On periodic solutions
of second–order differential equations with attractive–repulsive singularities
Robert Hakl, Pedro J. Torres
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
ω, λ, µ > 0 g, h ∈ L`[0, ω]; R+´ f ∈ L`[0, ω]; R´
Positive solution u : [0, ω] → ]0, +∞[
u, u0are absolutely continuous and satisfy (2), abbr. u ∈ AC1ω`[0, ω]; R´ u satisfies (1)
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
ω, λ, µ > 0
g, h ∈ L`[0, ω]; R+´ f ∈ L`[0, ω]; R´
Positive solution u : [0, ω] → ]0, +∞[
u, u0are absolutely continuous and satisfy (2), abbr. u ∈ AC1ω`[0, ω]; R´ u satisfies (1)
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
ω, λ, µ > 0 g, h ∈ L`[0, ω]; R+´
f ∈ L`[0, ω]; R´
Positive solution u : [0, ω] → ]0, +∞[
u, u0are absolutely continuous and satisfy (2), abbr. u ∈ AC1ω`[0, ω]; R´ u satisfies (1)
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
ω, λ, µ > 0 g, h ∈ L`[0, ω]; R+´ f ∈ L`[0, ω]; R´
Positive solution u : [0, ω] → ]0, +∞[
u, u0are absolutely continuous and satisfy (2), abbr. u ∈ AC1ω`[0, ω]; R´ u satisfies (1)
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
ω, λ, µ > 0 g, h ∈ L`[0, ω]; R+´ f ∈ L`[0, ω]; R´
Positive solution u : [0, ω] → ]0, +∞[
u, u0are absolutely continuous and satisfy (2), abbr. u ∈ AC1ω`[0, ω]; R´ u satisfies (1)
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
ω, λ, µ > 0 g, h ∈ L`[0, ω]; R+´ f ∈ L`[0, ω]; R´
Positive solution u : [0, ω] → ]0, +∞[
u, u0 are absolutely continuous and satisfy (2), abbr. u ∈ ACω1`[0, ω]; R´
u satisfies (1)
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
ω, λ, µ > 0 g, h ∈ L`[0, ω]; R+´ f ∈ L`[0, ω]; R´
Positive solution u : [0, ω] → ]0, +∞[
u, u0 are absolutely continuous and satisfy (2), abbr. u ∈ ACω1`[0, ω]; R´
u satisfies (1)
Special cases of (1)
u00(t) = g(t) uµ(t) − h(t)
uλ(t) for a. e. t ∈ [0, ω], (3)
u00(t) = − h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (4) u00(t) = g(t)
uµ(t)+ f (t) for a. e. t ∈ [0, ω], (5) Attractive singularity
x→0lim+q(t, x) = −∞ for a. e. t ∈ [0, ω]
Repulsive singularity
x→0lim+q(t, x) = +∞ for a. e. t ∈ [0, ω]
Special cases of (1)
u00(t) = g(t) uµ(t) − h(t)
uλ(t) for a. e. t ∈ [0, ω], (3)
u00(t) = − h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (4)
u00(t) = g(t)
uµ(t)+ f (t) for a. e. t ∈ [0, ω], (5) Attractive singularity
x→0lim+q(t, x) = −∞ for a. e. t ∈ [0, ω]
Repulsive singularity
x→0lim+q(t, x) = +∞ for a. e. t ∈ [0, ω]
Special cases of (1)
u00(t) = g(t) uµ(t) − h(t)
uλ(t) for a. e. t ∈ [0, ω], (3)
u00(t) = − h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (4) u00(t) = g(t)
uµ(t)+ f (t) for a. e. t ∈ [0, ω], (5)
Attractive singularity
x→0lim+q(t, x) = −∞ for a. e. t ∈ [0, ω]
Repulsive singularity
x→0lim+q(t, x) = +∞ for a. e. t ∈ [0, ω]
Special cases of (1)
u00(t) = g(t) uµ(t) − h(t)
uλ(t) for a. e. t ∈ [0, ω], (3)
u00(t) = − h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (4) u00(t) = g(t)
uµ(t)+ f (t) for a. e. t ∈ [0, ω], (5) Attractive singularity
x→0lim+q(t, x) = −∞ for a. e. t ∈ [0, ω]
Repulsive singularity
x→0lim+q(t, x) = +∞ for a. e. t ∈ [0, ω]
Special cases of (1)
u00(t) = g(t) uµ(t) − h(t)
uλ(t) for a. e. t ∈ [0, ω], (3)
u00(t) = − h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (4) u00(t) = g(t)
uµ(t)+ f (t) for a. e. t ∈ [0, ω], (5) Attractive singularity
x→0lim+q(t, x) = −∞ for a. e. t ∈ [0, ω]
Repulsive singularity
x→0lim+q(t, x) = +∞ for a. e. t ∈ [0, ω]
A. C. Lazer, S. Solimini (1987)
u00= − 1
uλ+ f (t); u(0) = u(ω), u0(0) = u0(ω) (6)
u00= 1
uµ+ f (t); u(0) = u(ω), u0(0) = u0(ω) (7) Theorem A. f continuous; (6) has a positive solution iffRω
0 f (s)ds > 0. Theorem B. f ∈ L`[0, ω]; R´, µ ≥ 1; (7) has a positive solution iffRω
0 f (s)ds < 0. I. Rach˚unkov´a, M. Tvrd´y, I. Vrkoˇc (2001)
Theorem C. f ∈ L`[0, ω]; R´, µ < 1; (7) has a positive solution ifRω
0 f (s)ds < 0 and
ess inf˘f (t) : t ∈ [0, ω]¯ > −
„ π2 ω2µ
«1+µµ (1 + µ).
A. C. Lazer, S. Solimini (1987)
u00= − 1
uλ+ f (t); u(0) = u(ω), u0(0) = u0(ω) (6)
u00= 1
uµ+ f (t); u(0) = u(ω), u0(0) = u0(ω) (7) Theorem A. f continuous; (6) has a positive solution iffRω
0 f (s)ds > 0. Theorem B. f ∈ L`[0, ω]; R´, µ ≥ 1; (7) has a positive solution iffRω
0 f (s)ds < 0. I. Rach˚unkov´a, M. Tvrd´y, I. Vrkoˇc (2001)
Theorem C. f ∈ L`[0, ω]; R´, µ < 1; (7) has a positive solution ifRω
0 f (s)ds < 0 and
ess inf˘f (t) : t ∈ [0, ω]¯ > −
„ π2 ω2µ
«1+µµ (1 + µ).
A. C. Lazer, S. Solimini (1987)
u00= − 1
uλ+ f (t); u(0) = u(ω), u0(0) = u0(ω) (6)
u00= 1
uµ+ f (t); u(0) = u(ω), u0(0) = u0(ω) (7)
Theorem A. f continuous; (6) has a positive solution iffRω
0 f (s)ds > 0. Theorem B. f ∈ L`[0, ω]; R´, µ ≥ 1; (7) has a positive solution iffRω
0 f (s)ds < 0. I. Rach˚unkov´a, M. Tvrd´y, I. Vrkoˇc (2001)
Theorem C. f ∈ L`[0, ω]; R´, µ < 1; (7) has a positive solution ifRω
0 f (s)ds < 0 and
ess inf˘f (t) : t ∈ [0, ω]¯ > −
„ π2 ω2µ
«1+µµ (1 + µ).
A. C. Lazer, S. Solimini (1987)
u00= − 1
uλ+ f (t); u(0) = u(ω), u0(0) = u0(ω) (6)
u00= 1
uµ+ f (t); u(0) = u(ω), u0(0) = u0(ω) (7) Theorem A. f continuous; (6) has a positive solution iffRω
0 f (s)ds > 0.
Theorem B. f ∈ L`[0, ω]; R´, µ ≥ 1; (7) has a positive solution iffRω
0 f (s)ds < 0. I. Rach˚unkov´a, M. Tvrd´y, I. Vrkoˇc (2001)
Theorem C. f ∈ L`[0, ω]; R´, µ < 1; (7) has a positive solution ifRω
0 f (s)ds < 0 and
ess inf˘f (t) : t ∈ [0, ω]¯ > −
„ π2 ω2µ
«1+µµ (1 + µ).
A. C. Lazer, S. Solimini (1987)
u00= − 1
uλ+ f (t); u(0) = u(ω), u0(0) = u0(ω) (6)
u00= 1
uµ+ f (t); u(0) = u(ω), u0(0) = u0(ω) (7) Theorem A. f continuous; (6) has a positive solution iffRω
0 f (s)ds > 0.
Theorem B. f ∈ L`[0, ω]; R´, µ ≥ 1; (7) has a positive solution iffRω
0 f (s)ds < 0.
I. Rach˚unkov´a, M. Tvrd´y, I. Vrkoˇc (2001)
Theorem C. f ∈ L`[0, ω]; R´, µ < 1; (7) has a positive solution ifRω
0 f (s)ds < 0 and
ess inf˘f (t) : t ∈ [0, ω]¯ > −
„ π2 ω2µ
«1+µµ (1 + µ).
A. C. Lazer, S. Solimini (1987)
u00= − 1
uλ+ f (t); u(0) = u(ω), u0(0) = u0(ω) (6)
u00= 1
uµ+ f (t); u(0) = u(ω), u0(0) = u0(ω) (7) Theorem A. f continuous; (6) has a positive solution iffRω
0 f (s)ds > 0.
Theorem B. f ∈ L`[0, ω]; R´, µ ≥ 1; (7) has a positive solution iffRω
0 f (s)ds < 0.
I. Rach˚unkov´a, M. Tvrd´y, I. Vrkoˇc (2001)
Theorem C. f ∈ L`[0, ω]; R´, µ < 1; (7) has a positive solution ifRω
0 f (s)ds < 0 and
ess inf˘f (t) : t ∈ [0, ω]¯ > −
„ π2 ω2µ
«1+µµ (1 + µ).
A. C. Lazer, S. Solimini (1987)
u00= − 1
uλ+ f (t); u(0) = u(ω), u0(0) = u0(ω) (6)
u00= 1
uµ+ f (t); u(0) = u(ω), u0(0) = u0(ω) (7) Theorem A. f continuous; (6) has a positive solution iffRω
0 f (s)ds > 0.
Theorem B. f ∈ L`[0, ω]; R´, µ ≥ 1; (7) has a positive solution iffRω
0 f (s)ds < 0.
I. Rach˚unkov´a, M. Tvrd´y, I. Vrkoˇc (2001)
Theorem C. f ∈ L`[0, ω]; R´, µ < 1; (7) has a positive solution ifRω
0 f (s)ds < 0 and
ess inf˘f (t) : t ∈ [0, ω]¯ > −
„ π2 ω2µ
«1+µµ (1 + µ).
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
Notation
For the sake of brevity we will use the following notation:
G = Z ω
0
g(s)ds, H = Z ω
0
h(s)ds,
F = Zω
0
f (s)ds, F+= Z ω
0
[f (s)]+ds, F−= Z ω
0
[f (s)]−ds.
Note that F = F+− F−.
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
Theorem
Let H > 0, F > 0, functions w, σ ∈ ACω1`[0, ω]; R´ be such that the equalities w00(t) = Hg(t) − Gh(t) for a. e. t ∈ [0, ω], σ00(t) = −F
Hh(t) + f (t) for a. e. t ∈ [0, ω], are fulfilled, and let there exist x0∈ ]0, +∞[ such that
x0`w(t) − mw´ + σ(t) − mσ≤
„ H
x0GH + F
«1/λ
−
„ 1 x0H
«1/µ
for t ∈ [0, ω],
where
mw= min˘w(t) : t ∈ [0, ω]¯, mσ= min˘σ(t) : t ∈ [0, ω]¯. Then the equation (1), (2) has at least one positive solution.
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
Theorem
Let H > 0, F > 0, functions w, σ ∈ ACω1`[0, ω]; R´ be such that the equalities w00(t) = Hg(t) − Gh(t) for a. e. t ∈ [0, ω], σ00(t) = −F
Hh(t) + f (t) for a. e. t ∈ [0, ω], are fulfilled, and let there exist x0∈ ]0, +∞[ such that
x0`w(t) − mw´ + σ(t) − mσ≤
„ H
x0GH + F
«1/λ
−
„ 1 x0H
«1/µ
for t ∈ [0, ω],
where
mw= min˘w(t) : t ∈ [0, ω]¯, mσ= min˘σ(t) : t ∈ [0, ω]¯.
Then the equation (1), (2) has at least one positive solution.
u00(t) = − h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (4)
u(0) = u(ω), u0(0) = u0(ω) (2)
Corollary
Let H > 0, F > 0, σ ∈ ACω1`[0, ω]; R´ be such that
σ00(t) = −F
Hh(t) + f (t) for a. e. t ∈ [0, ω], and let
`Mσ− mσ´λ
F < H, where
Mσ= max˘σ(t) : t ∈ [0, ω]¯, mσ= min˘σ(t) : t ∈ [0, ω]¯. Then the problem (4), (2) has at least one positive solution.
Corollary
Let H > 0, F > 0, and let
“ω 4F+
”λ
F ≤ H. Then the problem (4), (2) has at least one positive solution.
u00(t) = − h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (4)
u(0) = u(ω), u0(0) = u0(ω) (2)
Corollary
Let H > 0, F > 0, σ ∈ ACω1`[0, ω]; R´ be such that
σ00(t) = −F
Hh(t) + f (t) for a. e. t ∈ [0, ω], and let
`Mσ− mσ
´λ
F < H, where
Mσ= max˘σ(t) : t ∈ [0, ω]¯, mσ= min˘σ(t) : t ∈ [0, ω]¯.
Then the problem (4), (2) has at least one positive solution.
Corollary
Let H > 0, F > 0, and let
“ω 4F+
”λ
F ≤ H. Then the problem (4), (2) has at least one positive solution.
u00(t) = − h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (4)
u(0) = u(ω), u0(0) = u0(ω) (2)
Corollary
Let H > 0, F > 0, σ ∈ ACω1`[0, ω]; R´ be such that
σ00(t) = −F
Hh(t) + f (t) for a. e. t ∈ [0, ω], and let
`Mσ− mσ
´λ
F < H, where
Mσ= max˘σ(t) : t ∈ [0, ω]¯, mσ= min˘σ(t) : t ∈ [0, ω]¯.
Then the problem (4), (2) has at least one positive solution.
Corollary
Let H > 0, F > 0, and let
“ω 4F+
”λ
F ≤ H.
Then the problem (4), (2) has at least one positive solution.
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
Lemma
Let there exist positive functions α, β ∈ ACω1`[0, ω]; R´ such that
α00(t) ≥ g(t) αµ(t)− h(t)
αλ(t)+ f (t) for a. e. t ∈ [0, ω], (8) β00(t) ≤ g(t)
βµ(t)− h(t)
βλ(t)+ f (t) for a. e. t ∈ [0, ω], (9) α(t) ≤ β(t) for t ∈ [0, ω].
Then there exists at least one positive solution to (1), (2).
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
Lemma
Let there exist positive functions α, β ∈ ACω1`[0, ω]; R´ such that
α00(t) ≥ g(t) αµ(t)− h(t)
αλ(t)+ f (t) for a. e. t ∈ [0, ω], (8) β00(t) ≤ g(t)
βµ(t)− h(t)
βλ(t)+ f (t) for a. e. t ∈ [0, ω], (9) α(t) ≤ β(t) for t ∈ [0, ω].
Then there exists at least one positive solution to (1), (2).
w00(t) = Hg(t) − Gh(t), σ00(t) = −F
Hh(t) + f (t), x0`w(t) − mw´ + σ(t) − mσ≤
„ H
x0GH + F
«1/λ
−
„ 1 x0H
«1/µ
Proof.
α(t) =
„ 1 x0H
«1/µ
+ x0`w(t) − mw´ + σ(t) − mσ for t ∈ [0, ω].
α00(t) = x0Hg(t) −
„ x0G +F
H
«
h(t) + f (t) for a. e. t ∈ [0, ω],
„ 1 x0H
«1/µ
≤ α(t) ≤
„ H
x0GH + F
«1/λ
for t ∈ [0, ω].
α00(t) ≥ g(t) αµ(t)− h(t)
αλ(t)+ f (t) for a. e. t ∈ [0, ω].
w00(t) = Hg(t) − Gh(t), σ00(t) = −F
Hh(t) + f (t), x0`w(t) − mw´ + σ(t) − mσ≤
„ H
x0GH + F
«1/λ
−
„ 1 x0H
«1/µ
Proof.
α(t) =
„ 1 x0H
«1/µ
+ x0`w(t) − mw´ + σ(t) − mσ for t ∈ [0, ω].
α00(t) = x0Hg(t) −
„ x0G +F
H
«
h(t) + f (t) for a. e. t ∈ [0, ω],
„ 1 x0H
«1/µ
≤ α(t) ≤
„ H
x0GH + F
«1/λ
for t ∈ [0, ω].
α00(t) ≥ g(t) αµ(t)− h(t)
αλ(t)+ f (t) for a. e. t ∈ [0, ω].
w00(t) = Hg(t) − Gh(t), σ00(t) = −F
Hh(t) + f (t), x0`w(t) − mw´ + σ(t) − mσ≤
„ H
x0GH + F
«1/λ
−
„ 1 x0H
«1/µ
Proof.
α(t) =
„ 1 x0H
«1/µ
+ x0`w(t) − mw´ + σ(t) − mσ for t ∈ [0, ω].
α00(t) = x0Hg(t) −
„ x0G +F
H
«
h(t) + f (t) for a. e. t ∈ [0, ω],
„ 1 x0H
«1/µ
≤ α(t) ≤
„ H
x0GH + F
«1/λ
for t ∈ [0, ω].
α00(t) ≥ g(t) αµ(t)− h(t)
αλ(t)+ f (t) for a. e. t ∈ [0, ω].
w00(t) = Hg(t) − Gh(t), σ00(t) = −F
Hh(t) + f (t), x0`w(t) − mw´ + σ(t) − mσ≤
„ H
x0GH + F
«1/λ
−
„ 1 x0H
«1/µ
Proof.
α(t) =
„ 1 x0H
«1/µ
+ x0`w(t) − mw´ + σ(t) − mσ for t ∈ [0, ω].
α00(t) = x0Hg(t) −
„ x0G +F
H
«
h(t) + f (t) for a. e. t ∈ [0, ω],
„ 1 x0H
«1/µ
≤ α(t) ≤
„ H
x0GH + F
«1/λ
for t ∈ [0, ω].
α00(t) ≥ g(t) αµ(t)− h(t)
αλ(t)+ f (t) for a. e. t ∈ [0, ω].
w00(t) = Hg(t) − Gh(t), σ00(t) = −F
Hh(t) + f (t), x0`w(t) − mw´ + σ(t) − mσ≤
„ H
x0GH + F
«1/λ
−
„ 1 x0H
«1/µ
Proof.
α(t) =
„ 1 x0H
«1/µ
+ x0`w(t) − mw´ + σ(t) − mσ for t ∈ [0, ω].
α00(t) = x0Hg(t) −
„ x0G +F
H
«
h(t) + f (t) for a. e. t ∈ [0, ω],
„ 1 x0H
«1/µ
≤ α(t) ≤
„ H
x0GH + F
«1/λ
for t ∈ [0, ω].
α00(t) ≥ g(t) αµ(t)− h(t)
αλ(t)+ f (t) for a. e. t ∈ [0, ω].
w00(t) = Hg(t) − Gh(t), σ00(t) = −F
Hh(t) + f (t),
„ 1 x0H
«1/µ
≤ α(t) ≤
„ H
x0GH + F
«1/λ
.
Proof.
x1∈ ]0, x0] : x1`w(t) − mw´ + σ(t) − mσ≤
„ 1 x1H
«1/µ
−
„ H
x1GH + F
«1/λ
for t ∈ [0, ω]
β(t) =
„ H
x1GH + F
«1/λ
+ x1`w(t) − mw´ + σ(t) − mσ for t ∈ [0, ω].
β00(t) = x1Hg(t) −
„ x1G +F
H
«
h(t) + f (t) for a. e. t ∈ [0, ω],
„ H
x1GH + F
«1/λ
≤ β(t) ≤
„ 1 x1H
«1/µ
for t ∈ [0, ω].
β00(t) ≤ g(t) βµ(t)− h(t)
βλ(t)+ f (t) for a. e. t ∈ [0, ω]. α(t) ≤ β(t) for t ∈ [0, ω].
w00(t) = Hg(t) − Gh(t), σ00(t) = −F
Hh(t) + f (t),
„ 1 x0H
«1/µ
≤ α(t) ≤
„ H
x0GH + F
«1/λ
.
Proof.
x1∈ ]0, x0] : x1`w(t) − mw´ + σ(t) − mσ≤
„ 1 x1H
«1/µ
−
„ H
x1GH + F
«1/λ
for t ∈ [0, ω]
β(t) =
„ H
x1GH + F
«1/λ
+ x1`w(t) − mw´ + σ(t) − mσ for t ∈ [0, ω].
β00(t) = x1Hg(t) −
„ x1G +F
H
«
h(t) + f (t) for a. e. t ∈ [0, ω],
„ H
x1GH + F
«1/λ
≤ β(t) ≤
„ 1 x1H
«1/µ
for t ∈ [0, ω].
β00(t) ≤ g(t) βµ(t)− h(t)
βλ(t)+ f (t) for a. e. t ∈ [0, ω]. α(t) ≤ β(t) for t ∈ [0, ω].
w00(t) = Hg(t) − Gh(t), σ00(t) = −F
Hh(t) + f (t),
„ 1 x0H
«1/µ
≤ α(t) ≤
„ H
x0GH + F
«1/λ
.
Proof.
x1∈ ]0, x0] : x1`w(t) − mw´ + σ(t) − mσ≤
„ 1 x1H
«1/µ
−
„ H
x1GH + F
«1/λ
for t ∈ [0, ω]
β(t) =
„ H
x1GH + F
«1/λ
+ x1`w(t) − mw´ + σ(t) − mσ for t ∈ [0, ω].
β00(t) = x1Hg(t) −
„ x1G +F
H
«
h(t) + f (t) for a. e. t ∈ [0, ω],
„ H
x1GH + F
«1/λ
≤ β(t) ≤
„ 1 x1H
«1/µ
for t ∈ [0, ω].
β00(t) ≤ g(t) βµ(t)− h(t)
βλ(t)+ f (t) for a. e. t ∈ [0, ω]. α(t) ≤ β(t) for t ∈ [0, ω].
w00(t) = Hg(t) − Gh(t), σ00(t) = −F
Hh(t) + f (t),
„ 1 x0H
«1/µ
≤ α(t) ≤
„ H
x0GH + F
«1/λ
.
Proof.
x1∈ ]0, x0] : x1`w(t) − mw´ + σ(t) − mσ≤
„ 1 x1H
«1/µ
−
„ H
x1GH + F
«1/λ
for t ∈ [0, ω]
β(t) =
„ H
x1GH + F
«1/λ
+ x1`w(t) − mw´ + σ(t) − mσ for t ∈ [0, ω].
β00(t) = x1Hg(t) −
„ x1G +F
H
«
h(t) + f (t) for a. e. t ∈ [0, ω],
„ H
x1GH + F
«1/λ
≤ β(t) ≤
„ 1 x1H
«1/µ
for t ∈ [0, ω].
β00(t) ≤ g(t) βµ(t)− h(t)
βλ(t)+ f (t) for a. e. t ∈ [0, ω]. α(t) ≤ β(t) for t ∈ [0, ω].
w00(t) = Hg(t) − Gh(t), σ00(t) = −F
Hh(t) + f (t),
„ 1 x0H
«1/µ
≤ α(t) ≤
„ H
x0GH + F
«1/λ
.
Proof.
x1∈ ]0, x0] : x1`w(t) − mw´ + σ(t) − mσ≤
„ 1 x1H
«1/µ
−
„ H
x1GH + F
«1/λ
for t ∈ [0, ω]
β(t) =
„ H
x1GH + F
«1/λ
+ x1`w(t) − mw´ + σ(t) − mσ for t ∈ [0, ω].
β00(t) = x1Hg(t) −
„ x1G +F
H
«
h(t) + f (t) for a. e. t ∈ [0, ω],
„ H
x1GH + F
«1/λ
≤ β(t) ≤
„ 1 x1H
«1/µ
for t ∈ [0, ω].
β00(t) ≤ g(t) βµ(t)− h(t)
βλ(t)+ f (t) for a. e. t ∈ [0, ω]. α(t) ≤ β(t) for t ∈ [0, ω].
w00(t) = Hg(t) − Gh(t), σ00(t) = −F
Hh(t) + f (t),
„ 1 x0H
«1/µ
≤ α(t) ≤
„ H
x0GH + F
«1/λ
.
Proof.
x1∈ ]0, x0] : x1`w(t) − mw´ + σ(t) − mσ≤
„ 1 x1H
«1/µ
−
„ H
x1GH + F
«1/λ
for t ∈ [0, ω]
β(t) =
„ H
x1GH + F
«1/λ
+ x1`w(t) − mw´ + σ(t) − mσ for t ∈ [0, ω].
β00(t) = x1Hg(t) −
„ x1G +F
H
«
h(t) + f (t) for a. e. t ∈ [0, ω],
„ H
x1GH + F
«1/λ
≤ β(t) ≤
„ 1 x1H
«1/µ
for t ∈ [0, ω].
β00(t) ≤ g(t) βµ(t)− h(t)
βλ(t)+ f (t) for a. e. t ∈ [0, ω].
α(t) ≤ β(t) for t ∈ [0, ω].
w00(t) = Hg(t) − Gh(t), σ00(t) = −F
Hh(t) + f (t),
„ 1 x0H
«1/µ
≤ α(t) ≤
„ H
x0GH + F
«1/λ
.
Proof.
x1∈ ]0, x0] : x1`w(t) − mw´ + σ(t) − mσ≤
„ 1 x1H
«1/µ
−
„ H
x1GH + F
«1/λ
for t ∈ [0, ω]
β(t) =
„ H
x1GH + F
«1/λ
+ x1`w(t) − mw´ + σ(t) − mσ for t ∈ [0, ω].
β00(t) = x1Hg(t) −
„ x1G +F
H
«
h(t) + f (t) for a. e. t ∈ [0, ω],
„ H
x1GH + F
«1/λ
≤ β(t) ≤
„ 1 x1H
«1/µ
for t ∈ [0, ω].
β00(t) ≤ g(t) βµ(t)− h(t)
βλ(t)+ f (t) for a. e. t ∈ [0, ω].
α(t) ≤ β(t) for t ∈ [0, ω].
Definition
A function ϕ ∈ L`[0, ω]; R+´ is said to verify the property (P ) if the implication u ∈ ACω1`[0, ω]; R´
u00(t) + ϕ(t)u(t) ≥ 0 for a. e. t ∈ [0, ω]
)
=⇒ u(t) ≥ 0 for t ∈ [0, ω]
holds.
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
Theorem
Let G > 0, F < 0, functions w, σ ∈ ACω1`[0, ω]; R´ be such that the equalities
w00(t) = Hg(t) − Gh(t), σ00(t) = |F |
Gg(t) + f (t) for a. e. t ∈ [0, ω] are fulfilled, and let there exist x0∈ ]0, +∞[ such that
x0`w(t) − mw´ + σ(t) − mσ≤
„ G
x0GH + |F |
«1/µ
−
„ 1 x0G
«1/λ
for t ∈ [0, ω].
Moreover, let us define
β(t) =
„ 1 x0G
«1/λ
+ x0`w(t) − mw´ + σ(t) − mσ for t ∈ [0, ω]
and assume that ϕ(t) =βµg(t)1+µ(t) verifies the property (P ). Then the problem (1), (2) has at least one positive solution.
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
Theorem
Let G > 0, F < 0, functions w, σ ∈ ACω1`[0, ω]; R´ be such that the equalities
w00(t) = Hg(t) − Gh(t), σ00(t) = |F |
Gg(t) + f (t) for a. e. t ∈ [0, ω]
are fulfilled, and let there exist x0∈ ]0, +∞[ such that
x0`w(t) − mw´ + σ(t) − mσ≤
„ G
x0GH + |F |
«1/µ
−
„ 1 x0G
«1/λ
for t ∈ [0, ω].
Moreover, let us define
β(t) =
„ 1 x0G
«1/λ
+ x0`w(t) − mw´ + σ(t) − mσ for t ∈ [0, ω]
and assume that ϕ(t) =βµg(t)1+µ(t) verifies the property (P ). Then the problem (1), (2) has at least one positive solution.
u00(t) = g(t)
uµ(t)+ f (t) for a. e. t ∈ [0, ω], (5)
u(0) = u(ω), u0(0) = u0(ω) (2)
Corollary
Let G > 0, F < 0, σ ∈ ACω1`[0, ω]; R´ be such that
σ00(t) =|F |
Gg(t) + f (t) for a. e. t ∈ [0, ω], and let
`Mσ− mσ´µ
|F | < G. Let, moreover, either
µ|F |
1+µ µ g(t)
`G1/µ− |F |1/µ`Mσ− σ(t)´´1+µ≤“π ω
”2
for a. e. t ∈ [0, ω],
or
µ|F |
1+µ µ
Z ω 0
g(s)
`G1/µ− |F |1/µ`Mσ− σ(s)´´1+µds ≤ 4 ω. Then the problem (5), (2) has at least one positive solution.
u00(t) = g(t)
uµ(t)+ f (t) for a. e. t ∈ [0, ω], (5)
u(0) = u(ω), u0(0) = u0(ω) (2)
Corollary
Let G > 0, F < 0, σ ∈ ACω1`[0, ω]; R´ be such that
σ00(t) =|F |
Gg(t) + f (t) for a. e. t ∈ [0, ω], and let
`Mσ− mσ´µ
|F | < G.
Let, moreover, either µ|F |
1+µ µ g(t)
`G1/µ− |F |1/µ`Mσ− σ(t)´´1+µ≤“π ω
”2
for a. e. t ∈ [0, ω],
or
µ|F |1+µµ Z ω
0
g(s)
`G1/µ− |F |1/µ`Mσ− σ(s)´´1+µds ≤ 4 ω. Then the problem (5), (2) has at least one positive solution.
u00(t) = g(t)
uµ(t)+ f (t) for a. e. t ∈ [0, ω], (5)
u(0) = u(ω), u0(0) = u0(ω) (2)
Corollary
Let G > 0 and F < 0. Let, moreover,
“ω 4µG”1+µ1
|F |1/µ+ω
4F−|F |1/µ≤ G1/µ. Then the problem (5), (2) has at least one positive solution.
u00(t) = g(t)
uµ(t)+ f (t) for a. e. t ∈ [0, ω], (5)
u(0) = u(ω), u0(0) = u0(ω) (2)
Corollary
Let G > 0 and F < 0. Let, moreover,
“ω 4µG”1+µ1
|F |1/µ+ω
4F−|F |1/µ≤ G1/µ. Then the problem (5), (2) has at least one positive solution.
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
Lemma
Let there exist positive functions α, β ∈ ACω1`[0, ω]; R´ such that
α00(t) ≥ g(t) αµ(t)− h(t)
αλ(t)+ f (t) for a. e. t ∈ [0, ω], β00(t) ≤ g(t)
βµ(t)− h(t)
βλ(t)+ f (t) for a. e. t ∈ [0, ω], β(t) ≤ α(t) for t ∈ [0, ω].
Let, moreover, there exists ϕ ∈ L`[0, ω]; R+´ with the property (P ) and such that g(t)
uµ(t)− h(t) uλ(t)−
„ g(t) vµ(t)− h(t)
vλ(t)
«
≤ ϕ(t)`v(t) − u(t)´ for a. e. t ∈ [0, ω],
whenever β(t) ≤ u(t) ≤ v(t) ≤ α(t) for t ∈ [0, ω]. Then there exists at least one positive solution to (1), (2).
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
Lemma
Let there exist positive functions α, β ∈ ACω1`[0, ω]; R´ such that
α00(t) ≥ g(t) αµ(t)− h(t)
αλ(t)+ f (t) for a. e. t ∈ [0, ω], β00(t) ≤ g(t)
βµ(t)− h(t)
βλ(t)+ f (t) for a. e. t ∈ [0, ω], β(t) ≤ α(t) for t ∈ [0, ω].
Let, moreover, there exists ϕ ∈ L`[0, ω]; R+´ with the property (P ) and such that g(t)
uµ(t)− h(t) uλ(t)−
„ g(t) vµ(t)− h(t)
vλ(t)
«
≤ ϕ(t)`v(t) − u(t)´ for a. e. t ∈ [0, ω],
whenever β(t) ≤ u(t) ≤ v(t) ≤ α(t) for t ∈ [0, ω]. Then there exists at least one positive solution to (1), (2).
ϕ(t) = µg(t)
β1+µ(t) for t ∈ [0, ω], g(t)
uµ(t)− h(t) uλ(t)−
„ g(t) vµ(t)− h(t)
vλ(t)
«
≤ ϕ(t)`v(t) − u(t)´ for a. e. t ∈ [0, ω],
whenever β(t) ≤ u(t) ≤ v(t) ≤ α(t) for t ∈ [0, ω],
Proof.
ψ(y) = µ β1+µy + 1
yµ is nondecreasing for y ≥ β.
g(t)
„ µ
β1+µ(t)u(t) + 1 uµ(t)
«
− h(t) uλ(t)≤ g(t)
„ µ
β1+µ(t)v(t) + 1 vµ(t)
«
− h(t)
vλ(t) for t ∈ [0, ω] whenever β(t) ≤ u(t) ≤ v(t) for t ∈ [0, ω].
g(t) uµ(t)− h(t)
uλ(t)−
„ g(t) vµ(t)− h(t)
vλ(t)
«
≤ µg(t)
β1+µ(t)(v(t) − u(t)).
ϕ(t) = µg(t)
β1+µ(t) for t ∈ [0, ω], g(t)
uµ(t)− h(t) uλ(t)−
„ g(t) vµ(t)− h(t)
vλ(t)
«
≤ ϕ(t)`v(t) − u(t)´ for a. e. t ∈ [0, ω],
whenever β(t) ≤ u(t) ≤ v(t) ≤ α(t) for t ∈ [0, ω], Proof.
ψ(y) = µ β1+µy + 1
yµ is nondecreasing for y ≥ β.
g(t)
„ µ
β1+µ(t)u(t) + 1 uµ(t)
«
− h(t) uλ(t)≤ g(t)
„ µ
β1+µ(t)v(t) + 1 vµ(t)
«
− h(t)
vλ(t) for t ∈ [0, ω] whenever β(t) ≤ u(t) ≤ v(t) for t ∈ [0, ω].
g(t) uµ(t)− h(t)
uλ(t)−
„ g(t) vµ(t)− h(t)
vλ(t)
«
≤ µg(t)
β1+µ(t)(v(t) − u(t)).
ϕ(t) = µg(t)
β1+µ(t) for t ∈ [0, ω], g(t)
uµ(t)− h(t) uλ(t)−
„ g(t) vµ(t)− h(t)
vλ(t)
«
≤ ϕ(t)`v(t) − u(t)´ for a. e. t ∈ [0, ω],
whenever β(t) ≤ u(t) ≤ v(t) ≤ α(t) for t ∈ [0, ω], Proof.
ψ(y) = µ β1+µy + 1
yµ is nondecreasing for y ≥ β.
g(t)
„ µ
β1+µ(t)u(t) + 1 uµ(t)
«
− h(t) uλ(t)≤ g(t)
„ µ
β1+µ(t)v(t) + 1 vµ(t)
«
− h(t)
vλ(t) for t ∈ [0, ω]
whenever β(t) ≤ u(t) ≤ v(t) for t ∈ [0, ω].
g(t) uµ(t)− h(t)
uλ(t)−
„ g(t) vµ(t)− h(t)
vλ(t)
«
≤ µg(t)
β1+µ(t)(v(t) − u(t)).
ϕ(t) = µg(t)
β1+µ(t) for t ∈ [0, ω], g(t)
uµ(t)− h(t) uλ(t)−
„ g(t) vµ(t)− h(t)
vλ(t)
«
≤ ϕ(t)`v(t) − u(t)´ for a. e. t ∈ [0, ω],
whenever β(t) ≤ u(t) ≤ v(t) ≤ α(t) for t ∈ [0, ω], Proof.
ψ(y) = µ β1+µy + 1
yµ is nondecreasing for y ≥ β.
g(t)
„ µ
β1+µ(t)u(t) + 1 uµ(t)
«
− h(t) uλ(t)≤ g(t)
„ µ
β1+µ(t)v(t) + 1 vµ(t)
«
− h(t)
vλ(t) for t ∈ [0, ω]
whenever β(t) ≤ u(t) ≤ v(t) for t ∈ [0, ω].
g(t) uµ(t)− h(t)
uλ(t)−
„ g(t) vµ(t)− h(t)
vλ(t)
«
≤ µg(t)
β1+µ(t)(v(t) − u(t)).
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
Theorem
Let λ > µ, H > 0, G > 0, F = 0, functions w, σ ∈ AC1ω`[0, ω]; R´ be such that w00(t) = Hg(t) − Gh(t) for a. e. t ∈ [0, ω],
σ00(t) = f (t) for a. e. t ∈ [0, ω], and let there exist x0∈ ]0, +∞[ such that
x0`w(t) − mw´ + σ(t) − mσ≤
„ 1 x0G
«1/λ
−
„ 1 x0H
«1/µ
for t ∈ [0, ω],
where
mw= min˘w(t) : t ∈ [0, ω]¯, mσ= min˘σ(t) : t ∈ [0, ω]¯. Then the problem (1), (2) has at least one positive solution.
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
Theorem
Let λ > µ, H > 0, G > 0, F = 0, functions w, σ ∈ ACω1`[0, ω]; R´ be such that w00(t) = Hg(t) − Gh(t) for a. e. t ∈ [0, ω],
σ00(t) = f (t) for a. e. t ∈ [0, ω], and let there exist x0∈ ]0, +∞[ such that
x0`w(t) − mw´ + σ(t) − mσ≤
„ 1 x0G
«1/λ
−
„ 1 x0H
«1/µ
for t ∈ [0, ω],
where
mw= min˘w(t) : t ∈ [0, ω]¯, mσ= min˘σ(t) : t ∈ [0, ω]¯.
Then the problem (1), (2) has at least one positive solution.
u00(t) = g(t) uµ(t) − h(t)
uλ(t) for a. e. t ∈ [0, ω], (3)
u(0) = u(ω), u0(0) = u0(ω) (2)
Corollary
Let λ > µ, H > 0, G > 0, and let w ∈ ACω1`[0, ω]; R´ be such that w00(t) = Hg(t) − Gh(t) for a. e. t ∈ [0, ω] is fulfilled. Let, moreover,
Mw− mw≤Hλ−µ1+λ Gλ−µ1+µ
„ (1 + λ)µ (1 + µ)λ
«(1+λ)µ
λ−µ λ − µ (1 + µ)λ, where
Mw= max˘w(t) : t ∈ [0, ω]¯, mw= min˘w(t) : t ∈ [0, ω]¯. Then the problem (3), (2) has at least one positive solution.
u00(t) = g(t) uµ(t) − h(t)
uλ(t) for a. e. t ∈ [0, ω], (3)
u(0) = u(ω), u0(0) = u0(ω) (2)
Corollary
Let λ > µ, H > 0, G > 0, and let w ∈ ACω1`[0, ω]; R´ be such that w00(t) = Hg(t) − Gh(t) for a. e. t ∈ [0, ω]
is fulfilled. Let, moreover,
Mw− mw≤Hλ−µ1+λ G1+µλ−µ
„ (1 + λ)µ (1 + µ)λ
«(1+λ)µλ−µ λ − µ (1 + µ)λ, where
Mw= max˘w(t) : t ∈ [0, ω]¯, mw= min˘w(t) : t ∈ [0, ω]¯.
Then the problem (3), (2) has at least one positive solution.
u00(t) = g(t) uµ(t) − h(t)
uλ(t) for a. e. t ∈ [0, ω], (3)
u(0) = u(ω), u0(0) = u0(ω) (2)
Corollary
Let λ > µ, H > 0, and G > 0. Let, moreover, G1+λ
H1+µ ≤„ 4 ω
«λ−µ„ (1 + λ)µ (1 + µ)λ
«(1+λ)µ„ λ − µ (1 + µ)λ
«λ−µ
.
Then the problem (3), (2) has at least one positive solution.
u00(t) = g(t) uµ(t) − h(t)
uλ(t) for a. e. t ∈ [0, ω], (3)
u(0) = u(ω), u0(0) = u0(ω) (2)
Corollary
Let λ > µ, H > 0, and G > 0. Let, moreover, G1+λ
H1+µ ≤„ 4 ω
«λ−µ„ (1 + λ)µ (1 + µ)λ
«(1+λ)µ„ λ − µ (1 + µ)λ
«λ−µ
.
Then the problem (3), (2) has at least one positive solution.
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
Theorem
Let µ > λ, H > 0, G > 0, F = 0, functions w, σ ∈ AC1ω`[0, ω]; R´ be such that w00(t) = Hg(t) − Gh(t) for a. e. t ∈ [0, ω],
σ00(t) = f (t) for a. e. t ∈ [0, ω], and let there exist x0∈ ]0, +∞[ such that
x0`w(t) − mw´ + σ(t) − mσ≤
„ 1 x0H
«1/µ
−
„ 1 x0G
«1/λ
for t ∈ [0, ω].
Moreover, assume that ϕ(t) = µg(t)
β1+µ(t) verifies the property (P ), where β is given by
β(t) =
„ 1 x0G
«1/λ
+ x0`w(t) − mw´ + σ(t) − mσ for t ∈ [0, ω].
Then the problem (1), (2) has at least one positive solution.
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1)
u(0) = u(ω), u0(0) = u0(ω) (2)
Theorem
Let µ > λ, H > 0, G > 0, F = 0, functions w, σ ∈ ACω1`[0, ω]; R´ be such that w00(t) = Hg(t) − Gh(t) for a. e. t ∈ [0, ω],
σ00(t) = f (t) for a. e. t ∈ [0, ω], and let there exist x0∈ ]0, +∞[ such that
x0`w(t) − mw´ + σ(t) − mσ≤
„ 1 x0H
«1/µ
−
„ 1 x0G
«1/λ
for t ∈ [0, ω].
Moreover, assume that ϕ(t) = µg(t)
β1+µ(t) verifies the property (P ), where β is given by
β(t) =
„ 1 x0G
«1/λ
+ x0`w(t) − mw´ + σ(t) − mσ for t ∈ [0, ω].
Then the problem (1), (2) has at least one positive solution.
u00(t) = g(t) uµ(t) − h(t)
uλ(t) for a. e. t ∈ [0, ω], (3)
u(0) = u(ω), u0(0) = u0(ω) (2)
Corollary
Let µ > λ, H > 0, G > 0, w ∈ ACω1`[0, ω]; R´ be such that
w00(t) = Hg(t) − Gh(t) for a. e. t ∈ [0, ω] is fulfilled, and let
Mw− mw≤ Gµ−λ1+µ H
1+λ µ−λ
„ (1 + µ)λ (1 + λ)µ
«(1+µ)λµ−λ µ − λ (1 + λ)µ. Moreover, let us define
β(t) =„ (1 + µ)λ (1 + λ)µ
«µ−λµ „ G H
«µ−λ1
+„ (1 + λ)µ (1 + µ)λ
«µ−λλµ Hµ−λλ G
µ µ−λ
`w(t) − mw´
for t ∈ [0, ω],
and assume that ϕ(t) =βµg(t)1+µ(t) verifies the property (P ). Then the problem (3), (2) has at least one positive solution.
u00(t) = g(t) uµ(t) − h(t)
uλ(t) for a. e. t ∈ [0, ω], (3)
u(0) = u(ω), u0(0) = u0(ω) (2)
Corollary
Let µ > λ, H > 0, G > 0, w ∈ ACω1`[0, ω]; R´ be such that
w00(t) = Hg(t) − Gh(t) for a. e. t ∈ [0, ω]
is fulfilled, and let
Mw− mw≤ G1+µµ−λ H
1+λ µ−λ
„ (1 + µ)λ (1 + λ)µ
«(1+µ)λµ−λ µ − λ (1 + λ)µ. Moreover, let us define
β(t) =„ (1 + µ)λ (1 + λ)µ
«µ−λµ „ G H
«µ−λ1
+„ (1 + λ)µ (1 + µ)λ
«µ−λλµ Hµ−λλ G
µ µ−λ
`w(t) − mw´
for t ∈ [0, ω],
and assume that ϕ(t) =βµg(t)1+µ(t) verifies the property (P ). Then the problem (3), (2) has at least one positive solution.
u00(t) = g(t) uµ(t) − h(t)
uλ(t) for a. e. t ∈ [0, ω], (3)
u(0) = u(ω), u0(0) = u0(ω) (2)
Corollary
Let µ > λ, H > 0, and G > 0. Let, moreover,
H1+µ G1+λ ≤„ 4
ω
«µ−λ„ (1 + µ)λ (1 + λ)µ
«(1+µ)λ„ µ − λ (1 + λ)µ
«µ−λ
×
× min (
1,„ 1 + λ µ − λ
«µ−λ„ (1 + µ)λ (1 + λ)µ
«(µ−λ)(1+µ)) .
Then the problem (3), (2) has at least one solution.
u00(t) = g(t) uµ(t) − h(t)
uλ(t) for a. e. t ∈ [0, ω], (3)
u(0) = u(ω), u0(0) = u0(ω) (2)
Corollary
Let µ > λ, H > 0, and G > 0. Let, moreover,
H1+µ G1+λ ≤„ 4
ω
«µ−λ„ (1 + µ)λ (1 + λ)µ
«(1+µ)λ„ µ − λ (1 + λ)µ
«µ−λ
×
× min (
1,„ 1 + λ µ − λ
«µ−λ„ (1 + µ)λ (1 + λ)µ
«(µ−λ)(1+µ)) .
Then the problem (3), (2) has at least one solution.
Open problem:
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1) λ = µ and F = 0.
J. L. Bravo, P. J. Torres (preprint)
u00=a(t)
u3 ; u(0) = u(ω), u0(0) = u0(ω) (8)
a(t) =
(a+ for t ∈ [0, t0[
−a− for t ∈ [t0, ω] with a+, a−> 0.
Theorem. (8) has a positive solution iffRω
0 a(s)ds < 0.
Open problem:
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1) λ = µ and F = 0.
J. L. Bravo, P. J. Torres (preprint)
u00=a(t)
u3 ; u(0) = u(ω), u0(0) = u0(ω) (8)
a(t) =
(a+ for t ∈ [0, t0[
−a− for t ∈ [t0, ω]
with a+, a−> 0.
Theorem. (8) has a positive solution iffRω
0 a(s)ds < 0.
Open problem:
u00(t) = g(t) uµ(t)− h(t)
uλ(t)+ f (t) for a. e. t ∈ [0, ω], (1) λ = µ and F = 0.
J. L. Bravo, P. J. Torres (preprint)
u00=a(t)
u3 ; u(0) = u(ω), u0(0) = u0(ω) (8)
a(t) =
(a+ for t ∈ [0, t0[
−a− for t ∈ [t0, ω]
with a+, a−> 0.
Theorem. (8) has a positive solution iffRω
0 a(s)ds < 0.