Sobolev spaces for planar domains

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 H^k
₀(Ω) is defined as the closure of C^∞
_c(Ω) in the standard Sobolev space H^k(T²).

• H⁰
  ₀(Ω) = L²(Ω).
• Vanishing properties on boundary: For k > 0 the elements of H^k
  ₀(Ω) are referred to as "L² functions on Ω which vanish with their first k − 1 derivatives on ∂Ω."[1] In fact if  f  ∈ C^k(Ω) agrees with a function in H^k
  ₀(Ω), then g = ∂^αf  is in C¹. Let  f_n ∈ C^∞
  _c(Ω) be such that  f_n → f  in the Sobolev norm, and set g_n = ∂^αf_n . Thus g_n → g in H¹
  ₀(Ω). Hence for h ∈ C^∞(T²) and D = a∂_x + b∂_y,

${\displaystyle \iint _{\Omega }\left(g(Dh)+(Dg)h\right)\,dx\,dy=\lim _{n\to 0}\iint _{\Omega }\left(g(Dh_{n})+(Dg)h_{n}\right)\,dx\,dy=0.}$

By Green's theorem this implies

${\displaystyle \int _{\partial \Omega }gk=0,}$

where

${\displaystyle k=h\cos \left(\mathbf {n} \cdot (a,b)\right),}$

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.

• Support properties: Let Ω^c be the complement of Ω and define restricted Sobolev spaces analogously for Ω^c. Both sets of spaces have a natural pairing with C^∞(T²). The Sobolev space for Ω is the annihilator in the Sobolev space for T² of C^∞
  _c(Ω^c) and that for Ω^c is the annihilator of C^∞
  _c(Ω).[2] In fact this is proved by locally applying a small translation to move the domain inside itself and then smoothing by a smooth convolution operator.

Suppose g in H^k(T²) annihilates C^∞
_c(Ω^c). By compactness, there are finitely many open sets U₀, U₁, ... , U_N covering Ω such that the closure of U₀ is disjoint from ∂Ω and each U_i is an open disc about a boundary point z_i such that in U_i small translations in the direction of the normal vector n_i carry Ω into Ω. Add an open U_N+1 with closure in Ω^c to produce a cover of T² and let ψ_i be a partition of unity subordinate to this cover. If translation by n is denoted by λ_n, then the functions

${\displaystyle g_{t}=\psi _{0}g+\sum _{i=1}^{N}\psi _{n}\lambda _{tn_{i}}g}$

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 Ω^c, 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.

• Further vanishing properties on the boundary: The characterization in terms of annihilators shows that  f  ∈ C^k(Ω) lies in H^k
  ₀(Ω) if (and only if) it and its derivatives of order less than k vanish on ∂Ω.[3] In fact  f  can be extended to T² by setting it to be 0 on Ω^c. This extension F defines an element in H^k(T²) using the formula for the norm

${\displaystyle \|h\|_{(k)}^{2}=\sum _{j=0}^{k}{k \choose j}\left\|\partial _{x}^{j}\partial _{y}^{k-j}h\right\|^{2}.}$

Moreover F satisfies (F, g) = 0 for g in C^∞
_c(Ω^c).

• Duality: For k ≥ 0, define H^−k(Ω) to be the orthogonal complement of H^−k
  ₀(Ω^c) in H^−k(T²). Let P_k be the orthogonal projection onto H^−k(Ω), so that Q_k = I − P_k is the orthogonal projection onto H^−k
  ₀(Ω^c). When k = 0, this just gives H⁰(Ω) = L²(Ω). If  f  ∈ H^k
  ₀(Ω^c) and g ∈ H^−k(T²), then

${\displaystyle (f,g)=(f,P_{k}g).}$

This implies that under the pairing between H^k(T²) and H^−k(T²), H^k
₀(Ω^c) and H^−k(Ω) are each other's duals.

• Approximation by smooth functions: The image of C^∞
  _c(Ω) is dense in H^−k(Ω) for k ≤ 0. This is obvious for k = 0 since the sum C^∞
  _c(Ω) + C^∞
  _c(Ω^c) is dense in L²(T²). Density for k < 0 follows because the image of L²(T²) is dense in H^−k(T²) and P_k annihilates C^∞
  _c(Ω^c).
• Canonical isometries: The operator (I + ∆)^k gives an isometry of H^2k
  ₀(Ω) into H⁰(Ω) and of H^k
  ₀(Ω) onto H^−k(Ω). In fact the first statement follows because it is true on T². That (I + ∆)^k is an isometry on H^k
  ₀(Ω) follows using the density of C^∞
  _c(Ω) in H^−k(Ω): for  f, g ∈ C^∞
  _c(Ω) we have:

${\displaystyle {\begin{aligned}\left\|P_{k}(I+\Delta )^{k}f\right\|_{(-k)}&=\sup _{\|g\|_{(-k)}=1}\left|\left((I+\Delta )^{k}f,g\right)_{(-k)}\right|\\&=\sup _{\|g\|_{(-k)}=1}|(f,g)|\\&=\|f\|_{(k)}.\end{aligned}}}$

Since the adjoint map between the duals can by identified with this map, it follows that (I + ∆)^k is a unitary map.

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

${\displaystyle \Delta f=u,\qquad u\in H^{-1}(\Omega ),f\in H_{0}^{1}(\Omega ),}$

enlargements of the Sobolev spaces H^k
₀(Ω) have to be considered. Let C^∞(Ω⁻) 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 H^k(Ω) is defined to the Hilbert space completion of this space for the norm

${\displaystyle \|f\|_{(k)}^{2}=\sum _{j=0}^{k}{k \choose j}\left\|\partial _{x}^{j}\partial _{y}^{k-j}f\right\|^{2}.}$

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

• Invariance under diffeomorphism: Any diffeomorphism between the closures of two smooth domains induces an isomorphism between the Sobolev space. This is a simple consequence of the chain rule for derivatives.
• Extension theorem: The restriction of ρ_k to the orthogonal complement of its kernel defines an isomorphism onto H^k(Ω). The extension map E_k is defined to be the inverse of this map: it is an isomorphism (not necessarily norm preserving) of H^k(Ω) onto the orthogonal complement of H^k
  ₀(Ω^c) such that ρ_k ∘ E_k = I. On C^∞
  _c(Ω), it agrees with the natural inclusion map. Bounded extension maps E_k of this kind from H^k(Ω) to H^k(T²) were constructed first constructed by Hestenes and Lions. For smooth curves the Seeley extension theorem provides an extension which is continuous in all the Sobolev norms. A version of the extension which applies in the case where the boundary is just a Lipschitz curve was constructed by Calderón using singular integral operators and generalized by Stein (1970).

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 C^k( I ) in C^k(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

${\displaystyle E(f)=\sum _{m=0}^{k}a_{m}f\left(-{\frac {x}{m+1}}\right).}$

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

${\displaystyle \sum _{m=0}^{k}(-m-1)^{-k}a_{m}=1.}$

This matrix equation is solvable because the determinant is non-zero by Vandermonde's formula. It is straightforward to check that the formula for E( f ), when appropriately modified with bump functions, leads to an extension which is continuous in the above Sobolev norm.[4]

• Restriction theorem: The restriction map ρ_k is surjective with ker ρ_k = H^k
  ₀(Ω^c). This is an immediate consequence of the extension theorem and the support properties for Sobolev spaces with boundary condition.
• Duality: H^k(Ω) is naturally the dual of H^−k₀(Ω). Again this is an immediate consequence of the restriction theorem. Thus the Sobolev spaces form a chain:

${\displaystyle \cdots \subset H^{2}(\Omega )\subset H^{1}(\Omega )\subset H^{0}(\Omega )\subset H_{0}^{-1}(\Omega )\subset H_{0}^{-2}(\Omega )\subset \cdots }$

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

• Sobolev embedding theorem: H^k+2(Ω) is contained in C^k(Ω⁻). This is an immediate consequence of the extension theorem and the Sobolev embedding theorem for H^k+2(T²).
• Characterization: H^k(Ω) consists of  f  in L²(Ω) = H⁰(Ω) such that all the derivatives ∂^αf lie in L²(Ω) for |α| ≤ k.Here the derivatives are taken within the chain of Sobolev spaces above.[5] Since C^∞
  _c(Ω) is weakly dense in H^k(Ω), this condition is equivalent to the existence of L² functions f_α such that

${\displaystyle \left(f_{\alpha },\varphi \right)=(-1)^{|\alpha |}\left(f,\partial ^{\alpha }\varphi \right),\qquad |\alpha |\leq k,\varphi \in C_{c}^{\infty }(\Omega ).}$

To prove the characterization, note that if  f  is in H^k(Ω), then ∂^αf  lies in H^k−|α|(Ω) and hence in H⁰(Ω) = L²(Ω). Conversely the result is well known for the Sobolev spaces H^k(T²): the assumption implies that the (∂_x − i∂_y)^k f  is in L²(T²) and the corresponding condition on the Fourier coefficients of  f  shows that  f  lies in H^k(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 H^k: equivalently the restriction of the function  f_R (x, y) = f (Rx, y) lies in H^k for R > 1. On the other hand ∂^α f_R → ∂^α f in L² as R → 1, so that  f  must lie in H^k. 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.

• Regularity theorem: If  f  in L²(Ω) has both derivatives ∂_x f and ∂_y f in H^k(Ω) then  f  lies in H^k+1(Ω). This is an immediate consequence of the characterization of H^k(Ω) above. In fact if this is true even when satisfied at the level of distributions: if there are functions g, h in H^k(Ω) such that (g,φ) = (f, φ_x) and (h,φ) = (f,φ_y) for φ in C^∞
  _c(Ω), then  f  is in H^k+1(Ω).
• Rotations on an annulus: For an annulus I × T, the extension map to T² is by construction equivariant with respect to rotations in the second variable,

${\displaystyle R_{t}f(x,y)=f(x,y+t).}$

On T² it is known that if  f  is in H^k, then the difference quotient δ_h f = h⁻¹(R_h f − f ) → ∂_y f  in H^k−1; if the difference quotients are bounded in H^k then ∂_yf lies in H^k. Both assertions are consequences of the formula:

${\displaystyle {\widehat {\delta _{h}f}}(m,n)=h^{-1}(e^{-ihn}-1){\widehat {f}}(m,n)=-\int _{0}^{1}ine^{-inht}\,dt\,\,{\widehat {f}}(m,n).}$

These results on T² imply analogous results on the annulus using the extension.

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:

• g is harmonic in the interior of Ω;
• On ∂Ω we have: ∂_ng = κ − Ke^G, where κ is the curvature of the boundary curve, ∂_n is the derivative in the direction normal to ∂Ω and K is constant on each boundary component.

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:

${\displaystyle {\begin{aligned}U_{x}&=-V_{y},\\U_{y}&=V_{x}.\end{aligned}}}$

The solution is given by

${\displaystyle V(z)=\int _{0}^{z}-U_{y}dx+V_{x}dy,}$

where the integral is taken over any path in Ω. It is easily verified that V_x and V_y 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

${\displaystyle F(z)=e^{G(z)+iH(z)}=ze^{f(z)}}$

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 e^iH, 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| = e^G. 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 : Ω → D 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 C_i (the inner curve) and C_o (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 −u_ydx + u_ydx 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

${\displaystyle \int _{C}\left(-U_{y}\,dx+U_{x}\,dy\right)=2\pi }$

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

${\displaystyle V(z)=\int _{1}^{z}-u_{y}\,dx+u_{x}\,dy}$

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

${\displaystyle \displaystyle {F(z)=e^{U(z)+iV(z)}}}$

is smooth on Ω and holomorphic in Ω. On the outer curve |F| = 1 and on the inner curve |F| = e^−c = 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.

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

${\displaystyle \displaystyle {{\widehat {\tau f}}(n)=\sum _{m}{\widehat {f}}(m,n),}}$

so the Cauchy–Schwarz inequality yields

${\displaystyle {\begin{aligned}\left|{\widehat {\tau f}}(n)\right|^{2}\left(1+n^{2}\right)^{k-{\frac {1}{2}}}&\leq \left(\sum _{m}{\frac {\left(1+n^{2}\right)^{k-{\frac {1}{2}}}}{\left(1+m^{2}+n^{2}\right)^{k}}}\right)\left(\sum _{m}\left|{\widehat {f}}(m,n)\right|^{2}\left(1+m^{2}+n^{2}\right)^{k}\right)\\&\leq C_{k}\sum _{m}\left|{\widehat {f}}(m,n)\right|^{2}\left(1+m^{2}+n^{2}\right)^{k},\end{aligned}}}$

where, by the integral test,

${\displaystyle {\begin{aligned}C_{k}&=\sup _{n}\sum _{m}{\frac {\left(1+n^{2}\right)^{k-{\frac {1}{2}}}}{\left(1+m^{2}+n^{2}\right)^{k}}}<\infty ,\\c_{k}&=\inf _{n}\sum _{m}{\frac {\left(1+n^{2}\right)^{k-{\frac {1}{2}}}}{\left(1+m^{2}+n^{2}\right)^{k}}}>0.\end{aligned}}}$

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

${\displaystyle {\widehat {Eg}}(m,n)=\lambda _{n}^{-1}{\widehat {g}}(n){\frac {\left(1+n^{2}\right)^{k-{\frac {1}{2}}}}{\left(1+n^{2}+m^{2}\right)^{k}}},}$

where

${\displaystyle \lambda _{n}=\sum _{m}{\frac {\left(1+n^{2}\right)^{k-{\frac {1}{2}}}}{\left(1+m^{2}+n^{2}\right)^{k}}}.}$

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

${\displaystyle {\begin{aligned}\|Eg\|_{(k)}^{2}&=\sum _{m,n}\left|{\widehat {Eg}}(m,n)\right|^{2}\left(1+m^{2}+n^{2}\right)\\&\leq c_{k}^{-2}\sum _{m,n}\left|{\widehat {g}}(n)\right|^{2}{\frac {\left(1+n^{2}\right)^{2k-1}}{\left(1+m^{2}+n^{2}\right)^{k}}}\\&\leq c_{k}^{-2}C_{k}\|g\|_{k-{\frac {1}{2}}}^{2}.\end{aligned}}}$

It follows that there is a trace map τ of H^k(Ω) onto H^{k − 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 H^k of the collar, which can be identified with H^k 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 H^{k + ½}(T) is given by[12]

${\displaystyle \|f\|_{[k+{\frac {1}{2}}]}^{2}=\|f\|_{(k)}^{2}+\int _{0}^{2\pi }\int _{0}^{2\pi }{\frac {\left|f^{(k)}(s)-f^{(k)}(t)\right|^{2}}{\left|e^{is}-e^{it}\right|^{2}}}\,ds\,dt.}$

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 H^k(T²) under the induced map F* therefore implies invariance of H^{k − ½}(T) under f*, since f* = τ ∘ F* ∘ E.

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

${\displaystyle (0)\to H_{0}^{1}(\Omega )\to H^{1}(\Omega )\to H^{\frac {1}{2}}(\partial \Omega )\to (0)}$

and

${\displaystyle (0)\to H_{0}^{2}(\Omega )\to H^{2}(\Omega )\to H^{\frac {3}{2}}(\partial \Omega )\oplus H^{\frac {1}{2}}(\partial \Omega )\to (0),}$

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

${\displaystyle (0)\to H_{0}^{k}(\Omega )\to H^{k}(\Omega )\to \bigoplus _{j=1}^{k}H^{j-{\frac {1}{2}}}(\partial \Omega )\to (0).}$

The trace map to H^{j − ½}(∂Ω) takes f to ∂^{k − j}
_nf |_∂Ω

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 closed subspace H¹
  ₀(Ω) ⊆ H ⊆ H¹(Ω).
• A Dirichlet form for Δ given by a bounded Hermitian bilinear form D( f, g) defined for f, g ∈ H¹(Ω) such that D( f, g) = (∆f, g) for f, g ∈ H¹
  ₀(Ω).
• D is coercive, i.e. there is a positive constant C and a non-negative constant λ such that D( f, f ) ≥ C ( f, f )₍₁₎ − λ( f, f ).

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

${\displaystyle D(f,g)=(u,g)}$

for all g.

For both the Dirichlet and Neumann problem

${\displaystyle D(f,g)=\left(f_{x},g_{x}\right)+\left(f_{y},g_{y}\right).}$

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

${\displaystyle D(f,g)=(\Delta f,g),\qquad f,g\in H.}$

By the trace theorem the solution satisfies u|_Ω = 0 in H^½(∂Ω).

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

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

${\displaystyle {\begin{cases}\Delta u=f,&f,u\in C^{\infty }(\Omega ^{-}),\\\partial _{n}u=0&{\text{on }}\partial \Omega \end{cases}}}$

Green's theorem implies that for u, v ∈ C^∞(Ω⁻)

${\displaystyle (\Delta u,v)=(u_{x},v_{x})+(u_{y},v_{y})-(\partial _{n}u,v)_{\partial \Omega }.}$

Thus if Δu = 0 in Ω and satisfies the Neumann boundary conditions, u_x = u_y = 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

${\displaystyle \displaystyle {D(f,g)=(u_{x},v_{x})+(u_{y},v_{y}).}}$

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

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

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

In fact

${\displaystyle ((L+I)u,v)=(u,v)_{(1)}.}$

So

${\displaystyle \|(L+I)u\|_{(-1)}=\sup _{\|v\|_{(1)}=1}|((L+I)u,v)|=\sup _{\|v\|_{(1)}=1}|(u,v)_{(1)}|=\|u\|_{(1)}.}$

On the other hand, any  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

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

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

${\displaystyle T=R_{1}(I+L)^{-1}R_{0},}$

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:

• T is a contraction since it is a composition of contractions
• T is compact, since R₀ and R₁ are compact by Rellich's theorem
• T is self-adjoint, since if  f, g ∈ L²(Ω), they can be written  f = (L + I)u, g = (L + I)v with u, v ∈ H¹(Ω) so

${\displaystyle (Tf,g)=(u,(I+L)v)=(u,v)_{(1)}=((I+L)u,v)=(f,Tg).}$

• T has positive spectrum and kernel (0), for

${\displaystyle (Tf,f)=(u,u)_{(1)}\geq 0,}$

and Tf = 0 implies u = 0 and hence f = 0.

• There is a complete orthonormal basis  f_n of L²(Ω) consisting of eigenfunctions of T. Thus

${\displaystyle Tf_{n}=\mu _{n}f_{n}}$

with 0 < μ_n ≤ 1 and μ_n decreasing to 0.

• The eigenfunctions all lie in H¹(Ω) since the image of T lies in H¹(Ω).
• The  f_n are eigenfunctions of L with

${\displaystyle \displaystyle {Lf_{n}=\lambda _{n}f_{n},\qquad \lambda _{n}=\mu _{n}^{-1}-1.}}$

Thus λ_n are non-negative and increase to ∞.

• The eigenvalue 0 occurs with multiplicity one and corresponds to the constant function. For if u ∈ H¹(Ω) satisfies Lu = 0, then

${\displaystyle (u_{x},u_{x})+(u_{y},u_{y})=(Lu,u)=0,}$

so u is constant.

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