A uniformly convergent numerical method for a coupled system of two singularly perturbed linear reaction diffusion problems

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1 IMA Journal of Numerical Analysis ( , A uniformly convergent numerical method for a coupled system of two singularly perturbed linear reaction diffusion problems NIALL MADDEN Department of Mathematics, National University of Ireland Galway, Ireland AND MARTIN STYNES Department of Mathematics, National University of Ireland Cork, Ireland [Received on 9 June 2002; revised on December 2002] A coupled system of two singularly perturbed linear reaction diffusion two-point boundary value problems is examined. The leading term of each equation is multiplied by a small positive parameter, but these parameters may have different magnitudes. The solutions to the system have boundary layers that overlap and interact. The structure of these layers is analysed, and this leads to the construction of a piecewise-uniform mesh that is a variant of the usual Shishkin mesh. On this mesh central differencing is proved to be almost first-order accurate, uniformly in both small parameters. Supporting numerical results are presented for a test problem. Keywords: singularly perturbed; finite difference method; reaction diffusion; coupled system; Shishkin mesh.. Introduction The modelling of turbulence in water waves when they interact with currents is a complex and poorly understood physical phenomenon. In hydraulics, a standard way of dealing with the effects of turbulence is the well-known k-ɛ two-equation model (see Rodi, 993. Thomas (998 recently developed a hierarchical set of systems of equations that uses such a model to describe the effects of turbulence in the context of wave current interactions. Each system in this set comprises two singularly perturbed secondorder two-point boundary value problems. The system cannot be solved analytically, so numerical methods must be used to determine its solution. Experimental data for wavecurrent interactions and an understanding of the physical mechanisms involved lead one to expect that the solution of each system will exhibit two distinct but interacting layers that correspond to a turbulent layer and a viscous sublayer in the water velocities. The first system in Thomas s hierarchy is already quite complex. In this paper we consider a linearized version of that first system that discards a coupling between its second-derivative terms. Despite this simplification, the solution of our system retains the niall.madden@nuigalway.ie m.stynes@ucc.ie c The Institute of Mathematics and its Applications 2003

2 628 N. MADDEN AND M. STYNES main feature of Thomas s system: the presence of two interacting boundary layers. We shall develop a detailed knowledge of their properties that enables us to construct a suitable mesh for their numerical calculation and to prove convergence of a standard finite-difference method on that mesh. Thus, consider the following system of two linear second-order singularly perturbed two-point boundary value problems coupled by their zero-order terms: ( ε d 2 0 L u := dx 2 u + A u = f for x Ω := (0,, (.a 0 µ d2 dx 2 where the parameters ε and µ lie in (0, ], ( a (x a A = 2 (x a 2 (x a 22 (x and the boundary conditions at Ω are and f (x = ( f (x, (.b f 2 (x u(0 = γ 0, u( = γ, (.c where the data ε, µ, A, f, γ 0 and γ are given. The unknown solution of (. is the vector u(x := (u (x, u 2 (x T.Weassume that and that for some constant α we have a 2 (x 0 and a 2 (x 0 for all x Ω, (.2a 0 <α<min{a (x + a 2 (x, a 2 (x + a 22 (x}. (.2b Ω Consequently a (x > a 2 (x and a 22 (x > a 2 (x on Ω. Furthermore, assume that a 2, a 22, f, f 2 C 2 ( Ω, and a, a 2 C 3 ( Ω, where the bounds on the derivatives of these functions are independent of ε and of µ. The stronger assumptions for a and a 2 are needed when ε<µ/4inthe proof of Lemma 5. For existence and uniqueness of the solution of (., see Section 2. Without loss of generality, throughout the paper we shall take 0 <ε µ. (.3 Although the numerical solution of singularly perturbed differential equations has received much attention (see Roos et al., 996, few papers deal with the numerical analysis of systems of this type. Shishkin (995 examined a system of two parabolic partial differential equations that is analogous to (., posed on a strip that is infinite in one of the space variables. He considered the three cases (i 0 <ε= µ, (ii 0 <ε, µ=, and (iii 0 <ε µ, but observed that in cases (i and (ii the structure of the solution of (. is simpler than in case (iii, and that consequently his a priori bounds for the solutions of the former [(i and (ii] are more accurate and give a higher order of convergence of the special difference

3 AUNIFORMLY CONVERGENT METHOD FOR COUPLED SYSTEMS 629 schemes than for the latter [(iii]. For the general case (iii, the analysis in Shishkin (995, p. 43 shows that his difference scheme is convergent of order /4; in contrast, we shall obtain almost first-order convergence for our difference scheme (for the simpler problem (.. Matthews et al. (2000a, 2002 studied the problem (. but restricted their attention to the cases 0 < ε = µ and 0 < ε µ =. They gave numerical results and an outline of the analysis required to show convergence in the maximum norm of the numerical method. Further details and extensive numerical results appear in Matthews (2000, where numerical results for the case 0 <ε µ are also given. Shanthi & Ramanujan (2002 consider the numerical solution of fourth-order singularly perturbed reaction diffusion two-point boundary value problems by rewriting the differential equation as a coupled system of two second-order problems. This yields a system that corresponds to the special (and simpler case µ = of(., which they solve using a combination of asymptotic and numerical techniques. The analysis of Matthews et al. (2000a, 2002 and Shanthi & Ramanujan (2002 concentrates on special cases that yield significant simplifications in the structure of the solution u of (.. In contrast, we shall prove bounds for u and its derivatives over the full range of parameters 0 <ε µ. These bounds will enable us to construct a special piecewise-uniform mesh on which we can prove that the standard central differencing scheme is almost first-order accurate, uniformly in ε and µ. Section 2 contains some preliminary bounds on u and u 2, the components of the solution u of (.. In Section 3 these components are decomposed into smooth and layer functions and bounds are obtained on the derivatives of these functions. The mesh and finite-difference operator that approximate (. are constructed in Section 4. We show in Section 5 that the numerical solution computed by this method is almost first-order accurate in the maximum norm, uniformly in the parameters ε and µ. Finally, supporting numerical results are presented in Section 6. We shall use C to denote a generic constant that is independent of ε, µ and the mesh. Note that C can take different values in different places. We occasionally use a subscripted C (e.g. C ; this a fixed constant that is independent of ε, µ and the mesh and does not vary in value. 2. Norms and preliminary results On each closed set S Ω, define the maximum norm for each real-valued function φ C(S by φ S = max x Ω φ(x. For vector-valued functions v(x = ( v (x, v 2 (x T C(S C(S, set v(x =( v (x, v 2 (x T, and v S = max{ v S, v 2 S }. When S = Ω,wesimply write. Given two vector-valued functions v(x and w(x, wewrite v(x w(x if v (x w (x and v 2 (x w 2 (x for all x [0, ]. The notation v t, for some scalar t, is shorthand for v t(, T. A mesh Ω N := {x i } i=0 N is a set of points satisfying 0 = x 0 < x < < x N =. A mesh function V := {V (x i } i=0 N is a real-valued function defined on Ω N. Define the discrete maximum norm for such functions by V Ω N = max i=0,...,n V (x i. Where

4 630 N. MADDEN AND M. STYNES vector mesh functions V := (V, V 2 T = { (V (x i, V 2 (x i T } N are used, we define i=0 V Ω N = max{ V Ω N, V 2 Ω N }. Suppose that the parameter ν satisfies 0 <ν and the function b( satisfies b(x > b 0 > 0 for all x Ω, where b 0 is some positive constant. For each y C 2 ( Ω define the scalar differential operator L ν y(x := νy (x + b(xy(x. This operator has the following properties. LEMMA (Comparison principle for scalar reaction diffusion problem If y(0 z(0, y( z( and L ν y(x L ν z(x for all x Ω, then y(x z(x for all x Ω. Proof. This follows from a maximum principle; see Protter & Weinberger (967. In the context of Lemma, we say that y is a barrier function for z. We shall have frequent recourse to the scalar differential operators L ε u(x := εu (x + a u(x and L µ u(x := µu (x + a 22 u(x that come from the components of (.a. Note that Lemma can be applied to each of these operators. The operator L of the coupled system (. has a property analogous to that of Lemma, as the next result shows. LEMMA 2 (Comparison principle for reaction diffusion system If ψ(0 φ(0, ψ( φ( and L ψ L φ on Ω, then ψ(x φ(x for all x Ω. Proof. The proof is essentially the same as that of Theorem in Matthews et al. (2002. In the context of Lemma 2, we say that ψ is a barrier function for φ. Lemma 2 and standard arguments (see, e.g. Ladyzhenskaya & Ural tseva, 968 show that under the assumptions (.2, the reaction diffusion system (. has a solution u C 4 ( Ω C 4 ( Ω and that solution is unique. 3. Structure of u and u 2 To gain some insight into the behaviour of the solution u of (., we briefly consider a numerical example. EXAMPLE Take ( ( 2 A =, f (x = 2 ( 0 and γ 0 = γ = 0 (3. with ε = 0 7 and µ = 0 4. One expects the solution to this problem to display layer behaviour near the boundary of Ω. It is much less obvious that while both solutions have exponential layers of width O( µ /2 ln µ, only u (x has an additional sublayer of width O( ε /2 ln ε. This is demonstrated in Fig. ; in particular note the difference in behaviour of the two curves in the rightmost diagram.

5 AUNIFORMLY CONVERGENT METHOD FOR COUPLED SYSTEMS u (x u 2 (x u (x u 2 (x u (x u 2 (x FIG.. Solution of Example Analyses of fitted mesh methods for singularly perturbed two-point boundary value problems depending on a single parameter typically involve a decomposition of the solution into (i a smooth part that is close to the full solution over most of the domain and whose low-order derivatives can be bounded independently of the parameter and (ii a layer part that has large first-order derivatives at the boundary but decays exponentially as one moves into the interior of the domain. This splitting is often called a Shishkin decomposition ; see Miller et al. (996 for an exposition of the idea in a simple setting. As Fig. suggests, we are forced to develop a more intricate analysis that splits u (x into a smooth part, a layer part that depends on µ and a layer part that depends on ε. The interaction between these two layers makes the analysis more difficult than for problems with layers depending on a single small parameter. We write u = v + w, (3.2a where v = (v,v 2 T is the solution to L v = f on Ω and v = A f on Ω (3.2b (by (.2 the matrix A is diagonally dominant and hence invertible and w = (w,w 2 T satisfies L w = 0onΩ and w = u v on Ω. (3.2c (Later, a further decomposition of w is required. Here v(x is smooth, in the sense that its derivatives up to second order are bounded, uniformly for 0 <ε,µ. The function w(x is the layer part of the solution u. LEMMA 3 There exists a constant C such that v (k C for k = 0,, 2, with v Cε /2 and v 2 Cµ /2.

6 632 N. MADDEN AND M. STYNES Proof. Set C = max{ A f (0, A f ( }+ f /α. Then using the constant barrier function C (, T, one sees easily by means of Lemma 2 that v C. (3.3 Observe that the two equations of (3.2b together imply that εv = µv 2 = 0on Ω, i.e. v = 0 on Ω. The hypotheses on the regularity of the a ij and f imply that v C (4 (Ω C (4 (Ω. Differentiating L v = f twice yields ( L( v = f a v + 2a v + a 2 v 2 + 2a 2 v 2 a 2 v + 2a 2 v + a 22 v 2 + 2a 22 v 2 on Ω. (3.4 Set C 2 = 2 max{ a + a 2, a 2 + a 22 }. Then L( v C + C 2 v by (3.3 and (3.4. Invoking Lemma 2 with the barrier function [C/α + C 2 v /α](, T,weget v C/α + C 2 v /α C + C 2 v /α. (3.5 Let x [0, ] be such that max{ v (x, v 2 (x } = v.for definiteness, suppose that v (x = v. Set C 3 = min{,α/(2c 2 }. Choose an interval I =[a, a + C 3 ], for some a 0, such that x I [0, ]. Bythe mean value theorem, for some ζ I we have v (ζ =C 3 v (a + C 3 v (a 2C 3 v C by (3.3. Hence v = v (x = v (ζ + x ζ v (t dt C + x ζ v C + C 3 v. (3.6 Substituting this inequality into (3.5 yields v C + (C 2 C 3 /α v C + v /2. Rearranging, we obtain Now from (3.6 we see that v C. (3.7 v C. (3.8 Finally, to bound v and v 2,werecall (3.4. The first equation of this system can be written in the form L ε z = g, where z = v and g = f (a v + 2a v + a 2 v 2 + 2a 2 v 2 + a 2v 2. Our earlier bounds show that g C. Also, we know that z = 0on Ω. Thus Lemma 2.3 of O Riordan & Stynes (986 can be applied and yields z Cε /2, i.e. v Cε /2.Asimilar argument based on the second equation of (3.4 proves that v 2 Cµ /2. We now find bounds on the layer part w(x of the Shishkin decomposition of u. For this the following layer functions defined on [0, ] will be helpful: B ε (x = exp ( x α/ε + exp x α/ε, B µ (x = exp ( x α/µ + exp x α/µ

7 AUNIFORMLY CONVERGENT METHOD FOR COUPLED SYSTEMS 633 LEMMA 4 Let w(x = (w (x, w 2 (x T be the solution to (3.2c. Then there exists a constant C such that for all x Ω we have w (x CB µ (x, w 2 (x CB µ (x, (3.9a w (x C( ε /2 B ε (x + µ /2 B µ (x, w 2 (x Cµ /2 B µ (x, (3.9b w (x C( ε B ε (x + µ B µ (x, w 2 (x Cµ B µ (x, (3.9c w (x C( ε 3/2 B ε (x + µ 3/2 B µ (x, w 2 (x Cµ ( ε /2 B ε (x + µ /2 B µ (x (3.9d Proof. By (3.2c and Lemma 3 we have w C on Ω. Define the function ψ(x = ( CC Bµ (x with C chosen sufficiently large so that ψ w on Ω;as L ψ(x = CB µ (x ( a (x + a 2 (x εµ α 0 = L w(x on Ω, a 2 (x + a 22 (x α the comparison principle of Lemma 2 implies that ψ(x w(x on Ω, and (3.9a is proved. The second equation of (3.2c is µw 2 + a 22w 2 + a 2 w = 0. This identity and (3.9a imply that there is a constant C such that w 2 (x Cµ B µ (x for x Ω. (3.0 To bound w 2 (x we use a technique similar to that of Matthews (2000. Let x [0, /2] be arbitrary. Without loss of generality one can take µ /4. Choose an interval I =[a, a + µ /2 ], for some a 0, such that x I [0, /2]. Note that B µ (a B µ (a + µ /2.Bythe mean value theorem and (3.9a, for some ζ I we have w 2 (ζ =µ /2 w 2 (a + µ /2 w 2 (a Cµ /2 B µ (a. Hence, applying the mean value theorem to w 2, for some ξ I one has w 2 (x = w 2 (ζ + (x ζw 2 (ξ w 2 (ζ +µ /2 w 2 (ξ Cµ /2 B µ (a, where we used (3.0. But x I implies that so B µ (a = e (x a α/µ e x α/µ + e a e α/µ e α e x α/µ + e x e α/µ CB µ (x, w 2 (x Cµ /2 B µ (x for all x [0, /2]. A similar argument can be applied for x [/2, ]. This gives w 2 (x Cµ /2 B µ (x for all x Ω, (3. where C is some fixed constant. Applying the same techniques to the first equation of (3.2c, we obtain w (k (x Cε k/2 B µ (x for k =, 2 and x Ω. This bound is satisfactory on Ω but isnot sharp

8 634 N. MADDEN AND M. STYNES inside Ω, for if ε µ, then B µ (x decays much more slowly than B ε (x. Toimprove it, differentiate the first equation of (3.2c, which yields L ε w (x = ( εw + a w (x = (a w + a 2 w 2 + a 2 w 2 (x Cµ /2 B µ (x on Ω, (3.2 by (3.9a and (3.. We have just seen that w (x Cε /2 on Ω. Define a barrier ( function by setting Ψ(x = C 4 ε /2 B ε (x + µ /2 B µ (x, where the constant C 4 is chosen so large that Ψ(x w (x for x = 0, and that L ε Ψ(x = C 4 ε /2 B ε (x(a α + C 4 µ /2 B µ (x(a εµ α C 4 (a αµ /2 B µ (x L ε w (x for x Ω, where we used (.2 and (3.2. The comparison principle of Lemma now implies that w (x Φ(x on Ω.This inequality and (3. constitute (3.9b. To bound w (z, the argument resembles the one used to bound w (z. Wealready have w (x Cε on Ω. Differentiate the identity in (3.2 to get L ε w (x = (a w + a 2 w 2 + a 2 w 2 (x a (xw (x C ( ε /2 B ε (x + µ B µ (x x Ω, (3.3 by our previous bounds. Now a suitable barrier function yields w (x C(ε B ε (x + µ B µ (x on Ω, for some constant C. This inequality and (3.0 show that (3.9c holds true. Differentiating the second equation of (3.2c gives µw 2 = (a 2w + (a 22 w 2. The bounds (3.9a (3.9c now imply that w 2 (x Cµ (ε /2 B ε (x + µ /2 B µ (x on Ω. Finally, to bound w (x,from the identity in (3.2 and the previous bounds we infer that w (x Cε 3/2 on Ω. The hypotheses on the regularity of the a ij allow us to differentiate the identity in (3.3, which yields L ε (w = ( a w + a 2 w 2 + a 2 w 2 (a w a w. Now once again imitate the derivation of the bound on w (x ; this leads to the bound on (x in (3.9d and completes the proof. w When in Section 5 we prove an error estimate for our finite-difference method, the cases µ/4 ε µ and ε<µ/4 are considered separately. In the former case the bounds of Lemma 4 suffice, but in the latter case, the decomposition of w(x stated in the following lemma is also required. LEMMA 5 Suppose that ε<µ. Then there are functions w,ε (x, w,µ (x, w 2,ε (x and w 2,µ (x such that w (x = w,ε (x + w,µ (x, w 2 (x = w 2,ε (x + w 2,µ (x,

9 AUNIFORMLY CONVERGENT METHOD FOR COUPLED SYSTEMS 635 and w,ε (x Cε B ε (x, w,µ (x Cµ 3/2 B µ (x, (3.4a w 2,ε (x Cµ B ε (x, w 2,µ (x Cµ 3/2 B µ (x, (3.4b for all x [0, ]. Proof. Since ε < µ, there is a unique point x = x (ε, µ (0, /2 such that ε 3/2 B ε (x = µ 3/2 B µ (x. Also, ε 3/2 B ε ( x = µ 3/2 B µ ( x.on[0, x ( x, ] we have ε 3/2 B ε (x >µ 3/2 B µ (x and on (x, x we have ε 3/2 B ε (x < µ 3/2 B µ (x. Define a function w,µ on [0,] as follows: w (x for x [x, x ], w,µ (x = 3k=0 [(x x k /k!] w (k (x for x [0, x, 3k=0 [(x + x k /k!] w (k ( x for x ( x, ]. Then w,µ C 3 [0, ]. Forx [0, x,the construction of w,µ, Lemma 4 and the choice of x yield w,µ (x = w (x Cµ 3/2 B µ (x Cµ 3/2 B µ (x. A similar result holds true on ( x, ]. And on [x, x ] we have w,µ (x = w (x Cµ 3/2 B µ (x by Lemma 4 and the choice of x.insummary, w,µ = w on [x, x ] and w,µ (x Cµ 3/2 B µ (x on [0, ], (3.5 which is the second inequality in (3.4a. The construction of w,µ was inspired by Linß (200, p Set w,ε = w w,µ. Then w,ε 0on[x, x ].On[0, x ( x, ] one has ε 3/2 B ε (x >µ 3/2 B µ (x,so w,ε (x w (x + w,µ (x Cε 3/2 B ε (x. Thus for any x [0, ],asw,ε (/2 = 0itfollows that w,ε (x = x /2 w,ε (t dt x /2 Cε 3/2 B ε (t dt Cε [ B ε (/2 + B ε (x ] Cε B ε (x, (3.6 since B ε (/2 B ε (x for all x [0, ]. This proves the first inequality in (3.4a. Asomewhat similar analysis can be applied to w 2. Motivated by the bound on w 2 (x in Lemma 4, one starts by choosing the point ˆx (0, /2 such that ε /2 B ε ( ˆx = µ /2 B µ ( ˆx. Define a function w 2,µ on [0,] by w 2 (x for x [ˆx, ˆx], w 2,µ (x = 3k=0 [(x ˆx k /k!] w (k 2 ( ˆx for x [0, ˆx, 3k=0 [(x +ˆx k /k!] w (k 2 ( ˆx for x ( ˆx, ].

10 636 N. MADDEN AND M. STYNES Continuing as for w,µ leads to w 2,µ (x Cµ 3/2 B µ (x for all x [0, ]. (3.7 Set w 2,ε = w 2 w 2,µ. Then w 2,ε 0on[ˆx, ˆx], and on [0, ˆx ( ˆx, ] one has w 2,ε (x w 2 (x + w 2,µ (x Cµ ε /2 B ε (x by Lemma 4, the definition of ˆx and (3.7. Imitating the derivation of (3.6 now yields w 2,ε (x Cµ B ε (x on [0, ]. (3.8 The inequalities (3.7 and (3.8 together form (3.4b. 4. Discretization of the problem In this section the finite-difference operator that we shall use is first presented on an arbitrary mesh, then we specify a piecewise-uniform mesh that is particularly suited to solving (.. Let Ω N ={x i } i=0 N be an arbitrary mesh with 0 = x 0 < x < x 2 < x N =. Set h i = x i x i for each i. Given a mesh function {Y (x i } i=0 N, define the standard second-order central differencing operator D 2 Y (x i := 2 h ( Y (xi+ Y (x i Y (x i Y (x i for i =,...,N, i h i+ h i where h i = (h i+ + h i /2. It is well known that for any ψ(x C 3 ( Ω, ( D 2 d2 ψ(xi dx 2 3 (x i+ x i ψ [xi,x i+ ], and D 2 ψ(x i ψ [xi,x i+ ]. Thus (D 2 d2 { } dx 2 ψ(x i min 3 (h i + h i+ ψ [xi,x i+ ], 2 ψ [xi,x i+ ]. (4. To solve numerically the coupled problem (. a discrete vector operator is required: set ( (L N εd 2 0 U(x i = 0 µd 2 U(x i + A(x i U(x i, for i =,...N. We show now that the difference operator satisfies a discrete maximum principle. LEMMA 6 (Discrete maximum principle Let { Y (x i } i=0 N be any vector mesh function such that Y (x 0 0 and Y (x N 0. Suppose that L N Y (x i 0 for i N. Then Y (x i 0 for 0 i N.

11 AUNIFORMLY CONVERGENT METHOD FOR COUPLED SYSTEMS 637 Proof. If the conclusion of the lemma is false, one can choose k such that min Y [ j(x k = min min Y j (x i ] < 0. j=,2 j=,2 i Without loss of generality, it can be assumed that Y (x k Y 2 (x k. Clearly k {0, N}. Also D 2 Y (x k = 2 h ( Y (x k+ Y (x k Y (x k Y (x k 0. k h k+ h k But a (x k Y (x k < a 2 (x k Y 2 (x k,soa (x k Y (x k + a 2 (x k Y 2 (x k < 0. Hence D 2 Y (x k + a (x k Y (x k + a 2 (x k Y 2 (x k <0, which contradicts the hypothesis that L N Y (x k 0. The next lemma is a discrete stability result for the operator L N. LEMMA 7 Let { Y (x i } be any vector mesh function such that Y (x 0 = 0 and Y (x N = 0. Then Y Ω N α L N Y Ω N for 0 i N. Proof. Set M = α L N Y Ω.Clearly M( ± Y (x 0 0 and M ( ± Y (x N 0. Also, for j =,...,N, we have L N ( M ( ± Y (x j ( a + a = M 2 ± L N Y (x a 2 + a j 22 ( (a + a 2 /α L N Y (a 2 + a 22 /α Ω N L N Y (x j 0 The discrete maximum principle of Lemma 6 implies that M ( ± Y (x j 0 for j =,...,N, and the desired result follows. Now we move on the precise choice of mesh for solving (.. This mesh will be piecewise uniform and is closely related to the Shishkin meshes used in many recent papers. For a general discussion of these meshes see Farrell et al. (2000, Miller et al. (996 and Roos et al. (996. Shishkin meshes for a single reaction diffusion two-point boundary value problem divide the interval [0, ] into three subintervals, on each of which the mesh is uniform. These subintervals correspond to the region where the solution is smooth and to the boundary layers at x = 0 and x =. When 0 <ε<µ, the solution to (. has however two overlapping boundary layers at both x = 0 and x =. This necessitates the construction of a mesh that is uniform on each of five subintervals. On the main subinterval, where the solution is smooth, the mesh is coarse; on the other four subintervals it is very fine. We define τ ε and τ µ as { } µ τ µ = min /4, α and { } ε τ ε = min /8,τ µ /2, α. (4.2 A piecewise-uniform mesh Ω N = {x i } N 0 is constructed by dividing [0, ] into five subintervals [0,τ ε ], [τ ε,τ µ ], [τ µ, τ µ ], [ τ µ, τ ε ] and [ τ ε, ]. Then subdivide

12 638 N. MADDEN AND M. STYNES N/8 N/8 N/2 N/8 N/8 0 τ ε τ µ τ µ τ ε FIG.2.Piecewise-uniform mesh for two coupled reaction diffusion equations. [τ µ, τ µ ] into N/2 mesh intervals, and subdivide each of the other four subintervals into N/8 mesh intervals. Figure 2 is a sketch of such a mesh for the case N = 32. This mesh is quite similar to that proposed by Shishkin (995 and Matthews (2000, and differs only in how τ µ is chosen. Those authors take τ µ = min { /4, (ε + µ/α 0 }, where α 0 = min{a + a 2, a 2 + a 22 }. 5. Error estimates for the numerical method Let U = (U, U 2 T,avector mesh function on Ω N,bethe solution to the discrete problem ( (L N εd 2 0 U(x i = 0 µd 2 U(x i + A(x i U(x i = f (x i, for i =,...N, (5.a with U(x 0 = u(0 and U(x N = u(, (5.b where D 2 is the second-order finite-difference operator defined in Section 4. It follows from Lemma 6 that U is well defined. The next theorem, our main result, shows that U is a parameter-robust numerical approximation to the solution of (.. THEOREM 8 Let u(x be the solution to (. and U the solution to the discrete problem (5. on the mesh Ω N defined in Section 4. Then there exists a constant C such that u U Ω N CN. Proof. Recall the Shishkin decomposition (3.2a (3.2c. We similarly decompose the discrete solution as U = V + W, where V ={ V (x i } is the solution to V (x 0 = v(0, V (x N = v( and ( L N V (x i = f (x i for i =,...N, and W ={ W (x i } is the solution to W (x 0 = w(0, W (x N = w( and ( L N W (x i = 0 for i =,...N. Then the error satisfies U u Ω N = ( V v+( W w Ω N V v Ω N + W w Ω N.

13 AUNIFORMLY CONVERGENT METHOD FOR COUPLED SYSTEMS 639 First, consider V v Ω N.Fori =,...,N, L N ( V v (x i = f (x i L N v(x i = (L L N v(x i (ε(d = 2 d2 0 dx 2 0 µ(d 2 d v(x i dx ( 2 ε(d 2 d2 v = dx 2 (x i µ(d 2 d2 v dx 2 2 (x i It follows from (4. and the bounds on v (x given in Lemma 3 that L N ( V v (x i ( ε v 3 (h i + h i+ ( N µ v 2 C N for i =,...,N, since h i + h i+ 4N for all i. Recall that ( V v (x j = 0 for j = 0, N. Lemma 7 implies that V v Ω N CN. (5.3 Next, W w is estimated. Consider only the case where Ω N is not simply a uniform mesh, as otherwise N is exponentially large relative to µ, which simplifies the analysis. Note that L N ( W w ε D (x i = 0 L N w(x i = (L L N 2 d2 w (x w(x i = dx 2 i µ ( D 2, d2 w2 (x dx 2 i and ( W w (x j = 0 for j = 0, N. Wewish to show that ε D 2 ( d2 w (x dx 2 i N µ ( D 2 C d2 w2 (x dx 2 i N for i =,...,N, (5.4 so that Lemma 7 may be used to estimate W w Ω N. There are three cases to be considered: Case, x i [τ µ, τ µ ]; Case 2, x i (0,τ ε ( τ ε, and Case 3, x i [τ ε,τ µ ( τ µ, τ ε ]. Case : x i [τ µ, τ µ ].Suppose first that x i [τ µ, /2]. Then by (4. and Lemma 4, ε D 2 d2 w (x dx 2 i µ ( D 2 2 d2 w2 (x dx 2 i ( ε w [x i,x i+ ] µ w 2 C [x i,x i+ ] ( Bµ [xi,x i+ ] B µ [xi,x i+ ] Now x i [τ µ, /2] implies that B µ [xi,x i+ ] = B µ (x i and x i τ µ 8τ µ /N. Hence B µ [xi,x i+ ] 2e ( τ µ+8τ µ /N α/µ = 2e τ µ α/µ e (8τ µ/n α/µ. = 2e e 8N CN.

14 640 N. MADDEN AND M. STYNES Thus ε D 2 ( d2 w (x dx 2 i N µ ( D 2 C d2 w2 (x dx 2 i N. A similar argument shows that this inequality also holds true when x i [/2, τ µ ],so ε D 2 ( d2 w (x dx 2 i N µ ( D 2 C d2 w2 (x dx 2 i N for all x i [τ µ, τ µ ]. (5.5 Case 2: x i (0,τ ε ( τ ε,. By(4. and Lemma 4, ε D 2 ( d2 w (x dx 2 i µ ( D 2 ε w d2 3 w2 (x dx 2 i (h i + h i+ ( N µ w 2 C N, (5.6 where we used the inequality h i + h i+ 6 ε/α N. Case 3: x i [τ ε,τ µ ( τ µ, τ ε ]. Suppose first that µ/4 ε µ. Then from the definitions of τ ε and τ µ in (4.2, τ µ = µ/2lnn 2 ε/α. Thus for i such that x i [τ ε,τ µ ( τ µ, τ ε ], h i + h i+ = 8N τ µ 6 ε/α N. By (4. and Lemma 4, ε D 2 ( d2 w (x dx 2 i µ ( D 2 ε w d2 3 w2 (x dx 2 i (h i + h i+ µ w 2 C ( N N. (5.7 Next, suppose instead that ε<µ/4. Recalling the decomposition of Lemma 5, ε D 2 d2 w (x dx 2 i ε D 2 µ ( d2 w,ε (x D 2 dx 2 i ε D 2 d2 w2 (x dx 2 i µ ( d2 w,µ (x D 2 + dx 2 i d2 w2,ε (x dx 2 i µ ( D 2. d2 w2,µ (x dx 2 i (5.8 Using the bounds of Lemma 5, one can imitate the analysis of Case to show that ε D 2 ( d2 w,ε (x dx 2 i ε w µ ( D 2 2,ε ( ( [x i,x i+ ] Bε d2 w2,ε (x dx 2 i µ w 2,ε C [xi,x i+ ] N C [x i,x i+ ] B ε [xi,x i+ ] N and imitate the analysis of Case 2 to show that ε D 2 d2 w,µ (x dx 2 i µ ( D 2 d2 w2,µ (x dx 2 i ( ε w 3 (h i + h i+,µ µ w 2,µ C ( N N, where we note that x i [τ ε,τ µ ( τ µ,τ ε ] implies that h i +h i+ 6 µ/α N. Combining these two inequalities with (5.8 gives ε D 2 ( d2 w (x dx 2 i N µ ( D 2 C d2 w2 (x dx 2 i N for all x i [τ ε,τ µ ( τ µ,τ ε ]. (5.9

15 AUNIFORMLY CONVERGENT METHOD FOR COUPLED SYSTEMS 64 The inequalities (5.5 (5.7, and (5.9 together yield (5.4. Invoking the discrete maximum principle of Lemma 6, we get W w Ω N CN. Recalling (5.3, the proof is complete. COROLLARY 9 Let U be the solution to (5.. Define U h (x to be the piecewise linear interpolant to U on [0,]. Then there is a constant C such that ( u U h (x CN for all x [0, ]. Proof. Recall the decomposition (3.2a of u(x and the bounds on v (x and w (x given in Lemmas 3 and 4. Then, using Theorem 8, the analysis of Miller et al. (996, pp may easily be adapted to complete the proof. 6. Numerical results We consider the following test problem from Matthews et al. (2002. EXAMPLE 2 εu + 2(x + 2 u ( + x 3 u 2 = 2e x ( π x µu 2 2 cos u + 2 2e ( x u 2 = 0x +, 4 with the boundary conditions u(0 = 0, u( = 0. Let U N be the solution to (5. on the piecewise-uniform mesh Ω N with N intervals. The maximum pointwise error is U N u Ω N. The true solution u to Example 2 is unavailable, so this error must be estimated numerically. Let Ω 5N be the piecewise-uniform Shishkin-type mesh having 5N intervals, but with the same transition points as Ω N. Write U 5N for the solution to the discrete problem computed on Ω 5N.Asanestimate of U N u Ω N, let E N ε,µ = U N U 5N Ω N. If ε = 0 j for some non-negative integer j, set Eε N = max{eε, N, E N, ε,0 E N,...,E N }. Then the parameter-uniform error is computed as E N = ε,0 2 ε,0 j max{e N, E N,...,E N } Table shows, for various values of N and ε, the computed values of Eε N. Each number in the last row of the table is the maximum entry in its column. From these values we can see that the error is robust with respect to the parameters ε and µ, and is converging to zero as N is increased. Let U h 2N (x be the piecewise linear interpolant to U 2N on [0, ]. Tocompute the rate of convergence of the numerical method, we follow the practice suggested in Chapter 8 of Farrell et al. (2000 by computing the two-mesh difference Dε,µ N = U N U h 2N Ω N and the parameter-uniform two-mesh error D N := max ε,µ {Dε,µ N }, where ε and µ take the

16 642 N. MADDEN AND M. STYNES TABLE Errors E N ε in the computed solution to Example 2 N ε e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e 04 49e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e 02 8 e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e 04 E N 82e e e e e e e 04 TABLE 2 Rates of convergence for Example 2 N D N 382e e e e e e e 04 p values, 0, 0 2,...,0 6. Then the rate of convergence of our method is estimated by D N p = log 2 D 2N. Table 2 lists values of D N and p, and shows that the rate of convergence is at least first-order, confirming Theorem 8. REMARK 0 In this paper we obtained a priori bounds on the derivatives of the solution of a system of two singularly perturbed reaction diffusion equations. To derive them the solution had to be split into into a sum of two overlapping layers. The bounds were used to analyse the convergence of a difference scheme used to solve this problem numerically, and are the key to the analysis of any numerical method for this problem. Theorem 8 predicts convergence of order N and is in harmony with results like that stated for a single reaction diffusion equation in Miller et al. (996, Chapter 6, Theorem 4. The numerical results of Table 2 suggest that in fact one obtains convergence of order N 2 ln 2 N, and this higher order of convergence has been proved in Matthews et al. (2002 for special cases of (.; nevertheless, the derivative bounds of the present

17 AUNIFORMLY CONVERGENT METHOD FOR COUPLED SYSTEMS 643 paper do not permit a proof of almost-second-order convergence to do this for the general problem (. requires further investigation of the derivatives of its solution together with a change in the transition points of (4.2 to { } µ τ µ = min /4, 2 α and and this is the subject of ongoing research. { } ε τ ε = min /8,τ µ /2, 2 α, Acknowledgements This work was conducted as part of the project The Kinematics and Dynamics of Wave Current Interactions and financial support was provided by the Commission of the European Communities, under contract MAS3-CT95-00 of the MAST Programme. REFERENCES FARRELL, P.A.,HEGARTY, A.F.,MILLER, J.J.H., O RIORDAN, E.&SHISHKIN, G.I.(2000 Robust computational techniques for boundary layers Vol. 6 of Applied Mathematics and Mathematical Computation. Boca Raton, FL: Chapman & Hall/CRC Press. LADYZHENSKAYA, O. A. & URAL TSEVA, N. N. (968 Linear and Quasilinear Elliptic Equations. New York: Academic. LINSS, T.(200 The necessity of Shishkin decompositions. Appl. Math. Lett., 4, MATTHEWS, S.(March 2000 Parameter robust numerical methods for a system of two coupled singularly perturbed reaction diffusion equations. Master s Thesis, School of Mathematics Sciences, Dublin City University. MATTHEWS, S., MILLER, J. J. H., O RIORDAN, E.& SHISHKIN, G. I.(2000a A parameter robust numerical method for a system of singularly perturbed ordinary differential equations. Analytical and Numerical Methods for Convection-Dominated and Singularly Perturbed Problems. (J. J. H. Miller, G. I. Shishkin & L. Vulkov, eds. New York: Nova Science, pp MATTHEWS, S., O RIORDAN, E. & SHISHKIN, G. I. (2002 Numerical reaction diffusion equations. J. Comp. Appl. Math., 45, MILLER, J. J. H., O RIORDAN, E.& SHISHKIN, G. I.(996 Fitted Numerical Methods For Singular Perturbation Problems Error Estimates In The Maximum Norm For Linear Problems In One And Two Dimensions. Singapore: World Scientific. O RIORDAN, E.& STYNES, M.(986 A uniformly accurate finite element method for a singularly perturbed one-dimensional reaction diffusion problem. Math. Comput., 47, PROTTER, M. H.& WEINBERGER, H. F.(967 Maximum Principles in Differential Equations. Englewood Cliffs, NJ: Prentice-Hall. RODI, W.(993 Turbulence Models and Their Applications in Hydraulics (A. A. Balkema, ed.. IAHR Monograph Series. Rotterdam: 3rd edition. ROOS, H.-G., STYNES, M. & TOBISKA, L. (996 Numerical Methods for Singularly Perturbed Differential Equations, (Springer Series in Computational Mathematics, Vol. 24. Berlin: Springer. SHANTHI, V.& RAMANUJAN, N.(2002 A numerical method for boundary value problems for singularly perturbed fourth-order ordinary differential equations. Appl. Math. Comput., 29,

18 644 N. MADDEN AND M. STYNES SHISHKIN, G. I.(995 Mesh approximation of singularly perturbed boundary-value problems for systems of elliptic and parabolic equations. Comp. Maths. Math. Phys., 35, THOMAS, G. P. (998 Towards an improved turbulence model for wave-current interactions. In 2nd Annual Report to EU MAST-III Project The Kinematics and Dynamics of Wave Current Interactions, Contract No MAS3-CT95-00.