Commun Nonlinear Sci Numer Simulat 19 (2014) 2632–2643

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Commun Nonlinear Sci Numer Simulat

journal homepage: www.elsevier .com/locate /cnsns

Canard solution and its asymptotic approximationin a second-order nonlinear singularly perturbed boundaryvalue problem with a turning point

1007-5704/$ - see front matter � 2014 Elsevier B.V. All rights reserved.http://dx.doi.org/10.1016/j.cnsns.2013.12.033

⇑ Corresponding author. Tel.: +86 189 3094 0369.E-mail address: mahan@shnu.edu.cn (M. Han).

Jianhe Shen a, Maoan Han b,⇑a School of Mathematics and Computer Science, Fujian Normal University, 350007 Fuzhou, PR Chinab Department of Mathematics, Shanghai Normal University, 200234 Shanghai, PR China

a r t i c l e i n f o a b s t r a c t

Article history:Received 12 March 2013Received in revised form 26 November 2013Accepted 30 December 2013Available online 10 January 2014

Keywords:Singular perturbationNonlinear differential equationsTurning pointCanard solutionAsymptotic approximation

This paper considers the existence and uniformly valid asymptotic approximation ofcanard solutions in a second-order nonlinear singularly perturbed boundary value problemwith a turning point. We get the main results by constructing the asymptotic solution firstand then defining a couple of upper and lower solutions suitably on the basis of the asymp-totic solution. Two examples are carried out to illustrate and verify the theoretical results.

� 2014 Elsevier B.V. All rights reserved.

1. Introduction

Singularly perturbed boundary value problems for ordinary differential equations with turning point(s) may displayspecial dynamical behaviors like delayed loss of stability, which leads to the appearance of canard solutions. On the studyof this issue, one can trace back to Ackerberg and O’Malley’s work [1], where singularly perturbed linear two-point boundaryvalue problem was considered,

�y00 þ aðxÞy0 þ bðxÞy ¼ 0; x 2 ð�1;1Þ;yð�1Þ ¼ A; yð1Þ ¼ B;

�

in which, � is a small parameter, A and B are constants, aðxÞ and bðxÞ are sufficiently smooth functions on ½�1;1�. Ackerbergand O’Malley [1] detected a new phenomenon, called boundary layer resonance. Since the work of Ackerberg and O’Malley,depending on the properties of aðxÞ and bðxÞ near the turning point, many limiting behaviors of yðx; �Þ with �! 0 in theabove singularly perturbed problem were revealed, see Wong and Yang [2–4] for example. Fruchard and Schaefke [5,6]pointed out that, in fact, boundary layer resonance is equivalent to the appearance of canard solutions. On the advance ofthe study on this linear singularly perturbed boundary value problem, one can refer to O’Malley [7].

Recently, with the discoveries of applications of canards in some real world problems [8,9], more and more attention hasbeen paid to the study of canards, and several methods including non-standard analysis [10], asymptotic analysis[2–4,11,12], geometric method [13,14] and upper and lower solutions method [15] etc., had been developed. To revealthe feature of canards, O’Malley [16] provided a simple example as follows,

J. Shen, M. Han / Commun Nonlinear Sci Numer Simulat 19 (2014) 2632–2643 2633

�y0 ¼ ðx� 1Þy; yð0Þ ¼ 1;

where x ¼ 1 is the turning point. We plot the solution of this problem in Fig. 1 with � ¼ 0:05 and x 2 ½0;2:01�. Fig. 1 showsthat, near x ¼ 0, the flow first exhibits a fast motion (i.e., a boundary layer), and then, with �! 0, it stays near y ¼ 0; afterx ¼ 1, the flow stays still near y ¼ 0 on fxj1 < x < 2g despite of the repelling nature until x ¼ 2 where there is a blowup. Theflow displays delayed loss of stability. A solution with this feature is named canard [16].

But nevertheless, just as pointed out by O’Malley [7], whether canards can occur in nonlinear differential equations re-mains largely unexplored. This motivates us to consider the following second-order nonlinear singularly perturbed problem:

�y00 � yy0 þ f ðxÞy ¼ �gðxÞ; a < x < b; ð1Þ

yðaÞ � �py0ðaÞ ¼ A; ð2Þ

yðbÞ þ �py0ðbÞ ¼ B; ð3Þ

in which, 0 < �� 1; p > 0 is a constant, a; b;A and B are constants with the following symmetric conditions:

a < 0 < b; aþ b ¼ 0 and Aþ B ¼ 0; ð4Þ

f and g are sufficiently smooth functions on ½a; b�.Let � ¼ 0 in (1), one gets the reduced equation

yy0 ¼ f ðxÞy ð5Þ

admitting a trivial reduced solution

upðxÞ � 0 ð6Þ

and a family of non-trivial reduced solutions

uðxÞ ¼ FðxÞ þ C; ð7Þ

where C is a constant of integration and

FðxÞ ¼Z x

0f ðuÞdu: ð8Þ

Let y ¼ y1; y0 ¼ y01 ¼ y2, we can rewrite (1) in the form of ordinary differential systems of first-order,

y01 ¼ y2;

�y02 ¼ y1y2 � f ðxÞy1 þ �gðxÞ:

�

Thus, by geometric singular perturbation theory [17], one knows that the trivial reduced solution (6) is non-hyperbolic. Sowe do not expect that there are solutions of (1)–(3) to be attracted by (6). On the other hand, the non-trivial reduced solu-tions (7) are normally attracting and repelling respectively when FðxÞ þ C < 0 and FðxÞ þ C > 0. In this paper, we will provethat, with �! 0, there are solutions of (1)–(3), after x ¼ a, they stay near one of this family of non-trivial reduced solutions(7) (C in (7) will be determined later) not only on the attracting part fx jFðxÞ þ C < 0g but also on the repelling onefx jFðxÞ þ C > 0g.

Fig. 1. Canard in O’Malley’s example.

2634 J. Shen, M. Han / Commun Nonlinear Sci Numer Simulat 19 (2014) 2632–2643

The paper is arranged as follows. In Section 2, we construct the asymptotic solution of (1)–(3) through the method ofmatched asymptotic expansions. On the technique of matched asymptotic expansions and its modern developments, onecan refer to Fruchard and Schäfke [18]. In Section 3, on the basis of the asymptotic solution, we define a couple of upperand lower solutions suitably and prove the existence of canards and give the asymptotic approximation to the canard solu-tion. We also derive the error estimate between the asymptotic approximation and the canard. In Section 4, two examplesare carried out to illustrate the main results.

At the end of this section, we want to remark that:

Remark 1. By following the linearizing viewpoint of Howes [19], the turning points of nonlinear ordinary differential Eq. (1)is defined by fxðCÞ ¼ jFðxÞ þ C ¼ 0g. This way of definition is identical with which derived from geometric singularperturbation theory [17].

Remark 2. Although Howes [19] had considered the following more general equation with turning points,

�y00 ¼ Fðx; y; y0; �Þ;

Howes’s results can not cover the results of the present paper. On the one hand, Howes’s method is unable to provide the

uniformly valid asymptotic solution. On the other hand, since the upper and lower solutions constructed by Howes [19] isnot accurate enough (compare the upper and lower solutions defined in Howes [19] and in the present paper), thus, from thequantitative point of view, the solutions estimated by Howes’s method is relatively rough, especially on the boundaries ofthe interval where the layers occur.

2. The asymptotic solution

In this section, we construct the asymptotic solution of (1)–(3), which plays an important role in defining the desiredupper and lower solutions in the next section.

Let us introduce s1 ¼ ðx� aÞ=�, by which, (1) and (2) becomes respectively

d2yds2

1

� ydyds1þ �f ð�s1 þ aÞy ¼ �2gð�s1 þ aÞ ð9Þ

and

yð0Þ � pdyds1ð0Þ ¼ A: ð10Þ

Making �! 0 in (9) yields

d2yds2

1

� ydyds1¼ 0: ð11Þ

The solutions of (11) satisfying (10), denoted by VLðs1Þ, is called the zeroth-order approximation to the solutions of (9) and(10).

Let dVL=ds1 ¼ P, then d2VL=ds21 ¼ dP=ds1 ¼ PdP=dVL, so (11) reduces

PdP

dVL� PVL ¼ 0

admitting a particular solution P � 0 (discarded) and

dPdVL¼ VL:

Thus we have

P ¼ V2L

2þ K1;

i.e.,

dVL

ds1¼ V2

L

2þ K1; ð12Þ

where K1 is a nonzero constant of integration.Let K1 ¼ �a2

1=2 ða1 – 0Þ, then it can be solved from (12) that

VL � a1

VL þ a1¼ �M1ea1s1 ða1 – 0Þ; ð13Þ

J. Shen, M. Han / Commun Nonlinear Sci Numer Simulat 19 (2014) 2632–2643 2635

in which, M1 is a positive constant. There are two cases to be discussed.Case I: jVLj > ja1j.In this case, we have by direct calculations that

VLðs1Þ ¼ a11þM1ea1s1

1�M1ea1s1¼ a1

1þ ea1ðs1þd1Þ

1� ea1ðs1þd1Þ¼ �a1 coth

a1

2ðs1 þ d1Þ

h i

and

dVLðs1Þds1

¼ 2a21csch2 a1

2ðs1 þ d1Þ

h i> 0; ð14Þ

where d1 is a constant determined through

M1 ¼ ea1d1 : ð15Þ

It can be seen from (14) that VLðs1Þ is monotone increasing in s1, i.e., in x.Case II: jVLj < ja1j.In this case, it turns out that

VLðs1Þ ¼ a11�M1ea1s1

1þM1ea1s1¼ �a1 tanh

a1

2ðs1 þ d1Þ

h i

and

dVLðs1Þds1

¼ �2a21

M1ea1s1

ð1þM1ea1s1 Þ2¼ �2a2

1sech2 a1

2ðs1 þ d1Þ

h i< 0; ð16Þ

where d1 is also defined by (15). In the present case, VLðs1Þ is monotone decreasing.By the matching principle [20], it must be that

VLðþ1Þ ¼ uðaÞ:

Consequently, for case I, if a1 < 0, since

VLðþ1Þ ¼ lims1!þ1

� a1 cotha1

2ðs1 þ d1Þ

h i¼ a1;

then

a1 ¼ uðaÞ: ð17Þ

If a1 > 0, since

VLðþ1Þ ¼ lims1!þ1

� a1 cotha1

2ðs1 þ d1Þ

h i¼ �a1;

then

�a1 ¼ uðaÞ: ð18Þ

Both (17) and (18) imply that uðaÞ must be negative. Otherwise, the matching can not be carried out.Similarly, for case II, if a1 < 0, then

VLðþ1Þ ¼ lims1!þ1

� a1 tanha1

2ðs1 þ d1Þ

h i¼ a1;

which yields Eq. (17), and if a1 > 0, then

VLðþ1Þ ¼ lims1!þ1

� a1 tanha1

2ðs1 þ d1Þ

h i¼ �a1;

which gives Eq. (18).Therefore, if

uðaÞ ¼ FðaÞ þ C < 0 ð19Þ

for both cases I and II, the a1 < 0 and a1 > 0 cases are both possible for matching. Without loss of generality, we consider thea1 < 0 case for definiteness. Hence we have a1 ¼ uðaÞ ¼ FðaÞ þ C. Accordingly, we get the hyperbolic coth and tanh functions

VLðs1Þ ¼ uðaÞ1þM1euðaÞs1

1�M1euðaÞs1¼ �uðaÞ coth

uðaÞ2ðs1 þ d1Þ

� �ð20Þ

and

VLðs1Þ ¼ uðaÞ1�M1euðaÞs1

1þM1euðaÞs1¼ �uðaÞ tanh

uðaÞ2ðs1 þ d1Þ

� �: ð21Þ

2636 J. Shen, M. Han / Commun Nonlinear Sci Numer Simulat 19 (2014) 2632–2643

Taking (10) into account, it can be deduced from (20) and (21) that

A ¼ uðaÞ1þM1

1�M1� 2pu2ðaÞ M1

ð1�M1Þ2ð22Þ

and

A ¼ uðaÞ1�M1

1þM1þ 2pu2ðaÞ M1

ð1þM1Þ2: ð23Þ

By (22) or (23), M1, i.e., equivalently, d1 in (15) can be determined. Till now, VLðs1Þ in (20) and (21) are determinedcompletely.

Similarly, by another fast time scale s2 ¼ ðx� bÞ=� 2 ð�1;0�, we transform (1) and (3) respectively into

d2yds2

2

� ydyds2þ �f ð�s2 þ aÞy ¼ �2gð�s2 þ aÞ;

which turns out to be Eq. (10) as �! 0, and

yð0Þ þ pdyds2ð0Þ ¼ B: ð24Þ

The solutions of (10) and (24), denoted by VRðs2Þ, can be viewed as the zeroth-order approximation to the right boundarylayer.

Like the deductions of VLðs1Þ in (20) and (21), two functions possible to be the right boundary layer are obtained, namely,

VRðs2Þ ¼ uðbÞ1þM2euðbÞs2

1�M2euðbÞs2¼ �uðbÞ coth

uðbÞ2ðs2 þ d2Þ

� �ð25Þ

and

VRðs2Þ ¼ uðbÞ1�M2euðbÞs2

1þM2euðbÞs2¼ �uðbÞ tanh

uðbÞ2ðs2 þ d2Þ

� �; ð26Þ

where d2 is determined by

M2 ¼ euðbÞd2 :

In the deductions, we have used the following matching principle

VRð�1Þ ¼ uðbÞ;

in which, it is required that

uðbÞ ¼ FðbÞ þ C > 0: ð27Þ

In the same way, by taking (24) into account, we have

B ¼ uðbÞ1þM2

1�M2þ 2pu2ðbÞ M2

ð1�M2Þ2ð28Þ

and

B ¼ uðbÞ1�M2

1þM2� 2pu2ðbÞ M2

ð1þM2Þ2: ð29Þ

By (28) or (29), M2, i.e., d2 is determined. So far VRðs2Þ defined in (25) and (26) are fully known.Now we have (20), (21) and (25), (26), respectively possible to be the left and right boundary layers. Which pair of the

hyperbolic functions in (20), (21) and (25), (26) are chosen depending on the practical situations like the boundary condi-tions. In the next section, we will show that the hyperbolic coth functions defined in (20) and (25) should be selected. Con-sequently, M1 and M2 are determined through (22) and (28).

Now we have obtained the formal asymptotic solution

yasyðx; �Þ ¼ uðxÞ þ VLðs1Þ þ VRðs2Þ � uðaÞ � uðbÞ; ð30Þ

where s1 ¼ ðx� aÞ=�; s2 ¼ ðx� bÞ=�;uðxÞ is a non-trivial reduced solution contained in (7), in which, C will be determinedlater, VLðs1Þ and VRðs2Þ are defined in (20) and (25) respectively.

3. The main results

Theorem 1. If nonlinear singularly perturbed problem (1)–(3) with (4) satisfies

J. Shen, M. Han / Commun Nonlinear Sci Numer Simulat 19 (2014) 2632–2643 2637

(a) gð0Þ ¼ 0 ¼ F 00ð0Þ; FðaÞ þ FðbÞ ¼ 0 and

FðxÞ < 0; x 2 ½a;0Þ; FðxÞ > 0; x 2 ð0; b�;

(b) the boundary values A and B are chosen such that M1 and M2 determined respectively by (22) and (28) satisfy

1� ðM2 þM1Þ þM2M1 � pFðaÞ þ pFðaÞM2M1 – 0;

(c) when x 2 ða;0Þ [ ð0; bÞ,

�4cx2ð3FðxÞ þ xf ðxÞÞ þ f 00ðxÞ � g0ðxÞ– 0;

where c > 0 is a constant satisfying

c > max �pf ðaÞa4 ;�pf ðbÞ

b4

� �;

then (1)–(3) has at least one solution yðx; �Þ 2 C2½a; b� with

jyðx; �Þ � yasyðx; �Þj 6 c�; x 2 ½a; b�; 0 < �� 1; ð31Þ

where yasyðx; �Þ is given in (30), in which,

uðxÞ ¼ FðxÞ;

i.e., C ¼ 0 in (7), and c > 0 is a constant selected with c P b4cða4cÞ.The proof of Theorem 1 relies on the method of upper and lower solutions. Thus, the following lemma is needed.

Lemma 1 [21]. Consider

y00 ¼ f ðx; y; y0Þ; x 2 ða; bÞ;yðaÞ � py0ðaÞ ¼ A; yðbÞ þ qy0ðbÞ ¼ B;

�ð32Þ

where p; q; a; b;A and B are constants with p > 0; q > 0. If the following conditions hold,

(1) there are functions bðxÞ;aðxÞ 2 C2½a; b� with aðxÞ 6 bðxÞ satisfying

b00 6 f ðx; b;b0Þ; a00 P f ðx;a;a0Þ; x 2 ½a; b�

and

aðaÞ � pa0ðaÞ 6 A 6 bðaÞ � pb0ðaÞ; aðbÞ þ qa0ðbÞ 6 B 6 bðbÞ þ qb0ðbÞ;

(2) f ðx; y; y0Þ satisfies the Nagumo condition with respect to bðtÞ and aðtÞ, i.e., there is a positive continuous function uðsÞ on½0;þ1Þ such that f ðx; y; y0Þ 6 uðjy0jÞ for all a 6 x 6 b;aðxÞ 6 y 6 bðxÞ; jy0j 6 þ1 and

Z þ1

k

sdsuðsÞ > max

x2½a;b�bðxÞ � min

x2½a;b�aðxÞ;

where kðb� aÞ ¼maxfjaðaÞ � bðbÞj; jaðbÞ � bðaÞjg; then (32) has at least one solution yðxÞ 2 C2½a; b� with

aðxÞ 6 yðxÞ 6 bðxÞ:

Remark 3. In fact, the Nagumo condition guarantees that if the right-hand sided function f ðx; y; y0Þ in (32) increases in y0

quadratically at most, then the derivative of solutions, i.e., y0 is bounded provided aðxÞ 6 y 6 bðxÞ. Under the Nagumo con-dition, one can get the existence of solutions as well as the estimate of solutions for (32).

Proof of Theorem 1. Based on the asymptotic solution (30), we introduce an error function term and define a couple ofupper and lower solutions as follows,

bðx; �Þ ¼ uðxÞ þ VLðs1Þ þ VRðs2Þ � uðaÞ � uðbÞ þ x4c� ð33Þ

and

aðx; �Þ ¼ uðxÞ þ VLðs1Þ þ VRðs2Þ � uðaÞ � uðbÞ � x4c�; ð34Þ

in which, c is a positive constant to be determined.Obviously, aðx; �Þ 6 bðx; �Þ for x 2 ½a; b�.According to the condition (2) in Lemma 1, i.e., the definition of the Nagumo condition, we know that Eq. (1) satisfies the

Nagumo condition. So to prove Theorem 1, it is left to verify that the upper and lower solutions (33) and (34) satisfies thecondition (1) in Lemma 1. For this purpose, let us first claim

2638 J. Shen, M. Han / Commun Nonlinear Sci Numer Simulat 19 (2014) 2632–2643

�b00 � bb0 þ f ðxÞb 6 �gðxÞ: ð35Þ

By recalling the governing equations of VLðs1Þ;VRðs2Þ and uðxÞ, we have the following equalities by direct calculations,

�b00 � bb0 þ f ðxÞb� �gðxÞ ¼ � u00ðxÞ þ 1�2

d2VL

ds21

þ 1�2

d2VR

ds22

þ 12x2c�

!� ðuðxÞ þ VL þ VR � uðaÞ � uðbÞ þ x4c�Þ

� u0ðxÞ þ 1�

dVL

ds1þ 1�

dVR

ds2þ 4cx3�

� �þ f ðxÞðuðxÞ þ VL þ VR � uðaÞ � uðbÞ þ x4c�Þ � �gðxÞ

¼ �1�ðuðxÞ þ VR � uðaÞ � uðbÞ þ x4c�ÞdVL

ds1� 1�ðuðxÞ þ VL � uðaÞ � uðbÞ þ x4c�ÞdVR

ds2

þ � �4cx3ðuðxÞ þ VL þ VR � uðaÞ � uðbÞ þ x4c�Þ þ u00ðxÞ � gðxÞ þ 12cx2��

: ð36Þ

Denote

H1ðx; �Þ ¼ �1�ðuðxÞ þ VR � uðaÞ � uðbÞ þ x4c�ÞdVL

ds1;

H2ðx; �Þ ¼ �1�ðuðxÞ þ VL � uðaÞ � uðbÞ þ x4c�ÞdVR

ds2;

H3ðx; �Þ ¼ �½�4cx3ðuðxÞ þ VL þ VR � uðaÞ � uðbÞ þ x4c�Þ þ u00ðxÞ � gðxÞ þ 12cx2��

and

Hðx; �Þ ¼ H1ðx; �Þ þ H2ðx; �Þ þ H3ðx; �Þ;

then to prove the inequality (35), it is equivalently to prove Hðx; �Þ � 0.Firstly, we have from (20) that

VLðs1Þjx¼a ¼ uðaÞ1þM1euðaÞx�a�

1�M1euðaÞx�a�

x¼a

¼ uðaÞ1þM1

1�M1; ð37Þ

which is a constant and

dVL

ds1ðs1Þ

x¼a

¼ 2u2ðaÞM1

1�M1euðaÞx�a�

� �2 euðaÞx�a�

x¼a

¼ 2u2ðaÞM1

ð1�M1Þ2; ð38Þ

which is a positive constant. Similarly, it can be derived from (25) that, when x is near x ¼ a and � is sufficiently small,

VRðxÞ ¼ uðbÞ þ uðbÞ 2M2euðbÞx�b�

1�M2euðbÞx�b�¼ uðbÞ þ O euðbÞx�b

�

�; ð39Þ

in which and in the following of the paper, OðeuðbÞx�b� Þ always denotes an exponentially small quantity, and

dVR

ds2ðxÞ ¼ 2u2ðbÞM2

1�M2euðbÞx�b�

�2 euðbÞx�b� ¼ O euðbÞx�b

�

�: ð40Þ

The substitution of (37)–(40) into (36) gives

Hða; �Þ ¼ H1ða; �Þ þ Oð�Þ;

where O stands for a Landau order symbol, and

H1ða; �Þ ¼ �a4c2u2ðaÞM1

ð1�M1Þ2þ O euðbÞa�b

�

�< 0:

Thus, when � is sufficiently small, we have by continuity that Hða; �Þ < 0. Accordingly, there is a constant d > 0 sufficientlysmall and independent of � such that

Hðx; �Þ < 0; x 2 ½a; dÞ:

So the inequality (35) holds on ½a; aþ dÞ.Similarly, we have H2ðb; �Þ < 0, which means Hðb; �Þ < 0 for � sufficiently small. So there is a constant d > 0 sufficiently

small and independent of � such that (35) holds when x 2 ðb� d; b� and � sufficiently small.Now we consider (35) for x 2 ½aþ d; b� d�. Let us consider Hð0; �Þ firstly. By recalling the following conditions:

Fð0Þ ¼ F 00ð0Þ ¼ gð0Þ ¼ 0;uðaÞ ¼ FðaÞ þ C, uðbÞ ¼ FðbÞ þ C and FðaÞ þ FðbÞ ¼ 0, one gets

J. Shen, M. Han / Commun Nonlinear Sci Numer Simulat 19 (2014) 2632–2643 2639

Hð0; �Þ ¼ �1�

VR�b�

� �� C

� �dVL

ds1

�a�

�� 1�

VL�a�

�� C

� dVR

ds2

�b�

� �: ð41Þ

At this position, instead of continuing to study (41), let us turn to consider another differential inequality, namely,

�a00 � aa0 þ f ðxÞa P �gðxÞ; ð42Þ

in which, aðx; �Þ is defined in (34). It can be deduced similarly that

�a00 � aa0 þ f ðxÞa� �gðxÞ ¼ �1�ðuðxÞ þ VR � uðaÞ � uðbÞ � x4c�ÞdVL

ds1� 1�ðuðxÞ þ VL � uðaÞ � uðbÞ � x4c�ÞdVR

ds2

þ �½4cx3ðuðxÞ þ VL þ VR � uðaÞ � uðbÞ � x4c�Þ þ u00ðxÞ � gðxÞ � 12cx2��

¼ fH1ðx; �Þ þ fH2ðx; �Þ þ fH3ðx; �Þ ¼ eHðx; �Þ;

where fHiðx; �Þ; i ¼ 1;2;3 and eHðx; �Þ are defined like Hiðx; �Þ; i ¼ 1;2;3 and Hðx; �Þ respectively. Now let us consider eHð0; �Þ. Itcan be found easily that

Hð0; �Þ ¼ eHð0; �Þ:

Thus, to verify the inequalities (35) and (42) simultaneously, it must be that

Hð0; �Þ ¼ eHð0; �Þ ¼ 0:

Now the substitution of VL;VR and their derivatives into (41) yields

��Hð0; �Þ ¼ FðbÞ þ ðFðbÞ þ CÞ 2M2e�buðbÞ�

1�M2e�uðbÞb�

" #2u2ðaÞM1

1�M1e�uðaÞa�

�2 e�uðaÞa�

þ FðaÞ þ ðFðaÞ þ CÞ 2M1e�uðaÞa�

1�M1e�uðaÞa�

" #2u2ðaÞM2

1�M2e�uðbÞb�

�2 e�uðbÞb� :

By multiplying both sides of the above equality by the eaFðaÞ� term and recalling the following conditions:

a ¼ �b; FðaÞ ¼ �FðbÞ;uðaÞ ¼ FðaÞ þ C and uðbÞ ¼ FðbÞ þ C, one gets

��eaFðaÞ� Hð0; �Þ ¼ FðbÞ þ ðFðbÞ þ CÞ 2M2e

�buðbÞ�

1�M2e�uðbÞb�

" #2u2ðaÞM1

ð1�M1e�uðaÞa� Þ

2 e�aC�

þ FðaÞ þ ðFðaÞ þ CÞ 2M1e�uðaÞa�

1�M1e�uðaÞa�

" #2u2ðaÞM2

1�M2e�uðbÞb�

�2 eaC� :

At this position, Hð0; �Þ ¼ 0 means that the quantity on right-hand sided of the above equation vanishes, which is impossibleif C – 0 (Note: the first and second lines in the above equation are respectively exponentially small and large, or vice versa,for � sufficiently small). Thus, to verify (35) and (42) simultaneously, we must set

C ¼ 0: ð43Þ

By this choice, the non-trivial reduced solution turns out to be

uðxÞ ¼ FðxÞ:

Thus, by the symmetry conditions Aþ B ¼ 0 and FðaÞ þ FðbÞ ¼ 0, one can derive from (22) and (28) that

FðaÞ1þM1

1�M1� 2pF2ðaÞ M1

ð1�M1Þ2¼ FðaÞ1þM2

1�M2� 2pF2ðaÞ M2

ð1�M2Þ2:

Accordingly, under the assumption (b) in Theorem 1, we get by direct calculations that

M1 ¼ M2: ð44Þ

Thus, under (43) and (44) as well as the symmetry conditions (4), Hð0; �Þ ¼ eHð0; �Þ ¼ 0 holds.So far the upper and lower solutions become

bðx; �Þ ¼ uðxÞ þ VLðs1Þ þ VRðs2Þ þ x4c� ð45Þ

and

aðx; �Þ ¼ uðxÞ þ VLðs1Þ þ VRðs2Þ � x4c�: ð46Þ

2640 J. Shen, M. Han / Commun Nonlinear Sci Numer Simulat 19 (2014) 2632–2643

Now it is a good position to continue to prove (35) and (42) on the rest of interval. Since Hðx; �Þ < 0 for x 2 ½a; aþ dÞ andHð0; �Þ ¼ 0, thus, if (35) does not hold uniformly on ½aþ d;0Þ, then there must be some x0 2 ½aþ d;0Þ such that Hðx0; �Þ > 0.Hence, Hðx; �Þ has its positive maximum, say at x 2 ðaþ d;0Þ. Thus,

Hðx; �Þ > 0; H0xðx; �Þ ¼ 0 and H00xðx; �Þ � 0: ð47Þ

On the other hand, direct calculations give

H0xðx; �Þ ¼ � � 1�2 ðuðxÞ þ VR þ x4c�ÞdVL

ds1� 1�2 ðuðxÞ þ VL þ x4c�ÞdVR

ds2

� �x

�þ �4cx3ðuðxÞ þ VL þ VR þ x4c�Þ þ u00ðxÞ � gðxÞ þ 12x2��

x

�¼ �½�4cx2ð3FðxÞ þ xf ðxÞÞ þ f 00ðxÞ � g0ðxÞ þ Oð�Þ�;

where the subscript x means the derivative with respect to x. So if f ðxÞ and gðxÞ satisfy

�4cx2ð3FðxÞ þ xf ðxÞÞ þ f 00ðxÞ � g0ðxÞ– 0; x 2 ða;0Þ; ð48Þ

then

�4cx2ð3FðxÞ þ xf ðxÞÞ þ f 00ðxÞ � g0ðxÞ – 0;

which further means that, when � is sufficiently small,

H0xðx; �Þ – 0:

This is a contradiction. Hence, (35) holds uniformly for x 2 ½aþ d;0Þ.Similarly, since Hð0; �Þ ¼ 0 and Hðb; �Þ < 0, by the same idea and similar derivations, one can prove that when � is

sufficiently small, (35) holds on ð0; b� d� under the following condition:

�4cx2ð3FðxÞ þ xf ðxÞÞ þ f 00ðxÞ � g0ðxÞ– 0; x 2 ð0; bÞ: ð49Þ

By recalling eHð0; �Þ ¼ 0, eHða; �Þ > 0 and eHðb; �Þ > 0 and using the same idea in the above, one can prove (42) onða;0Þ [ ð0; bÞ under the following condition:

4cx2ð3FðxÞ þ xf ðxÞÞ þ f 00ðxÞ � g0ðxÞ – 0; x 2 ða;0Þ [ ð0; bÞ: ð50Þ

We now turn to prove the inequalities on the boundaries. Firstly, for � sufficiently small, it can be deduced that

bða; �Þ � �pb0ða; �Þ ¼ uðaÞ þ VLð0Þ þ VRa� b�

� �þ a4c�� �p u0ðaÞ þ 1

�dVL

ds1ð0Þ þ 1

�dVR

ds2

a� b�

� �þ 4a3c�

� �¼ FðaÞ þ FðbÞ þ O e�FðbÞa�b

�

�þ a4c�� �pf ðaÞ þ VLð0Þ � p

dVL

ds1ð0Þ � p

dVR

ds2

a� b�

� �� 4a3pc�2

¼ Aþ a4c�� �pf ðaÞ þ O e�FðbÞða�bÞ

�

�� p

dVR

ds2

a� b�

� �� 2apc�2 ¼ Aþ a4c�� �pf ðaÞ þ Oð�2Þ > A;

if c is chosen such that

c >pf ðaÞ

a4 :

We have similarly that

aða; �Þ � �pa0ða; �Þ < A;

if

c > � pf ðaÞa4 :

Thus, the following inequality

aða; �Þ � �pa0ða; �Þ < A < bða; �Þ � �pb0ða; �Þ

holds under

c > max �pf ðaÞa4

� �: ð51Þ

Similarly, one gets

aðb; �Þ þ �qa0ðb; �Þ 6 B 6 bðb; �Þ þ �qb0ðb; �Þ;

J. Shen, M. Han / Commun Nonlinear Sci Numer Simulat 19 (2014) 2632–2643 2641

under

c > max �pf ðbÞb4

� �: ð52Þ

Till now, by Lemma 1, the proof of Theorem 1 is completed.On the basis of the behaviors of aðx; �Þ and bðx; �Þ, we can know the dynamical behaviors of the solutions of (1)–(3). That

is, we have the following corollaries.

Corollary 1. Under the conditions (a)–(c) listed in Theorem 1, there exists at least one solution of (1)–(3) such that

lim�!0

yðx; �Þ ¼ FðxÞ; x 2 ða; bÞ; ð53Þ

in which, FðxÞ is defined in (8).

Corollary 2. Under the conditions (a)–(c) listed in Theorem 1, nonlinear singularly perturbed boundary value problem (1)–(3) hasat least one canard solution, whose explicit and uniformly valid zeroth-order approximation is given by (30).

Remark 4. Under the symmetric conditions (4), the canard solutions appearing in (1)–(3) display the feature of symmetry.

4. Two illustrating examples

Example 1. Consider

�y00 � yy0 þ y ¼ 0; �1 < x < 1; 0 < �� 1yð�1Þ � �y0ð�1Þ ¼ �3;yð1Þ þ �y0ð1Þ ¼ 3:

8><>: ð54Þ

Here p ¼ 1; a ¼ �1; b ¼ 1;A ¼ �3;B ¼ 3; gðxÞ ¼ 0; f ðxÞ ¼ 1 and FðxÞ ¼ x. In this example, x ¼ 0 is the turning point.According to Section 2, we have

uðxÞ ¼ x;

VLðs1Þ ¼ �1þM1e�s1

1�M1e�s1¼ �1þM1e�

xþ1�

1�M1e�xþ1�;

VRðs2Þ ¼1þM2es2

1�M2es2¼ 1þM2e

x�1�

1�M2ex�1�:

By (22), M1 ¼ 1�ffiffiffi2p

=2 is determined through

�3 ¼ �1þM1

1�M1� 2

M1

ð1�M1Þ2; ð55Þ

where another root M1 ¼ 1þffiffiffi2p

=2 should be discarded. Similarly, M2 ¼ 1�ffiffiffi2p

=2 is solved from

3 ¼ 1þM2

1�M2þ 2

M2

ð1�M2Þ2: ð56Þ

Substituting M1 ¼ M2 ¼ 1�ffiffiffi2p

=2; p ¼ 1; Fð�1Þ ¼ �1 into the assumption (b) in Theorem 1 yields

1� ðM2 þM1Þ þM2M1 � pFðaÞ þ pFðaÞM2M1 ¼ffiffiffi2p

– 0:

The condition (c) in Theorem 1 turns out to be

�16cx3 – 0; for x 2 ð�1;0Þ [ ð0;1Þ;

where c > 1 is chosen. Thereby, according to Theorem 1, (54) has at least one solution yðx; �Þ 2 C2½�1;1� satisfies

yasyðx; �Þ � x4c� 6 yðx; �Þ 6 yasyðx; �Þ þ x4c�: ð57Þ

Furthermore, if c P c > 1 is set, then we have

jyðx; �Þ � yasyðx; �Þj 6 c�;

2642 J. Shen, M. Han / Commun Nonlinear Sci Numer Simulat 19 (2014) 2632–2643

where

Fig. 2.solution

yasyðx; �Þ ¼ x�1þ 1�

ffiffi2p

2

�e�

xþ1�

1� 1�ffiffi2p

2

�e�xþ1

�

þ1þ 1�

ffiffi2p

2

�e

x�1�

1� 1�ffiffi2p

2

�ex�1�

: ð58Þ

The asymptotic solution (58) is plotted in Fig. 2 with different values of �. In Fig. 2, the solid, dashing and dotted linesrepresent respectively the reduced solution, the asymptotic solutions with � ¼ 0:1 and � ¼ 0:01. We can concludefrom Fig. 2 that: canards exist in (54) and the explicit and uniformly valid zeroth-order approximation of canards is givenby (58).

Example 2. Consider

�y00 � yy0 þ ðcos xÞy ¼ �� sin x; �1 < x < 1; 0 < �� 1;y � p

2

� �� �y0 � p

2

� �¼ �4;

y p2

� �þ �y0 p

2

� �¼ 4:

8><>: ð59Þ

Here p ¼ 1; a ¼ � p2 ; b ¼ p

2 ;A ¼ �4; B ¼ 4; gðxÞ ¼ � sin x, f ðxÞ ¼ cos x and FðxÞ ¼ sin x. Thus, x ¼ 0 is the turning point. It turnsout that

uðxÞ ¼ sin x;

VLðs1Þ ¼ �1þM1e�

xþp2�

1�M1e�xþp

2�

;

VRðs2Þ ¼1þM2e

x�p2�

1�M2ex�p2�

:

Like the deductions in the Example 1, M1 ¼ M2 ¼ 1�ffiffiffiffiffiffi10p

=5 can be solved and the condition b) in Theorem 1 is

1� ðM2 þM1Þ þM2M1 � pF �p2

�þ pF �p

2

�M2M1 ¼

2ffiffiffiffiffiffi10p

5– 0:

The condition (c) in Theorem 1 turns out to be

�4cx2ð3 sin xþ x cos xÞ – 0; x 2 ð�1;0Þ [ ð0;1Þ; ð60Þ

which can be proved by the technique of calculus.According to Theorem 1, singularly perturbed boundary value problem (59) has at least one solution yðx; �Þ 2 C2½�1;1�

satisfies

yasyðx; �Þ � cx4� 6 yðx; �Þ 6 yasyðx; �Þ þ cx4�; ð61Þ

The reduced solution and the asymptotic solutions of (54). the reduced solution; , the asymptotic solution when � ¼ 0:01; - - -, the asymptoticwhen � ¼ 0:1.

Fig. 3. The reduced solution and the asymptotic solutions of (59). the reduced solution; , the asymptotic solution when � ¼ 0:01; - - -, the asymptoticsolution when � ¼ 0:1.

J. Shen, M. Han / Commun Nonlinear Sci Numer Simulat 19 (2014) 2632–2643 2643

in which

yasyðx; �Þ ¼ sin x�1þ 1�

ffiffiffiffi10p

5

�e�

xþp2�

1� 1�ffiffiffiffi10p

5

�e�

xþp2�

þ1þ 1�

ffiffiffiffi10p

5

�e

x�p2�

1� 1�ffiffiffiffi10p

5

�e

x�p2�

: ð62Þ

The asymptotic solution (62) is plotted in Fig. 3 with different �. By Fig. 3 and (61), the existence of canards in (59) isknown. This canard is approximated uniformly and explicitly by (62) with the accuracy of zeroth-order.

Acknowledgements

Financial supports from the National Natural Science Foundations of China (11201072, 11102041) are gratefully acknowl-edged. The authors would like to express their appreciation and thanks to the referees for the proposed improvements.

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