GEORGIAN MATHEMATICAL JOURNAL: Vol. 2, No. 1, 1995, 93-98
EXISTENCE OF CONJUGATE POINTS FOR
SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
A. LOMTATIDZE
Abstract. The sufficient conditions are established under which the
second-order linear differential equation is conjugate.
1. Consider the differential equation
u00 = p(t)u,
(1)
where −∞ < a < b < +∞ and the function p :]a, b[→ R is Lebesgue
integrable on each compact subset of ]a, b[.
Definition 1. A function p :]a, b[→ R belongs to the class O(]a, b[) if the
condition
Z b
(2)
(s − a)(b − s)|p(s)|ds < +∞
a
is satisfied and the solution of equation (1), satisfying the initial conditions
u(a+) = 0,
u0 (a+) = 1,
(3)
has at least one zero on ]a, b[.
It is known [1] that if condition (2) is satisfied then any solution u of (1)
has the finite left- and right-hand side limits u(a+) and u(b−) and problem
(1),(3) is uniquely solvable.
Definition 2. A function p :]a, b[→ R belongs to the class O0 (]a, b[) if
Z b
(4)
(s − a)|p(s)|ds < +∞
a
and the derivative of the solution of problem (1),(3) has at least one zero
on ]a, b].
1991 Mathematics Subject Classification. 34C10.
Key words and phrases. Second-order linear differential equation, conjugate point.
93
c 1995 Plenum Publishing Corporation
1072-947X/95/0100-0093$07.50/0 
94
A. LOMTATIDZE
While investigating two-point singular boundary-value problems there
arises a question as to an effective description of the classes O(]a, b[) and
O0 (]a, b[) (see, for example, [2]). Earlier attempts in this direction were
undertaken in [3-7]. The statements given below complete the results of
these papers.
2. On the Class O(]a, b[). In this section the function p :]a, b[→ R is
assumed to satisfy (2).
Theorem 1. Let there exist a point t0 ∈]a, b[ and absolutely continuous
functions f1 : [a, t0 ] → [0, +∞[ and f2 : [t0 , b] → [0, +∞[ such that f1 (a) =
f2 (b) = 0, f1 (t) > 0 for a < t ≤ t0 , f2 (t) > 0 for t0 ≤ t < b,
Z b
Z t0
[f2 0 (s)]2
[f1 0 (s)]2
ds < +∞, g2 (t0 ) =
ds < +∞ (5)
g1 (t0 ) =
f1 (s)
f2 (s)
t0
a
and
f2 (t0 )
Z
t0
f1 (s)p(s)ds + f1 (t0 )
a
≤−
Then p ∈ O(]a, b[).
Z
b
t0
f2 (s)p(s)ds ≤

1€
g1 (t0 )f2 (t0 ) + g2 (t0 )f1 (t0 ) .
4
(6)
Corollary 1. Let there exist t0 ∈]a, b[ and α ∈]1, +∞[ such that
Z t0
Z b
α
α
α
(s − a) p(s)ds + (t0 − a)
(b − t0 )
(b − s)α p(s)ds ≤
a
Then p ∈ O(]a, b[).
t0
α2
[(t0 − a)(b − t0 )]α−1 .
≤−
4(α − 1)
and α = 2 this proposition implies the result of
When t0 = a+b
2
R. Putnam ([4], p. 177).
Corollary 2. Let one of the following three conditions hold:
Z b
3
9π(b − a)2
,
[(s − a)(b − s)] 2 p(s)ds ≤ −
32
a
Z b
n!2
[(s − a)(b − s)]n p(s)ds ≤ −
(b − a)2n−1 ,
2(n − 1)(2n − 1)!
a
n ≥ 2, n ∈ N,
Z b
€ π(s − a) 
π2
p(s)ds ≤ −
sin2
.
b−a
2(b − a)
a
Then p ∈ O(]a, b[).
EXISTENCE OF CONJUGATE POINTS
95
Corollary 3. Let the inequality
(b − t0 )
Z
a
t0
1
3
(s − a) 2 (b − s) 2 p(s)ds + (t0 − a)
Z
b
t0
1
3
(s − a) 2 (b − s) 2 p(s)ds ≤
1
5
≤ − [(t0 − a)(b − t0 )] 2 (b − a)
4
hold for some t0 ∈]a, b[. Then p ∈ O(]a, b[).
Theorem 2. Let the inequality
+(t − a)
Z
(b − t)
b
t
Z
a
t
3
1
(s − a) 2 (b − s) 2 [p(s)]− ds +
3
1
2
1
(s − a) (b − s) 2 [p(s)]− ds < [(t − a)(b − t)] 2 (b − a)
for a < t < b
hold, where [p(t)]− =
|p(t)|−p(t)
.
2
(7)
Then p 6∈ O(]a, b[).
3. On the Class O0 (]a, b[). In this section the function p :]a, b[→ R is
assumed to satisfy (4).
Theorem 3. Let there exist t0 ∈]a, b[ and absolutely continuous functions f1 : [a, t0 ] → [0, +∞[ and f2 : [t0 , b] → [0, +∞[ such that f1 (a) = 0,
f1 (t) > 0 for a < t < t0 , f2 (t) > 0 for t0 < t < b and conditions (5) and (6)
are satisfied. Then p ∈ O0 (]a, b[).
Corollary 4. Let there exist t0 ∈]a, b[ and α ∈]1, +∞[ such that
Z
t0
a
(s − a)α p(s)ds + (t0 − a)α
Z
b
t0
p(s)ds ≤ −
α2
(t0 − a)α−1 .
4(α − 1)
Then p ∈ O0 (]a, b[).
Corollary 5. Let the inequality
Z
a
t0
3
(s − a) 2 p(s)ds + (t0 − a)
Z
b
t0
1
t0 − a
5
≤ − (t0 − a) 2 +
1
4
8(b − a) 2
hold for some t0 ∈]a, b[. Then p ∈ O0 (]a, b[).
1
(s − a) 2 p(s)ds ≤
96
A. LOMTATIDZE
4. Proof of the Main results.
Proof of Theorem 1 (Theorem 3). Admit on the contrary that p 6∈ O(]a, b[)
(p 6∈ O0 (]a, b[)). Then equation (1) has the solution u satisfying
u(a+) = 0, u0 (a+) = 1, u(t) > 0 for a < t < b.
’
u(a+) = 1, u0 (b−) = 0, u(t) > 0 for a ≤ t ≤ b
Denote
ρ(t) =
“
.
u0 (t)
for a < t < b.
u(t)
It is clear that
ρ0 (t) = p(t) − ρ2 (t) for a < t < b.
(8)
Multiplying both sides of this equality by f1 and integrating from a + ε
to t0 where ε ∈]0, t0 − a[, we have
Z t0
f1 (s)p(s)ds + f1 (t0 )ρ(t0 ) − f1 (a + ε)ρ(a + ε) =
−
=
Z
t0
a+ε
”
a+ε
•
Z
1
1 t0 [f1 0 (s)]2
ds < g1 (t0 ).
f1 0 (s)ρ(s) − f1 (s)ρ2 (s) ds <
4 a+ε f1 (s)
4
From (5) it easily follows that
lim f1 0 (t) = 0.
t→a+
Therefore
lim f1 (t)ρ(t) = 0.
t→a+
In view of this the last inequality can be rewritten as
Z t0
1
−
f1 (s)p(s)ds + f1 (t0 )ρ(t0 ) < g1 (t0 ).
4
a
(9)
Now multiplying both sides of (8) by f2 and integrating from t0 to b − ε
where ε ∈]0, b − t0 [, we have
Z b−ε
−
f2 (s)p(s)ds − f2 (t0 )ρ(t0 ) + f2 (b − ε)ρ(b − ε) =
=
Z
b−ε
t0
”
t0
•
1
f2 (s)ρ(s) − f2 (s)ρ (s) ds <
4
0
2
Z
b−ε
t0
[f2 0 (s)]2
1
ds < g2 (t0 ).
f2 (s)
4
Taking into account
lim f2 0 (t) = 0 and
t→b−
lim (b − t)u0 (t) = 0,
t→b−
EXISTENCE OF CONJUGATE POINTS
from the last inequality we obtain
Z b
1
−
f2 (s)p(s)ds − f2 (t0 )ρ(t0 ) < g2 (t0 ).
4
t0
From (9) and (10) we have
”
Z
Z t0
f1 (s)p(s)ds + f1 (t0 )
− f2 (t0 )
b
t0
a
97
(10)
•
f2 (s)p(s)ds <
i
1h
< f2 (t0 )g1 (t0 ) + f1 (t0 )g2 (t0 ) ,
4
which contradicts (6).
Proof of Theorem 2. Admit on the contrary that p ∈ O(]a, b[). Then
equation (1) has the solution u satisfying
u(a+) = u(b1 ) = 0, u(t) > 0 for a < t < b1 ≤ b.
According to the Green formula
u(t) =
”
1
(b1 − t)
=−
b1 − a
Z
a
t
(s − a)p(s)u(s)ds + (t − a)
for a ≤ t ≤ b1 .
Z
t
b1
•
(b1 − s)p(s)u(s)ds
Hence we easily obtain
”
Z t
1
(b1 − t)
(s − a)[p(s)]− u(s)ds +
b1 − a
a
•
Z b1
+(t − a)
(b1 − s)[p(s)]− u(s)ds for a ≤ t ≤ b1 ,
u(t) ≤
t
i.e.,
v(t) ≤
Z t
1
3
1
b1 − t
≤
(s − a) 2 (b − s) 2 [p(s)]− v(s)ds +
1
a
−
b
2
[(t − a)(b − t)]
1
a
•
Z b
1
1
t−a
+
(s − a) 2 (b1 − s)(b − s) 2 [p(s)]− v(s)ds <
b1 − a t
”
Z
3
1
b−t t
1
(s − a) 2 (b − s) 2 [p(s)]− )ds +
<λ
1
b
−
a
2
[(t − a)(b − t)]
a
•
Z
1
3
t−a b
(s − a) 2 (b − s) 2 [p(s)]− ds for a < t < b1 ,
+
b−a t
”
98
A. LOMTATIDZE
where
v(t) =
and
u(t)
1
[(t − a)(b − t)] 2
for a < t < b1
λ = sup{v(t) t ∈]a, b1 [}.
Taking into account (7), we obtain the contradiction λ < λ.
Corollaries 1-5 are obtained from Theorems 1 and 3 by an appropriate
choice of functions f1 and f2 .
References
1. I. T. Kiguradze and A. G. Lomtatidze, On certain boundary-value
problems for second-order linear ordinary differential equations with singularities. J. Math. Anal. Appl. 101(1984), No. 2, 325-347.
2. A. G. Lomtatidze, On positive solutions of boundary-value problems
for second-order ordinary differential equations with singularities. (Russian)
Differentsial’nye Uravneniya 23(1987), 1685-1692.
3. F. Hartman, Ordinary differential equations. Wiley, New York/London/Sydney, 1964.
4. M. A. Krasnosel’ski, Vector fields on the plane. (Russian) Moscow,
Nauka, 1963.
5. N. l. Korshikova, On zeros of solutions of high-order linear equations.
(Russian) Differential equations and their applications. (Russian) 143-148,
Moscow University Press, Moscow, 1984.
6. A. G. Lomtatidze, On oscillatory properties of solutions of secondorder linear differential equations. (Russian) Sem. I. Vekua Inst. Appl.
Math. Tbiliss. St. Univ. Reports. 19(1985), 39-53.
7. O. Došlý, The multiplicity criteria for zero points of second-order
differential equations. Math. Slovaca 42(1992), No. 2, 181-193.
(Received 01.12.1993)
Author’s address:
N.Muskhelishvili Institute of Computational Mathematics
Georgian Academy of Sciences
8, Akuri St., Tbilisi 380093
Republic of Georgia