A coupled system of singularly perturbed semilinear reaction-diffusion equations

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1 A coupled system of singularly perturbed semilinear reaction-diffusion equations J.L. Gracia, F.J. Lisbona, M. Madaune-Tort E. O Riordan September 9, 008 Abstract In this paper singularly perturbed semilinear reaction-diffusion systems are examined. A numerical method is constructed for these systems which involves an appropriate layer adapted piecewise-uniform mesh. The numerical approximations generated from this method are shown to be uniformly convergent with respect to the singular perturbation parameters. Numerical experiments supporting the theoretical results are given. Key words: Singular perturbation, semilinear problem, reaction-diffusion problems, uniform convergence, coupled system, Shishkin mesh. 000 Mathematics Subject Classification : 65N06, 65N, 65N30 Introduction In this paper we consider semilinear systems of the form Tu := Eu + b(x,u) = 0, x Ω = (0,), u(0) = a, u() = b, (a) b(x,u) = (b (x,u),...,b m (x,u)) T C 4 ( Ω R m ), and (x,u) Ω R m we assume that the nonlinear terms satisfy b i u i β ii > 0, β ij b i u j 0, if i j, m β ij > β > 0, j= i =,...,m. (b) (c) where E = diag{ε,...,ε m} is a diagonal matrix, 0 < ε... ε m and u = (u,...,u m ) T. This research was partially supported by the Diputacion General de Aragon. Departamento de Matemática Aplicada, Universidad de Zaragoza, Spain. Departamento de Matemática Aplicada, Universidad de Zaragoza, Spain Laboratoire de Mathématiques Appliquées, Université de Pau et des Pays de l Adour, France School of Mathematical Sciences, Dublin City University, Ireland

2 Reaction-diffusion systems arise for example in catalytic reaction theory. The simplified physical problem involves an isothermal reaction which is catalyzed in a pellet. The unknowns u are the normalized concentrations, b represents the nonlinear kinetics, x is the normalized distance and the diffusion term is the reciprocal of the Thiele modulus (see [] for additional details). In the scalar case, the continuous solutions of semilinear and quasilinear problems have been analyzed using the method of matched asymptotic expansions (see for example [6], [0], [], [3]). In [5] and [9] information about the layer structure for linear singularly perturbed reaction diffusion systems, was obtained via linear decompositions of the solution into regular and singular components. Here we show that these techniques are applicable to a nonlinear system. In we give the asymptotic behaviour of the solution by defining an appropriate decomposition of the solution into a regular and singular components. In 3 we present the nonlinear finite difference scheme and we prove that the resulting approximations are first order uniformly convergent. In 4, some numerical results are presented which indicate that a uniform rate closer to second order may be possible. In the appendices, we show that second order (up to logarithmic factors) can be established in the special cases of two equations (in Appendix ) and in the case of ε i = ε, i =,...m (in Appendix ). For any v,w R m, we write v w if v i w i, i and v := ( v,..., v m ) T ; f := max x f(x) and f := max i f i ; C := C(,...,) T is a constant vector and C denotes a generic positive constant independent of (ε,...,ε m ) and the discretization parameter. Singularly perturbed semilinear systems Conditions (b), (c) and the implicit function theorem ensure that there exists a unique solution u (C 4 ( Ω)) m and the corresponding reduced problem b(x,r) = 0, x Ω, also has a unique solution in r (C 4 ( Ω)) m. We also note that the conditions given in (c) for the Jacobian matrix J where ( ) bi J(x,y) := (x,y). u j are the natural extension of the linear case [9] for the coupling matrix. These conditions guarantee that J is an M matrix for all (x,y) Ω R m. To deduce the asymptotic behaviour of the solution, we consider the following decomposition u = v + w + w R, where the regular component v is the solution of the problem Ev + b(x,v) = 0, x Ω, v(0) = r(0), v() = r(), ()

3 and the singular components w,w R are the respective solutions of the problems Ew + (b(x,v + w) b(x,v)) = 0, x Ω, w(0) = (u v)(0), w() = 0, Ew R + (b(x,v + w + w R) b(x,v + w)) = 0, x Ω, w R (0) = 0, w R () = (u v)(). (3) (4) Note (c) guarantees existence and uniqueness of v,w,w R and it will also be used below to establish existence and uniqueness for several further decompositions of these components. Below we state bounds on the derivatives of the left layer component w. The corresponding bounds on the right layer component w R are obtained by simply replacing x with x. Lemma. The regular component v satisfies dk v dx k C, k = 0,,, dk v i dx k Cε k i, k = 3,4, i =,...,m. (5) Proof. We consider the secondary decomposition of v = m q [i], where d q [m] E m dx + b(x,q [m] ) = 0, x Ω, (q [m] ) m (0) = r m (0), (q [m] ) m () = r m (), (6) d q [j] m m m E j dx + b(x, q [i] ) b(x, q [i] ) = ε j (q [i] ) j e j, x Ω, i=j i=j+ i=j+ (q [j] ) i (0) = (q [j] ) i () = 0, j i m, j < m, (7) with the matrix E i is the zero matrix except that on the main diagonal (E i ) jj = ε j,j i, (note that in this notation E = E) and e i is the i th vector of the canonical base. Conditions (c) imply that q [m] (0) = r(0), q [m] () = r() and q [j] (0) = q [j] () = 0 for j < m. To obtain estimates for the component q [m], we introduce the function z = q [m] r. Note that it is the solution of the problem i= E m z + b(x,r + z) b(x,r) = E m r, z(0) = z() = 0, which can be written as E m z + s=0 J(x,r + sz)ds z = E m r. The conditions (c) ensure that a maximum principle holds for this system, z Cε m and hence z m C. Using a Taylor expansion and these bounds, it follows that z m C. Using the fact that q [m] = z + r, where r is the solution of the reduced problem, we conclude that dk (q [m] ) m dx k C, k = 0,,, and q [m] C. 3

4 In addition, from the nonlinear system b (x,q [m] ) = = b m (x,q [m] ) = 0, we have that dk (q [m] ) i dx k C, k =,, i < m. Differentiating the m th equation of (6) twice and using the bounds for the lower derivatives, we conclude that d 4 (q [m] ) m /dx 4 Cε m. Hence we have d 3 (q [m] ) m /dx 3 Cε m and, using the first m equations of (6), we have that dk (q [m] ) i dx k Cεm k, k = 3,4, i < m. Now we consider the component q [j] with j < m. It is the solution of the problem E j d q [j] dx + 0 J(x, m i=j+ q [i] + sq [j] )ds q [j] = ε j m (q [i] ) j e j, x Ω, i=j+ (q [j] ) i (0) = (q [j] ) i () = 0, j i m. (8) The maximum principle proves q [j] Cε j, and then d (q [j] ) i /dx C(ε j /ε i ) C, j i m. Then, d(q [j] ) i /dx C, j i m, and hence (if j > ) we have d(q [j] ) i /dx C, d (q [j] ) i /dx C, i j. Differentiating twice the differential equation (7) and using the bounds for q [j] and its derivatives up to second order, we deduce that d 4 (q [j] ) i /dx 4 Cε i, i =,...,m. Hence d 3 (q [j] ) i /dx 3 Cε i, i =,...,m. To establish first order error bounds in the case of an arbitrary number of equations, we consider a further decomposition of the singular component w, which is similar to that used in [5] for linear systems. For simplicity, we present the main ideas for the particular case of two equations and these decompositions can be extended to the general case of m semilinear equations. Consider the following decomposition for the left singular component where w [] () = w [] () = 0, w = w [] + w [], (9) E d w [] dx + (b(x,v + w [] ) b(x,v)) = 0, x Ω, b (0,v(0) + w [] (0)) b (0,v(0)) = 0, w [] (0) = w (0), (0) and E d w [] dx + (b(x,v + w) b(x,v + w [] )) = 0, x Ω, () w [] (0) = w (0) w [] (0), w[] (0) = 0. Below we will see that the components w [] depend weakly on ε and the appearance of w [] requires that w (0) w [] (0) 0. Moreover, if ε = ε, it is not 4

5 necessary to decompose w into these subcomponents to perform the numerical analysis. For the sake of simplicity, we introduce the following notation where β is defined by (c). B ε (x) = e xβ/ε, Lemma. The component w [] satisfies for any x Ω Proof. Note that dk w [] dx k (x) Cε k (x), k = 0,,, (a) d3 w [] dx 3 (x) C(ε,ε )T ε (x). (b) E d w [] dx + s=0 J(x,v + sw [] )ds w [] = 0, from which it follows that w [] (x) CB ε (x). Then, from the second equation in (0) we can deduce that dk w [] dx k (x) Cε k (x), k = 0,,. (3) To obtain bounds for the first component, we consider the decomposition where ε w [] = p [] + r [], r [] 0 b (x,v + p [] ) b (x,v) = 0, (4a) d r [] dx + b (x,v + p [] + r [] ) b (x,v + p [] ) = ε d p [] dx. (4b) As p [] w [] this is simply a decomposition of the first component w [] that the condition on the coefficients (c) means that p [] p [] () = w[] (). Therefore r[] Writing (4a) in the form i= 0 (0) = r[] (0) = 0. b u i (x,v + sp [] )p [] i ds = 0, (0) = w[]. Note (0) and and using (b) and (3), we deduce that p [] (x) CB ε (x) for any x Ω. Differentiating (4a) and grouping terms, we have ( b (x,v + p [] ) b (x,v) ) + ( u b (x,v + p [] ) u b (x,v) ) dv x dx + u b (x,v + p [] ) dp[] dx = 0, 5

6 where u b := ( b u, b u ) T. Note if b (x,u + v) b (x,u) = Q(x), then [ b (x,u + v) b (x,u) ] = [ x x which implies that i= 0 b u i (x,u + tv)v i dt ], v C Q + C v + C v + C v. x x x From these expressions, (b) and (3), we have that dp[] (x) Cε dx B ε (x). Use the same argument to prove d p [] (x) Cε dx (x). The remainder is the solution of the following problem ε r [] d r [] dx + ( (0) = r[] b (x,v + p [] + sr [] 0 u )ds) r [] = ε () = 0. d p [] dx, The maximum principle proves that r [] (x) Cε ε (x). Hence, dk r [] Cε k (x), k =,. For the third derivatives, differentiating (0), we have E d3 w [] dx 3 dx k (x) + x (b(x,v + w[] ) b(x,v)) + (J(x,v + w [] )) J(x,v))) dv dx +J(x,v + w [] )) dw[] dx = 0. Therefore, ε w [] i d3 i dx 3 Cε (x), i =,. Lemma 3. The component w [] satisfies for any x Ω and i =, [] w (x) C(Bε (x) + ε B ε (x)), [] w (x) ε C B ε (x), dw[] i ε i ε ε (5a) dx (x) C(ε B ε (x) + ε (x)), (5b) d w [] i dx (x) C(Bε (x) + ε B ε (x)), (5c) ε ε d3 w [] i i dx 3 (x) C(ε B ε (x) + ε (x)). (5d) 6

7 Proof. Decompose w [] further into the following sum w [] = z [] + s [], where z [] = (z,z ) T, z [] (0) = w [] (0), z [] () = w [] () = s [] (0) = s [] () = 0 and for x Ω, ε ε d z [] dx + b (x,v + w [] + (z [],0)T ) b (x,v + w [] ) = 0, d z [] dx + b (x,v + w [] + z [] ) b (x,v + w [] ) = 0, (6) ε ε d s [] dx + b (x,v + w [] + z [] + s [] ) b (x,v + w [] + (z [],0)T ) = 0, d s [] dx + b (x,v + w [] + z [] + s [] ) b (x,v + w [] + z [] ) = 0, Note, that the first equation of (6) can be written as ε d z [] ( b ) dx + (x,v + w [] + s(z [] 0 u,0)t )ds z [] = 0. (7) Then, from the maximum principle, we have dk z [] dx k Cε k B ε (x), k = 0,,. We write the second differential equation of (6) as follows ε d z [] ( dx + b ) ( (x,v+w [] +tz [] )dt z [] b ) = (x,v+w [] +tz [] )dt 0 u 0 u (8) If ε ε, then the maximum principle proves z [] (x) CB ε (x). For the remaining case of ε ε to obtain appropriate bounds of the function z [], we observe that b (x,v + w [] + tz [] )dtz [] 0 u C B ε (x), and we consider the barrier function [5] which is the solution of the problem ε Z + β Z = C B ε (x), Z(0) = Z() = 0. (9) This allows one show that, if ε ε, then Thus, for all ε ε, we have z [] (x) Z(x) Cε ε (x). z [] (x) Cε ε (x). z []. 7

8 Using the differential equation (8) Hence, ε d z [] dx (x) C(B ε (x) + ε ε (x)) CB ε (x). dz[] dx (x) Cε ε (x). To obtain bounds for the remainder s [], we rewrite problem (7) in matrix vector form E d s [] dx + J(x,v + w [] + z [] + ts [] )dt s [] 0 = (b (x,v + w [] + (z [],0)T ) b (x,v + w [] + z [] ),0) T s [] (0) = s [] () = 0. Note that the first component of the right hand side can be written as b (x,v + w [] + (z [],0)T ) b (x,v + w [] + z [] ) b = (x,v + w [] + (z [] u,tz[] )T )dtz []. Then, the maximum principle proves Hence, 0 s [] (x) Cε ε (x). dk s [] (x) C(ε k dxk,ε k )T ε ε For the third derivatives, differentiating (), we have E d3 w [] dx 3 + x (b(x,v + w) b(x,v + w[] )) +(J(x,v + w) J(x,v + w [] )) d(v + w[] ) dx B ε (x), k = 0,,. Therefore, ε w [] i d3 i dx 3 C(ε B ε (x) + ε (x)), i =,. + J(x,v + w) dw[] dx = 0. 3 Numerical scheme and analysis of its uniform convergence The bounds established in the previous section show that in the case of m =, the solution has two overlapping boundary layers at x = 0 and x = with a width of O(ε i ), i =,. This information allows one design and analyze the uniform convergence of the numerical scheme described below. 8

9 The layer adapted piecewise uniform Shishkin mesh is defined as in the linear case [9]: Consider the transition parameters τ = min {/4,ε /β lnn}, τ = min {τ /,ε /β lnn}. (0) Distribute N/8 mesh points uniformly in each of the subintervals [0,τ ε ), [τ ε,τ ε ), [ τ ε, τ ε ), [ τ ε,) and N/ mesh points uniformly in [τ ε, τ ε ). We denote the mesh points by Ω N = {x i } N i=0, ΩN = Ω N Ω and the local steps size by h j = x j x j, j =,...,N. On the mesh Ω N = {x i } N i=0, consider the following nonlinear finite difference scheme (T N U)(x j ) := (Eδ U)(x j ) + b(x j,u(x j )) = 0, x j Ω N = Ω N Ω, () with U(0) = u(0), U() = u(), and δ Y(x j ) := h j+ + h j ( Y(xj+ ) Y(x j ) h j+ Y(x j) Y(x j ) h j To analyze the convergence of this nonlinear difference scheme we study its stability and we obtain appropriate bounds of the local error. Similarly to [5], we write the operator T N as follows T N = A + B where the linear mapping A is given by the diagonal block matrix A = diag{ ε A, ε A } and A i are tridiagonal matrices of the form ε i 0 0 r r c r + r r c r + A i = , r N rn c r + N 0 0 ε i and r j = h j (h j + h j+ ), r+ j = h j+ (h j + h j+ ), rc j = (r j + r+ j ), and B = diag{b,b } is a diagonal block matrix with B i = diag{0,b i (x,u(x )),b i (x,u(x )),...,b i (x N,U(x N ),0)}, i =,. Lemma 4. The mapping T N : (R N+ R N+ ) (R N+ R N+ ) is continuously differentiable. Also, the Frechet derivative T N satisfies (T N) min{,β }. Therefore, the problem () has a solution and it is an homeomorphism of R N+ onto R N+. ). 9

10 Proof. From (c), the Frechet derivative T N is an M matrix with all of its rows satisfying (T N) ii (T N) ij min{,β } > 0. i j Then from [6], (T N ) exists and (T N) min{,β }. () In addition, in [] the existence of solution is established from (). The Hadamard theorem proves that T N is an homeomorphism. Lemma 5. For any Y and Z mesh functions with Y(0) = Z(0) and Y() = Z(), it holds Y Z Proof. For any Y,Z (R N+ R N+ ), we have min{,β } T NY T N Z. (3) T N Y T N Z = Eδ (Y Z)+ ( J(x,Z+s(Y Z))ds ) (Y Z) = (T N)(Y Z). Then, Y Z 0 min{,β } T NY T N Z. An immediately consequence of Lemma 5 is the uniqueness of the solution to problem (). Lemma 6. The local error associated to the scheme () satisfies Proof. Note that we must bound the same terms T N u CN. (4) T N u(x) = T N u(x) Tu(x) = E (δ u u )(x) E (δ v v )(x) + E (δ w w )(x) than for the linear problem [9] and the derivatives of both the regular and singular components, given in Lemmas, and 3, have a similar behaviour to their linear counterparts. Theorem 7. Let u be the solution of the problem () with m = and U the solution of problem () on the Shishkin mesh Ω N. Then, U u CN. 0

11 Proof. Note that Lemma 5 proves T N u = T N u T N U min{,β } U u. Using Lemma 6 the results follows. Remark 8. Combining the decomposition given in this paper with the ideas developed in [4], it is possible to extend the convergence result in Theorem 7 to problem () with an arbitrary number of equations. In the case of m equations, the piecewise uniform mesh is defined as follows: the domain is divided into the subintervals [0,τ ε ],[τ ε,τ ε ],...,[τ εm, τ εm ],...,[ τ ε,]. Distribute half the mesh points uniformly in the main subinterval (τ εm, τ εm ] and the other half in the remaining intervals, distributing N/(4m) + mesh points uniformly in each (τ εi,τ εi+ ], using the following transition points τ εm = min {0.5,ε m /β lnn}, τ εi = min { 0.5τ εi+,ε i /β lnn }, i < m. (5) 4 Numerical experiments We consider a problem of type () where the nonlinear reaction term is b (x,u) = (u ( u ) 3 )+e u u, b (x,u) = (u 0.5 (0.5 u ) 5 )+e u u, and zero boundary conditions. The corresponding nonlinear systems of equations associated with the discrete problem are solved using Newton s method with zero as an initial guess. We iteratively compute U k (x j ), for k =,,...,K until U K (x j ) U K (x j ) N. To estimate the pointwise errors U K (x j ) u(x j ) we calculate a new approximation {ÛK (x j )} on the mesh {ˆx j } that contains the mesh points of the original mesh and its midpoints, i.e., ˆx j = x j, j = 0,...,N, ˆx j+ = (x j + x j+ )/, j = 0,...,N. At the coarse mesh points we calculate the uniform two-mesh differences and the orders of convergence d N,K i = max S ε max 0 j N UK i (x j ) ÛK i (x j ), p N,K i,uni = log (d N,K i /d N,K i ), i =,, where the singular perturbation parameters take values in the set S ε = {(ε,ε ) ε = 0,,..., 30, ε = ε, ε,..., 59, 60 }. The singular perturbation parameters vary over different ranges to allow the differences to stabilize. In Table we display the uniform two-mesh differences and the approximate orders of convergence for both components u and u. We observe uniform convergence of the finite difference approximations. Finally, we would like to report that K 4 for all (ε,ε ) S ε and all N = j, j = 5,...,.

12 Table : Uniform two-mesh differences d N,K and orders of convergence p N,K uni (ε, ε ) S ε N=3 N=64 N=8 N=56 N=5 N=04 N=048 N=4096 d N,K 6.86E-3 6.E E-3.33E E-4.43E E-5.9E-5 p N,K,uni d N,K 8.30E E-3.53E E-4.736E E-5.644E E-6 p N,K,uni References [] N.S. Bakhvalov, On the optimization of methods for boundary-value problems with boundary layers. J. Numer. Meth. Math. Phys., 9 (969) (in Russian). [] K.W. Chang, F.A. Howes, Nonlinear singular perturbation phenomena: Theory and applications, Springer Verlag, Berlin, 984. [3] P.A. Farrell, A.F. Hegarty, J.J.H. Miller, E. O Riordan and G.I. Shishkin, Robust Computational Techniques for Boundary Layers, Chapman and Hall/CRC Press, Boca Raton, U.S.A., 000. [4] J.L. Gracia, F.J. Lisbona, E. O Riordan, A system of singularly perturbed reaction-diffusion equations, DCU School of Mathematical Sciences preprint MS-07-0, (007). [5] J.L. Gracia, F.J. Lisbona, E. O Riordan, A coupled system of singularly perturbed parabolic reaction-diffusion equations, Adv. Comp. Math., (to appear). [6] F.A. Howes, R.E. O Malley Jr., Singular perturbations of semilinear second order systems, Lectures Notes in Mathematics, 87 Springer Verlag, Berlin, 980, [7] T. Linß, N. Madden, Accurate solution of a system of coupled singularly perturbed reaction-diffusion equations, Computing, 73 (004), 33. [8] T. Linß, N. Madden, Layer-adapted meshes for a system of coupled singularly perturbed reaction-diffusion equations, IMA J. Numer. Anal. (to appear) [9] N. Madden, M. Stynes, A uniformly convergent numerical method for a coupled system of two singularly perturbed linear reaction-diffusion problems, IMA J. Numer. Anal., 3 (003) [0] R.E. O Malley Jr., Introduction to singular perturbations, Academic Press, New York, 974.

13 [] R.E. O Malley Jr., Singular perturbation methods for ordinary differential equations, Springer, Berlin, 99. [] J.M. Ortega, C. Rheinboldt, Iterative solution on nonlinear equations in several variables, Academic Press, New York and London, 970. [3] H.G. Roos, M. Stynes, L. Tobiska, Numerical Methods for Singularly Perturbed Differential Equations. Convection Diffusion and Flow Problems, Springer Verlag, New York, 996. [4] G. I. Shishkin, Discrete approximation of singularly perturbed elliptic and parabolic equations, Russian Academy of Sciences, Ural section, Ekaterinburg, 99. (in Russian) [5] G. Sun, M. Stynes, An almost fourth order uniformly convergent difference scheme for a semilinear singularly perturbed reaction diffusion problem, Numer. Math. 70 (995) [6] R.S. Varga, On diagonal dominance arguments for bounding A, Linear Algebra Appl., 4 (976) 7. [7] R. Vulanović, On a numerical solution of a type of singularly perturbed boundary value problem by using a special discretization mesh, Rev. Res. Faculty Sci. Uni. Novi-Sad, 3 (983) Appendix : Special case of two equations In this appendix we show that it is possible to achieve almost second order convergence in the particular case of a system of two semilinear equations. Assume throughout this appendix that b is sufficiently smooth for the analysis to be valid. To obtain similar bounds of the error to those given in [7], we employ a decomposition of the solution u which is based on the decomposition in [9]. First, similarly to [7], we obtain an additional bound on the third derivative of the first component of the regular component. Note that the decomposition used in the proof of Lemma 9 is different to what was used in establishing the corresponding result in [7]. Lemma 9. For m =, the regular component satisfies v Cε. Proof. Let where v = q [] + t + s diag{0,ε } d q [] dx + b(x,q [] ) = 0, x Ω, (q [] ) (0) = r (0), (q [] ) () = r (), diag{0,ε }t + b(x,q [] + t) b(x,q [] ) = (ε d (q [] ) dx,0) T, x Ω, t (0) = 0 = t () Es + b(x,v) b(x,q [] + t) = (ε t,0) T, x Ω, s(0) = s() = 0. 3

14 As in the proof of Lemma we have that Note also that and dk q [] dx k C,k = 0,,, dk q [] dx k Cε k,k = 3,4,5. t Cε, t Cε ε, t Cε ε, t Cε ε, t Cε ε, t Cε ε 4, t Cε ε 3, t Cε ε 3 s Cε 4 ε, s Cε ε, s Cε 4 ε 4, s Cε 3 ε, s Cε 4 ε 3, s Cε ε 3 + Cε ε + Cε ε 3, s Cε 3 ε 4 + Cε 4 ε 5 Cε. In the next result we show that it is possible to have similar bounds to those in [7] and [9] for the left singular component. Thus in the particular case of two equations, it is not necessary to establish a further decomposition of the component w. Lemma 0. For m = the singular component w satisfies w i (x) CB ε (x), i =, (6a) w (k) i (x) C ε k l B εl (x), k =,, i =, (6b) l=i w (x) C(ε 3 B ε (x) + ε 3 (x)), (6c) w (x) Cε (ε B ε (x) + ε (x)), (6d) w (4) (x) C(ε 4 B ε (x) + ε 4 (x)), (6e) w (4) (x) Cε (ε B ε (x) + ε (x)). (6f) Proof. We define the following auxiliary linear operator L := E d ( dx + B, B = b i (x,v + sw)ds u j 0 ). i,j The singular component w is the solution of the boundary value problem Lw = 0, w(0) = (u v)(0), w() = (u v)(). (7) Applying the maximum principle and using the diagonal dominance assumption in (c), we deduce that w(x) CB ε (x). Then, we easily deduce (see [],[9]) that w (k) i (x) Cε k i B ε (x), k =,, i =,. (8) 4

15 We need to sharpen the bounds associated with the first component. We follow the argument in [9] for the linear problem. Differentiating the first equation ε w + (b (x,v + w) b (x,v)) = 0 in the system, we obtain the following problem for w L w := ε (w ) + b (x,v + w)w = h(x,v,w), x Ω, u (9) w (x) Cε, x {0,}, where the bounds on the boundary are obtained using the bound (8) and ( b h(x,v,w) = x (x,v) b ) ( (x,v + w) + v b (x,v) b ) (x,v + w) ( x u u v b (x,v) b ) (x,v + w) b (x,v + w)w u u u. Writing the three terms in brackets in integral form g(x,v + w) g(x,v) = i= 0 g u i (x,v + sw)ds w i, and using the hypothesis on b, the bounds for v, w and w, we deduce that h(x,v,w) Cε (x). Then, the maximum principle for problem (9) with the barrier function ψ(x) = C(ε B ε (x) + ε (x)), proves the estimate (6b) for k =. Using a similar argument, the estimate (6b) for the second order derivative can also be deduced. Now we consider the third order derivatives. Differentiating the equation (7) and using the estimates on the lower derivatives of w it follows that Ew C(ε B ε (x) + ε (x)). This establishes the appropriate bounds on w. For the first component w, use the same argument as for the first and second order derivatives. The previous ideas can be also applied to obtain the required bounds for the fourth derivatives. To obtain second order convergence, instead of applying Lemma 5, we use the following comparison principle. Lemma. Given three mesh functions Y, Z, Y such that T N Y T N Z T N Y, where T N is defined in 3, then Y Z Y. Proof. Note that Y Z = (T N) (T N Y T N Z) 0, since T N is an M matrix. Use the same argument to prove that Y Z. 5

16 Theorem. Let u be the solution of the problem () with m = and U the solution of problem () on the Shishkin mesh Ω N. Then, U u CN ln N. Proof. From the bounds given in Lemmas, 9 and 0, it follows from the arguments in [7] for the corresponding linear problem, that CN ε ε (ε,ε ) T + CN, if x i = τ, τ, T N u CN ε (ε,ε ) T + CN, if x i = τ, τ C(N lnn), otherwise. Consider the barrier function Z (x) = C(N ln N)(θ (x)+θ (x)+)+u N (x), where for i =, x, if x [0,τ i ], τ i θ i (x) =, if x [τ i, τ i ], x, if x [ τ i,]. τ i These functions satisfy CN δ θ (x) ε ε (ln N), if x = τ, τ, 0, otherwise, CN δ θ (x) ε (ln N), if x = τ, τ, 0, otherwise,, Therefore, we conclude that (u U N )(x i ) CN ln N. Remark 3. Note that in the case of a linear system of m equations Linß and Madden [8] have established second order convergence in the case of arbitrary ε i, under the assumption that the elements in the coefficient matrix B(x) of the zero order terms satisfy the conditions b ii (x) > 0, m b ik(x) <, i m. b ii (x) k i 6

17 Appendix : Second order in the special case of ε i = ε for all i =,...,m,. For an arbitrary number of equations and in the special case of ε =... = ε m = ε in () it is possible to prove essentially second order convergence, i.e., U u CN ln N. Assume throughout this appendix that b is sufficiently smooth for the analysis to be valid. To prove second order convergence, we use the following auxiliary result Lemma 4. The regular component v defined in () satisfies for i =,...,m, dk v dx k C, k = 0,,, dk v i dx k Cε k, k = 3,4. (30) The left singular component w defined in (3) satisfies for i =,...,m, dk w i dx k (x) Cε k B ε (x), k = 0,,,3,4. (3) Proof. See Lemma for (30). Follow the ideas given in [7] and in this paper to complete the argument. Theorem 5. Assume that ε i = ε for all i =,...,m,. Let u be the solution of the problem () and U the solution of problem () on the Shishkin mesh (5). Then, U u CN ln N. Proof. Now the local error satisfies C(N lnn), if x i < τ, x i > τ, T N u CN ε, if x i = τ, τ, CN, if x i (τ, τ). where τ and τ are the only transition point of the piecewise uniform Shishkin mesh. Consider the barrier function Z (x i ) = C(N lnn) (θ(x i )+)+U N (x i ), where Then, we have θ(x) = x τ, if x [0,τ],, if x [τ, τ], x, if x [ τ,]. τ T N Z = C(N lnn) ε δ θ ε δ U N + b(x,z ) = C(N lnn) ε δ θ + b(x,z ) b(x,u N ) = C(N lnn) ε δ θ + J(U N + s(z U N ))(Z U N ) C(N lnn) ε δ θ + β (Z U N ) T N u. For the barrier function Z (x i ) = C(N lnn) (θ(x i )+)+U N (x i ), it holds that T N Z T N u. Hence Z u Z and therefore (u U N )(x i ) C(N lnn). 7