Sharp H¹-norm error estimate of a cosine pseudo-spectral scheme for 2D reaction-subdiffusion equations

Abstract

A finite difference cosine pseudo-spectral scheme is presented for solving a linear reaction-subdiffusion problem with Neumann boundary conditions. The nonuniform version of L1 formula is employed for approximating the Caputo fractional derivative, and a cosine pseudo-spectral approximation is utilized in spatial discretization. With the help of discrete fractional Grönwall inequality and global consistency analysis, sharp H¹-norm error estimate reflecting the regularity of solution is verified for the proposed method. A fast algorithm is implemented in computation and numerical results confirm the sharpness of our analysis.

Appendices

Appendix A: Proof of Lemma 2.3

For simplicity, we substantiate the assertion in the x-direction. The other one in the y-direction can be derived similarly. The interpolation basis function in (2.12) gives

$$ \begin{array}{@{}rcl@{}} \overline{\text{X}}_{j+\frac{1}{2}}(x)&=&\frac{2}{N_{x}}\sum\limits_{l=1}^{N_{x}-1}\cos\left( \mu_{x}l(x_{j+\frac{1}{2}}-a)\right)\cos\left( \mu_{x}l(x-a)\right)+\frac{1}{N_{x}}. \end{array} $$

Denote $\theta _{x}=\mu _{x}\left (x-x_{j+\frac {1}{2}}\right )$ and $\theta _{x}^{\prime }=\mu _{x}\left (x+x_{j+\frac {1}{2}}-2a\right )$, the prosthaphaeresis formula yields:

$$ \begin{array}{@{}rcl@{}} \overline{\text{X}}_{j+\frac{1}{2}}(x)&=&\frac{1}{2N_{x}}\sum\limits_{l=1}^{N_{x}-1}\left[1+2\cos(\theta_{x}l)+\left( 1+2\cos(\theta_{x}^{\prime}l)\right)\right]\\ &=&\frac{1}{2N_{x}}\left[\frac{\sin\left( (N_{x}-\frac{1}{2})\theta_{x}\right)}{\sin(\frac{1}{2}\theta_{x})}+\frac{\sin\left( (N_{x}-\frac{1}{2})\theta_{x}^{\prime}\right)}{\sin(\frac{1}{2}\theta_{x}^{\prime})}\right], \end{array} $$

where the identity $1+2{\sum }_{k=1}^{N}\cos kx=\frac {\sin ((N+1/2)x)}{\sin (x/2)}$ is used. For convenience of presentation, we denote:

$$ \mathfrak{C}(N,\mu,\theta):=\frac{1}{2N}\frac{\sin\left( (N-\frac{1}{2})\theta\right)}{\sin\left( \frac{\theta}{2}\right)} =\frac{1}{2N}\left[\sin(N\theta)\cot\frac{\theta}{2}-\cos(N\theta)\right]. $$

Then, $\overline {\text {X}}_{j+\frac {1}{2}}(x)=\mathfrak {C}(N_{x},\mu _{x},\theta _{x})+\mathfrak {C}(N_{x},\mu _{x},\theta _{x}^{\prime })$, and the second derivative $\overline {\text {X}}_{j+\frac {1}{2}}^{\prime \prime }(x)$ can be indicated directly as:

$$ \begin{array}{@{}rcl@{}} \overline{\text{X}}_{j+\frac{1}{2}}^{\prime\prime}(x)= \mathfrak{\ddot{C}}(N_{x},\mu_{x},\theta_{x})+\mathfrak{\ddot{C}}(N_{x},\mu_{x},\theta_{x}^{\prime}), \end{array} $$

with

$$ \begin{array}{@{}rcl@{}} \mathfrak{\ddot{C}}(N,\mu,\theta)&=&-\frac{N\mu^{2}}{2}\sin(N\theta)\cot\frac{\theta}{2}-\frac{\mu^{2}}{2}\cos(N\theta)\csc^{2}\frac{\theta}{2}\\&&+ \frac{\mu^{2}}{4N}\sin(N\theta)\frac{\cos\frac{\theta}{2}}{\sin^{3}\frac{\theta}{2}}+\frac{N\mu^{2}}{2}\cos(N\theta). \end{array} $$

To obtain the second-order CSDM ${M_{2}^{x}}$, we divide the proof into two cases:

• (i) If $x=x_{p+\frac {1}{2}} \neq x_{j+\frac {1}{2}}$, the definitions of 𝜃_x and $\theta _{x}^{\prime }$ yield:

  $$ \begin{array}{@{}rcl@{}} \sin(N_{x}\theta_{x}) = \sin(N_{x}\theta_{x}^{\prime}) = 0,\ \ \cos(N_{x}\theta_{x}) = (-1)^{j+p},\ \ \cos(N_{x}\theta_{x}^{\prime}) = (-1)^{j+p+1}. \end{array} $$

  Substituting these results into $\overline {\text {X}}_{j+\frac {1}{2}}^{\prime \prime }(x)$ and utilizing the definition of μ_x, we have:

  $$ \begin{array}{@{}rcl@{}} \overline{\text{X}}_{j+\frac{1}{2}}^{\prime\prime}\left( x_{p+\frac{1}{2}}\right)&=&(-1)^{j+p}\frac{{\mu_{x}^{2}}}{2}\left( \csc^{2}\frac{\theta_{x}^{\prime}}{2}-\csc^{2}\frac{\theta_{x}}{2}\right)\\ &=&(-1)^{j+p}\frac{{\mu_{x}^{2}}}{2}\left[\csc^{2}\left( (j+p+1)\frac{\pi}{2N_{x}}\right)\right.\\ && \qquad\qquad\qquad \left. -\csc^{2}\left( (j-p)\frac{\pi}{2N_{x}}\right)\right], \quad j\neq p. \end{array} $$

• (ii) If $x=x_{p+\frac {1}{2}}=x_{j+\frac {1}{2}}$, the first part of $\overline {\text {X}}_{j+\frac {1}{2}}^{\prime \prime }(x)$, namely $\mathfrak {\ddot {C}}(N_{x},\mu _{x},\theta _{x})$ equals to:

  $$ \begin{array}{@{}rcl@{}} && \!\!\!\!\! \frac{\frac{{\mu_{x}^{2}}}{4N_{x}}\sin(N_{x}\theta_{x})\cos\frac{\theta_{x}}{2}-\frac{N_{x}{\mu_{x}^{2}}}{2}\sin(N_{x}\theta_{x})\cos\frac{\theta_{x}}{2}\sin^{2} \frac{\theta_{x}}{2}-\frac{{\mu_{x}^{2}}}{2}\cos(N_{x}\theta_{x})\cos\frac{\theta_{x}}{2} }{\sin^{3}\frac{\theta_{x}}{2}}\\ && \!\!\!\!\! +\frac{N_{x}{\mu_{x}^{2}}}{2}\cos(N_{x}\theta_{x}). \end{array} $$

Applying the Taylor’s expansion to each part of the molecule yields:

$$ \begin{array}{@{}rcl@{}} \sin(N_{x}\theta_{x})\cos\frac{\theta_{x}}{2}&=&\left( N_{x}\theta_{x} -\frac{(N_{x}\theta_{x})^{3}}{3!}+\cdots\right)\left( 1-\frac{(\frac{\theta_{x}}{2})^{2}}{2!}+\cdots\right)\\ &=&N_{x}\theta_{x}-N_{x}\left( \frac{\theta_{x}}{2}\right)^{3}-\frac{4{N_{x}^{3}}}{3}\left( \frac{\theta_{x}}{2}\right)^{3}+O({\theta_{x}^{5}}),\\ \sin(N_{x}\theta_{x})\cos\frac{\theta_{x}}{2}\sin^{2}\frac{\theta_{x}}{2}&=&\left( N_{x}\theta_{x} -\frac{(N_{x}\theta_{x})^{3}}{3!}+\cdots\right)\left( 1-\frac{\left( \frac{\theta_{x}}{2}\right)^{2}}{2!}+\cdots\right)\\ &&\times \left( \frac{\theta_{x}}{2}-\frac{\left( \frac{\theta_{x}}{2}\right)^{3}}{3!}+\cdots\right)^{2}\\ &=& 2N_{x}\left( \frac{\theta_{x}}{2}\right)^{3}+O({\theta_{x}^{5}}),\\ \cos(N_{x}\theta_{x})\sin\frac{\theta_{x}}{2}&=&\left( 1-\frac{(N_{x}\theta_{x})^{2}}{2!}+\cdots\right)\left( \frac{\theta_{x}}{2}-\frac{\left( \frac{\theta_{x}}{2}\right)^{3}}{3!}+\cdots\right)^{2}\\ &=&\frac{\theta_{x}}{2}-\frac{1}{6}\left( \frac{\theta_{x}}{2}\right)^{3}-2{N_{x}^{2}}\left( \frac{\theta_{x}}{2}\right)^{3}+O({\theta_{x}^{5}}). \end{array} $$

Since 𝜃_x → 0 as $x_{p+\frac {1}{2}} \rightarrow x_{j+\frac {1}{2}}$. We combine these above four equalities and obtain:

$$ \begin{array}{@{}rcl@{}} \mathfrak{\ddot{C}}(N_{x},\mu_{x},\theta_{x})\rightarrow-\frac{{\mu_{x}^{2}}}{6}+\frac{N_{x}{\mu_{x}^{2}}}{2}-\frac{{N_{x}^{2}}{\mu_{x}^{2}}}{3}, \quad\text{as } x_{p+\frac{1}{2}} \rightarrow x_{j+\frac{1}{2}}. \end{array} $$

Due to $\sin \frac {\theta _{x}^{\prime }}{2}\neq 0$, the second part of $\overline {\text {X}}_{j+\frac {1}{2}}^{\prime \prime }(x)$, namely $\mathfrak {\ddot {C}}(N_{x},\mu _{x},\theta _{x}^{\prime })$ gives:

$$ \begin{array}{@{}rcl@{}} \mathfrak{\ddot{C}}(N_{x},\mu_{x},\theta_{x}^{\prime})&=&-\frac{{\mu_{x}^{2}}}{2}(-1)^{j+p+1}\csc^{2}\frac{\theta_{x}^{\prime}}{2}+\frac{N_{x}{\mu_{x}^{2}}}{2}(-1)^{j+p+1}\\ &=&\frac{{\mu_{x}^{2}}}{2}\csc^{2}\left( (2j+1)\frac{\pi}{2N_{x}}\right)-\frac{N_{x}{\mu_{x}^{2}}}{2}. \end{array} $$

Combining these two parts and recalling $\overline {\text {X}}_{j+\frac {1}{2}}^{\prime \prime }(x)$, we have:

$$ \begin{array}{@{}rcl@{}} \overline{\text{X}}_{j+\frac{1}{2}}^{\prime\prime}\left( x_{p+\frac{1}{2}}\right)=\frac{{\mu_{x}^{2}}}{2}\csc^{2}\left( (2j+1)\frac{\pi}{2N_{x}}\right)-\frac{{\mu_{x}^{2}}}{6}-\frac{{N_{x}^{2}}{\mu_{x}^{2}}}{3}, \quad j=p. \end{array} $$

Cases (i) and (ii) complete the proof and obtain the claimed results in Lemma 2.3.

Appendix B: Proof of Lemma 2.4

We always demonstrate the case in the x-direction, and the other case is equally acceptable. Differentiating the interpolation basis function in (2.12) and taking $x=x_{p+\frac {1}{2}}$ yield:

$$ \begin{array}{@{}rcl@{}} ({M_{2}^{x}})_{j,p}=\frac{2}{N_{x}}\sum\limits_{l=0}^{N_{x}-1}\left[-\frac{(\mu_{x}l)^{2}}{Z_{l}} \cos\left( \mu_{x}l(x_{j+\frac{1}{2}}-a)\right)\cos\left( \mu_{x}l(x_{p+\frac{1}{2}}-a)\right)\right]. \end{array} $$

In view of spatial discretization, it is easy to find:

$$ \begin{array}{@{}rcl@{}} ({M_{2}^{x}})_{j,p}&=&\frac{2}{N_{x}}\sum\limits_{l=0}^{N_{x}-1}\left[-\frac{(\mu_{x}l)^{2}}{Z_{l}} \cos\left( \mu_{x}l\cdot (j+\frac{1}{2})h_{x}\right) \cos\left( \mu_{x}l\cdot (p+\frac{1}{2})h_{x}\right)\right]\\ &=&\frac{2}{N_{x}}\sum\limits_{l=1}^{N_{x}}\left[-\frac{\left( (l-1)\mu_{x}\right)^{2}}{Z_{l-1}} \cos \frac{\pi (j+\frac{1}{2})(l-1)}{N_{x}} \cos \frac{\pi (l-1)(p+\frac{1}{2})}{N_{x}}\right]. \end{array} $$

Denote the matrix D_x by its elements:

$$ (D_{x})_{\alpha,p}=\left( {D_{x}^{T}}\right)_{p,\alpha}=\sqrt{\frac{2}{N_{x}Z_{\alpha-1}}}\cos \frac{\pi (\alpha-1)\left( p+\frac{1}{2}\right)}{N_{x}}. $$

Then, we obtain:

$$ \begin{array}{@{}rcl@{}} ({\Lambda}_{x}D_{x})_{\alpha,p}=\sum\limits_{l=1}^{N_{x}}({\Lambda}_{x})_{\alpha,l}(D_{x})_{l,p}=({\Lambda}_{x})_{\alpha,\alpha}\sqrt{\frac{2}{N_{x}Z_{\alpha-1}}}\cos \frac{\pi (\alpha-1)\left( p+\frac{1}{2}\right)}{N_{x}}, \end{array} $$

and

$$ \begin{array}{@{}rcl@{}} ({D_{x}^{T}}{\Lambda}_{x}D_{x})_{j,p}&=&\sum\limits_{l=1}^{N_{x}}({D_{x}^{T}})_{j,l}({\Lambda}_{x}D_{x})_{l,p}\\ &=&\sum\limits_{l=1}^{N_{x}}\sqrt{\frac{2}{N_{x}Z_{l-1}}}\cos \frac{\pi (j+\frac{1}{2})(l-1)}{N_{x}}({\Lambda}_{x})_{l,l}\sqrt{\frac{2}{N_{x}Z_{l-1}}}\\ &&\times \cos \frac{\pi (l-1)\left( p+\frac{1}{2}\right)}{N_{x}}\\ &=&\frac{2}{N_{x}}\sum\limits_{l=1}^{N_{x}}\left[\frac{({\Lambda}_{x})_{l,l}}{Z_{l-1}}\cos \frac{\pi \left( j+\frac{1}{2}\right)(l-1)}{N_{x}}\cos \frac{\pi (l-1)\left( p+\frac{1}{2}\right)}{N_{x}}\right]\\&=&({M_{2}^{x}})_{j,p}. \end{array} $$

This completes the proof of x-direction and gets the claimed results (2.14) and the first assertion of (2.16). Equation (2.15) and the second assertion of (2.16) in the y-direction can be proved similarly.

Appendix C: Proof of Lemma 3.1

Similarly, we just validate the assertion in the x-direction. For convenience of proof, we multiply $-{h_{x}^{2}}$ on both two sides of the decomposition formula in Lemma 3.1. According to the definition of D_x in Lemma 2.4, we use the same way as Lemma 2.4 and obtain:

$$ \begin{array}{@{}rcl@{}} (D_{{x}}^{T}\bar{{\Lambda} }_{x}D_{x})_{j,p}&=&\sum\limits_{l=1}^{N_{x}}\sqrt{\frac{2}{N_{x}Z_{l-1}}}\cos \frac{\pi (j+\frac{1}{2})(l-1)}{N_{x}}({\Lambda}_{x})_{l,l}\sqrt{\frac{2}{N_{x}Z_{l-1}}}\\ &&\times \cos \frac{\pi (l-1)(p+\frac{1}{2})}{N_{x}}\\ &=&\frac{2}{N_{x}}\sum\limits_{l=1}^{N_{x}}\left[\frac{1}{Z_{l-1}}\cos \frac{\pi (j+\frac{1}{2})(l-1)}{N_{x}}\cos \frac{\pi (l-1)(p+\frac{1}{2})}{N_{x}}\right.\\ && \times \left. \left( 2-2\text{cos}\frac{(l-1)\pi}{N_{x}}\right)\right]. \end{array} $$

Since Z_ν = 1 + δ_0ν (δ_0ν is the Kronecker delta symbol) for any integer $\nu \in \mathbb {N}$, it is easy to check:

$$ \begin{array}{@{}rcl@{}} (D_{{x}}^{T}\bar{{\Lambda} }_{x}D_{x})_{j,p}=\frac{2}{N_{x}}\sum\limits_{l=1}^{N_{x}-1}\left[\cos \frac{\pi (j+\frac{1}{2})l}{N_{x}}\cos \frac{\pi l(p+\frac{1}{2})}{N_{x}}\left( 2-2\cos\frac{l\pi}{N_{x}}\right)\right]. \end{array} $$

We apply the prosthaphaeresis formula on the right side of the above equality and divide the results into two parts:

$$ \begin{array}{@{}rcl@{}} \vartheta_{1}:=\frac{2}{N_{x}}\sum\limits_{l=1}^{N_{x}-1}\left[\cos \frac{\pi (j+p+1)l}{N_{x}}+\cos \frac{\pi (j-p)l}{N_{x}}\right],\quad \vartheta_{2}:=\vartheta_{1}\cos\frac{l\pi}{N_{x}}. \end{array} $$

Plugging 𝜗₁ into 𝜗₂ and dividing 𝜗₂ into two parts equally, the prosthaphaeresis formula yields:

$$ \begin{array}{@{}rcl@{}} &&\vartheta_{21}:=\frac{1}{N_{x}}\sum\limits_{l=1}^{N_{x}-1}\left[\cos \frac{\pi \left( j+(p+1)+1\right)l}{N_{x}}+\cos \frac{\pi \left( j-(p+1)\right)l}{N_{x}}\right],\\ &&\vartheta_{22}:=\frac{1}{N_{x}}\sum\limits_{l=1}^{N_{x}-1}\left[\cos \frac{\pi \left( j+(p-1)+1\right)l}{N_{x}}+\cos \frac{\pi \left( j-(p-1)\right)l}{N_{x}}\right]. \end{array} $$

As a result, the equality can be reformulated as:

$$ \begin{array}{@{}rcl@{}} (D_{{x}}^{T}\bar{{\Lambda} }_{x}D_{x})_{j,p}=\vartheta_{1}-\vartheta_{21}-\vartheta_{22}, \end{array} $$

where three parts on the right side contain the similar structures. Invoking the identity:

$$ \sum\limits_{l=1}^{N_{x}-1}\cos \frac{\pi ml}{N_{x}}=\left\{ \begin{aligned} N_{x}&-1, \ \ \ \ \ \ m=0 \ \text{or} \ 2N_{x}\\ -&1,\ \ \ \ \ \ \ \ \ \ m \in (0, 2N_{x})\ \text{and is even}\\ &0,\ \ \ \ \ \ \ \ \ \ m \in (0, 2N_{x})\ \text{and is odd} \end{aligned} \right., $$

the following observations can be easily obtained:

$$ \vartheta_{1} = \left\{ \begin{aligned} &\frac{2(N_{x}-1)}{N_{x}}, \quad &j=p\\ &-\frac{2}{N_{x}},\quad &j\neq p \end{aligned} \right.,\ \ \vartheta_{21} = \left\{ \begin{aligned} &\frac{N_{x}-1}{N_{x}}, \ \ j = p + 1 \ \text{or} \ j = p = N_{x} - 1\\ & - \frac{1}{N_{x}},\ \ \ j \!\neq\! p + 1 \ \text{and} \ j\!\neq\! p = N_{x} - 1 \end{aligned} \right., $$

$$ \vartheta_{22}=\left\{ \begin{aligned} &\frac{N_{x}-1}{N_{x}}, \ \ j=p-1 \ \text{or} \ j=p=0\\ &-\frac{1}{N_{x}},\ \ \ j \neq p-1 \ \text{and} \ j\neq p=0 \end{aligned} \right.. $$

Substituting these results into the equality, we obtain the assertion in the x-direction. The assertion of y-direction can be derived equivalently. This completes the proof of Lemma 3.1.