Uniqueness and comparison principles for semilinear equations and inequalities in Carnot groups

Definition 3.1.

Let f∈Lloc1⁢(Ω). A distributional solution of the inequality

is a function u∈Lloc1⁢(Ω) such that for any nonnegative ϕ∈𝒞02⁢(Ω) we have that

Theorem 3.2 (Kato inequality).

Let u,f∈Lloc1⁢(Ω) be such that

Then

and

Lemma 3.3.

Let γ∈C2⁢(R) be a convex function with bounded first derivative. Let u,f∈Lloc1⁢(Ω) be such that

Then γ⁢(u)∈Lloc1⁢(Ω) and

Proof.

We need to prove that for any nonnegative ϕ∈𝒞02⁢(Ω) we have

and that the following inequality holds:

Fix ϕ∈𝒞02⁢(Ω). Let (mη)η be a family of symmetric mollifiers associated to a fixed homogeneous norm S. Set uη:=u⋆𝔾mη in Ωη:={x∈Ω:dist⁢(x,∂⁡Ω)>η}, that is,

For η small enough, it follows that supp⁡(ϕ)⊂Ωη.

Let x∈Ωη. Since mη(x∘⋅-1) is a nonnegative test function in Ω, by using the Fubini–Tonelli theorem we obtain

that is,^[1]

On the other hand, by the convexity of γ it follows that

which implies

The convergence of γ⁢(uη)→γ⁢(u) in Lloc1⁢(Ω) is assured by the convergence of uη→u in Lloc1⁢(Ω) and the fact that γ is a Lipschitz function (since γ′ is bounded). By observing that

it suffices to prove that

To this end, we first claim that

Indeed, since γ′ is continuous and uη→u a.e. in Ω (if necessary by passing to a subsequence), it follows that γ′⁢(uη)⁢ϕ→γ′⁢(u)⁢ϕ a.e. in Ω. Now, by γ′ being bounded, an application of the Lebesgue dominated convergence gives γ′⁢(uη)⁢ϕ→γ′⁢(u)⁢ϕ in L1⁢(Ω). Moreover,

Next, if necessary by passing to a subsequence, we may suppose that the convergence in (3.3) is a.e. on Ω.

Now, since γ′⁢(uη)⁢ϕ is uniformly bounded by M:=|γ′|∞⁢|ϕ|∞, we deduce that

Noticing that mη⋆𝔾(γ′⁢(uη)⁢ϕ) has compact support contained in suptϕ+Bη⊂suptϕ+B1=:K, it follows that

Finally, by the Lebesgue theorem we have

This completes the proof. ∎

Proof of Theorem 3.2.

The idea is first to approximate the function sign+ with a family of convex functions γϵ having bounded derivatives, and then apply Lemma 3.3 above.

Let m∈𝒞⁢(ℝ) be nonnegative with supt⁡(m)⊂[-1,1] and ∫m=1. For ϵ>0, set mϵ:=1ϵ⁢m⁢(t-ϵϵ) and consider γϵ as the solution of the problem

Clearly, we have γϵ⁢(t)=γϵ′⁢(t)=0 for t≤0. In addition, γϵ′⁢(t)=1 for t>2⁢ϵ and 0≤γϵ⁢(t)≤t+, 0≤γϵ′⁢(t)≤1. This implies the pointwise convergence, as ϵ→0, of γϵ⁢(t)→t+ and γϵ′⁢(t)→sign+⁡t. Finally, by Lemma 3.3 we have

and by the Lebesgue theorem we obtain

The proof of (3.2) follows from a similar argument as above, so we shall omit it. ∎

Remark 3.4.

Theorem 3.2 holds if we replace the functions sign+ and u+ respectively with

and uh+:=(u-h)+, where h∈ℝ. To this end, we can argue as in the proof of Theorem 3.2, replacing γϵ⁢(t) by γϵ⁢(t-h).

Remark 3.5.

Theorem 3.2 deals with Lloc1⁢(Ω) solutions of the inequality

while [8, Theorem 2.1] allows to consider Wloc1,2⁢(Ω) solutions.

One may try to prove (3.1) by mollifying the solution and then applying [8, Theorem 2.1]. In this case, one would obtain

Clearly, in order to prove (3.1) we need to know that

at least a.e. This is not always possible. Indeed, we can construct a function u (even continuous) such that each mollification uη has sign+⁡(uη)≡1, while sign+⁡(u)≢1. We shall prove this when Ω=]0,1[.

Let {qn}n≥1 be the set of rational numbers contained in ]0,1[. Fix 1>ϵ>0 and set

The set I is open and dense in [0,1]. Moreover, 0<|I|≤ϵ<1, thus |S|>0.

Next, for each n≥1 let ϕn:[0,1]→ℝ be a continuous nonnegative function such that ϕn⁢(x)>0 if and only if x∈In and ∥ϕn∥∞≤1. Set

Since the above series is uniformly convergent, the function u is continuous. Moreover, u⁢(x)>0 if and only if there exists n≥1 such that ϕn⁢(x)>0. This is obviously equivalent to the fact that x∈In. In other words, u vanishes on S and it is positive on I.

Let η>0 and let uη be a mollification of u, that is, u⋆mη, where (mη)η is a standard family of mollifiers. We claim that uη⁢(x0)>0 for any x0∈]0,1[. Indeed, let x0∈]0,1[. By our choice of {qn}n there exists n≥1 such that |qn-x0|<η. Hence In∩]x0-η,x0+η[≠∅ and

Definition 4.1.

Let f∈𝒞⁢(ℝ) and h∈Lloc1⁢(Ω). A weak solution of

is a function

with f⁢(u)∈Lloc1⁢(Ω), such that for any nonnegative ϕ∈𝒞01⁢(Ω) we have

If L=ΔG is a sub-Laplacian on a Carnot group, then a distributional solution of (4.1) is a function u∈Lloc1⁢(Ω) such that f⁢(u)∈Lloc1⁢(Ω), and for any nonnegative ϕ∈𝒞02⁢(Ω) we have

Theorem 4.2.

Let X be a subspace of Lloc1⁢(Ω) such that if u∈X, then u+∈X. Let b:[0,∞[→[0,+∞[ be a continuous function such that b⁢(0)=0 and the problem

has no nontrivial weak [distributional] solution belonging to X.

Let h∈Lloc1⁢(Ω) and let f∈C⁢(R) be such that

Then the equation

has at most one weak [distributional] solution belonging to X.

Proof.

Let h∈Lloc1⁢(Ω) and let u,v∈X be solutions of (4.3). The function u-v∈X is a weak solution of

An application of the appropriate Kato inequality (3.1) or [8, Theorem 2.1] yields

which in turn implies that the function w:=(u-v)+ is a weak (or distributional) solution of

In other words, w solves (4.2). Hence w≡0 a.e. on Ω, that is, u≤v a.e. on Ω. Inverting the role of u and v, the claim follows. ∎

Theorem 4.3.

Let f∈C⁢(R) be such that

where b:[0,+∞[→[0,+∞[ is a continuous function satisfying the following assumptions:

1. b⁢(0)=0, b⁢(t)>0 for t>0;

2. it holds that

   (4.5)∫1+∞(∫1tb⁢(s)⁢𝑑s)-12⁢𝑑t<+∞;

3. b is convex.

Let h∈Lloc1⁢(RN). Then the problem

has at most one distributional solution u∈Lloc1⁢(RN). Moreover, if h≥0, then u≤0 a.e. on RN.

Proof.

The obvious idea is to apply Theorem 4.2. To this end, it is enough to check that the inequality

has only the trivial solution. Indeed, let us assume that v∈Lloc1⁢(ℝN) is a solution of (4.6). By a mollification argument (as in the proof of Lemma 3.3) we have

Next, by the convexity of b and the Jensen inequality, it follows that

Now vη is smooth and solves (4.7) with the function b nondecreasing (indeed, it satisfies (i) and it is convex) and satisfying (4.5), thus we are in the position to apply [7, Theorem 3.10] (by changing u:=-vη), so we deduce that vη≡0. Thus, by letting η→0 we obtain v≡0. ∎

Remark 4.4.

When dealing with 𝒞1 solutions, hypothesis (iii) can be relaxed by assuming that b is nonincreasing; see [7].

Corollary 4.5.

Let q>1 and let h∈Lloc1⁢(RN). The problem

has at most one solution u∈Llocq⁢(RN). Moreover, if h≥0, then u≤0 a.e. on RN.

Remark 4.6.

The above result, as far it is concerned with uniqueness and nonpositivity of the possible solutions, is the analog on Carnot groups of [3, Theorem 2].

Remark 4.7.

All the above results still hold when one replaces the function h∈Lloc1 with a distribution h∈𝒟′.

Example 4.8.

Let f be defined by

where q1,q2>1. Theorem 4.3 applies to such f. Indeed, for t≥0 define g⁢(t):=min⁡{tq1,tq2}. The function b that we need is the convexification of cg for a small constant c>0.

We claim that there exists a constant c>0 such that for any t>s we have

Assume that q1≤q2. By the well-known inequality

we have the following three cases:

1. Let t>s>0. Then

   f⁢(t)-f⁢(s)=tq1-sq1≥cq1⁢(t-s)q1≥cq1⁢g⁢(t-s).

2. Let 0>t>s. Then

   f⁢(t)-f⁢(s)=-|t|q2+|s|q2≥cq2⁢(|s|-|t|)q2≥cq2⁢g⁢(|s|-|t|)=cq2⁢g⁢(t-s).

3. Let t>0>-s. The proof of the claim will follow if we prove that

   tq1+sq2≥c⁢g⁢(t+s) for any ⁢t,s>0.

By using the inequality

and distinguishing three different cases, we have the following:

1. Let s≥1 and t>0. Then

   tq1+sq2≥tq1+sq1≥21-q1⁢(t+s)q1≥21-q1⁢g⁢(t+s).

2. Let 1>t>0 and 1>s>0. Then

   tq1+sq2≥tq2+sq2≥21-q2⁢(t+s)q2≥21-q2⁢g⁢(t+s).

3. Let t≥1 and 1>s>0. Then

   tq1+sq2≥tq1≥2-q1⁢(t+1)q1≥2-q1⁢(t+s)q1≥2-q1⁢g⁢(t+s).

Next, by choosing

we get the claim.

By defining b:=conv⁡(c⁢g), it follows that assumptions (4.4) and Theorem 4.3 (i) and (iii) are fulfilled. Notice that Theorem 4.3 (ii) is satisfied since at infinity the function b behaves like tq1 with q1>1.

We point out that f does not satisfy the Brezis condition f′⁢(t)≥|t|q-1 for any t∈ℝ unless q1=q2. The interested reader may compare this with [3].

Theorem 5.1.

Let X be a subspace of Lloc1⁢(Ω) such that if u∈X, then u+∈X. Let b:[0,∞[→[0,+∞[ be a continuous function such that b⁢(0)=0 and the problem

has no nontrivial weak [distributional] solutions belonging to X. Let f∈C⁢(R) be such that

Let (u,v)∈X×X be a weak [distributional] solution of the system of inequalities

Then the following assertions hold:

1. u+v≤0 a.e. on Ω.

2. Let C≥1 and assume that the function f¯⁢(t):=-C⁢f⁢(-t) satisfies (5.2). Let (u,v)∈X×X be a weak [distributional] solution of the system

   (5.4){C⁢f⁢(u)≥L⁢v≥f⁢(u),C⁢f⁢(v)≥L⁢u≥f⁢(v) [{C⁢f⁢(u)≥ΔG⁢v≥f⁢(u),C⁢f⁢(v)≥ΔG⁢u≥f⁢(v)] on Ω.

   Then u=-v a.e. on Ω . Therefore, u satisfies

   (5.5)Cf(u)≥-Lu≥f(u) [Cf(u)≥-ΔGu≥f(u)]  on Ω,

   and the function f must be odd on the range of u, that is, for any t∈u⁢(Ω) the condition f⁢(t)=-f⁢(-t) holds.