Local truncation error

4.4 Convergence Analysis

4.4.1 Local truncation error

The local truncation error associated with (4.3.1), at an internal mesh point x_i is given by

τi =(

∆tLx,ε^N −I)

(z(xi))−(∆tLx,ε−I) (z(xi)), i= 1, 2, · · · , N −1.

We denote the remainder of the Taylor’s expression as R_n(a, p, F) = F⁽ⁿ⁺¹⁾(η)(p−a)ⁿ⁺¹

(n+ 1)! , a < η < p, (4.4.22) or

R_n(a, p, F) = 1 n!

∫p a

(p−s)ⁿF⁽ⁿ⁺¹⁾(s) ds. (4.4.23) Lemma 4.4.3. The local truncation error associated with (4.3.1) satisfies the following bound

|τi| ≤C∆thi+1

[1 +ε⁻¹exp(−mxi−1/√ ε)]

, 0< xi < σ, (4.4.24) and for σ ≤x_i <1

|τ_i| ≤





C∆th_i+1[1 +ε⁻¹exp(−mx_i−1/√

ε)], h_i+1 ≤√ ε, C∆t[

h_i+1+h⁻_i+1¹ exp(−mx_i−1/√ ε)]

, h_i+1 >√ ε.

(4.4.25)

Proof. Using the Taylor series expansion, one can obtain the local truncation error as τ_i = hi+1

2 ∆tx^p_iba₁(x_i, y)z^′′(x_i) + 2ε∆t

(h_i+h_i+1)h_iR₂(x_i, x_i−1, z) +

[ 2ε∆t

(h_i+h_i+1)h_i+1 + 1

h_i+1∆tx^p_iba₁(x_i, y) ]

R₂(x_i, x_i+1, z), (4.4.26) for 1< i < N.

Case 1. We will first consider the domain where 0 < x_i < σ. In this case, we observe

|τ_i| ≤C∆th_i+1[

1 +ε⁻¹exp(−mx_i−1/√ ε)]

, 0< x_i < σ, (4.4.27) by using Lemma 4.4.2 and the remainder (4.4.22) in the local truncation error (4.4.26).

Case 2. Now, we consider the domain σ < x_i < 1. Here we divide this case into two subcases, one for h_i+1 ≤√

ε and another for h_i+1 >√ ε.

Subcase (a). Ifhi+1 ≤√

ε, then by applying the estimates as given in Lemma 4.4.2 in the local truncation error (4.4.26), we get

|τ_i| ≤C∆th_i+1[

1 +ε⁻¹exp(−mx_i−1/√ ε)]

, σ < x_i <1, (4.4.28) whereR₂ is of the form (4.4.22).

Subcase (b). Forhi+1 >√

ε, we decompose the local truncation error τi as τ_i =τ_i+bτ_i,

where τ_i is the local truncation error associated to w(x) and bτ_i is the local truncation error associated to w_l(x). Therefore, using the estimates as given in Lemma 4.4.2 and the remainder (4.4.23), we deduce that

h_i+1

2 ∆tx^p_iba₁(x_i, y)w^′′(x_i)

≤C∆th_i+1[

1 +ε^−1/2exp(−mx_i/√ ε)]

, (4.4.29) 2ε∆tR₂(x_i, x_i−1, w)

(hi+hi+1)hi

≤ Cε∆t (hi+hi+1)hi

xi

∫

x_i−1

(x_i−1−s)²[1 +ε⁻¹exp(−ms/√ ε)] ds

≤ C∆th_i+1+C∆texp(−mx_i−1/√

ε), (4.4.30)

and [

2ε∆t

(h_i+h_i+1)h_i+1 + 1

h_i+1∆tx^p_iba₁(x_i, y) ]

R₂(x_i, x_i+1, w)

≤

[ 2ε∆t

(h_i+h_i+1)h_i+1 + 1

h_i+1∆tx^p_iba1(xi, y)]

x∫i+1 xi

(xi+1−s)²[1 +ε⁻¹exp(−ms/√ ε)] ds

≤C∆thi+1

[hi+1+ε^−1/2exp(−mxi/√ ε)]

. (4.4.31)

Therefore, by using (4.4.29), (4.4.30) and (4.4.31) in (4.4.26), we can say thatτ_i satisfies the following bound

|τ_i| ≤C∆th_i+1[

1 +ε^−1/2exp(−mx_i/√ ε)]

+C∆texp(−mx_i−1/√

ε). (4.4.32) Now, by using the expression of wl(x) as given in Lemma 4.4.2, we can express the local truncation error associated to w_l(x) as

b

τ_i = w_l(x_i)

[( 2ε∆t

(h_i+h_i+1)h_i+1 + 1

h_i+1∆tx^p_iba₁(x_i, y)) (

1−exp(−µh_i+1/√ ε))

+ 2ε∆t

(h_i+h_i+1)h_i

(1−exp(µh_i/√ ε))

− 2∆th_iµ²

(h_i+h_i+1)+ ∆tx^p_iba₁(x_i, y) µ

√ε ]

. Therefore,bτ_i satisfies the following bound

|bτ_i| ≤C ∆t

h_i+1 exp(−mx_i−1/√

ε) +C∆tε^−1/2exp(−mx_i/√

ε) +C∆texp(−mx_i/√ ε).

(4.4.33)

Now, from (4.4.32), (4.4.33) and hi =hi+1, we get

|τ_i| ≤ |τ_i|+|bτ_i|

≤C∆t[

h_i+1+h⁻_i+1¹ exp(−mx_i−1/√

ε) +ε^−1/2exp(−mx_i/√

ε) + exp(−mx_i/√ ε)]

≤C∆t [

hi+1+h⁻_i+1¹ exp(−mxi−1/√

ε) + ∆t h_i

hi

√εexp(−mhi/√

ε) exp(−mxi−1/√ ε)

]

≤C∆t [

h_i+1+ 1

h_i+1 exp(−mx_i−1/√ ε)

] .

Case 3. Here, we consider the case where x_i =σ. Again, we divide this case into two subcases, one for h_i+1 ≤√

ε and another for h_i+1 >√ ε.

Subcase (a). Forh_i+1 ≤√

ε, we proceed the same way as in Case 2 (Subcase (a)) and we get

|τ_i| ≤C∆th_i+1[

1 +ε⁻¹exp(−mx_i−1/√ ε)]

, x_i =σ. (4.4.34) Subcase (b). Forh_i+1 >√

ε, we decomposed the truncation error as in Case 2 (Subcase (b)) and using the bounds ofr_i⁻, r⁺_i , we can conclude that

|τ_i| ≤C∆t [

h_i+1+ 1

h_i+1 exp(−mx_i−1/√ ε)

]

, x_i =σ, (4.4.35) this completes the proof.

Now, we will discuss some technical lemmas which will be required to prove the error estimate.

Lemma 4.4.4. Let S_i be the mesh function defined by S_i =

∏i j=1

(

1 + mh_j

√ε )₋1

, for 1≤i≤N,

(with the usual convention that S₀ = 1). Then there exists a positive constant C such that

(∆tLx,ε^N −I)S_i ≤ − C∆t max{√

ε, h_i+1}S_i, for 1≤i≤N −1.

Proof. It is easy to obtain that S_i −S_i−1 = −S_i(mh_i/√

ε). Now, by applying the discrete operator (∆tLx,ε^N −I) on S_i, we get

(∆tLx,ε^N −I)S_i = 2∆tε h_i+h_i+1

(S_i+1−S_i

h_i+1 −S_i−S_i−1 h_i

)

+ ∆tx^p_iba₁(x_i, y)

(S_i+1−S_i h_i+1

)

−(∆tb1(xi, y) + 1)Si

≤ − m∆tS_i

√ε+mh_i+1 [

x^p_iba₁(x_i, y)− 2m√ εh_i+1

h_i+h_i+1 + (∆tb₁(x_i, y) + 1)

√ε+mh_i+1 m∆t

]

≤ − C∆t max{√

ε, hi+1}S_i,

which completes the proof.

Lemma 4.4.5. The mesh function S_i satisfies the following inequalities exp(−mx_i/√

ε)≤S_i, for 1≤i≤N, (4.4.36) and

exp(−mx_i−1/√

ε)≤CS_i, for 1≤i≤N and h_i+1 ≤√

ε. (4.4.37)

Proof. One can easily obtain (4.4.36), from the following inequality exp(−mx)≤ 1

1 +mx. Now, from (4.4.36), we have

exp(−mx_i−1/√

ε) ≤ⁱ∏⁻¹

j=1

(

1 + mh_j

√ε )₋1

=

∏i j=1

(

1 + mh_j

√ε

)₋1(

1 + mh_i

√ε )

≤ ∏ⁱ

j=1

(

1 + mh_j

√ε

)₋1(

1 + mh_i+1

√ε )

≤CS_i.

Theorem 4.4.6. For 1≤ i ≤N −1, there exists a positive constant C which is inde- pendent of y, ε, N, such that

|z(x_i)−Z_i| ≤CN⁻¹lnN, (4.4.38) where z(x) is the exact solution of (4.4.1) and {Z_i} is the numerical solution of the problem (4.3.1) obtained on the piecewise meshI_x^N.

Proof. We distinguish two situations depending on the value of h_i+1 and √ ε.

Case 1. For hi+1 ≤√

ε, we define two discrete functions as ψ^±₁(x_i) =ϕ₁(x_i)±χ

1(x_i), for 0≤i≤N, (4.4.39) where

ϕ₁(x_i) =C [

h_i+1(1−x_i) + h_i+1

√ε S_i ]

, for 0≤i≤N, (4.4.40) and

χ1(x_i) =





z(x_i)−Z_i, for 1 ≤i≤N −1,

0, for i= 0, N.

(4.4.41)

Now, by using Lemma 4.4.4 and Lemma 4.4.5, we have (∆tLx,ε^N −I)ϕ₁(x_i) ≤ −C∆t

(

h_i+1+h_i+1ε⁻¹S_i )

≤ −C∆thi+1

(

1 +ε⁻¹exp(−mxi−1/√ ε)

)

. (4.4.42) Therefore, from (4.4.39) and (4.4.42), one can show that for sufficiently large values of C,

(∆tLx,ε^N −I)ψ₁^±(x_i)≤0, for 1≤i≤N −1,

and also by (4.4.39), ψ^±₁(x_i) ≥ 0, for i = 0 and N. By using the discrete minimum principle (Lemma 4.3.1), we obtain

|z(xi)−Zi| ≤CN⁻¹lnN, for 1 ≤i≤N −1. (4.4.43) Case 2. Consider the case, when h_i+1 >√

ε. Here, we choose the discrete functions as ψ^±₂(x_i) =ϕ₂(x_i)±χ

1(x_i), for 0≤i≤N, (4.4.44) where

ϕ₂(x_i) = Ch_i+1(1−x_i) +CS_i−1,

and χ₁ is defined as in (4.4.41). Now, by applying Lemma 4.4.4 and Lemma 4.4.5, we get

(∆tLx,ε^N −I)ϕ₂(x_i) ≤ −C∆t (

h_i+1+h⁻_i+1¹S_i−1 )

≤ −C∆t (

hi+1+h⁻_i+1¹ exp(−mxi−1/√ ε)

)

. (4.4.45) From (4.4.44), it is easy to show that ψ₂^±(x_i) ≥ 0, for i = 0, N and also from (4.4.44) and (4.4.45), we have

(∆tL_x,ε^N −I)ψ₂^±(x_i)≤0, for 1≤i≤N −1,

for sufficiently large values ofC. Therefore, by applying the discrete minimum principle (Lemma 4.3.1), we obtain the required bound.

Note that, if we take N^−ν ≤C∆t with 0< ν <1, from Theorem 4.4.6, we can get

|z(xi)−Zi| ≤C∆tN^−1+νlnN, for 1 ≤i≤N −1. (4.4.46) Sincez(x_i) = bu^n+1/2(x_i, y) and Z_i =Ub_i,y^n+1/2, wherey∈I_y^N, we can write (4.4.46) as

bu^n+1/2(x_i, y)−Ub_i,y^n+1/2≤C∆tN^−1+νlnN, for 1 ≤i≤N −1. (4.4.47)

Now, for each x∈I_x^N, Ub_x,jⁿ⁺¹ satisfies the following problem













(∆tL_y,ε^N −I)Ub_{x, j}ⁿ⁺¹ =[

∆t(

εδ_y²+y^qba₂(x, y_j)D_y⁺−b₂(x, y_j))

−I] bU_{x, j}ⁿ⁺¹

=−Ub_x,j^n+1/2+ ∆tf₂(x, y_j, t_n+1), j = 1,2, · · · , N −1, Ub_x,ⁿ⁺¹₀ = 0, Ub_{x, N}ⁿ⁺¹ = 0.

(4.4.48)

Here, we cannot directly follow the same procedure as used in Theorem 4.4.6 to find the bound of

(buⁿ⁺¹(x, y_j)−Ub_x,jⁿ⁺¹ )

, because the RHS vector of (4.3.1) contains the exact solution of (4.1.1) whereas the RHS vector of (4.4.48) contains the numerical value of the previous time level. Therefore, for eachx∈I_x^N, we consider the following problem





(∆tLy,ε^N −I) ¯U_{x, j}ⁿ⁺¹ =−bu^n+1/2(x, yj) + ∆tf2(x, yj, tn+1), j = 1, · · ·, N −1, U¯_x,ⁿ⁺¹₀ = 0, U¯_{x, N}ⁿ⁺¹ = 0.

(4.4.49) Now, by following the same approach as used in Theorem 4.4.6, we can conclude that

buⁿ⁺¹(x, y_j)−U¯_x,jⁿ⁺¹≤C∆tN^−1+νlnN, for 1 ≤j ≤N −1, x∈I_x^N. (4.4.50) Now, from (4.3.4) and [91, Theorem 1], we can have

(

∆tLy,ε^N −I)₋1

∞≤1, (4.4.51)

and also, from (4.4.48) and (4.4.49), we can deduce U¯_{x, j}ⁿ⁺¹−Ub_{x, j}ⁿ⁺¹ =(

∆tLy,ε^N −I)₋1(

Ub_x,j^n+1/2−ub^n+1/2(x, y_j) )

, for 1≤j ≤N−1, (4.4.52) wherex∈I_x^N. Therefore, from (4.4.47), (4.4.50), (4.4.51), (4.4.52) and

b

uⁿ⁺¹(x, y_j)−Ub_x,jⁿ⁺¹ =buⁿ⁺¹(x, y_j)−U¯_{x, j}ⁿ⁺¹+ ¯U_{x, j}ⁿ⁺¹−Ub_x,jⁿ⁺¹, we get

buⁿ⁺¹(x, y_j)−Ub_x,jⁿ⁺¹≤C∆tN^−1+νlnN, for 1≤j ≤N −1, (4.4.53) wherex∈I_x^N.

Now, we prove the uniform convergence of fully discrete scheme, using (4.4.47) and (4.4.53).

Theorem 4.4.7. Let u be the exact solution of (4.1.1) and {Uⁿ} be the numerical solution of the fully discrete scheme (4.3.5) at time levelt_n=n∆t. For someν,∆t such that0< ν <1 andN^−ν ≤C∆t, there exists a positive constantC which is independent of ε, N, such that

u(x_i, y_j, t_n)−U_{i, j}ⁿ

∞≤C(∆t+N^−1+νlnN), for 1≤i, j ≤N−1. (4.4.54)

Proof. The global error at time level tn =n∆t can be written as u(x_i, y_j, t_n)−U_{i, j}ⁿ

∞ ≤u(x_i, y_j, t_n)−buⁿ_{i, j}

∞+buⁿ_{i, j} −Ub_{i, j}ⁿ

∞+ bU_{i, j}ⁿ −U_{i, j}ⁿ

∞. (4.4.55) By using Lemma 4.2.2 and the estimates given by (4.4.47) and (4.4.53), we get

u(x_i, y_j, t_n)−U_{i, j}ⁿ

∞ ≤C∆t(∆t+N^−1+νlnN) + bU_{i, j}ⁿ −U_{i, j}ⁿ

∞. (4.4.56) Now, from (4.3.1) and (4.4.48), we can write

(∆tLx,ε^N −I) (

∆tLy,ε^N −I) bU_{i, j}ⁿ⁺¹ = u(x_i, y_j, t_n+1)−∆tf₁(x_i, y_j, t_n+1) +∆t(

∆tL_x,ε^N −I)

f₂(x_i, y_j, t_n+1). (4.4.57) From (4.3.5), we have

(∆tLx,ε^N −I) (

∆tLy,ε^N −I)

U_{i, j}ⁿ⁺¹ = U_{i, j}ⁿ −∆tf₁(x_i, y_j, t_n+1) +∆t(

∆tL_x,ε^N −I)

f₂(x_i, y_j, t_n+1). (4.4.58) Subtracting (4.4.58) from (4.4.57) and taking the inverse operators(

∆tLx,ε^N −I)₋₁ ( and

∆tLy,ε^N −I)₋1

, we get Ub_{i, j}ⁿ⁺¹−U_{i, j}ⁿ⁺¹ =(

∆tLy,ε^N −I)₋1(

∆tLx,ε^N −I)₋1(

u(xi, yj, tn)−U_{i, j}ⁿ )

. (4.4.59) Now, taking maximum norm in (4.4.59) and by using the fact that

(

∆tLx,ε^N −I)₋1

∞≤1 and (

∆tLy,ε^N −I)₋1

∞ ≤1, we can deduce

bU_{i, j}ⁿ⁺¹−U_{i, j}ⁿ⁺¹

∞ ≤u(x_i, y_j, t_n)−U_{i, j}ⁿ

∞. (4.4.60)

Now, for the n-th time level, we get bU_{i, j}ⁿ −U_{i, j}ⁿ

∞≤u(x_i, y_j, t_n−1)−U_{i, j}ⁿ⁻¹

∞. (4.4.61)

We replace bU_{i, j}ⁿ −U_{i, j}ⁿ

∞ in (4.4.56) by the bound given in (4.4.61) and we obtain u(x_i, y_j, t_n)−U_{i, j}ⁿ

∞ ≤C(∆t+N^−1+νlnN), (4.4.62) for 1≤i, j ≤N −1, which is the required result.