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Physical constraints and manifold-valued inversion — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

A negative finite-difference PDE state is rejected before solving

Example

Discretize $- u^{''} = f$ on $(0, 1)$ with $u (0) = u (1) = 0$ using grid points $x_{i} = i / 3$ for $i = 0, 1, 2, 3$ and the two interior values $u_{i} = u (x_{i})$ , $z = (u_{1}, u_{2})$ . Put $h = 1 / 3$ and define the centered stencil at $i = 1, 2$ by $D_{h} u_{i} = (2 u_{i} - u_{i - 1} - u_{i + 1}) / h^{2}$ , with boundary values $u_{0} = u_{3} = 0$ . The resulting finite-difference system is $A_{h} z = f_{h}$ , where $A_{h} = h^{- 2} (\begin{array}{cc} 2 & - 1 \\ - 1 & 2 \end{array})$ and $f_{h} = (f (x_{1}), f (x_{2}))$ . Use parameter and data spaces $X = Y = R^{2}$ with their Euclidean metrics, take $U = X$ , and use the linear forward map $A_{h} ∣_{U} : U \to Y$ . Impose the physical constraint $C = {(z_{1}, z_{2}) \in R^{2} : z_{1} \geq 0, z_{2} \geq 0}$ for nonnegative nodal concentration. An input validator that checks both coordinates before invoking the linear solver rejects the candidate $z = (- 1, 1)$ before solving because its first coordinate is negative.

Facts & Assumptions

Given: The two-node centered finite-difference model, $X = Y = R^{2}$ with Euclidean metrics, $U = X$ , forward map $A_{h} ∣_{U}$ , and the nonnegative nodal constraint in the Example, with candidate state $z = (- 1, 1)$ .

[F1]

Admissible constraint classes and restricted identifiability: a constrained truth class is a nonempty subset of the parameter space, and its forward map is the restriction to that class.

[F2]

Linear map between vector spaces over the same field: a linear map satisfies $A (a u + b v) = a A (u) + b A (v)$ for scalar coefficients and vectors.

Verification

1.1givenF2algebra

By the definition of $D_{h}$ in the Example and the boundary values $u_{0} = u_{3} = 0$ , $D_{h} u_{1} = (2 u_{1} - u_{2}) / h^{2}$ and $D_{h} u_{2} = (2 u_{2} - u_{1}) / h^{2}$ , which gives the displayed matrix $A_{h}$ . For any vectors $v, w \in R^{2}$ and scalars $a, b$ , matrix multiplication gives $A_{h} (a v + b w) = a A_{h} v + b A_{h} w$ , so $A_{h}$ is linear.

2.1givenF1algebra∎

The set $C$ is nonempty because $(0, 0) \in C$ . The candidate has $z_{1} = - 1 < 0$ , so $z \notin C$ and the validator rejects it before the step that solves $A_{h} z = f_{h}$ . This verifies only the stated discrete admissibility check and makes no continuum PDE or solver-convergence claim.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

A line constraint removes one kernel direction

Example

Use $X = R^{2}$ with its Euclidean metric, $Y = R$ with its absolute-value metric, and $U = X$ . Let $A : X \to Y$ be $A (s, t) = s$ , and restrict the truth class to $C_{0} = {(t, 0) : t \in R}$ . Every scalar datum $y \in R$ has exactly one admissible preimage, and $(C_{0} - C_{0}) \cap \ker A = {0}$ ; this constraint removes the vertical null direction.

Facts & Assumptions

Given: $X = R^{2}$ and $Y = R$ with the stated Euclidean and absolute-value metrics, $U = X$ , $A (s, t) = s$ and the nonempty constraint class $C_{0} = {(t, 0) : t \in R}$ .

[F1]

Admissible constraint classes and restricted identifiability: restricted identifiability means injectivity of $A ∣_{C}$ , and $C - C = {x - x^{'} : x, x^{'} \in C}$ .

[F2]

A constraint removes kernel ambiguity exactly when its differences avoid the kernel: $A ∣_{C}$ is injective if and only if $(C - C) \cap \ker A = {0}$ .

Verification

1.1givenF1algebra

From $A (s, t) = s$ , $\ker A = {(0, t) : t \in R}$ . Also $C_{0} - C_{0} = {(t, 0) : t \in R}$ , whose intersection with this vertical kernel is ${(0, 0)}$ .

2.1givenF1F2step 1.1constructalgebra∎

By [F2] and step 1.1, $A ∣_{C_{0}}$ is injective. For any $y \in R$ , $(y, 0) \in C_{0}$ and $A (y, 0) = y$ , so it is the unique admissible preimage of $y$ .

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

A half-space constraint leaves a continuous kernel ambiguity

Example

Use $X = R^{2}$ with its Euclidean metric, $Y = R$ with its absolute-value metric, and $U = X$ . Let $A : X \to Y$ be $A (s, t) = s$ , and restrict the truth class to the half-space $C_{+} = {(s, t) \in R^{2} : t \geq 0}$ . For every datum $y \in R$ , the constrained fiber is ${(y, t) : t \geq 0}$ , a continuous half-line of admissible truths; moreover $(C_{+} - C_{+}) \cap \ker A = {(0, t) : t \in R} \neq {0}$ .

Facts & Assumptions

Given: $X = R^{2}$ and $Y = R$ with the stated Euclidean and absolute-value metrics, $U = X$ , $A (s, t) = s$ and the nonempty constraint class $C_{+} = {(s, t) \in R^{2} : t \geq 0}$ .

[F1]

Admissible constraint classes and restricted identifiability: $C - C = {x - x^{'} : x, x^{'} \in C}$ , $\ker A = {x : A x = 0}$ , and restricted identifiability is injectivity on $C$ .

[F2]

A constraint removes kernel ambiguity exactly when its differences avoid the kernel: injectivity on $C$ implies $(C - C) \cap \ker A = {0}$ .

Verification

1.1givenF1constructalgebra

We have $\ker A = {(0, t) : t \in R}$ . If $(a, b) \in R^{2}$ with $b \geq 0$ , then $(a, b) = (a, b) - (0, 0)$ with both points in $C_{+}$ . If $b < 0$ , then $(a, b) = (a, 0) - (0, - b)$ with both points in $C_{+}$ . Thus $C_{+} - C_{+} = R^{2}$ , so its intersection with $\ker A$ is the nonzero vertical axis.

2.1givenF1F2step 1.1algebra∎

Since this intersection is nontrivial, the forward implication in [F2] rules out injectivity of $A ∣_{C_{+}}$ . Directly, $A (s, t) = y$ and $t \geq 0$ hold exactly when $(s, t) = (y, t)$ with $t \geq 0$ ; distinct such $t$ give distinct admissible truths with the same datum.