NORM OF THE HIGHER-ORDER WENTE PROBLEM

NORM OF THE HIGHER-ORDER WENTE PROBLEM

SAMI BARAKET AND MAKKIA DAMMAK Received 1 June 2004

We study the best constant involving theL²norm of thep-derivative solution of Wente’s problem inR^2p. We prove that this best constant is achieved by the choice of some func- tionu. We give also explicitly the expression of this constant in the special casep=2.

1. Introduction and statement of the results

The Wente problem arises in the study of constant mean curvature immersions (see [6]).

LetΩbe a smooth and bounded domain inR². Givenu=(a,b) be function defined on Ω. Consider the following problem:

−∆ψ=det∇u=ax1bx2−ax2bx1 inΩ,

ψ=0 on∂Ω, (1.1)

wherex=(x1,x2) andaxi denote the partial derivative with respect to the variablexi, for i₌1, 2. IfΩ=R², we consider the limit condition lim_|x|→+_∞ψ(x)₌0, where|x_{| =}r₌ (x²₁+x²₂)^1/2. Whenu=(a,b)∈H¹(Ω,R²), it is proven in [7] and [3] thatψ, the solution of (1.1) is inL^∞(Ω). In particular, this provides control of∇ψinL²(Ω) and continuity of ψby simple arguments. We also have

ψ∞+∇ψ2≤C0(Ω)∇a2∇b2. (1.2) Denote

C_∞(Ω)= sup

∇a,∇b=0

ψ∞

∇a2∇b2, C1(Ω)= sup

∇a,∇b=0

∇ψ2

∇a2∇b2.

(1.3)

It is proved in [1,5,7] thatC_∞(Ω)=1/2πand in [4] thatC1(Ω)=

(3/16π).

Copyright©2005 Hindawi Publishing Corporation Abstract and Applied Analysis 2005:6 (2005) 599–606 DOI:10.1155/AAA.2005.599

Here, we are interested to study a generalization of problem (1.1) in higher dimen- sions. More precisely, letp∈N^∗andu∈W^1,2p(R^2p,R^2p). Consider the following prob- lem:

(−∆)^pϕ=det∇u inR^2p,

|xlim|→+_∞ϕ(x)=0. (1.4)

It was proved in [2] that the solutionϕof (1.4) is inL^∞(R^2p) and ˜∆^k/2ϕis inL^2p/k(R^2p) for 1≤k≤p, with the following estimates:

ϕ∞+∆˜^k/2ϕ_2p/k≤C∇u^2p_2p, (1.5) where

∆˜^k/2ϕ_2p/k=





∆^k/2ϕ_2p/k ifkis even, ∇

∆^(k−1)/2ϕ_2p/k ifkis odd. (1.6) Moreover, the best constant involving theL^∞norm was determined. Here, we will focus our attention to the quantity∆˜^p/2ϕ2. We will introduce some notations, denote byB^2p the unit ball inR^2p,S^2p the unit sphere inR^2p+1 and σ_2p+1=vol(S^2p). DenoteΨ the function defined on (0, +∞) by

Ψ(s)= 1 s^p R^2p

s|∇ϕ|²+|∇u|²p

2p+1

= 1 s^p



^p

k=0

C^k_p|∇ϕ|^k|∇u|^p−k2

2s^k





2p+1

. (1.7) Then, there exists a uniqueα=α(∇ϕ,∇u)∈(0, +∞) such that

Ψ(α)= inf

s∈(0,+∞)Ψ(s) (1.8)

satisfying

p k=0

(2p+ 1)k−pC^k_p|∇ϕ|^k|∇u|^p−k2

2α^k=0. (1.9)

Finally, let

Cp= sup

∇u≡0

∆˜^p/2ϕ²₂

Ψ^1/(2p)(α). (1.10)

Our main result is the following theorem.

Theorem1.1. There exists

C_p= 1

(2p+ 1)(2p)^(2p+1)/2σ_2p+1^1/(2p). (1.11) Moreover, the best constantCpis achieved by a family of one parameter of functionsϕ¯andu¯ given by

¯

ϕ(x)= 2

(2p)!(1 +cr²), u¯= 2^√cx

1 +cr², (1.12)

wherec >0is some arbitrary positive constant.

We can give for example more explicit expression of the best constant in the case where p=2. Letu∈W^1,4(R⁴,R⁴) andξis the solution of

∆²ξ=det∇u inR⁴,

|xlim|→+∞ξ(x)=0. (1.13)

We get that

Ψ(α)=

5⁵∇u¹²4

5|∇ξ||∇u|²

2+9|∇ξ||∇u|⁴

2+ 16∇ξ⁴4∇u⁴4

1/25

8⁴

3|∇ξ||∇u|²

2+9|∇ξ||∇u|⁴

2+ 16∇ξ⁴4∇u⁴4

1/23 .

(1.14) Corollary1.2. Letξbe a solution of (1.13), then

sup

∇u≡0

∆ξ²2

3|∇ξ||∇u|²

2+9|∇ξ||∇u|⁴

2+ 16∇ξ⁴4∇u⁴4

1/23/4

∇u³4

5|∇ξ||∇u|²

2+9|∇ξ||∇u|⁴

2+ 16∇ξ⁴4∇u⁴4

1/25/4

= 1 2⁸

15 8π²

1/4

,

(1.15)

and the supremum is achieved byξ¯andu¯given by ξ(x)¯ = 1

121 +cr², u(x)¯ = 2^√cx

1 +cr², (1.16)

wherecis some arbitrary positive constant.

2. Proof of results

First, we introduce some notations which we will use later. LetΩbe a bounded subset ofRⁿand letW:Ω→Rⁿ⁺¹be a regular function. DenoteW=(w¹,w²,. . .,wⁿ,wⁿ⁺¹) and Wi=(w¹,. . .,wⁱ⁻¹,wⁱ⁺¹,. . .,wⁿ,wⁿ⁺¹), fori=1,. . .,n+ 1. LetV be the algebric volume of the image ofWinRⁿ⁺¹and denote byAthe volume of the boundary ofV. Then, we have

V= 1

n+ 1 ΩW·Wx1×Wx2× ··· ×Wxn, (2.1) A=

Ω

Wx1×Wx2× ··· ×Wxn, (2.2)

whereWx1×Wx2× ··· ×Wxnis some vector ofRⁿ⁺¹given by

Wx1×Wx2× ··· ×Wxn=

e1 w¹_x1 ··· w_x¹_n e2 w²_x1 ··· w_x²_n

· · ··· ·

· · ··· ·

· · ··· · e_n+1 wⁿ⁺¹_x1 ··· w_xⁿ⁺¹_n

=

n+1

i=1

(−1)ⁱ⁻¹det(∇Wi)ei.

(2.3) Here (ei)1_≤i≤n+1is the canonic base ofRⁿ⁺¹. We need the following Lemma.

Lemma2.1. LetW:Ω→Rⁿ⁺¹defined as above. Suppose that there exist1≤i0≤nsuch thatwⁱ⁰=0on∂Ω, then

Ωwⁱdet∇W_i=(−1)ⁿ

Ωw^jdet∇W_j, (2.4) for1≤i < j≤n.

2.1. Proof ofTheorem 1.1. We will suppose thatu∈C^∞(R^2p,R^2p)∩W^1,2p(R^2p,R^2p).

The general case can be obtained by approximatinguby regular functions. Then we de- fineWinR^2p+1as follows:

W(x)=

u(x),tϕ(x), (2.5)

wheretis a reel parameter which will be chosen later. Using (2.4) the algebric volume closed by the image ofWinR^2p+1is

V=

R^2pw^2p+1det∇W2p+1

dx=t

R^2pϕdet∇udx=t

R^2pϕ(−∆)^pϕdx. (2.6) Then we have

V=t∆˜^p/2ϕ²₂. (2.7)

Next, we will estimateA. We have by (2.2) A≤

R^2p

Wx1Wx2···Wx2pdx=

R^2p

2p i=1

uxi²+t²ϕ²_xi^1/2. (2.8)

As (ⁿ_i=1αi)^1/n≤1/nⁿ_i=1αi, we have A≤ 1

(2p)^p R^2p

_2p

i=1

uxi²+t²ϕ²_xi

p

= 1 (2p)^p R^2p

|∇u|²+t²|∇ϕ|²p

. (2.9) Recall the isoperimetric inequality on a domainsΩofR^2p+1. Denote byV=Vol(Ω) andA=Vol(∂Ω), respectively, the volume ofΩand∂Ω, then

(2p+ 1)^2pσ_2p+1V^2p≤A^2p+1. (2.10) By (2.7) and (2.9), we have

(2p+ 1)^2pσ_2p+1t^2p∆˜^p/2ϕ^4p₂ ≤ 1

(2p)^p(2p+1) R^2p

|∇u|²+t²|∇ϕ|²p2p+1

. (2.11) We conclude that

∆˜^p/2ϕ²₂≤ 1

(2p+ 1)(2p)^(2p+1)/2σ_2p+1^1/2pΨ(t²)^1/2p. (2.12) Then we obtain

Cp≤ 1

(2p+ 1)(2p)^(2p+1)/2σ_2p+1^1/(2p). (2.13) Next, we will show thatC_pis achieved. We will consider a special case

u(x)=g|x|

x, (2.14)

whereg:R+→Ris a regular function which will be chosen later. Since det∇u= 1

2pr^2p−1 d dr

r^2pg^2p(r), (2.15)

then, the solutionϕof (1.4) is a radial function. Letχa general radial function onR^2p andW(x)=(g(|x|)x,tχ(|x|)). After a computation, we can show easily that in this case

Wx1×Wx2× ··· ×Wx2p²=g^4p−2(r)g²(r) + 2rg(r)g(r) +r²g²(r) +t²χ²(r) (2.16)

and for 1≤i≤2p,

W_xi²₌g²(r) +2rg(r)g(r) +r²g²(r) +t²χ²(r)x_i²

r². (2.17)

Next, we will suppose thatχandgsatisfy

2rg(r)g(r) +r²g²(r) +t²χ²(r)=0. (2.18) If we choseχas the solutionϕof (1.4) whenu=g(|x|)x, then by (2.16), (2.17) and under the hypothesis (2.18), the inequality (2.9) becomes an equality. Let now

¯

u(x)=g(¯ |x|)x with ¯g(r)= 2^√c

1 +cr², (2.19)

wherec >0 is some positive constant. Then the solution ¯ϕof (1.4) is given by

¯

ϕ(x)= 1 (2p)!

2

1 +cr². (2.20)

Indeed, the expression of∆^kϕ, for 1≤k≤pis

∆^kϕ(r)¯ = 2^2k+1(−1)^kk!c^k (2p)!1 +cr^22k+1

×



^k−1

l=0

(p+l) +

k−1 l=0

(p−2−l)c^kr^2k+

k−1 j=1

C_k^j

k−1 l=j

(p+l)

k−1

q=k−j

(p−2−q)c^jr^2j



. (2.21) Remark that all the coefficients ofr^2jfor 2≤j≤kin the expression of∆^kϕ¯have the term (p−k). Also, since

det∇u¯= 1 2pr^2p−1

d dr

r^2pg¯^2p(r)=2^2pc^p 1−cr²

1 +cr^22p+1, (2.22) so, we have

(−∆)^pϕ¯=det∇u¯ onR^2p. (2.23) If we choose ¯t=(2p)! and ¯χ(r)=ϕ(r)¯ −1/(2p)!, we remark that ¯t, ¯χand ¯gsatisfy (2.18).

Since ¯W₌( ¯u, ¯tχ) :¯ _R^2p_→S^2pand that the isoperimetric inequality (2.10) becomes equal- ity, then we have

∆˜^p/2ϕ¯²₂ Ψt¯^21/(2p)=

1

(2p+ 1)(2p)^(2p+1)/2σ_2p+1^1/(2p). (2.24) We conclude that ¯α=α(∇ϕ,¯ ∇u) defined by (1.8) in this case is just ¯¯ α=

(2p)!².

2.2. Proof ofCorollary 1.2. Following step by step the proof ofTheorem 1.1, we have A=

R⁴

W_x1×W_x2···W_x4≤ 1 16

t⁴∇ξ⁴4+ 2t²|∇ξ||∇u|²

2+∇u⁴4

. (2.25)

Choosing

t²=α= 2∇u⁴4

3|∇ξ||∇u|²

2+9|∇ξ||∇u|⁴

2+ 16∇ξ⁴4∇u⁴4

1/2, (2.26)

and using the fact that

4∇ξ⁴4α²+ 3|∇ξ||∇u|²

2α− ∇u⁴4=0, (2.27) we have

Ψ(α)=

5⁵∇u¹²4

5|∇ξ||∇u|²

2+9|∇ξ||∇u|⁴

2+ 16∇ξ⁴4∇u⁴4

1/25

8⁴

3|∇ξ||∇u|²

2+9|∇ξ||∇u|⁴

2+ 16∇ξ⁴4∇u⁴4

1/23 , (2.28) and then

sup

∇u≡0

∆ξ²2

3|∇ξ||∇u|²

2+9|∇ξ||∇u|⁴

2+ 16∇ξ⁴4∇u⁴4

1/23/4

∇u³4

5|∇ξ||∇u|²

2+9|∇ξ||∇u|⁴

2+ 16∇ξ⁴4∇u⁴4

1/25/4≤ 1 2⁸

15 8π²

1/4

.

(2.29) By taking

ξ(x)¯ = 1

121 +cr², u(x)¯ = 2^√cx

1 +cr², (2.30)

we find

∇u¯⁴4=2⁶×3×π²

7 ,

∆ξ¯²2= π²

3²×5, ∇ξ¯⁴4= π²

2⁶×3⁴×5×7, |∇ξ¯||∇u¯|²

2= 11π² 3³×5×7.

(2.31) Finally (1.15) follows.

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Sami Baraket: D´epartment de Math´ematiques, Facult´e des Sciences de Tunis, Campus Universi- taire, 2092 Tunis, Tunisie

E-mail address:[email protected]

Makkia Dammak: D´epartment de Math´ematiques, Facult´e des Sciences de Tunis, Campus Univer- sitaire, 2092 Tunis, Tunisie

E-mail address:[email protected]