ON A NESTED BOUNDARY-LAYER PROBLEM. X. Liang. R. Wong. Dedicated to Professor Philippe G. Ciarlet on the occasion of his 70th birthday

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1 COMMUNICATIONS ON doi:.3934/cpaa PURE AND APPLIED ANALYSIS Volume 8, Number, January 9 pp ON A NESTED BOUNDARY-LAYER PROBLEM X. Liang Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA 39, USA. R. Wong Department of Mathematics, City University of Hong Kong, Tat Chee Avenue, Kowloon, Hong Kong. Dedicated to Professor Philippe G. Ciarlet on the occasion of his 7th birthday Abstract. Nested boundary layers mean that one boundary layer lies inside another one. In this paper, we consider one such problem, namely, { 3 xy x + x y x x 3 + yx =, < x <, y =, y = e. An asymptotic solution, which holds uniformly for x [, ], is constructed rigorously. This result also provides an explicit formula for the exponentially small leading term in the interval where the exact solution exhibits such behavior. This phenomenon has never been mentioned in the existing literature.. Introduction. An interesting problem in singular perturbation theory is to consider the nested boundary layers, which mean that one boundary layer lies inside another one. A well-known boundary value problem with such a phenomenon is given by 3 xy x + x y x x 3 + yx =. with boundary conditions y =, y = e;. see Bender Orszag [, p.453]. By using the method of matched asymptotics, the authors of [] constructed the one-term asymptotic solution y unif = x x K + e x x + e,.3 where the first three terms represent the leading terms in the expansions of the inner-inner solution, inner solution outer solution, respectively, the fourth term results from the matching of the inner solution outer solution. The two boundary layers arise from the two corresponding distinguished limits. Distinguished limits are those proper stretchings which produce a nontrivial balance between two or more terms of the equation. To determine the distinguished limits, we let yx = Y X X = x/δ. Mathematics Subject Classification. Primary: 34E; Secondary: 34B5. Key words phrases. Nested boundary layers, exponentially small terms, modified Bessel function, triple deck. 49

2 4 X. LIANG AND R. WONG With these new variables,. becomes 3 δ X d Y dy + δx dx dx δ3 X 3 Y Y =..4 For the scaling δ =, the second fourth terms of.4 are of comparable size while the first third are smaller. Regarding the other choice δ =, it is clear that the first fourth terms are of same order while the rest are smaller. As we all know, for single boundary layer problems, the inner solution matches the outer solution also satisfies one boundary condition. Unfortunately, the inner solution for the boundary value problem.. Y in X; = e [ /X + 3X 3 + ] 4X 4 +,.5 where X = x/, does not satisfy the boundary condition y= since Y in X; vanishes exponentially fast as X +. Intuition suggests that an additional boundary layer very near x = might be required in order to satisfy the condition y =. Since δ = is the only other distinguished limit for., Bender Orszag obtained the inner-inner solution Ȳ X; = XK X +,.6 where X = x/. Combining.5.6 with the outer solution, one obtains the uniformly valid one-term asymptotic solution.3. However, in lieu of matching the inner-inner solution Ȳ X; with the inner solution Y in X,, Bender Orszag claimed that the two solutions match automatically because they both vanish exponentially. Indeed, they are exponentially small, but they may not be of the same order. Thus, the solution.3 is only valid up to O cannot give the correct leading term in the matching region where the exact solution may be exponentially small. The treatment of the nested boundary layers problem is based on Prtl s principle of matched asymptotics. However, the lack of rigor in Prtl s boundary layer theory does raise concern from mathematicians who believe that arguments based on purely heuristic reasoning may lead to incorrect results. Examples of this nature can be found in [7, pp ] [9]. With the same purpose, we are going to reinvestigate the problem.. to derive a uniformly valid asymptotic solution by a mathematically rigorous argument. Moreover, we can find explicitly the exponentially small leading term in the interval where the exact solution exhibits such behavior. Before proceeding, let us look at one more example: a triple-deck problem. For a nested boundary-layer problem, the domains of validity of the outer solution, inner solution inner-inner solution are called outer region, inner region inner-inner region, accordingly. Furthermore, these layers are also known as decks, especially by physicists; in other words, they are called right deck, middle deck left deck, respectively. Hence, a nested boundary-layer problem is also referred to as a triple-deck problem. Several typical examples are given in [4] [5]. In [4, pp ], Nayfeh considered the boundary-value problem { 3 y x + x 3 y x + x 3 yx =.7 y = α, y = β,

3 ON A NESTED BOUNDARY-LAYER PROBLEM 4 gave the composite expansion y c x = βe x + βe /x + αe x/ βe +..8 To illustrate that y c x is close to the exact solution, Nayfeh presented a figure in [4, p.37] to compare y c x with the numerical solution. In Figure, we plot the numerical solution, Nayfeh s solution y c x our rigorously established approximate solution Y x given in Numerical Solution.4. Our Solution solution y Nayfeh s Matched Asymptotic Solution variable x Figure. Comparison of the composite solution y c x the asymptotic solution Y x with the exact numerical solution when α =., β =. =.5.. Preliminary transformations. For the general singularly perturbed two-point boundary value problem B.V.P. with the boundary conditions y x + axy x + bxyx =. y = A y = B,. it is now well-known that if ax is positive, then the asymptotic solution which holds uniformly in the interval [, ] is given by { bt } y unif x = Bexp at dt bt + A Bexp at dt e a x..3 x This formula can be easily derived by using Prtl s boundary layer theory; see, e.g., [, p.45]. Moreover, y unif in.3 approximates the true solution of. in the sense that yx = y unif x + O, where the O-term is uniform with respect to x [, ]; cf. [, p.479]. As we have mentioned before, heuristic reasoning may lead to incorrect results. For instance, if the boundary value B in. is zero, then.3 becomes y unif x = Ae a x,

4 4 X. LIANG AND R. WONG which is exponentially small for x > asymptotically equal to zero with respect to the error estimate O. A more accurate formula for the B.V.P... is bt yx = B exp x at dt [ + O] + a { } bt x A B exp ax at dt bt exp at dt exp atdt [ + O]. One can establish this result by using the Liouville-Green WKB approximation, which provides good asymptotic solutions to differential equations of the type y x + λ px + qxyx =,.4 where λ is a positive large parameter px is positive. The basic idea in obtaining an asymptotic solution to.4 is to introduce the Liouville transformation ξ = p / xdx w = p /4 xyx..5 When the coefficient function px in.4 has a zero turning point, say x = x, then there is an ambiguity in taking the square root of the function px, hence the Liouville transformation.5 is not well defined. However, a good approximation can be obtained by introducing the Langer transformation [3], which can also be used to obtain asymptotic solutions to internal boundary-layer problems e.g., when ax in. has a zero in the interval, ; see [8]. A third situation arises when px in.4 has a simple pole at x. Again, the transformation for this case was first introduced by Langer, extended popularized by Olver; see [6]. Our treatment of. is based on this transformation. The key idea behind this transformation is to get a good approximate equation for., to estimate the error term by using the method of successive approximation. Motivated by the inner-inner solution in.3, we will try to derive an equation which is a perturbation of Weber s equation for modified Bessel functions. As we all know, the ordinary differential equation d w dz = 4z w.6 has two linearly independent solutions zk z zi z. Hence, we will seek a transformation which converts. into a perturbed equation of.6. To this end, we introduce new independent dependent variables Straightforward calculation gives ζ = ζx yx = AxUζx..7 dy dx = A xuζx + Ax du dζ dζ dx d y dx = A xuζ + A x du dζ dζ dx + Ax du d ζ dζ dx + d U dζ dζ. dx

5 ON A NESTED BOUNDARY-LAYER PROBLEM 43 With the new variables,. becomes dζ 3 d U Ax dx dζ + 3 A x dζ dx + 3 Ax d ζ du dx + xaxdζ dx dζ + 3 A x + xa x x + x Ax U =..8 To eliminate the first order derivative of Uζ, we set 3 A x dζ dx + 3 Ax d ζ dx + xaxdζ =..9 dx Upon solving.9, we have { } Ax = ζ x exp x Substituting. into.8 yields d U dζ U { 4ζ = x 3 ζ x x + + x { ζx + ζ ζ x 3 x 3ζ x ζ x Again, set x 3 ζ x x + + x Solving., we obtain ζ x = 4ζ 4ζ } U 4ζ. } U. =.. t + t t dt..3 As a consequence, ζ = ζx is a well-defined one-to-one mapping from [, ] into [, ζ + ], where = ζ =, ζ + = ζ. Let us summarize the result in the following lemma. Lemma. Under the transformations.7,..3,. is converted into d U dζ U 4ζ = φζ U 4ζ,.4 where φζ = ζx ζ ζ x 3 x 3ζ x ζ..5 x 3. Conversion to an integral equation. We construct two linearly independent solutions W ζ W ζ to the associated homogeneous equation of.4, namely, d U dζ U =, 3. 4ζ such that W =, W ζ + =, W =, W ζ + =. By straightforward calculation, we have W ζ = ζk ζ P ζi ζ,

6 44 X. LIANG AND R. WONG where Moreover, ζi ζ W ζ =, Q P = K ζ + I ζ +, Q = ζ + I ζ +. W{W ζ, W ζ} = Q. By considering the right-h side of.4 as an inhomogeneous term of the homogeneous equation 3., we can conver.4 into the integral equation Uζ = λw ζ + µw ζ + Gζ, sφs Us ds, 3. 4s where λ, µ are arbitrary constants, Gζ, s is the Green s function defined by W ζw s Gζ, s = W{W s, W s} = QW ζw s, s ζ ζ +, 3.3 W sw ζ W{W s, W s} = QW sw ζ, ζ s ζ Construction of the Solution. Lemma. Let ρ be a positive constant. We have the order estimates φx = O x for x ρ4/3, 4. φx = O 6 x 4 for ρ4/3 x. 4. Proof. For convenience, we put ft = t + t t for < t. 4.3 From.3,.5 4.3, it follows that where { 3 φ x = f 6 x 6 φζ = φ x + φ x, φ x = 3 4 ζx x x x x } Let us first consider the case < t x ρ 4/3. Assume temporarily that x is much smaller than 4/3 i.e., x 4/3, so that ft = { + t t t3 [ + O 4, t } ], 4.6 x ζ x = { + x x [ + Ox3 4, x } ], 4.7

7 ON A NESTED BOUNDARY-LAYER PROBLEM 45 φ x = 3 6 x + 3x [ + Ox3 3 4, x ], 4.8 where Oh, k is used to indicate that the error term is of the order Oh + Ok. Coupling yields Hence, φ x = 3 6x + 7x 64 [ + O x3 3 4, x, 3 ]. 4.9 x φx = φ x + φ x = 3 [ + Ox3 7 4, x, 3 ] = O x. 4. Since we are only interested in the leading order estimate for φx, our earlier assumption x 4/3 can now be dropped because x 3 / 4 is still bounded. For the second case ρ 4/3 x, imply that Thus, we have ζ / x = ftdt φ x = 3 4 ζx 6 x 4 t x dt = 4 3. = O 6 x 4. In a similar manner, it can be shown that φ x = O 6 /x 4. Lemma implies that φx is much smaller than, thus we can use the solution of the homogeneous equation 3. to approximate the solution of the inhomogeneous equation.4. In the following lemma, we are going to estimate the error term caused by discarding the inhomogeneous term of.4. Lemma 3. As, we have Proof. Define Gζ, s φs 4s W sds = W ζo /3, 4. Gζ, s φs 4s W sds = W ζo /3. 4. M ζ, := M ζ, := N ζ, := N ζ, := ζ Gζ, s φs 4s W sds, ζ ζ Gζ, s φs 4s W sds, Gζ, s φs 4s W sds, ζ Gζ, s φs 4s W sds. First of all, we recall some asymptotic formulas of K z I z: K z z I z z as z, 4.3 as z, 4.4

8 46 X. LIANG AND R. WONG K z I z π z e z as z, 4.5 ez πz as z. 4.6 By making the change of variable given in.3 with z s replacing x ζ, we obtain M ζ, = OW ζ I szk sz φsz sz z + z z dz. 4.7 To proceed further, we divide our discussion into three cases. Let k, k be positive constants. Case I: < x k. For < z < x, we have from 4.6, fz = [ + O z3 z 4, z ], 4.8 s / z = z [ + Oz3 4, z ], 4.9 φz = 3z [ + Oz3 7 4, z, 3 ]. 4. A combination of 4.3, 4.4, 4.7, 4.8, gives s 3 z M ζ, = OW ζ 4 z s 7 dz = W ζo. 4. z M ζ, = W ζo + OW ζ k Case II: k x k 4 3. On account of 4.5, , 4. 4., we have e s π e s πs 4 s 4 3 z 7 s z dz = W ζo Case III: k 4/3 x. In a similar manner, we obtain M ζ, = W ζo 3 + OW ζ k 3 4 e s π e s 6 πs 4 s 4 z 4 s z dz = W ζo Therefore, M ζ, = W ζo 3 for ζ ζ ζ +.

9 ON A NESTED BOUNDARY-LAYER PROBLEM 47 For real positive z, K z is a monotonically decreasing function, while I z is a strictly increasing function. Making use of this property, we get Hence, as, M ζ, = W ζ W ζ ζ ζ = W ζo 3. I ζ φs [K s PI s] K ζ PI ζ ds I s φs K s PI sds Gζ, s φs 4s W sds = M ζ, + M ζ, = W ζo /3 4.4 uniformly for all ζ [, ζ + ]. Similar arguments show that Thus, 4. follows. N ζ, = W ζo 3 N ζ, = W ζo 3. Remark. Lemma 3 infers that there exist positive constants γ such that for <, Gζ, s φs 4s W sds W ζγ /3 4.5 Gζ, s φs 4s W sds W ζγ / Theorem. There exists a constant > such that for <, equation.4 has two linearly independent solutions w ζ = ζk ζ + O w ζ = ζi ζ + O 3, 4.8 where the O-symbols hold uniformly with respect to ζ [, ζ + ]. Proof. In 3., we let λ = µ =. Then Uζ = W ζ + We define a sequence U j ζ, j =,,..., by U ζ = U j ζ = W ζ + Gζ, s φs Usds s Gζ, s φs 4s U j sds j. 4.3 Clearly, U ζ = W ζ. Now suppose that for a particular value j, we have U j ζ U j ζ W ζγ 3 j, 4.3 as indeed is the case when j =. From 4.3, we get U j+ ζ U j ζ = Gζ, s φs 4s U js U j s ds j. 4.3

10 48 X. LIANG AND R. WONG Hence, give U j+ ζ U j ζ = γ 3 j Gζ, s φs 4s W sγ 3 j ds W ζγ 3 j. By induction, 4.3 holds for all j, the series Ũ ζ := [U j+ ζ U j ζ] j= converges. From 4.33, it follows that Ũ ζ U ζ + W ζ + Gζ, s φs 4s W sds U j+ ζ U j ζ j= W ζγ 3 j j= = W ζ + O Since Ũζ = lim U n nζ = lim n n j= [U j+ζ U j ζ], taking limits as j on both sides of 4.3 shows that equation.4 has a solution Ũζ satisfying Ũ ζ =W ζ + O 3 = ζk ζ P ζi ζ + O Similar arguments show that equation.4 has another solution Ũ ζ = W ζ + O ζi ζ 3 = + O Q Since P Q depend only on, w ζ w ζ are just linear combinations of Ũζ Ũζ, the result stated in the theorem is proved. Now, we are ready to construct the solution to B.V.P.... Coupling..3, we have Ax = x + x } + 4 x exp { x x By Theorem, it is straightforward to show that B.V.P... has two linearly independent solutions y x = x + x } + 4 x exp { x 4 3 ζ 4 xk ζ + O 3, 4.38 y x = x + x } + 4 x exp { x 4 3 ζ 4 xi ζ + O

11 ON A NESTED BOUNDARY-LAYER PROBLEM 49 In view of 4.7, 4.3, , we obtain π y = lim y x = x + O 3, lim y x =, 4.4 x 4 exp { } { exp } ζ + O The last estimate indicates that y is exponentially small as. Let us define Y x = y x, 4.4 ey x Y x = y The solution to the B.V.P... is given by Y x = Y x + Y Y x Y Note that Y /Y = y / e is exponentially small as tends to. In summary, we have the following theorem. Theorem. There exists a constant > such that for <, the boundaryvalue problem.. has the unique solution Y x = Y x + Y x, 4.45 where Y x Y x are given by , respectively. 5. Simplification of our solution. To simplify our solution 4.45, we have to study the behavior of ζ / x. Theorem 3. There is a positive number ρ such that ζ x x = + O x3 4, 3 for x ρ ζ c x = + x log x x log O 4 x x 3, 3 + O 3 log for ρ 43 x. Proof. By the definition of ζ / x in.3, we have ζ x x = t dt. 5.3 t t It is easy to verify that + t = O 3 for < t. 5.4 t

12 43 X. LIANG AND R. WONG By the binomial expansion, we have / 4/3 ζ x = t t dt t / 4/3 ] dt [ + O 3. ] dt [ + O t 5.5 The first two integrals on the right-h side of 5.5 can be evaluated by symbolic calculation Mathematica 5., the result is ζ x x = F, 6 ; 7 6 ; x arcsinh x 3 [ + O 3 ] t dt[ + O 3 ]. 5.6 This result can also be verified by using integral representations of the above two special functions. Now the theorem follows from 5.6 the lemmas 4, 5 6. Lemma 4. x F, 6 ; 7 6 ; x3 4 4 x = x + O x3 4 F, 6 ; 7 6 ; x3 4 4 = c + x O 4 x x 3 where c = Γ 3 Γ Γ.64996, ρ = 3. Proof. We know that the hypergeometric series F a, b; c; z = n= a n b n c n z n n! for x ρ for ρ 4 3 x, is absolutely convergent for z, provided that Rec a b >. In particular, F, 6 ; 7 6 ; x3 4 is convergent for x follows. 4 Referring to [, p.559], we have the following connection formula ΓcΓb a F a, b; c; z = ΓbΓc a z a F a, c + a; b + a; z + ΓcΓa b ΓaΓc b z b F b, c + b; a + b; z for arg z < π. With a =, b = 6, c = 7 x3 6 z = 4, equation 5. becomes 4 F, 6 ; 7 6 ; x3 4 4 = Γ7 6 Γ 3 x 3 Γ 6 Γ Γ7 6 Γ 3 Γ Γ x F, 3 ; 3 ; 44 6 x 3 F 6, ; 5 3 ; 44 x

13 ON A NESTED BOUNDARY-LAYER PROBLEM 43 Furthermore, it can be simplified to F, 6 ; 7 6 ; x3 4 4 = x 3 8 F, 3 ; 3 ; 44 x c, 5. x where c = Γ 7 6 Γ Γ Multiplying both sides of 5. by x making use of 5.9, we obtain 5.8. Lemma 5. 3 arcsinh x 3 = x3 6 + O x3 4 3 arcsinh x 3 = log x + 3 log 4 3x 3 + O 4 x 3 Lemma 6. x t t dt = x7 7 + O x3 4 for x ρ 4 3, for ρ 4 3 x. for x ρ 4 3, dt = x + O 8 3 for ρ 4 3 x. Lemmas 5 6 can be proved by using some basic calculus the asymptotic formulas of arcsinh z [, p.88]. From Theorem 3 the asymptotic formulas of K z I z, the solution Y x in 4.45 can be simplified to { x Y x = K } { x + o exp x 4 } + o for < x O 4 3, { 3 x exp x } + o + o for O 4 3 x. 5.3 We claim that, for O x O, the solution Y x is exponentially small. To see this, we note from 5.3 that the terms K x exp x 4 for O x 3 O 4/3 x exp x for O 4/3 x O are both exponentially small. To compare our solution with Bender Orszag s solution, we evaluate the solution at some particular points. For example, at x = 4/8, Bender Orszag s solution.3 gives y unif 4/8 = 4/8 K 4/8 + e 4/8 + e 4/8 4/4, while our solution yields { π Y 4/8 4/8 4 exp 4/8 4 3 } 4/8 π exp{ 5/ }. 5/56 4/4

14 43 X. LIANG AND R. WONG Moreover, at x = 7/6, Bender Orszag s solution.3 gives y unif 7/6 = 7/6 K 7/6 + e 7/6 + e 7/6 7/3, while our solution gives Y 7/6 exp { 7/6 } + o 7/6 exp{ }. /6 In conclusion, for O x O Bender Orszag s asymptotic solution.3 seems to be incorrect since it shows that the leading term in the asymptotic approximation of the exact solution is of algebraic order, whereas our solution 4.45 shows that the true solution decays exponentially; this phenomenon has never been mentioned in the existing literature. 6. Triple-deck problem. Based on the experience from studying problem.., we believe that our method also works for problem.7. In fact, problem.7 is much simpler than problem.. due to the lack of singularity at the origin. Since the techniques are the same, we present only a brief outline of the derivation of an asymptotic solution to the triple-deck problem.7. In parallel to Lemma, we have Lemma 7. Under the transformations ζx = t t 3 dt 6..7 is converted to where } yx = ζ x exp { x4 8 3 Uζx, 6. d U U = φζu, 6.3 dζ φζx = ζ x ζ x 3 3 ζ x 4 ζ x The estimate φζ = O /3 implies that d U dζ U = 6.5 can be used as a perturbed equation to 6.3, guarantees success of the successive approximation method used in Sect. 4. As before, one can show that equation.7 has two linearly independent solutions y x = 3 x x 4x 3 } 4 exp { x4 8 { 3 exp 6.6 t t 3 dt } + O 3

15 ON A NESTED BOUNDARY-LAYER PROBLEM 433 y x = 3 x x 4x 3 } 4 exp { x4 8 { 3 exp t t 3 dt } + O 3. The solution which satisfies the boundary conditions of.7 is given by Y x = α y x + where y x y x are given in , respectively. REFERENCES β y y x, 6.8 [] M. Abramowitz I. A. Stegun, Hbook of Mathematical Functions with Formulas, Graphs, Mathematical Tables, Dover, New York, 99. [] C. M. Bender S. A. Orszag, Advanced Mathematical Methods for Scientists Engineers, McGraw-Hill, New York, 978. [3] R. E. Langer, The asymptotic solutions of ordinary linear differential equations of the second order, with special reference to a turning point, Trans. Amer. Math. Soc., , [4] A. H. Nayfeh, Introduction to Perturbation Techniques, John Wiley Sons, New York, 98. [5] A. H. Nayfeh, Problems in Perturbation, John Wiley Sons, New York, 985. [6] F. W. J. Olver, Asymptotics Special Functions, Academic Press, New York, 974. [7] R. Wong, Five Lectures on Asymptotics, in Differential Equations & Asymptotic Theory in Mathematical Physics Eds. Chen Hua R. Wong, World Scientific, Singapore, 4, [8] R. Wong H. Yang, On an internal layer problem, J. Comput. Appl. Math., 44, [9] R. Wong H. Yang, On a boundary-layer problem, Stud. Appl. Math., 8, Received March 8; revised September 8. address: xdliang@math.mit.edu address: mawong@cityu.edu.hk