positive solutions for parametric equidiffusive p-laplacian equations

Acta Mathematica Scientia 2014,34B(3):610–618

http://actams.wipm.ac.cn

POSITIVE SOLUTIONS FOR PARAMETRIC

EQUIDIFFUSIVE p-LAPLACIAN EQUATIONS∗

Leszek GASINSKI

Jagiellonian University, Faculty of Mathematics and Computer Science, ul. Lojasiewicza 6, 30-348

Krakow, Poland

E-mail : [email protected]

Nikolaos S. PAPAGEORGIOU

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

E-mail : [email protected]

Abstract We consider a parametric Dirichlet problem driven by the p-Laplacian with a

Caratheodory reaction of equidiffusive type. Our hypotheses incorporate as a special case

the equidiffusive p-logistic equation. We show that if λ1 > 0 is the principal eigenvalue of

the Dirichlet negative p-Laplacian and λ > λ1 (λ being the parameter), the problem has a

unique positive solution, while for λ ∈ (0, λ1], the problem has no positive solution.

Key words p-Laplacian; p-logistic equation; first eigenvalue; equidiffusive reaction; maxi-

mum principle; nonlinear regularity

2010 MR Subject Classification 35J25; 35J70

1 Introduction

Let Ω ⊆ RN be a bounded domain with a C2-boundary ∂Ω. We study the following

parametric Dirichlet problem:

(P )λ

−∆pu(z) = λu(z)p−1 − f(

z, u(z))

in Ω,

u|∂Ω = 0, λ > 0, u > 0,

with 1 < p < +∞. Here, ∆p denotes the p-Laplace differential operator, defined by

∆pu(z) = div(∥

∥∇u(z)∥

∥

p−2∇u(z)

)

∀u ∈W 1,p0 (Ω).

Also, f : Ω×R −→ R is a Caratheodory function, that is, for all ζ ∈ R, the function z 7−→ f(z, ζ)

is measurable and for almost all z ∈ Ω, the function ζ 7−→ f(z, ζ) is continuous. If p = 2 and

f(z, ζ) = f(ζ) = ζq−1 for ζ > 0 with q > 2, then problem (P )λ is the equidiffusive logistic

∗Received September 4, 2012. The first author is supported by the Marie Curie International Research

Staff Exchange Scheme Fellowship within the 7th European Community Framework Programme under Grant

Agreement No. 295118, the National Science Center of Poland under grant No. N N201 604640, the International

Project co-financed by the Ministry of Science and Higher Education of Republic of Poland under grant No.

W111/7.PR/2012, and the National Science Center of Poland under Maestro Advanced Project No. DEC-

2012/06/A/ST1/00262.

No.3 L. Gasinski & N.S. Papageorgiou: PARAMETRIC EQUIDIFFUSIVE p-LAPLACIAN EQUATIONS 611

equation, which is important in mathematical biology [12]. The case of the p-Laplacian and

with reaction (right hand side) equal to gλ(ζ) = λζp−1 − ζq−1 for ζ > 0, with q > p > 1,

was investigated by Guedda-Veron [9] (for N = 1, that is, ordinary differential equations) and

by Kamin-Veron [11] (for N > 2, that is, partial differential equations), who showed that for

λ > 0 sufficiently large, the solution uλ of the problem exhibits a flat core, that is, the set

Uλ = z ∈ Ω : uλ(z) = 1 is nonempty. Recently, the equidiffusive p-logistic equation was

investigated by Papageorgiou-Papalini [14] who focused on the existence of nodal (that is, sign

changing) solutions and by Gasinski-Papageorgiou [8] who deal with the Neumann boundary

condition problems.

In this article, we are interested on positive solutions for problem (P )λ and prove a re-

sult which describes precisely the positive solutions of (P )λ as the parameter λ > 0 varies.

Recently, the existence, nonexistence, and multiplicity of positive solutions for problems de-

pending on the parameter λ were studied by Gasinski-Papageorgiou [6, 7]. We also refer to

other article on boundary value problems (both equations and inclusions) such as Denkowski-

Gasinski-Papageorgiou [1, 2] and Gasinski-Papageorgiou [3–5], where particular eigenvalues λ

occur. Our hypotheses on the perturbation f(z, ζ) are such that incorporate as a special case the

equidiffusive p-logistic equation. Our approach is variational and also uses suitable truncation

and comparison techniques.

2 Mathematical Background and Hypotheses

The following simple lemma will be useful in the sequel.

Lemma 2.1 IfX is an ordered Banach space with order (positive) coneK and x0 ∈ intK,

then for every u ∈ K, we can find tu > 0, such that tux0 − u ∈ K (that is, u 6 tux0).

Proof As x0 ∈ intK, we can find r > 0, such that

Br(x0) ⊆ K,

where Br(x0) = x ∈ X : ‖x‖ < r. Let u ∈ K and tu >‖u‖r

. We have

x0 −1

tuu ∈ Br(x0) ⊆ K,

so tux0 − u ∈ K.

Consider the following nonlinear eigenvalue problem:

−∆pu(z) = λ∣

∣u(z)∣

∣

p−2u(z) in Ω,

u|∂Ω = 0.(2.1)

By an eigenvalue, we mean a real number λ ∈ R for which problem (2.1) has a nontrivial weak

solution u known as an eigenfunction corresponding to λ. The linear space spanned by the

eigenfunctions corresponding to the eigenvalue λ is the eigenspace corresponding to λ. It is well

known that there is a smallest eigenvalue λ1 > 0, which is simple (that is, the corresponding

eigenspace is one-dimensional) and isolated. It is the only eigenvalue with eigenfunctions of

constant sign. By u1 ∈ W 1,p0 (Ω) we denote the Lp-normalized (that is, ‖u1‖p = 1) eigenfunction

corresponding to λ1. As we already mentioned, we know that u1 has constant sign and we

612 ACTA MATHEMATICA SCIENTIA Vol.34 Ser.B

assume that u1 > 0. The eigenvalue λ1 > 0 admits the following variational characterization:

λ1 = inf

‖∇u‖pp

‖u‖pp

: u ∈W 1,p0 (Ω), u 6= 0

and the infimum is realized on the eigenspace corresponding to λ1 > 0.

Let

C10 (Ω) =

u ∈ C1(Ω) : u|∂Ω = 0

.

This is an ordered Banach space with positive cone

C+ =

u ∈ C10 (Ω) : u(z) > 0 for all z ∈ Ω

.

This cone has a nonempty interior, given by

intC+ =

u ∈ C+ : u(z) > 0 for all z ∈ Ω,∂u

∂n(z) < 0 for all z ∈ ∂Ω

,

where n(·) denotes the outward unit normal on ∂Ω.

Nonlinear regularity theory (see, for example, Papageorgiou-Kyritsi [13, pp. 309-312]), im-

plies that u1 ∈ C+ \ 0. Moreover, via the nonlinear maximum principle of Vazquez [15],

we have u1 ∈ intC+. For more details on the spectrum of(

− ∆p,W1,p0 (Ω)

)

, we infer to

Papageorgiou-Kyritsi [13]. Here, we only mention that for N > 2 and p 6= 2 (nonlinear eigen-

value problem), the spectrum of(

− ∆p,W1,p0 (Ω)

)

is far from completely known.

The hypotheses on the perturbation f(z, ζ) are the following:

H : f : Ω × R −→ R is a Caratheodory function, such that f(z, 0) = 0 and f(z, ζ) > 0 for

almost all z ∈ Ω, all ζ > 0 and

(i) there exist a ∈ L∞(Ω)+, c > 0, and r ∈ (p, p∗), with

p∗ =

Np

N − pif N > p,

+∞ if N ≤ p,

such that

f(z, ζ) 6 a(z) + cζr−1 for almost all z ∈ Ω, all ζ > 0;

(ii) we have

limζ→+∞

f(z, ζ)

ζp−1= +∞ uniformly for almost all z ∈ Ω;

(iii) we have

limζ→0+

f(z, ζ)

ζp−1= 0 uniformly for almost all z ∈ Ω;

(iv) for every > 0, there exist γ > 0 and τ = τ() > p, such that for almost all z ∈ Ω,

the map [0, ] ∋ ζ 7−→ γζτ−1 − f(z, ζ) is nondecreasing;

(v) for almost all z ∈ Ω,

the map (0,+∞) ∋ ζ 7−→f(z, ζ)

ζp−1is strictly increasing.

Remark 2.2 As we are interested in positive solutions and the above hypotheses concern

only the positive semiaxis R+ = [0,+∞), we may (and will) assume that

f(z, ζ) = 0 for almost all z ∈ Ω, all ζ 6 0.


Example 2.3 The following functions satisfy hypotheses H (for the sake of simplicity,

we drop the z-dependence):

f1(ζ) = ζr−1 ∀ζ > 0,

f2(ζ) =

cζϑ−1 − ζs−1 if ζ ∈ [0, 1],

ζp−1 ln ζ + (c− 1)ζ if 1 < ζ,

with r > p, p < ϑ < s, and c >s−pϑ−p

> 1.

Note that f1 corresponds to the standard equidiffusive p-logistic equation.

By a positive solution of problem (P )λ, we mean a function u ∈ W 1,p0 (Ω), u > 0, u 6= 0

which is a weak solution of (P )λ. Then, we have (see Papageorgiou-Kyritsi [13]):

−∆pu(z) = λu(z)p−1 − f(

z, u(z))

for almost all z ∈ Ω,

so u ∈ C+ \ 0 (by the nonlinear regularity theory; see Papageorgiou-Kyritsi [13]).

Let = ‖u‖∞ and let γ > 0 and τ = τ() > 0 be as postulated by hypothesis H(iv).

Then,

−∆pu(z) + γu(z)τ−1 = λu(z)p−1 − f(

z, u(z))

+ γu(z)τ−1

> 0 for almost all z ∈ Ω,

so

∆u(z) 6 (γτ−p)u(z)p−1 for almost all z ∈ Ω

and thus u ∈ intC+ (see Vazquez [15]).

So, if by S+(λ) we denote the set of positive solutions of problem (P )λ, then we have just

seen that S+(λ) ⊆ intC+.

In what follows, for any given measurable function h : Ω × R −→ R, set

Nh(u)(·) = h(

·, u(·))

∀u : Ω −→ R measurable.

Also, for any u ∈W 1,p0 (Ω), set

u+ = maxu, 0, u− = max−u, 0.

We know that u+, u− ∈ W 1,p0 (Ω) and |u| = u+ + u−. By | · |N we denote the Lebesgue measure

on RN . For every u ∈ W 1,p

0 (Ω), ‖u‖ = ‖∇u‖p (by the Poincare inequality). The same symbol

‖ · ‖ will be used to denote the norm in RN . However, no confusion is possible, since it will

always be clear from the context which one is in use.

By 〈·, ·〉 we denote the duality for the pair(

W−1,p′

(Ω) = W 1,p0 (Ω)∗,W 1,p

0 (Ω))

(where 1p

+1p′

= 1). Let A : W 1,p0 (Ω) −→ W−1,p′

(Ω) be the nonlinear map, defined by

⟨

A(u), y⟩

=

∫

Ω

‖∇u‖p−2(∇u,∇y)RN dz ∀u, y ∈W 1,p0 (Ω).

This map is continuous, bounded (that is, maps bounded sets to bounded sets), and maximal

monotone.

3 Positive Solutions

Let

Y =

λ > 0 : problem (P )λ has a positive solution

.


Proposition 3.1 If hypotheses H hold, then Y 6= ∅ and in fact (λ1,+∞) ⊆ Y.

Proof By virtue of hypotheses H , for a given ξ > λ1 and q ∈ (1, p), we can find c1 =

c1(ξ, q) > 0, such that

f(z, ζ) > ξζp−1 − c1ζq−1 for almost all z ∈ Ω, all ζ > 0. (3.1)

Let λ ∈ (λ1, ξ) and consider the following auxiliary Dirichlet problem:

−∆pu(z) = (λ− ξ)u+(z)p−1 + c1u+(z)q−1 in Ω,

u|∂Ω = 0.(3.2)

The energy functional ψλ : W 1,p0 (Ω) −→ R for problem (3.2) is defined by

ψλ(u) =1

p‖∇u‖p

p +ξ − λ

p‖u+‖p

p −c1q‖u+‖q

q ∀u ∈ W 1,p0 (Ω).

Evidently, ψλ ∈ C1(

W 1,p0 (Ω)

)

. As λ < ξ, we have

ψλ(u) >1

p‖u‖p − c2‖u‖

q ∀u ∈W 1,p0 (Ω),

for some c2 > 0. Since q < p, it follows that ψλ is coercive. Also, exploiting the compactness of

the embedding of W 1,p0 (Ω) into Lp(Ω), we can easily check that ψλ is sequentially weakly lower

semicontinuous. So, by the Weierstrass theorem, we can find uλ ∈ W 1,p0 (Ω), such that

ψλ(uλ) = inf

ψλ(u) : u ∈ W 1,p0 (Ω)

= mλ. (3.3)

Let u ∈ intC+ and t > 0. Then,

ψλ(tu) =tp

p‖∇u‖p

p +tp(ξ − λ)

p‖u‖p

p −c1t

q

q‖u‖q

q.

As q < p, by choosing t ∈ (0, 1) small, we see that

ψλ(tu) < 0,

so

ψλ(uλ) = mλ < 0 = ψλ(0)

(see (3.3)), that is, uλ 6= 0. From (3.3), we have

ψ′λ(uλ) = 0,

so

A(uλ) = (λ− ξ)(u+λ )p−1 + c1(u

+λ )q−1. (3.4)

Acting on (3.4) with −u−λ ∈W 1,p0 (Ω), we obtain

‖∇u−λ ‖pp = 0,

that is, uλ > 0, uλ 6= 0. Then, (3.4) becomes

A(uλ) = (λ− ξ)up−1λ + c1u

q−1λ ,

so

−∆puλ(z) = (λ− ξ)uλ(z)p−1 + c1uλ(z)q−1 for almost all z ∈ Ω,


thus,

−∆puλ(z) > λuλ(z)p−1 − f(

z, uλ(z))

for almost all z ∈ Ω (3.5)

(see (3.1)).

We consider the following truncation of the reaction term in problem (P )λ:

hλ(z, ζ) =

λζp−1 − f(z, ζ) if ζ 6 uλ(z),

λuλ(z)p−1 − f(z, uλ(z)) if uλ(z) < ζ.(3.6)

Evidently, this is a Caratheodory function. We set

Hλ(z, ζ) =

∫ ζ

0

hλ(z, s)ds

and consider the C1-functional ϕλ : W 1,p0 (Ω) −→ R, defined by

ϕλ(u) =1

p‖∇u‖p

p −

∫

Ω

Hλ

(

z, u(z))

dz ∀u ∈W 1,p0 (Ω).

From (3.6) it is clear that ϕλ is coercive. Also, it is sequentially weakly lower semicontinuous.

Therefore, by the Weierstrass theorem, we can find u0 ∈ W 1,p0 (Ω), such that

ϕλ(u0) = inf

ϕλ(u) : u ∈ W 1,p0 (Ω)

= mλ, (3.7)

so

ϕ′λ(u0) = 0,

thus

A(u0) = Nhλ(u0) (3.8)

and hence

−∆u0(z) = hλ

(

z, u0))

for almost all z ∈ Ω.

From nonlinear regularity theory (see Papageorgiou-Kyritsi [13]), we have u0 ∈ C10 (Ω). On

(3.8) we act with (u0 − uλ)+ ∈W 1,p0 (Ω). Then,

⟨

A(u0), (u0 − uλ)+⟩

=

∫

Ω

hλ(z, u0)(u0 − uλ)+dz

=

∫

Ω

(

λup−1λ − f(z, uλ)

)

(u0 − uλ)+dz

6⟨

A(uλ), (u0 − uλ)+〉

(see (3.6) and (3.5)), so∫

u0>uλ

(

‖∇u0‖p−2∇u0 − ‖∇uλ‖

p−2∇uλ, ∇u0 −∇uλ

)

RNdz 6 0. (3.9)

If 1 < p 6 2, we use the inequality which says that

c3(1 + ‖ξ‖ + ‖ξ′‖)2−p

‖ξ − ξ′‖26

(

‖ξ‖p−2ξ − ‖ξ′‖p−2‖ξ′, ξ − ξ′)

RN∀ξ, ξ′ ∈ R

N

for some c3 > 0. So, from (3.9), we have

c4∥

∥∇(u0 − uλ)+∥

∥

2

26 0,

for some c4 > 0 (recall that u0, uλ ∈ C10 (Ω)), so

u0 6 uλ.


If p > 2, then use the inequality which says that

c5‖ξ − ξ′‖p6

(

‖ξ‖p−2ξ − ‖ξ′‖p−2ξ′, ξ − ξ′)

RN, ∀ξ, ξ′ ∈ R

N

and some c5 > 0.

So, from (3.9), we have

c6∥

∥∇(u0 − uλ)+∥

∥

p

p6 0,

for some c6 > 0, thus

u0 6 uλ.

Finally, on (3.8) we act with −u−0 ∈ W 1,p0 (Ω) and obtain

‖∇u−0 ‖pp = 0

(see (3.6)), hence u0 > 0.

Therefore, we have proved that

u0 ∈ [0, uλ] =

u ∈W 1,p0 (Ω) : 0 6 u(z) 6 uλ(z) for almost all z ∈ Ω

.

By virtue of (3.6), the operator equation (3.8) becomes

A(u0) = λup−10 −Nf (u0),

so

−∆pu0(z) = λu0(z)p−1 − f

(

z, u0(z))

for almost all z ∈ Ω,

that is, u0 solves (P )λ.

It remains to show that u0 6= 0 and then λ ∈ Y. By virtue of hypothesis H(iii), for a given

ε > 0, we can find δ = δ(ε) > 0, such that

F (z, ζ) 6ε

pζp for almost all z ∈ Ω, all ζ ∈ [0, δ]. (3.10)

For t ∈ (0, 1) small, we have

tu1(z) ∈ [0, δ] ∀z ∈ Ω

and

tu1 6 uλ

(recall that u1, uλ ∈ intC+ and use Lemma 2.1). Then,

ϕλ(tu1) =tp

p‖∇u1‖

pp −

tpλ

p‖u1‖

pp +

∫

Ω

F(

z, tu1

)

dz 6tp

p

(

λ1 − λ+ ε)

(3.11)

(see (3.6), recall that ‖∇u1‖pp = λ1‖u1‖

pp, ‖u1‖p = 1 and use (3.10)).

Choose ε ∈ (0, λ− λ1) (recall that λ > λ1). Then, from (3.11), we have

ϕλ(tu1) < 0,

so

ϕλ(u0) = mλ < 0 = ϕλ(0)

(see (3.7)), thus,

u0 6= 0 and hence u0 ∈ intC+ is a positive solution of (P )λ.

We proved that λ ∈ Y and so Y 6= 0 and (λ1,+∞) ⊆ Y.

Let λ∗ = inf Y.


Proposition 3.2 If hypotheses H hold, then,

λ∗ = λ1 > 0 and λ∗ = λ1 6∈ Y.

Proof Let λ 6 λ1 and suppose that λ ∈ Y. Then, there exists uλ ∈ intC+, solution of

problem (P )λ. As u1 ∈ intC+, by Lemma 2.1, we can find t > 0, such that

tuλ 6 u1.

Let t > 0 be the largest such real number. We have

−∆p

(

tuλ(z))

= tp−1(

− ∆uλ(z))

= tp−1(

λuλ(z)p−1 − f(

z, uλ(z)))

< λ(

tuλ(z))p−1

6 λ1u1(z)p−1

= −∆pu1(z) for almost all z ∈ Ω

(since f(

z, uλ(z))

> 0, λ 6 λ1 and tuλ 6 u1), thus

u1 − tuλ ∈ intC+

(see Guedda-Veron [10, Proposition 2.2]), which contradicts the maximality of t > 0. Therefore,

λ 6∈ Y and so λ∗ = λ1 6∈ Y.

Proposition 3.3 If hypotheses H hold and λ > λ∗ = λ1, then problem (P )λ has a unique

positive solution.

Proof From Proposition 3.1, we already have the existence of a positive solution for (P )λ

for all λ > λ∗ = λ1. We need to show the uniqueness of this positive solution.

Let u, v be two positive solutions of (P )λ (λ > λ1). We know that u, v ∈ intC+. Let t > 0

be the largest real number, such that tu 6 v (see Lemma 2.1). Suppose that t ∈ (0, 1). For

= max

‖u‖∞, ‖v‖∞

, let γ > 0 and τ = τ > p be as postulated by hypothesis H(iv).

Then,

−∆p

(

tu(z))

+ γ

(

tu(z))τ−1

= tp−1(

λu(z)p−1 − f(

z, u(z)))

+ γ

(

tu(z))τ−1

(3.12)

for almost all z ∈ Ω. As we have assumed that t ∈ (0, 1), by hypothesis H(v), we have

f(z, tu(z))

(tu(z))p−1<f(z, u(z))

u(z)p−1for almost all z ∈ Ω,

so

f(

z, tu(z))

< tp−1f(

z, u(z))

for almost all z ∈ Ω. (3.13)

Using (3.13) in (3.12), we obtain

−∆p

(

tu(z))

+ γ

(

tu(z))τ−1

< λ(

tu(z))p−1

− f(

z, tu(z))

+ γ

(

tu(z))p−1

6 λv(z)p−1 − f(

z, v(z))

+ γv(z)τ−1

= −∆pv(z) + γv(z)τ−1 for almost all z ∈ Ω

(as tu 6 v; see hypothesis H(iv)), so

v − tu ∈ intC+

(see Papageorgiou-Papalini [14, Lemma 2]). This contradicts the maximality of t > 0. There-

fore, t > 1 and so u 6 v. Reversing the roles of u and v in the above argument, we also show


that v 6 u. Hence, u = v, which proves the uniqueness of the positive solution for problem

(P )λ, λ > λ1.

The next theorem summarizes the precise dependence of the positive solutions of (P )λ on

the parameter λ > 0. We stress that we identify explicitly the critical parameter λ∗ > 0.

Theorem 3.4 If hypotheses H hold, then,

(a) for all λ > λ∗ = λ1 > 0, problem (P )λ has a unique solution uλ ∈ intC+;

(b) for all λ ∈ (0, λ1], problem (P )λ has no positive solution.

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positive solutions for parametric equidiffusive p-laplacian equations

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