# On a New Characterization of Some Class Nonlinear Eigenvalue Problem

Authors: Lutfi Akin
Journal: Advances in Mathematical Physics
 (2019)

---

## Abstract

A normal mode analysis of a vibrating mechanical or electrical system gives rise to an eigenvalue problem. Faber made a fairly complete study of the existence and asymptotic behavior of eigenvalues and eigenfunctions, Green’s function, and expansion properties. We will investigate a new characterization of some class nonlinear eigenvalue problem.


---

## Body

## 1. Introduction

In this paper, we derive a new boundedness and compactness result for the Hardy operator in variable exponent Lebesgue spaces (VELS)L
p
.
0
,
l. A maximally weak condition is assumed on the exponent function. The last time such a study was carried out was in [1–12]. For a study of the Dirichlet problem of some class nonlinear eigenvalue problem with nonstandard growth condition the obtained results are applied. Such equations arise in the studies of the so-called Winslow effect physical phenomena [13] in the smart materials. In this connection, we mention recent studies for the multidimensional cases with application of Ambrosetti-Rabinowitz’s Mountain Pass theorem approaches (see, e.g., [1, 14, 15]).Theorem A.
Letq
,
p
:
0
,
l
→
(
1
,
∞
) be measurable functions with q
(
x
)
≥
p
(
x
) on 0
,
l. Assume p is monotonically increasing and the function x
-
1
/
p
′
(
x
)
+
δ is almost decreasing on 0
,
l
.. Then operator H boundedly acts the space L
p
(
0
,
l
) into L
q
.
,
-
1
/
p
′
-
1
/
q
.
(
0
,
l
). Moreover, the norm of mapping depends on p
-
,
p
+
,
δ
,
β.Theorem B.
Letq
,
p
:
0
,
l
→
(
1
,
∞
) be measurable functions such that ∞
>
q
+
≥
q
(
x
)
≥
p
(
x
)
≥
p
-
>
1 for all x
∈
(
0
,
l
). Assume that p is monotonically increasing and x
-
1
/
p
′
+
ε is almost decreasing. Then the identity operator maps boundedly the space W
p
.
1
.
(
0
,
l
) into L
q
.
,
-
1
/
p
′
-
1
/
q
.
(
0
,
l
). Moreover, the norm of mapping is estimated by a constant depending on p
-
,
p
+
,
q
,
ε
,
β
.Notice that Theorem B states the inequality(1)
y
x
-
1
/
p
′
-
1
/
q
.
L
q
.
0
,
l
≤
y
′
L
p
.
0
,
lfor any absolutely continuous function y
:
(
0
,
l
)
→
R with y
0
=
0
.In the given assertions,L
p
,
α
(
0
,
l
) denotes the space of measurable functions with finite norm y
x
α
L
p
.
(
0
,
l
), while W
p
.
,
α
1
.
(
0
,
l
) stands for the space of absolutely continuous functions y with y
0
=
0 and finite norm (2)
y
W
p
.
1
.
=
y
′
L
p
.
.We say that the function α
:
0
,
l
→
(
0
,
∞
) is almost increasing (decreasing) if there exists a constant C
>
0 such that for any 0
<
t
1
<
t
2
<
l it holds α
(
t
1
)
≤
C
α
(
t
2
)
α
(
t
1
)
≥
C
α
(
t
2
).We need the following assertion.Lemma 1.
Letp
(
x
) be increasing for x
∈
0
,
l. Let t
∈
A
n
x
=
(
2
-
n
-
1
x
,
2
-
n
x, where n is the natural number. Then it holds(3)
t
-
1
/
p
′
t
≤
C
t
-
1
/
p
x
,
n
-
′
,where p
x
,
n
=
inf
t
∈
A
n
(
x
)
⁡
p
(
t
).We will be inspired by [6–8] while proving Lemma 1.Proof.
Lety
∈
A
n
(
x
) be a point with t
-
1
/
p
′
(
y
)
≤
2
t
-
1
/
p
x
,
n
-
′. Let y
<
t and both lie in A
n
(
x
). Then using the almost decrease of x
-
1
/
p
′
+
ε it follows that (4)
t
-
1
/
p
′
t
+
ε
≤
c
y
-
1
/
p
′
y
+
εUsing t
,
y
∈
A
n
x
,
p
x
,
n
-
′
>
1 it follows that(5)
t
-
1
/
p
′
t
≤
2
+
ε
C
y
-
1
/
p
′
y
≤
2
+
2
+
ε
C
t
-
1
/
p
x
,
n
-
′Now let y
>
t; then, using the increase of p, 1
/
p
′ also will be increasing. Since 1
/
p
′
(
t
)
<
1
/
p
′
(
y
), it follows that(6)
1
t
1
/
p
′
t
≤
C
1
t
1
/
p
′
y
≤
2
C
t
-
1
/
p
x
,
n
-
′
,where c
=
l
1
/
p
-
′
+
l
1
/
p
+
′.
Lemma1 has been proved.In the light of the information given above, we can give proof of Theorem A.Proof of Theorem A.
Letf
:
0
,
l
→
(
0
,
∞
) be a positive measurable function. It holds the identity(7)
H
f
x
=
∑
n
=
1
∞
∫
2
-
n
-
1
x
2
-
n
x
f
t
d
tAssume f
p
=
1
. Using the triangle property of p
(
.
)
- norms(8)
x
α
H
f
q
.
≤
∑
n
=
1
∞
x
α
∫
A
n
x
f
t
d
t
q
.
,with α
(
x
)
=
-
1
/
p
′
(
x
)
-
1
/
q
x (recall A
n
x
=
2
-
n
-
1
x
,
2
-
n
x).
Derive estimation for every summand in (8). In this purpose get estimation for the proper modular(9)
I
q
.
x
α
.
∫
A
n
x
f
t
d
t
≔
∫
0
l
x
α
.
∫
A
n
x
f
t
d
t
q
x
d
x
.Applying the assumption on p (decreasing of x
-
1
/
p
′
+
ε), and using the expression for q
(
x
)
=
1
/
-
α
-
1
/
p
′
(
x
), we have (10)
I
q
x
-
1
/
p
′
-
1
/
q
∫
A
n
x
f
t
d
t
=
∫
0
l
x
-
1
/
p
′
+
ε
∫
A
n
x
f
t
d
t
q
x
d
x
x
1
+
ε
q
x
≤
C
q
+
2
-
n
ε
q
-
.
.
∫
0
l
d
x
x
∫
A
n
x
f
t
t
-
1
/
p
′
t
d
t
q
xNotice that we have used x
-
1
/
p
′
(
x
)
+
ε
≤
c
t
-
1
/
p
′
(
t
)
+
ε for any 0
<
x
<
l and 2
-
n
-
1
x
<
t
≤
2
-
n
x by using the almost decrease of x
-
1
/
p
′
(
x
)
+
ε.
Therefore, from (8), using Hölder’s inequality, it follows that(11)
I
q
x
α
.
∫
A
n
x
f
t
d
t
≤
C
q
+
2
-
n
ε
q
-
∫
0
l
d
x
x
∫
A
n
x
f
t
p
x
,
n
-
d
t
q
x
/
p
x
,
n
-
∫
A
n
x
t
-
p
x
,
n
-
′
/
p
′
t
d
t
q
x
/
p
x
,
n
-Applying Lemma 1 and estimate (3) it follows from (11) that(12)
I
q
x
α
.
∫
A
n
x
f
t
d
t
≤
c
ln
⁡
2
q
+
/
p
-
2
-
n
ε
q
-
C
q
+
∫
0
l
d
x
x
∫
A
n
x
f
t
p
x
,
n
-
d
t
q
x
/
p
x
,
n
-Since (13)
∫
A
n
x
f
t
p
x
,
n
-
d
t
≤
∫
A
n
x
f
t
p
t
d
t
+
∫
A
n
x
d
t
≤
1
+
2
-
n
x
≤
1
+
2
-
n
l
≤
l
+
1
,It follows that(14)
I
q
x
α
∫
A
n
x
f
t
d
t
≤
c
ln
⁡
2
q
+
2
-
n
ε
q
-
∫
0
l
d
x
x
1
l
+
1
∫
A
n
x
f
t
p
x
,
n
-
d
t
q
x
/
p
x
,
n
-
l
+
1
q
+
≤
c
ln
⁡
2
l
+
1
q
+
∫
0
l
1
l
+
1
∫
A
n
x
f
t
p
t
+
1
d
t
p
x
/
p
x
,
n
-
d
x
x
≤
2
-
n
ε
q
-
C
q
+
c
ln
⁡
2
q
+
l
+
1
q
+
-
1
∫
0
l
d
x
x
∫
A
n
x
f
t
p
t
+
1
d
t
.Hence,(15)
I
q
x
α
.
∫
A
n
x
f
t
d
t
≤
c
3
2
-
n
ε
q
-
C
q
+
∫
0
l
∫
A
n
x
f
t
p
t
+
1
d
t
d
x
x
≤
c
3
∫
0
2
-
n
l
f
t
p
t
+
1
∫
2
n
t
2
n
+
1
t
d
x
x
=
C
q
+
c
3
2
-
n
ε
q
-
ln
⁡
2
∫
0
2
-
n
l
f
t
p
t
+
1
d
t
≤
c
q
+
c
3
2
-
n
ε
q
-
ln
⁡
2
1
+
2
-
n
l
=
c
4
2
-
n
ε
q
-
.Therefore, it has been proved that (16)
I
q
x
-
1
/
p
′
-
1
/
q
∫
A
n
x
f
t
d
t
≤
c
4
2
-
n
ε
q
-
,which implies (17)
x
-
1
/
p
′
-
1
/
q
∫
A
n
x
f
t
d
t
q
.
;
0
,
l
≤
c
4
1
/
q
+
2
-
n
ε
q
-
/
q
+Inserting (17) into (8), we get (18)
x
-
1
/
p
′
-
1
/
q
H
f
q
.
;
0
,
l
≤
c
4
1
/
q
+
∑
n
=
1
∞
2
-
n
ε
q
-
/
q
+
=
c
5Theorem A has been proved.Theorem C.
Letq
,
p
:
0
,
l
→
(
1
,
∞
) be measurable functions such that ∞
>
q
+
≥
q
(
x
)
≥
q
-
>
p
+
≥
p
(
x
)
≥
p
-
>
1.
Assume that the functionp increases on 0
,
l and x
-
1
/
p
′
+
ε is almost decreasing in 0
,
l
. Then operator H acts compactly the space L
p
,
β
(
0
,
l
) into L
q
,
-
1
/
p
′
-
1
-
δ
/
q
(
0
,
l
) for any δ
∈
0,1.Proof.
In order to proof Theorem C, we may apply the approaches from [3–5]. In this way, insert the operators(19)
P
1
f
x
=
X
0
,
a
x
x
-
1
/
p
′
-
1
-
δ
/
q
∫
0
x
f
t
d
t
;
P
2
f
x
=
X
a
,
l
x
x
-
1
/
p
′
-
1
-
δ
/
q
∫
0
a
f
t
d
t
;
P
3
f
x
=
X
a
,
l
x
x
-
1
/
p
′
-
1
-
δ
/
q
∫
a
x
f
t
d
t
;As it was stated in [3], P
3 is a limit of finite rank operators, while P
2 is a finite rank operator. From the condition lim
t
→
0
⁡
B
t
=
0 it follows that (20)
H
f
-
P
2
f
-
P
3
f
L
q
.
0
,
l
≤
P
1
f
L
q
.
0
,
l
≤
c
a
δ
/
p
+
f
por8
H
-
P
2
-
P
3
L
p
→
L
q
,
-
1
/
p
′
-
1
-
δ
/
q
≤
P
1
L
p
,
β
→
L
q
,
-
1
/
p
′
-
1
-
δ
/
q
≤
c
a
δ
/
p
→
0as a
→
0
. To show the last estimation we shall use the arguments of Theorem A. Repeating all constructions there, we get the following estimates:(21)
I
q
x
-
1
/
p
′
-
1
-
δ
/
q
∫
A
n
x
f
t
d
t
=
∫
0
l
d
x
x
1
-
δ
+
ε
q
x
-
1
/
p
′
+
ε
∫
A
n
x
f
t
d
t
q
x
≤
C
q
+
2
-
n
ε
q
-
∫
0
l
d
x
x
1
-
δ
∫
A
n
x
t
-
1
/
p
′
t
f
t
d
t
q
x
.Notice that we have used x
-
1
/
p
′
+
ε
≤
c
t
-
1
/
p
′
(
t
)
+
ε for any t
∈
A
n
(
x
), where n belongs to the natural number.
Therefore, using Hölder’s inequality,(22)
I
q
x
-
1
/
p
′
-
1
-
δ
/
q
∫
A
n
x
f
t
d
t
≤
C
q
+
2
-
n
ε
q
-
∫
0
l
d
x
x
1
-
δ
∫
A
n
x
f
t
p
x
,
n
-
d
t
q
x
/
p
x
,
n
-
∫
A
n
x
t
-
p
x
,
n
-
′
/
p
′
t
d
t
q
x
/
p
x
,
n
-
′
.Applying Lemma 1 and the arguments above, we attain the estimates(23)
I
q
x
-
1
/
p
′
-
1
-
δ
/
q
∫
A
n
x
f
t
d
t
≤
C
q
+
2
-
n
ε
q
-
c
ln
⁡
2
q
+
l
+
1
q
+
-
1
∫
0
l
d
x
x
1
-
δ
∫
A
n
x
f
t
p
t
+
1
d
t
≤
C
3
2
-
n
ε
q
-
∫
0
-
2
-
n
f
t
p
t
+
1
∫
2
n
t
2
n
+
1
t
d
x
x
1
-
δ
d
t
≤
2
-
n
ε
q
-
1
+
2
-
n
l
C
3
C
q
+
l
δTherefore, it has been shown that (24)
I
q
x
-
1
/
p
′
-
1
-
δ
/
q
∫
A
n
x
f
t
d
t
≤
c
l
δ
2
-
n
ε
q
-if f
p
≤
1. This implies (25)
x
-
1
/
p
′
-
1
-
δ
/
q
∫
A
n
x
f
t
d
t
q
.
;
0
,
l
≤
C
1
/
q
+
l
δ
/
q
+
2
-
n
ε
q
-
/
q
+Inserting these estimates over n
=
1,2
,
… in the expression (26)
x
-
1
/
p
′
-
1
-
δ
/
q
H
f
q
.
;
0
,
l
≤
c
1
/
q
+
l
δ
/
q
+
∑
n
=
1
∞
2
-
n
ε
q
-
/
q
+
=
c
5
l
δ
/
q
+The last estimate is a needed estimation which completes the proof of Theorem C.
Consider the problem(27)
d
d
x
d
y
d
x
p
x
-
2
d
y
d
x
=
λ
y
q
x
-
1
l
x
-
x
2
q
-
1
/
p
′
-
1
-
δ
/
q
y
0
=
y
l
=
0
,
y
x
>
0where δ
∈
0,1
,
x
∈
0
,
l
.Theorem D.
Let1
<
p
-
<
p
x
≤
p
+
<
q
-
≤
q
x
≤
q
+
<
∞ be measurable functions on 0
,
l such that p
x is increasing and the function x
-
1
/
p
′
+
ε is almost decreasing on 0
,
l. Then for any fixed λ
>
0 there exists a nontrivial solution of the problem (27).Proof.
To prove this assertion, we shall use the well-known Mountain Pass theorem approaches. Define the functional(28)
I
λ
y
=
∫
0
l
1
p
x
y
′
p
x
d
x
-
λ
∫
0
l
1
q
x
x
l
-
x
2
-
1
/
p
′
-
1
-
δ
/
q
y
+
q
x
d
xDefine the space E
=
W
¯
p
1
0
,
l a closure of absolutely continuous functions on 0
,
l, such that y
0
=
y
l
=
0 with a norm(29)
y
W
¯
p
1
0
,
l
=
d
y
d
x
p
.
;
0
,
l
,Define also the space L
¯
p
,
β
0
,
l as a space of measurable functions with finite norm(30)
f
L
¯
p
0
,
l
=
inf
⁡
λ
>
0
:
∫
0
l
f
λ
p
x
d
x
≤
1
.Applying the standard approaches (see, e.g., [13]), it is not difficult to see that the functional I
λ
∈
C
1
E
,
R
. Further, I
λ
′
∈
E
∗ and (31)
I
λ
′
y
,
v
=
∫
0
l
y
′
p
x
-
λ
∫
0
l
x
l
-
x
2
-
1
/
p
′
-
1
-
δ
/
q
q
x
y
+
q
x
d
xBy applying Theorem B, we get the implication f
E
=
0
→
f
=
0
. Show that Palaisce-Smale (PS) condition is satisfied for the problem (27). Let y
n
∈
E be a sequence such that it is satisfied by the following two conditions:(1)
I
λ
y
n
≤
M;
(2)
I
′
(
y
n
)
E
∗
→
0 as n
→
∞
.
To prove the PS condition, we must prove that such a sequence is compact; that is, it contains a subsequencey
n
k converging in E to a function ∈
E. In order to show it, establish the boundedness of y
n.
From (1) it follows that(32)
∫
0
l
y
n
′
p
x
p
x
d
x
-
λ
∫
0
l
λ
q
x
x
l
-
x
2
-
1
/
p
′
-
1
-
δ
/
q
y
n
+
q
x
d
x
≤
MThen (33)
1
p
+
∫
0
l
y
n
′
p
x
d
x
≤
1
q
-
∫
0
l
x
l
-
x
2
-
1
/
p
′
-
1
-
δ
/
q
y
n
+
q
x
d
x
+
MUsing condition (2) I
λ
′
y
n
=
0
(
1
) for n
→
∞, it follows that I
λ
′
y
n
,
y
n
=
O
1
y
n; that is,(34)
λ
∫
0
l
x
l
-
x
2
-
1
/
p
′
-
1
-
δ
/
q
y
n
+
q
x
d
x
=
O
1
y
n
+
∫
0
l
y
n
′
p
x
d
xInserting this into (27), it follows that (35)
1
p
+
-
1
q
-
∫
0
l
y
n
′
p
x
d
x
≤
M
+
O
1
y
n
.From this, since q
-
>
p
+, it follows that(36)
∫
0
l
y
n
′
p
x
d
x
≤
2
M
p
+
q
-
q
-
-
p
+
+
0
1
y
nor (37)
y
n
p
-
≤
O
1
+
O
1
y
n
.Using Young’s inequality from here it follows that (38)
y
n
′
p
.
≤
C
M
.This completes the boundedness of y
n in E. Applying well-known fact, there exists a weak convergent subsequence y
n
k
→
y in E
. Denote it again as y
n
∙It follows from the compact embedding
Theorem C that a strong convergencey
n
→
y in L
q
,
-
1
/
p
′
-
1
-
δ
/
q
0
,
l holds; that is,(39)
y
n
-
y
L
q
,
-
1
/
p
′
-
1
-
δ
/
q
0
,
l
→
0Now, we are ready to show the strong convergence y
n
→
y in E. For this purpose, insert v
=
y
n
-
y into the equality(40)
I
λ
′
y
n
,
v
=
O
1
v
.Then (41)
∫
0
l
y
n
′
p
x
-
2
y
′
n
y
n
′
-
y
′
-
λ
∫
0
l
x
l
-
x
2
-
1
/
p
′
-
1
-
δ
/
q
q
x
y
n
+
q
x
-
1
y
n
-
y
=
O
1
y
n
-
y
.From this, since y
n
→
y in L
q
,
-
1
/
p
′
-
1
-
δ
/
q
0
,
l and using Holder’s inequality, it follows that(42)
∫
0
l
l
x
-
x
2
-
1
/
p
′
-
1
-
δ
/
q
q
x
y
n
+
q
x
-
1
y
n
-
y
≤
y
n
-
y
L
q
,
-
1
/
p
′
-
1
-
δ
/
q
0
,
l
y
n
+
l
x
-
x
2
-
1
/
p
′
-
1
-
δ
/
q
q
x
-
1
L
q
′
=
O
1
y
n
l
x
-
x
2
-
1
/
p
′
-
1
-
δ
/
q
q
q
+
-
1
→
0
,since y
n is bounded in L
q
,
-
1
/
p
′
-
1
-
δ
/
q
0
,
l.
Therefore,(43)
∫
0
l
y
n
′
p
x
-
2
y
n
′
y
n
′
-
y
′
=
O
1
+
O
1
y
n
-
y
.
From this we infer that(44)
∫
0
l
y
n
′
p
x
-
2
y
n
′
-
y
′
p
x
-
2
y
′
y
n
′
-
y
′
d
x
+
∫
0
l
y
′
p
x
-
2
y
′
y
n
′
-
y
′
d
x
=
0
1
+
0
1
y
n
-
ySince y
n
→
y weakly in E it holds that(45)
∫
0
l
y
′
p
x
-
2
y
′
y
n
′
-
y
′
d
x
→
0
.This ensures that (46)
∫
0
l
y
n
′
p
x
-
2
y
n
′
-
y
′
p
x
-
2
y
′
y
n
′
-
y
′
d
x
=
0
1
+
0
1
y
n
-
y
.We will apply the following two inequalities:(47)
y
n
′
p
x
-
2
y
n
′
-
y
′
p
x
-
2
y
′
,
y
n
′
-
y
′
≥
γ
1
p
y
n
′
-
y
′
p
xfor p
(
x
)
≥
2 and (48)
y
n
′
p
x
-
2
y
n
′
-
y
′
p
x
-
2
y
′
,
y
n
′
-
y
′
≥
γ
2
p
y
n
′
-
y
′
2
y
n
′
2
-
p
+
y
′
2
-
pfor 1
<
p
(
x
)
<
2
. Applying (47), for the case p
(
x
)
≥
2, we get (49)
∫
0
l
y
n
′
-
y
′
p
x
d
x
=
O
1
+
O
1
y
n
-
y
.As to the case 1
<
p
(
x
)
<
2, we have the inequality (50)
∫
0
l
y
n
′
-
y
′
2
y
n
′
2
-
p
+
y
′
2
-
p
d
x
≤
0
1
+
0
1
y
n
-
yUsing Young’s inequality from here it follows that (51)
∫
0
l
y
n
′
-
y
′
p
d
x
≤
ε
∫
0
l
y
n
′
+
y
′
p
x
x
β
p
d
x
+
C
ε
∫
0
l
y
n
′
-
y
′
2
y
n
′
+
y
′
2
-
p
d
x
≤
M
ε
+
C
ε
0
1
+
0
1
y
n
-
y
,
∀
ε
>
0
.Therefore, (52)
y
n
-
y
p
-
≤
M
ε
+
c
ε
0
1
+
0
1
y
n
-
ywhere M does not depend on n
∈
N. This and the above inequality and Young’s inequality give(53)
y
n
-
y
p
-
≤
M
ε
+
0
1This inequality yields y
n
→
y in E.
Now, we are ready to apply the Mountain Pass theorem. Ify
n
-
y
↛
0 from preceding equality one gets(54)
y
n
′
-
y
′
p
∙
p
-
=
O
1
+
O
1
y
n
′
-
y
′
p
∙Therefore, using assumption p
-
>
1 and Young’s inequality we have (55)
y
n
′
-
y
′
p
∙
=
O
1
,that is, y
n
→
y in E strongly.
The proof of PS property has been completed.Now, apply the Mountain Pass theorem in order to show the existence of solution for the problem (27).Fory
≤
1 we have(56)
I
λ
y
≥
1
p
+
∫
0
l
y
′
y
p
x
y
p
+
d
x
-
λ
q
-
∫
0
l
x
l
-
x
2
-
1
/
p
′
-
1
-
δ
/
q
y
+
y
q
x
y
q
-By using Theorem A, it follows that (57)
y
L
q
,
-
1
/
p
′
-
1
/
q
0
,
l
≤
c
y
L
p
0
,
lThen (56) implies that(58)
I
λ
y
≥
1
p
+
y
p
+
-
λ
q
-
∫
0
l
x
l
-
x
2
-
1
/
p
′
-
1
/
q
c
y
L
q
,
-
1
/
p
′
-
1
/
q
0
,
l
q
x
y
q
-
≥
1
p
+
y
p
+
-
λ
c
q
+
q
-
y
q
-Hence for y
≤
1 it follows that(59)
I
λ
y
≥
1
p
+
y
p
+
-
λ
c
q
+
q
-
y
q
-
=
y
p
+
1
p
+
-
λ
c
q
+
p
+
q
-
y
q
-
-
p
+Therefore,(60)
I
λ
y
≥
y
p
+
λ
c
q
+
q
-
q
-
λ
c
q
+
p
+
-
y
q
-
-
p
+
.If we choose y
=
m
i
n
1
,
q
-
/
2
λ
c
q
+
p
+
1
/
q
-
-
p
+, it follows that(61)
I
λ
y
≥
q
-
2
λ
c
q
+
p
+
p
+
/
q
-
-
p
+
1
2
q
-
2
λ
c
q
+
p
+Choose R
=
1
/
2
λ
c
q
+
p
+
1
/
q
-
-
p
+ to apply the Mountain Pass theorem.Now, what remains is to find a pointy
0
∈
E where I
λ
y
0
<
0. To show this, apply the fibering method; for y
∈
E to be fixed and sufficiently large t
>
1 it holds that(62)
I
λ
t
y
=
∫
0
l
t
p
x
p
x
y
′
p
x
d
x
-
∫
0
l
t
q
x
q
x
x
l
-
x
2
-
1
/
p
′
-
1
-
δ
/
q
y
+
q
x
d
x
≤
t
p
+
p
-
∫
0
l
y
′
p
x
d
x
-
t
q
-
q
+
∫
0
l
x
l
-
x
β
-
1
/
p
′
-
1
-
δ
/
q
y
+
q
x
d
x
<
0
.Insert y
0
=
t
y in order to get some point laid out of the ball B
0
,
R in E, such that I
λ
y
0
<
0. Applying Mountain Pass theorem, there exists a point y
0
∈
E with I
λ
y
0
=
c such that I
λ
′
y
0
=
0, where (63)
c
=
inf
sup
γ
t
∈
Γ
⁡
I
λ
γ
t
:  with
γ
:
0,1
→
E
,
γ
∈
C
1
0,1
;
E
,
and
γ
0
=
0
,
γ
1
=
0
.Therefore,I
λ
y
0
>
0, I
′
y
0
=
0. To show that y
0 is a positive solution of (27) insert in I
′
y
0
,
v
=
0, v
=
y
0
-
:
(64)
∫
0
l
y
′
p
x
-
2
y
′
y
-
′
-
λ
∫
0
l
x
l
-
x
2
-
1
/
p
′
-
1
-
δ
/
q
q
y
+
q
x
-
1
y
-
d
x
=
0Since the second integral is zero y
+
q
x
-
1
y
-
≡
0, we have(65)
0
=
∫
0
l
y
-
′
p
x
-
2
y
′
-
2
d
xTherefore, y
-
′
≡
0; using imbedding Theorem B we infer y
-
≡
0, which implies that y
0
(
x
)
>
0.We have proved the existence of problem (27) for any λ
>
0.



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*Source: 7865303-2019-02-07.xml*