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The feasible tangent cone for nonnegative stoichiometric amounts

Statement

Let $s, r \geq 1$ , let $ν \in R^{s \times r}$ be a stoichiometric matrix (column $k$ gives the signed species-amount change per unit extent of reaction $k$ ), and let $n^{0} \in [0, \infty)^{s}$ be initial species amounts. Extents $ξ \in R^{r}$ are measured in one fixed amount unit, so $n (ξ) = n^{0} + ν ξ$ has species amounts in that same unit; $ν$ is dimensionless. The feasible extent set is the polyhedron

$F = {ξ \in R^{r} : n^{0} + ν ξ \in [0, \infty)^{s}} .$

At $ξ_{*} \in F$ , let $J = {j : n_{j}^{0} + (ν ξ_{*})_{j} = 0}$ be the active species set. Define the Bouligand tangent cone $T_{F} (ξ_{*})$ to consist of all $h \in R^{r}$ for which there are $t_{k} > 0$ with $t_{k} \to 0$ and $h_{k} \to h$ such that $ξ_{*} + t_{k} h_{k} \in F$ . Then

$T_{F} (ξ_{*}) = {h \in R^{r} : (ν h)_{j} \geq 0 for every j \in J} .$

If $U \subseteq R^{r}$ is open with $ξ_{*} \in U$ , and $f : U \to R$ is differentiable with a local minimum at $ξ_{*}$ relative to $F \cap U$ , then $D f (ξ_{*}) h \geq 0$ for every $h \in T_{F} (ξ_{*})$ . Conversely, if $U$ is convex and $f$ is convex and differentiable on $U$ , this directional inequality for every $h \in T_{F} (ξ_{*})$ implies $f (ξ) \geq f (ξ_{*})$ for every $ξ \in F \cap U$ . Suppose $V \subseteq R^{s}$ is open, $n_{*} : = n^{0} + ν ξ_{*} \in V$ , $Φ : V \to R$ is totally differentiable at $n_{*}$ , and $f (ξ) = Φ (n^{0} + ν ξ)$ near $ξ_{*}$ . For each $j$ , let $e_{j} \in R^{s}$ have coordinate $1$ in position $j$ and $0$ in every other position; define $μ_{j} : = D Φ (n_{*}) e_{j}$ . Then

$D f (ξ_{*}) h = \sum_{j = 1}^{s} μ_{j} (ν h)_{j} = (ν^{T} μ) \cdot h .$

The tangent-cone and optimality claims are mathematical deductions. They do not assert that a particular physical system obeys this model or is at equilibrium.

Facts & Assumptions

Given: The finite stoichiometric matrix, initial amounts, feasible extent set, Euclidean convergence, and differentiability/convexity hypotheses in the Statement.

[F1]

The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder: differentiability at $ξ_{*}$ gives $f (ξ_{*} + w) = f (ξ_{*}) + D f (ξ_{*}) w + o (∥ w ∥_{2})$ as $w \to 0$ .

[F2]

A convex subset of $R^{m}$ contains every line segment between two of its points: convexity of $U$ and the affine inequalities defining $F$ keep every segment between feasible points in $F \cap U$ .

[F3]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : the derivative of a composition is the composition of its total derivatives.

[F4]

Convex and strictly convex functions on Euclidean convex sets: a real-valued function is convex when $f ((1 - t) x + t y) \leq (1 - t) f (x) + t f (y)$ for points in its convex domain and $t \in [0, 1]$ .

[F5]

Every Euclidean linear map has a unique matrix and satisfies $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for some $K \geq 0$ : a finite-dimensional linear map is bounded and therefore continuous.

Proof

technique · direct

1.1F5givenalgebra

Suppose $h \in T_{F} (ξ_{*})$ , and choose sequences $t_{k}, h_{k}$ from its definition. For each active $j \in J$ , feasibility gives $0 + t_{k} (ν h_{k})_{j} \geq 0$ , hence $(ν h_{k})_{j} \geq 0$ . Since [F5] makes the finite matrix map continuous, $h_{k} \to h$ implies $ν h_{k} \to ν h$ , so $(ν h)_{j} \geq 0$ . This proves one inclusion.

1.2givenchoosealgebra

Conversely suppose $(ν h)_{j} \geq 0$ for all $j \in J$ . For each inactive $j$ , put $b_{j} = n_{j}^{0} + (ν ξ_{*})_{j} > 0$ . If $(ν h)_{j} < 0$ , restrict $0 \leq t \leq b_{j} / [- (ν h)_{j}]$ ; if $(ν h)_{j} \geq 0$ , no restriction is needed. There are finitely many species, so choose a common $ϵ > 0$ no larger than all positive bounds (or $ϵ = 1$ if there are none). For $j \in J$ , $n_{j}^{0} + (ν (ξ_{*} + t h))_{j} = t (ν h)_{j} \geq 0$ ; for inactive $j$ , the choice of $ϵ$ gives $b_{j} + t (ν h)_{j} \geq 0$ for $0 \leq t \leq ϵ$ . Thus $ξ_{*} + t h \in F$ throughout that interval. Taking $h_{k} = h$ and any positive $t_{k} \leq ϵ$ tending to zero proves $h \in T_{F} (ξ_{*})$ .

2.1F1step 1.1step 1.2given

By steps 1.1–1.2, every tangent $h$ has a feasible ray $ξ_{*} + t h$ for all sufficiently small $t \geq 0$ . Openness of $U$ keeps the initial part of this ray in $U$ . At a relative local minimum, $[f (ξ_{*} + t h) - f (ξ_{*})] / t \geq 0$ for all sufficiently small $t > 0$ ; differentiability [F1] makes its limit $D f (ξ_{*}) h$ . Hence $D f (ξ_{*}) h \geq 0$ .

2.2F1F2F4step 1.2givenalgebra

Now assume $U$ is convex and $f$ convex and differentiable. For any $ξ \in F \cap U$ , put $h = ξ - ξ_{*}$ . For active $j$ , $(ν h)_{j} = n_{j} (ξ) - n_{j} (ξ_{*}) = n_{j} (ξ) \geq 0$ , so step 1.2 makes $h$ a tangent direction; the segment $ξ_{*} + t h$ also lies in $F \cap U$ by [F2]. Let $g (t) = f (ξ_{*} + t h)$ . [F4] gives $(g (t) - g (0)) / t \leq g (1) - g (0)$ for $0 < t < 1$ ; as $t ↓ 0$ , the left side tends to $D f (ξ_{*}) h$ by [F1]. If every tangent derivative is nonnegative, then $f (ξ) - f (ξ_{*}) = g (1) - g (0) \geq D f (ξ_{*}) h \geq 0$ .

3.1F3givenalgebra∎

The affine map $ξ \mapsto n^{0} + ν ξ$ has derivative $h \mapsto ν h$ . By [F3], $D f (ξ_{*}) h = D Φ (n_{*}) (ν h)$ . Since $ν h = \sum_{j = 1}^{s} (ν h)_{j} e_{j}$ and $D Φ (n_{*})$ is linear, this equals $\sum_{j = 1}^{s} (ν h)_{j} D Φ (n_{*}) e_{j} = \sum_{j = 1}^{s} μ_{j} (ν h)_{j} = (ν^{T} μ) \cdot h$ , as claimed.

The feasible tangent cone for nonnegative stoichiometric amounts

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources