
We investigate the boundary Harnack principle for uniformly elliptic operators in divergence form in Hölder domains of exponent α>0. We also deal with operators in nondivergence form with coefficient that remain constant in the graph direction.
Citation: Daniela De Silva, Ovidiu Savin. On the boundary Harnack principle in Hölder domains[J]. Mathematics in Engineering, 2022, 4(1): 1-12. doi: 10.3934/mine.2022004
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We investigate the boundary Harnack principle for uniformly elliptic operators in divergence form in Hölder domains of exponent α>0. We also deal with operators in nondivergence form with coefficient that remain constant in the graph direction.
In this paper we continue the study of the boundary Harnack principle for solutions to elliptic equations, based on the method developed in [7]. The classical boundary Harnack principle states that two positive harmonic functions that vanish on a portion of the boundary of a Lipschitz domain must be comparable up to a multiplicative constant, see for example [1,6,11,14]. Further extensions to more general operators and more general domains were obtained in several subsequent works [2,3,5,10,12,13].
In particular, Bass and Burdzy [3,4] and Banuelos, Bass and Burdzy [2] provided sharp versions using probabilistic methods. They established the boundary Harnack principle for nondivergence elliptic operators in Hölder domains (or more general twisted Hölder domains) of exponent α>12, and for divergence operators in Hölder domains of arbitrary exponent α>0. For the case of divergence operators, an analytical proof based on Green's function was given by Ferrari in [8].
To state precisely the boundary Harnack principle in Hölder domains, first we introduce some notation. Let g:¯B′1→R be a Cα Hölder function of n−1 variables with g(0)=0, and α∈(0,1). Denote by Γ⊂Rn the graph of g,
Γ:={xn=g(x′)},0∈Γ, |
and by Cr the cylinder above B′r and at height r on top of Γ
Cr:={x′∈B′r,g(x′)<xn<g(x′)+r}. |
We say that C1 is a Cα-Holder domain in a neighborhood of Γ.
The following version of the boundary Harnack principle is due to Banuelos, Bass and Burdzy, see [2].
Theorem 1.1. Let Lu=div(A(x)∇u) be a uniformly elliptic linear operator, and assume that u,v are two positive solutions to
Lu=Lv=0,in C1, |
which vanish on Γ. Then
uv (x)≤Cuv(12en)for all x∈C1/2, |
with C depending on n, α, ‖g‖Cα and the ellipticity constants of L.
The assumption that u=v=0 in Γ is understood in the H1 sense, i.e., u,v∈H10(C1) in a neighborhood of Γ.
Recently, in [7] we found a direct analytical method of proof of the boundary Harnack principle based on an iteration scheme and Harnack inequality. In particular we established the corresponding results in Hölder domains of exponent α>12 for general equations either in divergence or nondivergence form.
In the present paper we discuss further the case of Hölder domains of arbitrary exponent α>0, and give a proof of Theorem 1.1 using the same ideas from [7]. We also consider some novel extensions of Theorem 1.1 to non-divergence equations whose coefficients remain constant in the vertical direction (see Section 4).
The paper is self-contained and is organized as follows. In Section 2 we give two lemmas concerning Harnack inequality outside domains of small capacity. In Section 3 we use these lemmas and employ the arguments from [7] to prove Theorem 1.1. Finally in Section 4 we provide some extensions of Theorem 1.1 to more general divergence operators, and certain non-divergence or fully nonlinear operators.
In this section we present two lemmas concerning solutions to divergence equations in domains whose complement in the unit cube Q1 has small capacity.
Given a domain Ω and a compact set K⊂Ω, we say that two functions u,v∈H1(Ω) agree on K, and write u=v in K, if u−v∈H10,loc(Kc). Here Kc denotes the complement of K in Rn.
In particular, if L is a uniformly elliptic operator in divergence form
Lu=div(A(x)∇u), |
with
A(x) measurable, Λ|ξ|2≥ξTA(x)ξ≥λ|ξ|2,λ>0, |
then the statement that u solves
Lu=0in Ω∖K, and u=0 on ∂Ω, u=1 in K, | (2.1) |
means that Lu=0 in the open set Ω∖K, and
u−η∈H10(Ω∖K), |
where η∈C∞0(Ω), and η=1 in a neighborhood of K.
Notice that the solution u to (2.1) is a supersolution in Ω, i.e., Lu≤0 in Ω.
Let Q1 denote the unit cube in Rn centered at 0, and E⊂Q1 a closed set. Set,
cap3/4(E):=capQ1(E∩Q3/4)=infw∈A∫Q1|∇w|2dx, |
where
A:={w∈H10(Q1),w=1in E∩¯Q3/4}. |
The first lemma states that a solution to Lv=0 in Q1∖E satisfies the Harnack inequality in measure if E has small capacity. Positive constants depending on the dimension n and the ellipticity constants λ, Λ are called universal.
Lemma 2.1. Assume v≥0 is defined in Q1∖E and satisfies
Lv=0. |
Let
Qi:=Q1/8(xi)⊂Q1/2,i=1,2 |
be two cubes of size 1/8 included in Q1/2. Assume that
cap3/4(E)≤δand|{v≥1}∩Q1||Q1|≥1/2, |
for some δ small, universal. Then
|{v≥c0}∩Q2||Q2|≥1/2 |
for some c0 small.
The second lemma is standard and states that the weak Harnack inequality holds for a subsolution v≥0 which vanishes on a set E of positive capacity.
Lemma 2.2. Assume that v≥0 in Q1, and
Lv≥0 in Q1, v=0 in E∩¯Q3/4. |
If
cap3/4(E)≥δ, |
then
v(0)≤(1−c(δ))‖v‖L∞. |
Proof of Lemma 2.1. Let ψ be the solution to
Lψ=0 in Q3/4∖K, ψ=1 in K, ψ=0 on ∂Q3/4, |
where K is a compact subset of {v≥1}∩(Q1∖E) with |K|≥14|Q1| (see Figure 1). By hypothesis and weak Harnack inequality (see Theorem 8.18 and Theorem 9.22 in [9]) we find
ψ≥c0 inQ1/2, | (2.2) |
for some small c0.
Similarly as above we define ϕ to be the solution to
Lϕ=0 in Q1∖(E∩Q3/4), ϕ=1 in E∩Q3/4, ϕ=0 on ∂Q1. | (2.3) |
We claim that if δ is chosen sufficiently small then,
|{ϕ>14c0}∩Q2|≤12|Q2|. | (2.4) |
For this we let w be the solution to (2.3) when L=△. The Dirichlet energies of ϕ and w are comparable since
∫(∇(ϕ−w))TA∇ϕdx=0, |
hence
cap3/4(E)≤∫|∇ϕ|2dx≤C∫(∇ϕ)TA∇ϕdx≤C∫|∇w|2dx=Ccap3/4(E). |
By Poincaré inequality we find
∫ϕ2dx≤C∫|∇ϕ|2dx≤Cδ, |
which gives the claim (2.4).
Next we compare 2v with ψ−ϕ in Q3/4∖E.
They satisfy the same equation in Q3/4∖(E∪K), and in a neighborhood of K by the continuity of v we have
2v≥1≥ψ−ϕ. |
On the other hand
ψ−ϕ≤0on∂(Q3/4∖E) |
in the sense that (ψ−ϕ)+∈H10(Q3/4∖E). Since v≥0, the maximum principle gives
2v≥ψ−ϕ, |
which by (2.2), (2.4) yields the desired conclusion.
Proof of Lemma 2.2. Assume that ‖v‖L∞=1. Then, by the maximum principle we have
1−v≥ϕ, |
with ϕ as in (2.3) above. It suffices to show that ϕ≥c(δ) on ∂Q7/8 which by the maximum principle implies the desired conclusion ϕ(0)≥c(δ) small. Since all the values of ϕ are comparable near ∂Q7/8 by the Harnack inequality, we need to show that ϕ≥c′(δ) at some point on ∂Q7/8.
Assume by contradiction that |ϕ|≤μ is very close to 0 on ∂Q7/8. The Caccioppoli inequality (we think that ϕ is extended to 0 outside Q1) implies
‖∇ϕ‖L2(Q1∖Q15/16)≤C‖ϕ‖L2(Q1∖Q7/8)≤Cμ. | (2.5) |
On the other hand if η∈C∞0(Q1) with η=1 in Q15/16 then
∫∇[η2(1−ϕ)]A∇ϕdx=0, |
hence
∫η2∇ϕA∇ϕdx≤C∫|∇ϕ|2|∇η|2dx≤Cμ2. |
This together with (2.5) implies that the Dirichlet energy of ϕ in Q1 is bounded above by Cμ2, and we reach a contradiction.
This section is devoted to the proof of Theorem 1.1. We recall that Γ denotes the graph of a Cα function g, with α∈(0,1),
Γ:={xn=g(x′)},0∈Γ, |
and Cr denotes the cylinders of size r on top of Γ
Cr:={x′∈B′r,g(x′)<xn<g(x′)+r}. |
The main idea of the proof is to show through an iterative procedure that a solution w which vanishes on Γ and is mostly positive in Cr, becomes positive near the origin.
We denote by
Ar:={x∈B′r|g(x′)+rβ≤xn<g(x′)+r}, |
the points in the cylinder Cr at height greater than rβ on top of Γ, for some β>1.
We divide the proof in three steps.
Step 1. We show that, there exist C0,β>1 depending on n, α, ‖g‖Cα, and the ellipticity constants of L, such that if w is a solution to Lw=0 in Cr (possibly changing sign) which vanishes on Γ,
w≥f(r)onAr, |
and
w≥−1on Cr, |
where
f(r):=eC0rγ,γ:=β(1−1α)<0, |
then,
w≥f(r2)aon Ar2, | (3.1) |
and
w≥−aon Cr2, | (3.2) |
for some small a=a(r)>0, as long as r≤r0 universal.
The conclusion can be iterated and we obtain that if the hypotheses are satisfied in Cr0 then
w>0on the line segment {ten,0<t<r0}. |
Since g is Hölder continuous, we can apply interior Harnack inequality to w+1 in a chain of balls and need
C(rβ)1−1α=Crγballs |
to connect a point in Ar/2 with a point in Ar. We conclude that
w≥(f(r)+1)e−C1rγ−1in Ar/2, | (3.3) |
for some C1 universal, hence w≥1 in Ar/2 if C0 is sufficiently large.
Next we take a point on Γ:={xn=g(x′)}, say 0 for simplicity, and consider the cubes of size rβ/α centered on the en axis, i.e., Qrβ/α(ten) (see Figure 2).
When t>Crβ the cube is in the interior of the domain and when t<−Crβ the cube is in the complement. There are Crγ stacked cubes which connect the domain with its complement. The graph property of the domain implies that the capacity of the complement
E={xn≤g(x′)} |
in Qrβ/α(ten) is decreasing with t. By continuity we can find a cube centered at t0en such that, after a rescaling of factor r−β/α, cap3/4(E)=δ in that cube, with δ as in Lemma 2.1.
For all cubes centered at ten with t≥t0 we can apply Lemma 2.1 repeatedly for w+1 and obtain an inequality in measure as in (3.3),
|{w≥1}∩Qi(ten)|≥12|Qi|,Qi(ten):=Q18rβ/α(ten). |
Thus {w−=0} has positive density in all cubes centered at ten with t≥t0.
Now we notice that we can apply weak Harnack inequality for w− in all cubes, in the top cubes with t≥t0 because of the density property, and in the bottom ones with t≤t0 because of Lemma 2.2.
Hence w− decreased by a fixed factor on the en axis passing through a point on Γ, with respect to its maximum over all cubes of size rβ/α centered on that axis. As we move each time a rβ/α distance inside the domain from the sides of Cr, supw− decreases geometrically hence,
w≥−e−c0r1−β/αin Cr/2, |
for c0 small universal. We choose
a(r):=e−c0r1−β/α, |
and in view of (3.3), our claim
w≥1≥a(r)f(r2), |
is satisfied for all r small.
Step 2. [Carleson estimate] We show that,
u,v≤C2in C1/2, |
with C2 universal. We apply an iterative argument similar to the one in Step 1. Since u(en/2)=1, the interior Harnack inequality gives that
u≤eC1h1−1/αΓin C3/4,hΓ(x):=xn−g(x′), | (3.4) |
with C1 universal. With the same notation as Step 1, we wish to prove that if r is smaller than a universal r0 and
u(y)≥f(r), |
for some y∈C1/2, then we can find
z∈S:={|y′−z′|=r,0<hΓ(z)<rβ}, |
such that
u(z)≥f(r2). |
Since |z−y|≤Crα, we see that for r small enough, we can build a convergent sequence of points yk∈C3/4 with u(yk)≥f(2−kr)→∞. On the other hand the extension of u by 0 below Γ is a subsolution in a neighborhood of Γ. Therefore u is bounded above, and we reach a contradiction.
To show the existence of the point z we let
w:=(u−12eC0rγ)+,withC0≫C1. |
By (3.4) we know that
w=0 whenhΓ(x)≥rβ. |
By Lemma 2.1 this estimate can be extended in measure for the cubes of size rβ/α with t≥t0 since the capacity of the complement is bounded above. More precisely, as in Step 1, in each cube of size rβ/α we have that either {w=0} has positive density (for the cubes with t≥t0), or positive capacity (for the cubes with t≤t0).
Moreover, if our claim is not satisfied then we apply Weak Harnack inequality for w repeatedly as in Step 1 above. As we move inside the domain from the sides of Cr(y′,g(y′)) we obtain
w≤f(r2)e−c0r1−β/αin Cr/2(y′,g(y′)). |
In particular
12f(r)≤w(y)≤f(r2)e−c0r1−β/α, |
and we reach a contradiction.
Step 3. We prove the theorem using the Steps 1 and 2 above. After multiplication by a constant we may assume that u=v=1 at 12en. It suffices to show that for a large constant C3>0 universal,
w:=C3u−C−13v≥0in C1/2. |
By Step 2 we know that v≤C3 hence w≥−1 in C3/4. Moreover, since u(en/2)=1, we conclude by interior Harnack for u that
w≥f(r0)inC3/4∩{xn≥rβ0}, |
provided that C3 is chosen sufficiently large. Here f, r0 and β are as in Step 1.
We conclude by Step 1 that w≥0 on the line {ten,0<t<3/4}. We can repeat the argument at all points on Γ∩¯C1/2, and the theorem is proved.
In this section we state a few variants of the Theorem 1.1. First we remark that the proof applies to operators involving lower order terms.
Theorem 4.1. The statement of Theorem 1.1 holds for general uniformly elliptic operators
Lu=div(A(x)∇u)+b(x)⋅∇u+d(x)u,b∈Lq,d∈Lq/2,q>n, |
with the constant C depending also on q, ‖b‖Lq, ‖d‖Lq/2.
Indeed, we only need to check that the statements of Section 2 continue to hold in the small cubes of size rβ/α. After a dilation this corresponds to proving Lemmas 2.1 and 2.2 for operators L as above with ‖b‖Lq, ‖d‖Lq/2 sufficiently small. The proofs are identical since the presence of such lower order terms does not affect the energy estimates.
A counterexample of Bass and Burdzy in [4] shows that Theorem 1.1 does not hold in general for nondivergence equations when α<12. Here we remark that Theorem 1.1 remains valid with α>0 for nondivergence linear operators which are translation invariant in the vertical direction.
Theorem 4.2. The statement of Theorem 1.1 holds for linear nondivergence uniformly elliptic operators of the form
Lu=tr(A(x′)D2u). |
In this theorem we assume that the coefficient matrix A depends continuously on its argument, although the estimates do not depend on its modulus of continuity. Since u, v might not be continuous at all points on Γ, the hypothesis that u, v vanish on the boundary is understood in the sense that their extensions with 0 below Γ are bounded subsolutions for L, see [7].
In this case we provide the corresponding lemmas of Section 2 by defining the capacity (with respect to L) as
cap3/4(E)=infQ1/4ϕ, |
where ϕ solves
Lϕ=0 in Q1∖(E∩Q3/4), ϕ=1 in E∩Q3/4, ϕ=0 on ∂Q1. |
Then Lemma 2.2 follows directly from the definition of the capacity, with c(δ)=δ. For Lemma 2.1 we see that (2.4) is satisfied since by the Weak Harnack inequality the set {ϕ>c0} must have small measure in Q1 if δ is sufficiently small. The rest of the proof is the same.
The arguments of Section 3 can be repeated in the same way. The invariance of the operator L with respect to the vertical direction and the graph property of the boundary imply that the capacity of the complement E in the cubes Qrβ/α(ten) is monotone in t. We can apply again Lemma 2.1 for the top cubes with t≥t0 and Lemma 2.2 for the bottom cubes with t≤t0, and carry on as before.
We also discuss the case of fully nonlinear operators
F(D2u)=0inC1, | (4.1) |
with F uniformly elliptic with constants λ,Λ, and homogenous of degree 1.
We can prove the lemmas of Section 2 for the operator F by using as capacity the definition above with Lϕ=F(−D2ϕ). Then Lemma 2.2 follows again directly from the definition. For the proof of Lemma 2.1 we choose the function ψ to satisfy M−λ/n,Λ(ψ)=0. Here as usual, M− denotes the extremal Pucci operator,
M−λ,Λ(M)=infAtr(AM), |
with A a symmetric matrix whose eigenvalues belong to [λ,Λ].
Then ψ−ϕ is a subsolution
F(D2(ψ−ϕ))≥F(D2(−ϕ))+M−λ/n,Λ(ψ)≥0, |
and the rest of proof remains as before. However in the proof of the main Theorem 1.1 only Step 2, the Carleson estimate, can be carried out in this setting, since for Step 1 we need the lemmas of Section 2 to hold not only for solutions of the operator F but for the difference of two such solutions as well.
Theorem 4.3 (Carleson estimate). Assume that u≥0 satisfies (4.1) and u vanishes on Γ. Then
u≤Cu(12en)inC1/2, |
with C depending on n, α, ‖g‖Cα,λ and Λ.
Finally we mention that in R2 Theorem 1.1 holds under very mild assumptions on the domain and the operator. Here we state a version for L∞ graphs and linear operators.
Theorem 4.4. Assume Γ⊂R2 is the closure of the graph of a function g with ‖g‖L∞≤1/4. Then the statement of Theorem 1.1 holds for uniformly elliptic linear operators L in divergence or nondivergence form with constant C depending only on the ellipticity constants of L.
We only sketch Steps 1 and 2 of Section 3 in this setting which can be adapted to more general situations. They are based on topological considerations and do not require an iterative argument.
Step 1. If Lw=0 and w≥−1 in C1, and w vanishes continuously on Γ, then w>0 in C1/2 provided that u(12e2) is large.
To prove this, we assume by contradiction that there is a connected component U of {w<0} which intersects C1/2. This connected component must exit C1 and since u(12e2)≫1, U must stay close to Γ. Thus we can find a nonintersecting polygonal line ℓ, say included in
l⊂{12≤x1≤34}∩U, |
which connects the two lateral sides x1=12 and x1=34 (see Figure 3). The line ℓ splits the cylinder
D:=(12,34)×(12,12) |
into two disjoint sets, and we define ˜w to be equal to w on the set "above" ℓ and ˜w=min{w,0} on the set "below" ℓ. Then ˜w is a supersolution of L in D. Since ˜w is sufficiently large in a ball above ℓ, and ˜w≥−1 in D we find that ˜w≥0 on the segment
{x1=58}∩{|x2|≤38}. |
We reached a contradiction at the point where ℓ intersects this segment.
Step 2. Assume Lu=0 and u≥0 in C1, and u vanishes continuously on Γ, with u(12e2)=1. Then u≤C in C1/2, for some large C.
This follows similarly as in Step 1. If {u>C} has a connected component that intersects C1/2, then we can find a polygonal line ℓ as above where u is large. Thus min{u,C} extended by C below ℓ is a supersolution for L in D, and the maximum principle implies that u is large at the point (5/8,1/2). Therefore by Harnack inequality u(0,1/2) is large as well, and we reach a contradiction.
The authors declare no conflict of interest.
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