Allen-Cahn equation for the truncated Laplacian: unusual phenomena

We study entire viscosity solutions of the Allen-Cahn type equation for the truncated Laplacian that are either one dimensional or radial, in order to shed some light on a possible extension of the Gibbons conjecture in this degenerate elliptic setting.


Introduction
In this paper we consider some entire solutions of a degenerate elliptic equation with non linear forcing term, where the behaviour of the solutions is quite different from the uniformly elliptic case.
We begin by recalling the linear uniformly elliptic model problem we have in mind in order to underline the differences with the degenerate elliptic case that we shall treat here. Consider the entire solutions of the Allen-Cahn equation ∆u + u − u 3 = 0 in R N (1) where N ≥ 2.
Writing the variable as x = (x , x N ), with x N ∈ R, we are interested in solutions that satisfy the following conditions: |u| < 1 and lim Any solution of (1) satisfying (2) is necessarily a function of x N , i.e. it is independent of x . This was referred to as Gibbons conjecture and it has been proved by several authors [3,4,17] and then it has been generalized to different contexts [14,18]. It is well known that this conjecture is somehow related to a conjecture of De Giorgi, see [2], but it would be too long to dwell on the subject. The function f (u) = u − u 3 can be replaced by a much more general class of functions and the result is still valid. On the other hand, if instead of the Laplacian one consider some degenerate elliptic operators there are a few results, let us mention e.g. the works in the Heisenberg group and in Carnot groups [13,15].
In this work we shall mainly focus on equations whose leading term is the operator P − k which is sometimes referred to as the truncated Laplacian. The operator P − k is defined, for any N × N symmetric matrix X, by the partial sum for a general class of functions f modelled on f (u) = u − u 3 . Clearly P − k (D 2 u) corresponds to the Laplacian when k = N , hence in the whole paper we shall suppose that k = 1, . . . , N − 1.
These operators are degenerate elliptic in the sense that In order to emphasize the strong degeneracy of these operators, let us mention that for any matrix X there exists M ≥ 0, not identically zero, such that P − k (X) = P − k (X + M ). It is immediate to see that, for example, one can take M = v N ⊗ v N where v N is an eigenvector corresponding to the largest eigenvalue of X.
In previous works we have, together with Hitoshi Ishii and/or Fabiana Leoni, encountered a certain number of surprising results, related to these degenerate operators. It would be too long to recall them all here but we shall briefly recall some of those closer to the results in this paper.
The classical Liouville result states that any harmonic function which is bounded from below is a constant. This is not true for bounded from below entire solutions of P − k (D 2 u) = 0.
Similarly, concerning the semi-linear Liouville theorem, the existence of entire non negative solutions of F ( Indeed, for the Laplacian there is a threshold for p between existence and non existence. This threshold is different for solutions and supersolutions and it depends on the dimension of the space. Instead, for solutions of the equation Interestingly, if one considers instead non positive solutions, the non existence results are very similar to those of the Laplacian in R k . These Liouville theorems were proved in [12]. Hence this lead us to wonder what happens to bounded entire solutions of the equation which satisfy |u| < 1 and which, a priori, may change sign. Can one expect solutions to be one dimensional? Precisely, can we extend Gibbons conjecture to this degenerate case? In order to answer these questions the first step is to study the one dimensional solutions, which is what we do in Section 2. Interestingly the results are completely different from the uniformly elliptic case and this leads to different conjectures. Precisely, we consider only one dimensional solutions u(x , x N ) = v(x N ) of (5) that satisfy |u| < 1. The results can be summarized in the following way: 1. The only classical solution is u ≡ 0;

Any viscosity subsolution is non negative and there exists a non trivial viscosity solution that satisfies
3. There are no solutions that are strictly monotone or positive Observe that, even though u 3 − u is the derivative of the double well potential F (u) = 1 4 (1−u 2 ) 2 , the lack of ellipticity does not allow the solutions to go from −1 to 1. This is the first surprising result.
Hence Gibbons conjecture should be reformulated in the following way: If the answer was positive, this would imply that there are no solutions of (5) that satisfies (2), since such one dimensional solutions don't exist in view of point 2 above. Or, equivalently, if such a solution of (5) exists then the answer to Question 1 is negative.
This is more similar in nature to the uniformly elliptic case. Nonetheless, classical proofs of these symmetry results rely heavily on the strong maximum principle, or strong comparison principle, and on the sliding method [6], which in general don't hold for the truncated Laplacian (see [8,11]). And in particular they are not true here since we construct ordered solutions that touch but don't coincide.
Another remark we wish to make is that, even though the solutions we consider are one dimensional, since they are viscosity solutions, the test functions are not necessarily one dimensional. Hence the proofs in are not of ODE type.
Other surprising results concern Liouville type theorems for (5). Aronson and Weinberger in [1] and more explicitly, Berestycki, Hamel and Nadirashvili in [5] have proved the following Liouville type result.
If v is a bounded non negative classical solution of Once again this result fails if one replaces the Laplacian with the truncated Laplacian. Indeed we prove that there exists infinitely many bounded non negative smooth solutions of for a general class of nonlinearities that include f (u) = u − u 3 . This is done by constructing infinitely many radial solutions of P − k (D 2 u) + f (u) = 0 which are positive in R N but tend to zero at infinity.
Finally in Section we show a different surprising phenomena related to the so called principal eigenvalue of P − k . This is somehow different in nature but we believe that it sheds some light to these extremal degenerate operators.

One dimensional solutions
We consider one dimensional viscosity solutions u, i.e. u(x) = v(x N ) for x = (x 1 , . . . , x N ), of the problem The main result is the following.
Proposition 1. Concerning problem (7), the following hold: i) If u ∈ U SC(R N ) is a viscosity one dimensional subsolution then u ≥ 0.
ii) The only classical one dimensional solution is u ≡ 0.
iii) There exist nontrivial viscosity one dimensional solutions, e.g.
or lim iv) There are no positive viscosity one dimensional supersolutions.
v) If u ≥ 0 is a viscosity one dimensional supersolution e.g. u(x) = v(x N ) and it is nondecreasing in the x N -direction then there exists t 0 ∈ R such that Remark 1. A consequence of i) and v) is that there are no viscosity one dimensional solutions increasing in the x N -direction.
is a viscosity solution of (7). This is obvious for x N = 0. Now takex = (x 1 , . . . ,x N −1 , 0). Since there are no test functions ϕ touchingũ by above at x, automaticallyũ is a subsolution. Let us prove theũ is also a supersolution.
Let ϕ ∈ C 2 (R N ) such that ϕ(x) =ũ(x) = 0 and ϕ ≤ u in B δ (x). Our aim is to show that λ N −1 D 2 ϕ(x) ≤ 0, from which the conclusion follows. Let W 0 = w ∈ R N : w N = 0 and for any w ∈ W 0 such that |w| = 1 let Since ϕ touchesũ from below atx andũ = 0 when x N = 0, we deduce that g w (t) has a maximum point at t = 0. Then Using the Courant-Fischer formula as we wanted to show. As above one can check that, for any c ≥ 0, the one dimensional functioñ is non monotone in the x N -direction and is solutions of (7).

Fig: The functionũ
iv) By contradiction suppose that u(x) = v(x N ) is a positive viscosity supersolution of (7). We claim that v is strictly concave, leading to a contradiction with v > 0 in R. We first prove that v satisfies the inequality v (t) < 0 for any t ∈ R in the viscosity sense. For this let ϕ ∈ C 2 (R) be test function touching v from below at t 0 . If we consider ϕ as a function of N variables just by settingφ(x) = ϕ(x N ), thenφ is a test function touching u from below at (x 1 , . . . , x N −1 , t 0 ), for any (x 1 , . . . , x N −1 ) ∈ R N −1 . Hence and then necessarily ϕ (t 0 ) < 0.
If v was not strictly concave, then there would exist t 1 <t < t 2 ∈ R such that as a test function in (11) we obtain a contradiction. v) By iv) there exists t 0 ∈ R such that v(t 0 ) = 0. By the monotonicity assumption we get v(t) = 0 for any t ≤ t 0 .

Radial solutions
This section is concerned with the existence of entire radial solutions of the equation where f : R → R satisfies the following assumptions: there exists δ > 0 such that Prototypes of such nonlinearities are f (u) = αu + β|u| γ−1 u with α > 0, γ > 1 and any β ∈ R.
Proposition 2. Under the assumptions (13) there exist infinitely many positive and bounded radial (classical) solutions of the equation (12).
In the model case f (u) = u − u 3 the assumptions (13) are satisfied with δ = 1 √ 3 . By a straightforward computations, it is easy to see that for any α ∈ (0, 1
Another feature of µ − k is that it is the upper bound for the validity of the maximum principle. Precisely if µ < µ − k and It is well known that for uniformly elliptic operators, the principal eigenvalue goes to infinity when the domain decreases to a domain with zero Lebesgue measure.
In this section we shall construct a sequence of domains Q n ⊂ R 2 that collapse to a segment such that µ − 1 (Q n ) the principal eigenvalue of P − 1 stays bounded above by 1. Hence not only they collapse to a zero measure set, but they are narrower and narrower.
Obviously w n < 0 in Q n and w n = 0 on ∂Ω. We shall prove that This will imply that µ − 1 (Q n ) ≤ 1. Indeed if 1 < µ − 1 (Q n ) the maximum principle would imply that w n ≥ 0 which is a contradiction. Let us show (17). Clearly D 2 w n = n 2 sin(nx) 0 0 sin y .
This ends the proof.