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ScienceDirect J. Differential Equations 259 (2015) 4356–4375 www.elsevier.com/locate/jde

A note on the strong maximum principle Michiel Bertsch a,b , Flavia Smarrazzo c , Alberto Tesei b,∗ a Dipartimento di Matematica, Università di Roma “Tor Vergata”, Via della Ricerca Scientifica, 00133 Roma, Italy b Istituto per le Applicazioni del Calcolo “M. Picone”, CNR, Roma, Italy c Università Campus Bio-Medico di Roma, Via Alvaro del Portillo 21, 00128 Roma, Italy

Received 28 July 2014; revised 23 April 2015 Available online 30 June 2015

Abstract We give a necessary and sufficient condition for the validity of the strong maximum principle in one space dimension. © 2015 Elsevier Inc. All rights reserved. MSC: 35B50; 35D30; 35J15 Keywords: Strong maximum principle; Singular elliptic equations; Weak solutions; Differential inequalities

1. Introduction The maximum principle is an important feature of second-order elliptic equations. Although its weak form is sufficient for many purposes (e.g., see [12,19]), the strong maximum principle, which excludes the assumption of a nontrivial interior maximum, is often important. Its classical form can be stated as follows. Let ⊆ RN be a connected, open set with smooth boundary ∂. Consider a uniformly elliptic operator of the form Lu ≡ −

N

aij

i,j =1

N ∂ 2u ∂u + bi + cu ∂xi ∂xj ∂xi i=1

* Corresponding author.

E-mail addresses: [email protected] (M. Bertsch), [email protected] (F. Smarrazzo), [email protected] (A. Tesei). http://dx.doi.org/10.1016/j.jde.2015.05.022 0022-0396/© 2015 Elsevier Inc. All rights reserved.

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with coefficients aij = aj i , bi , c ∈ C() ∩ L∞ () (i, j = 1, . . . , N ). Let c ≥ 0 in , and let u ∈ C 2 () ∩ C() satisfy the inequality Lu ≤ 0 in . Then the strong maximum principle asserts that, if there exists x0 ∈ such that u(x0 ) = sup u =: M ≥ 0, then u ≡ M in . It follows that, if Lu ≥ 0 in and there exists x0 ∈ such that u(x0 ) = inf u =: m ≤ 0, then u ≡ m in . In particular, by the strong maximum principle, if u ≥ 0 and Lu ≥ 0 in , then either u > 0 in , or u ≡ 0 in . Much effort has been made to extend the above result in two main directions: (i) addressing degenerate elliptic operators, for which the assumptions of uniform ellipticity, and/or regularity and boundedness of the coefficients aij , bi , c are not satisfied; (ii) relaxing the assumptions concerning the boundary ∂. In connection with (i), weak or distributional formulations of the differential inequality Lu ≤ 0 have been used. Let us recall some results concerning degenerate elliptic operators. Let aij ∈ C 2 (), bi ∈ 1 C (), D α aij ∈ L∞ () for |α| ≤ 2, D α bi ∈ L∞ () for |α| ≤ 1; let u ∈ C 2 () satisfy Lu ≤ 0 in . For any x0 ∈ such that u(x0 ) = sup u > 0, consider the propagation set P(x0 ) := {x ∈ | u(x) = u(x0 )}. As proven in [24], P(x0 ) contains the closure (in the relative topology) of the set P (x0 ) consisting of points, which can be joined to x0 by a finite number of subunitary and/or drift trajectories (see [6,9,16,21] for the proof in particular cases; see also [1]). Remarkably, the set P (x0 ) coincides with the support of the Markov process corresponding to the operator L – namely, with the closure of the collection of all trajectories of a Markovian particle, starting at x0 , with generator L (see [23,24]). Therefore, only the attainable points of the boundary ∂ (see [10,11]) belong to P (x0 ). A similar idea underlies the so-called refined maximum principle proven in [3], where the operator L is uniformly elliptic, but no smoothness of the boundary ∂ is assumed. Consider the subset of E ⊆ ∂ consisting of attracting boundary points, where the minimal positive solution U0 of the first exit time equation −

N i,j =1

N ∂ 2u ∂u aij + bi = 1 in ∂xi ∂xj ∂xi

(1.1)

i=1

can be prolonged to zero (see [13,14]). It was proven in [3] that, if u is bounded from below in , u ≥ 0 in E and Lu ≥ 0 in , then u ≥ 0 in (in particular, in the usual situation the weak maximum principle is recovered). In the above literature boundedness of the coefficients aij , bi , c is always assumed. Concerning unbounded coefficients, while referring the reader to [8,15,20] and the references therein, let us focus on the following result. It is known that the strong maximum principle holds for √ p the operator Lu = −u + cu, if c ≥ 0 is measurable and c ∈ Lloc () for some p > N (see [22, Corollary 8.1]). More generally, the following holds (see [2], [7, Theorem 1], [17, Proposition 5.20]): Theorem 1.1. Let ⊆ RN be connected and open. Let c ∈ L1 (), c ≥ 0 a.e. in , and let u ∈ C() ∩ L1 () be nonnegative in . If −u + cu ≥ 0 in D () and u = 0 in a set of positive Newtonian capacity, then u ≡ 0 in . √ It was asked in [7] whether the condition c ∈ L1 () can be relaxed, e.g. by assuming c ∈ L1loc (). It is proven below that this is not the case (see Remark 2.2). This follows from the main

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result of this note (see Theorem 2.1): a necessary and sufficient condition on c for the validity of the strong maximum principle for the operator Lu = −u

+ cu, if N = 1. Apart from the intrinsic interest of the problem, the present study is motivated by a nonlinear evolution problem considered in [4,5], where a sharp version of the strong maximum principle is needed. Our proof is based on the classical Jacobi method of multiplicative variation, which relies on the construction of a proper solution U of the equation U

+ c = 0 in I ,

(1.2)

where I ⊆ is a bounded interval. This solution is a supersolution of the problem

−u

+ cu = 0 in I u=0 on ∂I

(see Subsection 3.2, where U = −ψ). Similar methods have been used to study the wellposedness of boundary value problems for elliptic and parabolic equations with singular coefficients (see [18,20] and the references therein). Observe that, by the connection between the operator −u

+ cu, with c unbounded, and the degenerate elliptic operator − 1c u

, equation (1.2) is the counterpart of the first exit time equation (1.1). Let us stress that both Theorem 2.1 and its proof heavily depend on assuming N = 1. If N > 1, the problem of the optimal condition on the coefficient c, which ensures the strong maximum principle for the operator −u + cu to hold, is to our knowledge completely open. 2. The main result In the sequel we shall always assume that

⊆ R is a nonempty open interval , c : → R is measurable, and c ≥ 0 a.e. in .

(H )

Let us state the following definitions. Definition 2.1. Let assumption (H ) be satisfied. A function u ∈ C(), u ≥ 0 is said to satisfy the inequality −u

+ cu ≥ 0 in D () ,

(2.1)

if cu ∈ L1loc () and

uρ dx ≤

cuρ dx

for any ρ ∈ Cc∞ (), ρ ≥ 0 .

(2.2)

Definition 2.2. Let assumption (H ) be satisfied. The strong maximum principle is said to hold in for the operator Lu = −u

+ cu, if for any nonnegative function u ∈ C() inequality (2.1) implies that either u ≡ 0, or u > 0 in .

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For any measurable function c : → R, c ≥ 0 set N := x0 ∈ | δ > 0 such that c ∈ L1 (Iδ (x0 )) ,

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(2.3)

where Iδ (x0 ) ≡ (x0 − δ, x0 + δ) with δ > 0 sufficiently small. Observe that \ N is open, and N = ∅ if and only if c ∈ L1loc (). Let us denote by dN (x) the distance dist(x, N ), for any x ∈ . The main result of this note is the content of the following Theorem 2.1. Let assumption (H ) be satisfied, and let the set N be defined by (2.3). The following statements are equivalent: (i) the strong maximum principle holds in for the operator Lu = −u

+ cu; (ii) either N = ∅, or N = ∅ and for any connected component (α, β) of \ N the map ⎧ 1 if β ∈ ∂ , ⎪ ⎨ Lloc ([α, β)) 1 x → dN (x)c(x) belongs to Lloc ((α, β]) if α ∈ ∂ , ⎪ ⎩ 1 1 Lloc ([α, β)) ∪ Lloc ((α, β]) if [α, β] ⊂ .

(2.4)

Remark 2.1. It follows from the proof of Theorem 2.1 in Section 4 that if N = ∅ and condition (2.4) is satisfied for any connected component of \ N , not only the strong maximum principle holds, but every function u ∈ C(), u ≥ 0 satisfying inequality (2.1) is identically zero in . −λ Remark 2.2. Let = (−1, 1) and c(x) = x −2 log |x| . Then N = {0} and \ N = (−1, 0) ∪ (0, 1). If λ > 1, the map x → |x|c(x) belongs to L1loc () whence, by Theorem 2.1, the strong √ maximum principle holds. On the other hand, c ∈ L1loc () if and only if λ > 2. Therefore, for √ λ ∈ (1, 2) the strong maximum principle holds, yet the function c does not belong to L1loc (). √ Hence the condition c ∈ L1loc () is not necessary for the validity of the strong maximum principle. √ The condition c ∈ L1loc () is not sufficient either. In fact, the coefficient n4 if n ≤ y ≤ n + n14 (n ∈ N) b log |x| c(x) = , where b(y) := 1 otherwise, x2 1+y 4 satisfies

√ c ∈ L1loc () with = (−1, 1), since 1 2

1

c(x) dx =

−1

∞ 0

∞ 1 dy b(y) dy ≤ + < ∞. 2 n 1 + y4 n=1 ∞

0

On the other hand, by Theorem 2.1 the strong maximum principle fails, since the map x → |x|c(x) does not belong to L1loc ([0, 1)): 1/e ∞ xc(x) dx = b(y) dy = ∞ . 0

1

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Remark 2.3. In Theorem 2.1 the inequality −u

+ cu ≥ 0 is meant in the sense of distributions (see Definition 2.1), but in principle its definition could be weakened. In particular, one could avoid the condition cu ∈ L1loc () by saying that the inequality −u

+ cu ≥ 0 is satisfied, if for any function ρ ∈ Cc∞ (), ρ ≥ 0 either cuρ ∈ / L1 (), or cuρ ∈ L1 () and uρ

dx ≤ cuρ dx .

(2.5)

To check that (2.5) is weaker than Definition 2.1, choose = (−1, 1), c(x) = |x|−1−α and u(x) = |x|α for some α ∈ (0, 1). Therefore cu ∈ / L1loc (), thus inequality (2.1) is not satisfied in the sense of Definition 2.1. On the other hand, condition (2.5) is satisfied: if cuρ = |x|−1 ρ ∈ L1 (), then ρ(0) = 0 and, since ρ ≥ 0 and ρ ∈ Cc∞ (), there holds ρ(x) ≤ Cx 2 for some constant C > 0. Therefore, 1

|x| [−ρ (x) + |x| α

−1−α

1 ρ(x)] dx = α

−1

x|x|

1

α−2

ρ (x) dx +

−1

|x|−1 ρ(x) dx =

−1

1 = α(1 − α)

1 |x|

α−2

ρ(x) dx +

−1

|x|−1 ρ(x) dx ≥ 0 .

−1

It also follows from the above example that Theorem 2.1 is false if we replace (2.1) by (2.5) in Definition 2.2 . In fact, the function u(x) = |x|α satisfies u(0) = 0 and u ≡ 0 in , whereas |x|c(x) = |x|−α with α ∈ (0, 1), hence condition (2.4) is satisfied with N = {0}. In view of Remark 2.3, we introduce another condition on u: for any compact interval K ⊂ , u(x) ≤ C|x − y| if u(y) = 0 and x, y ∈ K .

(L)

Let us also state the following definition. Definition 2.3. A nonempty subset N ⊂ is called locally finite, if its intersection N ∩ K with any compact interval K ⊂ is a finite set. Then Theorem 2.1 can be modified as follows. Theorem 2.2. Let (H ) be satisfied, and let N be defined by (2.3). The following statements are equivalent: (i) for all u ∈ C(), u ≥ 0 which satisfy (2.5) and condition (L), either u ≡ 0, or u > 0 in ;

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(ii) either N = ∅, or N is locally finite and the map x → dN (x)c(x) belongs to L1loc () .

(2.6)

Observe that: a) condition (2.6) is stronger than (2.4); b) condition (L) excludes the counterexample u(x) = |x|α (α ∈ (0, 1)) discussed in Remark 2.3. 3. Preliminaries 3.1. Notation We denote by M() the space of finite Radon measures on . For every μ ∈ M(), we denote by μs the singular part of μ, and by μr the density of the absolutely continuous part μac of μ – namely, according to the Radon–Nikodym Theorem, μr is the unique function in L1 () such that μac (E) = μr dx for all Borel sets E ⊆ . E

We denote by ·, · the duality map between M() and the space Cc () of continuous functions with compact support. If μ ∈ M() and ρ ∈ L1 (, μ) we set, by abuse of notation, μ, ρ :=

ρ(x) dμ(x) .

(3.1)

For any Borel set E ⊆ , we denote by L1loc (E) the set of functions f : E → R such that f ∈ L1 (K) for any compact subset K ⊆ E. 3.2. Some auxiliary functions Set I = (−1, 1), I0 = I \ {0}. Let the function a : I → R satisfy 1 a ≥ 0,

x → |x|a(x) ∈ L (I ),

|x|a(x) dx < 1 .

1

(A)

−1

Observe that assumption (A) implies that a ∈ L1loc (I0 ). Consider the problem ⎧

⎨ −ψ + a = 0 in I0 ψ(0) = 1 ⎩ ψ(±1) = 0 .

(3.2)

2,1 Definition 3.1. A function ψ ∈ C(I¯) ∩ Wloc (I0 ) is called a solution of problem (3.2) if it satisfies (3.2) in the strong sense.

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Lemma 3.1. Let assumption (A) be satisfied. Then the unique solution of problem (3.2) is the function ⎧ 1 ⎪ ⎪ ⎪ ⎪ ⎪ (y − x)a(y) dy if x ∈ [0, 1] , C(1 − x) + ⎪ ⎪ ⎨ x ψ(x) := x ⎪ ⎪ ⎪ ⎪ ⎪ D(1 + x) + (x − y)a(y) dy if x ∈ [−1, 0) , ⎪ ⎪ ⎩ −1

with 1 C := 1 −

0 D := 1 +

ya(y) dy ,

ya(y) dy .

−1

0

In particular, 1

ψ (x) = −C −

a(y) dy

(x ∈ (0, 1)) ,

x

xψ (x) → 0 as x → 0 , ψ (x) → −∞

as x → 0+ if and only if a ∈ / L1 (I ) .

(3.3)

Proof. By condition (A), C and D are positive constants, so ψ < 0 in (0, 1) and ψ > 0 in (−1, 0). Below we show that 1 a(y) dy = 0 ,

lim x

x→0+

(3.4)

x

thus xψ (x) → 0 as x → 0. The remainder of the proof is straightforward, thus we omit it. Consider the function 1 φ(x) :=

(y − x)a(y) dy

(x ∈ (0, 1)) .

x

Since

1

φ (x) = −

a(y) dy ≤ 0 in (0, 1) , x

the function φ is decreasing, hence there exists limx→0+ φ(x) =: φ(0). Moreover, by (3.5), 1 0 ≤ φ(0) ≤

ya(y) dy < ∞ . 0

(3.5)

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Letting x → 0+ in equality (3.5), we find that 1 k := lim x x→0+

1 a(y) dy = −φ(0) +

x

ya(y) dy 0

is well-defined and finite. By contradiction, let us suppose that k > 0. Then there exists ε > 0 such that 1 a(y) dy ≥

k 2x

for every x ∈ (0, ε) ,

x

whence ε

1 a(y) dy ≥

dz x

k ε log 2 x

(x ∈ (0, ε)) .

z

By letting x → 0+ in the above inequality we obtain ε xa(x) dx = ∞ , 0

which contradicts (Aε ). Hence equality (3.4) follows.

2

Let ψ the solution of problem (3.2), and let η ∈ (0, 1) be fixed. Consider the problem

ψ 2ξ + χ = 0 ξ(±η) = 0 ,

in (−η, η)

(3.6)

where χ ∈ C(I ), χ ≥ 0 and supp χ ⊂ (−η, η). The unique classical solution of problem (3.6) is given by x ξ(x) := −η

⎫ ⎧ y ⎬ dy ⎨ χ(z) dz + B − ⎭ ψ 2 (y) ⎩

(x ∈ [−η, η]) ,

(3.7)

−η

where η B :=

dy y −η ψ 2 (y) −η χ(z) dz η dy −η ψ 2 (y)

(3.8)

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(observe that both ξ and ψ 2 ξ belong to C 1 ([−η, η])). In particular, η 2

(ψ ξ )(−η) = B > 0 ,

2

(ψ ξ )(η) = −

dy η −η ψ 2 (y) y χ(z) dz η dy −η ψ 2 (y)

=: −B˜ < 0 .

(3.9)

Now let us define ζ : I → R by setting ζ :=

ξ

in [−η, η]

0

in I \ [−η, η] .

(3.10)

By the properties of ξ , the function ψ 2 ζ is continuously differentiable in the set I \ {±η} with derivative −χ , whereas it has a jump discontinuity at ±η. Therefore, its derivative (ψ 2 ζ ) belongs to M(I ), and by equalities (3.9) there holds (ψ 2 ζ ) = −χ + B δ−η + B˜ δη

in M(I )

(3.11)

(here δx denotes the Dirac measure concentrated at x ∈ I ; by abuse of notation, we have identified the absolutely continuous part of (ψ 2 ζ ) with its density). Proposition 3.2. Let assumption (A) be satisfied, and let w ∈ C(I ), w ≥ 0. Given η ∈ (0, 1), let ζ be defined by (3.10). If (ψ 2 ζ ) , w ≤ 0 ,

(3.12)

I

then w(η)

w(x) ≥ W (x) :=

x

η dy dy −η ψ 2 (y) + w(−η) x ψ 2 (y) η dy −η ψ 2 (y)

for any x ∈ [−η, η] .

(3.13)

Proof. The function W in the right-hand side of inequality (3.13) is a classical solution of the problem

ψ 2 W = 0 in (−η, η) W (±η) = w(±η) .

By (3.6), (3.9) and (3.11), (ψ 2 ζ ) , W = − χ W dx + (ψ 2 ξ )(−η) W (−η) − (ψ 2 ξ )(η) W (η) = η

I

−η

η = −η

ψ 2 ξ W dx + (ψ 2 ξ )(−η) W (−η) − (ψ 2 ξ )(η) W (η) =

M. Bertsch et al. / J. Differential Equations 259 (2015) 4356–4375

η =−

w(−η) − w(η) ψ ξ W dx = η dy 2

−η

−η

ψ 2 (y)

η

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ξ dx = 0 .

−η

On the other hand,

2

(ψ ζ ) , w − W

I

η =−

˜ χ (w − W ) dx + B[w(−η) − W (−η)] + B[w(η) − W (η)] =

−η

η =−

χ (w − W ) dx .

−η

In view of (3.12), we have that − (ψ 2 ζ ) , w = χ (w − W ) dx ≥ 0 for all χ ∈ Cc (−η, η), χ ≥ 0 . η

I

−η

Then the conclusion follows from the arbitrariness of χ . 2 We conclude this subsection with the proof of the following result. Lemma 3.3. Assume that a ∈ L1loc ((0, 1]) ,

a ≥ 0,

x → xa(x) ∈ / L1 ((0, 1)) .

(B)

Then there exists u ∈ C 1 ([0, 1]) ∩ W 2,1 ((0, 1)) such that −u

+ au = 0 a.e. in (0, 1), u > 0 in (0, 1) and u(0) = u (0+ ) = 0. Proof. Let uk ∈ C 1 ((0, 1]) be the solution of

in (0, 1) u

= au u(1) = 1 , u (1) = k (k > 0) .

(3.14)

If k ≤ 1, there holds uk (x) ≥ x > 0 for any x ∈ (0, 1]. On the other hand, it is easily seen that ˜ = 0 for some x˜ ∈ (0, 1) if we choose k > 1 sufficiently large. Hence, by setting uk (x) A0 := {k > 1 | ∃ x˜ ∈ (0, 1) such that uk (x) ˜ = 0, u k (x) ˜ > 0} , we have that k ∗ := inf A0 ∈ (1, ∞). Since uk depends continuously on k (locally in (0, 1]), the / A0 . It follows at once from the definition of k ∗ that uk ∗ (0) = 0 and set A0 is open and k ∗ ∈

+ uk ∗ (0 ) ≥ 0. It remains to prove that u k ∗ (0+ ) = 0. By contradiction, suppose that u k ∗ (0+ ) > 0. Since uk ∗ is convex, u

k ∗ (x) = a(x)uk ∗ (x) ≥ u k ∗ (0+ )xa(x) for x ∈ (0, 1) .

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By integrating between x and 1, we find that

u k ∗ (x) ≤ k ∗

− u k ∗ (0+ )

1 ya(y) dy

for x ∈ (0, 1) .

x

Hence u k ∗ (0+ ) = −∞, which gives a contradiction. 2 3.3. Approximation of a in L1 () Let = I , and let the function a satisfy assumption (A). Since the solution ψ of problem / L1 (I ) (see (3.3)), we consider the solutions ψn of the ap(3.2) satisfies ψ (0± ) = ∓∞ if a ∈ proximating problems

−ψn

+ an = 0 in I0 ψn (0) = 1 , ψn (±1) = 0

(n ∈ N) ,

(3.15)

where {an } ⊂ L1 (I ) is any sequence such that 0 ≤ an ≤ a in I, and the maps x → xan (x) converge to x → xa(x) in L1 (I ) .

(An )

An example is an = aχ(−1,− 1 )∪( 1 ,1) n

(n ∈ N) ,

n

where χE is the characteristic function of a Borel set E. Since Lemma 3.1 can be applied to problem (3.15), we have that ψn ∈ C(I¯) ∩ C 1 (I¯ \ {0}) and ψn (0+ ) − ψn (0− ) := lim ψn (x) − lim ψn (x) = x→0+

x→0−

(1 − |y|) an (y) dy =: −jn < 0 .

=− I

In particular, ψn

∈ M(I ) and

ψn

s

= −jn δ0 ,

ψn

r

= an ∈ L1 (I ) .

If we identify, by abuse of notation, the absolutely continuous part of ψn

with its density, this implies that ψn

= an − jn δ0

in M(I ) .

(3.16)

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Observe that, by assumption (A) and Lemma 3.1, sup |ψn (x) − ψ(x)| = x∈I

1 1 = sup −(1 − x) y(a − an )(y) dy + (y − x)(a − an )(y) dy ≤ x∈(0,1) x

0

1 ≤2

y(a − an )(y) dy → 0 as n → ∞ . 0

Therefore, ψn → ψ

in C(I ) .

(3.17)

Finally, let η ∈ (0, 1), let χ be as in (3.6), and let ζn : I → R be defined by ζn :=

ξn 0

in [−η, η] in I \ [−η, η]

(3.18)

(n ∈ N) ,

where ξn is the solution of the problem

ψn2 ξ + χ = 0

in (−η, η)

ξ(±η) = 0 . By (3.7) and (3.8), x ξn (x) = −η

⎫ ⎧ y ⎬ ⎨ dy χ(z) dz + B − n ⎭ ψn2 (y) ⎩

for x ∈ [−η, η]

−η

with dy y −η ψn2 (y) −η χ(z) dz η dy −η ψn2 (y)

η Bn :=

,

and dy η −η ψn2 (y) y χ(z) dz η dy −η ψn2 (y)

η (ψn2 ξn )(−η) = Bn > 0 ,

(ψn2 ξn )(η) = −

=: −B˜ n < 0 .

Clearly, the function ζn has the same regularity as ζ , and (ψn2 ζn ) = −χ + Bn δ−η + B˜ n δη

in M(I ) .

(3.19)

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By (3.17) there holds ζn → ζ in C(I¯) ,

(3.20)

and Bn → B ,

B˜ n → B˜ .

(3.21)

Proposition 3.4. Let the function c : (0, 1) → R satisfy 1 x → xc(x) ∈ L ([0, 1)),

c ≥ 0,

1

xc(x) dx < 1 .

(A )

0

Let u ∈ C([0, 1)), u ≥ 0, satisfy cu ∈ L1loc ((0, 1)) and −u

+ cu ≥ 0 in D ((0, 1)) .

(3.22)

If u(0) = u (0+ ) = 0, then u ≡ 0 in (0, 1). Proof. By contradiction, suppose that u(η) > 0 for some η ∈ (0, 1). We extend u and c to I = (−1, 1) by setting

u in [0, 1) u˜ := 0 in (−1, 0) ,

a :=

c 0

in [0, 1) in (−1, 0) .

Then u˜ ∈ C([−1, 1]), and the function a satisfies assumption (A) (see Subsection 3.2). Since u (0+ ) = 0, it follows from a standard approximation procedure and (3.22) that 1

1

uρ dx ≤ 0

cuρ dx

for any ρ ∈ Cc2 ([0, 1)), ρ ≥ 0 .

0

Therefore, a u˜ ∈ L1loc (I ) and −u˜

+ a u˜ ≥ 0 in D (I ) .

(3.23)

Set w :=

u˜ ψ

in I ,

(3.24)

where ψ is the solution of problem (3.2). Since by assumption u(η) > 0, we also have that w(η) > 0. Moreover, if w satisfies (3.12) for any ζ defined by (3.10),

(3.25)

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it follows from Proposition 3.2 that w(x) > 0 for any x ∈ [−η, η]. In particular, there holds u(0) ˜ = u(0) > 0, and we have found a contradiction. To prove (3.25) we set wn :=

u˜ ψn

in I ,

where ψn is the solution of problem (3.15). First we prove that for any n ∈ N 2

(ψn ζn ) , wn ≤ (a − an )u ζn dx for any n ∈ N , I

(3.26)

I

where ζn is defined by (3.18). To this purpose, observe that (ψn ζn )

, u˜ I = ψn

ζn + 2ψn ζn + ψn ζn

, u˜ I = = an u˜ ζn dx + 2ψn ψn ζn + ψn2 ζn

, wn = I

I

an u˜ ζn dx + (ψn2 ζn ) , wn ;

=

I

(3.27)

I

here we have used the equality

ψn

ζn , u˜ I =

an u˜ ζn dx − jn δ0 , u ˜ I=

I

an u˜ ζn dx , I

which follows from (3.16) since, by assumption, u(0) ˜ = u(0) = 0. Recall that, by inequality (3.23),

u(−ρ ˜ + aρ) dx ≥ 0 for any ρ ∈ Cc∞ (I ), ρ ≥ 0 . I

It follows by a standard approximation procedure that we can relax the regularity of ρ:

− ρ , u˜ I + a uρ ˜ dx ≥ 0 for any ρ ∈ Cc (I ), ρ ≥ 0, with ρ ∈ BVloc (I ) .

(3.28)

I

In fact, let j ∈ Cc∞ (I ), j ≥ 0, denote a standard mollifier ( > 0). Then for any ρ as in (3.28) (j ∗ ρ)

= j

∗ ρ → ρ

in D () as → 0+ , and, by (2.1),

u(−j ˜ ∗ ρ + aj ∗ ρ) dx ≥ 0 for any > 0 .

By letting → 0+ we obtain (3.28).

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In particular, we can choose ρ = ψn ζn in (3.28) (n ∈ N). By (3.27), this gives 0 ≤ − (ψn ζn )

, u˜ I + a uψ ˜ n ζn dx ≤ − (ψn2 ζn ) , wn + (a − an )u˜ ζn dx , I

I

I

since ψn ≤ 1 in I . Hence we have proven (3.26). By (3.11), (3.17), (3.19), (3.21) and the definition of u, ˜ ⎫ ⎧ ⎨ χ u˜ u(η) ⎬ lim (ψn2 ζn ) , wn = lim − dx + B˜ n = n→∞ n→∞ ⎩ I ψn ψn (η) ⎭ =−

I

χ u˜ u(η) 2 dx + B˜ = (ψ ζ ) , w . I ψ ψ(η)

I

On the other hand, by assumption (An ) and since a u˜ ∈ L1loc (I ), it follows from the Monotone Convergence Theorem and (3.20) that 0 ≤ (a − an )u˜ ζn dx ≤ (a − an )ζ u˜ dx + a|ζn − ζ |u˜ dx → 0 as n → ∞ . I

I

I

Then letting n → ∞ in (3.26) gives (3.12). This proves the result.

2

Remark 3.1. We shall use Proposition 3.4 in the proof of Theorem 2.1, where 0 can be thought of as a point of the set N and I ⊂ as a neighbourhood of 0 such that I ∩ N = {0} (a similar result can be stated and proven in the interval (−1, 0]). In this connection observe that the choice of the interval [0, 1) in Proposition 3.4, of (−1, 1) in Lemma 3.1, and the integral conditions in (A), (A ) are not restrictive. In fact, if a ≥ 0 and the map x → |x|a(x) is integrable in some neighbourhood of 0, we can always choose δ > 0 such that a satisfies the condition δ x → |x|a(x) ∈ L ((−δ, δ)),

a ≥ 0,

|x|a(x) dx < 1 .

1

−δ

Then the unique solution of

−ψ

+ a = 0 in (−δ, 0) ∪ (0, δ) ψ(0) = 1, ψ(±δ) = 0

is given by

ψδ (x) :=

⎧ δ ⎪ ⎪ ⎪ ⎪ ⎪ Cδ (δ − x) + (y − x)a(y) dy ⎪ ⎪ ⎨

if x ∈ [0, δ]

x

x ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ Dδ (δ + x) + (x − y)a(y) dy ⎪ ⎩ −δ

if x ∈ [−δ, 0) ,

(Aδ )

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where ⎞ ⎛ δ 1⎝ 1 − ya(y) dy ⎠ , Cδ = δ 0

⎞ ⎛ 0 1⎝ Dδ = 1 + ya(y) dy ⎠ . δ −δ

4. Proof of the main result To prove Theorem 2.1 we shall make use of the following lemma. Lemma 4.1. Let u ∈ C(), u ≥ 0 satisfy inequality (2.1). If u(x) ¯ = 0 for some x¯ ∈ , the derivative u (x) ¯ exists, and there holds u (x) ¯ = 0. Proof. By inequality (2.1) the distribution −u

+ cu is nonnegative, hence it is a nonnegative Radon measure over . Since cu ∈ L1loc () (see Definition 2.1), it follows that u

is a Radon measure in , and its singular part (u

)s is nonpositive. Since u

is a Radon measure in , there holds u ∈ BVloc (). Let u(x) ¯ = 0 for some x¯ ∈ . Since by assumption u ≥ 0 in , and u ∈ BVloc (), both derivatives u (x¯ + ) ≥ 0 and u (x¯ − ) ≤ 0 exist and are finite. Hence there holds (u

)s ({x}) ¯ ≥ 0. By the above remarks, it follows that (u

)s ({x}) ¯ = 0. This implies that u (x¯ + ) = u (x¯ − ) = 0, hence u (x) ¯ exists and u (x) ¯ = 0. 2 Proof of Theorem 2.1. (ii) ⇒ (i): Let us show that, if either N = ∅, or N = ∅ and for any connected component of \ N condition (2.4) is satisfied, the strong maximum principle holds in . Let N = ∅. Let u(x0 ) > 0 for some x0 ∈ , and let (x1 , x2 ) be the connected component of the open set {x ∈ |u(x) > 0} containing x0 . We must show that u > 0 in , namely (x1 , x2 ) = . By contradiction, suppose that x1 ∈ (the case x2 ∈ is similar). Then u(x1 ) = 0, whence by Lemma 4.1 u (x1 ) = 0. Moreover, since N = ∅, there holds c ∈ L1 (Iδ (x1 )) for some δ > 0, thus the assumptions of Proposition 3.4 (with [0, 1) replaced by [x1 , x1 + δ)) are satisfied. It follows that u ≡ 0 in [x1 , x1 + δ), which contradicts the definition of x1 . This proves the claim in this case. Now suppose that N = ∅, and for any connected component of \ N condition (2.4) is satisfied. Let us show that every nonnegative function u ∈ C() satisfying inequality (2.1) is identically zero in , thus in particular the strong maximum principle holds. In fact, let u ∈ C(), u ≥ 0 satisfy inequality (2.1), thus by Definition 2.1 there holds cu ∈ L1loc (). By the continuity of u, this implies that u(x) ¯ = 0 for every x¯ ∈ N (if u(x) ¯ > 0 for some ¯ where u > 0 strictly, whence c ∈ L1 (Iδ (x)), ¯ x¯ ∈ N , there would exist a neighbourhood Iδ (x) a contradiction). By Lemma 4.1, it follows that u (x) ¯ = 0. On the other hand, by (2.4) and the symmetry between α and β, we may assume that the map x → dN (x)c(x) belongs to L1 ([x, ¯ x¯ + δ)) for some δ > 0. Then by Proposition 3.4 (with [0, 1) replaced by [x, ¯ x¯ + δ)) there holds u ≡ 0 in [x1 , x1 + δ), whence the claim follows as in the case N = ∅. (i) ⇒ (ii): Suppose that the set N does not satisfy (2.4). Then there exists a connected component (α, β) of \ N such that one of the following holds:

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(1) α, β ∈ , (x − α)c(x) ∈ / L1loc ([α, β)) and (β − x)c(x) ∈ / L1loc ((α, β]); 1 (2) α ∈ ∂, β ∈ and (β − x)c(x) ∈ / Lloc ((α, β]); / L1loc ([α, β)). (3) α ∈ , β ∈ ∂ and (x − α)c(x) ∈ Let us prove that in this case the strong maximum principle does not hold. Consider case (1), the argument being analogous in the other cases. Let η1 , η2 > 0 satisfy / L1loc ([α, β)), by Lemma 3.3 there exists u1 ∈ C 1 ([α, α + η1 + η2 < β − α. Since (x − α)c(x) ∈ 2,1 η1 ]) ∩ Wloc ((α, α + η1 )) such that −u

1 + cu1 = 0 a.e. in (α, α + η1 ), u1 > 0 in (α, α + η1 ) and u1 (α) = u 1 (α + ) = 0. 2,1 Similarly, since (β − x)c(x) ∈ / L1loc ((α, β]), there exists u2 ∈ C 1 ([β − η2 , β]) ∩ Wloc ((β − η2 , β)) such that

−u

2 + cu2 = 0 a.e. in (β − η2 , β), u2 > 0 in (β − η2 , β) and u2 (β) = u 2 (β − ) = 0. Without loss of generality we may assume that u1 (α + η1 ) = u2 (β − η2 ) =: l. Set ⎧ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ u1 (x) u(x) = l ⎪ ⎪ ⎪ u2 (x) ⎪ ⎪ ⎪ ⎩ 0

if x < α , if α ≤ x ≤ α + η1 , if α + η1 < x < β − η2 , if β − η2 ≤ x ≤ β , if x > β .

It is easily seen that u ∈ C(), u ≥ 0 in , u > 0 in (α, β) in cu ∈ L1 (), (u

)s is nonpositive and −(u

)r + cu ≥ 0 a.e. in . Hence inequality (2.1) is satisfied, and the conclusion follows. 2 Proof of Theorem 2.2. (ii) ⇒ (i): Let N be nonempty, locally finite and satisfying (2.6). Let u ∈ C(), u ≥ 0 satisfy (2.5) and condition (L), and assume that u(x0 ) > 0 for some x0 ∈ . Let (x1 , x2 ) be the connected component of the open set {x ∈ |u(x) > 0} containing x0 . We must show that u > 0 in , namely (x1 , x2 ) = . By contradiction, suppose that x1 ∈ (the case x2 ∈ is similar). Then u(x1 ) = 0, and by condition (L) there holds u(x) ≤ C|x − x1 | for any x ∈ . By the local finiteness of N and condition (2.6), this implies that there exists a neighbourhood Iδ (x1 ) ≡ (x1 − δ, x1 + δ) such that cu ∈ L1 (Iδ (x1 )). In fact, 0≤ cu dx ≤ C |x − x1 |c(x) dx < ∞ . Iδ (x1 )

Iδ (x1 )

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It follows that for any function ρ ∈ Cc∞ (), ρ ≥ 0 such that supp ρ ⊆ Iδ (x1 )

cuρ dx =

cuρ dx ≤ ρ∞

Iδ (x1 )

cu dx < ∞ .

Iδ (x1 )

Then by (2.5) for any ρ as above

uρ

dx ≤

cuρ dx ,

namely −u

+ cu ≥ 0 in D (Iδ (x1 )) .

(4.1)

By the above remarks, we can adapt the proof of Proposition 3.4 to the present case. Hence we obtain that u ≡ 0 in the interval (x1 , x1 + δ), which contradicts the definition of x1 . This completes the proof of the claim in this case. If N is empty, we can make use of the above argument at any point of where u vanishes; hence the claim follows. (i) ⇒ (ii): Let (i) be satisfied, and let N = ∅. We argue by contradiction, by distinguishing two cases: either (C1 ) N is locally finite, and there exists a compact interval K = [α, β] ⊂ such that dN c ∈ / L1 (K), or (C2 ) N is not locally finite. In case (C1 ), there exist ξ ∈ K and η0 > 0 such that |x − ξ |c(x) ∈ / L1 (Uη0 ) and c ∈ \ {ξ }), where Uη0 = (ξ − η0 , ξ + η0 ). Without loss of generality we may assume that |x − ξ |c(x) ∈ / L1 ((ξ, ξ + η)). By Lemma 3.3, with I replaced by Uη0 , there exists u ∈ 1 C ([ξ, ξ + η0 ]) ∩ W 2,1 ((ξ, ξ + η0 )) such that u > 0 in (ξ, ξ + η0 ), u(ξ ) = u (ξ ) = 0, and −u

+ cu = 0 in D (Uη0 ). Extending u to by setting u(x) = 0 for x < ξ and u(x) = u(ξ + η0 ) for x > ξ + η0 we obtain a contradiction (it is enough to follow the proof of part (ii) of Theorem 2.1, without the request that cu ∈ L1loc ()). Next we consider case (C2 ). Without loss of generality we may assume that for some ξ ∈

L1loc (Uη0

c∈ / L1loc (Vη ) for each right neighbourhood Vη = (ξ, ξ + η) of ξ .

(4.2)

We fix η > 0. Then there exists x1 ∈ (ξ, ξ + η) such that c ∈ / L1 ((x1 , ξ + η)). By (4.2) with 1 Vη = (ξ, x1 ), there exists x2 ∈ (ξ, x1 ) such that c ∈ / L ((x2 , x1 )). By iteration, we find a strictly decreasing sequence {xn } converging to ξ such that xn−1

c(x)dx = ∞ for any n ∈ N . xn

Then for all n ∈ N there exists kn > 0 such that

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M. Bertsch et al. / J. Differential Equations 259 (2015) 4356–4375 xn−1

(Tkn c)(x)dx ≥ xn

1 , n(xn − ξ )

where (Tk c)(x) = min{c(x), k} for x ∈ (ξ, ξ + η). Let c(x) :=

∞ (Tkn c)(x)χ(xn ,xn−1 ) (x) for a.e. x ∈ (ξ, ξ + η) . n=1

Then c ∈ L1loc ((ξ, ξ + η)), 0 ≤ c ≤ c a.e. in (ξ, ξ + η) and (x − ξ )c ∈ / L1 (ξ, ξ + η): x ξ +η ∞ n−1 (x − ξ )c(x)dx = (x − ξ )(Tkn c)(x)dx ≥ n=1 xn

ξ

xn−1 ∞ ∞ 1 (xn − ξ ) (Tkn c)(x)dx ≥ ≥ = ∞. n n=1

n=1

xn

Arguing as in the proof of Lemma 3.3 (with the interval (0, 1) replaced by (ξ, ξ + η) and c by c) shows that there exists u ∈ C 1 ([ξ, ξ + η]) such that u > 0 in (ξ, ξ + η) and u is a strong solution of the problem ⎧

in (ξ, ξ + η) ⎨ u = cu u(ξ + η) = 1 ⎩ u(ξ ) = u (ξ ) = 0 . We extend the definition of u to \ [ξ, ξ + η]: u(x) =

0 if x ∈ , x < ξ 1 if x ∈ , x > ξ + η .

Then u ∈ C(), u

∈ M() and u

= c χ(ξ,ξ +η) − u ((ξ + η)− ) δη

in M() .

Therefore, for any ρ ∈ Cc∞ (), ρ ≥ 0,

u(−ρ

+ cρ) dx ≥ − u

, ρ +

c χ(ξ,ξ +η) ρu dx = u ((ξ + η)− )ρ(ξ + η) ≥ 0 .

To summarise, we have constructed a function u which is continuous and nonnegative in and satisfies (2.5) and condition (L), for which u = 0 for x ≤ ξ and u > 0 for x > ξ . In view of (i) we have found a contradiction. 2

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