Open access peer-reviewed chapter

# Perturbed Differential Equations with Singular Points

By Keldibay Alymkulov and Dilmurat Adbillajanovich Tursunov

Submitted: November 4th 2016Reviewed: February 13th 2017Published: June 14th 2017

DOI: 10.5772/67856

## Abstract

Here, we generalize the boundary layer functions method (or composite asymptotic expansion) for bisingular perturbed differential equations (BPDE that is perturbed differential equations with singular point). We will construct a uniform valid asymptotic solution of the singularly perturbed first-order equation with a turning point, for BPDE of the Airy type and for BPDE of the second-order with a regularly singular point, and for the boundary value problem of Cole equation with a weak singularity.A uniform valid expansion of solution of Lighthill model equation by the method of uniformization and the explicit solution—this one by the generalization method of the boundary layer function—is constructed. Furthermore, we construct a uniformly convergent solution of the Lagerstrom model equation by the method of fictitious parameter.

### Keywords

• turning point
• singularly perturbed
• bisingularly perturbed
• Cauchy problem
• Dirichlet problem
• Lagerstrom model equation
• Lighthill model equation
• Cole equation
• generalization boundary layer functions

## 1. Preliminary

### 1.1. Symbols O,o,~. Asymptotic expansions of functions

Let a function f(x)and ϕ(x)be defined in a neighborhood of x=0.

Definition1. If limx0f(x)ϕ(x)=M, then write f(x)=O(ϕ(x)),x0, and Mis constant.

If limx0f(x)ϕ(x)=0, then write f(x)=o(ϕ(x)),x0.

If limx0f(x)ϕ(x)=1, then write f(x)~ϕ(x),x0.

Definition2. The sequence {δn(ε)}, where δn(ε)defined in some neighborhood of zero, is called the asymptotic sequence in ε0, if

limε0δn+1(ε)δn(ε)=0,n=1,2,

For example.

{εn},{(1/ln(1/ε))n},{(εln(1/ε))n}.

Note 1. Everywhere below εdenotes a small parameter.

Definition3. We say that f(x)function can be expanded in an asymptotic series by the asymptotic sequence {ϕn(x)},x0, if there exists a sequence of numbers {fn}and has the relation

f(x)=k=0nfkϕk(x)+O(ϕn+1(x)),x0,

and write

f(x)~k=0fkϕk(x),x0.

### 1.2. The asymptotic expansion of infinitely differentiable functions

Theorem (Taylor (1715) and Maclaurin (1742)). If the function f(x)Cin some neighborhood of x=0, then it can be expanded in an asymptotic series for the asymptotic sequence {xn}, i.e.,

f(x)~n=1fnxn, where fn=f(n)(0)/n!.

Thus, the concept of an asymptotic expansion was given for the first time by Taylor and Maclaurin,although an explicit definition was given by Poincaré in 1886.

### 1.3. The asymptotic expansion of the solution of the ordinary differential equation

Consider the Cauchy problem for a normal ordinary differential equation

y'(x)=f(x,y,ε),y(0)=0.E1

The function f(x,y,ε)is infinitely differentiable on the variables x,y,εin some neighborhood O(0,0,0). It is correct next.

Theorem 1. The solution y=y(x,ε)of problem (1) exists and unique in some neighborhood point O(0,0,0)and y(x,ε)C, for small x,ε.

Corollary. The solution of problem (1) can be expanded in an asymptotic series by the small parameter ε, i.e.,

y(x,ε)=k=1εkyk(x).E2

Here and below, the equality is understood in an asymptotic sense.

Note 2. Theorem 1 for the case when f(x,y,ε)is analytical was given in [1] by Duboshin.

Note 3. This theorem 1 is not true if f(x,y,ε)is not smooth at ε. For example, the solution of a singularly perturbed equation

εy'(x)=y(x),y(0)=a

function y(x)=aex/εand is not expanded in an asymptotic series in powers of ε, because here f(x,y,ε)=y(x)/εand fhave a pole of the first order with respect to ε.

Note 4. The series 2 is a uniform asymptotic expansion of the function y(x)in a neighborhood of x=0.

For example. Series

y(x,ε)=1+εx1+(εx1)2++(εx1)n+

It is not uniform valid asymptotic series on the interval [0, 1], but it is a uniform valid asymptotic expansion of the segment [εα,1], where 0<α<1.

### 1.4. Singularly perturbed ordinary differential equations

We divide such equations into three types:

1. Singular perturbations of ordinary differential equations such as the Prandtl-Tikhonov [256], i.e., perturbed equations that contain a small parameter at the highest derivative, i.e., equations of the form

y'(x)=f(x,y,ε),y(0)=0,εz'(x)=g(x,y,ε),z(0)=0,

where f,gare infinitely differentiable in the variables x,y,εin the neighborhood of O(0,0,0). It is obvious that unperturbed equation (ε=0)

y0'(x)=f(x,y,0),0=g(x,y,0)

is a first order.

Definition 4. Singularly perturbed equation will be called bisingulary perturbed if the corresponding unperturbed differential equation has a singular point, or this one is an unbounded solution in the considering domain.

For example

1. Equation εy'(x)=y(x)is a singularly perturbed ordinary differential equation.

2. Equation Vander Pol

εy''(x)+(1y2(x))y'(x)+y(x)=0.

It is a bisingularly perturbed ordinary differential equation with singular points, if y(x)=±1.

2. εy'(x)xy(x)=1,x[0,1]is a bisingularly perturbed equation, because the unperturbed equation has an unbounded solution y0(x)=x1.

3. εy''(x)xy(x)=1,x[0,1]is a bisingularly perturbed equation also.

• Singularly perturbed differential equations such as the Lighthill’s type [5769], in which the order of the corresponding unperturbed equation is not reduced, but has a singular point in the considering domain.

For example, a Lighthill model equation

(x+εy(x))y'(x)+p(x)y(x)=r(x),y(1)=a

where x[0,1], p(x),r(x)C[0,1]. For unperturbed equation

xy'0(x)+p(x)y0(x)=r(x),

point x=0is a regular singular point.

• A singularly perturbed equation with a small parameter is considered on an infinite interval. For example, the Lagerstrom equation [7081]

y''(x)+nx1y'(x)+y(x)y'(x)=β(y'(x))2,y(ε)=0,y()=1.

where 0<βis a given number and nis the dimension space.

Remark. The division into such classes is conditional, because singularly perturbed equation of Van der Pol in the neighborhood of points y=±1leads to an equation of Lighthill type [2, 3].

• ### 1.5. Methods of construction of asymptotic expansions of solutions of singularly perturbed differential equations

1. The method of matching of outer and inner expansions [13, 19, 28, 29, 37, 49] is the most common method for constructing asymptotic expansions of solutions of singularly perturbed differential equations. Justification for this method is given by Il’in [22]. However, this method is relatively complex for applied scientists.

2. The boundary layer function method (or composite asymptotic expansion)dates back to the work of many mathematicians. For the first time, this method for a singularly perturbed differential equations in partial derivatives is developed by Vishik and Lyusternik [52] and for nonlinear integral-differential equations (thus for the ordinary differential equations) Imanaliev [24], O’Malley (1971) [38], and Hoppenstedt (1971) [42].

It should be noted that, for the first time, the uniform valid asymptotic expansion of the solution of Eq. (5) is constructed by Vasil’eva (1960) [50] after Wasow [69] and Sibuya in 1963 [68] by the method of matching.

This method is constructive and understandable for the applied scientists.

3. The method of Lomov or regularization method [33] is applied for the construction of uniformly valid solutions of a singularly perturbed equation and will apply Fredholm ideas.

4. The method WKB or Liouville-Green method is used for the second-order differential equations.

5. The method of multiple scales.

6. The averaging method is applicable to the construction of solutions of a singularly perturbed equation on a large but finite interval.

Here, we consider a bisingularly perturbed differential equations and types of equations of Lighthill and Lagerstrom.

Here, we generalize the boundary layer function method for bisingular perturbed equations. We will construct a uniform asymptotic solution of the Lighthill model equation by the method of uniformization and construct the explicit solution of this one by the generalized method of the boundary layer functions.

Furthermore, we construct a uniformly convergent solution of the Lagerstrom model equation by the method of fictitious parameter.

## 2. Bisingularly perturbed ordinary differential equations

### 2.1. Singularly perturbed of the first-order equation with a turning point

Consider the Cauchy problem [5]

εy'(x)+xy(x)=f(x),0<x1,y(0)=a,E3

where f(x)C[0,1], f(x)=k=0fkxk, fk=f(k)(0)/k!,f00;ais the constant

Explicit solution of the problem (3) has the form: y(x)=aex2/2ε+1ε0xe(s2x2)/2εf(s)ds.

The corresponding unperturbed equation (ε=0)

xy˜(x)+f(x)=0,

has a solution y˜(x)=f(x)/x, which is unbounded at x=0.

If you seek a solution to problem (1) in the form

y(x)=y0(x)+εy1(x)+ε2y2(x)+,E4

then

y0(x)=f(x)x~f0x1,x0,E9000
y1(x)=x1y'0(x)~f0x3,x0,
y2(x)=x1y'1(x)~3f0x5,x0,
y3(x)=x1y'2(x)~35f0x5,x0,
yn(x)=x1y'n1(x)~35(2n1)f0x(2n+1),x0,

and a series of Eq. (4) is asymptotic in the segment (ε,1], and the point x0=ε=μis singular point of the asymptotic series of Eq. (4). Therefore, the solution of problem (3) we will seek in the form

y(x)=μ1π1(t)+Y0(x)+π0(t)+μ(Y1(x)+π1(t))+μ2(Y2(x)+π2(t))+,μ0,E5

where Yk(x)C()[0,1],πk(t)C()[0,μ1],x=μtand boundary layer functions πk(t)decreasing by power law as t, that is,πk(t)=O(tm),t,mN.

Substituting Eq. (5) into Eq. (3), we obtain

π'1(t)+μ2Y'0(x)+μπ'0(t)+μ3Y'1(x)+μ2π'1(t)+μ4Y'2(x)+μ3π'2(t)+μ5Y'3(x)+μ4π'3(t)++xY0(x)+μxY1(x)+μ2xY2(x)+μ3xY3(x)++tπ1(t)+μtπ0(t)+μ2tπ1(t)+μ3tπ2(t)+μ4tπ3(t)+=f(x).E6

The initial conditions for the functions πk1(t),k=0,1,we take in the next form

π1(0)=0,π0(0)=aY0(0),πk(0)=Yk(0),k=1,2,

From Eq. (6), we have

μ0:π'1(t)+tπ1(t)+xY0(x)=f(x),E7.-1
μ1:π'0(t)+tπ0(t)+xY1(x)=0,E7.0
μk+1:π'k(t)+tπk(t)+xYk+1(x)+Y'k1(x)=0,k=1,2,E7.k

To Y0(x)function has been smooth, and we define it from the equation

xY0(x)=f(x)f0Y0(x)=(f(x)f0)/x,

and then from Eq. (7.−1), we have obtained the equation

π'1(t)+tπ1(t)=f0.

Therefore

π1(t)=f0et2/20tes2/2dsC[0,μ1],

Obviously, this function bounded and is infinitely differentiable on the segment [0,μ1], and

π1(t)=f0t(1+1t2+3t4+),t.

This asymptotic expression can be obtained by integration by parts the integral expression for π1(t).

Eq. (7.0) define Y1(x)and π0(t). Let Y1(x)0, then

π'0(t)+tπ0(t)=0,π0(0)=af1.

Hence, we find

π0(t)=(af1)et2/2.

From Eq. (7c) for k=1, we have

π'1(t)+tπ1(t)+xY2(x)+Y'0(x)=0.

Let xY2(x)=Y'0(0)Y'0(x), then π'1(t)+tπ1(t)=Y'0(0).

From these, we get

Y2(x)=(Y'0(0)Y'0(x))/x,π1(t)=f2et2/20tes2/2dsC[0,μ1],

and

π1(t)=f2t(1+1t2+3t4+),t∞.

From Eq. (7c) for k=2, we have

π'2(t)+tπ2(t)+xY3(x)+Y'1(x)=0orπ'2(t)+tπ2(t)+xY3(x)=0.

Let Y3(x)0, then

π'2(t)+tπ2(t)=0,π2(0)=Y2(0)=2f3.

From this, we get

π2(t)=2f3et2/2.

Analogously continuing this process, we determine the others of the functions Yk(x),πk(t).

In order to show that the constructed series of [Eq. (5)] is asymptotic series, we consider remainder term Rm(x)=y(x)ym(x),

where ym(x)=1μπ1(t)+Y0(x)+π0(t)+μ(Y1(x)+π1(t))++μm(Ym(x)+πm(t)).

For the remainder term Rm(x), we obtain a problem:

εRm(x)+xRm(x)=μm+2Y'm(x),0<x1,Rm(0)=0.E8

We note that if mis odd, then Y'm(x)0.

The problem (8) has a unique solution

Rm(x)=μmex2/2ε0xY'm(s)es2/2εds,

and from this, we have Rm(x)=O(μm),μ0,x[0,1].

### 2.2. Bisingularly perturbed in a homogenous differential equation of the Airy type

Consider the boundary value problem for the second-order ordinary in a homogenous differential equation with a turning point

εy''(x)xy(x)=f(x),x(0,1),E9
y(0)=0,y(1)=0.E10

where f(x)=k=0fkxk,x0,fk=f(k)(0)/k!,f00.

Note 5. It is the general case of this one was considered in Ref. [8, 4547].

Without loss of generality, we consider the homogeneous boundary conditions, since y(0)=a,y(1)=b,a2+b20,using transformation

y(x)=a+(ba)x+z(x),

If the asymptotic solution of the problems (9)–(10) we seek in the form

y(x)=y0(x)+εy1(x)+ε2y2(x)+…,E11

then we have

y0(x)=f(x)x~f0x1,x0,
y1(x)=x1y''0(x)~12f0x4,x0,
y2(x)=x1y''1(x)~1245f0x7,x0,
y3(x)=x1y''2(x)~124578f0x10,x0,
yn(x)=x1y''n1(x)~124578(3n2)(3n1)f0x(3n+1),0<n,x0,

and the series (11) is asymptotic in the segment (ε3,1]. The point x0= ε3=μis singular point of asymptotic series (11).

The solution of problems (9) and (10) will be sought in the form

y(x)=μ1π1(t)+k=0μk(Yk(x)+πk(t))+k=0λkwk(η),E12

where t=x/μ,μ=ε3, η=(1x)/λ,λ=ε. Here, Yk(x)C[0,1], πk(t)C[0,1/μ]is boundary layer function in a neighborhood of t=0and decreases by the power law as t, and the function wk(t)C[0,1/λ]is boundary function in a neighborhood of η=0and decreases exponentially as η.

Substituting Eq. (12) in Eq. (9), we get

k=0μk(πk1(t)tπk1(t))+k=0μk+3Yk(x)xk=0μkYk(x)=f(x)E13
k=0λk(w''k(η)(1λη)wk(η))=0.E14

From Eq. (13), we have

μ0:π1(t)tπ1(t)xY0(x)=f(x),E15.-1
μ1:π0(t)tπ0(t)xY1(x)=0,E15.0
μ2:π1(t)tπ1(t)xY2(x)=0,E15.1
μ3:π2(t)tπ2(t)+Y0(x)xY3(x)=0,E15.2
μk:πk1(t)tπk1(t)+Yk3(x)xYk(x)=0,k>3,E15.k

Boundary conditions for functions πk1(t),k=0,1,we take next form

π1(0)=0,πk(0)=Yk(0),limμ0πk1(1/μ)=0,k=0,1,2,

To Y0(x)function has been smooth; therefore, we define it from the equation

xY0(x)=f(x)f0Y0(x)=(f(x)f0)/x,

then from Eq. (15.1), we have the equation

π''1(t)tπ1(t)=f0.

Let us prove an auxiliary lemma.

Lemma 1. Next boundary value problem

z''(t)tz(t)=b,0<t<1/μ,herebistheconstant,E16
z(0)=z0,z(1/μ)0,μ0E17

will have the unique solution and this one have next form

z(t)=z0Ai(t)Ai(0)πb(Ai(t)0tBi(s)ds+Bi(t)t1/μAi(s)dsAi(t)301/μAi(s)ds),

and z(t)C[0,μ1].

Proof. We verify the boundary conditions:

z(0)=z0πb(Bi(0)01/μAi(s)dsAi(0)301/μAi(s)ds),

as Bi(0)=Ai(0)3, so z(0)=z0.

z(1/μ)=z0Ai(1/μ)Ai(0)πb(13)Ai(1/μ)01/μBi(s)ds,

as Ai(t)~t1/4e23t3/2,Bi(t)~t1/4e23t3/2,t, so z(1/μ)=O(μ),μ0.

Now we show that z(t) satisfies Eq. (16). For this, we compute derivatives:

z'(t)=z0Ai'(t)Ai(0)πb(Ai'(t)0tBi(s)ds+Bi'(t)t1/μAi(s)dsAi'(t)301/μAi(s)ds)
z''(t)=z0Ai''(t)Ai(0)πb(Ai''(t)0tBi(s)ds+Bi''(t)t1/μAi(s)ds1πAi''(t)301/μAi(s)ds)

Substituting the expressions for z''(t)and z(t)in Eq. (17), and given that Ai''(t)tAi(t)0and Bi''(t)tBi(t)0, we get: bb.

The uniqueness of z(t)the solution is proved by contradiction. Let u(t)also be a solution of problems (16) and (17), z(t)u(t). Considering the function r(t)=z(t)u(t), for the function r(t), we obtain the problem

r''(t)tr(t)=0,0<t<1/μ,r(0)=0,r(1/μ)0,μ0.

The general solution of the homogeneous equation is

r(t)=c1Ai(t)+c2Bi(t);c1,2is the constant.

Considering the boundary condition r(1/μ)0,μ0, we have c2=0;r(t)=c1Ai(t). And the second condition r(0)=0,c1=0follows. This implies that r(t)0.

Therefore, z(t)u(t). It is obvious that z(t)C[0,μ1]. Lemma 1 is proved.

This Lemma 1 implies the existence and uniqueness of π1(t)C[0,μ1]solution of the problem:

π''1(t)tπ1(t)=f0,0<t<1/μ,π1(0)=0,π1(1/μ)0,μ0.

This function bounded and is infinitely differentiable on the segment [0,μ1], and as t:

π1(t)=f0t(1+12t3+1245t6+).

This asymptotic expression can be obtained by integration by parts the integral expression for π1(t).

From Eq. (15.0), we define Y1(x)and π0(t). Let Y1(x)0, then

π''0(t)tπ0(t)=0,π0(0)=f1,π0(1/μ)0,μ0,

And by Lemma 1, we have

π0(t)=f1Ai(t)/Ai(0).

Analogously, from Eq. (15.1), we define Y2(x)and π1(t). Let Y2(x)0, then

π''1(t)tπ1(t)=0,π1(0)=0,π0(1/μ)0,μ0.

In view of Lemma 1, we have π1(t)0.

To Y3(x)function has been smooth; as above, we define it from the equation

xY3(x)=Y0(x)Y0(0)Y3(x)=(Y0(x)Y0(0))/x,(Y0(0)=2f3),

then Eq. (15.2) to π2(t)hase the problem

π2(t)tπ2(t)=2f3,π2(0)=0,π2(1/μ)0,μ0.

By Lemma 1, we can write an explicit solution to this problem, and this solution bounded and is infinitely differentiable on the segment [0,μ1], and as t:

π2(t)=2f3t(1+12t3+1245t6+).

Analogously continuing this process, we determine the rest of the functions Yk(x),πk(t).

Now we will define functions wk(η)from the equality (14) by using the boundary conditions y(1)=0We state problems

Lw0w0(η)w0(η)=0,w0(0)=Y0(1),limηw0(η)=0E18.0
Lwk=ηwk1(η),w2i(0)=Y3i(1),w2i1(0)=0,limηwk(η)=0,k,iN.E18.k

One can easily make sure that all these problems (18.0) and (18.k) have unique solutions such that wk(η)C[0,), wk(η)=O(eη)with η.

Thus, all functions Yk(x),wk(η), and πk(t)in equality (12) are defined, i.e., a formally asymptotic expansion is constructed. Let us justify the constructed expansion. Let

ym(x)=μ1π1(t)+k=03mμk(Yk(x)+πk(t))+k=02mλkwk(η),rm(x)=y(x)ym(x).

Then for the remainder term, we state the following problem:

εrm(x)xrm(x)=O(εm+1/2),ε0,x(0,1).E19
rm(0)=O(e1/ε),rm(1)=O(εm+1),ε0.E20

Let rm(x)=(2x2)Rm(x)/2, and then problems (19) and (20) take the form

εRm(x)4xε2x2R'm(x)(2ε2x2+x)Rm(x)=O(εm+1/2),ε0,
Rm(0)=O(e1/ε),Rm(1)=O(εm+1),ε0.

According to the maximum principle [23, p. 117, 82], we have Rm(x)=O(εm1/2),ε0,x[0,1].

Hence, we get rm(x)=O(εm1/2),ε0,x[0,1].

Thus, we have proved.

Theorem 2. Let f(0)0, then the solution to problem (9) and (10) will have next form

y(x)=1ε3π1(xε3)+k=0εk3(yk(x)+πk(xε3))+k=0εkwk(1xε).

Example. Consider the problem

εy''(x)xy(x)=1+x,x(0,1),y(0)=0,y(1)=0.

The asymptotic solution this problem we can represent in the form y(x)=μ1π1(t)+k=03μk(Yk(x)+πk(t))+w0(η)+λw1(η)+λ2w2(η)+R(x).

We have got Y0(x)=(1+x1)/x=1,Y1,2,3(x)0,

π1(t)=π(Ai(t)0tBi(s)ds+Bi(t)t1/μAi(s)dsAi(t)301/μAi(s)ds),
π0(t)=Ai(t)/Ai(0),π1,2,3(t)0,w0(η)=2eη,wk(η)=O(eη),k=1,2.
εR''(x)xR(x)=O(ε3/2),0<x<1,R(0)=O(e1/ε),R(1)=O(ε2),ε0.

We have

y(x)=ε1/3π1(t)1+2e(1x)/ε+π0(t)+εw1(η)+εw2(η)+O(ε),ε0.

### 2.3. Bisingularly perturbed equation of the second order with a regularly singular point

Consider the boundary value problem [6, 7]

Lεyεy+xyq(x)y=f(x),x[0,1],E21
y(0)=0,y(1)=0,E22

where q(x),f(x)C[0,1].

Here, for simplicity, we consider the case q(0)=1,q(x)1.

The solution of the unperturbed problem

Myxyq(x)y=f(x),

represented as

y0(x)xp(x)1xr(s)s2ds,E23

where

r(x)=p1(x)f(x),p(x)=exp{1x(q(x)1)s1ds}.

Extracting in Eq. (23), the main part of the integral in the sense of Hadamard [34], it can be represented as

y0(x)=a(x)+r1xp(x)lnx,E24

where

a(x)=xp(x)1x(r(s)r0r1s)s2ds+r0p(x)[x1],r0=r(0),r1=r(0)=p(0)1[f(0)q(0)f(0)].E25

Function a(x)C[0,1].

Theorem 3. Suppose that the conditions referred to the above with respect to q(x)and f(x). Then the asymptotic behavior of the solution of the problems (21) and (22) can be written as:

k=0μk(zk(x)+πk(t)),ε=μ2,x=μt,E26

where zk(x)C[0,1],πk(t)C[0,μ1].

Function z0(x)is a solution of equation

Mz0=f(x)c0xp(x),

wherec0=p(0)1[f(0)q(0)f(0)].

The coefficients zk(x)of the series (26) will be determined as the solution of equations

Mzk=zk1(x)ckxp(x),

where ck=p(0)1[zk1(0)+zk1(0)q(0)], with boundary conditions zk(1)=0,k1.

Functions πk(t)is the solution of the equations

Lπkπk(t)+tπk(t)q(μt)πk(t)ckμtp(μt)

with boundary conditions πk(0)=zk(0),πk(μ1)=0.

Next, we use the following lemma.

Lemma 2. The problem

My=f(x)r1xp(x)

It has a unique solution y(x)C[0,1].

The proof of Lemma 2 follows from Eqs. (24) and (25).

Lemma 3. A boundary value problem

L0vv+tvv(t)=0,v(0)=a,v(1/μ)=0,

has solution v(t)=aX(t), where

X(t)=ttμ1s2exp(s22)ds,0X(t)1,X(0)=1.

The proof of Lemma 3 is obvious.

Lemma 4. In order to solve the boundary value problem

L0W=μt,W(0)=W(μ1)=0,

we have the estimate

0W(μ,t)e1lnμ1.

Proof. This follows from the fact that the solution of this problem existsuniquely by the maximum principle [23, 82] and will be represented in the form

W(μ,t)=μttμ1y2exp(y22)0ys2exp(s22)dsdy.

Lemma 5. The estimate

|πk(μ,t)|<Bk,

where 0<Bkis constant.

Proof. Consider the function

V±(μ,t)=γ1W(μ,t)+γ2X(t)±πk(μ,t),

where γ1and γ2are positive constants such that

γ1>max[0,1]|p(x)|,γ2>|zk(0)|.

It is obvious that

V±(μ,0)>0,V±(μ,μ1)>0,L0V±V±(t)+tV±(t)V±(t)<0

From the maximum principle, it follows that |πk(μ,t)|<γ1W(μ,t)+γ2X(t).

Now the proof of the lemma 5 follows from estimates of W(μ,t)and X(t).

If we introduce the notation

Yn(x,ε)=k=0nεk(zk(x)+πk(μ,t)),

where zk(x),πk(μ,t)are constructed above functions, then

LεYn(x,ε)=f(x)+εn+1zn.

Let y(x,ε)be the solution of the problems (21) and (22). Then

|Lε(Yn(x,ε)y(x,ε))|<Bnεn+1,Yn(0,ε)y(0,ε)=Yn(1,ε)y(1,ε)=0.

Therefore, |Yn(x,ε)y(x,ε)|<Bnεn+1.

### 2.4. The bisingular problem of Cole equation with a weak singularity

The following problem is considered [9, 13, 28, 29],

εy''(x)+xy'(x)y(x)=0,0<x<1,E27
y(0)=a,y(1)=bE28

where x[0,1];a,bare the given constants.

The unperturbed equation xy(x)y(x)=0,0<x<1,

has the general solution

y0(x)=ce2x,cconst.

This is a nonsmooth function in [0,1].

We seek asymptotic representation of the solution of the problems (27) and (28) in the form:

y(x)=k=0nεkyk(x)+k=03(n+1)μkπk(t)+R(x,ε),E29

where t=x/μ2,ε=μ3,yk(x)C[0,1],πk(t)C[0,1/μ2],R(x,ε)is the reminder term.

Substituting Eq. (29) into Eq. (27), we have

k=0nεk(εyk(x)+xyk(x)yk(x))+1μ(π0(t)+tπ0(t))+k=13(n+1)μk1(πk(t)+tπk(t)πk1(t))μ3(n+1)π3(n+1)(t)+εR''(x,ε)+xR'(x,ε)R(x,ε)h(x,ε)+h(x,ε)=0E30

By the method of generalized boundary layer function, we put the term h(x,ε)=k=0n1εkhk(x)into the equation. We choose functions hk(x)so that yk(x)C[0,1].

Taking into account the boundary condition (28), from Eq. (30), we obtain

xy0(x)y0(x)=0,0<x<1,y0(1)=b.E31
xyk(x)yk(x)=hk1(x)yk1(x),0<x<1,kN,yk(1)=0.E32

The solution of the problems (31) and (32) exists. It is unique and has the form

y0(x)=be2(x1),yk(x)=e2x1xhk1(s)yk1(s)se2sds,kN.

We choose indefinite functions hk(x) as follows: y''k1(x)hk1(x)C[0,1]. We can represent

y0(x)=be2(1+2x+(2x)22!+(2x)33!+(2x)44!++(2x)nn!+).

Let h1(x)=be2(2x+(2x)33!)=be2(12x31x).

Then

y''0(x)h0(x)C[0,1],μ3h1(tμ2)=c1(12t3μ2t),c1=be2,
y1(x)=c1e2x1x(12s2+1s+12s2e2s1s3e2s)e2sds.

We can rewrite y1(x) in the form:

y1(x)=y1,0+y1,1(2x)+y1,2(2x)2+y1,3(2x)3+,

where y1,0=(32+12e2)c1,y1,1=(16+12e2)c1,y1,2=(16+14e2)c1,y1,3=(110+112e2)c1.

Analogously, we have obtained

h1(x)=(y1,1(2x)+y1,3(2x)3)=y1,12x3+6y1,3x.

Then

y''2(x)h2(x)C[0,1],μ6h2(tμ2)=μ3y1,12t3+μ5y1,3t.

Continuing this process, we have

hk1(x)=yk1,12x3+6yk1,3x,k=4,,n,

where yk1,1,yk1,3are corresponding coefficients of the expansion of yk1,1(x)in powers of (2 x).

From Eq. (30), we have the following equations for the boundary functions πk(t):

Lπ0π0(t)+tπ0(t)=0,0<t<μ˜,π0(0)=ay0(0),π0(μ˜)=0,μ˜=1/μ2,E33
Lπ3k+1(t)=π3k(t)+yk,12t3,0<t<μ˜,π3k+1(0)=0,π3k+1(μ˜)=0, k=0,1,,nE34
Lπ3k+2(t)=π3k+1(t),0<t<μ˜,π3k+2(0)=0,π3k+2(μ˜)=0, k=0,1,,nE35
Lπ3k+3=π3k+2(t)yk,3t,0<t<μ˜,π3k(0)=yk(0),π3k(μ˜)=0,k=0,1,,n1E36
Lπ3(n+1)(t)=π3n+2(t)yn,3t,0<t<μ˜,π3n(0)=0,π3n(μ˜)=0E37

The solution of problem (33) is represented in the form

π0(t)=(abe2)Atμ˜e23s3/2ds,A=(0μ˜e23s3/2ds)1.

We note that π0(t)will exponentially decrease as tμ˜.

Lemma 6. The general solution of this equation Lz(t)=0will have z(t)=c1Y(t)+c2X(t); here c1,c2are constants, and

Y(t)=1X(t),X(t)=αtμ˜e23s3/2ds(α0μ˜e23s3/2ds=1).

Two linearly independent solutions and Y(t)=O(t),t0,0<X(t)1,

X(t)=t12e23t3/2(112t32++(1)n2nΠk=1n14(3k2)t3n2+),tμ˜.E38

Lemma 7. The boundary problem Lz(t)=0,z(0)=z(μ˜)=0will have only trivial solution.

The proofs of Lemmas 6 and 7 are evident.

Theorem 4. The problem

Lz(t)=f(t),z(0)=0,z(μ˜)=0,

will have the unique solution and this one has the next form

z(t)=0μ˜G(t,s)e23s3/2f(s)ds,

and G(t,s)={Y(t)X(s),0ts,Y(s)X(t),stμ˜,

is the function of Green andf(t)C(0,μ˜].

Theorem 4 implies the existence and uniqueness of the solution of problem (34)–(37): |πk(t)|<l=const,t[0,μ˜].

Lemma 8. Asymptotical expansions of functions πk(t),tμ˜(k=1,2,) will have the next forms

π1(t)=y0,12t(1+45t3+74t3+4211t9+392t7+),
π2(t)=y0,1t(1+2340t3+1732t3+),π3(t)=23y0,160t3+O(1t3),
π3k+1(t)=t1j=0l3k+1,jt32j,π3k+2(t)=t1/2j=0l3k+2,jt32j,π3k(t)=j=1l3k,jt32j.

Proof for Lemma 8.

Firs proof. We can prove this lemma by applying formulas (38) and Theorem 4.

Second proof. We can receive these representations from Eqs. (34)–(37) directly.

Now we will prove the boundedness of the reminder function R(x,ε). This function will satisfy the next equation:

εR''(x,ε)+xR'(x,ε)R(x,ε)=μ3(n+1)π3(n+1)(t)+εn+1(hn(x)yn(x)),
R(0,ε)=0,R(1,ε)=0.

Applying to this problem theorem [23, p.117, 82], we obtained

|R(x,ε)|εn+1Cmax0x10tμ˜|π3(n+1)(t)+hn(x)yn(x)|.

Therefore, we have R(x,ε)=O(εn+1),ε0,x[0,1].

We prove next.

Theorem 5. The asymptotical expansion of the solution of the problems (27) and (28) and will have the next form

y(x)=k=0nεkyk(x)+k=03(n+1)μkπk(t)+O(εn+1),ε0.

## 3. Singularly perturbed differential equations Lighthill type

### 3.1. The idea of the method of Poincare

Consider the equation

My(x):=y''(x)+y(x)εy3(x)=0.E39

Unperturbed equation has solutions y0(x)=a1cosx+b1sinx(where a1,b1are arbitrary constants) with period 2π. We are looking for the periodic solution of the equation y(x,ε)with a period of ω(ε)=ω(0)=2π.

Note that the operator Mtransforms Fourier series k=1akcoskxand k=1aksinkxin itself. Poincare’s method reduces the existence of periodic solutions of differential equations to the existence of the solution of an algebraic equation.

We will seek a periodic solution of Eq. (39) with the initial condition

y(0)=1,y'(0)=0.

If we seek the solution in the form

y(x)=y0(x)+εy1(x)+ε2y2(x)+

with the initial conditions

y0(0)=1,y'0(1)=0,yk(0)=y'k(1)=0,k=1,2,

then for ys(x),s=0,1,we have next equations

Ly0:=y''0(x)+y0(x)=0y0(x)=cosx
Ly1=cos3x=34cosx+14cos3xy1(x)=38xsinx132cos3x+132cosx,

Thus, y(x)=cosx+ε8(3xsinx14cos3x+14cosx)+it is not a uniform expansion of the y(x) on the segment [,], since the term εxsinxis present here.

If these secular terms do not appear in Eq. (39), it is necessary to make the substitution

x=t(1+εα1+ε2α2+)

where the constant αkshould be selected so as not to have secular terms in t.

Thus, the solution of Eq. (39) must be sought in the form

y(t)=y0(t)+εy1(t)+ε2y2(t)+x=t(1+εα1+ε2α2+)E40

Then Eq. (39) has the form

z''(t)+(1+α1ε+α2ε2+)z(t)=ε(1+α1ε+α2ε2+)z3(t)

where y(w(ε)t)=z(t).

We will seek the 2π periodic solution of this equation in the form

z(t)=z0(t)+εz1(t)+ε2z2(t)+

Then

Lz0:=z''0(t)+z0(t)=0z0(t)=cost.
Lz1(t)=α1cost+34cost+14cos3t.

The function Z1(t) will have the periodical solution we take α1=3/4. Then z1(t)=132cos3t.

Similarly, from equations

αzn(t)=αncost+g(α1,α2,,αn1)cost+m=12n+1βncosmt

αnand etc. are uniquely determined.

Theorem 6. Equation (39) has a unique 2π/ωperiodic solution, and it can be represented in the form (40).

### 3.2. The idea of the Lighthill method

Lighthill in 1949 [67] reported an important generalization of the method of Poincare.

He considered the model equation [67, 82]:

(x+εy(x))y'(x)+q(x)y(x)=r(x),y(1)=aE41

where x[0,1]q(x),r(x)C[0,1].

Lighthill proposed to seek the solution of Eq. (41) in the form

y(ξ)=y0(ξ)+εy1(ξ)+ε2y2(ξ)+x=ξ+εx1(ξ)+ε2x2(ξ)+E42

It is obvious that Eq. (42) has generalized the Poincare ideas (see, the transformation Eq. (40)).

At first, we consider the example

(x+εy(x))y'(x)+y(x)=0,y(1)=b.E43

It has exact solution

y(x)=(x2+2bε+ε2b2x)/ε.E44

It is obvious that for b>0, the solution (43) exists on the interval [0,1]and

y(0)=2b+εb2/ε.

The solution of Eq. (43) is obtained by the method of small parameter that can be obtained from Eq. (44). For this purpose, we write Eq. (44) in the form

y(x)=xε(1+1+2bεx+b2(εx)2)

and considering x2>2εb, this expression can be expanded in powers of ε, and then we have

y(x)=bx+b22xεx2(x21)++O(1x(εx2)n)+E45

The series (45) is uniformly convergent asymptotic series only on the segment [εα,1],0<α<1/2.

First, we write Eq. (43) in the form

(x+εy(ξ))y'(ξ)+y(ξ)x'(ξ)=0E46

Substituting Eq. (42) into Eq. (46):

(ξ+ε(y0(ξ)+x1(ξ))++εn(yn1(ξ)+xn(ξ))+)(y'0(ξ)+εy'1(ξ)+++εny'n(ξ)+)+(y0(ξ)+εy'1(ξ)+εny'n(ξ)+)(1+εx'1(ξ)++εnx'n(ξ)+)=0

and equating coefficients of the same powers ε,we have

ξy'0(ξ)+y0(ξ)=0E47
ξy'n(ξ)+yn(ξ)+i=0n1((yi(ξ)+xi+1(ξ))y'n1i(ξ)+yi(ξ)x'ni(ξ))=0,yn(1)=0,n=1,2,E48

From Eq. (47), we have

y0(ξ)=bξ1.

Using Eq. (47), Eq. (48) for n=1can be written as

ξy'1(ξ)+y1(ξ)=(ξx'1(ξ)x1(ξ)+y0(ξ))y'0(ξ)=0,y1(1)=0.E49

If we put x1(ξ)=0in Eq. (49), we obtain

ξy'1(ξ)+y1(ξ)=b2ξ3,y1(1)=0.

Hence, solving this equation, we have

y1(ξ)=b2(2ξ)1b2(2ξ3)1.

Since differentiation increased singularity of nonsmooth function, we select x1(ξ)so that the expression in the right side of Eq. (49) is equal to zero, i.e.,

ξx'1(ξ)x1(ξ)+y0(ξ)=0,x1(1)=0.

Hence, we have

x1(ξ)=21bξ(2ξ)1b.

Then Eq. (49) takes the form

ξy'1(ξ)+y1(ξ)=0,y1(1)=0.

Hence, we obtain y1(ξ)=0.

Now Eq. (48) for n=2takes the form

ξy'2(ξ)+y2(ξ)=(ξx'2(ξ)x2(ξ))y'0(ξ)=0,y2(1)=0.

Let x2(ξ)=0, and then y1(ξ)=0. Further also choose xi(ξ)=yi(ξ)=0(i=3,4,), as they also satisfy the initial conditions. Thus, we have found that

y(ξ)=bξ1E50
x(ξ)=ξ+b2(ξ1ξ)ε.E51

Putting in Eq. (51) x=0, we have

η=bε/(2+bε).E52

For b>0, the point x=0is achieved. Moreover, the except in variable ξfrom Eq. (50) and to Eq. (51) setting ξ, we obtain the exact solution (44).

Now we will present the main idea of the Lighthill method to Eq. (41) under conditions:q(x),r(x)C[0,1]and q0=q(0)>0. We will write it in the form of

(x(ξ)+εy(ξ))y'(ξ)=[r(x(ξ))q(x(ξ))y(ξ)]x'(ξ),y(1)=y0.E53

It is obvious that we have one equation for two unknown functions, y(ξ), x(ξ). Now we substitute the series (42) to Eq. (53):

(ξ+k=0εk(yk(ξ)+xk(ξ)))k=0εky'k(ξ)==(j=0rj(ξ)(k=0xk(ξ)εk)jj=0qj(ξ)(k=0xk(ξ)εk)j)(1+k=0x'k(ξ)εk),

where qj=qj(ξ)=1j!q(j)(ξ),rj=rj(ξ)=1j!r(j)(ξ).

Hence, equating the coefficients of equal powers has ε

Lu0ξy'0(ξ)+q(ξ)y0(ξ)=r(ξ),y0(1)=y0,E54
Ly1=[ξy'0x'1y'0x1y0y'0]+(r1q1y0)x1,y1(1)=0,E55
Ly2=[ξy'0x'2(y0+x1)y'1(y1+x2)y'0+((r1q1y0)x1qy1)x'1]++{r1x2+r2x12q1x1y1(q1x2+q2x12)y0},y2(1)=0,E56

Lyn=[y'0x'ny'0xn+fn(y0,,yn1,x1,,xn1,y'0,,y'n1,x'1,,x'n1)]++{gn(y0,,yn1,x1,,xn1)},yn(1)=0;E57

where q=q0,r=r0,

fn=(y0+x1)y'n1(y1+x2)y'n2(yn2+xn1)y'1yn1y'0++(r1x1qy1q1x1y0)x'n1+(r1x2+r2x12qy2q1x1y1(q1x2+q2x12)y0)x'n2++(r1xn1+2r2x1xn2+2r2x2xn3++rn1x1n1q1y1xn2(q1xn1+2q2x1xn2++qn1x1n1)y'0)x'1,
gn=r1xn+2r2x1xn1++rnx1nq1x1yn1(q1x2+q2x12)yn2(q1xn+2q2x1xn1++qnx1n)y0.

In these equations, the coefficient r(ξ)q(ξ)y0(ξ)of the derivative x'n(ξ)(n=1,2,)was replaced by Eq. (54) on ξy'0(ξ).

From Eq. (57) for n= 1,2,…, it follows that if we want to define functions xn(ξ)(n=1,2,)from this differential equations, then we must assume that

ξy'0(ξ)=r(ξ)q(ξ)y0(ξ)0,ξ(0,1].E58

And this condition cannot be avoided by applying the Lighthill method to Eq. (41). Condition (58) first appeared in [69], justifying Lighthill method, then in the works Habets [66] and Sibuya, Takahashi [68]. Comstock [65] on the example shows that the condition (58) is not necessary for the existence of solutions on the interval [0,1]. Further assume that the condition (58) holds. Note that the right-handside of Eq. (57) is linear with respect to xn(ξ), and fnfunction depends from y'0,,y'n1,x'1,,x'n1only.

The solution of Eq. (54) can be written as

y0(ξ)=ξq0g(ξ)(y0+1ξsq01r(s)g1(s)ds):=ξq0w(ξ),E59

where g(ξ)=exp(1ξ(q0q(s))s1ds).

Let

w0=y001sq01r(s)g1(s)ds0w0=w(0)0.

Hence, we have

y0(ξ)~ξq0w0,ξ0.E60

Since the differentiation of y0(ξ)increased of its singularity at the point ξ=0, it is better to choose such that the first brace in Eq. (55) is equal to zero, i.e.,

ξx'1=x1+y0,x(1)=0.

Hence, using Eq. (60), we obtain

x1(ξ)=ξ+ξ1ξs2y0(s)ds~w01+q0ξq0.E61

Then Eq. (55) takes the form

Ly1=(r1q1y0)x1~a˜1ξ2q0,

where a˜1=const. Hence, we have

y1(ξ)~a1ξ2q0(a1=const),ξ0.E62

Now equating to zero the expression in the first brace in the right-hand side of Eq. (56), we have

ξx'2x2=y1+((y0+x1)y'1((r1q1y0)x1qy1)x'1)(y'0)1~b˜2ξ2q0,b˜2=const.

From this, we get

x2(ξ)~b2ξ2q0,b2=const,ξ0.E63

Now Eq. (56) takes the form

Ly2=g2(y0,y1,x1,x2)~a˜2ξ3q0,a˜2=const,ξ0

Solving this equation, we have

y2(ξ)~a2ξ3q0,a2=const,ξ0E64

Next, the method of induction, it is easy to show that

xj(ξ)~bjξjq0,yj(ξ)~ajξ(j+1)q0,j=1,2,….E65

Thus, the series (42) has the asymptotic

y(ξ)~ξq0(w0+a1εξq0++an(εξq0)n+),ξ0,E66
x~ξw01+q0ξq0ε+b2(εξq0)2++bn(εξq0)n+E67

From Eq. (67), it follows that the point x=0corresponds to the root of the equation

η+εx1(η)+ε2x2(η)+=0E68

Moreover, this equation should have a positive root and if the solution of Eq. (41) exists on the interval (0,1]. Solving Eq. (68), we obtain

η~(w0ε/1+q0)1/(1+q0),ε0.E69

And, under the conditionw0>0,η0will be positive. It is obvious that on the interval [ξ0,1]series (42) or (66) and (67) remains asymptotic. Substituting Eq. (69) into Eq. (66), we have

y(0)~w0(w0ε1+q0)q0/(1+q0),ε0.

If w0<0the point x=0does not have the positive root of Eq. (68), so that the solution of Eq. (41) goes to infinity, before reaching the point x=0.

We have the

Theorem 7. Suppose that the conditions (1) q(x),r(x)C[0,1]; (2) q0>0; (3) w0>0; (4) ξy'00,ξ[0,1]. Then the solution of problem (41) exists on the interval [0,1], and it can be represented in the asymptotic series (42), (66) and (67).

Theorem 7 proved by Wasow [69], Sibuya and Takahashi [68] in the case where q(x),r(x)are analytic functions on [0,1]; proved by Habets [66] in the case q(x),r(x)C2[0,1]. Moreover, instead of the condition (3) Wasow impose a stronger condition: a>>1.

In the proof of Theorem 7, we will not stop because it is held by Majorant method.

From the foregoing, it follows that Wasow condition y'0(ξ)0,ξ(0,1]is essential in the Lighthill method.

Comment 2. Prytula and later Martin [65] proposed the following variant of the Lighthill method. At first direct expansion determined using by the method of small parameter

y(x)=y0(x)+εy1(x)+ε2y2(x)+E70

and further at second they will make transformation

x=ξ+εx1(ξ)+ε2x2(ξ)+E71

Here unknowns xj(ξ)are determined from the condition that function yj(ξ)was less singular function yj1(ξ). We show that using the method Prytula or Martin, also cannot avoid Wasow conditions. Really, substituting Eq. (71) into Eq. (70) and expanding in a Taylor series in powers of ε, we have

y(ξ)=y0(ξ)+ε{y1(ξ)+y'0(ξ)x1(ξ)}+O(ε2).

Hence, to obtain a uniform representation of the solution to the second order by ε, we must to put to zero the expression in the curly brackets, i.e., x1(ξ)=y1(ξ)/y'0(ξ). Therefore, y(ξ)=y0(ξ)+O(ε2). Hence, it is clear that we must make the condition of Wasow: y'0(ξ)0in the method of Prytula or Martin also.

### 3.3. Uniformization method for a Lighthill model equation

We will consider the problem (41) again [3, 5860], i.e.,

(x+εy(x))y'(x)=r(x)q(x)y(x),y(1)=a,E72

Theorem 8. Suppose that the problem (72) has a parametric representation of the solution y=y(ξ),x=x(ξ), where ξ[η,1],η=η(ε)>0, then the problem (72) is equivalent to the problem

{ξy'(ξ)=r(x(ξ))q(x(ξ))y(ξ),y(1)=y0,ξx'(ξ)=x(ξ)+εy(ξ),x(1)=1,ξ[η,1],E73

where η=η(ε)is the root equation x(η)=0and if the root η=η(ε)>0and x(ξ)+εy(ξ)0on the interval [η,1].

Proof. Sufficiency. Let the solution of the problem (72) exists and x(ξ),y(ξ)are a parametric representation of the solution of the problem (72). Then introducing the variable-parameter ξ, we obtain the problem (73).

Necessity. Let it fulfill the conditions of Theorem 8. Then dividing the first equation by second one, we get Eq. (72). Theorem 8 is proved.

Equation (73) on the proposal of the Temple [43], we will call uniformizing equation for the problem (72).

We have the following

Theorem 9. Suppose that the first three conditions of Theorem 8. i.e.,(1) q(x),r(x)C[0,1]; (2) q0>0; (3) w0>0. Then the solution of problem (72) is represented in the form of an asymptotic series (42) and its solution can be obtained from uniformizing equation (73).

The proof of this theorem is completely analogous to the proof of Theorem 8, even more easily.

Only it remains to show that under the conditions of Theorem 9 we can get an explicit solution y=y(x,ε). Really, since

x~ξw01+q0ξq0ε,ξ0.

Let

F(x,ξ,ε)=xξ+w01+q0ξq0ε+O((εξq0)2),ξ0,η=w01+q0εq0+1,ε0.

then

F(x,ξ,ε)ξ|ξ=η(ε)=1q0+O(ε1/(1+q0))0,ξ[η,1].

Therefore, by the implicit function theorem, we can express ξ:ξ=ϕ(x,ε).

Then when we put it in first equality (42), we obtain an explicit solution y=y(x,ε).

Comment 3. Explicit asymptotic solution that this problem obtained in Section 3.4.

Example 43. Uniformized equation is

{ξy'(ξ)=y(ξ),y(1)=b,ξx'(ξ)=x(ξ)+εy(ξ),x(1)=1,ξ[η,1],

It is easy to integrate this system, and we obtain

y(ξ)=bξ1,x(ξ)=(1+21bε)ξ(2ξ)1bε,

Hence, excluding variable ξ, we have an exact solution (44).

Example 2 [37, 43])

(x+εy(x))y'(x)+(2+x)y(x)=0,y(1)=e1.

Uniformized equation is

{ξx'(ξ)=x+εy(ξ),x(1)=1,ξy'(ξ)=(2+x(ξ))y(ξ),y(1)=e1,ξ[η,1],E74

Let

{x(ξ)=x0(ξ)+εx1(ξ)+O(ε2),y(ξ)=y0(ξ)+εy1(ξ)+O(ε2),E75

Substituting Eq. (75) into Eq. (74), we have

x0(ξ)=ξ,x1(ξ)=ξ1ξess4ds,y0(ξ)=eξξ2,y1(ξ)=eξξ21ξess4ds,

Hence if ξ0, we obtain

x0(ξ)=ξ,x1(ξ)=13ξ2+,y0(ξ)=ξ2+,y1(ξ)=16ξ4+

From the equation x(η)=0, we find η:η~ε/33.

We prove that x(ξ)+εy(ξ)0on the interval [η,1].

Really,

x(ξ)+εy(ξ)~ξ+εξ20,ξ[η,1].

### 3.4. It is construction explicit form of the solution of the model Lighthill equation

We will consider the problem [57], i.e., (41) again

(x+εy(x))y'(x)+q(x)y(x)=r(x),y(1)=bE76

wherebis given constant, x[0,1],y'(x)=dy/dx. Given functions are subjected to the conditions U: q(x),r(x)C()[0,1].

Here, we consider the case q0=1; this is done to provide a detailed illustration of the idea of the application of the method. We search for the solution of problem (76) in the form

y(x)=μ1π1(t)+k=0(πk(t)+uk(x))μk,E77

where t=x/μ,ε=μ2,uk(x)C()[0,1]and πk(t)C()[0,μ0],μ0=1/μ.

Note that πk(t)=πk(t,μ), i.e., πk(t)depends also on μ, but this dependence is not indicated.

The initial conditions for the functions πj(t)are taken as

π1(1/μ)=bμ,b=u0k=0μkuk(1),πk(μ0)=0,k=0,1,E78

Substituting Eq. (77) into Eq. (76), we obtain to determine the functions πk(t),k=1,0,1,,un(x),n=0,1,,

we have the following equations:

(t+π1(t))π1(t)=q(μ t)π1(t),π1(μ0)=bμ,E79.-1
Lu0(x):=xu0(x)q(x)u0(x)=r(x),u0(x)C()[0,1]E80.0
Dπ0(t):=(t+π1(t))π0(t)+(π1(t)q(μ t))π0(t)=u0(tμ)π1(t),π0(μ0)=0E79.0
Lu1(x)=0,u1(x)C()[0,1],E80.1
Dπ1(t)=u0(tμ)π0(t)+π0(t)π0(t)u1(tμ)π1(t),π1(μ0)=0E79.1
Lu2(x):=u0(x)u0(x),u2(x)C()[0,1]E80.2
Dπ2(t):=u0(tμ)π1(t)π0(t)π1(t)u1(tμ)π0(t)π1(t)π0(t)u2(tμ)π1(t),π2(μ0)=0E79.2
Lu3(x):=u0(x)u1(x)u0(x)u1(x),u3(x)C()[0,1],E80.3
Dπ3(t)=i+j=2i0,j2ui(μ t)πj(t)+i+j=2i,j0πi(t)πj(t),π3(μ0)=0,E79.3

We solve these problems successively. We write problem (79.1) as

tz(t)q(μt)z(t)=z(t)z(t),z(μ0)=bμ,

where

z=π1(t),μ0=μ1.

The fundamental solution of the homogeneous equation corresponding to this equation is of the form

z0(t)=exp{μ0tq(μs)dss}=exp{μ0t(q(μs)+1)dssμ0tdss}=p(t,μ)μt,

where

p(t,μ)=exp{μ0t(q(μs)+1)dss}.

Using the expression for z0(t), the solution of the inhomogeneous equation for z(t)can be written as

z(t)=p(t,μ)μt[z(μ0)+μμ0 tp1(s,μ)z(s)z(s)ds] ,

Or tz(t)=p(t,μ)bp(t,μ)μ0 tp1(s,μ)z(s)z(s)ds.

After integrating by parts, we reduce the last expression to the following equation:

tz(t)=p(t,μ)bz2(t)2+p(t,μ)b2μ22+p(t,μ)2μ0 t1+q(μs)sp1(s,μ)z2(s)ds

or

z2(t)+2tz(t)p(t,μ)b0=p(t,μ)μ0tφ(s,μ)p1(s,μ)z2(s)ds:=p(t,μ)T(t,z2)E81

where ϕ(s,μ)=(1+q(μs))/s,b0=2b+b2μ2.

Let b0>0. Let us introduce the notation z0(t)=t+t2+b0p(t,μ). This function satisfies the inequality 0<z0(t)Mt1(t>0)and is a strictly decreasing bounded function on the closed interval [0,μ0]. Here and elsewhere, all constants independent of the small parameter μare denoted by M. Let Sμbe the set of functions z(t)satisfying the condition

zz0Mμ,where z=max0tμ0|z(t)|,

Theorem 10. Ifb0>0, then there exists a unique constraint of the solution of problem(79.-1) from the setSμ.

Proof. Equation (81) is equivalent to the equation z=F[t,z], where

F[t,z]=t+t2+bp(t,μ)+p(t,μ)T(t,z2).

Suppose that ϕ(t,μ)Mμ,0<mp(t,μ)M,p1(t)M.First, let us estimate T(t,z2)on the set Sμ. We have

|T(t,z2)|tμ0|ϕ(s,μ)||p1(s,μ)||z(s)|2dsMμtμ0|z(s)|2dsMμ0μ0|z(s)|2dsMμ01|z(s)|2ds+Mμ1μ0|z(s)|2dsMμ.

Here, we have used the triangle inequality

|z(t)||z(t)z0(t)|+|z0(t)|,

as well as the inequality

|z0(t)|Mt1(t>0).

The Fréchet derivative of the operator F(t,z)with respect to zat the point z0(t)is a linear operator:

Fz(t,z0)h=p(t,μ)tμ0ϕ(s,μ)p1(s,μ)z0(s)h(s)dst2+p(t,μ)(b+T(t,z2)),

where h(t)is a continuous function on the closed interval [0,μ0]. Note that, in view of T(t,z02)=O(μ),the denominator of this expression is strictly positive on the closed interval [0,μ0]. For Fz(t,z0), we can obtain the estimate Fz(t,z0)Mμlnμ1in the same way as the estimate for T(t,z2). Hence, in turn, it follows from the Lagrange inequality that the operator is a contraction operator in the set Sμ. Therefore, by the fixed-point principle, Eq. (81) has a unique solution from the class Sμ. The theorem is proved.

Corollary. The following inequalities hold:

1. z(t)=π1(t)M>0for allt[0,μ0];

2. π1(t)Mt1(t>0).

The other function πj(t),uj(x),j=0,1,2,is determined from the inhomogeneous linear equations; therefore, the following lemmas are needed.

Lemma 9. For any functionf(x)C()[0,1], the equationLξ=f(x)has a unique bounded solutionξ(x)C()[0,1]expressible as

ξ(x)=Q(x) 0 xQ1(s)f(s)dsx,Q(x)=exp{ 1 x(q(s)+1)dss}.

Proof. The proof follows from the fact that the general solution of the equation under consideration is expressed as

ξ(x)=Q(x)x1[ξ(1)+ 1 xQ1(s)f(s)ds].

If we choose

ξ(1)= 0 1Q1(s)f(s)ds.

then we obtain the required result.

This lemma implies that all the functions uk(x),k=0,1,are uniquely determined and belong to the class C[0,1].

Lemma 10. The problem

(t+π1(t))η(t)+(π1(t)q(μt))η(t)=k(t),η(μ0)=0,E82

where the functionk(t)belongs toC[0,1]is continuous and bounded, and if|k(t)|Mt2,t, has a unique uniformly bounded solutionη(t)=η(t,μ)on the closed intervalt[0,μ0]for a smallμ.

Proof. The fundamental solution of the homogeneous equation (82) is of the form

Ф(t)=(1+μ2b)g(t,μ)μ(t+π1(t)),g(t,μ)=exp{ t μ0(1+q(μs))dss+π1(s)}.

Obviously, g(t,μ)Mand g1(t,μ)Mfor t[0,μ0]and μaresmall. The solution of problem (82) can be expressed as

η(t)=g(t,μ)t+π1(t) μ0 tg1(s,μ)k(s)ds.E83

The estimate of the integral term in Eq. (83) shows that it is bounded by the constant M. Hence, it also follows that |η(t)|Mt1(t>0). The solution of problem (79.0) is defined by the integral Eq. (83), where

k(t)=u0(tμ)π1(t)=u0(tμ)q(μt)π1(t)t+π1(t),

satisfies the assumptions of the lemma. Therefore, the function π0(t)is bounded on [0,μ0]. The boundedness of the other functions πk(t),k=1,2,is proved in a similar way, because the right-hand sides of the equations defining these functions satisfy the assumptions of Lemma 10. The estimate of the asymptotic behavior of the series (77) is also carried out using Lemma 10.

Let us introduce the notation

y(x)=μ1π1(t)+k=0nμk(πk(t)+uk(x))+μn+1Rn+1(x,μ).E84

The following statement holds.

Theorem 11. Letb0>0(for this, it suffices that the conditionb0:=by0(1)>0holds). Then the solution of problem(76) exists on the closed interval[0,1]and its asymptotics can be expressed as Eq. (84) and|Rn+1(x,μ)|Mfor allx[0,1].

Example. Consider the equation

(x+εy(x))y'(x)+y(x)=1,y(1)=b,

This equation is integrated exactly

y(x)=ε1[x+x2+2b0ε+ε2(y(0))2+2εx],

where b0=b1. If b0>0, then the solution of problem (1) exists on the closed interval [0,1], which is confirmed by Theorem 11. The equation for π1(t)is of the form

(t+π1(t))π1(t)+π1(t)=0,π1(μ0)=bμ.

The solution of this problem can be expressed as

π1(t)=t+t2+2b+b2μ2.

The equation for u0(x) has the solution y0(x)=1C[0,1]. Further,

π0(t)=π1(t)+bμt+π1(t),uk(x)=0,k=1,2,,

where b=b0. The asymptotics of the solutions of problem (76) can be expressed as

y(x)=μ1π1(x/μ)+1+π0(x/μ)+o(μ)for all x[0,1],μ0.

## 4. Lagerstrom model problem

The problem [32]

v''(r)+krv'(r)+v(r)v'(r)=β[v'(r)]2,v(ε)=0,v()=1,E85

where 0<βis constant, kN.

It has been proposed as a model for Lagerstrom Navier-Stokes equations at low Reynolds numbers. It can be interpreted as a problem of distribution of a stationary temperature v(r).

The first two terms in Eq. (1) is (k+1)dimensional Laplacian depending only on the radius, and the other two members—some nonlinear heat loss.

It turns out that not only the asymptotic solution but also convergent solutions of Eq. (1) can be easily constructed by a fictitious parameter [70]. The basic idea of this method is as follows. The initial problem is entered fictitious parameter λ[0,1]with the following properties:

1. λ=0, the solution of the equation satisfies all initial and boundary conditions;

2. The solution of the problem can be expanded in integral powers of the parameter λfor all λ[0,1].

It is convenient in Eq. (85) to make setting r=εx,v=1u, then

u''(x)+(kx1+ε)u'(x)λεu(x)u'(x)=[u'(x)]2,u(1)=1,u()=0.E86

We have the following

Theorem 12. For small ε>0, the solution of problem (86) can be represented in the form of absolutely and uniformly convergent series

u(x)=u0(x,ε)+vk(ε)u1(x,ε)++vkn(ε)un(x,ε)+,

for the sufficiently small parameter ε, where

v1(ε)(ln1ε)1,v2εln1ε,vkk1k2ε(j>2);uk(x,ε)=O(1),x[1,)

Note that the function un(x,ε)also depends on k, but for simplicity, this dependence is not specified.

Proof. We introduce Eq. (86) parameter λ, i.e., consider the problem

u''(x)+(kx1+ε)u'(x)β[u'(x)]2=λεu(x)u'(x),u(1)=1,u()=0E87

Here, we will prove this Theorem 12 in the caseβ=0only for simplicity.

Setting λ=0in Eq. (87), we have

u0+(x1k+ε)u0=0,u0(1)=1,u0()=0.E88

It has a unique solution

u0=X(x,ε):=1X1(X,ε),X1=C01xskeεsds,C01=1skeεsds.

Therefore, Eq. (88) with zero boundary conditions is the Green’s function

K(x,s,ε)={C01X1(x,ε)X(s,ε),1xs,C01X1(s,ε)X(x,ε),s<x<.

Hence, the problem (87) is reduced to the system of integral equations

u(x)=X(x,ε)+λε1G(x,s,ε)u(s)u(s)ds,u(x)=X(x,ε)+λε1Gx(x,s,ε)u(s)u(s)ds,E89

where

G(x,s,ε)={X1(x,ε)X(s,ε)/X(s,ε),1xs,X1(s,ε)X(x,ε)/X(s,ε),s<x<.

In Eq. (89), we make the substitution u=X(x,ε)ϕ(x),u=X(x,ε)ψ(x), and then we have

ϕ(x)=1+λε1Q1(x,s,ε)ϕ(s)ψ(s)ds:=1+λεQ1(ϕψ),ψ(x)=1+λε1Q2(x,s,ε)ϕ(s)ψ(s)ds:=1+λεQ2(λψ),E90

where

Q1=X1(x,ε)G(x,s,ε)X(s,ε)X(s,ε),Q2=Xx1(x,ε)Gx(x,s,ε)X(s,ε)X(s,ε).

To prove the theorem, we need next

Lemma 11. The following estimate holds

1|Qj(x,s,ε)|ds1X(s,ε)ds(j=1,2)E91

Given that, we have 0X1(x,ε)1,|X(x,ε)|=X(x,ε),X(x,ε)0,x[1,), we have

1|Q1(x,s,ε)|ds1xX1(s,ε)X1(s,ε)|X(s,ε)|X(s,ε)ds++xX1(x,ε)X2(s,ε)|X(s,ε)|X1(s,ε)ds1xX(s,ε)ds+xX(s,ε)ds=1X(s,ε)ds.

Inequality Eq. (91) for j=2is proved similarly.

Further, by integrating by parts, we have

1X(s,ε)ds=1+C01sk+1eεsds1sk+1eεsds/1skeεsds:=vk(ε)ε.

Consequently,

ε1X(x,ε)dsvk(ε).E92

It is from integral expressing of vk(ε)we can obtain the asymptotic behavior such as indicated in the theorem.

With the solution of Eq. (90), we can expand in series

ϕ(x)=1+ϕ1(x,ε)λ+ϕ2(x,ε)λ2+,Ψ(x)=1+Ψ1(x,ε)λ+Ψ2(x,ε)λ2+….

The coefficients of this series are uniquely determined from the equations ϕ0=Ψ0=1,ϕ1=εQ1(1),Ψ1=Q2(1),ϕn=εQ1(ϕn1)+εQ1(Ψn1)+εQ1(ϕ1Ψn2)++εQ1(ϕn2Ψ1),Ψn=εQ2(ϕn1)+εQ2(Ψn1)+εQ2(ϕ1Ψn2)++εQ2(ϕn2Ψ1),(n=2,3,).

Let z=sup1x<{|ϕ(x)|,|Ψ(x)|},then by using Eq. (92) we have a Majorant equation: z=1+λvk(ε)z2. The solution of this equation can be expanded in powers λ the under condition 8vk(ε)1for all λ[0,1].

If we call un(x,ε)=X(x,ε)ϕn(x,ε)vkn(ε),we get the proof of the theorem.

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Keldibay Alymkulov and Dilmurat Adbillajanovich Tursunov (June 14th 2017). Perturbed Differential Equations with Singular Points, Recent Studies in Perturbation Theory, Dimo I. Uzunov, IntechOpen, DOI: 10.5772/67856. Available from:

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