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Elliptic H1 error estimate from the source coefficient bound

Statement

Assume Countable Choice. Let $U \subset R^{d}$ be bounded Lipschitz. Let $A = (a_{i j})$ be a bounded measurable real symmetric matrix field, uniformly elliptic with eigenvalues at least $λ_{0} > 0$ , and let $b, c, d$ be bounded measurable real coefficient fields satisfying $λ_{0}^{- 2} (∥ b (x) ∥^{2} + ∥ c (x) ∥^{2}) + λ_{0}^{- 1} ∣ d (x) ∣ \leq v^{2} (x \in U) .$ For the divergence operator, with all divergence derivatives interpreted distributionally, $L [e] = \sum_{i, j} D_{i} (a_{i j} D_{j} e) + \sum_{i} D_{i} (b_{i} e) + \sum_{i} c_{i} D_{i} e + d e,$ every real $e \in H_{0}^{1} (U) \cap C^{2} (U) \cap C^{0} (\overset{ˉ}{U})$ for which the distribution $L [e]$ is represented by an $L^{2} (U)$ function satisfies $∥ e ∥_{H^{1} (U)}^{2} \leq C_{E} (∥ L [e] ∥_{L^{2} (U)}^{2} + ∥ e ∥_{L^{2} (U)}^{2}),$ where $C_{E}$ depends only on $λ_{0}$ and $v$ . The zero-trace condition is required for the integration by parts.

Statement references

Shin et al., Appendix E: source elliptic H1 error estimate and coefficient bound

Shin et al., Appendix E: energy estimate, with the lower-order coefficient absorption written out here

Facts & Assumptions

Given: The bounded Lipschitz domain, uniformly elliptic symmetric matrix, coefficient bound, zero-trace function, and divergence operator in the Statement.

[A1]

Countable Choice is the explicit premise of the zero-boundary Sobolev closure convention (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Zero-boundary Sobolev space as a norm closure).

[F1]

$H^{1} (U) = W^{1, 2} (U)$ has norm squared $∥ e ∥_{L^{2} (U)}^{2} + ∥ \nabla e ∥_{L^{2} (U)}^{2}$ (Integer-order Sobolev spaces and their norms).

[F2]

The divergence operator uses the positive-principal-part convention displayed in Shin's positive divergence-form operator.

[F3]

$H_{0}^{1} (U)$ is the $H^{1}$ -closure of $C_{c}^{\infty} (U)$ , and the $H^{1}$ norm contains the $L^{2}$ norm and weak-gradient norm (Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms).

[F4]

Hölder's integral inequality at exponents $p = q = 2$ bounds the $L^{2}$ pairings used below (Holder's inequality for integrals, including the endpoint cases).

Proof

Proof strategy: integrate against the zero-trace error and absorb lower-order terms by explicit square inequalities.

1.1givenA1F2F3F4algebra

By [F3], choose $ϕ_{n} \in C_{c}^{\infty} (U)$ with $ϕ_{n} \to e$ in $H^{1} (U)$ . The distributional definition of each divergence term and the ordinary test-function integration by parts give $\int_{U} ϕ_{n} L [e] d x = - \int_{U} a_{i j} D_{j} e D_{i} ϕ_{n} d x - \int_{U} b_{i} e D_{i} ϕ_{n} d x + \int_{U} c_{i} D_{i} e ϕ_{n} d x + \int_{U} d e ϕ_{n} d x .$ All coefficients are bounded and $e, \nabla e, L [e] \in L^{2}$ , so [F4] makes each term continuous under $H^{1}$ convergence. Taking $n \to \infty$ and using [F2] yields $- \int_{U} e L [e] d x = \int_{U} a_{i j} D_{j} e D_{i} e d x + \int_{U} b_{i} e D_{i} e d x - \int_{U} c_{i} e D_{i} e d x - \int_{U} d e^{2} d x .$ The principal term is at least $λ_{0} ∥ \nabla e ∥_{2}^{2}$ by ellipticity.

2.1givenF4step 1.1algebra

The coefficient bound gives $∥ b ∥^{2} + ∥ c ∥^{2} \leq λ_{0}^{2} v^{2}$ and $∣ d ∣ \leq λ_{0} v^{2}$ . Hence the absolute value of the two first-order terms is at most $\sqrt{2} λ_{0} ∣ v ∣ ∥ e ∥_{2} ∥ \nabla e ∥_{2}$ by [F4], and the zero-order term is at most $λ_{0} v^{2} ∥ e ∥_{2}^{2}$ . The scalar bound $\sqrt{2} λ_{0} ∣ v ∣ x y \leq \frac{λ_{0}}{2} y^{2} + λ_{0} v^{2} x^{2}$ follows by expanding $(\sqrt{λ_{0} / 2} y - \sqrt{λ_{0}} ∣ v ∣ x)^{2} \geq 0$ . Therefore $- \int_{U} e L [e] d x \geq \frac{λ_{0}}{2} ∥ \nabla e ∥_{2}^{2} - 2 λ_{0} v^{2} ∥ e ∥_{2}^{2} .$

3.1F1F4step 2.1algebra∎

Since $L [e] \in L^{2} (U)$ , [F4] and step 2.1 imply $\frac{λ_{0}}{2} ∥ \nabla e ∥_{2}^{2} \leq ∥ L [e] ∥_{2} ∥ e ∥_{2} + 2 λ_{0} v^{2} ∥ e ∥_{2}^{2} .$ The elementary square inequality $x y \leq x^{2} / (2 λ_{0}) + λ_{0} y^{2} / 2$ follows from $(x / \sqrt{λ_{0}} - \sqrt{λ_{0}} y)^{2} \geq 0$ and gives $\frac{λ_{0}}{2} ∥ \nabla e ∥_{2}^{2} \leq \frac{1}{2 λ_{0}} ∥ L [e] ∥_{2}^{2} + (\frac{λ_{0}}{2} + 2 λ_{0} v^{2}) ∥ e ∥_{2}^{2} .$ Dividing by $λ_{0} / 2$ yields $∥ \nabla e ∥_{2}^{2} \leq λ_{0}^{- 2} ∥ L [e] ∥_{2}^{2} + (1 + 4 v^{2}) ∥ e ∥_{2}^{2} .$ By [F1], adding $∥ e ∥_{2}^{2}$ gives the claimed $H^{1}$ estimate with $C_{E} = \max {λ_{0}^{- 2}, 2 + 4 v^{2}}$ .

Depends on

Used by

Elliptic PINN minimizers converge to the Dirichlet solution Theorem

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Sources

Shin, Darbon, and Karniadakis, On the Convergence of Physics-Informed Neural Networks for Linear Second-Order Elliptic and Parabolic Type PDEs (standard reference, not scraped)