Sobolev spaces for planar domains

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 Ω ⊂ R2 be a bounded domain with smooth boundary. Since Ω is contained in a large square in R2, it can be regarded as a domain in T2 by identifying opposite sides of the square. The theory of Sobolev spaces on T2 can be found in Bers, John & Schechter (1979), an account which is followed in several later textbooks such as Warner (1983) and Griffiths & Harris (1994).

For k an integer, the (restricted) Sobolev space Hk
0
(Ω)
is defined as the closure of C
c
(Ω)
in the standard Sobolev space Hk(T2).

By Green's theorem this implies
where
with n the unit normal to the boundary. Since such k form a dense subspace of L2(Ω), it follows that g = 0 on ∂Ω.
Suppose g in Hk(T2) annihilates C
c
c)
. By compactness, there are finitely many opens U0, U1, ... , UN covering Ω such that the closure of U0 is disjoint from ∂Ω and each Ui is an open disc about a boundary point zi such that in Ui small translations in the direction of the normal vector ni carry Ω into Ω. Add an open UN+1 with closure in Ωc to produce a cover of T2 and let ψi be a partition of unity subordinate to this cover. If translation by n is denoted by λn, then the functions
tend to g as t decreases to 0 and still lie in the annihilator, indeed they are in the annihilator for a larger domain than Ωc, the complement of which lies in Ω. Convolving by smooth functions of small support produces smooth approximations in the annihilator of a slightly smaller domain still with complement in Ω. These are necessarily smooth functions of compact support in Ω.
Moreover F satisfies (F, g) = 0 for g in C
c
c)
.
This implies that under the pairing between Hk(T2) and Hk(T2), Hk
0
c)
and Hk(Ω) are each other's duals.
Since the adjoint map between the duals can by identified with this map, it follows that (I + ∆)k is a unitary map.

Application to Dirichlet problem

Invertibility of

The operator defines an isomorphism between H1
0
(Ω)
and H−1(Ω). In fact it is a Fredholm operator of index 0. The kernel of in H1(T2) consists of constant functions and none of these except zero vanish on the boundary of Ω. Hence the kernel of H1
0
(Ω)
is (0) and is invertible.

In particular the equation f = g has a unique solution in H1
0
(Ω)
for g in H−1(Ω).

Eigenvalue problem

Let T be the operator on L2(Ω) defined by

where R0 is the inclusion of L2(Ω) in H−1(Ω) and R1 of H1
0
(Ω)
in L2(Ω), both compact operators by Rellich's theorem. The operator T is compact and self-adjoint with (Tf, f) > 0 for all f. By the spectral theorem, there is a complete orthonormal set of eigenfunctions fn in L2(Ω) with

Since μn > 0, fn lies in H1
0
(Ω)
. Setting λn = μn, the fn are eigenfunctions of the Laplacian:

Sobolev spaces without boundary condition

To determine the regularity properties of the eigenfunctions  fn  and solutions of

enlargements of the Sobolev spaces Hk
0
(Ω)
have to be considered. Let C) be the space of smooth functions on Ω which with their derivatives extend continuously to Ω. By Borel's lemma, these are precisely the restrictions of smooth functions on T2. The Sobolev space Hk(Ω) is defined to the Hilbert space completion of this space for the norm

This norm agrees with the Sobolev norm on C
c
(Ω)
so that Hk
0
(Ω)
can be regarded as a closed subspace of Hk(Ω). Unlike Hk
0
(Ω)
, Hk(Ω) is not naturally a subspace of Hk(T2), but the map restricting smooth functions from T2 to Ω is continuous for the Sobolev norm so extends by continuity to a map ρk : Hk(T2) → Hk(Ω).

It is sufficient to construct an extension E for a neighbourhood of a closed annulus, since a collar around the boundary is diffeomorphic to an annulus I × T with I a closed interval in T. Taking a smooth bump function ψ with 0 ≤ ψ ≤ 1, equal to 1 near the boundary and 0 outside the collar, E(ψf) + (1 − ψ)f will provide an extension on Ω. On the annulus, the problem reduces to finding an extension for Ck(I) in Ck(T). Using a partition of unity the task of extending reduces to a neighbourhood of the end points of I. Assuming 0 is the left end point, an extension is given locally by
Matching the first derivatives of order k or less at 0, gives
This matrix equation is solvable because the determinant is non-zero by Vandermonde's formula. It is straightforward to check that the formula for E(f), when appropriately modified with bump functions, leads to an extension which is continuous in the above Sobolev norm.[4]
The differentiation operators x, ∂y carry each Sobolev space into the larger one with index 1 less.
To prove the characterization, note that if f is in Hk(Ω), then αf lies in Hk−|α|(Ω) and hence in H0(Ω) = L2(Ω). Conversely the result is well known for the Sobolev spaces Hk(T2): the assumption implies that the (∂xiy)kf is in L2(T2) and the corresponding condition on the Fourier coefficients of f shows that f lies in Hk(T2). Similarly the result an be proved directly for an annulus [−δ, δ] × T. In fact by the argument on T2 the restriction of f to any smaller annulus [−δ',δ'] × T lies in Hk: equivalently the restriction of the function fR(x, y) = f(Rx, y) lies in Hk for R > 1. On the other hand αfR → ∂αf in L2 as R → 1, so that f must lie in Hk. The case for a general domain Ω reduces to these two cases since f can be written as f = ψf + (1 − ψ)f with ψ a bump function supported in Ω such that 1 − ψ is supported in a collar of the boundary.
On T2 it is known that if f is in Hk, then the difference quotient δhf = h−1(Rhff) → ∂yf in Hk−1; if the difference quotients are bounded in Hk then ∂yf lies in Hk. Both assertions are consequences of the formula:
These results on T2 imply analogous results on the annulus using the extension.

Regularity for Dirichlet problem

Regularity for dual Dirichlet problem

If u = f with u in H1
0
(Ω)
and f in Hk−1(Ω) with k ≥ 0, then u lies in Hk+1(Ω).

Take a decomposition u = ψu + (1 − ψ)u with ψ supported in Ω and 1 − ψ supported in a collar of the boundary. Standard Sobolev theory for T2 can be applied to ψu: elliptic regularity implies that it lies in Hk+1(T2) and hence Hk+1(Ω). v = (1 − ψ)u lies in H1
0
of a collar, diffeomorphic to an annulus, so it suffices to prove the result with Ω a collar and replaced by

The proof[6] proceeds by induction on k, proving simultaneously the inequality

for some constant C depending only on k. It is straightforward to establish this inequality for k = 0, where by density u can be taken to be smooth of compact support in Ω:

The collar is diffeomorphic to an annulus. The rotational flow Rt on the annulus induces a flow St on the collar with corresponding vector field Y = rx + sy. Thus Y corresponds to the vector field θ. The radial vector field on the annulus rr is a commuting vector field which on the collar gives a vector field Z = px + qy proportional to the normal vector field. The vector fields Y and Z commute.

The difference quotients δhu can be formed for the flow St. The commutators [δh, ∆1] are second order differential operators from Hk+1(Ω) to Hk−1(Ω). Their operators norms are uniformly bounded for h near 0; for the computation can be carried out on the annulus where the commutator just replaces the coefficients of 1 by their difference quotients composed with Sh. On the other hand, v = δhu lies in H1
0
(Ω)
, so the inequalities for u apply equally well for v:

The uniform boundedness of the difference quotients δhu implies that Yu lies in Hk+1(Ω) with

It follows that Vu lies in Hk+1(Ω) where V is the vector field

Moreover, Vu satisfies a similar inequality to Yu.

Let W be the orthogonal vector field

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

It suffices to show that Wu lies in Hk+1(Ω). For then

so that xu and yu lie in Hk+1(Ω) and u must lie in Hk+2(Ω).

To check the result on Wu, it is enough to show that VWu and W2u lie in Hk(Ω). Note that

are vector fields. But then

with all terms on the right hand side in Hk(Ω). Moreover, the inequalities for Vu show that

Hence

Smoothness of eigenfunctions

It follows by induction from the regularity theorem for the dual Dirichlet problem that the eigenfunctions of in H1
0
(Ω)
lie in C). Moreover, any solution of u = f with f in C) and u in H1
0
(Ω)
must have u in C). In both cases by the vanishing properties, the eigenfunctions and u vanish on the boundary of Ω.

Solving the Dirichlet problem

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

By Borel's lemma g is the restriction of a function G in C). Let F be the smooth solution of F = ∆G with F = 0 on ∂Ω. Then f = GF solves the Dirichlet problem. By the maximal principle, the solution is unique.[7]

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.[8] For multiply connected regions with smooth boundary Schiffer & Hawley (1962) 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 g such that:

Taylor (2011) gives a proof of the Riemann mapping theorem for a simply connected domain Ω with smooth boundary. Translating if necessary, it can be assumed that 0 ∈ Ω. The solution of the Dirichlet problem shows that there is a unique smooth function U(z) on Ω which is harmonic in Ω and equals −log|z| on ∂Ω. Define the Green's function by G(z) = log|z| + U(z). It vanishes on ∂Ω and is harmonic on Ω away from 0. The harmonic conjugate V of U is the unique real function on Ω such that U + iV is holomorphic. As such it must satisfy the Cauchy-Riemann equations:

The solution is given by

where the integral is taken over any path in Ω. It is easily verified that Vx and Vy exist and are given by the corresponding derivatives of U. Thus V is a smooth function on Ω, vanishing at 0. By the Cauchy-Riemann f = U + iV is smooth on Ω, holomorphic on Ω and f(0) = 0. The function H = arg z + V(z) is only defined up to multiples of 2π, but the function

is a holomorphic on Ω and smooth on Ω. By construction, F(0) = 0 and |F(z)| = 1 for z ∈ ∂Ω. Since z has winding number 1, so too does F(z). On the other hand, F(z) = 0 only for z = 0 where there is a simple zero. So by the argument principle F assumes every value in the unit disc, D, exactly once and F′ does not vanish inside Ω. To check that the derivative on the boundary curve is non-zero amounts to computing the derivative of eiH, i.e. the derivative of H 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 G vanishes on the boundary and is strictly negative in Ω since |F| = eG. The Hopf lemma implies that the directional derivative of G in the direction of the outward normal is strictly positive. So on the boundary curve, F has nowhere vanishing derivative. Since the boundary curve has winding number one, F defines a diffeomorphism of the boundary curve onto the unit circle. Accordingly, F : ΩD is a smooth diffeomorphism, which restricts to a holomorphic map Ω → D and a smooth diffeomorphism between the boundaries.

Similar arguments can be applied to prove the Riemann mapping theorem for a doubly connected domain Ω bounded by simple smooth curves Ci (the inner curve) and Co (the outer curve). By translating we can assume 1 lies on the outer boundary. Let u be the smooth solution of the Dirichlet problem with U = 0 on the outer curve and −1 on the inner curve. By the maximum principle 0 < u(z) < 1 for z in Ω and so by the Hopf lemma the normal derivatives of u are negative on the outer curve and positive on the inner curve. The integral of uydx + uydx 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 c > 0 such that U = cu satisfies

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

and is well-defined up to multiples of 2π. The function

is smooth on Ω and holomorphic in Ω. On the outer curve |F| = 1 and on the inner curve |F| = ec = r < 1. 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 F restricts to a diffeomorphism between the boundary curves and the two concentric circles. Since the images of outer and inner curve have winding number 1 and 0 about any point in the annulus, an application of the argument principle implies that F assumes every value within the annulus r < |z| < 1 exactly once; since that includes multiplicities, the complex derivative of F is nowhere vanishing in Ω. This F is a smooth diffeomorphism of Ω onto the closed annulus r ≤ |z| ≤ 1, restricting to a holomorphic map in the interior and a smooth diffeomorphism on both boundary curves.

Trace map

The restriction map τ : C(T2) → C(T) = C(1 × T) extends to a continuous map Hk(T2) → Hk − ½(T) for k ≥ 1.[9] In fact

so the Cauchy–Schwarz inequality yields

where, by the integral test,

The map τ is onto since a continuous extension map E can be constructed from Hk − ½(T) to Hk(T2).[10][11] In fact set

where

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

It follows that there is a trace map τ of Hk(Ω) onto Hk − ½(∂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 diffeomorphims (or coordinate change) of the half-integer Sobolev spaces on the circle follows from the fact that an equivalent norm on Hk + ½(T) is given by[12]

It is also a consequence of the properties of τ and E (the "trace theorem").[13] 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 − ½(T) under f*, since f* = τ ∘ F* ∘ E.

Further consequences of the trace theorem are the two exact sequences[14][15]

and

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:

The trace map to Hj − ½(∂Ω) takes f to kj
n
f|∂Ω

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 f in H1(Ω), the normal derivative nf|∂Ω cannot be a priori defined at the level of Sobolev spaces. Instead an alternative formulation of boundary value problems for the Laplacian Δ on a bounded region Ω in the plane is used. It employs Dirichlet forms, sesqulinear bilinear forms on H1(Ω), H1
0
(Ω)
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 H2(Ω), 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 f|∂Ω is already defined on H1(Ω), Dirichlet forms do not need to be mentioned explicitly and the operator formulation is more direct. A unified discussion is given in Folland (1995) 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 Taylor (2011).

The Hilbert space formulation of boundary value problems for the Laplacian Δ on a bounded region Ω in the plane proceeds from the following data:[16]

A weak solution of the boundary value problem given initial data f in L2(Ω) is a function u satisfying

for all g.

For both the Dirichlet and Neumann problem

For the Dirichlet problem H = H1
0
(Ω)
. In this case

By the trace theorem the solution satisfies u|Ω = 0 in H½(∂Ω).

For the Neumann problem H is taken to be H1(Ω).

Application to Neumann problem

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

Green's theorem implies that for u, v ∈ C)

Thus if Δu = 0 in Ω and satisfies the Neumann boundary conditions, ux = uy = 0, and so u is constant in Ω.

Hence the Neumann problem has a unique solution up to adding constants.[17]

Consider the Hermitian form on H1(Ω) defined by

Since H1(Ω) is in duality with H−1
0
(Ω)
, there is a unique element Lu in H−1
0
(Ω)
such that

The map I + L is an isometry of H1(Ω) onto H−1
0
(Ω)
, so in particular L is bounded.

In fact

So

On the other hand, any f in H−1
0
(Ω)
defines a bounded conjugate-linear form on H1(Ω) sending v to (f, v). By the Riesz–Fischer theorem, there exists u ∈ H1(Ω) such that

Hence (L + I)u = f and so L + I is surjective. Define a bounded linear operator T on L2(Ω) by

where R1 is the map H1(Ω) → L2(Ω), a compact operator, and R0 is the map L2(Ω) → H−1
0
(Ω)
, its adjoint, so also compact.

The operator T has the following properties:

and Tf = 0 implies u = 0 and hence f = 0.
with 0 < μn ≤ 1 and μn decreasing to 0.
Thus λn are non-negative and increase to .
so u 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 L and the Dirichlet form D is a strong solution in the classical sense, expressed in terms of the Laplacian Δ and the Neumann boundary conditions. Thus if u = Tf with u ∈ H1(Ω), f ∈ L2(Ω), then u ∈ H2(Ω), satisfies Δu + u = f and nu|∂Ω = 0. Moreover, for some constant C independent of u,

Note that

since

Take a decomposition u = ψu + (1 − ψ)u with ψ supported in Ω and 1 − ψ supported in a collar of the boundary.

The operator L is characterized by

Then

so that

The function v = ψu and w = (1 − ψ)u are treated separately, v being essentially subject to usual elliptic regularity considerations for interior points while w 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 v = ψu lies in H1
0
1)
where Ω1 is a region with closure in Ω. If f ∈ C
c
(Ω)
and g ∈ C)

By continuity the same holds with f replaced by v and hence Lv = ∆v. So

Hence regarding v as an element of H1(T2), v ∈ L2(T2). Hence v ∈ H2(T2). Since v = φv for φ ∈ C
c
(Ω)
, we have v ∈ H2
0
(Ω)
. Moreover,

so that

Boundary estimates

The function w = (1 − ψ)u is supported in a collar contained in a tubular neighbourhood of the boundary. The difference quotients δhw can be formed for the flow St and lie in H1(Ω), so the first inequality is applicable:

The commutators [L, δh] are uniformly bounded as operators from H1(Ω) to H−1
0
(Ω)
. This is equivalent to checking the inequality

for g, h 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 δh with a differential operator is obtained by applying the difference operator to the coefficients after having applied Rh to the function:[18]

Hence the difference quotients δhw are uniformly bounded, and therefore Yw ∈ H1(Ω) with

Hence Vw ∈ H1(Ω) and Vw satisfies a similar inequality to Yw:

Let W be the orthogonal vector field. As for the Dirichlet problem, to show that w ∈ H2(Ω), it suffices to show that Ww ∈ H1(Ω).

To check this, it is enough to show that VWw, W2u ∈ L2(Ω). As before

are vector fields. On the other hand, (Lw, φ) = (∆w, φ) for φ ∈ C
c
(Ω)
, so that Lw and w define the same distribution on Ω. Hence

Since the terms on the right hand side are pairings with functions in L2(Ω), the regularity criterion shows that Ww ∈ H2(Ω). Hence Lw = ∆w since both terms lie in L2(Ω) and have the same inner products with φ's.

Moreover, the inequalities for Vw show that

Hence

It follows that u = v + w ∈ H2(Ω). Moreover,

Neumann boundary conditions

Since u ∈ H2(Ω), Green's theorem is applicable by continuity. Thus for v ∈ H1(Ω),

Hence the Neumann boundary conditions are satisfied:

where the left hand side is regarded as an element of H½(∂Ω) and hence L2(∂Ω).

Regularity of strong solutions

The main result here states that if u ∈ Hk+1 (k ≥ 1), ∆u ∈ Hk and nu|∂Ω = 0, then u ∈ Hk+2 and

for some constant independent of u.

Like the corresponding result for the Dirichlet problem, this is proved by induction on k ≥ 1. For k = 1, u is also a weak solution of the Neumann problem so satisfies the estimate above for k = 0. The Neumann boundary condition can be written

Since Z commutes with the vector field Y corresponding to the period flow St, the inductive method of proof used for the Dirichlet problem works equally well in this case: for the difference quotients δh preserve the boundary condition when expressed in terms of Z.[19]

Smoothness of eigenfunctions

It follows by induction from the regularity theorem for the Neumann problem that the eigenfunctions of D in H1(Ω) lie in C). Moreover, any solution of Du = f with f in C) and u in H1(Ω) must have u in C). In both cases by the vanishing properties, the normal derivatives of the eigenfunctions and u vanish on ∂Ω.

Solving the associated Neumann problem

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

By Borel's lemma g is the restriction of a function G ∈ C). Let F be a smooth function such that nF = G near the boundary. Let u be the solution of u = −∆F with nu = 0. Then f = u + F solves the boundary value problem.[20]

Notes

References

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