MULTIDIMENSIONAL KOLMOGOROV-PETROVSKY TEST FOR THE BOUNDARY REGULARITY AND IRREGULARITY OF SOLUTIONS docx

Petrovsky This paper establishes necessary and sufficient condition for the regularity of a charac-teristic top boundary point of an arbitrary open subset ofRN+1N ≥2 for the diffusion or he

FOR THE BOUNDARY REGULARITY AND IRREGULARITY

OF SOLUTIONS TO THE HEAT EQUATION

UGUR G ABDULLA

Received 25 August 2004

Dedicated to I G Petrovsky

This paper establishes necessary and sufficient condition for the regularity of a charac-teristic top boundary point of an arbitrary open subset ofRN+1(N ≥2) for the diffusion (or heat) equation The result implies asymptotic probability law for the standard

N-dimensional Brownian motion

1 Introduction and main result

Consider the domain

Ωδ =
(x,t) ∈ R N+1:| x | < h(t), − δ < t < 0

whereδ > 0, N ≥2,x =(x1, ,x N)∈ R N,t ∈ R,h ∈ C[ − δ,0], h > 0 for t < 0 and h(t) ↓0

ast ↑0

Foru ∈ C2,1x,t(Ωδ), we define the diffusion (or heat) operator

Du = u t − ∆u = u t −

N



i =1

u x i x i, (x,t) ∈Ωδ (1.2)

A functionu ∈ C x,t2,1(Ωδ) is called parabolic inΩδifDu =0 for (x,t) ∈Ωδ Let f : ∂Ω →

Rbe a bounded function First boundary value problem (FBVP) may be formulated as follows

Find a functionu which is parabolic in Ω δand satisfies the conditions

f ∗ ≤ u ∗ ≤ u ∗ ≤ f ∗ forz ∈ ∂Ω δ, (1.3) where f ∗,u ∗(or f ∗,u ∗) are lower (or upper) limit functions of f and u, respectively.

Assume thatu is the generalized solution of the FBVP constructed by Perron’s

su-persolutions or subsolutions method (see [1,6]) It is well known that, in general, the generalized solution does not satisfy (1.3) We say that a point (x0,t0)∈ ∂Ω δis regular if, for any bounded function f : ∂Ω → R, the generalized solution of the FBVP constructed

by Perron’s method satisfies (1.3) at the point (x0,t0) If (1.3) is violated for some f , then

(x0,t0) is called irregular point.

Boundary Value Problems 2005:2 (2005) 181–199

DOI: 10.1155/BVP.2005.181

The principal result of this paper is the characterization of the regularity and irregu-larity of the origin (ᏻ) in terms of the asymptotic behavior of h as t ↑0

We writeh(t) =2(t logρ(t))1/2, and assume thatρ ∈ C[ − δ,0], ρ(t) > 0 for − δ ≤ t < 0; ρ(t) ↓0 ast ↑0 and

logρ(t) = o

log

(seeRemark 1.2concerning this condition) The main result of this paper reads as follows

Theorem 1.1 The origin (ᏻ) is regular or irregular according as

 0− ρ(t)logρ(t)N/2

diverges or converges.

For example, (1.5) diverges for each of the following functions

ρ(t) =log| t |−1

, ρ(t) =
log| t |log(N+2)/2log| t |−1

,

ρ(t) =



log| t |log(N+2)/2log| t | n

k =3 logk | t |

−1 , n =3, 4, , (1.6)

where we use the following notation:

log2| t | =loglog| t |, log

n | t | =log logn −1 | t |, n ≥3. (1.7) From another side, (1.5) converges for each function

ρ(t) =log| t |−(1+
)

, ρ(t) =
log| t |log(N+2)/2+
log| t |−1

,

ρ(t) =
log| t |log(N+2)/2log| t |log1+

3 | t |−1,

ρ(t) =
log| t |log(N+2)/2log| t |log

3| t |log1+
4 | t |−1,

(1.8)

and so forth, where
> 0 is sufficiently small number.

If we takeN =1, thenTheorem 1.1coincides with the result of Petrovsky’s celebrated paper [6] From the proof ofTheorem 1.1, it follows that if (1.5) converges (in particular, for any example from (1.8)), then the functionu(x,t) which is parabolic in Ω δ, vanishes

on the lateral boundary ofΩδand is positive on its bottom, cannot be continuous at the pointᏻ, and its upper limit at ᏻ must be positive

It should be mentioned that Wiener-type necessary and sufficient condition for boundary regularity is proved in [2] However, it seems impossible to deriveTheorem 1.1

from Wiener condition

As in [6], a particular motivation for the consideration of the domainΩδis the prob-lem about the local asymptotic behavior of the Brownian motion trajectories for the

diffusion processes We briefly describe the probabilistic counterpart ofTheorem 1.1in

the context of the multidimensional Brownian motion Consider the standard

N-dimen-sional Brownian motion

Ᏸ=ξ(t) =x1(t),x2(t), ,x N(t)

:t ≥0,P •

in which the coordinates of the sample path are independent standard 1-dimensional Brownian motions andP •( B) is the probability of B as a function of the starting point ξ(0)

of theN-dimensional Brownian path (see [3]) Consider the radial partr(t) =(x2(t) +

x2(t) + ···+x2N(t))1/2:t ≥0 of the standardN-dimensional Brownian path

Blumen-thal’s 01 law implies thatP0[r(t) < h(t), t ↓0]=0 or 1;h is said to belong to the upper

class if this probability is 1 and to the lower class otherwise The probabilistic analog of

Theorem 1.1states that ifh ∈↑and ift −1/2 h ∈↓for smallt > 0, then h belongs to the upper

class or to the lower class according as

 0+t − N/2 −1h N(t)exp − h2

2t



converges or diverges WhenN =1, this is well-known Kolmogorov-Petrovsky test Note that the integral (1.10) reduces to (1.5) (with coefficient 2N/2) if we replaceh2(t) with

−2t logρ( − t) By adapting the examples (1.6) and (1.8), we easily derive that for any positive integern > 1, the function

h(t) = 2t



log21

t+

N + 2

2 log3

1

t+ log4

1

t+···+ logn −11

t+ (1 +
) logn1

t

 1/2

(1.11)

belongs to the upper or to the lower class according as
> 0 or
≤0

Obviously, one can replace the integral (1.10) with the simpler one for the function

h1(t) = t −1/2 h(t):

 0+t −1 h N1(t)exp − h2

2



It should be mentioned that the described probabilistic counterpart ofTheorem 1.1

is well known (see survey article [5, page 181]) and there are various known proofs of the N-dimensional Kolmogorov-Petrovsky test in the probabilistic literature (see [3]) Recently in [4], a martingale proof of theN-dimensional Kolmogorov-Petrovsky test for

Wiener processes is given

Remark 1.2 It should be mentioned that we do not need the condition (1.4) for the proof of the irregularity assertion ofTheorem 1.1and it may be replaced with the weaker assumption thatt log(ρ(t)) →0 ast ↓0 The latter is needed just to makeᏻ the top bound-ary point ofΩδ For the regularity assertion ofTheorem 1.1, the assumption (1.4) makes almost no loss of generality First of all, this condition is satisfied for all examples from (1.6) and (1.8) Secondly, note that the class of functions satisfying (1.4) contains the class

of functions satisfying the following inequality:

ρ(t) ≥ ρ M C =log(Ct)− M (1.13)

for all small| t |and for someC < 0, M > 1 Since the integral (1.5) is divergent, the func-tionρ(t) may not satisfy (1.13) with reversed inequality and for all small| t |because (1.5)

is convergent for each functionρ M C(t) Accordingly, the condition (1.13), together with divergence of (1.5), excludes only pathological functions with the property that in any small interval−
< t < 0 they intersect infinitely many times all the functions ρ M

C, with

C < 0, M > 1 We handle this kind of pathological functions inSection 3within the proof

of the irregularity assertion Finally, we have to mention that the assumption (1.4) (or even (1.13)) makes no loss of generality in the probabilistic context Indeed, since (1.10)

is divergent, any functionh(t) =(−2t logρ M C(− t))1/2withC < 0 and M > 1 belongs to the

lower class Hence, to get improved lower functions, it is enough to stay in the class of functionsh(t) =(−2t logρ( − t))1/2withρ satisfying (1.13) (or (1.4))

We present some preliminaries inSection 2 The proof of the cheap irregularity part of

Theorem 1.1is presented inSection 3, while a regularity assertion is proved inSection 4

2 Preliminary results

LetΩ⊂ R N+1(N ≥2) denote any bounded open subset and∂Ω its topological boundary.

For a given pointz0=(x0,t0) and a positive number
, define the cylinder

Q

z0,
=
z =(x,t) :x − x0<
,t0−
< t < t0

For the definition of the parabolic boundaryᏼΩ, lateral boundary ᏿Ω, and basic facts about Perron’s solution, super- and subsolutions of the FBVP, we refer to the paper in [1]

It is a standard fact in the classical potential theory that the boundary pointz0∈᏿Ω is regular if there exists a so-called “regularity barrier”u with the following properties:

(a)u is superparabolic in U = Q(z0,
)∩Ω for some
> 0;

(b)u is continuous and nonnegative in U, vanishing only at z0.

It is also a well-known fact in the classical potential theory that in order to prove the irregularity of the boundary pointz0∈᏿Ω, it is essential to construct a so-called irregu-larity barrieru with the following properties:

(a)u is subparabolic in U = Q(z0,
)∩Ω for some
> 0;

(b)u is continuous on the boundary of U, possibly except at z0, where it has a re-movable singularity;

(c)u is continuous in U \{ z0}and

lim sup

z → z0 ,z ∈ U u > limsup

z → z0 ,z ∈ ∂U

Obviously, we have

ᏼΩδ = ∂Ω δ, ᏿Ωδ =
z : | x | = h(t), − δ < t ≤0

Assume that all the boundary pointsz ∈᏿Ω\{ᏻ}are regular points For example, this is the case ifρ(t) is differentiable for t < 0 Then concerning the regularity or irregularity of

ᏻ, we have the following

Lemma 2.1 The origin (ᏻ) is regular for Ω δ if and only if there exists a regularity barrier u for ᏻ regarded as a boundary point of Ω δ for su fficiently small δ.

The proof is similar to the proof of Lemma 2.1 of [1]

Lemma 2.2 The origin ( ᏻ) is irregular for Ω δ if and only if there exists an irregularity barrier u for ᏻ regarded as a boundary point of Ω δ for sufficiently small δ.

Proof The proof of the “if ” part is standard (see [6]) Take a boundary function f = u

at the points nearᏻ (at ᏻ define it by continuity) and f = c at the rest of the boundary

withc > sup | u | Letu = HΩδ

f be Perron’s solution Applying the maximum principle to

u − u in domains Ω δ ∩ { t <
< 0 }and passing to limit as
↑0, we derive thatu ≥ u in

Ωδ In view of property (c) of the irregularity barrier, we have discontinuity ofu at ᏻ.

To prove the “only if ” part, take f = − t and let u = HΩδ

f be Perron’s solution Since all the boundary pointsz0∈ᏼΩδ,z0 ᏻ are regular points, u is continuous in Ω δ\ᏻ and in view of the maximum principle, it is positive inΩδ Therefore,u must be discontinuous

atᏻ Otherwise, it is a regularity barrier and we have a contradiction withLemma 2.1

The next lemma immediately follows from Lemmas2.1and2.2

Lemma 2.3 Let Ω be a given open set inRN+1 andᏻ∈ ᏼΩ, Ω − , whereΩ− = { z ∈

Ω : t < 0 } IfΩ− ⊂Ωδ , then from the regularity of ᏻ for Ω δ , it follows that ᏻ is regular for

Ω Otherwise speaking, from the irregularity of ᏻ for Ω or Ω − , it follows that ᏻ is irregular forΩδ

Obviously, “if ” parts of both Lemmas2.1and2.2are true without assuming that the boundary pointsz ∈᏿Ω\{ᏻ}are regular points

3 Proof of the irregularity

First, we prove the irregularity assertion ofTheorem 1.1by assuming thatρ(t) is

differ-entiable fort < 0 and

tρ (t)

Under these conditions, we construct an irregularity barrieru, exactly as it was done in

[6] for the caseN =1 Consider the function

v(x,t) = − ρ(t)exp − | x |2

4t



which is positive inΩδand vanishes on᏿Ωδ Since 0≤ v ≤1 inΩδ, we have

lim

t ↑0 v(0,t) =lim sup

Hence,v satisfies all the conditions of the irregularity barrier besides subparabolicity We

have

Dv = − ρ (t) − Nρ(t)

2t

 exp − | x |2

4t



Sinceρ(t) ↓0 ast ↑0,Dv > 0 and accordingly, it is a superparabolic function We consider

a functionw with the following properties

w(0,t) ≤1

Clearly, the functionu(x,t) = w(x,t) + v(x,t) would be a required irregularity barrier As

a functionw, we choose a particular solution of the equation from (3.5):

w(x,t) = − 1

(4π) N/2



Ωδ \Ω t

exp

− | x − y |2/4(t − τ) (t − τ) N/2 Dv(y,τ)dy dτ. (3.7) SinceDv > 0 in Ω δ,w is negative and we only need to check that for sufficiently small δ,

(3.6) is satisfied From (3.1) it follows that

| Dv | < C1



ρ(t) t exp − | x |2

4t



whereC1= C + N/2 and C is a constant due to (3.1) Hence,

w(0,t)< C1

(4π) N/2

t

− δ

ρ(τ)

| τ |(t − τ) N/2



B((4τ logρ(τ))1/2)exp − | y |2t

4(t − τ)τ



dy dτ, (3.9)

whereB(R) = { y ∈ RN:| y | < R } Changing the variable in the second integral, we have

w(0,t)< C1

(4π) N/2

t

− δ

ρ(τ)

| τ |(t − τ) N/2

τ t

N/2

B((4t logρ(τ))1/2)exp − | y |2

4(t − τ)



dy dτ.

(3.10)

We splitt

− δinto two parts ast

2t+ 2t

− δand estimate the first part as follows:



t

2t

ρ(τ) τ(t − τ) N/2

τ t

N/2

B((4t logρ(τ))1/2)exp − | y |2

4(t − τ)



dy dτ



< 2 N

t

2t

ρ(τ) τ

τ t

N/2

RNexp

− | y |2 

dy dτ



< 2 N(2π) N/2

t

2t

ρ(τ)

| τ | dτ.

(3.11)

From the convergence of the integral (1.5), it follows that the right-hand side of (3.11) converges to zero ast ↑0 We also have



−2t

δ

ρ(τ) τ(t − τ) N/2

τ t

N/2

B((4t logρ(τ))1/2)exp − | y |2

4(t − τ)



dy dτ



< ω N



−2t

δ

ρ(τ) τ

τ t

N/2 2

− τ

N/2

4t logρ(τ)N/2

dτ



=23N/2 ω N

2t

− δ

ρ(τ)logρ(τ)N/2

(3.12)

whereω Nis the volume of the unit ball inRN Hence, from the convergence of the integral (1.5), it follows that| w(0,t) | < 1/2 for − δ < t < 0 if δ is sufficiently small.

Now we need to remove the additional assumptions imposed on ρ To remove the

differentiability assumption, consider a function ρ1( t) such that ρ1isC1fort < 0, ρ1↓0 as

t ↑0 andρ(t) < ρ1(t) < 2ρ(t) for − δ ≤ t < 0 Then we consider a domain Ω1

δby replacingρ

withρ1inΩδ Since the integral (1.5) converges forρ, it also converges for ρ1 Therefore,

ᏻ is irregular point regarded as a boundary point of Ω1

δ SinceΩ1

δ ⊂ΩδfromLemma 2.3,

it follows thatᏻ is irregular point regarded as a boundary point of Ωδ

We now prove that the assumption (3.1) imposed onρ may be also removed In fact,

exactly this question was considered in [6] However, there is a point which is not clearly justified in [6] and for that reason, we present a slightly modified proof of this assertion Consider a one-parameter family of curves

ρ C(t) =log(Ct)−3

, C < 0, C −1 < t < 0. (3.13) Obviously, for each point (ρ(t),t) on the quarter plane, there exists a unique value

C = C(t) = t −1exp

− ρ −1/3(t)

such thatρ C(t) passes through the point (ρ(t),t) One cannot say anything about the

behavior ofC(t) as t ↑0 But it is clear thattC(t) ↓0 ast ↑0 It is also clear that ifC1<

C2< 0, then ρ C1(t) > ρ C2(t) for C −1< t < 0 It may be easily checked that for any C < 0,

the functionρ C(t) satisfies all the conditions which we used to prove the irregularity of

ᏻ Accordingly, ᏻ is irregular point regarded as a boundary point of Ωδwithρ replaced

byρ C By usingLemma 2.3, we conclude that if for someC < 0 and t0< 0,

thenᏻ must be irregular regarded as a boundary point of Ωδ Hence, we need only to consider the functionρ with the property that for arbitrary C < 0 and t0< 0, the

inequal-ity (3.15) is never satisfied Sinceρ C(C −1+ 0)=+∞, it follows that within the interval (− δ,0), our function ρ(t) must intersect all the functions ρ C(t) with C ≤ − δ −1 There-fore, at least for some sequence { t n }, we haveC(t n)→ −∞ as t n ↑0 In [6], Petrovsky

introduced the setM formed by all values of t (0 > t > − δ) with the following property

(which is called “conditionC” in [6]): the curveρ C(t) which passes through the point

(ρ(t),t) cannot intersect the curve ρ = ρ(t) for any smaller value of t > − δ Denote by

M the closure of M It is claimed in [6] that “C(t) monotonically decreases as t ↑0 and

t ∈ M; moreover, C(t) takes equal values at the end points of every interval forming the

complement ofM.”

We construct a functionρ which shows that this assertion is, in general, not true

Con-sider two arbitrary negative and strictly monotone sequences{ C(1n) },{ C(2n) },n =0, 1, 2,

such thatC10= C20> − δ −1and

C(1n) ↓ −∞, C2(n) ↑0 asn ↑ ∞ (3.16)

We form by induction a new sequence{ C n}via sequences{ C1(n) }and{ C2(n) }:

C0= C2= C(0)1 , C1= C1(1), C3= C(1)2 ,

C4n = C4n −3, C4n+1 = C(1n+1), C4n+2 = C4n −1,

C4n+3 = C2(n+1), n =1, 2,

(3.17)

The sequence{ C n}has arbitrarily large oscillations between −∞and 0 asn ↑ ∞ Our purpose is to construct a functionρ(t), − δ < t < 0 in such a way that the related function

C = C(t) will satisfy

C

a n

at some pointsa n We now construct the sequence{ a n }by induction:

a0= − δ, 0> a n+1 > max a n; C n

C n+1 a n;

1 (n + 1)C n+1

 ,n =0, 1, 2, (3.19)

Having{ a n}, we define the values of the functionρ at the end-points of intervals (a n,

a n+1),n =0, 1, 2, , as

ρ

a n

=log

C n a n−3

From (3.19) it follows thatρ(a n)↓0 asn ↑ ∞ Having the values{ ρ(a n)}, we construct monotonically decreasing functionρ(t) as follows: ρ is C1for− δ ≤ t < 0 and if C n+1 < C n

(resp.,C n+1 > C n) then within the interval [a n;a n+1], ρ(t) intersects each function x =

ρ C(t) with C n+1 ≤ C ≤ C n (resp., withC n ≤ C ≤ C n+1) just once, and moreover at the intersection point, we have

ρ (t) ≥(resp., ≤)ρ  C(t). (3.21) Obviously, it is possible to make this construction Clearly, the related functionC = C(t)

satisfies (3.18) It has infinitely large oscillations near 0 and for arbitraryC satisfying

−∞ ≤ C ≤0, there exists a sequencet n ↑0 asn ↑ ∞such thatC(t n)→ C One can easily

see that according to the definition of the setM given in [6], we have

+∞



n =0



a4n,a4n+1

∪a4n+2,a4n+3



In view of our definition, we haveC4n+1 < C4n,C4n+3 > C4n+2,n =0, 1, 2, Accordingly, C(t) is neither monotonically increasing nor monotonically decreasing function as t ↑0 andt ∈ M.

We now give a modified definition of the setM It is easier to define the set M in terms

of the functionC(t):

M =
t ∈[− δ,0) : C1(t) = C(t)

whereC1(t) =min− δ ≤ τ ≤ t C(t) Denote by (M) cthe complement ofM Since (M) cis open set, we have



Mc

n



t2n −1, t2n



From the definition, it follows thatC(t) monotonically decreases for t ∈ M and,

more-over, we have

C

t2n −1

= C

t2n



Indeed, we taket ,t ∈ M with t  < t SinceC1(t )= C(t ) andC1(t )= C(t ), it follows thatC(t )≤ C(t ) Fort ,t ∈ M, the same conclusion follows in view of continuity of C(t) To prove (3.25), first note that sincet2n −1,t2n ∈ M, we have C1(t2n −1)= C(t2n −1) andC1(t2n)= C(t2n) If (3.25) is not satisfied, then we haveC1(t2n −1) > C1(t2n) SinceC1

is continuous function, there exists
∈(0,t2n − t2n −1) such that C1(t2n −
)< C1(t2n −1).

LetC1(t2n −
)= C(θ) Obviously, θ ∈(t2n −1,t2n −
] andC1(θ) = C(θ) But this is the

contradiction with the fact that (t2n −1, t2n)∈(M) c Hence, (3.25) is proved

If we apply the modified definition ofM to the example constructed above, then one

can easily see that

+∞



n =0

a4n,a4n+1

Mc

=

+∞



n =0



a4n+1,a4(n+1)

,

C

a4n+1



= C1(n+1) = C

a4(n+1)



= C4(n+1) −3 = C4n+1 ↓ −∞ asn ↑ ∞

(3.26)

Now we define the new functionρ1(t) as follows:

(a)ρ1(t) = ρ(t) for t ∈ M;

(b)ρ1(t) = |log(C(t2n −1)t) | −3fort2n −1< t < t2n

Equivalent definition might be given simply by takingρ1(t) = |log(C1(t)t) | −3,− δ ≤ t < 0.

Otherwise speaking, the functionC(t) defined for ρ1(t) via (3.14) coincides withC1(t).

Obviously,ρ1is continuous function satisfyingρ1(t) ≥ ρ(t) and possibly ρ1(t) ρ(t) on a

numerate number of intervals (t2n −1, t2n) This new function may be nondifferentiable at the pointst = t2n −1, t2n Therefore, we consider another functionρ2(t) with the following

(a)ρ2isC1fort < 0;

(b)ρ2(t) ≥ ρ1(t);

(c)ρ2(t) satisfies everywhere weak condition C: the curve x = ρ C(t) which passes

through the point (ρ2(t),t) may not satisfy the condition ρ C(t) < ρ2(t) for any

smaller value oft > − δ;

(d) for arbitrary
with− δ <
< 0, we have



−

δ

1

t



ρ1(t)logρ1(t)N/2 − ρ2(t)logρ2(t)N/2

dt

< 1. (3.27)

Obviously, this function may be constructed Again, it is easier to express this construc-tion in terms of the related funcconstruc-tionC(t) Having a function C1(t), we consider a function

C2(t) which is C1fort < 0, monotonically decreasing, C2(t) ≤ C1(t) for all − δ ≤ t ≤0 and

tC2(t) →0 ast ↑0 Then we consider a functionρ2(t) as

ρ2(t) =log

C2(t)t−3

Monotonicity ofC2(t) is equivalent to the property (c) of ρ2 Finally, (d) will be achieved

by choosingC2(t) close to C1(t) The rest of the proof coincides with Petrovsky’s proof

from [6] First, it is easy to show thatρ2(t) satisfies (3.1) We have



tρ  C(t)

ρ C(t)



 =log(3Ct), (3.29) and the right-hand side is arbitrarily small for sufficiently small Ct From the property

(c) of the functionρ2(t), it follows that

ρ 

2(t) ≤ ρ 

C(t) = 3

t log4(Ct)



provided thatC = C2(t) or equivalently (3.28) is satisfied Hence, we have



tρ 2(t)

ρ2(t)



 =log3

C2t 

SincetC2→0 ast ↑0, the right-hand side is arbitrarily small for small| t |

Consider a domainΩ2

δby replacingρ with ρ2inΩδ Sinceρ2(t) ≥ ρ1(t), we have Ω2

δ ⊂

Ωδ FromLemma 2.3, it follows that ifᏻ is an irregular point regarded as a boundary point ofΩ2

δ, then it is also irregular point regarded as a boundary point ofΩδ

It remains only to show that the convergence of the integral (1.5) withρ implies the

convergence of the integral (1.5) withρ = ρ2 In view of the property (d) ofρ2, it is enough

to show the convergence of the integral (1.5) withρ = ρ1 Having a modified definition

of the setM, the elegant proof given in [6] applies with almost no change The proof of the irregularity assertion is completed

For the definition of the parabolic boundary? ??Ω, lateral boundary ᏿Ω, and basic facts about Perron’s solution, super- and subsolutions of the FBVP, we refer to the paper in [1]... similar to the proof of Lemma 2.1 of [1]

Lemma 2.2 The origin ( ᏻ) is irregular for Ω δ if and only if there exists an irregularity barrier u for ᏻ regarded as a boundary