Boundary Hölder continuity of stable solutions to semilinear elliptic problems in 𝐶^1,1 domains

A bounded domain Ω ⊂ R n is of class C 1 , 1 if, at each point x 0 ∈ ∂ Ω , there are a ball B = B ρ ⁢ ( x 0 ) and a one-to-one mapping Ψ of 𝐵 onto U ⊂ R n such that

1. Ψ ⁢ ( B ∩ Ω ) ⊂ R + n ;

2. Ψ ⁢ ( B ∩ ∂ Ω ) ⊂ ∂ R + n ;

3. Ψ ∈ C 1 , 1 ⁢ ( B ̄ ) and Ψ − 1 ∈ C 1 , 1 ⁢ ( U ̄ ) .

Let Ω ⊂ R n be a bounded domain of class C 1 , 1 . Then there is a function Φ ∈ C 1 , 1 ⁢ ( R n ) such that Ω = { Φ > 0 } , Φ = 0 on ∂ Ω , and ∇ Φ ⁢ ( x ) ≠ 0 for all x ∈ ∂ Ω .

Proof

By compactness, ∂ Ω may be covered by finitely many balls { B j = B ρ j ⁢ ( x j ) } j = 1 N , with x j ∈ ∂ Ω , ρ j > 0 , such that there are flattening maps Ψ j ∈ C 1 , 1 ⁢ ( B j ̄ ) as in Definition A.1.

Let ρ > 0 be sufficiently small so that the set B 0 : = { x ∈ Ω : dist ( x , ∂ Ω ) > ρ } satisfies Ω ̄ ⊂ ⋃ j = 0 N B j , and consider a partition of unity { η j } j = 0 N subordinated to the covering { B j } j = 0 N . The function

now satisfies the desired properties. ∎

Notice that, by construction, the function Φ above is compactly supported and takes negative values in a bounded neighborhood of ∂ Ω outside Ω ̄ .

Suppose that Ω = { Φ > 0 } ⊂ R n is a bounded domain of class C 1 , 1 , with Φ ∈ C 1 , 1 ⁢ ( R n ) as in Lemma A.2 above. Then there is an exhaustion of Ω by smooth sets Ω k = { Φ k > 0 } ,^[5] where the functions Φ k ∈ C ∞ ⁢ ( R n ) satisfy

for some constant 𝐶 depending only on Φ and Ω. Moreover, we have that ∂ Ω k → ∂ Ω in the sense of the Hausdorff distance.^[6]

Proof

Since Ω = { Φ > 0 } and ∇ Φ ≠ 0 on ∂ Ω , by continuity of the gradient, Φ is comparable to the distance function d ( x ) : = dist ( x , ∂ Ω ) in Ω ̄ , i.e.,

for some L ≥ 1 . For 𝑥 outside Ω, we define d ( x ) : = − dist ( x , ∂ Ω ) < 0 .

Consider a mollifying sequence ( η ε ) ε > 0 with supp ⁡ η ε ⊂ B ε . We will take the functions

for an appropriate sequence ε k ↓ 0 . Since ∥ ∇ Φ k ∥ C 1 = ∥ ∇ Φ k ∥ L ∞ + ∥ D 2 ⁢ Φ k ∥ L ∞ , recalling that ∥ D 2 ⁢ Φ k ∥ L ∞ ⁢ ( R n ) = [ ∇ Φ k ] C 0 , 1 ⁢ ( R n ) (since Φ k is C ∞ ) and Φ ∈ C 1 , 1 , the uniform bounds for ∥ ∇ Φ k ∥ C 1 stated in the lemma hold by the standard properties of convolutions.

Given δ > 0 , from (A.1), we see that

and taking the convolution with η ε , it follows that

Similarly, for δ ̃ > 0 , we have that

and regularizing, we obtain

Letting δ ̃ = δ / L 2 , since L ⁢ δ ̃ = δ / L , by (A.3) and (A.4), we have that

and the choice δ = 2 ⁢ L 2 ⁢ ε now yields the inclusions

Next, we construct the sequence ε k in (A.2) inductively. Fix ε 1 > 0 small. For k ≥ 1 , we define ε k + 1 : = ε k / { 2 ( 2 L 2 + 1 ) } = ε 1 / { 2 ( 2 L 2 + 1 ) } k . Hence, by (A.5), the sets

satisfy

They clearly exhaust Ω since ε k ↓ 0 and thus Ω = ⋃ k { d ≥ ε k − 1 } ⊂ ⋃ k Ω k . Furthermore, the inclusions (A.6) show that ∂ Ω k is at a Hausdorff distance of at most ε k ⁢ ( 2 ⁢ L 2 + 1 ) from ∂ Ω , and hence ∂ Ω k → ∂ Ω with respect to this distance.

It remains to prove the lower bound for | ∇ Φ k | on ∂ Ω k , which will also show the boundary ∂ Ω k to be smooth. Let x ∈ R n with | d ⁢ ( x ) | = dist ⁡ ( x , ∂ Ω ) = | x − x 0 | for some x 0 ∈ ∂ Ω . Since

we have the lower bound

where ρ > 0 depends only on ∥ | ∇ Φ | − 1 ∥ L ∞ ⁢ ( ∂ Ω ) and [ ∇ Φ ] C 0 , 1 ⁢ ( R n ) . Taking the convolution with η ε k in this last inequality, we obtain

By (A.6), the boundary ∂ Ω k is at a distance of at most ε k ⁢ ( 2 ⁢ L 2 + 1 ) from ∂ Ω . Hence, choosing ε 1 > 0 sufficiently small such that ρ − ε 1 > ε 1 ⁢ ( 2 ⁢ L 2 + 1 ) , from (A.7) and the definition of ε k = ε 1 / { 2 ⁢ ( 2 ⁢ L 2 + 1 ) } k − 1 , we deduce

and therefore ∥ | ∇ Φ k | − 1 ∥ L ∞ ⁢ ( ∂ Ω k ) ≤ 2 ⁢ ∥ | ∇ Φ | − 1 ∥ L ∞ ⁢ ( ∂ Ω ) , which concludes the proof. ∎

There are (small) numbers 0 < R 1 < R 2 and ρ > 0 depending only on ∥ ∇ Φ ∥ C 0 , 1 ⁢ ( R n ) and ∥ | ∇ Φ | − 1 ∥ L ∞ ⁢ ( ∂ Ω ) such that

Proof

By translation, we may assume that x 0 = 0 ∈ ∂ Ω . Since the map Ψ satisfies Ψ ⁢ ( 0 ) = 0 and D ⁢ Ψ ⁢ ( 0 ) = I , choosing R 2 > 0 small such that

by [20, Chapter XIV, Lemma 1.3], we deduce that, for all y ∈ B R 2 / 2 , there is a unique x ∈ B R 2 such that Ψ ⁢ ( x ) = y . Thus, we obtain the second inclusion

Using that [ D ⁢ Ψ ] C 0 , 1 ⁢ ( R n ) ≤ [ ∇ Φ ] C 0 , 1 ⁢ ( R n ) ⁢ ∥ | ∇ Φ | − 1 ∥ L ∞ ⁢ ( ∂ Ω ) , it is easy to check that condition (A.9) is fulfilled if

To show the first inclusion in (A.8), we proceed as above but considering the inverse map Ψ − 1 instead of Ψ. If R 1 > 0 is such that

then, again by [20, Chapter XIV, Lemma 1.3], we deduce

For (A.12) to hold, it suffices to take R 1 > 0 sufficiently small such that

Hence, to quantify the smallness of R 1 , we have to estimate [ D ⁢ Ψ − 1 ] C 0 , 1 ⁢ ( B 2 ⁢ R 1 ) .

Let x ̃ = Ψ ⁢ ( x ) and y ̃ = Ψ ⁢ ( y ) in B 2 ⁢ R 1 . If R 1 ≤ R 2 , with R 2 satisfying (A.11), then

and hence

Moreover, since

by the above, we conclude

for all R 1 ≤ R 2 satisfying (A.11). Therefore, if

then (A.14) holds and (A.13) follows.

Finally, choosing R 1 , R 2 > 0 satisfying (A.11), (A.15), and 8 ⁢ R 1 ≤ R 2 , we may take ρ = 4 ⁢ R 1 , and the inclusions (A.10) and (A.13) then yield the claim (A.8). ∎

Given Ω ⊂ R n a bounded domain, let u 1 , u 2 ∈ C 0 ⁢ ( Ω ̄ ) ∩ W loc 2 , n ⁢ ( Ω ) be two stable solutions of the equation − L ⁢ u = f ⁢ ( u ) in Ω, with u = 0 on ∂ Ω . Assume that f ∈ C 1 ⁢ ( R ) is convex. Then either u 1 = u 2 or f ⁢ ( u ) = μ 1 ⁢ [ L , Ω ] ⁢ u on the ranges of u 1 and u 2 .

Here μ 1 ⁢ [ L , Ω ] denotes the principal (or smallest) eigenvalue of 𝐿 in Ω, with the sign convention − L ⁢ φ = μ 1 ⁢ φ . It is characterized by (see [2])

Moreover, 𝐿 can be any uniformly elliptic second order operator. In particular, we allow it to have zero order terms.

In the proof of Theorem 1.1 above, we only need a weaker version of Proposition B.1. Namely, we could assume additionally that u 1 ≤ u 2 , which admits a shorter proof. However, the present statement might be more useful in applications.

Proof

Assume that u 1 ≠ u 2 and consider the difference w : = u 2 − u 1 . Let

and assume that Ω + ≠ 0 (otherwise, we exchange the roles of u 1 and u 2 ). By convexity, we have − L ⁢ w = f ⁢ ( u 2 ) − f ⁢ ( u 1 ) ≤ f ′ ⁢ ( u 2 ) ⁢ w , and hence

By the monotonicity of the principal eigenvalue with respect to the domain, since u 2 is stable, we have that

Since w > 0 in Ω + , by (B.1) and (B.2), it follows that

Applying [2, Corollary 2.2], we deduce that 𝑤 is a positive principal eigenfunction of J u 2 in Ω + , that is, satisfying

Using the equation − L ⁢ w = f ⁢ ( u 2 ) − f ⁢ ( u 1 ) in Ω + , from (B.3), we see that

By convexity, we also have the opposite inequality f ⁢ ( u 2 ) − f ⁢ ( u 1 ) − f ′ ⁢ ( u 2 ) ⁢ ( u 2 − u 1 ) ≤ 0 . Hence, by (B.4), we conclude μ 1 ⁢ [ J u 2 , Ω + ] = 0 , and the nonlinearity 𝑓 is affine in the union of intervals [ u 1 ⁢ ( x ) , u 2 ⁢ ( x ) ] with x ∈ Ω + .

If 𝑓 is of the form f ⁢ ( u ) = a ⁢ u + b in the ranges above, then − L ⁢ w = a ⁢ w in Ω + , with w > 0 , and therefore a = μ 1 ⁢ [ L , Ω + ] . Now, to have nontrivial solutions of

the Fredholm alternative forces b = 0 . We conclude that f ⁢ ( u ) = μ 1 ⁢ [ L , Ω + ] ⁢ u in the ranges of u 1 and u 2 in Ω + .

If Ω − : = { w < 0 } ≠ ∅ , then arguing as above with − w in place of 𝑤, we deduce that

with μ 1 ⁢ [ J u 1 , Ω − ] = 0 and f ⁢ ( u ) = μ 1 ⁢ [ L , Ω − ] ⁢ u in the ranges of u 2 and u 1 in Ω − .

The regularity of 𝑓 and the continuity of the solutions now forces

Now, on the one hand, by the stability of u 2 , we have that

On the other hand, since Ω + ⊊ Ω by assumption, the strict monotonicity of μ 1 yields^[7]

This contradiction forces either Ω + or Ω − to be empty, and therefore f ⁢ ( u ) = μ 1 ⁢ [ L , Ω ] ⁢ u in the ranges of u 1 and u 2 , which was the claim. ∎