Let and be two bounded smooth domains of , , such that . Moreover, let and . We consider weak solutions to the problem
i.e., we consider such that
When , actually we deal with the case of positive solutions. Necessary and sufficient conditions for the existence of solutions to (1.1) are provided by the classical results of Keller  and Osserman , under suitable assumptions on the nonlinearity. The literature regarding boundary blow-up solutions is really wide (see, for example, [1, 2, 3, 4, 6, 5, 7, 13, 14, 12, 15, 16, 17, 20, 21]). Here we exploit an adaptation of the celebrated moving plane technique (see ) in order to obtain monotonicity properties of the solutions to (1.1). The domain that we consider is not convex and the solutions are not in as in the classical case. This is the same difficulty that occurs when dealing with the study of the uniqueness, symmetry and monotonicity properties of solutions to singular semilinear elliptic equations, see [9, 10, 11]. These problems exhibit in fact some similarities with problem (1.1) although the proofs cannot be adapted to our case.
In the second part of the paper, we prove the radial symmetry of the solutions on annular domains, under suitable assumptions. In our setting, this cannot be done just using the moving plane method, since the domain is not convex, and we prove that the solution is radially symmetric showing directly that the angular derivative is zero. The technique is based on a refined maximum principle for the linearized equation.
Let us introduce some notations. Let ν be a direction in , with . Given a real number λ, we set
that is, the reflection of x trough the hyperplane . We will make the following assumption throughout the paper:
and are strictly convex with respect to the ν-direction and symmetric with respect to .
Moreover, we set
Also we set
Observe that, by assumption (A), it follows that is nonempty and for any . We define
and for any .
We are now ready to state our main results.
Let be a weak solution to (1.1). Then
Consequently, it follows that u is strictly increasing with respect to the ν-direction in the set
In order to get symmetry results for the solution to (1.1), we restrict our attention to annular domains. We denote by the open ball of center 0 and radius in . By Theorem 1.1, we immediately deduce the following.
Let and be a weak solution to (1.1) in . Then in .
In the following we state sufficient conditions in order to deduce the radial symmetry of the solution, once we prove the monotonicity. We set
and we denote by the angular derivative of u.
Let and let be a weak solution to (1.1) in that satisfies
If , then u is radially symmetric and radially decreasing in .
2 Proof of Theorem 1.1
Let . We define
We need to prove that in . We have
so that weakly satisfies
Observe that and for small. Moreover, since on , it follows that on and on a neighborhood of , thanks to the fact that
Let be sufficiently small so that the weak maximum principle in small domains works for the operator in , recalling that for small. We get in and, by the strong maximum principle, we obtain
We need to prove that . On the contrary, suppose . By continuity, it follows that in . Moreover, it is possible to apply the same argument of the first part of the proof to obtain, by the strong maximum principle, that
Take now sufficiently small such that . Given , we fix a compact set so that . Since in K, by the continuity of u, it follows that there exists such that in K. Therefore, we can find such that
whenever . Moreover, choosing sufficiently small such that and using as test function in the weak formulation of (1.1), since f is locally Lipschitz continuous, an application of the Poincaré inequality gives
Choosing δ sufficiently small, we have so that in . Hence, in . By the strong maximum principle, we get
This gives a contradiction with the definition of μ and shows that , namely, (1.2) is proved.
Now, let x and such that . Setting , we get and
Therefore, u is strictly increasing with respect to the ν-direction. To conclude the proof, fix and let be such that . We have that , in and in . By the Hopf lemma, we obtain
3 Proof of Theorem 1.3
We start by proving the following proposition.
Let and be a weak solution to (1.1) in . If , then
Let . Set for each . We have
Since in and , we obtain
so that . ∎
In the following we will also exploit the fact that
as it follows by direct computation.
Proof of Theorem 1.3.
We shall actually show that in . By assumption (1.3), choosing sufficiently small, we have in .
Furthermore, by Corollary 1.2, since v is continuous, we have
where with . Moreover, on , by the Hopf lemma. Then in . Since is bounded in , for t sufficiently large, we have in . Hence,
We need to prove that . Conversely, suppose . By the definition of and Proposition 3.1, we obtain
By the strong maximum principle, since and on , we get
Since (1.3) is in force, there exists such that
Moreover, we have
By continuity, for small, we have that
Resuming, we have that
This contradicts the definition of , showing that actually and in . This is possible only if in , namely, if the solution is radial. We conclude that the solution is also radially decreasing by Theorem 1.1. ∎
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About the article
Published Online: 2018-07-07
Published in Print: 2019-03-01
Citation Information: Advances in Nonlinear Analysis, Volume 9, Issue 1, Pages 1–6, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496, DOI: https://doi.org/10.1515/anona-2017-0221.
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