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The inhomogeneous weak Dirichlet problem by a trace lifting

Statement

Assume the Axiom of Choice (through the published Sobolev trace results) together with Countable Choice. Let Ω⊂Rn, n≥2, be a bounded C1 domain, let g∈H1/2(∂Ω)=W1/2,2(∂Ω) and F∈H−1(Ω), and let a be the divergence form of Uniformly elliptic divergence-form operators and their sesquilinear forms satisfying the coercivity condition of Lax--Milgram solvability for coercive divergence-form equations on H01(Ω); write α:=α0/(1+CP2)>0 for its coercivity constant, where α0=θ−n CPMb−CP2Mc. Fix a bounded right inverse R:H1/2(∂Ω)→H1(Ω) of the trace, T∘R=id, as in A bounded right inverse of the trace, supported in a prescribed collar. Then there is a unique u∈H1(Ω) with Tu=ganda(u,v)=F(v)for every v∈H01(Ω), and with Ca=nMa+nMb+Mc the bound of The elliptic form is well defined and bounded on H1 on all of H1, ∥u∥H1≤∥Rg∥H1+∥F∥H−1+Ca∥Rg∥H1α≤C(Ω,a,R)(∥g∥W1/2,2+∥F∥H−1). The solution is independent of the choice of lifting; the displayed estimate depends on the fixed right inverse R. Boundary data outside the trace range H1/2(∂Ω) are not admissible: no H1 function has such a trace, the trace range being exactly H1/2(∂Ω) (The sharp trace theorem: boundedness and range in the fractional space).

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; a bounded C1 domain Ω⊂Rn, n≥2; boundary data g∈H1/2(∂Ω)=W1/2,2(∂Ω); F∈H−1(Ω); a divergence form a on H1(Ω) whose restriction to H01(Ω) satisfies the coercivity condition of Lax--Milgram solvability for coercive divergence-form equations with α=α0/(1+CP2)>0, α0=θ−n CPMb−CP2Mc, and bounded with constant Ca; and a bounded right inverse R of the trace T with T∘R=id.

[F1]

The trace operator T:H1(Ω)→H1/2(∂Ω) is bounded and its kernel is exactly H01(Ω) (The kernel of the trace is the closure of the test functions, The fractional Sobolev space on a compact C1 boundary, Bounded C^k domains and boundary charts).

[F2]

The right inverse satisfies T(Rg)=g and ∥Rg∥H1≤∥R∥ ∥g∥W1/2,2 (A bounded right inverse of the trace, supported in a prescribed collar, Integer-order Sobolev spaces and their norms, Zero-boundary Sobolev space as a norm closure).

[F3]

The divergence-form solvability theorem: for every datum in H−1(Ω) there is a unique w∈H01(Ω) with a(w,v)=G(v) for all v∈H01(Ω), satisfying ∥w∥H01≤∥G∥H−1/α (Lax--Milgram solvability for coercive divergence-form equations, Weak Dirichlet solutions for a divergence-form operator).

[F4]

Boundedness on all H1 slots: with Ca=nMa+nMb+Mc, ∣a(u,v)∣≤Ca∥u∥H1∥v∥H1 for every u,v∈H1(Ω) (The elliptic form is well defined and bounded on H1, The negative Sobolev space H−1(Ω), Complex Lp classes and Euclidean test-function conventions, Holder's inequality for integrals, including the endpoint cases).

[F5]

Admissibility is exactly trace-range membership: the trace operator T:H1(Ω)→H1/2(∂Ω) has range exactly H1/2(∂Ω) (The Lp trace operator on a bounded C1 domain, The sharp trace theorem: boundedness and range in the fractional space, The fractional Sobolev space on a compact C1 boundary), so a datum outside H1/2(∂Ω) is the trace of no H1 function and the inhomogeneous problem admits no solution for it.

Proof

1.1F2F4algebra

Lift and shift: put u0:=Rg, so Tu0=g and ∥u0∥H1≤∥R∥ ∥g∥; define F~(v):=F(v)−a(u0,v) for v∈H01(Ω). Then F~ is conjugate-linear, and by [F4] ∥F~∥H−1≤∥F∥H−1+Ca∥u0∥H1.

2.1F3step 1.1

Zero-boundary correction: by [F3] applied to F~ there is a unique w∈H01(Ω) with a(w,v)=F~(v) for all v∈H01(Ω), and ∥w∥H1≤∥F~∥H−1/α.

3.1F1step 2.1algebra

The sum solves the inhomogeneous problem: let u:=u0+w. Since w∈H01(Ω)=ker⁡T by [F1], Tu=Tu0+Tw=g. For v∈H01(Ω), additivity of a in the first slot gives a(u,v)=a(u0,v)+a(w,v)=a(u0,v)+F~(v)=F(v).

3.2F2step 1.1step 2.1algebra

Estimate: ∥u∥H1≤∥u0∥H1+∥w∥H1≤∥Rg∥H1+(∥F∥H−1+Ca∥Rg∥H1)/α, and the right-inverse bound ∥Rg∥H1≤∥R∥ ∥g∥W1/2,2 makes the right-hand side at most C(Ω,a,R)(∥g∥W1/2,2+∥F∥H−1) for an explicit constant depending only on Ω, a and R.

4.1F1step 3.1

Uniqueness independent of the lifting: if u1,u2 are solutions, then z:=u1−u2 has Tz=0, so z∈H01(Ω) by [F1], and a(z,v)=0 for every v∈H01(Ω). Testing v=z and using coercivity gives α∥z∥H12≤Re⁡a(z,z)=0, so z=0.

5.1F5given∎

Admissibility: the construction needs g in the trace range; by [F5] a datum outside H1/2(∂Ω) is the trace of no H1 function, so the inhomogeneous problem has no solution for it and the trace-range hypothesis cannot be dropped.

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Sources