# Singular Initial Value Problem for Certain Classes of Systems of Ordinary Differential Equations

Authors: Josef Diblík; Josef Rebenda; Zdeněk Šmarda
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The paper is devoted to the study of the solvability of a singular initial value problem for systems of ordinary differential equations. The main results give sufficient conditions for the existence of solutions in the right-hand neighbourhood of a singular point. In addition, the
dimension of the set of initial data generating such solutions is estimated.
An asymptotic behavior of solutions is determined as well and relevant asymptotic formulas are derived. The method of functions defined implicitly and the topological method (Ważewski's method) are used in the proofs. The results generalize some previous ones on singular initial value problems for differential equations.


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## Body

## 1. Introduction

Letx0>0 and y0>0 be given constants. Define auxiliary set of points
(1)Dn:=(0,x0]×∏i=1n‍(0,y0i],wherey0i=y0.In the paper, we consider a system of ordinary differential equations in the form(2)gi(x,y)yi′=αi(x,y),wherei=1,…,n and functions gi,αi:Dn→(0,∞) can satisfy
(3)gr(0+,θ)=αr(0+,θ)=0for some indicesr∈{1,2,…,n} or
(4)1gs(0+,θ)=1αs(0+,θ)=0for some indicess∈{1,2,…,n}, where θ=(0,0,…,0) is the n-dimensional zero vector. Together with system (2) we consider the initial problem
(5)yi(0+)=0,i=1,…,n.Because of the above properties, the initial problem (2), (5) is called a singular initial problem.Definition 1.
Denote byM(x0,y0) a class of vector-functions φ:(0,x0]→ℝn having the following properties:  (1)
φis continuously differentiable on (0,x0];(2)
φi(0+)=0, i=1,…,n;(3)
0<φi(j)(x) for x∈(0,x0], j=0,1, i=1,…,n;(4)
φi(x0)<y0, i=1,…,n.Forφ∈M(x0,y0), define an auxiliary vector-function
(6)G(x,φ(x))=(G1(x,φ(x)),…,Gn(x,φ(x))),where(7)Gi(x,φ(x))≔-φi′(x)gi(x,φ(x))+αi(x,φ(x)),Gi(x,φ(x))≔-φi′(x)gi(x,φ(x))ddi=1,…,n.In the paper, sufficient conditions which guarantee the existence of a parametric class of solutions of initial value problem (2), (5) are given and asymptotic formulas
(8)yi(x)=φi(x)(1+o(1)),x⟶0+,i=1,…,nare derived, whereφi are the coordinates of a function φ∈M(x0,y0) and the symbol o(1) is the well-known Landau order symbol.There are numerous papers and books dealing with singular initial value problems (see, e.g., [1–16] and the references therein). Among others, we should mention pioneering results on the solvability of singular problems for ordinary differential equations achieved by Chechyk [15] and Kiguradze [13]. The results of the paper generalize previous investigation of the first author on the solvability of singular problems [5–7]. The main differences are as follows. In [5], a scalar singular differential equation was studied for the case that a function similar to the function G above does not change the sign for x→0+. In [6], system (2) is investigated under the assumption that the ith right-hand side of the system is bounded by the product of two functions, with the first depending only on the variable x while the second one only  depends on the variable yi, i=1,…,n. In comparison with the results of [7], we cannot expect that a first approximation of system (2) consists of equations with separable variables.The structure of the paper is the following. In Section2, auxiliary results on implicit functions are given. We refer to Corollary 4 where formula (27) is crucial for the proofs of the asymptotic behavior of solutions. The main results of the paper are formulated in Section 3. New results are proved and a progress is achieved by implicit construction of funnels, where solutions of the singular problem are expected. To prove the existence of such solutions, the topological method of Ważewski (see, e.g., [17–19]) is used. A simple illustrative example is shown here as well. A generalization of the results derived is discussed in Section 4.


## 2. Auxiliary Results on Implicit Functions

First, we give some properties of implicit functions used in the following proofs.Lemma 2.
Assume that a functionω:D1→ℝ satisfies the following conditions:(1)
ω(x,y) is continuously differentiable with respect to x and y;(2)
ω(0+,0+)=0;(3)
for everyy∈(0,y0], there exists a finite limit ω(0+,y), for every x∈(0,x0], there exists a finite limit ω(x,0+), and ω(0+,y)ω(x,0+)<0;(4)
ωx′(x,y)ωy′(x,y)<0, where (x,y)∈D1.
Then,(9)ω(x,y)=0
defines a unique implicit functiony=y(x) on some interval (0,x00], 0<x00≤x0 such that y(x)∈M(x00,y0).Proof.
Analysing assumptions (1)–(4), we deduce that only the following two cases can occur: either(10)ω(0+,y)<0,ωy′(x,y)<0,ω(x,0+)>0,ωx′(x,y)>0
or(11)ω(0+,y)>0,ωy′(x,y)>0,ω(x,0+)<0,ωx′(x,y)<0
while the remaining two cases(12)ω(0+,y)<0,ωy′(x,y)>0,ω(x,0+)>0,ωx′(x,y)<0,ω(0+,y)>0,ωy′(x,y)<0,ω(x,0+)<0,ωx′(x,y)>0
are in contradiction with assumptions(1) and (2). The rest of the proof is analogous to the proofs of the well-known implicit-function theorems and, therefore, we leave it out.To formulate the second lemma, we need some auxiliary notions. Define, for a giveny00>0 and ɛ*>0 satisfying the inequality 0<y00(1+ɛ*)<y0, the set
(13)D*(y00,ɛ*):={(y,ɛ):y∈[0,y00],ɛ∈(-ɛ*,ɛ*)}.Moreover, for a given continuously differentiable functionR:(0,y0)→(0,∞), let
(14)F(y,ɛ):=R(y(1+ɛ))R(y),where argumenty(1+ɛ) is assumed to be positive. In addition, we define
(15)F(0,ɛ):=limy→0+F(y,ɛ),Fɛ′(0,ɛ):=limy→0+Fɛ′(y,ɛ)provided that the limits exist and are finite.Lemma 3.
Assume that functions(16)R:(0,y0)⟶(0,∞),F:D*(y00,ɛ*)⟶[0,∞)
satisfy the following conditions:(1)
R(y) is continuously differentiable and R′(y)<0;(2)
F(y,ɛ) is continuous with respect to y and continuously differentiable with respect to ɛ;(3)
Fɛ′(y,0)≠0.
Then, for an arbitraryy∈[0,y00] and ɛ1∈(-ɛ~1,ɛ~1), where ɛ~1 is a positive and sufficiently small constant, there exists a unique continuous solution
(17)ɛ10=ɛ10(y,ɛ1)
of the equation(18)1+ɛ1-F(y,ɛ10)=0,
where(19)ɛ10(y,ɛ1):[0,y00]×(-ɛ~1,ɛ~1)⟶{(-ɛ~10,0)ifɛ1∈(0,ɛ~1),0ifɛ1=0,(0,ɛ~10)ifɛ1∈(-ɛ~1,0),
andɛ~10 is a sufficiently small positive constant, ɛ~10≤ɛ*.Proof.
Define an auxiliary function(20)ω:D*(y00,ɛ*)×(-ɛ~1,ɛ~1)⟶ℝ
as(21)ω(y,ɛ10,ɛ1):=1+ɛ1-F(y,ɛ10)
and consider implicit equation (18) in the form
(22)ω(y,ɛ10,ɛ1)=0
with respect toɛ10. In what follows, we will assume that y∈[0,y00] is a parameter. Since
(23)ω(y,0,0)=0,
ω
(
y
,
ɛ
1
0
,
ɛ
1
) is continuous with respect to all y, ɛ10, and ɛ1 and continuously differentiable with respect to ɛ10, and
(24)ωɛ10′(y,0,0)=-Fɛ10′(y,0)>0
for arbitraryy∈[0,y00]. Hence we can apply the classical implicit-function theorem. As a result, we state that (18) is uniquely solvable with respect to ɛ10. Thus
(25)ɛ10=ɛ10(y,ɛ1),
whereɛ10:[0,y00]×(-ɛ~1,ɛ~1)→ℝ is a continuous function with respect to both y and ɛ1 and ɛ~1 is a sufficiently small positive number. The sign of the function ɛ10(y,ɛ1) can be specified. In particular, since R is a decreasing function, the function F is decreasing with respect to ɛ,
(26)ɛ10=ɛ10(y,ɛ1):[0,y00]×(-ɛ~1,ɛ~1)⟶{(-ɛ~10,0)ifɛ1∈(0,ɛ~1),0ifɛ1=0,(0,ɛ~10)ifɛ1∈(-ɛ~1,0),
andɛ~10 is a sufficiently small positive constant satisfying ɛ~10≤ɛ*.Corollary 4.
It is possible to reformulate the statement of Lemma3 as follows. Since (18) is uniquely solvable, one can use the definition of F(y,ɛ) given by (14) to get
(27)R(y)(1+ɛ1)≡R(y(1-ɛ100(y,ɛ1))),
whereɛ100(y,ɛ1):=-ɛ10(y,ɛ1).


## 3. Main Results

In this part, the main results related to the solvability of problem (2), (5) are proved. We will discuss the dimension of the set of initial conditions generating solutions of this problem as well.Usingφ∈M(x0,y0), define the sets
(28)Nφi:={(x,y):(x,y)∈Dn,φi(x)<yi<φi(x0)},Nφi:={(x,y):(x,y)∈Dn,yi<φi(x)},wherei=1,2,…,n. To formulate the results we need auxiliary functions
(29)Wi:Nφi∪Nφi⟶ℝndefined as follows:(30)Wi(x,y)≔gi(φi-1(yi),φ(φi-1(yi)))αi(x,y)gi(x,y)Wi(x,y)≔-αi(x,φ(x))fori=1,2,…,n.Theorem 5.
Letgi:Dn→ℝn and αi:Dn→ℝn, i=1,…,n, be continuous functions. Let, moreover, for a function φ∈M(x0,y0), the following conditions be true:(
1
)
(31)
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,wherex00∈(0,x0) is a sufficiently small constant, x∈(0,x00], xki=φi-1(ψki), and ψki, k=1,2, i=1,…,n, are constants such that
(34)ψ1i<φi(x00)<ψ2i<y0,x2i<x0;(
4
)  there is an integer n1∈{0,1,…,n} such that(a)
Wi(x,y)>0 if (x,y)∈Nφi and i=1,…,n1;(b)
Wi(x,y)<0 if (x,y)∈Nφi and i=1,…,n1;(c)
Wi(x,y)<0 if (x,y)∈Nφi and i=n1+1,…,n;(d)
Wi(x,y)>0 if (x,y)∈Nφi and i=n1+1,…,n.Here, ifn1=0, conditions (a), (b) are omitted and, if n1=n, conditions (c), (d) are omitted.Then, problem (2), (5) has at least n1-parametric class of solutions y(x)=(y1(x),…,yn(x)) such that (x,y(x))∈Dn for x→0+.Proof.
The proof is divided into two parts. First, implicit curves are constructed and their properties are derived. Then, Ważewski’s method is applied to special domains having the shape of funnels with sides constructed using implicitly defined hypersurfaces. In this construction, we use implicit curves from the first part of the proof.
Implicitly Defined Curves and Their Properties. Let φ∈M(x0,y0) be fixed. Define auxiliary functions
(35)zki,z~ki:(0,x00]×(0,y0]⟶ℝ,k=1,2,i=1,…,n
as(36)zki(x,yki)≔∫ψkiyki‍gi(φi-1(t),φ(φi-1(t)))dtzki(x,yki)≔-∫x00x‍αi(t,φ(t))dt,z~ki(x,yki)≔[∫ψkiyki‍gi(φi-1(t),φ(φi-1(t)))dt]-1z~ki(x,yki)≔-[∫x00x‍αi(t,φ(t))dt]-1.
We prove that(37)zki(x,yki)=0,k=1,2,i=1,…,n,
define unique implicit functions(38)yki=yki(x)≡ψki(x)∈M(x00,y0)
on the interval(0,x00]. Observe that the function ψki(x) is a solution of
(39)z~ki(x,yki)=0,k=1,2,i=1,…,n,
on the interval(0,x00) as well. Therefore, we consider the latter equation and investigate its solvability using Lemma 2. Set
(40)ω(x,y):=z~ki(x,y),
wherek∈{1,2} and i∈{1,…,n}. We show that the function ω(x,y) satisfies all assumptions (1)–(4) of Lemma 2, where, instead of the region D1, we assume the region (0,x00*]×(0,y0*] with sufficiently small x00*, y0* such that 0<x00*<x00, 0<y0*<y0.
(a) It is easy to see (in view of the above assumptions) that the function(41)ω:(0,x00*]×(0,y0*]⟶ℝ
defined by (39) is continuously differentiable with respect to x and y and assumption (1) of Lemma 2, holds.
(b) Compute the limitω(0+,0+). We get
(42)ω(0+,0+)≡z~ki(0+,0+)=limx→0+,y→0+[[∫x00x‍αi(t,φ(t))dt]-1[∫ψkiy‍gi(φi-1(t),φ(φi-1(t)))dt]-1=limx→0+y→0+D-[∫x00x‍αi(t,φ(t))dt]-1]=limy→0+[∫ψkiy‍gi(φi-1(t),φ(φi-1(t)))dt]-1-limx→0+[∫x00x‍αi(t,φ(t))dt]-1
if the last two limits exist and are finite. Substitutingt=φi(x) into the first integral of the last expression and using condition (2), we get
(43)ω(0+,0+)≡z~ki(0+,0+)ω(0+,0+)=limy→0+[∫φi-1(ψki)φi-1(y)‍gi(x,φ(x))φi′(x)dx]-1ω(0+,0+)=-limx→+0[∫x00x‍αi(t,φ(t))dt]-1ω(0+,0+)=[∫xki0+‍gi(x,φ(x))φi′(x)dx]-1ω(0+,0+)=-[∫x000+‍αi(x,φ(x))dx]-1=0
and assumption(2) of Lemma 2 holds.
(c) Now we consider the existence of the productω(0+,y)ω(x,0+) for y∈(0,y0*], x∈(0,x00*] and determine its sign. We get
(44)ω(0+,y)ω(x,0+)≡z~ki(0+,y)z~ki(x,0+)=limx→0+[[∫x00x‍αi(t,φ(t))dt]-1[∫ψkiy‍gi(φi-1(t),φ(φi-1(t)))dt]-1-[∫x00x‍αi(t,φ(t))dt]-1]×limy→0+[[∫ψkiy‍gi(φi-1(t),φ(φi-1(t)))dt]-1-[∫x00x‍αi(t,φ(t))dt]-1].
Substitutingt=φi(x) into the first integrals in the square brackets, using conditions (1) and (2) of Theorem 5 and the property x00*<x00, and assuming y0*<mink=1,2;i=1,…,n{ψki}, we have
(45)ω(0+,y)ω(x,0+)≡z~ki(0+,y)z~ki(x,0+)=[[∫φi-1(ψki)φi-1(y)‍gi(x,φ(x))φi′(x)dx]-1-[∫x000+‍αi(x,φ(x))dx]-1[∫φi-1(ψki)φi-1(y)‍gi(x,φ(x))φi′(x)dx]-1]×[[∫φi-1(ψki)0+‍gi(x,φ(x))φi′(x)dx]-1-[∫x00x‍αi(x,φ(x))dx]-1]=[∫xkiφi-1(y)‍gi(x,φ(x))φi′(x)dx]-1×[∫x00x‍αi(x,φ(x))dx]-1·(-1)<0.
(d) Determine the sign ofωx′(x,y)ωy′(x,y). We get
(46)ωx′(x,y)·ωy′(x,y)≡(z~ki(x,y))x′·(z~ki(x,y))y′=(-1)·αi(x,φ(x))[∫x00x‍αi(t,φ(t))dt]-2×[∫ψkiy‍gi(φi-1(t),φ(φi-1(t)))dt]-2×gi(φi-1(y),φ(φi-1(y))).
From condition(1), it follows that
(47)αi(x,φ(x))>0,gi(φi-1(y),φ(φi-1(y)))>0.
Hence(48)ωx′(x,y)·ωy′(x,y)≡(z~ki(x,y))x′·(z~ki(x,y))y′<0.
Because of (a)–(d) all assumptions (1)–(4) of Lemma2 on (0,x00*]×(0,y0*] hold and (39) defines an implicit function
(49)y(x)=yki(x)=ψki(x)
on some interval(0,x00**],  where x00**≤x00*.
Now we turn to (37) and show that its solution given by formula (49) can be extended beyond x00**.
We show thatx00**=x00. On the contrary, assume x00**<x00. Then, (after a proper transformation of variables) we can apply Lemma 2 to the point (x00**,ψki(x00**))  again and, by well-known procedure, implicit function can be continued up to the boundary of the region (0,x00]×(0,y0].
Ifψki(x00**)=y0, then z~ki(x00**,y0)=zki(x00**,y0)=0. Moreover, we have
(50)zki(x00**,y0)=∫ψkiy0‍gi(φi-1(t),φ(φi-1(t)))dtzki(x00**,y0)=-∫x00x00**‍αi(t,φ(t))dt>0
sincex00**<x00 and ψki<y0 by condition (3). This is a contradiction and x00**=x00. Therefore, the implicit function yki=ψki(x) can be continued on the whole interval (0,x00]. Similarly we will show that the inequality
(51)ψki(x00)<y0
holds. We have(52)zki(x00,y0)=∫ψkiy0‍gi(φi-1(t),φ(φi-1(t)))dt-∫x00x00‍αi(t,φ(t))dt=gi(φi-1(t),φ(φi-1(t)))dt>0,
because, by condition(3), ψki<y0 holds. It is obvious that
(53)zki(x00,ψki(x00))=0,(zki(x00,y))y′=gi(φi-1(y),φ(φi-1(y)))>0.
Because the functionzki(x00,y) is monotonously increasing and
(54)0=zki(x00,ψki(x00))<zki(x00,y0),
we getψki(x00)<y0. Hence inequality (51) is proved.
Now we will investigate the behavior of implicit curves in a neighborhood of the functionφ. Since
(55)ψ1i<φi(x00)<ψ2i,
we have (by condition (8))
(56)φi(x1i)<φi(x00)<φi(x2i),x1i<x00<x2i.
Thus, (in the first integral we substitutet=φi(s))
(57)zki(x,φi(x))=∫ψkiφi(x)‍gi(φi-1(t),φ(φi-1(t)))dt-∫x00x‍αi(t,φ(t))dt=∫φi-1(ψki)x‍gi(s,φ(s))φi′(s)ds-∫x00x‍αi(t,φ(t))dt=∫x00x‍[gi(t,φ(t))φi′(t)-αi(t,φ(t))]dt+∫xkix00‍gi(t,φ(t))φi′(t)dt=-∫x00x‍Gi(t,φ(t))dt-∫x00xki‍gi(t,φ(t))φi′(t)dt,
wherek=1,2 and i=1,…,n. Consequently, we deduce that
(58)z1i(x,φi(x))>0,z2i(x,φi(x))<0,x∈(0,x00].
Since functionsz1i, z2i increase with respect to their second co-ordinates and
(59)z1i(x,ψ1i(x))≡0,z2i(x,ψ2i(x))≡0
on(0,x00], we get
(60)ψ1i(x)<φi(x)<ψ2i(x),x∈(0,x00]
for eachi∈{1,…,n}. Finally, we recall that
(61)ψki(0+)=φi(0+)=0,k=1,2,i=1,…,n.
Application of Ważewski’s Method to an Implicitly Defined Domain. In the next part of the proof we will apply the topological method of Ważewski. We use the above mentioned functions given implicitly to define an open set
(62)Ω0≔{(x,y):(x,y)∈Dn,vk(x,y)<0,uj(x,y)k=0,…,n1,uj(x,y)<0,j=n1+1,…,n},
where(63)v0(x,y)≡v0(x)≔x-x^,0<x^<x00,x^isaconstant,vk(x,y)≡vk(x,yk)≔(yk-ψ1k(x))(yk-ψ2k(x)),k=1,…,n1,uj(x,y)≡uj(x,yj)≔(yj-ψ1j(x))(yj-ψ2j(x)),j=n1+1,…,n.
Now we start to investigate the behavior of the integral curves of system (2) with respect to the boundary of the set Ω0, that is, on the sets
(64)Vβ={(x,y):(x,y)∈Dn,vβ(x,y)=0,vk(x,y)≤0,k=0,…,n1,k≠β,uj(x,y)≤0,j=n1+1,…,n},β=0,…,n1,Uα={(x,y):(x,y)∈Dn,uα(x,y)=0,vk(x,y)≤0,k=0,…,n1,uj(x,y)≤0,j=n1+1,…,n,j≠α},α=n1+1,…,n.
First, we calculate the full derivativev˙β(x,y) of the function vβ(x,y) along trajectories of system (2) on the set Vβ, β=0,…,n1. It is clear that v˙0(x)=1>0. Further, for β=1,…,n1, we have
(65)v˙β(x,y)=ddx[(yβ-ψ1β(x))(yβ-ψ2β(x))]v˙β(x,y)=[ddx(yβ-ψ1β(x))](yβ-ψ2β(x))v˙β(x,y)=+(yβ-ψ1β(x))[ddx(yβ-ψ2β(x))]v˙β(x,y)=[αβ(x,y)gβ(x,y)-ψ1β′(x)](yβ-ψ2β(x))v˙β(x,y)=+[αβ(x,y)gβ(x,y)-ψ2β′(x)](yβ-ψ1β(x)).
On the setVβ, as it follows from the condition vβ(x,y)=0, we have either yβ=ψ1β(x) or yβ=ψ2β(x).
(1) Letyβ=ψ1β(x). Then, we can see that
(66)v˙β(x,ψ1β(x))=[αβ(x,y1,…,yβ-1,ψ1β(x),yβ+1,…,yn)gβ(x,y1,…,yβ-1,ψ1β(x),yβ+1,…,yn)-ψ1β′(x)]×(ψ1β(x)-ψ2β(x)).
The derivativeψ1β′(x) of the function ψ1β(x) can be calculated using the well-known rules for differentiation of implicit functions given by identities
(67)z1β(x,ψ1β(x))≡0.
We get(68)ψ1β′(x)≡αβ(x,φ(x))gβ[φβ-1(ψ1β(x)),φ(φβ-1(ψ1β(x)))].
Using that relation, we have(69)v˙β(x,ψ1β(x))=[αβ(x,y1,…,yβ-1,ψ1β(x),yβ+1,…,yn)gβ(x,y1,…,yβ-1,ψ1β(x),yβ+1,…,yn)αβ(x,φ(x))gβ[φβ-1(ψ1β(x)),φ(φβ-1(ψ1β(x)))]+(-1)αβ(x,φ(x))gβ[φβ-1(ψ1β(x)),φ(φβ-1(ψ1β(x)))]]×(ψ1β(x)-ψ2β(x))=ψ1β(x)-ψ2β(x)gβ[φβ-1(ψ1β(x)),φ(φβ-1(ψ1β(x)))]×[αβ(x,y1,…,yβ-1,ψ1β(x),yβ+1,…,yn)gβ(x,y1,…,yβ-1,ψ1β(x),yβ+1,…,yn)gβ[φβ-1(ψ1β(x)),φ(φβ-1(ψ1β(x)))]×αβ(x,y1,…,yβ-1,ψ1β(x),yβ+1,…,yn)gβ(x,y1,…,yβ-1,ψ1β(x),yβ+1,…,yn)-αβ(x,φ(x))αβ(x,y1,…,yβ-1,ψ1β(x),yβ+1,…,yn)gβ(x,y1,…,yβ-1,ψ1β(x),yβ+1,…,yn)].
Since, by (60), ψ1β(x)<ψ2β(x) and ψ1β(x)<φβ(x)<ψ2β(x), assumption (1) of the theorem yields
(70)gβ[φβ-1(ψ1β(x)),φ(φβ-1(ψ1β(x)))]>0.
In view of assumption (4b) (Wβ(x,y)<0 when (x,y)∈Nφβ), we get
(71)gβ[φβ-1(ψ1β(x)),φ(φβ-1(ψ1β(x)))]×αβ(x,y1,…,yβ-1,ψ1β(x),yβ+1,…,yn)gβ(x,y1,…,yβ-1,ψ1β(x),yβ+1,…,yn)-αβ(x,φ(x))<0
and, consequently,(72)v˙β(x,ψ1β(x))>0.
(2) Letyβ=ψ2β(x). Then, by similar calculations and using assumption (4a) (Wβ(x,y)>0 when (x,y)∈Nφβ), we obtain
(73)v˙β(x,ψ2β(x))>0.
Hencev˙β(x,y)>0 for all β=1,…,n1.
Now we will calculate the full derivativeu˙α(x,y) of the function uα(x,y) along trajectories of system (2) on the set Uα, where α=n1+1,…,n. As above, we get
(74)u˙α(x,y)=ddx[(yα-ψ1α(x))(yα-ψ2α(x))]u˙α(x,y)=[ddx(yα-ψ1α(x))](yα-ψ2α(x))u˙α(x,y)=+(yα-ψ1α(x))[ddx(yα-ψ2α(x))]u˙α(x,y)=[αα(x,y)gα(x,y)-ψ1α′(x)](yα-ψ2α(x))u˙α(x,y)=+[αα(x,y)gα(x,y)-ψ2α′(x)](yα-ψ1α(x)).
On the setUα, as it follows from the condition uα(x,y)=0, we have either yα=ψ1α(x) or yα=ψ2α(x).
(1) Letyα=ψ1α(x). Then, we get (proceeding like in the previous part of the proof)
(75)u˙α(x,ψ1α(x))=[αα(x,y1,…,yα-1,ψ1α(x),yα+1,…,yn)gα(x,y1,…,yα-1,ψ1α(x),yα+1,…,yn)+(-1)αα(x,φ(x))gα[φα-1(ψ1α(x)),φ(φα-1(ψ1α(x)))]]×(ψ1α(x)-ψ2α(x))=ψ1α(x)-ψ2α(x)gα[φα-1(ψ1α(x)),φ(φα-1(ψ1α(x)))]×[αα(x,y1,…,yα-1,ψ1α(x),yα+1,…,yn)gα(x,y1,…,yα-1,ψ1α(x),yα+1,…,yn)gα[φα-1(ψ1α(x)),φ(φα-1(ψ1α(x)))]×αα(x,y1,…,yα-1,ψ1α(x),yα+1,…,yn)gα(x,y1,…,yα-1,ψ1α(x),yα+1,…,yn)[φα-1(ψ1α(x)),φ(φα-1(ψ1α(x)))]-αα(x,φ(x))αα(x,y1,…,yα-1,ψ1α(x),yα+1,…,yn)gα(x,y1,…,yα-1,ψ1α(x),yα+1,…,yn)].
Since, by (60), ψ1α(x)<ψ1β(x) and ψ1α(x)<φα(x)<ψ2α(x), applying assumption (1) of the theorem, we have
(76)gα[φα-1(ψ1α(x)),φ(φα-1(ψ1α(x)))]>0.
In view of assumption (4d) (Wα(x,y)>0 when (x,y)∈Nφα), we have
(77)gα[φα-1(ψ1α(x)),φ(φα-1(ψ1α(x)))]×αα(x,y1,…,yα-1,ψ1α(x),yα+1,…,yn)gα(x,y1,…,yα-1,ψ1α(x),yα+1,…,yn)-αα(x,φ(x))>0
and, consequently(78)u˙α(x,ψ1α(x))<0.
(2) Letyα=ψ2α(x). Then, by similar calculations and using assumption (4)-(c) (Wα(x,y)<0 when (x,y)∈Nφα), we obtain
(79)u˙α(x,ψ2α(x))<0.
Thus,u˙α(x,y)<0 for all α=n1+1,…,n.
By [18, Lemma  3.1, page  281], for decreasing values of the variable x, the set of all egress points Ωe0, Ωe0⊂Ω0, from the set Ω0 equals the set of all strict egress points Ωse0, Ωse0⊂Ω0 from the set Ω0 (for definitions of  Ωe0 and Ωse0 see, for example, [18, page  37 and page 278]); that is,
(80)Ωe0=Ωse0=⋃α=n1+1n‍Uα∖⋃β=0n1‍Vβ.
LetS be a subset of Ω0∪Ωe0 defined as
(81)S={(x~,yk0)(x~,y10,…,yn10,yn1+1,…,yn):uj(x~,yj)≤0,j=n1+1,…,n,v0(x~)<0,vk(x~,yk0)<0,k=1,…,n1(x~,yk0)},
whereyk0, k=1,…,n1, are fixed. Then,
(82)S∩Ωe0={(x~,y10,…,yn10,yn1+1,…,yn):uα(x~,yα)S∩Ωe0=0,uj(x~,yj)≤0,α,j=n1+1,…,n,S∩Ωe0α≠j,vk(x~,yk0)<0,k=1,…,n1}.
We can see that the setS∩Ωe0 is a subset of the boundary ∂S of the set S, but it is not a retract of S. The explanation is simple and is based on the well-known fact that the boundary of an (n-n1)-dimensional ball is not its retract [20], and the set S is topologically equivalent to an (n-n1)-dimensional ball.
We show thatS∩Ωe0 is a retract of Ωe0. Define a mapping
(83)π:Ωe0∋(x,y)⟼(x~,y10,…,yn10,y~n1+1,…,y~n)∈S∩Ωe0,
where(84)y~j=ψ1j(x~)+(yj-ψ1j(x))ψ2j(x~)-ψ1j(x~)ψ2j(x)-ψ1j(x),j=n1+1,…,n.
With respect to the behavior of functionsψkj(x), k=1,2, j=n1+1,…,n, the mapping π is continuous. From the definition of the mapping π, we get that S∩Ωe0 is a retract of Ωe0 and, furthermore, S is a compact set.
By seeing Corollary  3.1, [18, page 282] of Ważewski’s theorem [18, Theorem 3.1, page 282] there exists a solution of system (2) with the initial conditions in the set S∩Ωe0 and it is contained in Ω0 for x∈(0,x~]. This solution satisfies (5) since the set Ωe0 is contracted to the initial point for x→0+.
As we can change the constantsyk0, k=1,…,n1 within the inequality vk(x~,yk0)<0, we can repeat the above-mentioned construction for every admissible fixed set (y10,…,yn10). Then, there exists a class of solutions depending on n1 parameters and lying in Ω0 for x∈(0,x~]. If n1=n, the assertion of theorem remains true as well. The proof is complete.Remark 6.
Ifφ∈M(x0,y0) is a solution of initial problem (2), (5), then
(85)Gi(x,φ(x))=-φi′(x)gi(x,φ(x))+αi(x,φ(x))≡0,
wherex∈(0,x0] and i=1,…,n, and
(86)G(x,φ(x))≡θ,x∈(0,x0].
Then, condition(3) of Theorem 5 is satisfied and, in this case, we obtain the result on the dimension of the set of initial data generating solutions of initial problem (2), (5).Denote(87)Ri(y)≔∫yy0‍gi(φi-1(t),φ(φi-1(t)))dt,Fi(y,ɛ)≔Ri(y(1+ɛ))Ri(y),Fi(0,ɛ)≔limy→0+Fi(y,ɛ),(Fi)ɛ′(0,ɛ):=limy→0+(Fi)ɛ′(y,ɛ),i
=
1
,
…
,
n, provided that the limits exist and are finite.Theorem 7.
Let all assumptions of Theorem5 hold and, moreover, the functions
(88)Fi:D*(y00,ɛ*)⟶[0,∞),i=1,…,n,
satisfy  the following:(1)
Fi(y,ɛ) is continuous with respect to y and continuously differentiable with respect to ɛ;(2)
(Fi)ɛ′(y,0)≠0.
Then arbitrary solutiony(x)=(y1(x),…,yn(x)) of initial problem (2), (5) mentioned in Theorem 5 has the asymptotic form
(89)yi(x)=φi(x)(1+o(1)),x⟶0+,i=1,…,n.Proof.
From the proof of Theorem5, it follows that, for coordinates yi(x), i=1,…,n, of the solution y(x) of initial problem  (2), (5), the inequalities
(90)ψ1i(x)<yi(x)<ψ2i(x),i=1,…,n
are valid on an interval(0,x~] and we can assume that the inequalities
(91)ψ1i(x)<φi(x)<ψ2i(x),i=1,…,n
are valid on(0,x~] as well. Thus, to prove (89), it is sufficient to prove that
(92)limx→0+ψ2i(x)ψ1i(x)=1,i=1,…,n.
Applying L’Hospital’s rule to the limits (92), we do not obtain the desired result. Therefore we will apply L’Hospital’s rule to the limits
(93)limx→0+Ri(ψ2i(x))Ri(ψ1i(x)),i=1,…,n.
This is possible because, in view of condition(2) of the theorem, we obviously have
(94)limx→0+Ri(ψki(x))=∞,i=1,…,n,k=1,2.
Then, we use the auxiliary results for implicit functions. Applying the above-mentioned procedure, we get (fori∈{1,…,n})
(95)limx→0+Ri(ψ2i(x))Ri(ψ1i(x))=limx→0+[Ri(ψ2i(x))]′[Ri(ψ1i(x))]′=limx→0+-gi[φi-1(ψ2i(x)),φ(φi-1(ψ2i(x)))]ψ2i′(x)-gi[φi-1(ψ1i(x)),φ(φi-1(ψ1i(x)))]ψ1i′(x)=limx→0+gi[φi-1(ψ2i(x)),φ(φi-1(ψ2i(x)))]gi[φi-1(ψ1i(x)),φ(φi-1(ψ1i(x)))]×αi(x,φ(x))/gi[φi-1(ψ2i(x)),φ(φi-1(ψ2i(x)))]αi(x,φ(x))/gi[φi-1(ψ1i(x)),φ(φi-1(ψ1i(x)))]=limx→0+gi[φi-1(ψ2i(x)),φ(φi-1(ψ2i(x)))]gi[φi-1(ψ2i(x)),φ(φi-1(ψ2i(x)))]×gi[φi-1(ψ1i(x)),φ(φi-1(ψ1i(x)))]gi[φi-1(ψ1i(x)),φ(φi-1(ψ1i(x)))]×αi(x,φ(x))αi(x,φ(x))=1.
Hence it follows that there exists a sufficiently small constantɛi>0 such that the inequalities
(96)Ri(ψ1i(x))(1-ɛi)<Ri(ψ2i(x))<Ri(ψ1i(x))(1+ɛi),i=1,…,n,
hold on some interval(0,x*], 0<x*≤x~, where x* is a sufficiently small constant. Applying Lemma 3 and Corollary 4 (formula (27)) we show that inequalities (96) can be written as
(97)Ri[ψ1i(x)(1+ɛi2(ψ1i(x),ɛi))]<Ri(ψ2i(x))<Ri[ψ1i(x)(1-ɛi1(ψ1i(x),ɛi))],i=1,…,n,
whereɛij(ψ1i(x),ɛi), j=1,2, are constants depending on ψ1i(x) and ɛi such that
(98)0<ɛij(ψ1i(x),ɛi),limɛi→0+ɛij(ψ1i(x),ɛi)=0,j=1,2.
In Lemma3 we set, for every fixed index i=1,…,n,
(99)R(y):=Ri(y)=∫yy0‍gi(φi-1(t),φ(φi-1(t)))dt,F(y,ɛ):=Fi(y,ɛ)=Ri(y(1+ɛ))Ri(y).
Then, we haveR:(0,y0)→(0,∞), F:D*(y00,ɛ*)→[0,∞) and  we have the following:(a)
R(y) is continuously differentiable, R′(y)=-gi(φi-1(y),φ(φi-1(y)))<0;(b)
F(y,ɛ) is continuous with respect to y and continuously differentiable with respect to ɛ;(c)
Fɛ′(y,0)≠0 by condition (2) of the theorem.
Hence, Lemma3 holds. By Corollary 4, we can write inequalities (96) in the form of  (97). From (97), we get
(100)ψ1i(x)(1-ɛi1(ψ1i(x),ɛi))<ψ2i(x)<ψ1i(x)(1+ɛi2(ψ1i(x),ɛi))
for(0,x*]. The last inequalities are equivalent to
(101)limx→+0ψ2i(x)ψ1i(x)=1,i=1,…,n.
The proof is complete.Example 8.
Consider the following simple particular case of initial problem   (2), (5):
(102)x2(1+x2y6)y′=y2(1+xy7),(103)y(0+)=0.
This problem is a singular one. To apply Theorem5, we rewrite (102), (103) as
(104)1y2y′=1+xy7x2(1+x2y6),y(0+)=0.
Setn=1 (below we omit this index since we deal with a scalar problem), φ(x)=x, and
(105)g(x,y)=1y2,α(x,y)=1+xy7x2(1+x2y6).
Without loss of generality, we assume thatx0 and y0 in definition of D1 are positive and sufficiently small (from Definition (1), property (4), it follows x0<y0). Condition (1), obviously holds. Condition (2), is valid as well because
(106)∫0+‍g(x,φ(x))φ′(x)dx=∫0+‍1x2dx=∞,∫0+‍α(x,φ(x))dx=∫0+‍1+x8x2(1+x8)dx=∫0+‍1x2dx=∞.
Sinceφ is a solution of (104), we have G(x,φ(x))≡0 and Condition (3) holds (see Remark 6 as well). Next,
(107)W(x,y)=g(φ-1(y),φ(φ-1(y)))α(x,y)g(x,y)-α(x,φ(x))=g(y,y)α(x,y)g(x,y)-α(x,x)=1y2(1+xy7)/(x2(1+x2y6))1/y2-1x2=y6(y-x)x(1+x2y6).
Setn1=1 (i.e., conditions (4c) and (4d) are omitted). For x and y sufficiently small and positive, it is easy to see that W(x,y)>0 if y>x and W(x,y)<0 if y<x. Hence, conditions (4a) and (4b) hold. According to Theorem 5, problem (102), (103) has at least one-parametric class of solutions.
Now we apply Theorem7. We get
(108)R(y)=∫yy0‍g(φ-1(t),φ(φ-1(t)))dtR(y)=∫yy0‍g(t,t)dt=∫yy0‍1t2dt=1y-1y0,F(y,ɛ)=R(y(1+ɛ))R(y)=1/y(1+ɛ)-1/y01/y-1/y0.
Now we see that condition(1) holds since F is continuous with respect to y and continuously differentiable with respect to ɛ for (y,ɛ)∈D*(y00,ɛ*). The values F(0,ɛ) and Fɛ′(0,ɛ) are computed by above given formulas. Moreover, as
(109)∂F∂ɛ(y,0)=-1/y(1+ɛ)21/y-1/y0|ɛ=0=y0y-y0≠0
fory∈(0,y00] with y00<y0  and
(110)∂F∂ɛ(0,0)=limy→0+∂F∂ɛ(y,0)=-limy→0+1/y1/y-1/y0∂F∂ɛ(0,0)=limy→0+y0y-y0=-1≠0.
condition(2) holds too. Theorem 7 is valid and every solution of problem (102), (103) mentioned in Theorem 5 satisfies (89); that is,
(111)y(x)=x(1+o(1))
forx→0+.
Finally, we remark that, instead of (102), the same investigation can be performed, for example, for
(112)x2(1+x2y6)y′=y2(1+xy7+y20),
where the functionφ is no longer a solution and condition (3) of Theorem 5 holds (the moment x00 can be sufficiently small).


## 4. Generalization

The following theorem improves Theorems5 and 7. In particular, comparing assumptions of Theorems 5 and 7 with Theorem 9, we see that the corresponding assumptions in Theorem 9 are assumed to be valid on some sets that are reductions of sets used in Theorems 5 and 7.For given constantsCi*>1 and 0<Ci**<1, i=1,…,n, and for φ∈M(x0,y0), define sets
(113)Nφi(Ci*):={(x,y):(x,y)∈Dn,φi(x)<yi<Ci*φi(x)},Nφi(Ci**):={(x,y):(x,y)∈Dn,Ci**φi(x)<yi<φi(x)},N≔{(x,y):(x,y)∈Dn,Ci**φi(x)<yi<Ci*φi(x),i=1,…,n}.Theorem 9.
Let, for some constantsCi*, Ci**, Ci*>1, 0<Ci**<1, i=1,…,n, and some function φ∈M(x0,y0), the following conditions be fulfilled.  (a)
Considergi(x,y)∈C(N), αi(x,y)∈C(N), i=1,…,n.(b)
Assumptions (1)–(3) of Theorem5 hold.(c)
Fori=1,…,n1,
(114)Wi(x,y)>0if(x,y)∈Nφi(Ci*)∩N,Wi(x,y)<0if(x,y)∈Nφi(Ci**)∩N.
We omit this assumption ifn1=0. (d)
Fori=n1+1,…,n,
(115)Wi(x,y)<0if(x,y)∈Nφi(Ci*)∩N,Wi(x,y)>0if(x,y)∈Nφi(Ci**)∩N.
We omit this assumption ifn1=n. (e)
FunctionsFi(y,ɛ), i=1,…,n, satisfy the assumptions of Theorem 7.
Then, there exists at least ann1-parametric family of solutions
(116)y(x)=(y1(x),…,yn(x))
such that(117)yi(x)=φi(x)(1+o(1)),i=1,…,n,
holds forx→0+ and (x,y(x))∈Dn.Proof.
The proof is omitted since it is similar to the proofs of Theorems5 and 7.



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*Source: 207352-2013-12-10.xml*