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A global H² bound for the Poisson adjoint on a C² domain

Statement

Assume Countable Choice. Let $n \geq 2$ , let $Ω \subset R^{n}$ be a bounded $C^{2}$ domain, let $g \in L^{2} (Ω)$ , and let $z \in H_{0}^{1} (Ω)$ be a weak solution of $- Δ z = g$ with zero Dirichlet data. Suppose a Poincaré constant $C_{P}$ is specified such that $∥ v ∥_{L^{2} (Ω)} \leq C_{P} ∥ \nabla v ∥_{L^{2} (Ω)} (v \in H_{0}^{1} (Ω)) .$ Then $z \in H^{2} (Ω)$ and $∥ z ∥_{H^{2} (Ω)} \leq C_{r e g} (1 + C_{P}^{2}) ∥ g ∥_{L^{2} (Ω)},$ where $C_{r e g}$ is the constant from the global H² Dirichlet regularity theorem for the coefficient matrix $I$ and zero lower-order coefficients. The Poincaré inequality is an explicit hypothesis here; this item does not import a stronger domain or axiom branch for it.

Facts & Assumptions

Given: Countable Choice; the bounded $C^{2}$ domain; $g \in L^{2} (Ω)$ ; the zero-trace weak equation $\int_{Ω} \nabla z \cdot \overline{\nabla v} d x = \int_{Ω} g \overline{v} d x (v \in H_{0}^{1} (Ω));$ and the displayed Poincaré inequality with constant $C_{P}$ .

[F1]

A zero-trace weak solution of a divergence-form problem is an $H_{0}^{1}$ class satisfying the variational equation for every $H_{0}^{1}$ test function (Weak Dirichlet solutions for a divergence-form operator).

[F2]

For a bounded $C^{2}$ domain, coefficients $a^{i j} \in W^{1, \infty}$ , $b^{i}, c \in L^{\infty}$ , ellipticity, $g \in L^{2}$ , and a zero-trace weak solution, the global theorem gives $∥ z ∥_{H^{2}} \leq C (∥ g ∥_{L^{2}} + ∥ z ∥_{L^{2}})$ ; its statement retains the second term (Global $H^{2}$ Dirichlet regularity).

[F3]

On real or complex $L^{2}$ , the corresponding integral inner product is well defined and satisfies Cauchy–Schwarz, so $∣ \int g \overline{z} ∣ \leq ∥ g ∥_{L^{2}} ∥ z ∥_{L^{2}}$ (The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz).

[A1]

The Poincaré inequality with constant $C_{P}$ is assumed exactly as displayed in Given.

Proof

1.1F1F3A1algebra

Test the weak equation with $v = z$ and take real parts. By [F3] and [A1], $∥ \nabla z ∥_{L^{2}}^{2} = Re \int_{Ω} g \overline{z} d x \leq ∥ g ∥_{L^{2}} ∥ z ∥_{L^{2}} \leq C_{P} ∥ g ∥_{L^{2}} ∥ \nabla z ∥_{L^{2}} .$ If $∥ \nabla z ∥_{L^{2}} = 0$ , [A1] gives $z = 0$ in $L^{2}$ ; otherwise divide by $∥ \nabla z ∥_{L^{2}}$ . In either case $∥ \nabla z ∥_{L^{2}} \leq C_{P} ∥ g ∥_{L^{2}}$ and $∥ z ∥_{L^{2}} \leq C_{P}^{2} ∥ g ∥_{L^{2}}$ .

2.1F2step 1.1algebra∎

The operator $- Δ$ has principal matrix $a^{i j} = δ_{i j}$ , ellipticity constant $1$ , coefficients in $W^{1, \infty}$ , and zero lower order coefficients. Thus [F2] applies to the given weak solution and yields $∥ z ∥_{H^{2}} \leq C_{r e g} (∥ g ∥_{L^{2}} + ∥ z ∥_{L^{2}}) .$ Insert the bound from step 1.1 to obtain the asserted estimate.

Remarks

The estimate uses the global theorem only for the symmetric Poisson operator on its stated bounded $C^{2}$ domain branch. It does not extend that supplier to general Lipschitz domains or nonsymmetric adjoints.

Depends on

Used by

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Sources

John K. Hunter, Notes on Partial Differential Equations (standard reference, not scraped)
Richard S. Laugesen, Linear Analysis and Partial Differential Equations (standard reference, not scraped)