Positive Solution to a Nonlinear Elliptic Problem

AbstractLet L be a second order elliptic operator with smooth coefficients satisfying L1 = 0 defined in a domain Ω that is Greenian for L. Under fairly general hypotheses on the function φ, we solve the following problem:
Lu+φ(·,u)=0,in the sense of distributions inΩ;u>0,inΩ;u=0,on∂Ω.$$\left\{ \begin{array}{ll} Lu+\varphi(\cdot,u)=0, & \text{in the sense of distributions in \({\Omega}\);} \\ u>0, & \text{in \({\Omega}\) ;} \\ u=0, & \text{on \(\partial {\Omega}\).} \end{array} \right. $$


Introduction
Let L be a second order elliptic operator with smooth coefficients satisfying L1 = 0 defined in a domain that is Greenian for L. 1 In particular, boundedness of is not assumed here. 1 See the definition in Section 2. with various ϕ arise in a large number of mathematical models in physics, mechanics, chemistry and astronomy. In particular, they describe population dynamics, chemical reactions and morphogenesis. Therefore, positive solutions are often of main interest. Moreover, solutions to Eq. 1.3 can be interpreted as stationary solutions to the associated parabolic problem. When ϕ(x, u) = g(x)u γ , Eq. 1.3 is known as the generalized Emden-Fowler equation [23] and has been extensively studied since the beginning of the 20th century. For a good overview we refer to [14] and [5]. Let G be the Green function for L in . In this paper the function ϕ is required to satisfy the following hypotheses: • (H 1 ) ϕ is continuous and nonincreasing with respect to the second variable.
Notice that this allows the existence of a singularity at 0 of the type lim s→0 + ϕ(x, s) = ∞ as well as some growth of ϕ(·, c) at the boundary of . The idea of taking such ϕ comes from [7,18], where problems (1.1), (1.2) were studied for L = . Our aim here is to generalize the results of both papers. We do it in many ways, not only taking an arbitrary elliptic operator but also by weakening hypotheses on ϕ and . The following main theorem says something new also in the case of L = . With further assumptions on regularity of ϕ, we get more regularity of u (see Theorem 19).
Notice that if h has a well defined boundary value then so does u. In particular, in a Dirichlet domain i.e. a bounded domain satisfying an exterior sphere condition, for h just continuous on ∂ , we obtain solution u ∈ C(¯ ) under very mild assumptions both on regularity and growth of ϕ. Dirichlet domains are called here regular. 2 To have a feeling of H 3 , it is worth mentioning that Theorem 1 gives solution to the following problem for a uniformly elliptic operator L in a bounded domain with C 1,1 boundary,  [16] and to use the estimates for G proved in [26]. Then for a fixed x, there exist a constant M > 0 such that for every y outside a compact neighbourhood of x which is integrable. Problem (1.5) for L = has been recently considered by Díaz, Hernández and Rakotoson in [5] in bounded domains without restrictions on the boundary. So we get here a partial generalization of their results to uniformly elliptic operators.
As far as we know, there is no result about existence of solutions to Eq. 1.1 for elliptic operators in unbounded . There are some papers concerning (1.1) but the domain is always bounded and there are much stronger regularity assumptions on ϕ and ∂ [3,14,20,22] although monotonicity of ϕ with respect to the second variable is not always required [14,22]. The strength of our approach relies on a very mild regularity of ϕ as well as practically no assumptions on the domain except of being Greenian which is perfectly natural provided solution to Eq. 1.1, if it exists, is of the form However, we need to keep monotonicity with respect to the second variable which fits very well into the potential theoretical approach developed in [7] for the problem nonincreasing. The method of El Mabrouk works perfectly here. In fact, a possible generalization of Eq. 1.6 to elliptic operators and p being in the Kato class is mentioned in [7]. Combining ideas both of [7] and [18], we do it here with weaker hypotheses on ϕ than in [7,18].
Problems (1.1) and (1.2) for L = have been recently considered under variety of hypotheses on ϕ. Monotonicity with respect to the second variable is crucial in [4,6,8,12,17,24] but it is not longer required in very recent papers [9,11,13,17,21,25]. However, there are always more regularity assumptions on ϕ than in this work like Hölder continuity or the product form ϕ(x, s) = p(x)ψ(s). Also, the problem is usually considered either in a bounded regular domain or in = R d . The approach via potential theory initiated in [7] and used here allows us to go beyond these restrictions.
Theorem 1 is a result of a few steps. First, we prove that for every regular bounded domain D such thatD ⊂ and for every f ∈ C + (∂D) the problem: has a unique solution u ∈ C + (D). Moreover, where H D f is the solution of the classical Dirichlet problem for L with boundary values f .
After that, in a Greenian domain, we establish one-to-one correspondence between nonnegative continuous solutions u of the equation Lu + ϕ(·, u) = 0 in the sense of distributions in , (1.8) and nonnegative L-harmonic functions h: It turns out that the solution preserves the same regularity as in regular bounded domains under the same hypotheses. The remainder of this paper is organized as follows. In Section 2, we introduce the some notations and tools needed for the sequel. Next in Section 3, we solve the problem (1.7) in a regular domain and we investigate the regularity of the solution. Then, in Section 4, we establish a 1-to-1 correspondence between nonnegative L-harmonic functions and nonnegative continuous solutions of Eq. 1.8 in Greenian domain and we also address their regularity. In Section 5, we focus on boundary conditions and we prove Theorem 1. Finally in Appendix, we recall, for readers convenience, some basic tools of the potential theory used here.
The author is grateful to her advisors Ewa Damek and Mohamed Sifi for their work, constant encouragement and precious feedback. Besides, she'd like to thank Mohamed Selmi and Pawel Glowacki for their helpful suggestions. She also expresses her gratitude to Alano Ancona and Wolfhard Hansen who kindly responded to Ewa Damek's enquiries about potential theory.

Preliminaries
For every open set of R d with (d ≥ 3) let B( )(resp. C( )) be the set of all real valued Borel measurable (resp. continuous) functions on . We are also going to consider C 1 ( ) -the space of continuously differentiable functions on , C ∞ c ( ) -the space of infinitely differentiable functions on with compact support, C 2,α ( ) -the space of functions with the second derivative being α-Hölder continuous. Finally, for every set F of numerical functions, we denote by F + the set of all functions in F which are nonnegative.
A bounded set D satisfying D ⊂ is called regular (for L) if each function f ∈ C(∂D) admits a continuous extension H D f on D such that H D f is L-harmonic in D, in other words, the function h = H D f is the unique solution to the classical Dirichlet problem i.e.
An open subset of R d is called Greenian set if possesses a Green function (for L) which will be denoted by G i.e. for every y ∈ G (·, y) is a potential on and we have L(G (·, y)) = −ε y , in the sense of distributions, where ε y denotes the Dirac measure at the point y.
In this paper, by a solution to a partial differential equation we shall mean a continuous solution in the sense of distributions. In particular, a solution to Eq. 1.8 in an open set D ⊂ will be a function u ∈ C + (D) such that ϕ(·, u) is locally integrable on D and for all In particular, if is a bounded domain, then ψ ∈ L 1 ( ).
The following proposition was proved in [18] for the Green function corresponding to the Laplace operator in R d . But due to the estimate (2) the proof is the same. [18] and [16]]

Proposition 4 [see
Let D be a bounded regular domain in R d (d ≥ 3) and ψ ∈ K d (D), then Definition 5 A Borel measurable function ψ on belongs to K loc d ( ) if for every bounded subset D in , ψ ∈ K d (D).
Proof Let D a bounded domain satisfyingD ⊂ then there exists a constant a d depending only on the dimension d such that for every x ∈ D and α > 0. We can deduce the result.
Proof Let x ∈ and D be a bounded regular domain such that x 0 ∈ D andD ⊂ .
∩ D c , so using Harnack inequality (see Theorem 28) we can deduce that the first part is finite continuous in D. Also We can deduce by Harnack inequality that G φ is continuous in .

Solution of Eq. 1.7 in a Regular Domain
In this section, we solve the problem (1.7) in an arbitrary regular bounded set D and a given f ∈ C + (∂D) : Theorem 8 (Solution of Eq. 1.7 in a regular domain) Let D be a bounded regular domain such that D ⊂ and let L be a second order elliptic operator with smooth coefficients satisfying L1 ≤ 0. Suppose that f ∈ C + (∂D) non identically zero and ϕ : . Furthermore, we have: If in addition, ϕ satisfies (H 4 ), then the statement remains true for f being the zero function.
Before dealing with the proof, we start with a lemma that allows us to compare solutions to Eq. 1.1. For L = this result is stated and proved in [7]. The proof goes along the same lines -only properties of L-superharmonic functions in the sense of abstract potential theory are used. Then: Therefore we can conclude: and so V is empty.
Now we are ready to prove the main theorem of this section.
Proof First, we suppose that: inf In C we consider the topology of uniform convergence, C is nonempty bounded closed convex set. Also for every u ∈ C and every Using the Schauder theorem we will prove that T has a fixed point in C. We start with equicontinuity of the set T (C). Let > 0. First, observe that H D f is uniformly continuous in D. Therefore, only equicontinuity of {G D (ϕ(·, u)), u ∈ C} remains to be proved.
If x ∈ ∂D, then G D (ϕ(·, u))(x) = 0 for every u ∈ C. In addition, G D (ϕ(·, α)) ∈ C 0 (D), hence there exists a neighbourhood V x of x such that for every y ∈ V x ∩ D and u ∈ C Therefore, given , we can choose ν sufficiently small such that for all u ∈ C Secondly, T is continuous. Indeed, Let u n tend to u ∈ C. Then the function inside the integral tends to zero and it is dominated by By the dominated convergence theorem, we can conclude the pointwise convergence, by equicontinuity we can deduce the uniform convergence. Now in view of the Schauder theorem, there is a fixed point u ∈ C(D) of T i.e. , u)).
Secondly, let f be a nontrivial nonnegative continuous function on ∂D such that inf and let u k be the solution to Eq. 1.7 with the boundary value f k . Then In addition, We denote: Now we turn to prove that u is continuous in D. On one hand, (u n ) is a decreasing sequence of continuous function, so the limit u is upper semi-continuous. On the other hand, (u n − H D f n ) is an increasing sequence of continuous functions, so the limit u − H D f is lower semi-continuous.
Let's suppose that there exists But, a nonconstant L-superharmonic function cannot take a negative minimum inside D, so u n − H D f n = 0 which implies that This is a contradiction with H 4 and hence u > 0 in D.
Uniqueness of the solution follows immediately from Lemma 9.
Remark 10 Notice that in view of Proposition 4, H 3 is always satisfied in D. However, H 4 is needed in order that ϕ can take zero on ×]0, +∞[. The above theorem generalizes both Lemma 4.3 in [7] and Corollary 2 in [18].
A careful observation of the proof of the above theorem allows us to conclude the following upper bound for the solution which depends only on the boundary value and the size of the potential, not on the domain itself i.e. as far as D ⊂ D 0 the bound depends only on f, ϕ and G D 0 .

4)
Proof By Theorem 8 the solution u is the limit of a decreasing sequence (u k ) and u k = H D f k + G D (ϕ(·, u k )). Clearly, Hence (3.4) follows.
With further hypotheses on ϕ we may conclude more regularity of u.

Theorem 12
Suppose that the assumptions of Theorem 8 are satisfied and additionally that for every c > 0, ϕ(·, c) ∈ L ∞ loc (D), then the unique solution u of problem (1.7) Remark 13 Regularity of solutions is not mentioned in [7] and in [18] it is proved only for the case where the boundary value is a constant function.
Proof Let x 0 ∈ D and r > 0 be such that B(x 0 , r) ⊂ D. Let f 0 = u /∂B(x 0 ,r) be continuous and strictly positive. We denote B = B(x 0 , r). By uniqueness, u is also given by: Moreover, ϕ(·, u) is bounded in B. Therefore, and we may deduce that u is differentiable on x 0 and then in all D. Now, if ϕ ∈ C α loc (D×]0, ∞[) then we first deduce that u ∈ C 1 (D) and then we take D 1 to be a regular bounded domain with C 1,1 boundary such thatD 1 ⊂ D. We denote u L the solution of problem (1.7) so, ϕ(·, u L ) ∈ C α (D 1 ) bounded in D 1 . Then by what has been said above, ϕ(·, u L ) ∈ C α (D 1 ) and u L | ∂D 1 ∈ C(∂D 1 ). We denote ψ = u L/∂D 1 which is continuous on ∂D 1 . Hence by [[10] p. 101] we deduce that u L ∈ C 2,α loc (D) ∩C(D).

Solution in a Greenian Domain without Boundary Condition
In this section, we establish one-to-one correspondence between nonnegative L-harmonic functions and nonnegative continuous solutions of the Eq.

ϕ(·, u)) = h, u is a minimal solution satisfying u > h and h is a maximal L-harmonic function dominated by u .
Before we prove the main result, we need two lemmas analogous to Lemmas 5.1 and 5.2 in [7].
If h 1 − h 2 is L-superharmonic positive function in then: Proof We are going to apply Proposition 44. Let By assumption, K is closed and non empty. Let v = ϕ(·, u 1 ) − ϕ(·, u 2 ). Then the latter being continuous. So by Proposition 45

upper semi-continuous because G (v) is continuous and G (−|v|) is upper semicontinuous.
Finally, We can conclude by Proposition 44 holds everywhere which implies that u 1 − u 2 ≥ 0 in . Consequently, by the monotone convergence theorem, we get: which is finite at least on one point x 0 . Also, one one hand, u is a limit of an increasing sequence of continuous functions so it's lower semi-continuous and on the other hand, u − H D u is the limit of u n − H D u n = G D (ϕ(·, u n )) which is a decreasing sequence of continuous functions then u − H D u is upper-semi-continuous. Since H D u is continuous we can conclude the continuity of u as well as G D (ϕ(·, u)).
In addition, u 1 is a continuous, positive function onD which compact, so α = inf D u 1 > 0.

Now we are ready to prove Theorem 14:
Proof Let u be a nonnegative continuous solution of Eq. 1.8. Let (D n ) be a sequence of bounded regular domains exhausting i.e. − D n ⊂ D n+1 and ∪ D n = .
Since by Theorem 8, u ∈ C + (∂D n ), we have: On one hand, (H D n (u)) is nonincreasing. Indeed,
On the other hand, by the monotone convergence theorem we have, (G D n (ϕ(·, u))) G (ϕ (·, u)). Hence we may conclude that G (ϕ(·, u)). Now we turn our attention to prove that h is the maximal L-harmonic solution dominated by u: suppose that there is another L-harmonic nonnegative function h 1 such that: Or h 1 ≤ H D n (u) in D n because: When n tends to +∞, we get: By Theorem 8, we know that there is a unique continuous solution u n of: Lu + ϕ(·, u) = 0, in D n ; u = h, on ∂D n , and u n = h + G D n (ϕ(·, u n )) in D n . In addition the sequence (u n ) is not decreasing. In fact, by Lemma 9: Lu n + ϕ(·, u n ) = Lu n+1 + ϕ(·, u n+1 ) = 0, in D n ; u n − u n+1 = h − u n+1 ≤ 0, on ∂D n , and so u n ≤ u n+1 in D n . So by Lemma 16 sup n u n = u can be +∞ almost everywhere or a solution of the equation. Therefore we have to prove that u is finite in .
We can conclude that u is finite which implies that u is a continuous solution satisfying: ϕ(·, u)).
Second case: Let h be just a nonnegative L-harmonic function: Then we take Using L1 = 0, h k is L-harmonic too so by the first step we get a continuous solution u k such that : Or h k ≥ h k+1 so by Lemma 15 u k is a nonincreasing sequence in . We denote u = lim u k . As before, by upper and lower semi-continuity, we deduce that u is continuous in . Now, we turn to prove that: Notice here if h is a nonnegative nontrivial L-harmonic function in then h > 0 and so u > 0 in . Otherwise, h ≡ 0, then we suppose that there exists x 0 ∈ such that u( Hence L(u n − h n ) = −ϕ(·, u n ) ≤ 0 in ; (u n − h n )(x 0 ) = 0, However, any L-superharmonic function nonconstant cannot attain its negative minimum inside , so u n = h n = 1 n which implies that G D n ϕ(·, 1 n )(x 0 ) = 0. This is a contradiction with H 4 and hence (4.1) holds. Following this, lim n→+∞ ϕ(·, u n ) = ϕ(·, u) < ∞ so by the monotone convergence theorem, we get: Further, for every compact set K in , there exists α > 0 such that u(x) > α for x ∈ K. Therefore, ϕ(·, u) ≤ ϕ(·, α) ∈ L 1 (K). Also, G (ϕ(·, u)) = u − h is continuous, by Corollary 39 we may conclude L(G (ϕ(·, u)))= − ϕ(·, u), in , in the sense of distributions. The above proof suggests also the following corollary about bounded solutions. c > 0 such that G (ϕ(·, c)) ∈ L ∞ ( ). Then the continuous solution of Eq. 1.8 in given by u = h + G (ϕ(·, u)) is bounded in . Following this, there is one-to-one correspondence between L-harmonic nonnegative bounded functions and nonnegative bounded continuous solutions of Eq. 1.8.

Corollary 18 Let h be a nonnegative bounded L-harmonic function in . Suppose that the assumptions of Theorem 14, and in addition that there is
Proof By the proof of Theorem 14, u is the limit of a decreasing sequence of solution of Eq. 1.8 given by u n = h + 1 n + G (ϕ (·, u n )). Let u c be the solution of Eq. 1.8 given by u c = h + c + G (ϕ(·, u c )). Then by Lemma 15 u c ≥ u n for n big enough. So, 0 < u ≤ u n ≤ u c ≤ h + c + G (ϕ(·, c)). Moreover, by assumption h and G (ϕ(·, c)) are bounded in . Hence we may conclude that u is bounded too.
We obtain also a statement about regularity of solutions in a Greenian domain analogous to that for a bounded regular domain. Theorem 14, and assume in addition that ϕ(·, c) ∈ L ∞ loc ( ), for every c > 0. Then for every continuous solution of Eq. 1.8 in we have u ∈ C 1 ( ). Furthermore, if we suppose that ϕ ∈ C α loc ( ×]0, ∞[) then u ∈ C 2,α loc ( ).

Theorem 19 Suppose the same hypotheses as in
Proof Let u be a continuous solution of Eq. 1.8 in . Let D a bounded regular domain such thatD ⊂ . We denote f = u ∂D . By Theorem 8 ϕ(·, u)).

Boundary Condition
In this section, for a given a nonnegative L-harmonic function h in a Greenian domain , we give a sufficient and necessary conditions in order that the corresponding solution of the Eq. Thanks to (H 4 ), u > h because u − h = G (ϕ(·, u)) > 0. Furthermore, we denote h k = h + 1 k which L-harmonic then there exists u k a positive continuous solution of Eq. 1.8 such that u k = h k + G (ϕ(·, u k )). Hence 0 < u k − h k = G (ϕ(·, u k )) ≤ G (ϕ(·, 1 k )) which vanishes at ∂ . In addition, by Lemma 15 By tending k to ∞ we obtain u − h = 0 on ∂ . Moreover, by using Lemma 15, we conclude the uniqueness of solution.

Remark 21
The theorem remains true if we replace ∂ by ∂ ∪{∞} provided that for every c > 0, G (ϕ(·, c)) ∈ C 0 ( ). Unfortunately we cannot prove the converse statement. We have only the following theorem.  G (ϕ(·, c)) vanishes at the boundary.

Remark 23
As we have seen until now G (ϕ(·, c)) vanishing at the boundary for every c > 0 is just a sufficient condition for the existence of solution. However, in the special case when ϕ(x, y) = p(x)ψ(y), p ∈ K loc d ( ) positive and ψ is a positive continuous decreasing function (as in [7]) one can easily formulate a necessary and sufficient condition for the solution of Eq. 1.1. The following theorem generalizes Theorem 6.1 in [7].

Appendix
Let be a domain in R d , d ≥ 3 and let L be a second order elliptic operator with smooth coefficients defined in i.e. Remark 25 Notice that L is locally uniformly elliptic in .
Throughout all this section, we suppose that L1 ≤ 0 in and there exists s ∈ C ∞ ( ) such that s > 0 and Ls < 0 in . Such function can always exist if is bounded and L1 ≤ 0.
We consider at first D a bounded domain contained with its closure in with smooth boundary (C 1,α boundary is enough) where we represent basic notions of potential theory and most important properties of the corresponding Green function G D . Afterwards, we justify the existence of Green function in denoted G and we discuss her properties.

In a Regular Bounded Domain
The operator L satisfies the following properties in D provided that L1 ≤ 0 in D and D is a bounded domain with smooth boundary contained with its closure in :

Basic Properties
Theorem 26 ( Strong maximum and minimum principle). (see [10] p.34 or [1]  Remark 30 For f ∈ C(D), we can use a standard approximation of f which is explained e.g. in [2]. denoted G D f . Then we can define the Green operator by: Theorem 32 (Existence of Green function). (See [2] p.295 or [19] p.20) There exists a function G D (x, y) called a Green function C ∞ outside the diagonal such that for every f ∈ C(D) Furthermore, for every y ∈ D: • L(G D (·, y)) = − y , in the sense of distributions in D.
• lim x→∂D G D (·, y) = 0. Now using the weak maximum principle, we can easily prove that for ψ ∈ C ∞ c (D) G D (Lψ) = −ψ, in the sense of distributions in D, In other words, G D commute with L.

The Adjoint Operator
We denote L * the adjoint operator of L.
Now, we denote It is clear that L * s * = − 1 D , in the sense of distributions.
where 1 is the characteristic function of D. So s * is a smooth function satisfying s * > 0 and L * s * < 0. Such s * can be constructed in any bounded regular domain which allows us to conjugate L * and to obtain the corresponding preceding properties i.e. we define a new operator L * 1 by . We have L * 1 1 < 0, so all the preceding properties are true for L * 1 . Following this we can also obtain for L * the weak maximum principle, solvability of the Dirichlet problem and the Green function G * D in D. Preceding as in [2] we prove that: In what follows we recall some estimations of the Green function in D which facilitate the generalisation from the case of Laplace operator to the general elliptic operator.

Properties of the Green Function in Bounded Regular Domain
Proposition 36 • There is C > 0 such that for every x, y ∈ D, (see [16]). • Let x ∈ D and D 1 a compact set in D then there is C > 0 such that for every y ∈ D 1 .
By the Fubini theorem we get Secondly, let ψ ∈ C ∞ c (D). As before,

In a Greenian Domain
We consider (D n ) an increasing sequence of regular bounded domains exhausting i.e. D n ⊂ D n+1 and ∪ D n = .

Existence of the Green Function in
Using weak maximum principle we may easily justify: Proposition 40 • (G D n ) is an increasing sequence.
• G D n (−Ls) ≤ s for every n ∈ N.
Proposition 41 Let f ∈ C + c ( ). Then (G D n f ) is convergent.
Proof It is clear that (G D n f ) is increasing, so it is enough to prove that (G D n f ) is bounded.
We denote k f the support of f . The function x → −Ls(x) is continuous on k f so it is bounded on k f i.e. there exists a constant c k > 0 such that for every x ∈ k f c k ≤ −Ls(x), and then It follows by Riez theorem that there exists a function G such that lim n→+∞ G D n (x, y) = G (x, y). It follows that G * (x, y) = G (y, x) where G * is the Green function corresponding to L * in . We can check easily by monotone convergence theorem that • G (Lψ) = −ψ, in the sense of distributions in for every ψ ∈ C ∞ c ( ). • L(G (·, y)) = − y in the sense of distributions in for every y ∈ .
• L * (G (x, ·)) = − x in the sense of distributions in for every x ∈ .

Properties in a Greenian Domain
In a Greenian domain , we can obtain a generalized version of the maximum principle as follows: Proposition 42 Let f ∈ L 1 loc ( ), Lf ∈ L 1 loc ( ), Lf ≤ 0 as a distributions and lim inf which implies that f is lower semi-continuous and satisfies the super mean value property. The result follows by the minimum principle for so called L-superharmonic functions in the sense of the classical potential theory (see [14] p.427-8).
Now we focus on properties of the Green function G . First we recall the definition of potential in .

Definition 43
We say that a function p is a potential if p is L-superharmonic positive and if 0 ≤ h ≤ p, in , Lh = 0 then h = 0. We denote p ∈ P( ).
As examples, we can mention that G D n and G are potentials. So G f −G D f is L-harmonic in D and G D f is continuous which allows us to conclude.