In this section, we will show that the scheme (2.7) preserves the discrete maximum principle.
Theorem 1
Assume the initial value satisfies \({\max_{\mathbf{x} \in \bar{\Omega}} |u_{0} (\mathbf{x})|} \le 1\). There exist \(\delta >0\), the fully discrete scheme (2.7) preserves the maximum principle in the sense that \(\|U^{n}\|_{\infty}\le 1\) for all \(n\geq 1\) provided that the time stepsize satisfies
$$\begin{aligned}& 0< \tau \le \frac{1}{(2+\delta )\beta -1}, \qquad \gamma \leq 0, \qquad \beta \geq \frac{3}{2}-3\gamma , \qquad \delta \geq \frac{2d\epsilon ^{2}}{\beta h^{\alpha}}, \\& 0< \tau \le \frac{1}{(2+\delta )\beta +2\gamma -1},\qquad 0< \gamma \leq \frac{1}{2},\qquad \beta \geq \frac{3}{2}-\gamma , \qquad \delta \geq \frac{2d\epsilon ^{2}}{\beta h^{\alpha}}, \\& 0< \tau \le \frac{1}{(2+\delta )\beta +2\gamma -1},\qquad \gamma > \frac{1}{2},\qquad \beta \geq 2\gamma ,\qquad \delta \geq \frac{2d\epsilon ^{2}}{\beta h^{\alpha}}, \end{aligned}$$
where d is the dimension number.
Proof
First, it follows from the assumption on \(u_{0}\) that \(\|U^{0}\|_{\infty}\le 1\). Then, as in Theorem 1 in [16], when the time stepsize τ satisfies \(0<\tau \le \min \{\frac{1}{2}, \frac{h^{\alpha}}{2d\epsilon ^{2}} \} \), we have \(\|U^{1}\|_{\infty}\leq 1\). We will prove our theorem by induction. We now assume that the result holds for \(n=m-1\) and \(n=m\), i.e., \(\|U^{m-1}\|_{\infty}\le 1\) and \(\|U^{m}\|_{\infty}\le 1\). Below, we will check that this upper bound is also true for \(n=m+1\). Next, we divide the proof into three cases:
Case I: \(\gamma \leq 0\)
It follows from the scheme (2.7) that
$$\begin{aligned} &(1-\tau )U^{m+1}+2\tau \beta U^{m+1}+2\tau \gamma \bigl(U^{m}\bigr)^{.2}U^{m+1}- \tau \epsilon ^{2}D_{h}U^{m+1} \\ &\quad =\bigl(1+\tau -(2+\delta )\tau \beta \bigr)U^{m-1}-2\tau \gamma \bigl(U^{m}\bigr)^{.2}U^{m-1}+ \bigl( \delta \tau \beta I+\tau \epsilon ^{2}D_{h} \bigr)U^{m-1} \\ &\qquad {}+ \bigl(4\tau \beta +3(4\gamma -2)\tau \bigr)U^{m}+ \tau (4 \gamma -2 ) \bigl(\bigl(U^{m}\bigr)^{.3}-3U^{m} \bigr). \end{aligned}$$
(3.1)
Suppose \(\|U^{m+1}\|_{\infty}=U^{m+1}_{p}\). The pth component of (3.1) is
$$\begin{aligned} &(1-\tau +2\tau \beta )U^{m+1}_{p}+2\gamma \tau \bigl(U^{m}_{p}\bigr)^{2}U^{m+1}_{p}- \tau \epsilon ^{2} \Biggl(\sum_{j=1}^{N}b_{pj}U^{m+1}_{j} \Biggr) \\ &\quad =\bigl(1+\tau -(2+\delta )\tau \beta \bigr)U^{m-1}_{p}-2 \tau \gamma \bigl(U^{m}_{p}\bigr)^{2}U^{m-1}_{p}+ \delta \tau \beta U^{m-1}_{p}+\tau \epsilon ^{2} \Biggl(\sum_{j=1}^{N}b_{pj}U^{m-1}_{j} \Biggr) \\ &\qquad {}+ \bigl(4\tau \beta +3(4\gamma -2)\tau \bigr)U^{m}_{p}+ \tau (4 \gamma -2 ) \bigl(\bigl(U^{m}_{p} \bigr)^{3}-3U^{m}_{p} \bigr). \end{aligned}$$
(3.2)
If \(\beta \geq \frac{1}{2}-\gamma \), we deduce that \((1-\tau +2\tau \beta )U^{m+1}_{p}+2\gamma \tau (U^{m}_{p})^{2}U^{m+1}_{p}\) and \(-\tau \epsilon ^{2} (\sum_{j=1}^{N}b_{pj}U^{m+1}_{j} )\) are non-positive or non-negative simultaneously. Then, we notice that
$$\begin{aligned} & \Biggl\vert (1-\tau +2\tau \beta )U^{m+1}_{p}+2 \gamma \tau \bigl(U^{m}_{p}\bigr)^{2}U^{m+1}_{p}- \tau \epsilon ^{2} \Biggl(\sum_{j=1}^{N}b_{pj}U^{m+1}_{j} \Biggr) \Biggr\vert \\ &\quad \geq (1-\tau +2\tau \beta ) \bigl\vert U^{m+1}_{p} \bigr\vert +2\gamma \tau \bigl(U^{m}_{p} \bigr)^{2} \bigl\vert U^{m+1}_{p} \bigr\vert . \end{aligned}$$
(3.3)
Taking the absolute value of (3.2) and using (3.3), we see that
$$\begin{aligned} &(1-\tau +2\tau \beta ) \bigl\vert U^{m+1}_{p} \bigr\vert +2\gamma \tau \bigl(U^{m}_{p} \bigr)^{2} \bigl\vert U^{m+1}_{p} \bigr\vert \\ &\quad \leq \bigl\vert \bigl(1+\tau -(2+\delta )\tau \beta \bigr)U^{m-1}_{p}-2\tau \gamma \bigl(U^{m}_{p} \bigr)^{2} U^{m-1}_{p} \bigr\vert + \Biggl\vert \delta \tau \beta U^{m-1}_{p}+ \tau \epsilon ^{2} \Biggl(\sum_{j=1}^{N}b_{pj}U^{m-1}_{j} \Biggr) \Biggr\vert \\ &\qquad {}+ \bigl\vert \bigl(4\tau \beta +3(4\gamma -2)\tau \bigr)U^{m}_{p} \bigr\vert + \bigl\vert \tau (4 \gamma -2 ) \bigl(\bigl(U^{m}_{p}\bigr)^{3}-3U^{m}_{p} \bigr) \bigr\vert . \end{aligned}$$
(3.4)
If \(\tau \leq \frac{1}{(2+\delta )\beta -1}\) and \(\beta \geq \-\frac{1}{2+\delta}\), using \(|U^{m-1}_{p}|\leq \|U^{m-1}\|_{\infty}\leq 1\), we know that
$$ \bigl\vert \bigl(1+\tau -(2+\delta )\tau \beta \bigr)U^{m-1}_{p}-2\tau \gamma \bigl(U^{m}_{p} \bigr)^{2} U^{m-1}_{p} \bigr\vert \leq 1+ \tau -(2+\delta )\tau \beta -2\tau \gamma \bigl(U^{m}_{p} \bigr)^{2}. $$
(3.5)
Let \(H=\delta \tau \beta I+\tau \epsilon ^{2}D_{h}\). If \(\delta \geq \frac{2d\epsilon ^{2}}{\beta h^{\alpha}}\), then we know from Theorem 1 in [16] that
$$ \Vert H \Vert _{\infty}\le \delta \tau \beta . $$
(3.6)
Consequently, using (3.6) and \(\|U^{m-1}\|_{\infty}\leq 1\), we can obtain
$$ \Biggl\vert \delta \tau \beta U^{m-1}_{p}+ \tau \epsilon ^{2} \Biggl(\sum_{j=1}^{N}b_{pj}U^{m-1}_{j} \Biggr) \Biggr\vert \leq \Vert H \Vert _{\infty} \bigl\Vert U^{m-1} \bigr\Vert _{\infty}\leq \delta \tau \beta . $$
(3.7)
Then, using \(|U^{m}_{p}|\leq \|U^{m}\|_{\infty}\leq 1\), if \(\beta \geq \frac{3}{2}-3\gamma \), we know that
$$ \bigl\vert \bigl(4\tau \beta +3(4\gamma -2)\tau \bigr)U^{m}_{p} \bigr\vert \leq 4\tau \beta +3(4\gamma -2)\tau . $$
(3.8)
Let \(g(x)=x^{3}-3x\). It is easy to see that \(|g(x)|\leq 2\) for \(|x|\leq 1\). Since \(\gamma \leq 0\), we deduce that
$$ \bigl\vert \tau (4\gamma -2 ) \bigl(\bigl(U^{m}_{p} \bigr)^{3}-3U^{m}_{p} \bigr) \bigr\vert \leq 4\tau -8\tau \gamma . $$
(3.9)
It follows from (3.4)–(3.9) that
$$ \bigl(1-\tau +2\tau \beta +2\gamma \tau \bigl(U^{m}_{p} \bigr)^{2}\bigr) \bigl\vert U^{m+1}_{p} \bigr\vert \le 1-\tau +2\tau \beta -2\tau \gamma \bigl(U^{m}_{p} \bigr)^{2}+4\tau \gamma , $$
(3.10)
namely,
$$ \bigl\vert U^{m+1}_{p} \bigr\vert \le 1+ \frac{4\tau \gamma (1-(U^{m}_{p})^{2})}{1-\tau +2\tau \beta +2\gamma \tau (U^{m}_{p})^{2}}. $$
(3.11)
Since \(\gamma \leq 0\) and \(|U^{m}_{p}|\leq \|U^{m}\|_{\infty}\leq 1\), we can get \(\vert U^{m+1}_{p} \vert = \Vert U^{m+1} \Vert _{\infty}\leq 1\).
Case II: \(0<\gamma \leq \frac{1}{2}\)
The scheme is the same as (3.1). If \(\tau \leq \frac{1}{(2+\delta )\beta +2\gamma -1}\) and \(\beta \geq \frac{1}{2}\), we know that
$$ \bigl\vert \bigl(1+\tau -(2+\delta )\tau \beta \bigr)U^{m-1}_{p}-2\tau \gamma \bigl(U^{m}_{p} \bigr)^{2} U^{m-1}_{p} \bigr\vert \leq 1+ \tau -(2+\delta )\tau \beta -2\tau \gamma \bigl(U^{m}_{p} \bigr)^{2}. $$
(3.12)
Then, we reestimate (3.8) as
$$ \bigl\vert \bigl(4\tau \beta +3(4\gamma -2)\tau \bigr)U^{m}_{p} \bigr\vert \leq (4\tau \beta +12\tau \gamma -6\tau ) \bigl\vert U^{m}_{p} \bigr\vert ,\quad \beta \geq \frac{3}{2}-3\gamma . $$
(3.13)
Combining (3.4), (3.7), (3.9), (3.12) with (3.13) yields
$$\begin{aligned} (1-\tau +2\tau \beta ) \bigl\vert U^{m+1}_{p} \bigr\vert +2\gamma \tau \bigl(U^{m}_{p} \bigr)^{2} \bigl\vert U^{m+1}_{p} \bigr\vert \leq {}& 1+5\tau -2\tau \beta -2\tau \gamma \bigl(U^{m}_{p} \bigr)^{2}-8 \tau \gamma \\ &{}+(4\tau \beta +12\tau \gamma -6\tau ) \bigl\vert U^{m}_{p} \bigr\vert . \end{aligned}$$
Assume that \(\Vert U^{m+1} \Vert _{\infty}>1\), then we have
$$ 4\tau \gamma \bigl\vert U^{m}_{p} \bigr\vert ^{2}-(4\tau \beta +12\tau \gamma -6\tau ) \bigl\vert U^{m}_{p} \bigr\vert -6\tau +4\tau \beta +8\tau \gamma < 0. $$
Let
$$ h(x)=4\tau \gamma x^{2}-(4\tau \beta +12\tau \gamma -6\tau )x-6\tau +4 \tau \beta +8\tau \gamma .$$
It is easy to see that \(h(1)=0\). And if \(\beta \geq \frac{3}{2}-\gamma \), we can get \(\frac{4\tau \beta +12\tau \gamma -6\tau}{8\tau \gamma}\geq 1\), which contradicts \(\|U^{m}\|_{\infty}\leq 1\). Thus, we have \(\Vert U^{m+1} \Vert _{\infty}\leq 1\).
Case III: \(\gamma >\frac{1}{2}\)
We rewrite (3.1) as
$$\begin{aligned} &(1-\tau )U^{m+1}+2\tau \beta U^{m+1}+2\tau \gamma \bigl(U^{m}\bigr)^{.2}U^{m+1}- \tau \epsilon ^{2}D_{h}U^{m+1} \\ &\quad =\bigl(1+\tau -(2+\delta )\tau \beta \bigr)U^{m-1}-2\tau \gamma \bigl(U^{m}\bigr)^{.2}U^{m-1}+ \bigl( \delta \tau \beta I+\tau \epsilon ^{2}D_{h} \bigr)U^{m-1} \\ &\qquad {}+4\tau \beta U^{m}+\tau (4\gamma -2 ) \bigl(U^{m} \bigr)^{.3}. \end{aligned}$$
Using the same technique as in Case I, if \(\beta \geq \frac{1}{2}\), \(\delta \geq \frac{2d\epsilon ^{2}}{\beta h^{\alpha}}\) and \(\tau \leq \frac{1}{(2+\delta )\beta +2\gamma -1}\), we have
$$\begin{aligned} &(1-\tau +2\tau \beta ) \bigl\vert U^{m+1}_{p} \bigr\vert +2\gamma \tau \bigl(U^{m}_{p} \bigr)^{2} \bigl\vert U^{m+1}_{p} \bigr\vert \\ &\quad \leq 1+\tau -2\tau \beta -2\tau \gamma \bigl(U^{m}_{p} \bigr)^{2} + \bigl\vert 4 \tau \beta U^{m}_{p}+ \tau (4\gamma -2) \bigl(U^{m}_{p} \bigr)^{3} \bigr\vert . \end{aligned}$$
(3.14)
Since \(\beta \geq \frac{1}{2}\) and \(\gamma >\frac{1}{2}\), using \(|U^{m}_{p}|\leq \|U^{m}\|_{\infty}\leq 1\), it is easy to obtain that
$$ \bigl\vert 4\tau \beta U^{m}_{p}+ \tau (4\gamma -2) \bigl(U^{m}_{p} \bigr)^{3} \bigr\vert \leq 4\tau \beta \bigl\vert U^{m}_{p} \bigr\vert +\tau (4 \gamma -2 ). $$
(3.15)
Following (3.14) and (3.15) immediately yields
$$\begin{aligned} &(1-\tau +2\tau \beta ) \bigl\vert U^{m+1}_{p} \bigr\vert +2\gamma \tau \bigl(U^{m}_{p} \bigr)^{2} \bigl\vert U^{m+1}_{p} \bigr\vert \\ &\quad \le 1-\tau -2\tau \beta +4\tau \gamma -2\tau \gamma \bigl(U^{m}_{p}\bigr)^{2}+4 \tau \beta \bigl\vert U^{m}_{p} \bigr\vert . \end{aligned}$$
(3.16)
Assume that \(\Vert U^{m+1} \Vert _{\infty}>1\), then (3.16) becomes
$$ \gamma \bigl\vert U^{m}_{p} \bigr\vert ^{2} -\beta \bigl\vert U^{m}_{p} \bigr\vert + \beta -\gamma < 0. $$
Let \(z(x)=\gamma x^{2} -\beta x+\beta -\gamma \). If \(\beta \geq 2\gamma \), we have \(z(x)\geq 0\), which contradicts \(\|U^{m}\|_{\infty}\leq 1\). Thus, we have \(\Vert U^{m+1} \Vert _{\infty}\leq 1\).
This completes the proof of Theorem 1. □