# ON SINGULAR BOUNDARY VALUE PROBLEMS FOR FUNCTIONAL DIFFERENTIAL EQUATIONS OF HIGHER ORDER ```Georgian Mathematical Journal
Volume 8 (2001), Number 4, 791–814
ON SINGULAR BOUNDARY VALUE PROBLEMS FOR
FUNCTIONAL DIFFERENTIAL EQUATIONS OF HIGHER
ORDER
I. KIGURADZE, B. PŮŽA, AND I. P. STAVROULAKIS
Dedicated to the memory of Professor N. Muskhelishvili
on the occasion of his 110th birthday
Abstract. Sufficient conditions are established for the solvability of the
boundary value problem
x(n) (t) = f (x)(t), hi (x) = 0 (i = 1, . . . , n),
where f is an operator (hi (i = 1, . . . , n) are operators) acting from some
subspace of the space of (n − 1)-times differentiable on the interval ]a, b[
m-dimensional vector functions into the space of locally integrable on ]a, b[
m-dimensional vector functions (into the space Rm ).
2000 Mathematics Subject Classification: 34K10.
Key words and phrases: singular functional differential equation, boundary value problem, Fredholm property, a priori boundedness principle.
1. Formulation of the Main Results
1.1. Formulation of the problem and a brief survey of literature. Consider the functional differential equation of n-th order
x(n) (t) = f (x)(t)
(1.1)
hi (x) = 0 (i = 1, . . . , n).
(1.2)
with the boundary conditions
When the operators f : C n−1 ([a, b]; Rm ) → L([a, b]; Rm ) and hi : C n−1 ([a, b]; Rm )
→ Rm (i = 1, . . . , n) are continuous, problem (1.1), (1.2) is called regular. If the
operator f (operators hi (i = 1, . . . , n)) acts from some subspace of the space
C n−1 (]a, b[ ; Rm ) into the space Lloc (]a, b[ ; Rm ) (into the space Rm ), problem
(1.1), (1.2) is called singular.
The basic principles of the theory of a wide enough class of regular problems
of form (1.1), (1.2) are constructed in the monographs , , . Optimal
sufficient conditions for such problems to be solvable and uniquely solvable are
given in , , –, , , –, .
c Heldermann Verlag www.heldermann.de
ISSN 1072-947X / \$8.00 / &deg;
792
I. KIGURADZE, B. PŮŽA, AND I. P. STAVROULAKIS
As to singular problems of form (1.1), (1.2), they have been studied with
sufficient completeness in the case with the operator f having the form
&sup3;
&acute;
f (x)(t) = g t, x(t), . . . , x(n−1) (t)
(see , , –, –, ,  and the references cited therein). For
the singular functional differential equation (1.1), the weighted initial problem
is studied in , , two-point problems in , , , , , –,
whereas the multi-point Vallée-Poussin problem in . In the general case the
singular problem (1.1), (1.2) remains studied but little. An attempt is made in
this paper to fill up this gap to some extent.
Throughout the paper the following notation will be used.
R = ] − ∞, +∞[ , R+ = [0, +∞[ .
Rm is the space of m-dimensional column vectors x = (xi )m
i=1 with the components xi ∈ R (i = 1, . . . , m) and the norm
m
X
kxk =
|xi |.
i=1
m
Rm
+ = {x = (xi )i=1 : xi ∈ R+ (i = 1, . . . , m)}.
m&times;m
R
is the space of m &times; m matrices X = (xik )m
i,k=1 with the components
xik ∈ R (i, k = 1, . . . , m) and the norm
kXk =
m
X
|xik |.
i,k=1
m
m
m&times;m
If x = (xi )m
, then
i=1 ∈ R and X = (xik )i,k=1 ∈ R
m
|x| = (|xi |)m
i=1 and |X| = (|xik |)i,k=1 .
Rm&times;m
= {X = (xik )m
+
i,k=1 : xik ∈ R+ (i, k = 1, . . . , m)}.
r(X) is the spectral radius of the matrix X ∈ Rm&times;m .
Inequalities between matrices and vectors are understood componentwise,
m
m
m
i.e., for x = (xi )m
i=1 , y = (yi )i=1 , X = (xik )i,k=1 and Y = (yik )i,k=1 we have
x ≤ y ⇐⇒ xi ≤ yi (i = 1, . . . , m)
and
X ≤ Y ⇐⇒ xik ≤ yik (i, k = 1, . . . , m).
If k is a natural number and ε ∈ ]0, 1[ , then
(k − ε)! =
k
Y
(i − ε).
i=1
If m and n are natural numbers, −∞ &lt; a &lt; b &lt; +∞, α ∈ R and β ∈ R, then
n−1
Cα,β
(]a, b[ ; Rm ) is the Banach space of (n − 1)-times continuously differentiable
vector functions x : ]a, b[ → Rm having limits
lim(t − a)αi x(i−1) (t),
t→a
lim(b − t)βi x(i−1) (t) (i = 1, . . . , n),
t→b
(1.3)
ON SINGULAR BOUNDARY VALUE PROBLEMS
793
where
αi =
α + i − n + |α + i − n|
β + i − n + |β + i − n|
, βi =
2
2
(i = 1, . . . , n).
(1.4)
The norm of an arbitrary element x of this space is defined by the equality
kxkC n−1 = sup
α,β
&frac12;X
n
&frac34;
(t − a)αi (b − t)βi kx(i−1) (t)k : a &lt; t &lt; b .
k=1
n−1
n−1
Ceα,β
(]a, b[ ; Rm ) is the set of x ∈ Cα,β
(]a, b[ ; Rm ) for which x(n−1) is locally
absolutely continuous on ]a, b[ , i.e., absolutely continuous on [a + ε, b − ε] for
arbitrarily small positive ε.
Lα,β (]a, b[ ; Rm ) and Lα,β (]a, b[ ; Rm&times;m ) are respectively the Banach space of
vector functions y : ]a, b[ → Rm and the Banach space of matrix functions Y :
]a, b[ → Rm&times;m whose components are summable with weight (t − a)α (b − t)β .
The norms in these spaces are defined by the equalities
Zb
Zb
α
kykLα,β =
β
(t − a) (b − t) ky(t)k dt,
(t − a)α (b − t)β kY (t)k dt.
kY kLα,β =
a
a
m
m
Lα,β (]a, b[ ; Rm
+ ) = {y ∈ Lα,β (]a, b[ ; R ) : y(t) ∈ R+ for t ∈ ]a, b[ }.
m&times;m
m&times;m
Lα,β (]a, b[ ; R+ ) = {Y ∈ Lα,β (]a, b[ ; R
) : Y (t) ∈ Rm&times;m
for t ∈ ]a, b[ }.
+
In the sequel it will always be assumed that −∞ &lt; a &lt; b &lt; +∞,
α ∈ [0, n − 1], β ∈ [0, n − 1],
(1.5)
n−1
n−1
whereas f : Cα,β
(]a, b[ ; Rm ) → Lα,β (]a, b[ ; Rm ) and hi : Cα,β
(]a, b[ ; Rm ) → Rm
(i = 1, . . . , n) are continuous operators which, for each ρ ∈ ]0, +∞[, satisfy the
conditions
n
o
sup kf (x)(&middot;)k : kxkC n−1 ≤ ρ ∈ Lα,β (]a, b[ ; R+ ),
α,β
n
o
sup khi (x)k : kxkC n−1 ≤ ρ &lt; +∞ (i = 1, . . . , n).
α,β
(1.6)
(1.7)
By a solution of the functional differential equation (1.1) is understood a vector
n−1
function x ∈ Ceα,β
(]a, b[ ; Rm ) satisfying (1.1) almost everywhere on ]a, b[ . A
solution of (1.1) satisfying (1.2) is called a solution of problem (1.1), (1.2).
1.2. Theorem on the Fredholm property of a linear boundary value
problem. We begin by introducing
n−1
Definition 1.1. A linear operator p : Cα,β
(]a, b[ ; Rm ) → Rm is called strongly bounded if there exists ζ ∈ Lα,β (]a, b[ ; R+ ) such that
n−1
kp(x)(t)k ≤ ζ(t)kxkC n−1 for a &lt; t &lt; b, x ∈ Cα,β
(]a, b[ ; Rm ).
α,β
(1.8)
794
I. KIGURADZE, B. PŮŽA, AND I. P. STAVROULAKIS
Consider the boundary value problem
x(n) (t) = p(x)(t) + q(t),
`i (x) = c0i (i = 1, . . . , n),
(1.9)
(1.10)
n−1
where p : Cα,β
(]a, b[ ; Rm ) → Lα,β (]a, b[ ; Rm ) is a linear, strongly bounded opn−1
(]a, b[ ; Rm ) → Rm (i = 1, . . . , m) are linear bounded operators,
erator, `i : Cα,β
q ∈ Lα,β (]a, b[ ; Rm ), c0i ∈ Rm (i = 1, . . . , m).
Theorem 1.1. For problem (1.9), (1.10) to be uniquely solvable it is necessary and sufficient that the corresponding homogeneous problem
x(n) (t) = p(x)(t),
`i (x) = 0 (i = 1, . . . , n)
(1.90 )
(1.100 )
have only a trivial solution. Moreover, if problem (1.90 ) (1.100 ) has only a
trivial solution, then there exists a positive constant γ such that for any q ∈
Lα,β (]a, b[ ; Rm ) and c0i ∈ Rm (i = 1, . . . , m), a solution x of problem (1.9), (1.10)
kxk
≤γ
n−1
Cα,β
&micro;X
n
&para;
kc0i k + kqkLα,β .
(1.11)
i=1
The vector differential equation with deviating arguments
x(n) (t) =
n
X
Pi (t)x(i−1) (τi (t)) + q(t),
(1.12)
i=1
where τi : [a, b] → [a, b] (i = 1, . . . , n) are measurable functions, Pi : ]a, b[ →
Rm&times;m (i = 1, . . . , n) are matrix functions with measurable components and
q ∈ Lα,β (]a, b[ ; Rm ), is a particular case of equation (1.9). Along with (1.12),
consider the corresponding homogeneous equation
x(n) (t) =
n
X
Pi (t)x(i−1) (τi (t)).
(1.120 )
i=1
From Theorem 1.1 follows
Corollary 1.1. Let almost everywhere on ]a, b[ the inequalities
τi (t) &gt; a for i &gt; n − α, τj (t) &lt; b for j &gt; n − β
(1.13)
be fulfilled. Moreover,
Zb
&sup3;
&acute;−αi &sup3;
(t − a)α (b − t)β τi (t) − a
&acute;−βi
b − τi (t)
kPi (t)k dt
a
&lt; +∞ (i = 1, . . . , n).∗)
(1.14)
∗)
Here and in the sequel it will be assumed that if αi = 0 (βi = 0), then (τi (t) − a)−αi ≡ 1
((τi (t) − b)−βi ≡ 1).
ON SINGULAR BOUNDARY VALUE PROBLEMS
795
Then for problem (1.12), (1.2) to be uniquely solvable, it is necessary and sufficient that the corresponding homogeneous problem (1.120 ), (1.20 ) have only a
trivial solution. Moreover, if problem (1.120 ) (1.20 ) has only a trivial solution,
then there exists a positive constant γ such that for any q ∈ Lα,β (]a, b[ ; Rm )
and c0i ∈ Rm (i = 1, . . . , m), a solution x of problem (1.12), (1.2) admits estimate (1.11).
1.3. A priori boundedness principle for the nonlinear problem (1.1),
(1.2). To formulate this principle we have to introduce
Definition 1.2. Let γ be a positive number. The pair (p, (`i )ni=1 ) of continn−1
n−1
uous operators p : Cα,β
(]a, b[ ; Rm ) &times; Cα,β
(]a, b[ ; Rm ) → Lα,β (]a, b[ ; Rm ) and
n−1
n−1
(`i )ni=1 : Cα,β
(]a, b[ ; Rm ) &times; Cα,β
(]a, b[ ; Rm ) → Rmn is said to be γ-consistent if:
n−1
(i) the operators p(x, &middot;) : Cα,β
(]a, b[ ; Rm ) → Lα,β (]a, b[ ; Rm ) and `i (x, &middot;) :
n−1
n−1
Cα,β
(]a, b[ ; Rm ) → Rm are linear for any fixed x ∈ Cα,β
(]a, b[ ; Rm ) and i ∈
{1, . . . , n};
n−1
(ii) for any x and y ∈ Cα,β
(]a, b[ ; Rm ) and for almost all t ∈ ]a, b[ we have
inequalities
&sup3;
&acute;
kp(x, y)(t)k ≤ δ t, kxkC n−1 kykC n−1 ,
α,β
α,β
n
X
&sup3;
&acute;
k`i (x, y)k ≤ δ0 kxkC n−1 kykC n−1 ,
α,β
i=1
α,β
where δ0 : R+ → R+ is nondecreasing, δ(&middot;, ρ) ∈ Lα,β (]a, b[ ; R+ ) for every ρ ∈ R+ ,
and δ(t, &middot;) : R+ → R+ is nondecreasing for every t ∈ ]a, b[ ;
n−1
(iii) for any x ∈ Cα,β
(]a, b[ ; Rm ), q ∈ Lα,β (]a, b[ ; Rm ) and ci ∈ Rm (i =
1, . . . , n), an arbitrary solution y of the boundary value problem
y (n) (t) = p(x, y)(t) + q(t), `i (x, y) = ci (i = 1, . . . , n)
(1.15)
kykC n−1 ≤ γ
α,β
&micro;X
n
&para;
kci k + kqkLα,β .
(1.16)
i=1
Definition 1.20 .
The pair (p, (`i )ni=1 ) of continuous operators
n−1
n−1
m
p : Cα,β (]a, b[ ; R ) &times; Cα,β
(]a, b[ ; Rm ) → Lα,β (]a, b[ ; Rm ) and (`i )ni=1 :
n−1
n−1
Cα,β
(]a, b[ ; Rm )&times;Cα,β
(]a, b[ ; Rm ) → Rmn is said to be consistent if there exists
γ &gt; 0 such that this pair is γ-consistent.
Theorem 1.2. Let there exist a positive number ρ0 and a consistent pair
n−1
n−1
(]a, b[ ; Rm ) &times; Cα,β
(]a, b[ ; Rm ) →
(p, (`i )ni=1 ) of continuous operators p : Cα,β
n−1
n−1
(]a, b[ ; Rm ) &times; Cα,β
(]a, b[ ; Rm ) → Rmn such that for any
Rmn and (`i )ni=1 : Cα,β
λ ∈ ]0, 1[ an arbitrary solution of the problem
x(n) (t) = (1 − λ)p(x, x)(t) + λf (x)(t),
(λ − 1)`i (x, x) = λhi (x) (i = 1, . . . , n)
(1.17)
(1.18)
kxkC n−1 ≤ ρ0 .
(1.19)
α,β
796
I. KIGURADZE, B. PŮŽA, AND I. P. STAVROULAKIS
Then problem (1.1), (1.2) is solvable.
For n = 1 and α = β = 0, Theorem 1.2 implies Theorem 1 from .
Corollary 1.2. Let there exist a positive number γ, a γ-consistent pair
n−1
n−1
(p, (`i )ni=1 ) of continuous operators p : Cα,β
(]a, b[ ; Rm ) &times; Cα,β
(]a, b[ ; Rm ) →
n−1
n−1
Lα,β (]a, b[ ; Rm ), (`i )ni=1 : Cα,β
(]a, b[ ; Rm ) &times; Cα,β
(]a, b[ ; Rm ) → Rmn and functions η : ]a, b[ &times;R+ → R+ and η0 : R+ → R+ such that the inequalities
&deg;
&deg;
&sup3;
&acute;
&deg;
&deg;
&deg;f (x)(t) − p(x, x)(t)&deg; ≤ η t, kxkC n−1 ,
(1.20)
n &deg;
&deg;
&sup3;
&acute;
X
&deg;
&deg;
&deg;hi (x) − `i (x, x)&deg; ≤ η0 kxkC n−1
α,β
(1.21)
α,β
i=1
n−1
are fulfilled for any x ∈ Cα,β
(]a, b[ ; Rm ) and almost all t ∈ ]a, b[ . Moreover,
η(&middot;, ρ) ∈ Lα,β (]a, b[ ; R+ ) for ρ ∈ R+ and
&Atilde;
!
b
η0 (ρ) 1 Z
1
lim sup
+
(s − a)α (b − s)β η(s, ρ) ds &lt; .
ρ
ρa
γ
ρ→+∞
(1.22)
Then problem (1.1), (1.2) is solvable.
n−1
As an example, in Cα,0
(]a, b[ ; Rm ) consider the boundary value problem
&sup3;
&acute;
x(n) (t) = g t, x(τ1 (t)), . . . , x(n−1) (τn (t)) ,
lim x(i−1) (t) = ci (x) (i = 1, . . . , k),
t→a
(i−1)
lim x
t→b
(t) = ci (x) (i = k + 1, . . . , n).
(1.23)
(1.24)
Here k ∈ {1, . . . , n − 1}, α ∈ [0, n − k], τi : [a, b] → [a, b] (i = 1, . . . , n)
n−1
are measurable functions, ci : Ceα,0
(]a, b[ ; Rm ) → Rm (i = 1, . . . , m) are continuous operators, and g : ]a, b[ &times;Rmn → Rm is a vector function such that
g(&middot;, x1 , . . . , xn ) : ]a, b[ → Rm is measurable for any xi ∈ Rm (i = 1, . . . , n) and
g(t, &middot;, . . . , &middot;) : Rmn → Rm is continuous for almost all t ∈ ]a, b[ . We will also
suppose that for i &gt; n − α the inequality
τi (t) &gt; a
holds almost everywhere on ]a, b[ .
The following statement is valid.
Corollary 1.3. Let there exist η0 : R+ → R+ , Pi ∈ Lα,0 (]a, b[ ; Rmn
+ ) (i =
such
that
1, . . . , n) and q : ]a, b[ &times;R+ → Rm
+
n
X
i=1
&sup3;
kci (x)k ≤ η0 kxkC n−1
α,0
&acute;
n−1
for x ∈ Cα,0
(]a, b[ ; Rm )
(1.25)
ON SINGULAR BOUNDARY VALUE PROBLEMS
797
and on ]a, b[ &times;Rmn the inequality
≤
n &sup3;
X
&acute; αi
τi (t) − a
&macr;
&macr;
&macr;
&macr;
&macr;g(t, x1 , . . . , xn )&macr;
&micro; X
n &sup3;
Pi (t)|xi | + q t,
τi (t) − a
&acute;αi
&para;
kxi k
(1.26)
i=1
i=1
holds. Let, moreover, q(&middot;, ρ) ∈ Lα,0 (]a, b[ ; Rm
+ ) for every ρ ∈ R+ , the components
of q(t, ρ) are nondecreasing with respect to ρ,
&micro;
lim
ρ→+∞
&para;
b
η0 (ρ) 1 Z
+
(s − a)α kq(s, ρ)k ds = 0
ρ
ρa
(1.27)
and
r(P) &lt; 1,
(1.28)
where
b
Z
&sup3;
&acute;k+1−i−αk+1
(b − a)n−k−1−α+αk+1
Pi (s) ds
(s − a)α τi (s) − a
P=
i=1 (n − k − 1)!(k + 1 − i − αk+1 )! a
k
X
+
b
n
X
(b − a)n−i−α+αi Z
i=k+1
(n − i)!
(s − a)α Pi (s) ds.
a
Then problem (1.23), (1.24) is solvable.
Before passing to the formulation of the next corollary we introduce
n−1
Definition 1.3. An operator p : Cα,β
(]a, b[ ; Rm ) → Lα,β (]a, b[ ; Rm ) (an opn−1
erator ` : Cα,β
(]a, b[ ; Rm ) → Rm ) is called positive homogeneous if the equality
p(λx)(t) = λp(x)(t)
&sup3;
`(λx) = λ`(x)
&acute;
n−1
is fulfilled for all x ∈ Cα,β
(]a, b[ ; Rm ), λ ∈ R+ and almost all t ∈ ]a, b[ .
n−1
Definition 1.4. A positive homogeneous operator p : Cα,β
(]a, b[ ; Rm ) →
n−1
Lα,β (]a, b[ ; Rm ) (a positive homogeneous operator ` : Cα,β
(]a, b[ ; Rm ) → Rm ) is
called strongly bounded (bounded) if there exists a function ζ ∈ Lα,β (]a, b[ ; R+ )
(a positive number ζ0 ) such that the inequality
kp(x)(t)k ≤ ζ(t)kxkC n−1
α,β
&sup3;
k`(x)k ≤ ζ0 kxkC n−1
&acute;
α,β
n−1
holds for all x ∈ Cα,β
(]a, b[ ; Rm ) and almost all t ∈ ]a, b[.
Corollary 1.4. Let there exist a linear, strongly bounded operator
n−1
p : Cα,β
(]a, b[ ; Rm ) → Lα,β (]a, b[ ; Rm ), a positive homogeneous, continuous,
n−1
strongly bounded operator p : Cα,β
(]a, b[ ; Rm ) → Lα,β (]a, b[ ; Rm ), linear bounded
798
I. KIGURADZE, B. PŮŽA, AND I. P. STAVROULAKIS
n−1
operators `i : Cα,β
(]a, b[ ; Rm ) → Rm (i = 1, . . . , n), positive homogeneous, conn−1
tinuous, bounded operators `i : Cα,β
(]a, b[ ; Rm ) → Rm (i = 1, . . . , m), and
functions η :]a, b[ &times;R+ and η0 : R+ → R+ such that the inequalities
&deg;
&deg;
&sup3;
&acute;
&deg;
&deg;
&deg;f (x)(t) − p(x)(t) − p(x)(t)&deg; ≤ η t, kxkC n−1 ,
α,β
n &deg;
&deg;
&sup3;
&acute;
X
&deg;
&deg;
&deg;hi (x) − `i (x) − `i (x)&deg; ≤ η0 kxkC n−1
α,β
(1.29)
(1.30)
i=1
n−1
hold for any x ∈ Cα,β
(]a, b[ ; Rm ) and for almost all t ∈ ]a, b[ . Moreover,
η(&middot;, ρ) ∈ Lα,β (]a, b[ ; R+ ) for any ρ ∈ R+ ,
&Atilde;
lim
ρ→+∞
!
b
η0 (ρ) 1 Z
+
(s − a)α (b − s)β η(s, ρ) ds = 0
ρ
ρa
(1.31)
and for any λ ∈ [0, 1] the problem
x(n) (t) = p(x)(t) + λp(x)(t), `i (x) + λ`i (x) = 0 (i = 1, . . . , n) (1.32)
has only a trivial solution. Then problem (1.1), (1.2) is solvable.
As an example, for the second order singular half-linear differential equation
u00 (t) = p1 (t)|u(t)|&micro; |u0 (t)|1−&micro; sgn u(t) + p2 (t)u0 (t) + p0 (t)
(1.33)
let us consider the two-point boundary value problems
lim u(t) = c1 , lim u(t) = c2
(1.341 )
lim u0 (t) = c2 .
(1.342 )
t→a
t→b
and
lim u(t) = c1 ,
t→a
t→b
We are interested in the case where &micro; ∈ [0, 1] and pi : ]a, b[ → R (i = 0, 1, 2)
are measurable functions satisfying either the conditions
Zb
Zb
(t − a)(b − t)|pi (t)| dt &lt; +∞ (i = 0, 1),
a
&middot;
1+&micro;
p1 (t) ≥ −λ1 [σ(t)]
,
|p2 (t)| dt &lt; +∞, (1.351 )
a
&cedil;
σ 0 (t)
p2 (t) −
sgn(t0 − t) ≥ −λ2 σ(t)
σ(t)
for a &lt; t &lt; b,
(1.361 )
or the conditions
Zb
Zb
(t − a)|p0 (t)| dt &lt; +∞ (i = 0, 1),
a
p1 (t) ≥ −λ1 [σ(t)]1+&micro; ,
|p2 (t)| dt &lt; +∞,
(1.352 )
a
0
p2 (t) −
σ (t)
≥ −λ2 σ(t) for a &lt; t &lt; b. (1.362 )
σ(t)
ON SINGULAR BOUNDARY VALUE PROBLEMS
799
Here t0 ∈ ]a, b[ , λi ∈ R+ (i = 1, 2), and σ : ]a, b[ → R+ is a locally absolutely
continuous function such that either
+∞
Z
0
σ (t) sgn(t0 − t) ≤ 0 for a &lt; t &lt; b,
0
&gt;
&micro;
2
&quot; Zb
ds
λ1 + λ2 s + s(1+&micro;)/&micro;
&macr; Zt0
&macr;#
Zb
&macr;
&macr;
σ(s) ds + &macr;&macr; σ(s) ds − σ(s) ds&macr;&macr; ,
a
a
(1.371 )
t0
or
+∞
Z
0
b
Z
ds
&gt;
&micro;
σ(s) ds.
λ1 + λ2 s + s(1+&micro;)/&micro;
a
(1.372 )
By virtue of Theorems 3.1 and 3.2 from  Corollary 1.4 implies
Corollary 1.5. Let conditions (1.35i ), (1.36i ) and (1.37i ) be fulfilled for
some i ∈ {1, 2}. Then problem (1.33), (1.34i ) has at least one solution.
This corollary is a generalization of the classical result of Ch. de la ValléePoussin  for equation (1.33).
2. Auxiliary Propositions
Lemma 2.1. Let ρ &gt; 0, η ∈ Lα,β (]a, b[ ; R+ ), t0 ∈ ]a, b[ , and S be the set of
(n − 1)-times continuously differentiable vector functions x : ]a, b[ → Rm satisfying the conditions
&deg;
&deg;
&deg; (i−1)
&deg;x
(t0 )&deg;&deg; ≤ ρ (i = 1, . . . , n),
(2.1)
Zt
&deg;
&deg;
&deg; (n−1)
&deg;
(n−1)
&deg;x
(t) − x
(s)&deg; ≤ η(ξ) dξ for a &lt; s ≤ t &lt; b.
(2.2)
s
n−1
n−1
Then S ⊂ Ceα,β
(]a, b[ ; Rm ) and S is a compact set of the space Cα,β
(]a, b[ ; Rm ).
Proof. Let x be an arbitrary element of the set S. Then by (2.2) the function
x(n−1) is locally absolutely continuous on ]a, b[ and
kx(n) (t)k ≤ η(t) for almost all t ∈ ]a, b[ .
(2.3)
x(n) ∈ Lα,β (]a, b[ ; Rm ),
(2.4)
Therefore
x(i−1) (t) =
n
X
(t − t0 )j−i
j=i
+
1
(n − i)!
(j − i)!
x(j−1) (t0 )
Zt
(t − s)n−i x(n) (s) ds for a &lt; t &lt; b (i = 1, . . . , n),
t0
(2.5)
800
I. KIGURADZE, B. PŮŽA, AND I. P. STAVROULAKIS
and
&deg;
&deg;
&deg; (i−1) &deg;
&deg;x
(t)&deg; ≤ εi (t) for a &lt; t &lt; b (i = 1, . . . , n),
(2.6)
where
εi (t) = ρ
&macr; t
j=i
&macr;
&macr;
1 &macr;&macr; Z
n−i
+
η(s) ds&macr;&macr; (i = 1, . . . , n). (2.7)
&macr; (t − s)
(j − i)!
(n − i)!
n
X
(b − a)j−i
t0
Let
i1 = max{i : αi = 0}, i2 = max{i : βi = 0}.
Then
n − i ≥ α, αi = 0 for i ≤ i1 , αi = α + i − n &gt; 0 for i &gt; i1 ,
n − i ≥ β, βi = 0 for i ≤ i2 , βi = β + i − n &gt; 0 for i &gt; i2 .
(2.81 )
(2.82 )
Therefore
εi (t) ≤ εi (a+) &lt; +∞ for i ≤ i1 , a &lt; t ≤ t0 ,
(2.9)
Zt0
ε1+i1 (s) ds &lt; +∞ if i1 &lt; n − 1,
(2.10)
a
εi (t) ≤ εi (b−) &lt; +∞ for i ≤ i2 , t0 ≤ t &lt; b,
(2.11)
Zb
ε1+i2 (s) ds &lt; +∞ if i2 &lt; n − 1.
(2.12)
t0
If i &gt; i1 , then, with (2.7) and (2.81 ) taken into account, for any δ ∈ ]0, t0 − a[
we find
#
a+δ
(t − a)α+i−n Z
n−i
(s − t) η(s) ds
lim sup (t − a) εi (t) = lim sup
(n − i)!
t→a
t→a
h
&quot;
i
αi
t
≤
1
(n − i)!
a+δ
Z
(s − a)α η(s) ds.
a
Hence, because of the arbitrariness of δ, it follows that
h
i
lim (t − a)αi εi (t) = 0 for i &gt; i1 .
t→a
(2.13)
Analogously, it can be shown that
h
i
lim (b − t)βi εi (t) = 0 for i &gt; i2 .
t→b
(2.14)
ON SINGULAR BOUNDARY VALUE PROBLEMS
801
If i ≤ i1 (if i ≤ i2 ), then by virtue of conditions (2.3) and (2.81 ) (conditions
(2.3) and (2.82 )) we have
&Atilde; Zb
Zt0
n−i
(s − a)
kx
(n)
!
n−i
(s)k ds &lt; +∞
(b − s)
a
kx
(n)
(s)k ds &lt; +∞ .
t0
Hence (2.5) implies the existence of the limit
lim x(i−1) (t)
t→a
&sup3;
&acute;
lim x(i−1) (t) .
t→b
If however i &gt; i1 (i &gt; i2 ), then from (2.6) and (2.13) (from (2.6) and (2.14)) we
have
&sup3;
&acute;
lim(t − a)αi x(i−1) (t) = 0
lim(b − t)βi x(i−1) (t) = 0 .
t→a
t→b
We have thereby proved the existence of limit (1.3). Therefore S ⊂
n−1
Ceα,β
(]a, b[ ; Rm ).
By the Arzela–Ascoli lemma, from estimates (2.3), (2.6) and conditions (2.9)–
n−1
(2.14) it follows that S is a compact set of the space Cα,β
(]a, b[ ; Rm ).
Let (p, (`i )ni=1 ) be a γ-consistent pair of continuous operators p :
n−1
n−1
Cα,β (]a, b[ ; Rm ) &times; Cα,β
(]a, b[ ; Rm ) → Lα,β (]a, b[ ; Rm ) and (`i )ni=1 :
n−1
n−1
n−1
Cα,β
(]a, b[ ; Rm ) &times; Cα,β
(]a, b[ ; Rm ) → Rmn , and q : Cα,β
(]a, b[ ; Rm ) →
n−1
Lα,β (]a, b[ ; Rm ), c0i : Cα,β
(]a, b[ ; Rm ) → Rm (i = 1, . . . , n) be continuous operan−1
tors. For any x ∈ Cα,β
(]a, b[ ; Rm ), consider the linear boundary value problem
y (n) (t) = p(x, y)(t) + q(x)(t), `i (x, y) = c0i (x) (i = 1, . . . , n).
(2.15)
By condition (iii) of Definition 1.2, the homogeneous problem
y (n) (t) = p(x, y)(t),
`i (x, y) = 0 (i = 1, . . . , n)
(2.150 )
has only a trivial solution. By Theorem 1.1 this fact guarantees the existence
of a unique solution y of problem (2.15). We write
u(x)(t) = y(t).
n−1
n−1
Lemma 2.2. u : Cα,β
(]a, b[ ; Rm ) → Cα,β
(]a, b[ ; Rm ) is a continuous operator.
Proof. Let
n−1
xi ∈ Cα,β
(]a, b[ ; Rm ), yi (t) = u(xi )(t) (i = 1, 2)
and
y(t) = y2 (t) − y1 (t).
Then
y (n) (t) = p2 (x2 , y)(t) + q0 (x1 , x2 )(t),
`i (x2 , y) = ci (x1 , x2 ) (i = 1, . . . , n),
802
I. KIGURADZE, B. PŮŽA, AND I. P. STAVROULAKIS
where
q0 (x1 , x2 )(t) = p(x1 , y1 )(t) − p(x2 , y1 )(t) + q(x2 )(t) − q(x1 )(t),
ci (x1 , x2 ) = `i (x1 , x2 ) − `i (x2 , y1 ) + c0i (x2 ) − c0i (x1 ) (i = 1, . . . , n).
Hence, by condition (iii) of Definition 1.2 we have
&deg;
&deg;
&deg;
&deg;
&deg;u(x2 ) − u(x1 )&deg;
n−1
Cα,β
≤γ
&micro;X
n
&para;
kci (x1 , x2 )k + kq0 (x1 , x2 )kLα,β .
i=1
Since the operators p, q, `i and c0i (i = 1, . . . , n) are continuous, this estimate
implies the continuity of the operator u.
n−1
Lemma 2.3. Let k ∈ {1, . . . , n − 1}, α ∈ [0, n − k], and x ∈ Cα,0
(]a, b[ ; Rm )
be a vector function satisfying conditions (1.24). Then on ]a, b[ the following
inequalities are fulfilled:
|x
(i−1)
(t)| ≤
n
X
(b − a)j−i |cj (x)|
j=i
1
+
(b − a)n−i−α+αi (t − a)−αi y(x) (i = k + 1, . . . , n),
(n − i)!
|x
(i−1)
(t)| ≤
(2.16)
n
X
(b − a)j−i |cj (x)|
j=i
+
n−k−1−α+αk+1
(b − a)
(t − a)k+1−i−αk+1 y(x) (i = 1, . . . , k), (2.17)
(n − k − 1)!(k + 1 − i − αk+1 )!
where
Zb
(s − a)α |x(n) (s)| ds.
y(x) =
(2.18)
a
Proof. Let x0 (t) be a polynomial of degree not higher than n − 1 satisfying the
conditions
(i−1)
x0
(a) = ci (x) (i = 1, . . . , k),
(i−1)
x0
(b) = ci (x) (i = k + 1, . . . , n).
Then
(i−1)
|x0
(t)| ≤
n
X
(b − a)j−i |cj (x)| for a ≤ t ≤ b (i = 1, . . . , n).
(2.19)
b
(−1)n−i Z
−
(s − t)n−i x(n) (s) ds
(n − i)!
(2.20)
j=i
On the other hand,
x
(i−1)
(t) =
(i−1)
x0 (t)
t
(i = k + 1, . . . , n),
Zt
x
(i−1)
x(i) (s) ds (i = 1, . . . , k).
(t) = ci (x) +
a
(2.21)
ON SINGULAR BOUNDARY VALUE PROBLEMS
803
By (1.4)
n − i − α − αi ≥ 0 (i = 1, . . . , n).
Therefore
(s − t)n−i ≤ (s − a)n−i−α+αi (s − a)−αi (s − a)α
≤ (b − a)n−i−α+αi (t − a)−αi (s − a)α for t ≤ s &lt; b (i = 1, . . . , n).
If along with this we take into account inequality (2.19), then from (2.20) we
obtain estimates (2.16).
It is clear that
αk+1 ≤ 1,
since α ≤ n − k. If αk+1 &lt; 1, then by virtue of (2.16) and (2.19), from (2.21)
To complete the proof of the lemma it remains to consider the case where
αk+1 = 1. Then α = n − k and thus from (2.19)–(2.21) we find
|x
(k−1)
n
X
(t)| ≤
(b − a)j−k |cj (x)|
j=k
1
+
(n − k − 1)!
Zt
&Atilde; Zb
!
n−k−1
(s − τ )
a
=
|x
(n)
(s)| ds dτ
τ
n
X
(b − a)j−k |cj (x)|
j=k
&quot;
b
t
#
Z
Z
1
n−k−1 (n)
(t − a) (s − a)
|x (s)| ds + (s − a)n−k |x(n) (s)| ds
+
(n − k − 1)!
a
t
n
X
≤
(b − a)j−k |cj (x)| +
j=k
1
y(x)
(n − k − 1)!
and
|x(i−1) (t)| ≤
n
X
(b − a)j−i |cj (x)| +
j=i
1
(t − a)k−i y(x)
(n − k − 1)!(k − i)!
(i = 1, . . . , k).
Therefore estimates (2.17) are valid.
3. Proof of the Main Results
n−1
Proof of Theorem 1.1. Let B = Cα,β
(]a, b[ ; Rm )&times;Rmn be a Banach space with
n−1
elements u = (x; c1 , . . . , cn ), where x ∈ Cα,β
(]a, b[ ; Rm ), ci ∈ Rm (i = 1, . . . , n),
and the norm
kukB = kukC n−1 +
α,β
n
X
i=1
kci k.
804
I. KIGURADZE, B. PŮŽA, AND I. P. STAVROULAKIS
Fix arbitrarily t0 ∈ ]a, b[ and, for any u = (x; c1 , . . . , cn ), set
e
p(u)(t)
=
&Atilde; n
X (t − t0 )i−1 &sup3;
(i − 1)!
i=1
ci + x(i−1) (t0 )
&acute;
!
t
Z
1
+
(t − s)n−1 p(x)(s) ds; c1 − `1 (x), . . . , cn − `n (x) ,
(n − 1)!
t0
&Atilde;
e
q(t)
=
!
t
Z
1
(t − s)n−1 q(s) ds; c01 , . . . , c0n .
(n − 1)!
t0
Problem (1.9), (1.10) is equivalent to the operator equation
e
u = p(u)
+ qe
(3.1)
in the space B since u = (x; c1 , . . . , cn ) is a solution of equation (3.1) if and only
if ci = 0 (i = 1, . . . , n) and x is a solution of problem (1.9), (1.10). As for the
homogeneous equation
e
u = p(u)
(3.10 )
it is equivalent to the homogeneous problem (1.90 ), (1.100 ).
From condition (1.8) and Lemma 2.1 it immediately follows that the linear
operator pe : B → Be is compact. By this fact and the Fredholm alternative for
operator equations (, Ch. XIII, &sect; 5, Theorem 1), equation (3.1) is uniquely
solvable if and only if equation (3.10 ) has only a trivial solution. Moreover, if
equation (3.10 ) has only a trivial solution, then the operator I − pe is invertible
e −1 : B → B is a linear bounded operator, where I : B → B is an
and (I − p)
identical operator. Therefore there exists γ0 &gt; 0 such that for any qe ∈ B the
solution u of equation (3.1) admits the estimate
e B.
kukB ≤ γ0 kqk
However,
e B ≤
kqk
n
X
kc0i k + γ1 kqkLα,β ,
i=1
where γ1 &gt; 1 is a constant depending only on α, β, a, b, t0 and n. Hence
kukB ≤ γ
&micro;X
n
&para;
kc0i k + kqkLα,β ,
(3.2)
i=1
where γ = γ0 γ1 .
Since problem (1.9), (1.10) is equivalent to equation (3.1), it is clear that
problem (1.9), (1.10) is uniquely solvable if and only if problem (1.90 ), (1.100 )
has only a trivial solution. Moreover, if (1.90 ), (1.100 ) has only a trivial solution,
then by virtue of (3.2) the solution x of problem (1.9), (1.10) admits estimate
(1.11).
ON SINGULAR BOUNDARY VALUE PROBLEMS
805
Proof of Corollary 1.1. We set
p(x)(t) =
n
X
Pi (t)x(i−1) (τi (t))
i=1
n−1
Cα,β
(]a, b[; Rm ).
for any x ∈
Then equations (1.12) and (1.120 ) take respectively
forms (1.9) and (1.90 ). On the other hand, in view of (1.13) and (1.14)
n−1
p : Cα,β
(]a, b[ ; Rm ) → Lα,β (]a, b[ ; Rm )
is a strongly bounded linear operator. Therefore the conditions of Theorem 1.1
are fulfilled.
Proof of Theorem 1.2. Let δ, δ0 and γ be the functions and numbers appearing
in Definitions 1.2 and 1.20 . We set
n
o
η(t) = 2ρ0 δ(t, 2ρ0 ) + sup kf (x)(t)k : kxkC n−1 ≤ 2ρ0 ,
α,β
η0 = 2ρ0 δ0 (2ρ0 ) +
n
X
o
sup khi (x)k : kxkC n−1 ≤ 2ρ0 ,
i=1
&sup3;
n
α,β
&acute;
ρ1 = γ η0 + kηkLα,β , η ∗ (t) = δ(t, ρ1 )ρ0 + η(t),
(3.3)
n−1
B0 = x ∈ Cα,β
(]a, b[ ; Rm ) : kxkC n−1 ≤ ρ1 ,
(3.4)
n
χ(s) =
o
α,β



1



2 − s/ρ0
0
&sup3;
&acute;h
for 0 ≤ s ≤ ρ0
for ρ0 &lt; s &lt; 2ρ0 ,
for s ≥ 2ρ0
(3.5)
i
q(x)(t) = χ kxkC n−1 f (x)(t) − p(x, x)(t) ,
(3.6)
c0i (x) = χ kxkC n−1 `i (x, x) − hi (x) (i = 1, . . . , n).
(3.7)
&sup3;
&acute;h
α,β
i
α,β
By (1.6) and (1.7)
η ∗ ∈ Lα,β (]a, b[ ; R+ )
η0 &lt; +∞, η ∈ Lα,β (]a, b[ ; R+ ),
n−1
and for every x ∈ Cα,β
(]a, b[ ; Rm ) and almost all t ∈ ]a, b[ we have the inequalities
kq(x)(t)k ≤ η(t),
n
X
kc0i (x)k ≤ η0 .
(3.8)
i=1
n−1
n−1
Let u : Cα,β
(]a, b[ ; Rm ) → Cα,β
(]a, b[ ; Rm ) be an operator which to every
n−1
x ∈ Cα,β
(]a, b[ ; Rm ) assigns the solution y of problem (2.15). By Lemma 2.1,
u is a continuous operator. On the other hand, by conditions (ii) and (iii) of
Definition 1.2, notations (3.3), (3.4) and inequalities (3.8), the vector function
y = u(x) satisfies, for each x ∈ B0 , the conditions
kykC n−1 ≤ ρ1 ,
α,β
Zt
&deg;
&deg;
&deg; (n−1)
&deg;
(n−1)
&deg;y
(t) − y
(s)&deg; ≤ η ∗ (ξ) dξ for a &lt; s ≤ t &lt; b.
s
806
I. KIGURADZE, B. PŮŽA, AND I. P. STAVROULAKIS
By Lemma 2.2 this implies that the operator u maps the ball B0 into its own
compact subset. Therefore, owing to Schauder’s principle, there exists x ∈ B0
such that
x(t) = u(x)(t) for a &lt; t &lt; b.
By notations (3.6), (3.7) the function x is a solution of problem (1.17), (1.18),
where
&acute;
&sup3;
λ = χ kxkC n−1 .
(3.9)
α,β
Let us show that x admits estimate (1.19). Assume the contrary. Then either
ρ0 &lt; kxkC n−1 &lt; 2ρ0 ,
(3.10)
kxkC n−1 ≥ 2ρ0 .
(3.11)
α,β
or
α,β
If condition (3.10) is fulfilled, then by virtue of (3.5) and (3.9)
λ ∈ ]0, 1[ ,
which, by one of the conditions of the theorem, guarantees the validity of estimate (1.19). But this contradicts condition (3.10).
Assume now that inequality (3.11) is fulfilled. Then by virtue of (3.5) and
(3.9)
λ=0
and therefore x is a solution of problem (2.150 ). Thus x(t) ≡ 0 since problem
(2.150 ) has only a trivial solution. But this contradicts inequality (3.11). The
contradiction obtained proves the validity of estimate (1.19).
By (1.19), (3.5)–(3.7) and (3.9), it clearly follows from (1.17), (1.18) that
λ = 1 and x is a solution of problem (1.1), (1.2).
Proof of Corollary 1.2. By (1.22) there is ρ0 &gt; 0 such that
&Atilde;
!
Zb
α
γ η0 (ρ) +
β
(s − a) (b − s) η(s, ρ) ds &lt; ρ for ρ &gt; ρ0 .
(3.12)
a
Let x be a solution of problem (1.17), (1.18) for some λ ∈ ]0, 1[ . Then y = x
is also a solution of problem (1.15) where
&sup3;
&acute;
q(t) = λ f (x)(t) − p(x, x)(t) ,
&sup3;
&acute;
c0i (x) = λ `i (x, x) − hi (x)
Assume that
ρ = kxkC n−1 .
α,β
(i = 1, . . . , n).
ON SINGULAR BOUNDARY VALUE PROBLEMS
807
By the γ-consistency of the pair (p, (`i )ni=1 ) and inequalities (1.20), (1.21) we
have
ρ≤γ
&micro;X
n
&para;
kc0i (x)k + kqkLα,β
i=1
&micro;
&para;
Zb
α
≤ γ η0 (ρ) +
β
(s − a) (b − s) η(s, ρ) ds .
a
Hence by (3.12) it follows that ρ ≤ ρ0 . Therefore estimate (1.19) is valid, which
due to Theorem 1.2 guarantees the solvability of problem (1.1), (1.2).
Proof of Corollary 1.3. Problem (1.23), (1.24) is obtained from problem (1.1),
(1.2) when
&sup3;
&acute;
f (x)(t) ≡ g t, x(τ1 (t)), . . . , x(n−1) (τn (t)) ,
hi (x) = lim x(i−1) (t) − ci (x) (i = 1, . . . , k),
t→a
hi (x) = lim x(i−1) (t) − ci (x) (i = k + 1, . . . , n).
(3.13)
(3.14)
t→b
By virtue of the restrictions imposed on g, τi , ci (i = 1, . . . , n) and the inequaln−1
ity α ≤ n − k it is obvious that f : Cα,0
(]a, b[ ; Rm ) → Lα,0 (]a, b[ ; Rm ) and
n−1
hi : Cα,0 (]a, b[ ; Rm ) → Rm (i = 1, . . . , m) are continuous operators satisfying
conditions (1.6) and (1.7), where β = 0.
n−1
Assume for any x, y ∈ Cα,0
(]a, b[ ; Rm ) and t ∈ ]a, b[ that
p(x, y)(t) = 0,
`i (x, y) = lim y (i−1) (t) (i = 1, . . . , k),
t→a
`i (x, y) = lim y
t→b
(i−1)
(t) (i = k + 1, . . . , n).
(3.15)
According to Definition 1.20 and Theorem 1.2 the pair (p, (`i )ni=1 ) of continuous
n−1
n−1
operators p : Cα,0
(]a, b[ ; Rm ) &times; Cα,0
(]a, b[ ; Rm ) → Lα,0 (]a, b[ ; Rm ), (`i )ni=1 :
n−1
n−1
Cα,0
(]a, b[ ; Rm ) &times; Cα,0
(]a, b[ ; Rm ) → Rmn , is consistent.
To prove Corollary 1.3, by Theorem 1.2 it is sufficient to show that for each
λ ∈ ]0, 1[ an arbitrary solution x of problem (1.17), (1.18) admits the estimate
kxkC n−1 ≤ ρ0 ,
(3.16)
α,0
where ρ0 is a non-negative constant not depending on λ and x.
By virtue of (3.13)–(3.15) problem (1.17), (1.18) takes the form
&sup3;
&acute;
x(n) (t) = λg t, x(τ1 (t)), . . . , x(n−1) (τn (t)) ,
lim x(i−1) (t) = λci (x) (i = 1, . . . , k),
t→a
lim x(i−1) (t) = λci (x) (i = k + 1, . . . , n).
(3.17)
(3.18)
t→b
Let x be a solution of problem (3.17), (3.18) for some λ ∈ ]0, 1[ . Then by
virtue of Lemma 2.3 we conclude that estimates (2.16), (2.17), where y(x) is
808
I. KIGURADZE, B. PŮŽA, AND I. P. STAVROULAKIS
the vector given by equality (2.18), are true. On the other hand, on account of
(1.26) we have
b
y(x) ≤
n Z
X
&sup3;
&acute; αi
(s − a)α τi (s) − a
Pi (s)|x(i−1) (s)| ds
i=1 a
Zb
&sup3;
&acute;
(s − a)α q s, kxkC n−1 ds.
+
α,0
a
If, along with (2.16) and (2.17), we take into account that αi = 0 (i = 1, . . . , k),
then from the latter inequality we obtain
y(x) ≤ Py(x) + y0 (x)
and therefore
(E − P)y(x) ≤ y0 (x),
(3.19)
where E is the unique m &times; m matrix and
y0 (x) =
n &micro; Zb &sup3;
X
i=1
&acute; αi
τi (s) − a
Pi (s) ds
&para;X
n
(b − a)j−i |cj (x)|
j=i
a
Zb
&sup3;
&acute;
(s − a)α q s, kxkC n−1 ds.
+
(3.20)
α,0
a
By the nonnegativeness of the matrix P and inequality (1.28), from (3.19) it
follows that
y(x) ≤ (E − P)−1 y0 (x).
If along with this we take into account condition (1.25) and equality (3.20),
then (2.16) and (2.17) imply that
&sup3;
&acute;
kxkC n−1 ≤ η1 kxkC n−1 ,
α,0
where
(3.21)
α,0
&Atilde;
!
Zb
α
η1 (ρ) = &micro; η0 (ρ) +
(s − a) kq(s, ρ)k ds ,
a
and &micro; is a positive constant depending only on αi , Pi , τi (i = 1, . . . , n), a and
b. On the other hand, due to condition (1.27) we have
lim
ρ→+∞
η1 (ρ)
=0
ρ
and therefore
η1 (ρ) &lt; ρ for ρ &gt; ρ0 ,
where
&frac12;
ρ0 = inf ρ &gt; 0 :
&frac34;
η(s)
&lt; 1 for s ∈ [ρ, +∞[ .
s
ON SINGULAR BOUNDARY VALUE PROBLEMS
809
Therefore from (3.21) we obtain estimate (3.16). On the other hand, it is
obvious that the constant ρ0 does not depend on λ and x.
Proof of Corollary 1.4. The strong boundedness of the operators p and p and
the boundedness of the operators (`i )ni=1 and (`i )ni=1 guarantee the existence of
ζ ∈ Lα,β (]a, b[ ; R+ ) and ζ0 ∈ R+ such that the inequalities
kp(x)(t)k + kp(x)(t)k ≤ ζ(t)kxkC n−1 ,
α,β
n &sup3;
X
&acute;
(3.22)
k`i (x)k + k`i (x)k ≤ ζ0 kxkC n−1
α,β
i=1
n−1
hold for each x ∈ Cα,β
(]a, b[ ; Rm ) and almost all t ∈ ]a, b[ .
By Theorem 1.1 and Definition 1.20 the pair (p, (`i )ni=1 ) is consistent since for
λ = 0 problem (1.32) has only a trivial solution.
Let us consider for arbitrary λ ∈ [0, 1], q ∈ Lα,β (]a, b[ ; Rm ) and c0i ∈ Rm
(i = 1, . . . , n) the boundary value problem
x(n) (t) = p(x)(t) + λp(x)(t) + q(t),
(3.23)
`i (x) + λ`i (x) = c0i (i = 1, . . . , n)
(3.24)
and prove that every solution x of this problem admits the estimate
kxkC n−1 ≤ γ
&micro;X
n
α,β
&para;
kc0i k + kqkLα,β ,
(3.25)
i=1
where γ is a positive constant not depending on λ, q, c0i (i = 1, . . . , n) and x.
Assume the contrary that this is not so. Then for each natural k there are
λk ∈ [0, 1], qk ∈ Lα,β (]a, b[ ; Rm ), cki ∈ Rm (i = 1, . . . , n)
such that the problem
x(n) (t) = p(x)(t) + λk p(x)(t) + qk (t),
`i (x) = λk `i (x) + cki (i = 1, . . . , n)
has a solution xk admitting the estimate
def
ρk = kxk kC n−1 &gt; k
&micro;X
n
α,β
&para;
kcki k + kqk kLα,β .
i=1
If we assume that
−1
xk (t) = ρ−1
k xk (t), q k (t) = ρk qk (t),
cki = ρ−1
k cki (i = 1, . . . , n),
then we have
kxk kC n−1 = 1,
(3.26)
α,β
kq k kC n−1 &lt;
α,β
(n)
1
,
k
n
X
i=1
kcki k &lt;
1
,
k
xk (t) = p(xk )(t) + λk p(xk )(t) + q k (t),
(3.27)
(3.28)
810
I. KIGURADZE, B. PŮŽA, AND I. P. STAVROULAKIS
`i (xk ) = λk `i (xk ) + cki (i = 1, . . . , n).
Let t0 =
a+b
2
(3.29)
. Then (3.28) implies
xk (t) = yk (t) + zk (t),
(3.30)
where
yk (t) =
n
X
(t − t0 )i−1
i=1
(i − 1)!
(i−1)
xk
(t0 )
t
Z
&sup3;
&acute;
1
+
(t − s)n−1 p(xk )(s) + λk p(xk )(s) ds,
(n − 1)!
(3.31)
t0
t
Z
1
zk (t) =
(t − s)n−1 q k (s) ds.
(n − 1)!
t0
By (3.27) we have
lim kzk kC n−1 = 0,
k→+∞
α,β
lim cki = 0 (i = 1, . . . , n).
k→+∞
(3.32)
On the other hand, with (3.22) and (3.26) taken into account, from (3.31) we
find
(i−1)
kyk
(t0 )k ≤ ρ∗ (i = 1, . . . , n),
Zt
&deg;
&deg;
(n−1)
&deg; (n−1)
&deg;
&deg;yk
(t) − yk
(s)&deg; ≤ ζ(ξ) dξ for a &lt; s &lt; t &lt; b,
s
where ρ∗ is a positive constant not depending on k. By virtue of these inequalities and Lemma 1.1, we can assume without loss of generality that the sequence
n−1
m
(yk )+∞
k=1 is converging in the norm of the space Cα,β (]a, b[ ; R ). It can also
+∞
be assumed without loss of generality that the sequence (λk )k=1 is converging.
Assume that
λ = lim λk , x(t) = lim yk (t).
k→+∞
k→+∞
Then by (3.29)–(3.32) we have
lim kxk − xkC n−1 = lim kyk − xkC n−1 = 0
k→+∞
α,β
k→+∞
α,β
(3.33)
and
`i (x) = λ`i (x) (i = 1, . . . , n),
x(t) =
n
X
(t − t0 )i−1
i=1
(i − 1)!
t
x
(i−1)
Z
&sup3;
&acute;
1
(t0 ) +
(t − s)n−1 p(x)(s) + λp(x)(s) ds.
(n − 1)!
t0
Therefore x is a solution of problem (1.32). On the other hand, from (3.26) and
(3.33) it clearly follows that
kxkC n−1 = 1.
α,β
ON SINGULAR BOUNDARY VALUE PROBLEMS
811
But this is impossible because for each λ ∈ [0, 1] problem (1.32) has only a
trivial solution. The contradiction obtained proves the existence of a positive
number γ that possesses the above-mentioned property.
By condition (1.31) there is ρ0 &gt; 0 such that inequality (3.12) is fulfilled.
To prove Corollary 1.4, it is sufficient due to Theorem 1.2 to establish that
for each λ ∈ ]0, 1[ an arbitrary solution x of the problem
h
i
x(n) (t) = p(x)(t) + λ f (x)(t) − p(x)(t) ,
&sup3;
&acute;
`i (x) = λ `i (x) − hi (x)
(3.34)
(i = 1, . . . , n)
(3.35)
It is obvious that each solution x of problem (3.34), (3.35) is a solution of
problem (3.23), (3.24), where
&sup3;
&acute;
q(t) = λ f (x)(t) − p(x)(t) − p(x)(t) ,
&sup3;
&acute;
c0i = λ `i (x) + `i (x) − hi (x)
(3.36)
(i = 1, . . . , n).
According to the above proof, x admits estimate (3.25) from which, with (1.29),
(1.30) and (3.36) taken into account, we find
&Atilde;
&sup3;
&acute;
Zb
α
kxkC n−1 ≤ γ η0 kxkC n−1 +
α,β
α,β
β
&sup3;
&acute;
!
(s − a) (b − s) η s, kxkC n−1 ds .
a
α,β
Hence, by virtue of (3.12), we obtain estimate (1.19).
Acknowledgements
This work was supported by the Research Grant of the Greek Ministry of Development in the framework of Bilateral S&amp;T Cooperation between the Hellenic
Republic and the Republic of Georgia.
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1, M. Aleksidze St., Tbilisi 380093
Georgia
E-mail: [email protected]
B. Půža
Masaryk University
Faculty of Science
Department of Mathematical Analysis
Janáčkovo nám. 2a, 662 95 Brno
Czech Republic
E-mail: [email protected]
I. P. Stavroulakis
Department of Mathematics
University of Ioannina
451 10 Ioannina
Greece
E-mail: [email protected]
```