Existence and multiplicity results for a doubly anisotropic problem with

Existence and multiplicity results for a doubly anisotropic problem with

sign-changing nonlinearity

Ahmed R´ eda Leggat

Laboratoire d’Analyse Non Lin´ eaire et Math´ ematiques Appliqu´ ees, Tlemcen University, BP 119 Tlemcen. Algeria

leggat2000@yahoo.fr

Sofiane El-Hadi Miri

Laboratoire d’Analyse Non Lin´ eaire et Math´ ematiques Appliqu´ ees, Tlemcen University, BP 119 Tlemcen. Algeria

mirisofiane@yahoo.fr

Received: 8.5.2018; accepted: 24.6.2019.

Abstract. We consider in this paper the following problem







−

N

P

i=1

∂

i

|∂

i

u|

^pi−2

∂

i

u − P

^N

i=1

∂

i

|∂

i

u|

^qi−2

∂

i

u = λf (u) in Ω,

u = 0 on ∂Ω.

Where Ω is a bounded regular domain in R

^N

, 1 < p

1

≤ p

2

≤ ... ≤ p

N

and 1 < q

1

≤ q

2

≤ ... ≤ q

N

, we will also assume that f is a continuous function, that have a finite number of zeroes, changing sign between them.

Keywords: Anisotropic problem, exitence and mutiplicity, variational methods.

MSC 2000 classification: primary 35J66, secondary 35J35

1 Introduction







−

N

P

i=1

∂ i

h

|∂ _i u| ^p

ⁱ

⁻² ∂ i u i

−

N

P

i=1

∂ i

h

|∂ _i u| ^q

ⁱ

⁻² ∂ i u i

= λf (u) in Ω,

u = 0 on ∂Ω.

(1.1)

Where Ω is a bounded regular domain in R ^N , we will assume that f fulfill some suitable hypotheses, 1 < p ₁ ≤ p ₂ ≤ ... ≤ p _N and 1 < q ₁ ≤ q ₂ ≤ ... ≤ q _N .

We will often use the notation

http://siba-ese.unisalento.it/ c 2019 Universit` a del Salento

L _(p

_i

₎ u =

N

X

i=1

∂ _i h

|∂ _i u| ^p

ⁱ

⁻² ∂ _i u i ,

There is a huge literature related to the anisotropic operator, when con- sidered with a linear, non-linear or singular terms we invite the reader to see [1, 4, 5, 6, 7, 9, 13, 25]

f is supposed to be such that

(H1)f is a continuous function such that f (0) ≥ 0, and there are 0 < a ₁ <

b ₁ < a ₂ < ... < b _m−1 < a _m the zeroes of f such that

 f ≤ 0 in (a k , b k ) f ≥ 0 in (b _k , a _k+1 )

(H2)

a

k+1

R

a

_k

f (t)dt > 0; ∀k = 1, 2, .., m − 1.

These kind of hypotheses, was introduced by different authors in the some early works [3, 8, 11], with the aim to study the problem

 −4u = λf (u) in Ω,

u = 0 on ∂Ω.

More recently, the results obtained there was generalized in [2] for the p&q- laplacian, that is

 −4 _p u − 4 _q u = λf (u) in Ω,

u = 0 on ∂Ω,

and in the case of the φ-laplacian in [18], where the considered problem

 −div(φ(|∇u|)∇u) = λf (u) in Ω,

u = 0 on ∂Ω,

φ being a function fulfilling some suitable conditions.

Observe that the anisotropic operator, and the doubly anisotropic operator considered in this paper cannot be obtained as a particular case of the previous cited, and his its own structure, as we will present in this paper.

In the whole paper C will denote a constant that may change from line to

line.

2 Preliminary results

Problem (1.1) is associated to the following anisotropic Sobolev spaces W ^1,(p

ⁱ

⁾ (Ω) = v ∈ W ^1,1 (Ω) ; ∂ i v ∈ L ^p

ⁱ

(Ω)

and

W ₀ ^1,(p

ⁱ

⁾ (Ω) = W ^1,(p

ⁱ

⁾ (Ω) ∩ W ₀ ^1,1 (Ω) endowed by the usual norm

kvk W

₀^1,(pi)

(Ω) =

N

X

i=1

k∂ _i vk _L

_pi

_(Ω) .

As we are dealing with a doubly anisotropic operator, the natural functional space is

X = W ₀ ^1,(p

ⁱ

⁾ (Ω) ∩ W ₀ ^1,(q

ⁱ

⁾ (Ω) endowed with the norm

kvk _X = kvk

W

₀^1,(pi)

(Ω) + kvk

W

₀^1,(qi)

(Ω) .

Definition 1. We will say that u ∈ W ₀ ^1,(p

ⁱ

⁾ (Ω) is a weak solution to (1.1) if and only if

N

X

i=1

Z

Ω

|∂ _i u| ^p

ⁱ

⁻² ∂ _i u∂ _i ϕ+

N

X

i=1

Z

Ω

|∂ _i u| ^q

ⁱ

⁻² ∂ _i u∂ _i ϕ = λ Z

Ω

f (u)ϕ ∀ϕ ∈ W ₀ ^1,(p

ⁱ

⁾ (Ω) .

We will also use very often the following indices 1

p = 1 N

N

X

i=1

1 p _i and

p ^∗ = N p

N − p , p ∞ = max {p _N , p ^∗ } without loss of generality we will assume that p ^∗ ≤ q ^∗

The following Sobolev type inequalities will be often used in this paper, we

refer to the early works [23], [16] and [20].

Theorem 1. There exists a positive constant C, depending only on Ω, such that for every v ∈ W ₀ ^1,(p

ⁱ

⁾ (Ω) , we have

kvk ^p

^N

L

^p∗

(Ω) ≤ C

N

X

i=1

k∂ _i vk ^p _L

ⁱ_pi

_(Ω) , (2.1)

kvk _L

r

(Ω) ≤ C

N

X

i=1

k∂ _i vk _L

_pi

_(Ω) ∀r ∈ [1, p ^∗ ] (2.2)

kvk _L

r

(Ω) ≤ C

N

Y

i=1

k∂ _i vk

1 N

L

^pi

(Ω) ∀r ∈ [1, p ^∗ ] (2.3) and ∀v ∈ W ₀ ^1,(p

ⁱ

⁾ (Ω) ∩ L ^∞ (Ω) , p < N



 Z

Ω

|v| ^r





N p

−1

≤ C

N

Y

i=1



 Z

Ω

|∂ _i v| ^p

ⁱ

|v| ^t

ⁱ

^p

ⁱ





1 pi

, (2.4)

for every r and t j chosen such a way to have



 

 

1

r = ^γ

ⁱ

^{(N −1)−1+}

1 pi

t

i

+1 N

P

i=1

γ _i = 1.

We also have the following algebraic inequalities :

• There exists a C > 0 not depending on ρ ∈ (0, 1) such that for given σ i > 0, i = 1, 2...N we have

N

X

i=1

σ i = ρ =⇒

N

X

i=1

σ _i ^p

ⁱ

p i

≥ Cρ ^p

^N

(2.5)

• For p _i ≥ 2

C |a − b| ^p

ⁱ

≤ 

|a| ^p

ⁱ

⁻² a − |b| ^p

ⁱ

⁻² b 

(a − b) (2.6)

• For 1 < p _i ≤ 2

C |a − b| ²

(|a| + |b|) ^2−p

ⁱ

≤ 

|a| ^p

ⁱ

⁻² a − |b| ^p

ⁱ

⁻² b 

(a − b) (2.7)

In view of applying the above inequalities, allthrough this pper we will

suppose that all the p i are neither p i ≥ 2 nor 1 < p _i ≤ 2 and the same for

the q _i for i = 1, ..., N.

Lemma 1. Let g ∈ C(R) be a continuous function and s 0 > 0 be such that g(s) ≥ 0 if s ∈ (−∞, 0)

g(s) ≤ 0 if s ∈ [s ₀ , +∞) then if u is a solution of







−

N

P

i=1

∂ i

h

|∂ _i u| ^p

ⁱ

⁻² ∂ i u i

−

N

P

i=1

∂ i

h

|∂ _i u| ^q

ⁱ

⁻² ∂ i u i

= λg(u)

u = 0 on ∂Ω

(2.8)

it verifies u ≥ 0 a.e.in Ω, u ∈ L ^∞ (Ω) and kuk _L

∞

< s ₀ .

Proof. We recall that u = u ⁺ −u ⁻ where u ⁻ = max(−u, 0) and u ⁺ = max(0, u);

as ∂ _i u ⁻ =

 −∂ _i u if u < 0

0 if u ≥ 0 we have that u ⁻ ∈ W ₀ ^1,(p

ⁱ

⁾ (Ω) whenever u ∈ W ₀ ^1,(p

ⁱ

⁾ (Ω) . Using u ⁻ as a test function in (2.8) we obtain

N

X

i=1

Z

Ω

|∂ _i u| ^p

ⁱ

⁻² ∂ _i u∂ _i u ⁻ +

N

X

i=1

Z

Ω

|∂ _i u| ^q

ⁱ

⁻² ∂ _i u∂ _i u ⁻ = Z

Ω

g(u)u ⁻

that is

N

X

i=1

Z

Ω∩[u<0]

|∂ _i u| ^p

ⁱ

+

N

X

i=1

Z

Ω∩[u<0]

|∂ _i u| ^q

ⁱ

= Z

Ω∩[u<0]

g(u)u

by the definition of g, g(u)u ≤ 0 when u < 0 so

N

X

i=1

Z

Ω∩[u<0]

|∂ _i u| ^p

ⁱ

+

N

X

i=1

Z

Ω∩[u<0]

|∂ _i u| ^q

ⁱ

≤ 0

and thus necessarily the set (Ω ∩ [u < 0]) is a null measure set, and so u = u ⁺ ≥ 0.

On the other hand observe that ∂ i (u − s 0 ) ⁺ =

 +∂ i u if u > s 0

0 if u ≤ s ₀ we have that (u − s ₀ ) ∈ X whenever u ∈ X. Using (u − s ₀ ) ⁺ as test function in (2.8) we obtain

N

X

i=1

Z

Ω

|∂ _i u| ^p

ⁱ

⁻² ∂ i u∂ i (u − s 0 ) ⁺ +

N

X

i=1

Z

Ω

|∂ _i u| ^q

ⁱ

⁻² ∂ i u∂ i (u − s 0 ) ⁺ = Z

Ω

g(u) (u − s 0 ) ⁺

that is

N

X

i=1

Z

Ω∩[u>s

0

]

|∂ _i u| ^p

ⁱ

+

N

X

i=1

Z

Ω∩[u>s

0

]

|∂ _i u| ^q

ⁱ

= Z

Ω∩[u>s

0

]

g(u) (u − s 0 )

by the definition of g, g(u) (u − s 0 ) ≤ 0

N

X

i=1

Z

Ω∩[u>s

0

]

|∂ _i u| ^p

ⁱ

+

N

X

i=1

Z

Ω∩[u>s

0

]

|∂ _i u| ^q

ⁱ

≤ 0

and thus necessarily the set (Ω ∩ [u > s 0 ]) is a null measure set, and so u ≤

s 0 .

^QED

3 Existence and multiplicity results

For each k = 1, 2, .., m − 1, consider the following problem







−

N

P

i=1

∂ i

h

|∂ _i u| ^p

ⁱ

⁻² ∂ i u i

−

N

P

i=1

∂ i

h

|∂ _i u| ^p

ⁱ

⁻² ∂ i u i

= λf _k (u)

u = 0 on ∂Ω

(3.1)

where

f _k (s) =







f (0) if s ≤ 0 f (s) if 0 ≤ s ≤ a _k

0 s > a k

Proposition 1. There exists λ > 0 such that for every λ ∈ (λ, +∞), prob- lem (3.1) posses a nonnegative solution u = u _k,λ such that ku _k k _L

∞

≤ a _k . Proof. As a direct consequence of Lemma 1 we have ku _k k _L

∞

≤ a _k .

Let

Φ _k,λ (u) :=

N

X

i=1

1 p i

Z

Ω

|∂ _i u| ^p

ⁱ

+

N

X

i=1

1 q i

Z

Ω

|∂ _i u| ^q

ⁱ

− λ Z

Ω

F _k (u),

where F _k (t) =

t

R

0

f _k (s)ds, the set C _k,λ of the critical points of Φ _k,λ (u) corre-

sponds to the set of solution to (3.1). Observe that as f k is a bounded function

we have

m _k |t| ≤

t

R

0

m _k ds ≤ F _k (t) ≤

t

R

0

M _k ds ≤ M _k |t| , thus

Φ k,λ (u) =

N

X

i=1

1 p i

Z

Ω

|∂ _i u| ^p

ⁱ

+

N

X

i=1

1 q i

Z

Ω

|∂ _i u| ^q

ⁱ

− λ Z

Ω

F k (u)

≥

N

X

i=1

1 p i

Z

Ω

|∂ _i u| ^p

ⁱ

+

N

X

i=1

1 q i

Z

Ω

|∂ _i u| ^q

ⁱ

− λM _k Z

Ω

|u| ,

by H¨ older inequality we obtain that

Φ _k,λ (u) ≥

N

X

i=1

1 p _i

Z

Ω

|∂ _i u| ^p

ⁱ

+

N

X

i=1

1 q _i

Z

Ω

|∂ _i u| ^q

ⁱ

− λM _k kuk _L

p∗

,

by Sobolev inequality

Φ _k,λ (u) ≥

N

X

i=1

1 p i

Z

Ω

|∂ _i u| ^p

ⁱ

+

N

X

i=1

1 q i

Z

Ω

|∂ _i u| ^q

ⁱ

− λM _k C

N

X

i=1

k∂ _i uk _L

_pi

,

as p _N ≥ p _i for every i

Φ _k,λ (u) ≥ 1 p _N

N

X

i=1

k∂ _i uk ^p _L

ⁱ_pi

+ 1 q _N

N

X

i=1

k∂ _i uk ^q _L

ⁱ_qi

− λM _k C

N

X

i=1

k∂ _i uk _L

_pi

,

by the fact that

kuk _X → +∞ ⇒ k∂ _i uk _L

pi

→ +∞ or k∂ _i uk _L

qi

→ +∞ for some i, we obtain the coercivity of Φ _k,λ (u) that is

Φ _k,λ (u) → +∞ when kuk _X → +∞.

On the other hand, as Φ _k,λ (u) continuous it is also lower semi continuous, and thus by Weirstrass theorem, it is also possible to show that a Palais Smail s´ equence {u n } _n converges strongly, indeed as {u n } _n is bounded in X

u n * u weakly in X, thus

u _n → u strongly in L ^r (Ω) for evry 1 ≤ r < p ^∗ ,

and in particular

Z

Ω

|u _n | → Z

Ω

|u| ; using (u n − u) as test function in (3.1) we obtain

N

X

i=1

Z

Ω

|∂ _i u| ^p

ⁱ

⁻² ∂ i u∂ i (u n −u)+

N

X

i=1

Z

Ω

|∂ _i u| ^q

ⁱ

⁻² ∂ i u∂ i (u n −u) = λ Z

Ω

f k (u)(u n −u)

which gives

N

X

i=1



 Z

Ω

 |∂ _i u _n | ^p

ⁱ

⁻² ∂ _i u _n − |∂ _i u| ^p

ⁱ

⁻² ∂ _i u 

∂ _i (u _n − u) + ∂ _i u _n |∂ _i u| ^p

ⁱ

⁻² ∂ _i u − |∂ _i u| ^p

ⁱ







+

N

X

i=1



 Z

Ω

 |∂ _i u _n | ^q

ⁱ

⁻² ∂ _i u _n − |∂ _i u| ^q

ⁱ

⁻² ∂ _i u 

∂ _i (u _n − u) + ∂ _i u _n |∂ _i u| ^q

ⁱ

⁻² ∂ _i u − |∂ _i u| ^q

ⁱ







= λ

Z

Ω

f _k (u)(u _n − u),

that is

N

X

i=1



 Z

Ω



|∂ _i u n | ^p

ⁱ

⁻² ∂ i u n − |∂ _i u| ^p

ⁱ

⁻² ∂ i u



∂ i (u n − u) + Z

Ω



∂ i u n |∂ _i u| ^p

ⁱ

⁻² ∂ i u − |∂ i u| ^p

ⁱ







+

N

X

i=1



 Z

Ω



|∂ _i u n | ^q

ⁱ

⁻² ∂ i u n − |∂ _i u| ^q

ⁱ

⁻² ∂ i u



∂ i (u n − u) + Z

Ω



∂ i u n |∂ _i u| ^q

ⁱ

⁻² ∂ i u − |∂ i u| ^q

ⁱ







= λ

Z

Ω

f k (u)(u n − u)

by inequality (2.6) for p _i , q _i > 2

N

X

i=1

Z

Ω

|∂ _i (u n − u)| ^p

ⁱ

+

N

X

i=1

Z

Ω



∂ i u n |∂ _i u| ^p

ⁱ

⁻² ∂ i u − |∂ i u| ^p

ⁱ

 +

N

X

i=1

Z

Ω

|∂ _i (u n − u)| ^q

ⁱ

+

+

N

X

i=1

Z

Ω



∂ i u n |∂ _i u| ^q

ⁱ

⁻² ∂ i u − |∂ i u| ^q

ⁱ



≤ λC Z

Ω

f k (u)(u n − u)

N

X

i=1

Z

Ω

|∂ _i (u n − u)| ^p

ⁱ

+

N

X

i=1

Z

Ω

|∂ _i (u n − u)| ^q

ⁱ

≤ λC Z

Ω

f _k (u)(u n − u) −

N

X

i=1

Z

Ω



∂ i u n |∂ _i u| ^p

ⁱ

⁻² ∂ i u − |∂ i u| ^p

ⁱ



−

N

X

i=1

Z

Ω



∂ i u n |∂ _i u| ^q

ⁱ

⁻² ∂ i u − |∂ i u| ^q

ⁱ

 ,

by the weak convergence of {u _n } _n

−

N

X

i=1

Z

Ω



∂ _i u _n |∂ _i u| ^p

ⁱ

⁻² ∂ _i u − |∂ _i u| ^p

ⁱ



−

N

X

i=1

Z

Ω



∂ _i u _n |∂ _i u| ^q

ⁱ

⁻² ∂ _i u − |∂ _i u| ^q

ⁱ



= o(1)

thus

N

X

i=1

Z

Ω

|∂ _i (u n − u)| ^p

ⁱ

+

N

X

i=1

Z

Ω

|∂ _i (u n − u)| ^q

ⁱ

≤ λC Z

Ω

f _k (u)(u n − u) + o(1)

≤ λCM _k Z

Ω

(u n − u) + o(1)

as u _n → u strongly in L ^r (Ω) for every 1 ≤ r < p ^∗ , we conclude that ku _n − uk _X → 0.

The same result can be obtained for the cases (p i < 2 and q i > 2) and

(p _i < 2 and q _i < 2) by using in a smililar way inequality (2.7) instead of (2.6),

which ends the proof.

^QED

Theorem 2. There exists λ > 0 such that for every λ ∈ (λ, +∞), problem (1.1) posses at least (m − 1) nonnegative solutions u _i such that u _i ∈ X and a _i ≤ ku _i k _L

∞

≤ a _i+1 .

Proof. Let u be a solution of (3.1), so by lemma1 it is necessarily such that , u ∈ L ^∞ (Ω) and 0 ≤ u < a k−1 a.e.in Ω thus f k−1 (u) = f (u) and then u is also a solution to (1.1). To prove the last part of the theorem we claim that for each k

∈ {2, ...m} there is λ _k > 0,such that for all λ > λ _k we have u _k,λ ∈ C / _k−1,λ where Φ k,λ (u k,λ ) = min

v∈X Φ k,λ (v), first let δ > 0 and consider

Ω _δ = {x ∈ Ω, dist(x, ∂Ω) < δ} ,

and

α _k = F (a _k ) − max

0<s<a

k−1

|F (s)| = F (a _k ) − C _k , by hypothesis (H2) α _k > 0. Consider w _δ ∈ C ₀ ^∞ (Ω) such that

0 ≤ w _δ ≤ a _k , and

w δ = a k , when x ∈ Ω r Ω δ ,

we have Z

Ω

F (w _δ ) ≥ Z

Ω

F (a _k ) − 2C _k |Ω _δ | , which yields to

Z

Ω

F (w _δ ) − Z

Ω

F (u) ≥ α _k |Ω| − 2C _k |Ω _δ | , since |Ω _δ | → 0 as δ → 0 there must exit a δ such that

β _k = α _k |Ω| − 2C _k |Ω _δ | > 0 for that δ we put w _δ = w, we have

Φ _k,λ (w) − Φ _k−1,λ (u _k−1,λ ) =

N

X

i=1

1 p _i

Z

Ω

|∂ _i w| ^p

ⁱ

+

N

X

i=1

1 q _i

Z

Ω

|∂ _i w| ^q

ⁱ

− λ Z

Ω

F _k (w)

−

N

X

i=1

1 p _i

Z

Ω

|∂ _i u _k−1,λ | ^p

ⁱ

−

N

X

i=1

1 q _i

Z

Ω

|∂ _i u _k−1,λ | ^q

ⁱ

+ λ Z

Ω

F _k (u _k−1,λ )

≤

N

X

i=1

1 p i

Z

Ω

|∂ _i w| ^p

ⁱ

+

N

X

i=1

1 q i

Z

Ω

|∂ _i w| ^q

ⁱ

− λ Z

Ω

(F k (w) − F k (u k−1,λ ))

≤

N

X

i=1

1 p i

Z

Ω

|∂ _i w| ^p

ⁱ

+

N

X

i=1

1 q i

Z

Ω

|∂ _i w| ^q

ⁱ

− λβ _k ,

for λ large enough we have

Φ k,λ (w) − Φ k−1,λ (u k−1,λ ) < 0 that is

Φ _k,λ (w) < Φ _k−1,λ (u _k−1,λ )

so

Φ _k,λ (u _k,λ ) ≤ Φ _k,λ (w) < Φ _k−1,λ (u _k−1,λ )

so we have proved that u _k,λ and u _k−1,λ are two distinct solutions to (1.1). Now suppose by contradiction that

0 ≤ u _k,λ < a _k−1 we necessarily would have

Φ k−1,λ (u k−1,λ ) ≤ Φ k−1,λ (u k,λ ) = Φ k,λ (u k,λ ) wich is a contradiction, and in conclusion

a k−1 < ku k,λ k _L

∞

≤ a _k .

which ends the proof.

^QED

Remark 1. Obviously, and under the same conditions on f, all the results obtained here are still valid for the following simply anisotropic problem:







− P ^N

i=1

∂ _i h

|∂ _i u| ^p

ⁱ

⁻² ∂ _i u i

= λf (u) in Ω,

u = 0 on ∂Ω.

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