A SINGULARLY PERTURBED CONVECTION DIFFUSION PROBLEM WITH A MOVING INTERIOR LAYER

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1 INTERNATIONAL JOURNAL OF NUMERICAL ANALYSIS AND MODELING Volume 9, Number 4, Pages c 01 Institute for Scientific Computing and Information A SINGULARLY PERTURBED CONVECTION DIFFUSION PROBLEM WITH A MOVING INTERIOR LAYER J.L. GRACIA AND E. O RIORDAN Abstract. A singularly perturbed parabolic equation of convection-diffusion type with an interior layer in the initial condition is studied. The solution is decomposed into a discontinuous regular component, a continuous outflow boundary layer component and a discontinuous interior layer component. A priori parameter-explicit bounds are derived on the derivatives of these three components. Based on these bounds, a parameter-uniform Shishkin mesh is constructed for this problem. Numerical analysis is presented for the associated numerical method, which concludes by showing that the numerical method is a parameter-uniform numerical method. Numerical results are presented to illustrate the theoretical bounds on the error established in the paper. Key Words. Singular perturbation, interior layer, Shishkin mesh. 1. Introduction The solutions of singularly perturbed parabolic equations of convection-diffusion type typically contain boundary layers[1, 6, 16], which can appear along the boundary corresponding to the outflow boundary of the problem. Additional interior layers can form in the solution if either the coefficients, the inhomogenous term or the boundary/initial conditions are not sufficiently smooth []. In this paper, we examine a linear singularly perturbed parabolic problem with smooth data and an interior layer in the solution, which is created by artificially inserting a layer into the initial condition. This problem is motivated from studying a singularly perturbed parabolic problem of convection-diffusion type, with a singularity generated by a discontinuity between the boundary and initial conditions at the inflow corner. At some distance from this inflow corner, the solution is characterized by the presence of an interior layer moving in time along the characteristic of the reduced problem, which passes through the inflow corner. This paper formulates a related problem which captures this effect of an interior layer being transported in a convection-diffusion parabolic problem. Our interest is to design and analyse a parameter-uniform numerical method [6] for such a problem. Parameter-uniform numerical methods for several classes of singularly perturbed parabolic problems of the form εu xx +au x +bu+u t = f, x,t) 0,1) 0,T], with discontinuous coefficients or discontinuous inhomogeneous term) have been constructed and analysed in [4, 15]. These methods are based on upwind discretizations combined with appropriate piecewise-uniform Shishkin meshes [6], which are aligned to the trajectory of the point where the data are discontinuous [15]. In Received by the editors May 6, 010 and, in revised form, March 4, Mathematics Subject Classification. 65L11,65L1. 83

2 84 J.L. GRACIA AND E. O RIORDAN [4, 15], it is assumed that the boundary/initial conditions are sufficiently smooth and compatible at the corners. The nature and width of any interior layers appearing in the solutions of these problems, is dictated mainly by the sign of the convective coefficient a either side of a discontinuity. In this paper, we examine a problem where the coefficient a is smooth and always positive, but the solution contains a strong interior layer, generated solely from the fact that the initial condition contains an internal layer. Parameter-uniform numerical methods for singularly perturbed parabolic problems with a discontinuous initial condition have been examined by Shishkin et al. in a series of papers see [7, 10, 11] and the references therein). Rather than using simple upwind finite difference operators on a piecewise-uniform mesh, Shishkin et al. use suitable fitted operator methods to capture the singularity in the neighbourhood of the discontinuity. In this paper, the initial solution is smooth, but contains an interiorlayer. We will see that it is not necessaryto use a fitted operatormethod here, as a suitable Shishkin mesh combined with a standard upwind finite difference operator suffices to generate a parameter-uniform method. Thispaperisstructuredasfollows: In,westatetheproblemtobeinvestigated. In 3, we employ a mapping [5] which is used to align the mesh to the location of the interior layer. The solution is decomposed into a sum of a regular component, a boundary layer component and an interior layer component. In 4, we examine the regular component and deduce parameter-explicit bounds on its derivatives. In 5 parameter-explicit bounds on the derivatives of the layer components are established, which are central to the design of a piecewise-uniform Shishkin mesh, given in 6. In 7, the associated numerical analysis is presented and in the final 8 some numerical results are given to illustrate the theoretical error bounds established in 7. Notation. The space C 0+γ D), where D R is an open set, is the set of all functions that are Hölder continuous of degree γ with respect to the metric, where for all u = u 1,u ), v = v 1,v ) R, u v = u 1 v 1 ) + u v. For f to be in C 0+γ D) the following semi-norm needs to be finite f 0+γ,D = The space C n+γ D) is defined by C n+γ D) = {z : fu) fv) sup u v, u,v D u v γ. i+j z x i y j C0+γ D), 0 i+j n}, and n+γ, n+γ are the associated norms and semi-norms. Throughout the paper, c or C denotes a generic constant that is independent of the singular perturbation parameter ε and of all discretization parameters.. Continuous problem In this paper, we examine the following singularly perturbed parabolic problem: Find û such that 1a) 1b) 1c) 1d) ˆL ε û := εû ss +ât)û s +û t = ˆfs,t), s,t) Q := 0,1) 0,T], ûs,0) = φs;ε),0 s 1, û0,t) = φ L t), û1,t) = φ R t), 0 < t T, ât) > α > 0,

3 SINGULARLY PERTURBED PROBLEM WITH MOVING INTERIOR LAYER 85 where the initial condition φ is smooth, but contains an interior layer in the vicinity of a point s = d,0 < d < 1 and d is independent of ε. The function φ is defined as the solution of the singularly perturbed two point boundary value problem εφ α s d) 1e) +bs)φ = f 1 s) := qs)+atanh ), ε 1f) bs) β > α, α α, φ0) = φ 0 0), φ1) = φ 0 1), where φ 0 s) is the reduced discontinuous initial condition defined by bs)φ 0 s) := qs) A, s < d; bs)φ 0 s) := qs)+a, s > d. Our focus will be on the influence this interior layer has on the solution of the time dependent problem. Hence, the boundary values 1f) for the initial condition have been chosen to dampen boundary layers appearing in the initial condition. We also design q so that φ 0 0) = φ 0 1) = 0, which further reduces the amplitude of any boundary layers in the initial condition. The zero level of compatibility required such that û is continuous on the boundary Q\Q is specified by 1g) φ L 0) = φ0), φ R 0) = φ1). We can increase the level of compatibility by assuming that 1h) d φ ε ds +âdφ ds + dφ L) 0,0) = ˆf0,0), dt and a similar condition is assumed to hold at the corner 1,0). Differentiate 1a) with respect to time and let ŷ = û t then εŷ ss +âŷ s +ŷ t = ˆf t â t û s, s,t) Q, ŷs,0) = φ 1 s) := ˆfs,0)+εφ s) â0)φ s),0 s 1, ŷ0,t) = φ L t), ŷ1,t) = φ R t), 0 < t T. Imposing the first level compatibility conditions on this problem leads to the second level compatibility conditions for problem 1). At the corner 0, 0), we require that 1i) ε d φ 1 ds +âdφ 1 ds + d φ L dt = ˆf t â t dφ ds, and analogously at the corner 1, 0). With all these compatibility conditions 1g), 1h),1i), at both corners and if â, ˆf C +γ Q) then the solution of problem 1) û C 4+γ Q) [1, pg. 30]. The characteristic curve associated with the reduced differential equation formally set ε = 0 in 1a)) can be described by the set of points Γ := {dt),t) d t) = ât), 0 < d0) = d < 1}. We also define the two subdomains of Q either side of Γ by Q := {s,t) s < dt) < 1} and Q + := {s,t) s > dt) > 0}. The solution of problem 1) will have an interior layer of width O ε) emanating from the initial condition) which travels along Γ. In general a boundary layer of width Oε) will also appear in the vicinity of the edge x = 1. We restrict the size of the final time T so that the interior layer does not interact with this boundary

4 86 J.L. GRACIA AND E. O RIORDAN layer. Since â > 0, the function dt) is monotonically increasing. Thus, we limit the final time T such that 0 < c < 1 dt). We define the parameter ) δ := 1 dt) > 0, 1 d which plays an important role throughout this paper. In later sections, we construct a piecewise-uniform mesh, which is designed to be refined in the neighbourhood of the curve Γ. To analyse the parameter-uniform convergence of the resulting numerical approximations on such a mesh, it is more convenient to perform the analysis in a transformed domain where the location of the interior layer is fixed in time. As most of the paper deals with this transformed domain, we have adopted the notation ûs,t) for the solution in the original domain and we use the simpler notation of ux,t) for the solution in the transformed domain. 3. Mapping to fix the location of interior layer Consider the map X : s,t) x,t) given by d dt) s, s dt), 3) xs,t) = 1 1 d 1 s), s dt). 1 dt) Note that x = s at t = 0 and x = d for all t such that s = dt). This maps 4) Q Ω := [0,d] [0,T], Q+ Ω + := [d,1] [0,T]. Remark 3.1. We employ the following notation in subsequent sections: ux,t) := ûxs,t),t). Noting that d t) = ât),d0) = d, we have for s < dt) or x < d that û t = at) dt) x u x + u t, û s = d dt) u x. Using this map, the differential equation 1a) transforms into 5a) L ε u := εu xx +κx,t)u x ) +gx,t)ut = gx,t)fx,t), x d, 5b) ux,0) = φx;ε), 0 < x 1,u0,t) = φ L t), u1,t) = φ R t), 0 < t T, dt)at) x) 1, x < d, d d 5c) κx,t) := 1 dt))at) 1 x) 1, x > d, 1 d 1 d dt)), x < d, d 5d) gx,t) := 1 dt)), x > d. 1 d Note that for any t such dt) d, then gd,t) gd +,t). Also for x > d, gx,t) δ > 0, αδ x d κx,t) a x d 1 d 1 d. In this transformed problem, the coefficient κx, t) of the first derivative in space is positive, except along the internal line x = d where it is zero. To have a well-posed problem for u, we seek a solution u C 1 Ω) satisfying 5). As û C 4+γ Q) then u C 4+γ Ω ) C 4+γ Ω + ) ) C 1 Ω).

5 SINGULARLY PERTURBED PROBLEM WITH MOVING INTERIOR LAYER 87 Note that, by using a comparison principle for the steady state differential operator εw + bw)x), x 0,1); wx), x {0,1}, we deduce that φx) C, where φ is defined in 1e). We also associate the following differential operator L εωx,t) := ωx,t), x = 0,1, t 0, εω xx x,0)+bx)ωx,0), x = 0,1), εω xx +κx,t)ω x +gx,t)ω t, x = 0,d) d,1), t > 0, [ω x ], x = d, t 0, with the problem 5) where [z]d,t) := zd +,t) zd,t). For the operator L ε a comparison principle holds. Theorem 1. Assume that a function ω C 0 Ω Ω + ) C Ω Ω + ) satisfies L ε ωx,t) 0, for all x,t) Ω Ω + then ωx,t) 0, for all x,t) Ω Ω +. Proof. Consider the continuous function νx, t), defined by { e t εδ ωx,t) = e x d ε νx,t), x,t) Ω, e t εδ e x d 4ε νx,t), x,t) Ω +. Apply a standard proof by contradiction argument, where x d and d are treated seperately. From this, we deduce that the solution of problem 1) satisfies û Q C. 4. Discontinuous regular component The initial function φ defined in 1e) is assumed to be sufficiently smooth and can be split into the sum of two discontinuous functions φ v,φ w, where φ = φ v +φ w and εφ v +bφ v = bφ 0, s d, φ v 0) = φ0), bd)φ v d ± ) = bd)φ 0 d ± )+εφ 0d ± ), φ v 1) = φ1), εφ w +bφ w = f 1 s) bφ 0, s d. Thediscontinuouscomponentφ w satisfiesalinearsecondorderdifferentialequation on the two subintervals 0,d) and d,1). Hence, φ w is well defined when we impose the additional four conditions: [φ w ]d) = [φ v ]d), [φ w]d) = [φ v]d), φ w 0) = φ w 1) = 0. We can further decompose the regular component φ v such that φ v s) = φ 0 s)+εφ 1 s)+ε φ s), s d, where bφ 1 = φ 0, εφ +bφ = φ 1,s d,φ 0) = φ 1) = φ d) = 0. For s d, one can establish the following parameter-explicit bounds on the derivatives of the individual components φ i,i = 0,1, φ 0 s) k C, φ 1 s) k C, φ s) k C1+ε k/ e α ε s d ). For the layer component φ w, we see that on 0,d) εφ w +bφ w = Ae α ε d s) 1+e α ε d s), φ wd ) C.

6 88 J.L. GRACIA AND E. O RIORDAN Use the barrier function Ce α ε d x) and a maximum principle to bound φ w x) on the interval [0,d]. Then following the arguments in [3], one can deduce that at all points s d and for all k 7 6a) 6b) φ v s) k C1+ε k/ ), φ w s) k Cε k/ e α ε s d. On the domain Q we will define the regular left component ˆv s,t) such that ˆL εˆv = ˆf, s,t) Q,ˆv 0,t) = φ L t), ˆv s,0) = φ v s)+φ c s). In the next theorem, we construct a function rt) = ˆv dt),t) satisfying second order compatibility at the corner d,0), so that there are no layers in ˆv in the vicinity of dt),t). Recall that φ,φ L and ˆf satisfy compatibility up to second order at the inflow corner 0,0). The function φ c s) is a polynomial designed so that φ v,φ L and ˆf satisfy compatibility up to second order and φ c s) 0,s 0.5d. From the bounds on φ w in 6a) one can then deduce that φ c s) k Ce α ε d, k 7. Theorem. There exist functions r 0 t),r 1 t),r t) such that the solutions ˆv, ˆv + of the problems ˆL εˆv = ˆf, s,t) Q, ˆv 0,t) = φ L t), ˆv s,0) = φ v s)+φ c s), ˆL εˆv + = ˆf, s,t) Q +, ˆv + s,0) = φ v s), ˆv dt),t) = r 0 t), ˆv + dt),t) = r 1 t), ˆv + 1,t) = r t), satisfy ˆv C 4+γ Q ), ˆv + C 4+γ Q + ) and the bounds 7a) 7b) 7c) ˆv Q C, ˆv + Q + C, j+mˆv Q s j t m C1+ε j+m) ), 1 j +m 4, j+mˆv + Q s j t m + C1+ε j+m) ), 1 j +m 4. Proof. The main argument is similar to the proofs given in [13, Theorem 3.3] and [15, Theorem 3.3]. Using smooth extensions of the data, the problem ˆL εˆv = ˆf,s,t) Q can be extended to the domain 0,) 0,T] so that ˆL εˆv = ˆf, ˆv 0,t) = φ L t), ˆv s,0) = φ v +φ c )s), ˆv,t) = r t), where φ v,φ c are extensions of φ v,φ c to the interval [0,] so that the bounds 6a) are applicable for all s [0,1]. Then boundary data r t) can be specified so that ˆv C 4+γ [0,] [0,T]). Using the expansion ˆv = ˆv 0 + εˆv 1 + εˆv, where ˆv 0 is the solution of the extended reduced problem ˆL 0ˆv 0 := â s + t )v 0 = ˆf,ˆv 00,t) = ˆv 0,t),ˆv 0s,0) = ˆv s,0), and ˆL 0ˆv 1 = ˆv 0) then the bounds in 7a)-7b) hold. We now choose r 0 t) := ˆv dt),t). In the case of ˆv +, we again extend the problem to the domain,) 0,T] so that ˆL ε ˆv = ˆf, ˆv s,0) = φ v s).

7 SINGULARLY PERTURBED PROBLEM WITH MOVING INTERIOR LAYER 89 Boundary data can be chosen on the boundaries s =,s = so that the bounds 6a) apply to the extension ˆv. Then we choose r 1 t) := ˆv dt),t), r t) := ˆv 1,t). The fact that the initial condition φ depends on ε introduces additional issues into the proof of this result. These technical issues are addressed in the appendix. Note that for ˆv we use a transformation in space of the form [0,d] [0,1] and for ˆv +, [d,1] [0,1]) before using the arguments from the appendix. 5. Layer Components The numerical method presented below in 6 is specified in the transformed domains Ω Ω +. In the subsequent numerical analysis section, it is more convenient to have bounds on the derivatives of the layer components in the transformed variables. Note that u,v + C 4+γ Ω + ) and so L ε u v + ) = 0, and u v + C 4+γ Ω + ). Hence, u v + ) and φ φ v ) satisfy second order compatibility conditions at the corner 1,0). We further decompose the function u v + into two subcomponents u v + = w+z, where L ε w = L ε z = 0, and wd,t) = z1,t) = 0. The boundary layer function w is the solution of 8a) 8b) L ε w = 0, x,t) Ω + ; w 0, x,t) Ω ; wd,t) = 0, wx,0) = φ T x)e αδ ε 1 x), w1,t) = u1,t) v + 1,t), where φ T x) is a polynomial function such that w satisfies the second order compatibility conditions at the corners d,0),1,0). At the corner 1,0), we specify the values of φ k) T 1),0 k 4 so that φ T1) = 0, εφ w 1)+a0)φ w 1) = εφ Tx)e αδ ε 1 x) ) 1)+a0)φ T x)e αδ ε 1 x) ) 1), and, in addition, second order compatibility is also satisfied. For example, by setting φ k) T d) = 0,0 k, φ T 1) = φ w 1) and φ T 1) = φ w αδ ε φ w )1) we ensure that the first order compatibility condition is satisfied at both corners. Using the bounds on the derivatives of φ w at s = 1, we can deduce that φ k) T 1) Ce 3α ε 1 d),0 k 5. Hence, we can have that φ T x) k Ce 3α ε 1 d), 0 k 5. Theorem 3. The solution of 8) w C 0 Ω) C 4+γ Ω + ) and wx,t) Ce αδ ε 1 s=x s d 1 d ds e αt 1 d)δ, x,t) Ω +, j w x j x,t) Cε j e αδ1 x) m w t m x,t) Cε 1 m, x,t) Ω +, m = 1,. Proof. Consider the barrier function Note first that for x d, Φ 1 x,t) := Ce αδ ε ε, 1 j 4, x,t) Ω +, 1 s=x wx,0) Ce αδ ε 1 x) Ce αδ ε s d 1 d ds e αt 1 d)δ. 1 s=x s d 1 d ds,

8 830 J.L. GRACIA AND E. O RIORDAN and for x,t) Ω +, L ε Φ 1 αδ κ x d ε αδ x d) ) Φ 1 1 d)ε 1 d 0. This yields the first bound on wx,t). Introduce the stretched variables ζ = 1 x)/ε, τ = t/ε. Then wζ,t) satisfies w ζζ κ w ζ + g w τ = 0, ζ,τ) 0, 1 d ) 0, T ε ε ), wζ,0) = φ T ζ)e αδζ, w 1 d,τ) = 0, w0,τ) = u1,τ) v + 1,τ). ε We employ the a priori interior estimates in [8, pg. 13] or [1, pg. 35], where we also split the region between ζ 1 and ζ 1 see also the argument in [13, Theorem 4]). Transforming back to the original variables and using wx,t) Ce αδ1 x) ε, i wζ,0) ζ i Ce α ε 1 d) e αδζ, 0 i 5, we deduce that 9) j+m w x j t mx,t) Cε j ε m e αδ1 x) ε, 1 j +m 4, x,t) Ω +. To obtain the final bound on the time derivatives, we follow the argument in [14, Lemma 3.9]. Note that wx,t) = wx,0)+w1,t) w1,0))φx,t) +εrx,t), where εφ xx + κ1,t)φ x = 0, φd,t) = 0,φ1,t) = 1. As in [14], one can deduce that ε R t C,ε R tt C. The discontinuous multi-valued interior layer function z is defined as the solution of 10a) 10b) 10c) L ε z = 0, x,t) Ω Ω +, [z]d,t) = [v]d,t), [z x ]d,t) = [v x +w x ]d,t), t 0, z0,t) = z1,t) = 0, zx,0) = φ w x) φ c x) wx,0), x d. Theorem 4. The solution of 10) z C 4+γ Ω ) C 4+γ Ω + ) and we have 11a) 11b) zx,t) Ce α ε d x),x,t) Ω, a zx,t) Ce e δ1 d) T α ε x d), x,t) Ω +. If d1 d dt) ) < x d and 0 t T, then j z x jx,t) Cε j/ e α 11c) ε d x), 1 j 4, m z 11d) t mx,t) C, m = 1,. If d x < d+1 dt)) and 0 t T, then j z x jx,t) Cε j/ e 1 dt) α 11e) 1 d ε x d), 1 j 4, m z 11f) t mx,t) C, m = 1,.

9 SINGULARLY PERTURBED PROBLEM WITH MOVING INTERIOR LAYER 831 Proof. Recall that z = u v+w) and so z C 4+γ Ω ) C 4+γ Ω + ). Since u,v,w are bounded, we have that z C. Note that zx,0) Ce α ε d x),0 < x < d and α L ε e ε d x)dt) d e αt ) = 0. Thus we have established 1) zx,t) Ce α ε d x)dt) d e αt Ce α ε d x),x,t) Ω. To the right of the interface x = d we introduce the barrier function, Φ x,t) := Ce e C t α ε x d) e C t δ, x,t) Ω +. Note first that zx,0) Φ x,0),d < x < 1 and for δc 1 d) a, L ε Φ α δc αe C t 1 dt)) + ε 1 d) 1 dt))c a )x d)e C t ) Φ α δc αe C t 1 dt)) + ε 1 d) δ1 d)c a )x d)e C t ) Φ 0. Hence, zx,t) Φ x,t), x,t) Ω +. We now again use the interior estimates from [8, pg. 13] or [1, pg. 35]. Let us first determine bounds on the time derivatives of the solution u of 5) along the line x = d. We introduce the time dependent stretched variable 13) ζ := α d x)dt) d ε Note that with ũζ,t) := ûs,t) then û α s = ũ and ε ζ 14) u x = α dt) d ε ũ ζ = α s dt) ε. û t = α at) ũ ε ζ + ũ t, u and t = at) dt) ζ ũ ζ + ũ t. Hence the function ũζ, t) satisfies the Poisson equation αũ ζζ +ũ t = f. On the rectangle R := α, α) 0,T] use Ladyzhenskaya estimates [1, pg. 35]. Hence ũζ,t) 4+γ C, ζ,t) R. Transforming back to the variables x, t) we have 15) m u t m d,t) = m ũ t m d,t) C, m +γ. On the left domain Ω, we use the same time dependent stretched variable 13). The function zζ,t) := zx,t) satisfies the heat equation α z ζζ + z t = 0. Consider the rectangular region ζ,t) S := 0,K) 0,T], where K α d ε. Recall that from 1) we have that Using 7b), 15) and zζ,t) Ce ζ, ζ,t) S. m z m z mˆv tmd,t) = t m0,t), t m dt),t) = m ṽ 0,t), m +γ, tm

10 83 J.L. GRACIA AND E. O RIORDAN we have 16) Note further that 17) m z t m0,t) C, 0 t T, m +γ. j z ζ jζ,0) Ce ζ +Ce α ε d, j 4+γ, where the second term is due to the presence of the function φ c. If 0 ζ 1, from 16) and 17), [1, 10.5)] yields j+m z ζ j t mζ,t) Ce ζ +C Ce ζ, 0 ζ 1, 0 j +m 4, since ζ 1. If 1 < ζ K, use [1, 10.5)] again but now j+m z ζ j t mζ,t) Ce ζ, 1 < ζ K, t 0, 0 j +m 4. Hence j+m z ζ j t mζ,t) Ce ζ, 0 ζ K, t 0, 0 j +m 4. Thus, in particular, the space derivatives satisfy j z x jx,t) Cε j/ e α ε d x), d1 d ) < x d, t 0, 0 j 4. dt) For the time derivatives, noting 14) for the first order time derivative and z t = z ζ ) z ζ + ζ t ζ t t + z ζ ζ t + z t, for the second order time derivative, we deduce that z t x,t) Cζ +1)e ζ C, z t x,t) Cζ +ζ +1)e ζ C, for d1 d dt) ) < x d and 0 t T. An analogous argument applies on the region d,d+1 dt))) 0,T] Ω Numerical method Let N and M be two positive integers. To approximate the solution of problem 5) we use a uniform mesh in time {t j = j t, t = T/M} and a piecewise uniform mesh of Shishkin type in space {x i } N i=1 described below) in the transformed variables x,t). The grid is given by Ω N,M = {t j } M j=0 {x i } N i=0, Γ N,M = Ω N,M Ω\Ω), Ω N,M = Ω N,M \Γ N,M. The local spatial mesh sizes are denoted by h i = x i x i 1, 1 i N. To describe the numerical method we use the following notation for the finite difference approximations of the derivatives Dt Υx i,t j ) := Υx i,t j ) Υx i,t j 1 ), D t xυx i,t j ) := Υx i,t j ) Υx i 1,t j ), h i D xυx + i,t j ) := Υx i+1,t j ) Υx i,t j ), h i+1 δ x Υx i,t j ) := h i +h i+1 D + x Υx i,t j ) D x Υx i,t j )).

11 SINGULARLY PERTURBED PROBLEM WITH MOVING INTERIOR LAYER 833 Discretize problem 5) using an Euler method to approximate the time variable and an upwind finite difference operator to approximate in space. The finite difference equation associated with each grid point is given by 18a) εδxu +κx i,t j )Dx U +gx i,t j )Dt U = gx i,t j )fx i,t j ), x i d, t j > 0, 18b) DxUd,t j ) = D xud,t + j ),t j > 0, 18c) εδ xux i,0)+bx i )Ux i,0) = f 1 x i ), 0 < x i < 1, 18d) U0,t j ) = φ L t j ),U1,t j ) = φ R t j ), t j 0. The space domain is discretized using a piecewise uniform mesh which splits the space domain [0, 1] into four subintervals 18e) [0,d τ 1 ] [d τ 1,d+τ ] [d+τ,1 σ] [1 σ,1], where τ 1 = min{ d ε dt), α lnn}, τ = min{1 dt),e C T ε 18f) α lnn}, 18g) σ = min{ 1 d+τ ), 4ε αδ lnn}, C = a δ1 d). The grid points are uniformly distributed within each subinterval such that x 0 = 0, x N/4 = d τ 1, x N/ = d, x 5N/8 = d+τ, x 7N/8 = 1 σ, x N = 1. In the next section, the error analysis will concentrate on the case when ε 19) τ 1 = α lnn, τ = e C T ε 4ε lnn, σ = α αδ lnn. The other possibilities for the mesh parameters will be dealt with using a classical argument. 7. Numerical Analysis Define the following finite difference operator L N,M by Υx i,t j ), x i = 0,1, t j 0, L N,M Υx i,t j ) = εδx Υx i,0)+bx i )Υx i,0), x i 0,1), εδ x Υ+κDxΥ+gD t Υ ) x i,t j ), x i 0,d) d,1), t j > 0, Dx D+ x )Υx i,t j ), x i = d, t j > 0. Theorem 5. Let Υ be any mesh function defined on Ω N,M. If L N,M Υx i,t j ) 0, x i,t j ) Ω N,M then Υx i,t j ) 0, x i,t j ) Ω N,M. Proof. It is a standard proof by contradiction argument. First establish that Υx i,0) 0 and then apply the standard argument for any t j > 0. The cases of x i = d and x i d are treated separately. Noting that Ux i,0) C, we can conclude from the discrete maximum principle that U C. Let U = V +W+Z, where V, W, Z arethe discrete counterparts

12 834 J.L. GRACIA AND E. O RIORDAN to the continuous components v, w, z. The discrete regular component V is multivalued and defined separately on Ω and Ω + by εδ x +κdx ) +gd t V = f, x i,t j ) Ω, εδx V x i,0)+bx i )V x i,0) = qx i ) A, x i d, Note that V 0,t j ) = φ L t j ), V d,t j ) = v d,t j ), t j 0, εδ x +κdx ) +gd t V + = f, x i,t j ) Ω +, εδx V + x i,0)+bx i )V + x i,0) = qx i )+A, x i d, V + d,t j ) = v + d +,t j ), V + 1,t j ) = v + 1,t j ), t j 0. V x i,t j ) C, V + x i,t j ) C. The discrete boundary layer function W is defined by 0a) 0b) 0c) W 0, x i,t j ) Ω Ω N,M, εδ x +κdx +gdt ) W = 0, xi,t j ) Ω + Ω N,M, Wd,t j ) = 0, Wx i,0) = 0, W1,t j ) = w1,t j ). Observe that wx i,0) Ce α ε 1 d) Ce α ε τ1 CN 1. The discrete interior layer function Z is also multi-valued and defined by: εδ x +κdx ) 1a) +gd t Z = 0, x i < d,t j > 0, Z 0,t j ) = 0, εδ x +κdx ) 1b) +gd t Z + = 0, x i > d,t j > 0, Z + 1,t j ) = 0, εδx Z x i,0)+bx i )Z α x i d) 1c) x i,0) = Atanh )+1), x i < d, ε εδx Z+ x i,0)+bx i )Z + α x i d) 1d) x i,0) = Atanh ) 1), x i > d, ε and for all t j 0 1e) 1f) Z + +V + )d,t j ) = Z +V )d,t j ), D + x Z+ D x Z )d,t j ) = D x V D + x V + +W))d,t j ). Theorem 6. For sufficiently large M, the solution of 0) satisfies the bound Wx i,t j ) C Πi k=n/ 1+ αδx k d)h k ) ε1 d) Πk=N/ N 1+ αδx k d)h k ), N/ i N. ε1 d) Proof. Consider the following discrete barrier function Φ 3 x i,t j ) := C1 C 1T δ M ) jπi k=n/ 1+ C 1x k d)h k ) ε Πk=N/ N 1+ C 1x k d)h k ), C 1 := αδ 1 d, where M is sufficiently large so that ) 0 < c < 1 C 1T δ M. Note first that Φ 3 x i,0) 0, Φ 3 d,t j ) 0, Φ 3 1,t j ) C1 C 1T δ M ) M C > 0. ε

13 SINGULARLY PERTURBED PROBLEM WITH MOVING INTERIOR LAYER 835 We also observe that εd x + Φ 3x i,t j ) = C 1 x i+1 d)φ 3 x i,t j ), Dt Φ 3x i,t j ) = C 1 δ Φ 3x i,t j ), ε 1+ C 1x i d)h i ) D ε x Φ 3 x i,t j ) = C 1 x i d)φ 3 x i,t j ), εδxφ C 1 h i+1 3 x i,t j ) = + C 1x i d) h i ) Φ3 x i,t j ). h i +h i+1 4ε+C 1 x i d)h i From this we deduce that εδ xφ 3 +C 1 x i d)d x Φ 3 +δ D t Φ 3 0. Hence, εδ x +κdx ) +gd t Φ3 ±W) κ C 1 x i d) ) Dx Φ 3 +g δ )Dt Φ 3 0. Using a discrete comparison principle over the region Ω N,M Ω + will complete the proof. We remark that x i d) 1 d) 1 4, if x i 1 σ, and so, from the bound in Theorem 6, we have that W1 σ,t j ) C 1+ αδσ ) N/8 CN 1. Nε In passing we note that W Cx i d), x i d, and so D xwd,t + j ) C. Since U, W and V are all bounded we have that Z d,t j ), Z + d,t j ) C. Theorem 7. For sufficiently large M OlnN)), the solution of 1) satisfies the bounds a) Zx i,t j ) C Πi k=1 1+ ε αhk ) 1+ ε αhk ), x i d, Π N/ k=1 b) Zx i,t j ) CΠ i n=n/ 1+ e C T αh n ε ) 1 +CN 1 lnn, x i d. Proof. a) For x i d, consider the following barrier function Φ 4 x i,t j ) := C1 αt δ M ) jπi k=1 1+ ε αhk ) 1+ ε αhk ), where M is sufficiently large so that Π N/ k=1 3) 0 < c < 1 αt δ M. Note that εd + x Φ 4x i,t j ) = αφ 4 x i,t j ), D t Φ 4x i,t j ) = α δ Φ 4x i,t j ), ε 1+ αhi ε ) D x Φ 4 x i,t j ) = αφ 4 x i,t j ), εδx Φ 4x i,t j ) = αh i αhi ) 1Φ4 1+ x i,t j ) αφ 4 x i,t j ), h i +h i+1 ε

14 836 J.L. GRACIA AND E. O RIORDAN and so, it follows that, εδ x +κd x +gd t )Φ 4 x i,t j ) 0. Note also that Φ 4 0,t j ) 0,Φ 4 d,t j ) C > 0 and εδ xφ 4 x i,0)+bφ 4 x i,0) b α)φ 4 x i,0). Also, we have Φ 4 x i,0) = C Πi k=1 1+ ε αhk ) 1+ ε αhk ) Ce αx i d)/ ε Ce α ε x i d)/. Thus, Π N/ k=1 εδ xφ 4 x i,0)+bφ 4 x i,0) εδ xzx i,0)+bzx i,0). Finish using a discrete comparison principle. b) For x i d, consider the following barrier function where Bx i,t j ) := CΦ 5 x i )Ψ 1 t j )+CN 1 lnn)t j, Φ 5 x i ) := Π i n=n/ 1+ ρhn ) 1 θt lnn, Ψ 1 t j ) := 1 ε δ M ) j, and MN) is chosen sufficiently large so that θt lnn 4) 0 < c < 1 δ M ). The parameters ρ,θ are specified below. Note first that θt lnn Bd,t j ) CΨ 1 t j ) C1 δ M ) j C > 0, B1,t j ) 0. In addition we have that εd x Φ 5 x i ) = ρφ 5 x i ), Φ 5 d) = 1, εδx Φ 5x i ) ρφ 5 x i ), and D t Ψ 1t j ) = θlnn δ Ψ 1 t j ), Ψ 1 0) = 1. Initially at t j = 0, Φ 5 x i ) Z + x i,0), when ρ α. For d < x i d+τ, εδ x +κdx +gd t )Φ 5 x i )Ψ 1 t j ) ρ a ρx i d) 1 d) +θlnn ) Φ 5 x i )Ψ 1 t j ) ε ρ a ρe C T lnn 1 d) α +θlnn ) Φ 5 x i )Ψ 1 t j ). By defining, ρ := e C T α α, θ := 1 + a 1 d) 1, we have that for N sufficiently large independently of ε) εδ x +κd x +gd t )Φ 5 x i )Ψ 1 t j ) Φ 5 x i )Ψ 1 t j ), d < x i x i +τ. If d+τ < x i 1 σ, then using the inequality nt 1+t) n,t 0, x i d) ε Φ 5 x i ) = τ Φ 5 x i )+Φ 5 d+τ ) x i τ d) 1+ ε ε CN 1 lnn, ρh ε ) i 5N/8)

15 SINGULARLY PERTURBED PROBLEM WITH MOVING INTERIOR LAYER 837 where NH := 81 σ d+τ )). If 1 σ < x i < 1, then x i d) 1 σ d Φ 5 x i ) Φ 5 1 σ)+ x i 1 σ) Φ 5 x i ) CN 1. ε ε ε Then for sufficiently large N and d+τ < x i < 1, εδ x +κdx +gdt )Φ 5 x i )Ψ 1 t j ) ρ a ρx i d) 1 d) +θlnn ) Φ 5 x i )Ψ 1 t j ) ε ρ+θlnn ) Φ 5 x i )Ψ 1 t j ) CN 1 lnn. Then CΦ 5 x i )Ψ 1 t j )+CN 1 lnn)t j is a suitable barrier function for Z. Note by assuming 4) then the other constraints ), 3) are automatically satisfied. Constraint 4) requires M sufficiently large so that 5) M > lnn) a +1 d)t 1 d)δ, which will be more restrictive as T increases and δ decreases. Theorem 8. Assume 19) applies. For M sufficiently large so that 5) is satisfied, a) V v CN 1 +CM 1, b) W w CN 1 lnn +M 1 )lnn, c) U u CN 1 lnn +M 1 )lnn. Proof. a) For the regular component, we derive the necessary bound on Ω,Ω + separately, noting that For all other time levels, L N,M V v ) Ω L N,M V ± v ±) x i,0) CN 1. CεN 1 v xxx +N 1 v xx +M 1 v tt ) CN 1 +M 1 ). Use the barrier function CN 1 +M 1 )1+t j ) to bound V ± v ±. b) Firstly, for d x i 1 σ, use the exponential character of this component and the bound in Theorem 6 to get that Wx i,t j ) W1 σ,t j ) CN 1. Hence, W w)x i,t j ) Wx i,t j ) + wx i,t j ) CN 1, x i 1 σ. Secondly, for 1 σ < x i < 1 and t j > 0, we have L N,M W w)x i,t j ) CN 1 lnn +M 1 )ε 1. Initially, wehavewx i,0) = 0and wx i,0) Ce αδ ε 1 x) CN 1.Useadiscrete maximum principle on [1 σ,1] with the barrier function N 1 lnn +M 1 )lnnx i 1 σ))σ 1 to prove the result. c) From Theorem 4 and Theorem 7, we have that zx i,t j ) CN 1, Zx i,t j ) CN 1, for x i d τ 1. From Theorem 4, zx i,t j ) CN 1, for x i d + τ and from Theorem 7 we deduce that Zx i,t j ) CN 1 lnn, for x i d+τ. Then U u)x i,t j ) CN 1 lnn +M 1 )lnn, for x i [0,d τ 1 ] [d+τ,1].

16 838 J.L. GRACIA AND E. O RIORDAN We now examine the truncation error L N,M U u) in the layer region d τ 1,d+ τ ) [0,T]. Initially, we have L N,M U u)x i,0) Cεh i u xxx x,0) CN 1 lnn. Standard discrete comparison principle yields Note that Ux i,0) ux i,0) CN 1 lnn, x i d τ 1,d+τ ). u x d +,t j ) D + xud,t j ) v x d +,t j ) D + xvd,t j ) + w x d +,t j ) D + xwd,t j ) + z x d +,t j ) D + xzd,t j ) CN 1 τ +CN 1 τ +C N 1 τ, ε with a similar bound to the left of d. Hence, D + x D x)u u)d,t j ) = [u x ] D + x D x)u C N 1 τ 1 +τ ) ε C N 1 lnn ε. In the fine mesh region around the points x = d the truncation error will be of the form εn τ u xxxx +κx i,t j )N 1 τ u xx +M 1 u tt, τ := max{τ 1,τ }. From 9), w t [d τ1,d+τ ] C, w tt [d τ1,d+τ ] C. Here we also use the bounds in Theorem 4 and the fact that 19) allows the use of the derivative estimates 11c) and 11e) on the component z in the region d τ 1,d+τ ). So the truncation error in the fine mesh for x i d) is bounded by CN τ ε 1 +CN 1 τ ε 1 +CM 1 CN 1 lnn) +M 1. To complete the proof use the barrier function Φ 6 x i,t j ) = CN 1 lnn +M 1 )lnnt j +1)+CN 1 lnn) Φ x i,t j ), where Φ is a piecewise linear function defined by Note that Φ d τ 1,t j ) = 0 = Φ d+τ,t j ), Φ d,t j ) = 1. D x + Dx)Φ d,t j ) = τ +τ 1 α1 e C T ) =. τ τ 1 εlnn Remark 7.1. When εlnn C., we can establish that 6) U u CN 1 +M 1 )lnn) 5. Applying the argument given in the appendix to the original formulation of the problem 1), one can deduce that for i =,3, i u Ω x i i+1) Ω Cε, u Ω + t Ω Cε 5/. + Combining these bounds with a standard truncation error argument yields the required bound 6). Note we do not employ a decomposition of the solution u in this case where εlnn C.

17 SINGULARLY PERTURBED PROBLEM WITH MOVING INTERIOR LAYER 839 The nodal error estimate established in Theorem 8 and in 6)) is easily extended to a global error estimate using simple linear interpolation. Define Ūx,t) := N,M i=0,j=1 Ux i,t j )ϕ i x)ψ j t), where ϕ i x) is the standard hat function centered at x = x i and ψ j t) = Mt t j 1 ),t [t j 1,t j ). Theorem 9. For M sufficiently large so that 5) is satisfied, Ū û Q CN 1 +M 1 )lnn) 5, where U is the discrete solution of 18) and û is the solution of the continuous problem 1). Proof. Combine the arguments in [6, Theorem 3.1] with the interpolation bounds in [17, Lemma 4.1] and the bounds on the derivatives of the components v,w,z established in 4,5. Note that from [17, Lemma 4.1], we only require the first time derivative of u to be uniformly bounded. In the interior layer we use [17, Lemma 4.1] and outside of the layer that the continuous singular component z satisfies zx,t) CN Numerical experiments 7a) Consider the following test problem εû ss +û s +û t = 4s1 s), s,t) 0,1) 0,0.5], ûs,0) = φs;ε),0 < s 1, û0,t) = 0, û1,t) =, 0 < t 1, where the initial condition φ is the solution of the following problem 7b) εφ +φ = 1+tanh s 0.5 ), φ0) = 0,φ1) =. ε For this problem, the parameters in the numerical method 18) are taken to be α = 1 and δ = 1/3. As the exact solution is unknown, we estimate the errors using a variant of the double mesh principle, where the finest mesh contains the grid points of the original mesh and its midpoints. Denoting by U and Υ the solutions associated with the coarsermesh Ω N,M andthetwicefinermeshrespectively, wecomputethemaximum two mesh differences d N,M ε and the uniform differences d N,M from d N,M ε := max max U Υ)x i,t j ), d N,M := maxd N,M ε, 0 j M 0 i N S ε where S ε = { 0, 1,..., 30 }. From these values we calculate the corresponding computed orders of convergence qε N,M and the computed orders of uniform convergence q N,M using qε N,M := log d N,M ε /dε N,M ), q N,M := log d N,M /d N,M). The uniform differences and the computed orders of uniform convergence for test problem 7) are given in Table 1. We observe uniform convergence of the finite difference approximations, which is in agreement with the theory. In Figure 1 we display the solution and the corresponding two mesh differences for ε = 10 8 with N = 64, and M = 64, observing clearly the interior layer and that the largest differences appear in the interior layer region. Note that we compute the solution in the transformed domain, but we display the computed solution and differences in the original domain.

18 840 J.L. GRACIA AND E. O RIORDAN "eps=1.e-8,n=m=64" "eps=1.e-8,n=m=64" time space time space 1 Figure 1. Test problem 7): Solution U and two mesh differences U Υ)x i,t j ) for ε = 10 8 and N = M = 64. The final time is T = 0.5. Table 1. The maximum two mesh differences d N,M ε, the uniform differences d N,M, the computed orders of convergence qε N,M and the computed ordersof uniform convergenceq N,M for test problem 7) N=64 N=18 N=56 N=51 N=104 N=048 N=4096 N=819 M=64 M=18 M=56 M=51 M=104 M=048 M=4096 M=819 ε = E E E E E E E E ε = 0.88E E E E E E E E ε = E E E E E E E E ε = E E E E E E E E ε = E E E E E-0 0.0E E E ε = E E E E E E E E ε = E E E E E E E E ε = E E E E E E E E ε = E E E E E E E E ε = E E E E E E E E d N,M 0.59E E E E E E E E-0 q N,M uni In Table we examine if the restriction 5) is needed in practice. In this table, we display the numerical results for problem 7) associated with different values of the final time T and ε = Here, M denotes the minimum value of M such

19 SINGULARLY PERTURBED PROBLEM WITH MOVING INTERIOR LAYER 841 that 5) holds. From these results we conclude that this theoretical constraint appears to be required in practice. More extensive numerical experiments for other test problems are given in [9]. Table. Test problem 7): Maximum two mesh differences and computed orders of convergence for ε = 10 8 and different values of T. M denotes the minimum value of M satisfying 5) T = 0.5 N,M=64 N,M=18 N,M=56 N,M=51 N,M=104 N,M=048 M = 44) M = 51) M = 59) M = 66) M = 73) M = 81) d N,M ε 0.59E E E E E E-01 qε N,M T = 0.53 N,M=64 N,M=18 N,M=56 N,M=51 N,M=104 N,M=048 M = 60) M = 70) M = 80) M = 90) M = 100) M = 110) d N,M ε 0.35E E E E E E-01 qε N,M T = 0.6 N,M=64 N,M=18 N,M=56 N,M=51 N,M=104 N,M=048 M = 146) M = 170) M = 195) M = 19) M = 43) M = 67) d N,M ε 0.370E E E E E E-01 qε N,M T = 0.7 N,M=64 N,M=18 N,M=56 N,M=51 N,M=104 N,M=048 M = 159) M = 1784) M = 038) M = 93) M = 548) M = 803) d N,M ε 0.379E E E E E E-01 qε N,M Appendix. A priori bounds on the solution of the continuous problem In this appendix, we restate the standard arguments see for example [11]) to derive a priori parameter explicit bounds on the solution of a singularly perturbed parabolic problem of convection-diffusion type. However, in this appendix, we allow the initial condition to depend on the singular perturbation parameter and we track the influence of this parameter dependence in the derivation of the bounds on the derivatives of the solution. In addition, we also examine the effect such an initial condition has on the bounds of the regular component in a standard Shishkin decomposition of the solution into regular and layer components. In this appendix, let L ε denote the differential operator defined by L ε ωx,t) := εω xx +ax,t)ω x +bx,t)ω +ω t, where ax,t) α > 0, b 0. We also define L 0 ωx,t) := ax,t)ω x +bx,t)ω +ω t. Consider the following parabolic problem 8a) 8b) 8c) L ε u = f, x,t) 0,1) 0,T] =: Ω, u0,t) = φ L t),u1,t) = φ R t), ux,0) = φx;ε), φ i C1+ε k i ),i n+3. Note that the initial condition for the above problem is not as in the main part of the paper, but it does correspond to the initial condition used in the definition of the regular component v defined in Theorem after transforming in space from [0, d] to [0, 1]. Throughout this appendix we assume sufficient compatibility and regularity so that the solution of 8) u C n+γ Ω). We can write z = u φx) 1 x)φ L t) φ L 0)) xφ R t) φ R 0)). Then, z has zero initial/boundary conditions and L ε z = F := f L ε φx) +1 x)φ L t) φ L 0))+xφ R t) φ R 0))).

20 84 J.L. GRACIA AND E. O RIORDAN Note that F i C1+ε k i+1) ), i n+1. From the maximum principle, we have that z C1+ε k 1 ). Introduce the stretched variables ς := x/ε, τ := t/ε, gς,τ) := gx,t). Then z ςς +ã z ς ε b z d z = ε F. Using [8, pg. 65] or [1, pg. 30], we have z 1+γ C z +Cε F 0+γ, z n+γ C z +Cε F n +γ, n = i+j. Observe that z n z n+γ, n = i+j, and F n = n F i = i=0 n ε i F i. Note that F n C, for k 0.5. In the unstretched variables, for k 0.5, z 1 Cε 1, z n Cε n +Cε n 1) F n +γ Cε n. It easily follows that, for k 0.5, u n Cε n. Let the regular component v of the solution to problem 8) be further decomposed by v := v 0 +εv 1 +ε v. Here v 0 is the solution i=0 L 0 v 0 = f, v 0 0,t) = φ L t), v 0 x,0) = φx;ε). From the explicit closed form solution representation given in [1, pg.396] v 0 m C1+ε k m ), m = i+j n+3. The second term v 1 in the expansion of v is defined by L 0 v 1 = v 0 ) xx, v 1 x,0) = 0, v 1 0,t) = 0. Then using [1, pg.396] again v 1 m C1+ε k 1 m ), m = i+j n+1. Finally, define v as the solution of L ε v = v 1 ) xx, v x,0) = v 0,t) = v 1,t) = 0. Note that v 1 ) xx m C1 + ε k m ), m = i + j n 1. Hence, using the arguments for the parabolic problem given above, v Cε k, v 1 Cε k 3, v m Cε k ε m 3+m k +Cε, m = i+j n. To conclude, we have established the following bounds v m C1+ε k m 1) +ε k m ) C1+ε k m ), m = i+j n. Acknowledgments. This research was partially supported by the project MEC/FEDER MTM and the Diputación General de Aragón.

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