Helms L.L. Potential Theory

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368 9 Elliptic Operators

On U

∩ Ω,



g(x

+ k(j)w



≤ L



g(x



− k(j)

≤ L



g(x



−



f − L



g(x





≤ L



g(x



+ f − L



g(x



= f.

It follows that the function g(x

)+(1/j)+k(j)w is a superfunction on U

∩Ω

for all f ∈Fand g by Lemma 9.7.4. Let



min {g(x

+ k(j)w, g(x

)+u} on U

∩ Ω

g(x

)+u on Ω

−

∼ U

To show that w

is a superfunction on Ω, according to Lemma 9.7.4 it

suﬃces to show that Lw

≤ f for all f ∈Fat each point of Ω since

∈ C

(Ω

−

) ∩ C

(Ω). By Lemma 9.7.7, w

is a superfunction on U

∩ Ω

and on Ω ∼ U

−

.Thus,Lw

≤ f on each of these two sets by Lemma 9.7.4.

It remains only to show that Lw

≤ f at each point z ∈ ∂U

∩ Ω.Since

g(x

)+(1/j)+k(j)w(z) >g(x

)+u(z), there is a neighborhood V of z such

that g(x

)+(1/j)+k(j)w>g(x

)+u on V .Sincew

= g(x

)+u on V

and the latter is a superfunction on Ω, Lw

≤ f on V and, in particular,

(z) ≤ f. This completes the proof that w

is a superfunction on Ω.

On ∂Ω ∩ U

,g(x

)+(1/j)+k(j)w ≥ g(x

)+(1/j) >gso that w

≥ g on

∂Ω ∩U

.On∂Ω ∼ U

= g(x

)+u ≥ g since g(x

)+u is assumed to be

a supersolution relative to g.Thus,w

≥ g on ∂Ω and w

is a supersolution

on Ω for all f ∈Frelative to g.Atx

)=min{g(x

,g(x

)+u(x

)}

and

lim

j→∞

)=min{g(x

),g(x

)+u(x

)} = g(x

This completes the proof that {w

} is an upper approximate barrier at x

simultaneously for all f ∈F. Only minor modiﬁcations of the above argument

are needed to construct a lower approximate barrier {w

−

} at x

Theorem 9.8.9 Let Ω be a L-regular bounded open subset of R

and let

L be a strictly elliptic operator with coeﬃcients in H

(Ω), c ≤ 0,andlet

f ∈ C

(Ω)∩H

(2+b)

(Ω; d),α ∈ (0, 1),b∈ (−1, 0).Ifg ∈ C

(∂Ω),theDirichlet

problem

Lu = f on Ω, u = g on ∂Ω

has a unique solution u ∈ H

2+α,loc

(Ω).

9.8 Barriers 369

Proof: The requirement of regularity means that at each point x of ∂Ω,

lim

y→x,y∈Ω

f,g

(y)=g(x) for all x ∈ ∂Ω. Redeﬁning H

−

f,g

, if necessary, to

be equal to g on ∂Ω,H

−

f,g

∈ U(f, g)sothatH

f,g

≤H

−

f,g

and the two are

equal. Letting H

f,g

= H

f,g

= H

−

f,g

, LH

f,g

= f on Ω and H

f,g

= g on ∂Ω.

Uniqueness follows from Theorem 9.5.7.

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Chapter 10

Apriori Bounds

10.1 Introduction

As is well known, the mere fact that a function of a complex variable is

diﬀerentiable on a domain in the complex plane implies that the function

has many other properties; for example, the real and imaginary parts satisfy

the Cauchy-Riemann equations, the function is inﬁnitely diﬀerentiable on the

domain, etc. In addition, if a function on a domain Ω in R

has continuous

second partials and satisﬁes Laplace’s equation thereon, then other properties

follow as a consequence, such as the averaging property, etc. It will be shown

in this chapter that there are limitations on the size of the H

(b)

2+α

(Ω)normof

a solution of the oblique boundary derivative problem for elliptic equations.

These inequalities are known as apriori inequalities since it is assumed that

a function is a solution without actually knowing that there are solutions. The

establishment of such inequalities will pave the way for proving the existence

of solutions in the next chapter.

Eventually, very strong conditions will have to be imposed on Ω. So strong,

in fact, that the reader might reasonably conclude that only a spherical chip,

or some topological equivalent, will satisfy all the conditions. As the ultimate

application will involve spherical chips, the reader might beneﬁt from as-

suming that Ω is a spherical chip. But as some of the inequalities require

less stringent conditions on Ω, the conditions on Ω will be spelled out at

the beginning of each section. Since these conditions will not be repeated in

the formal statements, it is important to refer back to the beginning of each

section for a description of Ω.

L.L. Helms, Potential Theory, Universitext, 371

 Springer-Verlag London Limited 2009

372 10 Apriori Bounds

10.2 Green Function for a Half-space

Let Ω = R

and let γ,β

,...,β

be constants with γ ≤ 0inthen ≥ 3case,

γ<0inthen =2case,β

> 0, and |β| =1whereβ =(β

,...,β

). For

x ∈ R

, deﬁne a boundary operator M

by the equation

u(x)=γu(x)+



i=1

u(x)=γu(x)+β ·∇u(x),

which can be written M

u = γu+ D

u,whereD

u is the directional deriva-

tive in the β direction. Although M

is primarily of interest as an operator

on functions on R

, it can be regarded as an operator on functions with

domain Ω.

To solve the oblique derivative problem

Δu = f on Ω, M

u =0 onR

a Green function G

1/2

(x, y)onR

× R

will be constructed so that the

solution is given by

u(x)=



1/2

(x, y)f(y) dy.

The function G

1/2

(x, y) is deﬁned by the equation

1/2

(x, y)=u(|x − y|) − u(|x − y

|) − 2β



∞

γs

∂

∂x

u(|x − y

+ sβ|) ds,

where u(|x − y|) is the fundamental harmonic function with pole at x ∈ R

It will be shown below that the same equation can be used to deﬁne G

1/2

on R

−

× R

, recalling that R

−

denotes the closure of R

. The following

calculations will be carried out in the n ≥ 3case,then =2casebeing

essentially the same assuming that γ<0.

The integral part of G

1/2

(x, y) suggests deﬁning a kernel h(x, y)onR

−

by the equation

h(x, y)=−2β



∞

γs

∂

∂x

u(|x − y

+ sβ|) ds

=2β

(n − 2)



∞

γs

|x − y

+ sβ|

−n

+ y

+ sβ

) ds.

(10.1)

The integral can be seen to be ﬁnite for all (x, y) ∈ R

−

× R

as follows.

Letting ξ =(x − y

)/|x − y

10.2 Green Function for a Half-space 373

|h(x, y)|≤2β

(n − 2)



∞

γs

|x − y

+ sβ|

−n+1

=2β

(n − 2)|x − y

−n+2



∞

γ|x−y

|ξ + sβ|

−n+1

≤ 2β

(n − 2)|x − y

−n+2



∞

|ξ + sβ|

−n+1

ds.

Putting β

=(β

,...,β

n−1

, 0) and β

=(0,...,0,β

),ξ· β = ξ · β

+ ξ · β

= ξ · β

+ ξ

≥ ξ · β

≥−|β

| = −



1 − β

for all x ∈ R

−

,y ∈ R

,and

|ξ + sβ|

=1+2sξ · β + s

≥ 1 − 2s



1 − β

+ s

Since the latter function has the minimum value β

, |ξ + sβ|

−n+1

≤ 1/β

n−1

Noting that |ξ + sβ|≥||sβ|−|ξ|| = |s − 1|,

|ξ + sβ|

−n+1

≤ 1/|s − 1|

n−1

The integrand of the above integral is therefore dominated on [0, +∞)by

the integrable function min (1/β

n−1

, 1/|s − 1|

n−1

). This shows that h(x, y)is

ﬁnite for all x ∈ R

−

,y ∈ R

. Moreover, there is a constant C(n, β

)such

that

|h(x, y)|≤C(n, β

)|x − y

−n+2

x ∈ R

−

,y ∈ R

. (10.2)

If (x

) is a sequence in R

−

×R

that converges to (x, y) ∈ R

−

×R

,it

follows in the same way from the Lebesgue dominated convergence theorem

that lim

n→∞

h(x

)=h(x, y); i.e., h is continuous on R

−

×R

.Itiseasy

to see by a simple geometrical argument that |x − y|≤|x − y

|,x,y ∈ R

Therefore,

|h(x, y)|≤C(n, β

)

|x − y|

n−2

,x∈ R

−

,y ∈ R

. (10.3)

It will be shown now that M

1/2

(·,y)=0on(x

=0)foreachy ∈ R

For ﬁxed y ∈ R



i=1

∂

∂x



u(|x − y|) − u(|x − y



= β

(−n +2)



|x − y|

−n

− y

) −|x − y

−n

+ y

)



so that when x

=0,



u(|x − y|) − u(|x − y



=2(n − 2)|x − y|

−n

. (10.4)

374 10 Apriori Bounds

Moreover,



i=1

∂

∂x



− 2β



∞

γs

∂

∂x

u(|x − y

+ sβ|) ds



= −2nβ

(n − 2)



∞

γs

|x − y

+ sβ|

−n−2





i=1

− (y

)

+ sβ

)



× (x

+ y

+ sβ

) ds

+2β

(n − 2)



∞

γs

|x − y

+ sβ|

−n

ds.

Using the fact that

(|x − y

+ sβ|

−n

)=−n|x − y

+ sβ|

−n−2



i=1

− (y

)

+ sβ

integrating by parts, and simplifying,



i=1

∂

∂x



− 2β



∞

γs

∂

∂x

u(|x − y

+ sβ|) ds



=2β

(n − 2)



∞

γs

(|x − y

+ sβ|

−n

)(x

+ y

+ sβ

) ds

+2β

(n − 2)



∞

γs

|x − y

+ sβ|

−n

= −2β

(n − 2)|x − y

−n

+ y

)

− 2β

γ(n − 2)



∞

γs

|x − y

+ sβ|

−n

+ y

+ sβ

) ds

= −2β

(n − 2)|x − y

−n

+ y

)

+2γβ



∞

γs

∂

∂x

|x − y

+ sβ|

−n+2

= −2β

(n − 2)|x − y

−n

+ y

) − γh(x, y).

Putting x

= 0 in this equation, using Equation (10.4), and using the fact

that G

1/2

(x, y)=h(x, y)whenx

=0,

1/2

(x, y)=γG

1/2

(x, y)+2(n − 2)|x − y|

−n

− 2(n − 2)|x − y|

−n

− γG

1/2

(x, y)=0.

Thus, M

1/2

(·,y)=0onR

10.2 Green Function for a Half-space 375

Lemma 10.2.1 There is a constant C(n, β

) such that

(i) D

h(x, y)=−D

h(x, y), 1 ≤ i ≤ n −1,

(ii) D

h(x, y)=D

h(x, y),

(iii) |D

(x)

h(x, y)|≤C(n, β

)|x − y|

−n+2−k

if |β| = k, 0 ≤ k ≤ 3,

for all x ∈ R

−

,y ∈ R

Proof: The ﬁrst two assertions are straightforward and the k =0caseof

(iii) was proved above. For 1 ≤ i ≤ n − 1,

h(x, y)|≤2β

n(n − 2)



∞

γs

|x − y

+ sβ|

−n−2

− y

+ sβ

×|x

+ y

+ sβ

|ds

≤ 2β

n(n − 2)



∞

γs

|x − y

+ sβ|

−n

ds.

Letting ξ =(x − y

)/|x − y

h(x, y)|≤2β

n(n − 2)|x − y

−n+1



∞

γ|x−y

|ξ + sβ|

−n

ds.

As above, the integrand of the above integral is dominated on (0, +∞)by

the integrable function min (1/β

, 1/|s − 1|

). Thus,

h(x, y)|≤2β

n(n − 2)|x − y

−n+1



∞

min (1/β

, 1/|s − 1|

) ds

and there is a constant C = C(n, β

) such that

h(x, y)|≤C(n, β

)|x − y

−n+1

, 1 ≤ i ≤ n − 1,x∈ R

−

,y ∈ R

When i = n,

h(x, y)=2β

(n − 2)



∞

γs



− n|x − y

+ sβ|

−n−2

+ y

+ sβ

)

+ |x − y

+ sβ|

−n



so that

h(x, y)|≤4β

n(n − 2)



∞

γs

|x − y

+ sβ|

−n

ds.

The integral is the same as in the preceding step so that the same bound

is obtained except for a factor of 2 which can be absorbed into the con-

stant C(n, β

). Estimates for |D

h(x, y)|, |β| = k, follow in exactly the same

way using the integrable function min (1/β

n−1+k

, 1/|s − 1|

n−1+k

). This es-

tablishes (iii) with |x − y

| in place of |x −y|.Since|x − y|≤|x −y

| for all

x, y ∈ R

, |x − y

−n+2−k

≤|x − y|

−n+2−k

and (iii) holds.

376 10 Apriori Bounds

This lemma shows that the kernel h(x, y)onR

× R

satisﬁes Inequali-

ties (8.1), (8.2), (8.3), and (8.4) of Section 8.2. All of the results of that section

therefore apply to the kernel h(x, y). It was shown in Section 8.3 that the

same results apply to the Newtonian kernel u(|x−y|). Since |x−y|≤|x−y

the kernel u(|x−y

|) also satisﬁes the conditions of Section 8.2. It follows that

1/2

(x, y) satisﬁes the same conditions and all of the results of Section 8.2

are applicable to G

1/2

(x, y).

Deﬁnition 10.2.2 If f is a bounded measurable function with compact sup-

port in R

, deﬁne G

1/2

f corresponding to the kernel G

1/2

by the equation

1/2

f(x)=

(2 − n)σ



1/2

(x, y)f(y) dy, x ∈ R

Theorem 10.2.3 If f is a bounded measurable function with compact sup-

port in R

, f is locally H¨older continuous with exponent α ∈ (0, 1],and

u = G

1/2

f,thenu ∈ C

) ∩ C

∪ R

),Δu = f on R

,andM

u =0

on R

Proof: To show that Δu = f on R

,consider

1/2

f(x)=

(2 − n)σ



u(|x − y|)f(y) dy −

(2 − n)σ



u(|x − y

|)f(y) dy

−

2β

(2 − n)σ



∞

γs

∂

∂x

u(|x − y

+ sβ|)f (y) dy.

By Theorem 8.3.1, the ﬁrst term is in C

) and its Laplacian is just f .

It therefore suﬃces to show that the second and third terms are harmonic

functions on D

= {x ∈ R

; x

>δ} for each δ>0. Apply Theorem 1.7.13 to

show that these functions are harmonic. Thus, Δu = f on R

.Theassertion

that u ∈ C

∪ R

) follows from Theorem 8.2.7. The fact that M

u =0

on R

follows from the fact that MG

1/2

(·,y)=0onR

Remark 10.2.4 This proof includes the fact that G

1/2

f has the represen-

tation

1/2

f =

(2 − n)σ

Uf + h

where Uf is the Newtonian potential of f and h is harmonic on R

10.3 Mixed Boundary Conditions for Laplacian

This section will be limited to regions Ω that are spherical chips with

Σ = int(∂Ω ∩ R

)orballs in R

with Σ = ∅. Such regions are convex along

with the

and are stratiﬁable. An integral representation of a solution to

Poisson’s equation Δu = f on a spherical chip using a kernel

10.3 Mixed Boundary Conditions for Laplacian 377

(x, y)=

2σ

(n − 2)



|x − y|

n−2

−

|x − y

n−2



,x,y∈ R

will be derived next. If x

∈ Ω and ρ>0, let B

= B

,tρ

= B

∩ R

and B

= B

∩ R

. Consider only those ρ>0forwhich∂B

∼ B

⊂ Ω.

The integral representation requires the use of a cutoﬀ function ψ

,de-

pending upon ρ, which can be constructed as follows. Let ψ be a function in

∞

(0, ∞) satisfying (i)0 ≤ ψ ≤ 1, (ii) ψ =1on(0, 5/4], and (iii) ψ =0on

(7/4, ∞); for example, take ψ = J

1/4

χ where χ is the indicator function of

the interval (0, 3/2) and J

1/4

is the operator deﬁned in Section 2.5. For ρ>0,

as restricted above, let ψ

(y)=ψ(

),y ∈ B

,wherer = |x

−y| for y ∈ R

It is easily seen that ψ



0,B

≤ 1, Dψ



0,B

≤ C/ρ, D



0,B

≤ C/ρ

and [ψ

]

2+α,B

≤ C/ρ

2+α

. The following lemma is taken from [25]. The proof

requires the use of Green’s representation theorem, Theorem 1.4.2, with the

ball B replaced by B

. Green’s theorem is valid for regions with a Lipschitz

boundary (c.f.[48], p.121).

Lemma 10.3.1 If Ω is a spherical chip with Σ = int(∂Ω ∩ R

) and u ∈

(Ω ∪ Σ) ∩ C

(Ω),then

u(x)=



(x, y)Δ(uψ

) dy, x ∈ B

Proof: For x, y ∈ B

,letr

(x, y)=|x − y|

−n+2

/σ

(n − 2) and r

(x, y)=

|x − y

−n+2

/σ

(n − 2) so that K

=(r

−r

)/2. By Green’s representation

theorem, Theorem 1.4.2,

u(x)ψ

(x)=



∂B

(uψ

)−uψ

) dσ(y)−



Δ(uψ

) dy, x ∈ B

Since ψ

=0andD

(uψ

)=0on∂B

∼ B

and D

= D

on B

u(x)ψ

(x)=



(uψ

) − uψ

) dσ(y)

−



Δ(uψ

) dy, x ∈ B

(10.5)

Similarly, if x, y

∈ B

u(x)ψ

(x)=



)

(uψ

) − uψ

)

) dσ(y

)

−



Δ(uψ

) dy

,x∈ B