Existence and asymptotic behavior of non-radially

Nonlinear Analysis 65 (2006) 719–727 www.elsevier.com/locate/na

Existence and asymptotic behavior of non-radially symmetric ground states of semilinear singular elliptic equations✩ J.V. Goncalves a,∗ , C.A. Santos b a Universidade de Bras´ılia, Departamento de Matemática, 70910-900 Bras´ılia, DF, Brazil b Universidade Federal de Goiás, Departamento de Matemática, 75706-220 Catalão, GO, Brazil

Received 1 September 2005; accepted 29 September 2005

Abstract This paper deals with the existence and asymptotic behavior of entire positive solutions of the semilinear elliptic equation −u = ρ(x) f (u) in R N (N ≥ 3), where ρ is nonnegative and locally Hölder continuous r→0

and f is a positive locally Lipschitz continuous function, singular at 0 in the sense that f (r ) −→ ∞. No symmetry is required from ρ and no monotonicity condition is imposed on f . Arguments for lower and upper solutions are exploited. c 2005 Elsevier Ltd. All rights reserved. Keywords: Semilinear elliptic equations; Ground states; Asymptotic behavior; Lower–upper solutions

1. Introduction In this work we study the existence and asymptotic behavior of entire solutions of the problem −u = ρ(x) f (u) in R N , (1.1) |x|→∞ u > 0, u(x) −→ 0, where N ≥ 3, ρ : R N → [0, +∞) is continuous and f : (0, +∞) → (0, +∞) belongs to r→0

Liploc ((0, ∞)) with f singular at 0 in the sense that f (r ) −→ ∞. No symmetry is required for ✩ Research partially supported by CNPq, PADCT/UFG 620039/2004-8, PRONEX/UnB. ∗ Corresponding author. Tel.: +55 61 273 3356; fax: +55 61 273 2737.

E-mail addresses: [email protected] (J.V. Goncalves), [email protected] (C.A. Santos). c 2005 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2005.09.036

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J.V. Goncalves, C.A. Santos / Nonlinear Analysis 65 (2006) 719–727

ρ and no monotonicity condition is imposed on f . This sort of problem appears in the physical sciences and has been exhaustively investigated in the last few years. As regards corresponding problems on bounded domains, we refer the reader to the pioneering work [5] by Crandall, Rabinowitz and Tartar, where both perturbation and global bifurcation methods are exploited and even linear operators more general than are considered. We also refer the reader to the work of Lazer and McKenna [13], where singular problems involving the Monge–Ampére operator rather than are investigated. Returning to problems on the whole space, as a brief historical account of problems closer to our interests in the present paper, we recall that Edelson in [9] and Shaker in [14] addressed the cases f (r ) = r −λ with λ ∈ (0, 1) and λ ∈ (0, ∞) respectively, and additionally assumed ∞ s N−1+λ(N−2) ϕ(r )dr < ∞, (1.2) 1

where ϕ(r ) := max|x|=r ρ(x) while Lair and Shaker in [12] assumed λ ∈ (0, ∞) and ∞ r ϕ(r )dr < ∞.

(1.3)

1

Zhang in [16] improved on the results above by exploiting C 1 -nonlinearities f , singular at 0, satisfying the condition f < 0 which includes pure powers as a special case. Cirstea and Radulescu in [4] showed the existence of entire solutions of (1.1) in the case of a C 1 -term f , singular at 0, satisfying the following additional conditions: lim

r→0

f (r ) f (r ) = ∞, f bounded at ∞, decreasing for some b > 0. r r +b

There is now a broad literature on singular problems. We refer the reader to the recent work [8] by Dinu, where the condition “ f (r )/r is decreasing in (0, ∞)” which is weaker than the related one in [4] mentioned above is applied. We also refer the reader to Cirstea, Ghergu and Radulescu [3], Ghergu and Radulescu [10], Dinu [7] and their references. The main result of this paper is the following. 0,α Theorem 1.1. Assume ρ ∈ Cloc for some α ∈ (0, 1). Assume also (1.3) and

(i)

f (r ) f (r ) f (r ) is decreasing, (ii) lim = ∞, (iii) lim = 0. r→+∞ r→0 r r r

(1.4)

2,α Then (1.1) admits a solution u ∈ Cloc . If, in addition,

lim |x|µ ϕ(|x|) < ∞,

(1.5)

|x|→∞

for some µ ∈ (2, N), then u(x)2 [ f (u(x)/4)]−1 = O(|x|(2−µ) ) as |x| → ∞.

(1.6)

Remark 1. (i) The class of functions f (r ) = r −λ + r γ ,

r > 0,

where λ > 0 and 0 ≤ γ ≤ 1 provides examples of singular nonlinearities, covered by Theorem 1.1, but not by the results referred to above. To the best of our knowledge a decay rate such as (1.6) was not shown earlier in the study of (1.1).


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r→∞

(ii) If f is nonincreasing and f (r ) −→ 0, (1.6) turns into u(x)[ f (u(x)/4)]−1 = O(|x|(2−µ) ) In particular, when f is of the u(x) = O(|x|

2−µ 1+λ

)

form r −λ

as |x| → ∞.

where λ > 0, the condition above turns into

as |x| → ∞.

(iii) The technique of this paper (follow our proofs) applies to the more general problem |x|→∞

−u = p(x) f (u) + q(x)g(u) in R N , u > 0 in R N , u(x) −→ 0, 0,α where f + g satisfies the conditions of Theorem 1.1, p + q ∈ Cloc satisfies (1.3). Setting ϕ(r ) := max|x|=r p(|x|) + max|x|=r q(|x|) in (1.5), condition (1.6) turns into

u(x)2 [ f (u(x)/4) + g(u(x)/4)]−1 = O(|x|(2−µ) )

as |x| → ∞.

u −λ

A special case of the problem above, with f (u) = and g(u) = u γ where 0 < λ, γ < 1, was treated in the inspiring article [15] by Sun and Li. (iv) Theorem 1.1 holds true even for terms ρ vanishing at subsets of R N with zero measure. Our approach is based on arguments for upper and lower solutions. In this regard, we construct in the next section a positive upper solution decaying to zero at infinity. 2. Existence of an upper solution decaying to zero Theorem 1.2. Assume (1.3) and (1.4). Then, there is a radially symmetric function v ∈ C 2 such that −v(x) ≥ ρ(x) f (v) in R N ,

v > 0,

Proof of Theorem 1.2. Set r t 1−N N−1 I (r ) := t s ϕ(s)ds dt, 0

|x|→∞

v(x) −→ 0.

(2.1)

r > 0.

0

We claim (details appear in the Appendix) that ∞ t 1−N N−1 0 < a := t s ϕ(s)ds dt < ∞. 0

(2.2)

0

Now setting w(r ) := a − I (r ), it is easily shown that w ∈ C 2 ([0, ∞)) and, in fact, w is the unique solution of the initial value problem −(r N−1 w ) = r N−1 ϕ(r ) in (0, ∞), w > 0 in [0, ∞), w(0) = a, w (0) = 0.

(2.3)

Now, it is an easy matter to check, since f ∈ Liploc ((0, ∞)), that f defined by r2 f (r ) := r t

0 f (t ) dt

,

r > 0,

is a ∈ C 1 ((0, ∞)) function satisfying (i) f (t) ≥ f (t),

(ii)

f (t) is decreasing, t

(iii) lim

t →∞

f (t) = 0. t

(2.4)

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Claim. There are R > 0 and a function v := v R ∈ C 2 ([0, ∞)) such that 1 v(r) t dt R 0 f (t) v satisfies (2.1). w(r ) =

(2.5) (2.6)

Verification of the Claim. Notice first that R t dt for some R > 0. Ra ≤ f (t) 0

(2.7)

Indeed, by (2.4)(iii), r t lim dt = ∞ r→∞ 0 f (t) and hence lim

r→∞

r

t dt 0 f (t )

r

= lim

r→∞

showing (2.7). Now set 1 s t F(s) := dt, R 0 f (t)

r = ∞, f (r )

s ≥ 0. s→∞

Using (2.4)(iii) it follows that F(0) = 0, F is increasing and F(s) → ∞. By a suitable application of the Implicit Function Theorem there is some function v ∈ C 2 ((0, ∞)) satisfying (2.5). It remains to show (2.6) and v ∈ C 2 ([0, ∞)). Indeed, by construction, v is nonincreasing. Hence using the fact that w(0) = a, (2.5) and (2.7), we get v(r) v(0) R t t t dt ≤ dt = Rw(0) = Ra ≤ dt, f (t) f (t) f (t) 0 0 0 and in particular v(r ) ≤ R for r ≥ 0. Now, differentiating in (2.5) with respect to r and computing, we get v v 1 1 d (r N−1 w (r )) = (2.8) (r N−1 v (r )) + r N−1 |v |2 . R R dv f (v) f (v) Now using (2.3), (2.4)(ii) and the fact that v ≤ R in (2.8), we have R N−1 (r r N−1 v (r )) ≤ − f (v(r ))ϕ(r ) ≤ −r N−1 ϕ(r ) f (v(r )). v Making r → 0 in (2.5) it follows that v (0) = 0 and as a further consequence of (2.8) r→∞ v ∈ C 2 ([0, ∞)). Setting r = |x| and v(x) := v(r ) and noting that v(r ) −→ 0 ends the proof of Theorem 1.2.


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3. Results on bounded domains Let Ω ⊂ R N be a bounded smooth domain and consider the family of problems −u = ρ(x) f (u + ε) in Ω , u > 0, u = 0 on ∂Ω ,

(3.1)

where > 0 is a parameter. A few auxiliary results and remarks will be established. 0,α Lemma 3.1. Assume (1.4). If ρ ∈ Cloc , there is, for each > 0 small enough, a solution 2,α u ∈ C (Ω ) of (3.1) such that

(i) u + ≤ u δ + δ in Ω provided < δ, (ii) there is c0 := c0,Ω > 0 with c0 ϕ1 ≤ u ε + ε in Ω , where ϕ1 := ϕ1,Ω is the λ1 -eigenfunction of the eigenvalue problem −u = λρ(x)u in Ω , u=0 on ∂Ω . We will use some notation, results and arguments inspired by Brézis and Oswald [2] and D´ıaz and Saa [6]. To begin with, the proof of Lemma 3.1 uses the following result. Lemma 3.2. Let i, j ∈ {1, 2} and let O ⊂ R N be an open set. If wi ∈ L ∞ (Ω ) satisfy wi > 0 1/2 1/2 a.e. in O, w1 = w2 on ∂O, wi ∈ H 1(O), wi ∈ L ∞ (O) and wi /w j ∈ L ∞ (O), then

1/2 1/2 −w1 w2 + 1/2 (w1 − w2 )dx ≥ 0. 1/2 O w1 w2 Now, let a : Ω → R ∪ {∞} be a measurable function; set λ1 (−v − a(x)v) := inf |∇v|2 dx − v∈H01 (Ω ),|v| L 2 (Ω ) =1

and ρ (x, r ) := ρ(x) f (r + ε) so that

Ω

v =0

a(x)|v|2 dx

f (r + ε) ρ (x, r ) ε = ρ(x) 1+ . r r +ε r

(3.2)

Setting a0ε (x) := lim

r→0

ρ (x, r ) r

and

ε a∞ (x) = lim

r→∞

ρ (x, r ) r

ε = 0. Hence it follows by (3.2) and (1.4)(ii)(iii) that a0ε = ∞ and a∞

λ1 (−v − a0 (x)v) < 0

and

λ1 (−v − a∞ (x)v) > 0.

Proof of Lemma 3.1. By the results of [2] and [6] ∇u ε ∇ϕdx = ρ(x) f (u ε (x) + ε)ϕdx, Ω

Ω

ϕ ∈ H01(Ω ),

∩ L ∞ (Ω ). Applying the regularity theory for elliptic equations we find for some u ∈ 2, p that u ∈ W (Ω ) for p ∈ (1, ∞) and thus u ∈ C 1,α (Ω) for α ∈ (0, 1). Moreover, the outward H01(Ω )

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0,α normal derivative ∂ν u is negative. Since f ∈ Liploc and ρ ∈ Cloc it follows by the Schauder theory that u ∈ C 2,α (Ω ) and, in fact, (3.1) holds in the classical sense.

Verification of (i). Set Bε,δ = {x ∈ Ω | u ε (x) + ε > u δ (x) + δ}. It suffices to show that Bε,δ = φ. To this end, notice that Bε,δ ⊂ Ω is open and, in fact, Bε,δ ⊂⊂ Ω . Assume, to the contrary, that Bε,δ = φ. Let w1 := (u ε + ε)2 and w2 := (u δ + δ)2 . Applying Lemma 3.2 we have

1/2 1/2 w2 w1 − 1/2 − − 1/2 (w1 − w2 )dx 0≤ Bε,δ w1 w2 −(u ε + ε) (u δ + δ) + = ((u ε + ε)2 − (u δ + δ)2 )dx (u ε + ε) (u δ + δ) Bε,δ f (u δ + δ) f (u ε + ε) − = ρ(x) ((u ε + ε)2 − (u δ + δ)2 )dx < 0, u + ε u + δ ε δ Bε,δ which is impossible, due to (1.4)(i). This proves (i). Verification of (ii). Let c > 0 be a parameter and set Bc,ε := {x ∈ Ω | cϕ1(x) > u ε (x) + ε}. It suffices to show that Bc0 ,ε = φ for some c0 > 0. Of course, Bc,ε ⊂ Ω is open and, in fact, Bc,ε ⊂⊂ Ω . Set w1 := (cϕ1 )2 and w2 := (u ε + ε)2 in Lemma 3.2. By (1.4)(i) and the properties of ϕ1 and λ1 , we get, by estimating as above, f (cϕ1 L ∞ ) ((cϕ1 )2 − (u ε + ε)2 )dx. 0≤ ρ(x) λ1 − (3.3) cϕ1 L ∞ Bc,ε By (1.4)(ii) there a positive constant c0 := c0,Ω (not depending on ) such that λ1 −

f (c0 |ϕ1 | L ∞ ) < 0. c0 |ϕ1 | L ∞

This inequality, (3.3) and the definition of Bc,ε show that Bc0 ,ε = φ. This shows (ii). Lemma 3.1 is proved. 4. Proof of Theorem 1.1 Pick an integer k ≥ 1. Set := 1/n, u n := u 1/n and Ω := Bk with n > k in (3.1). By Lemma 3.1 there are both a solution u n ∈ C 2,α (B k ) of (3.1) satisfying u n ≥ c0,Bk ϕ1,Bk and some function u k ∈ C(B k ) such that n

(i) u n → u k a.e. in Bk , (ii) u k ≥ c0,k ϕ1 a.e. in Bk , where ϕ1 := ϕ1,Bk . Letting ϕ ∈ C0∞ (Bk ), it follows from −u n = ρ(x) f (u n + 1/n)

in Bk

(4.1)


that

725

−

u n ϕdx =

ρ(x) f (u n + 1/n)ϕdx.

Bk

Bk

Passing to the limit in n, we have − u k ϕdx = ρ(x) f (u k )ϕdx. Bk

(4.2)

Bk 2, p

By the elliptic regularity theory (cf. [1, Theorem 7.1] and [11]), u k ∈ Wloc (Bk ) for p ∈ (1, ∞) and, in fact, −u k = ρ(x) f (u k ) a.e. in Bk .

(4.3)

1,α Cloc (Bk ),

∈ and invoking the interior Schauder estimates, By the Sobolev embeddings 2,α k k u ∈ Cloc (Bk ). Extending u , as zero, outside Bk we claim that uk

(i) u k ≤ u k+1

and

(ii) u k ≤ v.

(4.4)

Indeed, to show (i), consider the open subset of R N , namely Bk,k+1 = {x ∈ R N | u k (x) > u k+1 (x)} ⊂⊂ Bk . Setting w1 := (u k )2 and w2 := (u k+1 )2 we get, as earlier,

1/2 1/2 w2 −w1 + 1/2 (w1 − w2 ) 0≤ 1/2 Bk,k+1 w2 w2 k k+1 u −u = + k+1 ((u k )2 − (u k+1 )2 ) uk u Bk,k+1 f (u k ) f (u k+1 ) ρ(x) − = ((u k )2 − (u k+1 )2 ) < 0, uk u k+1 Bk,k+1 which is impossible. Hence, Bk,k+1 = φ, showing (i). In order to show (ii), set Bk,v := {x ∈ R N | u k (x) > v(x)} and assume, by way of contradiction, that Bk,v = φ. Applying Lemma 3.2 with w1 := (u k )2 and w2 := v 2 and estimating as above we get f (u k ) f (v) 0≤ ρ(x) − ((u k )2 − v 2 ) < 0, uk v Bk,v which is impossible. So Bk,v = φ and hence u k ≤ v. Now, as a consequence of (4.4)(i), u k → u pointwise for some u > 0. Passing to the limit in 2,α (4.2) and using the elliptic regularity theory again we find that u ∈ Cloc and −u = ρ(x) f (u),

x ∈ RN .

In order to show (1.6), from (2.5) we have 1 v(r) t 1 v(r) 1 v(r) t h 1 (t)dt dt ≥ dt := w(r ) = R 0 R v(r)/2 R v(r)/2 f (t) f (t)

726


where h 1 (t) := t / f (t), t > 0. Using (2.4)(i)–(iii) and the definition of f we find that v(r)/2 1 t 1 w(r ) ≥ h 1 (v(r )/2)(v(r )/2) ≥ dt. R R v(r)/4 f (t) Using (1.4)(i), w(r ) ≥

1 (v(r )/4)2 [ f ((v(r )/4))]−1 . R

(4.5)

From (4.5) and (1.4)(i) and the fact that u ≤ v we get u(x)2 [ f (u(x)/4)]−1 ≤ 16Rw(|x|),

x ∈ RN .

On the other hand, using (2.1), w(r ) w (r ) 1 1 lim 1−µ = lim lim 2−µ = r→∞ r 2 − µ r→∞ r µ − 2 r→∞ 1 lim |x|µ ϕ(|x|) < ∞. = µ − 2 r→∞

(4.6)

r 0

s N−1 ϕ(s)ds r N−µ

Hence w(|x|) = O(|x|2−µ ) as |x| → ∞. So, (1.6) follows from the above and (4.6). Setting 1 r 1 r dt, and F(r ) := f (r ) := r 1 R 0 f (t) 0 f (t ) dt remark (ii) follows by similar arguments. This ends the proof of Theorem 1.1.

Appendix Verification of (2.2). Set t H (t) := s N−1 ϕ(s)ds,

t ≥0

0

and notice that either H (t) < 1 for each t > 0 or H (δ0) = 1 for some δ0 > 0. In the first case, we get r t 1−N dt I (r ) ≤ C1 + 1

for some positive constant C1 > 0 and so (2.2) holds because N ≥ 3. In the second case, we get by taking C1 larger, if necessary, r I (r ) ≤ C1 + t 1−N H (t)dt, 1

for all r > δ0 . Integrating by parts and estimating we get r tϕ(t)dt I (r ) ≤ C1 + C2 1

and using (1.3) we infer once more that (2.2) holds true. This ends the verification of (2.2).


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