HAL Id: hal-01348953
https://hal.archives-ouvertes.fr/hal-01348953v3
Preprint submitted on 17 Jan 2018
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An inequality of type sup+inf on S_4 for the Paneitz operator.
Samy Skander Bahoura
To cite this version:
Samy Skander Bahoura. An inequality of type sup+inf on S_4 for the Paneitz operator.. 2018.
�hal-01348953v3�
An inequality of type sup+inf on S
4
for the Paneitz operator.
Samy Skander Bahoura
Equipe d’Analyse Complexe et Géométrie
Université Pierre et Marie Curie, 4 Place Jussieu, 75005, Paris, France e-mails: samybahoura@yahoo.fr, samybahoura@gmail.com
Abstract
We give an inequality of typesup + infonS4for the Paneitz equation.
1 Introduction and Main Result.
We set∆ =−∇i(∇i)the Laplace-Beltrami operator.
OnS4 we consider the following Paneitz equation:
∆2u+ 2∆u+ 6 =V e4u (E) with,0≤V(x)≤b, x∈S4andb >0.
Here, we try to derive a minoration of the sumsup + inf with minimal con- dition onV.
Not that, Paneitz (see [7]) had introduced this equation which is invariant under conformal change of metrics.
This equation is similar to the Gaussian curvature equation in dimension 2, which is:
∆u+K= ˜Ke2u (E′)
whereK is the Gaussian curvature andK˜ =V /2is the prescribed Gaussian curvature. (V is the prescribed scalar curvature).
We refer to [2,7], for some examples where this equation where considered and studied.
Some authors where interested by this equation and established some Sobolev inequalities, see [2-5,7].
For the corresponding equation in the heigher dimensionnal case to (E) (dimension n≥5), see [5,7].
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Our main result is:
Theorem. For all b > 0, there is a constant c which depends only on b, c=c(b), such that,
sup
S4
u+ inf
S4 u≥c, for all solutions uof (E).
2 Proof of the Theorem.
We multiply the equation byuand integrate:
Z
S4
(∆u)2+ 2 Z
S4
|∇u|2+ 6 Z
S4
u= Z
S4
uV e4u≤6|S4|sup
S4
u. (1)
Let’s considerG1 andG2 the following two Green functions:
∆G1+ 2G1=δ, ∆G2=δ− 1
|S4|. G1 andG2are symmetric.
We refer to [1,6] for the construction of the Green function of 2nd order operators.
We set:
G=G1∗G2= Z
S4
G1(x, z)G2(z, y)dVg(z).
It is easy to see that:
∆2G+ 2∆G=δ− 1
|S4|. It is clear that,G1≥0. We chooseG2such that:
G2≥0, Z
S4
G2≡cte >0 We write:
infS
4
u=u(x) = 1
|S4| Z
S4
u+ Z
S4
(V e4uG−6G)≥ 1
|S4| Z
S4
u−C. (2) We use(1)and(2)to obtain:
sup
S4
u+ inf
S4
u≥ 2
|S4| Z
S4
u+ 2 6|S4|
Z
S4
|∇u|2+ 1 6|S4|
Z
S4
(∆u)2−C. (3) We use the biharmonic version of the Moser-Trudinger inequality:
2
log 1
|S4| Z
S4
e4u
≤ 1 3|S4|
Z
S4
(∆u)2+ 2 3|S4|
Z
S4
|∇u|2+ 4
|S4| Z
S4
u. (4) We combine(3)and(4)to obtain:
sup
S4
u+ inf
S4
u≥1 2log
1
|S4| Z
S4
e4u
−C, We integrate the eqaution(E), we obtain:
1
|S4| Z
S4
V e4u= 6, thus,
1
|S4| Z
S4
e4u≥ 6 b. hence,
sup
S4
u+ inf
S4
u≥ 1 2log
6 b
−C=log 6 2 −logb
2 −C. (5)
Remark that:
C≡6 Z
S4
G(x, y)dVg(y) = 6 Z
S4
Z
S4
G1(x, z)G2(z, y)dVg(z)dVg(y) = 3 Z
S4
G2(x, y)dVg(y).
because,R
S4G1(x, y)≡ 1 2.
References
[1] T. Aubin. Some Nonlinear Problems in Riemannian Geometry. Springer- Verlag 1998
[2] Beckner. Sharp Sobolev inequalities on the Sphere and Moser-Trudinger inequality. Ann. of Math. (2) 138, (1993), 213-242.
[3] T.P. Branson, S.Y.A. Chang, P.C. Yang. Estimates and extremals for zeta function determinants on four manifolds, Comm. Math. Phys. 149 (1992), 241-262.
[4] S.Y.A.Chang, P.C.Yang. Extremal metrics of zeta function determinants on 4-manifolds. Ann. of Math. (2) 142 (1995), 171-212.
[5] Z. Djadli, E. Hebey, M. Ledoux. Paneitz-type operators and Applications.
Duke. Math. J, (104), 1, (2000) 129-169.
[6] O. Druet, E. Hebey, F.Robert, Blow-up theory in Riemannian Geometry, Princeton University Press 2004.
[7] S. Paneitz. A quartic conformally covariant differential operator for arbi- trary pseudo-Riemannian manifolds, preprint, 1983.
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