A convection-diffusion problem with a small variable diffusion coefficient
aa r X i v : . [ m a t h . NA ] J a n A convection-diffusion problem with a small variablediffusion coefficient
Hans-G. Roos and Martin SchopfJanuary 14, 2020
Abstract
Consider a singularly perturbed convection-diffusion problem with a small, variable dif-fusion. Based on certain a priori estimates for the solution we prove robustness of a finiteelement method on a Duran-Shishkin mesh.
Key words: singular perturbation, finite element method, layer-adapted mesh
MSC (2000)
Consider the one dimensional boundary value problem L ε u := − ( εu ′ ) ′ − bu ′ + cu = f in (0 , ,u (0) = 0 ,u (1) = 0 , (1.1)with smooth functions ε, b, c, f : [0 , → R , satisfying0 < β < b ( x )0 < ε ≤ ε ( x ) ≤ ε ≪ x ∈ [0 , . (1.2)Moreover we assume c ≥ , c + b ′ / ≥ γ > , (1.3)which can be ensured using the assumptions (1.2) and the transformation u = ˆ u e δx with suitablychosen constant δ , see, for instance, [6].We do not know any results concerning robust numerical methods for such problems, theonly exceptions are [2, 3], where ε ( x ) has piecewise the special form ε i p i ( x ) in Ω i with differentparameters ε i .Assuming additionally ε ′ > − β , we have an outflow boundary layer at x = 0. It is relativelytechnical to prove a priori estimates for derivatives of u and to prove the existence of a solutiondecomposition into a smooth part and a layer part. But this can be done with well knowntechniques (Kellogg/Tsan; use of extended domains), see the Appendix.Under additional assumptions ( ε ′ is nonnegative and bounded; moreover conditions on ε ′′ , seeTheorem 10 and Remark 4) we have: There exists a solution decomposition u = S + E with | S ( k ) ( x ) | ≤ C for k = 0 , , , (1.4a)1nd (cid:12)(cid:12) E ( k ) ( x ) (cid:12)(cid:12) ≤ C ε ( x ) k e − βe ( x ) for k = 0 , , , (1.4b)here e ( x ) = Z x ε ( t ) dt. Based on the solution decomposition we are going to analyze the finite element method on aspecial mesh. The weak formulation of the problem uses the bilinear form a ( v, w ) := ( εv ′ , w ′ ) − ( bv ′ , w ) + ( cv, w ) . (1.5)Let V h ∈ H (0 ,
1) be the space of linear finite elements. We look for u h ∈ V h such that a ( u h , v h ) = ( f, v h ) for all v h ∈ V h . (1.6)Define an energy norm by k v k ε := k ε / v ′ k + k v k . Then we ask: on which layer adapted mesh can we prove an (almost) robust error estimate for ourfinite element method in that energy norm?
Near the layer we use a fine graded mesh, otherwise an equidistant mesh with the step size h .First we introduce a point τ ∗ satisfying e ( τ ∗ ) = − β ln h. (2.1)Observe that as e (0) = 0 and e is strictly increasing (2.1) has a unique solution.Since e is strictly increasing the choice (2.1) also impliese − βe ( x ) ≤ e − βe ( τ ∗ ) ≤ h for x ≥ τ ∗ . (2.2)Moreover, τ ∗ satisfies − β ε ln h ≤ τ ∗ ≤ − β ε ln h. (2.3)Following [1], we introduce near x = 0 the graded mesh (D-L mesh) x = 0 ,x = hδε,x i +1 = x i + hx i , for 1 ≤ i ≤ N ∗ . (2.4)We choose N ∗ in such a way that τ = x N ∗ +1 is the first point with τ ≥ τ ∗ . Then, τ has similarproperties as τ ∗ . In the subinterval [ τ,
1] we use an equidistant mesh with a mesh size of order O ( h ).To simplify the notation, we introduce the symbol (cid:22) and note A (cid:22) B , if there exists a constant C independent of ε , such that A ≤ CB .Because the smooth part S satisfies | S ′′ | ≤ C , we have for the interpolation error of thepiecewise linear interpolant k S − S I k (cid:22) h , | S − S I | (cid:22) h. On [ τ,
1] we obtain for the layer component k E − E I k , [ τ, (cid:22) k E k ∞ , [ τ, (cid:22) h . k ε / ( E − E I ) ′ k , [ τ, (cid:22) Z τ ε ( x ) e − βe ( x ) + 1 h k ε / E I k , [ τ, (cid:22) h . (2.5)Next we study the interpolation error on the fine subinterval [0 , τ ], using the definition of themesh, the estimate of E ′′ and x ≤ e ( x ) ε ( x ): k ε − / ( E − E I ) k , [0 ,τ ] = Z x ε − ( E − E I ) + N ∗ X Z x i +1 x i ε − ( E − E I ) (cid:22) h + N ∗ X Z x i +1 x i ε − ( hx i ) e − βe ( x ) (cid:22) h + h Z τ ε − x e − βe ( x ) ≤ h (cid:18) Z τ ε − ( e ( x )) e − βe ( x ) (cid:19) (cid:22) h (cid:18) Z ∞ s e − βs ) (cid:19) (cid:22) h . (2.6)Thus we obtain k ε − / ( E − E I ) k , [0 ,τ ] (cid:22) h and k E − E I k , [0 ,τ ] (cid:22) h k ε / k , [0 ,τ ] . (2.7)Similarly we get k ε / ( E − E I ) ′ k , [0 ,τ ] = Z x ε (( E − E I ) ′ ) + N ∗ X Z x i +1 x i ε ( E − E I ) ′ ) (cid:22) h (cid:18) Z ∞ s e − βs (cid:19) , (2.8)resulting in k ε / ( E − E I ) ′ k , [0 ,τ ] (cid:22) h. (2.9) So far we proved k u − u I k ε (cid:22) h and start now to estimate k u h − u I k ε . As usual, we have basedon the coercivity of our bilinear form in the given norm k u I − u h k ε (cid:22) a ( u I − u h , u I − u h )= a ( u I − u, u I − u h )= ( ε ( u I − u ) ′ , v ′ h ) − ( b ( u I − u ) ′ , v h ) + ( c ( u I − u ) , v h ) (3.1)with v h = u I − u h . The first and the third term can be easily estimated, only the convection termwith respect to the layer part E needs some care. We use integration by parts and on the finepart of the mesh | ( E − E I , ( v h ) ′ ) | ≤ k ε − / ( E − E I ) k k v h k ε , while on the coarse part an inverse inequality yields | ( E − E I , ( v h ) ′ ) | (cid:22) h k E − E I k k v h k ≤ h k E − E I k k v h k ε . Using (2.7), we get finally
Theorem 1.
If there exists a solution decomposition with the properties (1.4) , then the finiteelement approximation with linear elements on our DL-Shishkin mesh satisfies k u h − u k ε (cid:22) h. (3.2)Remark that our result is not fully robust: the number of mesh points used is of order O ( ψ ( ε, h ) h ), where ψ ( ε, h ) can be estimate by ln(( ε ) / ( ε )) + ln(( − ln h ) /h ).3 Appendix
Consider the one dimensional boundary value problem L ε u := − ( εu ′ ) ′ − bu ′ + cu = f in (0 , ,u (0) = 0 ,u (1) = 0 . (4.1)Assume (1.2) and (1.3).The differential equation in (4.1) can be rewritten in the equivalent form − εu ′′ − ( b + ε ′ ) u ′ + cu = f. (4.2)Thus the first derivative of ε has a crucial influence on the behavior of the exact solution: If forinstance ε ′ < − b then the outflow boundary will shift to the point x = 1 leading to the formationof an exponential boundary layer at that point. We shall consider the case ε ′ > − β ≥ − b leavingthe outflow boundary point at the origin of the unit interval. Lemma 2.
Let u be the solution of (4.1) and T be the coordinate transformation ξ = T ( x ) = Z x r εε ( t ) d t, (4.3) mapping the domain (0 , to (cid:0) , T (1) (cid:1) . Then in the transformed variable ξ it holds | ˜ u ( k ) ( ξ ) | ≤ C (cid:18) ε − k e − σ +2 β √ εε ξ (cid:19) (4.4) with ˜ u := u ◦ T − and σ := min z ∈ [0 , ε ′ ( z ) > − β .Proof. Let T be the coordinate transformation defined by (4.3). As strict monotone mapping T is injective and therefore T − : [0 , T (1)] → [0 , ξ x exists. The chain rule yields for˜ u (cid:0) T ( x ) (cid:1) = u ( x ) and x ∈ (0 , u ′ ( x ) = dd x ˜ u (cid:0) T ( x ) (cid:1) = ˜ u ′ (cid:0) T ( x ) (cid:1) T ′ ( x ) ,u ′′ ( x ) = ˜ u ′′ (cid:0) T ( x ) (cid:1)(cid:0) T ′ ( x ) (cid:1) + ˜ u ′ (cid:0) T ( x ) (cid:1) T ′′ ( x ) . Thus the differential equation (4.2) is transformed into ε ˜ u ′′ ( T ( x )) − (cid:0) b ( x ) + ε ′ ( x ) (cid:1)r εε ( x ) − ε ( x ) εε ′ ( x )2 q εε ( x ) (cid:0) ε ( x ) (cid:1) ˜ u ′ ( T ( x )) + c ( x )˜ u ( T ( x )) = f ( x ) . Note that the coefficient of the highest derivative of ˜ u is the constant ε and that the functions˜ c := c ◦ T − and ˜ f := f ◦ T − remain bounded. Therefore rewriting (4.2) in the new variable ξ yields: ε ˜ u ′′ ( ξ ) − ˜ b ( ξ )˜ u ′ ( ξ ) + ˜ c ( ξ )˜ u ( ξ ) = ˜ f ( ξ ) , ξ ∈ [0 , T (1)]˜ b = 2 b ◦ T − + ε ′ ◦ T − r εε ◦ T − . In order to obtain bounds on the derivatives of ˜ u we need an estimate ˜ β ≤ ˜ b ( ξ ) for ξ ∈ [0 , T (1)]with a constant ˜ β >
0. Equivalently, we provide an estimate ˜ β ≤ ˜ b (cid:0) T ( x ) (cid:1) for x ∈ [0 , b (cid:0) T ( x ) (cid:1) = 2 b ( x ) + ε ′ ( x )2 r εε ( x ) ≥ r εε ( x ) σ + 2 β ≥ r εε σ + 2 β β > . ε/ε ( t ) ≤ T ( x ) ≤ x and hence T (1) ≤ T ( x ) = Z x r εε ( t ) d t ≤ Z x d t = x. Thus, we can apply well known a-priory estimates for the case when ε is a constant to obtain | ˜ u ( k ) ( ξ ) | ≤ C (cid:18) ε − k e − ˜ βε ξ (cid:19) ≤ C (cid:18) ε − k e − σ +2 β √ εε ξ (cid:19) Lemma 3.
Set ˜ e ( x ) := Z x p εε ( t ) d t (4.5) The solution u of Problem (4.1) satisfies | u ( x ) | ≤ C (cid:16) − σ +2 β ˜ e ( x ) (cid:17) , (4.6a) | u ′ ( x ) | ≤ C s εε ( x ) (cid:16) ε − e − σ +2 β ˜ e ( x ) (cid:17) , (4.6b) | u ′′ ( x ) | ≤ C εε ( x ) (cid:16) ε − e − σ +2 β ˜ e ( x ) (cid:17) + εε ′ ( x )2 (cid:0) ε ( x ) (cid:1) (cid:16) ε − e − σ +2 β ˜ e ( x ) (cid:17)! . (4.6c) with σ := min z ∈ [0 , ε ′ ( z ) > − β .Proof. Lemma 2 yields (cid:12)(cid:12)(cid:12) ( u ◦ T − ) ( k ) ( ξ ) (cid:12)(cid:12)(cid:12) = | ˜ u ( k ) ( ξ ) | ≤ C (cid:18) ε − k e − σ +2 β √ εε ξ (cid:19) . The transformation ξ = T ( x ) gives (cid:12)(cid:12)(cid:12) ( u ◦ T − ) ( k ) (cid:0) T ( x ) (cid:1)(cid:12)(cid:12)(cid:12) ≤ C (cid:18) ε − k e − σ +2 β √ εε T ( x ) (cid:19) = C (cid:16) ε − k e − σ +2 β ˜ e ( x ) (cid:17) . (4.7)We use (4.7) to deduce our proposition. First (4.6a) is an immediate consequence of (4.7) for k = 0. Next we want to verify (4.6b). A simple calculation yields( u ◦ T − ) ′ ( ξ ) = u ′ (cid:0) T − ( ξ ) (cid:1) (cid:0) T − (cid:1) ′ ( ξ ) = u ′ (cid:0) T − ( ξ ) (cid:1) T ′ (cid:0) T − ( ξ ) (cid:1) . With ξ = T ( x ) we conclude ( u ◦ T − ) ′ (cid:0) T ( x ) (cid:1) = u ′ ( x ) 1 T ′ ( x ) . (4.8)Collecting (4.7) with k = 1 and (4.8) the estimate (4.6b) follows. Same techniques yield (cid:12)(cid:12) ( u ◦ T − ) ′′ (cid:0) T ( x ) (cid:1)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u ′′ ( x ) 1 (cid:0) T ′ ( x ) (cid:1) − u ′ ( x ) T ′′ ( x ) (cid:0) T ′ ( x ) (cid:1) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ | u ′′ ( x ) | (cid:0) T ′ ( x ) (cid:1) − | u ′ ( x ) | | T ′′ ( x ) | (cid:0) T ′ ( x ) (cid:1) (4.9)Combining (4.7) with k = 2 and (4.9) we obtain | u ′′ ( x ) | ≤ C (cid:0) T ′ ( x ) (cid:1) (cid:16) ε − e − σ +2 β ˜ e ( x ) (cid:17) + | u ′ ( x ) || T ′′ ( x ) |≤ C εε ( x ) (cid:16) ε − e − σ +2 β ˜ e ( x ) (cid:17) + | u ′ ( x ) | εε ′ ( x )2 q εε ( x ) (cid:0) ε ( x ) (cid:1) . Using (4.7) with k = 1 for the second term the proof is complete.5 emark 1. In the classical constant setting ε ≡ ε = ε the formulas (4.6) reduce to the well-knownform (cid:12)(cid:12)(cid:12) u ( k ) ( x ) (cid:12)(cid:12)(cid:12) ≤ C (cid:16) ε − k e − βε x (cid:17) , k = 0 , , . (4.10)Unfortunately, all summands of the right hand side of the bounds (4.6b) and (4.6c) have alarge multiplier if ε changes on a huge scale. Moreover the exponential decay in the estimates(4.6) appears to be suboptimal. In order to provide better bounds we will use the followingLemmas. Lemma 4.
The differential operator L ε obeys the following maximum principle: For any function v ∈ C ( a, b ) ∩ C [ a, b ] L ε v ≤ a, b ) ,v ( a ) ≤ ,v ( b ) ≤ = ⇒ v ≤ on [ a, b ] .Proof. A proof can be found e.g. in [5].The maximum principle applied to v − v also yields a comparison principle. Lemma 5.
Let a < x , ℓ ∈ N , suppose ε ′ ≥ σ ≥ on [ a, x ] and set e a ( t ) := R ta /ε ( z )d z . Then Z xa ε ( t ) ℓ e γe a ( t ) d t ≤ γ + ( ℓ + 1) σ (cid:0) ε ( x ) ℓ +1 e γe a ( x ) − ε ( a ) ℓ +1 (cid:1) . (4.11) for γ > − ( ℓ + 1) σ .Proof. Since σ ≤ ε ′ ( t ) for t ∈ [ a, x ] multiplication with ε ( t ) ℓ exp (cid:0) γe a ( t ) (cid:1) > σ Z xa ε ( t ) ℓ e γe a ( t ) d t ≤ Z xa e γe a ( t ) ε ( t ) ℓ ε ′ ( t )d t. Integration by parts gives σ Z xa ε ( t ) ℓ e γe a ( t ) d t ≤ ℓ + 1 (cid:18) ε ( x ) ℓ +1 e γe a ( x ) − ε ( a ) ℓ +1 − Z xa ε ( t ) ℓ +1 γ e γe a ( t ) e ′ a ( t )d t (cid:19) . (4.12)Inserting − Z xa ε ( t ) ℓ +1 γ e γe a ( t ) e ′ a ( t )d t = − γ Z xa ε ( t ) ℓ e γe a ( t ) d t into (4.12) we obtain (cid:18) γℓ + 1 + σ (cid:19) Z xa ε ( t ) ℓ e γe a ( t ) d t ≤ ℓ + 1 (cid:16) ε ( x ) ℓ +1 e γe a ( x ) − ε ( a ) ℓ +1 (cid:17) and (4.11) follows.Next, we want to proof some pointwise bounds for the solution of the following problem in apossibly extended domain ( a,
1) with a ≤ − ( ε ∗ w ′ ) ′ − b ∗ w ′ + c ∗ w = f ∗ in ( a, , w ( a ) = 0 , w (1) = u , (4.13)with smooth functions ε ∗ , b ∗ , c ∗ and f ∗ defined on ( a,
1) and satisfying Cε ≤ ε ∗ ( x ) ≤ εβ ≤ b ∗ ( x )0 ≤ c ∗ ( x ) for x ∈ [ a, . (4.14)6 emma 6. Suppose ≤ ( ε ∗ ) ′ on [ a, . Then the solution w of problem (4.13) satisfies | w ( x ) | ≤ C, x ∈ [ a, . (4.15) Proof.
Using the comparison principle induced by Lemma 4 with the barrier functions ψ ± definedby ψ ± ( x ) := ± β k f ∗ k ∞ (1 − x ) ± | u | one obtains the result, because (cid:0) L ε ψ + (cid:1) ( x ) = b ∗ ( x ) + ( ε ∗ ) ′ ( x ) β k f ∗ k ∞ + c ∗ ( x ) (cid:18) β (1 − x ) + | u | (cid:19) ≥ k f ∗ k ∞ ≥ ( L ε w ) ( x ) in ( a, ,ψ + ( a ) = 1 β k f ∗ k ∞ (1 − a ) + | u | ≥ w ( a ) ,ψ + (1) = | u | ≥ u = w (1) . Hence w ≤ ψ + ≤ C on [ a, ψ − .The following argument is an extension of [4]. Lemma 7.
Suppose ≤ ( ε ∗ ) ′ on [ a, and set e a ( t ) := R ta /ε ∗ ( z )d z . Then the solution w ofproblem (4.13) satisfies | w ′ ( x ) | ≤ C (cid:18) ε ∗ ( x ) e − βe a ( x ) (cid:19) , x ∈ [ a, . (4.16) Proof.
For the sake of readability, we drop the star from the notation of the functions ε ∗ , b ∗ , c ∗ and f ∗ within this proof. Set h := f − cw . The problem − w ′′ ( x ) − b ( x ) + ε ′ ( x ) ε ( x ) w ′ ( x ) = h ( x ) ε ( x ) , w ( a ) = 0 , w (1) = u is equivalent to problem (4.13). It’s solution w admits the representation w ( x ) = w p ( x ) + K + K Z xa e − (cid:0) B ( t ) − B ( a ) (cid:1) d t, where w p ( x ) := − Z xa z ( t )d t, z ( x ) := Z xa h ( t ) ε ( t ) e − (cid:0) B ( x ) − B ( t ) (cid:1) d t,B ( x ) := Z xa b ( t ) + ε ′ ( t ) ε ( t ) d t = Z xa b ( t ) ε ( t ) d t + ln (cid:0) ε ( x ) (cid:1) − ln (cid:0) ε ( a ) (cid:1) , i.e. B is an indefinite integral of ( b + ε ′ ) /ε . The constants K and K may depend on ε . Theboundary condition w ( a ) = 0 yields K = 0 whereas the other boundary condition w (1) = u gives u − w p (1) = K Z a e − (cid:0) B ( t ) − B ( a ) (cid:1) d t = K Z a e − R ta b ( z ) ε ( z ) d z +ln ( ε ( a ) ε ( t ) )d t = K ε ( a ) Z a ε ( t ) e − R ta b ( z ) ε ( z ) d z d t. (4.17)7ecause of Lemma 6 we know k w k ∞ ≤ C . Thus | z ( x ) | ≤ C Z xa ε ( t ) e − (cid:0) B ( x ) − B ( t ) (cid:1) d t. (4.18)For t ≤ x a simple calculation yields1 ε ( t ) e − (cid:0) B ( x ) − B ( t ) (cid:1) = e − B ( x )+ B ( t ) − ln (cid:0) ε ( t ) (cid:1) = e − R xa b ( z ) ε ( z ) d z + R ta b ( z ) ε ( z ) d z − ln (cid:0) ε ( x ) (cid:1) = 1 ε ( x ) e − R xt b ( z ) ε ( z ) d z ≤ ε ( x ) e − β R xt ε ( z ) d z = 1 ε ( x ) e − β (cid:0) e a ( x ) − e a ( t ) (cid:1) . Inserting this estimate into (4.18) and applying Lemma 5 with ℓ = 0 we obtain | z ( x ) | ≤ Cε ( x ) Z xa e − β (cid:0) e a ( x ) − e a ( t ) (cid:1) d t ≤ Cε ( x ) e − βe a ( x ) Z xa e βe a ( t ) d t ≤ Cε ( x ) e − βe a ( x ) β + σ ε ( x )e βe a ( x ) ≤ C. Moreover | z ( x ) | ≤ C for all x ∈ [0 ,
1] implies | w p (1) | ≤ C . We still need to estimate Z a ε ( t ) e − R ta b ( z ) ε ( z ) d z d t ≥ Z a ε ( t ) e −k b k ∞ e a ( t ) d t = Z e a (1)0 e −k b k ∞ s d s = 1 k b k ∞ (cid:16) − e −k b k ∞ e a (1) (cid:17) ≥ C. Here we used the substitution s = e a ( t ) with d s/ d t = e ′ a ( t ) = 1 /ε ( t ). Thus, with (4.17) we get | K | ≤ C ε ( a ) . Combining this with w ′ ( x ) = − z ( x ) + K e − (cid:0) B ( x ) − B ( a ) (cid:1) , (4.19)we obtain | w ′ ( x ) | ≤ | z ( x ) | + C ε ( x ) e − βe a ( x ) ≤ C (cid:18) ε ( x ) e − βe a ( x ) (cid:19) and (4.16) is verified. Remark 2.
Again in the classical case where ε ∗ is constant the formula (4.16) reduces to (4.10) which is known to be optimal. Remark 3.
An inspection of the proof of (4.16) shows that the assumption ε ≪ can be droppedprovided w remains uniformly bounded and e a (1) is sufficiently large — Remark that e a is a strictlyincreasing function. With (4.16) we readily obtain a pointwise estimate for w ′′ . Lemma 8.
Let ≤ ( ε ∗ ) ′ on [ a, and set e a ( t ) := R ta /ε ∗ ( z )d z . Then the solution w of problem (4.13) satisfies | w ′′ ( x ) | ≤ C ε ∗ ) ′ ( x ) ε ∗ ( x ) (cid:18) ε ∗ ( x ) e − βe a ( x ) (cid:19) , x ∈ [ a, . (4.20) Proof.
The result is an immediate consequence of (4.13), k w k ∞ ≤ C and (4.16).8 emma 9. Suppose ≤ ( ε ∗ ) ′ on [ a, . Then for the solution w of Problem (4.13) | w ′′ ( x ) | ≤ C (cid:18) K ( x ) + 1 ε ∗ ( x ) (cid:0) ε ∗ ) ′ ( a ) + k ( ε ∗ ) ′′ k L ( a,x ) (cid:1) e − βe a ( x ) (cid:19) , x ∈ [ a,
1] (4.21) holds with e a ( t ) := R ta /ε ∗ ( z )d z and K ( x ) ≤ C min {k ( ε ∗ ) ′′ k ∞ , ( a,x ) , √ ε ∗ ( x ) k ( ε ∗ ) ′′ k , ( a,x ) } .Proof. In order to simplify the illustration we again drop the star from the notation of the functions ε ∗ , b ∗ , c ∗ and f ∗ within this proof. A differentiation of (4.13) yields − w (3) − b + 2 ε ′ ε w ′′ = f ′ + ( b ′ + ε ′′ − c ) w ′ − c ′ wε =: g. Thus, we obtain a differential equations for ω := w ′′ , indeed − ω ′ − ( b + 2 ε ′ ) /ε ω = g . Setting˜ B ( x ) := Z xa b ( t ) + 2 ε ′ ( t ) ε ( t ) d t = Z xa b ( t ) ε ( t ) d t + 2 ln (cid:0) ε ( x ) (cid:1) − (cid:0) ε ( a ) (cid:1) (i.e. ˜ B is an indefinite integral of ( b + 2 ε ′ ) /ε ) the function ω can be represented as ω ( x ) = K e − (cid:0) ˜ B ( x ) − ˜ B ( a ) (cid:1) − Z xa g ( t )e − (cid:0) ˜ B ( x ) − ˜ B ( t ) (cid:1) d t. (4.22)Here the constant K may depend on ε . Because of the identitye − (cid:0) ˜ B ( x ) − ˜ B ( t ) (cid:1) = (cid:18) ε ( t ) ε ( x ) (cid:19) e − R xt b ( z ) ε ( z ) d z and K = ω ( a ) = w ′′ ( a ) the representation (4.22) implies | ω ( x ) | ≤ | w ′′ ( a ) | (cid:18) ε ( a ) ε ( x ) (cid:19) e − βe a ( x ) + Z xa | g ( t ) | (cid:18) ε ( t ) ε ( x ) (cid:19) e − β (cid:0) e a ( x ) − e a ( t ) (cid:1) d t. (4.23)Because | g ( t ) | ≤ | b ′ ( t )+ ε ′′ ( t ) − c ( t ) | ε ( t ) | w ′ ( t ) | + | c ′ ( t ) | ε ( t ) | w ( t ) | + | f ′ ( t ) | ε ( t ) the integral in (4.23) is dominated bythe sum of the two integrals I ( x ) and I ( x ) with I ( x ) := Z xa (cid:18) | f ′ ( t ) | ε ( t ) + | c ′ ( t ) | ε ( t ) | w ( t ) | (cid:19) (cid:18) ε ( t ) ε ( x ) (cid:19) e − β (cid:0) e a ( x ) − e a ( t ) (cid:1) d t,I ( x ) := Z xa | b ′ ( t ) + ε ′′ ( t ) − c ( t ) | ε ( t ) | w ′ ( t ) | (cid:18) ε ( t ) ε ( x ) (cid:19) e − β (cid:0) e a ( x ) − e a ( t ) (cid:1) d t. Using k w k ∞ ≤ C and applying Lemma 5 with ℓ = 1 we see that I ( x ) ≤ C ε ( x ) e − βe a ( x ) Z xa ε ( t )e βe a ( t ) d t ≤ C ε ( x ) e − βe a ( x ) ε ( x ) e βe a ( x ) ≤ C. (4.24)For I the bound (4.16) yields with Lemma 5 ( ℓ = 1) I ( x ) ≤ C ε ( x ) e − βe a ( x ) Z xa (cid:0) | ε ′′ ( t ) | (cid:1) (cid:18) ε ( t ) e − βe a ( t ) (cid:19) ε ( t )e βe a ( t ) d t ≤ C ε ( x ) e − βe a ( x ) Z xa ε ( t )e βe a ( t ) + | ε ′′ ( t ) | ε ( t )e βe a ( t ) + 1 + | ε ′′ ( t ) | d t ≤ C ε ( x ) e − βe a ( x ) (cid:16) ε ( x ) e βe a ( x ) + K ( x ) + (cid:0) k ε ′′ k L ( a,x ) (cid:1)(cid:17) ≤ C (cid:18) K ( x ) + 1 ε ( x ) (cid:0) k ε ′′ k L ( a,x ) (cid:1) e − βe a ( x ) (cid:19) (4.25)9ith K ( x ) := R xa | ε ′′ ( t ) | ε ( t )e βe a ( t ) d t . H¨older’s inequality and the Cauchy-Schwarz inequality yieldfor K ( x ) with Lemma 5 K ( x ) ≤ C ε ( x ) e − βe a ( x ) k ε ′′ k ∞ , ( a,x ) Z xa ε ( t )e βe a ( t ) d t ≤ C k ε ′′ k ∞ , ( a,x ) (4.26a) K ( x ) ≤ C ε ( x ) e − βe a ( x ) k ε ′′ k , ( a,x ) (cid:18)Z xa ε ( t ) e βe a ( t ) d t (cid:19) / ≤ C p ε ( x ) k ε ′′ k , ( a,x ) (4.26b)From (4.20) we deduce the bound | w ′′ ( a ) | ≤ C ε ′ ( a ) ε ( a ) . (4.27)We conclude our proposition by collecting (4.23), (4.24), (4.25), (4.26) and (4.27). Theorem 10 (Solution Decomposition) . Suppose ≤ ε ′ on [0 , and define e ( t ) := R t /ε ( z )d z .Then there exists a constant S with | S | ≤ C such that the solution u of (4.1) can be decomposedinto the sum of a smooth part S and an exponential boundary layer component E BL , i.e. u = S + E BL such that S and E solve the boundary-value problems L ε S = f in (0 , , S (0) = S , S (1) = 0 , (4.28a) L ε E BL = 0 in (0 , , E BL (0) = − S , E BL (1) = 0 . (4.28b) Moreover there exists a constant C such that for x ∈ [0 , | S ( k ) ( x ) | ≤ C for k = 0 , , (4.28c) | S ′′ ( x ) | ≤ C ε ′ ( x ) ε ( x ) , (4.28d) | ε ( x ) S ′′ ( x ) + ε ′ ( x ) S ′ ( x ) | ≤ C, (4.28e) and (cid:12)(cid:12) ( E BL ) ( k ) ( x ) (cid:12)(cid:12) ≤ C ε ( x ) k e − βe ( x ) for k = 0 , , (4.28f) (cid:12)(cid:12) ( E BL ) ′′ ( x ) (cid:12)(cid:12) ≤ C ε ′ ( x ) ε ( x ) e − βe ( x ) . (4.28g) Proof.
We start off with the regular solution component S : Fix a < − β ε ln ε <
0. On the theinterval ( a,
1) choose smooth extension ε ∗ , b ∗ , c ∗ and f ∗ of ε , b , c and f in such a way that ε ∗ isnon-decreasing and the assumptions (4.14) are met. Thus we can apply Lemma 6 and Lemma 7to the boundary value problem − (cid:0) ε ∗ ( S ∗ ) ′ (cid:1) ′ − b ∗ ( S ∗ ) ′ + c ∗ S ∗ = f ∗ in ( a, , S ∗ ( a ) = 0 , S ∗ (1) = u to obtain | S ∗ ( x ) | ≤ C and | ( S ∗ ) ′ ( x ) | ≤ C (cid:18) ε ∗ ( x ) e − βe a ( x ) (cid:19) for x ∈ [ a, . (4.29)Since ε is non-decreasing we can set ε := ε (0) = ε ∗ (0). This implies for all x ∈ ( a,
0] that Cε ≤ ε ∗ ( x ) ≤ ε . Hence e − βe a (0) = e − β R a ε ∗ ( z ) d z ≤ e βaε < ε. The fact that e a is a strictly increasing function implies e a ( x ) > e a (0). Thus we arrive ate − βe a ( x ) ≤ e − βe a (0) < ε for x ∈ [0 , . (cid:12)(cid:12) ( S ∗ ) ( k ) ( x ) (cid:12)(cid:12) ≤ C for k = 0 , x ∈ [0 , ε ∗ | (0 , = ε ≥ ε . Setting S := S ∗ | (0 , it satisfies the boundary value problem (4.28a)because b ∗ | (0 , = b , c ∗ | (0 , = c as well as f ∗ | (0 , = f . The bound on S ∗ and ( S ∗ ) ′ yields(4.28c), in particular S (0) = S := S ∗ (0) with | S | ≤ C . Since (4.28d) and (4.28e) are immediateconsequences of (4.28a) and (4.28c) all propositions for the regular part S are verified.To bound the layer component E BL we use the barrier functions φ ± defined by φ ± ( x ) = ± (cid:12)(cid:12) E BL (0) (cid:12)(cid:12) e − βe ( x ) . A simple calculation yields (cid:0) L ε φ + (cid:1) ( x ) = (cid:12)(cid:12) E BL (0) (cid:12)(cid:12) (cid:18) β b ( x ) + ε ′ ( x ) − β − ε ′ ( x ) ε ( x ) + c ( x ) (cid:19) e − βe ( x ) ≥ (cid:0) L ε E BL (cid:1) ( x ) ,φ + (0) = (cid:12)(cid:12) E BL (0) (cid:12)(cid:12) ≥ E BL (0) ,φ + (1) = (cid:12)(cid:12) E BL (0) (cid:12)(cid:12) e − βE (1) ≥ E BL (1) . Hence E BL ( x ) ≤ φ + ( x ) ≤ C e − βe ( x ) . Because φ − = − φ + the estimate (4.28f) for k = 0 follows.For the first derivative of the boundary layer term E BL we use the representation E BL ( x ) = Z x z ( s )d s + K Z x e − B ( s ) d s with z ( x ) = − Z x b ( s ) + ε ′ ( s ) ε ( s ) E BL ( s )e − (cid:0) B ( x ) − B ( s ) (cid:1) d s,B ( x ) = Z x b ( s ) + ε ′ ( s ) ε ( s ) d s = Z x b ( s ) ε ( s ) d s + ln (cid:0) ε ( x ) (cid:1) − ln (cid:0) ε (0) (cid:1) . The estimate (4.28f) with k = 0 yields for z : | z ( x ) | ≤ C Z x b ( s ) + ε ′ ( s ) ε ( s ) e − βE ( s ) ε ( s ) ε ( x ) e − β (cid:0) e ( x ) − e ( s ) (cid:1) d s ≤ C ε ( x ) e − βe ( x ) Z x (cid:0) ε ′ ( s ) (cid:1) d s ≤ C ε ( x ) e − βe ( x ) . The constant K is governed by the boundary condition E BL (0) = u − S : K Z e − B ( s ) d s = u − S − Z z ( s )d s. (4.30)Using the substitution t = e ( s ) with d t d s = e ′ ( s ) = ε ( s ) we obtain Z e − B ( s ) d s = Z ε (0) ε ( s ) e − R s b ( x ) ε ( x ) d x d s ≥ Z ε (0) ε ( s ) e −k b k ∞ e ( s ) d s = ε (0) Z e (1)0 e −k b k ∞ t d t ≥ Cε (0)and Z z ( s )d s ≤ Z | z ( s ) | d s ≤ C Z ε ( s ) e − βe ( s ) d s ≤ C Z e (1)0 e − βt d t ≤ C Z ∞ e − βt d t ≤ C. Hence (4.30) gives | K | ≤ Cε (0) and because (cid:0) E BL (cid:1) ′ ( x ) = − z ( x ) − K e − B ( x )
11e can estimate (cid:12)(cid:12)(cid:12)(cid:0) E BL (cid:1) ′ ( x ) (cid:12)(cid:12)(cid:12) ≤ | z ( x ) | + | K | e − B ( x ) ≤ C ε ( x ) e − βe ( x ) + Cε (0) ε (0) ε ( x ) e − βe ( x ) which is (4.28f) for k = 1. For the remaining result (4.28g) we use the differential equation (4.28b)and the bounds (4.28f): (cid:12)(cid:12)(cid:12)(cid:0) E BL (cid:1) ′′ ( x ) (cid:12)(cid:12)(cid:12) ≤ | b ( x ) | + ε ′ ( x ) ε ( x ) (cid:12)(cid:12)(cid:12)(cid:0) E BL (cid:1) ′ ( x ) (cid:12)(cid:12)(cid:12) + | c ( x ) | ε ( x ) (cid:12)(cid:12)(cid:0) E BL (cid:1) ( x ) (cid:12)(cid:12) ≤ C ε ′ ( x ) ε ( x ) e − βe ( x ) . Remark 4.