Helms L.L. Potential Theory

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11.3 Curved Boundaries 407

x ∈ Σ,letN(x) be the set of neighborhoods U of x associated with x in the

deﬁnition. For x ∈ Ω, N(x) will denote the set of balls B

x,ρ

with closure in

Ω;theτ

associated with B

x,ρ

will be the identity map.

Consider any z ∈ Ω ∪ Σ, U ∈N(z), and let τ be the diﬀeomorphism

associated with z satisfying the above conditions mapping U onto the ball

B ⊂ R

.Lety = τ(x)=(τ

(x),...,τ

(x)) and ˜u(y)=u(x). Under the map

τ, the equation Lu = f on Ω ∩U becomes the equation

L˜u =

f on B

where

L˜u(y)=



p,q=1

˜a

(y)D

˜u(y)+



p=1

˜u(y)+˜c(y)˜u(y)

and

˜a

(y)=



i,j=1

∂τ

∂x

∂τ

∂x

(x) p, q =1,...,n

(y)=



i,j=1

∂

∂x

(x)+



i=1

∂τ

∂x

(x) p =1,...,n

˜c(y)=c(x),

f(y)=f(x).

Moreover, the boundary condition Mu = g on U ∩∂Ω becomes the condition

M˜u =˜g on B ∩R

where

M˜u(y





p=1



˜u(y)+˜γ(y



)˜u(y



, 0)

and





i=1

(x)

∂τ

∂x

, ˜γ(y)=γ(x).

In order to show that

L is strongly elliptic on B

, a few facts related to

quadratic forms will be reviewed. Consider the quadratic function

Q(y)=



p,q=1

where the n ×n matrix [c

] is a real symmetric matrix. The quadratic form

Q is said to be positive deﬁnite if Q(y) ≥ 0 for all y ∈ R

and Q(y)=0

only when y =0.TheformQ is positive deﬁnite if and only if there is a

constant m such that Q(y) ≥ m|y|

for all y ∈ R

(cf. [43], p. 210). Consider

the quadratic form

408 11 Oblique Derivative Problem

Q(y)=



p,q=1

˜a



p,q=1





i,j=1

∂τ

∂x

∂τ

∂x

(x)





i,j=1





p,q=1

∂τ

∂x

∂τ

∂x



(x)



i,j=1





p=1

∂τ

∂x





q=1

∂τ

∂x



(x).

Since the matrix [a

] is positive deﬁnite, Q(y) ≥ 0andQ(y) = 0 implies that



p=1

∂τ

∂x

=0,i=1,...,n;

but since the Jacobian of the map τ does not vanish on U ∩Ω by the multi-

plication theorem for Jacobians (c.f. [1]), this system of linear equations can

only have the trivial solution y =0.Thus,Q(y) is positive deﬁnite and

L is

strongly elliptic on B

Suppose N(x) contains a convex U with associated τ

mapping U onto

aballB = B

y,ρ

⊂ R

. Letting K denote the uniform bound on the H

2+α

norms of the τ

and τ

−1

and using the convexity of U, it is easily seen that

|y −z|≤|τ

(y) − τ

(z)|≤K|y − z|,y,z∈ U. (11.13)

Each U ∈N(x)containsaV ∈N(x), not necessarily convex, for which V ∩Ω

is the inverse image under τ

of an admissible spherical chip and satisﬁes

Inequalities (11.13). This can be seen as follows. Suppose τ

(U)=B

y,ρ

,let

x,

be a ball with B

x,

⊂ U,letW = τ

x,

), and let B



,ρ



be a ball

for which B



,ρ



⊂ W and B



,ρ



∩ R

is an admissible spherical chip. Let

V = τ

−1



) ⊂ B

x,

⊂ U . The convexity of B

x,

canbeusedtoshowthat

V satisﬁes Inequalities (11.13) even though V need not be convex.

Deﬁnition 11.3.4 If x ∈ Σ,U ∈N(x)andτ

is the associated diﬀeomor-

phism, U is an admissible neighborhood of x if it is the inverse image

under τ

of a ball deﬁning an admissible spherical chip and satisﬁes Inequal-

ity (11.13). The nonempty collection of admissible neighborhoods of x ∈ Σ

will be denoted by A(x)withA(x)=N(x)forx ∈ Ω.

It can be assumed that each admissible neighborhood is the inverse image of

a spherical chip for which the conclusion of Theorem 11.2.8 is valid as the

functions f and g will satisfy global weighted H¨older norms.

Consider any x ∈ Σ and U ∈A(x) with associated diﬀeomorphism τ

mapping U onto B. Letting

U∩Ω

(y)=d(y,∂(U ∩ Ω) ∼ (U ∩ Σ)) and

+ (z)=d(z,∂B

∼ B

), if z = τ

(y), then

U∩Ω

(y) ≤

(z) ≤ K

U∩Ω

(y).

11.3 Curved Boundaries 409

Moreover, if u ∈ H

(b)

2+α

(U ∩Ω)and˜u = u ◦τ

−1

,thenasinSection10.3there

is a constant C>0 such that

|u|

(b)

2+α,U∩Ω

≤|*u|

(b)

2+α,B

≤ C|u|

(b)

2+α,U∩Ω

Since Theorem 7.6.7 and Inequality (7.33) apply to spherical chips Ω with

Σ = int(Ω ∩ R

), if 0 ≤ a



≤ a, b



≤ b<0,a



+ b



≥ 0, and b, b



are not

negative integers or 0, then

|u|

(b)



,U∩Ω

≤ C|u|



)

a,U∩Ω

(11.14)

where C = C(a, a



,b,b



,K,A,d(Ω)).

Although no mention is made of an interior sphere condition in the fol-

lowing versions of the Boundary Point Lemma and the Strong Maximum

Principle, a version of the condition is implicit in the requirement that Σ be

of class C

2+α

Lemma 11.3.5 (Boundary Point Lemma) If x ∈ Σ ∈C

2+α

,U ∈N(x),

u ∈ C

(cl(U ∩ Ω)) ∩ C

(U ∩Ω), Lu ≥ 0 on U ∩ Ω, y

∈ U ∩ Σ,u(y

) >u(y)

for all y ∈ U ∩Ω,thenMu(y

) < 0, if deﬁned, whenever u(y

) ≥ 0.

Proof: Using the above notation, the function ˜u on B

induced by the

diﬀeomorphism τ

associated with x and U will satisfy the conditions

L˜u ≥ 0

on B

, ˜u(z

) > ˜u(z) for all z ∈ B

where z

= τ

) ∈ B

and z =

(y),y ∈ U ∩Ω. Since the interior sphere condition is satisﬁed at z

, it follows

that

M˜u(z

) < 0 whenever ˜u(z

) ≥ 0, and consequently that Mu(y

) < 0

whenever u(y

) ≥ 0.

The following corollary is proved in the same way Corollary 11.2.2 is

proved, using the above boundary point lemma.

Corollary 11.3.6 (Strong Maximum Principle) If Σ ∈C

2+α

,u ∈ C

(Ω

−

) ∩C

(Ω) satisﬁes Lu ≥ 0 on Ω,Mu ≥ 0 on 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.

It should be noted that the f and g of the following theorem satisfy global

weighted H¨older norms so that the conditions of Theorems 11.2.5 and 11.2.8

are satisﬁed.

Theorem 11.3.7 If b ∈ (−1, 0),α ∈ (0, 1),Ω is a bounded open subset of

,Σ is a relatively open subset of ∂Ω of class C

2+α

,f ∈ H

(2+b)

(Ω),and

g ∈ H

(1+b)

1+α

(Σ), then for each x ∈ Ω ∪Σ, U ∈A(x),andh ∈ C

(∂(U ∩Ω) ∼

(U ∩ Σ)) there is a unique function u ∈ C

(cl(U ∩ Ω)) ∩ C

(U ∩ (Ω ∪ Σ))

satisfying

Lu = f on U ∩Ω, Mu = g on U ∩Σ, and u = h on ∂(U ∩Ω) ∼ (U ∩ Σ).

410 11 Oblique Derivative Problem

Moreover, there is a constant C = C(a

,c,α,b,β

,γ,d(Ω)) such that

|u|

(b)

2+α,U∩Ω

≤ C



|f|

(2+b)

α,U∩Ω

+ |g|

(1+b)

1+α,U∩Σ

+ |h|

0,∂(U∩Ω)∼(U∩Σ)



. (11.15)

Proof: If x ∈ Ω and U is a ball, the result is covered by Theorem 11.2.8.

Consider any x ∈ Σ,U ∈A(x),τ

the diﬀeomorphism associated with x and

U,andtheballB = τ

(U). Under the map τ

, the ﬁrst two of the above

equations become

L˜u =

f on B

and

M˜u =˜g on B

Since the components of τ

and τ

−1

are in H

2+α

(U)andH

2+α

(B), respec-

tively, the maps τ

and τ

−1

have continuous extensions, denoted by the same

symbols, to U

−

and B

−

, respectively. It is easily seen that the extensions

are inverses to each other and are one-to-one. It is also easily seen that the

regions ∂(U ∩Ω) ∼ (U ∩Σ)and∂B

∼ B

are mapped onto each other. For

z ∈ ∂B

∼ B

,let

h(z)=h(τ

−1

(z)). Under the map τ

, the equation u = h

on ∂(U ∩Ω) ∼ (U ∩Σ) becomes ˜u =

h on ∂B

∼ B

. By Theorem 11.2.8, the

transformed equations have a unique solution ˜u ∈ C

(cl(B

) ∩C

∪B

The function u(y)=˜u(τ

(y)),y ∈ U ∩ Ω, satisﬁes the above equations. The

inequality follows from Inequality (11.8).

11.4 Superfunctions for Elliptic Operators

Just as superharmonic functions and superfunctions played a role in proving

the existence of a solution to the Dirichlet problems for the Laplacian and

elliptic operators, respectively, the concept of superfunction can be used to

prove the existence of a solution to the oblique derivative boundary problem.

As usual Ω will denote a bounded open subset of R

,α and b will be param-

eters with α ∈ (0, 1),b ∈ (−1, 0),Σ will denote a nonempty, relatively open

subset of ∂Ω of class C

2+α

. No further conditions will be imposed on Ω and

Σ in this section. Throughout this section f, g,andh will denote functions

in H

(2+b)

(Ω),H

(1+b)

1+α

(Σ), and C

(∂Ω ∼ Σ), respectively.

By Theorem 11.3.7, for each x ∈ Σ and U ∈A(x) there is a unique

function w ∈ C

(cl(U ∩ Ω)) ∩ C

(U ∩(Ω ∪ Σ)) satisfying the equations

Lw = f on U ∩ Ω, Mw = g on U ∩ Σ, and w = φ on ∂(U ∩ Ω) ∼ (U ∩ Σ)

for φ ∈ C

(∂(U ∩ Ω) ∼ (U ∩ Σ)). The function w will be called the local

solution associated with x, U ,andφ and will be denoted by LS(x, U, φ). It

will be understood that if the domain of the function φ contains ∂(U ∩Ω) ∼

(U ∩Σ), then the third argument of LS(x, U, φ) is the restriction of φ to the

latter set. It will be seen that LS(x, U, φ)playsthesameroleasthePoisson

integral and might justiﬁably be denoted by PI(x, U, φ).

11.4 Superfunctions for Elliptic Operators 411

Lemma 11.4.1 (i) If x ∈ Ω ∪ Σ,U ∈A(x),u ∈ C

(cl(U ∩ Ω)) ∩ C

(U ∩

Ω), Lu =0on U ∩Ω, Mu =0on U ∩Σ,andu ≥ 0 on ∂(U ∩Ω) ∼ (U ∩Σ),

then u ≥ 0 on U ∩ Ω. (ii) If u

∈ C

(cl(U ∩ Ω)) ∩ C

(U ∩ Ω), Lu

= f on

U ∩Ω, Mu

= g on U ∩Σ,andu

= h

on ∂(U ∩Ω) ∼ (U ∩Σ),i=1, 2,with

≥ h

,thenu

≥ u

on U ∩ Ω.

Proof: The second assertion is an immediate consequence of the ﬁrst. To

prove (i), let u satisfy the above conditions and assume that u attains a

strictly negative minimum at a point x

in cl(U ∩Ω). If x

is an interior point

of U ∩Ω, then it must be a constant by Theorem 9.5.7; that is, u = k<0on

U ∩Ω and must be equal to k on ∂(U ∩Ω) ∼ (U ∩Σ), contradicting the fact

that u ≥ 0on∂(U ∩ Ω) ∼ (U ∩ Σ). By the Boundary Point Lemma, x

can

not be a point of U ∩Σ. Therefore, x

∈ ∂(U ∩Ω) ∼ (U ∩Σ)withu(x

) < 0,

contradicting the fact that u ≥ 0on∂(U ∩ Ω) ∼ (U ∩ Σ). Therefore, u ≥ 0

on U ∩ Ω.

Deﬁnition 11.4.2 A function u ∈ C

(Ω)isasuperfunction on Ω if for

each x ∈ Ω ∪ Σ there is a V

x,u

∈A(x) such that for all U ∈A(x),U ⊂

x,u

,v ∈ C

(cl(U ∩ Ω)) ∩ C

(U ∩Ω)

Lv = f on U ∩Ω, Mv = g on U ∩Σ, and u ≥ v on ∂(U ∩Ω) ∼ (U ∩ Σ)

(11.16)

implies that u ≥ v on U ∩Ω. The set of all such superfunctions will be denoted

by U(f, g).

There is a corresponding deﬁnition of subfunction and L(f,g).

Remark 11.4.3 Using the above notation, u ∈ C

(Ω) is a superfunction on

Ω if and only if there is a V

x,u

∈A(x) such that

u ≥ LS(x, U, u)onU ∩Ω for all U ⊂ V

x,u

,U ∈A(x). (11.17)

This can be seen as follows. Assume u is a superfunction on Ω,x ∈ Ω ∪

Σ,U ⊂ V

x,u

,U ∈A(x), and let v = LS(x, U, u),U ⊂ V

x,u

.ThenLv = f

on U ∩ Ω, Mv = g on U ∩ Σ,andv = u on ∂(U ∩ Ω) ∼ (U ∩ Σ). Thus,

u ≥ v = LS(x, U, u)onU ∩Ω for all U ⊂ V

x,u

,U ∈A(x). Suppose now that

(11.17) holds. Consider any v satisfying (11.16). By (ii) of Lemma 11.4.1,

u ≥ LS(x, U, u) ≥ LS(x, U, v)=v on U ∩Ω so that u is a superfunction on Ω.

Lemma 11.4.4 If u ∈ C

(Ω

−

) ∩ C

(Ω ∪ Σ) satisﬁes Lu ≤ f on Ω and

Mu ≤ g on Σ and k>0,thenu and u + k are superfunctions on Ω.

Proof: Suppose Lu ≤ f on Ω and Mu ≤ g on Σ.Consideranyy ∈ Ω∪Σ,U ∈

A(y), and a function v such that Lv = f on U ∩Ω and Mv = g on U ∩Σ with

u ≥ v on ∂(Ω ∩U ) ∼ (U ∩Σ). Then L(u −v) ≤ 0onU ∩Ω, M(u −v) ≤ 0on

U ∩Σ,andu −v ≥ 0on∂(U ∩Ω) ∼ (U ∩Σ). It follows from Corollary 11.3.6

412 11 Oblique Derivative Problem

that u ≥ v on U ∩ Ω,provingthatu is a superfunction. If k>0, then

L(u + k) ≤ Lu ≤ f and M(u + k) ≤ Mu ≤ g, and it follows from the ﬁrst

part of the proof that u + k is a superfunction.

Lemma 11.4.5 (Lieberman [39]) (i) If u and v are superfunctions on

Ω,thenmin (u, v) is a superfunction on Ω.

(ii) If u is a superfunction on Ω,y ∈ Ω ∪ Σ,andU ⊂ V

y,u

,U ∈A(y),the

function



u on Ω ∼ U

LS(y, U,u) on U ∩Ω

is a superfunction on Ω.Ifu, v are superfunctions on Ω and u ≥ v on Ω,

then u

≥ v

on Ω.

(iii)If Σ = ∂Ω,eitherc<0 on Ω or γ<0 on Σ, u ∈ L(f, g),andv ∈ U(f, g),

then u ≤ v on Ω.

Proof: The ﬁrst assertion (i) follows from the deﬁnition of a superfunction.

Suppose u and u

are as described in (ii). Since u ≥ LS(y, U, u)=u

U ∩ Ω and u

= u on Ω ∼ U, u ≥ u

on Ω.Consideranyz ∈ Ω ∪ Σ and

W ⊂ V

z,u

,W ∈A(z) and the functions

v =



u on (∂(U ∩ Ω) ∼ (U ∩ Σ)) ∩ W

on (∂(W ∩ Ω) ∼ (W ∩ Σ)) ∩ U

−

w =

⎧

⎨

⎩

LS(z, W, u

)on(∂(U ∩ Ω) ∼ (U ∩Σ)) ∩W

on (∂(W ∩ Ω) ∼ (W ∩Σ)) ∩ U

−

Since u ≥ LS(z,W, u) ≥ LS(z,W,u

)onW ∩ Ω, v ≥ w on (∂(U ∩ Ω) ∼

(U ∩ Σ)) ∩ W ). Since LS(y,U,u)=u = v on (∂(U ∩ Ω) ∼ (u ∩ Σ)) ∩ W

and LS(y, U,u)=u

= v on (∂(W ∩ Ω) ∼ (W ∩ Σ)) ∩ U

−

, LS(y,U, u)

is the unique local solution on U ∩ W ∩ Ω corresponding to the boundary

function v on ∂(U ∩ W ∩ Ω) ∼ (U ∩ W ∩ Σ). Since LS(z,W,u

)=w on

(∂(U ∩ Ω) ∼ (U ∩ Σ)) ∩ W and LS(z,W,u

)=u

= w on (∂(W ∩ Ω) ∼

(W ∩ Σ)) ∩ U

−

, LS(z,W, u

) is the unique local solution on U ∩ W ∩ Ω

corresponding to the boundary function w on ∂(U ∩W ∩Ω) ∼ (U ∩W ∩Σ).

Note that v = u ≥ LS(z,W,u) ≥ LS(z, W, u

)=w on (∂(U ∩ Ω) ∼ (U ∩

Σ)) ∩W and v = u

= w on (∂(W ∩Ω) ∼ (W ∩Σ)) ∩U

−

so that v ≥ w on

∂(U ∩ W ∩ Ω) ∼ (U ∩ W ∩ Σ). Thus,

= LS(y, U, u) ≥ LS(z,W,u

)onU ∩ W ∩ Ω.

On (W ∼ U) ∩ Ω,

= u ≥ LS(z, W, u) ≥ LS(z,W,u

11.4 Superfunctions for Elliptic Operators 413

Therefore,

≥ LS(z, W, u

)onW ∩ Ω for all W ∈A(z),

proving that u

is a superfunction. To prove (iii), consider any u ∈ L(f,g)

and any v ∈ U(f,g). Letting

ˆu(x) = lim sup

y→x,y∈Ω

u(y)andˆv(x) = lim inf

y→x,y∈Ω

v(y),

for x ∈ Ω

−

, ˆu and ˆv are u.s.c. and l.s.c. on Ω

−

, respectively. If x ∈ Ω ∪ Σ,

let V

x,u

and V

x,v

be the neighborhoods associated with the deﬁnitions of u

and v, respectively. Let

μ =sup{ˆu(x) − ˆv(x); x ∈ Ω

−

assume μ>0, and let

Γ = {y ∈ Ω

−

;ˆu(y) − ˆv(y)=μ}.

Since both V

x,u

and V

x,v

contain arbitrarily small neighborhoods of x in A(x),

for each x ∈ Ω ∪Σ there is a W

⊂ V

x,u

∩V

x,v

∈A(x). It will be shown

ﬁrst that W

∩ ∂Ω = ∅ for some x ∈ Γ . Suppose W

∩∂Ω = ∅ for all x ∈ Γ .

In this case, Γ is a compact subset of Ω and there is a point y ∈ Γ nearest

to ∂Ω.NotethatW

−

⊂ Ω by choice of V

y,u

and V

y,v

when y ∈ Ω.Let

= LS(y,W

,u) − LS(y,W

,v)onW

.ThenLw

=0onW

= u − v

on ∂W

,and

(y)=LS(y, W

,u)(y) − LS(y, W

,v)(y) ≥ u(y) − v(y)=μ,

the latter inequality holding since u is a subfunction and v is a superfunction.

Since w

= u − v ≤ μ on ∂W

and w

(y) ≥ μ, w

attains a nonnegative

maximum value at an interior point of W

and must be equal to a constant

on W

,namelyμ, by the Strong Maximum Principle, Theorem 9.5.6. Thus,

there are points of Γ that are closer to ∂Ω than y, a contradiction. Therefore,

there is a point x

∈ Γ such that W

∩Ω = ∅. Letting w

= LS(x

,u)−

LS(x

,v), as above, Lw

=0onW

∩ Ω, Mw

=0onW

∩ Σ,

= u−v on ∂(W

∩Ω) ∼ (W

∩Σ), and w

) ≥ u(x

)−v

)=μ.Ifw

attains its nonnegative maximum value at an interior point of W

∩Ω,thenit

must be a constant, namely μ,onW

∩Ω.Ifc<0, then 0 = Lw

= cμ < 0

on W

∩ Ω, a contradiction. Therefore, w

cannot attain its nonnegative

maximum value on W

∩ Ω and must do so at x

. By the Boundary Point

Lemma, Lemma 11.2.1, Mw

) < 0, a contradiction. The assumption that

μ>0 is untenable and so u ≤ v on Ω.

The function u

of (ii) in the preceding lemma will be called the lower

or lowering of u on U. The corresponding v

for v a subfunction on Ω will

be called the lift or lifting of v on U.

414 11 Oblique Derivative Problem

Deﬁnition 11.4.6 A function u ∈ C

(Ω)isasupersolution for the

problem

Lv = f on Ω, Mv = g on Σ, and v = h on ∂Ω ∼ Σ

if u is a superfunction for f and g and lim inf

y→x,y∈Ω

u(y) ≥ h(x)forx ∈

∂Ω ∼ Σ. The set of all such supersolutions will be denoted by U(f,g,h).

There is a corresponding deﬁnition of subsolution and L(f,g,h). When Σ =

∂Ω, the concept of supersolution is the same as that of superfunction. The

following theorem eliminates the requirement that γ ≤−γ

< 0onΣ in [40].

Lemma 11.4.7 (Σ = ∂Ω) (i)If u, v∈U(f,g,h),thenmin (u, v) ∈ U(f, g,h).

(ii) If u ∈ U(f,g,h),y ∈ Ω ∪ Σ,andU ∈A(y),thenu

∈ U(f,g,h).

(iii) If u ∈ L(f,g,h) and v ∈ U(f,g, h),thenu ≤ v on Ω.

Proof: To prove (i), it need only be noted that lim inf

y→x,y∈Ω

min (u, v)(y) ≥

h(x)on∂Ω ∼ Σ.Toprove(ii), it suﬃces to show that lim inf

y→x,y∈Ω

(y) ≥

h(x)on∂Ω ∼ Σ since u

is a superfunction according to Lemma 11.4.5.

Since ∂(U ∩ Ω) ⊂ Ω ∪ Σ by (ii) of Deﬁnition 11.3.2, (U ∩ Ω)

−

⊂ Ω ∪ Σ.

This implies that for x ∈ ∂Ω ∼ Σ, there is a neightborhood V of x such

that V ∩ (U ∩ Ω)

−

= ∅, which in turn implies that u

= u on V and

therefore lim inf

y→x,y∈Ω

(y) = lim inf

y→x,y∈Ω

u(y) ≥ h(x), completing the

proofof(ii). Consider any u ∈ L(f,g,h)andv ∈ U(f,g,h). Letting ˆu(x)=

lim sup

y→x,y∈Ω

u(y)andˆv(x) = lim inf

y→x,y∈Ω

v(y)forx ∈ Ω

−

, ˆu and ˆv are

u.s.c. and l.s.c. on Ω

−

, respectively, ˆu(x) ≤ h(x) ≤ ˆv(x)forx ∈ ∂Ω ∼ Σ,

and ˆu − ˆv ≤ 0on∂Ω ∼ Σ.Toprove(iii), it suﬃces to show that u ≤ v on

each component of Ω. It therefore can be assumed that Ω is connected. Let

μ =sup{ˆu(x) − ˆv(x); x ∈ Ω

−

assume μ>0, and let

Γ = {x ∈ Ω

−

;ˆu(x) − ˆv(x)=μ} = ∅.

If x

∈ Ω∩Γ and W

∈A(x

), then W

−

⊂ Ω. Letting w

= LS(x

,u)−

LS(x

,v)onW

∩ Ω, Lw

=0onW

∩ Ω, w

= u − v ≤ μ on

∂(W

∩ Ω), and

)=LS(x

,u)(x

) − LS(x

,v)(x

) ≥ u(x

) − v(x

)=μ.

Thus, w

attains its nonnegative maximum value at an interior point of W

By the Strong Maximum Principle, Theorem 9.5.6, w

is constant on W

∩Ω

and since w

≤ u − v ≤ μ on ∂(W

∩ Ω),w

= μ on W

∩ Ω. Therefore,

= W

∩ Ω ⊂ Γ . This shows that Ω ∩Γ is an open subset of Ω.Since

Ω ∼ Γ is an open subset of Ω,eitherΩ ∩ Γ = ∅ or Ω ∼ Γ = ∅ by the

connectedness of Ω. Suppose Ω ∼ Γ = ∅.ThenΓ ⊃ Ω so that u − v = μ on

Ω or u = v + μ on Ω.Ifx is a point in ∂Ω ∼ Σ,then

h(x) ≥ lim sup

y→x

u(y) = lim sup

y→x

(v(y)+μ) ≥ lim inf

y→x

v(y)+μ ≥ h(x)+μ,

11.4 Superfunctions for Elliptic Operators 415

a contradiction. Thus, under the assumption that μ>0,Ω ∼ Γ = ∅. Suppose

now that Ω ∩ Γ = ∅ and consider any y

∈ Γ ⊂ ∂Ω,W

∈A(y

), and let

= LS(y

,u) − LS(y

,v). Then Lw

=0onW

∩ Ω, Mw

on W

∩ Σ,w

= u − v on ∂(W

∩ Ω) ∼ (W

∩ Σ), and w

) ≥ u(y

) −

v(y

)=μ as above. If w

attains its nonnegative maximum value at an

interior point of W

, then it must be a constant and equal to μ on W

∩Ω.

Since w

= u − v on ∂(W

∩ Ω) ∼ (W

∩ Σ), there would be points of Ω

where u −v = μ, a contradiction. Thus, w

<μ= w

)onW

∩Ω.Bythe

Boundary Point Lemma, Lemma 9.5.5, Mw

) < 0, contradicting the fact

that Mw

) = 0. Thus, under the assumption μ>0,Ω ∩ Γ = ∅ and, as

noted above, Ω ∼ Γ = ∅; but this contradicts the fact that Ω is connected.

Thus, the assumption that μ>0 is untenable and u ≤ v on Ω.

Lemma 11.4.8 If x ∈ Ω∪Σ,U ∈A(x),and{v

} is a sequence in C

(cl(U ∩

Ω)) ∩ C

(U ∩ (Ω ∪ Σ)) with sup

v



0,∂(U∩Ω)∼(U∩Σ)

< +∞ satisfying the

equations

= f on U ∩Ω, Mv

= g on U ∩ Σ, k ≥ 1, (11.18)

for f ∈ H

(2+b)

(Ω) and g ∈ H

(1+b)

1+α

(Σ), then there is a subsequence {v



} that

converges uniformly on U ∩(Ω ∪Σ) to v ∈ C

(cl(U ∩Ω)) ∩C

(U ∩(Ω ∪Σ))

satisfying

Lv = f on U ∩ Ω, Mv = g on U ∩ Σ. (11.19)

Proof: Letting K

=sup

v



0,∂(U∩Ω)∼(U∩Σ)

and using Theorem 11.3.7,

(b)

2+α,U∩Ω

≤ C



+ |f |

(2+b)

α,U∩Ω

+ |g|

(1+b)

1+α,U∩Σ



. (11.20)

For 0 <<d(U ∩ Ω),



2+b+α

2+α,



(U∩ Ω)



≤ C



+ |f |

(2+b)

α,Ω

+ |g|

(1+b)

1+α,Σ



. (11.21)

It follows from this inequality that the sequences {v

}, {D

},and{D

}

are uniformly bounded on (



U ∩ Ω)



. Since the term [v

]

2+α,(



U∩Ω)



is in-

cluded in the expression on the left side, the {D

} sequences are uniformly

equicontinuous on (



U ∩ Ω)



and therefore on cl(



U ∩ Ω)



.Since|v

(b)

1+α,U∩Ω

≤

C|v

(b)

2+α,U∩Ω

by Inequality (11.14), the same argument is applicable to the

sequences {D

}; that is, the sequences {D

} are uniformly equicontinu-

ous on cl(



U ∩ Ω)



.Since|v

(b)

α,U∩Ω

need not be deﬁned, the same argument

cannot be applied to the {v

} sequence. By considering the images *v

the v

under the map τ

, the mean value theorem can be applied to show

that the diﬀerence quotients |v

(y) − v

(z)|/|y − z| are uniformly bounded

on (



U ∩ Ω)



so that the sequence {v

} is also uniformly equicontinuous on

cl(



U ∩ Ω)



. Therefore, the sequences {v

}, {D

},and{D

} are uniformly

416 11 Oblique Derivative Problem

bounded and equicontinuous on the compact set cl(



U ∩ Ω)



.Considerase-

quence {

} strictly decreasing to zero. It follows from the Arzela-Ascoli

Theorem, Theorem 0.2.3, that there is a subsequence {v

(1)

} of the {v

} se-

quence such that the sequences {v

(1)

}, {D

(1)

},and{D

(1)

} converge uni-

formly to continuous functions on cl



(U ∩Ω)



. Applying the same argument

to the {v

(1)

} sequence, there is a subsequence {v

(2)

} of the {v

(1)

} sequence

such that the sequences {v

(2)

}, {D

(2)

},and{D

(2)

} converge uniformly

to continuous functions on cl



(U ∩ Ω)



, etc. Using the Cantor diagonaliza-

tion method, the sequence {v



} with v



= v

()



has the property that the

sequences {v



}, {D



},and{D



} converge to continuous functions on

∞

m=1



(U ∩Ω)



= U ∩ (Ω ∪ Σ).

Letting v = lim

→∞



, replacing v

in Inequality (11.21) by v



, letting

 →∞and then taking the supremum over 0 <<d(U ∩ Ω), v satisﬁes

Inequality (11.20). Since |v|

−b,U∩Ω

= |v|

(b)

−b,U∩Ω

≤|v|

(b)

2+α,U∩Ω

≤ C(K

|f|

(2+b)

α,U∩Ω

+ |g|

(1+b)

1+α,U∩Σ

), |v(x) −v(y)|≤K|x −y|

−b

from which it follows that

v has a continuous extension to cl(U ∩Ω)andthatv ∈ C

(cl(U ∩Ω)). Since

the D

converge to D

v on each cl



(U ∩ Ω)



,v ∈ C

(U ∩ (Ω ∪ Σ)).

Equation (11.19) follows easily by taking limits in Equation (11.18).

Lemma 11.4.9 u ∈ U(f, g,h) if and only if −u ∈ L(−f,−g, −h).

Proof: For y ∈ Ω ∪ Σ,letLS

(y,U, u)andLS

−

(y,U, u) denote local so-

lutions relative to (f,g,h)and(−f,−g, −h), respectively. Also, let V

⊕

y,u

and



y,u

be the elements of A(y)relativeto(f,g,h)and(−f,−g, −h) described

in Deﬁnition 11.4.2. Consider any W

⊂ V

⊕

y,u

∩ V



y,−u

∈A(y). It will be

shown ﬁrst that

(y,W

,u)=−LS

−

(y,W

, −u). (11.22)

Since L(LS

(y,U, u)+LS

−

(y,U, −u)) = f −f =0onW

∩Ω,M(LS

(y,U, u)

+LS

−

(y,U, −u)) = g−g =0onW

∩Σ,andLS

(y,U, u)+LS

−

(y,U, −u)=

u − u =0on∂W

∩ Ω, LS

(y,U, u)+LS

−

(y,U, −u)=0onW

∩ Ω

by Corollary 11.2.2, establishing Equation 11.22. If u ∈ U(f,g, h), then

u ≥ LS

(y,W

,u)onW

∩ Ω.Thus,

−u ≤−LS

(y,W

,u)=LS

−

(y,W

, −u)onW

∩ Ω.