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

Downloaded: 1077


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.


  • 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


For example.


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


and write


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


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.,


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


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


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


    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)


    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


    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


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


    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]


    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]


    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)


    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




    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


    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


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


    From Eq. (6), we have


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


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




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


    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


    Hence, we find


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


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

    From these, we get




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


    Let Y3(x)0, then


    From this, we get


    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:


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

    The problem (8) has a unique solution


    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


    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


    can lead to conditions (10).

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


    then we have


    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


    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


    From Eq. (13), we have


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


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


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


    Let us prove an auxiliary lemma.

    Lemma 1. Next boundary value problem


    will have the unique solution and this one have next form


    and z(t)C[0,μ1].

    Proof. We verify the boundary conditions:


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


    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:


    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


    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:


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


    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


    And by Lemma 1, we have


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


    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


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


    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:


    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


    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


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


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


    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


    Example. Consider the problem


    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,


    We have


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

    Consider the boundary value problem [6, 7]


    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


    represented as




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




    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:


    where zk(x)C[0,1],πk(t)C[0,μ1].

    Function z0(x)is a solution of equation



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


    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


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

    Next, we use the following lemma.

    Lemma 2. The problem


    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


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


    The proof of Lemma 3 is obvious.

    Lemma 4. In order to solve the boundary value problem


    we have the estimate


    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


    Lemma 5. The estimate


    where 0<Bkis constant.

    Proof. Consider the function


    where γ1and γ2are positive constants such that


    It is obvious that


    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


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


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


    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],


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

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

    has the general solution


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

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


    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


    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


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


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


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



    We can rewrite y1(x) in the form:


    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




    Continuing this process, we have


    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):

    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

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


    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


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


    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


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


    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


    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:


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


    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


    3. Singularly perturbed differential equations Lighthill type

    3.1. The idea of the method of Poincare

    Consider the equation


    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


    If we seek the solution in the form


    with the initial conditions


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


    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


    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


    Then Eq. (39) has the form


    where y(w(ε)t)=z(t).

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




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

    Similarly, from equations


    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]:


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

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


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

    At first, we consider the example


    It has exact solution


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


    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


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


    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


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


    and equating coefficients of the same powers ε,we have


    From Eq. (47), we have


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


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


    Hence, solving this equation, we have


    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.,


    Hence, we have


    Then Eq. (49) takes the form


    Hence, we obtain y1(ξ)=0.

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


    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


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


    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


    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):


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

    Hence, equating the coefficients of equal powers has ε



    where q=q0,r=r0,


    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


    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


    where g(ξ)=exp(1ξ(q0q(s))s1ds).



    Hence, we have


    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.,


    Hence, using Eq. (60), we obtain


    Then Eq. (55) takes the form


    where a˜1=const. Hence, we have


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


    From this, we get


    Now Eq. (56) takes the form


    Solving this equation, we have


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


    Thus, the series (42) has the asymptotic


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


    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


    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


    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


    and further at second they will make transformation


    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


    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.,


    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


    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






    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


    It is easy to integrate this system, and we obtain


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

    Example 2 [37, 43])


    Uniformized equation is




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


    Hence if ξ0, we obtain


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

    We prove that x(ξ)+εy(ξ)0on the interval [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


    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


    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


    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
    Dπ0(t):=(t+π1(t))π0(t)+(π1(t)q(μ t))π0(t)=u0(tμ)π1(t),π0(μ0)=0E79.0
    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




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




    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



    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


    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


    Here, we have used the triangle inequality


    as well as the inequality


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


    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


    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


    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


    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


    This equation is integrated exactly


    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


    The solution of this problem can be expressed as


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


    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]


    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


    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


    for the sufficiently small parameter ε, where


    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


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

    Setting λ=0in Eq. (87), we have


    It has a unique solution


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


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




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




    To prove the theorem, we need next

    Lemma 11. The following estimate holds


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


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

    Further, by integrating by parts, we have




    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


    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.

    © 2017 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution 3.0 License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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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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