Appl. Math. Mech. -Engl. Ed.     2014, Vol. 35 Issue (5) : 655–666     PDF       
http: //dx. doi. org/10.1007/s10483-014-1819-7
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Article Information

Ai- feng WANG, Ming-kang NI 2014.
Contrast structure for singular singularly perturbed boundary value problem
Appl. Math. Mech. -Engl. Ed., 35 (5) : 655–666
http: //dx. doi. org/10.1007/s10483-014-1819-7

Article History

Received 2013-08-27;
in final form 2013-11-16
Contrast structure for singular singularly perturbed boundary value problem
Ai- feng WANG1 , Ming-kang NI2       
1 School of Mathematical Science, Huaiyin Normal University, Huaian 223001, Jiangsu Province, P. R. China;
2 Department of Mathematics, East China Normal University, Shanghai 200062, P. R. China
ABSTRACT:The step-type contrast structure for a singular singularly perturbed problem is shown. By use of the method of boundary function, the formal asymptotic expansion is constructed. At the same time, based on sewing orbit smooth, the existence of the steptype solution and the uniform validity of the asymptotic expansion are proved. Finally, an example is given to demonstrate the effectiveness of the present results.
Keywordscontrast structure        singular singularly perturbation        asymptotic expansion        boundary function       

1 Introduction

Singular singularly perturbed problems can be found in chemical kinetics[1] ,semiconductor device[2, 3] ,Poisson-Nernst-Planck system[4],etc.,and have gained many mathematicians’ attention. In the work by Vasil'eva and Butuzov[3],the initial value problem and a special structure boundary value problem of singular singularly perturbed systems are considered by the boundary function method. A general singular singularly perturbed boundary value problem was examined by Schmeiser[2] using the same method as Vasil’eva. Within the framework of geometric singulary perturbation theory,Liu[4] studied the steady-state Poisson-Nernst-Planck system. There were many other studies on the singular singularly perturbed boundary value problem[5, 6, 7, 8] ,which considered the phenomenon of the boundary layer by the boundary function method or numerical method. However,the study on the contrast structure for the singular singularly perturbed problem is seldomly discussed in a relevant literature. When the reduced system has several isolate roots and the solution of the original problem approaches different reduced roots in different time,the contrast structure occurs. The contrast structure in singularly perturbed problems is mainly classified as a step-type contrast structure[9, 10, 11, 12, 13, 14, 15, 16, 17] or a spike-type contrast structure[18, 19] ,which corresponds to the heteroclinic orbit or homoclinic orbit in their phase space. A step-type contrast structure problem is only concerned in this paper . Its fundamental characteristic is that there is a t*(or multiple t*) within the domain of interest,which is called as an internal transition point. The position of t* is unknown in advance and needs to be determined thereafter. In the neighborhood of t*,the solution y(t,u) will have an abrupt structure change.

A singular perturbed boundary value problem

is studied[5] ,where a(x,y,t) keeps the sign unchanged.

Let μdy/dt = z. Then,(1) can be written as an equivalent system of the form

Denote u = (x,y,z)T,v = (y,z)T,u( ∓)(t) = (x( ∓)(t),y( ∓)(t),z( ∓)(t))T,where z( ∓)(t) ≡ 0,and F(u,t,μ) = (z,a(x,y,t)z + μg(x,y,t)T. The Jacobian matrix of F with respect to v is

where a( ∓)(t) = a(x( ∓)(t),y( ∓)(t),t). This shows that (1) is a singular singularly perturbed problem with a slow variable x. Furthermore,there exists a matrix

such that

Further study shows that,if a(x,y,t) has different signs as t ∈ [0,1],this problem will have a construct structure. In this paper,we discuss the contrast structure for (1) with the boundary conditions given by

2 Construction of asymptotic expansion

H1 Suppose that a(x,y,t),g(x,y,t),and f(x,y,t) are sufficiently smooth for (x,y,t) in D,where D is a bounded domain in the (x,y,t) space.

H2 For each t ∈ [0,1],suppose that the reduced equation

has a solution x ( − ) (t),y ( − ) (t) which is satisfied with x ( − ) (0) = x0,y ( − )(0) = y0 and a solution x ( + ) (t),y ( + ) (t) which is satisfied with x ( + ) (1) = x ( − ) (1),y ( + ) (1) = y1.

Moreover, a(x ( − ) (t),y( − )(t),t) > 0,a(x ( + ) (t),y ( + ) (t),t) < 0,t ∈ [0,1].

Remark 1 If H1 and H2 hold,there maybe exists a contrast structure for Problems (1)-(2), but the phenomena of boundary layers cannot occur.

Let t* be the internal transition point which has the formal asymptotic expansion of the following form:

where tj (j = 0,1,· · · ) are constants to be determined. Consider the following two problems.

The left problem (t ∈ [0,t*]) is

The right problem (t ∈ [t*,1]) is

where and x(+) (1,μ) = x(−) (1,t*,μ).

Suppose that the asymptotic solution of (1) and (2) is composed of these two parts. In terms of the remark and the boundary function method,the formal asymptotic expressions of the left problem and the right problem are given by

and

respectively,where are the coefficients of regular terms,are the coefficients of internal transition terms,and

By (6) and (8),we know that the component y(t,μ) is continuous at t*. In order to get a smooth solution with a step-type contrast structure from y(− )(t) to y(+)(t),the following connection condition

would hold.

In terms of the boundary function method,the zeroth-order regular terms satisfy the following system:

and

By H2,. Furthermore,.

For the high order terms (j = 1,2,· · · ),we have the equations and their boundary conditions as follows:

where

are determined functions. Here ,

Obviously,(12) and (13) are initial value problems. Therefore, exist.

Next,we give the equations and their conditions for determining u as follows:

By x(∓∞) = 0,we obtain x(τ) ≡ 0. (14) can be transformed into the following system:

Let and . Then,(16) and (15) can be transformed into

By (17),we get

Integrating it from 0 to τ,we obtain

Let τ → ∓∞. Then,

Let

H3 Suppose that I(t0) = 0 is solvable with respect to t0 and I'(t0) ≠ 0.

To solve (16),we rewrite (16) as an equivalent system of the form

where

which is satisfied with

By using the linear transformation

(21)∓ can be written as the diagonal form

where

We obtain that (23) satisfying ξ(∓) ( ∓ ∞) = 0 and η(∓) ( ∓ ∞) = 0 is equivalent to the integral equations

Replaced by y and z,(24) becomes

It follows from (22) and the successive approximation method that there is a unique solution

to (25) fo r s ufficiently s ma ll Q (0∓)z(0). Thus,

(27) is an analytical representation of a stable invariant manifold of (21) ,and (27)+ is an analytical representation of a stable invariant manifold of (21)+. Obviously,

By the inverse function theorem,(27) ∓ has a unique inverse function z(0) = P( ∓) ( y(0)) in the neighborhood of y(0) = 0. Let Ω⊂ R be the maximum domain of P(−) and Ω+ ⊂ R be the maximum domain of P(+) .

H4 Suppose that ,and.

Lemma 1 Suppose that H1−H4 hold. Then,u(τ) exist and satisfy the following in equalities:

where C and γ are positive constants.

For the high order terms (j = 1,2,· · · ),we have the equations and their boundary conditions as follows:

and

where ,and take their values at,and take their values at . are the known functions depending on (i = 0,1,· · · ,j − 1),and they satisfy the exponential decay. Integrating (28),we can obtain . are known functions depending only on (i = 0,1,· · · ,j) and z.

Problems (29) and (30) are the linear boundary value problems. Therefore,it is not difficult to verify that their solutions are

where

Φ(∓) (τ) and Ψ(∓) (τ) are the solutions of the matrix initial value problems

and

respectively. It is easy to prove H(∓) (τ) →a(∓) (t0) as τ → ∓∞. Moreover,H(∓) (τ) satisfy the Reccati equations,i.e.,

which plays a key role in matrix diagonalizations. It is therefore to get the estimate

where k is a positive constant.

Lemma 2 Suppose that H1−H4 hold. Then, (j = 1,2,· · · ) exist and satisfy the following in equalities:

where C and γ are positive constants.

So far,we have already determined ,but only contained the unknown tj (j = 1,2,· · · ).

By ,we get

Substituting them into

by H3,we have the coefficient of t1

Then,t1 is solvable. Similarly,tj can be calculated. So far,all the coefficients of the internal transition terms are determined. 3 Existence of step - type contrast structure

From Ref. [1],we obtain the existence of the solution for the left problem. It has the following asymptotic expansion:

Similarly,the solution to the right problem exists,and it has the following asymptotic expansion:

where

It is noted that t* is regarded as a parameter this time. It is not necessarily like the expansion (4) of μ,where satisfy the equations and the boundary value conditions similar to (14),(16),and (28)-(30),respectively. Here,we need to change t0 into t*.

Let

and

Then,

In (3 3 ),if we ta ke δ sufficiently large,μ sufficiently small,then,when δ takes different signs, the right side of (33) has different signs. Thus,there exists such that H(t,μ) = 0. That is,

Thus,we have obtained a step-type contrast structure at the neighborhood of t with an internal transition layer as follows:

In summary,we have the following theorem.

Theorem 1 Suppose that H1−H4 hold. Then,there exists μ0 > 0 such that Problem s (1) and (2) have a step- type contrast structure as 0 < μ < μ0. Moreover,the following uniformly valid asymptotic expansion holds:

4 Example

In this section,an example is presented for how to construct a zeroth-order asymptotic solution with a step-like contrast structure.

The system is given by

According to the algorithm of this paper,we obtain that the reduced system

has a solution x( −)(t) = 12t2 − t + 1,y( −)(t) = t − 1,t ∈ [0,1],and the reduced system

has a solution x(+)(t) = 12t2 − t0 + 1,y(+)(t) = t,t ∈ [0,1].

Thus,

Obviously,−y (−)0(t) = 1 − t > 0,−y(+)0 (t) = −t <0. Thus,H2 holds. By (20),we get

which is the equation to determine t0. It is easy to obtain that t0 = 1/2. Then,H3 holds.

Moreover,x(τ) ≡ 0. For the zeroth-order terms y(τ) and z(τ),we have the equations and their boundary conditions as follows:

and

Thus,

where

Obviously,as long as y0* ≠ ± 1/2,y(τ) and z(τ) converge to zero exponentially.
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