Sharp Liouville Theorems

Conjecture.

Let u∈C2⁢(ℝN) be a bounded solution of the Allen–Cahn equation -Δ⁢u=u-u3 which is monotone in one direction (for instance, ∂⁡u/∂xN>0 in ℝN). Then u is a one-dimensional function (or equivalently, all level sets {u=s} of u are hyperplanes), at least if N≤8.

Theorem 1.1.

Let φ∈Lloc∞⁢(RN) be a positive function. Assume that σ∈Hloc1⁢(RN) satisfies σ⁢div⁡(φ2⁢∇⁡σ)≥0 in RN in the distributional sense. For every R>0, let BR={|x|<R} and assume that there exists a constant independent of R such that

Then σ is constant.

Property (P).

We say that a function 0<Ψ∈C⁢([1,∞)) satisfies (P) if it has the following property: if φ∈Lloc∞⁢(ℝN) is a positive function, σ∈Hloc1⁢(ℝN) satisfies

in the distributional sense and

then σ is necessarily constant.

Property (P’).

We say that a function 0<Ψ∈C⁢([1,∞)) satisfies (P’) if it has the following property: if φ∈Lloc∞⁢(ℝN) is a positive function, σ∈Hloc1⁢(ℝN) satisfies

in the distributional sense and

then σ is necessarily constant.

Theorem 1.2.

Suppose 0<Ψ∈C⁢([1,∞)). The following conditions are equivalent:

1. Ψ satisfies (P).

2. Ψ satisfies (P’).

3. ∫ 1 ∞ 1 h ′ = ∞ for every nondecreasing function 0 ≤ h ∈ C ⁢ ( [ 1 , ∞ ) ) satisfying h ≤ Ψ in [ 1 , ∞ ) .

Theorem 1.3.

Suppose 0<Ψ∈C1⁢([1,∞)) satisfies Ψ′>0 in [1,+∞) and Ψ is convex in [R0,+∞) for some R0>1. The following conditions are equivalent:

1. Ψ satisfies (P).

2. Ψ satisfies (P’).

3. ∫ 1 ∞ 1 Ψ ′ = ∞ .

Remark 1.4.

For general 0<Ψ∈C⁢([1,∞)), it is possible to prove that if lim infx→∞⁡Ψ⁢(x)/x2<+∞, then Ψ satisfies (P) and (P’). Therefore, we can restrict our attention to the case limx→∞⁡Ψ⁢(x)/x2=+∞. Thus, the condition of convexity of Ψ in Theorem 1.3 seems natural and not too restrictive.

To see that lim infx→∞⁡Ψ⁢(x)/x2<+∞ implies that Ψ satisfies (P) and (P’), we will apply Theorem 1.2. Suppose that there exists a divergent sequence {Rn} and a real number C>0 such that Ψ⁢(Rn)≤C⁢Rn2, n≥1, and take a nondecreasing function 0≤h∈C⁢([1,∞)) satisfying h≤Ψ in [1,∞). Our purpose is to obtain ∫1∞1/h′=∞. To this end, take an arbitrary R>1 and consider n0∈ℕ such that Rn>R for every n≥n0. Then

for every n≥n0. Hence

for n≥n0. Taking limit as n tends to infinity, we deduce

Since R>1 is arbitrary we conclude ∫1∞1/h′=∞, which is the desired conclusion.

Remark 1.5.

As said before, if 0<Ψ∈C1⁢([1,∞)) satisfies Ψ′>0 in [1,+∞), it is obvious from Theorem 1.2 that the condition ∫1∞1/Ψ′=∞ is necessary to have (i) and (ii). To show that this is not sufficient, it suffices to construct functions Ψ,h∈C∞⁢([1,∞)) satisfying 0<h<Ψ and 0<Ψ′, h′ in [1,∞), such that

To do this, for every integer n≥1 define the function fn:[n,n+12]→ℝ by

Clearly,

Hence, there exists 0<Ψ∈C∞([1,+∞) satisfying Ψ′>0 and Ψ⁢(x)=fn⁢(x) for every n≤x≤n+12 and n≥1.

It follows that

Then ∫1∞1/Ψ′=∞. On the other hand, take an arbitrary x≥1. Then there exists an integer n≥1 such that n≤x<n+1. Thus

Therefore, taking h⁢(x)=x3, we have ∫1∞1/h′<∞, which is our claim.

Proof of Theorem 1.2.

It is evident that (i)⟹(ii). Therefore, we shall have established the theorem if we prove (iii)⟹(i) and (ii)⟹(iii).

Proof of (iii)⟹(i). Suppose that 0<Ψ∈C⁢([1,∞)) satisfies (iii) and ∫BR(φ⁢σ)2≤Ψ⁢(R) for every R≥1, where φ∈Lloc∞⁢(ℝN) is a positive function and σ∈Hloc1⁢(ℝN) satisfies (1.1) in the distributional sense. Our purpose is to obtain that σ is constant.

If inf⁡Ψ=0, then there exists a divergent sequence {Rn} such that Ψ⁢(Rn) tends to 0 as n tends to ∞. Thus ∫BRn(φ⁢σ)2 also tends to 0, which implies σ=0.

Otherwise, let 0<m:=inf⁡Ψ and consider the function

Clearly, h≤12⁢Ψ+12⁢m≤Ψ in [1,∞) and h is a positive continuous and nondecreasing function satisfying

for almost every r>1. From this, we obtain ∫1∞1/h′=∞. Taking into account that 1/h′⁢(r)≤2⁢er/m for almost every r>1, we have 1/h′∈Lloc∞⁢([1,∞)). Thus

Now, for arbitrary 1<R1<R2 define in the ball BR2 the radial function η by

for every r=|x|≤R2. Multiplying (1.1) by η2 and integrating by parts in BR2, we obtain

Therefore,

Thus

Fix R1>1. Applying (2.1) and taking limit in the above inequality as R2 tends to ∞, we obtain

Since R1>1 is arbitrary, σ is constant, which is the desired conclusion.

Proof of (ii)⟹(iii). Now, suppose that (iii) does not hold. That is, there exists a nondecreasing function 0≤h∈C⁢([1,∞)) satisfying h≤Ψ in [1,∞) and ∫1∞1h′<∞. The proof is completed by constructing a positive function φ∈Lloc∞⁢(ℝN) and a nonconstant function σ∈Hloc1⁢(ℝN) satisfying (1.2) in the distributional sense and ∫BR(φ⁢σ)2≤Ψ⁢(R) for every R≥1.

First of all, note that

Since Ψ>0 in [1,∞), we have that 0<m:=inf⁡Ψ. Consider the odd function μ:ℝ→ℝ such that

Clearly, μ is continuous and increasing in ℝ and satisfies, almost everywhere, that

Hence

Therefore, 1/μ′∈L1⁢(1,∞), and it follows immediately that 1/μ′∈L1⁢(ℝ). For this reason, taking any 0<H∈C∞⁢(ℝN-1) satisfying ∫ℝN-1H2=12, we can define the functions φ,σ:ℝN→ℝ by

if N=1, then we define

and we apply the same reasoning as in the case N>1.

It is easy to check that

From the above it follows that 0<φ∈Lloc∞⁢(ℝN) and |∇⁡σ|∈Lloc∞⁢(ℝN). Thus σ∈Hloc1⁢(ℝN). Moreover, an easy computation shows that

which implies div⁡(φ2⁢∇⁡σ)=0 in ℝN.

Finally, taking into account that BR⊂ℝN-1×(-R,R), for every R≥1 we obtain

which completes the proof. ∎

Proposition 3.1.

Let ϕ∈C1⁢([a,b]) be a convex function satisfying ϕ′>0 in [a,b]. Then

for every nondecreasing function g∈C⁢([a,b]) satisfying g⁢(a)=ϕ⁢(a) and g≤ϕ in [a,b].

Moreover, equality holds if and only if g=ϕ.

Lemma 3.2.

Let g∈C⁢([a,b]) be a nondecreasing function. Let p⁢(x)=A⁢x+B, A>0 and B∈R be such that g⁢(a)=p⁢(a) and g⁢(b)≤p⁢(b). Then

Moreover, equality holds if and only if g=p.

Proof.

If ∫ab1/g′=∞, the lemma is trivial. Otherwise, applying the Cauchy–Schwarz inequality, we obtain

Hence

On the other hand, if equality holds, then all previous inequalities become equalities. This implies that g⁢(b)=p⁢(b) and that g′ is a real multiple of 1/g′. That is, g′ is constant and, since g⁢(a)=p⁢(a) and g⁢(b)=p⁢(b), we obtain g=p. ∎

Lemma 3.3.

Let g∈C⁢([a,b]) be a nondecreasing function. For 1≤i≤m consider pi⁢(x)=Ai⁢x+Bi, Ai>0, Bi∈R, such that pi⁢(a)≤g⁢(a). Define

Then

Moreover, if ∫ab1/g′<∞, then equality holds if and only if g=g¯.

Proof.

Note that g¯ is a nondecreasing continuous function in [a,b]. Therefore, the statement of the lemma makes sense. If

the lemma is trivial. Hence, we will suppose in the rest of the proof that

The proof is by induction on m.

We first prove the lemma for m=1. To do this, consider the open set G={x∈(a,b):p1⁢(x)>g⁢(x)}. If G=∅, then g¯=g and the lemma follows. Otherwise, G is the countable (possible finite) disjoint union of open intervals. That is, G=⋃n∈X(an,bn), where X⊂ℕ and p1⁢(an)=g⁢(an) and p1⁢(bn)≥g⁢(bn) for every n∈X. Then

Applying Lemma 3.2 in each interval (an,bn), we conclude the lemma for the case m=1.

We now proceed by induction. We suppose that the lemma holds for m-1≥1 and we will prove that it holds for m. Define

By the induction hypothesis, we have

On the other hand, note that

It is easily seen that h is a continuous nondecreasing function satisfying pm⁢(a)≤g⁢(a)=h⁢(a). Therefore, applying the case of m=1 (which is already proved) to the functions h⁢(x) and pm⁢(x), we obtain

Combining inequalities (3.3) and (3.4), we obtain the desired inequality (3.2). Finally, if equality holds in (3.2), then equalities also hold in (3.3) and (3.4). This gives g=h=g¯ and the proof is completed. ∎

Proof of Proposition 3.1.

We first prove (3.1) in the case g⁢(x)<ϕ⁢(x) for every x∈(a,b). To do this, for every positive integer n, consider a partition of the interval (a,b] in 2n subintervals of the same length. That is,

Consider now the 2n lines which are tangent to the graphic of the function y=ϕ⁢(x) at xk,n, 1≤k≤2n. That is,

Define

Note that the convexity of ϕ gives gn⁢(x)≤ϕ⁢(x) for every a≤x≤b, n≥1.

We claim that gn→ϕ in L∞⁢(a,b) as n→∞. To do this, take an arbitrary x∈(a,b). Then, for fixed n≥1, there exists 1≤k≤2n such that xk-1,n<x≤xk,n. Using the convexity and monotonicity of ϕ, we deduce

This gives

and the claim is proved.

Now fix n0>1 and consider a0=a+(b-a)/2n0 and b0=b-(b-a)/2n0. Note that a0=x2n-n0,n and b0=x2n-2n-n0,n for every n≥n0. Since [a0,b0]⊂(a,b) and g<ϕ in (a,b), we deduce that there exists ε0>0 (depending on n0) such that g⁢(x)<ϕ⁢(x)-ε0 for every x∈[a0,b0]. Using gn→ϕ in L∞⁢(a0,b0), we can assert that there exists n1≥n0 (depending on ε0) such that g⁢(x)<gn⁢(x) for every x∈[a0,b0] and n1≥n0. Then

Consider n≥n1 and 2n-n0<k≤2n-2n-n0. Take x∈[xk-1,n,xk,n]. The convexity of ϕ yields

and consequently gn′⁢(x)≤ϕ′⁢(xk,n). This gives

Therefore, applying Lemma 3.3 in the interval [a,b], we obtain

for every n≥n1. Since 1/ϕ′ is continuous in [a0,b0] and xk,n-xk-1,n=(b-a)/2n, we deduce that the right term of the last inequality tends to ∫a0b01/ϕ′ as n tends to ∞. Thus,

Finally, since n0>1 is arbitrary, we conclude (3.1) for the case g<ϕ in (a,b).

We now turn out to the general case g≤ϕ in (a,b) and we proceed to show (3.1). For this purpose, consider the open set G={x∈(a,b):ϕ⁢(x)>g⁢(x)}. If G=∅, then (3.1) is trivial. Otherwise, G is the countable (possible finite) disjoint union of open intervals. That is, G=⋃n∈X(an,bn), where X⊂ℕ, ϕ⁢(an)=g⁢(an), ϕ⁢(bn)≥g⁢(bn) and ϕ>g in (an,bn) for every n∈X. Applying the previous case in each interval (an,bn), we conclude

It remains to prove that equality holds in (3.1) if and only if g=ϕ. To this end, suppose that we have equality in (3.1) for some g. Take an arbitrary x0∈[a,b] and consider the function

Clearly, gx0 is nondecreasing and satisfies g≤gx0≤ϕ in [a,b] and gx0⁢(a)=g⁢(a)=ϕ⁢(a). Hence

Applying Lemma 3.3 yields g=gx0 in [a,b]. In particular, g⁢(x0)=gx0⁢(x0)=max⁡{g⁢(x0),ϕ⁢(x0)}=ϕ⁢(x0). Since x0∈[a,b] is arbitrary, we conclude that g=ϕ in [a,b] and the proposition follows. ∎