Singularly perturbed differential equations with

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In this paper a singularly perturbed reaction-diffusion equation with a dis- continuous ... with a discontinuous source term is considered on the unit interval Ω = (0,1). A ... where φ1(x),φ2(x) are the solutions of the boundary value problems.
Singularly perturbed differential equations with discontinuous source terms. ∗ P. A. Farrell



J. J. H. Miller



E. O’Riordan

§

G. I. Shishkin



Abstract In this paper a singularly perturbed reaction-diffusion equation with a discontinuous source term is examined. A numerical method is constructed for this problem which involves an appropriate piecewise-uniform mesh. The method is shown to be uniformly convergent with respect to the singular perturbation parameter. Numerical results are presented which validate the theoretical results.

1

Introduction

In this paper a singularly perturbed reaction - diffusion equation in one dimension with a discontinuous source term is considered on the unit interval Ω = (0, 1). A single discontinuity in the source term is assumed to occur at a point d ∈ Ω. It is convenient to introduce the notation Ω− = (0, d) and Ω+ = (d, 1) and to denote the jump at d in any function with [ω](d) = ω(d+) − ω(d−). The corresponding self-adjoint two point boundary value problem is     

¯ ∩ C 2 (Ω− ∪ Ω+ ) such that Find uε ∈ C 1 (Ω) −εu00ε + a(x)uε = f for all x ∈ Ω− ∪ Ω+ (Pε )  uε (0) = u0 , uε (1) = u1    f (d−) 6= f (d+), a(x) ≥ α > 0 ∗

This research was supported by the National Science Foundation under Grant DMS-9627244 and partially by the Russian Foundation for Basic Research under Grant N 98-01-00362. † Department of Mathematics and Computer Science, Kent State University, Kent, Ohio 44242, U.S.A. The work of this author was supported in part by e-mail: [email protected] ‡ Department of Mathematics, Trinity College, Dublin 2, Ireland. e-mail: [email protected] § School of Mathematical Sciences, Dublin City University, Glasnevin, Dublin 9, Ireland. e-mail: [email protected] ¶ Institute of Mathematics and Mechanics, Russian Academy of Sciences, Ekaterinburg, Russia. e-mail: [email protected] 0 To appear in Proceedings of Workshop’98, Lozenetz, Bulgaria, Aug 27–31,1998.

1

¯ \ {d}. Because f is discontinuous at d where a and f are sufficiently smooth on Ω the solution uε of (Pε ) does not necessarily have a continuous second order derivative at the point d. Thus uε ∈ 6 C 2 (Ω), but the first derivative of the solution exists and is continuous. Theorem 1 The problem (Pε ) has a solution uε ∈ C 1 (Ω) ∩ C 2 (Ω− ∪ Ω+ ). Proof: The proof is by construction. Let y1 , y2 be particular solutions of the differential equations −εy100 + a(x)y1 = f, x ∈ Ω− ,

and

− εy200 + a(x)y2 = f, x ∈ Ω+ .

Consider the function (

y(x) =

y1 (x) + (uε (0) − y1 (0))φ1 (x) + Aφ2 (x), y2 (x) + Bφ1 (x) + (uε (1) − y2 (1))φ2(x),

x ∈ Ω− x ∈ Ω+

where φ1 (x), φ2 (x) are the solutions of the boundary value problems −εφ001 + a(x)φ1 = 0, x ∈ Ω, −εφ002 + a(x)φ2 = 0, x ∈ Ω,

φ1 (0) = 1, φ1 (1) = 0 φ2 (0) = 0, φ2 (1) = 1

and A, B are constants to be chosen so that y ∈ C 1 (Ω). Note that on the open interval (0, 1), 0 < φi < 1, i = 1, 2. Thus φ1 , φ2 cannot have an internal maximum or minimum and hence φ01 < 0, φ02 > 0 x ∈ (0, 1). We wish to choose the constants A, B so that y ∈ C 1 (Ω). That is we impose y(d−) = y(d+),

and y 0 (d−) = y 0 (d+).

For the constants A, B to exist we require

φ2 (d) −φ1 (d) φ02 (d) −φ01 (d)



6= 0

This follows from observing that φ02 (d)φ1(d) − φ2 (d)φ01 (d) > 0.

2

A priori bounds on the solution and its derivatives

Let Lε denote the differential operator occurring in (Pε ), which is defined by Lε ω = −εω 00 + a(x)ω ¯ Then Lε satisfies the following minimum principle on Ω. 2

Lemma 2

¯ ∩ C 2 (Ω− ∪ Ω+ ) satisfies Suppose that a function ω ∈ C 0 (Ω) ω(0) Lε ω(x) [ω] (d) then ω(x)

≥ ≥ = ≥

0, 0, 0, 0,

ω(1) ≥ 0 for all x ∈ Ω− ∪ Ω+ [ω 0 ](d) ≤ 0 ¯ for all x ∈ Ω

¯ If ω(p) ≥ Proof: Let p be any point at which ω attains its minimum value in Ω. 0, there is nothing to prove. Suppose therefore that ω(p) < 0, then the proof is completed by showing that this leads to a contradiction. With the above assumption on the boundary values, either p ∈ Ω− ∪ Ω+ or p = d. In the first case ω 00 (p) ≥ 0 and so Lε ω(p) = −εω 00 (p) + a(p)ω(p) < 0 which is false. In the second case the argument depends on whether or not ω is differentiable at d. If ω 0 (d) does not exist, then [ω 0 ](d) 6= 0 and because ω 0 (d−) ≤ 0, ω 0 (d+) ≥ 0 it is clear that [ω 0 ](d) > 0, which is a contradiction. On the other hand if ω is differentiable at d, then ω 0 (d) = 0 and ω ∈ C 1 (Ω). Recalling that ω(d) < 0 it follows that there exists a neighbourhood Nh = (d − h, d) such that ω(x) < 0 for all x ∈ Nh . Now choose a point x1 6= d, x1 ∈ Nh such that ω(x1 ) > ω(d). It follows from the mean values theorem that, for some x2 ∈ Nh , ω 0 (x2 ) =

ω(d) − ω(x1 ) 0 d − x2 d − x2

Note also that ω(x3 ) < 0, since x3 ∈ Nh . Thus Lε ω(x3 ) = −εω 00 (x3 ) + a(x3 )ω(x3 ) < 0 which is the required contradiction. An immediate consequence of the minimum principle is the following stability result. Theorem 3 Let uε be a solution of (Pε ), then 1 kuε kΩ¯ ≤ max{|u0 |, |u1|, kf kΩ−∪Ω+ } α Proof: Put Ψ± (x) = M ± uε (x), where M = max{|u0 |, |u1|, α1 kf kΩ− ∪Ω+ }. Then clearly Ψ± (0) ≥ 0, Ψ± (1) ≥ 0 and for each x ∈ Ω− ∪ Ω+ Lε Ψ± (x) = a(x)M ± Lε uε (x) ≥ αM ± f (x) ≥ 0 3

Furthermore, since uε ∈ C 1 (Ω) [Ψ± ](d) = ±[uε ](d) = 0,

h

and

i

Ψ0± (d) = ±[u0ε ](d) = 0.

¯ which leads at It follows from the minimum principle that Ψ±(x) ≥ 0 for all x ∈ Ω, once to the desired bound on uε . An immediate consequence of this result is that the solution uε of (Pε ) is unique. To establish the parameter-robust properties of the numerical methods involved in this paper, the following decomposition of uε into smooth vε and singular wε components is required. The smooth component vε is defined as the solution of Lε vε = f vε (0) =

f (0) , a(0)

vε (d−) =

on Ω− ∪ Ω+

f (d−) , a(d)

vε (d+) =

f (d+) , a(d)

vε (1) =

f (1) a(1)

and the singular component wε is given by Lε wε = 0 on Ω− ∪ Ω+ [wε0 (d)] = −[vε0 (d)].

[wε (d)] = −[vε (d)], wε (0) = uε (0) − vε (0),

wε (1) = uε (1) − vε (1)

As in Theorem 1, the singular component wε is well defined and is given by (

wε (x) =

x ∈ Ω− x ∈ Ω+

wε (0)ψ1 (x) + A1 ψ2 (x), B1 ψ3 (x) + wε (1)ψ4 (x),

where ψi (x), i = 1, 2, 3, 4 are the solutions of the boundary value problems −εψ100 + a(x)ψ1 −εψ200 + a(x)ψ2 −εψ300 + a(x)ψ3 −εψ400 + a(x)ψ4

= 0, = 0, = 0, = 0,

x ∈ Ω− , x ∈ Ω− , x ∈ Ω+ , x ∈ Ω+ ,

ψ1 (0) = 1, ψ2 (0) = 0, ψ3 (d) = 1, ψ4 (d) = 0,

ψ1 (d) = 0 ψ2 (d) = 1 ψ3 (1) = 0 ψ4 (1) = 1

and A1 , B1 are constants to be chosen so that the jump conditions at x = d are satisfied. One can easily show that |A1 |, |B1 | ≤ C, where C is a constant independent of ε. Lemma 4 For each integer k, satisfying 0 ≤ k ≤ 4, the smooth and regular components vε and wε satisfy the bounds. (

|vε(k) (x)|



k

C(1 + ε1− 2 e1 (x)), k C(1 + ε1− 2 e2 (x)), 4

x ∈ Ω− x ∈ Ω+

(

|wε(k)(x)|



C(ε− 2 e1 (x)), k C(ε− 2 e2 (x)), k

x ∈ Ω− x ∈ Ω+

where C is a constant independent of ε and √ √ √ √ e1 (x) = e−x α/ε + e−(d−x) α/ε , e2 (x) = e−(x−d) α/ε + e−(1−x) α/ε Proof: This is found using the stability result in Theorem 2 above, and the techniques in [3]. Note that vε , wε 6∈ C 0 (Ω), but uε = vε + wε ∈ C 1 (Ω).

3

Discrete Problem

A fitted mesh method for problem (Pε ) is now introduced. On Ω− ∪ Ω+ a piecewise− uniform mesh of N mesh intervals is constructed as follows. The interval Ω is subdivided into the three subintervals [0, σ1 ],

[σ1 , d − σ1 ] and [d − σ1 , d].

for some σ1 that satisfies 0 < σ1 ≤ d4 . On [0, σ1 ] and [d − σ1 , d] a uniform mesh with N mesh-intervals is placed, while on [σ1 , d − σ1 ] has a uniform mesh with N4 mesh8 intervals. The subintervals [d, d+σ2 ], [d+σ2 , 1−σ2 ], [1−σ2 , 1] are treated analogously for some σ2 satisfying 0 < σ2 ≤ 1−d . The interior points of the mesh are denoted by 4 ΩN ε = {xi : 1 ≤ i ≤

N N − 1} ∪ {xi : + 1 ≤ i ≤ N − 1} 2 2

N

Clearly x N = d and Ωε = {xi }N 0 . 2 Note that this mesh is a uniform mesh when σ1 = d4 and σ2 = 1−d . It is fitted to 4 the singular perturbation problem (Pε ) by choosing σ1 and σ2 to be the following functions of N and ε (

) d q σ1 = min , 2 ε/α ln N 4

(

) 1−d q and σ2 = min , 2 ε/α ln N 4

N

On the piecewise-uniform mesh Ωε a standard centred finite difference operator is used. Then the fitted mesh method for (Pε ) is      

(PεN )     

Find a mesh function Uε such that 2 N LN ε Uε = −εδ Uε (xi ) + a(xi )Uε (xi ) = f (xi ) for all xi ∈ Ωε Uε (0) = u0 , Uε (1) = u1 D−Uε (x N ) = D+Uε (x N ) 2

2

5

where δ 2 Zi = (

Zi+1 − Zi Zi − Zi−1 1 − ) xi+1 − xi xi − xi−1 xi+1 − xi−1

The following lemma shows that the finite difference operator LN ε has properties analogous to those of the differential operator Lε . Lemma 5

Suppose that a mesh function W satisfies W (0) ≥ 0,

W (1) ≥ 0,

N LN ε W (xi ) ≥ 0 for all xi ∈ Ωε ,

and D+ W (x N ) − D− W (x N ) ≤ 0, 2

2

N

then W (xi ) ≥ 0

for all xi ∈ Ωε . N

Proof : Let xp be any point at which W (xp ) attains its minimum value on Ωε . If W (xp ) ≥ 0 there is nothing to prove. Suppose therefore that W (xp ) < 0, then the proof is completed by showing that this leads to a contradiction. Because W attains its minimum value at xp it is clear that D− W (xp ) ≤ 0 ≤ D+ W (xp ) and hence 2 LN ε W (xp ) = −εδ W (xp ) + a(xp )W (xp ) < 0

If xp ∈ ΩN ε this leads to a contradiction. Because of the boundary values, the only other possibility, is that xp = x N . Then 2

D− W (x N ) ≤ 0 ≤ D+ W (x N ) ≤ D− W (x N ) 2

2

2

and so W (x N −1 ) = W (x N ) = W (x N +1 ) < 0. 2

2

2

Then LN ε W (x N −1 ) < 0 2

which provides the desired contradiction.

4

Truncation error analysis

¯N The error at each mesh point xi ∈ Ω ε is denoted by e(xi ) = Uε (xi ) − uε (xi ). 6

Now, we examine the truncation error at the interior mesh points ΩN ε N N N LN ε e(xi ) = Lε (Uε (xi ) − uε (xi )) = f (xi ) − Lε uε (xi ) = (Lε − Lε )uε (xi ) d2 d2 = −ε( 2 − δ 2 )uε (xi ) = −ε( 2 − δ 2 )(vε + wε )(xi ) dx dx

Now, by classical estimates (see [2] p21), for all 1 ≤ i < | − ε(

N 2

√ −1 d2 ε 2 − δ )v (x )| ≤ − x )|v | ≤ C εN (x ε i i+1 i−1 ε 3 dx2 3

and by [2] p46 for (a) and [2] p47 for (b) we have    

ε(xi+1 − xi−1 )|wε |3 d 2 | − ε( 2 − δ )wε (xi )| ≤  dx  max |wε00 (x)|  2ε 2

x∈[xi−1 ,xi+1 ]

(a) (b)

Using (b) outside the layers and at x = σ, x = d − σ gives | − ε(

d2 − δ 2 )wε (xi )| ≤ 2εCε−1 max e1 (x) x∈[xi−1 ,xi+1 ] dx2 ≤ C max e1 (x) ≤ CN −1 x∈[xi−1 ,xi+1 ]

as on [2] p47

Using (a) inside the layers gives, as above for vε , | − ε(

1 8σ1 − 3 d2 − δ 2 )wε (xi )| ≤ Cε2 ε 2 e1 (xi ) ≤ Cσ1 ε− 2 N −1 ≤ CN −1 ln N. 2 dx N

Hence

d2 | − ε( 2 − δ 2 )wε (xi )| ≤ CN −1 ln N dx N for all xi , 0 < i < 2 . Similarly for all xi , N2 < i < N. Hence −1 |LN ln N, ε e(xi )| ≤ CN

i 6= N/2

At the point x = d, (D+ − D− )e(x N ) = (D+ − D− )(Uε − uε )(x N ) 2

2

= (D+ − D− )Uε (x N ) − (D+ − D− )uε (x N ) 2

2

Recall that (D+ − D− )Uε (x N ) = 0. Let h± be the mesh interval sizes on either side 2 of xd and h = max{h− , h+ }. Thus h− =

8σ1 , N

h+ = 7

8σ2 N

|(D+ − D− )e(x N )| = |(D+ − D−)uε (d)| 2

d d )uε (d)| + |(D− − )uε (d)| dx dx 1 + 1 − h |uε |2 + h |uε |2 ≤ 2 2 1 + C (h + h− ) ≤ 2 ε ≤ |(D+ −

5

Error analysis

We now state and proof the main result of this paper. Theorem 6 Let uε be the solution of problem (Pε ) and Uε the solution of (PεN ). Then, for N sufficiently large, max |Uε (xi ) − uε (xi )| ≤ CN −1 ln N

¯N xi ∈Ω ε

where C is a constant independent of ε and N. Proof: Consider the discrete barrier function Φd defined by −εδ 2 Φd (xi ) + αΦd (xi ) = 0 for all xi ∈ ΩN ε Φd (0) = 0, Φd (d) = 1, Φd (1) = 0 ¿From the discrete minimum principle on the separate intervals [0, d] and [d, 1], one easily derives that 0 ≤ Φd ≤ 1. Also, using induction and D+Φd (xi ) − D−Φd (xi ) = ε((xi+1 − xi−1 )/2)αΦd (xi ) one can deduce that D− Φd (xi ) ≥ 0, i ≤ N/2,

D+Φd (xi ) ≤ 0, i ≥ N/2

Note that LN ε Φd (xi ) = (a(xi ) − α)Φd (xi ) ≥ 0,

for all xi ∈ ΩN ε

Define the ancillary continuous functions u1 , u2 by −εu001 + αu1 = 0,

u1 (0) = 0, u1 (d) = 1

−εu002 + αu2 = 0,

u2 (d) = 1, u2 (1) = 0 8

Note that

√ √ sinh( αx/ ε) √ √ , u1 (x) = sinh( αd/ ε)

Hence −

D u1 (d) = = = ≥ Similarly

√ √ sinh( α(1 − x)/ ε) √ √ u2 (x) = sinh( α(1 − d)/ ε)

√ √ √ √ sinh( αd/ ε) − sinh( α(d − h− )/ ε) √ √ h− sinh( αd/ ε) √ √ √ √ 2 sinh( αh− /2 ε) cosh( α(d − h− /2)/ ε) √ √ h− sinh( αd/ ε) √ − √ √ √ √ − √ 1 − e− αh / ε 1 + e−2 αd/ ε e αh / ε √ √ ( )( ) −2 αd/ ε h−√ 1 − e − √ 1 − e− αh / ε C( ) h− √

1 − e− αh |D u2 (d)| ≥ C( h+



+/

+

ε

)

Hence D+ u2 (d) − D− u1 (d) = −[|D+ u2 (d)| + D− u1 (d)] C 1 − e−ρ C ≤ −√ ( ) ≤ −√ ε ρ ε √

−x

where ρ = √αh , since 1−ex is a decreasing function of x and ρ ≤ 16(N −1 ln N). Note ε that by applying the results from [3] on the intervals [0, d] and [d, 1] separately, it follows that |Φd (xi ) − u1 (xi )| ≤ C(N −1 ln N)2 , i ≤ N/2 |Φd (xi ) − u2 (xi )| ≤ C(N −1 ln N)2 ,

i ≥ N/2

For i = N/2 Φd (d + h+ ) − 1 1 − Φd (d + h− ) − h+ h− C(N −1 ln N)2 = D+ u2 (d) − D− u1 (d) ± min{h− , h+ } C C(N −1 ln N) √ ≤ −√ ± ε ε C1 ≤ −√ ε

D+ Φd (d) − D− Φd (d) =

for N sufficiently large. Consider the mesh function h W (xi ) = C2 N −1 ln N + C3 √ Φd (xi ) ± e(xi ) ε 9

where C2 and C3 are suitably large constants. Hence for i 6= N/2 h −1 LN ln N + C3 √ (a(xi ) − α)Φd (xi ) ± LN e(xi ) ≥ 0 ε W (xi ) = C2 a(xi )N ε for C2 suitably large. For i = N/2 D+ W (d) − D− W (d) ≤ −C3

C1 h Ch ± ε ε

Hence, for C3 suitably large, D+ W (d) − D− W (d) ≤ 0 Thus, for N sufficiently large, |e(xi )| ≤ CN −1 ln N which completes the proof.

6

Numerical results

Numerical results are presented in this section for the particular problem −εu00ε (x) + uε (x) = d(x), d(x) = .7, x < .5; d(x) = −.6, x > .5, uε (0) = uε (1) = 0 which validate the theoretical results established in the previous section. The nodal errors and orders of convergence are estimated using the double mesh principle (see [1]). Define the double mesh differences to be DεN ≡ maxN |UεN (xi ) − U¯ε2N (xi )| and xi ∈Ωε

DN = max DεN ε

where U¯ε2N is the piecewise linear interpolant of the mesh function Uε2N onto [0, 1]. From these quantities the parameter-robust orders of convergence are computed from pN = log2 (

DN ). D2N

The corresponding approximate maximum pointwise error is taken to be ¯ε65,536 (xi )| and E N = max EεN EεN = maxN |UεN (xi ) − U ε xi ∈Ωε

The computed maximum pointwise errors EεN , E N and the computed orders of convergence pN are given in Table 1. These numerical results are in agreement with the theoretical results presented in this paper. 10

Table 1: Computed maximum pointwise errors EεN , computed ε-uniform errors E N and the computed ε-uniform orders of convergence pN .

ε −1

2 2−2 2−3 2−4 2−5 2−6 2−7 2−8 2−9 2−10 2−11 2−12 2−13 2−14 2−15 2−16 2−17 EN pN

32 .008740 .015457 .025489 .039081 .056822 .080547 .113102 .157572 .216555 .290530 .375053 .425322 .425190 .425004 .424758 .424757 .424757 .425322 .25

64 .004368 .007727 .012746 .019557 .028476 .040484 .057174 .080544 .113038 .157489 .216442 .290371 .299367 .299181 .298936 .298936 .298936 .299367 .78

128 .002182 .003860 .006368 .009772 .014233 .020250 .028642 .040467 .057117 .080464 .112926 .157331 .188182 .187996 .187751 .187751 .187751 .188182 .53

Number 256 .001089 .001926 .003178 .004876 .007104 .010109 .014303 .020223 .028586 .040387 .057005 .080305 .110534 .110348 .110102 .110102 .110102 .110534 .91

11

of Mesh Points N 512 1024 .000542 .000269 .000959 .000476 .001583 .000785 .002429 .001205 .003538 .001755 .005035 .002498 .007125 .003535 .010076 .004999 .014247 .007069 .020144 .009996 .028474 .014135 .040229 .019985 .056780 .028249 .062424 .034513 .062179 .034267 .062179 .034267 .062179 .034267 .062424 .034513 .85 .78

2048 .000132 .000234 .000386 .000593 .000864 .001229 .001739 .002460 .003479 .004919 .006957 .009838 .013911 .018716 .018471 .018471 .018471 .018716 .86

4096 .000064 .000113 .000187 .000287 .000418 .000595 .000842 .001190 .001683 .002380 .003366 .004761 .006732 .009520 .009678 .009678 .009678 .009678 .94

8192 .000030 .000053 .000087 .000134 .000195 .000278 .000393 .000555 .000785 .001111 .001571 .002222 .003142 .004443 .004840 .004840 .004840 .004840 .88

References [1] P.A. Farrell, A.F.Hegarty, J.J.H. Miller, E. O’Riordan, G.I. Shishkin, On the design of piecewise uniform meshes for solving advection-dominated transport equations to a prescribed accuracy, Notes on Numerical Fluid Mechanics, 49, (Vieweg-Verlag), (1995), 86-95. [2] J.J.H. Miller, E. O’Riordan and G.I. Shishkin, Fitted numerical methods for singular perturbation problems, World-Scientific, (Singapore), 1996. [3] J.J.H. Miller, E. O’Riordan and G.I. Shishkin, Fitted mesh methods for the singularly perturbed reaction diffusion problem, in proc. of V-th International Colloquium on Numerical Analysis, Aug. 13-17, 1996, Plovdiv, Bulgaria, Academic Publications, ed. E. Minchev, 99- 105. [4] G.I. Shishkin, Discrete approximation of singularly perturbed elliptic and parabolic equations, Russian Academy of Sciences, Ural section, Ekaterinburg, 1992. (in Russian)

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