Temporal second-order difference methods for solving multi-term time fractional mixed diffusion and wave equations

Abstract

This article deals with an establishment and sharp theoretical analysis of a numerical scheme devised for solving the multi-dimensional multi-term time fractional mixed diffusion and wave equations. The governing equation contains both fractional diffusion term and fractional wave term which make the numerical analysis challenging. With the help of the method of order reduction, we convert the time multi-term fractional diffusion and wave terms into the time multi-term fractional integral and diffusion terms respectively, and then develop L2-1_σ formula for solving the latter problem. In addition, the formula is used to numerically solve the time distributed-order diffusion and wave equations. The stability and convergence of these numerical schemes are rigorously analyzed by the energy method. The convergence rates are of order two in both time and space. A difference scheme on nonuniform time grids is also constructed for solving the problem with weak regularity at the initial time. Finally, we illustrate our results with some examples.

Appendices

Appendix 1: The proof of Lemma 2.3

For simplicity, we just have to prove the property of $d_{k}^{(n,\beta _{q})}$ here.

(I) Proof of (2.12). It is easy to know that

$$d_{0}^{(n,\beta_{q})}=\frac{\sigma^{2-\beta_{q}}}{2-\beta_{q}}>0,\quad d_{1}^{(n,\beta_{q})}=\frac{\sigma^{2-\beta_{q}}}{2-\beta_{q}}\Big((1+\frac{1}{\sigma})^{2-\beta_{q}}-2\Big)\geq0, $$

and

$$ \begin{array}{@{}rcl@{}} d_{n+1}^{(n,\beta_{q})}=(n+\sigma)^{1-\beta_{q}}-\frac{1}{2-\beta_{q}}\left[(n+\sigma)^{2-\beta_{q}}- (n+\sigma-1)^{2-\beta_{q}}\right]\geq0. \end{array} $$

For 2 ≤ l ≤ n, we have

$$ \begin{array}{@{}rcl@{}} d_{l}^{(n,\beta_{q})}&=&\frac{1}{2-\beta_{q}}\left[(l+\sigma)^{2-\beta_{q}}-2(l+\sigma-1)^{2-\beta_{q}}+ (l+\sigma-2)^{2-\beta_{q}}\right]\\ &\geq&(1-\beta_{q})(l+\sigma)^{-\beta_{q}}\geq0. \end{array} $$

(II) The overall structure of the proof of (2.13) is as follows.

For 3 ≤ l ≤ n we have

$$ \begin{array}{@{}rcl@{}} d_{l}^{(n,\beta_{q})}-d_{l-1}^{(n,\beta_{q})} &=&\frac{1}{2-\beta_{q}}\left[(l+\sigma)^{2-\beta_{q}}-3(l+\sigma-1)^{2-\beta_{q}}+3(l+\sigma-2)^{2-\beta_{q}}\right.\\ &&\left.\qquad\quad\ - (l+\sigma-3)^{2-\beta_{q}}\right]\\ &\leq&-\beta_{q}(1-\beta_{q})(l+\sigma)^{-1-\beta_{q}}\leq0. \end{array} $$

In addition, we have

$$ \begin{array}{@{}rcl@{}} d_{2}^{(n,\beta_{q})}-2d_{1}^{(n,\beta_{q})}&=&\frac{\sigma^{2-\beta_{q}}}{2-\beta_{q}} \left[5+ (1+\frac{2}{\sigma})^{2-\beta_{q}}-4(1+\frac{1}{\sigma})^{2-\beta_{q}}\right] \\ &=&\frac{\sigma^{2-\beta_{q}}}{2-\beta_{q}}H(x,y), \end{array} $$

(A1.1)

where

$$H(x,y)=5+(1+2x)^{y}-4(1+x)^{y},\quad x=\frac{1}{\sigma}\in(1,2),~y=2-\beta_{q}\in(1,2).$$

By tedious calculation, one can find that H(x, y) < 0. Consequently, we have $d_{2}^{(n,\beta _{q})}-2d_{1}^{(n,\beta _{q})}<0$.

$$ \begin{array}{@{}rcl@{}} 2d_{1}^{(n,\beta_{q})}-20d_{0}^{(n,\beta_{q})}=\frac{2\sigma^{2-\beta_{q}}}{2-\beta_{q}} \left[\left( 1+\frac{1}{\sigma}\right)^{2-\beta_{q}}-12\right]<0. \end{array} $$

(III) The proof of (2.14) is as follows.

For 3 ≤ l ≤ n − 1 we have

$$ \begin{array}{@{}rcl@{}} &&d_{l+1}^{(n,\beta_{q})}-2d_{l}^{(n,\beta_{q})}+d_{l-1}^{(n,\beta_{q})} \\ &=&\frac{1}{2-\beta_{q}}\left[(l+\sigma+1)^{2-\beta_{q}}-4(l+\sigma)^{2-\beta_{q}}+6(l+\sigma-1)^{2- \beta_{q}}\right.\\ &&\left.\qquad\quad\ -4(l+\sigma-2)^{2-\beta_{q}}+(l+\sigma-3)^{2-\beta_{q}}\right]\\ &\geq&(1+\beta_{q})\beta_{q}(1-\beta_{q})(l+\sigma+1)^{-2-\beta_{q}}\geq0. \end{array} $$

(IV) Through similar processing technique with (A1.1), one can find that (2.15) is also valid.

(III) Proof of (2.16).

$$ \begin{array}{@{}rcl@{}} &&{}d_{s+1}^{(s,\beta_{q})}-8d_{0}^{(s,\beta_{q})}\\ &=&(s+\sigma)^{1-\beta_{q}}-\frac{1}{2-\beta_{q}}\left[(s+\sigma)^{2-\beta_{q}}-(s+\sigma-1)^{1-\beta_{q}} \right]-8 \frac{\sigma^{2-\beta_{q}}}{2-\beta_{q}}\\ &\leq&(s+\sigma)^{1-\beta_{q}}-(s+\sigma-1)^{1-\beta_{q}}-8 \frac{\sigma^{2-\beta_{q}}}{2-\beta_{q}}\\ &\leq&(1-\beta_{q}) \sigma^{-\beta_{q}}-8 \frac{\sigma^{2-\beta_{q}}}{2-\beta_{q}}=\frac{\sigma^{- \beta_{q}}}{2-\beta_{q}}\left( (1-\beta_{q})(2-\beta_{q})-8 \sigma^{2}\right)\leq0, \end{array} $$

and

$$ \begin{array}{@{}rcl@{}} \sum\limits_{s=1}^{n}d_{s+1}^{(s,\beta_{q})} &=&\sum\limits_{s=1}^{n}\left[(s+\sigma)^{1-\beta_{q}}-\frac{1}{2-\beta_{q}}\left( (s+\sigma)^{2-\beta_{q}}-(s+\sigma-1)^{2- \beta_{q}}\right)\right]\\ &\leq&\sum\limits_{s=1}^{n}\left[(s+\sigma)^{1-\beta_{q}}-(s+\sigma-1)^{1-\beta_{q}}\right]\leq(n+\sigma)^{1-\beta_{q}}. \end{array} $$

By the definition of $\hat {d}_{k}^{(n)}$, we can easily obtain the results. This completes the proof.

Appendix 2: The proof of Lemma 3.6

Notice

$$ \begin{array}{@{}rcl@{}} &&{}\sum\limits_{s=1}^{n}\left( \sum\limits_{k=0}^{s+1}\hat{d}^{(s)}_{s-k+1}v^{k},v^{s+\sigma}\right)\\ &=&\sigma\sum\limits_{s=1}^{n}\left( \sum\limits_{k=0}^{s}\hat{d}^{(s)}_{k}v^{s-k+1}+\hat{d}^{(s)}_{s+1}v^{0}, v^{s+1}\right)\\ &&+(1-\sigma)\sum\limits_{s=1}^{n}\left( \hat{d}^{(s)}_{0}v^{s+1}+\sum\limits_{k=0}^{s-1} \hat{d}^{(s)}_{k+1}v^{s-k}+\hat{d}^{(s)}_{s+1}v^{0},v^{s}\right). \end{array} $$

(A2.1)

For the first term on the right hand side of (A2.1), we get

$$ \begin{array}{@{}rcl@{}} \text{I}&\equiv&\sum\limits_{s=1}^{n}\left( \sum\limits_{k=0}^{s}\hat{d}^{(s)}_{k}v^{s-k+1}+\hat{d}^{(s)}_{s+1}v^{0},v^{s+1}\right)\\ &=&\sum\limits_{s=1}^{n}\left( 20\hat{d}^{(s)}_{0}v^{s+1}+2\hat{d}^{(s)}_{1}v^{s}+\sum\limits_{k=2}^{s}\hat{d}^{(s)}_{k}v^{s-k+1}-19 \hat{d}^{(s)}_{0}v^{s+1}-\hat{d}^{(s)}_{1}v^{s}\right. \\ &&\qquad\left.{\vphantom{\sum\limits_{k=2}^{s}}}+\hat{d}^{(s)}_{s+1}v^{0},v^{s+1}\right)\\ &=&\sum\limits_{s=0}^{n}\left( \sum\limits_{k=0}^{s}\tilde{g}^{(s)}_{k}v^{s-k+1},v^{s+1}\right)-19\sum\limits_{s=1}^{n}\hat{d}^{(s)}_{0}\|v^{s+1}\|^{2} \\ &&+\sum\limits_{s=1}^{n}\left( \hat{d}^{(s)}_{s+1}v^{0}-\hat{d}^{(s)}_{1}v^{s},v^{s+1}\right)-\tilde{g}^{(0)}_{0}\|v^{1}\|^{2}, \end{array} $$

(A2.2)

where

$$ \tilde{g}^{(s)}_{0}=20\hat{d}^{(s)}_{0},\quad \tilde{g}^{(s)}_{1}=2\hat{d}^{(s)}_{1},\quad \tilde{g}^{(s)}_{k}=\hat{d}^{(s)}_{k} (k\geq2). $$

In view of the properties of Lemma 2.3, we see that $\{\tilde {g}^{(s)}_{k}, k\geq 0\}$ satisfies the condition of Lemma 3.4. Therefore, applying Corollary 3.1, one gets that the first term on the right hand side of (A2.2) is non-negative. Further, with the help of the Cauchy-Schwarz inequality and recalling Lemma 2.3, we find that

$$ \begin{array}{@{}rcl@{}} \text{I}&\geq&-19\sum\limits_{s=1}^{n}\hat{d}^{(s)}_{0}\|v^{s+1}\|^{2}+\sum\limits_{s=1}^{n}(\hat{d}^{(s)}_{s+1}v^{0}-\hat{d}^{(s)}_{1}v^{s}, v^{s+1})-\tilde{g}^{(0)}_{0}\|v^{1}\|^{2}\\ &\geq&-19\hat{d}^{(s)}_{0}\sum\limits_{s=1}^{n}\|v^{s+1}\|^{2}-\frac{1}{2}\sum\limits_{s=1}^{n}\hat{d}^{(s)}_{s+1}(\|v^{0}\|^{2}+\|v^{s+1} \|^{2})\\ \end{array} $$

$$ \begin{array}{@{}rcl@{}} &&-\frac{1}{2}\sum\limits_{s=1}^{n}\hat{d}^{(s)}_{1}(\|v^{s}\|^{2}+\|v^{s+1}\|^{2})-20\hat{d}^{(0)}_{0}\|v^{1}\|^{2}\\ &\geq&-28\hat{d}^{(s)}_{0}\sum\limits_{s=2}^{n+1}\|v^{s}\|^{2}-5\hat{d}^{(s)}_{0}\sum\limits_{s=1}^{n}\|v^{s}\|^{2}-\frac{1}{2}\sum\limits_{q=0}^{Q} \mu_{q} \frac{T^{1-\beta_{q}}}{\Gamma(2-\beta_{q})}\|v^{0}\|^{2}\\ &&-20\hat{d}^{(0)}_{0}\|v^{1}\|^{2}. \end{array} $$

(A2.3)

On the other hand, for the second term on the right hand side of (A2.1), we have

$$ \begin{array}{@{}rcl@{}} \text{II}&=&\sum\limits_{s=1}^{n}\left( \hat{d}^{(s)}_{0}v^{s+1}+\sum\limits_{k=0}^{s-1}\hat{d}^{(s)}_{k+1}v^{s-k}+\hat{d}^{(s)}_{s+1}v^{0}, v^{s}\right)\\ &=&\sum\limits_{s=1}^{n}\left( \sum\limits_{k=0}^{s-1}\check{g}^{(s)}_{k}v^{s-k},v^{s}\right)+\sum\limits_{s=1}^{n}(\hat{d}^{(s)}_{0}v^{s+1}- \hat{d}^{(s)}_{1}v^{s}+\hat{d}^{(s)}_{s+1}v^{0},v^{s}), \end{array} $$

(A2.4)

where

$$ \check{g}^{(s)}_{0}=2\hat{d}^{(s)}_{1},\quad \check{g}^{(s)}_{k}=\hat{d}^{(s)}_{k+1} (k\geq1). $$

Recalling Lemmas 2.3 and 3.4 again, we see that the first term on the right hand side of (A2.4) is non-negative. Further, with the similar treatment to (A2.3), we get

$$ \begin{array}{@{}rcl@{}} \text{II}&\geq&\!-\frac{1}{2}\hat{d}^{(s)}_{0}\sum\limits_{s=2}^{n+1}\|v^{s}\|^{2} - \frac{29}{2}\hat{d}^{(s)}_{0} \sum\limits_{s=1}^{n}\|v^{s}\|^{2} - \frac{1}{2}\sum\limits_{q=0}^{Q}\mu_{q} \frac{T^{1-\beta_{q}}}{\Gamma(2 - \beta_{q})}\| v^{0}\|^{2}. \end{array} $$

(A2.5)

Substituting (A2.3) and (A2.5) into (A2.1), we obtain

$$ \begin{array}{@{}rcl@{}} \sum\limits_{s=1}^{n}\left( \sum\limits_{k=0}^{s+1}\hat{d}^{(s)}_{s-k+1}v^{k},v^{s+\sigma}\right)&\geq&-3(5+6\sigma)\sum\limits_{q=0}^{Q} \mu_{q} \frac{\sigma^{2-\beta_{q}}\tau^{1-\beta_{q}}}{\Gamma(3-\beta_{q})}\sum\limits_{s=1}^{n+1}\|v^{s}\|^{2}\\ &&-\sum\limits_{q=0}^{Q}\mu_{q} \frac{T^{1-\beta_{q}}}{2{\Gamma}(2-\beta_{q})}\|v^{0}\|^{2}\\ &&-20\sigma\sum\limits_{q=0}^{Q} \mu_{q} \frac{\sigma^{2-\beta_{q}}\tau^{1-\beta_{q}}}{\Gamma(3-\beta_{q})}\|v^{1}\|^{2}. \end{array} $$

This completes the proof.