2. Properties of the PDEs

In this section we will study the basic properties of the equations (1.5) and (1.8). We will develop explicit Poisson formulas among other properties.

3. Relation with Fractional Laplacian

In this section we will see how the equations (1.5) or (1.8) relate to the operator (− ▵) ^s . Namely, we will show that up to a constant factor

4. Reflection Extensions

In order to apply interior Harnack estimates to our PDE (1.5) (or (1.8)), we must show that if the operator (− ▵) ^s f = 0 in an open ball, then we can reflect the solution u and make it a solution of (1.5) (or (1.8)) across y = 0 (z = 0) in a suitable sense.

We first address the divergence case.

Lemma 4.1

Suppose that u: ℝ ⁿ  × [0, ∞) → ℝ is a solution to (1.5) such that for |x| ≤ r then the extension to the whole space is a solution to in the week sense in the (n + 1) dimensional ball of radius R (lcub;(x, y): |x|² + |y|² ≤ R ²}).

Proof

Let be a test function. We want to show that Let ϵ > 0, we separate a strip of width ϵ around y = 0 in the domain of the integral above

When we let ϵ → 0, the second term of the right hand side above clearly goes to zero because |y| ^a |∇ u|² is locally integrable, and the first term converges to zero if ϵ ^a (x, ϵ) → 0 as ϵ → 0.

Therefore, if ϵ ^a ũ _y (x, ϵ) → 0 as ϵ → 0, then div(|y| ^a ∇ u) = 0 in the whole ball B _R (accross y = 0).

We must clarify in what sense we take the limit in (4.1). In this particular case, since the limit vanishes, we could prove that actually u is C ^∞ in x and the limit holds in the uniform sense. In general it would be convenient to have a weak definition of (4.1). The equation (4.2) becomes the definition of (4.1) in the weak sense. In any case the limit (4.1) holds in the sense that, if we take any smooth test function ,

For a general class of functions g, the problem has a unique solution in the weighted Sobolev space H ^{1, 2} (B _R , |y| ^a ). This kind of problems and general properties of elliptic equations with A ₂ weights are studied in Fabes et al. (1982b).

We now address the nondivergence case.

Lemma 4.2

Given a continuous function g on ∂ B _R , such that g(x, z) = g(x, − z), there exists a unique function such that:

i.	u solves u xx  + |z|α u zz  = 0 in B R  ∪ {z ≠ 0} in the classical sense.
ii.	u ∈ C 1(B R )
iii.	u z (x, 0) = 0

Moreover, for this solution the Harnack inequality result of Caffarelli and Gutierrez (1997) applies.

Proof

We point out that any viscosity solution of (1.8) is C ² away from z = 0, so it would be a solution in the classical sense in B _R  ∪ {z ≠ 0}.

Let us prove uniqueness first. Let us suppose there were two solutions u and v satisfying (i), (ii), and (iii). For an arbitrary ϵ > 0, consider w = u − v + ϵ |z|. Then w ≤ ϵ R on ∂ B _R . Let us suppose that w has an interior maximum at a point x ∈ B _R . This point cannot be in B _R  ∩ {z ≠ 0}, since there w solves the non-degenerate elliptic equation (1.8) and thus it does not have an interior maximum. Therefore, if there is any interior maximum, it has to be on {z = 0} ∩ B _R . But clearly there cannot be a maximum since ∂ _z ⁺ w > ∂ _z ⁻ w. Therefore w < ϵ R in the whole ball B _R . Since ϵ is arbitrary, we conclude that u ≤ v in B _R . Similarly we can obtain that v ≤ u in B _R , so they must coincide. Notice that we only used (i) and (ii) for uniqueness.

Now, let us prove existence. The subtle point here is to show that there is a C ¹ solution. What we do is to prove a uniform C ^{1, α} estimate for solutions to the problem For any ϵ > 0, by the Schauder theory, this problem has classical solutions. If we have a C ^{1, η} estimate uniform in ϵ, we take limit as ϵ → 0 and obtain the desired solution.

The solutions u ^ϵ are uniformly bounded in L ^∞ due to the maximum principle for uniformilly elliptic equations. The equation (4.4) has constant coefficients with respect to x. The Hölder estimate of Caffarelli and Gutierrez (1997) holds independently of ϵ and we can apply it to u ^ϵ and to any differential quotient in the direction of x to obtain uniform estimates for derivatives of any order with respect to x (In terms of fractional Laplacian, this corresponds to the fact that functions such that (− ▵) ^s u = 0 are C ^∞). Therefore we have that ▵ _x u ^ϵ is bounded independently of ϵ in any smaller ball B _(1−δ/2)R .

Since u ^ϵ ∈ C ²(B _R ) and u is symmetric respect to the hyperplane z = 0, then .

From the equation (4.4):

Recall that and since a ∈ (− 1, 1), then α < 1. We can then integrate for any x, z such that |x| < 1 − 2δ and 0 < z < 1 − δ.

This shows is C ^η in B _(1−δ)R for η = min(1, 1 − α), independently of ϵ for any δ.

So, we can take ϵ → 0 and extract a subsequence that converges to the desired solution u.

Moreover, for any ϵ, u ^ϵ is smooth and satisfies the equation (4.4) that has smooth coefficients and for which the Harnack inequality from Caffarelli and Gutierrez (1997) applies. Therefore, the same estimate passes to the limit as ϵ → 0 and the solution u of the original problem satisfies Harnack inequality.

Remark 4.3

We have to be careful if we want to study the viscosity solutions to (1.8). The naive definition using C ² test functions would not suffice for uniqueness. If, for example, α > 0 and we consider the function u(x, z) = |z|, then for any C ² function φ touching u from below at a point (x, 0), ▵ _x φ(x, 0) + 0^α φ _zz (x, 0) = ▵ _x φ ≤ 0 and thus u(x, z) = |z| would be a (non-differentiable) viscosity solution. However if we test against less regular functions, like φ(x, z) = z ^2−α, then u would fail to be a supersolution. In the notation of Caffarelli et al. (1996), this corresponds to the distinction between C-viscosity solutions and L ^p -viscosity solutions.

5. Harnack and Boundary Harnack Type Estimates

As an application on how to apply the equations (1.5) or (1.8) to the study of fractional harmonic functions, we prove a Harnack inequality and a boundary Harnack inequality using (local) pde methods.

Harnack inequality is not a new result for the fractional Laplacian. It can actually be proved using direct classical potential methods like in Landkof (1972). A boundary Harnack estimate for the fractional Laplacian was first proved in Bogdan (1997) using potential methods.

In this article, we derive the Harnack and boundary Harnack inequality for the fractional Laplacian from the Harnack inequality for singular elliptic equations, either with A ₂ weights (see Fabes et al., 1982b;, 1983; Smith, 1982/83) or for certain classes of nondivergence problems (see Caffarelli and Gutierrez, 1997).

Theorem 5.1 (Harnack Inequality)

Let f:ℝ ⁿ  → ℝ be nonnegative such that (− ▵) ^s f = 0 in B _r . Then there is a constant C (depending only on s and dimension) such that

Proof using (1.5)

We consider the extension u of f that solves (1.5). This function u is going to be nonnegative since f is. We reflect it through the {y = 0} hyperplane. Since (− ▵) ^s f = 0 in B _r , by Lemma 4.1, u is a solution to in the n + 1 dimensional ball of radius r centered at the origin. We can then apply the result of Fabes et al. (1982b) to obtain Harnack inequality for u and thus also for f in half of the ball.

Proof using (1.8)

We consider the extension u of f that solves (1.8). This function u is going to be nonnegative since f is. We reflect it through the {z = 0} hyperplane. Since (− ▵) ^s f = 0 in B _r , then u satisfies the conditions of Lemma 4.2. We can then apply the result of Caffarelli and Gutierrez (1997) to obtain Harnack inequality for u and thus also for f in half of the ball.

Remark 5.2

In the two proofs above we can observe that what is needed for Harnack inequality is that the function u is nonnegative in an n + 1 dimensional ball. The condition f ≥ 0 is indeed sufficient for that, but it is not strictly necessary.

Theorem 5.3 (Boundary Harnack)

Let f, g: ℝ ⁿ  → ℝ be two nonnegative functions such that (− ▵) ^s f = (− ▵) ^s g = 0 in a domain Ω. Suppose that x ₀ ∈ ∂ Ω, f(x) = g(x) = 0 for any x ∈ B ₁\ Ω, and ∂ Ω ∩ B ₁ is a Lipschitz graph in the direction of x ₁ with Lipschitz constant less than 1. Then there is a constant C depending only on dimension such that for any x, y ∈ Ω ∩ B _1/2(x ₀).

Proof

We consider the extension u ⁽¹⁾ of f and u ⁽²⁾ of g that solve (1.5). As before, we reflect u ^(k) through {y = 0} for k = 1, 2 and obtain a solution across this hyperplane through Ω. What we want to do is to find a map that straightens up the domain in order to apply the result of Fabes et al. (1983).

First, since Ω is a Lipschitz domain, we can find a bilipschitz map ϕ₁: ℝ ⁿ  → ℝ ⁿ such that ϕ₁(x ₀) = 0 and ϕ₁(Ω) ∩ B _1/2 = B _1/2 ∩ {x ₁ > 0}. We can extend this map to ℝ ⁿ⁺¹ as constant in the variable y. Now, the functions are solutions of the equation where the matrix b _ij is given by . Since ϕ₁ is bilipschitz, then the eigen values of Dϕ₁ are bounded below and above and b _ij is uniformally elliptic.

Now we take the map ϕ₂ that maps ℝ ⁿ⁺¹\ {x ₁ ≤ 0 ∧ y = 0} into the half space ℝ ⁿ⁺¹ ∩ {x ₁ > 0}. This map is constant in the variables x ₁,…, x _n , and if we write the pair (x ₁, y) in polar coordinates (r, θ), it maps it to (r, θ/2). The eigen values of D ϕ₂ are all equal to 1 but the one in the direction of (i.e., the one in the direction (− y, 0,…, 0, x ₁)), that is 1/2. The functions satisfy the equation where c _ij is given by the matrix 2Dϕ₂(b _ij )Dϕ₂, and then it is uniformailly elliptic.

We can apply the boundary Harnack inequality of Fabes et al. (1983) to the functions and to obtain

Recalling that and , we obtain that for some universal r < 1. By a standard covering argument, (5.2) together with Theorem 5.1 imply (5.1).

As before, the condition in Theorem 5.3 that requires f and g to be nonnegative in the whole ℝ ⁿ is not sharp. In the proof we can see that what we need to apply the result of Fabes et al. (1983) is that the corresponding extensions u ⁽¹⁾ and u ⁽²⁾ are nonnegative in the unit ball of n + 1 dimensions. The advantage is that this is a local condition, therefore we can obtain the standard corollary of boundary Harnack saying that is C ^α in a neighborhood of the origin. In particular is C ^α, which is not strictly a corollary of Theorem 5.3 but of its proof.

Corollary 5.4

Let f and g be as in Theorem 5.3, then is a C ^α function in for some universal α.

Remark 5.5

In Fabes et al. (1983), the boundary Harnack principle is proven for A ₂ weights and Lipschitz domains. An extension of that result to nontangentiably accesible (NTA) domains (as in the last section of Athanasopoulos et al., to appear) would lead to a boudary Harnack principle for the fractional Laplacian to domains only requiring a uniform capacity condition.

6. Monotonicity Formulas

Monotonicity formulas are a very powerfull tool in the study of the regularity properties of elliptic PDEs. They have been used in a number of problems to exploit the local properties of the equations by giving information about the blowup configurations. Since the equation (1.5) represents a harmonic function in n + 1 + a dimensions, it is to be expected that any monotonicity formula known for harmonic functions will have its counterpart for solutions to (1.5). For example, the simplest one would be that if then where B _R ⁺ stands for the n + 1 dimensional half ball {(x, y): |x|² + y ² < 1 ∧ y > 0}.

The weight y ^a is the correct one if we think that the function u is radially symmetric in a + 1 variables and the y is the modulus of those variables as it is explained in Section 2.1.

More interestingly, we have Almgren's frequency formula

Theorem 6.1

If u is a solution to (1.5) in B _{R 0} ⁺ such that for any x in B _{R 0} ⁺, either u(x, 0) = 0 or lim _y→0 y ^a u _y (x, y) = 0, then Moreover, Φ is constant if and only if u is homogeneous.

The proof is essentially the same as for harmonic functions and it is mainly computational. The original proof can be found in Almgren (2000). Here we only need a few minor modifications for our case. First we need the following lemma.

Lemma 6.2

Let u be s solution of (1.5) in B _{R 0}  ∩ {y ≠ 0}, such that y ^a u _y (x, y) is bounded and for every x with |x|<R ₀ either u(x, 0) = 0 or lim _y→0 y ^a u _y (x, y) = 0. Then the following identity holds for any R ≤ R ₀. where u _τ stands for the gradient of u tangential to ∂ B _r .

Proof

The only thing we have to notice is that since u solves (1.5), then So we apply divergence theorem in the set B _R  ∩ {y > ϵ} to obtain The second term comes from the integration of the vector field at the bottom of the half sphere. We observe it goes to zero as ϵ → 0 since there we have that for each x either u(x, 0) = 0 or lim _y→0 y ^a u _y (x, y) = 0 and a < 1. So we can remove that term and extend the expresion to the whole ball B _R . Expanding |∇ u|² = |u _τ|² + |u _ν|² we obtain (6.1).

Proof of Theorem 6.1

We will show that log Φ(R) is nondecreasing. By the usual scaling argument, it is enough to check it at r = 1. We compute log Φ: and now we compute its derivative at R = 1. Let S ₁ = ∂ B ₁ ∩ {y > 0}, then

Now we observe that since div (y ^a ∇ u) = 0, we have div y ^a u ∇ u = y ^a |∇ u|², and then

From Lemma 6.2 we have that

Putting (6.3) and (6.4) together with (6.2) we obtain

Therefore (log Φ)^′ (1) ≥ 0 follows from Cauchy–Schwartz inequality since and the equality is achieved if and only if u _ν = λ u for some constant λ. Therefore, only when u is homogeneous of degree λ.

7. Other Operators

The results in this article suggest that many integro-differential operators can be thought in the same way. Let us suppose we have smooth uniformilly elliptic coefficients a _ij in ℝ ⁿ  × [0, ∞). Consider the following extension problem:

If we now consider the operator f ↦ u _y (x, 0), this is going to be some integro-differential operator of degree one that we can also obtain from a local type pde. We can push the situation even further by considering coefficients a _ij that are only measurable, and maybe singular near y = 0. It is not reasonable to expect that every integro-differential operator can be realized in this way. It looks difficult to characterize the ones that can. But certainly for many it is possible.

Let us consider the case when the operators are invariant under translation in x. We want to study the operator that maps the Dirichlet to the Neumann condition for the equation we take the Fourier transform in x and proceed as in Section 3.2 to obtain the differential equation

The operator T: u(x, 0) ↦ − u _z (x, 0) is then the pseudodifferential operator whose symbol s(ξ) is given by solving (7.3) and computing φ _z (0) for each value of ξ.

The question is what symbols s(ξ) can we obtain by the above procedure. We can see that s(ξ) is radially symmetric and monotone increasing in |ξ|. Indeed, if φ₁ is a solution of (7.3) for |ξ| = r ₀ and ϕ₁ for |ξ₁| = r ₁ ≥ r ₀ then Thus φ₁ < φ₀ by comparison principle.

As ξ → 0, the solution φ of (7.3) converges to φ ≡ 1, so s(0) = 0. We leave the question if every radially symmetric symbol s such that s(0) = 0 and it is monotone increasing with respect to |ξ| can be obtained by a suitable choice of a(z).