Sobolev spaces for planar domains explained

In mathematics, Sobolev spaces for planar domains are one of the principal techniques used in the theory of partial differential equations for solving the Dirichlet and Neumann boundary value problems for the Laplacian in a bounded domain in the plane with smooth boundary. The methods use the theory of bounded operators on Hilbert space. They can be used to deduce regularity properties of solutions and to solve the corresponding eigenvalue problems.

Sobolev spaces with boundary conditions

Let be a bounded domain with smooth boundary. Since is contained in a large square in, it can be regarded as a domain in by identifying opposite sides of the square. The theory of Sobolev spaces on can be found in, an account which is followed in several later textbooks such as and .

For an integer, the (restricted) Sobolev space is defined as the closure of in the standard Sobolev space .

\iint\Omega\left(g(Dh)+(Dg)h\right)dxdy=\limn\to\iint\Omega\left(g(Dhn)+(Dg)hn\right)dxdy=0.

By Green's theorem this implies

\int\partial\Omegagk=0,

where

k=h\cos\left(n(a,b)\right),

with the unit normal to the boundary. Since such form a dense subspace of, it follows that on .

Suppose in annihilates . By compactness, there are finitely many open sets covering such that the closure of is disjoint from and each is an open disc about a boundary point such that in small translations in the direction of the normal vector carry into . Add an open with closure in to produce a cover of and let be a partition of unity subordinate to this cover. If translation by is denoted by, then the functions

gt=\psi0g

N
+\sum
i=1

\psin

λ
tni

g

tend to as decreases to and still lie in the annihilator, indeed they are in the annihilator for a larger domain than, the complement of which lies in . Convolving by smooth functions of small support produces smooth approximations in the annihilator of a slightly smaller domain still with complement in . These are necessarily smooth functions of compact support in .

k
\|h\|
j=0

{k\choosej}\left

k-j
\|\partial
y

h\right\|2.

Moreover satisfies for g in .

(f,g)=(f,Pkg).

This implies that under the pairing between and, and are each other's duals.

\begin{align} \left\|Pk(I+\Delta)kf\right\|(-k)&=

\sup
\|g\|(-k)=1

\left|\left((I+\Delta)kf,g\right)(-k)\right|\\ &=

\sup
\|g\|(-k)=1

|(f,g)|\\ &=\|f\|(k). \end{align}

Since the adjoint map between the duals can by identified with this map, it follows that is a unitary map.

Application to Dirichlet problem

See also: Dirichlet problem.

Invertibility of

The operator defines an isomorphism between and . In fact it is a Fredholm operator of index . The kernel of in consists of constant functions and none of these except zero vanish on the boundary of . Hence the kernel of is and is invertible.

In particular the equation has a unique solution in for in .

Eigenvalue problem

Let be the operator on defined by

T=

-1
R
1\Delta

R0,

where is the inclusion of in and of in, both compact operators by Rellich's theorem. The operator is compact and self-adjoint with for all . By the spectral theorem, there is a complete orthonormal set of eigenfunctions in with

Tfn=\munfn,    \mun>0,\mun\to0.

Since, lies in . Setting, the are eigenfunctions of the Laplacian:

\Deltafn=λnfn,    λn>0,λn\toinfty.

Sobolev spaces without boundary condition

To determine the regularity properties of the eigenfunctions and solutions of

\Deltaf=u,    u\inH-1(\Omega),f\in

1
H
0(\Omega),

enlargements of the Sobolev spaces have to be considered. Let be the space of smooth functions on which with their derivatives extend continuously to . By Borel's lemma, these are precisely the restrictions of smooth functions on . The Sobolev space is defined to the Hilbert space completion of this space for the norm

2
\|f\|
(k)

=

k
\sum
j=0

{k\choosej}\left

k-j
\|\partial
y

f\right\|2.

This norm agrees with the Sobolev norm on so that can be regarded as a closed subspace of . Unlike, is not naturally a subspace of, but the map restricting smooth functions from to is continuous for the Sobolev norm so extends by continuity to a map .

It is sufficient to construct an extension for a neighbourhood of a closed annulus, since a collar around the boundary is diffeomorphic to an annulus with a closed interval in . Taking a smooth bump function with, equal to 1 near the boundary and 0 outside the collar, will provide an extension on . On the annulus, the problem reduces to finding an extension for in . Using a partition of unity the task of extending reduces to a neighbourhood of the end points of . Assuming 0 is the left end point, an extension is given locally by

E(f)=

k
\sum
m=0

amf\left(-

x
m+1

\right).

Matching the first derivatives of order k or less at 0, gives

k
\sum
m=0

(-m-1)-kam=1.

This matrix equation is solvable because the determinant is non-zero by Vandermonde's formula. It is straightforward to check that the formula for, when appropriately modified with bump functions, leads to an extension which is continuous in the above Sobolev norm.

\subsetH2(\Omega)\subsetH1(\Omega)\subsetH0(\Omega)\subset

-1
H
0(\Omega)

\subset

-2
H
0(\Omega)

\subset

The differentiation operators carry each Sobolev space into the larger one with index 1 less.

\left(f\alpha,\varphi\right)=(-1)|\alpha|\left(f,\partial\alpha\varphi\right),    |\alpha|\leqk,\varphi\in

infty
C
c(\Omega).

To prove the characterization, note that if is in, then lies in Hk−|α|(Ω) and hence in . Conversely the result is well known for the Sobolev spaces : the assumption implies that the is in and the corresponding condition on the Fourier coefficients of shows that lies in . Similarly the result can be proved directly for an annulus . In fact by the argument on the restriction of to any smaller annulus [−δ',δ'] × T lies in : equivalently the restriction of the function lies in for . On the other hand in as, so that must lie in . The case for a general domain reduces to these two cases since can be written as with ψ a bump function supported in such that is supported in a collar of the boundary.

Rtf(x,y)=f(x,y+t).

On it is known that if is in, then the difference quotient in ; if the difference quotients are bounded in Hk then ∂yf lies in . Both assertions are consequences of the formula:

\widehat{\deltahf}(m,n)=h-1(e-ihn

1
-1)\widehat{f}(m,n)=-\int
0

ine-inhtdt\widehat{f}(m,n).

These results on imply analogous results on the annulus using the extension.

Regularity for Dirichlet problem

Regularity for dual Dirichlet problem

If with in and in with, then lies in .

Take a decomposition with supported in and supported in a collar of the boundary. Standard Sobolev theory for can be applied to : elliptic regularity implies that it lies in and hence . lies in of a collar, diffeomorphic to an annulus, so it suffices to prove the result with a collar and replaced by

\begin{align} \Delta1&=\Delta-[\Delta,\psi]\\ &=\Delta+\left(p\partialx+q\partialy-\Delta\psi\right)\\ &=\Delta+X. \end{align}

The proof proceeds by induction on, proving simultaneously the inequality

\|u\|(k+1)\leC\|\Delta1u\|(k-1)+C\|u\|(k),

for some constant depending only on . It is straightforward to establish this inequality for, where by density can be taken to be smooth of compact support in :

2
\begin{align} \|u\|
(1)

&=|(\Deltau,u)|\\ &\le|(\Delta1u,u)|+|(Xu,u)|\\ &\le\|\Delta1u\|(-1)\|u\|(1)+C\prime\|u\|(1)\|u\|(0). \end{align}

The collar is diffeomorphic to an annulus. The rotational flow on the annulus induces a flow on the collar with corresponding vector field . Thus corresponds to the vector field . The radial vector field on the annulus is a commuting vector field which on the collar gives a vector field proportional to the normal vector field. The vector fields and commute.

The difference quotients can be formed for the flow . The commutators are second order differential operators from to . Their operators norms are uniformly bounded for near ; for the computation can be carried out on the annulus where the commutator just replaces the coefficients of by their difference quotients composed with . On the other hand, lies in, so the inequalities for apply equally well for :

\begin{align} \|\deltahu\|(k+1)&\leC\|\Delta1\deltahu\|(k-1)+C\|\deltahu\|(k)\\ &\leC\|\deltah\Delta1u\|(k-1)+C\|[\deltah,\Delta1]u\|(k-1)+C\|\deltahu\|(k)\\ &\leC\|\Delta1u\|(k)+C\prime\|u\|(k+1). \end{align}

The uniform boundedness of the difference quotients implies that lies in with

\|Yu\|(k+1)\leC\|\Delta1u\|(k)+C\prime\|u\|(k+1).

It follows that lies in where is the vector field

V=

Y
\sqrt{r2+s2
} = a\partial_x + b\partial_y, \qquad a^2 + b^2 = 1.

Moreover, satisfies a similar inequality to .

\|Vu\|(k+1)\leC\prime\prime\left(\|\Delta1u\|(k)+\|u\|(k+1)\right).

Let be the orthogonal vector field

W=-b\partialx+a\partialy.

It can also be written as for some smooth nowhere vanishing function on a neighbourhood of the collar.

It suffices to show that lies in . For then

(V\pmiW)u=(a\mpib)\left(\partialx\pmi\partialy\right)u,

so that and lie in and must lie in .

To check the result on, it is enough to show that and lie in . Note that

\begin{align} A&=\Delta-V2-W2,\\ B&=[V,W], \end{align}

are vector fields. But then

\begin{align} W2u&=\Deltau-V2u-Au,\\ VWu&=WVu+Bu, \end{align}

with all terms on the right hand side in . Moreover, the inequalities for show that

\begin{align} \|Wu\|(k+1)&\leC\left(\|VWu\|(k)+\left\|W2u\right\|(k)\right)\\ &\leC\left\|\left(\Delta-V2-A\right)u\right\|(k)+C\|(WV+B)u\|(k)\\ &\leC1\|\Delta1u\|(k)+C1\|u\|(k+1). \end{align}

Hence

\begin{align} \|u\|(k+2)&\leC\left(\|Vu\|(k+1)+\|Wu\|(k+1)\right)\\ &\leC\prime\|\Delta1u\|(k)+C\prime\|u\|(k+1). \end{align}

Smoothness of eigenfunctions

It follows by induction from the regularity theorem for the dual Dirichlet problem that the eigenfunctions of in lie in . Moreover, any solution of with in and in must have in . In both cases by the vanishing properties, the eigenfunctions and vanish on the boundary of .

Solving the Dirichlet problem

The dual Dirichlet problem can be used to solve the Dirichlet problem:

\begin{cases}\Deltaf|\Omega=0\f|\partial\Omega=g&g\inCinfty(\partial\Omega)\end{cases}

By Borel's lemma is the restriction of a function in . Let be the smooth solution of with on . Then solves the Dirichlet problem. By the maximal principle, the solution is unique.

Application to smooth Riemann mapping theorem

The solution to the Dirichlet problem can be used to prove a strong form of the Riemann mapping theorem for simply connected domains with smooth boundary. The method also applies to a region diffeomorphic to an annulus. For multiply connected regions with smooth boundary have given a method for mapping the region onto a disc with circular holes. Their method involves solving the Dirichlet problem with a non-linear boundary condition. They construct a function such that:

gives a proof of the Riemann mapping theorem for a simply connected domain with smooth boundary. Translating if necessary, it can be assumed that . The solution of the Dirichlet problem shows that there is a unique smooth function on which is harmonic in and equals on . Define the Green's function by . It vanishes on and is harmonic on away from . The harmonic conjugate of is the unique real function on such that is holomorphic. As such it must satisfy the Cauchy–Riemann equations:

\begin{align} Ux&=-Vy,\ Uy&=Vx. \end{align}

The solution is given by

z
V(z)=\int
0

-Uydx+Vxdy,

where the integral is taken over any path in . It is easily verified that and exist and are given by the corresponding derivatives of . Thus is a smooth function on, vanishing at . By the Cauchy-Riemann is smooth on, holomorphic on and . The function is only defined up to multiples of, but the function

F(z)=eG(z)+iH(z)=zef(z)

is a holomorphic on and smooth on . By construction, and for . Since has winding number, so too does . On the other hand, only for where there is a simple zero. So by the argument principle assumes every value in the unit disc,, exactly once and does not vanish inside . To check that the derivative on the boundary curve is non-zero amounts to computing the derivative of, i.e. the derivative of should not vanish on the boundary curve. By the Cauchy-Riemann equations these tangential derivative are up to a sign the directional derivative in the direction of the normal to the boundary. But vanishes on the boundary and is strictly negative in since . The Hopf lemma implies that the directional derivative of in the direction of the outward normal is strictly positive. So on the boundary curve, has nowhere vanishing derivative. Since the boundary curve has winding number one, defines a diffeomorphism of the boundary curve onto the unit circle. Accordingly, is a smooth diffeomorphism, which restricts to a holomorphic map and a smooth diffeomorphism between the boundaries.

Similar arguments can be applied to prove the Riemann mapping theorem for a doubly connected domain bounded by simple smooth curves (the inner curve) and (the outer curve). By translating we can assume 1 lies on the outer boundary. Let be the smooth solution of the Dirichlet problem with on the outer curve and on the inner curve. By the maximum principle for in and so by the Hopf lemma the normal derivatives of are negative on the outer curve and positive on the inner curve. The integral of over the boundary is zero by Stoke's theorem so the contributions from the boundary curves cancel. On the other hand, on each boundary curve the contribution is the integral of the normal derivative along the boundary. So there is a constant such that satisfies

\intC\left(-Uydx+Uxdy\right)=2\pi

on each boundary curve. The harmonic conjugate of can again be defined by

z
V(z)=\int
1

-uydx+uxdy

and is well-defined up to multiples of . The function

\displaystyle{F(z)=eU(z)

}

is smooth on and holomorphic in . On the outer curve and on the inner curve . The tangential derivatives on the outer curves are nowhere vanishing by the Cauchy-Riemann equations, since the normal derivatives are nowhere vanishing. The normalization of the integrals implies that restricts to a diffeomorphism between the boundary curves and the two concentric circles. Since the images of outer and inner curve have winding number and about any point in the annulus, an application of the argument principle implies that assumes every value within the annulus exactly once; since that includes multiplicities, the complex derivative of is nowhere vanishing in . This is a smooth diffeomorphism of onto the closed annulus, restricting to a holomorphic map in the interior and a smooth diffeomorphism on both boundary curves.

Trace map

The restriction map extends to a continuous map for . In fact

\displaystyle{\widehat{\tauf}(n)=\summ\widehat{f}(m,n),}

so the Cauchy–Schwarz inequality yields

\begin{align} \left|\widehat{\tauf}(n)\right|2\left(1+n2\right

k-1
2
)

&\le\left(\summ

\left(1+n2
k-1
2
\right)
\left(1+m2+n2\right)k

\right)\left(\summ\left|\widehat{f}(m,n)\right|2\left(1+m2+n2\right)k\right)\\ &\leCk\summ\left|\widehat{f}(m,n)\right|2\left(1+m2+n2\right)k, \end{align}

where, by the integral test,

\begin{align} Ck&=\supn\summ

\left(1+n2\right
k-1
2
)
\left(1+m2+n2\right)k

<infty,\ ck&=infn\summ

\left(1+n2\right
k-1
2
)
\left(1+m2+n2\right)k

>0. \end{align}

The map is onto since a continuous extension map can be constructed from to . In fact set

-1
\widehat{Eg}(m,n)
n

\widehat{g}(n)

\left(1+n2\right
k-1
2
)
\left(1+n2+m2\right)k

,

where

λn=\summ

\left(1+n2\right
k-1
2
)
\left(1+m2+n2\right)k

.

Thus . If g is smooth, then by construction Eg restricts to g on 1 × T. Moreover, E is a bounded linear map since

2
\begin{align} \|Eg\|
(k)

&=\summ,n\left|\widehat{Eg}(m,n)\right|2\left(1+m2+n2\right)\\ &\le

-2
c
k

\summ,n\left|\widehat{g}(n)\right|2

\left(1+n2\right)2k-1
\left(1+m2+n2\right)k

\\ &\le

-2
c
k

Ck\|g

2. \end{align}
\|
k-1
2

It follows that there is a trace map τ of Hk(Ω) onto Hk − 1/2(∂Ω). Indeed, take a tubular neighbourhood of the boundary and a smooth function ψ supported in the collar and equal to 1 near the boundary. Multiplication by ψ carries functions into Hk of the collar, which can be identified with Hk of an annulus for which there is a trace map. The invariance under diffeomorphisms (or coordinate change) of the half-integer Sobolev spaces on the circle follows from the fact that an equivalent norm on Hk + 1/2(T) is given by

2
\|f\|
[k+1]
2

=

2
\|f\|
(k)

+

2\pi
\int
0
2\pi
\int
0
\left|f(k)(s)-f(k)(t)\right|2
\left|eis-eit\right|2

dsdt.

It is also a consequence of the properties of τ and E (the "trace theorem"). In fact any diffeomorphism f of T induces a diffeomorphism F of T2 by acting only on the second factor. Invariance of Hk(T2) under the induced map F* therefore implies invariance of Hk − 1/2(T) under f*, since f* = τ ∘ F* ∘ E.

Further consequences of the trace theorem are the two exact sequences

(0)\to

1
H
0(\Omega)

\toH1(\Omega)\to

1
2
H

(\partial\Omega)\to(0)

and

(0)\to

2
H
0(\Omega)

\toH2(\Omega)\to

3
2
H

(\partial\Omega)

1
2
H

(\partial\Omega)\to(0),

where the last map takes f in H2(Ω) to f|∂Ω and ∂nf|∂Ω. There are generalizations of these sequences to Hk(Ω) involving higher powers of the normal derivative in the trace map:

(0)\to

k
H
0(\Omega)

\toHk(\Omega)\to

k
oplus
j=1
j-1
2
H

(\partial\Omega)\to(0).

The trace map to takes f to

Abstract formulation of boundary value problems

The Sobolev space approach to the Neumann problem cannot be phrased quite as directly as that for the Dirichlet problem. The main reason is that for a function in, the normal derivative cannot be a priori defined at the level of Sobolev spaces. Instead an alternative formulation of boundary value problems for the Laplacian on a bounded region in the plane is used. It employs Dirichlet forms, sesqulinear bilinear forms on, or an intermediate closed subspace. Integration over the boundary is not involved in defining the Dirichlet form. Instead, if the Dirichlet form satisfies a certain positivity condition, termed coerciveness, solution can be shown to exist in a weak sense, so-called "weak solutions". A general regularity theorem than implies that the solutions of the boundary value problem must lie in, so that they are strong solutions and satisfy boundary conditions involving the restriction of a function and its normal derivative to the boundary. The Dirichlet problem can equally well be phrased in these terms, but because the trace map is already defined on, Dirichlet forms do not need to be mentioned explicitly and the operator formulation is more direct. A unified discussion is given in and briefly summarised below. It is explained how the Dirichlet problem, as discussed above, fits into this framework. Then a detailed treatment of the Neumann problem from this point of view is given following .

The Hilbert space formulation of boundary value problems for the Laplacian on a bounded region in the plane proceeds from the following data:

A weak solution of the boundary value problem given initial data in is a function u satisfying

D(f,g)=(u,g)

for all g.

For both the Dirichlet and Neumann problem

D(f,g)=\left(fx,gx\right)+\left(fy,gy\right).

For the Dirichlet problem . In this case

D(f,g)=(\Deltaf,g),    f,g\inH.

By the trace theorem the solution satisfies in .

For the Neumann problem is taken to be .

Application to Neumann problem

The classical Neumann problem on consists in solving the boundary value problem

\begin{cases} \Deltau=f,&f,u\inCinfty

-),\\ \partial
(\Omega
n

u=0&on\partial\Omega \end{cases}

Green's theorem implies that for

(\Deltau,v)=(ux,vx)+(uy,vy)-(\partialnu,v)\partial.

Thus if in and satisfies the Neumann boundary conditions,, and so is constant in .

Hence the Neumann problem has a unique solution up to adding constants.

Consider the Hermitian form on defined by

\displaystyle{D(f,g)=(ux,vx)+(uy,vy).}

Since is in duality with, there is a unique element in such that

\displaystyle{D(u,v)=(Lu,v).}

The map is an isometry of onto, so in particular is bounded.

In fact

((L+I)u,v)=(u,v)(1).

So

\|(L+I)u\|(-1)

=\sup
\|v\|(1)=1

|((L+I)u,v)|=

\sup
\|v\|(1)=1

|(u,v)(1)|=\|u\|(1).

On the other hand, any in defines a bounded conjugate-linear form on sending to . By the Riesz–Fischer theorem, there exists such that

\displaystyle{(f,v)=(u,v)(1).}

Hence and so is surjective. Define a bounded linear operator on by

-1
T=R
1(I+L)

R0,

where is the map, a compact operator, and is the map, its adjoint, so also compact.

The operator has the following properties:

(Tf,g)=(u,(I+L)v)=(u,v)(1)=((I+L)u,v)=(f,Tg).

(Tf,f)=(u,u)(1)\ge0,

and implies and hence .

Tfn=\munfn

with and decreasing to .

\displaystyle{Lfnnfn,    λn=\mu

-1
n

-1.}

Thus are non-negative and increase to .

(ux,ux)+(uy,uy)=(Lu,u)=0,

so is constant.

Regularity for Neumann problem

Weak solutions are strong solutions

The first main regularity result shows that a weak solution expressed in terms of the operator and the Dirichlet form is a strong solution in the classical sense, expressed in terms of the Laplacian and the Neumann boundary conditions. Thus if with, then, satisfies and . Moreover, for some constant independent of,

\|u\|(2)\leC\|\Deltau\|(0)+C\|u\|(1).

Note that

\|u\|(1)\le\|Lu\|(-1)+\|u\|(0),

since

2
\begin{align} \|u\|
(1)

&=|(Lu,u)|

2
+\|u\|
(0)

\\ &\le\|Lu\|(-1)\|u\|(1)+\|u\|(0)\|u\|(1). \end{align}

Take a decomposition with supported in and supported in a collar of the boundary.

The operator is characterized by

(Lf,g)=\left(fx,gx\right)+\left(fy,gy\right)=(\Deltaf,g)\Omega-\left(\partialnf,g\right)\partial\Omega,    f,g\inCinfty(\Omega-).

Then

([L,\psi]f,g)=([\Delta,\psi]f,g),

so that

\displaystyle{[L,\psi]=-[L,1-\psi]=\Delta\psi+2\psix\partialx+2\psiy\partialy.}

The function and are treated separately, being essentially subject to usual elliptic regularity considerations for interior points while requires special treatment near the boundary using difference quotients. Once the strong properties are established in terms of and the Neumann boundary conditions, the "bootstrap" regularity results can be proved exactly as for the Dirichlet problem.

Interior estimates

The function lies in where is a region with closure in . If and

(Lf,g)=(\Deltaf,g)\Omega.

By continuity the same holds with replaced by and hence . So

\Deltav=Lv=L(\psiu)=\psiLu+[L,\psi]u=\psi(f-u)+[\Delta,\psi]u.

Hence regarding as an element of, . Hence . Since for, we have . Moreover,

2
\|v\|
(2)

=\|\Deltav\|2+

2,
2\|v\|
(1)

so that

\|v\|(2)\leC\left(\|\Delta(v)\|+\|v\|(1)\right).

Boundary estimates

The function is supported in a collar contained in a tubular neighbourhood of the boundary. The difference quotients can be formed for the flow and lie in, so the first inequality is applicable:

\begin{align} \|\deltahw\|(1)&\le\|L\deltahw\|(-1)+\|\deltahw\|(0)\\ &\le\|[L,\deltah]w\|(-1)+\|\deltahLw\|(-1)+\|\deltahw\|(0)\\ &\le\|[L,\deltah]w\|(-1)+C\|Lw\|(0)+C\|w\|(1). \end{align}

The commutators are uniformly bounded as operators from to . This is equivalent to checking the inequality

\left|\left(\left[L,\deltah\right]g,h\right)\right|\leA\|g\|(1)\|h\|(1),

for, smooth functions on a collar. This can be checked directly on an annulus, using invariance of Sobolev spaces under dffeomorphisms and the fact that for the annulus the commutator of with a differential operator is obtained by applying the difference operator to the coefficients after having applied to the function:

\left[\deltah,\suma\alpha\partial\alpha\right]=\left

h(a
(\delta
\alpha)\circ

Rh\right)\partial\alpha.

Hence the difference quotients are uniformly bounded, and therefore with

\|Yw\|(1)\leC\|Lw\|(0)+C\prime\|w\|(1).

Hence and satisfies a similar inequality to :

\|Vw\|(1)\leC\prime\prime\left(\|Lw\|(0)+\|w\|(1)\right).

Let be the orthogonal vector field. As for the Dirichlet problem, to show that, it suffices to show that .

To check this, it is enough to show that . As before

\begin{align} A&=\Delta-V2-W2\ B&=[V,W] \end{align}

are vector fields. On the other hand, for, so that and define the same distribution on . Hence

\begin{align}\left(W2w,\varphi\right)&=\left(Lw-V2w-Au,\varphi\right),\\ (VWw,\varphi)&=(WVw+Bw,\varphi). \end{align}

Since the terms on the right hand side are pairings with functions in, the regularity criterion shows that . Hence since both terms lie in and have the same inner products with 's.

Moreover, the inequalities for show that

\begin{align} \|Ww\|(1)&\leC\left(\|VWw\|(0)+\left\|W2w\right\|(0)\right)\\ &\leC\left\|\left(\Delta-V2-A\right)w\right\|(0)+C\|(WV+B)w\|(0)\\ &\leC1\|Lw\|(0)+C1\|w\|(1). \end{align}

Hence

\begin{align} \|w\|(2)&\leC\left(\|Vw\|(1)+\|Ww\|(1)\right)\\ &\leC\prime\|\Deltaw\|(0)+C\prime\|w\|(1). \end{align}

It follows that . Moreover,

\begin{align} \|u\|(2)&\leC\left(\|\Deltav\|+\|\Deltaw\|+\|v\|(1)+\|w\|(1)\right)\\ &\leC\prime\left(\|\psi\Deltau\|+\|(1-\psi)\Deltau\|+2\|[\Delta,\psi]u\|+\|u\|(1)\right)\\ &\leC\prime\prime\left(\|\Deltau\|+\|u\|(1)\right). \end{align}

Neumann boundary conditions

Since, Green's theorem is applicable by continuity. Thus for,

\begin{align} (f,v)&=(Lu,v)+(u,v)\\ &=(ux,vx)+(uy,vy)+(u,v)\\ &=((\Delta+I)u,v)+(\partialnu,v)\partial\Omega\\ &=(f,v)+(\partialnu,v)\partial\Omega. \end{align}

Hence the Neumann boundary conditions are satisfied:

\partialnu|\partial=0,

where the left hand side is regarded as an element of and hence .

Regularity of strong solutions

The main result here states that if and, then and

\|u\|(k+2)\leC\|\Deltau\|(k)+C\|u\|(k+1),

for some constant independent of .

Like the corresponding result for the Dirichlet problem, this is proved by induction on . For, is also a weak solution of the Neumann problem so satisfies the estimate above for . The Neumann boundary condition can be written

Zu|\partial\Omega=0.

Since commutes with the vector field corresponding to the period flow, the inductive method of proof used for the Dirichlet problem works equally well in this case: for the difference quotients preserve the boundary condition when expressed in terms of .

Smoothness of eigenfunctions

It follows by induction from the regularity theorem for the Neumann problem that the eigenfunctions of in lie in . Moreover, any solution of with in and in must have in . In both cases by the vanishing properties, the normal derivatives of the eigenfunctions and vanish on .

Solving the associated Neumann problem

The method above can be used to solve the associated Neumann boundary value problem:

\begin{cases}\Deltaf|\Omega=0\\partialnf|\partial\Omega=g&g\inCinfty(\partial\Omega)\end{cases}

By Borel's lemma is the restriction of a function . Let be a smooth function such that near the boundary. Let be the solution of with . Then solves the boundary value problem.

Notes and References

  1. See: