@@ -395,27 +395,26 @@ we find
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\mathbf{x}(t) = \mathbf{x}(0) + \sum_{k=1}^K \left(\int_0^t v_k(\mathbf{x})(s) \, ds\right) \mathbf{N}_k,
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```
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which demonstrates that `` \mathbf{x}(t) - \mathbf{x}(0) `` is always given by a
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- linear combination of the stochiometry vectors, i.e.
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+ linear combination of the stoichiometry vectors, i.e.
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``` math
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- \DeclareMathOperator{\span}{span}
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- \mathbf{x}(t) - \mathbf{x}(0) \in \span\{\mathbf{N}_k \}.
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+ \mathbf{x}(t) - \mathbf{x}(0) \in \operatorname{span}\{\mathbf{N}_k \}.
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```
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In particular, this says that `` \mathbf{x}(t) `` lives in the translation of the
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- `` \span\{\mathbf{N}_k \} `` by `` \mathbf{x}(0) `` which we write as
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- `` (\mathbf{x}(0) + \span\{\mathbf{N}_k\}) `` . In fact, since the solution should
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- stay non-negative, if we let $\bar{\mathbb{R}}_ +^{M}$ denote the subset of
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- vectors in $\mathbb{R}^{M}$ with non-negative components, the possible physical
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- values for the solution, `` \mathbf{x}(t) `` , must be in the set
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+ `` \operatorname{ span} \{\mathbf{N}_k \} `` by `` \mathbf{x}(0) `` which we write as
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+ `` (\mathbf{x}(0) + \operatorname{ span} \{\mathbf{N}_k\}) `` . In fact, since the
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+ solution should stay non-negative, if we let $\bar{\mathbb{R}}_ +^{M}$ denote the
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+ subset of vectors in $\mathbb{R}^{M}$ with non-negative components, the possible
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+ physical values for the solution, `` \mathbf{x}(t) `` , must be in the set
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``` math
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- (\mathbf{x}(0) + \span\{\mathbf{N}_k\}) \cap \bar{\mathbb{R}}_+^{M}.
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+ (\mathbf{x}(0) + \operatorname{ span} \{\mathbf{N}_k\}) \cap \bar{\mathbb{R}}_+^{M}.
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```
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This set is called the stoichiometric compatibility class of `` \mathbf{x}(t) `` .
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The key property of stoichiometric compatibility classes is that they are
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invariant under the RRE ODE's dynamics, i.e. a solution will always remain
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within the subspace given by the stoichiometric compatibility class. Finally, we
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note that the * positive* stoichiometric compatibility class generated by
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- $\mathbf{x}(0)$ is just ``(\mathbf{x}(0) + \span\{ \mathbf{N}_ k\} ) \cap
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- \mathbb{R}_ +^{M}`` , where `` \mathbb{R}_ +^{M}`` denotes the vectors in
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+ $\mathbf{x}(0)$ is just ``(\mathbf{x}(0) + \operatorname{ span} \{ \mathbf{N}_ k\} )
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+ \cap \ mathbb{R}_ +^{M}`` , where `` \mathbb{R}_ +^{M}`` denotes the vectors in
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`` \mathbb{R}^M `` with strictly positive components.
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With these definitions we can now see how knowing the deficiency and weak
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