1 Introduction and Main Results
We consider the problem of classifying solutions to
(1.1)
{
-
Δ
u
=
f
(
u
)
in
ℝ
+
2
,
u
≥
0
in
ℝ
+
2
,
u
=
0
on
∂
ℝ
+
2
.
under very general assumptions on the nonlinear function f. Here, by ℝ+2, we mean the open half-plane {(x,y)∈ℝ2:y>0}.
In our two previous works [9, 16], we proved the following result.
Theorem 1.1
Let u∈C2(R+2¯) be a solution to (1.1)with flocally Lipschitz continuous on [0,+∞).Then the following hold:
If
f
(
0
)
≥
0
, then either
u
vanishes identically or
u
is positive on
ℝ
+
2
with
∂
u
∂
y
>
0
in
ℝ
+
2
.
If
f
(
0
)
<
0
, then either
u
is positive on
ℝ
+
2
, with
∂
u
∂
y
>
0
in
ℝ
+
2
,or
u
is one-dimensional and periodic (and unique).
The case f(0)≥0 was first treated in [9] (in which also the case of the p-Laplace operator is considered), while the case f(0)<0 was carried out later, in the paper [16].Both of them are based on a refined version of the moving plane method [26] (see also [5, 21]). More precisely, the authors of [9] exploit arotating line techniqueand a sliding line technique, while in [16] we have used a refinement of these techniques combined with the unique continuation principle, needed to handle the new, challenging and difficult case of nonnegative solutions.The techniques developed in [9, 16] also provided an affirmative answer to a conjecture and to an open question posed by Berestycki, Caffarelli and Nirenberg in [2, 3].
In this paper we first give a (new) unique/unified proof to Theorem 1.1 and at the same time we make the effort to deal with the case of continuous nonlinearities f that fulfills very weak and general regularity assumptions, i.e., less regular than locally Lipschitz continuous.The assumptions are introduced and commented in the next section, where we also state our main results.
2 Assumptions and New Results
We start by discussing the main assumptions on the nonlinearity f.In Theorem 1.1 we assumed f to be locally Lipschitz continuous on [0,+∞).This allow us to deal with nonnegative solutions.In the following Theorem 2.1 we will restrict our attention to positive solutions weakening the assumptions on the nonlinearity.It is convenient to set the following:
f is continuous in [0,+∞).
f is locally Lipschitz continuous from above in [0,∞), i.e., for every b>0 there exists Lb>0 such that
f
(
u
)
-
f
(
v
)
≤
L
b
(
u
-
v
)
for any
0
≤
v
≤
u
≤
b
.
For any t¯∈[0,+∞), there exists δ=δ(t¯)>0 such that
(2.1)
f
(
s
)
-
f
(
t
)
≤
g
(
t
-
s
)
for all
s
≤
t
∈
[
t
¯
-
δ
2
,
t
¯
+
δ
2
]
∩
[
0
,
+
∞
)
,
with g∈C0[0,δ], g(0)=0 and g either vanishes identically in [0,δ] or it is positive and non-decreasing in [0,δ] with
(2.2)
∫
0
δ
d
s
G
(
s
)
=
∞
,
where G(s):=∫0sg(t)𝑑t.
Condition (A3) holds but only for any t¯∈(0,+∞).
Note that the hypothesis (A2) is very weak (actually it is not even enough to ensure the continuity of f) and it is clearly satisfied by any non-increasing function. Actually, all the nonlinearities of the form
f
(
s
)
:=
f
1
(
s
)
+
f
2
(
s
)
,
for some non-increasing continuous function f1(⋅) in [0,∞) and some f2(⋅) which is locally Lipschitz continuous in [0,∞), satisfy both (A1) and (A2).
We also observe that assumption (A3) is natural. Indeed, the conditions imposed on g are the well-known optimal assumptions ensuring the validity of the strong maximum principle and Hopf’s lemma (see, for instance, [25]).It is also clear that every locally Lipschitz continuous function on [0,∞) satisfies (A1), (A2) and (A3) with g(t)=Lδt, Lδ>0 being any constant larger than the Lipschitz constant of f on the interval [0,δ].On the other hand, the converse is not true.This is the case, e.g., when f(⋅) has the form
f
(
s
)
:=
g
(
s
)
+
c
,
where g(s)≡slog(s) in some interval (0,δ), δ>0 and g(⋅) of class C1 in (0,∞).It is easy to verify that such a nonlinearity fulfills (A3) but it is not Lipschitz continuous at zero.
Having in mind the assumptions above, we have the following theorem.
Theorem 2.1
Let u∈C2(R+2¯) be a solution to (1.1)and assume that either of the following holds:
f
(
0
)
=
0
and
f
fulfills
(A1), (A2) and (A3),
f
(
0
)
≠
0
and
f
fulfills
(A1), (A2) and (A4).
Then
∂
u
∂
y
>
0
in
ℝ
+
2
.
We observe that the conclusion of Theorem 2.1 is not true if we drop the assumption (A3) in item (i).This will be discussed in the last section of the paper.
Let us point out that, with the same technique, we can also prove a symmetry and monotonicity result in strips, for possibly unbounded solutions.More precisely, with the notation
Σ
2
b
:=
{
(
x
,
y
)
∈
ℝ
2
:
y
∈
(
0
,
2
b
)
}
,
b
>
0
,
we have the following result.
Theorem 2.2
Let u∈C2(R×[0,2b)) be a solution to
{
-
Δ
u
=
f
(
u
)
in
Σ
2
b
,
u
>
0
in
Σ
2
b
,
u
=
0
on
{
y
=
0
}
,
and assume that either of the following holds:
f
(
0
)
=
0
and
f
fulfills
(A1), (A2) and (A3),
f
(
0
)
≠
0
and
f
fulfills
(A1), (A2) and (A4).
Then
∂
u
∂
y
>
0
in
Σ
b
.
If u∈C2(Σ2b¯)and u=0 on ∂Σ2b, thenu is symmetric about {y=b}.
As already observed for Theorem 2.1, the conclusion of Theorem 2.2 is not true if we drop the assumption (A3) in item (i) (see Section 6).
The theorem above complements [16, Theorem 1.3] and also provides an affirmative answer to an (extended version of an) open question posed by Berestycki, Caffarelli and Nirenberg in [2].
Next we prove a symmetry result in the case of the half-plane.
Theorem 2.3
Let u∈C2(R+2¯) be a solution to (1.1), with f being locally Lipschitz continuous on [0,+∞) and having the property thatthere exists p>1 such that
lim
t
→
+
∞
f
(
t
)
t
p
=
l
∈
(
0
,
+
∞
)
.
Then u is bounded and one-dimensional, i.e.,
u
(
x
,
y
)
=
u
0
(
y
)
for all
(
x
,
y
)
∈
ℝ
+
2
,
for some bounded function u0∈C2([0,+∞)).
As a consequence of the results above we obtain the following corollary.
Corollary 2.4
Let f be locally Lipschitz continuous on [0,+∞) satisfying
(2.3)
f
(
t
)
>
0
for all
t
>
0
,
and having the property that there exists p>1 such that
lim
t
→
+
∞
f
(
t
)
t
p
=
l
∈
(
0
,
+
∞
)
.
Then the following hold:
If
f
(
0
)
=
0
, then the only solution of class
C
2
(
ℝ
+
2
¯
)
of (
1.1
) is
u
≡
0
.
If
f
(
0
)
>
0
, then problem (
1.1
) has no solution of class
C
2
(
ℝ
+
2
¯
)
.
We conclude this section with the following classification result.
Theorem 2.5
Let f be non-decreasing and locally Lipschitz continuous on [0,+∞) satisfying
(2.4)
f
(
0
)
≥
0
.
Then problem (1.1) has a nontrivial solution of class C2(R+2¯) if and only if f≡0.In the latter case u is necessarily linear, i.e., u(x,y)=cy for some constant c>0.Furthermore, when f≢0, we have that
If
f
(
0
)
=
0
, then the only solution of class
C
2
(
ℝ
+
2
¯
)
of (
1.1
) is
u
≡
0
.
If
f
(
0
)
>
0
, then problem (
1.1
) has no solution of class
C
2
(
ℝ
+
2
¯
)
.
Remark 2.6
The assumption (2.4) is sharp.Indeed, the function u(x,y)=1-cosy is a nontrivial solution of
{
-
Δ
u
=
u
-
1
in
ℝ
+
2
,
u
⩾
0
in
ℝ
+
2
,
u
=
0
on
∂
ℝ
+
2
,
and f(0)=-1<0.
Remark 2.7
Theorem 2.5 applies, for instance, to the functions f(u)=up+c, with p≥1 and c≥0.In particular, for c=0 (i.e., for f(u)=up), we obtain a new and different proof of a celebrated result of Gidas and Spruck [22] in dimension two (see also [9]).
Note also that one can apply item (i) of Corollary 2.4 to f(u)=up, p>1, to obtain another new and different proof of the above mentioned result of Gidas and Spruck [22].
In this work we focused on the two-dimensional case and we provided a precise description of the situation under very general assumptions (both on f and on u). Differently from the two-dimensional case, the situation is not yet well-understood for dimensions N≥3. For results in the higher dimensional case (and with additional assumptions on f and on u),we refer to [2, 3, 4, 1, 7, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22].
The rest of the paper is devoted to the proof of our results. Some arguments rely on the techniques exploited in [16] (see also [3, 9]). For the reader’s convenience and in order to make the proofs self contained, we provide all the details.
3 Preliminary Results
We start proving a weak comparison principle in domains of small measure.Namely, we have the following proposition.
Proposition 3.1
Proposition 3.1 (Weak Comparison Principle in Domains of Small Measure)
Let N≥1, assume that f fulfills (A2) and fix a real number k>0. Then there exists ϑ=ϑ(N,k,f)>0 such that for any domain D⊂RN with L(D)≤ϑ, and any u,v∈H1(D)∩C0(D¯) such that
(3.1)
{
-
Δ
u
-
f
(
u
)
≤
-
Δ
v
-
f
(
v
)
in
D
,
0
≤
u
,
v
≤
k
in
D
,
u
≤
v
on
∂
D
,
we have
u
≤
v
in
D
.
Proof.
We use (u-v)+∈H01(D) as test function in the weak formulation of (3.1) and get
∫
D
|
∇
(
u
-
v
)
+
|
2
d
x
≤
∫
D
(
f
(
u
)
-
f
(
v
)
)
(
u
-
v
)
+
d
x
⩽
L
k
∫
D
(
(
u
-
v
)
+
)
2
d
x
,
where Lk is the positive constant appearing in (A2) and corresponding to b=k>0. Note that Lk depends only on k and f.
An application of the Poincaré inequality gives
∫
D
|
∇
(
u
-
v
)
+
|
2
d
x
⩽
L
k
(
C
N
(
ℒ
(
D
′
)
)
2
N
)
∫
D
|
∇
(
u
-
v
)
+
|
2
d
x
,
where CN>0 is a constant depending only on the euclidean dimension N.
The desired conclusion then follows by choosing ϑ=1/(2LkN/2CN).Indeed, the latter implies that Lk(CN(ℒ(D))2/N)<1, and so we get that (u-v)+≡0 and the thesis.∎
Now we focus on the two-dimensional case and fix some notations.
Given x0∈ℝ, s>0 and θ∈(0,π2), let Lx0,s,θ be the line, with slope tan(θ), passing through (x0,s). Also, let Vθ be the vector orthogonal to Lx0,s,θ such that (Vθ,e2)>0 and ∥Vθ∥=1.
We denote by𝒯x0,s,θ,the (open) triangle delimited by Lx0,s,θ, {y=0} and {x=x0}, and we define
u
x
0
,
s
,
θ
(
x
)
=
u
(
T
x
0
,
s
,
θ
(
x
)
)
,
x
∈
𝒯
x
0
,
s
,
θ
,
where Tx0,s,θ(x) is the point symmetric to x with respect to Lx0,s,θ.It is immediate to see that ux0,s,θ still fulfills -Δux0,s,θ=f(ux0,s,θ) on the triangle 𝒯x0,s,θ.
We also consider
(3.2)
w
x
0
,
s
,
θ
=
u
-
u
x
0
,
s
,
θ
on
𝒯
x
0
,
s
,
θ
,
and observe that
(3.3)
w
x
0
,
s
,
θ
≤
0
on
𝒯
x
0
,
s
,
θ
⟹
either
w
x
0
,
s
,
θ
≡
0
or
w
x
0
,
s
,
θ
<
0
on
𝒯
x
0
,
s
,
θ
,
thanks to assumption (A3) (resp. to (A4), if u is supposed to be positive).Indeed, if x¯∈𝒯x0,s,θ is such that wx0,s,θ(x¯)=0, then, by the continuity of u and of ux0,s,θ, we can find an open ball centered at x¯, say Bx¯⊂𝒯x0,s,θ, such that
(
u
(
x
¯
)
-
δ
2
)
+
≤
u
(
x
)
≤
u
(
x
¯
)
+
δ
2
for all
x
∈
B
x
¯
,
(
u
(
x
¯
)
-
δ
2
)
+
≤
u
x
0
,
s
,
θ
(
x
)
≤
u
(
x
¯
)
+
δ
2
for all
x
∈
B
x
¯
,
where t¯=u(x¯)=ux0,s,θ(x¯) and δ=δ(t¯)>0 is the one provided by assumption (A3) (resp. A4).Now, since wx0,s,θ≤0 on 𝒯x0,s,θ, we can apply (2.1) to get
(3.4)
{
Δ
(
-
w
x
0
,
s
,
θ
)
≤
g
(
-
w
x
0
,
s
,
θ
)
in
B
x
¯
,
-
w
x
0
,
s
,
θ
≥
0
in
B
x
¯
,
-
w
x
0
,
s
,
θ
(
x
¯
)
=
0
,
and the strong maximum principle (see, for instance, [25]) yields wx0,s,θ≡0 on Bx¯.After that, a standard connectedness argument provides wx0,s,θ≡0 on the entire triangle 𝒯x0,s,θ.
In what follows we shall make repeated use of a refined version of the moving plane technique [26] (see also [5, 21]).Actually we will exploit a rotating plane technique and a sliding plane technique developed in [9, 16].
Let us give the following definition.
Definition 3.2
Given x0,s and θ as above, we say that the condition (ℋ𝒯x0,s,θ) holds in the triangle 𝒯x0,s,θ if the following hold:
w
x
0
,
s
,
θ
<
0
in 𝒯x0,s,θ,
w
x
0
,
s
,
θ
⩽
0
on ∂(𝒯x0,s,θ),
w
x
0
,
s
,
θ
is not identically zero on ∂(𝒯x0,s,θ),
with wx0,s,θ defined in (3.2).
We have the following lemma.
Lemma 3.3
Lemma 3.3 (Small Perturbations)
Let u∈C2(R+2¯) be a nonnegative solution to (1.1)and assume that f fulfills (A2) and (A3).Let (x0,s,θ) and Tx0,s,θ be as above and assume that(HTx0,s,θ) holds. Then there exists μ¯=μ¯(x0,s,θ)>0 such that
{
|
θ
-
θ
′
|
+
|
s
-
s
′
|
<
μ
¯
,
w
x
0
,
s
′
,
θ
′
⩽
0
on
∂
(
𝒯
x
0
,
s
′
,
θ
′
)
,
w
x
0
,
s
′
,
θ
′
is not identically zero on
∂
(
𝒯
x
0
,
s
′
,
θ
′
)
⟹
(
ℋ
𝒯
x
0
,
s
′
,
θ
′
)
holds.
If u is positive, the same result holds assuming only (A4) instead of (A3).
Proof.
Let R>0 and μ~ be fixed so that
⋃
|
θ
-
θ
′
|
+
|
s
-
s
′
|
<
μ
~
𝒯
x
0
,
s
′
,
θ
′
∪
T
x
0
,
s
′
,
θ
′
(
𝒯
x
0
,
s
′
,
θ
′
)
⊂
B
R
(
x
0
)
.
We set
k
:=
max
B
R
+
(
x
0
)
¯
u
and
ϑ
=
ϑ
(
N
,
k
,
f
)
,
where ϑ(N,k,f) is the one appearing in Proposition 3.1 and BR+(x0)=BR(x0)∩{y>0}.
We fix 0<μ^≤μ~ such that
(3.5)
ℒ
(
𝒯
x
0
,
s
+
μ
^
,
θ
-
μ
^
∖
𝒯
x
0
,
s
-
μ
^
,
θ
+
μ
^
)
<
ϑ
2
,
and note that
𝒯
x
0
,
s
-
μ
^
,
θ
+
μ
^
⊂
⋃
|
θ
-
θ
′
|
+
|
s
-
s
′
|
<
μ
^
𝒯
x
0
,
s
′
,
θ
′
⊂
𝒯
x
0
,
s
+
μ
^
,
θ
-
μ
^
.
Now we can consider a compact set K⊂𝒯x0,s-μ^,θ+μ^ such that
(3.6)
ℒ
(
𝒯
x
0
,
s
-
μ
^
,
θ
+
μ
^
∖
K
)
<
ϑ
2
.
By assumption, we know that wx0,s,θ<0 in 𝒯x0,s,θ and consequently in the compact set K.Therefore, by a uniform continuity argument, for some0<μ¯≤μ^,we can assume that
(3.7)
w
x
0
,
s
′
,
θ
′
<
0
in
K
for
|
θ
-
θ
′
|
+
|
s
-
s
′
|
<
μ
¯
.
By (3.5) and (3.6), we deduce that
ℒ
(
𝒯
x
0
,
s
′
,
θ
′
∖
K
)
<
θ
and,observing that wx0,s′,θ′⩽0 on ∂(𝒯x0,s′,θ′∖K) (see (3.7)), we can apply Proposition 3.1 to get that
w
x
0
,
s
′
,
θ
′
⩽
0
in
𝒯
x
0
,
s
′
,
θ
′
∖
K
,
and therefore in the triangle 𝒯x0,s′,θ′.Then the desired conclusion
w
x
0
,
s
′
,
θ
′
<
0
in
𝒯
x
0
,
s
′
,
θ
′
follows from (3.3).∎
Now, from small perturbations, we move to larger translations and rotations. We have the following lemma.
Lemma 3.4
Lemma 3.4 (The Sliding-Rotating Technique)
Let u∈C2(R+2¯) be a nonnegative solution to (1.1) and assume that f fulfills (A2) and (A3).Let (x0,s,θ) be as above and assume that (HTx0,s,θ) holds.Let (s^,θ^) be fixed and assume that there exists a continuous function g(t)=(s(t),θ(t)):[0,1]→(0,+∞)×(0,π2) such that g(0)=(s,θ) and g(1)=(s^,θ^).Assume that
w
x
0
,
s
(
t
)
,
θ
(
t
)
⩽
0
on
∂
(
𝒯
x
0
,
s
(
t
)
,
θ
(
t
)
)
for every
t
∈
[
0
,
1
)
and thatwx0,s(t),θ(t) is not identically zero on ∂(Tx0,s(t),θ(t)) for every t∈[0,1).Then(HTx0,s^,θ^) holds.If u is positive, the same result holds assuming only (A4) instead of (A3).
Proof.
By the assumptions and by exploiting Lemma 3.3, we obtain the existence of a small t~>0 such that (ℋ𝒯x0,s(t),θ(t)) holds for 0⩽t⩽t~.
We now set
T
¯
=
{
t
~
∈
[
0
,
1
]
such that
(
ℋ
𝒯
x
0
,
s
(
t
)
,
θ
(
t
)
)
holds for any
0
⩽
t
⩽
t
~
}
and
t
¯
=
sup
T
¯
.
We claim that actually t¯=1.To prove this, assume t¯<1 and note that in this case we have
w
x
0
,
s
(
t
¯
)
,
θ
(
t
¯
)
≤
0
in
𝒯
x
0
,
s
(
t
¯
)
,
θ
(
t
¯
)
,
w
x
0
,
s
(
t
¯
)
,
θ
(
t
¯
)
≤
0
on
∂
(
𝒯
x
0
,
s
(
t
¯
)
,
θ
(
t
¯
)
)
,
by continuity, and that wx0,s(t¯),θ(t¯) is not identically zero on ∂(𝒯x0,s(t¯),θ(t¯)), by assumption.
Hence, by (3.3), we see that
w
x
0
,
s
(
t
¯
)
,
θ
(
t
¯
)
<
0
in
𝒯
x
0
,
s
(
t
¯
)
,
θ
(
t
¯
)
.
Therefore, (ℋ𝒯x0,s(t¯),θ(t¯)) holds, and using once again Lemma 3.3, we can find a sufficiently small ε>0 such that (ℋ𝒯x0,s(t),θ(t)) holds for any0⩽t⩽t¯+ε, which contradicts the definition of t¯.∎
4 Further Preliminary Results
In this section we assume that u is a nonnegative and non-trivial solution of (1.1), i.e., u≢0. We also suppose that f fulfills (A1) and, when f(0)=0, also that f fulfills (A3).
Given x0∈ℝ, let us set
B
r
+
(
x
0
)
=
B
r
(
x
0
)
∩
{
y
>
0
}
,
where Br(x0) denotes the two-dimensional open ball centered at (x0,0) and of radius r>0.
We claim that for some r¯>0 and for some θ¯=θ¯(r¯)∈(0,π2),
(4.1)
∂
u
∂
V
θ
>
0
in
B
r
¯
+
(
x
0
)
for
-
θ
¯
≤
θ
≤
θ
¯
.
Case 1: f(0)<0.Since ∂xxu(x0,0)=0, we have that
-
∂
y
y
u
(
x
0
,
0
)
=
-
Δ
u
(
x
0
,
0
)
=
f
(
u
(
x
0
,
0
)
)
=
f
(
0
)
<
0
.
Recalling that u∈C2(ℝ+2¯), we conclude that we can take r¯>0 small enough such that
∂
y
y
u
>
0
in
B
r
¯
+
(
x
0
)
.
Exploiting again the fact that u∈C2(ℝ+2¯), we can consequently deduce that
(4.2)
∂
∂
V
θ
(
∂
u
∂
V
θ
)
>
0
in
B
r
¯
+
(
x
0
)
for
-
θ
¯
≤
θ
≤
θ
¯
.
Also, since we assumed that u is nonnegative in ℝ+2, it follows that
(4.3)
∂
u
∂
V
θ
(
x
,
0
)
⩾
0
for any
-
θ
¯
≤
θ
≤
θ
¯
and any
x
∈
ℝ
.
Combining (4.2) and (4.3), we deduce (4.1).Case 2: f(0)≥0 and u≢0.In this case we first observe that
(4.4)
{
u
>
0
in
ℝ
+
2
,
∂
u
∂
y
(
x
,
0
)
>
0
for all
x
∈
ℝ
.
Indeed, if f(0)>0 and u(x¯)=0, with x¯∈ℝ+2, then Δu≤0 in an open connected neighborhood of x¯, by the continuity of u and f.The classical maximum principle and a standard connectedness argument imply thatu≡0 on ℝ+2. The latter contradicts the non-triviality of u. Therefore, u>0 everywhere and the classical Hopf lemma provides the second claim in (4.4).
To treat the case f(0)=0, we follow the arguments leading to (3.4) and (3.3).More precisely, when f(0)=0 and u(x¯)=0, the continuity of u and (2.1) with t¯=0 yieldΔu=-f(u)=f(0)-f(u)≤g(u) in an open connected neighborhood of x¯.Then, since (A3) is in force, we can use the strong maximum principle and the Hopf boundary lemma (see, for instance, [25, Chapter 5]) to get (4.4), as before.
Remark 4.1
Note that, in the previous argument, we used assumption (A3) only for t¯=0 (and only in the case f(0)=0).
The desired conclusion (4.1) then follows immediately from (4.4) and the C2-regularity of u up to the boundary.
From the analysis above, we find the existence of (possible very small)
(4.5)
s
¯
=
s
¯
(
θ
¯
)
>
0
such that for any 0<s⩽s¯, the following hold:
Both the triangle 𝒯x0,s,θ¯ and its reflection with respect to Lx0,s,θ¯ are contained in Br+(x0) (as well as their reflections with respect to the axis {x=x0}).
Both the segment {(x0,y):0≤y≤s} and its reflection with respect to Lx0,s,θ are contained in Br+(x0) for every θ∈(0,θ¯].
u
<
u
x
0
,
s
,
θ
¯
in 𝒯x0,s,θ¯.
u
⩽
u
x
0
,
s
,
θ
on ∂(𝒯x0,s,θ) for every θ∈(0,θ¯].
u
<
u
x
0
,
s
,
θ
on the set {(x0,y):0<y<s} for every θ∈(0,θ¯].
Note that, from (iii)–(iv), we have that
(4.6)
(
ℋ
𝒯
x
0
,
s
,
θ
¯
)
holds for all
s
∈
(
0
,
s
¯
)
.
To continue the description of our results, we denote by p:=(x,y) a general point in the plane and, for a nonnegative solution u of (1.1), we say that u satisfies the property (𝒫μ) if there exists a real number μ>0 and a point p∈{y=μ} such that u(p)≠0.Equivalently,
(
𝒫
μ
)
holds if
{
y
=
μ
}
∩
{
u
≠
0
}
≠
∅
.
For a non-trivial u and under the assumptions stated at the beginning of this section, we have that the set
Λ
*
=
Λ
*
(
u
)
:=
{
λ
>
0
:
(
𝒫
μ
)
holds for every
0
<
μ
≤
λ
}
is not empty.The latter claim follows from (4.4) when f(0)≥0 and from [16, Theorem 6.1] when f(0)<0 (note that [16, Theorem 6.1] holds true for functions f which are only continuous on [0,+∞), and so it applies in our situation since (A1) is in force).
Therefore, we set
(4.7)
λ
*
=
λ
*
(
u
)
:=
sup
Λ
*
∈
(
0
,
+
∞
]
,
and also note that, by a continuity argument, if λ* is finite, we get that {y=λ*}⊆{u=0}.
Next we prove a result that allows to start the moving plane procedure.
Lemma 4.2
Lemma 4.2 (Monotonicity Near the Boundary)
Let u∈C2(R+2¯) be a nonnegative and non-trivial solution to (1.1) and assume that f is locally Lipschitz continuous on [0,+∞).Then there exists λ^>0 such that for any 0<λ≤λ^, we have
(4.8)
u
<
u
λ
in
Σ
λ
.
Furthermore,
(4.9)
∂
y
u
>
0
in
Σ
λ
^
.
If u is positive, the conclusions above hold when either of the following holds:
f
(
0
)
=
0
and
f
fulfills
(A1), (A2) and (A3),
f
(
0
)
≠
0
and
f
fulfills
(A1), (A2) and (A4).
Proof.
Let θ¯ given by (4.2) and s¯=s¯(θ¯)as in (4.5).We showed that (ℋ𝒯x0,s,θ¯) holds for any 0<s<s¯.
We use now Lemma 3.4 as follows.For any fixed s∈(0,s¯) and θ′∈(0,θ¯), we consider the rotation
g
(
t
)
=
(
s
(
t
)
,
θ
(
t
)
)
:=
(
s
,
t
θ
′
+
(
1
-
t
)
θ
¯
)
,
t
∈
[
0
,
1
]
.
Recalling that (ℋ𝒯x0,s,θ¯) holds by (4.6), we deduce that also (ℋ𝒯x0,s,θ′) holds.Therefore, by the fact that 0<θ′<θ¯ is arbitrary and by a continuity argument, we pass to the limit for θ′→0 and get
u
(
x
,
y
)
≤
u
s
(
x
,
y
)
in
Σ
s
∩
{
x
⩽
x
0
}
for
0
<
s
<
s
¯
.
The invariance of the considered problem with respect to the axis {x=x0} enables us to use the same argument to treat the case of negative θ, yielding
u
(
x
,
y
)
≤
u
s
(
x
,
y
)
in
Σ
s
∩
{
x
⩾
x
0
}
for
0
<
s
<
s
¯
,
possibly by reducing s¯.
Thus, u(x,y)≤us(x,y) in Σs for every s∈(0,s¯).The desired conclusion (4.8) then follows by taking λ^ such that 0<λ^<min{s¯,λ*2}.Here we have used, in a crucial way, that the property (𝒫λ) holds for every λ∈(0,λ^], so that the case u≡uλ in Σλ is not possible.
Moreover, when f is locally Lipschitz continuous on (0,+∞], the function uλ-u>0 solves a linear equation of the form Δ(uλ-u)=c(x)(uλ-u), with c locally bounded on Σλ. Therefore, by Hopf’s lemma, for every λ∈(0,λ^] and every x∈ℝ, we get
(4.10)
-
2
∂
y
u
(
x
,
λ
)
=
∂
(
u
λ
-
u
)
∂
y
(
x
,
λ
)
<
0
.
The latter proves (4.9) when f is locally Lipschitz continuous on (0,+∞].
If u is everywhere positive, (4.10) is still true since (A3) (resp. A4) is in force. Indeed, the arguments already used to prove (3.3) and (4.4) and the crucial fact that u(x,λ)>0 for every x∈ℝ lead to
{
Δ
(
u
λ
-
u
)
≤
g
(
u
λ
-
u
)
in
B
x
,
u
λ
-
u
>
0
in
B
x
,
u
λ
(
x
,
λ
)
-
u
(
x
,
λ
)
=
0
for all
x
∈
ℝ
,
where Bx⊂ℝ2 is an open ball centered at (x,λ). Therefore, since (A3) (resp. A4) is in force, the boundary lemma gives (4.10). This concludes the proof.∎
Remark 4.3
Note that when u is positive, we used only (A4).
Let λ* be defined as in (4.7).In the case λ*=∞, we set
Λ
=
{
λ
>
0
:
u
<
u
λ
′
in
Σ
λ
′
for all
λ
′
<
λ
}
.
If λ* is finite, we use the same notation but considering values of λ such that 0<λ<λ*2, namely,
Λ
=
{
λ
<
λ
*
2
:
u
<
u
λ
′
in
Σ
λ
′
for all
λ
′
<
λ
}
.
By Lemma 4.2, we know that Λ is not empty and we can define
λ
¯
=
sup
Λ
.
Now we assume that λ¯<+∞ when λ*=∞ (resp. λ¯<λ*2 when λ* is finite), and observe that, arguing as above and under the same assumptions of Lemma 4.2 (cf. the proof of (4.10)), we deduce that
u
<
u
λ
¯
on
Σ
λ
¯
,
(4.11)
∂
y
u
(
x
,
λ
)
>
0
for all
(
x
,
λ
)
∈
ℝ
×
(
0
,
λ
¯
]
.
Then we can prove the following lemma.
Lemma 4.4
Let u and f be as in Lemma 4.2.Let λ* and λ¯ be as above.Assume that there is a point x0∈R satisfying u(x0,2λ¯)>0.Then there exists δ¯>0 such that for any -δ¯⩽θ⩽δ¯ and any0<λ⩽λ¯+δ¯, we have
u
(
x
0
,
y
)
<
u
x
0
,
λ
,
θ
(
x
0
,
y
)
for
0
<
y
<
λ
.
Proof.
First we note that, by (4.11), we have ∂yu(x0,λ¯)>0.
We argue now by contradiction. If the lemma were false, then we could find a sequence of small δn→0 and-δn⩽θn⩽δn,0<λn⩽λ¯+δn,0<yn<λn such that
u
(
x
0
,
y
n
)
⩾
u
x
0
,
λ
n
,
θ
n
(
x
0
,
y
n
)
.
Possibly considering subsequences, we may and do assume that λn→λ~⩽λ¯.Also yn→y~ for some y~⩽λ~.Considering the construction of Br¯+(x0) as above and, in particular, taking into account (4.2) and (4.3), we deduce that λ~>0 and,by continuity, it follows that u(x0,y~)⩾uλ~(x0,y~).Consequently,yn→λ~=y~, since we know that u<uλ′ in Σλ′ for any λ′⩽λ¯, andwe assumed that u(x0,2λ¯)>0 so that, in particular,u(x0,0)=0<u(x0,2λ¯).By the mean value theorem, since u(x0,yn)⩾ux0,λn,θn(x0,yn), it follows
∂
u
∂
V
θ
n
(
x
n
,
y
n
)
⩽
0
at some point ξn≡(xn,yn) lying on the line from (x0,yn) to Tx0,λn,θn(x0,yn), recalling that the vector Vθn is orthogonal to the line Lx0,λn,θn.Since Vθn→e2 as θn→0,by taking the limit, it follows that∂yu(x0,λ~)⩽0,which is impossible by (4.11).∎
5 Proof of Theorem 1.1
Proof of Theorem 1.1.
Since we are assuming that λ¯<+∞ when λ*=∞ (resp. λ¯<λ*2 when λ* is finite), by the definition of λ*, we can find x0∈ℝ such that u(x0,2λ¯)>0.Let Br¯+(x0) be constructed as above and pick θ¯ given by (4.2).
Let also δ¯ be as in Lemma 4.4.Then fix θ0>0 with θ0⩽δ¯ and θ0⩽θ¯.Let us set
s
0
:=
s
0
(
θ
0
)
,
so that the triangle 𝒯x0,s0,θ0 and its reflection with respect to Lx0,s0,θ0 is contained in Br¯+(x0) and, consequently, (ℋ𝒯x0,s0,θ0) holds.It is convenient to assume that s0⩽λ^ with λ^ as in Lemma 4.2.For any
s
0
<
s
⩽
λ
¯
+
δ
¯
,
0
<
θ
<
θ
0
,
we carry out the sliding-rotating technique exploitingLemma 3.4 with
g
(
t
)
=
(
s
(
t
)
,
θ
(
t
)
)
:=
(
t
s
+
(
1
-
t
)
s
0
,
t
θ
+
(
1
-
t
)
θ
0
)
t
∈
[
0
,
1
]
.
By Lemma 4.4, we deduce that the boundary conditions required to apply Lemma 3.4 are fulfilled, and therefore, by Lemma 3.4, we get that (ℋ𝒯x0,s,θ) holds.We can now argue as in the proof of Lemma 4.2 and deduce that u(x,y)<uλ(x,y) in Σλ for any 0<λ⩽λ¯+δ¯.This provides a contradiction unless λ¯=+∞ (resp. λ¯=λ*2 if λ* is finite).Arguing as in the proof of Lemma 4.2, we deduce
∂
y
u
>
0
in
ℝ
+
2
if
λ
*
=
+
∞
,
while
∂
y
u
>
0
in
Σ
λ
*
/
2
if
λ
*
<
+
∞
.
As a consequence of the monotonicity result, we deduce that u is positive in ℝ+2 if λ*=+∞.
Let us now deal with the case when λ* is finite, which may occur only in the case f(0)<0.We deduce by continuity that
u
≤
u
λ
*
/
2
in
Σ
λ
*
/
2
.
By the strong comparison principle, we deduce that either u<uλ*/2 or u≡uλ*/2 in Σλ*/2.Note that, by the definition of λ*, we have that {y=λ*}⊆{u=0}, which also implies {y=λ*}⊆{∇u=0} since u is nonnegative.If u<uλ*/2 in Σλ*/2,we get, by Hopf’s boundary lemma (see [23]), that ∂y(uλ*/2-u)>0 on {y=0}. Since ∂y(uλ*/2)=0 on {y=0} (by the fact that {y=λ*}⊆{∇u=0}) this provides a contradiction with the fact that u is nonnegative.Therefore, u≡uλ*/2 in Σλ*/2.
Note now that since{y=λ*}⊆{u=0}∩{∇u=0},by symmetry, we deduce
{
y
=
0
}
⊆
{
u
=
0
}
∩
{
∇
u
=
0
}
.
Therefore, we deduce that u is one-dimensional by the unique continuation principle (see, for instance, [20, Theorem 1] and the references therein).Here we use, in a crucial way, the fact that f is locally Lipschitz continuous on [0,+∞).Indeed, for every t∈ℝ, the function ut(x,y):=u(x+t,y) is a nonnegative solution of (1.1) with ut=∇ut=0 on ∂ℝ+2, and the unique continuation principle implies that u≡ut on ℝ+2.This immediately gives that u depends only on the variable y, i.e.,
u
(
x
,
y
)
=
u
0
(
y
)
for all
(
x
,
y
)
∈
ℝ
+
2
,
where u0∈C2([0,+∞)) is the unique solution of u0′′+f(u0)=0 with u0′(0)=u0(0)=0.
The remaining part of the statement, namely, the properties of u0, follows by a simple ODE analysis.∎
6 Proof of Theorems 2.1 and 2.2
Proof of Theorem 2.1.
Since u is positive, we immediately have that λ*=∞.We now observe that when u is positive, the first part of the proof of Theorem 1.1 holds under the assumptions of Theorem 2.1 (see Lemmas 3.4, 4.2 and 4.4).Therefore, arguing as in Theorem 1.1, we get∂yu>0 in ℝ+2.∎
Proof of Theorem 2.2.
The proof follows by arguing exactly as in the proof of Theorem 2.1, with just observing that the translating rotating technique can be performed until we reach the maximal position in the middle of the strip.This provides the fact that u is strictly monotone increasing in Σb.To prove that∂u∂y>0 in Σb, we just argue again as in the proof of (4.10) (see also (3.3)).If u∈C2(Σ2b¯) and u=0 on ∂Σ2b, then the technique can be applied in the opposite direction, thus proving that u is symmetric about {y=b}.∎
7 Proof of Theorem 2.3, Corollary 2.4 and Theorem 2.5
Proof of Theorem 2.3.
Since f is locally Lipschitz continuous on [0,+∞), Theorem 1.1 implies that either u is one-dimensional and periodic (possibly identically equals zero), and in this case we are done, or u>0 and ∂u∂y>0 everywhere in ℝ+2.To conclude the proof it remains to consider the second case. First we observe that u is necessarily bounded on ℝ+2. Indeed, by [24, Theorem 2.1], there exists a positive constant C, depending only on p,f and the euclidean dimension, such that
(7.1)
u
(
x
,
y
)
≤
C
(
1
+
dist
-
2
/
(
p
-
1
)
(
(
x
,
y
)
;
∂
ℝ
+
2
)
)
=
C
(
1
+
y
-
2
/
(
p
-
1
)
)
for all
(
x
,
y
)
∈
ℝ
+
2
,
and therefore the boundedness of u follows by combining the monotonicity of u, i.e., ∂u∂y>0 on ℝ+2, together with estimate (7.1).Then, by standard elliptic estimates, we also get that |∇u| is bounded, and so we can apply Theorem 1.6 of our previous work [16] to get that u is one-dimensional. This concludes the proof.∎
Proof of Corollary 2.4.
If u is a solution to (1.1), by the strong maximum principle, we have that either u≡0 or u>0.Then, by proceeding as in the proof of Theorem 2.3, we get that either u≡0 or u is a positive, bounded, one-dimensional and monotonically increasing function, say u=u(y).The second case is impossible since l¯:=limy→+∞u(y) would be a positive zero of f, contradicting assumption (2.3).Therefore, if u is a solution, then necessarily u≡0, and so f(0)=0. This completes the proof.∎
Proof of Theorem 2.5.
By assumption, f≥0 on [0,+∞), and so either u≡0 or u>0, thanks to the strong maximum principle. Since the linear function u(x,y)=cy, c≥0, is harmonic, to conclude the proof of the first claim it is enough to show that u>0 implies u(x,y)=cy for some c>0 (which in turn implies that f≡0).To this end, we first observe that
(7.2)
v
:=
∂
y
u
>
0
on
ℝ
+
2
¯
.
Indeed, u>0 on ℝ+2 implies v>0 on ℝ+2 by Theorem 1.1, and v>0 on ∂ℝ+2 by Hopf’s Lemma, since u>0 on ℝ+2 and f(0)≥0 by assumption.
Then we remark that for every r>0, every p∈ℝ+2¯ and every open ball Br(p),
u
∈
H
3
(
B
r
(
p
)
∩
ℝ
+
2
)
,
by standard elliptic regularity (see, for instance, [23, Theorem 8.13]), and so we get that
(7.3)
v
∈
C
1
(
ℝ
+
2
¯
)
,
v
∈
H
2
(
B
r
(
p
)
∩
ℝ
+
2
)
for all
r
>
0
and all
p
∈
ℝ
+
2
¯
.
Now, since f is locally Lipschitz continuous, by differentiating the equation satisfied by u and using (7.2) and (7.3), we obtain that v satisfies
v
>
0
everywhere in
ℝ
+
2
¯
,
-
Δ
v
=
f
′
(
u
)
v
≥
0
a.e. in
ℝ
+
2
.
Then, for any ψ∈Cc0,1(ℝ2), we multiply the latter equation by ψ2v-1 and integrate by parts to get
0
≤
-
∫
ℝ
+
2
Δ
v
ψ
2
v
-
1
=
-
∫
ℝ
+
2
|
∇
v
|
2
v
2
ψ
2
+
∫
ℝ
+
2
2
v
-
1
ψ
∇
v
∇
ψ
+
∫
∂
ℝ
+
2
∂
v
∂
y
v
-
1
ψ
2
.
Now we observe that
∂
v
∂
y
=
∂
2
u
∂
y
2
=
Δ
u
=
-
f
(
0
)
≤
0
on
∂
ℝ
+
2
,
by assumption (2.4), and therefore we deduce from the latter that
∫
ℝ
+
2
|
∇
v
|
2
v
2
ψ
2
≤
∫
ℝ
+
2
2
v
-
1
ψ
∇
v
∇
ψ
.
Then
∫
ℝ
+
2
|
∇
v
|
2
v
2
ψ
2
≤
2
[
∫
ℝ
+
2
|
∇
v
|
2
v
2
ψ
2
]
1
/
2
[
∫
ℝ
+
2
|
∇
ψ
|
2
]
1
/
2
,
which gives
(7.4)
∫
ℝ
+
2
|
∇
v
|
2
v
2
ψ
2
≤
4
∫
ℝ
+
2
|
∇
ψ
|
2
.
Now, for every R>1, consider the functions ψR∈Cc0,1(ℝ2) given by
ψ
R
(
x
)
:=
𝟏
B
R
(
x
)
+
2
ln
(
R
/
|
x
|
)
ln
R
𝟏
B
R
∖
B
R
(
x
)
for all
x
∈
ℝ
2
,
and used them into (7.4) to get
(7.5)
∫
ℝ
+
2
|
∇
v
|
2
v
2
ψ
R
2
≤
4
∫
ℝ
+
2
|
∇
ψ
R
|
2
≤
C
log
R
,
where C is a positive constant independent of R.By letting R→+∞ in (7.5), we find ∫ℝ+2|∇v|2v2=0, and so v is a positive constant, say v≡c>0.This means that ∂yu≡c>0, and so u(x,y)=cy for every (x,y)∈ℝ+2.Therefore, 0=-Δ(cy)=f(cy) for every y>0, and thus f≡0, since c>0.This concludes the proof of the first claim.
In view of the discussion above, if f≢0, the only solution of (1.1) is u≡0, which immediately implies item (i) and item (ii).∎
8 A Counterexample
In this section we provide a counterexample showing that the conclusion of Theorem 2.1 (and of Theorem 2.2) fails if f satisfies (A1), (A2) but not (A3), i.e., the monotonicity property ∂u∂y>0 in ℝ+2 does not hold true if f satisfies (A1), (A2) but not (A3).To this end, we shall follow [13, Section 6].With the notation of [13, Example 6.3], the function
u
(
x
,
y
)
:=
{
u
1
(
x
,
y
)
,
y
≤
2
,
v
(
x
,
y
-
5
)
,
y
>
2
,
given by [13, formula (6.9)], with s=0 and x0=(0,5)∈ℝ2, is a smooth entire solution of the equation -Δu=h(u) in ℝ2, where h is given by [13, formula (6.8)]. Observe that u is identically zero on the closed affine half-plane {(x,y)∈ℝ2:y≤-1} and positive on the open affine half-plane {(x,y)∈ℝ2:y>-1}.Therefore, the function v(x,y):=u(x,y-1) is a solution of
{
-
Δ
v
=
h
(
v
)
in
ℝ
+
2
,
v
>
0
in
ℝ
+
2
,
v
=
0
on
∂
ℝ
+
2
,
which is neither monotone nor one-dimensional.On the other hand, h (extended to be equal to the constant h(2)=0 for t≥2) is a function satisfying (A1), (A2) but not (A3).Indeed, h is Holder continuous on [0,+∞), and so it satisfies (A1).Moreover, h fulfills (A2), since it is non-increasing in a neighborhood of the points 0,1 and 2, and smooth on [0,+∞)∖{0,1,2}.Finally, let us prove that h does not satisfy assumption (A3) at t¯=0.To this end, we first observe that h(t)=-192[t(1-t1/4)]1/2[1-54t1/4] for t∈[0,1] (cf. [13, Example 6.1]), and we suppose, for contradiction, that h satisfies (A3) at t¯=0.Hence, there exists a function g such that
(8.1)
h
(
s
)
-
h
(
t
)
≤
g
(
t
-
s
)
for all
s
≤
t
∈
[
0
,
δ
2
]
for some δ∈(0,1), fulfilling the integral condition (2.2).By choosing s=0 in (8.1), we have
-
h
(
t
)
≤
g
(
t
)
for all
t
∈
[
0
,
δ
2
]
and, in view of the explicit form of h near zero, we can find η∈(0,δ2) small enough such that
g
(
t
)
≥
-
h
(
t
)
=
|
h
(
t
)
|
≥
γ
t
1
/
2
for all
t
∈
[
0
,
η
]
,
for some γ>0.The latter yields G(s)≥23γs3/2 in [0,η] and so ∫0ηdsG(s)<∞, contradicting (2.2). So, assumption (A3) is not satisfied at t¯=0 (also note that a similar argument shows that h does not satisfy (A3) neither at 1 nor at 2). Clearly, the same example can be used as a counterexample for Theorem 2.2.
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and
Berardino Sciunzi
