3. Introduction
In 1836, the first comparison theorem was given by C. Strum [2],
which states that
Let r(t), p1(t) and p2(t) be real-valued continuous functions defined
on (a, b), a < b such that r(t) > 0 and p2(t) ≥ p1(t) for t ∈ (a, b).
Assume that x(t) and y(t) are solutions of
(r(t)x0
)0
+ p1(t)x = 0,
(r(t)y0
)0
+ p2(t)y = 0
respectively on (a, b). Then between any two consecutive zeros t1, t2
∈ (a, b) of x(t), there is at least one zero of y(t).
In 1909, the general form of Strum’s comparison theorem has been
given by M. Picone [6], which is known as Strum-Picone theorem and
when r1(t) = r2(t) = r(t), Strum-Picone theorem reducing to
Strum’s comparison theorem and hence Strum-Picone theorem is
more general.
Jyotiranjan Sahoo (Sambalpur university, India) Singular Strum-Picone Theorem September 16, 2019 3 / 20
4. Preliminaries
Definition 1 ([4])
If p(t) is any real function
p+
(t) = max{p(t), 0} and p−
(t) = max{−p(t), 0},
are said to be the positive and negative parts of p(t), respectively and
p(t) = p+(t) − p−(t); |p(t)| = p+(t) + p−(t); p+(t), p−(t) ≥ 0.
Jyotiranjan Sahoo (Sambalpur university, India) Singular Strum-Picone Theorem September 16, 2019 4 / 20
5. Theorem 2 ([2] Strum-Picone Comparison Theorem:)
Let ri , pi , i = 1, 2, be real valued continuous on the interval (a, b).
Assume that x(t) and y(t) are solutions of
r1(t)x0
0
+ p1(t)x = 0,
r2(t)y0
0
+ p2(t)y = 0,
respectively on (a, b). Further, let
r1(t) r2(t) 0, p2(t) p1(t), t ∈ (a, b). Then one of the following
properties holds.
(i) Between any two consucative zero t1, t2 ∈ (a, b) of x(t), there is al
least one zero of y(t); or
(ii) there exists a λ in R such that y(t) = λx(t).
Jyotiranjan Sahoo (Sambalpur university, India) Singular Strum-Picone Theorem September 16, 2019 5 / 20
6. Main Results
Consider a pair of second order non-linear differential equations
lsx ≡ r1(t)x0
0
+ p1(t)f1(x(t)) = 0, (1)
Lsy ≡ r2(t)y0
0
+ p2(t)f2(y(t)) = 0, (2)
where
t ∈ (t1, t2), r1, r2 ∈ C1 (t1, t2), (0, ∞)
, p1, p2 ∈ C (t1, t2), R
,
some of them or all r1, r2, p1, p2 may not be continuous at t1 or t2 or at
t1 and t2 both. Let f1, f2 ∈ (R, R), ls and Ls be differential operators or
mapping whose domain consists of all real valued functions x ∈ C1(t1, t2)
such that r1x0 ∈ C1(t1, t2) and r2x0 ∈ C1(t1, t2), respectively.
Jyotiranjan Sahoo (Sambalpur university, India) Singular Strum-Picone Theorem September 16, 2019 6 / 20
7. We need the following hypothesis on non-linearity f1 and f2 :
(H1) f2 ∈ C(R, R) and there exist α1, α2 0 such that
α3y2
≤ yf2(y) ≤ α2y2
, for all 0 6= y ∈ R,
that is,
f2(y)
y
is bounded for all 0 6= y ∈ R.
(H2) f1 ∈ C1(R, R) and there exist α1 0 such that
0 α1 ≤ f 0
1(y), for all 0 6= y ∈ R.
(H3) f1(y) 6= 0, for all 0 6= y ∈ R.
Jyotiranjan Sahoo (Sambalpur university, India) Singular Strum-Picone Theorem September 16, 2019 7 / 20
8. Consider the quadratic functionals corresponding to (1) and (2),
respectively. Let t1 ξ η t2 and let
jξη[u] =
Z η
ξ
r1(t)(u0
(t))2
− α1p1(t)(u(t))2
dt
and
Jξη[u] =
Z η
ξ
r2(t)(u0
(t))2
− (α2p+
2 (t) − α3p−
2 (t))(u(t))2
dt.
Let us define js[u] = limξ→t+
1 , η→t−
2
jξη[u], Js[u] = limξ→t+
1 , η→t−
2
Jξη[u],
whenever the limits exist.
Jyotiranjan Sahoo (Sambalpur university, India) Singular Strum-Picone Theorem September 16, 2019 8 / 20
9. The domain Djs of js and DJs of Js are defined to be the set of all real
valued continuous functions u ∈ C1(t1, t2) with u(t1) = 0 and u(t2) = 0
such that js[u] and Js[u] exist and p+
2 (t) = max{p2(t), 0} and
p−
2 (t) = max{−p2(t), 0}. Let us define
At1t2 [u, x] = lim
t→t−
2
(u(t))2p1(t)x0(t)
f1(x(t))
− lim
t→t+
1
(u(t))2p1(t)x0(t)
f1(x(t))
,
whenever the limit of the right side exist. The variation Vs[u] is defined as
Vs[u] = Js[u] − js[u].
Jyotiranjan Sahoo (Sambalpur university, India) Singular Strum-Picone Theorem September 16, 2019 9 / 20
10. Lemma 3 ([5])
Suppose there exists a function u ∈ Djs not identically zero in any open
subinterval (t1, t2) such that js[u] ≤ 0. Let x be any non-trivial solution of
(1) and At1t2 [u, x] ≥ 0, then under hypothesis (H2) − (H3), x has at least
one zero in (t1, t2) unless f1(x) is a constant multiple of u.
Proof. Let x(t) 6= 0 for every t ∈ (t1, t2). Then by (H3), f1(x(t)) 6= 0, for
all t ∈ (t1, t2). Now,
(u(t))2
f1(x(t))
r1(t)x0
(t)
0
=
(u(t))2
f1(x(t))
(r1(t)x0
(t))0
+ r1(t)x0
(t)
2f1(x(t))u(t)u0(t) − (u(t))2f 0
1(x(t))x0(t)
(f 0
1(x(t)))2
Jyotiranjan Sahoo (Sambalpur university, India) Singular Strum-Picone Theorem September 16, 2019 10 / 20
11. contd.
= −p1(t)(u(t))2
− r1(t)
u(t)x0(t)
p
f 0
1(x(t))
f1(x(t))
−
u0(t)
f 0
1(x(t))
2
+
r1(t)(u0(t))2
f 0
1(x(t))
≤ −p1(t)(u(t))2
− r1(t)
u(t)x0(t)
p
f 0
1(x(t))
f1(x(t))
−
u0(t)
f 0
1(x(t))
2
+
r1(t)(u0(t))2
α1
,
that is,
r1(u0
(t))2
− α1p1(t)(u(t))2
≥ α1
(u(t))2
f1(x(t))
r1(t)x0
(t)
0
+ α1r1(t)
u(t)x0(t)
p
f 0
1(x(t))
f1(x(t))
−
u0(t)
f 0
1(x(t))
2
.
Jyotiranjan Sahoo (Sambalpur university, India) Singular Strum-Picone Theorem September 16, 2019 11 / 20
12. contd.
Integrating both sides from ξ to η, we get
jξη[u] ≥ α1
(u(t))2
r1(t)x0
(t)
f1(x(t))
η
ξ
+ α1
Z η
ξ
r1(t)
u(t)x0
(t)
p
f 0
1 (x(t))
f1(x(t))
−
u0
(t)
f 0
1 (x(t))
2
dt. (3)
Letting ξ → t+
1 , η → t−
2 and using At1t2
≥ 0, we get
js[u] ≥ α1
Z t2
t1
r1(s)
u(s)x0
(s)
p
f 0
1 (x(s))
f1(x(s))
−
u0
(s)
f 0
1 (x(s))
2
ds ≥ 0.
Therefore,
f1(x(t)) = C1u(t), for all t ∈ (t1, t2). (4)
Now for t → t1 or t → t2, L. H. S. of (4) is non-zero while R. H. S. is zero.
Therefore, js[u] 0, which is a contradiction.
Jyotiranjan Sahoo (Sambalpur university, India) Singular Strum-Picone Theorem September 16, 2019 12 / 20
13. Corollary 4 ([5])
Let f1(x) = x in (1) and
lim
ξ→t+
1 , η→t−
2
Z η
ξ
[r1(t)(u0
(t))2
− p1(t)(u(t))2
] dt
exists and is non-positive, where u ∈ C1(t1, t2) not identically zero in any
open subinterval (t1, t2) with u(t1) = 0 and u(t2) = 0. Let x be any
non-trivial solutions of (1) and At1t2 [u, x] ≥ 0, then x vanishes at least
once in (t1, t2) unless x is a constant multiple of u.
Clearly, f1 satisfies (H2) − (H3), α1 = 1 and js[u] ≤ 0. Thus, the result
follows from Lemma 3.
Jyotiranjan Sahoo (Sambalpur university, India) Singular Strum-Picone Theorem September 16, 2019 13 / 20
14. Theorem 5 ([5])
Suppose there exists a non-trivial u of (2) in (t1, t2) such that u(t1) = 0
and u(t2) = 0. Let x be any non-trivial solution of (1). Let
At1t2 [u, x] ≥ 0, and
lim
t→t+
1
r2(t)u(t)u0
(t) ≥ 0, lim
t→t−
1
r2(t)u(t)u0
(t) ≤ 0. (5)
Let (H1) − (H2), holds and Vs[u] ≥ 0, then x has at least one zero in
(t1, t2) unless f1(x) is a constant multiple of u.
Proof. Multiplying (2) by y(t) and integrating by parts from ξ to η, we
obtain
Z η
ξ
r2(t)(y0
(t))2
− p2(t)f2(y(t))y(t)
dt =
r2(t)u(t)u0
(t)
η
ξ
. (6)
Since
Z η
ξ
p2(t)f2(y(t))y(t) − (α2p+
2 (t) − α3p−
2 (t))(y(t))2
dt ≤ 0. (7)
Jyotiranjan Sahoo (Sambalpur university, India) Singular Strum-Picone Theorem September 16, 2019 14 / 20
15. contd.
Letting ξ → t+
1 , η → t−
2 in (7) and by (5), we get Js[u] ≤ 0. Since Vs[u] ≥ 0,
that is, Js[u] − js[u] ≥ 0, and hence the result follows from Lemma 3.
Corollary 6 ([5])
Let us consider (1) and (2) with f1(u) = u and f2(u) = u. Let x be any
non-trivial solution of (1). Let
At1t2
[u, x] ≥ 0, lim
t→t+
1
r2(t)u(t)u0
(t) ≥ 0, lim
t→t−
1
r2(t)u(t)u0
(t) ≤ 0 (8)
and
Vs[u] =
Z t2
t1
(r2(s) − r1(s))(u0
(s))2
+ (p2(s) − p1(s))(u(s))2
ds ≥ 0. (9)
Suppose there exists a non-trivial solution u of (2) in (t1, t2) such that u(t1) = 0
and u(t2) = 0, then x has at least one zero in (t1, t2) unless x is a constant
multiple of u.
Jyotiranjan Sahoo (Sambalpur university, India) Singular Strum-Picone Theorem September 16, 2019 15 / 20
16. Theorem 7 ([5])
Suppose there exists a non-trivial solution u of (2) in (t1, t2) such that
u(t1) = 0 and u(t2) = 0. Let (H3) − (H5), hold. Suppose r2(t) ≥ r1(t).
Let x be any non-trivial solution of (1). Let
At1t2 [u, x] ≥ 0, lim
t→t+
1
r2(t)u(t)u0
(t) ≥ 0, lim
t→t−
1
r2(t)u(t)u0
(t) ≤ 0
and
α1p1(t) − (α2p2(t) − (α3 − α2)p−
2 (t)) ≥ 0, for all t ∈ (t1, t2), (10)
then every solution x of (1) has at least one zero in (t1, t2) unless f1(x) is
a constant multiple of u.
From (10), it is clear that Vs[u] ≥ 0 and proof of the theorem follows from
Theorem 5.
Jyotiranjan Sahoo (Sambalpur university, India) Singular Strum-Picone Theorem September 16, 2019 16 / 20
17. Corollary 8 ([5])
Consider (1) and (2) with f1(u) = u and f2(u) = u. Let r2(t) ≥ r1(t) and
p1(t) ≥ p2(t), for all t ∈ (t1, t2). Let At1t2 [u, x] ≥ 0 and
lim
t→t+
1
r2(t)u(t)u0
(t) ≥ 0, lim
t→t−
1
r2(t)u(t)u0
(t) ≤ 0.
Suppose there exists a non-trivial solution u of (2) in (t1, t2) such that
u(t1) = 0 and u(t2) = 0, then every solution x of (1) has at least one zero
in (t1, t2) unless x is a constant multiple of u.
Clearly, f1(u), f2(u) satisfies (H1) − (H3) and α1 = α2 = α3 = 1. Hence
the result follows from Theorem 7.
Jyotiranjan Sahoo (Sambalpur university, India) Singular Strum-Picone Theorem September 16, 2019 17 / 20
18. Appendix Reference
Reference I
1 A. Tiryaki,
Strum-Picone type theorems for second order non-linear differential
equations,
Electron. J. Differ. Eq., 2014 (2014), 1-11.
2 C. Sturm,
Sur les equations differentielles linearies du second ordere,
J. Math. Pures. Appl., 1 (1836), 106-186.
3 C. A. Swanson,
Comparison and Oscillation Theory of Linear Differential Equations,
Academic Press, New York, 1968.
4 G. de Barra,
Measure Theory and Integration,
New Age International, New Delhi, 2013.
Jyotiranjan Sahoo (Sambalpur university, India) Singular Strum-Picone Theorem September 16, 2019 18 / 20
19. Appendix Reference
Reference II
5 J. Tyagi,
Generalization of Strum-Picone theorem for second order non-linear
differential equations,
Taiwan. J. Math., 17 (2013), 361-378.
6 M. Picone,
Sui valori eccezionalle di un parametro da cui dipende un equatione
differenzialle lineare del secondo ordine,
Ann. Scuola Norm. Sup. Pisa., 11 (1910), 1-141.
7 W. Leighton,
Comparison theorems for linear differential equations of second order,
Proc. Amer. Math. Soc., 13 (1962), 603-610.
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