The Lane-Emden Function and Nonlinear Eigenvalues Problems

ANNALES

DE LA FACULTÉ DES SCIENCES

Mathématiques

OULDAHMEDIZIDBIHISSELKOU

The Lane-Emden Function and Nonlinear Eigenvalues Problems

Tome XVIII, n^o4 (2009), p. 635-650.

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Annales de la Facult´e des Sciences de Toulouse Vol. XVIII, n^◦4, 2009 pp. 635–650

The Lane-Emden Function and Nonlinear Eigenvalues Problems

Isselkou Ould Ahmed Izid Bih⁽¹⁾

R^´ESUM´E.—Nous consid´erons un probl`eme aux valeurs propres, semi- lin´eaire elliptique, sur une boule deR<sup>n</sup> et montrons que ces valeurs et fonctions propres peuvent s’obtenir `a partir de la fonction de Lane-Emden.

ABSTRACT.—We consider a semilinear elliptic eigenvalues problem on a ball ofRⁿand show that all the eigenfunctions and eigenvalues, can be obtained from the Lane-Emden function.

1. Introduction We consider the problem

(P_λ^α)





∆u+λ(1 +u)^α= 0, in B1

u >0, in B1

u= 0, on ∂B1

whereB₁is the unit ball of Rⁿ,n3,λ >0 andα >1.

This problem arises in many physical models like the nonlinear heat generation and the theory of gravitational equilibrium of polytropic stars(cf.

[2] and [11]). It is well known (cf. [2], [10], [12]) that there exists a criti- cal constant λ^∗(α), such that (P_λ^α) admits, at least, one solution if 0 <

λ < λ^∗(α) and no solution if λ > λ^∗(α). We deal here with these critical constants and the corresponding eigenfunctions.

(∗) Re¸cu le 31/07/07, accept´e le 20/01/08

(1) Facult´e des Sciences et Techniques, B.P. 5026 Nouakchott, Mauritanie.

isselkou@univ-nkc.mr

Letφbe the Lane-Emden function(cf. [1], [5], [6],[15]) in then-dimensional space andr₀ the first ”zero” ofφ, we show that

λ^∗(α) = max

r∈[0,r0[r²φ^α−1(r).

We use this formula to compute λ^∗(α), when α is the Critical Sobolev Exponent. We also extend, to the subcritical case, an estimate of λ^∗(α) given in [10] and show qualitative properties of the eigenfunctions.

In the Appendix, we show how to approximateφ, so one can use numerical approaches (Maple or Matlab) to get estimates ofλ^∗(α).

2. Scalings of the Lane-Emden function as solutions When 0< λλ^∗(α), it is known that any regular solution of (P_λ^α) is radial and the minimal one is stable and analytical (cf.[8], [12]).

Proposition 2.1. — Letube a regular solution of(P_λ^α), then u(r) = (1 +u(0))φ

√

λ(1 +u(0))^α−12 r

−1, ∀r∈[0,1]

whereφ is the Lane-Emden function, in then-dimensional space.

Proof. — The Lane-Emden function(cf. [1], [5], [6], [15]) is the solution of

(L−E)

φ”(r) +ⁿ⁻¹_r φ(r) +φ(r)|φ(r)|^α−1= 0, φ(0) = 1, φ(0) = 0.

The proof of the proposition is quite immediate.

3. The Subcritical Case

Let us consider the problem (P_λ^α), with 1 < α < ⁿ⁺²_n−2. Let φ be the Lane-Emden function.

Proposition 3.1. — There existsr₀>0, such thatφ(r₀) = 0, φ(r)>0,

∀r∈[0, r0[and

λ^∗(α) = max

ρ∈[0,r0]ρ²φ^α−1(ρ).

We also have

λ^∗(α) 2

(α−1)²(α(n−2)−n), if n

n−2 < α < n+ 2 n−2.

Proof. — Asφ(0)>0,we infer thatφ >0,on a maximal interval [0, r₀[.

The problem

∆u+u^α= 0, inRⁿ u >0, inRⁿ

does not admit a solution (cf.[4]), so we infer thatr₀<∞andφ(r₀) = 0.

Let us put

ψρ(r) =φ(ρr)−φ(ρ)

φ(ρ) , ∀r∈[0,1],

with 0 < ρ < r₀, then ψ_ρ is a solution of (P_λ^α), with λ= ρ²φ^α−1(ρ). We infer that

ρ∈[0,rmax0]ρ²φ^α−1(ρ)λ^∗(α).

Let us suppose that

ρ∈[0,rmax0]ρ²φ^α−1(ρ)< λ^∗(α),

ifu_λ∗(α)is the unique solution of (P_λ^α∗(α))(cf.[10]),one can use Proposition 1 to show that

u_λ∗(α)(r) =

1 +u_λ∗(α)(0) φ

(λ^∗(α))¹²

1 +u_λ∗(α)(0)

α−1 2 r

− 1

1 +u_λ∗(α)(0)

.

Let us put ρλ^∗(α) = (λ^∗(α))¹²

1 +uλ^∗(α)(0)

α−1

2 . As uλ^∗(α) 0, we infer that ρ_λ∗(α)< r₀.Asu_λ∗(α)(1) = 0, we infer that

1

1 +u_λ∗(α)(0) =φ

(λ^∗(α))¹²

1 +u_λ∗(α)(0)

α−1 2

. So we get

u_λ∗(α)(r) =φ(ρ_λ∗(α)r)−φ(ρ_λ∗(α))

φ(ρ_λ∗(α)) and λ^∗(α) = ρ_λ∗(α)

2φ^α−1(ρ_λ∗(α)).

The last equality leads to a contradiction.

To prove the last statement, we use the fact that the maximum here is achieved at a uniquer_α (see the next lemma). So we get

φ(r_α) =− 2

(α−1)r_αφ(r_α), and

φ^α−3(rα)

2φ²(rα) + 4rα(α−1)φ(rα)φ(rα) + (α−1)r²_α

(α−2)

φ(rα) ²+φ(rα)φ(rα)

^0.

We first replace φ(r_α) by its value from (L−E) and then φ(r_α), from the previous equality, to get

φ^α−1(rα)

−(α−1)λ^∗(α) + 2(n−4) + 4α−2 α−1

0.

Simplifying, one gets the estimate.

Remark 3.2. — The last statement in Proposition 2 is also true forα

n+2n−2, with the same proof, provided that sup_r∈R+r²φ^α−1(r) is attained (see the next Proposition 6); this has been proved in [10], using sophisticated arguments.

Lemma 3.3. — Let us put g(r) =r²φ^α−1(r), r∈[0, r0],there exists ρ0∈]0, r0[ such thatg is increasing on[0, ρ0]and decreasing on [ρ0, r0].

Proof. — Let ρbe an arbitrary positive constant with ρ < r₀, then, as we have already mentionedψ_ρ is a solution of

P_γ^α ,whereγ=g(ρ).As g(r) =rφ^α−2(r) (2φ(r) + (α−1)rφ(r)),we infer thatg is increasing on a maximal intervalI0⊂[0, r0] with 0∈I0.

Using Proposition 2, there exists ρ0 ∈]0, r0[, such that g(ρ0) = max_r∈[0,r0]g(r) = λ^∗(α). This ρ0 is unique, otherwise, if there exists λ ∈ [0, r0], such that g(λ) = max_r∈[0,r0]g(r) = λ^∗(α), then ψρ0 and ψλ

are both solutions of the problem

P_λ^α∗(α)

. As φ is decreasing on [0, r0], we infer that ψρ0(0) = ^1−φ(ρ_φ(ρ ⁰⁾

0) = ^1−φ(λ)_φ(λ) =ψλ(0). So we get two different solutions of the problem

P_λ^α_∗(α)

.This leads to a contradiction (cf. [10]).

Asg(r0) = 0, we infer thatI0= [0, r0].Let us putδ= supI0. The function g can’t be constant on a nontrivial intervalJ ⊂[δ, r0], for if g(r) =c in J, then for every λ ∈ J, ψλ is a solution of (P_c^α). As ψλ1(0) = ψλ2(0), if λ1, λ2 ∈J andλ1=λ2,we infer that the problem (P_c^α) admits an infinity of solutions. This leads again to a contradiction (cf. [10]).

So ifg is not decreasing on [δ, r₀], then there existsβ₁ andβ₂ with r₀ > β₂ > β₁ > δ, such that g is decreasing on [δ, β₁] and increasing on [β1, β2].Let us putc0=min(g(δ), g(β2)), thenc0> g(β1).Let us choosec∈ ]g(β1), c0[, so the problem g(t) =cadmits at least three different solutions λi∈]0, β2[, 1i3.Asψλi(0)=ψλj(0), if i=j, 1i, j 3,we obtain three solutions for the problem (P_c^α).So we get a contradiction.

We conclude thatgis increasing on [0, δ], decreasing on [δ, r0] andδ=ρ0. Proposition 3.4. — Ifλ=λ^∗(α), there exists a uniqueρ_λ∗(α)∈]0, r0[, such that

λ^∗(α) = ρ_λ∗(α)

2φ^α−1(ρ_λ∗(α)) and the unique solution u_λ∗(α) of (P_λ^α∗(α)) is

u_λ∗(α)(r) =φ(ρ_λ∗(α)r)−φ(ρ_λ∗(α))

φ(ρ_λ∗(α)) =ψρλ∗(α)(r), ∀ r∈[0,1].

When0< λ < λ^∗(α), there exist exactly two constantsrλ andρλ, such that 0< rλ< ρ_λ∗(α)< ρλ< r0,λ=r²_λφ^α−1(rλ) =ρ²_λφ^α−1(ρλ)and the only two solutions of(P_λ^α) are

uλ=ψrλ, vλ=ψρλ; the minimal one(cf.[2]) isu_λ,lim_λ→0u_λ= 0 in C⁰

B₁ and lim_λ→0v_λ(r) =∞, ∀r∈[0,1[.

Proof. — Using Proposition 2 and Lemma 1, one infers that the only solution of (P_λ^α_∗(α)) isψ_ρ0.We putρ_λ∗(α)=ρ₀.If 0< λ < λ^∗(α), using the lemma again, we infer that g(t) = λ admits exactly two solutions rλ and ρλ, with 0< rλ < ρλ^∗(α) < ρλ < r0.Let us put uλ =ψrλ and vλ =ψρλ, u_λ(0) = v_λ(0). These two functions u_λ and v_λ are solutions of the the problem (P_λ^α), which admits only two ones (cf. [10]).

As φis decreasing on [0, r0], one can verify that u_λ(0) < v_λ(0), so we infer that the minimal solution (cf.[2]) isuλ.

Asλ=r²_λφ^α−1(rλ) =ρ²_λφ^α−1(ρλ),0< rλ< ρ_λ∗(α)< ρλ< r0, we get lim_λ→0rλ = 0, lim_λ→0ρλ = r0, lim_λ→0uλ(r) = limrλ→0φ(rλr)

φ(rλ) −1 = 0, and limλ→0vλ(r) = limρλ→r0

φ(ρλr)−φ(ρλ)

φ(ρλ) =φ(r0r)

limρλ→r0

1 φ(ρλ)

=

∞, ∀r∈ [0,1[.

4. The Critical Sobolev Exponent Case In this section, we suppose thatα=ⁿ⁺²_n−2 andn3.

Let us consider the following problem (P^α)

∆u+u^α= 0, inRⁿ u >0, inRⁿ.

Remark 4.1. — Every radially symmetrical solution of (P^α) verifies lim_r→∞u(r) = 0 (cf. [9]).

Following the method of Pohozaev in [14], the problem (Q^α)

u”(r) +ⁿ⁻¹_r u(r) +u^α(r) = 0, ∀r >0 u >0, u(0) = 1, u(0) = 0

admits a solutionφ.

Lemma 4.2. — Letube a radially symmetrical regular solution of(P^α), then

u(r) =u(0)φ

u(0)^α−12 r

. Proof. — This proof is immediate.

Lemma 4.3. — Let us put g(r) =r²φ^α−1(r), r∈R+, then there exists r₀>0, such thatg is increasing on[0, r₀], decreasing on [r₀,∞[, with lim_r→∞g(r) = 0.

Proof. — As we have already mentioned, g is increasing near 0. Let us assume thatg is nondecreasing on [0,∞[, then we have two possibilities

r→∞lim g(r) =∞or lim

r→∞g(r) =c, 0< c <∞.

For everyρ >0,ψρ is a solution of (P_γ^α), with γ=ρ²φ^α−1(ρ) =g(ρ).We infer (cf. [2], [10]) that g(r)λ^∗(α), ∀ r > 0, so the first limit becomes impossible.

In the second case, we have two subcases:c is achieved or not.

Ifcis not achieved, then∀lsuch that 0< l < c, there existsrl>0 such that g(rl) =l.One can verify that ∀ 0< l < c, the problem (P_l^α) admits the solutionψrl, so we infer thatcλ^∗(α).Letube a radially symmetrical solution (cf. [2], [10] and [3]) of (P_c^α).As in the proof of Proposition 2, one can verify that

u=ψρ, ρ=√

c(1 +u(0))^α−12 and 1

1 +u(0) =φ(ρ).

Asc=ρ²φ^α−1(ρ) =g(ρ),we get a contradiction.

Let us suppose thatc is achieved, asg is assumed to be nondecreasing, there exists r0 such that g(r) = c, ∀ r r0. Let us choose, an arbitrary constant ρ > 0 such that ρ r₀. The function ψ_ρ is a solution of the problem

P_γ^α , where γ = ρ²φ^α−1(ρ) = g(ρ) = c,∀ ρ r₀ . This means that this problem, with such a γ, admits an infinity of solutions ψ_ρ; this leads to a contradiction (cf. [2], [10]). Sog is not nondecreasing on [0,∞[.

Asg can’t be constant on a nontrivial interval, we deduce that there exists positive constants r₁ and r₂, such that r₁ < r₂, with g is increasing on [0, r1] and decreasing on a maximal interval [r1, r2[.Let us suppose thatg increases again on [r2, r3], with r2 < r3. If γ ∈ ]g(r2), min(g(r1), g(r3)) [, theng(r) =γadmits, at least, three roots, so the problem

P_γ^α admits, at least, three solutions; this gives again a contradiction (cf. [10]).

Finally, we get the existence ofr₀>0, such thatgis increasing on [0, r₀] and decreasing on [r₀,∞[. As g >0, we infer that lim_r→∞g(r) =c₀ 0.

Ifc₀>0, then for everyc∈]0, c₀[, there exists a uniqueρ_c∈R+, verifying g(ρ_c) = c. As c < λ^∗(α), the problem (P_c^α) admits exactly two solutions (cf. [10]). One of these two solutions isψ_ρc. Letu_c be the other one, then, using Proposition 2 again, we get

uc(r) =ψγ, γ=c¹²(1 +uc(0))^α−12 =c¹²φ^1−α2

c¹²(1 +uc(0))^α−12

. So we infer that c =g(γ). As the two solutions are different, ρc = γ and γ is another root of g(r) = c. This gives a contradiction and proves that necessarilyc= 0.This ends the proof of the lemma.

Proposition 4.4. — Let us assume α= ⁿ⁺²_n−2, n3, then λ^∗(α) = max

r∈]0,∞[g(r).

Proof. — Let γ=g(ρ) =ρ²φ^α−1(ρ), ρ∈R^∗+, we have seen that ψρ is a solution of

P_γ^α .So we infer thatg(ρ)λ^∗(α), ∀ ρ ∈R+. Let us suppose that

r∈max]0,∞[g(r)< λ^∗(α)

and let u be the unique solution (cf. [10]) of (P_λ^α∗(α)). As in the proof of Proposition 2, we get that u=ψ_ρ andλ^∗(α) =g(ρ). This gives a contra- diction.

Proposition 4.5. — We have λ^∗(α) =ⁿ⁽ⁿ₄⁻²⁾. There exists a unique r_λ∗(α) =

n(n−2), such that λ^∗(α) = r_λ²∗(α)φ^α−1(r_λ∗(α)) and a unique solution of(P_λ^α∗(α))

u_λ∗(α)=ψrλ∗(α). If 0< λ < λ^∗(α), there exist exactly two constants

r_λ=

1−_n(n^2λ₋₂₎−

1−_n(n^4λ₋₂₎ (n(n−2))⁻¹√

2λ and ρ_λ=

1−_n(n^2λ₋₂₎+

1−_n(n^4λ₋₂₎ (n(n−2))⁻¹√

2λ such that 0 < r_λ < r_λ∗(α) < ρ_λ, λ = g(r_λ) = g(ρ_λ) and the only two solutions of(P_λ^α) are

u_λ=ψ_rλ and v_λ=ψ_ρλ,

the minimal one (cf. [2]) is uλ; limλ→0uλ= 0, in C⁰(B1)and limλ→0vλ(r) =r²⁻ⁿ−1, ∀ r∈]0,1].

Proof. — One can use Lemma 3 to get the existence (and the uniqueness) ofr_λ∗(α)=r₀, r_λ andρ_λ.It is then easy to verify thatψ_rλ∗(α) is a solution of (P_λ^α_∗(α)), u_λ = ψ_rλ and v_λ = ψ_ρλ are solutions of (P_λ^α). The problem (P_λ^α) admits only two solutions (cf. [10]), asφis decreasing onR^∗+, one can verify thatu_λ(0)< v_λ(0), sou_λ=v_λ. We conclude thatu_λ andv_λ are the only solutions of (P_λ^α) and the minimal one (cf. [2]) isu_λ.

Let us compute the constantsr_λ∗(α),rλ andρλ.

It is well known (cf. [13]) that, ifα= ⁿ⁺²_n−2, the problem (Q^α) admits the continuum of spherically symmetrical ”instantons”

u_γ(r) =γⁿ⁻²² (n(n−2))ⁿ⁻⁴²

γ²+r^{2 2−n2} , γ >0.

Let us fixγ > 0, sou_γ(0) =γ²⁻²ⁿ(n(n−2))ⁿ⁻²⁴ .Using Lemma 2, we get the expression of the Lane-Emden function

φ(r) = 1 u_γ(0)uγ

uγ(0)ⁿ⁻²⁻² r

=

1 + r² n(n−2)

²⁻ⁿ₂ . Asα−1 = ⁿ⁺²_n−2−1 = _n−2⁴ , we infer that

g(r) =r²φ^α−1(r) =r²

1 + r² n(n−2)

₋2

. Using Proposition 4, a direct calculation gives

λ^∗(α) = max

r>0 r²

1 + r² n(n−2)

−2

=r²

1 + r² n(n−2)

₋2

|_{r=rλ∗(α) =}√

n(n−2)

= n(n−2)

4 .

In [7], the previous constant has been computed, using the Pohozaev Iden- tity. If 0< λ < λ^∗(α),the equationg(r) =λadmits two positive roots

rλ=

1−_n(n−2)^2λ −

1−_n(n−2)^4λ (n(n−2))⁻¹√

2λ and ρλ=

1−_n(n−2)^2λ +

1−_n(n−2)^4λ (n(n−2))⁻¹√

2λ .

This gives usuλ=ψrλ andvλ=ψρλ; asrλ< ρλ, we getuλ(0)< vλ(0),so uλis the minimal solution.

Asλ=r_λ²φ^α−1(r_λ) =ρ²_λφ^α−1(ρ_λ),0< r_λ < r_λ∗(α) < ρ_λ <∞, one can verify that

lim_λ→0r_λ= 0, lim_λ→0ρ_λ=∞, lim_λ→0u_λ= 0, in C⁰

B₁ and lim_λ→0v_λ(0) = lim_ρλ→∞^φ(ρ_φ(ρ^λr)

λ) −1 =r²⁻ⁿ−1, ∀r∈]0,1].

5. The Supercritical Case

We consider here the caseα > ⁿ⁺²_n−2, n3.Let us put f(α) = 4α

α−1+ 4 α

α−1, ∀α >1.

Let’s first detail a condition, f(α)> n−2, used in [10].

Lemma 5.1. — If

3n10and α > ⁿ⁺²_n−2 or

n >10 and ⁿ⁺²_n−2 < α < _n−ⁿ⁻₂^√2√_nⁿ₋⁻₁₋¹₄

,

thenf(α)> n−2. If n >10and _n−2^n−√2√_n−1−4ⁿ⁻¹ α, thenf(α)n−2.

Proof. — Let us put p(t) = 4t²+ 4t and u=

α

α−1, so we get f(α) = p(u).The only positive root ofp(t) =n−2, ist0=^√n−₂¹⁻¹ and the equation u = ^√n−1−1₂ has the only solution α0 = _nⁿ⁻²₋₂√^√n−1

n−1−4. But α0 >0, if and only ifn >10.

For everyα > ⁿ⁺²_n−2, we haveα > 1 so we get

α

α−1 >1 > ^√n−1−1₂ , if 3n10. We infer thatf(α)> n−2, if 3n10.

Ifn >10, we haveα0> ⁿ⁺²_n−2 >1, one can verify that if ⁿ⁺²_n−2 < α < α0, thenf(α)> n−2 andf(α)n−2,ifαα0.

Proposition 5.2. — Let us putλs=_(α−1)² 2 (α(n−2)−n). If

3n10and ⁿ⁺²_n−2 < α

or

n >10and ⁿ⁺²_n−2 < α < _n−2ⁿ⁻²√^√n−1 n−1−4

then

λ^∗(α) = max

R^∗+

g(r), λ^∗(α)> λs and φ(r)∼λ

α−11

s r^1−α2 , as r→ ∞.

If (ρ_i) is an increasing sequence of positive reals, such that (ψ_ρi) are so- lutions of (P_λ^α_s) and limi→∞ρi = ∞, then limi→∞ψρi(r) = λ

α−11

s (r^1−α2 − 1), ∀ r∈]0,1].

Ifn >10and _n−2ⁿ⁻²√^√n−1

n−1−4 αthen λ^∗(α) = sup

R^∗+

g(r) =λs and φ(r)∼λ

1 α−1

s r^1−α2 , as r→ ∞.

If (λ_i) is an increasing positive sequence such that lim_i→∞λ_i = λ_s and

∀i, w_i is the unique solution of(P_λ^α_i), then lim_i→∞w_i(r) =λ

α−11

s (r^1−α2 −1), ∀ r∈]0,1].

Proof. — As in the proof of Proposition 4, one can verify that λ^∗(α) = sup_R∗

+g(r),where g(r) =r²φ^α−1(r).

If

3n10and ⁿ⁺²_n−2 < α

or

n >10and ⁿ⁺²_n−2 < α < _nⁿ₋⁻₂^√2√_nⁿ₋⁻₁₋¹₄

, using Lemma 4, we getf(α)> n−2. So we can use Theorem 1 in [10] to infer thatλ^∗(α)> λs,

P_λ^α∗(α)

admits a unique solution and

P_λ^α_s admits an infinity of solutions. Using the unique solution u_λ∗(α) of

P_λ^α∗(α)

, one can deduce from Proposition 1 thatu_λ∗(α)=ψ_ρ, whereρ∈R^∗+andg(ρ) = λ^∗(α).We conclude that the supremum is achieved andλ^∗(α) = max_R∗

+g(r).

Let us suppose that a= lim inf

r→∞ g(r)< A= lim sup

r→∞ g(r).

For everyλ∈]a, A[, the equationg(r) =λadmits a sequence of roots (r_i), with limi→∞ri =∞. As for every i, ψri is a solution of (P_λ^α), we get an infinity of solutions for this problem; but an infinity of solutions exists only whenλ=λs (cf. [10]). We get a contradiction and infer that

a=A=λs= lim

r→∞g(r), so φ(r)∼λ

α−11

s r^1−2α, as r→ ∞.

If (ρ_i) is an increasing sequence of positive constants, such that (ψ_ρi) are solutions of (P_λ^α_s) and lim_i→∞ρ_i=∞,then one can use the previous asymp- totic behavior ofφto get lim_i→∞ψ_ρi(r) =λ

α−11

s (r^1−α2 −1), ∀r∈]0,1].

Ifn >10 and _n−2ⁿ⁻²√^√n−1

n−1−4 α, we get from Lemma 4 thatf(α)n−2.

Using [10] again, we infer that λ^∗(α) =λ_s, (P_λ^α) admits a unique solution for everyλ∈]0, λ^∗(α)[.As the functiong is increasing nearr= 0, we infer thatg is increasing onR^∗₊.For, on one hand, ifg decreases on a nontrivial open intervalI⊂R^∗₊, then the equationg(r) =λadmits at least two roots r1< r2, ifλ∈] minIg(r),maxIg(r)[.Asψr1 andψr2 are solutions of (P_λ^α),

withψ_r1(0)=ψ_r2(0),this violates the uniqueness result of [10]. On another hand, the function g can’t be constant on a nontrivial interval, otherwise we get an infinity of solutions for someλ.One can then see that

r→∞lim g(r) = sup

R^∗+

g(r) =λ^∗(α); λ^∗(α) =λs(cf.[10]).

Soφ(r)∼λ

α−11

s r^1−α2 , as r→ ∞.

Using this asymptotic behavior, one can show the last statement of the proposition.

Let us put

(Q^α_λ)





∆u+λ(1 +u)^α= 0, in B_r0 u >0, in Br0

u= 0, on ∂Br0

where Br0 ={x∈ Rⁿ, x < r0}.For every solution u of (Q^α_λ), we put v(r) =u(r0r) for everyr ∈[0,1].Letλ^∗_r0(α), be the maximal eigenvalue of (Q^α_λ).

Lemma 5.3. — A function uis a solution of(Q^α_λ), if and only if v is a solution of(P_r^α2

0λ). In particular, we getλ^∗_r0(α) =r²₀λ^∗(α).

Proof. — The proof is easy.

Remark 5.4. — According to the previous lemma, the results obtained here for (P_λ^α) (on the unit ball B₁), can be easily stated for (Q^α_λ) (on any ballB_r0).

6. Appendix

LetS_kⁱ be the set of all the (k−i)−selections of {1, ..., i} ands(j) the multiplicity of the elementj, 1ji.Ifuis a analytical solution of (P_λ^α), with u(r) = Σ^∞_k=0akr^k near r= 0, r0 the convergence radius of this series, then

Proposition 6.1. —

∀k0, a_2k+1= 0, a₂= λ

n−2(1 +a₀)^α 1

n−1 2

and∀ k >1, a_2k = λ n−2

1

2k+n−2 − 1 2k

× Σ^k_i=1⁻¹(1 +a0)^α−i 1

i!Πⁱ_p=0⁻¹(α−p)Σ_s∈Sⁱ

k−1Πⁱ_j=1a2(1+s(j)).

Proof. — Let us choose 0< rρ < r₀,by standard integrations, we get u(r)−u(ρ) = λ

n−2×

(r²⁻ⁿ−ρ²⁻ⁿ) _r

0

tⁿ⁻¹(1 +u(t))^αdt+ _ρ

r

(t−ρ²⁻ⁿtⁿ⁻¹)(1 +u(t))^αdt

. Let us point out that

(1 +u(r))^α= (1 +u(0)−u(0) +u(r))^α

= (1 +u(0))^α

1 +u(r)−u(0) 1 +u(0)

α

= (1 +a0)^α

1 + Σ^∞_i=1 ai

1 +a0

rⁱ α

, u(0) =a0. By the Maximum Principle, we have∀r∈]0,1[,0< u(r)< u(0), so we get

u(0)−u(r) 1 +u(0)

<1, ∀r∈ [0,1],

we infer that

(1+u(r))^α= (1 +a₀)^α

1 + Σ^∞_j=1α(α−1)...(α−j+ 1) j!

Σ^∞_i=1 a_i 1 +a0

rⁱ j

. All these series are uniformly convergent on [0, ρ].If we put

(1 +u(r))^α= Σ^∞_j=0c_jr^j,we get u(r) = λ

n−2

(r²⁻ⁿ−ρ²⁻ⁿ) _r

0

tⁿ⁻¹Σ^∞_j=0c_jt^jdt+ _ρ

r

(t−ρ²⁻ⁿtⁿ⁻¹)Σ^∞_j=0c_jt^jdt

= λ

n−2

Σ^∞_j=0cj

r^2+j

j+n−Σ^∞_j=0cj

ρ²⁻ⁿr^j+n

j+n + Σ^∞_j=0cj

ρ^j+2

j+ 2 −Σ^∞_j=0cj

ρ^j+2 j+n

+ λ

n−2

−Σ^∞_j=0c_jr^j+2

j+ 2 + Σ^∞_j=0c_jρ²⁻ⁿr^j+n j+n

= λ

n−2

Σ^∞_j=2c_j−2 r^j

j+n−2+ Σ^∞_j=0c_jρ^j+2

j+ 2 −Σ^∞_j=0c_jρ^j+2

j+n−Σ^∞_j=2c_j−2r^j j

. We finally obtain

(2) u(r) = λ n−2

Σ^∞_j=2c_j−2( 1

j+n−2−1

j)r^j+ Σ^∞_j=0cjρ^j+2( 1

j+ 2− 1 j+n)

. Using the previous identity, we obtain

a₁= 0, ∀ k >1, a_k= λ n−2

1

k+n−2 −1 k

c_k−2.

Using (1), we get

c0= (1 +a0)^α, c1=α(1 +a0)^α−1a1= 0 and

∀k >1, c_k= (1 +a₀)^αΣ^k_j=11

j!Π^j_p=0⁻¹(α−p) 1

(1 +a₀)^jΣ_s∈S^j

kΠ^j_i=1a_1+s(i)

= Σ^k_j=11

j!Π^j_p=0⁻¹(α−p) (1 +a₀)^α−jΣ_s∈S^j

k

Π^j_i=1a_1+s(i).

Using the previous relation and the fact that a₁ = 0 , one can verify (by induction) that a_2k+1 = 0, ∀k > 0. We then obtain from (2) and the expression ofc_k

a_2k= λ n−2

1

2k+n−2 − 1 2k

c_2k−2

= λ

n−2

1

2k+n−2− 1 2k

Σ^k−1_j=1 1

j!Π^j−1_p=0(α−p) (1 +a0)^α−jΣ_s∈S^j

k−1Π^j_i=1a_2(1+s(i)).

∀j ∈ [1, k−1], Card(S_k−1^j ) =C_k−2^j−1. Let us put

d2= 1

2n and ∀k >1,

d2k= 1

(2k+n−2)(2k)Σ^k−1_i=1 1

i!Πⁱ⁻¹_p=0(α−p)Σ_s∈Sⁱ

k−1Πⁱ_j=1d_2(1+s(j)), then

Lemma 6.2. — a_2k = (−1)^kλ^k(1 +a₀)^k(α−1)+1d_2k, ∀k >1.

Proof. —

a₄= αλ²

(n−2)² = (1 +a₀)^2α−1 1

n+ 2 −1 4

1 n−1

2

=λ²(1 +a₀)^2α−1 1 4(n+ 2)

α

2n=λ²(1 +a₀)^2(α−1)+1 1 4(n+ 2)

α 2n. d4= 1

4(n+ 2)Σ¹_i=11

i!Πⁱ⁻¹_p=0(α−p)Σ_s∈Sⁱ

1Πⁱ_j=1d_2(1+s(j))

= α

4(n+ 2)d₂= 1 4(n+ 2)

α 2n,