The semidiscrete problem

7.2 Time Semidiscretization

7.2.1 The semidiscrete problem

0 < ε 1 is the singular perturbation parameter and τ > 0 is the delay parameter.

The coefficientsa= (a₁, a₂), bandcare sufficiently smooth and bounded functions such that a₁(x, y)≥α

x >0, a₂(x, y)≥α

y >0, b(x, y)≥0 and c(x, y) is nonzero on D.

Under sufficient smoothness and necessary compatibility conditions imposed on the functionsfandϕ_b, the delay parabolic IBVP (7.1.1) admits a unique solutionu(x, y, t), which exhibits a regular boundary layer of widthO(ε) along the sides x= 1 and y= 1, and a corner layer at (x, y) = (1,1).

The rest part of this chapter is organized as follows: In Section 7.2, the time semidis- cretization process through the fractional-step method has been discussed and the uni- form convergence of this scheme has been studied. Section 7.3 deals with the piecewise- uniform Shishkin mesh and the classical upwind scheme. The main proof of convergence has been provided in Section 7.4. Numerical results are presented in Section 7.5 and the chapter ends with Section 7.6, that summarizes the main conclusions.

Now, we split the differential operator Lε as Lε =Lx,ε+Ly,ε, where L_x,ε ≡ −ε ∂²

∂x² +a₁(x, y) ∂

∂x +b₁(x, y), L_y,ε ≡ −ε ∂²

∂y² +a₂(x, y) ∂

∂y +b₂(x, y),

withb(x, y) = b₁(x, y) +b₂(x, y),c(x, y) =c₁(x, y) +c₂(x, y) and f(x, y, t) = f₁(x, y, t) + f₂(x, y, t), such that

f1(x,0, t) = f1(x,1, t) =f2(0, y, t) =f2(1, y, t) = 0. (7.2.1) The operator L_x,ε can be considered as a family of one-dimensional differential op- erators with one parametery ∈(0,1) (similarly for Ly,ε).

Now, we introduce a time discretization process by means of the following fractional- step scheme:

u^−m =ϕ_b(x, y,−t_m), m= 0,· · · , p,

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(I+ ∆tL_x,ε)u^n+1/2 =uⁿ−∆tc₁(x, y)u^n+1−p+ ∆tf₁(x, y, t_n+1), y ∈(0,1), u^n+1/2(0, y) =u^n+1/2(1, y) = 0,

(7.2.2)

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(I+ ∆tLy,ε)uⁿ⁺¹ =u^n+1/2−∆tc2(x, y)u^n+1−p+ ∆tf2(x, y, tn+1), x∈(0,1), uⁿ⁺¹(x,0) = uⁿ⁺¹(x,1) = 0.

(7.2.3)

The above scheme provides the approximation uⁿ(x, y) to the solution u(x, y, t) of (7.1.1) at the time level t_n =n∆t.

The operators (I+ ∆tLi,ε), i=x, y, satisfy the following maximum principle.

Lemma 7.2.2. (Maximum principle) Let Ψ(x, y)≥0 on the boundary ∂D and (I+

∆tL_i,ε)Ψ(x, y)≥0, i=x, y on D. Then Ψ(x, y)≥0 on D.

Proof. First we prove the maximum principle for the operator (I + ∆tL_x,ε). Let (x^∗, y^∗) ∈ D be such that Ψ(x^∗, y^∗) = min_DΨ(x, y) < 0. It is clear that the point (x^∗, y^∗)∈/ ∂D, which implies (x^∗, y^∗)∈D. Since Ψ_x(x^∗, y^∗) = 0 and Ψ_xx(x^∗, y^∗)≥0, we get

(I+ ∆tL_x,ε)Ψ(x^∗, y^∗) = Ψ(x^∗, y^∗)−ε∆tΨ_xx(x^∗, y^∗) + ∆ta₁(x^∗, y^∗)Ψ_x(x^∗, y^∗) +∆tb₁(x^∗, y^∗)Ψ(x^∗, y^∗)<0,

which is a contradiction. Hence the required result follows for the operator (I+ ∆tL_x,ε).

Similarly, we can prove the maximum principle for the operator (I+ ∆tL_y,ε).

In order to analyze the convergence of the semidiscrete scheme (7.2.2)-(7.2.3), we introduce the local truncation errore_n+1 defined by e_n+1 =u(t_n+1)−buⁿ⁺¹, where buⁿ⁺¹ is the solution of the following system:

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(I+ ∆tL_x,ε)ub^n+1/2 =u(x, y, t_n)−∆tc₁(x, y)u(x, y, tb n+1−p) +∆tf₁(x, y, t_n+1), y ∈(0,1), ub^n+1/2(0, y) = bu^n+1/2(1, y) = 0,

(7.2.4)

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(I+ ∆tL_y,ε)ubⁿ⁺¹ =ub^n+1/2−∆tc₂(x, y)u(x, y, tb n+1−p) +∆tf₂(x, y, t_n+1), x∈(0,1), ubⁿ⁺¹(x,0) =buⁿ⁺¹(x,1) = 0,

(7.2.5)

whereu(x, y, tb −m) = u(x, y,−tm), form= 0,· · · , p.

Now, one can obtain the following consistency result.

Lemma 7.2.3. The local error corresponding to the scheme (7.2.4)-(7.2.5) satisfies ke_n+1k_∞ ≤C(∆t)².

Proof. Since the function bu^n+1/2 satisfies

(I+ ∆tLx,ε)ub^n+1/2 =u(x, y, tn)−∆tc1(x, y)bu(x, y, tn+1−p) + ∆tf1(x, y, tn+1), therefore, from (7.2.5), we obtain

u(x, y, t_n) = (I+ ∆tL_x,ε)(I + ∆tL_y,ε)ubⁿ⁺¹−∆tf₁(t_n+1) +c₁∆tu(x, y, tb n+1−p)

−∆t(I+ ∆tL_x,ε)f₂(t_n+1) +c₂∆t(I+ ∆tL_x,ε)u(x, y, tb n+1−p).

On the other hand, since the solution of (7.1.1) is smooth enough, we have u(t_n) = u(t_n+1−∆t) =u(t_n+1)−∆t∂u

∂t _tn+1 +

Z tn

tn+1

(t_n−s)∂²u

∂t²(s)ds

= u(t_n+1) + ∆t[(L_x,ε+L_y,ε)u(t_n+1)−(f₁(t_n+1) +f₂(t_n+1)) +c₁(x, y)u(x, y, tn+1−p) +c₂(x, y)u(x, y, tn+1−p)]

+ Z tn

tn+1

(t_n−s)∂²u

∂t²(s)ds

= (I+ ∆tL_x,ε) [(I+ ∆tL_y,ε)u(t_n+1)−∆tf₂(t_n+1)

+c2∆tu(x, y, tn+1−p)]−∆tf1(tn+1) +c1∆tu(x, y, tn+1−p) +O(∆t²).

Hence, en+1 satisfies

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(I+ ∆tL_x,ε)(I+ ∆tL_y,ε)e_n+1 =−c₂∆t(I+ ∆tL_x,ε)en+1−p−c₁∆ten+1−p+O(∆t²), e_n+1(0, y) =e_n+1(1, y) =e_n+1(x,0) =e_n+1(x,1) = 0.

(7.2.6) For the first time interval, i.e., when t_n ∈[0, τ], the delay terms of both the continuous and semidiscrete equations will take value from given initial condition. So, en+1−p = 0.

Therefore, the above equation reduces to

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(I+ ∆tL_x,ε)(I+ ∆tL_y,ε)e_n+1 =O(∆t²),

e_n+1(0, y) =e_n+1(1, y) = e_n+1(x,0) = e_n+1(x,1) = 0.

Finally, by applying the maximum principle given in Lemma 7.2.2 for the operators, (I + ∆tL_i,ε), i=x, y, we obtain

ke_n+1k_∞≤C(∆t)², (7.2.7)

fort_n∈[0, τ].

For the second time interval, i.e., when t_n ∈ (τ,2τ], we obtain the required bound by using (7.2.7) and the maximum principle (Lemma 7.2.2) on (7.2.6).

In a similar way, we can prove the required bound for tn>2τ.

Lemma 7.2.4. Under the hypothesis of Lemma 7.2.3, we have sup

n≤T /(∆t)

ku(tn)−uⁿk_∞≤C∆t.

Proof. LetE_n =u(t_n)−uⁿ. Now, subtracting and adding ubⁿ, we have

E_n=e_n+buⁿ−uⁿ=e_n+R₁En−1+R₁(−∆tc₁(x, y)en−p)+R₂(−∆tc₂(x, y)en−p), (7.2.8) whereR₁ ≡(I+ ∆tL_x,ε)⁻¹(I+ ∆tL_y,ε)⁻¹ and R₂ ≡(I+ ∆tL_y,ε)⁻¹. We knowe_n−p = 0, when the time discretization parametern varies from 0 to p. Therefore, we get

E_n =e_n+R₁En−1. Using the above recurrence, we get E_n=Pn

i=1Rⁿ⁻ⁱ₁ e_i. Now, by using the idea given in [12], we get

sup

0≤n≤p

ku(t_n)−uⁿk_∞≤C∆t. (7.2.9) For the next time interval, i.e., when n varies from p+ 1 to 2p, by using the bound given in Lemma 7.2.3 and (7.2.9) to (7.2.8), and proceeding in an analogous way as

done above, we can get the required bound. In a similar way one can get the required bound for the entire time domain Λ^M_t .

Therefore, the time semidiscretization process is uniformly convergent of first-order in time.

Before proceeding, we present a lemma which gives the bounds of the derivatives of the solution of (7.2.4). To prove that without loss of generality we assume that the initial condition of (7.2.4) is zero.

Lemma 7.2.5. The solution of (7.2.4) satisfies the following bound:

∂ⁱbu^n+1/2

∂xⁱ

≤C 1 +ε⁻ⁱexp −α

x(1−x)/ε

, (x, y)∈D, 0≤i≤4.

Proof. To prove the above bound, we will consider several cases.

Case (i). Here, we consider the case, when i = 0. By using the maximum principle (Lemma 7.2.2), it is easy to show that |bu^n+1/2| ≤C.

Case (ii). Let us take i= 1. To obtain the bound for the first derivative ofub^n+1/2 with respect tox, let

κ(x, y) = bu^n+1/2(x, y)−u(x, y, t_n)

∆t , (7.2.10)

be the solution of

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(I+ ∆tL_x,ε)κ=−L_x,εu(x, y, t_n) +f₁(x, y, t_n+1)−c₁(x, y)bu(x, y, tn+1−p), κ(0, y) =κ(1, y) = 0.

(7.2.11)

Since, the right hand side is containing a delay term, we will proceed step by step.

In the first time level, i.e., t ∈ (0, τ), since the initial condition is zero, therefore, (7.2.11) reduces to

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(I+ ∆tLx,ε)κ=−Lx,εu(x, y, tn) +f1(x, y, tn+1), κ(0, y) = κ(1, y) = 0.

Taking into account that|Lx,εu(x, y, tn)| ≤C [14, Appendix A], the maximum principle (Lemma 7.2.2) implies that |κ| ≤C.

Now, to find a bound for bu^n+1/2 we applyL_x,ε on (7.2.10). Simple calculation shows that bu^n+1/2 satisfies the following problem:

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L_x,εub^n+1/2 =−κ+f₁(x, y, t_n+1)−c₁(x, y)u(x, y, tb _n+1−p), ub^n+1/2(0, y) = bu^n+1/2(1, y) = 0.

(7.2.12)

Now, for the first time level, (0, τ), we know bu(x, y, tn+1−p) = 0. Therefore, (7.2.12) reduces to

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L_x,εbu^n+1/2 =−κ+f₁(x, y, t_n+1), ub^n+1/2(0, y) = bu^n+1/2(1, y) = 0.

By proceeding in an analogous way as in [41], we can prove that

∂ⁱub^n+1/2

∂xⁱ (0, y)

≤C, 0≤i≤1,

∂ⁱub^n+1/2

∂xⁱ (1, y)

≤Cε⁻ⁱ, 0≤i≤2,

and

∂ub^n+1/2

∂x

≤C

1 +ε⁻¹exp −α_x(1−x)/ε

. (7.2.13)

In the next time level,i.e., when t∈(τ,2τ), one can write (7.2.11) as

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(I+ ∆tL_x,ε)κ=−L_x,εu(x, y, t_n) +f₁(x, y, t_n+1)−c₁(x, y)ub_τ(x, y, tn+1−p), κ(0, y) =κ(1, y) = 0,

where,bu_τ(x, y, tn+1−p) is the semidiscrete solution in (0, τ). We know thatbu_τ(x, y, tn+1−p) is bounded. Therefore, using the same argument as done in previous time interval, we can prove that|κ| ≤C.

Fort ∈(τ,2τ), (7.2.12) can be written as

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L_x,εub^n+1/2 =−κ+f₁(x, y, t_n+1)−c₁(x, y)bu_τ(x, y, tn+1−p), bu^n+1/2(0, y) =ub^n+1/2(1, y) = 0.

Sinceub_τ(x, y, tn+1−p) is bounded, the right hand side of the above equation is bounded.

Now, by using the similar idea used in [41], one can obtain the required bound for the second time interval, i.e., t∈(τ,2τ).

In an analogous way, we can get the bound (7.2.13) for t ≥2τ as well.

Case (iii). Now, we consider the case for i= 2. To obtain the bounds for the second- order derivative of bu^n+1/2 with respect to x, we differentiate (7.2.12) with respect to x and obtain

L_x,ε

∂bu^n+1/2

∂x

= 1

∆t

∂u(x, y, t_n)

∂x −∂bu^n+1/2

∂x

+∂f₁

∂x − ∂a₁

∂x

∂ub^n+1/2

∂x

−∂b₁

∂xub^n+1/2− ∂c₁

∂xbu(x, y, tn+1−p)−c1

∂ub

∂x(x, y, tn+1−p).

By using the same idea as we applied for (7.2.13), we can obtain

∂²bu^n+1/2

∂x²

≤C

1 +ε⁻²exp −α_x(1−x)/ε . But, the above bound can only be obtained if we can prove that

1

∆t

∂u(x, y, t_n)

∂x −∂ub^n+1/2

∂x

≤C

1 +ε⁻¹exp −α

x(1−x)/ε

. (7.2.14) For that we assume, κ(x, y) =L_x,εκ, which satisfies

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(I+ ∆tL_x,ε)κ=L_x,εf₁− L²_x,εu(x, y, t_n)−c₁L_x,εbu(x, y, t_n+1−p), κ(0, y) = 1

∆t(f1(0, y, tn+1)− Lx,εu(0, y, tn)−c1bu(0, y, tn+1−p)), κ(1, y) = 1

∆t(f1(1, y, tn+1)− Lx,εu(1, y, tn)−c1bu(1, y, tn+1−p)).

(7.2.15)

In the first time level, i.e., t∈(0, τ), the IBVP (7.2.15) can be rewritten as

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(I+ ∆tL_x,ε)κ =L_x,εf₁− L²_x,εu(x, y, t_n), κ(0, y) = 1

∆t(f₁(0, y, t_n+1)− L_x,εu(0, y, t_n)), κ(1, y) = 1

∆t(f₁(1, y, t_n+1)− L_x,εu(1, y, t_n)). Now, by using the same idea used in [14], we can prove|κ| ≤C.

Now, one can verify that κ satisfies the following problem:

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Lx,εκ=κ,

κ(0, y) = κ(1, y) = 0.

So, again proceeding in a similar way as in [41], we can show that

∂κ

∂x

≤C

1 +ε⁻¹exp −α

x(1−x)/ε , and hence

∂²bu^n+1/2

∂x²

≤C

1 +ε⁻²exp −α

x(1−x)/ε .

In the next time level, i.e., whent ∈(τ,2τ), (7.2.15) can be expressed as

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(I+ ∆tLx,ε)κ=Lx,εf1− L²_x,εu(x, y, tn)−c1Lx,εubτ(x, y, tn+1−p), κ(0, y) = 1

∆t(f1(0, y, tn+1)− Lx,εu(0, y, tn)−c1buτ(0, y, tn+1−p)), κ(1, y) = 1

∆t(f₁(1, y, t_n+1)− L_x,εu(1, y, t_n)−c₁bu_τ(1, y, tn+1−p)).

Since, we know buτ(x, y, tn+1−p) is bounded, by using the same idea used in [14], it is easy to show that|κ| ≤C. By following the same way as done for previous time interval (0, τ), we can obtain the required bound.

In an analogous way, we can prove the required bound for t≥2τ.

To obtain the corresponding bounds for higher-order derivatives of ub^n+1/2 with re- spect to x, we can use the similar idea.

We decompose the solution ub^n+1/2 of (7.2.4) into the smooth and singular compo- nents, i.e., bu^n+1/2 = bv^n+1/2 + wb^n+1/2, similarly, we decompose u as u = v +w and bu(x, y, tn+1−p) as bu(x, y, tn+1−p) =bv(x, y, tn+1−p) +w(x, y, tb n+1−p).

We further express the smooth components in the form bv^n+1/2 = P3

k=0ε^kbv_k^n+1/2, v = P3

k=0ε^kvk and bv(x, y, tn+1−p) = P3

k=0ε^kbvk(x, y, tn+1−p), where they satisfy the following problem for y∈(0,1):

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(I+ ∆tL_x,ε)bv^n+1/2 =v(x, y, t_n)−∆tc₁(x, y)bv(x, y, tn+1−p) + ∆tf₁(x, y, t_n+1),

bv^n+1/2(0) = 0, bv^n+1/2(1) =

3

X

i=0

εⁱbv_i^n+1/2(1), bv(x, y, tn+1−p) =u(x, y, tb n+1−p), (7.2.16) and the singular components satisfy the following problem:

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(I+ ∆tL_x,ε)wb^n+1/2 =w(x, y, t_n)−∆tc₁(x, y)w(x, y, tb n+1−p), y∈(0,1), wb^n+1/2(0) = 0, wb^n+1/2(1) =bu^n+1/2|_x=1−bv^n+1/2(1), w(x, y, tb n+1−p) = 0.

(7.2.17) The following lemma provides the bounds for the derivatives of the smooth and singular components.

Lemma 7.2.6. The smooth componentbv^n+1/2 and the singular componentwb^n+1/2defined in (7.2.16) and (7.2.17), respectively, satisfy the following bounds:

∂ⁱbv^n+1/2

∂xⁱ _∞

≤C 1 +ε³⁻ⁱ ,

and

∂ⁱwb^n+1/2

∂xⁱ

≤Cε⁻ⁱexp(−α

x(1−x)/ε), 0≤i≤4.

Proof. Since, we have obtained the bounds of the derivatives ofub^n+1/2, one can easily obtain the required bounds by the same idea as given in Theorem 2.2.4 of Chapter 2.