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 (ngeq 1) provided that the time stepsize satisfies
$$begin{aligned}& 0< tau le frac{1}{(2+delta )beta -1}, qquad gamma leq 0, qquad beta geq frac{3}{2}-3gamma , qquad delta geq frac{2depsilon ^{2}}{beta h^{alpha}}, \& 0< tau le frac{1}{(2+delta )beta +2gamma -1},qquad 0< gamma leq frac{1}{2},qquad beta geq frac{3}{2}-gamma , qquad delta geq frac{2depsilon ^{2}}{beta h^{alpha}}, \& 0< tau le frac{1}{(2+delta )beta +2gamma -1},qquad gamma > frac{1}{2},qquad beta geq 2gamma ,qquad delta geq frac{2depsilon ^{2}}{beta h^{alpha}}, end{aligned}$$
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}}{2depsilon ^{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
$$begin{aligned} &(1-tau )U^{m+1}+2tau beta U^{m+1}+2tau gamma bigl(U^{m}bigr)^{.2}U^{m+1}- tau epsilon ^{2}D_{h}U^{m+1} \ &quad =bigl(1+tau -(2+delta )tau beta bigr)U^{m-1}-2tau gamma bigl(U^{m}bigr)^{.2}U^{m-1}+ bigl( delta tau beta I+tau epsilon ^{2}D_{h} bigr)U^{m-1} \ &qquad {}+ bigl(4tau beta +3(4gamma -2)tau bigr)U^{m}+ tau (4 gamma -2 ) bigl(bigl(U^{m}bigr)^{.3}-3U^{m} bigr). end{aligned}$$
(3.1)
Suppose (|U^{m+1}|_{infty}=U^{m+1}_{p}). The pth component of (3.1) is
$$begin{aligned} &(1-tau +2tau beta )U^{m+1}_{p}+2gamma tau bigl(U^{m}_{p}bigr)^{2}U^{m+1}_{p}- tau epsilon ^{2} Biggl(sum_{j=1}^{N}b_{pj}U^{m+1}_{j} Biggr) \ &quad =bigl(1+tau -(2+delta )tau beta bigr)U^{m-1}_{p}-2 tau gamma bigl(U^{m}_{p}bigr)^{2}U^{m-1}_{p}+ delta tau beta U^{m-1}_{p}+tau epsilon ^{2} Biggl(sum_{j=1}^{N}b_{pj}U^{m-1}_{j} Biggr) \ &qquad {}+ bigl(4tau beta +3(4gamma -2)tau bigr)U^{m}_{p}+ tau (4 gamma -2 ) bigl(bigl(U^{m}_{p} bigr)^{3}-3U^{m}_{p} bigr). end{aligned}$$
(3.2)
If (beta geq frac{1}{2}-gamma ), we deduce that ((1-tau +2tau beta )U^{m+1}_{p}+2gamma 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
$$begin{aligned} & Bigglvert (1-tau +2tau beta )U^{m+1}_{p}+2 gamma tau bigl(U^{m}_{p}bigr)^{2}U^{m+1}_{p}- tau epsilon ^{2} Biggl(sum_{j=1}^{N}b_{pj}U^{m+1}_{j} Biggr) Biggrvert \ &quad geq (1-tau +2tau beta ) biglvert U^{m+1}_{p} bigrvert +2gamma tau bigl(U^{m}_{p} bigr)^{2} biglvert U^{m+1}_{p} bigrvert . end{aligned}$$
(3.3)
Taking the absolute value of (3.2) and using (3.3), we see that
$$begin{aligned} &(1-tau +2tau beta ) biglvert U^{m+1}_{p} bigrvert +2gamma tau bigl(U^{m}_{p} bigr)^{2} biglvert U^{m+1}_{p} bigrvert \ &quad leq biglvert bigl(1+tau -(2+delta )tau beta bigr)U^{m-1}_{p}-2tau gamma bigl(U^{m}_{p} bigr)^{2} U^{m-1}_{p} bigrvert + Bigglvert delta tau beta U^{m-1}_{p}+ tau epsilon ^{2} Biggl(sum_{j=1}^{N}b_{pj}U^{m-1}_{j} Biggr) Biggrvert \ &qquad {}+ biglvert bigl(4tau beta +3(4gamma -2)tau bigr)U^{m}_{p} bigrvert + biglvert tau (4 gamma -2 ) bigl(bigl(U^{m}_{p}bigr)^{3}-3U^{m}_{p} bigr) bigrvert . end{aligned}$$
(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
$$ biglvert bigl(1+tau -(2+delta )tau beta bigr)U^{m-1}_{p}-2tau gamma bigl(U^{m}_{p} bigr)^{2} U^{m-1}_{p} bigrvert leq 1+ tau -(2+delta )tau beta -2tau gamma bigl(U^{m}_{p} bigr)^{2}. $$
(3.5)
Let (H=delta tau beta I+tau epsilon ^{2}D_{h}). If (delta geq frac{2depsilon ^{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
$$ Bigglvert delta tau beta U^{m-1}_{p}+ tau epsilon ^{2} Biggl(sum_{j=1}^{N}b_{pj}U^{m-1}_{j} Biggr) Biggrvert leq Vert H Vert _{infty} biglVert U^{m-1} bigrVert _{infty}leq delta tau beta . $$
(3.7)
Then, using (|U^{m}_{p}|leq |U^{m}|_{infty}leq 1), if (beta geq frac{3}{2}-3gamma ), we know that
$$ biglvert bigl(4tau beta +3(4gamma -2)tau bigr)U^{m}_{p} bigrvert leq 4tau beta +3(4gamma -2)tau . $$
(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
$$ biglvert tau (4gamma -2 ) bigl(bigl(U^{m}_{p} bigr)^{3}-3U^{m}_{p} bigr) bigrvert leq 4tau -8tau gamma . $$
(3.9)
It follows from (3.4)–(3.9) that
$$ bigl(1-tau +2tau beta +2gamma tau bigl(U^{m}_{p} bigr)^{2}bigr) biglvert U^{m+1}_{p} bigrvert le 1-tau +2tau beta -2tau gamma bigl(U^{m}_{p} bigr)^{2}+4tau gamma , $$
(3.10)
namely,
$$ biglvert U^{m+1}_{p} bigrvert le 1+ frac{4tau gamma (1-(U^{m}_{p})^{2})}{1-tau +2tau beta +2gamma tau (U^{m}_{p})^{2}}. $$
(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 +2gamma -1}) and (beta geq frac{1}{2}), we know that
$$ biglvert bigl(1+tau -(2+delta )tau beta bigr)U^{m-1}_{p}-2tau gamma bigl(U^{m}_{p} bigr)^{2} U^{m-1}_{p} bigrvert leq 1+ tau -(2+delta )tau beta -2tau gamma bigl(U^{m}_{p} bigr)^{2}. $$
(3.12)
Then, we reestimate (3.8) as
$$ biglvert bigl(4tau beta +3(4gamma -2)tau bigr)U^{m}_{p} bigrvert leq (4tau beta +12tau gamma -6tau ) biglvert U^{m}_{p} bigrvert ,quad beta geq frac{3}{2}-3gamma . $$
(3.13)
Combining (3.4), (3.7), (3.9), (3.12) with (3.13) yields
$$begin{aligned} (1-tau +2tau beta ) biglvert U^{m+1}_{p} bigrvert +2gamma tau bigl(U^{m}_{p} bigr)^{2} biglvert U^{m+1}_{p} bigrvert leq {}& 1+5tau -2tau beta -2tau gamma bigl(U^{m}_{p} bigr)^{2}-8 tau gamma \ &{}+(4tau beta +12tau gamma -6tau ) biglvert U^{m}_{p} bigrvert . end{aligned}$$
Assume that (Vert U^{m+1} Vert _{infty}>1), then we have
$$ 4tau gamma biglvert U^{m}_{p} bigrvert ^{2}-(4tau beta +12tau gamma -6tau ) biglvert U^{m}_{p} bigrvert -6tau +4tau beta +8tau gamma < 0. $$
Let
$$ h(x)=4tau gamma x^{2}-(4tau beta +12tau gamma -6tau )x-6tau +4 tau beta +8tau gamma .$$
It is easy to see that (h(1)=0). And if (beta geq frac{3}{2}-gamma ), we can get (frac{4tau beta +12tau gamma -6tau}{8tau 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
$$begin{aligned} &(1-tau )U^{m+1}+2tau beta U^{m+1}+2tau gamma bigl(U^{m}bigr)^{.2}U^{m+1}- tau epsilon ^{2}D_{h}U^{m+1} \ &quad =bigl(1+tau -(2+delta )tau beta bigr)U^{m-1}-2tau gamma bigl(U^{m}bigr)^{.2}U^{m-1}+ bigl( delta tau beta I+tau epsilon ^{2}D_{h} bigr)U^{m-1} \ &qquad {}+4tau beta U^{m}+tau (4gamma -2 ) bigl(U^{m} bigr)^{.3}. end{aligned}$$
Using the same technique as in Case I, if (beta geq frac{1}{2}), (delta geq frac{2depsilon ^{2}}{beta h^{alpha}}) and (tau leq frac{1}{(2+delta )beta +2gamma -1}), we have
$$begin{aligned} &(1-tau +2tau beta ) biglvert U^{m+1}_{p} bigrvert +2gamma tau bigl(U^{m}_{p} bigr)^{2} biglvert U^{m+1}_{p} bigrvert \ &quad leq 1+tau -2tau beta -2tau gamma bigl(U^{m}_{p} bigr)^{2} + biglvert 4 tau beta U^{m}_{p}+ tau (4gamma -2) bigl(U^{m}_{p} bigr)^{3} bigrvert . end{aligned}$$
(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
$$ biglvert 4tau beta U^{m}_{p}+ tau (4gamma -2) bigl(U^{m}_{p} bigr)^{3} bigrvert leq 4tau beta biglvert U^{m}_{p} bigrvert +tau (4 gamma -2 ). $$
(3.15)
Following (3.14) and (3.15) immediately yields
$$begin{aligned} &(1-tau +2tau beta ) biglvert U^{m+1}_{p} bigrvert +2gamma tau bigl(U^{m}_{p} bigr)^{2} biglvert U^{m+1}_{p} bigrvert \ &quad le 1-tau -2tau beta +4tau gamma -2tau gamma bigl(U^{m}_{p}bigr)^{2}+4 tau beta biglvert U^{m}_{p} bigrvert . end{aligned}$$
(3.16)
Assume that (Vert U^{m+1} Vert _{infty}>1), then (3.16) becomes
$$ gamma biglvert U^{m}_{p} bigrvert ^{2} -beta biglvert U^{m}_{p} bigrvert + beta -gamma < 0. $$
Let (z(x)=gamma x^{2} -beta x+beta -gamma ). If (beta geq 2gamma ), 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. □
Rights and permissions
Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.
Disclaimer:
This article is autogenerated using RSS feeds and has not been created or edited by OA JF.
Click here for Source link (https://www.springeropen.com/)
