diffusive logistic equations in R with indefinite weight

diffusive logistic equations in R with indefinite weight

Dimitri Mugnai ^∗

Dipartimento di Matematica e Informatica Universit` a di Perugia

Via Vanvitelli 1, 06123 Perugia - Italy tel. +39 075 5855043, fax. +39 075 5855024,

e-mail: dimitri.mugnai@unipg.it Nikolaos S. Papageorgiou

Department of Mathematics National Technical University Zografou Campus, Athens 15780 Greece

e-mail: npapg@math.ntua.gr

Abstract

We consider a diffusive p–logistic equation in the whole of R

with absorption and an indefinite weight. Using variational and truncation techniques we prove a bifurcation theorem and describe completely the bifurcation point. In the semilinear case p = 2, under an additional hypothesis on the absorption term, we show that the positive solution is unique.

Keywords: diffusive p–logistic equation, positive solution, Hardy’s inequal- ity, bifurcation theorem, nonlinear regularity, indefinite weight.

2000AMS Subject Classification: 35J20, 35J70, 35P30

1 Introduction

In this paper we are interested in non existence, existence and uniqueness (in the semilinear case p = 2) of positive solutions for the following p–logistic type equation with absorption defined in the whole of R ^N (N > p):

(P λ )

( −∆ p u(z) = λ β(z)u ^p−1 − f (z, u(z))

in R ^N , u(z) → 0 as kzk → ∞, u > 0,

∗

Research supported by the M.I.U.R. project Metodi Variazionali ed Equazioni Differen- ziali Nonlineari.

1

where λ > 0, k · k is the Euclidean norm in R ^N and ∆ p denotes the p–Laplace differential operator defined by ∆ p u = div(kDuk ^p−2 Du) for all u ∈ W ^1,p (R ^N ).

Moreover, β is a sign changing weight function and f : R ^N × R → R is a Carath´ eodory perturbation (i.e., for all x ∈ R the map z 7→ f (z, x) is measur- able, and for a.e. z ∈ R ^N the map x 7→ f (z, x) is continuous).

Problems of this type arise as the steady state equation of population dy- namics models (for example, see [3], [10], [11], [12], [13], [14], [26]). In this case the unknown u corresponds to the density of a population, the weight β corresponds to the birth rate of the population if self–limitation is ignored and the perturbation −f expresses the fact that the population is self–limiting. The function β is in general sign changing and the region where β is positive (resp.

negative) the population, ignoring self–limitation, has positive (resp. negative) birth rate. Since u describes the population density, we are interested only in positive solutions of problem (P λ ), such solutions corresponding to possible steady state distributions of the population.

The result we prove establishes a relation between the diffusion parameter λ > 0 (extent of diffusion) and the existence of a nontrivial steady state popula- tion density. Thus, the population will prosper in those regions where the birth rate β is positive, and if the diffusion is small (i.e. if λ > 0 is large) nontrivial steady state solutions are possible, even if the population is subject to some overall disadvantage (i.e. β is predominantly negative). On the other hand, if the diffusion is big (i.e. λ > 0 is small), the population is not safely protected in small regions of disadvantage and a steady state solution cannot exist.

Problem (P _λ ) settled in a bounded region with Neumann boundary condi- tion was studied by Cantrell–Cosner–Hutson [4] and Umezu [25], and exten- sions to the p–Laplacian Dirichlet case can be found in the works of Dong [7], Garcia Melian–Sabina de Lis [15], Guedda–Veron [17], Kamin–Veron [19] and Papageorgiou–Papalini [21]. A related Neumann problem can be found also in Cardinali–Papageorgiou–Rubbioni [5]. Diffusive logistic equations in the whole of R ^N were studied by Afrouzi–Brown [2], who dealt with the semilinear equa- tion (i.e. p = 2). More precisely, they considered the equation

−∆u = λβu − u ² 

in R ^N

with β : R ^N → R a smooth function bounded above. Our semilinear result here (see Theorem 9) extends the result of Afrouzi–Brown [2], since we do not assume that β is smooth and bounded above, and our absorption term −f (z, x) is more general than the one considered therein.

2 The bifurcation parameter

Following Szulkin–Willem [24], we impose the following conditions on the weight function β:

H(β): β ∈ L ^∞ _loc (R ^N ), β ⁺ = β ₁ + β ₂ 6= 0, with β 1 ∈ L ^N/p (R ^N ) ∩ L ¹ (R ^N ) (1 < p < N ) and

kzk ^p β 2 (z) → 0 as kzk → ∞, lim

z→ˆ z kz − ˆ zk ^p β 2 (z) = 0 for all ˆ z ∈ R ^N ,

kzk

β 2 (z) → 0 as z → 0.

Here, as usual, β ⁺ = max{β, 0} and β ⁻ = max{−β, 0}.

Moreover, in view of the Bifurcation Theorems 7 and 9, we also assume the following condition:

H(β) 2 :

( (rˆ a(r)) ^1/(p−1) ∈ L ¹ (1, ∞) if 1 < p ≤ 2, or r

ˆ a(r) ∈ L ¹ (1, ∞) if p ≥ 2, where we have set

ˆ

a(r) := sup

kzk=R

β ⁺ (z).

Now, for any r > 0, let B r = z ∈ R ^N : kzk < r . We introduce the following quantity:

λ ˆ ₁ (r) = inf n Z

B

kDuk ^p dz : z ∈ W ₀ ^1,p (B _r ), Z

B

β|u| ^p dz = 1 o

. (1)

From Szulkin–Willem [24] we know that ˆ λ ₁ (r) is the principal eigenvalue of the nonlinear eigenvalue problem

(E)

( −∆ p u = λβ|u| ^p−2 u in B r ,

u = 0 on ∂B r .

From (1) it is clear that the map r 7→ ˆ λ ₁ (r) is decreasing on (0, ∞) and so we can define

λ ˆ ^∗ = lim

r→∞

λ ˆ 1 (r). (2)

In the following we will need the so called Hardy’s inequality, an extension of the Poincar´ e inequality to the case of singular potentials:

Z

R

 |u(z)|

kzk

 ^p

dz ≤ p ^p (N − p) ^p

Z

R

kDuk ^p dz (3)

for all u ∈ D ^1,p (R ^N ) = C _c ^∞ (R ^N ), the closure being taken in the norm kuk = kDuk p .

Proposition 1. If hypotheses H(β) hold, then ˆ λ ^∗ > 0.

Proof. For r > 0, let u ∈ W ₀ ^1,p (B r ) be such that R

B

β|u| ^p dz = 1. We have 1 =

Z

B

β|u| ^p dz = Z

B

(β ⁺ − β ⁻ )|u| ^p dz

≤ Z

B

β ⁺ |u| ^p dz = Z

B

β 1 |u| ^p dz + Z

B

β 2 |u| ^p dz.

(4)

By hypotheses H(β), we have that β 1 ∈ L ^N/p (R ^N ), and by the Sobolev Embed- ding Theorem we have that u ∈ L ^p

(B r ), where p ^∗ = N p/(N − p) is the critical Sobolev exponent. Note that

p N + p

p ^∗ = p

N + N − p

N = 1. (5)

So, using H¨ older’s and Sobolev’s inequalities, we have Z

B

β ₁ |u| ^p dz ≤ kβ ₁ k _L

(B

) kuk ^p _L

_(B

r

) ≤ c ₁ kβ ₁ k _L

(B

) kDuk ^p _L

_(B

r

,R

) (6) for some c 1 > 0. From the assumptions on β 2 in H(β), given ε > 0, we can find r 0 = r 0 (ε) > δ = δ(ε) > 0 such that

kzk ^p β ₂ (z) ≤ ε for all z ∈ R ^N , kzk ≥ r ₀ (7) and

kzk

β 2 (z) ≤ ε for all z ∈ B δ . (8) Let r > r ₀ be such that 1/r < δ and set Ω ₁ = B _r \ B r

. We have

Z

Ω

β 2 |u| ^p dz ≤ ε Z

Ω

|u| ^p

kzk ^p dz (by (7))

≤

 p

N − p

 ^p ε

Z

Ω

kDuk ^p dz (by (3))

= c 2 ε Z

Ω

kDuk ^p dz

(9)

with c 2 =  _p

N −p

 p

> 0.

Now let Ω ₂ = B _δ − B 1/r , so that Z

Ω

β ₂ |u| ^p dz ≤ ε Z

Ω

|u| ^p kzk

dz (by (8))

≤ ε



 Z

Ω

1 kzk

! N/p

dz





p/N

kuk ^p _L

_(Ω

2

)

(by H¨ older’s inequality, see (5))

≤ c 3 ε Z δ

1/r

1

s ^{N −1} s ^{N −1} ω _N ds

! ^p/N

kDuk ^p _L

(B

,R

)

by Sobolev’s inequality and for some c 3 > 0

= c 4 ε

 δ − 1

r

 ^p/N Z

B

kDuk ^p dz for some c 4 > 0;

(10)

Here ω N denotes the measure of the unit sphere in R ^N .

Next let Ω 3 = B r

\ B δ . The compactness of Ω 3 and hypotheses H(β) imply that we can find closed balls B ρ

(z k ) m

k=1 such that in correspondence of ε > 0 chosen above, there holds

kz − z k k ≤ ρ k ⇒ kz − z k k

β 2 (z) ≤ ε. (11) From (11) it follows that we can find ρ > 0 such that for all k ∈ {1, . . . , m}

we have ρ < ρ k and

kz − z _k k ≤ ρ ⇒ kz − z _k k

β ₂ (z) ≤ ε

m . (12)

Setting U = ∪ ^m _k=1 B ρ (z k ), we have Z

B

(z

)

β 2 |u| ^p dz ≤ ε m

Z

B

(z

)

|u| ^p kz − z k k

dz (by (12))

≤ ε m

"

Z

B

(z

)

1 kz − z k k

!# ^p/N

kuk ^p _L

_(B

r

)

by H¨ older’s inequality and (5)

≤ c ₅ ε m

"

Z

B

1 kzk ^{N −1} dz

# ^p/N

kDuk ^p _L

(B

,R

)

by Sobolev’s inequality

≤ c ₆ ε

m kDuk ^p _L

(B

,R

) ,

(13)

for some c ₅ > 0 and with c ₆ = (ω _N ρ) ^p/N .

Adding these inequalities over k ∈ {1, . . . , m} we obtain Z

U

β 2 |u| ^p dz ≤ c 6 εkDuk ^p _L

(B

,R

) . (14)

Finally, if z ∈ Ω ₃ \ U , then we can find k ∈ {1, . . . , m} such that ρ < kz − z k k ≤ ρ k .

Hence from (11) it follows that

β 2 (z) ≤ ε ρ

. (15)

Therefore, we have Z

Ω

\U

β ₂ |u| ^p dz ≤ ε ρ

Z

Ω

\U

|u| ^p dz (see (15))

≤ c ₇ εkDuk ^p _L

_(B

r

,R

) ,

(16)

where c 7 > 0 is a constant depending on r 0 and ρ, but not on r.

Now, we return to (15) and use (6), (9), (10), (14) and (15), obtaining 1 ≤ h

c 1 kβ 1 k _L

(B

) + c 2 ε + c 4 ε

 δ − 1

r

 ^p/N

+ (c 6 + c 7 )ε i Z

B

kDuk ^p dz.

Since u ∈ W ₀ ^1,p (B r ) with R

B

β|u| ^p dz = 1 was arbitrary, from (1) it follows that

h

c 1 kβk _L

(B

) + c 2 ε + c 4 ε

 δ − 1

r

 ^p/N

+ (c 6 + c 7 )ε i −1

≤ ˆ λ 1 (r).

Passing to the limit as r → ∞ in the previous inequality, we obtain 0 < h

c 1 kβk _L

(R

) + c 2 ε + c 4 εδ ^p/N + (c 6 + c 7 )ε i −1

≤ ˆ λ ^∗ , as claimed.

Remark 1. For Proposition 1, the requirement β 2 ∈ L ^∞ _loc (R ^N ) is sufficient, but in the future, namely for Proposition 2 and its consequences, we shall use β ∈ L ^∞ _loc (R ^N ).

We end up this section recalling that, if Y is a Banach space, Y ^∗ is its topological dual and E : Y → Y ^∗ is a map, we say that E is of type (S) + if

y n * y in Y and lim sup

n→∞

hE(y n ), y n − yi Y

,Y ≤ 0

(by h , i Y

,Y we denote the duality brackets for the pair (Y ^∗ , Y )), then y n → y in Y (for example, see Gasinski–Papageorgiou [16]).

3 The bifurcation theorem

The hypotheses on the absorption term f are:

H(f ): f : R ^N × R → R is a Carath´eodory function such that for a.e. z ∈ R ^N there holds f (z, 0) = 0, f (z, x) > 0 for all x > 0 and

(i): for every ρ > 0 there exists a ρ ∈ L ^∞ _loc (R ^N ) ∩ L ^(p

⁾

(R ^N ) (1/p ^∗ + 1/(p ^∗ ) ⁰ = 1) such that

f (z, x) ≤ a ρ (z) for a.e. z ∈ R ^N and all 0 ≤ x ≤ ρ;

(ii): there exists η ∈ L ^∞ (R ^N ) such that ϑ = ess inf

R

(η − β) > 0 and

lim inf

x→∞

f (z, x)

x ^p−1 ≥ η(z) uniformly for a.e. x ∈ R ^N ; (iii): we have

lim

x→0

f (z, x)

x ^p−1 = 0 uniformly for a.e. z ∈ R ^N .

Remark 2. Since we are interested in positive solutions and hypotheses H(f ) concern the positive semiaxis R ⁺ = [0, ∞) only, by truncating f if necessary, we may (and will) suppose that f (z, x) = 0 for a.e. z ∈ R ^N and all x ≤ 0.

Examples of functions satisfying H(f ) are easy to construct; as an example, if β ⁺ ∈ L ^∞ (R ^N ) ∩ L ^(p

⁾

(R ^N ), let us present

f (z, x) =

( β ⁺ (z)x ^p−1 if x ≥ 1, β ⁺ (z)x ^℘−1 if x < 1, where ℘ is any number strictly greater than p.

In what follows a fundamental rˆ ole will be played by the suitably weighted space

V = n

u ∈ D ^1,p (R ^N ) : kuk ^p _V = kDuk ^p _L

(R

,R

) + Z

R

β ⁻ |u| ^p dz < ∞ o , which turns out to be a reflexive Banach space, and in which we will look for solutions to problem (P _λ ).

Proposition 2. If hypotheses H(β), H(β) 2 , H(f ) hold and λ > ˆ λ ^∗ > 0, then problem (P _λ ) has a positive solution u ₀ ∈ C ¹ (R ^N ) ∩ V such that

u ₀ (z) > 0 for all z ∈ R ^N and u ₀ (z) → 0 as kzk → ∞.

Proof. The proof is rather long, and will need several intermediate steps.

By virtue of hypothesis H(f )(ii), given ε > 0, there exists M = M (ε) > 0 such that

f (z, x) ≥ (η(z) − ε)x ^p−1 for a.e z ∈ R ^N and all x ≥ M . (17) Let ξ > M . For any r > 0 and any h ∈ W ₀ ^1,p (B r ), h ≥ 0, we have

Z

B

(β(z)ξ ^p−1 − f (z, ξ))h(z) dz

≤ Z

B

(β(z)ξ ^p−1 − η(z)ξ ^p−1 + εξ ^p−1 )h(z) dz (see (17))

≤ ξ ^p−1 (ε − ϑ) Z

B

h dz (see hypothesis H(f )(ii)).

(18)

Choosing ε ∈ (0, ϑ), from (18) we infer that Z

B

(β(z)ξ ^p−1 − f (z, ξ))h(z) dz ≤ 0 for all h ∈ W ₀ ^1,p (B r ), h ≥ 0. (19) Now, we consider the following truncation for the reaction term:

g(z, x) =



 

 

0 if x ≤ 0,

β(z)x ^p−1 − f (z, x) if 0 ≤ x ≤ ξ, β(z)ξ ^p−1 − f (x, ξ) if x > ξ.

(20)

Of course, g is a Carhath´ eodory function. We set G(z, x) = R x

0 g(z, s) ds for all (z, x) ∈ B r × R, and we consider the C ¹ functional ψ λ : W ₀ ^1,p (B r ) → R defined by

ψ _λ (u) = 1

p kDuk ^p _p − λ Z

B

G(z, u(z)) dz for all u ∈ W ₀ ^1,p (B r ).

It is clear from (20) that ψ λ is coercive. Moreover, exploiting the compact embedding of W ₀ ^1,p (B r ) into L ^p (B r ), we can easily check that ψ λ is sequentially weakly lower semicontinuous. Therefore, by the Weierstrass Theorem, we can find u ^r ₀ ∈ W ₀ ^1,p (B r ) such that

ψ λ (u ^r ₀ ) = inf n

ψ λ (u) : u ∈ W ₀ ^1,p (B r ) o

= m r . (21)

Lemma 3. u ^r ₀ 6= 0.

Proof. By virtue of hypothesis H(f )(iii), given ε > 0, we can find δ = δ(ε) > 0 such that

f (z, x) ≤ εx ^p−1 for a.e. z ∈ B r and all x ∈ [0, δ], so that

F (z, x) ≤ ε

p x ^p for a.e. z ∈ B r and all x ∈ [0, δ]. (22) The ordered Banach space C ₀ ¹ (B _r ) = n

u ∈ C ¹ (B _r ) : u = 0 on ∂Ω has a positive cone given by C ₊ ^r = n

u ∈ C ₀ ¹ (B r ) : u(z) ≥ 0 for all z ∈ B r

o

, and this cone has a nonempty interior given by

int C ₊ ^r = n

u ∈ C ₊ ^r : u(z) > 0 for all z ∈ B _r and ^∂u _∂ν < 0 for all z ∈ ∂B _r o , where ν(z) denotes the unit outward normal to ∂B r at z.

Let ˆ u r ∈ int C ₊ ^r be the principal eigenfunction of the eigenvalue problem (E) (see Szulkin–Willem [24] for the existence and Vazquez [27] for the regularity).

We can find t ∈ (0, 1) so small that tˆ u _r (z) ∈ [0, δ] for all z ∈ B _r . Then ψ λ (tˆ u r ) = t ^p

p kDˆ u r k ^p _L

(B

,R

) − λt ^p p

Z

B

β ˆ u ^p _r dz + λ Z

B

F (z, tˆ u r ) dz (see (20))

≤ (ˆ λ ₁ (r) − λ)t ^p p

Z

B

β ˆ u ^p _r dz + λt ^p

p εkˆ u _r k ^p _p (see (22)).

(23) But R

B

β ˆ u ^p _r dz = 1, and for r > 0 large enough we have ˆ λ 1 (r) < λ (see Proposi- tion 1 and recall that λ > ˆ λ ^∗ ). So, if in (23) we choose ε > 0 sufficiently small, then we have ψ _λ (tˆ u _r ) < 0, and by (21)

m r = ψ λ (u ^r ₀ ) < ψ λ (0) = 0,

and therefore u ^r ₀ 6= 0, as claimed.

Lemma 4. u ^r ₀ ∈ int C ₊ ^r , and in addition u ^r ₀ ∈ [0, ξ] = n

u ∈ W ₀ ^1,p (B _r ) : 0 ≤ u(z) ≤ ξ a.e. in B _r o . Proof. From (21) we have (ψ λ ) ⁰ (u 0 ) = 0, that is

A(u ^r ₀ ) = λN g (u ^r ₀ ), (24) where A : W ₀ ^1,p (B _r ) → W ^−1,p

(B _r ) is the nonlinear map defined by

hA(u), yi = Z

B

kDuk ^p−2 (Du, Dy) _R

dz for all u, y ∈ W ₀ ^1,p (B _r ),

and N g (u)(·) = g(·, u(·)) for all u ∈ W ₀ ^1,p (B r ).

On (24) we first act with −(u ^r ₀ ) ⁻ ∈ W ₀ ^1,p (B _r ), and recalling (20), we obtain Z

B

kD(u ^r ₀ ) ⁻ k ^p dz = 0,

that is u ^r ₀ ≥ 0 and u ^r ₀ 6= 0 (from above).

Next we act on (24) with (u ^r ₀ − ξ) ⁺ ∈ W ₀ ^1,p (B r ), obtaining hA(u ^r ₀ ), (u ^r ₀ − ξ) ⁺ i = λ

Z

B

g(z, u ^r ₀ )(u ^r ₀ − ξ) ⁺ dz

= λ Z

B

(βξ ^p−1 − f (z, ξ))(u ^r ₀ − ξ) ⁺ dz (by (20))

≤ 0 (by (19)), that is

Z

{u

>ξ}

kDu ^r ₀ k ^p dz ≤ 0.

Thus, either |{u ^r ₀ > ξ}| N = 0, where | · | N denotes the Lebesgue measure in R ^N , or u ^r ₀ is constant in the set {u ^r ₀ > ξ}. Nonlinear regularity theory (see, for example, Gasinski–Papageorgiou [16, pp. 737–738]) implies that u ^r ₀ ∈ C ₀ ¹ (B r ), so that the second possibility is neglected, and thus u ^r ₀ ≤ ξ.

Therefore, we have proved that

u ^r ₀ ∈ [0, ξ].

Then (20) implies that (24) becomes

A(u ^r ₀ ) = λβ(u ^r ₀ ) ^p−1 − λN f (u ^r ₀ ), with N f (u)(·) = f (·, u(·)) for all u ∈ W ₀ ^1,p (B r ), that is

( −∆ p u ^r ₀ (z) = λβ(u ^r ₀ (z)) ^p−1 − f (z, u ^r ₀ (z))) in B _r ,

u ^r ₀ = 0 on ∂B r . (25)

Then

∆ p u ^r ₀ (z) ≤ λ(kβ ⁺ k _L

_(B

₎ u ^r ₀ (z) ^p−1 + f (z, x)) a.e. in B r . (26) The assumptions on f say that there exists C, δ > 0 such that

0 ≤ f (z, x) ≤

 1 + C

δ



|x| ^p−1 for a.e. z ∈ B r and all x ≥ 0. (27) Then, from (26) and (27) we have

∆ _p u ^r ₀ (z) ≤ λ



kβ ⁺ k _L

(B

) +

 1 + C

δ



u ^r ₀ (z) ^p−1 a.e. in B _r , and so u ^r ₀ ∈ int C ₊ ^r , see Vazquez [27] and Pucci–Serrin [22].

Lemma 5. Problem (25) has a smallest solution u belonging to [0, ξ] ∩ int C ₊ ^r . Proof. Set

S _ξ ^r = n

u ∈ W ₀ ^1,p (B _r ) : u is a nontrivial solution of (25) with u ∈ [0, ξ] o . We have just shown above that S _ξ ^r 6= ∅ and that S _ξ ^r ⊂ int C ₊ ^r . Using Lemma 4.3 of Filippakis–Kristaly–Papageorgiou [9], we infer that S _ξ ^r is downward di- rected, i.e., if u, v ∈ S _ξ ^r , then there exists y ∈ S _ξ ^r such that y ≤ min{u, v}. We now show that S _ξ ^r has a minimal element. To this purpose, let C ⊂ S _ξ ^r be a chain (i.e., a totally ordered subset of S _ξ ^r ). From Dunford–Schwartz [8, p. 336]

we know that we can find a sequence {u _n } n≥1 ⊂ C such that inf

n≥1 {u n } n≥1 = inf C.

Evidently, {u n } n≥1 ⊂ W ₀ ^1,p (B r ) is bounded, and so we may assume without loss of generality that

u n * u in W ₀ ^1,p (B r ), u n → u in L ^p (B r ) and a.e. in B r as n → ∞. (28) Since u _n ∈ S _ξ ^r for any n ∈ N, we have

A(u n ) = λβu ^p−1 _n − λN f (u n ) for all n ≥ 1. (29) On (29) we act with u n − u ∈ W ₀ ^1,p (B r ) and then pass to the limit as n → ∞.

By (28) we immediately obtain

n→∞ lim hA(u n ), u n − ui = 0,

and so, A being of type (S) + (see, for example, [16]), we get that

u n → u in W ₀ ^1,p (B r ). (30)

Thus, if in (29) we pass to the limit as n → ∞ and use (30), we obtain

A(u) = λβu ^p−1 − λN _f (u). (31)

By (28) we have that u ∈ [0, ξ]; we now show that u 6= 0. If u = 0, then from (30) we would have that ku n k = ku n k _W

0

(B

) → 0 as n → ∞, and so, up to a subsequence, u n → 0 a.e. in B r .

Let y _n = u _n /ku _n k, n ≥ 1. Then ky n k = 1 for all n ≥ 1 and so we may assume that

y n * y in W ₀ ^1,p (B r ), y n → y in L ^p (B r ) and a.e. in B r as n → ∞. (32) From (29) we have

A(y n ) = λβy ^p−1 _n − λ N _f (u _n )

ku n k ^p−1 , n ≥ 1. (33)

Acting on (33) with y n − y ∈ W ₀ ^1,p (B r ) and passing to the limit as n → ∞, by H(f )(iii) and (32) we obtain

n→∞ lim hA(y n ), y n − yi = 0, and thus

y n → y in W ₀ ^1,p (B r ), so that kyk = 1 and y ≥ 0. (34) From H(f )(i) it is clear that n _N

f

(u

) ku

k

o

n≥1 ⊂ L ^p

(B _r ) is bounded, and so we may assume that _ku ^N

^(u

⁾

n

k

* h in L ^p

(B r ) as n → ∞. Moreover, hypothesis H(f )(iii) implies h = 0. Therefore, if in (33) we pass to the limit as n → ∞, we obtain

A(y) = λβy ^p−1 . Equivalently,

( −∆ p y = λβy ^p−1 in B _r ,

y = 0 on ∂B r ,

that is (λ, y) solve problem (E), and so y ∈ C ₀ ¹ (B r ) must be nodal, since λ >

ˆ λ ₁ (r) (for example, see [6, Theorem 3.2]).

This contradicts (34) and this proves that u 6= 0. As before, via the nonlinear maximum principle (see [22], [27]), we have that u ∈ int C ₊ ^r , and so u ∈ S _ξ ^r . Since C was an arbitrary chain in S _ξ ^r , from the Kuratowski–Zorn Lemma we infer that S _ξ ^r has a minimal element u _∗ ∈ S _ξ ^r ⊂ int C ₊ ^r . If u ∈ S _ξ ^r , since S _ξ ^r is downward directed and due to the minimality of u _∗ , we have u _∗ ≤ u, and so u _∗ ∈ int C ₊ ^r is the smallest element of S _ξ ^r . This proves the lemma.

Now, let r _n ↑ ∞ be such that ˆ λ ₁ (r ₁ ) < λ, and hence ˆ λ ₁ (r _n ) < λ for all n ≥ 1. Let u ⁿ ₀ = u ^r ₀

, n ≥ 1. Extending by zero outside B _r

, we obtain a function ¯ u ⁿ ₀ ∈ W ^1,p (R ^N ). From the claim above we know that

0 ≤ ¯ u ⁿ ₀ ≤ ξ in R ^N for all n ≥ 1. (35)

We also know that A(¯ u ² ₀ ) = λβ(¯ u ² ₀ ) ^p−1 − λN f (¯ u ² ₀ ) in W ₀ ^1,p (B r

) ^∗ . In view of the first part of the proof, we make the truncation

g(z, x) =



 

 

0 if x ≤ 0,

β(z)x ^p−1 − f (z, x) if 0 ≤ x ≤ ¯ u ² ₀ (z), β(z)¯ u ² ₀ (z) ^p−1 − f (x, ¯ u ² ₀ (z)) if x > ¯ u ² ₀ (z).

In this way we obtain a solution of (26) with r = r ₁ which is located in the order interval [0, ¯ u ² ₀ 

 _B

r1

]. But ¯ u ² _{0   B}

r1

= u ¹ ₀ is the smallest solution of (26) (with r = r 1 ) in [0, ξ] ⊃ [0, ¯ u ² ₀ 

 _B

r1

] by (35). Thus ¯ u ¹ ₀ ≤ ¯ u ² ₀ . Continuing this way, by induction it follows that

0 < ¯ u ⁿ ₀ ≤ ¯ u ⁿ⁺¹ ₀ in R ^N for all n ≥ 1, (36) and thus there exists

u ₀ = lim

n→∞ u ¯ ⁿ ₀ in R ^N .

For every open and bounded domain Ω ⊂ R ^N with C ² boundary, there exists n 0 ∈ N such that Ω ⊂ B r

for all n ≥ n 0 ; from the nonlinear regularity theory (see, for example [16, p. 738]) we know that there exists α ∈ (0, 1) and M 0 > 0 such that

¯ u ⁿ ₀ 

 _{Ω ¯} ∈ C ^1,α ( ¯ Ω) and k¯ u ⁿ ₀ 

 _{Ω ¯} k _C

( ¯ Ω) ≤ M 0 for all n ≥ n 0 . (37) From the compact embedding of C ^1,α ( ¯ Ω) into C ¹ ( ¯ Ω) and (36), (37), we have

¯ u ⁿ ₀ 

 _{Ω ¯} → u _{0| ¯ Ω} in C ¹ ( ¯ Ω), so that

u 0 ∈ C ¹ (R ^N ) and u 0 > 0 in R ^N . We also know that

A(u ⁿ ₀ ) = λβ(u ⁿ _n ) ^p−1 − λN f (u ⁿ ₀ ) in W ₀ ^1,p (B r

) ^∗ for all n ≥ 1. (38) In (38) we act with u ⁿ ₀ ∈ W ₀ ^1,p (B _r

) and obtain

Z

B

kDu ⁿ ₀ k ^p dz = λ Z

B

β(u ⁿ ₀ ) ^p dz − λ Z

B

f (z, u ⁿ ₀ )u ⁿ ₀ dz

≤ λ Z

B

β(u ⁿ ₀ ) ^p dz (see H(f ))

≤ λ Z

B

β ⁺ (u ⁿ ₀ ) ^p dz

= λ Z

B

(β 1 + β 2 )(u ⁿ ₀ ) ^p dz (see H(β)).

(39)

We have Z

B

β 1 (u ⁿ ₀ ) ^p dz ≤ ξ ^p Z

B

|β 1 |dz (by (35))

≤ ξ ^p Z

R

|β 1 |dz

= c 9 = c 9 (ξ, kβk 1 ) > 0 (see H(β)).

(40)

Moreover, from the estimates found for (13) in the proof of Proposition 1, we have

Z

B

β 2 (u ⁿ ₀ ) ^p dz ≤ c 10 ε Z

B

kDu ⁿ ₀ k ^p dz (41)

for some c ₁₀ > 0.

We return to (39) and use (40), (41), obtaining Z

B

kDu ⁿ ₀ k ^p dz ≤ c 9 + c 10 ε Z

B

kDu ⁿ ₀ k ^p dz for all n ≥ 1. (42)

Now choose ε ∈ (0, 1) so small that c ₁₀ ε < 1. Then from (42) it follows that Z

B

kDu ⁿ ₀ k ^p dz ≤ c 11 for some c 11 > 0 and all n ≥ 1. (43)

But we know that D ¯ u ⁿ _{0 |B}

rn

= Du ⁿ ₀ and D ¯ u ⁿ _{0 |R}

\B

= 0 for all n ≥ 1, see, for example, Kesavan [20, p. 69,70]. Therefore, from (43) we have

kD¯ u ⁿ ₀ k _L

(R

,R

) ≤ c 11 for all n ≥ 1. (44) We also have

0 ≤ Z

R

kD¯ u ⁿ ₀ k ^p dz = λ Z

R

β(¯ u ⁿ ₀ ) ^p dz − λ Z

R

f (z, ¯ u ⁿ ₀ )¯ u ⁿ ₀ dz ≤ λ Z

R

β(¯ u ⁿ ₀ ) ^p dz by H(f ), so that

λ Z

R

β ⁻ (u ⁿ ₀ ) ^p dz ≤ λ Z

R

β ⁺ (u ⁿ ₀ ) ^p dz. (45) From the estimates in the proof of Proposition 1 and (44), (45) it follows

λ Z

R

β ⁺ (u ⁿ ₀ ) ^p dz ≤ c ₁₂ for some c ₁₂ > 0 and all n ≥ 1. (46)

Then from (44) and (46), we infer that n

¯ u ⁿ ₀ o

n≥1 ⊂ V is bounded, and so we may assume that

¯

u ⁿ ₀ * u 0 in V. (47)

Let ¯ A : V → V ^∗ be the nonlinear map defined by h ¯ A(y), vi _V

_,V =

Z

R

kDyk ^p−2 (Dy, Dv) _R

dz for all y, v ∈ V.

From Szulkin–Willem [24] (see also Huang [18]) we know that ¯ A is of type (S) +

in V . By the Claim above, we know that

A(¯ ¯ u ⁿ ₀ ) = λβ(¯ u ⁿ ₀ ) ^p−1 − λN f (¯ u ⁿ ₀ ) in V ^∗ for all n ≥ 1. (48) Acting on (48) with ¯ u ⁿ ₀ − u 0 ∈ V , we obtain

h ¯ A(¯ u ⁿ ₀ ), ¯ u ⁿ ₀ − u 0 i = λ Z

R

β(¯ u ⁿ ₀ ) ^p−1 (¯ u ⁿ ₀ − u 0 ) dz − λ Z

R

f (z, ¯ u ⁿ ₀ )(¯ u ⁿ ₀ − u 0 ) dz.

(49) By Szulkin–Willem [24], Lemma 4.2, we have

Z

R

β(¯ u ⁿ ₀ ) ^p−1 (¯ u ⁿ ₀ − u 0 ) dz

= Z

R

β ⁺ (¯ u ⁿ ₀ ) ^p−1 (¯ u ⁿ ₀ − u 0 ) dz − Z

R

β ⁻ (¯ u ⁿ ₀ ) ^p−1 (¯ u ⁿ ₀ − u 0 ) dz → 0

(50)

as n → ∞. Moreover, for any r > 0 

   Z

R

f (z, ¯ u ⁿ ₀ )(¯ u ⁿ ₀ − u 0 )dz    

≤ Z

R

a ξ |¯ u ⁿ ₀ − u 0 | dz (see (35) and H(f )(i))

= Z

B

a ξ |¯ u ⁿ ₀ − u 0 | dz + Z

R

\B

a ξ |¯ u ⁿ ₀ − u 0 | dz.

(51)

Evidently, by H(f ))(i) and the compact embedding of V in L ¹ _loc (R ^N ), we have

n→∞ lim Z

B

a ξ |¯ u ⁿ ₀ − u 0 | dz = 0. (52) Moreover, given ε > 0, using H¨ older’s inequality, we get

Z

R

\B

a _ξ |¯ u ⁿ ₀ − u ₀ | dz ≤ k¯ u ⁿ ₀ − u ₀ k _L

(R

\B

)

Z

R

\B

a ^(p _ξ

⁾

dz

! 1/(p

)

≤ c ₁₃ ε (53) for r > 0 large enough and some c ₁₃ > 0.

Therefore, if in (51) we pass to the limit as n → ∞ and use (52) and (53), we obtain

n→∞ lim     Z

R

f (z, ¯ u ⁿ ₀ )(¯ u ⁿ ₀ − u 0 ) dz    

≤ c 13 ε, and since ε > 0 was arbitrary, we conclude that

n→∞ lim Z

R

f (z, ¯ u ⁿ ₀ )(¯ u ⁿ ₀ − u 0 ) dz = 0. (54)

We now return to (49), pass to the limit as n → ∞ and use (50) and (54), obtaining

n→∞ lim h ¯ A(¯ u ⁿ ₀ ), ¯ u ⁿ ₀ − u 0 i V

,V = 0, and since ¯ A is of type (S) + , we finally get

¯

u ⁿ ₀ → u 0 in V. (55)

Note that for h ∈ V we have

|hN _f (¯ u ⁿ ₀ ) − N _f (u ₀ ), hi _V

_,V |

≤     Z

B

[f (z, ¯ u ⁿ ₀ ) − f (z, u 0 )]h dz    

+      Z

R

\B

[f (z, ¯ u ⁿ ₀ ) − f (z, u 0 )]h dz     

. (56) From the continuity of the Nemitsky operator on the bounded domain B _r , we have

n→∞ lim Z

B

[f (z, ¯ u ⁿ ₀ ) − f (z, u ₀ )]h dz = 0. (57) On the other hand, via H¨ older’s inequality, as in (53), we obtain

     Z

R

\B

[f (z, ¯ u ⁿ ₀ ) − f (z, u ₀ )]h dz     

≤ c ₁₄ ε (58)

for r > 0 large enough, for all n ≥ 1 and some c 14 > 0.

Thus, if in (56) we let n → ∞ and we use (57), (58) and the fact that ε > 0 was arbitrary, we conclude that

n→∞ lim hN f (¯ u ⁿ ₀ ) − N f (u 0 ), hi V

,V = 0 for all h ∈ V, that is

N f (¯ u ⁿ ₀ ) * N ^∗ f (u 0 ) in V ^∗ . (59) From (55) it follows that

Z

R

β ⁻ (¯ u ⁿ ₀ ) ^p−1 hdz → Z

R

β ⁻ u ^p−1 ₀ hdz for all h ∈ V, while, invoking [24, Lemma 4.2], we have

Z

R

β ⁺ (¯ u ⁿ ₀ ) ^p−1 h dz → Z

R

β ⁺ u ^p−1 ₀ h dz for all h ∈ V, so that

Z

R

β(¯ u ⁿ ₀ ) ^p−1 h dz → Z

R

βu ^p−1 ₀ h dz for all h ∈ V. (60) Therefore, if in (48) we pass to the limit as n → ∞ and we use (55), (59) and (60), obtaining

A(u ¯ 0 ) = λβu ^p−1 ₀ − λN f (u 0 ) in V ^∗ ,

and so u 0 ∈ C ¹ (R ^N ) ∩ V is a solution of (P λ ) and

u 0 > 0 in R ^N . Finally, we show that lim

kzk→∞ u 0 (z) = 0. For this, it is enough to prove that the problem

( −∆ p v = λβ ⁺ u ^p−1 ₀ ∈ L ^∞ _loc (R ^N ) in R ^N ,

v > 0 and v(z) → 0 as kzk → ∞ (61)

has a bounded solution. Indeed, if this is the case, we find

−∆ p v = λβ ⁺ u ^p−1 ₀ ≥ λβ ⁺ (¯ u ⁿ ₀ ) ^p−1 ≥ λβ(¯ u ⁿ ₀ ) ^p−1 − f (z, ¯ u ⁿ ₀ ) = −∆ p u ¯ ⁿ ₀ . Moreover, by (36), we know that 0 ≤ ¯ u ⁿ ₀ ≤ ξ and ¯ u ⁿ ₀ = 0 at infinity for every n ∈ N, so that ¯ u ⁿ ₀ ≤ v by Theorem 3.5.1 and the following Remark in [22]; hence the claim follows by (55). Thus we are reduced to consider problem (61) and to show that it admits a solution vanishing at infinity. But finding a bounded solution of (61) is equivalent to finding a bounded one for

( −∆ p w = λβ ⁺ u ^p−1 ₀ w ^q−1 in R ^N ,

w > 0 and w(z) → 0 as kzk → ∞ (62)

with q < p, see Abdellaoui-Peral [1, Theorem 4.6]. Hence, let us prove that (62) actually admits a C ¹ solution vanishing at infinity, as required. For this, it is enough to apply a result of Santos [23, Theorem 1.1], whose assumptions (H 1 ) and (H 2 ) are satisfied with a = β ⁺ u ^p−1 ₀ and b ≡ 0, while hypothesis (H 3 ) is a consequence of H(β) 2 and of the fact that u 0 is bounded.

Next, we prove that for λ ∈ [0, ˆ λ ^∗ ] problem (P _λ ) has no positive solutions.

Proposition 6. If hypotheses H(β) and H(f ) hold and λ ∈ (0, ˆ λ ^∗ ], then prob- lem (P λ ) has no positive solutions.

Proof. Suppose that (P λ ) has a positive solution u. Then, as usual, u ∈ C ¹ (R ^N ) ∩ V and

−∆ p u(z) = λ β(z)u(z) ^p−1 − f (z, u(z))
in R ^N . By definition of solution and by assumption, we have

Z

R

kDuk ^p dz = λ Z

R

[βu ^p − f (z, u)u] dz < λ Z

R

βu ^p dz (see H(f )). (63)

Now, choose ζ ∈ C _c ¹ (R ^N ) such that 0 ≤ ζ ≤ 1, ζ 

 _{kzk≤1} = 1 and ζ 

 _{kzk≥2} = 0.

Let ζ _n (z) = ζ ^z _n
and define u n = ζ _n u. Then u _n (z) → u(z) for all z ∈ R ^N .

Since u ∈ D ^1,p (R ^N ), from the Sobolev Embedding theorem, we have that u ∈

L ^p

(R ^N ). Since 0 ≤ u n ≤ u for any n ≥ 1, from the Lebesgue Dominated Convergence Theorem we infer that

u _n → u in L ^p

(R ^N ). (64)

Consider the ring R n = n

z ∈ R ^N : n < kzk < 2n o

. We have kDu n − Duk _L

(R

,R

) ≤ k(ζ n − 1)Duk _L

(R

,R

) + kuDζ n k _L

(R

,R

)

≤ k(ζ n − 1)Duk _L

(R

,R

) + kuk ^p _L

(R

) kDζ n k ^{N −p} _L

_(R

₎ (65) by H¨ older’s inequality.

Since Du ∈ L ^p (R ^N , R ^N ) and ζ n (z) → 1 for all z ∈ R ^N as n → ∞, from the Lebesgue Dominated Convergence Theorem we have

k(ζ n − 1)Duk _L

(R

,R

) → 0 as n → ∞. (66) Moreover, we have

kDζ n k _L

(R

) = kDζ n k _L

(R

,R

) = kDζk _L

(R

,R

) < ∞ (67) and

kuk _L

(R

) → 0 as n → ∞, since u ∈ L ^p

(R ^N ). (68) From (67) and (68) it follows that

kuk ^p _L

_(R

n

) kDζ n k ^{N −p} _L

(R

) → 0 as n → ∞. (69) We return to (65), pass to limit as n → ∞ and use (66) and (69), obtaining Du _n → Du in L ^p (R ^N , R ^N ). (70) We know that β ⁻ u ^p ∈ L ¹ (R ^N ) (recall that u ∈ V ), while from the proof of Proposition 1 we have β ⁺ u ^p ∈ L ¹ (R ^N ). Moreover, 0 ≤ β ^± u ^p _n ≤ β ^± u ^p . So, again via the Lebesgue Dominated Convergence Theorem we have

Z

R

β ^± u ^p _n dz → Z

R

β ^± u ^p dz as n → ∞, and so

Z

R

βu ^p _n dz → Z

R

βu ^p dz as n → ∞. (71)

From (63), (70) and (71) it follows that we can find n 0 ≥ 1 such that Z

R

kDu n k ^p dz < λ Z

R

βu ^p _n dz for all n ≥ n 0 . (72) But from the definition of ˆ λ ^∗ , we have

Z

R

kDu n k ^p dz ≥ ˆ λ ^∗ Z

R

βu ^p _n dz ≥ λ Z

R

βu ^p _n dz, (73)

since 0 < λ ≤ ˆ λ ^∗ and R

R

βu ^p _n dz > 0 by (72). Comparing (72) and (73) we reach a contradiction and this proves that for λ ∈ (0, ˆ λ ^∗ ] there is no positive solution to problem (P λ ).

In conclusion, summarizing the situation for problem (P λ ), by Propositions 2 and 6, we can state the following bifurcation theorem:

Theorem 7. If hypotheses H(β), H(β) 2 and H(f ) hold, then there exists ˆ λ ^∗ > 0 such that

(a) for every λ > ˆ λ ^∗ problem (P _λ ) has at least one positive solution u ∈ C ¹ (R ^N ) ∩ V vanishing at infinity;

(b) for all λ ∈ (0, ˆ λ ^∗ ] problem (P λ ) has no positive solutions.

Remark 3. We emphasize the fact that the quantities introduced in (1) and (2) provide a complete variational description of the bifurcation point ˆ λ ^∗ .

In the semilinear case (i.e. p = 2) we can improve this theorem; so from now on we consider the following problem:

(S λ )

( −∆u(z) = λ β(z)u − f (z, u(z))

in Ω, u(z) → 0 as kzk → ∞, u > 0, λ > 0

and N > 2. By strengthening hypotheses H(f ) we will prove uniqueness of the positive solution for problem (S _λ ) when λ > ˆ λ ^∗ . Indeed, the assumptions on the perturbation f are now the following:

H(f ) ⁰ : f : R ^N × R → R is a Carath´eodory function such that for a.e. z ∈ R ^N f (z, 0) = 0 and f (z, x) > 0 for all x > 0, hypotheses H(f ) ⁰ (i), (ii), (iii) are the same as the corresponding hypotheses H(f )(i), (ii), (iii) with p = 2 and (iv) for a.e. z ∈ R ^N the function x 7→ ^{f (z,x)} _x is strictly increasing in (0, ∞).

Proposition 8. If hypotheses H(β), H(β) ₂ and H(f ) ⁰ hold and λ > ˆ λ ^∗ , then the positive solution of (S _λ ) is unique.

Proof. The existence of a positive solution in C ¹ (R ^N ) ∩ V follows from Propo- sition 2. Now, suppose u, v are two solutions of (S _λ ). As before, using Lemma 4.3 of [9] and the truncation technique introduced in the proof of Proposition 2, we see that we may assume u ≤ v. Then, for any r > 0, using Green’s identity, we have

Z

B

(Du, Dv) _R

dz − Z

∂B

∂u

∂n v dσ = λ Z

B

[βu − f (z, u)]v dz (74)

and

− Z

B

(Dv, Du) _R

dz + Z

∂B

∂v

∂n u dσ = −λ Z

B

[βv − f (z, v)]u dz. (75)

Adding (74) and (75) we obtain Z

∂B

 ∂v

∂n u − ∂u

∂n v

 dσ = λ

Z

B

 f (z, v)

v − f (z, u) u



uv dz. (76) If we show that

r→∞ lim Z

∂B

 ∂v

∂n u − ∂u

∂n v



dσ = 0, (77)

then, if in (76) we pass to the limit as r → ∞ and use (77), we have λ

Z

R

 f (z, v)

v − f (z, u) u



uv dz = 0, which implies u ≡ v by H(f ) ⁰ (iv).

Thus, we conclude proving (77), which does not hold a priori, since u, v are functions in V and not in W ^1,p (R ^N ). However, if in (74) and (75) we pass to the limit, we get

Z

R

(Du, Dv) _R

dz − λ Z

R

[βu − f (z, u)]v dz = lim

r→∞

Z

∂B

∂u

∂n v dσ and

Z

R

(Dv, Du) _R

dz − λ Z

R

[βv − f (z, v)]u dz = lim

r→∞

Z

∂B

∂v

∂n u dσ.

Since both u and v solve problem (S _λ ), we get that both limits in the previous two identities equal 0, and so (77) holds.

Therefore, for problem (S _λ ) we can state the following bifurcation theorem:

Theorem 9. If hypotheses H(β), H(β) 2 and H(f ) ⁰ hold, then there exists ˆ λ ^∗ >

0 such that

(a) for every λ > ˆ λ ^∗ problem (S λ ) has a unique positive solution u ∈ C ¹ (R ^N )∩

V vanishing at infinity;

(b) for all λ ∈ (0, ˆ λ ^∗ ] problem (S _λ ) has no positive solutions.

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