The Species-Reaction Graph

Abstract

The Species-Reaction Graph is a diagrammatic representation of a reaction network that closely resembles those commonly used to depict biochemical pathways. We shall see that a network’s Species-Reaction Graph often carries an extraordinary amount of far-from-obvious information about how the network might behave. In fact, the theorems in this chapter will tell us a great deal about behavior across the entire reaction network landscape, in particular about why dull, stable behavior is more prevalent than one might expect within a mathematical macrocosm so rife with nonlinearity.

Appendix 11.A Proof of Theorem 11.11.7

Here we give a proof of Theorem 11.11.7 , which is repeated below as Theorem 11.A.1. The central idea of the proof is largely the one (“symmetrization”) used in earlier Knot Graph work, but the style is different.

Theorem 11.A.1 ( [20, 54, 112, 113])

Consider a mass action system ${\{\mathcal {S},\mathcal {C},\mathcal {R},k\}}$ with stoichiometric subspace S and species-formation-rate function $f:{\mathbb {R}^{\mathcal {S}}} \to S$ . If the Knot Graph of the underlying network ${\{\mathcal {S},\mathcal {C},\mathcal {R}\}}$ has no cycles, then at each positive composition c ^∗ (not necessarily an equilibrium), all eigenvalues of the derivative of the species-formation-rate function, df(c ^∗) : S → S, are real and negative. Moreover, there exists for S a basis consisting of eigenvectors of df(c ^∗). If c ^∗ is an equilibrium, it is asymptotically stable.

Remark 11.A.2

Keep in mind that the base subnetwork of network ${\{\mathcal {S},\mathcal {C},\mathcal {R}\}}$ is presumed reversible. Keep in mind also the presumption that every reaction of the form is accompanied by another reaction of the form . If, in the mass action system of Theorem 11.A.1, there is a reaction of the form unaccompanied by a reaction of the form , we can still claim that all eigenvalues are real and nonpositive—see the proof—but an eigenvalue of 0 can occur. An example is provided by the simple network

$$\displaystyle \begin{aligned} 0 \to A \rightleftarrows B. \end{aligned} $$

(11.A.1)

This network is degenerate, so for every choice of mass action rate constants, we will have singularity of df(c ^∗) (and an eigenvalue of 0) at every positive composition.

By way of preparation for the proof of Theorem 11.A.1, we review some fundamental ideas of linear algebra. Suppose that V is a real vector space with a scalar product, denoted “◇”. A linear transformation T : V → V is symmetric relative to the scalar product ◇ if

$$\displaystyle \begin{aligned} u \diamond Tw = w \diamond Tu \quad \forall u, w \in V. \end{aligned} $$

(11.A.2)

Moreover, was say that T is negative-semi-definite relative to the scalar product ◇ if, for all v ∈ V ,

$$\displaystyle \begin{aligned} v \diamond Tv \leq 0 \end{aligned} $$

(11.A.3)

and negative-definite relative to that scalar product if equality holds only when v = 0.

Remark 11.A.3

Suppose that V is finite-dimensional. To establish the symmetry of a linear transformation T : V → V relative to a scalar product ◇, it is enough to test (11.A.2) against a basis for V . That is, if {v ₁, v ₂, …, v _n} is a basis for V , then T is symmetric if and only if

$$\displaystyle \begin{aligned} v_i \diamond Tv_j = v_j \diamond Tv_i, \quad i,j = 1,2,\dots,n. \end{aligned} $$

(11.A.4)

The information contained in the following theorem is well known [98, 120].

Theorem 11.A.4 (Properties of a symmetric linear transformation)

Let V be a finite-dimensional real vector space and let T : V → V be a linear transformation. If there exists for V a scalar product “◇” with respect to which T is symmetric, then all roots of the characteristic polynomial of T are real. Moreover, there exists for V a basis that is orthonormal relative to ◇ and consists entirely of eigenvectors of T. If T is negative-semi-definite [negative-definite] relative to ◇, then all eigenvalues of T are nonpositive [negative].

Proof (Theorem 11.A.1)

To begin the proof of Theorem 11.A.1, we first note that, in the theorem statement, the derivative at c ^∗ of the species-formation-rate function, df(c ^∗) : S → S, is regarded as a map from the stoichiometric subspace into itself. For the purposes of the proof, it will be convenient to consider instead (temporarily) the extension of df(c ^∗) to all of ${\mathbb {R}^{\mathcal {S}}}$, which we will denote by $Df(c^*):{\mathbb {R}^{\mathcal {S}}} \to {\mathbb {R}^{\mathcal {S}}}$. This is just the linear map given by

$$\displaystyle \begin{aligned} Df(c^*)\sigma = \left. \frac{df(c^* + \theta \sigma)}{d\theta} \right|{}_{\theta = 0}, \quad \forall \sigma \in {\mathbb{R}^{\mathcal{S}}}. \end{aligned} $$

(11.A.5)

For the mass action system ${\{\mathcal {S},\mathcal {C},\mathcal {R},k\}}$ of the theorem statement, the derivative Df(c ^∗) takes the following form:

$$\displaystyle \begin{aligned} D f(c^*)\sigma = \sum_{{y \to y'} \in {\mathcal{R}}}\eta_{{y \to y'}}(y*\sigma)(y'-y), \quad \forall \sigma \in {\mathbb{R}^{\mathcal{S}}}, \end{aligned} $$

(11.A.6)

where, for each ${y \to y'} \in {\mathcal {R}}$,

$$\displaystyle \begin{aligned} \eta_{{y \to y'}} := k_{{y \to y'}}(c^*)^y \end{aligned} $$

(11.A.7)

and “∗” indicates the scalar product in ${\mathbb {R}^{\mathcal {S}}}$ defined by

(11.A.8)

We denote by ${\mathcal {B}}$ the subset of ${\mathcal {R}}$ consisting of reactions of the base subnetwork. Because the base subnetwork is presumed reversible, we can choose for each reversible reaction pair a “forward” reaction, and we denote by ${\mathcal {F}}$ the subset of ${\mathcal {B}}$ consisting of all forward reactions. With this in mind, we can rewrite (11.A.6) in the following way: For all $\sigma \in {\mathbb {R}^{\mathcal {S}}}$

$$\displaystyle \begin{aligned} &D f(c^*)\sigma = \\ {} & \sum_{{y \to y'} \in {\mathcal{F}}}[\eta_{{y \to y'}}(y*\sigma) - \eta_{y' \to y}(y'*\sigma)](y'-y)\ \ + \sum_{{y \to y'} \in {\mathcal{R}} \setminus {\mathcal{B}}}\eta_{{y \to y'}}(y*\sigma)(y'-y). \end{aligned} $$

(11.A.9)

Our aim will be to show that, when the Knot Graph is acyclic, it is possible to choose $p \in {\mathbb {R}_+^{\mathcal {S}}}$ such that Df(c ^∗) is symmetric relative to the scalar product “◇” defined by

(11.A.10)

In fact, we will show that we can choose p to get symmetry term by term in (11.A.9). That is, we will show that we can choose p such that, for all σ and σ′ in ${\mathbb {R}^{\mathcal {S}}}$, we have, for each ${y \to y'} \in {\mathcal {F}}$,

$$\displaystyle \begin{aligned} &[\eta_{{y \to y'}}(y*\sigma) - \eta_{y' \to y}(y'*\sigma)](y'-y)\diamond \sigma'\\ =\; &[\eta_{{y \to y'}}(y*\sigma') - \eta_{y' \to y}(y'*\sigma')](y'-y)\diamond\sigma \end{aligned} $$

(11.A.11)

and, for each ${y \to y'} \in {\mathcal {R}} \setminus {\mathcal {B}}$,

$$\displaystyle \begin{aligned} (y*\sigma)(y' -y)\diamond \sigma' = (y*\sigma')(y' -y)\diamond \sigma. \end{aligned} $$

(11.A.12)

It is easy to see that the requirement (11.A.12) for reactions not in the base subnetwork is satisfied trivially no matter how $p \in {\mathbb {R}_+^{\mathcal {S}}}$ is chosen. Such reactions are either of the form or . In the first case, both sides of (11.A.12) reduce to zero. In the second case, both sides reduce to

(11.A.13)

We turn, then, to consideration of reactions in the base subnetwork. To begin, we will choose $p \in {\mathbb {R}_+^{\mathcal {S}}}$ to satisfy the requirement that when species and are in the same knot, then . (There will eventually be additional constraints on p.) When K is a knot, we denote by p _K the value of common to all . Moreover, when y is a complex, we denote by p _y the value of common to all . (All members of supp y are in the same knot.)

Now let y → y′ be a particular reaction in ${\mathcal {F}}$. In light of Remark 11.A.3 and because we can regard the species set ${\mathcal {S}}$ as a basis for ${\mathbb {R}^{\mathcal {S}}}$, the symmetry condition (11.A.11) will be satisfied if it is satisfied for all choices of and , where and are distinct species.

For the particular reaction ${y \to y'} \in {\mathcal {F}}\;$ under study, we consider various cases:

(i) Suppose that neither nor is in supp y′. In this case (11.A.11) is satisfied, for both sides reduce to

(11.A.14)

(ii) Suppose that neither nor is in supp y. Equation (11.A.11) is again satisfied, for both sides reduce to

(11.A.15)

(iii) Now suppose that is in supp y and that is in supp y′. In this case and are in different knots, denoted K and K′. (Were the two species in the same knot, the Knot Graph would have a loop connecting the knot to itself, in contradiction to the hypothesis of Theorem 11.A.1.) Equation (11.A.11) reduces to

This last equation will be satisfied if we choose

$$\displaystyle \begin{aligned} p_{K} = \frac{\eta_{{y \to y'}}}{\eta_{y' \to y}}p_{K'}. \end{aligned} $$

(11.A.16)

There is precisely one equation of the form (11.A.16) for each edge in the Knot Graph. The question then becomes whether there is, for each knot K, a choice of p _K > 0 such that the resulting system of equations is satisfied, regardless of the values of $\{\eta _{{y \to y'}}\}_{{y \to y'} \in {\mathcal {B}}}$. Because the Knot Graph has no cycles, the answer is yes: The Knot Graph is a forest, so in each tree we can choose a leaf—that is, a knot K ^∗ adjacent to precisely one edge—and then choose $p_{K^*} = 1$. The assignments of p _K for the remaining knots in the tree can be made by proceeding through the tree, applying (11.A.16) across each edge.

The resulting $p \in {\mathbb {R}_+^{\mathcal {S}}}$ then gives the scalar product ◇ in ${\mathbb {R}^{\mathcal {S}}}$ with respect to which $Df(c^*): {\mathbb {R}^{\mathcal {S}}} \to {\mathbb {R}^{\mathcal {S}}}$ is symmetric. Clearly, the stoichiometric subspace $S \subset {\mathbb {R}^{\mathcal {S}}}$ inherits the scalar product ◇ from ${\mathbb {R}^{\mathcal {S}}}$, and, with respect to ◇, the restriction of Df(c ^∗) to S—i.e., df(c ^∗) : S → S—is also symmetric. From Theorem 11.A.4 it follows that all roots of the characteristic polynomial of df(c ^∗) are real and that there exists for S a basis, orthonormal with respect to ◇, consisting entirely of eigenvectors of df(c ^∗).

It remains to be shown that the eigenvalues of df(c ^∗) are all negative. When the Species-Reaction Graph of the network ${\{\mathcal {S},\mathcal {C},\mathcal {R}\}}$ has no cycles, the negativity of the eigenvalues already follows from Theorem 11.5.1 and Corollary 10.7.3, provided that ${\{\mathcal {S},\mathcal {C},\mathcal {R}\}}$ is nondegenerate. Nevertheless, we are required to prove the negativity of the eigenvalues under the hypothesis of Theorem 11.11.7 (and the constraints imposed by Remark 11.11.1).

This we shall do by arguing that df(c ^∗) is negative-definite relative to the scalar product ◇ just constructed. (See Theorem 11.A.4.) Thus, for σ ∈ S we consider the sign of

$$\displaystyle \begin{aligned} &\sigma \diamond df(c^*)\sigma = \\ {} & \sum_{{y \to y'} \in {\mathcal{F}}}[\eta_{{y \to y'}}(y*\sigma) - \eta_{y' \to y}(y'*\sigma)](y'-y)\diamond \sigma\ + \sum_{{y \to y'} \in {\mathcal{R}} \setminus {\mathcal{B}}}\eta_{{y \to y'}}(y*\sigma)(y'-y) \diamond \sigma. \end{aligned} $$

(11.A.17)

We will argue that the various individual terms in the two sums on the right side of (11.A.17) are all nonpositive with at least one negative so long as σ is not zero.

First consider a reaction ${y \to y'} \in {\mathcal {F}}$ with both y and y′ nonzero. In this case, supp y and supp y′ must reside in different knots K and K′. (Were they in the same knot, in the Knot Graph that knot would have a self-loop.) By virtue of the way ◇ and p were constructed, the term in (11.A.17) corresponding to y → y′ takes the form

$$\displaystyle \begin{aligned}{}[\eta_{{y \to y'}}(y*\sigma)& - \eta_{y' \to y}(y'*\sigma)](y'-y)\diamond \sigma \\ &=[\eta_{{y \to y'}}(y*\sigma) - \eta_{y' \to y}(y'*\sigma)][p_{K'}(y'*\sigma) - p_K( y*\sigma)] \\ &= \frac{\eta_{{y \to y'}}}{p_K}[p_{K}(y*\sigma) - p_{K'}( y'*\sigma)][p_{K'}(y'*\sigma) - p_K( y*\sigma)]\\ &=-\frac{\eta_{{y \to y'}}}{p_K}[(y'-y)\diamond \sigma]^2. \end{aligned} $$

(11.A.18)

The result is clearly nonpositive and is zero only when, for the particular ${y \to y'} \in {\mathcal {F}}$ under study, σ is orthogonal to the corresponding reaction vector relative to the ◇ scalar product.

We do not preclude the possibility that there are in ${\mathcal {F}}$ reactions of the form y → 0 (such as A + B → 0) or 0 → y (such as 0 → A + B). In such cases, it is not difficult to see that the corresponding terms in (11.A.17) reduce to, respectively,

$$\displaystyle \begin{aligned} -\frac{\eta_{y \to 0}}{p_y}(y\diamond \sigma)^2 \quad \mathrm{or} \quad -\frac{\eta_{0 \to y}}{p_y}(y\diamond \sigma)^2. \end{aligned} $$

(11.A.19)

Again, each is nonpositive and is zero only when σ is ◇-orthogonal to the reaction vector for the corresponding reaction, either y → 0 or 0 → y.

We turn now to terms in (11.A.17) that correspond to reactions that do not reside in the base subnetwork. These reactions are of the form or . In the first case, the corresponding term in (11.A.17) reduces to zero. In the second case, the term reduces to

(11.A.20)

The result is again nonpositive and is zero only if is zero or, stated differently, only if σ is ◇-orthogonal to the reaction vector () corresponding the reaction .

In summary, then, we have established that the right side of (11.A.17) is negative and is zero precisely when σ ∈ S is ◇-orthogonal to all reaction vectors corresponding to reactions in the set

$$\displaystyle \begin{aligned} {\mathcal{R}}^* := {\mathcal{F}} \cup \{y \to y' \in {\mathcal{R}} \setminus {\mathcal{B}}\;\; \mathrm{with}\;\; y' = 0\}. \end{aligned} $$

(11.A.21)

From the presumptions of Remark 11.11.1, it follows easily that

$$\displaystyle \begin{aligned} S := {\mathrm{span} \,} \{y'-y: {y \to y'} \in {\mathcal{R}}\} = {\mathrm{span} \,} \{y'-y: {y \to y'} \in {\mathcal{R}}^*\}. \end{aligned} $$

(11.A.22)

Thus, the right side of (11.A.17) is zero only when σ ∈ S is in the orthogonal complement of S (relative to ◇). This can happen only if σ = 0. We have established, then, that df(c ^∗) : S → S is negative-definite relative to ◇, so all of its eigenvalues are negative.

When c ^∗ is an equilibrium, it is asymptotically stable. This follows from the negativity of the eigenvalues associated with df(c ^∗). [34, 104]. □

Remark 11.A.5

In Remark 11.11.1 we required that when, for a particular , there is a reaction of the form , there is also one of the form . In fact, this requirement is stronger than it needed to be. It is enough that , viewed as a vector of ${\mathbb {R}^{\mathcal {S}}}$, be a member of ${\mathrm {span} \,} \{y' - y : y \to y' \in {\mathcal {R}}^*\}$.