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For this purpose, we combine a blowup argument, the strong maximum principle, and Liouville-type theorems to obtain a priori estimates.. We will obtain a priori estimate to positive solu

Volume 2007, Article ID 47218, 12 pages

doi:10.1155/2007/47218

Research Article

Existence and Multiplicity Results for Degenerate Elliptic

Equations with Dependence on the Gradient

Leonelo Iturriaga and Sebastian Lorca

Received 17 October 2006; Revised 2 January 2007; Accepted 9 February 2007

Recommended by Shujie Li

We study the existence of positive solutions for a class of degenerate nonlinear elliptic equations with gradient dependence For this purpose, we combine a blowup argument, the strong maximum principle, and Liouville-type theorems to obtain a priori estimates Copyright © 2007 L Iturriaga and S Lorca This is an open access article distributed un-der the Creative Commons Attribution License, which permits unrestricted use, distri-bution, and reproduction in any medium, provided the original work is properly cited

1 Introduction

We consider the following nonvariational problem:

−Δm u = f (x, u, ∇ u) − a(x)g(u, ∇ u) + τ inΩ, u =0 on∂Ω, (P) τ

whereΩ is a bounded domain with smooth boundary ofRN,N ≥3.Δmdenotes the usual

m-Laplacian operators, 1 < m < N and τ ≥0 We will obtain a priori estimate to positive solutions of problem(P) τunder certain conditions on the functions f , g, a This result

implies nonexistence of positive solutions toτ large enough.

Also we are interested in the existence of a positive solutions to problem(P)0, which does not have a clear variational structure To avoid this difficulty, we make use of the blow-up method over the solutions to problem (P) τ, which have been employed very often to obtain a priori estimates (see, e.g., [1,2]) This analysis allows us to apply a result due to [3], which is a variant of a Rabinowitz bifurcation result Using this result, we obtain the existence of positive solutions

Throughout our work, we will assume that the nonlinearities f and g satisfy the

fol-lowing conditions

(H1) f :Ω× R × R N → Ris a nonnegative continuous function

(H2)g : R × R N → Ris a nonnegative continuous function

(H3) There existL > 0 and c0≥1 such thatu p − L | η | α ≤ f (x, u, η) ≤ c0u p+L | η | αfor all (x, u, η) ∈Ω× R × R N, wherep ∈(m −1,m ∗−1) andα ∈(m −1,mp/(p+ 1)).

Here, we denotem ∗ = m(N −1)/(N − m).

(H4) There existM > 0, c1≥1, q > p, and β ∈(m −1,mp/(p + 1)) such that | u | q −

M | η | β ≤ g(u, η) ≤ c1| u | q+M | η | βfor all (u, η) ∈ R × R N

We also assume the following hypotheses on the functiona.

(A1)a :Ω→ Ris a nonnegative continuous function

(A2) There is a subdomainΩ0 withC2-boundary so thatΩ0⊂ Ω, a ≡0 inΩ0, and

a(x) > 0 for x ∈Ω\Ω0

(A3) We assume that the functiona has the following behavior near to ∂Ω0:

a(x) = b(x)d

x, ∂Ω0

γ

x ∈Ω\Ω0, whereγ is positive constant and b(x) is a positive continuous

func-tion defined in a small neighborhood of∂Ω0

Observe that particular situations on the nonlinearities have been considered by many authors For instance, whena ≡0 and f verifies (H3), Ruiz has proved that the problem

(P)0has a bounded positive solution (see [2] and reference therein) On the other hand, when f (x, u, η) = u p and g(x, u, η) = u q,q > p and m < p, and a ≡1, a multiplicity of results was obtained by Takeuchi [4] under the restrictionm > 2 Later, Dong and Chen

[5] improve the result because they established the result for allm > 1 We notice that the

Laplacian case was studied by Rabinowitz by combining the critical point theory with the Leray-Schauder degree [6] Then, whenm ≥ p, since ( f (x, u) − g(x, u))/u m−1 becomes monotone decreasing for 0< u, we know that the solution to(P)0 is unique (as far as it exists) from the D´ıaz and Sa´a’s uniqueness result (see [7]) For more information about this type of logistic problems, see [1,8–13] and references cited therein

Our main results are the following

Theorem 1.1 Let u ∈ C1(Ω) be a positive solution of problem(P) τ Suppose that the condi-tions (H1)–(H4) and the hypotheses (A1)–(A3) are satisfied with γ m(q − p)/(1 − m + p) Then, there is a positive constant C, depending only on the function a and Ω, such that

for any x ∈ Ω.

Moreover, if γ = m(q − p)/(1 − m + p), then there exists a positive constant c1= c1(p, α,

β, N, c0) such that the conclusion of the theorem is true, provided that inf ∂Ω 0b(x) > c1.

Observe that this result implies in particular that there is no solution for 0< τ large

enough By using a variant of a Rabinowitz bifurcation result, we obtain an existence result for positive solutions

Theorem 1.2 Under the hypotheses of Theorem 1.1 , the problem(P)0has at least one pos-itive solution.

2 A priori estimates and proof of Theorem 1.1

We will use the following lemma which is an improvement of Lemma 2.4 by Serrin and Zou [14] and was proved in Ruiz [2]

Lemma 2.1 Let u be a nonnegative weak solution to the inequality

in a domainΩ⊂ R N , where p > m − 1 and m −1≤ α < mp/(p + 1) Take λ ∈(0,p) and let B( ·,R0) be a ball of radius R0such that B( ·, 2R0) is included in Ω.

Then, there exists a positive constant C = C(N, m, q, α, λ, R0) such that



B(·,R) u λ ≤ CR(N−mλ)/(p+1−m), (2.2)

for all R ∈(0,R0].

We will also make use of the following weak Harnack inequality, which was proved by Trudinger [15]

Lemma 2.2 Let u ≥ 0 be a weak solution to the inequalityΔm u ≤ 0 in Ω Take λ ∈[1,m ∗ −

1) and R > 0 such that B( ·, 2R) ⊂ Ω Then there exists C = C(N, m, λ) (independent of R) such that

inf

B(·,R) u ≥ CR −N/λ



B(·,2R) u λ

 1/λ

The following lemma allows us to control the parameterτ in the Blow-Up analysis.

(SeeSection 2.1.)

Lemma 2.3 Let u be a solution to the problem(P) τ Then there is a positive constant k0

which depends only onΩ0such that

τ ≤ k0



max

x∈Ωu

m−1

Proof Since u is a positive solution, the inequality holds if τ =0 Now ifτ > 0, then from

(H1) and (A2) we get

−Δm u = f (x, u, ∇ u) − a(x)g(u, ∇ u) + τ ≥ τ ∀ x ∈Ω0. (2.5) Letv be the positive solution to

−Δm v =1 inΩ0,

andw =(τ/2)1/(m−1)v inΩ0, then it follows that−Δm w = τ/2 < −Δm u inΩ0andu > w

on∂Ω0 Thus, using the comparison lemma (see [16]), we obtainu ≥ w inΩ0 Therefore,

there is a positive constantk0such that

2.1 A priori estimates We suppose that there is a sequence{(u n,τ n)}n∈Nwithu nbeing

aC1-solution of(P) τ n such that u n+τ n −−−→

n→∞ ∞ ByLemma 2.3, we can assume that there existsx n ∈ Ω such that u n(x n)= u n =:S n −−−→ n→∞ ∞ Letd n:= d(x n,∂Ω), we define

w n(y) = S −1

n u n(x), where x = S −θ n y + x nfor some positiveθ that will be defined later The

functionsw nare well defined at leastB(0, d n S θ

n), andw n(0)= w n =1 Easy computa-tions show that

−Δm w n(y) = S1n −(θ+1)m

f

S −θ n y + x n,S n w n(y), S1n −θ ∇ w n(y)

− a

S −θ n y + x n



g

S n w n(y), S1n −θ ∇ w n(y)

+τ n



From our conditions on the functions f and g, the right-hand side of (2.8) reads as

S1n −(θ+1)m

f

S −θ n y + x n,S n w n(y), S1n −θ ∇ w n(y)

− a

S −θ n y + x n

g

S n w n(y), S1−θ

n ∇ w n(y)

+τ n

≤ S1n −(θ+1)m+q

c0S p−q n w n(y) p+MS(1n −θ)α−q w n(y) α

− a

S −θ n y + x n

w n(y) q − g0S β(1−θ)−q n w n(y) β +S1−(θ+1)m

n τ n

(2.9)

We note that fromLemma 2.3we haveS1n −(θ+1)m τ n ≤ c0S1n −(θ+1)m S m−1

n −−−→ n→∞ 0

We split this section into the following three steps according to location of the limit pointx0of the sequence{ x n}n

(1)x0∈Ω\Ω0 Here, up to subsequence, we may assume that { x n}n ⊂Ω\Ω0 We de-fineδ n  =min{dist(x n,∂ Ω),dist(x n,∂Ω0)}andB = B(0, δ n  S θ

n) if dist(x0,∂ Ω) > 0, or δ 

n =

dist(x n,∂Ω0) andB = B(0, δ n  S θ

n)∩ Ω if dist(x0,∂Ω)=0 Then,w nis well defined inB and

satisfies

sup

Now, taking θ =(q + 1 − m)/m in (2.9) and applying regularity theorems for the

m-Laplacian operator, we can obtain estimates forw nsuch that for a subsequencew n → w,

locally uniformly, withw be a C1-function defined inRNor in a halfspace, if dist(x0,∂Ω)

is positive or zero, satisfying

−Δm w ≤ − a

x0 

which is a contradiction with the strong maximum principle (see [17])

(2)x0∈Ω0 In this case, up to subsequence we may assume that { x n}n ⊂Ω0 Letd n =

dist(x n,∂Ω0) andθ =(1 +p − m)/m Then, w nis well defined inB(0, d n S θ

n) and satisfies sup

y∈B(0,d n S θ)

On the other hand, for anyn ∈ N, we havea(S −θ n y + x n)=0 and

−Δm w n(y) = S1−(θ+1)m

n



f

S −θ n y + x n,S n w n(y), S1−θ

n ∇ w n(y)

+τ n

From the hypothesis (H4),

−Δm w n(y) = S1−(θ+1)m

n



f

S −θ n y + x n,S n w n(y), S1−θ

n ∇ w n(y)

+τ n

≥ w n(y) p − MS α(1−θ)+1−(θ+1)m

n w n(y) α+τ n S1−(θ+1)m

From our choice of the constantsα and θ, we have α(1 − θ) + 1 −(θ + 1)m = α(2m −(1 +

p))/m − p < 0, that is, S α(1−θ)+1− n (θ+1)m |∇ w n(y) | αandτ n S1n −(θ+1)mtend to 0 asn goes to

∞ This implies that for a subsequencew nconverges to a solution of−Δm v ≥ v p,v ≥0 in

RN,v(0) =maxv =1 This is a contradiction with [14, Theorem III]

(3)x0∈ ∂Ω0 Let δ n = d(x n,z n), wherez n ∈ ∂Ω0 Denote byν nthe unit normal of∂Ω0

atz npointing toΩ\Ω0

Up to subsequences, We may distinguish two cases:x n ∈ ∂Ω0for alln or x n ∈Ω\ ∂Ω0

for alln.

Case 1 (x n ∈ ∂Ω0for alln) In this case, x n = z n Forε sufficiently small but fixed take



x n = z n − εν n Then we have the following

Claim 1 For any large n we have

u n



x n

< S n

Proof of Claim 1 In other cases, define for all n sufficiently large, passing to a subsequence

if necessary, the following functions



w n(y) = S − n1u n





x n+S − n(p+1−m)/m y

which are well defined at least inB(0, εS(n p+1−m)/m),w n(0)≥1/4 and sup B(0,εS(p+1 − m)/m

n )wn ≤1 Arguing as in the previous casex0∈Ω0, we arrive to a contradiction 

Now, by continuity, for any largen there exist two points inΩ0x n ∗ = x n − t n ∗ ν n and

x n ∗∗ = x n − t ∗∗ n ν n, 0< t n ∗ < t ∗∗ n < ε such that

u n

x ∗ n

= S n



x ∗∗ n 

= S n

Claim 2 There exists a number δn ∈(0, min{ d(x n,x ∗

n),d(x n ∗,x ∗∗ n )}) such that S n /4 <

u n(x) < S n for allx ∈ B(x n ∗,δn) Moreover, there exists y nsatisfying d(x ∗

n,y n)=  δ n and eitheru n(y n)= S n /4 or else u n(y n)= S n

Proof of Claim 2 Define δn =sup{ δ > 0 : S n /4 < u n(x) < S nfor allx ∈ B(x ∗

n,δ) } It is easy

to prove thatδnis well defined Thus, the continuity ofu nensures the existence of y n.



Now we will obtain an estimate from below ofδn S(p+1−m)/m

Claim 3 There exists a positive constant c = c(p, α, β, N, c0) such that



for anyn sufficiently large

Proof of Claim 3 Assume, passing to a subsequence if necessary, that δn S(p+1−m)/m

n < 1 for

anyn We have that the functions wn(y) = S −1

n u n(x n ∗+S − n(p+1−m)/m y) are well defined in B(0, 1) for n sufficiently large and satisfy

−Δm wn ≤ c0wn p+ w n α+ w n β (2.19) Applying Lieberman’s regularity (see [18]), we obtain that there exists a positive con-stant k = k(p, α, β, N, c0) such that |∇  w n| ≤ k in B(0, 1) Assume for example that

u n(y n)= S n /4 By the generalized mean value theorem, we have

1

4=1

2−1

4=  w n(0)−  w n

S θ n



y n − x n ∗

≤ w n(ξ) δ n S θ



Claim 4 For any n su fficiently large, we have B(x ∗

n,δn)⊂ B( xn,ε).

Proof of Claim 4 Take x ∈ B(x ∗ n,δn), byClaim 2we get

d

x,x n

≤ d

x, x ∗ n

+d

x n ∗,xn

< δn+dx ∗

n,x n

≤ d

x n,x ∗ n

+d

x n ∗,xn

= d

x n,x n

So,x ∈ B(x n,ε).

Let λ be a number such that N(p + 1 − m)/m < λ < p (this is possible because p <

m ∗ −1) By Claims3and4, and byLemma 2.2, we get



inf

B( xn,ε/2) u n

λ

≥ cε −N



B( xn,ε) u λ n ≥



B(x ∗

n,δn)u λ n

≥ C δN

n S λ

n /4 ≥ C1S N(m− n 1−p)/m+λ −−−→ n→∞ ∞

(2.22)

Therefore, the last inequality tells us that



B(x n,ε/2) u λ

n −−−→

Now, we will analyze the other case

Case 2 (x n ∈Ω\ ∂Ω0for alln) Define 2d =dist(x0,∂ Ω) > 0 Since Ω0hasC2-boundary

as in [19], we have

d

x n+S −θ n y, ∂Ω0



= δ n+S −θ n ν n · y + o

S −θ n ,

a

x n+S −θ n y

=

⎧

⎪

⎪

b

x n+S −θ n y

S −γθ n δ n S θ

n+ν n · y + o(1) γ, ifx n+S −θ n y ∈Ω\Ω0,

(2.24)

We defineb n(x n+S −θ n y) = S γθ n a(x n+S −θ n y).

Forn large enough, w nis well defined inB(0, dS θ

n) and we get sup

y∈B(0,dS θ)

By (2.9), we obtain

−Δm w n(y) ≤ S1n −(θ+1)m+q

c0S n p−q w n(y) p+MS(1n −θ)α−q w n(y) α

− b n



x n+S −θ n y

S −γθ n

w n(y) q − g0S β(1−θ)−q n w n(y) β

+S1n −(θ+1)m τ n

(2.26) Now we need to consider the following cases

If 0< γ < m(q − p)/(1 − m + p), we choose θ =(1− m + q)/(γ + m).

We first assume that{ δ n S θ

n }n∈Nis bounded Up to subsequence, we may assume that

δ n S θ

n −−−→ n→∞ d0≥0, from (2.26) we get

−Δm w n(y) ≤ S γθ n

c0S n p−q w n(y) p+MS(1n −θ)α−q w n(y) α

− b n

x n+S −θ n y

S −γθ n

w n(y) q − g0S β(1−θ)−q n w n(y) β +S1−(θ+1)m

= c0S p−q+γθ n w n(y) p+MS γθ+(1−θ)α−q n w n(y) α

− b n

x n+S −θ n y

w n(y) q − g0S β(1−θ)−q n w n(y) β

+S1−(θ+1)m

n τ n

(2.27) Thus, up to a subsequence, we may assume thatw nconverges to aC1functionw defined

inRNand satisfyingw ≥0,w(0) =maxw =1 inRN, and

−Δm w(y) ≤

⎧

⎨

⎩−

b

x0 d0+ν0· y γ w q(y), ifν0· y > σ,

whereσ = − d0ifx n ∈Ω\Ω0orσ = d0ifx n ∈Ω0andν0is a unitary vector inRN This

is impossible by the strong maximum principles

Suppose now that { δ n S θ

n } is unbounded, we may assume that β n =(δ −1

n S −θ n )γ/m

−−−→ n→∞ 0 for anyr > 0 Let us introduce z = y/β n andv n(z) = w n(β n z), using (2.26) we see thatv nsatisfies

−Δm v n(z) ≤ β m

n S γθ n

c0S p−q n v n(z) p+MS(1n −θ)α−q β −α v n(z) α

− b n



x n+S −θ n β n z

S −γθ n

v n(z) q − g0S β(1−θ)−q n β −β n v n(z) β

+S1−(θ+1)m

= c0β m n S γθ+p−q n v n(z) p+MS γθ+(1−θ)α−q n β m−α n v n(z) α

− β m n b n

x n+S −θ n β n z

v n(z) q − g0S β(1−θ)−q n β m−β n v n(z) β

+S1n −(θ+1)m τ n

(2.29)

On the other hand,

β m n b n

x n+S −θ n β n z

= b

x n+S −θ n β n z

1 +β n(m+γ)/γ ν n · z + o

β n m/γγ

−−−→ n→∞ b

x0 

.

(2.30) Thus, sinceγ < m(q − p)/(1 − m + p) and our choice of θ and β n, it is easy to see that

S γθ+p−q n ,S γθ+(1−θ)α−q n β m−α n and S β(1−θ)−q n β m−β n tend to 0 asn goes to + ∞ Therefore, we obtain a limit functionv that satisfies −Δm v ≤ − b(x0)v q,v ≥0,v(0) =maxv =1 inRN

which is again impossible

Ifγ = m(q − p)/(1 − m + p), in this case, by our assumptions on the function b, we

obtain forθ =(1− m + p)/m

−Δm w n(y) ≤ c0w n(y) p+MS(1n −θ)α−p w n(y) α

− b n



x n+S −θ n y

w n(y) q − g0S β(1−θ)−q n w n(y) β

+S1n −(θ+1)m τ n

(2.31) Arguing as in the proof ofClaim 3in the above casex n ∈ ∂Ω0for alln, we may assume

thatδ n S n θ ≥ d0= d0(p, α, β, N, c0)> 0 Therefore, the limit w of the sequence w nsatisfies

−Δm w(y) ≤ c0w(y) p − b

x0 d0− ν0· y + o(1) γ w(y) q (2.32) Now, evaluating inx =0, the last inequality reads as

−Δm w(0) ≤ c0− b

x0 

provided thatb(x0)> c0/d γ0 This contradicts the strong maximum principle

Ifγ > m(q − p)/(1 − m + p), we choose θ =(p − m + 1)/m, then we get

−Δm w n(y) ≥ w n(y) p − MS(1n −θ)α−p w n(y) α

− S q−p−γθ n b n

x n+S −θ n y

g1w n(y) q+g2S β(1−θ)−q n w n(y) β

+S1−(θ+1)m

n τ n

(2.34)

Arguing as seen before, that is,{ δ n S −θ n }is whether bounded or unbounded, we obtain that the limit equation of the last inequality becomes

−Δm v ≥ v p, v ≥0 inRN,v(0) =maxv =1, (2.35) which is a contradiction with [14, Theorem III]

3 Proof of Theorem 1.2

The following result is due to Azizieh and Cl´ement (see [3])

Lemma 3.1 LetR +:=[0, +∞ ) and let ( E, · ) be a real Banach space Let G :R +× E → E

be continuous and map bounded subsets on relatively compact subsets Suppose moreover that G satisfies the following:

(a)G(0, 0) = 0,

(b) there exists R > 0 such that

(i)u ∈ E,  u ≤ R, and u = G(0, u) imply that u = 0,

(ii) deg(Id− G(0, ·),B(0, R), 0) = 1.

Let J denote the set of the solutions to the problem

inR +× E LetCdenote the component (closed connected maximal subset with respect to the inclusion) of J to which (0, 0) belongs Then if

C∩{0} × E

=(0, 0)

thenCis unbounded inR +× E.

Proof of Theorem 1.2 First, we consider the following problem:

−Δm u = f

x, u+,∇ u+ 

− a(x)g

u+,∇ u+ 

+τ inΩ,

and letu be a nontrivial solution to the problem above, then u is nonnegative and so is

solution for the problem(P) τ In fact, suppose thatU = { x ∈ Ω : u(x) < 0 }is nonempty Thenu is a weak solution to

−Δm u = τ ≥0 inU,

UsingLemma 2.3, we obtain thatu(x) ≥0, which is a contradiction with the definition

ofU.

ConsiderT : L ∞(Ω)→ C1(Ω) as the unique weak solution T(v) to the problem

−Δm T(v) = v inΩ,

It is well known that the functionT is continuous and compact (e.g., see [3, Lemma 1.1])

Next, denote byG(τ, u) : = T( f (x, u+,∇ u+)− a(x)g(u+,∇ u+) +τ), then G :R +× C1(Ω)

→ C1(Ω) is continuous and compact Now, we will verify the hypotheses ofLemma 3.1

It is clear thatG(0, 0) =0 On the other hand, consider the compact homotopyH(λ, u) :

[0, 1]× C1(Ω)→ C1(Ω) given by H(λ,u) = u − λG(0, u) We will show that

ifu is a nontrivial solution to H(λ, u) =0, then u  > R > 0. (3.4)

This fact implies that condition (i) of (b) holds Moreover, (3.4) also implies that deg(H(λ, ·)B(0, R), 0) is well defined since there is not solution on ∂B(0, R) By the

in-variance property of the degree, we have

deg

Id− λG(0, ·),B(0, R), 0

=deg

Id,B(0, R), 0

=1, ∀ λ ∈(0, 1] (3.5)

and (ii) of (b) holds

In order to prove (3.4), note thatH(λ, u) =0 implies thatu is a solution to the problem

−Δm u = λ

f

x, u+,∇ u+ 

− a(x)g

u+,∇ u+ 

inΩ,

Multiplying (3.6) byu, integrating overΩ the equation obtained, and applying H¨older’s and Poincare’s inequalities, we have that



Ω|∇ u | m ≤ c0



Ωu p+1+M1



Ω|∇ u | α u +



Ω|∇ u | β u



≤ C



Ω|∇ u | m

 (p+1)/m

+M1



Ω|∇ u | m

α/m

Ωu m/(m−α)

 (m−α)/m

+M1



Ω|∇ u | mβ/m

Ωu m/(m−β)

 (m−β)/m

≤ C



Ω|∇ u | m (p+1)/m

+C1



Ω|∇ u | m (α+1)/m

+C1



Ω|∇ u | m (β+1)/m

.

(3.7) This inequality implies that

Ω|∇ u | m > c > 0 Hence, we have  u  > R > 0.

Now, we note thatTheorem 1.1andC1,ρestimates imply that the componentCwhich contains (0, 0) is bounded So, applyingLemma 3.1, we obtain thatC∩({0} × C1(Ω)) (0, 0) Therefore, we have a positive solutionu to the problem(P)0 

Acknowledgments

The first author would like to thank the hospitality of Departamento de Matem´aticas, Universidad de Tarapac´a He also wants to thank Professors Heriberto Roman and Yurilev

which is a contradiction with the strong maximum principle (see [17])

(2)x0∈Ω0... known that the functionT is continuous and compact (e.g., see [3, Lemma 1.1])