Helms L.L. Potential Theory

Подождите немного. Документ загружается.

11.2 Boundary Maximum Principle 397

Lemma 11.2.4 If Ω is an admissible spherical chip with Σ = int(∂Ω ∩R

)

or a ball in R

with Σ = ∅,b∈ (−1, 0),α∈ (0, 1),u∈ C

(Ω

−

) ∩C

(Ω) satis-

ﬁes Lu = f on Ω for f ∈ H

(2+b)

(Ω), Mu = g on Σ for g ∈ H

(1+b)

1+α

(Σ),and

u =0on ∂Ω ∼ Σ, then there is a constant C = C(n, m, b

,b,α,β

,γ,d(Ω))

such that



u

0,Ω

≤ C



|f|

(2+b)

+ |g|

(1+b)

1+α



Proof: Since the inequality was proven above for a ball B ⊂ R

with Σ = ∅,

only the spherical chip case need be considered. Deﬁne t ∈ (0, 1),ρ

and w as

above. Consider any x ∈ Ω and any ϕ ∈ H

(1+b)

1+α

(Ω) satisfying the condition

ϕ = g on Σ.If

v = u −

|f|

(2−t)

+ |ϕ|

(1−t)

then by Inequality (11.5) and the fact that |f|

(2−t)

≥ (ρ − r)

2−t

|f|

ρ−r

Lv(x) ≥ f(x)+|f|

(2−t)

(ρ − r)

t−2

≥ f (x)+|f|

ρ−r

≥ f (x) − f (z)

for all z ∈

ρ−r

.Notingthatf is continuous at x and x is a limit point of

ρ−r

, Lv(x) ≥ f(x) − f(x), and therefore Lv ≥ 0onΩ. Consider now any



, 0) ∈ Σ. By Inequality (11.6)and a similar continuity argument,

Mv(x



, 0) ≥ ϕ(x



, 0) + |ϕ|

(1−t)

(ρ − r)

t−1

≥ ϕ(x



, 0) + (ρ − r)

1−t

|ϕ|

ρ−r

(ρ − r)

t−1

≥ ϕ(x



, 0) + |ϕ|

ρ−r

≥ 0.

Since v =0on∂Ω ∼ Σ,v ≤ 0onΩ by the Strong Maximum Principle,

Corollary 11.2.2; that is,

u(x) ≤

|f|

(2−t)

+ |ϕ|

(1−t)

(ρ

− r

)

Since −u also satisﬁes the hypotheses of the lemma with f and ϕ replaced

by −f and −ϕ, respectively,

|u(x)|≤

|f|

(2−t)

+ |ϕ|

(1−t)

(ρ

− r

)

≤

|f|

(2−t)

+ |ϕ|

(1−t)

(2ρ)

(ρ − r)

Using Lemma 7.6.7 and the fact that

d(x)=ρ − r,

d(x)

−t

|u(x)|≤C



|f|

(2−t)

+ |ϕ|

(1−t)



≤ C



|f|

(2−t)

+ |ϕ|

(1−t)

1+α



398 11 Oblique Derivative Problem

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



u

0,Ω

≤ C



|f|

(2+b)

+ |ϕ|

(1+b)

1+α



Taking the inﬁmum over ϕ ∈ H

(1+b)

1+α

(Ω), the conclusion is established.

The requirement in the hypotheses of the following theorem that the func-

tion f satisﬁes a weighted H¨older norm on a superset is not an essential

requirement in so far as future developments are concerned. In applying the

theorem, starting with an f satisfying a H¨older requirement on a spherical

chip, an admissible spherical chip satisfying the requirement can be con-

structed. Only the k = 0 case of the following theorem will be proved in

detail for spherical chips to illustrate the techniques. The theorem is stated

only for spherical chips but is also applicable for balls with Σ = ∅.

Theorem 11.2.5 If Ω = B

y,ρ

where y =(0,...,0, −ρ

), 0 <ρ

<ρ,is

a spherical chip with Σ = int(∂Ω ∩ R

) α ∈ (0, 1),b ∈ (−1, 0),andf ∈

(2−k+b)

k+α

(Ω



) where Ω



= B



,ρ

with y



=(0,...,0, −ρ



), 0 <ρ



<ρ

ρ, k =0, 1, then there is a sequence {f

} in H

k+α

(Ω ∪ Σ) that converges

uniformly to f on each

−

, 0 <δ<1,and|f

(2−k+b)

k+α,Ω

≤ C|f |

(2−k+b)

k+α,Ω

,j ≥ 1.

Proof: Let Ω = B

y,ρ

where y =(0,...,o,− ρ

), 0 <ρ

< 1, let m be the

molliﬁer of Section 2.5, and let f ∈ H

(2+b)

(Ω). Since δ

2+b+α

|f|

α,

< +∞

for each 0 <δ<1,f has a continuous extension to

−

, denoted by the

same symbol, and therefore to



0<δ<1

−

= Ω ∪ Σ.Consideranyx ∈ Ω

and the point x



= x − (x

/(x

− y

))(x − y) at which the line joining

y to x intersects R

. Letting r

= |x



− x| =(x

/(x

− y

))|x − y|,if

x ∈

= B

y,ρ− δ

,thenB



⊂

.NotealsothatifB is any ball in



,thenB ⊂

.Foreachx ∈ Ω,letB(x, r

) be a ball with center

at ξ = ξ(x)=x − (x

/j(x

− y

))(x − y)ofradiusr

= r

(x)=|ξ − x| =

−1

/(x

−y

))|x−y|.Ifj>2ρ/ρ

> 2, then ξ

= x

(1−j

−1

) >x

/2and

the nth coordinate of the lowest point on the boundary of the ball B(x, r

)

is ξ

−j

−1

/(x

−y

))|x −y| > 0, showing that the ball B(x, r

) ⊂ B



whenever j ≥ 2ρ/ρ

.Thus,B(x, r

) ⊂

if x ∈

.Forj ≥ 1, let ϕ

be the

truncation function ϕ

(x)=−j for x ≤−j, ϕ

(x)=x for −j<x<j,and

(x)=j for x ≥ j and let ψ

= J

1/j

be the smoothened version of ϕ

The latter function has the following properties:

(i) ψ

∈ C

∞

(R),

(ii) ψ

(x)=x if −j +

<x<j−

,ψ

(0) = 0, |ψ

|≤j, |ψ



|≤1, and

(iii) |ψ

(x) − ψ

(y)|≤|x − y| for all x, y ∈ R.

For j>2ρ/ρ

and x ∈ Ω ∪ Σ,let

(x)=

⎧

⎪

⎨

⎪

⎩



(f)(ξ + r

z)m(z) dz



(f)(z)m



(z − ξ)



dz.

(11.7)

11.2 Boundary Maximum Principle 399

Note that if x ∈

and j ≥ 2ρ/ρ

,thenf

(x) is the weighted average of

(f)overB(x, r

) ⊂

.Whenx

=0,f

(x)=ψ

(f)(x). It will be shown

now that the sequence {f

} converges uniformly to f on each

−

.Sincef

is uniformly continuous on

−

,given>0thereisanη>0 such that

|f(z) − f(z



)| <whenever |z −z



| < η,z,z



∈

−

.Choosej

≥ 2ρ/ρ

such

that |f |

−

and j>2ρ(ρ − ρ

)/ρ

η for all j ≥ j

.Ifz ∈ B(x, r

then |z − x| < 2r

<ηand it follows from the second form for f

(x)in

Equation (11.7) that

(x) − f(x)|≤



|ψ

(f)(z) − f (x)|m



(z − ξ)



≤



|f(z) − f(x)|m



(z − ξ)



dz < 

simultaneously for all x ∈

and j ≥ j

; that is, the sequence {f

} converges

uniformly to f on

−

.Itwillbeshownnextthateachf

∈ H

(Ω). Using the

ﬁrst form for f

(x) in Equation (11.7) and the fact that |ψ

|≤j, [f

]

0,Ω

+∞.Consideranyx, x



∈ Ω. Using the ﬁrst form for f

(x)andf



)and

the fact that |ψ(z) − ψ(z



)|≤|z − z



(x) − f



)|≤



|(ψ

(f)(ξ(x)+r

(x)z) − ψ

(f)(ξ(x



)+r



)z)|m(z) dz

≤



|f(ξ(x)+r

(x)z) − f(ξ(x



)+r



)z)|m(z) dz.

Using the easily checked fact that |ξ(x)+r

(x)z)−ξ(x



)−r



)z)|≤|x−x



(x) − f



|x − x



≤



[f]

α,(



)

ρ−ρ



|ξ(x)+r

(x)z) − ξ(x



) − r



)z)|

|x − x



m(z) dz

≤ C[f ]

α,(



)

ρ−ρ



< +∞.

Thus [f

]

α,Ω

< +∞. Combining this inequality with the inequality [f

]

0,Ω

+∞, |f

α,Ω

< +∞ and f

∈ H

(Ω). It remains in the k = 0 case to show

that |f

(2+b)

α,Ω

≤ C|f|

(2+b)

α,Ω

.Consideranyx ∈

.SinceB(x



,r) ⊂

,it

follows from the ﬁrst form for f

(x)that[f

]

≤ [f]

.Considerany

x, x



∈

. Using the fact that |ψ

(z) − ψ



)|≤|z − z



| for all z,z



∈ R,

(x) − f



)|≤



|ψ

(f)(ξ(x)+r

(x)z) − ψ

(f)(ξ(x



)+r



)z)|m(z) dz

≤



|f(ξ(x)+r

(x)z) − f(ξ(x



)+r



)z)|m(z) dz

400 11 Oblique Derivative Problem

≤ [f]

α,



|ξ(x)+r

(x)z − ξ(x



) − r



)z|

m(z) dz

≤ C[f]

α,

(|ξ(x) − ξ(x



)| + |r

(x) − r



)|)

≤ C[f]

α,

|x − x



and it follows that [f

]

α,

≤ C[f]

α,

. Combining this inequality with the

inequality [f

]

≤ [f]

, |f

α,

≤ C|f|

α,

. Multiplying by δ

2+b+α

and taking the supremum over 0 <δ<1, |f

(2+b)

α,Ω

≤ C|f |

(2+b)

α,Ω

.When

f ∈ H

(1+b)

1+α

(Ω), the f

(x) are deﬁned by putting f

(x)=



(ψ

(f)(ξ

(x)+

(x)z)m(z) dz, x ∈ Ω. As above, it can be seen that [f

]

0,Ω

≤ j, [f

]

1+0,Ω

≤ 1,

and [f

]

1+α,Ω

≤ 2C|m|

0,B

0,1

d(Ω)

1−α

so that |f

1+α,Ω

< +∞ and f

∈

1+α

(Ω). Also, [f

]

≤ [f]

1+α,

,and[f

]

1+α,

≤ C[ψ

(ψ

)]

2+0,R

[f]

≤ C[f]

1+α,

. Adding corresponding members, |f

1+α,

≤ C[f]

1+α,

Multiplying both sides by δ

2+b+α

and taking the supremum over 0 <δ<1,

(1+b)

1+α,Ω

≤ C|f |

(1+b)

1+α,Ω

. As above, the sequence {f

} converges uniformly to

f on each

−

,0<δ<1.

It is assumed in the following lemma that f and g satisfy the hypotheses

of the preceding theorem.

Lemma 11.2.6 If Ω is an admissible spherical chip with Σ = int(∂Ω ∩R

)

or a ball in R

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

(2+b)

(Ω),and

g ∈ H

(1+b)

1+α

(Σ), then there is a unique u ∈ C

(Ω

−

) ∩ C

(Ω ∪ Σ) ∩ C

(Ω)

satisfying u = f on Ω, D

u = g on Σ,andu =0on ∂Ω ∼ Σ with

|u|

(b)

2+α,Ω

≤ C(|f |

(2+b)

α,Ω

+ |g|

(1+b)

1+α,Σ

Proof: Assume ﬁrst that Ω is an admissible spherical chip with Σ = ∅.Let

} be the sequence in H

(Ω) of the preceding theorem that converges to

f uniformly on compact subsets of Ω with |f

(2+b)

α,Ω

≤ C|f|

(2+b)

α,Ω

. Similarly,

there is a sequence {g

} of functions on Σ such that each g

∈ H

1+α

(Σ), the

sequence {g

} converges uniformly to g on each compact subset of Σ,and

(1+b)

1+α,Σ

≤ C|g|

(1+b)

1+α,Σ

. By Theorem 8.5.7, for each j ≥ 1, there is a unique

∈ C

(Ω

−

) ∩C

(Ω ∪Σ) ∩C

(Ω) such that u

= f

on Ω, D

= g

Σ,andu

=0on∂Ω ∼ Σ. By Theorem 10.3.4 and Lemma 11.2.4,

(b)

2+α,Ω

≤ C



(2+b)

α,Ω

+ |g

(1+b)

1+α,Σ



≤ C



|f|

(2+b)

α,Ω

+ |g|

(1+b)

1+α,Σ



By the subsequence selection principle of Section 7.2, there is a subsequence

} that converges uniformly on compact subsets of Ω to a unique function

u ∈ H

(b)

2+α

(Ω) such that u = f on Ω, D

u = g on Σ,andu =0on∂Ω ∼ Σ.

In the case of a ball with Σ = ∅, the proof is the same except for using

Theorem 8.3.1 to assert the existence of a unique u

satisfying the equation

u

= f

subject to the condition u

=0on∂Ω.

11.2 Boundary Maximum Principle 401

Lemma 11.2.7 Let Ω be a bounded open subset of R

,letΩ



be an open

set containing Ω

−

,andletu ∈ C

) vanish outside Ω



.Ifη>0,α ∈

(0, 1) and b>−1,then|J

(b)

2+α,Ω

≤ C|u|

0,Ω



, |LJ

(2+b)

α,Ω

≤ C|u|

0,Ω



,and

|MJ

(1+b)

1+α,Σ

≤ C|u|

0,Ω



where C = C(a

,c,α,b,β

,γ,d(Ω)).

Proof: Using the function m ∈ C

∞

) deﬁned by Equation (2.4), if 0 <

δ<1,j ≥ 0andx, x



∈

,then

u(x) − D

u(x



|x − x



≤



|u(y)|



(x)



y−x



− D

(x)



y−x







|x − x



dy.

Letting z =(y −x)/η, z



=(y −x



)/η,forsomepointz



=(y −x



)/η on the

line segment joining z to z





(x)



y−x



− D

(x)



y−x







|x − x



≤

j+1

(z)

m(z) − D

(z)

m(z



|z − z



j+1

|∇

(z)

m(z





∇

(x)



y − x







so that

u(x) − D

u(x



|x − x



≤|u|

0,Ω



d(Ω)

1−α

(Ω)





∇

(x)



y −x







≤|u|

0,Ω



d(Ω)

1−α

[m]

j+1+0,R

≤ C|u|

0,Ω



where C=C(j, α, d(Ω)). Taking the supremum over x, x



∈

, [J

2+α,

α,

≤ C|u|

0,Ω



. Using interpolation as in Theorem 7.3.5, |J

2+α,

≤ C|u|

0,Ω



. Multiplying by δ

2+b+α

and taking the supremum over 0 <δ<

1, |J

(b)

2+α,Ω

≤ C|u|

0,Ω



. Similarly,

|LJ

α,





i,j

u +



u + cJ



α,

≤ A(|D

α,

+ |D

α,

+ |J

α,

)

≤ C|u|

0,Ω



Multiplying by δ

2+b+α

and taking the supremum over 0 <δ<1, |LJ

(2+b)

α,Ω

≤ C|u|

0,Ω



. A similar argument shows that |MJ

(1+b)

1+α,Σ

≤ C|u|

0,Ω



402 11 Oblique Derivative Problem

It is assumed in the following theorem that the f and g satisfy the hy-

potheses of Theorem 11.2.5.

Theorem 11.2.8 If Ω is an admissible spherical chip with Σ = int(∂Ω∩R

)

or a ball in R

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

(2+b)

(Ω),g ∈

(1+b)

1+α

(Σ),andh ∈ C

(∂Ω ∼ Σ), then there is a unique u ∈ C

(Ω

−

) ∩

(Ω ∪ Σ) satisfying Lu = f on Ω, Mu = g on Σ,andu = h on ∂Ω ∼ Σ.

Moreover, there is a constant C = C(a

,c,α,b,β

,γ,d(Ω)) such that

|u|

(b)

2+α,Ω

≤ C



|f|

(2+b)

α,Ω

+ |g|

(1+b)

1+α,Σ

+ |h|

0,∂Ω∼Σ



. (11.8)

Proof: The existence and uniqueness in the homogeneous case h = 0 will be

proved ﬁrst. Consider the Banach space

B = {u ∈ H

(b)

2+α

(Ω):u =0on∂Ω ∼ Σ},

with norm |u|

= |u|

(b)

2+α,Ω

, the normed linear space

X = H

(2+b)

(Ω) × H

(1+b)

1+α

(Σ)

with norm |(f,g)|

= |f |

(2+b)

α,Ω

+ |g|

(1+b)

1+α,Σ

, and the family of operators from B

to X deﬁned by

u =(u, D

u =(Lu, Mu)

u =(1−t)L

u + tL

u, 0 ≤ t ≤ 1.

As in the proof of Theorem 9.6.5, each of the operators (1−t)Δ+tL and (1−

t)D

+ tM satisﬁes the general conditions previously imposed on coeﬃcients.

Solvability of the oblique derivative problem Lu = f on Ω, Mu = g on Σ,

and u =0on∂Ω ∼ Σ is equivalent to showing that the map L

: B → X is

invertible. If u ∈ B and L

u =(f, g) ∈ X,then

(1 − t)u + tLu = f on Ω

(1 − t)D

u + tMu = g on Σ

u =0on∂Ω ∼ Σ.

By Theorem 10.4.1, for such u,

|u|

= |u|

(b)

2+α,Ω

≤ C





u

0,Ω

+ |f |

(2+b)

α,Ω

+ |g|

(1+b)

1+α,Σ



Note that u ∈ H

(b)

2+α

(Ω) implies that u ∈ H

2+α

(

)for0<δ<1which

in turn implies that u, Du,andD

u have continuous extensions to

−

that u, Du,andD

u have continuous extensions to Ω ∪ Σ.Inaddition,it

11.2 Boundary Maximum Principle 403

is also true that u ∈ C

(Ω

−

)sinceforx ∈

(x)|u(x)|≤δ

[u]

≤

|u|

(b)

0,Ω

≤|u|

(b)

2+α,Ω

< +∞; therefore,

(x)|u(x)|≤|u|

(b)

2+α,Ω

for all x ∈ Ω and

lim

x→x

∈∂Ω∼Σ

u(x) = 0 since lim

x→x

∈∂Ω∼Σ

(x)=+∞. This shows that

u ∈ C

(Ω

−

) ∩ C

(Ω ∪ Σ) so that Lemma 11.2.4 is applicable and

|u|

≤ C



|f|

(2+b)

+ |g|

(1+b)

1+α



= |(f,g)|

= C|L

where C is independent of t. By Theorem 9.2.2, L

maps B onto X if and

only if L

=(u, D

u)mapsB onto X. According to Lemma 11.2.6, there

is a unique u ∈ B satisfying u = f on Ω, D

u = g on Σ,andu =0on

∂Ω ∼ Σ.Thus,L

maps B onto X and so L

maps B onto X;thatis,there

is a u ∈ H

(b)

2+α

(Ω) such that

Lu = f on Ω, Mu = g on Σ, u =0on∂Ω ∼ Σ.

It follows from Corollary 11.2.2 that u is unique. As above, u ∈ H

(b)

2+α

(Ω)

implies that u ∈ C

(Ω

−

) ∩ C

(Ω ∪ Σ). It remains only to show that the

condition u =0on∂Ω ∼ Σ can be replaced by u = h on ∂Ω ∼ Σ.By

the Tietze Extension Theorem, there is a continuous function on R

that

agrees with h on ∂Ω ∼ Σ, denoted by the same symbol, with sup norm no

greater than that of h on ∂Ω ∼ Σ and vanishing outside the ball B

y,2ρ

where Ω = B

y,ρ

∩ R

or Ω = B

y,ρ

.Forj ≥ 1, let h

= J

1/j

h ∈ C

∞

The sequence { h

} converges uniformly to h on R

.Foreachj ≥ 1, consider

the problem

= f − Lh

on Ω, Mv

= g − Mh

on Σ, v

=0on∂Ω ∼ Σ.

(11.9)

By the ﬁrst part of the proof, this problem has a unique solution v

∈

(b)

2+α

(Ω) which implies that v

∈ C

(Ω

−

) ∩ C

(Ω ∪ Σ)asabove.Itwill

be shown now that the sequence {|v

(b)

2+α,Ω

} is bounded. By Lemma 11.2.4

and Lemma 11.2.7,



0,Ω

≤ C



|f − Lh

(2+b)

α,Ω

+ |g − Mh

(1+b)

1+α,Σ



≤ C



|f|

(2+b)

α,Ω

+ |g|

(1+b)

1+α,Σ

+ |h|

0,∂Ω∼Σ



It follows from Lemma 10.4.1 that

(b)

2+α,Ω

≤ C



|f|

(2+b)

α,Ω

+ |g|

(1+b)

1+α,Σ

+ |h|

0,∂Ω∼Σ



≤ K

< +∞,j≥ 1.

(11.10)

Letting u

= v

+ h

∈ H

(b)

2+α

(Ω) since this is true of h

and

= f on Ω, Mu

= g on Σ, u

= h

on ∂Ω ∼ Σ. (11.11)

404 11 Oblique Derivative Problem

Note that u

∈ C

(Ω

−

) ∩ C

(Ω ∪ Σ) since this is true of v

and h

.Forthe

diﬀerences u

− u

L(u

−u

)=0onΩ, M(u

−u

)=0onΣ, u

−u

= h

−h

on ∂Ω ∼ Σ.

Itcanbeassumedthatu

− u

attains a nonnegative, absolute maximum

value at a point x

of Ω

−

, for if not, replace it by its negative. If x

∈ Ω,then

−u

is a constant function by Lemma 9.5.6, and since u

− u

= h

− h

on ∂Ω ∼ Σ,u

−u



0,Ω

= h

−h



0,∂Ω∼Σ

.Ifx

is not a point of Ω,then

∈ ∂Ω ∼ Σ,forifx

∈ Σ, it would follow that M(u

−u

)(x

) < 0bythe

boundary point lemma, Lemma 9.5.5, a contradiction. In either case,

u

− u



0,Ω

≤h

− h



0,∂Ω∼Σ

As the sequence {h

} converges uniformly to h,itisCauchyinthe·

0,∂Ω∼Σ

norm and so it follows that the sequence {u

} is Cauchy in the ·

0,Ω

norm.

Thus there is a u ∈ C

(Ω

−

) which is the uniform limit of the {u

} sequence.

Using Inequality (11.10) and the fact that |h

(b)

2+α,Ω

≤ C|h|

0,∂Ω∼Σ

for all

j ≥ 1 according to Lemma 11.2.7, the sequence {|u

(b)

2+α,Ω

} is bounded by a

constant K.Considerany0<δ<1forwhich

= ∅.Then

2+b+α



u



+[Du

]

+[D

]

+[D

]

α,



≤ K.

As above, the u

,Du

,andD

have continuous extensions to

−

.More-

over, the sequences {u

}, {Du

},and{D

} are bounded and equicontin-

uous on

−

⊃

∪ Σ

and can be assumed to converge to u, Du,and

u, respectively, on

⊃

∪ Σ

by passing to a subsequence if neces-

sary. Thus, u ∈ C

(

∪Σ

) for all δ as described above, and it follows that

u ∈ C

(Ω

−

)∩C

(Ω ∪Σ). Letting j →∞in Equations (11.11), u satisﬁes the

equations Lu = f on Ω, Mu = g on Σ,andu = h on ∂Ω ∼ Σ. Uniqueness

follows from Corollary 11.2.2. By Inequality (11.10),

(b)

2+α,Ω

≤|v

(b)

2+α,Ω

+ |h

(b)

2+α,Ω

≤ C



|f|

(2+b)

α,Ω

+ |g|

(1+b)

1+α,Σ

+ |h|

0,∂Ω∼Σ



Thus for 0 <δ<1,

2+b+α

2+α,

≤ C



|f|

(2+b)

α,Ω

+ |g|

(1+b)

1+α,Σ

+ |h|

0,∂Ω∼Σ



By expanding the left side, letting j →∞, and then taking the supremum

over 0 <δ<1, Inequality (11.8) is established.

11.3 Curved Boundaries 405

11.3 Curved Boundaries

The procedure for dealing with curved boundaries involves the use of trans-

formations to “straighten the boundary” as follows. Let Ω be a bounded open

subset of R

and let Σ be a nonempty, relatively open subset of ∂Ω,possibly

equal to ∂Ω.Foreachx ∈ Σ,letU

be a neighborhood of x and let τ

be a

map of U

into R

with continuous partials on U

such that τ

(x)=0∈ R

If the map τ

has an inverse, then facts known for spherical chips can be

carried back to U

by means of τ

−1

. A suﬃcient condition for the existence

of an inverse for τ

is that the Jacobian of τ

not vanish at x;thatis,if

=(τ

,...,τ

), then

J(τ

)(x)=



∂τ

(x)

∂x

···

∂τ

(x)

∂x

∂τ

(x)

∂x

···

∂τ

(x)

∂x



=0.

If this condition is satisﬁed, then there is a neighborhood O of 0 ∈ R

and

a neighborhood V

of x such that τ

−1

is deﬁned on O, V

= τ

−1

(O),τ

one-to-one on V

,τ

−1

(τ

)=x,andτ

−1

has continuous ﬁrst partials on O

(c.f. [1]).

Deﬁnition 11.3.1 If τ is a mapping from Ω into R

,let

τ(x)=(τ

(x),...,τ

(x)) if x =(x

,...,x

) ∈ Ω.

The map τ is said to be diﬀerentiable of class C

on Ω if each component

∈ H

(Ω). If U and V are open subsets of R

and τ : U → V is a

homeomorphism such that τ and τ

−1

are of class C

,thenτ is said to be a

diﬀeomorphism of U onto V .

Let Ω be a bounded, convex, open subset of R

,letΛ be a neighborhood

of Ω

−

,andletτ be a C

k+α

diﬀeomorphism, k ≥ 1, 0 <α≤ 1, which maps Λ

onto a bounded, convex, open subset Λ



of R

. Using the mean value theorem,

it is easy to see that there is a constant K>0 such that

|x − y|≤|τ(x) − τ(y)|≤K|x − y|,x,y∈ Ω (11.12)

where K depends only upon τ and Ω.

Deﬁnition 11.3.2 A relatively open subset Σ of the boundary of the open

set Ω ⊂ R

is class C

,wherea = k + α, k ≥ 1, 0 <α≤ 1, if for each x ∈ Σ

there is a neighborhood U of x and a C

diﬀeomorphism τ

of U onto a ball

B = B

y,ρ

⊂ R

of radius ρ with center at y =(0,...,0, −ρ

) such that

406 11 Oblique Derivative Problem

(i) U ∩ ∂Ω ⊂ Σ,

(ii) ∂(U ∩Ω) ⊂ Ω ∪ Σ,

(iii) τ

(U ∩Σ)=B

,and

(iv) τ

(U ∩Ω)=B

The relation Σ ∈C

signiﬁes that the H

(Ω) norms of the τ

and τ

−1

,x∈ Σ,

are uniformly bounded.

Lemma 11.3.3 Let τ be a diﬀeomorphism of class C

k+α

,k ≥ 1, 0 <α≤ 1 of

Ω onto Ω

∗

.Ifu ∈ H

j+β

(Ω), ˜u(y)=u(τ

−1

(y)),y ∈ Ω

∗

,andj +β ≤ k+α, 1 ≤

j ≤ k, 0 <β≤ 1, then there is a constant C>0 such that

|u|

j+β,Ω

≤|˜u|

j+β,Ω

∗

≤ C|u|

j+β,Ω

where C depends only upon β, τ, Ω.

Proof: The proof will be carried out only in the j = 1 case. It follows from

the deﬁnition of ˜u that ˜u

0,Ω

∗

= u

0,Ω

≤|u|

1+β,Ω

. By the chain rule for

diﬀerentiation,

˜u(y)=D

u(τ

−1

(y)) =



m=1

u(τ

−1

(y))D

(τ

−1

(y)|.

Thus, [˜u]

1+0,Ω

∗

≤ C[u]

1+0,Ω

where C is a constant depending only upon τ.

By Inequality (11.12),

˜u(y) − D

˜u(x)|

|x − y|



m=1

u(τ

−1

(y))D

(τ

−1

(y)) −



m=1

u(τ

−1

(x))D

(τ

−1

(x))|

|x − y|

≤ C



m=1

|(D

u(τ

−1

(y)) − D

u(τ

−1

(x))D

(τ

−1

(y))|

|τ

−1

(y) − τ

−1

(x)|



m=1

|(D

u(τ

−1

(x))(D

(τ

−1

(y) − D

(τ

−1

(x))|

|x − y|

≤ C



[u]

1+β,Ω

[τ

−1

]

1+0,Ω

∗

+[u]

1+0,Ω

[τ

−1

]

1+β,Ω

∗



≤ C[u]

1+β,Ω

[τ

−1

1+β,Ω

∗

≤ C[u]

1+β,Ω

Therefore, |˜u|

1+β,Ω

∗

≤ C|u|

1+β,Ω

. Interchanging the roles of u, τ and ˜u, τ

−1

there is a constant K such that |u|

1+β,Ω

≤ K|˜u|

1+β,Ω

∗

. The assertion follows

by replacing C and K by the maximum of the two.

Consider a relatively open subset Σ of ∂Ω of class C

2+α

and the diﬀeo-

morphisms τ

,x∈ Σ, satisfying (i) through (iv) of Deﬁnition 11.3.2. For each