# Numerical analysis of a linear second-order finite difference scheme for space-fractional Allen–Cahn equations – Advances in Continuous and Discrete Models

#### ByKai Wang, Jundong Feng, Hongbo Chen and Changling Xu

Aug 20, 2022

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. □

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