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On bounded domains, the unshifted equation is an identity-minus-compact equation

Statement

Assume Countable Choice. Let Ω⊆Rn be open, let μ≥β and let Kμ be the shifted solution operator of The shifted elliptic solution operator for the form a of Uniformly elliptic divergence-form operators and their sesquilinear forms. For f∈L2(Ω) and u∈H01(Ω) the following are equivalent:

  1. a(u,v)=(f,v)L2 for every v∈H01(Ω), i.e. u is a weak Dirichlet solution in the sense of Weak Dirichlet solutions for a divergence-form operator;
  2. (I−μKμ)u=Kμf as elements of L2(Ω), where I is the identity of L2(Ω).

Both sides of (2) lie in H01(Ω). Thus for arbitrary open Ω the weak equation is equivalent to this identity-minus-bounded-operator equation. If Ω is bounded, also assume the Axiom of Choice; then Kμ on L2(Ω), and hence μKμ, is compact by The shifted solution operator is compact on L2, so (2) is an identity-minus-compact Fredholm equation. The algebraic equivalence applies to the general divergence-form operator, including nonsymmetric lower-order terms.

Facts & Assumptions

Given: Countable Choice; an open set Ω⊆Rn; a fixed μ≥β; the shifted solution operator Kμ of the general divergence-form operator; f∈L2(Ω) and u∈H01(Ω).

[F1]

Definition of Kμ: for g∈L2(Ω), Kμg is the unique class in H01(Ω) with aμ(Kμg,v)=(g,v)L2 for all v∈H01(Ω), where aμ=a+μ(⋅,⋅)L2; Kμ is linear and maps L2(Ω) into H01(Ω) (The shifted elliptic solution operator, Zero-boundary Sobolev space as a norm closure).

[F2]

Weak Dirichlet solutions: u is a weak solution of a(u,v)=(f,v)L2 for every v∈H01(Ω) exactly when the identity holds for all test classes v∈H01(Ω) with datum f∈L2(Ω) (Weak Dirichlet solutions for a divergence-form operator, Uniformly elliptic divergence-form operators and their sesquilinear forms).

[F3]

Compactness: if Ω is bounded and the Axiom of Choice holds, the L2 realization of Kμ is compact, hence so is μKμ, and I−μKμ is an identity-minus-compact operator on L2(Ω) (The shifted solution operator is compact on L2, The space Lp(μ) as the quotient by null functions, The Axiom of Choice).

Proof

technique · direct
1.1F1F2givenalgebra

Equivalence. Condition (1) says a(u,v)=(f,v)L2 for all v∈H01(Ω). Adding μ(u,v)L2 to both sides, this is equivalent to aμ(u,v)=(f+μu,v)L2 for all v∈H01(Ω), where f+μu∈L2(Ω). By the defining uniqueness clause of [F1] for the datum f+μu, this holds exactly when u=Kμ(f+μu) in H01(Ω). Rearranging the linear identity gives u−μKμu=Kμf, that is (I−μKμ)u=Kμf in L2(Ω), and conversely the same rearrangement recovers the defining identity for f+μu and hence condition (1). No symmetry of a and no sign condition on the lower-order coefficients is used.

1.2F1givenalgebra

Location of the two sides. Since u∈H01(Ω) by hypothesis and Kμ maps L2(Ω) into H01(Ω), both u and μKμu, hence both sides of (2), lie in H01(Ω) (μKμu∈H01 because H01 is a linear subspace); the equality itself is an equality of L2 classes.

2.1F3step 1.1step 1.2given∎

Compact case. If Ω is bounded and the Axiom of Choice is assumed, [F3] makes the L2 realization of μKμ compact, so (2) is the equation (I−C)u=Kμf with C:=μKμ compact, an identity-minus-compact equation; for unbounded Ω the equivalence of step 1.1 remains valid as an identity-minus-bounded-operator equation and no compactness or Fredholm claim is made.

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