The Gelfand problem for the 1-homogeneous p -Laplacian

: In this paper, we study the existence of viscosity solutions to the Gelfand problem for the 1-homo-geneous p -Laplacian in a bounded domain Ω ⊂ ℝ N , that is, we deal with


Introduction
This paper deals with the Gelfand problem corresponding to the 1-homogeneous p-Laplacian, { −∆ N p u = λe u in Ω, u = 0 on ∂Ω, (P λ,p ) where Ω ⊂ ℝ N is a regular bounded domain, p ∈ [2, ∞] and the operator ∆ N p is the 1-homogeneous p-Laplacian (it is also called the normalized p-Laplacian) defined, for p < ∞, by with α = 1/(p − 1) and β = (p − 2)/(p − 1), and for p = ∞, is the 1-homogeneous infinity Laplacian.These kinds of elliptic operators for 2 ≤ p < ∞ have 1 and 1/(p − 1) as ellipticity constants, hence there is a lack of uniform ellipticity when we let p → ∞.Therefore, the theory of uniformly elliptic operators can not be applied.Moreover, we remark the lack of variational structure and differentiability of this operator, in contrast to what happens with the classical p-Laplacian.This fact implies that the theory concerning "stable solutions" can not be applied to our problem.Note that the 1-homogeneous p-Laplacian is a convex combination of Laplacian and infinity Laplacian operators since α + β = 1.Moreover, α = 1, β = 0 if p = 2, and α → 0, β → 1 as p → ∞.This operator appears when one considers Tug-of-War games with noise; see [18,23,24], where the Poisson problem is studied.Moreover, the sublinear problem and the eigenvalue problem associated to the 1-homogeneous p-Laplacian, namely, the problem with right-hand side λu q for 0 < q ≤ 1 , has been studied in [20] and [19].In view of these two references it seems natural to deal with the superlinear case (which for this operator is challenging due to the fact that there is no variational structure and no Sobolev spaces framework).
Concerning the Gelfand problem, since it is a classical problem, there is a large number of references.We refer to [2,4,5,10,21] and the references therein for the Laplacian, and to [25] for the fractional Laplacian.
Our first result for this problem reads as follows.
One of our main tools for the proof of this result is a comparison principle (that we prove here) adapted to the particular structure of the 1-homogeneous p-Laplacian (see Theorem 3.3).This result generalizes previous ones in [3,20].We believe that this comparison principle is of independent interest.Using arguments from degree theory, we can obtain the following result concerning solutions that are not necessarily the minimal one.Remark that we even obtain a continuum of solutions for a fixed p using λ as parameter, or for fixed λ small taking p as parameter.More precisely, for fixed p we denote by S p the solution set Analogously, for fixed λ we denote by S λ the solution set Theorem 1.2.For every fixed p ∈ [2, ∞], there exists an unbounded continuum of solutions C ⊂ S p that emanates from λ = 0, u = 0, i.e., (0, 0) ∈ C.Moreover, for every fixed λ < λ 0 = min{λ * (Ω, N, 2), (2d We remark that, as a consequence of the previous theorem, there is a lower bound for the extremal parameter found in Theorem 1.1: The use of degree theory is new for this kind of operators.Here we perform homotopies both in the parameters λ and p.The deformation in p is needed in order to start the argument with the trivial solution u = 0 for the problem with p = 2 and λ = 0, ∆u = 0, which is known to have degree 1.Note that, due to the nonsmoothness of the operator, there is a nontrivial difficulty in the computation of the degree of the trivial solution to ∆ N p u = 0. Also note that the necessary compactness is nontrivial; we rely here on results from [7].
Remark 1.3.Our results can be generalized to handle the equation with a general continuous nonlinearity f that verifies To simplify the exposition we just write the details for f(s) = e s and we make a comment at the end of the paper on how to deal with this general case.The rest of this paper is organized as follows: In Section 2, we collect some preliminaries and state the definition of a viscosity solution to our equation.In Section 3, we prove our comparison result.Finally, in Sections 4 and 5 we prove our main results concerning the Gelfand problem.

Preliminaries
In this section, we introduce the notion of a viscosity solution for problem (P λ,p ).Actually, we give the definition for a more general family of nonlinearities and we consider the following boundary value problem: where f : Ω × ℝ → ℝ is a continuous function.
Since the normalized infinity Laplacian is not well defined at the points where |∇u(x)| = 0, we have to use the semicontinuous envelopes of the operator in order to define viscosity solutions for problem (2.1) (see [8,9]).To this end, we denote the largest and the smallest eigenvalue for A ∈  N by M(A) and m(A), respectively.That is, Let us denote by USC(ω) the set of upper semicontinuous functions u : ω ⊂ ℝ N → ℝ, and we denote by LSC(ω) the set of lower semicontinuous functions.Definition 2.1.(i) u ∈ USC(Ω) is a viscosity subsolution of the equation −∆ N p u = λf(x, u) if whenever x 0 ∈ Ω and φ ∈ C 2 (Ω) such that φ(x 0 ) = u(x 0 ) and φ − u > 0 in Ω \ {x 0 }, then If, in addition, u ∈ USC(Ω) and u ≤ 0 on ∂Ω, we say that u is a subsolution of (2.1).(ii) u ∈ LSC(Ω) is a viscosity supersolution of the equation −∆ If, in addition, u ∈ LSC(Ω) and u ≥ 0 on ∂Ω, we say that u is a supersolution of (2.1).(iii) A continuous function u : Ω → ℝ is a viscosity solution of (2.1) if it is both a viscosity supersolution and a viscosity subsolution.
In what follows, φ stands for test functions whose graph touches the graph of u from above, and ψ denotes test functions whose graph touches the graph of u from below.Notice that the inequalities φ − u > 0 and u − ψ > 0 have to be satisfied in a neighborhood of {x 0 } and not necessarily in the whole Ω \ {x 0 }.

Let us observe that
Thus, as in [9], we can characterize viscosity sub-and supersolutions by using the concept of upper and lower semijets in the sense of the following definition.
Remark 2.5.In [12], a parabolic equation of the form was studied using viscosity solutions.The definition of viscosity solutions given there (inspired by [22]) differs from ours.In fact, in [12] Imbert, Lin and Silvestre restrict the class of test functions in order to give sense to the equation when the gradient vanishes (note that this parabolic problem can be singular or degenerate according to the value of γ).In our definition we do not restrict the test functions but we give a meaning to ∆ N ∞ u in terms of the largest and the smallest eigenvalue of D 2 u at points where the gradient vanishes.With our definition we can prove a comparison principle in the following section.

Comparison principle and uniqueness
In this section, we start giving sufficient conditions on f to prove a comparison principle and hence obtain uniqueness for (2.1).Definition 3.1.Given a positive function h ∈ C 1 (0, +∞) such that h ∈ L 1 (0, 1) and h  (s)/h 2 (s) is nondecreasing, we say that f : Ω × ℝ → ℝ satisfies the h-decreasing condition if for every x ∈ Ω, h(s)f(x, s) is decreasing with respect to s. (3.1) Brought to you by | Universidad de Granada Authenticated Download Date | 2/14/20 8:35 AM Remark 3.2.Observe that if f(x, s) = f 0 (x) > 0, that is, f does not depend on s, then f satisfies the h-decreasing condition for h(s) = 1/s q for any 0 < q < 1.In addition, when f(x, s) = f 0 (x)s q > 0 for some 0 ≤ q < 1, then f satisfies the h-decreasing condition for h(s) = 1/s q+ε for any 0 < ε < 1 − q.Moreover, taking a decreasing function h, we obtain that any function 0 < f ∈ C 1 (Ω × ℝ) nonincreasing with respect to s also satisfies the h-decreasing condition (since h  (s)f(x, s) + h(s)f  s (x, s) < 0 in this case).
Proof.We argue by contradiction following closely the ideas in [9].Suppose that for s ≥ 0. By hypothesis, u ≤ u on ∂Ω.Using that u, u ∈ C(Ω), we have that there exists x ∈ Ω + with Since Ω + is an open set, we can take Ω, an open neighborhood of x , such that Ω ⊂ Ω + .Now, let w and w be the positive functions defined for x ∈ Ω by Clearly w, w ∈ C( Ω) and Now, we claim that w and w are a sub-and a supersolution (in the viscosity sense) of the equation Indeed, we proceed to show that w is a subsolution (the fact that w is a supersolution can be proved in the same way).For every x 0 ∈ Ω, we take φ ∈ C 2 ( Ω) with φ(x 0 ) = w(x 0 ) and φ(x) > w(x) for every x ∈ Ω \ {x 0 }.If ∇φ(x 0 ) ̸ = 0 and we take φ = H −1 (φ), then it is easy to check that Consequently, In the case ∇φ(x 0 ) = 0, since ∇ φ(x 0 ) = 0 and D 2 φ(x 0 ) = h( φ(x 0 ))D 2 φ(x 0 ), we have Therefore, we conclude that w is a subsolution of problem (Q), which was our claim.
Brought to you by | Universidad de Granada Authenticated Download Date | 2/14/20 8:35 AM Now, consider the sequence of functions For every n ∈ ℕ, let (x n , y n ) ∈ Ω × Ω be such that We note that Ψ n (x n , y n ) is finite since w − w is continuous and Ω is compact.Moreover, Furthermore, we can assume that x n , y n → x , ŷ , w(x n ) → w( x ) and w(y n ) → w( ŷ ) as n → ∞, and that x = ŷ (see [9, Lemma 3.1 and Proposition 3.7]).Next, by [9, Theorem 3.2], there exist Hence, if x n ̸ = y n , having in mind that w and w are sub-and supersolution of (Q) and using Remark 2.4, we obtain that Letting n → ∞, by the continuity of w, w, f , h, h  , and using that h  /h 2 is nondecreasing, we get This is a contradiction to (3.2) since it implies, by using (3.1), that If x n = y n for n ≥ n 0 , then η n = 0 and by (iii) we have arguing as above, this leads to a contradiction.
Let us extract easy consequences of this comparison principle.
Proposition 3.4 (Uniqueness).Assume that 0 < f ∈ C(Ω × ℝ) satisfies the h-decreasing condition.Then there exists at most one positive viscosity solution of Proof.Suppose that there exist two solutions u 1 , u 2 ≥ 0 of (P).Using Theorem 3.3 twice, we obtain that u 1 ≤ u 2 and u 2 ≤ u 1 , and we conclude that u 1 = u 2 .
Brought to you by | Universidad de Granada Authenticated Download Date | 2/14/20 8:35 AM The next result improves [20], where a starshaped condition on the domain Ω was required.

Corollary 3.5. As a particular case, we can assert that there exists a unique positive solution of
for every λ > 0 and 0 < q < 1.Moreover, for λ = 0, the problem admits as unique solution u = 0.
Proof.For λ > 0, the uniqueness is due to Proposition 3.4 and the existence due to [20, Theorem 3.1] (which can be extended to the case p = ∞ by using the same iterative procedure as in [20, Theorem 3.1]).For λ = 0, we observe that u is a solution of −∆ N p u = 0 if and only if −∆ p u = 0 in the viscosity sense, (this holds since it is enough to test the equation −∆ p u = 0 with test functions with ∇φ ̸ = 0; see [15]).Thus, the trivial solution u = 0 is the unique solution when λ = 0.

Existence of minimal solutions for the Gelfand problem
The first result of this section shows how one can pass to the limit in a sequence of viscosity solutions of a sequence of problems to obtain a viscosity solution of the limit problem.Proof.First, we prove that u is a subsolution.For every x 0 ∈ Ω, we take φ ∈ C 2 (Ω) such that φ(x 0 ) = u(x 0 ) and φ − u > 0 in Ω \ {x 0 }.Fix δ > 0 such that B δ (x 0 ) ⊂ Ω, and for every n ∈ ℕ we consider x n as the strict minimum point (not necessarily unique) of φ − u n in B δ (x 0 ), i.e., Up to a subsequence, we can assume that x n → x * ∈ B δ (x 0 ).Using that u n is continuous and that the sequence u n uniformly converges to u, we deduce that u n (x n ) → u(x * ).We obtain, taking limits in the above inequality, that and we can assert that x * = x 0 .We set It is easy to check that φ n satisfies in a neighborhood of x n .Thus, using that u n is a subsolution of (4.1) and taking φ n as test function, we obtain the following: Case 1: ∇φ(x 0 ) ̸ = 0.In this case, we can suppose that, up to a subsequence, ∇φ n (x n ) ̸ = 0 for n ≥ n 0 and, taking into account that φ ∈ C 2 and the continuity and uniform convergence of f n , we can pass to the limit in (4.2) as n → ∞ to obtain Case 2: ∇φ(x 0 ) = 0 and, up to a subsequence, ∇φ n (x n ) ̸ = 0 for n ≥ n 0 .In this case, since replacing in (4.2), we get (4.3).Taking limits, we obtain the desired inequality Case 3: ∇φ(x 0 ) = ∇φ n (x n ) = 0 for n ≥ n 0 .We obtain (4.4) directly from (4.3).
On the other hand, to prove that u is a supersolution, we argue in a similar way.To be more specific, for every x 0 ∈ Ω we take the test function ψ ∈ C 2 (Ω) satisfying that u − ψ has a strict minimum at x 0 with ψ(x 0 ) = u(x 0 ).Now, taking x n , the strict minimum of u n − ψ in B δ (x 0 ) ⊂ Ω, we set as the test function in (4.1) touching the graph of u n from below in x n .The rest of the proof runs as before.Now we can prove the existence of minimal solutions of (P λ,p ) for λ small and the nonexistence of solutions for λ large, that is, we prove Theorem 1.1.r) , r in (0, 1), with α = 1 p − 1 if p < +∞ and α = 0 in the other case.
Then u(x) := z(|x|) is a solution to the problem in the sense of Definition 2.1 (iii) (see also Remark 2.2).Due to [14], it is well known that there exists a positive number λ (B 1 ), depending only on p, N, such that problem (4.5) has no solution for λ > λ (B 1 ).Moreover, for every 0 ≤ λ < λ (B 1 ) there exists a classical solution z ∈ C 2 ([0, 1]) (see also [13] for a complete description of the multiplicity of solutions).Observe that for any classical solution z ∈ C 2 ([0, 1]), λ ≥ 0, of (4.5) it holds that λ ≤ λ (B 1 ) (we refer again to [13] for a complete description of the multiplicity of solutions).Note also that the relationship between classical solutions of (4.5) and viscosity radial solutions of (4.6) is bidirectional.Given a solution u ∈ C( B1 ) of (4.6) radially symmetric and decreasing, then z(r) = u(|x|) for some x ∈ Ω with |x| = r satisfies (4.5) in the weak sense (which is equivalent to be a classical solution in this case).
By taking into account Remark 2.2, u is also a solution to our problem in the viscosity sense.Now, for any fixed R > 0, we can rescale the problem and consider It is easy to check that we arrive at the ODE Brought to you by | Universidad de Granada Authenticated Download Date | 2/14/20 8:35 AM Summarizing, we have that there exists a positive value which is decreasing with respect to R, such that problem (P λ,p ) admits at least a solution for every λ < λ (B R ) in the ball of radius R, Ω = B R .Let now Ω be a bounded domain and R 1 > 0 given by We claim that there exists a solution of problem (P λ,p ) with λ = Λ.Indeed, to prove this fact we use a standard monotone iteration argument: let w 0 = 0, and for every n ≥ 1 we define the recurrent sequence {w n } by The sequence {w n } ∈ C( Ω) is well defined by [18,24]; see also [17].Note that we are solving a problem of the form −∆ N p w n = f in Ω, with f > 0 and w n = 0 on ∂Ω as boundary condition.Then the existence is a consequence of a limit procedure involving game theory (in this problem the right-hand side, f , enters into the problem as a running payoff and the boundary condition w n = 0 as the final payoff).The existence of such a solution can also be proved directly by using Perron's method thanks to our general comparison principle.
Moreover, the sequence {w n } is increasing with n.Indeed, taking into account that 0 < w 1 , we obtain λe w 0 ≤ λe w 1 , and by using the comparison principle in Theorem 3.3, it follows that w 1 ≤ w 2 .By an inductive argument, we get 0 < w 1 ≤ w 2 ≤ ⋅ ⋅ ⋅ ≤ w n for all n ≥ 1.From the fact that u R 1 is a supersolution of problem (P λ,p ), with a similar inductive argument, we prove that w n ≤ u R 1 for every n ∈ ℕ.
In addition, thanks to the subtle Krylov-Safonov C 0,α -estimates of w n for every p ∈ [2, ∞] (here we refer to [6,7]), we obtain that w n → w λ uniformly.Taking f n = λe w n−1 and p n = p in Lemma 4.1, we get that w λ is a solution of problem (P λ,p ).
To prove that the obtained solution w λ is minimal let v λ be a solution of problem (P λ,p ).By a similar argument, using the comparison principle and induction in n, we have w n ≤ v λ for all n ∈ ℕ.As w λ (x) = lim n→∞ w n (x) (we use again comparison here), we obtain w λ ≤ v λ .
we remark that without loss of generality we can assume that 0 ∈ Ω.In that way, taking w λ , the minimal solution in Ω, as a supersolution in B R 2 and applying the above argument again, with Ω replaced by B R 2 , we obtain that λ * (Ω, N, p) ≤ λ * (B R 2 , N, p).Note that in the case Ω = B r we can perform the previous argument starting with w 0 = 0 and obtain that the minimal solution is radial.In fact, by uniqueness, in this case w n is radial for every n.Remark that in this case the unique minimal solution leads to a solution of the ODE (4.7), and thus λ * (B R 2 , N, p) ≤ λ (B R 2 ).Remark 4.2.The arguments used in the previous proof show that the extremal parameter verifies λ * (Ω, N, p) = sup{λ > 0 : there exists a minimal solution of (P λ,p )} = sup{λ > 0 : there exists a solution of (P λ,p )} = sup{λ > 0 : there exists a nonnegative supersolution of (P λ,p )}.Now, we prove that K is completely continuous, which allows us to apply the Leray-Schauder degree techniques (see [16]), in order to study the existence of "continua of solutions" of (P λ,p ), i.e., connected and closed subsets in the solution set for every fixed p ∈ [2, +∞], or, if we fixed λ instead, in Proof.If λ n = 0, then u n = 0 is the unique solution (Corollary 3.5) and the proof is immediate.In the other case, since 0 < λ n e w n ≤ C for some positive constant, u n is a subsolution of the problem It is well known, by the theory of uniformly elliptic fully nonlinear equations, that, for every fixed n ∈ ℕ, u n ∈ C 0,ν(n) (Ω) whenever 2 ≤ p(t n ) ≤ M for some M sufficiently large (for instance, greater than the dimension N), with 0 < ν(n) < 1 (see [6,11]).We stress that this Hölder estimates depend on the ratio between the ellipticity constants, which in this case is p(t n ) − 1 and, consequently, it blows-up as p(t n ) → ∞.However, for p Thus, we can assert that the sequence u n ∈ C 0,γ (Ω), where γ = min{ν(n), ρ(n) : n ∈ ℕ}.Hence, the Ascolí-Arzelá theorem gives that u n possesses a subsequence converging in C(Ω), which concludes the first part of the lemma.Finally, the second part is a direct consequence of the uniqueness of solutions by Proposition 3.4 and Lemma 4.1.
Proof of Theorem 1.2.For fixed R > 0, let O R be the open ball of radius R of C(Ω), and we fix some λ R with 0 where d is the diameter of Ω.
By Lemma 5.4, we obtain that In fact, we argue by contradiction: Suppose that ‖u‖ ∞ = R and there exist t ∈ [0, 1] and λ ∈ [0, λ R ] such that u satisfies the equation In order to conclude this proof, we apply the continuation theorem of Leray-Schauder (Theorem 5.1) with T(λ, u) = K(t, λ, u) for every fixed t ∈ [0, 1], which is completely continuous (Lemma 5.4).Therefore, using that deg(I − T(0, ⋅ ), O R , 0) = 1 ̸ = 0, we can assert that there exists a maximal connected subset C of S p that contains (0, 0).Furthermore, C is not bounded since 0 is the unique solution for λ = 0. Finally, since for every λ such that there is a solution of (P λ,p ) we can construct a minimal solution, we can state that C ⊂ [0, λ * ] × C(Ω).
With the same arguments, using Theorem 5. Since f is assumed to be continuous, it is easy to check that T is completely continuous.Now, as T(0, u) = 0 for every u ∈ C([0, 1]), using Leray-Schauder's theorem, we obtain the existence of a continuum of solutions C ⊂ [0, ∞) × C([0, 1]) that is unbounded with (0, 0) ∈ C. In particular, there exist solutions for values of λ close to 0.

Lemma 4 . 1 .
Let u n , f n ∈ C(Ω) and p n ∈ [2, ∞] be three sequences satisfying− ∆ N p n u n = f n ,(4.1)in the viscosity sense, such that f n → f , u n → u uniformly for every ω ⋐ Ω and p n → p ∈ [2, ∞].Then u is a viscosity solution to the problem −∆ N p u = f.