C. R. Acad. Sci. Paris, t. 330, Série I, p. 785–790, 2000Équations aux dérivées partielles/Partial Differential Equations

On the asymptotic behavior of solutionsto a semilinear elliptic boundary problemin an unbounded domainYouri V. EGOROV a, Vladimir A. KONDRATIEV b

a Laboratoire des mathématiques pour l’industrie et la physique, UMR5640, Université Paul-Sabatier,UFR MIG, 118, route de Narbonne, 31062, Toulouse cedex 4, France

b Mehmat Faculty, Lomonossov University, Vorob’evy Gory, 119899 Moscow, Russia

(Reçu le 28 juin 1999, accepté après révision le 16 mars 2000)

Abstract. We consider solutions to the elliptic linear equation (1) of second order in an unboundeddomainQ in Rn supposing thatQ= x= (x′, xn) : 0< xn <∞, |x′|< γ(xn), where1 6 γ(t)6At+B, and thatu satisfy the boundary condition (2). We show that any suchsolutionu growing moderately at infinity is bounded and tending to0 asxn→∞. Earlierwe showed in our notes [3,4] this theorem for the caseγ(xn) = B, i.e., for a cylindricaldomainQ= Ω× (0,∞), Ω⊂ Rn−1. 2000 Académie des sciences/Éditions scientifiqueset médicales Elsevier SAS

Sur le comportement asymptotique des solutions d’un problème au bordelliptique semi-linéaire dans un domaine non borné

Résumé. On considère des solutions de l’équation elliptique linéaire(1) dans un domaine nonbornéQ de Rn en supposant queQ = x = (x′, xn) : 0 < xn < ∞, |x′| < γ(xn),où 1 6 γ(t) 6 At + B, et queu vérifie les conditions au bord(2). Nous montrons queles solutionsu qui ne croissent pas trop vite à l’infini sont bornées et tendent vers0 sixn→∞. Dans nos notes[3,4] nous avons consideré le casγ(xn) = B, i.e. le cas d’undomaine cylindriqueQ= Ω× (0,∞), Ω⊂ Rn−1. 2000 Académie des sciences/Éditionsscientifiques et médicales Elsevier SAS

Version française abrégée

Nous considérons des solutions de l’équation elliptique linéaire :

Lu :=

n∑i,j=1

∂

∂xi

(aij(x)

∂u

∂xj

)+ c(x)u= 0 (1)

Note présentée par Pierre-Louis LIONS.

S0764-4442(00)00264-0/FLA 2000 Académie des sciences/Éditions scientifiques et médicales Elsevier SAS. Tous droits réservés. 785

Y.V. Egorov, V.A. Kondratiev

dans un domaine non bornéQ deRn en supposant que

Q=x=

(x′, xn

): 0< xn <∞,

∣∣x′∣∣< γ(xn),

où16 γ(t)6At+B, et queu vérifie la condition au bord :

∂u

∂N+ b(x)

∣∣u(x)∣∣p−1

u(x) = 0 (2)

sur la surfaceS =x ∈Rn : |x′|= γ(xn), 0< xn <∞

, oùp > 0, b(x)> b0 > 0, γ ∈C1(0,∞) et

∂u

∂N=

n∑i,j=1

aij(x)∂u

∂xjcosθi,

θi est l’angle entre l’axexi and le vecteur normal extérieur. Nous supposons que

n∑i, j=1

aij(x)ξiξj > c0|ξ|2, c0 > 0, x∈Q,

et que|anj(x)|6C pourj = 1, . . . , n, 06 c(x) en tout pointx ∈Q.Nous montrons que toute fonctionu satisfaisant (1) et (2) qui ne croît pas trop vite à l’infini est bornée

et tend vers0 si xn →∞. Dans nos notes [3,4] nous avons consideré le casγ(xn) = B, i.e. le cas d’undomaine cylindriqueQ= Ω× (0,∞), Ω⊂Rn−1.

Nous considérons aussi les solutions de l’équation (1) satisfaisant la condition au bord :

∂u

∂N− b(x)

∣∣u(x)∣∣p−1

u(x) = 0.

1. We study the solutions to the elliptic second order linear equation:

Lu :=

n∑i,j=1

∂

∂xi

(aij(x)

∂u

∂xj

)+ c(x)u= 0 (1)

in an unbounded domainQ in Rn supposing

Q=x=

(x′, xn

):∣∣x′∣∣< γ(xn), 0<xn <∞

,

where16 γ(t)6At+B, and thatu satisfies the boundary condition:

∂u

∂N+ b(x)

∣∣u(x)∣∣p−1

u(x) = 0 (2)

on the lateral surfaceS =x ∈ Rn : |x′| = γ(xn), 0 < xn < ∞

, wherep > 0, b(x) > b0 > 0, γ ∈

C1(0,∞) and

∂u

∂N=

n∑i, j=1

aij(x)∂u

∂xjcosθi,

θi is the angle between the axisxi and the outer normal vector.Suppose that

n∑i,j=1

aij(x)ξiξj > c0|ξ|2, c0 > 0, x ∈Q,

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Semilinear elliptic boundary problems in unbounded domains

and that|anj(x)|6C for j = 1, . . . , n and06 c(x)6C for all x ∈Q. We don’t assume thataij andc arecontinuous.

Let us denoteΩT andΣT the sections of the domainQ and the boundaryS by the planexn = T , andQTandST the parts ofQ andS between the planesxn = 1 andxn = T .

We consider functionsu satisfying (1) and (2) weakly. It means thatu ∈H1loc(Q)∩ Lp+1

loc (S) and∫Q

[n∑

i,j=1

aij(x)∂u

∂xj

∂ϕ

∂xi+ c(x)uϕ

]dx+

∫S

b(x)∣∣u(x)

∣∣p−1u(x)ϕ(x) dS = 0 (3)

for all functionsϕ(x) ∈H1(Q), equal to0 asxn = 0 and in a neighborhood ofxn =∞.We will show that any solution of our problem growing moderately at infinity is bounded and tending

to 0 asxn→∞.In our notes [3,4] we showed such a theorem for the caseγ(xn) = B, i.e., for a cylindrical domain

Q= Ω× (0,∞), Ω⊂Rn−1.

2. Consider firstly conical domains corresponding to the functionγ(xn) =Axn +B.

THEOREM 1. –Let γ(xn) = Axn + B, p > 1. There exist constantsa0, A0 > 0 such that a functionu, satisfying(1) and (2) and the inequality|u(x)| 6 bxan in the domainQ = x ∈ Rn : |x′| < γ(xn),0< xn <∞, with some constantsb > 0, 0< a< a0, 0<A<A0, tends to0 asxn→∞ uniformly inQ.

Proof. –Let h(xn) be a smooth function such thath(xn) = 1 as1< xn < T , h(xn) = 0 for xn > 3T/2and forxn < 1/2. We can assume that|h′(xn)|6C/T and|h′′(xn)|6C/T 2 asxn > T . Set

J(T ) =

∫Q

[n∑

i, j=1

aij(x)∂u

∂xj

∂u

∂xi+ c(x)u2

]hx2−n

n dx+

∫S

hb(x)∣∣u(x)

∣∣p+1x2−nn dS.

Substituting in (3) the functionϕ(x) = h(xn)x2−nn u(x), we obtain that

J(T ) =−∫Q

n∑i,j=1

aij(x)∂u

∂xj

∂(x2−nn h(xn))

∂xiu(x) dx6C1 +C2J(T )1/2

(∫Q3T/2

|x|−nu(x)2 dx

)1/2

.

Therefore, if|u(x)|6 bxan, thenJ(T )6C3T2a.

Set now

I(T ) =

∫QT

[n∑

i,j=1

aij(x)∂u

∂xj

∂u

∂xi+ c(x)u2

]x2−nn dx+

∫ST

b(x)∣∣u(x)

∣∣p+1x2−nn dS.

If |u(x)| 6 bxan, thenI(T ) 6 J(T ) 6 C4T2a. Substituting in (3) the functionϕ(x) = u(x)x2−n

n hε(xn),wherehε(xn) is a smooth function, which is equal to1 asxn < T − ε and0 for xn > T , and passing to thelimit as ε→ 0, we obtain that

I(T ) = (n− 2)

∫QT

n∑i=1

ani(x)u∂u

∂xix1−nn dx+

∫ΩT

n∑i=1

ani(x)∂u

∂xiux2−n

n dx′ + c1,

where

c1 =−∫

Ω1

n∑i=1

ani(x)∂u

∂xiudx′.

We have

787

Y.V. Egorov, V.A. Kondratiev

(n− 2)

∫QT

n∑i=1

ani(x)u∂u

∂xix1−nn dx6C5

(∫QT

u2x−nn dx

)1/2(∫QT

|∇u|2x2−nn dx

)1/2

;

∫QT

u(x)2x−nn dx6C6

(A2

0

∫QT

∣∣∇u(x)∣∣2x2−n

n dx+

∫ST

u(x)2x1−nn dS

)6C6

[A2

0

∫QT

∣∣∇u(x)∣∣2x2−n

n dx+

∫ST

(A2

0

∣∣u(x)∣∣p+1

+A−4/(p−1)0 x−1−2/(p−1)

n

)x2−nn dS

]6C6A

20I(T ) +C7T

−2/(p−1),

sinceu(x)2 6A20xn|u(x)|p+1 +A

−4/(p−1)0 x

−2/(p−1)n asxn > 0. Using the Sobolev inequality we obtain∫

ΩT

u(x)2x2−nn dx′ 6C8

[A2

0T2

∫ΩT

∣∣∇u(x)∣∣2x2−n

n dx′ + T

∫ΣT

u(x)2x2−nn dS

]6C9

[A2

0T2

∫ΩT

∣∣∇u(x)∣∣2x2−n

n dx′ + T

∫ΣT

(T∣∣u(x)

∣∣p+1+ T−2/(p−1)

)x2−nn dS

],

sinceu(x)2 6 T |u(x)|p+1 + T−2/(p−1) asT > 0. If A0 is so small thatC6A20 6 1/2, then∫

ΩT

u(x)2x2−nn dx′ 6C10

[T 2I ′(T ) + T (p−3)/(p−1)

],

and therefore ∫ΩT

n∑i=1

ani(x)∂u

∂xiux2−n

n dx′ 6C11TI′(T ) +C12.

Thus

I(T )6C11TI′(T ) +C13.

Integrating this inequality we see that eitherI(T )6 C13, or I(T )> C13 +C14T1/C11 . Since the latter is

impossible when2aC11 < 1, we obtain thatI(T )6C13, i.e.,∫Q

|∇u|2 dx+

∫S

|u|p+1 dx <∞.

Now we can use the methods of [1] to prove our statement.

COROLLARY 1. –Two solutions of the equationLu= 0, satisfying boundary condition(2) on the lateralsurface of the domainQ and the estimate|u(x)|6 bxan with a constanta depending onQ, coincide inQ,if they coincide atxn = 0.

Example1. – SettingQ = (x1, x2) : 0 < x1 < x2 < ∞ and u(x1, x2) = r4 cos4ϕ + r16 sin16ϕ,wherer, ϕ are polar coordinates, then we obtain an example of a harmonic function of power growthin a cone. It is easy to see that∂u/∂ν + 16u|u|3 = 0 if ϕ= 0 or ϕ= π/4.

3. Let nowQ= Ω× (0,∞), whereΩ is a domain inRn−1, i.e.,Q is a cylinder inRn.

THEOREM 2. –Letp> 1. Suppose that the coefficientsc(x), aij(x) for i, j = 1, . . . , n−1 do not dependonxn. Letλ2 be the first eigenvalue of the Dirichlet problem inΩ, i.e.,

λ2 = infw∈C∞0 (Ω)

∫Ω

[∑n−1i, j=1 aij(x)∂w(x)/∂xj ∂w(x)/∂xi + c(x)w(x)2

]dx∫

Ωw(x)2 dx

.

788

Semilinear elliptic boundary problems in unbounded domains

Letγ1 > 0, γ2 > 0 be such constants that

γ21

∣∣∣∣∣n∑j=1

anj(x)ξj

∣∣∣∣∣2

6n∑

i,j=1

aij(x)ξiξj , γ22

n−1∑i,j=1

aij(x)ξiξj 6n∑

i,j=1

aij(x)ξiξj ,

for all ξj ∈ R. Setρ = γ1γ2. For any ε ∈ ]0, ε0[ there exists a positive constanta such that if|u(x)| 6a eρ(λ−ε)xn/2, thenu(x)→ 0 asxn→∞ uniformly inQ.

Remark1. – If the coefficientsaij(x), c(x) are depending onxn, then Theorem 1 is true in the followingsense: “there exist positive constantsa, b such that if a solution to the problem (1), (2) satisfies the inequality|u(x)|6 a ebxn , thenu(x)→ 0 asxn→∞”.

COROLLARY 2. –Two solutions to the equationLu = 0, satisfying the boundary condition(2) on thelateral surface of the cylinder and the estimate|u(x)|6 a eρ(λ−ε)xn/2 with a constanta, depending onΩ,coincide inQ, if they coincide asxn = 0.

Example2. – The following example shows that our hypothesis on the growth of solutions is essential.Let n= 2 andQ= (x1, x2) :−2π < x1 < 2π, 0< x2 <∞, u(x1, x2) = ex2 sinx1 − ex2/4 sin(x1/4). Itis easy to see that the functionu is harmonic inQ and∂u/∂ν + u|u|3 = 0 for x1 =±2π.

4. Let now|x′|6 γ(xn), 0< xn <∞, x= (x′, xn) ∈Q. LetF (t) =∫ t

0dsγ(s) .

Suppose thatγ(s) = o(s) ass→∞, γ(s)→∞ ass→∞, andγ(T + s)6 Cγ(T ) if 0< s6C1γ(T ).Note thatF (t)→∞ ast→∞.

THEOREM 3. – If u is a solution of equation(1) in Q, satisfying(2), and|u(x)|6 b eaF (xn) in Q with asmall enough constanta, thenu(x)→ 0 asxn→∞ uniformly inQ.

5. Let us consider now the case0< p< 1.

THEOREM 4. –Let 0 < p < 1 and0 < a < 1/(1− p). Let γ(xn) 6 Axn + B andA < A0 with smallenoughA0. If |u(x)|6 bxan with someb, thenu(x)→ 0 asxn→∞ uniformly inQ.

6. We consider now solutions to the equation

Lu :=

n∑i,j=1

∂

∂xi

(aij(x)

∂u

∂xj

)= 0 (4)

in the domainQ = x= (x′, xn) : |x′| < Axσn +B, 0 < xn <∞, 0 6 σ 6 1, in Rn. We study the weaksolutions of (4) satisfying the nonlinear boundary condition

∂u

∂N− b(x)

∣∣u(x)∣∣p−1

u(x) = 0 (5)

on the lateral surfaceS wherep > 0, b(x)> b0 > 0.We show that a global solution of the problem cannot exist for all values of parametersp, σ and indicate

these values.

THEOREM 5. –Letn> 3, Q = x= (x′, xn) : |x′|<Axσn +B, 1 < xn <∞, 1n−1 < σ 6 1. Suppose

thatv(x) satisfies(4), (5)andv(x)> 0 in Q. If 1< p6 1 + 2−σσ(n−2) , thenv(x)≡ 0.

THEOREM 6. –LetQ= x= (x′, xn) : |x′|<Axσn +B, 1< xn <∞, 06 σ(n−1)< 1. Suppose thatv(x) satisfies(4), (5)andv(x)> 0 in Q. If p > 1, thenv(x)≡ 0.

789

Y.V. Egorov, V.A. Kondratiev

The proofs of Theorems 5 and 6 are based on the following lemmas.

LEMMA 1. –Let σ(n − 1) > 1, σ 6 1, n > 3, Q = x = (x′, xn) : |x′| < Axσn + B, 1 < xn <∞,A> 0, B > 0. There exists a weak solution to the following problem:

LE = 0 in Q,∂E

∂N= 0 onS, E = 1 for xn = 1,

from the classH1loc(Q), such thatE(x)> cxσ(2−n)

n , c > 0. Moreover,

limxn→∞

E(x) = 0,

∣∣∣∣∣∫

Ω1

n∑j=1

anj(x)∂E

∂xjdx′

∣∣∣∣∣= c0 6= 0.

LEMMA 2. –Let 06 σ(n− 1)< 1, n > 2, Q= x= (x′, xn) : |x′|<Axσn +B, 1< xn <∞, A> 0,B > 0. There exists a weak solutionE of the problem:

LE = 0 in Q,∂E

∂N= 0 onS, E = 0 at xn = 1,

from the classH1loc(Q), satisfying the estimateE(x)> c > 0 for xn > 2.

THEOREM 7. –LetQ= x= (x′, xn) : |x′|<Axσn +B, 1< xn <∞, 06 σ(n−1)6 1. Suppose thatv(x) satisfies(4) andv(x)> 0 in Q. Suppose that

n∑j=1

anj(x)∂v(x)

∂xj6 0 asxn = 1,

and∂v(x)/∂N > 0 onS. Thenv(x)≡ 0.

Proof. –Let ε be a small positive number. Put in the definition of weak solutionsϕ(x) = h(xn)/(v(x) +ε), whereh(xn) = 1 for 1< xn < T , h(xn) = 0 for 2T < xn, h is a smooth function forxn > 1. We have

J(T )≡∫QT

h(xn)1

(v + ε)2

n∑i, j=1

aij(x)∂v(x)

∂xj

∂v(x)

∂xidx

6∫QT

h′(xn)1

v+ ε

n∑j=1

anj(x)∂v(x)

∂xjdx6C1

(∫QT

h′(xn)2 dx

)1/2

J(T )1/2,

and therefore,

J(T )6C∫QT

T−2 dx6C1T−2+1+σ(n−1).

We see tendingT →∞ thatv(x)≡ 0 if σ(n− 1)< 1. If σ(n− 1) = 1 the proof is slightly modified.

References

[1] Stampacchia G., Le problème de Dirichlet pour les équations elliptiques du second ordre à coefficients discontinus,Ann. Inst. Fourier Grenoble 15 (1965) 189–258.

[2] Hu B., Nonexistence of a positive solution of the Laplace operator with a nonlinear boundary condition, Differ. andInteg. Eq. 7 (1994) 301–313.

[3] Egorov Yu.V., Kondratiev V.A., On a Oleinik’s problem, Uspekhi Mat. Nauk 57 (1997) 159–160.[4] Egorov Yu.V., Kondratiev V.A., On asymptotic behavior in an infinite cylinder of solutions to an elliptic solution of

second order, Appl. Anal. 71 (1999) 25–41.

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