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Dropping the H1-zero boundary condition makes least squares nonunique

Statement

If zero boundary values are removed from the continuous first-order Poisson least-squares problem with zero load, every constant scalar $u$ with $σ = 0$ has zero residual; hence the unconstrained minimizer is not unique.

Facts & Assumptions

Given: The least-squares residual convention and $f = 0$ .

[F1]

The residual is $(σ + \nabla u, d i v σ - f)$ and the intended scalar space is $H_{0}^{1}$ . (First-order least-squares formulation of the Dirichlet Poisson problem)

Proof

1.1F1constructalgebra

Fix any $c \in R$ and set $u (x) = c$ , $σ (x) = 0$ , $f = 0$ . Then $\nabla u = 0$ and $d i v σ = 0$ , so both residual components in [F1] vanish.

2.1step 1.1givenalgebra∎

Distinct constants give distinct H1 functions, and every one has objective zero, the minimum of a sum of squares. They are all admissible after the stated replacement of H1_0 by H1, so the unconstrained minimizer is not unique.

Depends on

First-order least-squares formulation of the Dirichlet Poisson problem

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Peter Bochev and Max Gunzburger, Least-squares finite element methods (standard reference, not scraped)
J. Bond et al., A First-Order System Least-Squares Finite Element Method for the Poisson-Boltzmann Equation (standard reference, not scraped)