In this paper, both cases are dealt with and the transition from convection-diffusion to reaction-diffusion is examined ... O'Malley ([4],[5]), where the ratio of µ to. â.
Singularly Perturbed Problems Modelling Reaction-Convection-Diffusion Processes ∗ E. O’Riordan1 , M.L. Pickett1 , and G.I. Shishkin2 1 2
School of Mathematical Sciences, Dublin City University, Dublin, Ireland Institute for Mathematics and Mechanics, Russian Academy of Sciences, Ekaterinburg, Russia. Abstract In this paper, parameter - uniform numerical methods for singularly perturbed ordinary differential equations containing two small parameters are studied. Parameter-explicit theoretical bounds on the derivatives of the solutions are derived. A numerical algorithm based on an upwind finite difference operator and an appropriate piecewise uniform mesh is constructed. Parameter-uniform error bounds for the numerical approximations are established. Numerical results are given to illustrate the parameter-uniform convergence of the numerical approximations.
1
Introduction
Consider the following two-parameter singularly perturbed boundary value problem Lu = εu00 (x) + µa(x)u0 (x) − b(x)u(x) = u(0), u(1) given,
f (x),
x ∈ Ω = (0, 1)
(1.1)
where a, b, f ∈ C 4 (Ω), 0 < ε ≤ 1, 0 ≤ µ ≤ 1, 0 < α ≤ a(x) and 0 < β ≤ b(x). When the parameter µ = 1, the problem is the well-studied one-dimensional convection-diffusion problem ([3],[6]). In this case, a boundary layer of width O(ε) appears in a neighbourhood of the point x = 0. When the parameter √ µ = 0, the problem is called reaction-diffusion and boundary layers of width O( ε) appear at both x = 0 and x = 1. In this paper, both cases are dealt with and the transition from convection-diffusion to reaction-diffusion is examined. The asymptotic structure of the solutions to (1.1) was examined by √ O’Malley ([4],[5]), where the ratio of µ to ε was identified as significant. In this paper, a numerical method is constructed and analysed for this problem. The convergence of the numerical approximations to the exact solution is shown to be independent of the small parameters. We call such methods parameter– uniform numerical methods. Parameter–uniform methods for a two–parameter singularly perturbed problem of the form (1.1) were constructed in [10] on the basis of special fiited finite difference operators (similar to Il’in’s [2]) on uniform meshes. In [8] multi–parameter singularly perturbed parabolic reaction-diffusion equations on a strip and on a rectangle were examined. It is shown that no finite difference scheme from the natural class of fitted operator methods on a uniform mesh exists, whose numerical solutions converge uniformly with respect to the parameters. Using a condensing mesh technique, parameter–uniform finite difference schemes are constructed for this problem. In [9], it is shown that no parameter–uniform numerical method exists on a uniform rectangular mesh for a multi–parameter singularly perturbed parabolic convection-diffusion equation, whose solution contains a parabolic layer. The analysis in this paper is based on the principles laid down in [7] and in the books [3] and [1] for a single parameter singularly perturbed problem. The argument consists of establishing a maximum principle, a decomposition of the solution into regular and layer components and deriving sharp parameter-explicit bounds on these components and their derivatives. The discrete solution is decomposed into analagous components and the numerical error between the discrete and continuous components are analysed separately using discrete maximum principle, truncation error analysis and appropriate√barrier functions. √ The analysis for the two-parameter problem naturally splits into two cases: µ ≤ C ε and µ ≥ C ε. For the first case, the analysis follows closely the single parameter reaction-diffusion case. However, the second case is more intricate. We should also note the following notation ∂Ω = {0, 1},
ν (i) =
di ν , d xi
||u||Ω¯ = max |u(x)|, ¯ Ω
∗ This research was supported in part by the National Centre for Plasma Science and Technology Ireland, by the Albert College Fellowship scheme of Dublin City University, by the Enterprise Ireland grant SC–2000–070 and by the Russian Foundation for Basic Research under grant No. 01–01–01022.
and if the norm is not subscripted we can assume ||.|| = ||.||Ω¯ . Throughout this paper C (sometimes subscripted) will denote a generic constant independent of the parameters ε and µ and N (the dimension of the discrete problem). 2
Bounds on the Solution u and its Derivatives
In this section we will establish a priori bounds on the solution of (1.1) and it’s derivatives. These bounds will be used in the error analysis in later sections. We start by stating a continuous minimum principle for the differential operator in (1.1), the proof is standard. Minimum Principle. If w ∈ C 2 [0, 1] such that Lw |Ω ≤ 0 and w |∂Ω ≥ 0 then w |Ω¯ ≥ 0. Lemma 2.1. The solution u of the differential equation (1.1), satisfies the following bound ||u||Ω¯ ≤ max{|u(0)|, |u(1)|} +
1 ||f ||. β
(2.1)
Proof. Let us consider the following barrier functions 1 ||f || ± u(x). β
ψ ± (x) = max{|u(0)|, |u(1)|} +
Clearly the functions ψ ± are nonnegative at x = 0 and x = 1. Also since Lψ ± (x) = −b(x) max{|u(0)|, |u(1)|} −
b(x) ||f || ± f (x) ≤ 0, β
¯ The required we can apply the minimum principle stated above to show that ψ ± (x) ≥ 0 for all x ∈ Ω. result now follows. Lemma 2.2. The derivatives u(k) of the solution u of (1.1) satisfy the following bounds µ µ ¶k ¶ C µ ||u ||Ω¯ ≤ √ k 1 + √ max{||u||, ||f ||}, k = 1, 2 ( ε) ε µ µ ¶3 ¶ C µ (3) √ √ ||u ||Ω¯ ≤ 1+ max{||u||, ||f ||, ||f 0 ||} ( ε)3 ε where C depends only on ||a||, ||a0 ||, ||b|| and ||b0 ||. (k)
(2.2a) (2.2b)
Proof. Given any x ∈ (0, 1) we can construct a neighbourhood Nx = (p, p + r) (where r is some combination of ε and µ yet to be determined) such that x ∈ Nx and Nx ⊂ (0, 1). The mean value theorem implies that there exists y ∈ Nx such that u0 (y) =
u(p + r) − u(p) . r
From (2.1), it follows that ||u|| . r Z x 0 0 u00 (ξ) dξ u (x) = u (y) + |u0 (y)| ≤ 2
We have
y
and therefore from the original differential equation (1.1) and using integration by parts we obtain Z x Z x b(ξ)u(ξ) dξ f (ξ) dξ + ε−1 u0 (x) = u0 (y) + ε−1 y
y
µ −
ε−1
¯x Z ¯ µau¯¯ − µ y
x
¶ a0 (ξ)u(ξ) dξ .
y
Using the fact that x − y ≤ r, we have (after some calculations) the following bound µ ¶ µ ¶ 1 r µ r 1 r µ 0 |u (x)| ≤ C + + ||u|| + ||f || ≤ C + + max{||u||, ||f ||}. r ε ε ε r ε ε √ If we choose r = ε, then the right hand side of the above expression is minimised with respect to r and we obtain the required result for k = 1. Using (1.1) the differential equation for u we can obtain the required bounds for k = 2 and by differentiating (1.1) the result for k = 3 follows.
3
Decomposition of the Solution
In order to obtain parameter-uniform error estimates we will decompose the solution of (1.1) into regular and singular components. Firstly we want to show that there exists a function v (regular component) where the boundary conditions can be chosen such that ||v (i) || ≤ C for i = 0, 1, 2. √ We need to split the analysis √ into two cases depending on the ratio of µ to ε. Starting with µ ≤ C1 ε we consider the following differential equation Lv = f on (0, 1)
and
Lv = f on (0, 1).
(3.1)
We can decompose v as follows v = v0 +
√
εv1 + εv2
(3.2)
where −bv0
= f
(3.3a)
√ µ = − √ av00 − εv000 ε √ µ = − √ av10 − εv100 , ε
−bv1 Lv2 We know that
(i)
||v0 || ≤ C
if
√ and since µ ≤ C1 ε we also have (i)
||v1 || ≤ C
if
(3.3b) v2 (0) = v2 (1) = 0.
(3.3c)
¯ ¡ f ¢(i) ¯ ¯| ¯| ≤ C b
¯ ¡ f ¢(i+2) ¯ ¯| ¯| ≤ C, b
||a(i) || ≤ C,
||b(i) || ≤ C.
Hence, if we have f, b ∈ C 4 and a ∈ C 2 , we can use Lemma 2.2 in order to obtain (i) ||v2 ||
µ µ ¶i ¶µ ¶i µ ¶i µ 1 1 √ ≤C 1+ √ ≤C √ , ε ε ε
and therefore we conclude
i≤2
||v (i) || ≤ C for i = 0, 1, 2. √
In the second case µ ≥ C2 ε where C2 < C1 and C22 < α1 minΩ¯ { ab }, we consider the differential equation b v = fˆ on (0, d) Lˆ d≥1 vˆ(d) = 1, vˆ(0) chosen below (3.4) b coincides with L on the interval (0, 1) and a where the differential operator L ˆ, ˆb and fˆ are extensions of the functions a, b and f to the interval (0, d) (they have the same properties as a, b and f and also coincide with the functions on the interval (0, 1)). We extend the functions in such a way that the critical points of the functions remain unchanged i.e. ˆb ≥ β > 0, a ˆ ≥ α > 0, ||ˆ a|| = ||a||, ||ˆb|| = ||b|| and ˆ b b minΩˆ { aˆ } = minΩ { a }. Let us now decompose vˆ as follows vˆ = vˆ0 + εˆ v1 + ε2 vˆ2
(3.5)
where µˆ avˆ00 − ˆbˆ v0 0 µˆ avˆ − ˆbˆ v1 1
b v2 Lˆ We note that vˆ(0) = vˆ0 (0) + εˆ v1 (0).
= fˆ, = =
−ˆ v000 , −ˆ v100 ,
vˆ0 (d) = 1
(3.6a)
vˆ1 (d) = 0
(3.6b)
vˆ2 (0) = vˆ2 (d) = 0.
(3.6c)
Using Lemma A-2 from the appendix, we conclude that vˆ is bounded above away from x = d and imposing the condition that d > 1 we know ∃ vˆ ∈ C 3 (0, 1) such that Lˆ v = f and ||ˆ v (i) || ≤ C on (0, 1) for i = 0, 1, 2. In this case we define the regular component v as the solution to the following problem Lv = f
v(0) = vˆ(0)
v(1) = vˆ(1).
In both cases we now have the following decomposition of the solution u u = v + wL + wR
(3.7a)
v(0), v(1) chosen wL (0) = u(0) − v(0), wL (1) = 0, wR (0) = 0, wR (1) = u(1) − v(1).
(3.7b) (3.7c) (3.7d)
where Lv LwL LwR
= f, = 0, = 0,
The boundary conditions of v are chosen (as above) so that it satisfies the bounds ||v (i) || ≤ C
i = 0, 1, 2
and
||v (3) || ≤
C ε
(3.8)
and therefore we call v the regular component of the solution. The singular components wL and wR satisfy the bounds in Lemma 2.2. However, we also can obtain the following sharper bounds on the exponential character of the two components. Lemma 3.1. When the solution of (1.1) is decomposed as in (3.7a), the singular components wL and wR satisfy the following bounds |wL (x)| ≤ Ce−θ1 x (3.9a) |wR (x)| ≤ Ce−θ2 (1−x) where θ1 =
µα +
p
and θ2 =
(3.9b)
−µA +
µ2 α2 + 4εβ , 2ε
p µ2 A2 + 4εβ . 2ε
(A = ||a||Ω¯ and θ1 and θ2 are respectively the positive roots of the equations εθ12 − µαθ1 − β = 0 and εθ22 + µAθ2 − β = 0). Proof. Consider the following barrier functions ψ ± (x) = Ce−θ1 x ± wL (x), where θ1 is as stated. We find that for C large enough the functions are both nonnegative at x = 0 and x = 1 and after a simple calculation we also find that Lψ ± (x) ≤ 0. We therefore can apply the minimum principle in order to obtain |wL (x)| ≤ Ce−θ1 x . The proof in the case of wR is similar. Remark 1. The following properties of θ1 and θ2 can easily be established. They will be required in order to analyse the error in the numerical approximations to the solution. ¾ ½√ β αµ , (3.10a) θ1 ≥ max √ , ε ε √ C if µ ≤ C ε then θ2 ≥ √ , ε
√ C if µ ≥ C ε then θ2 ≥ . µ
(3.10b)
4
Discrete Problem
Consider the following upwind finite difference scheme LN U (xi ) = ²δ 2 U (xi ) + µa(xi )D+ U (xi ) − b(xi )U (xi ) = f (xi ), where D+ U (xi ) =
U (xi+1 ) − U (xi ) , xi+1 − xi
and δ 2 U (xi ) =
D− U (xi ) =
xi ∈ ΩN
(4.1)
U (xi ) − U (xi−1 ) , xi − xi−1
D+ U (xi ) − D− U (xi ) . (xi+1 − xi−1 )/2
The piecewise-uniform mesh, ΩN , on which we apply the above finite difference operator consists of two transition points ¾ ½ 1 2 , ln N σ1 = min 4 θ1 ½ ¾ 1 2 σ2 = min , ln N . 4 θ2 More specifically
( Ω
N
=
xi |xi =
4σ1 i N , σ1 + (i
1−
− N4 )H, 4σ2 σ2 + (i − 3N 4 ) N ,
) i ≤ N4 N 3N , 4 ≤i≤ 4 3N ≤ i ≤ N 4
(4.2)
where H = 1 − σ1 − σ2 . We now state a discrete comparison principle for (4.1), whose proof is standard. Discrete Minimum Principle. If W is any mesh function and LN W |ΩN ≤ 0 and W |∂ΩN ≥ 0, then W |Ω¯ N ≥ 0. We have the following discrete decomposition U = V + WL + WR
(4.3a)
where the components are the solutions of the following LN V LN W L LN WR
= f (xi ), V (0) = v(0), V (1) = v(1), = 0, WL (0) = wL (0), WL (1) = 0, = 0, WR (0) = 0, WR (1) = wR (1).
(4.3b) (4.3c) (4.3d)
We can prove the following bounds on the discrete counterparts of the singular components wL and wR . Theorem 1. We have the following bounds on WL and WR |WL (xj )| ≤ C
j Y
(1 + θL hi )−1 = ΨL,j ,
ΨL,0 = C
(4.4a)
i=1
|WR (xj )| ≤ C
N Y
(1 + θR hi )−1 = ΨR,j ,
ΨR,N = C
(4.4b)
i=j+1
where WL and WR are solutions of (4.3c) and (4.3d) respectively and hi = xi − xi−1 . The parameters θL and θR are defined to be the positive roots of the following equations. 2 2εθL − µαθL − β = 0,
2 2εθR + µAθR − β = 0,
(A = ||a||).
Proof. We start with WL . Consider ΦL,j = ΨL,j ± WL (xj ). Now LN ΦL,j = εδ 2 ΨL,j + µaD+ ΨL,j − bΨL,j ± 0, and using ΨL,j > 0,
D+ ΨL,j = −θL ΨL,j+1 < 0
and
δ 2 ΨL,j = θL 2 ΨL,j+1
hj+1 h¯j
> 0,
we obtain
hj+1 LN ΦL,j ≤ εθL 2 ΨL,j+1 ¯ − µαθL ΨL,j+1 − βΨL,j , hj
where h¯j = hj+1 + hj . Rewriting the right hand side of this equation we have µ µ ¶ ¶ hj+1 2 LN ΦL,j ≤ ΨL,j+1 2εθL 2 − 1 + (2εθ − µαθ − β) − βθ h ≤ 0. L L L j+1 2h¯j Now using the discrete minimum principle we obtain the required result. The same idea is applied to WR , we consider ΦR,j = ΨR,j ± WR (xj ). Now LN ΦR,j = εδ 2 ΨR,j + µaD+ ΨR,j − bΨR,j ± 0, and using ΨR,j ≤ ΨR,j+1 , ΨR,j > 0, D+ ΨR,j = θR ΨR,j and δ 2 ΨR,j =
hj θR 2 ΨR,j ¯ , (1 + θR hj ) hj
we obtain N
L ΦR,j
µ ¶ ΨR,j 2 hj 2 εθR ( ¯ − 2) + 2εθR + µAθR (1 + θR hj ) − β(1 + θR hj ) . ≤ (1 + θR hj ) hj
Rewriting the right hand side of this inequality we have ¶ µ ΨR,j 2 hj N 2 3 L ΦR,j ≤ εθR ( ¯ − 2) + (2εθR + µAθR − β)(1 + θR hj ) − 2εθR hj ≤ 0. (1 + θR hj ) hj
5
Error Analysis
We now wish to analyse the bounds on the error between the discrete solution and the continuous solution. ¯ N the regular component of the error satisfies the following Lemma 5.1. At each mesh point xi ∈ Ω estimate |(V − v)(xi )| ≤ CN −1 , (5.1) where v is the solution of (3.7b) and V is the solution of (4.3b). Proof. Using the usual truncation error argument and (3.8) we have |LN (V − v)(xi )| ≤ CH (ε||v 000 || + µ||v 00 ||) ≤ CH ≤ CN −1 , where H is the maximum step size. If we choose ψ ± (xi ) = C1 N −1 ± (V − v)(xi ) as our barrier functions we know that these functions are both nonnegative at x = 0 and x = 1. We also find that LN ψ ± ≤ 0 for C1 large enough and therefore we can apply the discrete minimum principle in order to obtain the required result. ¯ N the left singular component of the error satisfies the following Lemma 5.2. At each mesh point xi ∈ Ω estimate |(WL − wL )(xi )| ≤ CN −1 (ln N )2 , (5.2) where wL is the solution of (3.7c) and WL is the solution of (4.3c). Proof. We can use a classical argument in order to obtain the following truncation error bounds |LN (WL − wL )(xi )| ≤ C(hi+1 + hi ) (ε||w000 || + µ||w00 ||) . Since wL satisfies a similar equation to u, we can use Lemma 2.2 to obtain µ µ µ ³ ´2 ¶¶ ³ ´3 ¶ µ µ 1 N |L (WL − wL )(xi )| ≤ C(hi+1 + hi ) √ε 1 + √ε + ε 1 + √µε .
Simplifying the right hand side of this expression we have à |LN (WL − wL )(xi )| ≤ C(hi+1 + hi )
µ µ ¶3 ¶! 1 µ √ 1+ √ . ε ε
(5.3)
Starting with when σ1 = 14 we can show that in this case θ1 ≤ 8 ln N and therefore using (3.10a) our bound for the truncation error now becomes |LN (WL − wL )(xi )| ≤ CN −1 (ln N )2 ,
if σ1 =
1 . 4
If we choose ψ ± (xi ) = CN −1 (ln N )2 ± (WL − wL )(xi ) as our barrier functions we find that we can apply the discrete minimum principle in order to obtain |(WL − wL )(xi )| ≤ CN −1 (ln N )2 ,
if σ1 =
1 . 4
(5.4)
The next case to consider is σ1 < 14 . In this case the mesh is piecewise uniform. We firstly analyse the error in the course mesh region [σ1 , 1) and then we proceed to analyse the fine mesh on (0, σ1 ). With the course mesh region, instead of using the usual truncation error argument, we will use (3.9a) and (4.4a) to obtain the required error bounds. From (4.4a) we have N
|WL (x N )| ≤ C(1 + θL hL )− 4 4
where hL =
4σ1 N .
When σ1 < 14 , we can prove that θL hL ≥ 4N −1 ln N . We obtain the following N
|WL (x N )| ≤ C(1 + 4N −1 ln N )− 4 . 4
Using the standard inequality ln(1 + t) > t(1 − 2t ) and letting t = 4N −1 ln N , we can show that (1 + N 4N −1 ln N )− 4 ≤ 4N −1 and therefore we conclude that on the interval [σ1 , 1) we have |WL (xi )| ≤ CN −1 . Looking at the continuous solution in this region we have 2
|wL (x)| ≤ Ce−θ1 x ≤ Ce−θ1 ( θ1
ln N )
≤ CN −2 .
Combining these two results we now obtain the following error bounds |(WL − wL )(xi )| ≤ CN −1 ,
xi ∈ [σ1 , 1)
and
σ1
γ we can choose C1 such that L1 Φ± ≤ 0 and therefore we can apply (A-2) in order to obtain µ ¶ γ 1 |y(x)| ≤ C 1 + p e− µ (d−x) . (A-3) µ To derive the required bounds on the derivative of y, we decompose the solution as follows µ ¶ g(x, µ) g(d, µ) y=− + y(d) + s(x) + µz(x). k(x) k(d)
(A-4)
where L1 s = L1 z
=
0, s(d) = 1. µ ¶0 g , z(d) = 0. k
(A-5) (A-6)
γ
Starting with (A-5), we can use Φ± (x) = Ce− µ (d−x) ± s(x) as our barrier functions in order to obtain γ
L1 Φ± (x) = C(γ − k)e− µ (d−x) ± 0. Again since k > γ we find that the above expression is always negative and we therefore can apply (A-2) in order to obtain the following bound on s γ
|s(x)| ≤ Ce− µ (d−x) . Using the above bound and (A-5), we obtain |s0 (x)| ≤
C − µγ (d−x) e . µ
Next, since z satisfies a similar equation to y we have from (A-3) that µ ¶ 1 − µγ (d−x) |z(x)| ≤ C 1 + p+1 e . µ The bounds on the derivative of z can be derived using (A-6) and the above result. We obtain µ ¶ 1 − µγ (d−x) 0 |µz (x)| ≤ C 1 + p+1 e . µ Combining this with (A-4) we now have µ |y 0 (x)| ≤ C 1 +
Lemma A-2. If µ2 ≥ following bounds
γ∗ε α ,
¶ γ (d−x) −µ . e µp+1 1
(A-7)
γ ∗ < min{ ab } and fˆ, a ˆ, ˆb ∈ C 4 then the solution vˆ of (3.4) satisfies the
µ ¶ γ∗ 1 |ˆ v (i) (x)| ≤ C 1 + i e− 2µ (d−x) , µ
i = 0, 1, 2,
(A-8)
where C depends only on ||a||,||a0 ||, ||b|| and ||b0 ||. Proof. Note that vˆ = vˆ0 + εvˆ1 + ε2 vˆ2 . We first consider vˆ0 which is the solution of (3.6a). Since vˆ0 (d) = 1 ˆ and ||( faˆ )i || ≤ C for i = 0, 1 we apply Lemma A-1 with p = 0 in order to obtain µ ¶ 1 − γµ∗ (d−x) b (i) |ˆ v0 (x)| ≤ C 1 + i e , i = 0, 1, > γ∗. µ a Differentiating (3.6a) we have ˆb µ(ˆ v00 )0 − vˆ00 = a ˆ
µ ˆ¶0 f a ˆ
+
µ ˆ ¶0 b vˆ0 = g1 (x). a ˆ γ∗
(i)
In this case |ˆ v00 (d)| ≤ µc and we also know |g1 (x)| ≤ C(1 + µ1i e− µ (d−x) ) for i = 0, 1. We therefore can apply Lemma A-1 with p = 1 in order to obtain µ ¶ 1 − γµ∗ (d−x) 00 |ˆ v0 (x)| ≤ C 1 + 2 e . µ Continuing in this way (differentiating (3.6a) and applying lemma A-1 to differential equations involving derivatives of vˆ0 for the appropriate value of p) we obtain µ ¶ 1 − γµ∗ (d−x) (i) |ˆ v0 (x)| ≤ C 1 + i e i = 0, 1, 2, 3, 4. µ Next we consider vˆ1 which is the solution of (3.6b). Letting g2 (x) = − (i)
and |g2 (x)| ≤ C(1 + have the following
1 µ2+i e
∗ − γµ
(d−x)
v ˆ000 (x) a ˆ(x) ,
we find that vˆ1 (d) = 0
). We therefore start by applying Lemma A-1 with p = 2. We now
µ (i) |ˆ v1 (x)| ≤ C 1 +
1
e− µ2+i
γ∗ µ
¶ (d−x)
i = 0, 1.
As with vˆ0 , we differentiate (3.6b) in order to obtain µ(ˆ v10 )0
µ 00 ¶0 µ ˆ ¶0 ˆb vˆ b 0 − vˆ1 = − 0 + vˆ1 a ˆ a ˆ a ˆ
and applying the lemma with p = 3, we now have µ ¶ 1 − γµ∗ (d−x) (i) |ˆ v1 (x)| ≤ C 1 + 2+i e µ
i = 0, 1, 2.
µ ¶ γ∗ Finally we consider vˆ2 . Choosing Φ± (x) = C1 1 + µ14 e− 2µ (d−x) ± vˆ2 as our barrier functions we see that both are nonnegative at x = d. We also have µ ∗2 ¶ ˆγ ∗ ˆ − γ2µ∗ (d−x) ˆ ± (x) = −C1ˆb + C1 εγ + a LΦ − b e ± vˆ100 . µ4 4µ2 2 ∗
If we take µ2 ≥ γαε we can show that the above expression is negative if C1 is large enough (since γ ∗ < min{ ab }). We can therefore apply the minimum principle in order to obtain µ ¶ 1 − γ2µ∗ (d−x) |ˆ v2 (x)| ≤ C 1 + 4 e . (A-9) µ √ ˆ Given any x ∈ (0, d) we can construct a neighbourhood Nx = (p, p + ε), where x ∈ Nx and Nx ⊂ Ω. The mean value theorem implies there exists y ∈ Nx such that √ vˆ2 (p + ε) − pˆ2 (˜ a) 0 √ vˆ2 (y) = . ε Using (A-9) we now obtain µ ¶ µ ¶ √ √ γ∗ γ∗ C 1 1 C |ˆ v20 (y)| ≤ √ 1 + 4 e− 2µ (d−(p+ ε)) ≤ √ 1 + 4 e− 2µ (d−(x+2 ε)) . µ µ ε ε However this can be simplified to µ ¶ √ γ∗ ε γ∗ C 1 |ˆ v20 (y)| ≤ √ 1 + 4 e− 2µ (d−x) e µ . µ ε √ ∗ ε √ Since γ ∗ < min{ ab } and in this case µ ≥ C2 ε, we know that eγ µ ≤ C. We therefore obtain µ ¶ 1 − γ2µ∗ (d−x) C 0 |ˆ v2 (y)| ≤ √ 1 + 4 e . µ ε
From the original differential equation for vˆ2 , we have Z vˆ20 (x) = vˆ20 (y) + and using the fact that x − y ≤
y
x
vˆ200 (ξ) dξ
√
ε we find (as in the proof of Lemma 2.2) µ ¶ √ ¶µ µ+ ε 1 − γ2µ∗ (d−x) 0 |ˆ v2 (x)| ≤ C 1 + 4e . ε µ
√ Also given that µ ≥ C ε this can be simplified in order to obtain µ ¶µ ¶ γ∗ µ 1 |ˆ v20 (x)| ≤ C 1 + 4 e− 2µ (d−x) . ε µ Substituting (A-9) and (A-10) into (3.6c),we now have the following bounds for v200 ¶µ µ 2 ¶ γ 1 1 − 2µ µ (d−x) 00 1 + 4e |ˆ v2 (x)| ≤ C 2 + . ε ε µ
(A-10)
(A-11)
Finally we use the bounds for vˆ0 ,ˆ v1 and vˆ2 and their derivatives to obtain the required result. References [1] P. A. Farrell, A. F. Hegarty, J. J. H Miller, E O’Riordan and G. I. Shishkin (2000). Robust computational techniques for boundary layers. Applied Mathematics 16. [2] A. M. Il’in . Differencing scheme for a differential equation with a small parameter affecting the highest derivative. Math. Notes, 1969 6, (2), 596–602.
[3] J. J. H Miller, E O’Riordan and G. I. Shishkin (1996). Fitted Numerical Methods for Singular Perturbation Problems. World Scientific Publishing Co. Pte. Ltd.. [4] R. E. O’Malley Jr, Two–parameter singular perturbation problems for second order equations, J. Math. Mech. 16, (1967), pp. 1143-1164. [5] R. E. O’Malley Jr (1974). Introduction to singular perturbations, Academic Press, New York, . [6] H. -G. Roos, M. Stynes and L. Tobiska (1996). Numerical methods for singularly perturbed differential equations. Springer Series in Computational Mathematics 24. [7] G. I. Shishkin (1992). Discrete approximation of singularly perturbed elliptic and parabolic equations. Russian Academy of Sciences, Ural Section, Ekaterinburg. [8] G.I. Shishkin. Grid approximation of parabolic equations with small parameters multiplying the space and time derivatives. Reaction-diffusion equations. Math. Balkanica (N.S.), 1998, 12 (1-2), 179–214. [9] G.I. Shishkin. On finite difference schemes for singularly perturbed problems with an initial parabolic layer. Communications in Applied Analysis, 2001, 5 (1), 1-16. [10] G.I. Shishkin, V.A. Titov. A difference scheme for a differential equation with two small parameters at the derivatives. Chisl. Metody Meh. Sploshn. Sredy, 1976, 7 (2), 145–155. (in Russian)
