# Variational Integrals of a Class of Nonhomogeneous𝒜-Harmonic Equations

Authors: Guanfeng Li; Yong Wang; Gejun Bao
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We introduce a class of variational integrals whose Euler equations are nonhomogeneous𝒜-harmonic equations. We investigate the relationship between the minimization problem and the Euler equation and give a simple proof of the existence of some nonhomogeneous 𝒜-harmonic equations by applying direct methods of the calculus of variations. Besides, we establish some interesting results on variational integrals.


---

## Body

## 1. Introduction

In this paper, we study the variational integral of the form(1)
I
(
F
1
,
F
2
)
(
u
,
E
)
=
∫
E
‍
(
F
1
(
x
,
∇
u
(
x
)
)
+
F
2
(
x
,
u
(
x
)
)
)
d
x
,
whose Euler equations are nonhomogeneous 𝒜-harmonic equations
(2)
-
div
⁡
𝒜
(
x
∇
u
)
+
ℬ
(
x
,
u
)
=
0
,
where F
1
:
ℝ
n
×
ℝ
n
→
ℝ, F
2
:
ℝ
n
×
ℝ
→
ℝ, 𝒜
:
ℝ
n
×
ℝ
n
→
ℝ
n, and ℬ
:
ℝ
n
×
ℝ
→
ℝ are operators satisfying some assumptions. There are many literatures on (2) and a large of useful results have been established; see [1–3] and their references. We investigate the relationship between the minimization of I
(
F
1
,
F
2
)
(
u
,
E
) and solutions of the Euler equation. Based on that, we give a simple proof of the existence of some nonhomogeneous 𝒜-harmonic equations by applying direct methods of the calculus of variations. Besides, we establish some interesting results on variational integrals. The results of this paper make the theory on (2) easier to comprehend.We recall the weighted Sobolev spacesH
1
,
p
(
Ω
;
μ
) which are adopted in [4].Letℝ
n be the real Euclidean space with the dimension n, n
≥
2. Throughout this paper, Ω will denote an open subset of ℝ
n and 1
<
p
<
∞. Let w be a locally integrable, nonnegative function in ℝ
n. A Radon measure μ is canonically associated with the weight w,
(3)
μ
(
E
)
=
∫
E
‍
w
(
x
)
d
x
.
Thus d
μ
(
x
)
=
w
(
x
)
d
x, where d
x is the n-dimensional Lebesgue measure. In this paper, unless otherwise stated, we always assume that μ is a p-admissible measure and d
μ
(
x
)
=
w
(
x
)
d
x; see [4].LetL
p
(
Ω
;
μ
)
=
{
φ
:
Ω
→
ℝ
:
∫
Ω
‍
|
φ
|
p
d
μ
<
∞
} and L
p
(
Ω
;
μ
;
ℝ
n
)
=
{
φ
:
Ω
→
ℝ
n
:
∫
Ω
‍
|
φ
|
p
d
μ
<
∞
}. Denote the norm of L
p
(
Ω
;
μ
) and L
p
(
Ω
;
μ
;
ℝ
n
) by ∥
·
∥
p,
(4)
∥
ϕ
∥
p
=
(
∫
Ω
‍
|
ϕ
|
p
d
μ
)
1
/
p
,
where ϕ
∈
L
p
(
Ω
;
μ
) (or L
p
(
Ω
;
μ
;
ℝ
n
)).Forφ
∈
C
∞
(
Ω
), let
(5)
∥
φ
∥
1
,
p
=
(
∫
Ω
‍
|
φ
|
p
d
μ
)
1
/
p
+
(
∫
Ω
‍
|
∇
φ
|
p
d
μ
)
1
/
p
,
where ∇
φ
=
(
∂
1
φ
,
…
,
∂
n
φ
) is the gradient of φ. The Sobolev space H
1
,
p
(
Ω
;
μ
) is defined to be the completion of the set {
φ
∈
C
∞
(
Ω
)
:
∥
φ
∥
1
,
p
<
∞
} with respect to the norm ∥
·
∥
1
,
p. In other words, u
∈
H
1
,
p
(
Ω
;
μ
) if and only if u
∈
L
p
(
Ω
;
μ
) and there is a function v
∈
L
p
(
Ω
;
μ
;
ℝ
n
) and a sequence φ
i
∈
C
∞
(
Ω
), such that
(6)
∫
Ω
‍
|
φ
i
-
u
|
p
d
μ
⟶
0
,
∫
Ω
‍
|
∇
φ
i
-
v
|
p
d
μ
⟶
0
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
i
⟶
∞
.
We call v the gradient of u in H
1
,
p
(
Ω
;
μ
) and write v
=
∇
u.The spaceH
0
1
,
p
(
Ω
;
μ
) is the closure of C
0
∞
(
Ω
) in H
1
,
p
(
Ω
;
μ
). Obviously, H
1
,
p
(
Ω
;
μ
) and H
0
1
,
p
(
Ω
;
μ
) are Banach space with respect to the norm ∥
·
∥
1
,
p. Moreover, ∥
·
∥
1
,
p is uniformly convex and the Sobolev space H
1
,
p
(
Ω
;
μ
) and H
0
1
,
p
(
Ω
;
μ
) are reflexive; see [5] for details.The corresponding local Sobolev spaceH
loc
⁡
1
,
p
(
Ω
;
μ
) is defined in the obvious manner: a function u is in H
loc
⁡
1
,
p
(
Ω
;
μ
) if and only if u is in H
loc
⁡
1
,
p
(
Ω
′
;
μ
) each open set Ω
′
⋐
Ω.


## 2. Variational Integrals

Suppose thatE is a measurable set and that u
∈
H
loc
⁡
1
,
p
(
Ω
;
μ
) for an open neighborhood Ω of E. Then, we have the following variational integral:
(7)
I
(
F
1
,
F
2
)
(
u
,
E
)
=
∫
E
‍
(
F
1
(
x
,
∇
u
(
x
)
)
+
F
2
(
x
,
u
(
x
)
)
)
d
x
,
where F
1
:
ℝ
n
×
ℝ
n
→
ℝ is a variational kernel satisfying the following assumptions for some constants 0
<
γ
1
≤
δ
1
<
∞:
(8)
the
mapping
x
⟼
F
1
(
x
,
ξ
)
is
measurable
∀
ξ
∈
ℝ
n
;
for a.e. x
∈
ℝ
n;
(9)
γ
1
w
(
x
)
|
ξ
|
p
≤
F
1
(
x
,
ξ
)
≤
δ
1
w
(
x
)
|
ξ
|
p
,
ξ
∈
ℝ
n
,
(10)
the
mapping
ξ
⟼
F
1
(
x
,
ξ
)
is
strictly
convex
and
differentiable
,
(11)
F
1
(
x
,
λ
ξ
)
=
|
λ
|
p
F
1
(
x
,
ξ
)
,
λ
∈
ℝ
,
ξ
∈
ℝ
n
,
and F
2
:
ℝ
n
×
ℝ
→
ℝ is also a variational kernel satisfying the following assumptions for some constants 0
<
γ
2
≤
δ
2
<
∞:
(12)
the
mapping
x
⟼
F
2
(
x
,
t
)
is
measurable
∀
t
∈
ℝ
;
for a.e. x
∈
ℝ
n;
(13)
γ
2
w
(
x
)
|
t
|
p
≤
F
2
(
x
,
t
)
≤
δ
2
w
(
x
)
|
t
|
p
,
t
∈
ℝ
;
(14)
the
mapping
t
⟼
F
2
(
x
,
t
)
is
convex
and
differentiable
.Remark 1.
Note that a convex function is differential if and only if it is continuously differentiable; see [6]. Thus, by assumptions (10) and (14), mappings ξ
↦
F
1
(
x
,
ξ
) and t
↦
F
2
(
x
,
t
) are continuously differentiable for a.e. x. Denote by ∇
ξ
F
1
(
x
,
·
) the usual gradient of F
1 with respect to the second variable and by ∂
t
F
2
(
x
,
·
) the usual derivative of F
1 with respect to the second variable. Obviously, ∇
ξ
F
1
(
x
,
·
) and ∂
t
F
2
(
x
,
·
) exist for a.e. x
∈
ℝ
n.
The valueI
(
F
1
,
F
2
)
(
u
,
E
) lies in the interval [
0
,
∞
] and by assumptions (9) and (13), I
(
F
1
,
F
2
)
(
u
,
E
)
<
∞ if and only if u
∈
L
p
(
E
;
μ
) and ∇
u
∈
L
p
(
E
;
μ
); that is, u
∈
H
1
,
p
(
E
;
μ
).
The convexity assumptions (10) and (14) can imply the following useful inequalities.Lemma 2.
For a.e.x
∈
ℝ
n,
(15)
F
1
(
x
,
ξ
1
)
-
F
1
(
x
,
ξ
2
)
>
∇
ξ
F
1
(
x
,
ξ
2
)
·
(
ξ
1
-
ξ
2
)
,
(16)
F
2
(
x
,
t
1
)
-
F
2
(
x
,
t
2
)
≥
∂
t
F
2
(
x
,
t
2
)
(
t
1
-
t
2
)
,
whenever ξ
1
,
ξ
2
∈
ℝ
n, ξ
1
≠
ξ
2, and t
1
,
t
2
∈
ℝ.Proof.
The proof is based on assumptions (10) and (14) and the definition of directional derivative. Here, we only show the proof of the first inequality (15) and the other is similar.
Fixx
∈
ℝ
n such that the mapping ξ
↦
F
1
(
x
,
ξ
) is strictly convex and differentiable. Then, for 0
<
s
<
1,
(17)
F
1
(
x
,
ξ
2
+
s
(
ξ
1
-
ξ
2
)
)
=
F
1
(
x
,
(
1
-
s
)
ξ
2
+
s
ξ
1
)
<
(
1
-
s
)
F
1
(
x
,
ξ
2
)
+
s
F
1
(
x
,
ξ
1
)
.
Setting ξ
=
ξ
1
-
ξ
2, we can get
(18)
F
1
(
x
,
ξ
2
+
s
ξ
)
-
F
1
(
x
,
ξ
2
)
<
s
(
F
1
(
x
,
ξ
2
+
ξ
)
-
F
1
(
x
,
ξ
2
)
)
.
Dividing by s and subtracting ∇
ξ
F
1
(
x
,
ξ
2
)
·
ξ from both sides, we obtain that
(19)
F
1
(
x
,
ξ
2
+
s
ξ
)
-
F
1
(
x
,
ξ
2
)
s
-
∇
ξ
F
1
(
x
,
ξ
2
)
·
ξ
<
F
1
(
x
,
ξ
2
+
ξ
)
-
F
1
(
x
,
ξ
2
)
-
∇
ξ
F
1
(
x
,
ξ
2
)
·
ξ
.
By the definition of directional derivative, we have that
(20)
lim
⁡
s
→
0
+
F
1
(
x
,
ξ
2
+
s
ξ
)
-
F
1
(
x
,
ξ
2
)
s
=
∇
ξ
F
1
(
x
,
ξ
2
)
·
ξ
.
Then, we can get that F
1
(
x
,
ξ
1
)
-
F
1
(
x
,
ξ
2
)
≥
∇
ξ
F
1
(
x
,
ξ
2
)
·
(
ξ
1
-
ξ
2
).
Suppose that there existξ
1
,
ξ
2
∈
ℝ
n, ξ
1
≠
ξ
2, such that F
1
(
x
,
ξ
1
)
-
F
1
(
x
,
ξ
2
)
=
∇
ξ
F
1
(
x
,
ξ
2
)
·
(
ξ
1
-
ξ
2
), let ξ
=
(
1
/
2
)
(
ξ
1
+
ξ
2
), and then we can obtain that
(21)
F
1
(
x
,
ξ
)
=
F
1
(
x
,
1
2
(
ξ
1
+
ξ
2
)
)
<
1
2
(
F
1
(
x
,
ξ
1
)
+
F
1
(
x
,
ξ
2
)
)
=
F
1
(
x
,
ξ
2
)
+
1
2
∇
ξ
F
1
(
x
,
ξ
2
)
·
(
ξ
1
-
ξ
2
)
.
On the other hand, since ξ
≠
ξ
2, we have that
(22)
F
1
(
x
,
ξ
)
≥
F
1
(
x
,
ξ
2
)
+
∇
ξ
F
1
(
x
,
ξ
2
)
·
(
ξ
-
ξ
2
)
=
F
1
(
x
,
ξ
2
)
+
1
2
∇
ξ
F
1
(
x
,
ξ
2
)
·
(
ξ
1
-
ξ
2
)
.
Then, (22) contradicts (21) and the lemma follows.


## 3. Nonhomogeneous𝒜-Harmonic Equations and the Obstacles Problem

The following nonlinear elliptic equation:(23)
-
div
⁡
𝒜
(
x
,
∇
u
)
+
ℬ
(
x
,
u
)
=
0
is called the nonhomogeneous 𝒜-harmonic equation, where 𝒜
:
ℝ
n
×
ℝ
n
→
ℝ
n is an operator satisfying the following assumptions for some constants 0
<
α
≤
β
<
∞:
(24)
the
mapping
x
⟼
𝒜
(
x
,
ξ
)
is
measurable
∀
ξ
∈
ℝ
n
,
the
mapping
ξ
⟼
𝒜
(
x
,
ξ
)
is
continuous
for
a
.
e
.
x
∈
ℝ
n
;


for all ξ
∈
ℝ
n and almost all x
∈
ℝ
n,
(25)
𝒜
(
x
,
ξ
)
·
ξ
≥
α
w
(
x
)
|
ξ
|
p
,
(26)
|
𝒜
(
x
,
ξ
)
|
≤
β
w
(
x
)
|
ξ
|
p
-
1
,
(27)
(
𝒜
(
x
,
ξ
1
)
-
𝒜
(
x
,
ξ
2
)
)
·
(
ξ
1
-
ξ
2
)
>
0
,
whenever ξ
1
,
ξ
2
∈
ℝ
n, ξ
1
≠
ξ
2, and
(28)
𝒜
(
x
,
λ
ξ
)
=
λ
|
λ
|
p
-
2
𝒜
(
x
,
ξ
)
,
whenever λ
∈
ℝ, λ
≠
0, and ℬ
:
ℝ
n
×
ℝ
→
ℝ is also an operator satisfying the following similar assumptions for some constants 0
<
γ
≤
δ
<
∞:
(29)
the
mapping
x
⟼
ℬ
(
x
,
t
)
is
measurable
∀
t
∈
ℝ
,
the
mapping
t
⟼
𝒜
(
x
,
t
)
is
continuous
hhhhhhhhhhhhhhhhhhh
for
a
.
e
.
x
∈
ℝ
n
;
for all t
∈
ℝ and almost all x
∈
ℝ
n,
(30)
ℬ
(
x
,
t
)
t
≥
γ
w
(
x
)
|
t
|
p
,
(31)
|
ℬ
(
x
,
t
)
|
≤
δ
w
(
x
)
|
t
|
p
-
1
,
(32)
(
ℬ
(
x
,
t
1
)
-
ℬ
(
x
,
t
2
)
)
(
t
1
-
t
2
)
≥
0
,
whenever t
1
,
t
2
∈
ℝ.Definition 3.
A functionu
∈
H
loc
⁡
1
,
p
(
Ω
;
μ
) is a (weak) solution of (2) in Ω if
(33)
∫
Ω
‍
(
𝒜
(
x
,
∇
u
)
·
∇
φ
+
ℬ
(
x
,
u
)
φ
)
d
x
=
0
for all φ
∈
C
0
∞
(
Ω
). A function u
∈
H
loc
⁡
1
,
p
(
Ω
;
μ
) is a supersolution of (2) in Ω if
(34)
∫
Ω
‍
(
𝒜
(
x
,
∇
u
)
·
∇
φ
+
ℬ
(
x
,
u
)
φ
)
d
x
≥
0
,
whenever φ
∈
C
0
∞
(
Ω
) is nonnegative. A function u
∈
H
loc
⁡
1
,
p
(
Ω
;
μ
) is a subsolution of (2) in Ω if
(35)
∫
Ω
‍
(
𝒜
(
x
,
∇
u
)
·
∇
φ
+
ℬ
(
x
,
u
)
φ
)
d
x
≤
0
,
whenever φ
∈
C
0
∞
(
Ω
) is nonnegative.Next is the obstacles problem associated with nonhomogeneous𝒜-harmonic equations (2).Suppose thatΩ is bounded in ℝ
n
,
ψ
:
Ω
→
[
-
∞
,
∞
] is a function, and ϑ
∈
H
1
,
p
(
Ω
;
μ
). Let
(36)
𝒦
ψ
,
ϑ
=
𝒦
ψ
,
ϑ
(
Ω
)
=
{
v
∈
H
1
,
p
(
Ω
;
μ
)
:
v
≥
ψ
h
h
h
a
.
e
.
in
Ω
,
v
-
ϑ
∈
H
0
1
,
p
(
Ω
;
μ
)
}
.
If ψ
=
ϑ, write 𝒦
ψ
,
ψ
(
Ω
)
=
𝒦
ψ
(
Ω
).The problem is to find a functionu in 𝒦
ψ
,
ϑ such that
(37)
∫
Ω
‍
(
𝒜
(
x
,
∇
u
)
·
(
∇
v
-
∇
u
)
+
ℬ
(
x
,
u
)
(
v
-
u
)
)
d
x
≥
0
,
whenever v
∈
𝒦
ψ
,
ϑ. We call the function ψ an obstacle.Definition 4.
If a functionu
∈
𝒦
ψ
,
ϑ
(
Ω
) satisfies (37) for all v
∈
𝒦
ψ
,
ϑ
(
Ω
), we say that u is a solution to the obstacle problem with obstacle ψ and boundary values ϑ or a solution to the obstacle problem in 𝒦
ψ
,
ϑ
(
Ω
).
Ifu is a solution to the obstacle problem in 𝒦
ψ
,
u
(
Ω
), we say that u is a solution to the obstacle problem with obstacle ψ.Proposition 5.
(1) A solutionu to the obstacle problem is always a supersolution to (2) in Ω.
(2) Ifu is a supersolution to (2) in Ω, u is a solution to the obstacle problem in 𝒦
u
,
u
(
D
) for each open sets D
⋐
Ω. Moreover, if Ω is bounded and u
∈
H
1
,
p
(
Ω
;
μ
), u is a solution to the obstacle problem in 𝒦
u
,
u
(
Ω
).
(3) A solutionu to the obstacle problem in 𝒦
-
∞
,
u
(
Ω
) is a solution to (2) in Ω.
(4) Ifu is a solution to (2) in Ω, u is a solution to the obstacle problem in 𝒦
-
∞
,
u
(
D
) for each open set D
⋐
Ω. Moreover, if Ω is bounded u
∈
H
1
,
p
(
Ω
;
μ
), u is a solution to the obstacle problem in 𝒦
-
∞
,
u
(
Ω
).Proof.
By the definition of supersolution and solution to (2) and the definition of solution to the obstacle problem, it is easy to prove that Proposition 5 is true. Here, we only give a proof of (1).
Supposeu is a solution to the obstacle problem in 𝒦
ψ
,
ϑ
(
Ω
), and u is obviously in H
1
,
p
(
Ω
;
μ
). For all nonnegative φ
∈
C
0
∞
(
Ω
), u
+
φ
∈
H
1
,
p
(
Ω
;
μ
), u
+
φ
≥
u
≥
ϑ a.e. in Ω, and
(38)
u
+
φ
-
ϑ
=
(
u
-
ϑ
)
+
φ
∈
H
0
1
,
p
(
Ω
;
μ
)
.
Then, u
+
φ
∈
𝒦
ψ
,
ϑ
(
Ω
). By (37), we can get
(39)
∫
Ω
‍
(
𝒜
(
x
,
∇
u
)
·
∇
φ
+
ℬ
(
x
,
u
)
φ
)
d
x
=
∫
Ω
‍
(
(
u
+
φ
-
u
)
𝒜
(
x
,
∇
u
)
·
(
∇
u
+
∇
φ
-
∇
u
)
hhhhh
+
ℬ
(
x
,
u
)
(
u
+
φ
-
u
)
)
d
x
≥
0
.
Therefore, u is a supersolution to (2) in Ω.Theorem 6 (see [3]).
Supposeu is a solution to the obstacle problem in 𝒦
ψ
,
ϑ
(
Ω
). If v
∈
H
1
,
p
(
Ω
;
μ
) is a supersolution to (2) in Ω, such that min
⁡
{
u
,
v
}
∈
𝒦
ψ
,
ϑ
(
Ω
), then v
≥
u a.e. in Ω.


## 4. Relationship between the Minimization Problem and the Euler Equation

In this section, we establish that the variational integral (7) gives rise to an equation of the type (2) as its Euler equation, where the mappings 𝒜
(
x
,
ξ
)
=
∇
ξ
F
1
(
x
,
ξ
) and ℬ
(
x
,
t
)
=
∂
t
F
2
(
x
,
t
) satisfy the structural assumptions (24)–(32).Theorem 7.
Suppose thatF
1 and F
2 are two variational kernels satisfying (8)–(14) and let 𝒜
(
x
,
ξ
)
=
∇
ξ
F
1
(
x
,
ξ
) and ℬ
(
x
,
t
)
=
∂
t
F
2
(
x
,
t
). Then, 𝒜 and ℬ satisfy assumptions (24)–(32) with α
=
γ
1, β
=
2
p
δ
1, γ
=
γ
2, and δ
=
2
p
δ
2.Proof.
For pointsx for which (9), (10), (11), (13), and (14) do not hold, we are free to define 𝒜
(
x
,
ξ
) and ℬ
(
x
,
t
) arbitrarily. Fix x
∈
ℝ
n such that F
1 satisfies (9)–(11) and F
2 satisfies (13) and (14).
(i) By the definition of partial derivative, thekth coordinate of 𝒜
(
x
,
ξ
) equals
(40)
lim
⁡
i
→
∞
i
(
F
1
(
x
,
ξ
+
e
k
i
)
-
F
1
(
x
,
ξ
)
)
.
Then, the mapping x
↦
𝒜
(
x
,
ξ
) is measurable. Moreover, by (10), ξ
↦
F
1
(
x
,
ξ
) is continuously differentiable. Then, ξ
↦
𝒜
(
x
,
ξ
) is continuous and 𝒜 satisfies (24).
(ii) Ifξ
≠
0, then ξ
′
=
0
≠
ξ. By (15), we have that
(41)
F
1
(
x
,
ξ
′
)
-
F
1
(
x
,
ξ
)
>
∇
ξ
F
1
(
x
,
ξ
)
·
(
ξ
′
-
ξ
)
=
𝒜
(
x
,
ξ
)
·
(
ξ
′
-
ξ
)
.
Since ξ
′
=
0 and F
1
(
x
,
ξ
′
)
=
F
1
(
x
,
0
)
=
0, we can obtain that -
F
1
(
x
,
ξ
)
>
𝒜
(
x
,
ξ
)
·
(
-
ξ
). Then,
(42)
𝒜
(
x
,
ξ
)
·
ξ
>
F
1
(
x
,
ξ
)
≥
γ
1
w
(
x
)
|
ξ
|
p
.
If ξ
=
0, 𝒜
(
x
,
ξ
)
·
ξ
=
0
=
γ
1
w
(
x
)
|
ξ
|
p. Therefore, 𝒜 satisfies (25).
(iii) Ifξ
≠
0, then 𝒜
(
x
,
ξ
)
≠
0 by (25). Write
(43)
v
=
𝒜
(
x
,
ξ
)
|
𝒜
(
x
,
ξ
)
|
and ξ
1
=
ξ
+
|
ξ
|
v. Then, ξ
1
=
ξ
+
|
ξ
|
v
≠
ξ and |
ξ
1
|
≤
|
ξ
|
+
|
ξ
|
|
v
|
=
2
|
ξ
|. Applying (15) with ξ
1 and ξ, we can obtain that
(44)
|
𝒜
(
x
,
ξ
)
|
|
ξ
|
=
𝒜
(
x
,
ξ
)
·
|
ξ
|
𝒜
(
x
,
ξ
)
|
𝒜
(
x
,
ξ
)
|
=
𝒜
(
x
,
ξ
)
·
(
|
ξ
|
v
)
=
𝒜
(
x
,
ξ
)
·
(
ξ
1
-
ξ
)
<
F
1
(
x
,
ξ
1
)
-
F
1
(
x
,
ξ
)
≤
F
1
(
x
,
ξ
1
)
≤
δ
1
w
(
x
)
|
ξ
1
|
p
≤
2
p
δ
1
w
(
x
)
|
ξ
|
p
.
Since ξ
≠
0, we have that |
𝒜
(
x
,
ξ
)
|
<
2
p
δ
1
w
(
x
)
|
ξ
|
p
-
1.
Ifξ
=
0, we just need to verify that 𝒜
(
x
,
0
)
=
0. If not, for each k
∈
ℕ, write
(45)
v
k
=
𝒜
(
x
,
0
)
k
|
𝒜
(
x
,
0
)
|
,
and |
v
k
|
=
1
/
k
≠
0. By (15), F
1
(
x
,
v
k
)
=
F
1
(
x
,
v
k
)
-
F
1
(
x
,
0
)
>
𝒜
(
x
,
0
)
·
v
k. Therefore,
(46)
1
k
|
𝒜
(
x
,
0
)
|
=
𝒜
(
x
,
0
)
·
𝒜
(
x
,
0
)
k
|
𝒜
(
x
,
0
)
|
=
𝒜
(
x
,
0
)
·
v
k
<
F
1
(
x
,
v
k
)
≤
δ
1
w
(
x
)
|
v
k
|
p
=
δ
1
w
(
x
)
k
-
p
.
Thus, |
𝒜
(
x
,
0
)
|
≤
δ
1
w
(
x
)
k
1
-
p for k
∈
ℕ. The right hand side goes to zero as k
→
∞ and 𝒜
(
x
,
0
)
=
0. Then, 𝒜 satisfies (26).
(iv) Forξ
1
,
ξ
2
∈
ℝ
n, ξ
1
≠
ξ
2, by (15), we have that
(47)
F
1
(
x
,
ξ
1
)
-
F
1
(
x
,
ξ
2
)
>
∇
ξ
F
1
(
x
,
ξ
2
)
·
(
ξ
1
-
ξ
2
)
=
𝒜
(
x
,
ξ
2
)
·
(
ξ
1
-
ξ
2
)
,
(48)
F
1
(
x
,
ξ
2
)
-
F
1
(
x
,
ξ
1
)
>
∇
ξ
F
1
(
x
,
ξ
1
)
·
(
ξ
2
-
ξ
1
)
=
𝒜
(
x
,
ξ
1
)
·
(
ξ
2
-
ξ
1
)
.
Combining (47) with (48), we obtain that
(49)
0
>
𝒜
(
x
,
ξ
2
)
·
(
ξ
1
-
ξ
2
)
+
𝒜
(
x
,
ξ
1
)
·
(
ξ
2
-
ξ
1
)
=
-
(
𝒜
(
x
,
ξ
1
)
-
𝒜
(
x
,
ξ
2
)
)
·
(
ξ
1
-
ξ
2
)
.
Then, (
𝒜
(
x
,
ξ
1
)
-
𝒜
(
x
,
ξ
2
)
)
·
(
ξ
1
-
ξ
2
)
>
0 and 𝒜 satisfies (27).
(v) Forλ
∈
ℝ, λ
≠
0, F
1
(
x
,
λ
ξ
1
)
=
|
λ
|
p
F
1
(
x
,
ξ
1
). Taking partial derivative from both sides with respect to ξ yields
(50)
λ
𝒜
(
x
,
λ
ξ
)
=
λ
∇
ξ
F
(
x
,
λ
ξ
)
=
|
λ
|
p
∇
ξ
F
(
x
,
ξ
)
=
|
λ
|
p
𝒜
(
x
,
ξ
)
.
Then, 𝒜
(
x
,
λ
ξ
)
=
λ
|
λ
|
p
-
2
𝒜
(
x
,
ξ
).
By the similar argument, we can obtain thatℬ
(
x
,
t
) satisfies assumptions (29)–(32).The next theorem shows that minimizers of the variational integralI
(
F
1
,
F
2
)
(
u
,
Ω
) are solutions to the corresponding Euler equation and vice versa.Theorem 8.
Suppose thatK
⊂
H
1
,
p
(
Ω
;
μ
) is a convex set and u
∈
K. Then,
(51)
I
(
F
1
,
F
2
)
(
u
,
Ω
)
=
min
⁡
{
I
(
F
1
,
F
2
)
(
v
,
Ω
)
:
v
∈
K
}
if and only if
(52)
∫
Ω
‍
(
∇
ξ
F
1
(
x
,
∇
u
)
·
(
∇
v
-
∇
u
)
+
∂
t
F
2
(
x
,
u
)
(
v
-
u
)
)
d
x
≥
0
for all v
∈
K.Proof.
(i) By Lemma2, we have that
(53)
∇
ξ
F
1
(
x
,
∇
u
)
·
(
∇
v
-
∇
u
)
≤
F
1
(
x
,
∇
v
)
-
F
1
(
x
,
∇
u
)
∂
t
F
2
(
x
,
u
)
(
v
-
u
)
≤
F
2
(
x
,
v
)
-
F
2
(
x
,
u
)
for all v
∈
K
⊂
H
1
,
p
(
Ω
;
μ
). Then,
(54)
0
≤
∫
Ω
‍
(
∇
ξ
F
1
(
x
,
∇
u
)
·
(
∇
v
-
∇
u
)
+
∂
t
F
2
(
x
,
u
)
(
v
-
u
)
)
d
x
≤
∫
Ω
‍
(
F
1
(
x
,
∇
v
)
-
F
1
(
x
,
∇
u
)
+
F
2
(
x
,
v
)
-
F
2
(
x
,
u
)
)
d
x
≤
∫
Ω
‍
(
F
1
(
x
,
∇
v
)
+
F
2
(
x
,
v
)
)
d
x
-
∫
Ω
‍
(
F
1
(
x
,
∇
u
)
+
F
2
(
x
,
u
)
)
d
x
=
I
(
F
1
,
F
2
)
(
v
,
Ω
)
-
I
(
F
1
,
F
2
)
(
u
,
Ω
)
.
That is, I
(
F
1
,
F
2
)
(
u
,
Ω
)
≤
I
(
F
1
,
F
2
)
(
v
,
Ω
). Therefore, I
(
F
1
,
F
2
)
(
u
,
Ω
)
=
min
⁡
{
I
(
F
1
,
F
2
)
(
v
,
Ω
)
:
v
∈
K
}.
(ii) Fixv
∈
K and let φ
=
v
-
u. Then, since K is convex and by (51), we have that
(55)
u
+
ε
φ
=
(
1
-
ε
)
u
+
ε
v
∈
K
for all 0
<
ε
<
1 and I
(
F
1
,
F
2
)
(
u
,
Ω
)
≤
I
(
F
1
,
F
2
)
(
u
+
ε
φ
,
Ω
). Therefore,
(56)
∫
Ω
‍
(
F
1
(
x
,
∇
u
+
ε
∇
φ
)
-
F
1
(
x
,
∇
u
)
ε
h
h
h
h
h
+
F
2
(
x
,
u
+
ε
φ
)
-
F
2
(
x
,
u
)
ε
)
d
x
≥
0
.
By assumptions (10) and (14) and the definition of directional derivative,
(57)
lim
⁡
ε
→
0
F
1
(
x
,
∇
u
+
ε
∇
φ
)
-
F
1
(
x
,
∇
u
)
ε
=
∇
ξ
F
1
(
x
,
∇
u
)
·
∇
φ
,
lim
⁡
ε
→
0
F
2
(
x
,
u
+
ε
φ
)
-
F
2
(
x
,
u
)
ε
=
∂
t
F
2
(
x
,
u
)
φ
for a.e. x
∈
ℝ
n. By the mean value theorem, there exists a real number t
0
∈
(
0,1
), such that
(58)
F
1
(
x
,
∇
u
+
ε
∇
φ
)
+
F
2
(
x
,
u
+
ε
φ
)
-
F
1
(
x
,
∇
u
)
-
F
2
(
x
,
u
)
=
∇
ξ
F
1
(
x
,
∇
u
+
t
0
ε
∇
φ
)
·
ε
∇
φ
+
∂
t
F
2
(
x
,
u
+
t
0
ε
φ
)
ε
φ
.
Then, we have
(59)
F
1
(
x
,
∇
u
+
ε
∇
φ
)
-
F
1
(
x
,
∇
u
)
ε
+
F
2
(
x
,
u
+
ε
φ
)
-
F
2
(
x
,
u
)
ε
=
∇
ξ
F
1
(
x
,
∇
u
+
t
0
ε
∇
φ
)
·
∇
φ
+
∂
t
F
2
(
x
,
u
+
t
0
ε
φ
)
φ
.
By Theorem 7, the following inequalities hold:
(60)
|
∇
ξ
F
1
(
x
,
∇
u
+
t
0
ε
∇
φ
)
·
∇
φ
|
≤
|
∇
ξ
F
1
(
x
,
∇
u
+
t
0
ε
∇
φ
)
|
|
∇
φ
|
≤
2
p
δ
1
w
(
x
)
|
∇
u
+
t
0
ε
∇
φ
|
p
-
1
|
∇
φ
|
≤
2
2
p
-
1
δ
1
w
(
x
)
(
|
∇
u
|
p
-
1
+
|
t
0
ε
∇
φ
|
p
-
1
)
|
∇
φ
|
≤
2
2
p
-
1
δ
1
w
(
x
)
(
|
∇
u
|
p
-
1
|
∇
φ
|
+
|
∇
φ
|
p
)
,
|
∂
t
F
2
(
x
,
u
+
t
0
ε
φ
)
φ
|
≤
|
∂
t
F
2
(
x
,
u
+
t
0
ε
φ
)
|
|
φ
|
≤
2
p
δ
2
w
(
x
)
|
u
+
t
0
ε
φ
|
p
-
1
|
φ
|
≤
2
2
p
-
1
δ
2
w
(
x
)
(
|
u
|
p
-
1
|
φ
|
+
|
φ
|
p
)
,
|
∇
ξ
F
1
(
x
,
∇
u
)
·
∇
φ
|
≤
2
p
δ
1
w
(
x
)
|
∇
u
|
p
-
1
|
∇
φ
|
,
|
∂
t
F
2
(
x
,
u
)
φ
|
≤
2
p
δ
2
w
(
x
)
|
u
|
p
-
1
|
φ
|
.
Write g
(
x
)
=
2
2
p
-
1
δ
1
w
(
x
)
(
|
∇
u
|
p
-
1
|
∇
φ
|
+
|
∇
φ
|
p
)
+
2
2
p
-
1
δ
2
w
(
x
)
(
|
u
|
p
-
1
|
φ
|
+
|
φ
|
p
) and by u
,
φ
∈
H
1
,
p
(
Ω
;
μ
), we have that
(61)
∫
Ω
‍
g
(
x
)
d
x
≤
2
2
p
-
1
δ
1
∫
Ω
‍
w
(
x
)
(
|
∇
u
|
p
-
1
|
∇
φ
|
+
|
∇
φ
|
p
)
d
x
+
2
2
p
-
1
δ
2
∫
Ω
‍
w
(
x
)
(
|
u
|
p
-
1
|
φ
|
+
|
φ
|
p
)
d
x
=
2
2
p
-
1
δ
1
∫
Ω
‍
|
∇
u
|
p
-
1
|
∇
φ
|
d
μ
+
2
2
p
-
1
δ
1
×
∫
Ω
‍
|
∇
φ
|
p
d
μ
+
2
2
p
-
1
δ
2
×
∫
Ω
‍
|
u
|
p
-
1
|
φ
|
d
μ
+
2
2
p
-
1
δ
2
∫
Ω
‍
|
φ
|
p
d
μ
=
2
2
p
-
1
δ
1
(
∫
Ω
‍
|
∇
u
|
p
d
μ
)
1
-
1
/
p
(
∫
Ω
‍
|
∇
φ
|
p
d
μ
)
1
/
p
+
2
2
p
-
1
δ
1
∫
Ω
‍
|
∇
φ
|
p
d
μ
+
2
2
p
-
1
δ
2
(
∫
Ω
‍
|
u
|
p
d
μ
)
1
-
1
/
p
(
∫
Ω
‍
|
φ
|
p
d
μ
)
1
/
p
+
2
2
p
-
1
δ
2
∫
Ω
‍
|
φ
|
p
d
μ
<
∞
;
that is, g
∈
L
1
(
Ω
;
d
x
). Then, we can get the following conditions:
(62)
|
F
1
(
x
,
∇
u
+
ε
∇
φ
)
-
F
1
(
x
,
∇
u
)
ε
+
F
2
(
x
,
u
+
ε
φ
)
-
F
2
(
x
,
u
)
ε
|
≤
g
(
x
)
,
|
∇
ξ
F
1
(
x
,
∇
u
)
·
∇
φ
+
∂
t
F
2
(
x
,
u
)
φ
|
≤
g
(
x
)
,
F
1
(
x
,
∇
u
+
ε
∇
φ
)
-
F
1
(
x
,
∇
u
)
ε
+
F
2
(
x
,
u
+
ε
φ
)
-
F
2
(
x
,
u
)
ε
⟶
∇
ξ
F
1
(
x
,
∇
u
)
·
∇
φ
+
∂
t
F
2
(
x
,
u
)
φ
as ε
→
0. By the Lebesgue’s dominated convergence theorem, we obtain that
(63)
∫
Ω
‍
(
∇
ξ
F
1
(
x
,
∇
u
)
·
∇
φ
+
∂
t
F
2
(
x
,
u
)
φ
)
d
x
=
lim
⁡
ε
→
0
∫
Ω
‍
(
F
1
(
x
,
∇
u
+
ε
∇
φ
)
-
F
1
(
x
,
∇
u
)
ε
hhhhhhhhh
+
F
2
(
x
,
u
+
ε
φ
)
-
F
2
(
x
,
u
)
ε
)
d
x
≥
0
.
The theorem is proved.


## 5.(
F
1
,
F
2
)-Extremals and (
F
1
,
F
2
)-Superextremals with Obstacles

Definition 9.
A functionu
∈
H
1
,
p
(
Ω
;
μ
) is called an (
F
1
,
F
2
)-extremal in Ω with boundary values ϑ
∈
H
1
,
p
(
Ω
;
μ
) if u
-
ϑ
∈
H
0
1
,
p
(
Ω
;
μ
) and
(64)
I
(
F
1
,
F
2
)
(
u
,
Ω
)
≤
I
(
F
1
,
F
2
)
(
v
,
Ω
)
whenever v
-
ϑ
∈
H
0
1
,
p
(
Ω
;
μ
). A function u
∈
H
loc
⁡
1
,
p
(
Ω
;
μ
) is called a (free) (
F
1
,
F
2
)-extremal in Ω if u is an (
F
1
,
F
2
)-extremal with boundary values u in each open set D
⋐
Ω.It is immediate that an(
F
1
,
F
2
)-extremal with boundary values is a free (
F
1
,
F
2
)-extremal.Theorem 10.
Suppose thatu
∈
H
1
,
p
(
Ω
;
μ
) is a free (
F
1
,
F
2
)-extremal in Ω. Then,
(65)
I
(
F
1
,
F
2
)
(
u
,
Ω
)
≤
I
(
F
1
,
F
2
)
(
v
,
Ω
)
whenever v
-
u
∈
H
0
1
,
p
(
Ω
;
μ
).Proof.
Forφ
∈
C
0
∞
(
Ω
), φ has compact support. Choose an open set D
⋐
Ω such that spt
φ
⊂
D. Then, φ
∈
C
0
∞
(
D
) and u is an (
F
1
,
F
2
)-extremal with boundary values u in D. Since (
u
+
φ
)
-
u
∈
H
0
1
,
p
(
D
;
μ
), we have that
(66)
I
(
F
1
,
F
2
)
(
u
,
D
)
≤
I
(
F
1
,
F
2
)
(
u
+
φ
,
D
)
.
Since φ vanishes outside D, we can obtain that
(67)
I
(
F
1
,
F
2
)
(
u
,
Ω
)
=
∫
Ω
‍
(
F
1
(
x
,
∇
u
)
+
F
2
(
x
,
u
)
)
d
x
=
∫
D
‍
(
F
1
(
x
,
∇
u
)
+
F
2
(
x
,
u
)
)
d
x
+
∫
Ω
∖
D
‍
(
F
1
(
x
,
∇
u
)
+
F
2
(
x
,
u
)
)
d
x
=
I
(
F
1
,
F
2
)
(
u
,
D
)
+
∫
Ω
∖
D
‍
(
F
1
(
x
,
∇
u
)
+
F
2
(
x
,
u
)
)
d
x
≤
I
(
F
1
,
F
2
)
(
u
+
φ
,
D
)
+
∫
Ω
∖
D
‍
(
F
1
(
x
,
∇
u
)
+
F
2
(
x
,
u
)
)
d
x
=
∫
D
‍
(
F
1
(
x
,
∇
u
+
∇
φ
)
+
F
2
(
x
,
u
+
φ
)
)
d
x
+
∫
Ω
∖
D
‍
(
F
1
(
x
,
∇
u
)
+
F
2
(
x
,
u
)
)
d
x
=
∫
Ω
‍
(
F
1
(
x
,
∇
u
+
∇
φ
)
+
F
2
(
x
,
u
+
φ
)
)
d
x
=
I
(
F
1
,
F
2
)
(
u
+
φ
,
Ω
)
,
whenever φ
∈
C
0
∞
(
Ω
).
Fixv with v
-
u
∈
H
0
1
,
p
(
Ω
;
μ
) and let φ
j
∈
C
0
∞
(
Ω
) be a sequence with φ
j converging to v
-
u in H
1
,
p
(
Ω
;
μ
). By Lemma 2, we can get that
(68)
F
1
(
x
,
∇
u
+
∇
φ
j
)
≤
F
1
(
x
,
∇
v
)
+
∇
ξ
F
1
(
x
,
∇
u
+
∇
φ
j
)
·
(
∇
u
+
∇
φ
j
-
∇
v
)
,
F
2
(
x
,
u
+
φ
j
)
≤
F
2
(
x
,
v
)
+
∂
t
F
2
(
x
,
u
+
φ
j
)
(
u
+
φ
j
-
v
)
.
By the inequality (67), we obtain that
(69)
I
(
F
1
,
F
2
)
(
u
,
Ω
)
≤
I
(
F
1
,
F
2
)
(
u
+
φ
j
,
Ω
)
=
∫
Ω
‍
(
F
1
(
x
,
∇
u
+
∇
φ
)
+
F
2
(
x
,
u
+
φ
)
)
d
x
≤
∫
Ω
‍
(
F
1
(
x
,
∇
v
)
+
F
2
(
x
,
v
)
)
d
x
+
∫
Ω
‍
∇
ξ
F
1
(
x
,
∇
u
+
∇
φ
j
)
hhhhh
·
(
∇
u
+
∇
φ
j
-
∇
v
)
d
x
+
∫
Ω
‍
∂
t
F
2
(
x
,
u
+
φ
j
)
(
u
+
φ
j
-
v
)
d
x
≤
I
(
F
1
,
F
2
)
(
v
,
Ω
)
+
∫
Ω
‍
|
∇
ξ
F
1
(
x
,
∇
u
+
∇
φ
j
)
|
hhhhhhhhhhhhhhh
×
|
∇
u
+
∇
φ
j
-
∇
v
|
d
x
+
∫
Ω
‍
|
∂
t
F
2
(
x
,
u
+
φ
j
)
|
|
u
+
φ
j
-
v
|
d
x
≤
I
(
F
1
,
F
2
)
(
v
,
Ω
)
+
∫
Ω
‍
2
p
δ
1
w
(
x
)
|
∇
u
+
∇
φ
j
|
p
-
1
hhhhhhhhhhhhhhh
×
|
∇
u
+
∇
φ
j
-
∇
v
|
d
x
+
∫
Ω
‍
2
p
δ
2
w
(
x
)
|
u
+
φ
j
|
p
-
1
|
u
+
φ
j
-
v
|
d
x
≤
I
(
F
1
,
F
2
)
(
v
,
Ω
)
+
2
p
δ
1
(
∫
Ω
‍
|
∇
u
+
∇
φ
j
|
p
d
μ
)
1
-
1
/
p
×
(
∫
Ω
‍
|
∇
u
+
∇
φ
j
-
∇
v
|
p
d
μ
)
1
/
p
+
2
p
δ
2
(
∫
Ω
‍
|
u
+
φ
j
|
p
d
μ
)
1
-
1
/
p
×
(
∫
Ω
‍
|
u
+
φ
j
-
v
|
p
d
μ
)
1
/
p
≤
I
(
F
1
,
F
2
)
(
v
,
Ω
)
+
2
p
(
δ
1
+
δ
2
)
×
∥
u
+
φ
j
∥
1
,
p
p
-
1
∥
u
+
φ
j
-
v
∥
1
,
p
.
Since φ
j converges to v
-
u in H
1
,
p
(
Ω
;
μ
), letting j converge to ∞ in inequality (69), it follows that
(70)
I
(
F
1
,
F
2
)
(
u
,
Ω
)
≤
I
(
F
1
,
F
2
)
(
v
,
Ω
)
.
The theorem follows.Theorem 11.
A functionu
∈
H
loc
⁡
1
,
p
(
Ω
;
μ
) is an (free) (
F
1
,
F
2
)-extremal in Ω if and only if
(71)
-
div
⁡
∇
ξ
F
1
(
x
,
∇
u
)
+
∂
t
F
2
(
x
,
u
)
=
0
in Ω, that is,
(72)
∫
Ω
‍
(
∇
ξ
F
1
(
x
,
∇
u
)
·
∇
φ
+
∂
t
F
2
(
x
,
u
)
φ
)
d
x
=
0
for all φ
∈
C
0
∞
(
Ω
).Proof.
Write(73)
K
D
=
{
v
∈
H
1
,
p
(
D
;
μ
)
:
u
-
v
∈
H
0
1
,
p
(
D
;
μ
)
}
for each open set D
⋐
Ω.
(i) Fixφ
∈
C
0
∞
(
Ω
) and let D
⋐
Ω be an open set such that spt
φ
⊂
D. Then, u
-
(
u
+
φ
)
∈
H
0
1
,
p
(
D
;
μ
) and u
-
(
u
-
φ
)
∈
H
0
1
,
p
(
D
;
μ
). Thus, u
∈
K
D, u
+
φ
∈
K
D, u
-
φ
∈
K
D, and K
D
⊂
H
1
,
p
(
D
;
μ
)   is a convex set. Since u is an (free) (
F
1
,
F
2
)-extremal in Ω and D
⋐
Ω, we have that
(74)
I
(
F
1
,
F
2
)
(
u
,
D
)
=
min
⁡
{
I
(
F
1
,
F
2
)
(
v
,
D
)
:
v
∈
K
D
}
.
By Theorem 8, it follows that
(75)
∫
D
‍
(
∇
ξ
F
1
(
x
,
∇
u
)
·
(
∇
u
+
∇
φ
-
∇
u
)
h
h
h
+
∂
t
F
2
(
x
,
u
)
(
u
+
φ
-
u
)
∇
ξ
)
d
x
≥
0
,
∫
D
‍
(
∇
ξ
F
1
(
x
,
∇
u
)
·
(
∇
u
-
∇
φ
-
∇
u
)
h
h
h
+
∂
t
F
2
(
x
,
u
)
(
u
-
φ
-
u
)
∇
ξ
)
d
x
≥
0
.
Then,
(76)
∫
D
‍
(
∇
ξ
F
1
(
x
,
∇
u
)
·
∇
φ
+
∂
t
F
2
(
x
,
u
)
φ
)
d
x
=
0
.
Since φ and ∇
φ vanish outside D, it follows that
(77)
∫
Ω
‍
(
∇
ξ
F
1
(
x
,
∇
u
)
·
∇
φ
+
∂
t
F
2
(
x
,
u
)
φ
)
d
x
=
0
.
(ii) Fix the open setD
⋐
Ω and v
∈
K
D. Then, u
∈
H
1
,
p
(
D
;
μ
) and u
-
v
∈
H
0
1
,
p
(
D
;
μ
). Choose a sequence φ
j
∈
C
0
∞
(
D
) with φ
j converging to u
-
v in H
1
,
p
(
D
;
μ
). By Theorem 7, we obtain that
(78)
|
∫
Ω
‍
∇
ξ
F
1
(
x
,
∇
u
)
·
(
∇
v
-
∇
u
)
d
x
+
∫
Ω
‍
∇
ξ
F
1
(
x
,
∇
u
)
·
∇
φ
j
d
x
|
≤
∫
Ω
‍
|
∇
ξ
F
1
(
x
,
∇
u
)
|
|
∇
v
-
∇
u
+
∇
φ
j
|
d
x
≤
∫
Ω
‍
2
p
δ
1
w
(
x
)
|
∇
u
|
p
-
1
|
∇
v
-
∇
u
+
∇
φ
j
|
d
x
≤
2
p
δ
1
(
∫
Ω
‍
|
∇
u
|
p
d
μ
)
1
-
1
/
p
×
(
∫
Ω
‍
|
∇
v
-
∇
u
+
∇
φ
j
|
p
d
μ
)
1
/
p
⟶
0
,
|
∫
Ω
‍
∂
t
F
2
(
x
,
u
)
(
v
-
u
)
d
x
+
∫
Ω
‍
∂
t
F
2
(
x
,
u
)
φ
j
d
x
|
≤
∫
Ω
‍
|
∂
t
F
2
(
x
,
u
)
|
|
v
-
u
+
φ
j
|
d
x
≤
∫
Ω
‍
2
p
δ
2
w
(
x
)
|
u
|
p
-
1
|
v
-
u
+
φ
j
|
d
x
≤
2
p
δ
2
(
∫
Ω
‍
|
u
|
p
d
μ
)
1
-
1
/
p
×
(
∫
Ω
‍
|
v
-
u
+
φ
j
|
p
d
μ
)
1
/
p
⟶
0
as j
→
∞. Therefore, it follows that
(79)
∫
D
‍
(
∇
ξ
F
1
(
x
,
∇
u
)
·
(
∇
v
-
∇
u
)
+
∂
t
F
2
(
x
,
u
)
(
v
-
u
)
)
d
x
=
-
lim
⁡
j
→
∞
∫
D
‍
(
∇
ξ
F
1
(
x
,
∇
u
)
·
∇
φ
j
+
∂
t
F
2
(
x
,
u
)
φ
j
)
d
x
=
-
lim
⁡
j
→
∞
∫
Ω
‍
(
∇
ξ
F
1
(
x
,
∇
u
)
·
∇
φ
j
+
∂
t
F
2
(
x
,
u
)
φ
j
)
d
x
=
0
.
By Theorem 8, we have that I
(
F
1
,
F
2
)
(
u
,
D
)
≤
I
(
F
1
,
F
2
)
(
v
,
D
). Then, u is a free (
F
1
,
F
2
)-extremal in Ω.Based on the proof of Theorem11, we easily infer the following corollary.Corollary 12.
Suppose that a sequenceu
j converges to u in H
1
,
p
(
Ω
;
μ
), and then
(80)
I
(
F
1
,
F
2
)
(
u
,
Ω
)
=
lim
⁡
j
→
∞
I
(
F
1
,
F
2
)
(
u
j
,
Ω
)
.Next, we formulate the obstacle problem in terms of variational integrals. This makes the essence of the problem clearer.Definition 13.
Suppose thatΩ is bounded. Let ψ
:
Ω
→
[
-
∞
,
∞
] be an arbitrary function and call it an obstacle. For ϑ
∈
H
1
,
p
(
Ω
;
μ
), write
(81)
K
ψ
,
ϑ
(
Ω
)
=
{
v
∈
H
1
,
p
(
Ω
;
μ
)
:
v
-
ϑ
∈
H
0
1
,
p
(
Ω
;
μ
)
,
hhhhhhhh
v
≥
ψ
a
.
e
.
in
Ω
}
.
A function u
∈
K
ψ
,
ϑ
(
Ω
) is called an (
F
1
,
F
2
)-superextremal with obstacle ψ and boundary values ϑ if
(82)
I
(
F
1
,
F
2
)
(
u
,
Ω
)
≤
I
(
F
1
,
F
2
)
(
v
,
Ω
)
for all v
∈
K
ψ
,
ϑ
(
Ω
).
A functionu
∈
H
loc
⁡
1
,
p
(
Ω
;
μ
) is called a (free) (
F
1
,
F
2
)-superextremal in Ω if u is an (
F
1
,
F
2
)-superextremal with obstacle and boundary values u in each open set D
⋐
Ω.Remark 14.
(1) The(
F
1
,
F
2
)-superextremal u with obstacle ψ and boundary values ϑ minimizes the variational integral I
(
F
1
,
F
2
)
(
v
,
Ω
) among all functions v which, roughly speaking, coincide with ϑ on the boundary ∂
Ω and lie above the obstacle ψ. Naturally, this problem makes sense only if K
ψ
,
ϑ
(
Ω
) is nonempty. Moreover, we always assume that the notation K
ψ
,
ϑ
(
Ω
) that includes the assumptions Ω is bounded in this paper.
(2)(
F
1
,
F
2
)-extremal can be interpreted as (
F
1
,
F
2
)-superextremal with ψ identically -
∞.Theorem 15.
Suppose thatψ
:
Ω
→
[
-
∞
,
∞
] and ϑ
∈
H
1
,
p
(
Ω
;
μ
). Then, a function u
∈
K
ψ
,
ϑ
(
Ω
) is an (
F
1
,
F
2
)-superextremal with obstacle ψ and boundary values ϑ if and only if u is a solution to the obstacle problem in K
ψ
,
ϑ
(
Ω
) with 𝒜
=
∇
ξ
F
1 and ℬ
=
∂
t
F
2.Proof.
It is easy to see thatK
ψ
,
ϑ
(
Ω
) is a convex subset of H
1
,
p
(
Ω
;
μ
) and u
∈
H
1
,
p
(
Ω
;
μ
). Then, u is a (
F
1
,
F
2
)-superextremal with obstacle ψ and boundary values ϑ if and only if
(83)
I
(
F
1
,
F
2
)
(
u
,
Ω
)
≤
I
(
F
1
,
F
2
)
(
v
,
Ω
)
for all v
∈
K
ψ
,
ϑ
(
Ω
).
By Theorem8, we can get that I
(
F
1
,
F
2
)
(
u
,
Ω
)
≤
I
(
F
1
,
F
2
)
(
v
,
Ω
) for all v
∈
K
ψ
,
ϑ
(
Ω
) if and only if
(84)
∫
Ω
‍
(
𝒜
(
x
,
∇
u
)
·
(
∇
v
-
∇
u
)
+
ℬ
(
x
,
u
)
(
v
-
u
)
)
d
x
=
∫
Ω
‍
(
∇
ξ
F
1
(
x
,
∇
u
)
·
(
∇
v
-
∇
u
)
+
∂
t
F
2
(
x
,
u
)
(
v
-
u
)
)
d
x
≥
0
for all v
∈
K
ψ
,
ϑ
(
Ω
).
Thus, the theorem follows by the definition of the solution to the obstacle problem.


## 6. Existence of(
F
1
,
F
2
)-Superextremals

In this section, we establish the existence of(
F
1
,
F
2
)-superextremals by the direct methods of the calculus of variations.First, we show a lemma, which is a direct corollary of Mazur lemma.Lemma 16.
IfX is a normed space with the norm ∥
·
∥ and x
j converges weakly in X to x, then there exists a sequence x
~
j of convex combinations of x
j,
(85)
x
~
j
=
∑
k
=
j
l
‍
λ
j
,
k
x
k
,
λ
j
,
k
≥
0
,
∑
k
=
j
l
‍
λ
j
,
k
=
1
such that x
~
j converges to x in the norm topology of X.Proof.
Fixj
=
1,2
,
… and it is easily to see that the subsequence x
k, k
≥
j, of x
j converges weakly in X to x. By the Mazur lemma, we can get that y
k converges to x in the norm topology of X, where
(86)
y
k
=
∑
s
=
j
k
‍
λ
k
,
s
u
s
,
λ
k
,
s
≥
0
,
∑
s
=
j
k
‍
λ
k
,
s
=
1
.
Then, there exists a number k
j
∈
ℕ, such that
(87)
∥
y
k
-
x
∥
≤
1
j
for all k
≥
k
j. Let x
~
j
=
y
k
j and the lemma follows.Theorem 17.
Suppose thatK
⊂
H
1
,
p
(
Ω
;
μ
) is a nonempty convex closed set. Then there is u
∈
K such that
(88)
I
(
F
1
,
F
2
)
(
u
,
Ω
)
=
min
⁡
{
I
(
F
1
,
F
2
)
(
v
,
Ω
)
:
v
∈
K
}
.Proof.
Letu
j
∈
K be a minimizing sequence, that is,
(89)
I
(
F
1
,
F
2
)
(
u
j
,
Ω
)
⟶
I
0
=
min
⁡
{
I
(
F
1
,
F
2
)
(
v
,
Ω
)
:
v
∈
K
}
as j
→
∞. Since K
≠
∅, 0
≤
I
0
<
∞, and we can assume that
(90)
I
(
F
1
,
F
2
)
(
u
j
,
Ω
)
≤
I
0
+
1
for all j. By assumptions (9) and (13), we have that
(91)
γ
1
∫
Ω
‍
|
∇
u
j
|
p
d
μ
+
γ
2
∫
Ω
‍
|
u
j
|
p
d
μ
≤
I
(
F
1
,
F
2
)
(
u
j
,
Ω
)
≤
I
0
+
1
<
∞
.
Therefore, u
j is a bounded sequence in H
1
,
p
(
Ω
;
μ
). Thus a subsequence which we still denote by u
j converges weakly in H
1
,
p
(
Ω
;
μ
) to a function u
∈
H
1
,
p
(
Ω
;
μ
). By Lemma 16, there exists a sequence u
~
j of convex combinations of u
j,
(92)
u
~
j
=
∑
k
=
j
l
‍
λ
j
,
k
u
k
,
λ
j
,
k
≥
0
,
∑
k
=
j
l
‍
λ
j
,
k
=
1
such that u
~
j converges to u in H
1
,
p
(
Ω
;
μ
). Since K is closed and convex, u
∈
H
1
,
p
(
Ω
;
μ
). By Corollary 12, we have that
(93)
I
0
≤
I
(
F
1
,
F
2
)
(
u
,
Ω
)
=
lim
⁡
j
→
∞
I
(
F
1
,
F
2
)
(
u
~
j
,
Ω
)
.
For each ε
>
0, since I
(
F
1
,
F
2
)
(
u
j
,
Ω
)
→
I
0 as j
→
∞, there exists a number j
ε
∈
ℕ such that
(94)
max
⁡
{
I
0
-
ε
,
0
}
≤
I
(
F
1
,
F
2
)
(
u
j
,
Ω
)
<
I
0
+
ε
for all j
≥
j
ε. By assumptions (10) and (14), we obtain that
(95)
I
(
F
1
,
F
2
)
(
u
~
j
,
Ω
)
=
∫
Ω
‍
(
F
1
(
x
,
∇
u
~
j
)
+
F
2
(
x
,
u
~
j
)
)
d
x
=
∫
Ω
‍
(
F
1
(
x
,
∑
k
=
j
l
‍
λ
j
,
k
∇
u
k
)
hhhh
+
F
2
(
x
,
∑
k
=
j
l
‍
λ
j
,
k
u
k
)
)
d
x
≤
∫
Ω
‍
∑
k
=
j
l
‍
λ
j
,
k
(
F
1
(
x
,
∇
u
k
)
+
F
2
(
x
,
u
k
)
)
d
x
=
∑
k
=
j
l
‍
λ
j
,
k
∫
Ω
‍
(
F
1
(
x
,
∇
u
k
)
+
F
2
(
x
,
u
k
)
)
d
x
=
∑
k
=
j
l
‍
λ
j
,
k
I
(
F
1
,
F
2
)
(
u
k
,
Ω
)
≤
∑
k
=
j
l
‍
λ
j
,
k
(
I
0
+
ε
)
=
I
0
+
ε
whenever j
≥
j
ε. By (93) and (95), it follows that
(96)
I
0
≤
I
(
F
1
,
F
2
)
(
u
,
Ω
)
≤
I
0
+
ε
.
Then, I
0
=
I
(
F
1
,
F
2
)
(
u
,
Ω
) and u is the desired minimizer.Theorem 18.
Suppose thatΩ is bounded, that ψ
:
Ω
→
[
-
∞
,
∞
], and that ϑ
∈
H
1
,
p
(
Ω
;
μ
). If
(97)
K
ψ
,
ϑ
(
Ω
)
=
{
v
∈
H
1
,
p
(
Ω
)
:
v
≥
ψ
a
.
e
.
i
n
Ω
,
v
-
ϑ
∈
H
0
1
,
p
(
Ω
)
}
≠
∅
,
there exists a unique (
F
1
,
F
2
)-superextremal with obstacle ψ and boundary values ϑ.Proof.
Since the setK
ψ
,
ϑ
(
Ω
) is nonempty convex subset of H
1
,
p
(
Ω
;
μ
), the existence follows from Theorem 17 if we can show that K
ψ
,
ϑ
(
Ω
) is closed in H
1
,
p
(
Ω
;
μ
). To accomplish this, let u
j be a sequence such that u
j converges to a function u in H
1
,
p
(
Ω
;
μ
). Since u
j
-
ϑ
∈
H
0
1
,
p
(
Ω
;
μ
), u
-
ϑ
∈
H
0
1
,
p
(
Ω
;
μ
). Since u
j converges to u in H
1
,
p
(
Ω
;
μ
), there is subsequence of u
j that converges a.e. to u. Therefore, u
≥
ψ a.e. Ω. Then, u
∈
K
ψ
,
ϑ
(
Ω
) and the existence part is thereby established.
For the uniqueness, suppose thatu
1
,
u
2
∈
K
ψ
,
ϑ
(
Ω
) are two distinct minimizers. Since u
1
-
ϑ, u
1
-
ϑ
∈
H
0
1
,
p
(
Ω
;
μ
), the set {
∇
u
1
≠
∇
u
2
} has positive measure. By the strict convexity (10) of F
1, we can get that
(98)
F
1
(
x
,
∇
v
(
x
)
)
<
1
2
(
F
1
(
x
,
∇
u
1
(
x
)
)
+
F
1
(
x
,
∇
u
2
(
x
)
)
)
for each x
∈
{
∇
u
1
≠
∇
u
2
} and
(99)
∫
{
∇
u
1
≠
∇
u
2
}
‍
F
1
(
x
,
∇
v
)
d
x
<
1
2
(
∫
{
∇
u
1
≠
∇
u
2
}
‍
F
1
(
x
,
∇
u
1
)
d
x
h
h
h
h
h
h
h
+
∫
{
∇
u
1
≠
∇
u
2
}
‍
F
1
(
x
,
∇
u
2
)
d
x
)
.
Then,
(100)
∫
Ω
‍
F
1
(
x
,
∇
v
)
d
x
<
1
2
(
∫
Ω
‍
F
1
(
x
,
∇
u
1
)
d
x
+
∫
Ω
‍
F
1
(
x
,
∇
u
2
)
d
x
)
.
By the convexity (14) of F
2, we can obtain that
(101)
∫
Ω
‍
F
2
(
x
,
v
)
d
x
≤
1
2
(
∫
Ω
‍
F
2
(
x
,
u
1
)
d
x
+
∫
Ω
‍
F
2
(
x
,
u
2
)
d
x
)
.
Thus,
(102)
I
(
F
1
,
F
2
)
(
v
,
Ω
)
=
∫
Ω
‍
(
F
1
(
x
,
∇
v
)
+
F
2
(
x
,
v
)
)
d
x
<
1
2
(
∫
Ω
‍
F
1
(
x
,
∇
u
1
)
d
x
+
∫
Ω
‍
F
1
(
x
,
∇
u
2
)
d
x
)
+
1
2
(
∫
Ω
‍
F
2
(
x
,
u
1
)
d
x
+
∫
Ω
‍
F
2
(
x
,
u
2
)
d
x
)
=
1
2
(
I
(
F
1
,
F
2
)
(
u
1
,
Ω
)
+
I
(
F
1
,
F
2
)
(
u
2
,
Ω
)
)
=
min
⁡
{
I
(
F
1
,
F
2
)
(
u
,
Ω
)
:
u
∈
K
ψ
,
ϑ
(
Ω
)
}
.
This contradiction completes the proof.Similar to the proof of Theorem18, we can obtain the existence of (
F
1
,
F
2
)-extremals.Theorem 19.
Suppose thatϑ
∈
H
1
,
p
(
Ω
;
μ
), then there exists a unique (
F
1
,
F
2
)-extremals u in Ω with u
-
ϑ
∈
H
1
,
p
(
Ω
;
μ
).Remark 20.
In Theorem19, the open set Ω can be unbounded.



---

*Source: 697974-2014-02-27.xml*