Math. Z. (2006) DOI 10.1007/s00209-006-0947-5
Mathematische Zeitschrift
Enrico Valdinoci
Flatness of Bernoulli jets
Received: 6 May 2005 / Accepted: 12 December 2005 / Published online: © Springer-Verlag 2005
Abstract Level sets of minimizers of a possibly singular or degenerate elliptic variational problem related with fluid jets are shown to possess a Harnack-type inequality. A few flatness and one-dimensional results thence follow. “...e diedi un volto a quelle mie chimere le navi costruii di forma ardita, concavi navi dalle vele nere...” (Fancesco Guccini, Odysseus.)
Keywords p-Laplacian operator · Free boundary problems · Sliding methods · Geometric and qualitative properties of solutions · Variational models in fluid dynamics Subject Classification: 49J10 · 35R35 · 35B05 · 35J70 1 Introduction Given m ∈ (0, +∞), p ∈ (1, +∞) and a domain ⊆ R N , we define the following functional on W 1, p (): |∇u(x)| p F (u) = + mχ(−1,1) (u(x)) d x . p This work has been partially supported by MIUR Variational Methods and Nonlinear Differential Equations. E. Valdinoci Dipartimento di Matematica, Università di Roma Tor Vergata, Via della Ricerca Scientifica, 1, I-00133 Roma Italy E-mail:
[email protected] 00209 0947 B Jour. No.
Ms. No.
Dispatch: 14/2/2006 Total pages: 42 Disk Received ✓ Disk Used ✓
Journal: Math. Z. Not Used Corrupted Mismatch
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The functional above is a model for fluid jets and cavitation. The “kinetic term” given by “|∇u| p ” gives that critical points of the functional are p-harmonic when they take value in (−1, 1): this corresponds to a “continuity equation” driven by a p-Laplacian operator. From the physical point of view, for p = N = 2, a critical point u of the above functional represents the stream function of an ideal fluid1 and the particles of the fluid move along the level sets of u. The discontinuity provided by the “potential term” of the functional, that is, the term “χ(−1, 1) ”, gives a free boundary condition. Physically, such condition corresponds to a balance between the speed of the fluid and the external pressure and it is governed by the well-known Bernoulli’s law (thence the name of the model). We refer to [2], [3], [7], [8] and [15] for further details on such model. Our system is also related to the one studied in [13] and [4], which also deals with the interfaces of immiscible fluids. Related functionals also arise in cavitation and flame propagation (see again [7], [8] and references therein). In the pure mathematics setting, such functional is interesting for both the behavior of the free boundary (see [2], [3], [7] and [8]) and its relation with the surfaces of minimal perimeter (see, e.g., [5] and [20]). Throughout the paper, we will mainly deal with minimizers of the above functional. As customary, we say that u is a local minimizer for F in if F (u) ≤ F (u + ϕ), for any ϕ ∈ C0∞ (). If u is defined in the whole space and it is a local minimizer in any domain, we say that u is a Class A minimizer. We will prove a Harnack-type inequality for level sets of minimizers, stating that if one knows that the zero level set of a minimizer is trapped in a rectangle whose height is small enough, then, in a smaller neighborhood, it can be trapped in a rectangle with even smaller height. More precisely, Theorem 1 Let l > 0, θ > 0. Let u be a local minimizer for F in (x , x N ) ∈ R N −1 × R |x | < l , |x N | < l . 1 Here is a rough sketch of the physical motivation of the problem for p = N = 2, in relation with ideal fluid jets. Consider a homogeneous, incompressible, irrotational, inviscid fluid jet in the plane (subject to no gravity). Let v be its velocity and P the pressure. The equation of continuity simply reads here div v = 0. Bernoulli’s law (that can be easily derived from energy balance, or from the Euler equation) states that the motion of the particles of fluid preserves the quantity |v|2 /2 + P (up to dimensional constants). Since the fluid is irrotational, its velocity admits a potential ψ, i.e. v = ∇ψ. From the equation of continuity, ψ = 0. Hence, there exist a function u (usually called stream function) so that
∂1 u = −∂2 ψ
and
∂2 u = ∂1 ψ .
By construction, ∇u · v = 0, therefore the particles of fluid move along the level sets of u. Also, using the equation of continuity and Bernoulli’s law, one sees that u = 0 in the region occupied by the fluid jet, and that |∇u|2 /2 + P = const on the boundary of the jet. One can prove (see [2] or [4]) that the minimizers of the functional we deal with satisfy the above equations in a suitable weak sense.
Flatness of Bernoulli jets
3
Assume that u(0) = 0 and that {u = 0} ⊆ {|x N | < θ } . Then, there exists a universal constant c ∈ (0, 1) so that, for any θ0 > 0 there exists ε0 (θ0 ) > 0 such that, if θ ≤ ε0 (θ0 ) l
and
θ ≥ θ0 ,
then {u = 0} ∩ {|x | < cl} ⊆ {|x N | < (1 − c) θ } . Theorem 1 is much related to analogous results recently obtained by Savin for phase transition equations in connection with a famous conjecture of De Giorgi (see [17] and [21]). The proof of Theorem 1 will also employ some results from [5], [7], [15] and [20]. On a heuristic ground, we may say that results of this type should be expected at least for the following reason: if the zero level set of a minimizer u oscillated “too much”, then either the gradient nearby must be big, or there is a large region where u is close to zero. Then, either the kinetic or the potential term would be too large, in contradiction with the minimality of u. Moreover, one may think that the above result shows an interplay between the PDE structure of u and the one of the limiting minimal surface, which impose some “rigidity” in the behavior of the minimizers. A very rough idea of the proof of Theorem 1 might be the following. Near a given point x¯ ∈ {u = 0}, one expects u to “behave” like a plane of constant slope passing through x. ¯ Namely, if we slide such planes in the e N -direction (and we “bend” them a little, to avoid boundary touching points), we hope to be able to touch u at a “wide” portion of its level sets; moreover, by the standard Harnack inequality, one expects u and the bended planes to be close (if the bending is small). Then, if {u = 0} oscillated “too much” in the e N direction, one would find “many” bended planes (to which u is close) whose e N -projections overlap. These overlappings would give that {u = 0} is “too far” from having minimal perimeter, in contradiction with the prescription given by the limit case. However, to make such argument work, some fine measure estimates on the projection of these contact points are necessary. In particular, some covering lemmata is needed to relate the “local behavior” of u (to wit, u being close to be a plane), with the behavior “in the large” (where u looks “like having zero mean curvature” and thus it may be more and more easily touched by more and more bended planes). This paper, which has been largely inspired by many fundamental ideas of [17], is organized as follows. In § 2, following [17], we derive a few flatness and onedimensional symmetry results from Theorem 1. In § 3, we will confine our minimizer by using a barrier built in [20]; this will be also of some help in avoiding boundary contact points. The construction of the barriers (the “bended plane” in the above heuristics) needed in this paper is contained in § 4, while § 5 is devoted to some measure estimates on the touching point set with such barriers. Such estimates are then improved in § 7, via some measure theoretic arguments developed
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in § 6 (these lemmata are the ones linking the small and the large scales, as mentioned in the heuristics above); the proof of one of such arguments is deferred to the Appendix. The proof of Theorem 1 is then completed in § 8. Fixing some notation. Given two vectors v and w, we define (v, w) to be the angle between these vectors, i.e.,
(v, w) := arccos
v·w . |v| |w|
We will denote by q the dual exponent of p. Also, to make2 some computations easier, we assume throughout the paper that m =
p−1 1 = , p q
(1.1)
which can be obtained from the general case by a simple homogeneous rescaling in the space variables. As usual, ω N denotes the volume of the N -dimensional unit ball (with respect to the standard Lebesgue measure, which is denoted by L N ). Also, given X ∈ R N +1 we often denote x ∈ R N and x N +1 ∈ R in such a way X = (x, x N +1 ) ∈ R N × R . We will use the notation x = (x , x N ) ∈ R N −1 × R . Given a vector ξ ∈ S N , we denote by Pξ the hyperplane in R N +1 orthogonal to ξ , i.e. Pξ := {X ∈ R N +1 | ξ · X = 0} .
We also denote by πξ the projection onto Pξ , i.e. πξ (X ) := X − (ξ · X )ξ ,
∀X ∈ R N +1 .
With a slight abuse of notation, if ξ N +1 = 0, we will sometimes identify ξ with its N -dimensional projection by implicitly dropping the last zero-coordinate. Furthermore, for k = 1, . . . , N + 1, we let πk : R N +1 −→ {xk = 0} be the natural projection, i.e. πk (x1 , . . . , x N +1 ) := (x1 , . . . , xk−1 , 0, xk+1 , . . . , x N +1 ) . As standard, given A ⊆ B, the notation Per (A, B) will denote the perimeter of A in B (see, for instance, [11] for full details on such definition). 2 E.g., with respect to the notation in [20], we note that the quantity called there ω is 1 here, due to the choice in (1.1), which has to be compared to (5.2) in [20].
Flatness of Bernoulli jets
5
2 Consequences of Theorem 1 Following some ideas of [17], we manage to deduce several flatness results for our problem. First of all, we have the following “flatness improvement”: if a level set is trapped inside a flat cylinder, then, by possibly changing coordinates, it is trapped in a flatter cylinder in the interior. More precisely, Theorem 2 Let l > 0, θ > 0. Let u be a local minimizer for F in
(x , x N ) ∈ R N −1 × R |x | < l , |x N | < l .
Assume that u(0) = 0 and that {u = 0} ⊆ {|x N | < θ } . Then, there exist universal constants η1 , η2 > 0, with 0 < η1 < η2 < 1, such that, for any θ0 > 0, there exists ε1 (θ0 ) > 0 such that, if θ ≤ ε1 (θ0 ) l
and
θ ≥ θ0 ,
then {u = 0} ∩ {|πξ x| < η2 l} × {|(x · ξ )| < η2 l} ⊆ {|πξ x| < η2 l} × {|(x · ξ )| < η1 θ } for some unit vector ξ . We omit here the proof of Theorem 2, since it follows verbatim the proof of Theorem 1.2 in [21] (exploiting [20] instead of [16]). De Giorgi conjectured that the “typical” phase transition solution should have flat level sets, at least in low dimension: see [9] for a rigorous exposition of such conjecture and [11] for its relation with the minimal surface theory; see also [12], [1], [17], [21] and references therein for results in that direction. We now state that also the “typical” (low dimensional, or sub-linear) Bernoulli jet is flat. Indeed, repeating the proofs of Theorems 1.3-1.4 in [21], one deduces from an iteration of Theorem 2 the following Theorem 3 Let u be a non-constant Class A minimizer for F , with |u| ≤ 1. Assume one of the following conditions: • N ≤ 7, or • for any θ ∈ (−1, 1), {u = θ } is a graph in the e N -direction and either N ≤ 8 or the graphs {u = θ } have at most linear growth at infinity. Then, the level sets of u are hyperplanes.
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The above result also gives a one-dimensional symmetry property for u: namely, if
−1 u(t) := t 1
if t < −1 , if −1 ≤ t ≤ 1 , if t > 1 ,
then, it follows from Theorem 3 that u(x) = u(ν · x + x0 ), for some ν ∈ S N −1 and x0 ∈ R N . The results presented here also use and extend the regularity theory in bounded domain which has been performed in [15]. Namely, the optimal Lipschitz regularity for local minimizers of F has been proven in Theorem 4.1 of [15] (see also [7]). Moreover, if u ε is a local minimizer of |∇v| p v −→ εp + mχ(−1,1) (v) d x , p for an infinitesimal sequence of ε > 0, Theorem 2.3 of [15] give that, possibly taking subsequences, {u ε = 0} ∩ B1/2 lies in an arbitrarily small neighborhood of a hypersurface ∂ E with minimal perimeter. If one knows (as it happens in Theorem 3, due to the theory of minimal surfaces) that ∂ E is flat, one may conclude that {u ε = 0} ∩ B1/2 is uniformly close to being flat. In this sense, Theorem 3 straightens this property when one deals with global solutions. Very roughly speaking, the improved flatness of the level sets comes from the interplay between the small and the large scales of the problem, that is, the Bernoulli jets “look like” minimal surfaces at large scale, coupled with harmonic functions at small scale. Thus, the level sets of our solution “look nice” in the large, thanks to the theory of minimal surfaces (in the case the latter ones “look nice”) and then they also “look nice” in the small, by elliptic regularity. 3 Trapping a minimizer We first make a precise statement about the growth from the free boundary of a minimizer. From a physical viewpoint, one expects Bernoulli’s law to prescribe the speed of a fluid jet on its free boundary via a pressure balance. In our case, due to the choice in (1.1), these heuristic considerations would lead to “|∇u| = 1 on ∂{|u| < 1}”. Here below, we give a precise formulation of this fact, in terms of a growth from the free boundary “by a slope 1”: Proposition 4 Let u be a local minimizer for F . Then, if there is an open ball B+ ⊆ {|u| < 1} touching ∂{u < 1} at x+ , then u(x+ + t j ν0 ) ≥ 1 − t j − o(t j ) ,
(3.1)
for any infinitesimal positive sequence t j −→ 0+ , where ν0 is the interior normal of B+ at x+ . Analogously, if there is an open ball B− ⊆ {u = −1} touching ∂{u < −1} at x− , then u(x− − t j ν0 ) ≥ −1 + t j − o(t j ) ,
(3.2)
for any infinitesimal positive sequence t j −→ 0+ , where ν0 is the interior normal of B− at x− .
Flatness of Bernoulli jets
7
For the proof, see § 5 in [20] (with ω = 1). We now confine a minimizer of the type of the one in Theorem 1 by a barrier constructed in [20]. For this, we fix κ = 1/(cl) ¯ and κ > 0 suitably small (as requested in Lemma 2.1 of [20]) and we define ω± :=
(1 ± κ )2 ± 4κ
and
β(s) :=
min{ω+ s − κs 2 , 1} if s ≥ 0, max{ω− s − κs 2 , −1} if s < 0.
Then, we have: Lemma 5 Let l > 0, θ > 0. Let u be a local minimizer for F in
(x , x N ) ∈ R N −1 × R |x | < l , |x N | < l .
Assume that |u| ≤ 1, that u satisfies the free boundary growth conditions (3.1)(3.2), that θ/l is suitably small, that u(0) = 0, that u(x) > 0 if x N ≥ θ and that u(x) < 0 if x N ≤ −θ . Then, −β(θ − x N ) ≤ u(x) ≤ β(x N + θ ) , for any x ∈ [−l/10, l/10]. Proof We proof the latter inequality, the first one being analogous. We let g and y,l be the barriers defined in Lemma 2.1 of [20]. From the density estimates in [15] and the fact that {u = 0} ⊆ {|x N | ≤ θ }, it easily follows that u(x) = −1 for any x ∈ R N with x N ≤ l/16 and |x | ≤ l/2. Then, u(x) ≤ y,l/8 (x) , where y := (0, . . . , 0, −l/2), provided that u(x) = −1. Let l eN yˆ := x − x N + θ + 8 and e := yˆ − y. We consider the barrier y+te,l/8 . Since yˆ N + l/8 = −θ , { y+te,l/8 = 0} does not meet {u = 0} for t ∈ [0, 1]: thus, by Lemma 2.3 in [20], u ≤ y+te,l/8 for any t ∈ [0, 1]. In particular, u(x) ≤ y+e,l/8 (x) = yˆ ,l/8 (x) = g(|x − yˆ | − l/8) = g(x N + θ ) = β(x N + θ ) , which gives the desired inequality.
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4 Construction of barriers We fix a constant C 0 > 1, to be chosen conveniently large. We will also fix R, to be assumed suitably large (with respect to C 0 and some other universal constants). Let also |s0 | ≤ 1/4. We define −R + R 2 − 2C 0 R s0 ∓ 21 . σ± := s0 + C0 Note that σ± ∼ ±1/2 for large R. Let also 8C 8R R 0 + 1 1 − 1 − τ+ := σ+ + C0 (8R + C 0 )2 and τ− := σ− +
16R C0
−1
1 −
1+
16C 0 R (16R − C 0 )2
.
Note that τ± ∼ ±1 for large R. We define the function gs0 ,R (t) : [τ− , τ+ ] −→ [−1, 1] by −1 if t ≤ τ− , C C 1 2 0 0 − 2 + 1 − 16R (t − σ− ) − 32R (t − σ− ) if τ− < t < σ− , C0 2 t + 2R 0) (t − s 1 + 1 + C 0 (t − σ+ ) − 2 8R
if σ− ≤ t ≤ σ+ , C0 16R (t
− σ+
)2
if σ+ < t ≤ τ+ .
Note that gs0 ,R is continuous, increasing and smooth outside σ± and τ± , that gs0 ,R (σ± ) = ±1/2 and that gs0 ,R (τ± ) = ±1. What is more, lim g y N +1 ,R (t) > lim g y N +1 ,R (t) ,
t→σ−−
t→σ−+
lim g y N +1 ,R (t) > lim g y N +1 ,R (t)
(4.1)
lim g y N +1 ,R (t) < 1 < lim g y N +1 ,R (t) .
(4.2)
t→σ+−
t→σ++
and t→τ−+
t→τ+−
Given Y ∈ R N +1 with |y N +1 | ≤ 1/4, we also define gS(Y,R) (x) := g y N +1 ,R (y N +1 + |x − y| − R) and
S(Y, R) = x ∈ R N +1 x N +1 = gS(Y,R) (x) .
Flatness of Bernoulli jets
9
In the above definition, we will sometimes refer to Y as the “center” and to R as the “radius” of S. Note that, in the domain of gS(Y,R) , |x − y| ∼ R ,
(4.3)
for large R. We now show a p-superharmonicity property for gS(Y,R) : Proposition 6 Let Y ∈ R N +1 with |y N +1 | ≤ 41 . Then, gS(Y,R) is strictly p-superharmonic at any x ∈ R N for which gS(Y,R) (x) ∈ (−1, −1/2) ∩ (1/2, 1). Also, no smooth function can touch gS(Y,R) by below at x if |gS(Y,R) (x)| = 21 . Proof From (4.1), it follows that no smooth functions can touch the barrier at the ±1/2-level sets. The p-subharmonicity for level sets in (−1, −1/2) ∪ (1/2, 1) is a straightforward computation.
As a consequence of the above result, we show that non-trivial touching points between S(Y, R) and p-subharmonic functions with free boundary growth conditions may only occur on the (−1/2, 1/2)-level sets: Corollary 7 Let U be continuous in the closure of , C 1 and p-subharmonic in |U | < 1. Suppose that |U | ≤ 1 and that U satisfies the free boundary growth conditions in (3.1)-(3.2). Assume that U ≤ gS(Y,R) and that U (x ) = gS(Y,R) (x ) for some x in the interior of . Then, either x lies in the interior of {U = −1} or |gS(Y,R) (x )| < 1/2. Proof Let us assume that x is not in the interior of {U = −1}. That |gS(Y,R) (x )| = 1 is ruled out by (3.1)-(3.2) and (4.2). We now show that gS(Y,R) (x ) ∈ (−1, −1/2]∪ [1/2, 1). For this, we recall that gS(Y,R) is smooth with non-vanishing gradient in
1 0 := |gS(Y,R) (x)| ∈ ,1 , 2 and it is strictly p-superharmonic in 0 . This implies that U cannot coincide with gS(Y,R) in 0 . Therefore, by the Strong Comparison Principle (see, e.g., Corollary B.5 in [21] and references therein) we have that U < gS(Y,R) in 0 . Furthermore, |gS(Y,R) (x )| = 21 by the last claim in Proposition 6.
We now introduce another hypersurface in R N +1 , which will be denoted by S(Y, R) and we investigate its relation with S(Y, R). While S(Y, R) is continuous but not smooth, S(Y, R) is smooth, it coincides with S(Y, R) in {|x N +1 | ≤ 21 } and it always stays above outside. For any |t| ≤ 99/100, we define ρs0 ,R (t) := t +
C0 (t − s0 )2 . 2R
Also, given Y ∈ R N +1 with |y N +1 | ≤ 41 , we define g S(Y,R) (x) := ρ y N +1 ,R (y N +1 + |x − y| − R) ,
(4.4)
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and S(Y, R) =
x ∈ R N +1 x N +1 = g (x) . S(Y,R)
Note that, in the domain of g S(Y,R) , (4.3) holds. Moreover, C C0 2 g . + ∂i, j g p S(Y,R) S(Y,R) ≤ R
(4.5)
Also, g S(Y,R) (x) ≥ gS(Y,R) (x) 1 and, if |g S(Y,R) (x)| ≤ 2 , then
g S(Y,R) (x) = gS(Y,R) (x) . 5 The first contact points This section is close to Chapter 4 of [17] and § 3 of [21] (though the details of the calculations involved are different) and it deals with the geometric and measure theoretic properties of the set collecting the points of first touching (at a higher level than −1) between our solution and the barrier, when the latter is slided in some direction. In particular, the measure theoretic properties of an N -dimensional projection of the touching point set will be dealt with in Proposition 20 below: roughly speaking, we will prove there that the measure of the projection of the “first occurrence” touching points controls the measure of the projection of the centers of the corresponding surfaces. This gives that, if we consider a “large projection” set of centers, we obtain a “large projection” set of contact points. In all this section, we fix C to be a suitably large constant; also, l and R are fixed and suitably large (possibly in dependence of C ) and l/R is assumed to be conveniently small. We also assume that u is continuous in [−C l , C l] N , that |u| ≤ 1, that u is p-harmonic if |u| < 1 and that it satisfies the free boundary growth (3.1)-(3.2). We denote the graph of u by Gu := {x N +1 = u(x)}. We define the set of first contact points as follows. Given a compact set3 A ⊆ Pξ ⊂ R N , we set = A A := {Y ∈ R N +1 s.t. ∃Yˆ ∈ A, tˆ ∈ R s.t. Y = Yˆ + tˆξ , A gS(Yˆ +tξ,R) (ξ ) > u(ξ ) for any t < tˆ, provided that u(ξ ) = −1 gS(Y,R) ≥ u and ∃x ∈ R N +1 gS(Y,R) (x) = u(x) = −1} .
s.t.
may be seen as the set of the “first contact centers”. Moreover, The above set A we define A := {X = (x, x N +1 ) ∈ R N +1 s.t. ∃Y ∈ A s.t. = B B gS(Y,R) (x) = u(x) = x N +1 = −1} 3
For the notation about Pξ , recall page 4.
Flatness of Bernoulli jets
11
as the set of “first contact points”. From the geometric viewpoint, and we refer to B indeed, in the definition of the above sets, we are sliding our barriers until it touches collects Gu (recall Corollary 7) for the first time at a level higher than −1: the set B all such first occurrence contact points, while A collects the corresponding centers. We also define ) . B := πξ (B In this section, we will assume that A ⊆ [−C l/2, C l/2] N × [−1/4, 1/4]
(5.1)
∩ ∂[−C l, C l] N = ∅ , B
(5.2)
and that
that is, we rule out touching points on the boundary of the domain. Due to (5.2) and Corollary 7, we have that ⊆ {X ∈ R N +1 | |x N +1 | < 1/2} . B
(5.3)
Also, the compactness property of A is inherited by the other sets defined above: namely, by repeating the proof of Lemma 3.2 of [21] (verbatim but requiring the contact points to lie at levels higher than −1), one deduces that , B and B are compact (thence measurable) sets. Lemma 8 A S(Y, R) We now show how to reconstruct the center Y of the rotation barrier from the normal ν S(Y,R) at a given point X (in particular, at a contact point, see Corollary 10 below).
Lemma 9 For any X ∈ R N +1 with X ∈ S(Y, R) and ν ∈ S N , let − ∇g (x), 1 S(Y,R) ν S(Y,R) (X ) := ∈ R N +1 , 2 1 + |∇g S(Y,R) (x)| |(ν1 , . . . , ν N )|2 R 1− ∈ R, ω(X, ν) := ν N2 +1 2C 0 1 − 1 − 2C 0 ω(X,ν) R ∈ R, σ (X, ν) := R 1 − C0 (ν1 , . . . , ν N ) σ (X, ν), x N +1 + ω(X, ν) ∈ R N +1 . F(X, ν) := x + |(ν1 , . . . , ν N )| Then,
Y = F(X, ν S(Y,R) (X )) , S(Y, R). for any X ∈
(5.4)
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Proof Note that ν S(Y,R) (X ) is normal to S(Y, R) at the point X . For short, we will set here S := S(Y, R). Since S is a rotation surface, there exists σ ∈ R so that y−x =σ
(ν1S (X ), . . . , ν NS (X )) |(ν1S (X ), . . . , ν NS (X ))|
.
More precisely, from (4.3), we have that σ ∼ R, for R large. What is more, since x N +1 = g S (x), we have that y N +1 + |x − y| − R = x N +1 −
C0 (|x − y| − R)2 . 2R
Hence, if ω := y N +1 − x N +1 , we gather that 1− 1− σ = |x − y| = R 1 − C0
2C 0 ω R
,
which determines σ , as claimed. In order to determine ω, note that
|(ν1S (X ), . . . , ν NS (X ))|
ν NS +1 (X ))
= |∇g S (x)| = ρs0 ,R (y N +1 + |x − y| − R)
C0 C0 = 1+ (|x − y| − R) = 1 + (σ − R) R R 2C 0 ω , = 1− R thanks to the above computations on σ . This determines ω, as desired.
At contact points, where the gradient of u (and thus the normal) agrees with the one of the barrier, we thus deduce: Corollary 10 Let the notation of Lemma 9 holds. For X = (x, x N +1 ) ∈ R N , with x N +1 = u(x), let − ∇u(x), 1 ν u (X ) := ∈ R N +1 , (5.5) 2 1 + |∇u(x)| Let X, Y ∈ R N be so that gS(Y,R) (z) ≥ u(z) ∀z ∈ R N , gS(Y,R) (x) = u(x) = x N +1 = −1 . Then, Y = F(X, ν u (X )). Note that ν u (X ) here above is normal to the graph of u at the point X . We also point out an easy estimate on the slope of the normals.
Flatness of Bernoulli jets
13
, then Lemma 11 In the notation of Corollary 10, if X ∈ B y−x u |y−x| , 1 ≤ C C0 , ν (X ) − √ R 2 for a suitable universal constant C ≥ 1. Proof Since, by (5.3), |x N +1 | ≤ 1/2, we have that |∇u(x)| = |∇gS(Y,R) (x)| ∈
inf
t∈[σ− , σ+ ]
ρs 0 ,R (t),
sup
t∈[σ− , σ+ ]
ρs 0 ,R (t) ,
which implies the desired claim.
We now make a computation related to the Second Fundamental Form at touching points. Lemma 12 Let the notation of Lemma 9 and Corollary 10 hold. Let Y (X ) := F(X, ν u (X )) .
(5.6)
Then, there exists a positive constant C such that |D X Y (X )| ≤ C , . for any X ∈ B Proof By differentiating (5.6), D X Y (X ) = D X F(X, ν u (X )) + Dν F(X, ν u (X )) D X ν u (X ) .
(5.7)
On the other hand, by differentiating (5.4),
0 = D X F(X, ν S(Y,R) (X )) + Dν F(X, ν S(Y,R) (X )) D X ν S(Y,R) (X ) . Thus, from (5.7) and (5.8), we gather that D X Y (X ) = Dν F(X, ν S(Y,R) (X )) D X ν u (X ) − D X ν S(Y,R) (X ) , . By Lemmata 9 and 11, one obtains that for any X ∈ B |∂ν ω(X, ν)| ≤ const R and thus
|Dν F(X, ν S(Y,R) (X ))| ≤ const R . Therefore,
|D X Y (X )| ≤ const R D X ν u (X ) − D X ν S(Y,R) (X ) .
(5.8)
14
E. Valdinoci
Thence, from Lemma 11 and a direct computation on the Second Fundamental Form (see, e.g., Appendix B of [21]), we obtain that |D X Y (X )| ≤ const R D 2 u(x) − D 2 g (x) (5.9) . S(Y,R) Also, since g S(Y,R) touches u from above at X , we have that 2 D 2 g S(Y,R) (x) − D u(x) is a non-negative definite matrix
and so
(5.10)
const D 2 u(x) − D 2 g (x) ≤ g − u (x) . S(Y,R) S(Y,R)
Thus, we deduce from (5.9) that
|D X Y (X )| ≤ const R g − u (x) . S(Y,R)
(5.11)
On the other hand,
p−2 g p g S(Y,R) (x) − p u(x) = |∇g S(Y,R) (x)| S(Y,R) (x) − u(x) p−4 ∇g + ( p − 2) |∇g S(Y,R) (x)| S(Y,R) (x) 2 · D 2 g S(Y,R) (x) − D u(x) · ∇g S(Y,R) (x)
and therefore, by exploiting Lemma 11 and (5.10), we gather that (x) − u(x) ≥ const g (x) − u(x) . p g p S(Y,R) S(Y,R) The desired claim thus follows from (5.11), (5.12) and (4.5).
(5.12)
and B easily implies the following observation: The construction of A lies in Y (Gu ), which are Lipschitz surfaces. lies in Gu and A Lemma 13 B and Also, by repeating verbatim the proof of Lemma 3.8 in [21], we see that A are graphs with respect to the ξ -direction, that is: B and on B . Lemma 14 πξ is injective on A Given X ∈ R N +1 , we now define X := {Z | X ∈ S(Z , R)} .
(5.13)
Note that X is the surface containing all the centers of the surfaces S(·, R) to which X belongs. From Corollary 10, one sees that . Lemma 15 Let Y be as in Lemma 12. Then, Y (X ) ∈ X for any X ∈ B We denote by ν X (Z ) the unit normal vector to the surface X at a point Z ∈ X (in a fixed orientation): this is a bona fide definition, since X is a Lipschitz surface, as we are now going to show. Also, we can express ν X in terms of the normal to S(Y, R), according to the following result:
Flatness of Bernoulli jets
15
. Then, X is a Lipschitz rotaLemma 16 Let Y be as in Lemma 12 and let X ∈ B tion surface with axis parallel to e N +1 and passing through X . Also, ν X (Y (X )) belongs to the space spanned by e N +1 and ν S(Y (X ),R) (X ). Proof Let us first show that X is a Lipschitz rotation surface. By (4.4), we have that Z ∈ X if and only if x N +1 = z N +1 + |z − x| − R +
C0 (|z − x| − R)2 , 2R
which proves the rotational symmetry and the Lipschitz properties of X . The rotational symmetry also gives that ν X (Y (X )) is in the space spanned by (πe N +1 Y (X ) − x) and e N +1 . But ν S(Y (X ),R) (X ) is also in the space spanned by S(Y (X ), R). Moreover, (πe N +1 Y (X ) − x) and e N +1 , due to the radial symmetry of S ( Y (X ),R) ν (X ) and e N +1 cannot be parallel (because |x N +1 | ≤ 1/2 due to (5.3) and so ∇gS(Y (X ),R) (x) = 0). Therefore, ν X (Y (X )) belongs to the space spanned by ν S(Y (X ),R) (X ) and e N +1 .
Here is a relation between ν X and ν u : Lemma 17 There exists a positive constant C such that |ν X (Y (X )) · ξ | ≤ C |ν u (X ) · ξ | , . for any X ∈ B We omit the details of the proof of Lemma 17, since it follows exactly the proof of Lemma 3.11 in [21] (just exploit Lemmata 16 and 11 here instead of Lemmata 3.10 and 3.5 there). Lemma 18 Let X ∈ Gu , with |x N +1 | ≤ 1/2. Assume that g S(Y,R) ≥ u. Then, Y is above X with respect to the e N +1 -direction. ∗ Proof Assume that (y, y N∗ +1 ) ∈ X . Then, X ∈ S((y, y N +1 ), R), so
g S((y,y ∗
N +1 ),R)
(x) = x N +1 = u(x) ≤ g S(Y,R) (x)
and therefore y N∗ +1 ≤ y N +1 .
touches X from above at Y (X ). . Then A Lemma 19 Let X ∈ B Proof First, note that ) ⊆ A . Y (X ) ∈ Y (B On the other hand, Y (X ) ∈ X by Lemma 15. Thus, to complete the proof of this is above X with respect to the e N +1 -direction. For result we need to show that A . Then, by construction, g this purpose, take Y ∈ A S(Y,R) ≥ u (and equality holds
at some point). Thus, by Lemma 18, Y is above X . We now give an explicit measure estimate for the projection of the contact points:
16
E. Valdinoci
Proposition 20 Assume that for any Y ∈ A there exist t ∈ R and x ∈ R N , such that gS(Y +tξ,R) (x) ≤ u(x).
(5.14)
Then, L N (A) ≤ C L N (B ) ,
for a suitable positive universal constant C. To prove Proposition 20, repeat the proof of the analogous Proposition 3.14 of [21] (but use here Lemmata 8, 13, 14, 19, 17 and 12 instead of Lemmata 3.2, 3.7, 3.8, 3.13, 3.11 and 3.6 of [21], respectively). 6 Measure theoretic results This section collects some measure theory lemmata, which are extensions of analogous ones in [17] and [21]. We define, for l > 0, L := (x , 0, x N +1 ) ∈ R N −1 × R × R | |x N +1 | ≤ 1/2 , Q l := (x , 0, x N +1 ) ∈ L | |x | ≤ l . Then, the following results hold: Lemma 21 Let u be p-harmonic in {|x | < l} × {|x N | < l}, if |u| < 1. Suppose that u is continuous, that u satisfies (3.1)-(3.2), that |u| ≤ 1 and that S(Y, R) touches the graph of u by above at X 0 , with u(x0 ) = −1. Assume that ∇u(x0 ) π , eN ≤ . (6.1) |∇u(x0 )| 8 Then, there exists a universal a0 > 1 so that, for any a ≥ a0 , there exist a universal κ > 1 and a suitable C > 1, the latter depending only on a and on universal constants, such that the following holds. For any point Z ∈ L ∩ Ba (π N X 0 ) there exists x satisfying the following properties: • |x − x0 | ≤ κa, • Z = π N (x, u(x)), ∇u(x0 ) • (x − x0 ) · |∇u(x ≤ u(x) − u(x0 ) + 0 )|
C R,
provided that R is large enough (possibly in dependence of a). Lemma 22 Let u be p-harmonic in {|x | < l} × {|x N | < l} if |u| < 1. Assume that u is continuous, that u satisfies (3.1)-(3.2), that |u| ≤ 1, that S(Y0 , R) is above the graph of u and that S(Y0 , R) touches the graph of u at the point X 0 , with u(x0 ) = −1. Suppose that • |u(x0 )| < 1/2, |x0N | < l/4, q := |x0 | < l/4;
Flatness of Bernoulli jets
•
∇u(x0 ) |∇u(x0 )|
, eN
17
≤
π 8.
Then, there exist universal constants C1 , C2 > 1 > c > 0 such that, if √ 3 q ≥ C1 and 4 R ≤ l ≤ c R , the following holds. Let be the set of points (ξ, u(ξ )) ∈ R N × R satisfying the following properties: • |ξ | < q/15, |u(ξ )| < 1/2, |ξ − x0 | < 2q; • there exists Y ∈ R N × [−1/4, 1/4] such that S(Y, R/C2 ) is above u and it touches u at (ξ, u(ξ)); C1 q ∇u(ξ ) ∇u(x0 ) • |∇u(ξ )| , |∇u(x0 )| ≤ R ; • (ξ − x0 ) · Then,
∇u(x0 ) |∇u(x0 )|
≤
C1 q 2 R
+ u(ξ ) − u(x0 ).
L N π N () ≥ cq N −1 .
The proofs of the above lemmata will be given in the Appendix. For another general covering argument which will be of some use in this paper, see also Lemma 4.2.3 in [17] or Lemma 4.3 in [21]. 7 Further measure estimates on the projection of the contact set We now show how to use Proposition 20 and the measure theoretic lemmata stated in § 6 in order to deduce a measure estimate on the projection of the contact sets between barriers and minimal solutions. To this aim, we first need an estimate on the contact sets obtained by touching u by above “for the first time”, as dealt with by the following result. Lemma 23 Let C, C > 1 be suitably large constants. Let K l := {|x | < C l} × {|x N | < C l}. Let u ∈ W 1, p (K l ) be a local minimizer for F in K l . Assume that u(0) = 0 and that u(x) < 0 if x N < −θ , for some θ > 0. Define R0 := l 2 /(Cθ ). Let be the set of points (ξ, u(ξ )) ∈ K l satisfying: • |ξ | ≤ l, |u(ξ )| < 1/2; • there exists Y ∈ R N × [−1/4, 1/4] such that S(Y, R0 ) is above the graph of u in {|x | < C l} × {|x N | < C l} and it touches the graph of u at (ξ, u(ξ )), with u(ξ ) = −1; ∇u(ξ ) • ≤ R8l0 ; , e N |∇u(ξ )| • ξ N ≤ θ4 + u(ξ ). Then, there exists a universal constant c > 0 such that, for any θ0 > 0 there exists ε0 (θ0 ) > 0 for which, if θ ≤ ε0 (θ0 ) , l one has that
θ ≥ θ0 ,
L N π N () ≥ c l N −1 .
18
E. Valdinoci
Proof The density estimates in [15] gives that u(x) = −1 for any x so that |x | ≤ 9Cl/10 and x N ≤ −l/10.
(7.1)
be the set of (ξ, u(ξ ))’s described in the statement of Lemma 23. Let Let now and us also define := πe N O = Y = (y , 0, y N +1 ) ∈ R N +1 such that |y | ≤ l/C , |y N +1 | ≤ 1/4 , where C > C is a suitably large constant. Obviously, L N (O) ≥ const l N −1 .
(7.2)
If Y ∈ O, (7.1) implies that S(Y − te N , R0 ) is above the graph of u if t is large enough. By decreasing t, the fact that u(0) = 0 gives that there exists a suitable t ∗ for which S(Y − t ∗ e N , R0 ) touches for the first time the graph of u, say on point := Y − t ∗ e N . We denote by G the set of such touching points X . Let Y X ’s and we define G := πe N G . We claim that G ⊆ .
(7.3)
. We show that To prove this, take any X ∈G x is in the interior of {|x | ≤ C l} × {|x N | ≤ C l} .
(7.4)
First, we show that | x | ≤ l .
(7.5)
For this, we define R0 2C 0 y N +1 1− 1− r0 := R0 − R0 C0 and we make some considerations on the zero level set of gS(Y,R0 ) which, in turn, agrees with ∂ Br0 ( y). We observe that the first touching property of X implies that, y) is below the origin with respect to the e N -direction, ∂ Br0 (
(7.6)
and, consequently, y N + r0 ≤
l2 . C r0
Also, if |x | ≥ l, we gather from (4.3) that l2 l2 yN | 1 + ≥ |x − y | + . |x − y| ≥ |x N − N N 4|x N − y N |2 32r0
(7.7)
(7.8)
Flatness of Bernoulli jets
19
From (7.8) and (7.7), we deduce that, if |x | ≥ l, then l2 − R0 32r0 l2 ≥ y N +1 + x N − yN + − R0 32r0 l2 R0 2C 0 y N +1 1− 1− ≥ y N +1 + x N + − 64r0 R0 C0
y| − R0 ≥ y N +1 + |x N − yN | + y N +1 + |x −
l2 C0 − const 64r0 R0 2 l ≥ xN + ≥ x N + 3θ . 128 R0 ≥ xN +
(7.9)
On the other hand, Lemma 5 gives that u(x) ≤ β(x N + θ ). Thus, if x N + θ ≤ −2, then u(x) = −1, which excludes the possibility of touching gS(Y,R0 ) at x on a level higher than −1. Also, from (7.6), if |x | ≥ l, then x N ≤ −2θ . Therefore, if |x | ≥ l, we may consider the case in which x N + θ ∈ [−2, 0] and so, by the construction on page 7, we have that u(x) ≤ β(x N + θ ) ≤ ω− (x N + θ ) ≤ x N + θ + 8(κ + κ ) ≤ x N + 2θ .
(7.10)
We deduce from (7.9) and (7.10) that, if |x | ≥ l were a touching point between , R0 ) and the graph of u, one would have that S(Y u(x) < y N +1 + |x − y| − R0 ≤ gS(Y,R0 ) (x) , against the supposed touching property of x. This proves (7.5). Also, x N > −C l, because of (7.1). Finally, (7.6) and (4.3) imply that x N < C l, thus proving (7.4). Proposition 6 and (7.4) yield that |u( x )| < 1/2. On the other hand, | x − y | ≤ | x | + | y | ≤ 2l , thanks to (7.5). This and (4.3) give that x − y 8l . , eN ≤ | x − y| R0 Therefore, from (7.11) and the touching property of X , we get that ∇u( x) 8l . , eN ≤ |∇u( x )| R0
(7.11)
(7.12)
Let now x¯ be on the half line from y to x be so that gS(Y,R0 ) (x) ¯ = 0. From (7.7), x¯ N ≤
θ . 8
20
E. Valdinoci
We thus use the definition of g S(·,·) to get that const C 0 R0
u( x ) = g x ) − g ¯ ≥ | x − y| − |x¯ − y| − ,R0 ) ( ,R0 ) ( x) S(Y S(Y = ±| x − x| ¯ −
const C 0 , R0
x ) ≥ 0 (respectively, g where the sign + (respectively, −) holds for g ,R0 ) ( ,R0 ) S(Y S(Y ( x ) ≤ 0). Also, by (7.11) and some trigonometry, we have that x N − x¯ N | ≥ | x − x| ¯ ±( x N − x¯ N ) = |
const l 2 1− R02
,
with the same sign convention as above. Thus, the above estimates give that u( x) ≥ x N − x¯ N − ≥ xN −
const l 3 const C 0 − 2 R0 R0
θ const C 0 const l 3 − , − 2 8 R0 R0
that is x) + x N ≤ u(
θ . 4
(7.13)
and, therefore, Hence, in the light of (7.12) and (7.13), we have that X ∈ that πe N X ∈ , ending the proof of (7.3). We now exploit Proposition 20, applied4 to G and O: from that, (7.3) and (7.2), we get that L N () ≥ L N (G ) ≥ const L N (O) ≥ const l N −1 ,
ending the proof of Lemma 23.
Then, one deduces the following result: Proposition 24 Let C be a suitably large constant. Let K l := {|x | < C l} × {|x N | < C l}. Let u ∈ W 1, p (K l ) be a local minimizer for F in K l . Assume that u(0) = 0 and that u(x) < 0 if x N < −θ , for some θ > 0. Fix C¯ > 0 and k ∈ N. Let be the set of points (ξ, u(ξ )) satisfying the following properties: • |ξ | ≤ l, |ξ N +1 | ≤ 1/2; • ξ N ≤ C¯ k θ + u(ξ ). 4 Note that (5.14) is fulfilled because u(0) = 0 and R is large. Also, (5.2) is satisfied due 0 to (7.4).
Flatness of Bernoulli jets
21
Then, there exist positive universal constants c and cˆ for which the following holds. For any θ0 > 0, there exists ε0 (θ0 ) > 0, so that, if θ ≤ ε0 (θ0 ) , l then
θ ≥ θ0
and
C¯ k θ ≤ cˆ , l
L N π N () ≥ (1 − (1 − c)k ) L N (Q l ) .
The proof of such result makes use of the “abstract” covering argument in Lemma 4.3 of [21]. For the proof, repeat the analogous one of Proposition 5.2 in [21], with the following three modifications: the function t → H0 (t) there is replaced here by the identity; the use of Lemma 4.2 there is here replaced by the use of the analogous Lemma 22; the touching points are taken here to occur at a higher level than −1. 8 Proof of Theorem 1 The proof is much related to the one of the Harnack inequalities of [17] and [21]. We provide full details, since some computations differ from those in the above mentioned papers. First, note that u must attain both positive and negative values, due to the density estimates in [15]. Thus, possibly replacing l by C l, we may assume that u is a local minimizer for F in {|x | < C l} × {|x N | < C l}, that u(0) = 0 and that u(x) > 0 if x N > θ > 0 and u(x) < 0 if x N < −θ .
(8.1)
The proof of Theorem 1 is by contradiction and it deals with the fact that, if there existed a point in {u = 0} ∩ {|x | < l/4} close to x N = −θ , then the energy of u would be too large for being minimal. Thence, we assume, by contradiction, that {u = 0} ∩ {|x | < C¯ −k0 l/4} ∩ {x N < (−1 + C¯ −k0 /4)θ } = ∅ , with k0 ∈ N large and θ/l small (possibly in dependence of k0 ) and we take x ∗ in the above set. We define 0 := (x, u(x)) ∈ R N × R s.t. x N ≤ u(x) − θ/2 , |x − (x ∗ ) | ≤ l/2 , |u(x)| ≤ 1/2 . Then, by applying Proposition 24 to u(x + x ∗ ), we see that L N π N (0 ) ≥ (1 − (1 − c0 )k0 ) L N (Q l/2 ) , for a suitably small constant c0 > 0. Let 1 := (x, u(x)) ∈ R N × R s.t.
x N ≥ u(x) − θ/4 , |x | ≤ l/2 , |u(x)| ≤ 1/2 .
(8.2)
22
E. Valdinoci
Then, by applying Lemma 23 to −u(x , −x N ), we gather that L N π N (1 ) ≥ c1 L N (Q l/2 ) ,
(8.3)
for a suitably small constant c1 > 0. Furthermore, by construction, π N (0 ) ⊆ Q l + 2
therefore
L N π N (0 ) \ Q l/2 ≤
l 4C¯ k0
,
const L N (Q l/2 ) . (N ¯ C −1)k0
This, (8.2) and the assumption that k0 is large imply that const N L N π N (0 ) ∩ Q l/2 ≥ 1 − (1 − c0 )k0 − L (Q l/2 ) ¯ (N −1)k0 C c1 N ≥ 1− L (Q l/2 ) , 2 where c1 is the constant introduced in (8.3). Thus, c1 N L N Q l/2 \ π N (0 ) ∩ Q l/2 ≤ L (Q l/2 ) . 2 From this, (8.3) and the fact that π N (1 ) ⊆ Q l/2 , we obtain that c1 N L N π N (0 ) ∩ π N (1 ) ≥ L (Q l/2 ) . 2 On the other hand, we have that
θ θ = ∅. 0 ∩ 1 ⊆ − ≤ x N − u(x) ≤ − 4 2 Let now V :=
(8.4)
(8.5)
Z ∈ Q l/2 ∃x˜ = xˆ , s.t. Z = π N x, ˜ u(x) ˜ = π N x, ˆ u(x) ˆ .
By (8.5), we have that V ⊇ π N (0 ) ∩ π N (1 ) ,
thus, due to (8.4), L N (V ) ≥ const l N −1 .
(8.6)
With these inequalities, we are now in the position of estimating the energy of u, in order to deduce the desired contradiction. First of all, for any x ∈ R N with |x | ≤ l, let us define Tx (x N ) := u(x , x N ) Cx := {x N ∈ R | DTx (x N ) = 0} . and
Flatness of Bernoulli jets
23
By the regularity results in [10] and [18], we have that Tx is C 1 in {|Tx | < 1}, hence, by Sard’s Lemma, L N Tx (Cx ) = 0 . (8.7) Thus, using that Tx is locally invertible in {|Tx | < 1} \ Cx , we may write the latter set as Ja,x , in such a way Tx
a
Ja,x
is a diffeomorphism. Therefore, by Young’s inequality and
by changing variable x N +1 := Tx (x N ), |∇u(x , x N )| p + mχ(−1,1) (u(x , x N )) d x N p Ja,x |∂ N u(x , x N )| p ≥ + mχ(−1,1) (u(x , x N )) d x N p Ja,x |χ(−1,1) (Tx (x N ))|q |DTx (x N )| p = + dxN p q Ja,x ≥ χ(−1,1) (Tx (x N )) |DTx (x N )| d x N Ja,x
=
Tx (Ja,x )
χ(−1,1) (x N +1 ) d x N +1
= L1 Tx (Ja,x ) , therefore, ! a
≤
|x |≤l
! a
=
L1 Tx (Ja,x ) d x
|x |≤l
|x |≤l
Ja,x
|∇u(x , x N )| p + mχ(−1,1) (u(x , x N )) d x N d x p
[−Cl,Cl]\Cx
≤ F Al (u) ,
|∇u(x , x N )| p + mχ(−1,1) (u(x , x N )) d x N d x p (8.8)
where Al := {|x | < l} × {|x N | < Cl}. Now, we notice that V ⊆ (x , 0, x N +1 ) |x | ≤ l, x N +1 ∈ Tx (Ja,x ) ∩ Tx (Ja,x ˆ , ˆ ) for some a = a therefore, by (8.6), we have that L1 x N +1 ∈ Tx (Ja,x ) ∩ Tx (Ja,x d x ≥ L N (V ) ≥ c˜1 l N −1 , ˆ ) |x |≤l
a =aˆ
24
E. Valdinoci
for a suitably small positive constant c˜1 . Thence, ! L1 Tx (Ja,x ) d x |x |≤l
a
≥
|x |≤l
L1
a
+ ≥
|x |≤l
|x |≤l
Tx (Ja,x ) d x
L1
Tx (Ja,x ) ∩ Tx (Ja,x ) d x N +1 d x ˆ a =aˆ
L1 u(x , [−Cl, Cl] \ Cx ) d x + c˜1 l N −1 .
Thus, from (8.7), we have that ! 1 L Tx (Ja,x ) d x ≥ a
|x |≤l
|x |≤l
L1 u(x , [−l, l]) d x N +1 d x + c˜1 l N −1 .
On the other hand, from Lemma 5, we get that u x , [−l, l] ⊇ [−1, 1] , thus
! a
|x |≤l
L1 Tx (Ja,x ) d x ≥ 2ω N −1 l N −1 + c˜1 l N −1 .
So, from (8.8), we deduce that F Al (u) ≥ 2ω N −1 l N −1 + c˜1 l N −1 .
(8.9)
Let us now define the rescaled functional p−1 m ε |∇v(x)| p ε F (v) := + χ(−1,1) (v(x)) d x . p ε Such rescaling corresponds to a “blow down” solution: that is, if ε := 1/l and u ε (x) := u(x/ε), by scaling (8.9), we get that F Aε 1 (u ε ) ≥ 2ω N −1 + c˜1 .
(8.10)
By § 3 of [5], up to subsequences, we have that u ε converges almost everywhere and in L 1loc to χ E − χR N \E , for a suitable set E ⊆ R N , and that lim F Aε 1 (u ε ) = 2 Per (E, A1 ) .
ε→0+
(8.11)
Moreover, from (8.1), we see that there exists κ > 0 so that u ε (x) ≥ κ if |x | ≤ 1 and x N ≥ 2εθ and u(x) ≤ −κ if |x | ≤ 1 and x N ≤ −2εθ . In particular, for almost any x ∈ A1 , lim u ε (x) ≥ κ if x N > 0 and
ε→0+
lim u ε (x) ≤ −κ if x N < 0.
ε→0+
This implies that E = A1 ∩ {x N > 0} and thus, by (8.11), we get that lim F Aε 1 (u ε ) = 2ω N −1 ,
ε→0+
in contradiction with (8.10). This ends the proof of Theorem 1.
Flatness of Bernoulli jets
25
A Proof of the auxiliary measure theoretic results A.1 Proof of Lemma 21 Since (x0 − y) · e N ≥ const R , we have that x N − y N ≥ const R , S(Y, R), for any X ∈ B3a (X 0 ), if R is large enough and therefore, if X ∈ B3a (X 0 )∩ then
∂ N g S(Y,R) (x) ≥ const > 0 . By inspection, one also has that π N and π N +1 S(Y,R)∩B3a (X 0 )
S(Y,R)∩B3a (X 0 )
Thus, we define
T := π N +1
S(Y,R)∩B3a (X 0 )
are diffeomorphisms.
(A.1)
(A.2)
◦ π N−1
S(Y,R)∩B3a (X 0 )
and we introduce the domains O1 := T {x N = 0} ∩ {|x N +1 | < 3/4} ∩ Ba+2 (π N (X 0 )) O2 := T {x N = 0} ∩ {|x N +1 | < 5/8} ∩ Ba+1 (π N (X 0 )) . Of course, x0 ∈ O2 ⊂ O1 and, more precisely, by (A.2), dist (O2 , ∂ O1 ) ≥ const .
(A.3)
By (4.5), p g S(Y,R) − p u ≤
const , R
hence, from the p-Laplacian-Harnack-comparison inequality (see, for instance [6] and [19]), we get that sup(g S(Y,R) − u) ≤ O2
C , R
(A.4)
for a suitable C > 1, which may also depend on a. (1) Fix now Z ∈ L∩Ba (π N X 0 ). Then, from (A.2), there exists X (1) = (x (1) , x N +1 ) S(Y, R) ∩ B3a (X 0 ) so that ∈ π N (X (1) ) = Z , S(Y,R)∩B3a (X 0 )
26
E. Valdinoci
that is S(Y, R) X (1) = Z + t (1) e N
(A.5)
for some t (1) ∈ R. Moreover,
|X (1) − X 0 | ≤ π N−1 (Z − π N X 0 ) ≤ const |Z − π N X 0 | ≤ const a .
(A.6)
Furthermore, from (A.1), we have that, for any t ≥ 0, (1) (1) g + te N ) ≥ const t + g S(Y,R) (x S(Y,R) (x )
= const t + x N(1)+1 = const t + z N +1 . Therefore, from (A.4), u(x (1) + te N ) > z N +1 ,
(A.7)
provided that t ≥ C /R, for a suitable C > 1, which may also depend on a. Analogously, u(x (1) − te N ) < z N +1 ,
(A.8)
provided that t ≥ C /R. From (A.7) and (A.8), we deduce the existence of a suitable t (2) ∈ [−C /R, C /R] so that u(x (1) + t (2) e N ) = z N +1 . Let us define X (2) := X (1) + t (2) e N . The point x (2) = x (2) (Z ) will be the one satisfying the thesis of Lemma 21, as we are now going to show. Notice that, by construction, C (1) (A.9) and x N(2)+1 = z N +1 = u(x (2) ) X − X (2) ≤ R and thus
π N x (2) , u(x (2) ) = Z and |x (2) − x0 | ≤ |x (1) − x0 | +
C ≤ const a , R
thanks to (A.6). We now show that ∇u(x ) C 0 ≤ z N +1 − u(x0 ) + , x (1) − x0 · |∇u(x0 )| R
(A.10)
(A.11)
for some C > 0 which may depend on a. To prove (A.11), we use (A.5) to deduce that (1) z N +1 = x N(1)+1 = g S(Y,R) (x )
Flatness of Bernoulli jets
27
and so, by (A.6), (1) z N +1 − u(x0 ) = g S(Y,R) (x ) − g S(Y,R) (x 0 ) (1) (1) − x0 ) − const |D 2 g − x 0 |2 , ≥ ∇g S(Y,R) (x 0 ) · (x S(Y,R) (ξ )||x
from which (A.11) follows. In the light of (A.9) and (A.11), we deduce that ∇u(x ) ∇u(x ) 0 0 x (2) − x0 · ≤ x (1) − x0 · + x (1) − x (2) |∇u(x0 )| |∇u(x0 )| C ≤ z N +1 − u(x0 ) + R C (2) , = u(x ) − u(x0 ) + R for a suitable C > 1, which may depend on a. This, together with (A.6) and (A.10), completes the proof of Lemma 21. A.2 Proof of Lemma 22 The proof of Lemma 22 relies on an auxiliary result, namely Lemma 25 here below, which may be seen as a rotation of the desired claim (see (A.14) here below) plus a Lipschitz property on level sets (see also the Appendix in [21] for further comments and details). For this, we now state some further notation. Given R > 0 and Y = (y, y N +1 ) ∈ R N × [−1/4, 1/4], we define (Y, R) as the zero level set of S(Y, R), that is: (Y, R) := S(Y, R) ∩ {x N +1 = 0} = {gS(Y,R) = 0} . By construction, (Y, R) is the (N −1)-dimensional sphere centered at y of radius R 2C 0 −1 + 1 − (A.12) r = r (Y, R) := R + y N +1 . R C0 Notice that, if |s| ≤ 1/2, the signed distance between the s-level set of gS(Y,R) and (Y, R) is R 2C 0 2C 0 1− (y N +1 − s) − 1 − y N +1 . (A.13) R R C0 The analysis of the zero level sets of suitable barriers will be dealt with in the following pages. The arguments involved will often be based on the following heuristics: given two barriers, if their zero level sets are at some distance from each other in some region, then one expects that one barrier is above the other in a neighborhood of that region, since the s-level sets are “almost at exactly” the same distance.
28
E. Valdinoci
Given x ∈ R N , we denote by TY,R x = TY,R (x) the intersection point between (Y, R) and the half-line from y to x. With this notation, we now state and prove the above mentioned auxiliary result. Lemma 25 Let C˘ > 1 be a suitably large constant. Let u be p-harmonic in {|x | < l}×{|x N | < l}, if |u| < 1. Assume that u is continuous, that u satisfies (3.1)(3.2), that |u| ≤ 1, that S(Y, R) is above the graph of u in {|x | ≤ l} × {|x N | ≤ l/2} and that S(Y, R) touches the graph of u at the point (x0 , u(x0 )) with u(x0 ) = −1. Suppose that • |u(x |x0N | < l/4, |x0 | < l/4, 0 )| < 1/2, ∇u(x0 ) • |∇u(x , e N ≤ π8 . 0 )| Assume also that TY,R x0 ∈ {|x | = q} ∩ {x N = 0}
and
y = −e N
r 2 − q2 .
(A.14)
Then, there exist universal constants C1 , C2 > 1 > c > 0 such that, if C1 ≤ q ≤
l C1
and
√ 3 4 R ≤ l ≤ cR ,
(A.15)
the following holds. Let be the set of points (ξ, u(ξ )) ∈ R N × R satisfying the following properties: ˘ • |ξ | < q/15, |u(ξ )| < 1/2, |ξ − x0 | < Cq; N +1 ˆ ˆ • there exists Y ∈ R such that S(Y , R/C2 ) is above u and it touches u at (ξ, u(ξ )), with u(ξ ) = −1; C1 q ∇u(ξ ) ∇u(x0 ) • |∇u(ξ )| , |∇u(x0 )| ≤ R ; • (ξ − x0 ) ·
∇u(x0 ) |∇u(x0 )|
≤
C1 q 2 R
+ u(ξ ) − u(x0 ).
Then, L N π N () ≥ cq N −1 .
(A.16)
More precisely, for any s ∈ (−1/2, 1/2), there exists a set s ⊆ ∩ {x N +1 = s}, which is contained in a Lipschitz graph in the e N -direction, with Lipschitz constant less than 1, and so that, if ˘ :=
s , s∈(−1/2, 1/2)
we have that
˘ L N π N () ≥ cq N −1 .
(A.17)
Flatness of Bernoulli jets
29
Proof It follows from (A.13) that TY,R x0 − x0 ≤ const . By construction,
(A.18)
q x0 − y, e N ) = . r By means of (A.18) and (A.19), we gather that sin
(A.19)
const q . (A.20) R We now fix C∗ > 1, to be chosen conveniently large. The first step towards the proof of Lemma 25 consists in proving the existence of a suitable Y∗ ∈ R N +1 and R∗ > R/ const, so that πe N Y∗ = πe N Y , and S(Y∗ , R∗ ) touches the graph of u from ˘ and above at the point (x∗ , u(x∗ )), with |x∗ | ≤ Cq const q 2 q TY∗ ,R∗ x∗ ∈ x N ≤ . (A.21) × |x | < R C∗ |x0 | ≤ q +
The proof of (A.21) is by iteration. Namely, we will set Y0 := Y , R0 := R and, for any k ∈ N, we will find inductively Yk+1 ∈ R N +1 and Rk+1 > Rk /4, so that πe N Yk+1 = πe N Yk , and S(Yk+1 , Rk+1 ) touches the graph of u from above at the | ≤ Cq ˘ and point (xk+1 , u(xk+1 )), with |xk+1 const q 2 × |x | < ηq , TYk+1 ,Rk+1 xk+1 ∈ x N ≤ Rk
(A.22)
for some η ∈ (0, 1). Since (A.21) follows by iterating (A.22) a finite number of times, we only deal with the proof of (A.22). In fact, we will proof the first step in (A.22), i.e., the step with k = 0, the others being analogous. For this, we fix a small parameter ∈ (0, 1/2] and a large parameter γ > 1 and we define ω := 2−1/(γ +2) and η := 1 − (1 − ω). For any t > 0, set ˜ ψ(t) := and define
" G :=
1 γ
1 −1 tγ q2
X ∈ R | xN = ψ˜ r 2 − q2 N
(A.23)
# |x | q
.
Note that G touches (Y, R) from above when |x | = q. Analogously, if q2 1 γ +2 2 2 yω := − r − q eN and 1− γ + ω ω γ r 2 − q2 γ +2 rω := ω r 2 + q 2 (ω−2γ −2 − 1) ,
30
E. Valdinoci
one sees that G touches ∂ Brω (yω ) from above when |x | = ωq. We also define
and
1 := (Y, R) ∩ {x N < 0} 2 := G ∩ {|x | ∈ [ωq, q]} 3 := ∂ Brω (yω ) ∩ {|x | < ωq} ∩ {x N > 0} .
Let := 1 ∪ 2 ∪ 3 . By construction, is a C 1,1 closed hypersurface in R N . We denote by d the signed distance to , with the convention that d is positive in the exterior of and negative in the interior. Also, since R is much bigger than l (recall (A.15)), we will often look at ∩ [−l, l] N (that is, in the domain we are interested in) as a graph in the e N -direction. We now define g (x) := g y N +1 ,R d (x) + r − R + y N +1 , and we denote the graph of g by . Since d (x) ≤ |x − y| − r , we have that g ≤ gS(Y,R) ; also, if d (x) = d1 (x), then g (x) = gS(Y,R) (x).
(A.24)
Therefore, g (x) is strictly p-superharmonic in the viscosity sense at any x ∈ R N for which d (x) is attained on 1 and |g (x)| ∈ (1/2, 1).
(A.25)
Furthermore,5 in an appropriate system of coordinates, ∇g (x) = g y N +1 ,R e N
(A.26)
and D 2 g (x) is the N × N diagonal matrix with entries κ1 κ N −1 , g , g y N +1 ,R , . . . , g κ1 d (x) − 1 κ N −1 d (x) − 1 y N +1 ,R y N +1 ,R
(A.27)
where have we omitted, for short, to explicitly write the function evaluation at the point “d (x) + r − R + y N +1 ” and where κi denotes the ith principal curvature of at a point realizing d (x). By a direct computation, −κ12 = · · · = −κ N2−2 =
q γ +2 ≥0 |x | (r 2 − q 2 )|x |2γ +2 + q 2γ +4
(γ + 1) q γ +2 (r 2 − q 2 ) |x |2γ +1 κ N2−1 = 3/2 ≥ 0 . (A.28) (r 2 − q 2 ) |x |2γ +2 + q 2γ +4 In particular, since |x | ≥ ωq on 2 , −κi2 ≤ 5
3 , R
(A.29)
See, e.g., [16] for definitions and comments on p-Laplace viscosity sub/super/solutions.
Flatness of Bernoulli jets
31
for i = 1, . . . , N − 2, if R is large enough. Furthermore, since q ≥ |x | ≥ ωq on 2 , const
(γ + 1) (γ + 1) ≤ κ N2−1 ≤ const . R R
(A.30)
We now claim that γ p g (x) ≤ − const R
at any x ∈ R N for which |g (x)| = 1/2, ∇g (x) = 0 and d is attained on 2 .
(A.31)
To prove this, take x as requested here above: then, |d (x)| ≤ |d (x) − r + R + y N +1 | + |r − R| + 1 ≤ |r − R| + 3 ≤ 6 . (A.32) Therefore, by (A.29) and (A.30), we get that −2 N! const 2 2 κi d (x) ≤ √ κ N −1 d (x) + R i=1
and so, if R is large enough, N −1 ! i=1
−κi2
1 − κi2 d (x)
≤ −2
N −2 !
κi2 −
κ N2−1
i=1
2
.
From this, (A.29), (A.30), (A.26) and (A.27), we deduce that p g ≤ ( p − 1) (g y N +1 ,R ) p−2 g y N +1 ,R +
−κi2
(g y N +1 ,R ) p−1 2 i=1 1 − κi d N −2 ! κ N2−1 2 − 1) g y N +1 ,R g y N +1 ,R − κi − 2 i=1
≤
(g y N +1 ,R ) p−2
≤
const ( const − γ ) , R
(p
N −1 !
which proves (A.31), if γ is suitably large. We now prove that g is strictly p-superharmonic in the viscosity sense at any x ∈ R N for which |g (x)| ∈ (1/2, 1) and d (x) is attained on 3 . Indeed, since 3 is a portion of sphere, κ13 = · · · = κ N3−1 = −
1 < 0. rω
(A.33)
32
E. Valdinoci
Consequently, using also (A.32), we get that p g ≤ ( p − 1) g y N +1 ,R (g y N +1 ,R ) p−2 + (g y N +1 ,R ) p−1 ≤ (p
− 1) g y N +1 ,R
+
const g y N +1 ,R
( p − 1) C 0 const ≤ − + 8R R
i=1
R
N −1 !
κi3
κi3 d − 1
(g y N +1 ,R ) p−2
(g y N +1 ,R ) p−2 ,
which gives (A.33), if we choose C 0 large enough. Since, by Proposition 6, no smooth function can touch g from below at level ±1/2, we gather from (A.25) and (A.33), that g is strictly p-superharmonic in the viscosity sense outside the set |g | ∈ [0, 1/2) ∪ {1} ∩ d realized on 1 ∪ 3 . (A.34) Also, from (A.24), we have that ˘ then g (x) = gS(Y,R) (x). if |x | ≥ Cq,
(A.35)
We now define Yω := (yω , y N +1 ) , 5 C0 , R1 := rω + y N +1 + R R1 2C 0 y N +1 , r1 := R1 + −1 + 1 − R1 C0 (Yω , R1 ) := S(Yω , R1 ) ∩ {x N +1 = 0} . By definition, r1 − rω ∈
3C 0 5C 0 , R R
(A.36)
and (Yω , R1 ) = ∂ Br1 (yω ). Thus, we infer from (A.36) that (Yω , R1 ) stays at distance greater than 3C 0 /R outside ∂ Brω (yω ) ⊃ 3 .
(A.37)
By looking at and ∂ Brω (yω ) in [−l, l] N as graphs in the e N -direction, and by computing their “vertical distance” (see, e.g., (A.59) in [21] for further details), one sees that ∂ Brω (yω ) ∩ {|x | ≥ ηq} is at distance at least
const q 2 R
inside .
(A.38)
Flatness of Bernoulli jets
33
This implies that, for any a ∈ ∂ Brω (yω ) and any b ∈ 1 , |a − b| ≥ const
q2 . R
(A.39)
Also, from (A.38) and (A.36), we get that (Yω , R1 ) ∩ {|x | ≥ ηq} is at distance at least
3C 0 R
inside .
Also, yωN + r1 ≤ const q 2 /R, therefore
const q 2 (Yω , R1 ) ⊆ x N ≤ . R
(A.40)
(A.41)
Also, by inspection,
The region X ∈ where |x N +1 | < 1/2 and d (x) is realized on 3 is above S(Yω , R1 ) with respect to the e N +1 -direction.
(A.42)
We also claim that
The region X ∈ S(Yω , R1 ) where d(Yω ,R1 ) (x) is realized at a point z with |z | ≥ ηq is strictly above in the e N +1 -direction.
(A.43)
Indeed: let X ∈ S(Yω , R1 ) and assume that d(Yω ,R1 ) (x) is attained at z ∈ (Yω , R1 ) and |z | ≥ ηq. Let also y˘ ∈ on the half-line from yω towards z. Let also y˜ ∈ be the point realizing d (x). The geometry of these points and (A.40) imply that 3C 0 dist x, (Yω , R1 ) ≥ d (x) + , R which, in turn, yields (A.43). Furthermore, ˘ S(Yω , R1 ) ∩ {|x | ≥ Cq} is strictly above in the e N +1 -direction.
(A.44)
˘ and let z ∈ (Yω , R1 ) be realizing d(Yω ,R1 ) (x): To prove this, take x with |x | ≥ Cq by an elementary triangle similarity argument, |z | =
r1 |x | r1 C˘ q ˘ ≥ ηq , ≥ const ≥ const Cq |x − yω | l+R
if C˘ is large enough. Then, (A.44) follows from (A.43). We now consider the domain ˘ × {|x N | ≤ l/2} {|x | ≤ Cq}
34
E. Valdinoci
and we slide from −∞ in the e N direction, until we touch the graph of u at a level strictly larger than −1 for the first time in such domain (this must happen since |u(x0 )| < 1/2 and, by (3.1)-(3.2), it cannot happen on the free boundary). For fixing the notations, we take that for some β ∈ R, − βe N touches for the first time the graph of u by above at a point Z , with z N +1 > −1. Notice that, by the hypotheses of Lemma 25 and (A.24), we have that u(x0 ) = gS(Y,R) (x0 ) = g (x0 ) .
(A.45)
Therefore, we have that β ≥ 0. More precisely, it holds that β > 0.
(A.46)
Indeed: if, by contradiction, β = 0, (A.45) says that touches the graph of u from above at x0 (which is, by construction, an interior point): thence, by (A.34), d (x0 ) must be realized on 1 ∪ 3 , in contradiction with the fact that d (x0 ) is realized at TY,R (x0 ) ∈ 2 . This proves (A.46). By construction, the domain of g (and so, a fortiori, the domain of g−βe N ) is below the hyperplane x N = const (1 + log l) (A.47) with respect to the e N -direction. Also, we deduce from (A.46) that |z + βe N − y|2 = |z − y|2 + β 2 + 2β (z N − y N ) > |z − y|2 .
(A.48)
Moreover ˘ , |z | < Cq
(A.49)
otherwise (A.35) and (A.48) would yield that u(z) = g (z + βe N ) = gS(Y,R) (z + βe N ) > gS(Y,R) (z) ≥ u(z) , which is, of course, a contradiction. By means of (A.49) and (A.47), we have that z lies in the interior of ˘ × {|x N | ≤ l/2}. {|x | ≤ Cq} We now claim that |z N +1 | < 1/2 and d (z + βe N ) is realized on 3 .
(A.50)
Indeed, since touches by above u + βe N at z := z + βe N , (A.34) implies that 1/2 > |z N +1 | = |z N +1 | and that d (z) is realized on 1 ∪ 3 . To prove (A.50), we thus have to show that d (z) is not realized on 1 . If, by contradiction, d (z) were realized on 1 , it would follow from (A.24) and (A.48) that u(z) = g (z + βe N ) > gS(Y,R) (z) ≥ u(z) . This contradiction proves (A.50).
Flatness of Bernoulli jets
35
Let us also note that, in order not to go out of our domain, β ≤ 5l; thus, since by construction |Y − Yω | ≥ const r , we deduce that |Y − Yω | − β ≥ 0 .
(A.51)
For t ≥ 0, we now consider the surface S(Y + te N , R1 ). For t = 0, this surface is above S(Y, R), since R1 < R. In particular, S(Y + te N , R1 ) is, for t = 0, above ˘ × {|x N | ≤ l/2}. Recalling Corollary 7, we now the graph of u in {|x | ≤ Cq} increase t until we touch the graph of u in ˘ × {|x N | ≤ l/2} ∩ {|x N +1 | < 1} . {|x | ≤ Cq} Say this happen for t = t1 ≥ 0 and let X 1 be the above mentioned touching point. Set also Y1 := Y + t1 e N
and
r˜1 := R1 + σ− − y N +1 ,
so that
gS(Y1 ,R1 ) = −
1 = ∂ Br˜1 (y1 ) . 2
Thus, since u(x0 ) > −1/2, the first touching property of X 1 implies that x0 is above ∂ Br˜1 (y1 ). Then, recalling (A.18), we deduce that the domain of S(Y1 , R1 ) lies below the hyperplane x N = const (1 + log l) . (A.52) Notice also that, when Y + te N = Yω − βe N , (A.42) and (A.50) imply that gS(Y +te N ,R1 ) (z) = gS(Yω −βe N ,R1 ) (z) = gS(Yω ,R1 ) (z + βe N ) ≤ g (z + βe N ) = u(z) and so, by (A.51), we get that 0 ≤ t1 ≤ |Y − Yω | − β .
(A.53)
˘ . X 1 ∈ {|x | < Cq}
(A.54)
We now show that
Indeed, if (A.54) were false, set x := x1 + |Yω − Y1 | e N .
(A.55)
˘ so (A.44) yields that gS(Yω ,R ) (x ) > g (x ). Therefore, Thus, |x | ≥ Cq, 1 u(x1 ) = gS(Y1 ,R1 ) (x1 ) = gS(Yω −|Yω −Y1 |e N ,R1 ) (x1 ) ˆ N) (x1 ) = gS(Yω ,R1 ) (x1 + βe =g ˆ S(Yω −βe N ,R1 )
ˆ N ) ≥ u(x1 ) . = gS(Yω ,R1 ) (x ) > g (x ) = g (x1 + βe This contradiction proves (A.54).
(A.56)
36
E. Valdinoci
Since t ≥ 0, an inspection on the domain of S(Y, R1 ) gives that x1,N > −l/4. Therefore, recalling also (A.54) and (A.52), we have that x1 is in the interior ˘ × {|x N | ≤ l/2}; thence, by means of Corollary 7, of the domain {|x | ≤ Cq} |u(x1 )| < 1/2. We now claim that TY1 ,R1 (x1 ) ∈ {|x | < ηq} ∩ {x N < const q 2 /R} .
(A.57)
To prove (A.57), first observe that, by (A.53), y1 is below yω −βe N in the e N -direction, and so (Y1 , R1 ) is below (Yω −βe N , R1 ) and, a fortiori, below (Yω , R1 ). Thence, by (A.41), TY1 ,R1 (x1 ) ∈ (Y1 , R1 ) ⊆ {x N < const q 2 /R} .
(A.58)
This gives a first step towards the proof of (A.57); we now show that TY1 ,R1 (x1 ) ∈ {|x | < ηq} .
(A.59)
Assume, by contradiction, that (A.59) is false. Then, by translating in the e N -direction, we have that TYω ,R1 (x ) ∈ {|x | ≥ ηq} ,
(A.60)
where x was defined in (A.55). Let now βˆ := |Y − Yω | − t1 = |Yω − Y1 |. Then, ˆ N and, by (A.53), βˆ ≥ β. The latter inequality and the first touching x = x1 + βe ˆ N is above the graph of u, i.e. property of β yield that − βe ˆ N ) ≥ u(x) . g (x + βe
(A.61)
From (A.60) and (A.43), we get that gS(Yω ,R1 ) (x ) > g (x ) . Thence, using also the touching property of X 1 and (A.61), repeating the argument in (A.56) verbatim, one obtains the contradiction which ends the proof of (A.59). Thus, (A.59) and (A.58) end the proof of (A.57). The fact that R1 ≥ R0 /4, (A.54) and (A.57) end the proof of (A.22) (in case k = 0, the other steps being analogous) and, therefore, the proof of (A.21). With (A.21) in hand, we now complete the proof of Lemma 25. For this, let us make some estimates on the point x∗ found in (A.21). Since x∗ is a touching point between the barrier and u, we have that |u(x∗ )| < 1/2 thanks to Corollary 7 and, therefore, |∇u(x∗ )| = |∇gS(Y∗ ,R∗ ) (x∗ )| = 0 . We deduce from (A.13) that x∗ − TY∗ ,R∗ x∗ ≤ const .
(A.62)
Consequently, from a second order Taylor expansion of g S(Y∗ ,R∗ ) and (4.5), we get that const ∇u(x∗ ) u(x∗ ) ≤ . (A.63) TY∗ ,R∗ x∗ − x∗ + |∇u(x∗ )| R
Flatness of Bernoulli jets
37
Also, since |TY∗ ,R∗ x∗ − y∗ | ≥ const R∗ , we have that ∇u(x∗ ) const q , eN ≤ . |∇u(x∗ )| R
(A.64)
From (A.63) and (A.64), we have that ∇u(x∗ ) const · e N u(x∗ ) + |∇u(x∗ )| R const . (A.65) ≤ (TY∗ ,R∗ x∗ ) · e N + u(x∗ ) + R We now consider the set ∗ := Y¯ = ( y¯ , 0, y¯ N +1 ) ∈ R N +1 s.t. |( y¯ − TY∗ ,R∗ x∗ ) | ≤ c q , | y¯ N +1 | ≤ 1/4 x∗ · e N ≤ (TY∗ ,R∗ x∗ ) · e N +
and let R˘ := c R
(A.66)
with c > 0 suitably small. Recalling the definition of r (·, ·) given in (A.12), we ˘ also denote r˘ := r (Y˘ , R). ˘ from −∞ in the e N direction, until it For any Y¯ ∈ ∗ , we slide S(Y¯ , R) touches the graph of u by above for the first time at a level different from ±1 (recall again (3.1)-(3.2) and Corollary 7), and let Y˘ denote the center of the cor˘ responding barrier: more explicitly, Y˘ = Y¯ − te N , for some t ∈ R and S(Y˘ , R) touches the graph of u from above for the first time coming from −∞ in the e N ˘ the set collecting all these points direction at some point X˘ . We will denote by X˘ , when Y¯ varies in ∗ . We know from (A.21) that q2 TY∗ ,R∗ x∗ (A.67) ≤ C N R for some suitably large constant C . We now claim that, if Y¯ ∈ ∗ , then 2 ˘ TY∗ ,R∗ x∗ , 2C qR is outside (Y˘ , R). (A.68) The proof of (A.68) is by contradiction. If (A.68) were false, we would have that ˘ ∩ {|(x − TY∗ ,R∗ x∗ ) | ≤ cq} ∩ {x N ≥ y˘ N } (Y˘ , R) is above the hyperplane {x N = 3C q 2 /(2R)},
(A.69)
provided that c is small enough. Then, comparing with the “vertical distance” to ˘ and making use of (A.65), (A.67) and (A.69), we get that (Y˘ , R) q2 d(Y˘ , R) ˘ (x ∗ ) ≤ x ∗ − TY˘ , R˘ x ∗ · e N + const R q2 ≤ TY∗ ,R∗ x∗ − TY˘ , R˘ x∗ · e N + u(x∗ ) + const R C q 2 q2 ≤ u(x∗ ) + const − . R 2R
(A.70)
38
E. Valdinoci
On the other hand, in the light of (A.13), letting s∗ := u(x∗ ), we have that the ˘ ˘ signed distance between the s∗ -level set of gS(Y˘ , R) ˘ and (Y , R) is bigger than u(x∗ ) −
const . R
Thus, from (A.70), by taking q and C suitably large, we have that d(Y˘ , R) ˘ (ξ ) > d(Y˘ , R) ˘ (x ∗ ) , for any ξ so that gS(Y˘ , R) ˘ (ξ ) = s∗ . Therefore, x ∗ is strictly in the interior of the s∗ -level set of gS(Y˘ , R) , that is gS(Y˘ , R) ˘ ˘ (x ∗ ) < s∗ = u(x ∗ ). This is in contradiction with the fact that gS(Y˘ , R) touches u by above and yields the proof of (A.68). ˘ In the light of (A.68), some elementary trigonometry implies that y + r˘ e N is below the hyperplane {x N = 4C q 2 /R} and therefore ˘ is below {x N = 4C q 2 /R}. (Y˘ , R)
(A.71)
Let us now fix a small constant c∗ > 0. By (A.68), ˘ ∩ {|x | ≥ c∗ q} is below {x N = 0}. (Y˘ , R)
(A.72)
We now deduce from the above geometric facts that outside {|x | ≤ c∗ q} × {x N > 0}, ˘ is at distance greater than q 2 /(4R) (Y˘ , R) in the interior of (Y, R).
(A.73)
For the proof of (A.73), first notice that y˘ N − r˘ ≥ −const r˘ ≥ y N
(A.74)
if c in (A.66) is chosen suitably small. Let us now consider the surface ˘ := (Y˘ , R) ˘ + te N t (Y˘ , R) ˘ then ξ N ≥ y N and so with t ≥ 0. By (A.74), it follows that, given ξ ∈ (Y˘ , R), |(ξ + te N ) − y| ≥ |ξ − y| . Hence, d(Y,R) (ξ ) ≤ d(Y,R) (ξ + te N ) .
(A.75)
Recalling (A.72), we now select the first t ≥ 0 for which ˘ ∩ {|x | = c∗ q} ∩ {x N = 0} = ∅ . S := t (Y˘ , R)
(A.76)
˘ which is below {x N = 0} with We also denote by S − the portion of t (Y˘ , R) ˘ which is below S . The respect to the e N -direction, i.e. the portion of t (Y˘ , R) choice in (A.76) implies that y˘ N + t + r˘ ≥ 0
(A.77)
Flatness of Bernoulli jets
39
and, by the definition of ∗ and (A.21), that q |( y˘ + te N ) | = | y˘ | ≤ ( y˘ − TY∗ ,R∗ x∗ ) + (TY∗ ,R∗ x∗ ) ≤ cq + = c˜∗ q , C∗ where c˜∗ := c +
1 C∗
(A.78)
is a small positive constant. Therefore, from (A.66) and (A.78), y˘ + te N is above the cone C∗ defined by " # c∗ q N 2 2 C∗ := x ∈ R s.t. |x | = |x N + r − q | . r 2 − q2 This fact, (A.76) and some elementary geometric considerations yield that if x ∈ S − , and x ∈ S , then d(Y,R) (x) ≥ d(Y,R) (x ) . (A.79) By construction, if x ∈ S , then x lies inside (Y, R). Hence, d(Y,R) (x ) < 0 . Also, from (A.76) and the fact that r˘ < r , if x ∈ and therefore
(A.80)
S − , then
x is also inside (Y, R)
d(Y,R) (x) < 0 .
(A.81)
Thus, from (A.79), (A.80) and (A.81), d(Y,R) (x) ≤ d(Y,R) (x ) ,
(A.82)
S−
and x ∈ S . for x ∈ Also, since y = 0 by its definition, we gather that, if x ∈ S , then |x − y|2 = |y N |2 + |x |2 = r 2 − q 2 + c∗2 q 2 and therefore d(Y,R) (x ) = |x − y| − r =
r 2 − q 2 + c∗2 q 2 − r .
(A.83)
Thus, taking x ∈ S − and x ∈ S , making use of (A.82) and (A.83), we have that −d(Y,R) (x) ≥ −d(Y,R) (x ) ≥
q2 q2 q2 1 − c∗2 − const 2 > . (A.84) 2r R 4R
˘ be outside {|x | ≤ To complete the proof of (A.73), let now ξ ∈ (Y˘ , R) − c∗ q} × {x N > 0} and define x := ξ + te N . Then, x ∈ S by the choice in (A.76). Consequently, from (A.75) and (A.84), we gather that −d(Y,R) (ξ ) ≥ −d(Y,R) (x) >
q2 . 4R
40
E. Valdinoci
This ends the proof of (A.73). By inspection, one sees that, if C˜ is large enough, then ˜ at any x for which d(Y˘ , R) ˘ (x) − d(Y,R) (x) ≥ C C 0 /R, we have that gS(Y˘ , R) ˘ (x) > gS(Y,R) (x).
(A.85)
Moreover, thanks to (A.73) and (A.85), we have that ˘ ∩ ∩{|x | > c∗ q} is strictly above S(Y, R). S(Y˘ , R)
(A.86)
Hence, since u ≤ gS(Y,R) , we gather from (A.86) that ˘ ∩ {|x | > c∗ q} is strictly above the graph of u. S(Y˘ , R)
(A.87)
|x| ˘ ≤ c∗ q .
(A.88)
Thus,
Also, by our assumptions on x0 , it follows that |x˘ N |
