EXISTENCE OF SOLUTIONS FOR QUASILINEAR DIRICHLET PROBLEMS WITH GRADIENT TERMS

. In this paper we prove an existence theorem for positive solutions of a nonlinear Dirichlet problem involving the p -Laplacian operator on a smooth bounded domain when a nonlinearity depending on the gradient is considered. Our main theorem extends a previous result by Ruiz in [19], in which a slight modiﬁcation of the celebrated blowup technique due to Gidas and Spruck, [11] and [12], is introduced.


1.
Introduction. Motivated by [19], in this paper we prove an existence theorem for positive weak solutions of a nonlinear Dirichlet problem involving the p-Laplacian operator, namely ∆ p u + f (x, u, ∇u) = 0, x ∈ Ω u(x) = 0, x ∈ ∂Ω (1) where Ω ⊂ R N , N > 1, is a bounded smooth domain, ∆ p u = div |∇u| p−2 ∇u is the p-Laplacian operator with 1 < p < N and f : Ω × R + 0 × R N → R is a nonnegative continuous function such that for all (x, u, η) ∈ Ω×R + 0 ×R N , where c 0 and M are positive constants, with c 0 ≥ 1, while the exponents δ, s, and ϑ satisfy s ∈ [0, S) and ϑ ∈ p − s − 1, p(δ − s) δ + 1 with S = min p − 1, δ N , δ − ϑ p (δ + 1) . The novelty we introduce, respect to [19], is to consider a nonlinearity f involving an explicit dependence on the solution u in the gradient term. We point out that the presence of a factor depending both on a power of u and of |∇u| makes the analysis fairly delicate. Ruiz in [19], considers the subcase of (F ) when s = 0.
Observe that problem (1) does not have, in general, a variational structure because of the dependence on the gradient of the nonlinearity. Thus topological methods will be used to prove the existence of solutions. Similar problems have been intensively studied in literature, especially when p = 2 , for instance in the classical paper by Brezis and Turner [3], the authors assume conditions on f stronger than (F ) and while in the paper by De Figueiredo, Lions and Nussbaum [7], the nonlinearity f does not depend on the gradient and some other technical conditions are imposed. For further result relative to boundary Dirichlet problems when p = 2 and with a gradient term we refer to the pioneering paper by Ghergu and Radulescu [10], see also [8] where a competition between an anisotropic potential, a convection term |∇u| and a singular nonlinearity is taken under consideration, see also the recent paper [9] where also a diffusion term depending on u inside the divergence.
Concerning the p-Laplacian case, a first natural approach to solve problem (1) is to deal with radial solutions. This procedure has been carried out by Clèment, Manàsevich and Mitidieri in [4] for the case of systems, but without a dependence of the nonlinearity on the gradient. Later in [2], Azizieh and Clement, studied problem (1), under some additional conditions, indeed they consider the case 1 < p ≤ 2 and a nonlinear function f not depending on x and on ∇u, and when Ω is a convex domain. In [19], Ruiz removes all these conditions in order to treat problem (1), with a nonlinearity f satisfying (F ) in the subcase s = 0.
For further existence results for more general Dirichlet problems with gradient terms, we refer to [16] where Motreanu and Tanaka develope an approach based on approximate solutions and on a new strong maximum principle, and to the paper of Radulescu, Xiang and Zhang, [18], where existence of nonnegative solutions for a p-Kirchhoff type problem driven by a non-local integro-differential operator with homogeneous Dirichlet boundary data is investigated.
To prove the existence of positive solutions for (1), when s = 0, Ruiz uses a degree argument which was first utilized by Kranoselskii (see [13]). The natural approach to do that is to find a priori L ∞ estimates for positive solutions and then use the degree theory. The a priori estimates have been established by a blowup procedure, a method developed for the semilinear case by Gidas and Spruck in their celebrated papers, [11] and [12]. Roughly speaking, the blowup technique is based on suitable scaling arguments and allows to obtain existence of solutions of Dirichlet problems on bounded domains by combining a priori estimates and Liouville type theorems. Indeed, in [12], where generical bounded domains Ω are considered, not necessarily convex for instance, Gidas and Spruck reduce the question of the a priori bounds for positive solutions of ∆u = f (x, u) in Ω, u = 0 in ∂Ω to the nonexistence of positive classical solutions for the two problems and where the exponent δ is related to a growth condition on f (x, u) with respect to u and H + = {x ∈ R N | x N > 0} is a halfspace in R N . The latter problem, when the p-Laplacian operator is involved, is very delicate and still nowadays Liouville type theorems in general cases, are not available in literature. For this reason, to avoid problem (5) several methods in literature have been developed, for instance the convexity of the domain as in [4] and [2]. The technique used starts with the proof of a priori estimates on the pair (u, λ), with λ ≥ 0, solution of the parametric problem when λ is small. Indeed, in Proposition 1, we prove that (6) has no solutions at all when λ ≥ λ 0 , for a certain λ 0 positive. Thus we need to obtain uniform estimates (in L ∞ sense) on the weak solutions of (6), only when λ ∈ [0, λ 0 ). This uniform estimates are obtained by using suitable scaling arguments together with Liouville type theorems on the entire space R N . The idea is to suppose, by contradiction, that there exists a divergent sequence of positive solutions u n of (6), attaining their maxima in x n ∈ Ω. This procedure yields the existence of a nontrivial positive solutions of the analogous limit problems (4) or (5) for the p-Laplacian. In particular, Gidas and Spruck, in [12], show that this blowup method produces a solution of (5) in a halfspace, when the points x n approach sufficiently fast to the boundary of Ω. In order to avoid the boundary case, for instance in [2], the authors assume that Ω is convex and that 1 < p ≤ 2, this restriction is due to a simmetry result due to Damascelli and Pacella in [5] based on the moving plane method which guarantees that the sequence x n cannot approach to the boundary. In [19], Ruiz, to avoid problem (5), produces a slight modification of the blowup method using the same technique but centered on a certain fixed point y 0 ∈ Ω instead of x n . In order to do this, he needs some Harnack type inequalities, due to Trudinger [23], Serrin [20] and Serrin and Zou [21]. Hence, to use the above new version of the blowup technique, we have extended these Harnack type inequalities, following mainly the arguments in [20] and [21] to include the new case (F ), where s = 0. Thanks to this procedure, the corrisponding limit problem will be defined in all of R N so that we obtain a contradiction by a classical Liouville theorem. This paper is organized as follows. In Section 2, some Harnack type inequalities are proved together with some preliminaries, then Section 3 is devoted to prove the main a priori estimates for solution of the parametric problem (6). Finally, in Section 4 we present the main result of the paper, namely Theorem 4.2.
2. Harnack inequalities. Before stating and proving the main result, we need to give several preliminary lemmas, which extend the analogous ones in [19], given for the case s = 0. In what follows C is a positive constant which may vary from one expression to another, but it is always independent to u.
Let R 0 > 0 be fixed, and 0 < R < R 0 . Denote by B R a ball of radius R such that the corresponding ball B 2R of radius 2R is contained in the domain Ω. Then, there exists a positive constant C = C(N, p, δ, γ, R 0 ) such that for all γ ∈ (0, δ). Similarly, there exists a positive constant C = C(N, p, δ, γ, σ, µ, Proof. The proof basically uses the same ideas of the proof of Lemma 2.1 of [19], but adapted to the new case with s = 0. We can suppose that the ball B R is centered at zero. First we prove the integral estimate (9).
Let ξ be a radially symmetric C 2 cut-off function such that as a test function for the weak formulation of inequality (7). This gives at once We apply Young inequality to the first term on the right hand side of (11), with exponents µ and µ , and since Now apply again Young inequality to the second term on the right hand side of (11), with exponents p and p , and since Observe that by (iii) then |∇ξ |x|/R | ≤ C/R, therefore we have where (12) and (13) in (11) we where r = δ + 1 − p.
Let us focus on the case γ > r. Apply again Young inequality to the first integral of the right hand side, with exponents ν and ν , and we obtain Now we choose ν such that .
In particular, d < δ from the choice of d, while γ > r forces that

Consequently, it holds
. Now apply Young inequality to the second integral of the right hand side on (14), with exponents α and α , and since ROBERTA FILIPPUCCI AND CHIARA LINI and we impose and α > 1 follows from δ > p − 1 and γ > r, that is d < p − 1. Consequently, from 0 ≤ ξ ≤ 1 in A R and for k sufficiently large, say k > αp, and since γ = δ − d we have Replacing (15) and (16) in (14) and using that γ = δ − d we obtain hence for γ > r the following estimate holds with C 1 , C 2 > 0 and where we have chosen ε and τ such that .
Taking into account that namely the estimate (9) holds for γ > r. Now consider the case γ = r. Observe that we have since ϑ > p − 1 − s and furthermore < γ, indeed

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Thus replacing γ = r in (14), with ε given in (18), we have which, as above, yields From < r we can apply Young inequality to the first integral on the right hand side, with exponents r/ and r/(r − ) so that, thank to (21), we arrive to Thus also for γ = r inequality (9) holds. Now consider the case γ < r. We apply Hölder inequality, with exponents r γ and r/(r − γ) , to obtain and applying (9) with γ = r, we have This gives the required conclusion (9), since r = δ + 1 − p.
The following lemma is a result of Harnack type due to Serrin, see also Lemma 4.2 in [21].

Lemma 2.2. (Theorem 5, [20]) Let u be a nonnegative weak solution in a domain
Then, for every R such that B 2R ⊂ Ω , there exists constant C depending on Note that assumption q > N implies p > N/q , so that all the exponents of R in (24) are positive.
Combining Lemma 2.1 and Lemma 2.2 it follows another Harnack type inequality, which extends Theorem 2.3 in [19], which corresponds to the case s = 0.  [21], where only the case ϑ = p − 1 and s = 0 is considered.

Remark 2. As noted in [19] that Theorem 2.3 is a version of Theorem 4.1(b) in
Proof. We use Lemma 2.2 with thus we have to verify that c ∈ L q (B 2R ), for some q > N , and d, f ∈ L q (B 2R ) so that R with for certain C > 0. Consequently, the estimate (25) follows at once since f L q (B 2R ) = λCR N/q from which R p−N/q f L q (B 2R ) = λCR p .
To obtain the first estimate in (26) it is enough to apply inequality (10) in Lemma 2.1 with µ = q (ϑ + 1 − p) and σ = sq , for some q > N . Thus we have to verify σ < δ and µ < p(δ − σ)/(δ + 1). The first inequality follows choosing q > N close enough to N since sN < δ by assumption.
To obtain the second inequality it is enough to prove µ < p(δ − sN )/(δ + 1). Indeed, denoting with l 1 and l 2 we have that for q close to N, q > N , then thus choosing ε and τ small enough such that l 1 + ε < l 2 − τ the inequality follows. This obviously occurs if l 1 < l 2 , that is
It remains to prove the second inequality in (26). We have In the above expression we have used inequality (9) in Lemma 2.1 and to do that we need to verify that γ < δ. Again, as above since q > N implies that q > N/p, thus taking q > N/p as close as necessary to N/p, it is suffices to show that and this holds since δ < p * − 1. Hence (26) 2 is proved.
In the proof of the main theorem we will also make use of the following weak Harnack inequality, due to Trudinger, [23].
Theorem 2.4. Let u ≥ 0 be a weak solution of the inequality Then there exists

3.
A priori estimates. In order to prove the main result of the existence of positive solutions for (1), we will use a degree argument. First we need to obtain a priori bound for the parameter λ of the parametric problem (6) To find this bound, as in [19], we use the following consequence of Picone identity for the p-Laplacian, see also Theorem 1.1 in [1].
Lemma 3.1. Let u be a positive solution of the problem Then where λ 1 is the first eigenvalue of the p-Laplacian with Dirichlet boundary conditions, and φ 1 is the associated eigenfunction. Moreover, the equality holds if and only if u(x) = aφ 1 (x) for some a > 0.
Remark 3. Note that φ p 1 u p−1 belongs to W 1,p 0 (Ω) since u is positive in Ω and has nonzero outward derivative on the boundary because of the Hopf lemma (see [24]). Proposition 1. Consider problem (6) and let f satisfy condition (F ) with Then there exists λ 0 > 0 such that problem (6) has no positive solutions for any λ ≥ λ 0 .
Proof. Suppose that u is a positive solution of (6). Applying Lemma 3.1 with Consider the function Obviously, (λ) ≥ 0 for all λ > 0, and (λ) → ∞ if λ → ∞ since 1−(p−1)/δ > 0. Using property (F ), and the positivity of φ 1 and u, we get where in the last inequality we have used the definition of (λ). Combining the previous inequality with (30) we have that Obviously, the integral on the right hand side is bounded from below thanks to the positivity of (λ). Now we claim that it is bounded. If so, (31) provides a bound for (λ) from above and therefore the coercivity of (λ) forced the boundness of λ.
In order to prove the claim, we use the weak formulation of with test function ψ = φ p 1 /u p−1 which belongs to W 1,p 0 (Ω), as noted at the end of the statement of Lemma 3.1, namely

ROBERTA FILIPPUCCI AND CHIARA LINI
Now we estimate the first term on the right hand side of (32) by using Young inequality with exponents p and p and we obtain where C ε = ∇φ 1 p L p (Ω) /pε p . Substituting in (32) and choosing ε such that where C p = ∇φ 1 p L p (Ω) /ε p . Now we apply Young inequality to the second integral on the right hand side, with exponents µ and µ > 1, and we have Choosing µ such that ϑµ = p , that is µ = p/ϑ and µ = p/(p − ϑ), this can be done since µ > 1 from ϑ < p(δ − s)/(δ + 1) so that ϑ < p, we obtain Using again Young inequality to the last integral of the above inequality, with the exponents ν and ν , we obtain and inserting in (34) we arrive to having chosen ν such that this is possible since ϑ > p − s − 1 and δ > p − 1, and thus furthermore condition ν > 1 holds since ϑ < p(δ − s)/(δ + 1). Consequently, (33) yields where we have chosen ε and τ such that .
We have so obtained thanks to (35) and (36). The claim is proved.
Remark 4. Proposition 1 in the case s = 0 reduces to Proposition 3.2 in [19]. Furthermore, as a consequence of Proposition 1, we need to give a priori estimates on the solutions u of (6) only when λ is bounded.
To obtain the required estimates, Ruiz, in [19], uses a slight modification of the well know blowup technique due to Gidas and Spruck in [11], [12]. As described in the Introduction, usually the blowup technique is based on a contradiction argument, that is the construction of a divergent sequence (u n ) n of solutions of (6) and on a sequence of points x n in which u n attain their maxima. When x n → x ∈ Ω, then a Liouville theorem in the entire R N , produces the required contradiction. However, it may happen that d(x n , ∂Ω) → 0 and in this case the blowup method provides a solution in a halfspace when the points x n approach sufficiently fast (in comparison with the L ∞ norm of u n ) to the boundary of Ω. Thus in order to get a contradiction we need a Liouville-type theorem in the halfspace. In [19], Ruiz uses the same blowup technique but centered on a certain fixed point y 0 ∈ Ω, instead of x n . Proposition 2. Assume that (F ) holds and that λ < λ 0 for some λ 0 fixed.
Then, there exists C > 0 such that u ≤ C for any C 1 solution u of (6), where · denotes the uniform norm.
Proof. We argue by contradiction and we suppose that there exist a sequence of parameters (λ n ) n , λ n < λ 0 for all n ∈ N and a sequence of points (x n ) n in Ω such that, denoting with u n a solution of (6) with λ substituted by λ n , we have We proceed in several steps.
From now on, we use c to denote positive constants, which may vary from one expression to another, but always independent of n. To obtain (39) we make a change of variable and define the so called blowup functions Therefore, z n converges in the C 1 norm, up to a subsequence, to a certain function z 0 . Observe that z 0 (0) = 1, being ζ n (0) = 1. Applying (46) with |x| <M n −1 d 0 /2, we obtain that ℘ n (x, z n , ∇z n ) − z δ n ≥MS n −δS n ϑ δ+s+1 p C ϑ + λ nSn −δ → 0 n → ∞, so that arguing as in (49), we obtain that z 0 is a nonnegative weak solution of the problem ∆ p z 0 + z δ 0 ≤ 0, in B(0, R). The strong maximum principle of Vázquez [24], see also for further generalizations [17], yields that z 0 is actually positive in that ball. Since R is arbitrary, using a diagonal procedure, we can take a subsequence (still denoted by z n ) such that z n converge to z 0 in compact sets of R n (in the norm C 1 ) where z 0 is now defined in all R N , namely z 0 solves ∆ p z 0 + z δ 0 ≤ 0 in R N . Then we obtain that z 0 is a positive solution of problem (50) which is a contradiction with [15]. 4. Existence result. In this section we are ready to prove the main result of the paper. For this purpose we use a particular version of a theorem of Krasnoselskii, precisely Theorem 4.1 stated in [19]. The a priori estimates obtained in the previous section play a key role in verifying the conditions required in that fixed point theorem which we state for completeness.
Theorem 4.1. Let C be a cone in a Banach space and K : C → C a compact operator such that K(0) = 0 . Assume that there exists r > 0 , verifying a) u = tK(u) for all u = r, t ∈ [0, 1]. Assume also that there exist a compact homotopy H : [0, 1] × C → C, and R > r such that b1) K(u) = H(0, u) for all u ∈ C. b2) H(t, u) = u for any u = R, t ∈ [0, 1]. b3) H(1, u) = u for any u ≤ R. Let D = {u ∈ C | r < u < R}. Then K has a fixed point in D.
As in [19], we consider C(Ω) as a Banach space with the uniform norm · and C 1,τ (Ω) with the Hölder norm · C 1,τ . We define the continuous operator T v : C(Ω) → C 1,τ (Ω) such that T v (v) ∈ C 1,τ (Ω) denotes the unique weak solution u of the problem with zero Dirichlet conditions in the boundary of Ω. Clearly T v maps bounded sets into bounded sets. We define also the compact operator N : C 1,τ (Ω) → C(Ω), N (u) = f (x, u, ∇u) and consider K = T v • N : C 1,τ (Ω) → C 1,τ (Ω) which is also a compact operator. We apply Theorem 4.1 in the cone C = {u ∈ C 1,τ (Ω) | u is nonnegative}. (52) In particular, K maps C into C because of the maximum principle, furthermore positive solution of (1) are obviously fixed points of K in C.