• Tidak ada hasil yang ditemukan

Convergence Analysis

Dalam dokumen Anirban Majumdar (Halaman 45-54)

The following theorem ensures that the discrete operatorLNε satisfies the discrete mini- mum principle and therefore, the proposed scheme is ε-uniformly stable.

Theorem 2.6.1. Let us suppose that N ≥N0 8, where (∥b∥+ ∆t1)

2p1α

N0p1 and N0

(lnN0)2 4σ02∥a0. (2.6.1) Then, we have

ri>0, r+i >0, for 1≤i≤N 1,

|ri|+|r+i |<|ri0|, for 1< i < N 1,

|r1+|<|r01| and |rN1|<|r0N1|. Proof: From (2.5.4) and (2.5.5), it is clear that

ri>0, for N/2≤i < N−1, ri+>0, for 1≤i < N/2,

|ri|+|r+i |<|ri0| for 1 < i < N 1,

|r1+|<|r01| and |rN1|<|rN01|. Now, we have to prove

ri>0, for 1 ≤i < N/2, and r+i >0, for N/2≤i < N 1.

From (2.5.4), one can write ri = 2εt

ˆhh ait

ˆh tN2

16σ2ln2N ∥a0Nt

4 = Nt

16σ2 [ N

ln2N − ∥a04σ02 ]

. (2.6.2)

For 8≤N0 ≤N, it is easy to show that N0

ln2N0 N

ln2N. (2.6.3)

Therefore, by using (2.6.2) and (2.6.3), we get

ri >0, for 1 ≤i < N/2, when N0

ln2N0 ≥ ∥a04σ20. Now, forN/2≤i < N 1, from (2.5.5), we have

ri+ = 2εt

ˆhihi+1 + ai+1/2t

hi+1 −bi+1/2t

2 1

2

2εt ˆhihi+1

+ αt 2hi+1

(xpi +xpi+1)

bi+1/2t

2 1

2

2εt

ˆhihi+1 + αt 2hi+1

(

xpN/2+xpN/2+1

) bi+1/2t

2 1

2

2εt

ˆhihi+1 + αt

2hi+1xpN/2+1−bi+1/2t

2 1

2. (2.6.4)

ForN ≥N0, one can easily show that xpN/2+1≥xpN

0/2+1 and also xN0/2+1=σ+ 21−σ

N0 = 2 N0 +σ

( 1 2

N0 )

2

N0. (2.6.5)

Therefore, by using (2.6.5), we can conclude thatxpN

0/2+1 ( 2

N0 )p

and αt

2hi+1xpN

0/2+1 bi+1/2t

2 1

2 αt

4N01 ( 2

N0 )p

∥b∥t

2 1

2

t 2

[2p1α N0p1 (

∥b∥+ ∆t1)]

. (2.6.6) Hence, from (2.6.4) and (2.6.6), we have

r+i >0, for N/2≤i < N−1, when 2p1α N0p1 (

∥b∥+ ∆t1) , which completes the proof.

Now, we define the truncation error as τi,ubn+1 =LNε

(ubn+1i −Ubin+1 )

= ∆i,xubn+1, where

τi,xubn+1 =



LNcenbun+1i (Lx,εbun+1) (xi), for 1≤i < N/2, LNmidbun+1i (Lx,εbun+1)i+1/2, for N/2≤i≤N 1.

Therefore, using the Taylor’s formula one can easily obtain the following lemma.

Lemma 2.6.2. The local truncation error τi,ubn+1 at an internal mesh point satisfies

i,bun+1| ≤Ct(

h2+h2ε3/2exp(−mxi/√ ε))

, 0< xi < σ, (2.6.7)

i,bun+1| ≤Ct(

εhi+1+h2i+1+εh2i+1+ε1/2exp(−mxi1/√ ε))

, xi =σ, (2.6.8) and for σ < xi <1

i,bun+1| ≤



Ct(

εhi+1+h2i+1+εh2i+1+ε1/2exp(−mxi1/√ ε))

, hi+1 ≤√ ε, Ct(

εhi+1+h2i+1+εh2i+1+hi+11 exp(−mxi1/√ ε))

, hi+1 >√ ε.

(2.6.9) Proof. We divided the proof into two cases. Case 1 and Case 2 are corresponding to the proof of the truncation error in the inner and outer regions, respectively.

Case 1 (Inner region): Following the idea given in [42], one can easily show that τi,xbun+1≤Ch

xi+1

xi−1

[ ε

d4bun+1 dx4 (s)

+

d3bun+1 dx3 (s)

]

ds, for 1 < i < N/2. (2.6.10) Now, we decompose the local truncation errorτi,ubn+1 byτi,wbn+1 andτi,cw

ln+1, whereτi,wbn+1 is the local truncation error associated tow(x) and τi,wc

ln+1 is the local truncation error associated to wl(x). From (2.6.10) and the estimates given in Lemma 2.4.2, we obtain

i,wbn+1| ≤ Ct[

h2+1/2(exp(−mxi1/√

ε)exp(−mxi+1/√ ε))]

, (2.6.11) and

i,wc

ln+1| ≤Cthε1[

exp(−mxi1/√

ε)exp(−mxi+1/√ ε)]

. (2.6.12)

Therefore, by adding (2.6.11) and (2.6.12), we get

i,ubn+1| ≤Ct[h2+1(exp(−mxi1/√

ε)exp(−mxi+1/√ ε))]

≤Ct[h2+1exp(−mxi/√

ε) sinh(mh/√ ε)]. From (2.6.1), one can conclude thatmh/√

ε≤1, thus sinh(mh/√

ε)≤Cmh/√

ε. There- fore,

i,bun+1| ≤Ct[

h2+h2ε3/2exp(−mxi/√ ε)]

, for 0 < xi < σ.

Case 2 (Outer region): For N/2 i N 1, one can easily obtain the following truncation error expression

τi,xbun+1 = 2ε hi+hi+1

[h2i+1 3!

d3ubn+1

dx3 (ξ1) h2i 3!

d3bun+1 dx3 (ξ2)

]

+ai+1/2h2i+1 3!

d3bun+1 dx3 (ξ1)

−ε 2

[

hi+1d3bun+1

dx3 (xi) h2i+1 2!

d4ubn+1 dx4 (ξ1)

]

ai+1/2h2i+1 4

d3bun+1

dx3 (ξ1),(2.6.13) by using the Taylor’s expansion, wherexi1 < ξ2 < xi and xi < ξ1 < xi+1.

Now, we are going to divide this case into two subcases, one for hi+1 ε and another one forhi+1 >√

ε.

Subcase 1 (hi+1

ε): Here, also we decompose the local truncation error τi,bun+1 by τi,wbn+1 and τi,cw

ln+1. For N/2 i N 1, by using the estimates (2.4.2) in local truncation error expression (2.6.13), we deduce that

i,wbn+1| ≤ Ct[

εhi+1+h2i+1+εh2i+1+hi+1exp(−mxi−1/√

ε) + exp(−mxi/√ ε)]

, and

i,cw

ln+1| ≤C1/2exp(−mxi1/√ ε).

Therefore, we can conclude that

i,ubn+1| ≤Ct[

εhi+1+h2i+1+εh2i+1+ε1/2exp(−mxi1/√ ε)]

. Subcase 2 (hi+1 >√

ε): For N/2< i≤ N 1, by using the estimates (2.4.2) in local truncation error expression (2.6.13) and applying the inequalityskexp(−s)≤C,s 0, we obtain

i,wbn+1| ≤ Ct[

εhi+1+h2i+1+εh2i+1+hi+1exp(−mxi1/√

ε) + exp(−mxi1/√ ε)]

. Now,

τi,cw

ln+1 = ∆t[

LNmidbun+1(xi)(

Lx,εubn+1)

(xi1/2)]

= −εt

2 [wl′′(xi+1) +w′′l(xi)]t

2 [ai+1wl(xi+1) +aiwl(xi)]

+∆trmid,i[wl(xi1)−wl(xi)] + ∆tr+mid,i[wl(xi+1)−wl(xi)]

+∆tbi+1

2 [wl(xi+1)−wl(xi)]. (2.6.14) By using the expression of wl given in Lemma 2.4.2, we get

ε

2[w′′l(xi+1) +w′′l(xi)]≤Cexp(−µxi/√

ε), (2.6.15)

1

2[ai+1wl(xi+1) +aiwl(xi)]

≤Cε1/2exp(−µxi/√

ε), (2.6.16)

1

2bi+1[wl(xi+1)−wl(xi)]

≤Cexp(−µxi/√

ε). (2.6.17)

From (2.5.7), one can establish the following inequalities rmid,i= 2ε

hi(hi+hi+1) 1, for N/2< i≤N 1, (2.6.18) and

rmid,i+ 2ε

hi+1(hi+hi+1) +ai+1/2

hi+1 + bi+1/2

2 ≤Chi+11, for N/2≤i≤N 1. (2.6.19) Therefore, by applying (2.6.15)-(2.6.19) in (2.6.14), we get

i,cw

ln+1| ≤Cthi+11 exp(−µxi1/√

ε), forN/2< i≤N 1, and hence, forN/2< i≤N 1,

i,bun+1| ≤Ct(

εhi+1+h2i+1+εh2i+1+hi+11 exp(−mxi1/√ ε))

. From (2.5.7) and by using the inequality exp(s)1≤Cs , we have

rmid,N/2[

wl(xN/21)−wl(xN/2)]

≤C 2ε

h(h+H)

[exp(−µxN/21/√

ε)exp(−µxN/2/√ ε)]

≤C 2ε

h(h+H)exp(−µxN/2/√

ε)[exp(µh/√

ε)1]

≤Cexp(−µxN/2/√

ε), (2.6.20)

and from (2.6.19), we get rmid,N/2+ [

wl(xN/2+1)−wl(xN/2)] C

H

[exp(−µxN/2+1/√

ε)exp(−µxN/2/√ ε)]

C

H exp(−µxN/2/√

ε)[exp(−µH/√

ε)1]

CH1exp(−µxN/2/√

ε). (2.6.21)

Therefore, by applying (2.6.15)-(2.6.17), (2.6.20) and (2.6.21) in (2.6.14), we conclude

N/2,cw

ln+1| ≤C1/2exp(−µxN/2/√

ε). (2.6.22)

From the estimates given in Lemma 2.4.2 and the local truncation error expression (2.6.13), we can obtain

N/2,wbn+1| ≤ Ct[

εhN/2+1+h2N/2+1+εh2N/2+1+ε1h2N/2+1exp(−mxN/2/√ ε) +ε1/2h2N/2+1exp(−mxN/21/√

ε)]

, (2.6.23)

Hence, from (2.6.22) and (2.6.23), we get

N/2,ubn+1| ≤Ct[

εhN/2+1+h2N/2+1+εh2N/2+1+ε1/2exp(−mxN/21/√ ε)]

, which is the required result.

Now, we state some technical lemmas that will be used later.

Lemma 2.6.3. Let Si be the mesh function defined by Si =

i j=1

(

1 + mhj

√ε )1

, for 1≤i≤N, (with the usual convention that S0 = 1), where m =

β. Then there exists a positive constant C such that

LNε Si







−Ct

√ε Si, for 1≤i < N/2,

Ct max{√

ε, hi+1}Si, for N/2≤i≤N 1.

Proof. It is easy to obtain that Si−Si1 = −Simhi

√ε . Now, by applying the discrete operatorLNε on Si, for 1≤i < N/2, we get

LNε Si = 2εt hi+hi+1

(Si+1−Si

hi+1 Si−Si1 hi

)

+ ∆tai

(Si+1−Si1 hi+hi+1

)

tbiSi−Si

≤ − mt

√ε+mhSi [

−m√ ε+ ai

2 +bi

√ε+mh m

]

≤ −Ct

√ε Si,

and for N/2≤i≤N 1, we obtain LNε Si = 2εt

hi+hi+1

(Si+1−Si

hi+1 Si−Si−1 hi

)

+ ∆tai+1/2

(Si+1−Si hi+1

)

tbi+1/2Si+1/2−Si+1/2

= 2εt

hi+hi+1

(

−m

√εSi+1+ m

√εSi )

tai+1/2 m

√εSi+1tbi+1/2

2 Si

(∆tbi+1/2

2 Si+1+Si+1/2 )

≤ − mt

√ε+mhSi

[

ai+1/2+ bi+1/2 2

√ε+mhi+1

m 2m√

ε hi+1 hi+hi+1

]

≤ − Ct max{√

ε, hi+1}Si.

Lemma 2.6.4. The mesh function Si satisfies the following inequality exp(−mxi/√

ε)≤Si, for all i, (2.6.24) and there exists a positive constant C, such that

SN/2 ≤CN2. (2.6.25)

Proof. Since exp(−x)(1 +x)1, x≥0, for each j, we have exp

(

−mhj

√ε )

(

1 + mhj

√ε )1

. (2.6.26)

By multiplying the inequality (2.6.26), for j = 1, 2, · · · , i, we get exp

(

−m

√ε(h1 +h2+· · ·+hi) )

i

j=1

(

1 + mhj

√ε )1

. (2.6.27)

Therefore, the inequality (2.6.24) follows by using the fact that h1+h2 +· · ·+hi =xi, in (2.6.27). The second inequality (2.6.25) can be proved by following the idea as given in [89].

Theorem 2.6.5. There exists a positive constantC such that bun+1i −Ubin+1



CN2ln2N, for 1≤i < N/2, CN2, for N/2≤i≤N 1,

where bun+1(x) is the exact solution of (2.3.2) and {Ubin+1} is the numerical solution of the problem (2.5.2).

Proof. We define two discrete functions χ±1 by

χ±1(xi) =ϕ1(xi)±ψ1(xi), for 0 ≤i≤N, (2.6.28) where

ϕ1(xi) =C[(

εhi+1+εh2i+1+h2i+1)

(1−xi) +Si1]

, for 0≤i≤N, (2.6.29) and

ψ1(xi) =ubn+1i −Ubin+1, for 0≤i≤N. (2.6.30) Thus, from Lemma 2.6.3, we have

LNε χ±1(xi)0, for 1≤i≤N 1, χ±1(x0)>0 and χ±1(xN)>0,

for sufficiently large values of C. Hence, by applying the discrete minimum principle, we get

bun+1i −Ubin+1≤ϕ1(xi), for 0 ≤i≤N. (2.6.31) Now, from (2.6.31), we can write

bun+1i −Ubin+1 ≤C[

(εH+εH2+H2) +SN/21]

=C [

(εH+εH2+H2) + (

1 + mh

√ε )

SN/2 ]

, for N/2 i N 1. Therefore, by using H 2N1,

ε N1, mh/√

ε C and (2.6.25), we deduce that

bun+1i −Ubin+1≤CN2, for N/2≤i≤N−1. (2.6.32)

Next, for 0≤i≤N/2, we choose two discrete functions χ±2 as

χ±2(xi) = ϕ2(xi)±ψ2(xi), for 0≤i≤N/2, (2.6.33) where

ϕ2(xi) =C[

N2(1−xi) +h2ε1Si]

, for 0 ≤i≤N/2, (2.6.34) and

ψ2(xi) =



 b

un+1i −Ubin+1, for 1≤i≤N/21,

0, for i= 0, N/2.

(2.6.35) Thus, from Lemma 2.6.3, we have

LNε χ±2(xi)0, for 1≤i≤N/21, χ±2(x0)>0 and χ±2(xN/2)>0,

for sufficiently large value ofC and hence, by applying the discrete minimum principle, we get

bun+1i −Ubin+1≤ϕ2(xi)≤C [

N2 + ( h

√ε )2]

≤C[

N2+N2ln2N] , for 1≤i < N/2.

Theorem 2.6.6. Let u be the exact solution of (2.1.1) and {Un} be the numerical solution of the fully discrete scheme (2.5.8) at time level tn = nt. Taking ν,t such that0< ν <1 andNν ≤Ct, there exists a positive constantC which is independent of ε, N, such that

∥u(xi, tn)−Uin



C(∆t+N2+νln2N), for 1≤i < N/2, C(∆t+N2+ν), for N/2≤i≤N 1.

Proof. The global error at time level tn =nt can be written as

∥u(xi, tn)−Uin ≤ ∥u(xi, tn)buni+buni −Ubin

+ bUin−Uin

. (2.6.36) Now, by using Lemma 2.3.2 and Theorem 2.6.5 in (2.6.36), we get

∥u(xi, tn)−Uin



Ct(∆t+N2+νlnN) + bUin−Uin

, for 1≤i < N/2, Ct(∆t+N2+ν) + bUin−Uin

, for N/2≤i≤N 1.

(2.6.37) Subtracting (2.5.8) from (2.5.3) and taking the inverse operator(

LNε )1

, we get Ubin+1−Uin+1 =(

LNε )1

(u(xi, tn)−Uin), for 1 ≤i < N/2. (2.6.38) Now, from (2.5.4) and [91, Theorem 1], we can have (

LNε )1

1. Thus, by taking supremum norm in (2.6.38) and using the fact that (

LNε )1

1, we can deduce bUin+1−Uin+1

≤ ∥u(xi, tn)−Uin, for 1≤i < N/2. (2.6.39) For then-th time level, we get

bUin−Uin

≤u(xi, tn1)−Uin1

, for 1≤i < N/2. (2.6.40) Now, we replace bUin−Uin

in (2.6.37) by the bound given in (2.6.40) and we obtain

∥u(xi, tn)−Uin≤C(∆t+N2+νln2N), for 1≤i < N/2. (2.6.41) By following the same approach, we can get the required result forN/2≤i≤N−1.

Dalam dokumen Anirban Majumdar (Halaman 45-54)

Dokumen terkait