To prove the main results of the current paper, the following lemma is needed.

**Lemma 1**[

18] Consider second-order nonlinear boundary value problems with Dirichlet boundary conditions,

$\left\{\begin{array}{cc}\hfill {y}^{\u2033}=f\left(x,y,{y}^{\prime}\right),\hfill & \hfill \phantom{\rule{1em}{0ex}}x\in \left(a,b\right)\hfill \\ \hfill y\left(a\right)=A,\hfill & \hfill \phantom{\rule{1em}{0ex}}y\left(b\right)=B\hfill \end{array}\right.$

in which *a, b, A*, and *B* are constants.

For this boundary value problem, if the following conditions hold,

- (1)
there exist the upper and lower solutions, i.e., there are functions

*β*(

*x*),

*α*(

*x*) ∈

*C* ^{2}[

*a*,

*b*] with

*β*(

*x*) ≥

*α*(

*x*) such that

$\begin{array}{c}{\beta}^{\u2033}\le f\left(x,\beta ,{\beta}^{\prime}\right),\phantom{\rule{1em}{0ex}}x\in \left(a,b\right),\\ \beta \left(a\right)\ge A,\phantom{\rule{1em}{0ex}}\beta \left(b\right)\ge B\end{array}$

and

$\begin{array}{c}{\alpha}^{\u2033}\ge f\left(x,\alpha ,{\alpha}^{\prime}\right),\phantom{\rule{1em}{0ex}}x\in \left(a,b\right),\\ \alpha \left(a\right)\le A,\phantom{\rule{1em}{0ex}}\alpha \left(b\right)\le B,\end{array}$

- (2)
the function

*f*(

*x, y, y*') satisfies the Nagumo condition with respect to

*β*(

*t*) and

*α*(

*t*), then there exists at least one solution

*y*(

*x*) ∈

*C* ^{2}[

*a*,

*b*] with the following estimate:

$\alpha \left(x\right)\le y\left(x\right)\le \beta \left(x\right),\phantom{\rule{1em}{0ex}}x\in \left[a,b\right].$

Based on Lemma 1, we turn to prove the following theorems.

**Theorem 1** There exists at least one solution of boundary value problem (1-3) such that

$|y\left(x,\epsilon \right)-{y}_{asy}\left(x,\epsilon \right)|\le \gamma \epsilon ,\phantom{\rule{1em}{0ex}}x\in \left[a,b\right],$

(26)

where

*γ* is a positive constant,

*y*_{asy}(

*x*,

*ε*) is given by Equation (

25) in which

i.e., *C* = 0 in Equation (6) is determined.

By Theorem 1 and the dynamical behavior of *y*_{asy}(*x*, *ε*), the following Theorem 2 can be concluded directly.

**Theorem 2** There exist at least one solution of boundary value problem (1-3) with the following asymptotic behavior:

$\underset{\epsilon \to 0}{lim}y\left(x,\epsilon \right)=-x,\phantom{\rule{1em}{0ex}}x\in \left(a,b\right).$

Theorems 1 and 2 together mean Theorem 3 as follows.

**Theorem 3** Boundary value problem (1-3) has at least one canard solution, whose zero-order approximation is given by Equation (25).

**Proof of Theorem 1** Define the upper and lower solutions as follows:

$\beta \left(x,\epsilon \right)=u\left(x\right)+{V}_{L}\left({\tau}_{1}\right)+{V}_{R}\left({\tau}_{2}\right)+{x}^{4}\gamma \epsilon $

(27)

and

$\alpha \left(x,\epsilon \right)=u\left(x\right)+{V}_{L}\left({\tau}_{1}\right)+{V}_{R}\left({\tau}_{2}\right)-{x}^{4}\gamma \epsilon ,$

(28)

in which, *γ* is a positive constant.

Since the right-hand side function in Equation (1) satisfies the Nagumo condition, thus, to obtain Theorem 1, it is left to verify that the upper and lower solutions (31) and (32) satisfy the condition (1) in Lemma 1.

Firstly, we prove the following inequality:

$\epsilon {\beta}^{\u2033}+\beta {\beta}^{\prime}+\beta \le 0.$

(29)

In fact,

$\begin{array}{c}\epsilon {\beta}^{\u2033}+\beta {\beta}^{\prime}+\beta \\ =\epsilon \left(\frac{{\ddot{V}}_{L}}{{\epsilon}^{2}}+\frac{{\ddot{V}}_{R}}{{\epsilon}^{2}}+12{x}^{2}\gamma \epsilon \right)+\left(u\left(x\right)+{V}_{L}+{V}_{R}+{x}^{4}\gamma \epsilon \right)\left({u}^{\prime}\left(x\right)+\frac{{\dot{V}}_{L}}{\epsilon}+\frac{{\dot{V}}_{R}}{\epsilon}+4{x}^{3}\gamma \epsilon \right)\\ \phantom{\rule{1em}{0ex}}+u\left(x\right)+{V}_{L}+{V}_{R}+{x}^{4}\gamma \epsilon \\ =12{x}^{2}\gamma {\epsilon}^{2}+\left(u\left(x\right)+{V}_{R}+{x}^{4}\gamma \epsilon \right)\frac{{\dot{V}}_{L}}{\epsilon}+\left(u\left(x\right)+{V}_{L}+{x}^{4}\gamma \epsilon \right)\frac{{\dot{V}}_{R}}{\epsilon}\\ \phantom{\rule{1em}{0ex}}+4{x}^{3}\gamma \epsilon \left(u\left(x\right)+{V}_{L}+{V}_{R}+{x}^{4}\gamma \epsilon \right)\\ =\frac{1}{\epsilon}\left[\left(u\left(x\right)+{V}_{R}+{x}^{4}\gamma \epsilon \right){\dot{V}}_{L}+\left(u\left(x\right)+{V}_{L}+{x}^{4}\gamma \epsilon \right){\dot{V}}_{R}\right.\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\left.4{x}^{3}\gamma {\epsilon}^{2}\left(u\left(x\right)+{V}_{L}+{V}_{R}+{x}^{4}\gamma \epsilon \right)+12{x}^{2}\gamma {\epsilon}^{3}\right],\end{array}$

(30)

in which as well as in the following of the paper, the prime and the dot always denote the derivations with respect to the slow scale *x* and the fast scales *τ*_{1}, *τ*_{2}, respectively.

We want to prove that the quantity defined in Equation (30) is not positive. The proof is completed by dividing the interval [*a, b*] into five parts.

Part I. *x* ∈ [*a, a* + *δ*_{1}), where *δ*_{1} > 0 is a sufficiently small constant independent of *ε*.

In this case, it can be deduced from Equation (

17) that

${V}_{L}\left({\tau}_{1}\right){|}_{x=a}={V}_{L}\left(0\right)=A\phantom{\rule{2.77695pt}{0ex}}and\phantom{\rule{2.77695pt}{0ex}}{\dot{V}}_{L}\left({\tau}_{1}\right){|}_{x=a}={\dot{V}}_{L}\left(0\right)=-\frac{2{u}^{2}\left(a\right){M}_{1}}{{\left(1-{M}_{1}\right)}^{2}}$

(31)

which are both constants. Similarly, we can derive from Equation (

21) that

${V}_{R}\left(a\right)=u\left(b\right)\frac{1+{M}_{2}{e}^{-u\left(b\right)\frac{a-b}{\epsilon}}}{1-{M}_{2}{e}^{-u\left(b\right)\frac{a-b}{\epsilon}}}=u\left(b\right)+u\left(b\right)\frac{2{M}_{2}{e}^{-u\left(b\right)\frac{a-b}{\epsilon}}}{1-{M}_{2}{e}^{-u\left(b\right)\frac{a-b}{\epsilon}}}=u\left(b\right)+O\left({e}^{-u\left(b\right)\frac{a-b}{\epsilon}}\right),$

(32)

in which, for

*ε* sufficiently small,

$O\left({e}^{-u\left(b\right)\frac{a-b}{\epsilon}}\right)$ denotes a quantity that is exponentially small and negative, and

${\dot{V}}_{R}\left(a\right)=-\frac{2{u}^{2}\left(b\right){M}_{2}{e}^{-u\left(b\right)\frac{a-b}{\epsilon}}}{{\left(1-{M}_{2}{e}^{-u\left(b\right)\frac{a-b}{\epsilon}}\right)}^{2}},$

(33)

which is a exponential small quantity too.

Substituting Equations (

31-33) into Equation (

30) and taking

*a* +

*b* = 0 into account yields

$\begin{array}{c}\left[\left(u\left(x\right)+{V}_{R}+{x}^{4}\gamma \epsilon \right){\dot{V}}_{L}+\left(u\left(x\right)+{V}_{L}+{x}^{4}\gamma \epsilon \right){\dot{V}}_{R}\right.\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left.+4{x}^{3}\gamma {\epsilon}^{2}\left(u\left(x\right)+{V}_{L}+{V}_{R}+{x}^{4}\gamma \epsilon \right)+12{x}^{2}\gamma {\epsilon}^{3}\right]{|}_{x=a}\\ =-\left[2C+O\left({e}^{-u\left(b\right)\frac{a-b}{\epsilon}}\right)+{a}^{4}\gamma \epsilon \right]\frac{2{u}^{2}\left(a\right){M}_{1}}{{\left(1+{M}_{1}\right)}^{2}}-\left(-a+C+A+{a}^{4}\gamma \epsilon \right)\frac{2{u}^{2}\left(b\right){M}_{2}{e}^{-u\left(b\right)\frac{a-b}{\epsilon}}}{{\left(1-{M}_{2}{e}^{-u\left(b\right)\frac{a-b}{\epsilon}}\right)}^{2}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+4{a}^{3}\gamma {\epsilon}^{2}\left[2C+A+O\left({e}^{-u\left(b\right)\frac{a-b}{\epsilon}}\right)+{a}^{4}\gamma \epsilon \right]+12{a}^{2}\gamma {\epsilon}^{3}.\end{array}$

(34)

By comparing the order of the four parts in Equation (

34), we can find that, for

*ε* sufficiently small, the sign of Equation (

34) is determined by its first part, i.e.,

$-\left[2C+O\left({e}^{-u\left(b\right)\frac{a-b}{\epsilon}}\right)+{a}^{4}\gamma \epsilon \right]\frac{2{u}^{2}\left(a\right){M}_{1}}{{\left(1-{M}_{1}\right)}^{2}}.$

(35)

Hence, if the constant

*C* in Equation (

35) is chosen such that

then

$-\left[2C+O\left({e}^{-u\left(b\right)\frac{a-b}{\epsilon}}\right)+{a}^{4}\gamma \epsilon \right]\frac{2{u}^{2}\left(a\right){M}_{1}}{{\left(1-{M}_{1}\right)}^{2}}<0.$

Consequently, when *x* = *a*, the quantity defined in (30) is negative if the inequality (36) holds and *ε* is sufficiently small. Hence, there exists a sufficiently small constant *δ*_{1} > 0 independent of *ε* such that the quantity defined in (30) is negative for *x* ∈ [*a, a* + *δ*_{1}).

On the contrary, we can see that when the following differential inequality to be proved,

$\epsilon {\alpha}^{\u2033}+\alpha {\alpha}^{\prime}+\alpha \ge 0,$

(37)

in which,

*α*(

*x*,

*ε*) is defined in Equation (

28), it is required that

Accordingly, the inequalities (36) and (38) together yield

Therefore, in what follows, *C* = 0 is set in Equation (6). Thus, *u*(*x*) = -*x* turns out to be the reduced solution.

Part II. *x* = 0.

In this case, since the boundary values in (2-3) satisfy

it then follows from Equations (

19) and (

23) that

$0=-a\frac{1+{M}_{1}}{1-{M}_{1}}-b\frac{1+{M}_{2}}{1-{M}_{2}}=b\left(\frac{1+{M}_{1}}{1-{M}_{1}}-\frac{1+{M}_{2}}{1-{M}_{2}}\right)=2b\frac{{M}_{1}-{M}_{2}}{\left(1-{M}_{1}\right)\left(1-{M}_{2}\right)},$

in which,

*a* = -

*b* has been noted, which finally implies that

Consequently, by setting

*x* = 0 in Equation (

30), one gets

$\begin{array}{c}\frac{1}{\epsilon}\left[\left(u\left(0\right)+{V}_{R}\left(\frac{-b}{\epsilon}\right)\right){\dot{V}}_{L}\left(\frac{-a}{\epsilon}\right)+\left(u\left(0\right)+{V}_{L}\left(\frac{-a}{\epsilon}\right)\right){\dot{V}}_{R}\left(\frac{-b}{\epsilon}\right)\right]\\ =\frac{1}{\epsilon}\left[{V}_{R}\left(\frac{-b}{\epsilon}\right){\dot{V}}_{L}\left(\frac{-a}{\epsilon}\right)+{V}_{L}\left(\frac{-a}{\epsilon}\right){\dot{V}}_{R}\left(\frac{-b}{\epsilon}\right)\right]\\ =\frac{1}{\epsilon}\left[-2{a}^{2}b\frac{1+{M}_{2}{e}^{\frac{-{b}^{2}}{\epsilon}}}{1-{M}_{2}{e}^{\frac{-{b}^{2}}{\epsilon}}}\frac{{M}_{1}{e}^{\frac{-{a}^{2}}{\epsilon}}}{{\left(1+{M}_{1}{e}^{\frac{-{a}^{2}}{\epsilon}}\right)}^{2}}-2a{b}^{2}\frac{1+{M}_{1}{e}^{\frac{-{a}^{2}}{\epsilon}}}{1-{M}_{1}{e}^{\frac{-{a}^{2}}{\epsilon}}}\frac{{M}_{2}{e}^{\frac{-{b}^{2}}{\epsilon}}}{{\left(1+{M}_{2}{e}^{\frac{-{b}^{2}}{\epsilon}}\right)}^{2}}\right]\\ =0,\end{array}$

in which, *a* = -*b* and *M*_{1} = *M*_{2} have been taken into account. Thus, when *x* = 0, the inequality (29) holds.

Part III. *x* ∈ [*a* = *δ*_{1}, 0].

Taking the cases in Parts I and II into account, if the inequality (29) does not hold uniformly in this region, then there must be at least one point

*x** ∈ (

*a* =

*δ*_{1}, 0) such that

$H\left({x}^{*},\epsilon \right)>0,{H}^{\prime}\left({x}^{*},\epsilon \right)=0\phantom{\rule{0.3em}{0ex}}and{H}^{\u2033}\left({x}^{*},\epsilon \right)\le 0,$

in which

$\begin{array}{ll}\hfill H\left(x,\epsilon \right)& =\frac{1}{\epsilon}\left[\left(-x+{V}_{R}+{x}^{4}\gamma \epsilon \right){\dot{V}}_{L}+\left(-x+{V}_{L}+{x}^{4}\gamma \epsilon \right){\dot{V}}_{R}\right.\phantom{\rule{2em}{0ex}}\\ \left.\phantom{\rule{1em}{0ex}}+4{x}^{3}\gamma {\epsilon}^{2}\left(-x+{V}_{L}+{V}_{R}+{x}^{4}\gamma \epsilon \right)+12{x}^{2}\gamma {\epsilon}^{3}\right].\phantom{\rule{2em}{0ex}}\end{array}$

However, for

*ε* sufficiently small, since

${\dot{V}}_{L}$,

${\dot{V}}_{R}$ and

*V*_{
L
} +

*V*_{
R
} are exponentially small in this region, thus, it can be shown by direct calculations that

${H}^{\prime}\left(x*,\epsilon \right)=\epsilon \left[-12\gamma {\left({x}^{*}\right)}^{3}+O\left(\epsilon \right)\right]>0,$

which is a contradiction.

Part IV. *x* ∈ (*b* - *δ*_{2}, *b*], where *δ*_{2} > 0 is a sufficiently small constant independent of *ε*.

In this region, the proof of the inequality (29) is parallel to Part I completely. Like the deductions of Equations (31-33), the values of ${V}_{L}\left({\tau}_{1}\right){|}_{x=b}$, ${\dot{V}}_{L}\left({\tau}_{1}\right){|}_{x=b}$, ${V}_{R}\left({\tau}_{2}\right){|}_{x=b}$, and ${\dot{V}}_{R}\left({\tau}_{2}\right){|}_{x=b}$ can be calculated. Consequently, we can see that, when *x* = *b*, the other parts in Equation (30) are the higher-order small quantities compared with its second part. Thus, the sign of Equation (30) is determined by its second part, which is a negative quantity. Accordingly, the inequality (29) is proved.

Part V. *x* ∈ (0, *b* - *δ*_{2},], In this region, the proof of the inequality (29) is parallel to Part III completely.

So far the proof of the differential inequality (29) has been finished for *x* ∈ [*a*, *b*]. In the same way, the differential inequality (37) can be proved.

In what follows, we turn to prove the inequalities on the boundaries. For

*ε* sufficiently small, we have

$\begin{array}{ll}\hfill \beta \left(a,\epsilon \right)& =u\left(a\right)+{V}_{L}\left(0\right)+{V}_{R}\left(\frac{a-b}{\epsilon}\right)+{a}^{4}\gamma \epsilon \phantom{\rule{2em}{0ex}}\\ =u\left(a\right)+A+u\left(b\right)+O\left({e}^{-u\left(b\right)\frac{a-b}{\epsilon}}\right)+{a}^{4}\gamma \epsilon \phantom{\rule{2em}{0ex}}\\ =A+O\left({e}^{\frac{b\left(a-b\right)}{\epsilon}}\right)+{a}^{4}\gamma \epsilon \phantom{\rule{2em}{0ex}}\\ \ge A\ge \alpha \left(a,\epsilon \right)=u\left(a\right)+{V}_{L}\left(0\right)+{V}_{R}\left(\frac{a-b}{\epsilon}\right)-{a}^{4}\gamma \epsilon \phantom{\rule{2em}{0ex}}\end{array}$

in which, *u*(*a*) = -*a*, *u*(*b*) = -*b*, and *a* + *b* = 0 have been used.

Similarly, it can be proved that

$\beta \left(b,\epsilon \right)\ge B\ge \alpha \left(b,\epsilon \right).$

Therefore, according to Lemma 1, we have

$\alpha \left(x,\epsilon \right)\le y\left(x,\epsilon \right)\le \beta \left(x,\epsilon \right),\phantom{\rule{1em}{0ex}}x\in \left[a,b\right],$

and accordingly, Theorem 1 is derived.

**Remark 3**. From the proof of Theorem 1, we know that the construction of the upper and lower solutions defined in (27) and (28), respectively, is essential. The error term *x*^{4}*γε* introduced in (27) and (28) seems necessary for discussing the existence of canard solutions in singularly perturbed problems (1-3).