Directory UMM :Journals:Journal_of_mathematics:EJQTDE:
Electronic Journal of Qualitative Theory of Differential Equations
2010, No. 62, 1-14; http://www.math.u-szeged.hu/ejqtde/
On the Step-type Contrast Structure of a
Second-order Semilinear Differential Equation
with Integral Boundary Conditions
Feng Xie1,3∗, Zhaoyang Jin1 and Mingkang Ni2,3
1
Department of Applied Mathematics, Donghua University
Shanghai 201620, P.R. China
2
Department of Mathematics, East China Normal University
Shanghai 200241, P.R. China
3
Division of Computational Science, E-Institute of Shanghai Universities,
at SJTU, Shanghai 200030, P.R. China
Abstract
In this paper we investigate the step-type contrast structure of a second-order
semilinear differential equation with integral boundary conditions. The asymptotic
solution is constructed by the boundary function method, and the uniform validity
of the formal solution is proved by the theory of differential equalities.
2000 Mathematics Subject Classification. 34B15; 34E05; 34E20.
Key words and phrases. Singular perturbation; contrast structure; integral
boundary conditions.
1. Introduction
We shall consider the existence of contrast structure for the following singularly perturbed
differential equation with integral boundary conditions
µ2
y(0, µ) =
Z
d2 y
= f (t, y),
dt2
1
h1 (y(s, µ))ds,
0
0 < t < 1,
Z 1
y(1, µ) =
h2 (y(s, µ))ds,
(1.1)
(1.2)
0
where µ is a small and positive parameter, and f : [0, 1] × R → R, hi : R → R (i = 1, 2)
are C (2) -functions.
∗
Corresponding author. E-mail: [email protected].
EJQTDE, 2010 No. 62, p. 1
Singularly perturbed boundary value problems arise naturally in various applications,
and have received more and more attention in recent years. Contrast structures, namely
solutions which have internal transition layers, are initially investigated by using the boundary function method by Butuzov and Vasil’eva in 1987 [1], and have recently been one of
the hot topics in singular perturbation problems; see [2-8], for instance. The step-type
contrast structure of the equation (1.2) with two-point boundary conditions y(0, µ) = y 0
and y(1, µ) = y 1 was considered by Butuzov and Vasil’eva [1]. They gave the conditions
which ensure the existence of step-type contrast structure and applied the boundary function method to construct the corresponding asymptotic solution. Recently, Ni and Lin [7]
proved rigorously the uniform validity of asymptotic solution by using Nagumo’s Theorem.
In 2009, Ni and Wang [8] extended the equation (1.2) to higher dimension and studied
the following semilinear singularly perturbed system
µ2 y1′′ = f1 (y1 , y2 , . . . , yn , t),
µ2 y2′′ = f2 (y1 , y2 , . . . , yn , t),
..
.
µ2 yn′′ = fn (y1 , y2 , . . . , yn , t),
subject to the conditions
yk (0, µ) = yk0 ,
yj′ (0, µ) = zj1 ,
k = 1, 2, . . . , n,
k = 1, 2, . . . , n − 1,
yn′ (1, µ) = zn1 ′ .
The authors gave the conditions under which there exists an internal transition layer,
and constructed the uniformly valid asymptotic expansion of a solution with a step-type
contrast structure.
To our knowledge, contrast structures of singularly perturbed problems with integral
boundary conditions have not been investigated. Boundary value problems with integral
boundary conditions have significant applications in thermal conduction [9], semiconductor
problems [10], biomedical science [11], and so on. In [12], Cakir and Amiraliyev studied
the singularly perturbed nonlocal boundary value problem
ε2 y ′′ + εa(t)y ′ − b(t)y = f (t), 0 < t < l, 0 < ε ≪ 1,
Z l1
0
1
g(s)y(s)ds, 0 ≤ l0 < l1 ≤ l,
y(0) = y , y(1) = y +
l0
where y 0 , y 1 are given constants, and a(t) ≥ α > 0, b(t) ≥ β > 0, g(t) and f (t) are
sufficiently smooth functions in [0, 1]. The authors presented a finite difference method for
EJQTDE, 2010 No. 62, p. 2
numerical solutions of the problem which exhibited two boundary layers at t = 0 and t = l.
In [13], Xie and Zhang extended the above problem to general weakly nonlinear singular
perturbation problems with integral boundary conditions by using the boundary function
method.
The present paper is devoted to investigate the existence of step-type contrast structure for the problem (1.1)-(1.2). Integral boundary conditions (1.2) can be viewed as the
generalization of two-point and nonlocal boundary conditions. The boundary function
method and the theory of differential inequalities will be applied to obtain the uniformly
valid asymptotic solution of the problem (1.1) -(1.2). The main difficulty different from
the corresponding two-point boundary value problem lies in that the integral boundary
conditions of the two associated problems are coupled, which will be overcome with the aid
of the property that boundary layer functions decay exponentially.
The remainder of this paper is organized as follows. In section 2 we give some assumptions and construct the formal asymptotic solution of the original problem. In section 3
the uniform validity of formal solution is proved by the theory of differential equalities.
2. Basic Assumptions and Construction of Asymptotic Solution
Let us begin with two basic assumptions.
[H1 ] f : [0, 1] × R → R, hi : R → R (i = 1, 2) are C (2) -functions, and h′i (x) ≥ 0.
[H2 ] The reduced equation f (t, y) = 0 has three isolated solutions ϕi (t) (i = 1, 2, 3) on
[0, 1], satisfying
ϕ1 (t) < ϕ2 (t) < ϕ3 (t),
∂f
(t, ϕi (t)) > 0 (i = 1, 3),
∂y
∂f
(t, ϕ2 (t)) < 0.
∂y
Assumption [H2 ] is a so-called stability condition. It follows from the assumption [H2 ]
that in the phase plane (y, y ′), the equilibria (ϕ1,3 , 0) is a saddle point and (ϕ2 , 0) is a center.
We are interested in the solution of step-type which has a transition from the vicinity of
ϕ1 (t) to that of ϕ3 (t) at some point t = t∗ . That is, for some t∗ ∈ (0, 1) the following limit
holds:
ϕ (t), 0 < t < t∗ ,
1
lim y(t, µ) =
µ→0
ϕ3 (t), t∗ < t < 1.
t = t∗ is called the transition point.
We shall adopt the following strategy which is due to Butuzov and Vasil’eva [1]. We
divide the original problem into the two associated pure boundary layer problems, that is,
EJQTDE, 2010 No. 62, p. 3
the associated left problem
d2 y (−)
= f (t, y (−) ), 0 < t < t∗ ,
dt2
Z t∗
Z 1
(−)
(−)
y (0, µ) =
h1 (y (s, µ))ds +
h1 (y (+) (s, µ))ds,
µ2
0
t∗
y
(−)
∗
∗
(t , µ) = ϕ2 (t ),
and the associated right problem
d2 y (+)
= f (t, y (+) ), t∗ < t < 1,
2
dt
Z t∗
Z 1
(+)
(−)
y (1, µ) =
h2 (y (s, µ))ds +
h2 (y (+) (s, µ))ds,
µ2
0
t∗
y
(+)
∗
∗
(t , µ) = ϕ2 (t ).
Considering that the solution y(t, µ) is smooth at t = t∗ , it follows that
dy (+) ∗
dy (−) ∗
(t , µ) =
(t , µ),
dt
dt
(2.1)
which is the condition determining the position of transition point.
Here the main difficulty different from the corresponding two-point boundary value
problem lies in that the integral boundary conditions of the two associated problems are
coupled. In order to overcome this difficulty, we need to handle these two associated
problems simultaneously, and have the aid of properties of boundary layer functions to
uncouple the boundary conditions.
Let us describe the formal scheme of seeking an asymptotic solution of the problem
(1.1)-(1.2) of the form
y (−) (t, µ) + Π(−) y(τ , µ) + Q(−) y(τ, µ), 0 ≤ t ≤ t∗ ,
0
y(t, µ) =
(2.2)
y (+) (t, µ) + R(+) y(τ1 , µ) + Q(+) y(τ, µ), t∗ ≤ t ≤ 1,
where
(∓)
(∓)
(∓)
y (∓) (t, µ) = y 0 (t) + µy1 (t) + µ2 y 2 (t) + · · ·
are the regular parts of the left problem and the right problem, respectively,
(−)
(−)
(−)
Π(−) y(τ0 , µ) = Π0 y(τ0 ) + µΠ1 y(τ0) + µ2 Π2 y(τ0 ) + · · · ,
τ0 =
t
µ
is the left boundary layer part of the left problem,
(∓)
(∓)
(∓)
Q(∓) y(τ, µ) = Q0 y(τ ) + µQ1 y(τ ) + µ2 Q2 y(τ ) + · · · ,
τ=
t − t∗
µ
EJQTDE, 2010 No. 62, p. 4
are the right boundary layer part of the left problem and the left boundary layer part of
the right problem, respectively, and
(+)
(+)
(+)
R(+) y(τ1 , µ) = R0 y(τ1 ) + µR1 y(τ1 ) + µ2 R2 y(τ1 ) + · · · ,
τ1 =
t−1
µ
is the right boundary layer part of the right problem. We also seek the asymptotic expansion
of the transition point t∗ :
t∗ = t0 + µt1 + µ2 t2 + · · · .
Substituting (2.2) into (1.1) and equating the coefficients in like powers of µ, we get a
(∓)
recurrent sequence of algebraic equations for the functions y i (t) (i = 1, 2, . . .).
(∓)
f t, y 0 (t) = 0,
∂f (∓) (∓)
t, y 0 (t) y 2 (t),
∂y
(∓)
∂f (∓) (∓)
d2 y 2
(±)
t, y 0 (t) y 4 (t) + g4 ,
=
2
dt
∂y
···
(∓)
d2 y 2k−2
∂f (∓) (∓)
(±)
t, y 0 (t) y 2k (t) + g2k ,
=
2
dt
∂y
(∓)
d2 y 0
dt2
(±)
=
(∓)
where g2k are the determined functions of y i (0 ≤ i ≤ 2k − 2). From the assumption
(∓)
[H2 ], the coefficients y i can be obtained recurrently. In particular, we have
(−)
y0
= ϕ1 (t),
(+)
y0
= ϕ3 (t).
For simplicity, we only consider the approximation of first order for the boundary layer
(−)
(−)
series. The left boundary layer functions Π0 y(τ0 , µ) and Π1 y(τ0, µ) satisfy
(−)
d2 Π0 y
(−)
= f 0, ϕ1 (0) + Π0 y ,
2
dτ0
Z t0
Z 1
(−)
(2.3)
Π0 y(0) =
h1 (ϕ1 (s))ds +
h1 (ϕ3 (s))ds − ϕ1 (0),
0
t0
(−)
Π0 y(+∞) = 0,
and
(−)
d2 Π1 y
∂f
(−)
(−)
(−)
=
0,
ϕ
(0)
+
Π
y
Π1 y + ∆1 ,
1
0
dτ02
∂y
Z +∞
Z 0
(−)
(−)
(−)
′
Π1 y(0) =
h1 (ϕ1 (0))Π0 y(s)ds +
h′1 (ϕ1 (t0 ))Q0 y(s)ds
0
−∞
Z +∞
Z 0
(+)
(+)
′
+
h1 (ϕ3 (t0 ))Q0 y(s)ds +
h′1 (ϕ3 (1))R0 y(s)ds,
0
−∞
(−)
Π1 y(+∞)
= 0,
EJQTDE, 2010 No. 62, p. 5
respectively, where
∂f
∂f
(−)
(−)
(−)
(−)
(−)
∆1 =
0, ϕ1 (0) + Π0 y y 0 ′ (0)τ0 + y 1 (0) +
0, ϕ1 (0) + Π0 y τ0
∂y
∂t
∂f
∂f
(−)
(−)
(0, ϕ1 (0)) y 0 ′ (0)τ0 + y 1 (0) −
(0, ϕ1 (0)) τ0 .
−
∂y
∂t
(+)
(+)
Analogously, the right boundary layer functions R0 y(τ1 ) and R1 y(τ1 ) satisfy the
following boundary value problems
(+)
d2 R0 y
(+)
=
f
1,
ϕ
(1)
+
R
y
,
3
0
2
dτ1
Z t0
Z 1
(+)
(2.4)
R0 y(1) =
h2 (ϕ1 (s))ds +
h2 (ϕ3 (s))ds − ϕ3 (1),
0
t0
(+)
R0 y(−∞) = 0,
and
∂f
d2 R1 y
(+)
(+)
(+)
=
1,
ϕ
(1)
+
R
y
R1 y + ∆1 ,
3
0
2
dτ1
∂y
Z +∞
Z 0
(−)
(−)
(+)
′
R1 y (1) =
h2 (ϕ1 (0))Π0 y(s)ds +
h′2 (ϕ1 (t0 ))Q0 y(s)ds
0
−∞
Z +∞
Z 0
(+)
(+)
+
h′2 (ϕ3 (t0 ))Q0 y(s)ds +
h′2 (ϕ3 (1))R0 y(s)ds,
(+)
0
−∞
(+)
R1 y(−∞)
= 0,
respectively, where
∂f
∂f
(+)
(+)
(+)
(+)
(+)
∆1 =
1, ϕ3 (1) + R0 y y 0 ′ (1)τ1 + y 1 (1) +
1, ϕ3 (1) + R0 y τ1
∂y
∂t
∂f
∂f
(+)
(+)
−
(1, ϕ3 (1)) y 0 ′ (1)τ1 + y 1 (1) −
(1, ϕ3 (1)) τ1 .
∂y
∂t
In order to ensure the existence of solutions for the boundary value problems (2.3) and
(2.4), we need the following assumptions. See [3] for their geometrical interpretation.
!
Z t0
(−)
(−)
dΠ0 y
y
dΠ
(−)
0
, let the straight line
=
h1 (ϕ1 (s))ds+
[H3 ] In the phase plane Π0 y,
dτ0
dτ0
0
Z 1
h1 (ϕ3 (s))ds intersect the separatrix entering the saddle (0, ϕ1 (0)) as τ0 → +∞;
t0
[H4 ] In the phase plane
Z
1
!
Z t0
(+)
(+)
dR
y
dR0 y
(+)
0
R0 y,
, let the straight line
=
h2 (ϕ1 (s))ds+
dτ1
dτ1
0
h2 (ϕ3 (s))ds intersect the separatrix entering the saddle (0, ϕ3 (1)) as τ1 → −∞.
t0
EJQTDE, 2010 No. 62, p. 6
Note that t0 in the above assumptions is unknown, which is determined by (2.10).
From the assumptions [H1 ], [H2 ] and [H3 ], we have the following estimates of decay(−)
(−)
(+)
ing exponentially for the boundary layer functions Π0 y(τ0), Π1 y(τ0 ), R0 y(τ1 ) and
(+)
R1 y(τ1 ).
Lemma 2.1 Under the assumptions [H1 ], [H2 ] and [H3 ], the following estimates
dΠ(−) y
0
(−)
≤ c1 exp(−κ1 τ0 ), τ0 ≥ 0,
Π0 y ≤ c1 exp(−κ1 τ0 ),
dτ0
dR(+) y
0
≤ c2 exp(κ2 τ1 ), τ1 ≤ 0,
dτ1
(+)
τ0 ≥ 0; R1 y ≤ c4 exp(κ4 τ1 ), τ1 ≤ 0
(+)
R0 y ≤ c2 exp(κ2 τ1 ),
(−)
Π1 y ≤ c3 exp(−κ3 τ0 ),
hold, where ci and κi (i = 1, 2, 3, 4) are positive constants.
Proof. The proof is essential similar to that of [3], and we omit it here.
Let us now consider the right boundary layer of the left problem and the left boundary
layer of the right problem, that is, the interior layer of the original problem. We rewrite
the equation (1.1) into the equivalent system
µ
dz
= f (t, y),
dt
µ
dy
= z.
dt
(2.5)
Substituting (2.2) into (2.5), and separate the equations according to the scales t and τ ,
we obtain
µ
dz (∓)
= f t, y (∓) (t, µ) ,
dt
µ
dy(∓)
= z (∓) ,
dt
dQ(∓) z
= f t∗ + µτ, y (∓) (t∗ + µτ, µ) + Q(∓) y − f t∗ + µτ, y (∓) (t∗ + µτ, µ) ,
dτ
dQ(∓) y
= Q(∓) z.
dτ
(∓)
(∓)
Therefore, the coefficients Q0 y and Q0 z are determined by the following boundary
value problems
(∓)
dQ0 z
(∓)
=
f
t
,
ϕ
(t
)
+
Q
y
,
0
1,3
0
0
dτ
(∓)
dQ0 y
(∓)
= Q0 z,
(2.6)
dτ
(∓)
Q0 y(0) = ϕ2 (t0 ) − ϕ1,3 (t0 ),
Q(∓) y(∓∞) = Q(∓) z(∓∞) = 0.
0
0
EJQTDE, 2010 No. 62, p. 7
(∓)
(∓)
(∓)
(∓)
(∓)
By the transformations ye(∓) = y 0 (t0 ) + Q0 y, ze(∓) = z 0 (t0 ) + Q0 z = Q0 z, the
problem (2.6) becomes
de
z (∓)
(∓)
=
f
t
,
y
e
,
0
dτ
(∓)
de
y
= ze(∓) ,
dτ
(2.7)
ye(∓) (0) = ϕ2 (t0 ),
ye(∓) (∓∞) = ϕ1,3 (t0 ),
ze(∓) (∓∞) = 0.
It follows from the assumption [H2 ] that there exists a solution of the problem (2.7) for
given t0 . In what follows we will give the condition determining t0 . Integrating the first
two equations in (2.7) we have
Z ye(−) (τ )
Z ye(+) (τ )
(−) 2
(+) 2
ze (τ ) = 2
ze (τ ) = 2
f (t0 , y)dy,
f (t0 , y)dy.
(2.8)
ϕ1 (t0 )
ϕ3 (t0 )
Note that the zero order approximation of the smooth connection condition (2.1) becomes
It follows from (2.8) and (2.9) that
ze(−) (0) = ze(+) (0).
I(t0 ) ≡
Z
(2.9)
ϕ3 (t0 )
f (t0 , y)dy = 0,
(2.10)
ϕ1 (t0 )
which is the equation determining the dominant term t0 of t∗ .
[H4 ] Assume that the equation (2.10) has a root t = t0 with I ′ (t0 ) < 0.
In a similar way, we can also get the expression t1 which is closely related to the
(∓)
(∓)
equations for y (∓) , Q1 y and Q1 z, and the details are omitted here.
(∓)
(∓)
Similar to Lemma 2.1, for the boundary layer functions Qi y and Q1 z (i = 0, 1) we
have the following estimates of decaying exponentially.
Lemma 2.2 Under the assumptions [H1 ], [H2 ] and [H3 ], the following
(−)
(−)
Q0 y(τ ) ≤ c5 exp(κ5 τ ), Q0 z(τ ) ≤ c5 exp(κ5 τ ), τ
(−)
(−)
Q1 y(τ ) ≤ c6 exp(κ6 τ ), Q1 z(τ ) ≤ c5 exp(κ5 τ ), τ
(+)
(+)
Q0 y(τ ) ≤ c7 exp(−κ7 τ ), Q0 z(τ ) ≤ c7 exp(−κ7 τ ),
(+)
(+)
Q1 y(τ ) ≤ c8 exp(−κ8 τ ), Q1 z(τ ) ≤ c8 exp(−κ8 τ ),
estimates
≤ 0,
≤ 0,
τ ≥ 0,
τ ≥0
hold, where ci and κi (i = 5, 6, 7, 8) are positive constants.
EJQTDE, 2010 No. 62, p. 8
3. Existence of Step-type Solution
In this section we will prove the existence of step-type solution for the original problem
and give the estimate for the remainder term.
Theorem 3.3 Under the assumptions [H1 ]−[H4 ], there exists a step-type contrast structure
solution y(t, µ) of the problem (1.1)-(1.2) for sufficiently small µ > 0. Moreover, the
following asymptotic expansion holds
ϕ (t) + Π(−) y(τ ) + Q(−) y(τ ) + O(µ), 0 ≤ t < t + µt ;
1
0
0
1
0
0
y(t, µ) =
(3.1)
(+)
(+)
ϕ3 (t) + R y(τ1 ) + Q y(τ ) + O(µ), t0 + µt1 < t ≤ 1.
0
0
In order to prove Theorem 3.3 we need the following lemma which is a slight modification
of Theorem 2.2 in [14].
Lemma 3.4 Assume that the assumption [H1 ] holds and the continuous functions α(t, µ)
and β(t, µ) are of C (2) class on the intervals (0, tα ) ∪ (tα , 1) and (0, tβ ) ∪ (tβ , 1), respectively,
having the following properties
(1) α(t, µ) ≤ β(t, µ),
t ∈ [0, 1];
2
d2 α
2d β
≥
f
(t,
α),
t
∈
(0,
t
)
∪
(t
,
1);
µ
≤ f (t, β),
α
α
dt2
dt2
Z 1
Z 1
(3) α(0, µ) ≤
h1 (α(s, µ))ds, α(1, µ) ≤
h2 (α(s, µ))ds,
Z0 1
Z 01
β(0, µ) ≥
h1 (β(s, µ))ds, β(1, µ) ≥
h2 (β(s, µ))ds;
(2) µ2
0
(4)
dα
dα
(tα −) ≤
(tα +),
dt
dt
t ∈ (0, tβ ) ∪ (tβ , 1);
0
dβ
dβ
(tβ −) ≥
(tβ +),
dt
dt
where tα , tβ ∈ (0, 1). Then, there exists a solution y(t, µ) of the problem (1.1)-(1.2) such
that
α(t, µ) ≤ y(t, µ) ≤ β(t, µ), t ∈ [0, 1].
Remark 3.5 The functions α(t, µ) and β(t, µ) satisfying the above conditions are called
lower and upper solutions of the problem (1.1)-(1.2), respectively.
Remark 3.6 It is requested in [14] that the functions α(t, µ), β(t, µ) ∈ C (2) [0, 1]. Here we
only need the functions α(t, µ) and β(t, µ) to be piecewise C (2) - smooth and an additional
condition (4). It should be noted that the proof of Lemma 3.4 has no essential difference
from that of Theorem 2.2 in [14], but some slight modifications.
EJQTDE, 2010 No. 62, p. 9
Proof of Theorem 3.3. We select the auxiliary functions
ϕ (t) + Π(−) y(τ ) + µΠ(−) y(τ ) + Q(−) y(τ ) + µQ(−) y(τ ) − γµ, 0 ≤ t ≤ t ,
1
0
0
α
α
α
0
1
0α
1α
α(t, µ) =
(+)
(+)
(+)
(+)
ϕ3 (t) + R y(τ1 ) + µR y(τ1 ) + Q y(τα ) + µQ y(τα ) − γµ, tα ≤ t ≤ 1,
0
1
0α
1α
and
ϕ (t) + Π(−) y(τ ) + µΠ(−) y(τ ) + Q(−) y(τ ) + µQ(−) y(τ ) + γµ, 0 ≤ t ≤ t ,
1
0
0
β
β
β
0
1
0β
1β
β(t, µ) =
(+)
(+)
(+)
(+)
ϕ3 (t) + R y(τ1 ) + µR y(τ1 ) + Q y(τβ ) + µQ y(τβ ) + γµ, tβ ≤ t ≤ 1,
0
1
1β
0β
where
tα = t0 + µδ,
tβ = t0 − µδ,
τα =
t − tα
,
µ
τβ =
t − tβ
,
µ
(∓)
(∓)
while γ, δ are sufficiently large positive parameters. The functions Q0α y and Q1α y satisfy
respectively the following boundary value problems
d2 Q0α y
(∓)
=
f
τ
,
ϕ
(t
)
+
Q
y
,
α
1,3
α
0α
dτα2
(∓)
(∓)
Q0α y(0) = ϕ2 (tα ) − ϕ1,3 (tα ),
(∓)
Q0α y(∓∞) = 0,
and
(∓)
d2 Q1α y
∂f
(∓)
(∓)
(∓)
=
τ
,
ϕ
(t
)
+
Q
y
Q1α y + ∆1α − ω exp(±κ0 τα ),
α
1,3 α
0α
dτα2
∂y
(∓)
(∓)
Q1α y(0) = ϕ′2 (t0 ) − ϕ′1,3 (t0 ) t1 , Q1α y(∓∞) = 0,
where ω, κ0 are positive constants and
(∓)
∆1α =
∂f
∂f
(∓)
(∓)
τα , ϕ1,3 (tα ) + Q0α y ϕ′1,3 (tα )τα +
τα , ϕ1,3 (tα ) + Q0α y τα .
∂y
∂t
(∓)
(∓)
The functions Q0β y and Q1β y are also determined by the corresponding boundary value
problems.
To verify the conditions in Lemma 3.4 we divide the interval [0, 1] into five subintervals
[0, tβ /2], [tβ /2, tβ ], [tβ , tα ], [tα , (tα + 1)/2] and [(tα + 1)/2, 1].
Let us first check the condition (1). On the intervals [0, tβ /2] and [(tα +1)/2, 1], β(t, µ)−
α(t, µ) = 2γµ + EST > 0, where EST denotes exponentially small
terms. On the interval
(+)
(−)
(+)
(−)
[tβ , tα ], β(t, µ)−α(t, µ) = ϕ3 (t)−ϕ1 (t)+Q0β y(τβ )−Q0α y(τα )+µ Q1β y(τβ ) − Q1α y(τα ) +
2γµ + EST > 0. On the interval [tβ /2, tβ ],
(−)
(−)
(−)
(−)
β(t, µ) − α(t, µ) = 2γµ + Q0β y(τβ ) − Q0α y(τα ) + µ Q1β y(τβ ) − Q1α y(τα ) > 0,
EJQTDE, 2010 No. 62, p. 10
where we have used the following formula
(−)
(−)
(−)
Q0β y(τβ ) − Q0α y(τα ) =
dQ0 y ∗
(τ )2δ > 0,
dτ
τα ≤ τ ∗ ≤ τβ .
Similarly, on the interval [tα , (tα + 1)/2],
(+)
(+)
(+)
(+)
β(t, µ) − α(t, µ) = 2γµ + Q0β y(τβ ) − Q0α y(τα ) + µ Q1β y(τβ ) − Q1α y(τα ) > 0.
Next we check the condition (2). Here we only verify the condition (2) on the interval
[tβ /2, tβ ], which are similar on other subintervals.
(−)
(−)
d2 Q0β y
d2 Q1β y
β
2 ′′
µ 2 − f (t, β) = µ ϕ1 (t) +
+µ
dt
dτ 2
dτ 2
(−)
(−)
−f t, ϕ1 (t) + γµ + Q0β y + µQ1β y + EST.
2d
2
We rewrite f in f = fe(µ) + f (µ), where
(−)
(−)
fe(µ) = f τ µ, ϕ1 (τ µ) + γµ + Q0β y + µQ1β y − f (τ µ, ϕ1 (τ µ) + γµ)
!
(−)
(−)
d2 Q0β y
d2 Q1β y
2
=
+
µ
−
ω
exp(κ
τ
)
+
O
µ
,
0
dτ 2
dτ 2
(3.2)
(3.3)
and
f(µ) = f (t, ϕ1 (t) + γµ) =
Inserting (3.3) and (3.4) into (3.2) we have
µ2
∂f
(t, ϕ1 (t)) γµ + O µ2 .
∂y
(3.4)
d2 β
∂f
2
−
f
(t,
β)
=
−
(t,
ϕ
(t))
γµ
+
µω
exp(κ
τ
)
+
O
µ
< 0.
1
0
dt2
∂y
In a similar way we can show that
µ2
d2 α
≥ f (t, α),
dt2
We now show that
β(0, µ) ≥
t ∈ (0, tα ) ∪ (tα , 1).
Z
1
h1 (β(s, µ))ds.
0
It follows from the construction of asymptotic solution that
Z 1
β(0, µ) −
h1 (β(s, µ))ds
0
Z tβ
Z 1
(−)
(−)
= ϕ1 (0) + Π0 y(0) + µΠ1 y(0) + γµ − h1 (β(s, µ))ds − h1 (β(s, µ))ds + EST
0
tβ
EJQTDE, 2010 No. 62, p. 11
=
(−)
µΠ1 y(0)
−µ
Z
−µ
t0
(h1 (ϕ1 (s)) − h1 (ϕ3 (s))) ds − µ
t0 −δµ
0
t
δ− µ0
Z
Z
+ γµ +
(−)
h′1 (ϕ1 (t0 ))Q0 y(s)ds − µ
0
−
(+)
1−t0
−δ
µ
Z
t0 −δµ
µ
0
1−t0
−δ
µ
0
h′1 (ϕ3 (1))R0 y(s)ds + O µ2
= γµ + O (µ) > 0,
Z
(−)
h′1 (ϕ1 (0))Π0 y(s)ds
(+)
h′1 (ϕ3 (t0 ))Q0 y(s)ds
provided that γ is large enough.
Other inequalities in the condition (3) can be proved analogously.
Finally, let us check the condition (4).
(−)
dβ
dβ
dβ
µ (tβ −) − µ (tβ +) = µ
dt
dt
dt (+)
(−)
(−)
dQ1β y
dQ0β y
dQ0β y
′
= µy 0 (tβ ) +
+µ
=
+ O(µ).
dτβ
dτβ (+)
dτβ (+)
From the process similar to (2.6)-(2.10) we can obtain
(−) Z ϕ3 (tβ )
dQ0β y
=
f (tβ , y)dy = −I ′ (t0 )δµ + O µ2 > 0.
dτβ (+)
ϕ1 (tβ )
(3.5)
(3.6)
It follows from (3.5) and (3.6) that
dβ
dβ
(tβ −) ≥
(tβ +).
dt
dt
We can prove in a similar way that
dα
dα
(tα −) ≤
(tα +).
dt
dt
Thus from Lemma 3.4 there exists a step-type contrast structure solution y(t, µ) of the
problem (1.1)-(1.2) for sufficiently small µ > 0, and the asymptotic formula (3.1) holds.
The proof is completed.
Acknowledgment
F. Xie was supported by the National Natural Science Foundation of China (Nos 10701023,
10871040) and the Fundamental Research Funds for the Central Universities (No 2010B082-1); M. Ni was supported by the Natural Science Foundation of Shanghai (No 10ZR1409200),
and in part by E-Institutes of Shanghai Municipal Education Commission (No N.E03004).
EJQTDE, 2010 No. 62, p. 12
References
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structure type, Mathematical Notes, vol. 42, no. 6, pp. 956-961, 1987.
[2] V. F. Butuzov, A. B. Vasil’eva and N. N. Nefedov, Contrast structures: Theory and
applications,Computational Mathematics and Modeling, vol. 6, no. 4, pp. 231-234,
1995.
[3] A. B. Vasil’eva, V. F. Butuzov and L. V. Kalachev, The Boundary Function Method
for Singular Perturbation Problems, vol. 14 of SIAM Studies in Applied Mathematics,
SIAM, Philadelphia, Pa, USA, 1995.
[4] P. A. Farrell, A. F. Hegarty, J. J. H. Miller, E. O’Riordan and G. I. Shishkin, Singularly perturbed convection-diffusion problems with boundary and weak interior layers,
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[5] J. Grasman and S. Shih, A parabolic singular perturbation problem with an internal
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no. 6, Article ID 065301, 2007.
[7] M. Ni and W. Lin, Singular perturbation with step-type contrast structures, Applied
Mathematics: A Journal of Chinese Universities (Ser.A), vol. 24, no. 3, pp. 290-300,
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[9] J. R. Cannon, The solution of the heat equation subject to the specification of energy,
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[10] N. I. Ionkin, Solution of a boundary value problem in heat conduction theory with
nonlocal boundary conditions, Differential Equations, vol. 13, pp. 294-304, 1977.
[11] F. Nicoud, T. Sch¨onfeld, Integral boundary conditions for unsteady biomedical CFD
applications, Internat. J. Numer. Methods Fluids, vol. 40, pp. 457-465, 2002.
[12] M. Cakir and G. M. Amiraliyev, A finite difference method for the singularly perturbed
problem with nonlocal boundary condition Applied Mathematics and Computation,
vol. 160, pp. 539-549, 2005.
EJQTDE, 2010 No. 62, p. 13
[13] F. Xie and L. Zhang, Non-linear second-order singularly perturbed problem with integral boundary condition, Journal of East China Normal University (Natural Science),
vol. 2010, no. 1, pp. 1-5, 2010.
[14] R. A. Khan, The generalized method of quasilinearization and nonlinear boundary
value problems with integral boundary conditions, Electronic Journal of Qualitative
Theory of Differential Equations, vol. 2003, no. 10, pp. 1-15, 2003.
(Received July 14, 2010)
EJQTDE, 2010 No. 62, p. 14
2010, No. 62, 1-14; http://www.math.u-szeged.hu/ejqtde/
On the Step-type Contrast Structure of a
Second-order Semilinear Differential Equation
with Integral Boundary Conditions
Feng Xie1,3∗, Zhaoyang Jin1 and Mingkang Ni2,3
1
Department of Applied Mathematics, Donghua University
Shanghai 201620, P.R. China
2
Department of Mathematics, East China Normal University
Shanghai 200241, P.R. China
3
Division of Computational Science, E-Institute of Shanghai Universities,
at SJTU, Shanghai 200030, P.R. China
Abstract
In this paper we investigate the step-type contrast structure of a second-order
semilinear differential equation with integral boundary conditions. The asymptotic
solution is constructed by the boundary function method, and the uniform validity
of the formal solution is proved by the theory of differential equalities.
2000 Mathematics Subject Classification. 34B15; 34E05; 34E20.
Key words and phrases. Singular perturbation; contrast structure; integral
boundary conditions.
1. Introduction
We shall consider the existence of contrast structure for the following singularly perturbed
differential equation with integral boundary conditions
µ2
y(0, µ) =
Z
d2 y
= f (t, y),
dt2
1
h1 (y(s, µ))ds,
0
0 < t < 1,
Z 1
y(1, µ) =
h2 (y(s, µ))ds,
(1.1)
(1.2)
0
where µ is a small and positive parameter, and f : [0, 1] × R → R, hi : R → R (i = 1, 2)
are C (2) -functions.
∗
Corresponding author. E-mail: [email protected].
EJQTDE, 2010 No. 62, p. 1
Singularly perturbed boundary value problems arise naturally in various applications,
and have received more and more attention in recent years. Contrast structures, namely
solutions which have internal transition layers, are initially investigated by using the boundary function method by Butuzov and Vasil’eva in 1987 [1], and have recently been one of
the hot topics in singular perturbation problems; see [2-8], for instance. The step-type
contrast structure of the equation (1.2) with two-point boundary conditions y(0, µ) = y 0
and y(1, µ) = y 1 was considered by Butuzov and Vasil’eva [1]. They gave the conditions
which ensure the existence of step-type contrast structure and applied the boundary function method to construct the corresponding asymptotic solution. Recently, Ni and Lin [7]
proved rigorously the uniform validity of asymptotic solution by using Nagumo’s Theorem.
In 2009, Ni and Wang [8] extended the equation (1.2) to higher dimension and studied
the following semilinear singularly perturbed system
µ2 y1′′ = f1 (y1 , y2 , . . . , yn , t),
µ2 y2′′ = f2 (y1 , y2 , . . . , yn , t),
..
.
µ2 yn′′ = fn (y1 , y2 , . . . , yn , t),
subject to the conditions
yk (0, µ) = yk0 ,
yj′ (0, µ) = zj1 ,
k = 1, 2, . . . , n,
k = 1, 2, . . . , n − 1,
yn′ (1, µ) = zn1 ′ .
The authors gave the conditions under which there exists an internal transition layer,
and constructed the uniformly valid asymptotic expansion of a solution with a step-type
contrast structure.
To our knowledge, contrast structures of singularly perturbed problems with integral
boundary conditions have not been investigated. Boundary value problems with integral
boundary conditions have significant applications in thermal conduction [9], semiconductor
problems [10], biomedical science [11], and so on. In [12], Cakir and Amiraliyev studied
the singularly perturbed nonlocal boundary value problem
ε2 y ′′ + εa(t)y ′ − b(t)y = f (t), 0 < t < l, 0 < ε ≪ 1,
Z l1
0
1
g(s)y(s)ds, 0 ≤ l0 < l1 ≤ l,
y(0) = y , y(1) = y +
l0
where y 0 , y 1 are given constants, and a(t) ≥ α > 0, b(t) ≥ β > 0, g(t) and f (t) are
sufficiently smooth functions in [0, 1]. The authors presented a finite difference method for
EJQTDE, 2010 No. 62, p. 2
numerical solutions of the problem which exhibited two boundary layers at t = 0 and t = l.
In [13], Xie and Zhang extended the above problem to general weakly nonlinear singular
perturbation problems with integral boundary conditions by using the boundary function
method.
The present paper is devoted to investigate the existence of step-type contrast structure for the problem (1.1)-(1.2). Integral boundary conditions (1.2) can be viewed as the
generalization of two-point and nonlocal boundary conditions. The boundary function
method and the theory of differential inequalities will be applied to obtain the uniformly
valid asymptotic solution of the problem (1.1) -(1.2). The main difficulty different from
the corresponding two-point boundary value problem lies in that the integral boundary
conditions of the two associated problems are coupled, which will be overcome with the aid
of the property that boundary layer functions decay exponentially.
The remainder of this paper is organized as follows. In section 2 we give some assumptions and construct the formal asymptotic solution of the original problem. In section 3
the uniform validity of formal solution is proved by the theory of differential equalities.
2. Basic Assumptions and Construction of Asymptotic Solution
Let us begin with two basic assumptions.
[H1 ] f : [0, 1] × R → R, hi : R → R (i = 1, 2) are C (2) -functions, and h′i (x) ≥ 0.
[H2 ] The reduced equation f (t, y) = 0 has three isolated solutions ϕi (t) (i = 1, 2, 3) on
[0, 1], satisfying
ϕ1 (t) < ϕ2 (t) < ϕ3 (t),
∂f
(t, ϕi (t)) > 0 (i = 1, 3),
∂y
∂f
(t, ϕ2 (t)) < 0.
∂y
Assumption [H2 ] is a so-called stability condition. It follows from the assumption [H2 ]
that in the phase plane (y, y ′), the equilibria (ϕ1,3 , 0) is a saddle point and (ϕ2 , 0) is a center.
We are interested in the solution of step-type which has a transition from the vicinity of
ϕ1 (t) to that of ϕ3 (t) at some point t = t∗ . That is, for some t∗ ∈ (0, 1) the following limit
holds:
ϕ (t), 0 < t < t∗ ,
1
lim y(t, µ) =
µ→0
ϕ3 (t), t∗ < t < 1.
t = t∗ is called the transition point.
We shall adopt the following strategy which is due to Butuzov and Vasil’eva [1]. We
divide the original problem into the two associated pure boundary layer problems, that is,
EJQTDE, 2010 No. 62, p. 3
the associated left problem
d2 y (−)
= f (t, y (−) ), 0 < t < t∗ ,
dt2
Z t∗
Z 1
(−)
(−)
y (0, µ) =
h1 (y (s, µ))ds +
h1 (y (+) (s, µ))ds,
µ2
0
t∗
y
(−)
∗
∗
(t , µ) = ϕ2 (t ),
and the associated right problem
d2 y (+)
= f (t, y (+) ), t∗ < t < 1,
2
dt
Z t∗
Z 1
(+)
(−)
y (1, µ) =
h2 (y (s, µ))ds +
h2 (y (+) (s, µ))ds,
µ2
0
t∗
y
(+)
∗
∗
(t , µ) = ϕ2 (t ).
Considering that the solution y(t, µ) is smooth at t = t∗ , it follows that
dy (+) ∗
dy (−) ∗
(t , µ) =
(t , µ),
dt
dt
(2.1)
which is the condition determining the position of transition point.
Here the main difficulty different from the corresponding two-point boundary value
problem lies in that the integral boundary conditions of the two associated problems are
coupled. In order to overcome this difficulty, we need to handle these two associated
problems simultaneously, and have the aid of properties of boundary layer functions to
uncouple the boundary conditions.
Let us describe the formal scheme of seeking an asymptotic solution of the problem
(1.1)-(1.2) of the form
y (−) (t, µ) + Π(−) y(τ , µ) + Q(−) y(τ, µ), 0 ≤ t ≤ t∗ ,
0
y(t, µ) =
(2.2)
y (+) (t, µ) + R(+) y(τ1 , µ) + Q(+) y(τ, µ), t∗ ≤ t ≤ 1,
where
(∓)
(∓)
(∓)
y (∓) (t, µ) = y 0 (t) + µy1 (t) + µ2 y 2 (t) + · · ·
are the regular parts of the left problem and the right problem, respectively,
(−)
(−)
(−)
Π(−) y(τ0 , µ) = Π0 y(τ0 ) + µΠ1 y(τ0) + µ2 Π2 y(τ0 ) + · · · ,
τ0 =
t
µ
is the left boundary layer part of the left problem,
(∓)
(∓)
(∓)
Q(∓) y(τ, µ) = Q0 y(τ ) + µQ1 y(τ ) + µ2 Q2 y(τ ) + · · · ,
τ=
t − t∗
µ
EJQTDE, 2010 No. 62, p. 4
are the right boundary layer part of the left problem and the left boundary layer part of
the right problem, respectively, and
(+)
(+)
(+)
R(+) y(τ1 , µ) = R0 y(τ1 ) + µR1 y(τ1 ) + µ2 R2 y(τ1 ) + · · · ,
τ1 =
t−1
µ
is the right boundary layer part of the right problem. We also seek the asymptotic expansion
of the transition point t∗ :
t∗ = t0 + µt1 + µ2 t2 + · · · .
Substituting (2.2) into (1.1) and equating the coefficients in like powers of µ, we get a
(∓)
recurrent sequence of algebraic equations for the functions y i (t) (i = 1, 2, . . .).
(∓)
f t, y 0 (t) = 0,
∂f (∓) (∓)
t, y 0 (t) y 2 (t),
∂y
(∓)
∂f (∓) (∓)
d2 y 2
(±)
t, y 0 (t) y 4 (t) + g4 ,
=
2
dt
∂y
···
(∓)
d2 y 2k−2
∂f (∓) (∓)
(±)
t, y 0 (t) y 2k (t) + g2k ,
=
2
dt
∂y
(∓)
d2 y 0
dt2
(±)
=
(∓)
where g2k are the determined functions of y i (0 ≤ i ≤ 2k − 2). From the assumption
(∓)
[H2 ], the coefficients y i can be obtained recurrently. In particular, we have
(−)
y0
= ϕ1 (t),
(+)
y0
= ϕ3 (t).
For simplicity, we only consider the approximation of first order for the boundary layer
(−)
(−)
series. The left boundary layer functions Π0 y(τ0 , µ) and Π1 y(τ0, µ) satisfy
(−)
d2 Π0 y
(−)
= f 0, ϕ1 (0) + Π0 y ,
2
dτ0
Z t0
Z 1
(−)
(2.3)
Π0 y(0) =
h1 (ϕ1 (s))ds +
h1 (ϕ3 (s))ds − ϕ1 (0),
0
t0
(−)
Π0 y(+∞) = 0,
and
(−)
d2 Π1 y
∂f
(−)
(−)
(−)
=
0,
ϕ
(0)
+
Π
y
Π1 y + ∆1 ,
1
0
dτ02
∂y
Z +∞
Z 0
(−)
(−)
(−)
′
Π1 y(0) =
h1 (ϕ1 (0))Π0 y(s)ds +
h′1 (ϕ1 (t0 ))Q0 y(s)ds
0
−∞
Z +∞
Z 0
(+)
(+)
′
+
h1 (ϕ3 (t0 ))Q0 y(s)ds +
h′1 (ϕ3 (1))R0 y(s)ds,
0
−∞
(−)
Π1 y(+∞)
= 0,
EJQTDE, 2010 No. 62, p. 5
respectively, where
∂f
∂f
(−)
(−)
(−)
(−)
(−)
∆1 =
0, ϕ1 (0) + Π0 y y 0 ′ (0)τ0 + y 1 (0) +
0, ϕ1 (0) + Π0 y τ0
∂y
∂t
∂f
∂f
(−)
(−)
(0, ϕ1 (0)) y 0 ′ (0)τ0 + y 1 (0) −
(0, ϕ1 (0)) τ0 .
−
∂y
∂t
(+)
(+)
Analogously, the right boundary layer functions R0 y(τ1 ) and R1 y(τ1 ) satisfy the
following boundary value problems
(+)
d2 R0 y
(+)
=
f
1,
ϕ
(1)
+
R
y
,
3
0
2
dτ1
Z t0
Z 1
(+)
(2.4)
R0 y(1) =
h2 (ϕ1 (s))ds +
h2 (ϕ3 (s))ds − ϕ3 (1),
0
t0
(+)
R0 y(−∞) = 0,
and
∂f
d2 R1 y
(+)
(+)
(+)
=
1,
ϕ
(1)
+
R
y
R1 y + ∆1 ,
3
0
2
dτ1
∂y
Z +∞
Z 0
(−)
(−)
(+)
′
R1 y (1) =
h2 (ϕ1 (0))Π0 y(s)ds +
h′2 (ϕ1 (t0 ))Q0 y(s)ds
0
−∞
Z +∞
Z 0
(+)
(+)
+
h′2 (ϕ3 (t0 ))Q0 y(s)ds +
h′2 (ϕ3 (1))R0 y(s)ds,
(+)
0
−∞
(+)
R1 y(−∞)
= 0,
respectively, where
∂f
∂f
(+)
(+)
(+)
(+)
(+)
∆1 =
1, ϕ3 (1) + R0 y y 0 ′ (1)τ1 + y 1 (1) +
1, ϕ3 (1) + R0 y τ1
∂y
∂t
∂f
∂f
(+)
(+)
−
(1, ϕ3 (1)) y 0 ′ (1)τ1 + y 1 (1) −
(1, ϕ3 (1)) τ1 .
∂y
∂t
In order to ensure the existence of solutions for the boundary value problems (2.3) and
(2.4), we need the following assumptions. See [3] for their geometrical interpretation.
!
Z t0
(−)
(−)
dΠ0 y
y
dΠ
(−)
0
, let the straight line
=
h1 (ϕ1 (s))ds+
[H3 ] In the phase plane Π0 y,
dτ0
dτ0
0
Z 1
h1 (ϕ3 (s))ds intersect the separatrix entering the saddle (0, ϕ1 (0)) as τ0 → +∞;
t0
[H4 ] In the phase plane
Z
1
!
Z t0
(+)
(+)
dR
y
dR0 y
(+)
0
R0 y,
, let the straight line
=
h2 (ϕ1 (s))ds+
dτ1
dτ1
0
h2 (ϕ3 (s))ds intersect the separatrix entering the saddle (0, ϕ3 (1)) as τ1 → −∞.
t0
EJQTDE, 2010 No. 62, p. 6
Note that t0 in the above assumptions is unknown, which is determined by (2.10).
From the assumptions [H1 ], [H2 ] and [H3 ], we have the following estimates of decay(−)
(−)
(+)
ing exponentially for the boundary layer functions Π0 y(τ0), Π1 y(τ0 ), R0 y(τ1 ) and
(+)
R1 y(τ1 ).
Lemma 2.1 Under the assumptions [H1 ], [H2 ] and [H3 ], the following estimates
dΠ(−) y
0
(−)
≤ c1 exp(−κ1 τ0 ), τ0 ≥ 0,
Π0 y ≤ c1 exp(−κ1 τ0 ),
dτ0
dR(+) y
0
≤ c2 exp(κ2 τ1 ), τ1 ≤ 0,
dτ1
(+)
τ0 ≥ 0; R1 y ≤ c4 exp(κ4 τ1 ), τ1 ≤ 0
(+)
R0 y ≤ c2 exp(κ2 τ1 ),
(−)
Π1 y ≤ c3 exp(−κ3 τ0 ),
hold, where ci and κi (i = 1, 2, 3, 4) are positive constants.
Proof. The proof is essential similar to that of [3], and we omit it here.
Let us now consider the right boundary layer of the left problem and the left boundary
layer of the right problem, that is, the interior layer of the original problem. We rewrite
the equation (1.1) into the equivalent system
µ
dz
= f (t, y),
dt
µ
dy
= z.
dt
(2.5)
Substituting (2.2) into (2.5), and separate the equations according to the scales t and τ ,
we obtain
µ
dz (∓)
= f t, y (∓) (t, µ) ,
dt
µ
dy(∓)
= z (∓) ,
dt
dQ(∓) z
= f t∗ + µτ, y (∓) (t∗ + µτ, µ) + Q(∓) y − f t∗ + µτ, y (∓) (t∗ + µτ, µ) ,
dτ
dQ(∓) y
= Q(∓) z.
dτ
(∓)
(∓)
Therefore, the coefficients Q0 y and Q0 z are determined by the following boundary
value problems
(∓)
dQ0 z
(∓)
=
f
t
,
ϕ
(t
)
+
Q
y
,
0
1,3
0
0
dτ
(∓)
dQ0 y
(∓)
= Q0 z,
(2.6)
dτ
(∓)
Q0 y(0) = ϕ2 (t0 ) − ϕ1,3 (t0 ),
Q(∓) y(∓∞) = Q(∓) z(∓∞) = 0.
0
0
EJQTDE, 2010 No. 62, p. 7
(∓)
(∓)
(∓)
(∓)
(∓)
By the transformations ye(∓) = y 0 (t0 ) + Q0 y, ze(∓) = z 0 (t0 ) + Q0 z = Q0 z, the
problem (2.6) becomes
de
z (∓)
(∓)
=
f
t
,
y
e
,
0
dτ
(∓)
de
y
= ze(∓) ,
dτ
(2.7)
ye(∓) (0) = ϕ2 (t0 ),
ye(∓) (∓∞) = ϕ1,3 (t0 ),
ze(∓) (∓∞) = 0.
It follows from the assumption [H2 ] that there exists a solution of the problem (2.7) for
given t0 . In what follows we will give the condition determining t0 . Integrating the first
two equations in (2.7) we have
Z ye(−) (τ )
Z ye(+) (τ )
(−) 2
(+) 2
ze (τ ) = 2
ze (τ ) = 2
f (t0 , y)dy,
f (t0 , y)dy.
(2.8)
ϕ1 (t0 )
ϕ3 (t0 )
Note that the zero order approximation of the smooth connection condition (2.1) becomes
It follows from (2.8) and (2.9) that
ze(−) (0) = ze(+) (0).
I(t0 ) ≡
Z
(2.9)
ϕ3 (t0 )
f (t0 , y)dy = 0,
(2.10)
ϕ1 (t0 )
which is the equation determining the dominant term t0 of t∗ .
[H4 ] Assume that the equation (2.10) has a root t = t0 with I ′ (t0 ) < 0.
In a similar way, we can also get the expression t1 which is closely related to the
(∓)
(∓)
equations for y (∓) , Q1 y and Q1 z, and the details are omitted here.
(∓)
(∓)
Similar to Lemma 2.1, for the boundary layer functions Qi y and Q1 z (i = 0, 1) we
have the following estimates of decaying exponentially.
Lemma 2.2 Under the assumptions [H1 ], [H2 ] and [H3 ], the following
(−)
(−)
Q0 y(τ ) ≤ c5 exp(κ5 τ ), Q0 z(τ ) ≤ c5 exp(κ5 τ ), τ
(−)
(−)
Q1 y(τ ) ≤ c6 exp(κ6 τ ), Q1 z(τ ) ≤ c5 exp(κ5 τ ), τ
(+)
(+)
Q0 y(τ ) ≤ c7 exp(−κ7 τ ), Q0 z(τ ) ≤ c7 exp(−κ7 τ ),
(+)
(+)
Q1 y(τ ) ≤ c8 exp(−κ8 τ ), Q1 z(τ ) ≤ c8 exp(−κ8 τ ),
estimates
≤ 0,
≤ 0,
τ ≥ 0,
τ ≥0
hold, where ci and κi (i = 5, 6, 7, 8) are positive constants.
EJQTDE, 2010 No. 62, p. 8
3. Existence of Step-type Solution
In this section we will prove the existence of step-type solution for the original problem
and give the estimate for the remainder term.
Theorem 3.3 Under the assumptions [H1 ]−[H4 ], there exists a step-type contrast structure
solution y(t, µ) of the problem (1.1)-(1.2) for sufficiently small µ > 0. Moreover, the
following asymptotic expansion holds
ϕ (t) + Π(−) y(τ ) + Q(−) y(τ ) + O(µ), 0 ≤ t < t + µt ;
1
0
0
1
0
0
y(t, µ) =
(3.1)
(+)
(+)
ϕ3 (t) + R y(τ1 ) + Q y(τ ) + O(µ), t0 + µt1 < t ≤ 1.
0
0
In order to prove Theorem 3.3 we need the following lemma which is a slight modification
of Theorem 2.2 in [14].
Lemma 3.4 Assume that the assumption [H1 ] holds and the continuous functions α(t, µ)
and β(t, µ) are of C (2) class on the intervals (0, tα ) ∪ (tα , 1) and (0, tβ ) ∪ (tβ , 1), respectively,
having the following properties
(1) α(t, µ) ≤ β(t, µ),
t ∈ [0, 1];
2
d2 α
2d β
≥
f
(t,
α),
t
∈
(0,
t
)
∪
(t
,
1);
µ
≤ f (t, β),
α
α
dt2
dt2
Z 1
Z 1
(3) α(0, µ) ≤
h1 (α(s, µ))ds, α(1, µ) ≤
h2 (α(s, µ))ds,
Z0 1
Z 01
β(0, µ) ≥
h1 (β(s, µ))ds, β(1, µ) ≥
h2 (β(s, µ))ds;
(2) µ2
0
(4)
dα
dα
(tα −) ≤
(tα +),
dt
dt
t ∈ (0, tβ ) ∪ (tβ , 1);
0
dβ
dβ
(tβ −) ≥
(tβ +),
dt
dt
where tα , tβ ∈ (0, 1). Then, there exists a solution y(t, µ) of the problem (1.1)-(1.2) such
that
α(t, µ) ≤ y(t, µ) ≤ β(t, µ), t ∈ [0, 1].
Remark 3.5 The functions α(t, µ) and β(t, µ) satisfying the above conditions are called
lower and upper solutions of the problem (1.1)-(1.2), respectively.
Remark 3.6 It is requested in [14] that the functions α(t, µ), β(t, µ) ∈ C (2) [0, 1]. Here we
only need the functions α(t, µ) and β(t, µ) to be piecewise C (2) - smooth and an additional
condition (4). It should be noted that the proof of Lemma 3.4 has no essential difference
from that of Theorem 2.2 in [14], but some slight modifications.
EJQTDE, 2010 No. 62, p. 9
Proof of Theorem 3.3. We select the auxiliary functions
ϕ (t) + Π(−) y(τ ) + µΠ(−) y(τ ) + Q(−) y(τ ) + µQ(−) y(τ ) − γµ, 0 ≤ t ≤ t ,
1
0
0
α
α
α
0
1
0α
1α
α(t, µ) =
(+)
(+)
(+)
(+)
ϕ3 (t) + R y(τ1 ) + µR y(τ1 ) + Q y(τα ) + µQ y(τα ) − γµ, tα ≤ t ≤ 1,
0
1
0α
1α
and
ϕ (t) + Π(−) y(τ ) + µΠ(−) y(τ ) + Q(−) y(τ ) + µQ(−) y(τ ) + γµ, 0 ≤ t ≤ t ,
1
0
0
β
β
β
0
1
0β
1β
β(t, µ) =
(+)
(+)
(+)
(+)
ϕ3 (t) + R y(τ1 ) + µR y(τ1 ) + Q y(τβ ) + µQ y(τβ ) + γµ, tβ ≤ t ≤ 1,
0
1
1β
0β
where
tα = t0 + µδ,
tβ = t0 − µδ,
τα =
t − tα
,
µ
τβ =
t − tβ
,
µ
(∓)
(∓)
while γ, δ are sufficiently large positive parameters. The functions Q0α y and Q1α y satisfy
respectively the following boundary value problems
d2 Q0α y
(∓)
=
f
τ
,
ϕ
(t
)
+
Q
y
,
α
1,3
α
0α
dτα2
(∓)
(∓)
Q0α y(0) = ϕ2 (tα ) − ϕ1,3 (tα ),
(∓)
Q0α y(∓∞) = 0,
and
(∓)
d2 Q1α y
∂f
(∓)
(∓)
(∓)
=
τ
,
ϕ
(t
)
+
Q
y
Q1α y + ∆1α − ω exp(±κ0 τα ),
α
1,3 α
0α
dτα2
∂y
(∓)
(∓)
Q1α y(0) = ϕ′2 (t0 ) − ϕ′1,3 (t0 ) t1 , Q1α y(∓∞) = 0,
where ω, κ0 are positive constants and
(∓)
∆1α =
∂f
∂f
(∓)
(∓)
τα , ϕ1,3 (tα ) + Q0α y ϕ′1,3 (tα )τα +
τα , ϕ1,3 (tα ) + Q0α y τα .
∂y
∂t
(∓)
(∓)
The functions Q0β y and Q1β y are also determined by the corresponding boundary value
problems.
To verify the conditions in Lemma 3.4 we divide the interval [0, 1] into five subintervals
[0, tβ /2], [tβ /2, tβ ], [tβ , tα ], [tα , (tα + 1)/2] and [(tα + 1)/2, 1].
Let us first check the condition (1). On the intervals [0, tβ /2] and [(tα +1)/2, 1], β(t, µ)−
α(t, µ) = 2γµ + EST > 0, where EST denotes exponentially small
terms. On the interval
(+)
(−)
(+)
(−)
[tβ , tα ], β(t, µ)−α(t, µ) = ϕ3 (t)−ϕ1 (t)+Q0β y(τβ )−Q0α y(τα )+µ Q1β y(τβ ) − Q1α y(τα ) +
2γµ + EST > 0. On the interval [tβ /2, tβ ],
(−)
(−)
(−)
(−)
β(t, µ) − α(t, µ) = 2γµ + Q0β y(τβ ) − Q0α y(τα ) + µ Q1β y(τβ ) − Q1α y(τα ) > 0,
EJQTDE, 2010 No. 62, p. 10
where we have used the following formula
(−)
(−)
(−)
Q0β y(τβ ) − Q0α y(τα ) =
dQ0 y ∗
(τ )2δ > 0,
dτ
τα ≤ τ ∗ ≤ τβ .
Similarly, on the interval [tα , (tα + 1)/2],
(+)
(+)
(+)
(+)
β(t, µ) − α(t, µ) = 2γµ + Q0β y(τβ ) − Q0α y(τα ) + µ Q1β y(τβ ) − Q1α y(τα ) > 0.
Next we check the condition (2). Here we only verify the condition (2) on the interval
[tβ /2, tβ ], which are similar on other subintervals.
(−)
(−)
d2 Q0β y
d2 Q1β y
β
2 ′′
µ 2 − f (t, β) = µ ϕ1 (t) +
+µ
dt
dτ 2
dτ 2
(−)
(−)
−f t, ϕ1 (t) + γµ + Q0β y + µQ1β y + EST.
2d
2
We rewrite f in f = fe(µ) + f (µ), where
(−)
(−)
fe(µ) = f τ µ, ϕ1 (τ µ) + γµ + Q0β y + µQ1β y − f (τ µ, ϕ1 (τ µ) + γµ)
!
(−)
(−)
d2 Q0β y
d2 Q1β y
2
=
+
µ
−
ω
exp(κ
τ
)
+
O
µ
,
0
dτ 2
dτ 2
(3.2)
(3.3)
and
f(µ) = f (t, ϕ1 (t) + γµ) =
Inserting (3.3) and (3.4) into (3.2) we have
µ2
∂f
(t, ϕ1 (t)) γµ + O µ2 .
∂y
(3.4)
d2 β
∂f
2
−
f
(t,
β)
=
−
(t,
ϕ
(t))
γµ
+
µω
exp(κ
τ
)
+
O
µ
< 0.
1
0
dt2
∂y
In a similar way we can show that
µ2
d2 α
≥ f (t, α),
dt2
We now show that
β(0, µ) ≥
t ∈ (0, tα ) ∪ (tα , 1).
Z
1
h1 (β(s, µ))ds.
0
It follows from the construction of asymptotic solution that
Z 1
β(0, µ) −
h1 (β(s, µ))ds
0
Z tβ
Z 1
(−)
(−)
= ϕ1 (0) + Π0 y(0) + µΠ1 y(0) + γµ − h1 (β(s, µ))ds − h1 (β(s, µ))ds + EST
0
tβ
EJQTDE, 2010 No. 62, p. 11
=
(−)
µΠ1 y(0)
−µ
Z
−µ
t0
(h1 (ϕ1 (s)) − h1 (ϕ3 (s))) ds − µ
t0 −δµ
0
t
δ− µ0
Z
Z
+ γµ +
(−)
h′1 (ϕ1 (t0 ))Q0 y(s)ds − µ
0
−
(+)
1−t0
−δ
µ
Z
t0 −δµ
µ
0
1−t0
−δ
µ
0
h′1 (ϕ3 (1))R0 y(s)ds + O µ2
= γµ + O (µ) > 0,
Z
(−)
h′1 (ϕ1 (0))Π0 y(s)ds
(+)
h′1 (ϕ3 (t0 ))Q0 y(s)ds
provided that γ is large enough.
Other inequalities in the condition (3) can be proved analogously.
Finally, let us check the condition (4).
(−)
dβ
dβ
dβ
µ (tβ −) − µ (tβ +) = µ
dt
dt
dt (+)
(−)
(−)
dQ1β y
dQ0β y
dQ0β y
′
= µy 0 (tβ ) +
+µ
=
+ O(µ).
dτβ
dτβ (+)
dτβ (+)
From the process similar to (2.6)-(2.10) we can obtain
(−) Z ϕ3 (tβ )
dQ0β y
=
f (tβ , y)dy = −I ′ (t0 )δµ + O µ2 > 0.
dτβ (+)
ϕ1 (tβ )
(3.5)
(3.6)
It follows from (3.5) and (3.6) that
dβ
dβ
(tβ −) ≥
(tβ +).
dt
dt
We can prove in a similar way that
dα
dα
(tα −) ≤
(tα +).
dt
dt
Thus from Lemma 3.4 there exists a step-type contrast structure solution y(t, µ) of the
problem (1.1)-(1.2) for sufficiently small µ > 0, and the asymptotic formula (3.1) holds.
The proof is completed.
Acknowledgment
F. Xie was supported by the National Natural Science Foundation of China (Nos 10701023,
10871040) and the Fundamental Research Funds for the Central Universities (No 2010B082-1); M. Ni was supported by the Natural Science Foundation of Shanghai (No 10ZR1409200),
and in part by E-Institutes of Shanghai Municipal Education Commission (No N.E03004).
EJQTDE, 2010 No. 62, p. 12
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