Helms L.L. Potential Theory

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10.4 Nonconstant Coeﬃcients 387

Theorem 10.4.1 If Ω is any spherical chip with Σ = int(∂Ω∩R

) or a ball

B ⊂ R

with Σ = ∅, α ∈ (0, 1),b ∈ (−1, 0), u ∈ H

(b)

2+α

(Ω) satisﬁes Lu = f

for f ∈ H

(2+b)

(Ω) and Mu = g for g ∈ H

(1+b)

1+α

(Σ), then there is a constant

C = C(a

,α,b,β

,γ,d(Ω)) such that

|u|

(b)

2+α,Ω

≤ C





u

0,Ω

+ |g|

(1+b)

1+α,Σ

+ |f |

(2+b)

α,Ω



Proof: Consider any ρ>0, let μ<1/2 be a constant to be speciﬁed later,

let x

∈

2ρ

,andletB = B

,2μρ

.ThenB ∩Ω ⊂

since |y −x

| < 2μρ

implies that

≥

− 2μρ > 2ρ − 2μρ > ρ.Fory

∈ B

,μρ

∩Ω,

(2ρ)

2+b+α

u(x

) − D

u(y

− y

≤





2+b+α

(μρ)

2+b+α

|u|

2+α,



(Ω∩B)

μρ

≤





2+b+α

|u|

(b)

2+α,Ω∩B

(10.21)

Since |y

− x

|≥μρ for y

∈ B

,μρ

∩ Ω,

(2ρ)

2+b+α

u(x

) − D

u(y

− y

≤ (2ρ)

2+b+α

(|D

u(x

)| + |D

u(y

)|)

≤

α+1

|u|

(b)

2+0,Ω

By Theorem 7.6.1, there is a constant C

= C

(b, α, μ, d(Ω)) such that

|u|

(b)

2+0,Ω

≤ C



u

0,Ω

+ μ

2α

|u|

(b)

2+α,Ω

. (10.22)

Therefore, for y

∈ B

,μρ

∩ Ω

(2ρ)

2+b+α

u(x

) − D

u(y

− y

≤

α+1



u

0,Ω

α+1

|u|

(b)

2+α,Ω

(10.23)

Combining Inequalities (10.21) and (10.23) and using the fact that x

∈

2ρ

|u|

(b)

2+α,Ω

≤





2+b+α

|u|

(b)

2+α,Ω∩B

α+1



u

0,Ω

α+1

|u|

(b)

2+α,Ω

Further restricting μ so that 2

α+1

< 1/2,

|u|

(b)

2+α,Ω

≤ C



u

0,Ω





2+b+α

|u|

(b)

2+α,Ω∩B

(10.24)

388 10 Apriori Bounds

where C

= C

(b, α, μ). Consider the term |u|

(b)

2+α,Ω∩B

. Deﬁning operators

and M

as in Equations (10.17) and (10.18), the equations Lu = f and

Mu = g can be written

u = F =(L

u − Lu)+f

and

u = G =(M

u − Mu)+g

respectively. Applying Theorem 10.3.5 to L

and M

|u|

(b)

2+α,Ω∩B

≤ C





u

0,Ω∩B

+ |G|

(1+b)

1+α,Σ∩B

+ |F |

(2+b)

α,Ω∩B



where C

= C

(n, a

,b,α,β

,γ). Thus,

|u|

(b)

2+α,Ω∩B

≤ C





u

0,Ω

+ |g|

(1+b)

1+α,Σ

+ |f |

(2+b)

α,Ω

+ |M

u − Mu|

(1+b)

1+α,Σ∩B

+ |L

u − Lu|

(2+b)

α,Ω∩B



(10.25)

Note that

u − Lu =



i,j=1

) − a

u −



i=1

u − cu

and

− Mu =



i=1

(β



) − β

u +(γ(x



) − γ)u.

Consider |(a

)−a

(2+b)

α,Ω∩B

. By Inequality (10.19), there is a constant

C = C(b, α) such that

|(a

) − a

(2+b)

α,Ω∩B

≤ C|a

) − a

(0)

α,Ω∩B

(2+b)

α,Ω∩B

≤ C|a

) − a

(0)

α,Ω∩B

|u|

(b)

2+α,Ω∩B

Since

) − a

(0)

α,Ω∩B

=sup

>0



) − a

α,



(Ω∩B)





(Ω ∩ B)



= ∅ for >2μρ,and



(Ω ∩ B)



⊂ Ω ∩ B ⊂

) − a

(0)

α,Ω∩B

≤ 2

) − a

α,

≤ 2

) − a

(0)

α,Ω

Note also that

) − a

(0)

α,Ω

≤|a

(0)

α,Ω

+ |a

(0)

α,Ω

≤a



0,Ω

+ |a

(0)

α,Ω

10.4 Nonconstant Coeﬃcients 389

Thus,

|(a

) − a

(2+b)

α,Ω∩B

≤ Cμ



a



0,Ω

+ |a

(b)

α,Ω



|u|

(b)

2+α,Ω∩B

where C = C(b, α). Using Inequalities (7.25) and (7.26) and arguments sim-

ilar to the above,

(2+b)

α,Ω∩B

≤ Cμ

(1)

α,Ω

|u|

(b)

2+α,Ω∩B

≤ CAμ

|u|

(b)

2+α,Ω∩B

|cu|

(2+b)

α,Ω∩B

≤ Cμ

|c|

(2)

α,Ω

|u|

(b)

2+α,Ω∩B

≤ CAμ

|u|

(b)

2+α,Ω∩B

Since |u|

(b)

2+α,Ω

=sup

0<δ<1

2+b+α

|u|

2+α,

< +∞, u and its derivatives of

order two or less have continuous extensions to the closures of each

without

increasing norms and therefore |u|

(b)

2+α,Σ∩B

≤|u|

(b)

2+α,Ω∩B

. Using arguments

similar to the above,

|(β



) − β

(1+b)

1+α,Σ∩B

≤ Cμ



β



0,Σ

+ |β

(0)

1+α,Σ



|u|

(b)

2+α,Ω∩B

≤ CAμ

|u|

(b)

2+α,Ω∩B

|(γ(x



) − γ)u|

(1+b)

1+α,Σ∩B

≤ Cμ



γ

0,Σ

+ |γ|

(1)

1+α,Σ



|u|

(b)

2+α,Ω∩B

≤ CAμ

|u|

(b)

2+α,Ω∩B

It follows that

u − Lu|

(2+b)

α,Ω∩B

+ |M

u − M|

(1+b)

1+α,Σ∩B

≤ CAμ

|u|

((b)

2+α,Ω∩B

Returning to Inequality (10.25),

|u|

(b)

2+α,Ω∩B

≤ C





u

0,Ω

+ |g|

(1+b)

1+α,Σ

+ |f |

(2+b)

α,Ω

+ Aμ

|u|

(b)

2+α,Ω∩B



Further restricting μ so that C

d(Ω)

2+α

Aμ

< 1/2,

|u|

(b)

2+α,Ω∩B

≤ C





u

0,Ω

+ |g|

(1+b)

1+α,Σ

+ |f |

(2+b)

α,Ω



Choosing μ = μ

so that μ

< 1/2, 2

α+1

< 1/2, C

Aμ

< 1/2 and return-

ing to Inequality (10.24),

|u|

(b)

2+α,Ω

≤ C



u

0,Ω

+ C





2+b+α





u

0,Ω

+ |g|

(1+b)

1+α,Σ

+ |f |

(2+b)

α,Ω



≤ C





u

0,Ω

+ |g|

(1+b)

1+α,Σ

+ |f |

(2+b)

α,Ω



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Chapter 11

Oblique Derivative Problem

11.1 Introduction

The Dirichlet problem for an elliptic operator was solved in Chapter 9 by

morphing a solution of the Dirichlet problem for the Laplacian on a ball

into a solution of the Dirichlet problem for an elliptic operator on a ball. In

applying the method of continuity, it was necessary to show that an elliptic

operator with a suitably restricted domain and range is a bounded operator.

The ﬁrst step in mimicking this procedure is to show that the solution of a

mixed boundary value problem for the Laplacian on a spherical chip, as in

Theorem 8.5.7, can be morphed into a solution of the corresponding problem

for an elliptic operator with an oblique derivative condition rather than a

normal derivative condition. This results in the existence of local solutions

of the oblique derivative problem. By adapting the Perron-Wiener-Brelot

procedure, it will be shown that the oblique derivative problem has a solution

for regions satisfying appropriate geometric conditions.

11.2 Boundary Maximum Principle

Let L denote an elliptic operator on Ω ⊂ R

deﬁned by the equation

Lu(x)=



i,j=1

(x)

∂

∂x

(x)+



i=1

(x)

∂u

∂x

(x)+c(x)u(x),x∈ Ω,

with c ≤ 0and

m|x|

≤



i,j=1

(x)x

≤ M|x|

,x∈ Ω, (11.1)

L.L. Helms, Potential Theory, Universitext, 391

 Springer-Verlag London Limited 2009

392 11 Oblique Derivative Problem

for positive constants m, M .AlsoletM be a boundary operator deﬁned for

functions on the relatively open subset Σ of ∂Ω by the equation

Mu(x)=



i=1

(x)

∂u

∂x

(x)+γ(x)u(x),x∈ Σ,

where γ ≤ 0andβ is a vector with β · ν > 0atpointsofΣ and ν = −n

is the inner unit normal at points of Σ. It will be assumed throughout this

chapter that the coeﬃcients of L and M satisfy the condition

A =sup

1≤i,j≤n



a



0,Ω

+ |a

(0)

α,Ω

+ |b

0,Ω

+ |b

(1)

α,Ω

+ |c|

0,Ω

+ |c|

(2)

α,Ω

+ β



0,Σ

+ |β

(0)

1+α,Σ

+ γ

0,Σ

+ |γ|

(1)

1+α,Σ



< +∞.

(11.2)

It is customary to assume that the vector of coeﬃcients β(x)=(β

(x),...,

(x)) in the equation Mu(x



, 0) = g(x



) is normalized by requiring that

|β(x)| =1foreachx ∈ Σ. If this is not the case, the equation Mu(x



, 0) =

g(x



) can be normalized by multiplying both sides of the equation by 1/|β(x)|.

Justiﬁcation of this procedure requires showing that the coeﬃcients β

/|β|

and γ/|β| satisfy the above inequality and that the forcing function g/|β|

satisﬁes conditions required of the forcing function; namely, that g/|β|∈

(1+b)

1+α

(Σ)ifg ∈ H

(1+b)

1+α

(Σ). This can be done under the above additional

conditions on the β

and |β|. It will be assumed in this chapter that

|β(x)| =1for all x ∈ Σ.

In discussions of the oblique derivative problem

Lu = f on Ω and Mu = g on Σ

when Σ = ∂Ω and the problem

Lu = f on Ω, Mu = g on Σ, and u = h on ∂Ω ∼ Σ,

when ∅ = Σ = ∂Ω,thesymbolsf,g,andh will be used solely for functions

in H

(2+b)

(Ω),H

(1+b)

1+α

(Σ), and C

(∂Ω ∼ Σ), respectively. The norm symbols

will be simpliﬁed by omitting the domain of the functions, if clear from the

context; for example, |a

(0)

α,Ω

will be denoted simply by |a

(0)

and |γ|

(1)

1+α,Σ

will be denoted by |γ|

(1)

1+α

. It also will be assumed that b ∈ (−1, 0), α ∈ (0, 1),

c ≤ 0, and γ ≤ 0 without further mention in this chapter. Some sections of

this chapter may require additional restrictions on Ω and Σ which will be

prescribed at the beginning of each such section.

In order to proceed further, it is necessary to estimate the term 

u

0,Ω

Inequality (10.12). The ﬁrst lemma of this section, known as the boundary

point lemma, is valid if Ω is any bounded, convex, open subset of R

and

Σ is a nonempty, relatively open subset of ∂Ω having a unit normal vector

11.2 Boundary Maximum Principle 393

at each point of Σ. After the proof of this lemma and the following corollary,

it will be assumed that Ω is a spherical chip with Σ = int(∂Ω ∩ R

)ora

ball in R

with Σ = ∅.

Lemma 11.2.1 (Boundary Point Lemma, Hopf [33]) Let u ∈ C

(Ω ∪

Σ) ∩ C

(Ω) satisfy Lu ≥ 0 on Ω.Ifforx

∈ Σ,

(i) u(x

) >u(x) for all x ∈ Ω and

(ii) the interior sphere condition is satisﬁed at x

then Mu(x

) < 0, if deﬁned, whenever u(x

) ≥ 0.

Proof: Choose a local coordinate system (z

,...,z

)withoriginatx

that (0,...,0, 1) is the inner unit normal to ∂Ω at x

.LetB

y,ρ

be the ball

with center at y =(0,...,0,ρ

)ofradiusρ

specifying the interior sphere

condition at x

∈ ∂B

y,ρ

.Forz ∈ Ω,letr = |z − y|.Fix0<ρ

<ρ

and

deﬁne w on the closed annulus Γ = B

−

y,ρ

∼ B

y,ρ

by putting

w(z)=e

−kr

− e

−kρ

where k is a constant to be chosen as follows. As in the proof of Lemma 9.5.5

and using the bound given by (11.2),

Lw(z) ≥ e

−kr

(4k

mρ

− 2knA − 2k

√

nAρ

− A)

and k>0 can be chosen so that Lw ≥ 0onΓ. As in the same proof, there is

an >0 such that u − u(x

)+w ≤ 0onΓ .Sinceγ(x

)u(x

) ≤ 0 whenever

u(x

) ≥ 0andw(x

)=0,forx

+ tβ ∈ Γ

u(x

+ tβ) − u(x

)

+ γ(x

)u(x

) ≤

u(x

+ tβ) − u(x

)

≤−

w(x

+ tβ)

= −

w(x

+ tβ) − w(x

)

and it follows that Mu(x

) ≤−D

w(x

). If it can be shown that D

w(x

) >

0, it would follow that Mu(x

) < 0 whenever u(x

) ≥ 0. Letting β =

(β

,...,β

) relative to the local coordinate system,

w(x

)=D

z=0

= β ·∇w|

z=0



i=1

(−2k(z

− y

−kr

z=0

Since y

=0,i=1,...,n− 1andy

= ρ

,whenz =0

w(x

)=D

z=0

=2kρ

−kρ

β ·ν > 0.

394 11 Oblique Derivative Problem

Corollary 11.2.2 (Strong Maximum Principle) If u ∈ C

(Ω

−

)∩C

(Ω)

satisﬁes Lu ≥ 0 on Ω, Mu ≥ 0 on Σ, the interior sphere condition is satisﬁed

at each point of Σ, and either (i) u ≤ 0 on ∂Ω ∼ Σ when nonempty or (ii)

c ≡ 0 on Ω or γ ≡ 0 on Σ when Σ = ∂Ω, then either u =0on Ω or u<0

on Ω and, in particular, u ≤ 0 on Ω.

Proof: If u is a nonpositive constant, the conclusion is trivial in both the

∂Ω ∩ Σ = ∅ and Σ = ∂Ω cases; likewise, u cannot be a positive constant

if ∂Ω ∼ Σ = ∅ and if Σ = ∂Ω, for in the latter case Lu = cu < 0at

some point of Ω or Mu = γu < 0 at some point of Σ, contrary to the

hypotheses. It therefore can be assumed that u is not a constant function.

The two cases Σ = ∂Ω and Σ = ∂Ω will be considered separately. Consider

the latter case ﬁrst. If u attains a negative maximum value at a point x

∈ Ω,

then u ≤ u(x

) < 0; on the other hand, since u is nonconstant, it cannot

attain a nonnegative maximum value at a point of Ω by the strong maximum

principle, Lemma 9.5.6, and must do so at a point x

∈ ∂Ω, which implies that

Mu(x

) < 0 by the boundary point lemma, Lemma 11.2.1, a contradiction.

Therefore, u<0 in the nonconstant case. Suppose now that Σ = ∂Ω.As

above, if u attains a negative maximum at a point x

∈ Ω,thenu<0onΩ.

Since u is not a constant, it cannot attain a nonnegative maximum value

at a point of Ω andmustdosoatapointx

∈ ∂Ω.Thepointx

∈ Σ,

for otherwise Mu(x

) < 0, a contradiction. The nonnegative maximum must

occur at x

∈ ∂Ω ∼ Σ and so u<u(x

) ≤ 0onΩ.

It will be assumed throughout this section that Ω is a spherical chip with

Σ = int(∂Ω ∩ R

)oraballB in R

with Σ = ∅. A function w will be

constructed that will be used to estimate 

u

0,Ω

for Ω a ball or a particular

kind of spherical chip. Let Ω = B

y,ρ

with Σ = ∅ or Ω = B

y,ρ

∩ R

with

Σ = B

y,ρ

∩R

< 0, let K

be a positive constant satisfying



i=1

(|b

+ |β

) ≤ K

let t ∈ (0, 1), let r = |x − y| for x ∈ Ω,andletw =(ρ

− r

)

.Thenfor

x ∈ Ω,

Lw(x)=4t(t − 1)(ρ

− r

)

t−2



i,j=1

− y

)(x

− y

)

−2t(ρ

− r

)

t−1



i=1

+ b

− y

)) + c(ρ

− r

)

Since c ≤ 0, the right side will be increased if the last term is omitted. This

is also the case if the term involving the a

is omitted since the a

are

nonnegative. By Inequality (11.1),



i,j=1

− y

)(x

− y

) ≥ m|x − y|

= mr

11.2 Boundary Maximum Principle 395

Since |x

− y

|≤|x − y|,





i=1

− y

)



≤



i=1

− y

|≤K

from which it follows that



i=1

− y

) ≥−K

r.Thus,forx ∈ Ω

Lw(x) ≤ 4t(t − 1)(ρ

− r

)

t−2

+2t(ρ

− r

)

t−1

≤−t(ρ

− r

)

t−2

Q(r)

where Q(r)=r



4(1 −t)mr −2K

+2K



.SinceQ(r)=0on(0,ρ)when

r = r

−4(1 − t)m +



16(1 − t)

+16K

and Q(ρ)=4(1− t)mρ

> 0,Q(r) ≥ q

> 0forr ∈ [(ρ + r

)/2,ρ]where

= Q((ρ + r

)/2). For x ∈ Ω with r ∈ [(ρ + r

)/2,ρ],

Lw(x) ≤−q

t(ρ

−r

)

t−2

≤−q

(2ρ)

t−2

t(ρ−r)

t−2

≤−C

t(ρ−r)

t−2

(11.3)

where C

= C

(m, b

,β

,ρ)=q

(2ρ)

−2

min (1, 2ρ).

At this point, it is convenient to prove the next lemma in the case that

Ω is a ball with Σ = ∅.Inthiscase,

d = d,and

= Ω

. Suppose u ∈

(Ω

−

) ∩C

(Ω)satisﬁesLu = f on Ω for f ∈ H

(2+b)

(Ω)andu =0on∂Ω.

Consider any x ∈ Ω and r ∈ ((ρ + r

)/2,ρ)forwhichx ∈ Ω

ρ−r

. Letting

v = u −

|f|

(2−t)

Lv(x) ≥ f(x)+|f|

(2−t)

(ρ − r)

t−2

Since |f|

(2−t)

≥ (ρ − r)

2−t

|f|

0,Ω

ρ−r

, Lv(x) ≥ f(x)+|f|

0,Ω

ρ−r

≥ 0. Since x

can be any point of Ω, Lv(x) ≥ 0onΩ;sinceu =0on∂Ω by hypothesis,

v ≤ 0onΩ by the Strong Maximum Principle, Theorem 9.5.6; that is,

u(x) ≤

|f|

(2−t)

(ρ

− r

)

≤

|f|

(2−t)

(2ρ)

(ρ − r)

Since −u satisﬁes the same hypotheses as u,

|u(x)|≤

|f|

(2−t)

(2ρ)

(ρ − r)

Noting that

d(x)

−t

≤ (ρ − r)

−t

and using Inequality (7.26),

−t

(x)u(x)|≤C

|f|

(2−t)

≤ C

|f|

(2−t)

396 11 Oblique Derivative Problem

Letting b = −t and taking the supremum over x ∈ Ω,

u

0,Ω

≤ C

|f|

(2+b)

where C

= C

(m, b

,β

,b,d(Ω)). This is the conclusion of the next lemma

when Ω is a ball.

Returning to the case of a spherical chip, let B be a ball of radius ρ with

center at y =(0, −ρ

),ρ

∈ (0,ρ), let Ω = B

= ∅,andletΣ = B ∩ R

The constant ρ

will be speciﬁed later. As above Lw ≤−C

t(ρ − r)

t−2

for

x ∈ Ω, r ∈ [(ρ + r

)/2,ρ]. As before, the quantity on the right side of

Mw(x



, 0) =



i=1

(−2t)(ρ

− r

)

t−1

− y

)+γ(ρ

− r

)



∈ Σ,

is increased by omitting the last term since γ ≤ 0. Since y =(0,...,0, −ρ

Mw(x



, 0) ≤−2t(ρ

− r

)

t−1



n−1



i=1

+ β



Since |x

|≤|x



|≤



− ρ

, 1 ≤ i ≤ n − 1,

n−1



i=1

≥−K

− ρ

and

Mw(x



, 0) ≤−2t(ρ

− r

)

t−1



− K

− ρ

+ β



Since K

ρ/



+ K

<ρ, ρ

∈ (0,ρ) can be chosen so that

ρ>ρ

≥ max



√



+ K



. (11.4)

For this choice of ρ

, −K



− ρ

+ β

> 0andMw(x



, 0) ≤−C

t(ρ

−

)

t−1

for x ∈ Σ.Sincer<ρfor x =(x



, 0) ∈ Σ, with this choice of ρ

Lw(x) ≤−Ct(ρ − r)

t−2

x ∈ Ω, r ≥ ρ

(11.5)

Mw(x



, 0) ≤−Ct(ρ − r)

t−1



∈ Σ,r ≥ ρ

(11.6)

where C = C(m, b

,β

,d(Ω))

Following Lieberman [39], spherical chips fulﬁlling the above conditions

will play a signiﬁcant role in the following.

Deﬁnition 11.2.3 The spherical chip Ω is admissible if Ω = B

y,ρ

∩ R

where y =(0,...,−ρ

),ρ

∈ (0,ρ), and ρ

satisﬁes Inequality (11.4).

In the course of the preceding proof, it was shown that there is an admissible

spherical chip corresponding to each ρ>0. The requirement that ρ

>ρ/

√

in Inequality (11.4) validates the use of Theorem 8.5.7 for admissible chips.