_{n - 1 - p} )\left( {s_{n} - s_{n - 1 - p} } \right)^{{ - \alpha_{k + 1} }} } \hfill \\ & - \sum\limits_{p = 1}^{j} {u(s_{n - p} )\left( {s_{n} - s_{n - p} } \right)^{{ - \alpha_{k + 1} }} } + \sum\limits_{p = 1}^{j} {\frac{{u(s_{n - p} ) - u(t_{n - p - 1} )}}{{\tau_{n - p} }}\int\limits_{{s_{n - p - 1} }}^{{s_{n - p} }} {(s_{n} - \xi )^{{ - \alpha_{k + 1} }} } } d\xi - \left( {R_{1}^{k} } \right)^{n} \hfill \\ \end{aligned} \right) \hfill \\ &= \sum\limits_{p = 1}^{j} {\frac{{u(s_{n - p} ) - u(s_{n - p - 1} )}}{{\tau_{n - p} }}\int\limits_{{s_{n - p - 1} }}^{{s_{n - p} }} {(s_{n} - \xi )^{{ - \alpha_{k + 1} }} } } d\xi - \left( {R_{1}^{k} } \right)^{n} \hfill \\ \end{aligned} $$

(22)

Considering the linear interpolation, we can get

$$ \left( {R_{1}^{k} } \right)^{n} = \alpha_{k + 1} \sum\limits_{p = 1}^{j} {\int\limits_{{s_{n - p - 1} }}^{{s_{n - p} }} {\frac{1}{2}f^{^{\prime\prime}} (\xi_{p} )(\xi - s_{n - p} )(\xi - s_{n - p - 1} )(s_{n} - \xi )^{ - \alpha - 1} d\xi } } , \, \xi_{p} \in \left( {s_{n - p - 1} ,s_{n - p} } \right) $$

Then, we have

$$ \begin{aligned} \left| {\left( {R_{1}^{k} } \right)^{n} } \right| & \le \frac{{\alpha_{k + 1} }}{8}\sum\limits_{p = 1}^{j} {\tau_{n - p}^{2} } \left| {u^{\prime\prime}(s_{n - p} )} \right|\int\limits_{{s_{n - p - 1} }}^{{s_{n - p} }} {(s_{n} - \xi )^{{ - \alpha_{k + 1} - 1}} } d\xi \hfill \\ & \le \frac{{\alpha_{k + 1} \max \left( {\tau_{{^{n - p} }}^{2} } \right)}}{8}\mathop {\max }\limits_{{t_{k} \le s_{n - p} \le s_{n - 1} }} \left| {u^{\prime\prime}(s_{n - p} )} \right|\int\limits_{{t_{k} }}^{{s_{n - 1} }} {(s_{n} - \xi )^{{ - \alpha_{k + 1} - 1}} } d\xi \hfill \\ & = \frac{{\max \left( {\tau_{{^{n - p} }}^{2} } \right)}}{8}\mathop {\max }\limits_{{t_{k} \le s_{n - p} \le s_{n - 1} }} \left| {u^{\prime\prime}(s_{n - p} )} \right|(\tau_{n}^{{ - \alpha_{k + 1} }} - \left( {s_{n} - t_{k} } \right)^{{ - \alpha_{k + 1} }} ) \hfill \\ \end{aligned} $$

(23)

Using the Taylor series expansion, the error of the second term on the right-hand side of formula (19) is

$$ \begin{aligned} \left| {\left( {R_{2}^{k} } \right)^{n} } \right| &= \left| {\int\limits_{{s_{n - 1} }}^{{s_{n} }} {u^{^{\prime}} (\xi )(s_{n} - \xi )^{{ - \alpha_{k + 1} }} d\xi } - \int\limits_{{s_{n - 1} }}^{{s_{n} }} {\frac{{f(s_{n} ) - f(s_{n - 1} )}}{{\tau_{n} }}(s_{n} - \xi )^{{ - \alpha_{k + 1} }} d\xi } } \right| \hfill \\ & \le \frac{{\tau_{n}^{2} }}{2}\mathop {max}\limits_{{s_{n - 1} \le s \le s_{n} }} \left| {u^{^{\prime\prime}} (s)} \right|\frac{{\tau_{n}^{{ - \alpha_{k + 1} }} }}{{1 - \alpha_{k + 1} }} \hfill \\ & \le \frac{{\tau_{n}^{{2 - \alpha_{k + 1} }} }}{{2\left( {1 - \alpha_{k + 1} } \right)}}\mathop {\max }\limits_{{s_{n - 1} \le s \le s_{n} }} \left| {u^{^{\prime\prime}} (s)} \right| \hfill \\ \end{aligned} $$

(24)

By summing formulas (23) and (24), we complete the proof.$\square$

Appendix B

The inner products, L² norm, H¹ seminorm and H¹ norm, are introduced:

$$ \begin{gathered} \left\langle {m,n} \right\rangle = \int\limits_{\Omega } {mnd{\mathbf{x}}} \, \hfill \\ \left\| m \right\| = \sqrt {\left\langle {m,m} \right\rangle } \hfill \\ \left\| {\nabla m} \right\| = \sqrt {\left\langle {\nabla m,\nabla m} \right\rangle } \hfill \\ \left\| m \right\|_{1} = \sqrt {\left\| m \right\|^{2} { + }\left\| {\nabla m} \right\|^{2} } \hfill \\ \end{gathered} $$

Then the Green’s first formula can be written as:

$$ m,\Delta w + \nabla m,\nabla w = \mathop {\oint }\limits_{{\partial {\Omega }}}^{{}} m\left( {\nabla w \cdot \vec{n}} \right)dS $$

where $\Delta$ and $\nabla$ represent the Laplace and gradient operator, respectively; and $\overrightarrow {n}$ is the outer normal vector of the limited surface dS.

Lemma 2.2.

The difference scheme for the five collocation meshes for Eq. (11) is unconditionally stable to the initial condition $u_{{t_{k} }} + R_{ku}^{{t_{k} }}$ u₀ and the source part f:

$$ \left\| {\nabla \left( {u_{n + 1} + R_{k}^{n + 1} } \right)} \right\|^{2} \le \left\| {\nabla \left( {u_{{t_{k} }} + R_{ku}^{{t_{k} }} } \right)} \right\|^{2} + \frac{{\Gamma (1 - \alpha_{k + 1} )}}{{2\left( {t_{k + 1} - t_{k} } \right)^{{ - \alpha_{k + 1} }} }}\mathop {\max }\limits_{1 \le h \le n + 1} \left\| {f_{h + 1}^{k + 1} } \right\|^{2} ,{ 0} \le t_{n} \le T $$

Proof.

Multiplying both sides of Eq. (11) by $- 2\Delta \left( {u(s_{n + 1} ) + R_{k} (s_{n + 1} )} \right)$ and then applying Green’s first formula, we get:

$$ \begin{aligned} & 2\chi_{n + 1}^{n + 1} \left\| {\nabla \left( {u_{n + 1} + R_{k}^{j + 1} } \right)} \right\|^{2} + 2\left\| {\Delta \left( {u_{n + 1} + R_{k}^{j + 1} } \right)} \right\|^{{2}} \hfill \\ & = \left( \begin{aligned} & 2\sum\limits_{h = 0}^{j} {(\chi_{n + 1}^{h + 1} - \chi_{n + 1}^{h} )} \left\langle {\nabla (u^{h} + R_{k}^{h} ),\Delta (u_{n + 1} + R_{k}^{j + 1} )} \right\rangle \hfill \\ & + 2\chi_{n + 1}^{0} \left\langle {\nabla \left( {u_{{t_{k} }} + R_{k}^{{}} } \right),\Delta \left( {u_{n + 1} + R_{k}^{j + 1} } \right)} \right\rangle - 2\left\langle {f_{{^{n + 1} }}^{k + 1} ,\Delta \left( {u_{n + 1} + R_{k}^{j + 1} } \right)} \right\rangle \hfill \\ \end{aligned} \right) \hfill \\ & \le \left( \begin{aligned} & \sum\limits_{h = 0}^{j} {(\chi_{n + 1}^{h + 1} - \chi_{n + 1}^{h} )} \left( {\left\| {\nabla (u^{h} + R_{k}^{h} )} \right\|^{2} + \left\| {\Delta (u_{n + 1} + R_{k}^{j + 1} )} \right\|^{2} } \right) \hfill \\ & + \chi_{n + 1}^{0} \left( {\left\| {\nabla \left( {u_{{t_{k} }} + R_{k}^{{}} } \right)} \right\|^{2} + \left\| {\Delta \left( {u_{n + 1} + R_{k}^{j + 1} } \right)} \right\|^{2} } \right) + \frac{1}{2}\left\| {f_{{^{n + 1} }}^{k + 1} } \right\|^{2} + 2\left\| {\Delta \left( {u_{n + 1} + R_{k}^{j + 1} } \right)} \right\|^{2} \hfill \\ \end{aligned} \right) \hfill \\ & = \left( \begin{aligned} & \sum\limits_{h = 0}^{j} {(\chi_{n + 1}^{h + 1} - \chi_{n + 1}^{h} )} \left\| {\nabla (u^{h} + R_{k}^{h} )} \right\|^{2} + \chi_{{n{ + 1}}}^{{n{ + 1}}} \left\| {\nabla (u_{n + 1} + R_{k}^{j + 1} )} \right\|^{2} \hfill \\ & + \chi_{n + 1}^{0} \left\| {\nabla \left( {u_{{t_{k} }} + R_{k}^{{}} } \right)} \right\|^{2} + \frac{1}{2}\left\| {f_{{^{n + 1} }}^{k + 1} } \right\|^{2} + 2\left\| {\Delta \left( {u_{n + 1} + R_{k}^{j + 1} } \right)} \right\|^{2} \hfill \\ \end{aligned} \right) \hfill \\ \end{aligned} $$

where j is the j-th node of s_n within [t_k, t_k+1]. According to the monotonic increasing function $\chi_{n + 1 - p}^{k} > 0$, we have

$$ \chi_{n - j}^{k} > \frac{{\left( {s_{n + 1} - t_{k} } \right)^{{ - \alpha_{k + 1} }} }}{{\Gamma (1 - \alpha_{k + 1} )}} > \frac{{\left( {t_{k + 1} - t_{k} } \right)^{{ - \alpha_{k + 1} }} }}{{\Gamma (1 - \alpha_{k + 1} )}} $$

And we have $\left\| {\nabla (u^{h} + R_{k}^{h} )} \right\|^{2} \le \left\| {\nabla \left( {u_{{t_{k} }} + R_{k}^{{}} } \right)} \right\|^{2} + \frac{{\Gamma (1 - \alpha_{k + 1} )}}{{2\left( {t_{k + 1} - t_{k} } \right)^{{ - \alpha_{k + 1} }} }}\mathop {\max }\limits_{1 \le h \le n + 1} \left\| {f_{h + 1}^{k + 1} } \right\|^{2} h = 0,1 \cdots j + 1$

Then

$$ \begin{aligned} \chi_{n + 1}^{n + 1} \left\| {\nabla \left( {u_{n + 1} + R_{k}^{j + 1} } \right)} \right\|^{2}& \le \left( \begin{aligned} & \sum\limits_{h = 0}^{j} {(\chi_{n + 1}^{h + 1} - \chi_{n + 1}^{h} )} \left\| {\nabla (u^{h} + R_{k}^{h} )} \right\|^{2} \hfill \\ & + \chi_{n + 1}^{0} \left( {\left\| {\nabla \left( {u_{{t_{k} }} + R_{k}^{{}} } \right)} \right\|^{2} + \frac{{\Gamma (1 - \alpha_{k + 1} )}}{{2\left( {t_{k + 1} - t_{k} } \right)^{{ - \alpha_{k + 1} }} }}\left\| {f_{{^{n + 1} }}^{k + 1} } \right\|^{2} } \right) \hfill \\ \end{aligned} \right) \hfill \\ & \le \left( \begin{aligned} & \sum\limits_{h = 0}^{j} {(\chi_{n + 1}^{h + 1} - \chi_{n + 1}^{h} )} \left( {\left\| {\nabla \left( {u_{{t_{k} }} + R_{k}^{{}} } \right)} \right\|^{2} + \frac{{\Gamma (1 - \alpha_{k + 1} )}}{{2\left( {t_{k + 1} - t_{k} } \right)^{{ - \alpha_{k + 1} }} }}\left\| {f_{{^{n + 1} }}^{k + 1} } \right\|^{2} } \right) \hfill \\ & + \chi_{n + 1}^{0} \left( {\left\| {\nabla \left( {u_{{t_{k} }} + R_{k}^{{}} } \right)} \right\|^{2} + \frac{{\Gamma (1 - \alpha_{k + 1} )}}{{2\left( {t_{k + 1} - t_{k} } \right)^{{ - \alpha_{k + 1} }} }}\left\| {f_{{^{n + 1} }}^{k + 1} } \right\|^{2} } \right) \hfill \\ \end{aligned} \right) \hfill \\ & = \chi_{n + 1}^{n + 1} \left( {\left\| {\nabla \left( {u_{{t_{k} }} + R_{k}^{{}} } \right)} \right\|^{2} + \frac{{\Gamma (1 - \alpha_{k + 1} )}}{{2\left( {t_{k + 1} - t_{k} } \right)^{{ - \alpha_{k + 1} }} }}\left\| {f_{{^{n + 1} }}^{k + 1} } \right\|^{2} } \right) \hfill \\ \end{aligned} $$

According to Eq. (13) and $R_{0}^{{t_{0} }} = 0$, the value of $R_{k}^{n + 1}$ approaches zero with the increasing node number N by the recurrence relation.

The proof is completed.$\square$