On a Two-Point Boundary-Value Problem with Spurious Solutions

Transcription

1 On a Two-Point Boundary-Value Problem with Spurious Solutions By C. H. Ou and R. Wong The Carrier Pearson equation εü + u = with boundary conditions u ) = u) = 0 is studied from a rigorous point of view. Known solutions obtained from the method of matched asymptotics are shown to approximate true solutions within an exponentially small error estimate. The so-called spurious solutions turn out to be approximations of true solutions, when the locations of their spikes are properly assigned. An estimate is also given for the maximum number of spikes that these solutions can have.. Introduction Despite the fact that the boundary-layer theory has been around for nearly 00 years, much of the asymptotic analysis used in this theory still lacks mathematical rigor. Although this fact does not seem to bother most of the applied mathematicians using this theory, it does raise concern to some of the more rigor-oriented mathematicians. This is mainly because they believe that arguments based on purely heuristic reasoning may lead to incorrect results. As an illustration, let us consider the familiar singularly perturbed boundary-value Address for correspondence: R. Wong, Department of Mathematics, City University of Hong Kong, Tat Chee Avenue, Kowloon, Hong Kong. STUDIES IN APPLIED MATHEMATICS : C 00 by the Massachusetts Institute of Technology Published by Blackwell Publishing, 50 Main Street, Malden, MA 048, USA, and 9600 Garsington Road, Oxford, OX4 DQ, UK.

2 78 C. H. Ou and R. Wong problem BVP) { εy + ax)y + bx)y = 0, 0 < x <, ) y0) = A, y) = B, where A, B are prescribed constants, 0 <ε is a small positive number, the coefficients ax) and bx) are sufficiently smooth functions, and ax) is positive in the whole interval. Using the method of matched asymptotics, one easily derives the asymptotic approximation ) { bt) )} yx) = B exp x at) dt bt) + A B exp 0 at) dt e a0)x/ε + Oε), ) holding uniformly for all x [0, ]. This formula is given in at least 0 standard texts; see, e.g., [, p. 45], [, pp. 5 & 58], [, p. 59], [4, p. 68], [5, p. 4], [6, p. 89], [7, p. 94] and [8, p. 09]. If the boundary value B is zero, then the order estimate Oε) is of not much use because the leading term is exponentially small when x is bounded away from zero. Furthermore, the exponential function e a0)x/ε does not have much to do with the coefficient function ax) except for its value at the end point x = 0. Indeed, from the matching procedure, one readily finds that this exponential function can be replaced by e h0)x/ε for any hx) as long as h0) = a0). The correct asymptotic approximation should be yx) = B exp x + a0) ax) exp ε bt) at) dt ) { A B exp x 0 at) dt )} bt) x ) 0 at) dt bt) exp 0 at) dt ) [ + Oε)] ) for all x [0, ]. This result can be obtained by using the Liouville Green approximation given in Olver [9, Chapter 0], and is also given in Smith [0, Section 8.]. Another illustrative example is provided by the turning-point problem { εy + ax)y + bx)y = 0, x [, ], 4) y ) = A, y) = B, where ax) and bx) are sufficiently smooth functions with the behavior given by ax) αx and bx) β as x 0. When α<0and β/α 0,,,..., Bender and Orszag [] used a matching technique to derive the leading-order uniform asymptotic solution y unif x) = Ae a )x+)/ε + Be a) x)/ε. 5)

3 A Two-Point Boundary-Value Problem with Spurious Solutions 79 By applying 5) to the concrete problem εy xy + + x )y = 0, y ) =, y) =, 6) they obtained the approximation y unif x) = e x+)/ε + e x)/ε. 7) Taking x = 0 in 7) gives y unif 0) = e /ε. 8) This result is incorrect, as pointed out recently by Wong and Yang []. Using rigorous mathematics, they showed that the correct approximation should be y0) 6 π Ɣ ) e /4 ε /4 e /ε. 4 9) The mistakes that lead to ) and 5) are mainly due to the matching techniques used. In boundary-layer problems, exponentially small terms are expected to appear in leading-order approximations. Thus, Poincaré sdefinition of asymptotic expansion is inadequate to describe the behavior of solutions to such problems. This observation has already been made by Carrier and Pearson [, pp. 0 05], almost 40 years ago. They used the simple example { εu x) + u x) =, u ) = u) = 0 < x <, 0) to show that the method of matched asymptotics can lead to spurious solutions. More precisely, they first showed that an approximate solution to 0) is given by where ũ x,ε) = + e p p e + + e p ) + e, ) p ) p = ε + x) + ln + ) ) and p = ε x) + ln + ). ) Let x 0, ),ζ = x x 0 )/ ε, and They also noted that qζ ) = e ζ + e ζ ). 4) ũx,ε) = ũ x,ε) + qζ ) 5)

4 80 C. H. Ou and R. Wong appears to be a valid approximate solution if x 0 ε. 6) However, by using the phase-plane method, they demonstrated that ũx,ε) in 5) could not be an approximate solution unless x 0 = 0. Thus, for most values of x 0, the solutions given in 5) cannot be valid, and they are referred to as the spurious solutions. The function qζ ) in 4) behaves like a spike near x 0 for sufficiently small ε. In [], Lange considered a situation in which there are n internal boundary layers, i.e., n spikes. For any n points satisfying < x < x < < x n <, 7) the function n ) x x j ũx,ε) = ũ x,ε) + q 8) ε could be a valid approximate solution to 0), where ũ and q are given, respectively, by ) and 4). Lange provided an explanation why the method of matched asymptotics fails to locate these internal boundary layers, and presented a modification of the method. Instead of Poincaré-type perturbation series, he used expansions in powers of exponentially small functions in ε, and allowed the higher-order terms in these expansions to contain growing exponentials in the stretched variable, e.g., ζ in 4). More recently, MacGillivray [4] re-visited the problem 0) with a single spike. His idea is basically similar to that of Lange. By incorporating exponentially small terms in the matching, he was able to show directly that 5) could be a valid approximate solution only if x 0 = 0, without using the phase-plane method. All results obtained in [ 4] are based on heuristic arguments, and it is generally recognized that it is very difficult to establish them on mathematically rigorous grounds; see a comment in the first paragraph of []. In this paper, we are concerned with three mathematically more fundamental questions; namely, i) in what sense does the function ũ in ) approximate a solution of 0), ii) for a given ε>0, how many spikes can there be in the interval, ) so that the function ũ in 8) is a valid approximate solution to 0), iii) how well does ũ in 8) approximate a solution of 0)? An expected answer to question i) is that there is a solution u of 0) satisfying j= u x,ε) = ũ x,ε) + Oε) for all x [, ]; cf. [, p. 479, Prob. 9.]. It turns out that we have the much better result u x,ε) = ũ x,ε) + Oe /ε ) 9)

5 A Two-Point Boundary-Value Problem with Spurious Solutions 8 for all x [, ], the meaning of 9) being that there exist ε 0 > 0 and K > 0 such that for 0 <ε<ε 0, u x,ε) ũ x,ε) Ke /ε 0) for all x [, ]. As to question ii), i.e., the number N of internal) spikes, we have the estimate N ) ε Our answer to question iii) is similar to that of question i); namely, there exists a solution u of 0) satisfying n ) [ x x j ux,ε) = ũ x,ε) + q + O exp )] ) ε n + ε j= for all x [, ], where ũ and q are given in ) and 4), respectively.. The shooting method Our approach to the BVP 0) is through the shooting method. Thus, instead of 0), we consider the associated initial-value problem { εu x) + u x) =, u ) = 0, u ) ) = k, where k is a real number. It is readily verified that this problem has a unique solution, which can be extended to infinity. The results in the following lemmas will play an important role in the proof of our main theorems. LEMMA. Let ux, k) denote the solution of the initial-value problem ): a) If k < 4/ε, then ux, k) is decreasing for x and lim x ux,k) =. b) if k > 4/ε, then there exists a number η> such that uη, k) = 0, ux, k) is increasing in < x < η )/ and decreasing in η )/ < x <. Furthermore, lim x ux, k) =. c) If k =± 4/ε, then [ ux, k) = + sech x + + ln + ] ). 4) d) If k < 4/ε, then ux, k) is periodic and intersects the x-axis infinitely many times.

6 8 C. H. Ou and R. Wong Proof : As in [, p. 0], we let ux) = + vx) so that 0) is converted into the equivalent problem { εv v + v = 0, v ) = v) =. 5) The corresponding initial conditions in ) now become v ) =, v ) = k. 6) a) Multiplying both sides of the differential equation in 5) by v and integrating from to x, we obtain where ε v ) v + v = C, 7) C = ε k. 8) If k < 4/ε, we claim that v < 0 for all x >. Assume the contrary. Then, on one hand, there exists a number α> such that v < 0 for x, α), v α, ε, k) = 0, and vα, ε, k) <. On the other hand, because k > 4/ε and v / + v > 0 when v<, we have from 7) thus proving the claim, and v α, ε, k) 0, lim vx,ε,k) = l. x We wish to show that l =. Because k < 4/ε, there is a number β>0such that k > 4 ε + β. Again using the fact that v / + v > 0 when v<, we have from 7) for x >, which clearly implies v x,ε,β) < β, lim vx,ε,k) =. x b) When k > 4/ε, vx,ε,k) is increasing in the right neighborhood of x =. Because C > 0 in this case, it is readily seen from the graph of the

7 A Two-Point Boundary-Value Problem with Spurious Solutions 8 v Figure. Graph of Fv). function Fv) = v + v, depicted in Figure, that there exists a point x = η 0 > such that v η 0,ε,k) + v η 0,ε,k) = C and vη 0, ε, k) >. From 7), it follows that v η 0, ε, k) = 0 and v > 0 in the interval, η 0 ). From 5), we also have v η 0, ε, k) < 0. Thus, vx, ε, k) attains a local maximum at x = η 0. Using 7) and the graph in Figure, it can be further seen that there exists a number η such that v x, ε, k) < 0inη 0 < x <η, vη, ε, k) = and v η, ε, k) = k. With the initial values vη) = and v η) = k, the argument in a) can be repeated to show that v x, ε, k) < 0 for x >ηand lim x vx, ε, k) =. c) When k =± 4/ε, C = 0 in 7). Because the positive cubic term dominates the negative square term in that equation, we must have lim x vx, ε, k) = 0. With v ) = and v ) = 0, Equation 7) has the solutions [ v = sech x + + ln + ] ). 9) d) When k < 4/ε, wehave C < 0. First, we consider the case 4/ε <k < 0. Because F) =, from the graph of Fv) in Figure it is easily seen that Fv) = C has two positive roots v and v such that v 0, ) and v, ). From 7), it also follows that there are two points x = d and x = d with d < d such that v d,ε,k) = 0, vd,ε,k) = v 0, ) and vd,ε,k) =, v d,ε,k) = k > 0. Starting from x = d, one can use the argument in case b) to show that there exist two points d and d 4 such that

8 84 C. H. Ou and R. Wong and v d,ε,k) = 0, vd,ε,k) = v, ), vd 4,ε,k) =, v d 4,ε,k) = k < 0. At x = d 4, vx, ε, k) has the same initial values as those at the starting point x =. Hence, vx, ε, k) is periodic with period d 4 +. The case 0 < k < 4/ε is entirely similar, and can be handled in the same manner. When k = 0, we have C = in 7). From that equation, it is clear that we must have Fv), and hence v +. Therefore, it follows that there is a point d > such that v d,ε,k) = 0 and vd,ε,k) = +. With these initial values, one can show by using 5) that vx, ε, k) is decreasing for x > d, until x reaches the point d, where vd,ε,k) = and v d,ε,k) = 0. The last set of initial values are the same as the first set given in 6). Hence the values of v begin to repeat itself, and there are infinitely many points where we have v = and v = 0. Equivalently, the value of vx, ε, k) travels back and forth between and + infinitely many times. LEMMA. If k 4/ε, then u, ε, k) 0 for ε /. Proof : When k < 4/ε, we have shown in Lemma that ux, ε, k) is strictly decreasing for x >, and hence u, ε, k) 0. When k = 4/ε, direct differentiation of Equation 9) shows that v x, ε, k) is negative for x >. Hence, we again have u, ε, k) 0. When k = 4/ε, it follows from 9) that v + α, ε, k) = 0, v + α, ε, k) =, and vx, ε, k) is decreasing for x > + α, where α = ln + ). The derivation of the second equation makes use of the identity cosh z = lnz + z ). Because ln + ) < for ε /, one has + α < and v, ε, k) <. When k > 4/ε, C > 0 in 7) and the equation Fv) = C has only one real root v = a >, i.e., a + a = C; see Figure. Let η be the point where v η) = a. By Lemma, there is only one point x = η such that vη, ε, k) =. We will now estimate the distance L between x = and x = η. From 7), we have

9 A Two-Point Boundary-Value Problem with Spurious Solutions 85 ε dv = dx v v a + a for < x < η. Integration on both sides gives ε a dv η + =. v v a + a Similarly, we have for η <x <η, and ε η η = ε dv = dx v v a + a a dv v v a + a. Coupling the two results gives η + = a dv a v + v a = a dv 6ε a v)v + av v + a a). Note that the function f v) = v + av v + a a is increasing in the interval [, a], i.e., f v) > 0. Hence L a 6ε a a < 4 ε, which in turn implies u,ε,k) = + v,ε,k) < 0. From the above lemma, it is clear that to have u, ε, k) = 0, we must choose k in the interval 4/ε, 4/ε), in which case by Lemma the solution to the initial-value problem ) is periodic and intersects the x-axis infinitely many times. We consider only the case k 0, because the case k < 0 is entirely similar. Let x be the first point to the right of such that ux ) = 0 and ux) > 0 for < x < x, and x be the first point to the right of x such that ux ) = 0 and ux) < 0 for x < x < x. We denote by T and T, respectively, the lengths of the intervals [, x ] and [x, x ]; see Figure.

10 86 C. H. Ou and R. Wong _ T x T x T Figure. Lengths T and T ; k > 0. For 0 < a, we define a one-to-one mapping from k to a such that C = ε k = a a. 0) From 7), we have ε v ) = a a v + v where v = a, = v v )v v )v v), ) v = a 9 + 6a a ), v = a a a ). ) It can be easily verified that for 0 < a, we also have v < 0 and + v <. ) Note that v, v, and v all satisfy a a = v v = v v = v v. 4) From ), it follows as in the proof of Lemma that T a) = v dv v v a + a 5) and T a) = dv, 6) a v v a + a

11 A Two-Point Boundary-Value Problem with Spurious Solutions 87 where T and T are the lengths of the intervals, x ) and x, x ); see Figure. Obviously, T a) and T a) are continuous in a, and in fact differentiable with respect to a. LEMMA. T a) is increasing and T a) is decreasing for 0 < a <. In addition, we have dt + T ) da for 0 < a <. < 0, d T da > 0, and d T + T ) da > 0 7) Proof : In view of 4), we can replace a by v in 5). Expanding v v into a Taylor series at v = gives Similarly, we have v v = + v ) v ). v v = + v ) v ). Upon subtracting the second equation from the first, 5) becomes T a) = v { v v ) [ v ) v ) ]} / dv. Now we make the change of variable t = v )/v ). This yields T a) = { 0 t v + v ) / + t + t )} dt. Straightforward differentiation shows that Because dt dv < 0 and d T dv > 0. dv da = ) a + < 0 8) 9 + 6a a and d v /da is also negative for 0 < a <, we have dt da = dt dv dv da > 0

12 88 C. H. Ou and R. Wong and d T da = d T dv ) dv + dt d v da dv da > 0 for 0 < a <. In a similar manner, we make the change of variable t = v)/ a) in 6), and obtain T a) = 0 { t a a) / + t + t )} dt. 9) Again, straightforward differentiation shows that dt /da is negative for 0 < a <. From ), 5), and 6), it follows that T a) + T a) = 6ε v a dv v a)v v )v v). Set v = acos θ + v sin θ. Then v a = v a) sin θ, v v = v a) cos θ, and dv v a)v v )v v) = dθ. v v Adding the two equations in ) yields v + v = a. Hence, v v = a cos θ + v sin θ a v ). Changing the integration variable from v to θ gives T a) + T a) = π 6ε 0 dθ. + a + cos θ) + v + sin θ) For θ [0, π ] and a 0, ), it can be easily verified that + cos θ + dv da + sin θ) > 0, because / dv /da < 0; see 8). Now straightforward differentiation shows that dt + T ) da < 0 and d T + T ) da > 0 for a 0, ). In the last inequality, the result d v /da < 0 has been used for a 0, ). This completes the proof of the lemma.

13 A Two-Point Boundary-Value Problem with Spurious Solutions 89 From 5) and 6), we also have T 0) = dv 6ε v v = ln + ) ε.4 ε, 40) T ) = + 6ε dv v + )v ) + v) 4.88 ε, 4) T ) = 0 and lim T a) =. 4) a 0 To prove the second result in 4), we observe that the function a + a is decreasing in 0 < a < ; cf. Figure. Hence, from 6), we have T a) dv 6ε a v v. The integral on the right becomes infinite when a tends to 0. Remark : For small values of ε, T a) T ) 4.88 ε<. From this, it follows that the BVP 0) does not have a solution, which is positive in the entire interval, ). This explains why Carrier and Pearson chose the outer solution to be, instead of, in their matching process; see ). The discussion for the case 4/ε <k < 0 is similar, and vx, ε, k) is again oscillatory about the straight line v =, except that it now starts at x = with a negative slope. Hence, the graph of vx, ε, k) is below the horizontal line v = in the interval, x ), where x is the first point to the right of at which vx, ε, k) meets the line v = again. Equivalently, we have ux) < 0 in, x ). Let x denote the first point to the right of x such that ux ) = 0 and ux) > 0 for x < x < x. Unlike before, we now reverse the order, and denote by T and T the lengths of the intervals [, x ] and [x, x ], respectively. The values of T and T again depend on the slope k. Note that the relationship between a and k given in 0) still hold, and formulas 5) and 6) remain valid. Hence, T and T are actually dependent on the values of k.. Number of spikes and solutions We first recall that our objective is to use the shooting method to determine whether the BVP 0) has solutions. If it has, then we would like to give an estimate for the maximum number of spikes that solutions to this problem can have. This is equivalent to choosing appropriate values of k so that v,ε,k) =. 4)

14 90 C. H. Ou and R. Wong Because vx, ε, k) is periodic and oscillatory about the line v =, Equation 4) can be expressed in the form NT a) + MT a) =, 44) where N and M are nonnegative integers and represent, respectively, the number of intervals with lengths T and T. Now, we look at the arches bounded by the graph of vx, ε, k) and the line v =. Because every arch above v = is followed by an arch below v = and vice versa, the integers N and M are related in such a way that for every given N, there are only three possible values of M, namely M = N, N, orn +. From here on, our task is to choose appropriate values of kor a), and determine possible integers N and M so that 44) holds. For ε>0, we define [ ] [ 0.4 Nε) := ], 45) T ) ε where [x] denotes the largest integer less than or equal to x. The following results provide estimates for the number of spikes and solutions to the boundary-value problem 0). Here, by spikes we mean those, which lie above the line v =. LEMMA 4. IfN> Nε) +, then the boundary-value problem 0) has no solution with N spikes. For any integer 0 N Nε), there exists at least one solution for the problem 0) with N peaks including the boundary and internal spikes). If N = Nε) +, then either case can happen. Proof : When N > Nε) +, we have by the monotonicity of T + T NT a) + MT a) [Nε) + ]T a) + [Nε) + ]T a) = [Nε) + ][T a) + T a)] + T a) > [Nε) + ][T ) + T )]. Because T ) = 0, by 45) NT a) + MT a) > ; that is, 44) fails to hold and the BVP 0) has no solution. For 0 N Nε), we may choose M = N +. Then by 45), and NT ) + MT ) = NT ) Nε)T ) lim a 0 {NT a) + N + )T a)} =

15 A Two-Point Boundary-Value Problem with Spurious Solutions 9 by 4). In view of the continuity of the function NT a) + N + )T a), there exists at least one point a 0 0, ] such that NT a 0 ) + N + )T a 0 ) =, which means that there exists at least one solution to the BVP 0) with N arches with base length T and N + arches with base length T. When N = Nε) +, both functions NT a) + NT a) and NT a) + N + )T a) are decreasing in a, and their values at a = satisfy NT ) + NT ) = [Nε) + ]T ) > and NT + N + )T ) = [Nε) + ]T ) >. Thus it is not possible to expect that NT a) + NT a) = or NT a) + N + )T a) = for some a 0, ]. However, if we choose M = N, then the function NT a) + N )T a) = Nε)[T a) + T a)] + T a) is convex with NT 0) + MT 0) = and NT ) + MT ) >. Let m := min 0<a [NT a) + MT a)]. Clearly, m depends on ε. Ifm >, then NT a) + MT a) > for all a 0, ] and there is no solution to the BVP 0). When m =, there exists a 0 0, ] such that NT a 0 ) + MT a 0 ) =, i.e., there exists a solution to the BVP 0). When m <, then there exist two distinct points a i, i =,, such that NT a i ) + MT a i ) =. This completes the proof of the lemma. THEOREM. If Nε) /T ε), then the number nε) of solutions to the BVP 0) satisfies 4Nε) nε) 4Nε) +. 46) If Nε) = /T ), then it satisfies 4Nε) ) + nε) 4Nε) ) +. 47) Proof :Wefirst consider the case Nε) /T ). By Lemma 4, the BVP 0) has no solution with N spikes if N > Nε) +. Let us now count the number of solutions with N spikes, where N ranges from 0 to Nε) +.

16 9 C. H. Ou and R. Wong ) If N = 0, then we should look for solutions with arches whose base length is all T a). This can happen only if the initial slope k is negative. Because T 0) = and T ) = 0, by the monotonicity and continuity of T a), there exists one and only one value a 0, ) such that 44) holds with M =. This, in turn, means that the BVP 0) has a unique solution. This is the negative solution referred to in Carrier Pearson s book [].) ) If N =, then we should look for solutions with only one arch whose base length is T a). These arches are exactly the ones which lie above the line v =. Possible values of M are 0,, and. By Remark, the BVP 0) has no solution which is entirely above v =. Hence, M 0 and the only other possibilities are M = and M =. When N = M =, we have NT a) + MT a) = T a) + T a). The function on the right-hand side of this equation is decreasing in 0, ) by Lemma. Because T 0) + T 0) = and T ) + T ) = T ) <, there is one and only one) value a 0 0, ) such that T a 0 ) + T a 0 ) =. In view of 0), each value a 0 is associated with two values of k, namely, k 0 and k 0 such that 4) holds; that is, when N = M =, there are two solutions to the BVP 0). When N = and M =, we look for solutions with one arch above the line v = and two arches below v =. Because these arches alternate, the initial value k in 6) must be negative. Clearly, T 0) + T 0) = and T ) + T ) = T ) <. By Lemma, there is one and only one a 0 such that T a 0 ) + T a 0 ) =. In view of the restriction that k must be negative, we conclude that there is only one solution to the BVP 0). In summary, there are three solutions to 0) when N =. ) For N Nε), possible choices for M are M = N, N, and N +. If M = N, then the solution must start at x = with a positive slope k. Again, we have NT 0) + MT 0) = and NT ) + MT ) = NT ) <. Although NT a) + MT a) is not necessarily decreasing in 0, ), it is convex by Lemma. Hence, we can have only one value a 0 0, ) such that NT a 0 ) + MT a 0 ) =. If M = N, then the solution will have the same number of arches above and below the line v =, and it can start at x = with either a positive or a negative slope. The values of NT a) + NT a) at the endpoints 0 and are again and less than, respectively. Because this is a decreasing function, there exists only one value a 0 0, ) such that NT a 0 ) + NT a 0 ) =. In terms of k, there are two values k 0 and k 0 such that v, ε, k 0 ) = v, ε, k 0 ) = ; i.e., we have two solutions to the BVP 0). If M = N +, then the solution will have N + arches below v = and N arches above v =. Thus, it must start at x = with a negative slope k. As before, we have NT 0) + N + )T 0) = and NT ) + N + )T ) = NT ) <. Because NT a) + N + )T a) is decreasing in a by Lemma, there exists one and only one value a 0 0, ) such that NT a 0 ) +

17 A Two-Point Boundary-Value Problem with Spurious Solutions 9 N + )T a 0 ) =. Furthermore, because k must be negative, there is only one value k 0 such that v, ε, k 0 ) =. Therefore, we conclude that for each N in N Nε), there are four solutions to the BVP 0). Adding them up, we have 4Nε) ) solutions when N varies from to Nε). 4) For N = Nε) +, the only choice for M is M = N. To see this, we note that if M = N or N +, we have NT a) + MT a) NT a) + T a)) NT ) + T )) by Lemma. Because T ) = 0 and Nε) + > /T ) by the definition in 45), it follows that NT a) + MT a) > for all a [0, ]. Hence, we only need to find out how many solutions there are for the BVP 0) when N = Nε) + and M = Nε). Note that in the present case, the values of NT a) + MT a) at the endpoints a = 0 and a = are both greater than ; i.e., NT 0) + MT 0) > and NT ) + MT ) >. However, by Lemma, it is easily seen that this function is convex, i.e., d da [NT a) + MT a)] 0. Therefore, it follows that there are at most two solutions for the BVP 0). Adding up all the solutions together, we obtain + + 4Nε) ) nε) + + 4Nε) ) +, thus proving 46). Finally, we come to the proof of 47), when Nε) = /T ) is an integer. As before, we consider the function NT a) + MT a), and choose appropriate values of N, M, and a so that NT a) + MT a) =. First, we claim that N < Nε) +. Indeed, if N Nε) + then M Nε), and NT a) + MT a) Nε)T a) + T a)) + T a) Nε)T ) + T )) + T a) >, because T ) = 0 and T a) > 0. In a similar manner as in the earlier part of this proof, it can be shown that when N varies from 0 to Nε), there are exactly 4Nε) ) solutions for the BVP 0). For the last case N = Nε) = /T ), we also need to consider possible values of M and a such that NT a) + MT a) =. Here, again, M may be equal to N, N, orn +. When N = Nε) and M = N = Nε), the function NT a) + MT a) is convex by Lemma and its endpoint values are given by NT 0) + MT 0) = and NT ) + MT ) =. Hence, there are at most two values of a [0, ] such that NT a) + MT a) =. The right endpoint a = happens to correspond to a solution of the BVP

18 94 C. H. Ou and R. Wong 5), or equivalently 0). Let us denote this solution by v, and the other solution by v. Note that a = implies k = 0. Hence, v = vx, ε, 0). When N = M = Nε), the function NT a) + MT a) is decreasing in a with NT 0) + NT 0) = and NT ) + MT ) =. Hence, there is only one value of a, namely a =, that yields a solution to the BVP 5). We denote this solution by v, and note that v = vx, ε, 0). When N = Nε) and M = N + = Nε) +, the function NT a) + MT a) is also decreasing in a with NT 0) + MT 0) = and NT ) + MT ) =. Hence, there is again only one solution, corresponding to the value a =, which we denote by v 4. Note that v 4 = vx, ε, 0). Because v, v, and v 4 are the same solution vx, ε, 0), the number of solutions to 5) or 0), in the case when N = Nε) = /T ), is at most and at least. Recall that there are 4Nε) ) solutions for 0 N Nε). Adding the number 4Nε) ) to the number of solutions when N = Nε) = /T ), we conclude that the number of solutions to 0) or, equivalently, 5) satisfies 4Nε) ) + nε) 4Nε) ) +. This completes the proof of the theorem. Remark : From Theorem, it follows that nε) has the asymptotic formula nε).64 as ε ) ε 4. Proof of formula 9) First, we observe that the approximate solution ũ x,ε) given in ) can be expressed as x + ũ x,ε) = + sech + ln + ) ) x + sech + ln + ) ). 49) It is known that there are three other approximate solutions of a similar type; they are ũ x,ε) = + sech x + + ln + ) ) x + sech + ln + ) ), 50)

19 A Two-Point Boundary-Value Problem with Spurious Solutions 95 x + ũ x,ε) = + sech + ln + ) ) + sech x + ln + ) ), 5) and ũ 4 x,ε) = + sech x + + ln + ) ) + sech x + ln + ) ). 5) THEOREM. For small values of ε, there exist exactly four solutions u i x, ε), i =,...,4,to the BVP 0), which have boundary-layer behavior near the endpoints of the interval and no internal spikes. These solutions are uniformly approximated by the functions ũ i x,ε), i =,...,4, given in 49) 5) with an error estimate of the order Oe /ε ), that is, u i x,ε) = ũ i x,ε) + Oe /ε ), i =,...,4, 5) as ε 0 +, uniformly for x [, ]. Proof : a) We first consider the case i =. Recall that lim a 0 T a) =, T ) = 0, and T a) is decreasing for 0 < a <. Hence there is only one value a 0 0, ) such that T a 0 ) =, 54) which, in turn, means that if we choose k 0 = ε a 0 + a 0, then the solution ux, ε, k 0 ) will satisfy u, ε, k 0 ) = 0 and is negative in the entire interval < x <. First, we claim that ux, ε, k 0 ) or, equivalently, vx, ε, k 0 ) is symmetric with respect to the origin, i.e., vx, ε, k 0 ) = v x, ε, k 0 ) and u 0, ε, k 0 ) = v 0, ε, k 0 ) = 0. To see this, we note that if vx, ε, k 0 )is a solution of the BVP 0), then so is v x, ε, k 0 ). But, by the monotonicity of T a), there is only one negative solution to the BVP 0). Hence, vx, ε, k 0 ) = v x, ε, k 0 ) and v 0, ε, k 0 ) = 0. Next, we give an estimate for a 0, which will play a key role in the proof of the theorem. From 6) and 54), we have } {v v a 0 + a 0 dv = ε, 55) a 0

20 96 C. H. Ou and R. Wong which can be written as a 0 { v a 0 ) [ v a 0 v a 0 ]} ) dv = ε. Because the quantity inside the square brackets is greater than, the last equation gives and hence which, in turn, yields a 0 dv > v a0 cosh a 0 > ε, ε a 0 e /ε. 56) This is only a rough estimate of a 0. A more precise description of the behavior of a 0 can be obtained by using an asymptotic formula for the elliptic integral R F x, y, z) = dt. 57) t + x)t + y)t + z) 0 In [5], Carlson and Gustafson have shown that for real x, y and z, wehave θ )R F x, y, z) = z / log 4z / x / + y /, 0 <θ < x + y. 58) 4z Returning to 9), we now make the change of variable τ = t in the integral on the right-hand side. This gives where a 0 {v v a 0 + a 0 } dv = a 0 b = 6a 0 a 0 a 0 ). 0 t t t + b dt, Note that b 6a 0 and hence b 0asa 0 0. Now we make a further transformation t = b/b + y) in the last integral to obtain a 0 {v v a 0 + a 0 } dv = a 0 0 dy y + b)y + c)y + d), 59)

21 A Two-Point Boundary-Value Problem with Spurious Solutions 97 where and c = b + 4b + 9) = b + b 7 + ob ), as b 0, d = b + + 4b + 9) = + 4b b 7 + ob ), as b 0. Applying 58) to 59) gives } dv a 0 {v v a 0 + a 0 a 0 d log 4 d b + c as ε 0. Inserting the last formula into 55) yields ) 4 a exp, ε ) ε To prove 5) when i =, we first restrict x to the interval [, 0] and put x + v L x) := sech + ln + ) ). From the proof of case c) in Lemma, it is readily seen that v L satisfies and εv L v L + v L = 0 v L ) =, v L ) = 4 ε. By definition, v, ε, k 0 ) = k 0. Hence, it follows from Lemma a) that 4 v,ε,k 0 ) = k 0 > ε = v L ), 6) for otherwise one can deduce that v, ε, k 0 ). Now, let Rx) denote the difference between v L x) and the true solution vx, ε, k 0 ), i.e., In view of 60), we have We further claim that Rx) := vx,ε,k 0 ) v L x). 6) R ) = 0 and R ) > 0. 6) Rx) > 0 and R x) > 0 64)

22 98 C. H. Ou and R. Wong for all x, 0]. Indeed, if Rx) 0 for some x, 0], then there exists a point ξ, 0] such that Rξ) = 0 and Rx) > 0 for x, ξ). Because v v is an increasing function for <v<, we have R x) = ε [ vx,ε,k0 ) v x,ε,k 0 ) v L x) + v L x)] > 0 65) for x, ξ). Furthermore, from 6), it follows that ξ s Rξ) = R ) + R )ξ + ) + R τ) dτ ds > 0, which contradicts the assumption that Rξ) = 0. Hence, we conclude that Rx) > 0 for all x, 0] and by 65), R x) > 0 for all x, 0], thus proving 64). A combination of the results in 6) and 64) shows that Rx) is an increasing function in the interval [, 0] and bounded by R0) = v0,ε,k 0 ) sech + ln + ) ). Because v0, ε, k 0 ) = a 0 as shown in the first paragraph of this proof, we obtain ) R0) exp ε on account of 60). Thus, for x 0, Similarly, with 0 Rx) R0) exp ε ). x v R x) := sech + ln + ) ), one can show that vx, ε, k 0 ) v R x) is a decreasing function in [0, ] and bounded by R0). This also follows from the symmetry of the functions vx, ε, k 0 ) and v R x), thus completing the proof of 5) when i =. b) We next consider the case i =. Let u be a solution of the initial-value problem ), which is positive in the interval, x ) and negative in x, ), where u x ) = 0. The existence of such a solution is established in Lemma d). As in the case of i =, we need to estimate the value of a 0 0, ), which is related to k 0 via 0) and k 0 satisfies vx,ε,k 0 ) =, v,ε,k 0 ) =, vx,ε,k 0 ) > for < x < x,

23 A Two-Point Boundary-Value Problem with Spurious Solutions 99 and vx,ε,k 0 ) < for x < x < ; see Figure and Equation 54), and recall that v = + u. By definition, T a 0 ) is the length of the interval, x ), which is equal to x +. By Lemma, T a 0 ) < T ) = k ε, where k Integrating the equation ε v = v v a0 + a 0 from v = + x ) to gives a 0 v v a 0 + a 0 ) x dv = ε ε k ε ) ; see the proof of Lemma. Using the same argument as in the previous case, one can show that ) 4 0 < a ek / exp ε for small values of ε, i.e., )) a 0 = O exp, ε ε ) To estimate the value of x, we return to 5) T a 0 ) = v dv v v a0 + a 0 and obtain T a 0 ) = { 0 t v + v / + t + t )} dt; see the proof of Lemma. By ), we have v = a 0 + O a 0), ε 0 +. Coupling the last two results yields T a 0 ) = { t + } / + t + t ) dt + O εa0). 0

24 400 C. H. Ou and R. Wong The integral on the right can be evaluated explicitly, and its value is ln + ). Hence T a 0 ) = α + O εa0), ε 0 +, 67) where α = ln + ), 68) from which it follows that x = + T a 0 ) = + α + O εa0). 69) Now note that u x) is entirely negative in the interval x, ). By repeating the argument in case a), one can show that u x) = + sech x x + ln + ) x + sech + ln + ) ) + Oe /ε ) for x [x, ]. Using 69), we further obtain u x) = + sech x + + ln + ) ) x + sech + ln + ) ) + Oe /ε ), 70) because sech is an even function. In the interval [, x ], vx, k 0 ) attains the maximal value v at x = + x ). Consider the function x vx) = sech + x ) ). 7) It is easily verified that this function also attains its maximal value at x = + x ), and satisfies ε v v + v = 0. 7) From 69), we have ) x + α vx) = sech + O a ) 0 = sech x + + ln + ) ) + Oe /ε ). 7) )

25 A Two-Point Boundary-Value Problem with Spurious Solutions 40 v _ x x y 0 Figure. Graph of vy). We claim that for x [, x ], vx, k 0 ) = vx) + Oe /ε ). 74) First, we restrict x to the interval [, + x )], and observe that vx, k 0 ). From ), it follows that the point x, vx, k 0 )) satisfies x = + x ε v vx,k 0 ) ds; 75) s v )s v )v s) see, also, 5). Now, for any x [, + x ) ], we let y be the point in the same interval such that vy) = vx, k 0 ); 76) see Figure. From 7), we obtain as in 75) y = + x ε vy) s ds. 77) s In 75) and 77), we make respectively the change of variables s = vx, k 0 ) cos θ + v sin θ and s = vy) cos θ + sin θ; see the proof of Lemma. Using the facts: v, v + v = Oa 0 ), and v v = Oa 0 ), we have from 75) and 77) y x = where ε π/ 0 sin θ vx, k 0 ) cos θ + sin θ V x, k 0) dθ + O εa0 ), 78) V x, k 0 ) = vx, k 0 ) v vx, k 0 ). 79)

26 40 C. H. Ou and R. Wong Differentiating both sides of 77) with respect to y gives d v dy = ε vy) vy). 80) From 7), it can be shown that v y) is bounded by /ε in the interval [, ]. Because v = a 0 + Oa 0 ) and V x, k 0) = Oa 0 ), expanding vx) at x = y yields on account of 76). Note that V x, k 0 ) vy) = vx) = vx, k 0 ) + v y)x y) + O a0 ) v vx, k0 ) + v vx, k 0 ) vx, k0 ). 8) Because 0 <v vx, k 0 ) < vx, k 0 ), the right-hand side of this equation is less than v. Hence, substituting 78) and 79) into 80), we obtain vx) = vx, k 0 ) + O a 0) = vx, k0 ) + Oe /ε ) 8) as ε 0. This proves our claim in 74) for x [, + x )]. Coupling 7) and 8), we get ux, k 0 ) = + sech x + + ln + ) ) + Oe /ε ), 8) as ε 0, for x [, + x ) ]. By a similar argument, one can prove that formula 8) also holds for x [ + x ] ), x. The desired result 5), when i =, follows from 70) and 8). Observe that there are three terms in the approximation in 70), whereas in 8) we have only two terms. The explanation for this difference is that the third term on the right-hand side of 70) is exponentially small in the interval [, x ]. c) The proof of this case is essentially the same as that of the case i =. d) Similar to the case i =, we consider the function T a) + T a). As a 0, this function approaches infinity; as a, it is O ε). By Lemma, we also know that this function is convex. Hence, there is a unique value a 0 0, ) such that Furthermore, by Lemma we have T a 0 ) + T a 0 ) =. T a 0 ) T ) = 9.76 ε. Using an argument similar to that in a), we can also deduce that a 0 = Oe /ε );

27 A Two-Point Boundary-Value Problem with Spurious Solutions 40 see 60). Because T a 0 ) = x + = x, we obtain x = + α + O εa0 ) 84) and x = α + O εa0) ; 85) cf. 69). By the result in a), one can write x u 4 = + sech x + ln + ) ) + sech x x + ln + ) ) + Oe /ε ) 86) for x [x, x ]. Inserting 84) and 85) into 86) gives u 4 = + sech x + + ln + ) ) + sech x + ln + ) ) + Oe /ε ). 87) For x [, x ], we have by the result in case b) u 4 = + sech x + + ln + ) ) + Oe /ε ). 88) Similarly, for x [x, ], we have from case c) u 4 = + sech x + ln + ) ) + Oe /ε ). 89) Note that the third and the second term in 87) are Oe /ε ) for x in [, x ] and [x, ], respectively. Hence, upon comparing 87) with 88) and 89), we conclude that 87) actually holds for all x [, ]. This completes the proof of the theorem. Remark : In the above proof, we have on several occasions made use of a result corresponding to 5) with i = when the interval [, ] is replaced by [a, b]. The precise approximation is given by x a u x,ε) = + sech + ln + ) ) b x + sech + ln + ) { ) + O exp b a )} uniformly for x [a, b]. 90)

28 404 C. H. Ou and R. Wong 5. Solutions with spikes We are now ready to establish the result stated in ), i.e., the case with n internal spikes. In fact, we have the following formula corresponding to 5). THEOREM. For each i =,...,4and n =,,..., there exists a unique solution u i,n x, ε) satisfying n ) [ x xi, j u i,n x,ε) = ũ i x,ε) + q + O exp )] 9) ε n + ε for all x [, ], where j= x, j = ln + ) + x, j = + x, j = + j n + [ + ln + )], 9) j n + + ln + ), 9) j n + ln + ), 94) x 4, j = + ln + ) + j n + [ ln + )] 95) for j =,..., n. Proof : We consider only the case i = ; all other cases can be handled in similar manners. As in the proof of Theorem, by using Lemma, it can be shown that there exists a unique value a 0 0, ) such that nt a 0 ) + n + )T a 0 ) =. 96) Because T a 0 ) < T ) and T ) 4.88 ε,wehave T a 0 ) 4.88n ε. n + From 6), it follows that dv 4.88n ε. a 0 v v + a 0 n + a 0 The argument given in the proof of the case i = in Theorem then yields { a 0 = O exp )}. 97) n + ε

29 A Two-Point Boundary-Value Problem with Spurious Solutions 405 On account of 67) and 96), we obtain respectively { ε T a 0 ) = α + O exp )} n + ε and T a 0 ) = nα n + 98) { ε + O exp )}, 99) n + ε where α = ln + ). Let u x, k 0 ) be the solution of the initial value problem ) with k 0 related to a 0 via 0) and a 0 satisfying 96). Then u x, k 0 ) is also the unique solution to the BVP 0) with n internal spikes, i.e., u x, k 0 ) = u,n x, ε). Let x i, ), i =,...,n, denote the zeros of u x, k 0 ). The position of x i can be represented by x i = + T a 0 ) + T a 0 ) + i )[T a 0 ) + T a 0 )], i =,..., n. From 98) and 99), we have x i = ln + ) + { + O exp n + )}. ε i n + [ + ln + )] With the notation x,i defined in 9), the last equation can be written as { x i = x,i + O exp )}, i =,...,n. 00) n + ε Let β i and γ i, i =,..., n, denote the zeros of u,n x, ε), which are closest to x i so that x i β i,γ i ) and u,n x, ε) > 0 for x β i, γ i ); see Figure 4. Note that the length of each of the intervals β i, x i ) and x i,γ i )is T a 0 ). Hence, by virtue of 00), { β i = x,i α + O exp )} n + ε and { γ i = x,i + α + O exp )}. n + ε Using the argument given in the proof of case b) in Theorem, one can show that for x β i, γ i )

30 406 C. H. Ou and R. Wong β i _ x i γ i Figure 4. Zeros β i and γ i. ) { x u,n x,ε) = + sech xi + O exp ) { x = + sech x,i + O exp )} n + ε n + ε )} ; 0) see 7), 8), and 97). Similarly, by using the argument given in the proof of case a) in Theorem, we obtain for x γ i, β i+ ) x u,n x,ε) = + sech γi + ln + ) ) which in turn gives + sech βi+ x + ln + ) { ) + O exp )}, n + ε ) ) x u,n x,ε) = + sech x,i x + sech x,i+ { + O exp )} ; 0) n + ε see the remark following the proof of Theorem. Furthermore, we have x + u,n x,ε) = + sech + ln + ) ) ) { x + sech x, + O exp )} n + ε 0)

31 A Two-Point Boundary-Value Problem with Spurious Solutions 407 for x, β ), and x u,n x,ε) = + sech + ln + ) ) ) { x + sech x,n + O exp )} n + ε for x γ n, ). Let 04) and note that sech x xi ) = O δ := [T a 0 ) + T a 0 )], { exp n + for x / x i δ, x i + δ). Analogously, we have + x sech + ln + ) ) = O )}, i =,...,n, ε { exp n + )} ε for x + T a 0 ) and x sech + ln + ) { ) = O exp )} n + ε for x T a). The final result given in 9) with i = now follows from a combination of 0) 04). References. C. M. BENDER and S. A. ORSZAG, Advanced Mathematical Methods for Scientists and Engineers, McGraw-Hill, New York, J. KEVORKIAN and J. D. COLE, Perturbation Methods in Applied Mathematics, Springer Verlag, New York, 98.. P. A. LAGERSTROM, Matched Asymptotic Expansions, Springer Verlag, New York, J. D. LOGAN, Applied Mathematics A Contemporary Approach, John Wiley & Sons, New York, J. A. MURDOCH, Perturbations Theory and Methods, John Wiley & Sons, New York, A. H. NAYFEH, Introduction to Perturbation Techniques, John Wiley & Sons, New York, R. E. O MALLEY, Singular Perturbation Methods for Ordinary Differential Equations, Springer Verlag, New York, J. G. SIMMONDS and J. E. MANN, A First Look at Perturbation Theory, Robert E. Krieger Publishing Co., Malabar, FL, 986.

32 408 C. H. Ou and R. Wong 9. F. W. J. OLVER, Asymptotics and Special Functions, Academic Press, 974. Reprinted by A. K. Peters, Ltd., Wellesley, 997.) 0. D. R. SMITH, Singular Perturbation Theory, Cambridge University Press, Cambridge, R. WONG and H. YANG, On a boundary layer problem, Stud. Appl. Math. 08: ).. G. F. CARRIER and C. E. PEARSON, Ordinary Differential Equations, Blaisdell Publishing Co., Waltham, MA, 968. Reprinted in SIAM s Classics in Applied Mathematics series, vol. 6, SIAM, Philadelphia, 99.). C. G. LANGE, On spurious solutions of singular perturbation problems, Stud. Appl. Math. 68: ). 4. A. D. MACGILLIVRAY, A method for incorporating transcendentally small terms into the method of matched asymptotic expansions, Stud. Appl. Math. 99: ). 5. B. C. CARLSON and J. L. GUSTAFSON, Asymptotic expansion of the first elliptic integral, SIAM J. Math. Anal. 6: ). CITY UNIVERSITY OF HONG KONG Received September 6, 00)