Covei_Corectat. 132KB Jun 04 2011 12:08:34 AM

ACTA UNIVERSITATIS APULENSIS

No 10/2005

Proceedings of the International Conference on Theory and Application of
Mathematics and Informatics ICTAMI 2005 - Alba Iulia, Romania

A FEW RESULTS ABOUT THE P-LAPLACE’S OPERATOR

Covei C. Dragos-Patru
Abstract. The aim of this paper is to obtain few results for p-Laplace’s
operator and these representation an extension of the very know results for
laplacian.
2000 Mathematics Subject Classification: 3505,35J60.
1.The p-Laplace equation
The equation
△p u = 0 on Ω, 1 < p < ∞,

(1.1.)

is called p-Laplace’s equation.

Here, Ω ⊂ RN is an open set, u : Ω −→ R is the unknown, and △p is the
p-Laplace operator defined by
△p u := div(|∇u|p−2 ∇u),

(1.2.)

The previous investigations have led to the equation’s critical points
Z
Dp (u; Ω) = |∇u|p dx
(1.3.)
Ω

are weak solutions for (1.1.) , thus they can be named p-harmonic functions.

105

C. C. Dragos-Patru - A few results about the p-Laplace’s operator
2.Fundamental solutions for p-Laplace equation
We will first construct a simple radial solution of p-Laplace’s equation. To
look for radial solutions of p-Laplace’s equation on Ω = RN of the form

q
u(x) = v(r); r = |x| := 2 x21 + ... + x2N ,
(2.1.)
Here, v : [0, ∞) −→ R
We note that

u xi =

∂v(r)
xi
′
= v (r) ,
∂xi
r

(2.2.)

and

uxi xi =

x2i ′′
x2i ′
∂ 2 v(r)
1 ′
v
(r)
+
v (r), ∀1 ≤ i ≤ N,
=
v
(r)
−
∂x2i
r2
r
r3

(2.3.)

and summation yields
′′

∆2 u(x) = v (r) +

N −1 ′
v (r), r 6= 0.
r

(2.4.)

We have
r
2

and

∂u
∂x1

2

+ ... +
|∇u| =
p


′
′
2
(v (r))2 = v (r) ,



∂u
∂xN

2

=

q
2

v ′ (r) xr1


2

+ ... + v ′ (r) xrN


2

=
(2.5.)

q
p−2
 ′ p−2
2
2

′
∂
v (r)
= ∂xi
=
(v (r))
p−3 ′ x ′′
q
v (r) ri v (r)
,
(p − 2) 2 (v ′ (r))2
|v′ (r)|
∂
∂xi

(2.6.)

But (1.1.) equivalently

|∇u|p−2 ∆2 u + ∇ |∇u|p−2 · ∇u = 0.
106

(2.7.)

C. C. Dragos-Patru - A few results about the p-Laplace’s operator
We have
p−2


∇ |∇u|



 
 ′ p−2
· ∇u = ∇ v (r)
· ∇v(r) =


 

∂  ′ p−2
∂v(r)
∂v(r)
∂  ′ p−2
=
, ...,
·
v (r) , ...,
v (r)
∂x1
∂xN
∂x1
∂xN
 ′ p−3 ′
 ′ p−3 ′
′′
′′
′
′
(p − 2) v (r) v (r) xrN v (r) v (r)xN

(p − 2) v (r) v (r) x1
v (r) v (r)x1
r
+...+
=
|v ′ (r)|
r
|v ′ (r)|
r
 ′ p−3 ′
2 ′′


(p − 2) v (r)
v (r) v (r) 2
x1 + ... + x2N =
′
2
|v (r)| r
 ′ p−3 ′
2 ′′
(p − 2) v (r)
v (r) v (r)

.
(2.8.)
′
|v (r)|
So (2.7.) equivalently



 
N −1 ′
′′
 ′ p−2
v (r) = 0.
(p − 1)v (r) +
v (r)
r
 ′ 
Assume v (r) 6= 0.
Hence, we have
∆p u = 0 for x 6= 0

(2.9.)

if and only if
′′

(p − 1)v (r) +

N −1 ′
v (r) = 0,
r

(2.10.)

′

In the case (2.10.) note v = z, follows
(p − 1)z + N r−1 z = 0 ⇐⇒
= 1−N
dr ⇐⇒
(p − 1) dz
z
r
(p − 1) lnq
|z| = (1 − N ) ln r + ln |C|p−1 ⇐⇒
′

z(r) =

p−1

p−1

|C|
rN −1

=

|C|

N −1
r p−1

(2.11.)

.

We conclude that

C

′

v (r) =
r
107

N −1
p−1

,

(2.12.)

C. C. Dragos-Patru - A few results about the p-Laplace’s operator
for an arbitrary constant C ∈ R+ and thus
(
C ln r+C1 ,
if N = p
p−N
v(r) =
, r > 0,
p−1 p−1
+ C1 , if N ≥ p + 1
C p−N
r

(2.13.)

with constants C1 ∈ R.
3.Gauss-Green, Gauss-Ostrogradski and Green′ s formulas for
the p-Laplace′ operator
Definition 3.1. Let Ω ⊂ RN be open and bounded
i) We say that Ω has a C k -boundary, k ∈ N ∪ {∞}, if for any x ∈ ∂Ω
there exists r > 0 and a function β ∈ C k (RN ) such that
Ω ∩ B(x; r) = {y ∈ B(x; r) : yN > β(y1 , ..., yN −1 )} ,
ii)If ∂Ω is C k then we can define the unit outer normal field υ : ∂Ω −→
R , where, υ(x), |υ(x)| = 1, is the outward pointing unit normal of ∂Ω at x.
iii)Let ∂Ω be C k . We call the directional derivative
N

∂u
(x) := ν(x) · ∇u(x), x ∈ ∂Ω,
∂υ
the normal derivative of u.
In addition to C k (Ω) we define the function space


C k (Ω) := u ∈ C k (Ω) : Dα u can be continuously extended to ∂Ω for |α| ≤ k ,

where

N

Dα u =

X
∂ α1 +...+αN
u,
|α|
=
αi .
∂xα1 1 ...∂xαNN
i=1

We recall the Gauss-Green theorem.
Theorem 3.2. Let Ω ⊂ RN be open and bounded with C 1 -boundary.
Then for all u ∈ C 1 (Ω)
Z
Z
uxi (x)dx = u(x)υi (x)dσ(x).
Ω

∂Ω

108

C. C. Dragos-Patru - A few results about the p-Laplace’s operator
Remark(Gauss-Ostrogradscki): Let Ω ⊂ RN be open and bounded
−
→
−
→
with C 1 -boundary. Then for all f : Ω −→ RN such that f ∈ C(Ω) ∩ C 1 (Ω).
We have
Z
Z
−
→
−
→
div f dx = f · υdσ(x).
Ω

∂Ω

Theorem 3.3. If u ∈ C 2 (Ω) such that ∆p u ∈ C(Ω) then
Z
Z
∂u
|∇u|p−2 dσ(x).
∆p udx =
∂υ
Ω

(3.1.)

∂Ω

−
→
Proof. In theorem Gauss-Ostrogradscki let f = |∇u|p−2 ∇u.
We have




R
R
|∇u|p−2 ∇u · υdσ(x) = ∆p udx =
div |∇u|p−2 ∇u dx =
Ω
∂Ω


RΩ
R
|∇u|p−2 ∆2 udx + ∇ |∇u|p−2 · ∇udx =
Ω
Ω R


R ∂u
p−2
|∇u|
dσ(x)
−
∇ |∇u|p−2 · ∇udx+
∂υ
Ω R
∂Ω


R
|∇u|p−2 dσ(x)
∇ |∇u|p−2 · ∇udx = ∂u
∂ν
R

Ω

∂Ω

Moreover, we easily obtain Green’s formulas for the p-Laplace operator:
Theorem 3.4. Let Ω ⊂ RN be open and bounded with C 1 -boundary.
Then for all u, v ∈ C 2 (Ω) such that ∆p u ∈ C(Ω), we have


R
R
R
dσ(x) − ∇v · |∇u|p−2 ∇u dx
G1) (∆p u) vdx = v |∇u|p−2 ∂u
∂υ
Ω
∂Ω


R
RΩ
∂v
v |∇u|p−2 ∂u
G2) [(∆p u) v − (∆p v) u] dx =
− u |∇v|p−2 ∂υ
dσ(x).
∂υ
Ω

∂Ω



−
→
Proof. G1) Let f = v |∇u|p−2 ∇u .We have







div v |∇u|p−2 ∇u = vdiv |∇u|p−2 ∇u + ∇v · |∇u|p−2 ∇u .
So

Z

Ω




v∆p u + ∇v · |∇u|p−2 ∇u dx =
109

Z

∂Ω

v |∇u|p−2

∂u
dσ(x).
∂υ

(3.2.)

C. C. Dragos-Patru - A few results about the p-Laplace’s operator
Proof. G2) By G1) we have
Z
Z
Z


p−2 ∂u
(∆p u) vdx = v |∇u|
dσ(x) − ∇v · |∇u|p−2 ∇u dx
∂υ

(3.3.)

we inverse the role u and v,so
Z
Z
Z


p−2 ∂v
dσ(x) − ∇u · |∇v|p−2 ∇v dx
(∆p v) udx = u |∇v|
∂υ

(3.4.)

Ω

Ω

Ω

∂Ω

Ω

∂Ω

Using (3.3.) and (3.4.) we deduce G2)
4.Green function, Kelvin transform, or Poisson Kernel?
The following ideas are from [3]: From a physical standpoint equation
(1.1.), or rather its generalizations, arises naturally, e.g., in the steady rectilinear motion of incompressible non-Newtonian fluids or in phenomena of phase
transition. A glimpse at (1.1.) immediately reveals two unfavorable features:
(i) the operator is badly nonlinear;
(ii) ellipticity is lost at points where ∇u = 0.
The strong nonlinearity makes it impossible to develop a potential theory
along the lines of classical one. p-harmonic functions do not enjoy integral
representation formulas such as
I
I
u(x) =
udσ =
udy,
∂Br (x)

Br (x)

there is no Green function, or Kelvin transform, or Poisson Kernel. p-subharmonicity
is not preserved by the clasical mollification processes, as is the case for subharmonic functions. This makes it impossible to regularize p-subharmonic
functions. In retrospect, this obstruction is also deeply connected with (ii)
above. The lack of ellipticity results in loss of regularity of p-harmonic functions.
By results of Lewis [4], solutions to the p-Laplacian are C 1,α for some α > 0,
for instance the function
p
u(x) = |x| p−1
satisfies the equation
∆p u = const, but u ∈
/ C 2 , when p > 2.
110

C. C. Dragos-Patru - A few results about the p-Laplace’s operator
In particular |∇u| is C α in any region where u satisfies the p-Laplace equation
∆p u = 0.
However the operator Lu , defined above, may fail to have the maximum/comparison
principle. The weak maximum principle for the p-Laplace operator is well
known and can be find in standard literature in this filed; see [3], [5] and [1],
the latter treats the parabolic case.
5.The existence of positive solutions in C2 (RN ) for the problem
with p-Laplacian
Consider the problem

 −∆p u = p(x)f (u) in RN
u > 0 in RN

u(x) → 0 as | x |→ ∞ ,

(5.1.)

where N > 2, ∆p u (1 < p ≤ 2) is the p-Laplacian operator and
-the function p(x) fulfills the following hypotheses:
(p1) p(x) ∈ C(RN ) and p(x) > 0 in RN .
(p2) we have
Z

∞

1

1

r p−1 Φ p−1 (r)dr < ∞ if 1 < p ≤ 2

0

where Φ(r) := maxp(x).
|x|=r

-the function f ∈ C 1 ((0, ∞), (0, ∞)) such that lim f (u) = ∞ and satisfies
u→0
the following assumptions:
(u)
(f 1) mapping u −→ ufp−1
is decreasing on (0, ∞);
f (u)
(f 2) lim up−1 = +∞;
uց0

(f 3)lim inf f (u) > 0.
u→0
It easy to prove that
Theorem 5.1. If j : I ⊆ R −→ R is a integrable nonnegative function,
then

h
Z b
Z b
1
1
j(x)dx ≤
j h (x)dx
b−a a
b−a a
111

C. C. Dragos-Patru - A few results about the p-Laplace’s operator
∀ a, b ∈ I, a < b and 1 < h < +∞
Theorem 5.2.Under hypotheses (f 1) − (f 3), (p1), (p2), the problem (5.1.)
has a radially symmetric solution u ∈ C 2 (RN \{0}) ∩ C 1 (RN ).
Proof. By Theorem 1.3. in [2] the problem

−∆p U = p(x)f (U ), if |x| < k,
U = 0,
if
|x| = k.


has a radially symmetric solution in C B k ∩ C 1 (Bk ) ∩ C 2 (Bk \(0))


We now prove the existence of a positive function u ∈ C 2 RN . As in [2]
we construct first a positive radially symmetric function w such that −∆p w =
Φ(r), (r = |x|) on RN and lim w(r) = 0.
r−→∞
We obtain
1
 p−1
Z r
Z ξ
ξ 1−N
w(r) := K −
σ N −1 Φ(σ)dσ
dξ,
0

where
K=

Z

∞

0

0

1

 p−1
Z ξ
1−N
N −1
dξ.
ξ
σ
Φ(σ)dσ

0

We first show that (p2) implies that
Z

0

+∞

1
 p−1

Z ξ
N −1
1−N
σ
Φ(σ)dσ
ξ
dξ,

0

is finite.
Let 1 < p ≤ 2, so 0 < p − 1 ≤ 1, follows that 1 ≤
Using Theorem 5.1. for any r > 0, we have

1
p−1

1
i p−1
R r 1−N 1 h 1 R ξ N −1
p−1 ξ p−1
dξ
=
dξ ≤
ξ
σ
Φ(σ)dσ
ξ 0
ξ 0
0
0
R r 2−N 1 R ξ N −1 1
R r 2−N −1 R ξ N −1 1
ξ p−1 ξ 0 σ p−1 Φ p−1 (σ)dσdξ = 0 ξ p−1
σ p−1 Φ p−1 (σ)dσdξ =
0
0
R
R
N
−1
2−N
1
ξ
r d
ξ p−1 0 σ p−1 Φ p−1 (σ)dσdξ =
− Np−1
−2
dξ
h 0 2−N
i
Rr 1
R r N −1 1
Rr 1
1
1
p−1
p−1
p−1 Φ p−1 (σ)dσ +
p−1 Φ p−1 (ξ)dξ
−r
ξ p−1 Φ p−1 (ξ)dξ,
≤ Np−1
σ
ξ
N −2
−2 0
0
0

Rr

so

1−N

ξ p−1

h R
ξ ξ

σ N −1 Φ(σ)dσ

1
i p−1

< +∞.

1
 p−1
Z ξ
Z r
dξ < ∞
σ N −1 Φ(σ)dσ
ξ 1−N

0

0

112

C. C. Dragos-Patru - A few results about the p-Laplace’s operator
as r −→ ∞.
Then we obtain
p−1
K=
·
N −2

Z

∞

1

1

ξ p−1 Φ p−1 (ξ)dξ if 1 < p ≤ 2,

0

clearly, we have
p−1
w(r) ≤
·
N −2

Z

∞

1

1

ξ p−1 Φ p−1 (ξ)dξ if 1 < p ≤ 2.

0

An upper-solution to (5.1.) will be constructed.
Consider the function
1

f (t) = (f (t) + 1) p−1 , t > 0.
Note that

1

f (t) ≥ f (t) p−1
′
f (t)
, is decreasing, (f1 )
tp−1
′
= ∞,
(f2 )
lim f (t)
t
t−→0

Let v be a positive function such that
Z
1 v(r) tp−1
w(r) =
dt where C > 0
C 0
f (t)
will be chosen such that
1

KC ≤

Z

C p−1

0

tp−1
dt.
f (t)
′

We prove that we can find C > 0 with this property. By our hypothesis (f2 )
we obtain that
Z x p−1
t
lim
dt = +∞.
x−→+∞ 0 f (t)
Now using L’Hopital’s rule we have
R x tp−1
dt
x
0 f (t)
= lim
= +∞.
lim
p−1
x−→∞ (p − 1) f (x)
x−→∞
x
113

C. C. Dragos-Patru - A few results about the p-Laplace’s operator
From this we deduce that there exists x1 > 0 such that
Z x p−1
t
dt ≥ Kxp−1 , for all x ≥ x1 .
0 f (t)
It follows that for any C ≥ x1 we have
1

KC ≤

Z

C p−1

0

tp−1
dt.
f (t)

But w is a decreasing function, and this implies that v is a decreasing function
too.
Then
1

Z

v(r)

0

tp−1
dt ≤
f (t)

Z

v(0)

0

tp−1
dt = Cw(0) = CK ≤
f (t)

Z

C p−1

0

tp−1
dt.
f (t)

1

It follows that v(r) ≤ C p−1 for all r > 0. From w(r) −→ 0 as r −→ +∞
we deduce v(r) −→ 0 as r −→ +∞.
By the choice of v we have
∇w =

1 v p−1
∇v
·
C f (v)

follows that
∆p w =

1
C p−1



v p−1
f (v)

p−1

∆p v + (p − 1)

f (u)
up−1

1
C p−1

p

|∇v|



v p−1
f (v)

p−2 

v p−1
f (v)

′

.

(5.2.)
is a decreasing function on (0, +∞), we deduce

From (5.2.) and u −→
that

p−1
p−1

f (v)
f (v)
p−1
p−1
∆p v ≤ C
∆p w = −C
Φ(r) ≤ −f (v)Φ(r). (5.3.)
v p−1
v p−1
It follows that v is a radially symmetric solution of the problem:

 −∆p u ≥ p(x)f (u) in RN
u > 0 in RN

u(x) → 0 as | x |→ ∞ ,
114

(5.4.)

C. C. Dragos-Patru - A few results about the p-Laplace’s operator
By the proof of Theorem 1.1. in [2] the problem (5.1.) has positive solutions.
Now using
 1

Rr
u′ (r) = r1−N 0 σ N −1 p(σ)f (u(σ))dσ p−1
′′

u (r) = −

R
p(r)f (u(r))+(1−N )r −N 0r σ N −1 p(σ)f (u(σ))dσ
p−1

 2−p
 1−N R r N −1
σ
p(σ)f (u(σ))dσ p−1 ,
r
0

2−p
≥ 0 ⇐⇒ 1 < p ≤ 2
p−1
R r N −1
σ
p(σ)f (u(σ))dσ
lim 0
=0
rN
r−→0
R r N −1
σ
p(σ)f (u(σ))dσ
=0
lim 0
r N −1
r−→0
′′

we deduce lim u (r) is finite, so u(r) ∈ C 2 (RN ).
r−→0

References
[1] E. DiBenedetto, Degenerate and singular parabolic equations, Universitext, Springer-Verlag, New York, 1993.
[2] J.V. Goncalves and C.A. Santos, Positive solutions for a class of quasilinear singular equations, Journal of Differential Equations, Vol. 2004(2004),
No. 56, pp 1-15.
[3] J. Heinonen, T. Kilpel¨ainen, O. Martio, Nonlinear potential theory of
degenerate elliptic equations, Oxford Mathematical Monographs, Oxford Science Publications
[4] J.L. Lewis, Capacitary functions in convex rings, Arch. Rat. Mech.
Anal. 66 (1977), 201-224
[5] J. Maly, W. P. Ziemer, Fine regularity of solutions elliptic partial differential equations. Math. Surv. Monogr. 51, American Mathematical Society,
Providence, RI, 1997.
[6] V. Radulescu, Partial Differential Equations , Repografia Univ. Craiova
(http://inf.ucv.ro/˜radulescu), Craiova, 2005.
Dragos-Patru Covei
Department of Science and General Culture
University of Targu-Jiu ”Constantin Brancusi”
Bd. Republicii, No. 1, Tg. Jiu, Romania
dragoscovei77@yahoo.com
115