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Blog - network theory (part 24) (Rev #1)

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Review

The storm is brewing for the final climax, so let’s remember where we are. We start with a stochastic reaction network:

This consists of:

• finite sets of transitions TT, complexes KK and species SS,

• a map r:T(0,)r: T \to (0,\infty) giving a rate constant for each transition,

source and target maps s,t:TKs,t : T \to K saying where each transition starts and ends,

• a map Y:K SY : K \to \mathbb{N}^S saying how each complex is made of species.

Then we extend s,ts, t and YY to linear maps:

Then we put inner products on these vector spaces as described last time, which lets us ‘turn around’ the maps ss and tt by taking their adjoints:

s ,t : K T s^\dagger, t^\dagger : \mathbb{R}^K \to \mathbb{R}^T

More surprisingly, we can ‘turn around’ YY and get a nonlinear map using ‘matrix exponentiation’:

S K x x Y \begin{array}{ccc} \mathbb{R}^S &\to& \mathbb{R}^K \\ x &\mapsto& x^Y \end{array}

This is most easily understood by thinking of xx as a row vector and YY as a matrix:

x Y = (x 1, x 2, , x k) (Y 11 Y 12 Y 1 Y 21 Y 22 Y 2 Y k1 Y k2 Y k) = (x 1 Y 11x k Y k1,,x 1 Y 1x k Y k) \begin{array}{ccl} x^Y &=& {\left( \begin{array}{cccc} x_1 , & x_2 , & \dots, & x_k \end{array} \right)}^{ \left( \begin{array}{cccc} Y_{11} & Y_{12} & \cdots & Y_{1 \ell} \\ Y_{21} & Y_{22} & \cdots & Y_{2 \ell} \\ \vdots & \vdots & \ddots & \vdots \\ Y_{k1} & Y_{k2} & \cdots & Y_{k \ell} \end{array} \right)} \\ \\ &=& \left( x_1^{Y_{11}} \cdots x_k^{Y_{k1}} ,\; \dots, \; x_1^{Y_{1 \ell}} \cdots x_k^{Y_{k \ell}} \right) \end{array}

So, we get these maps:

The boundary operator

: T K \partial : \mathbb{R}^T \to \mathbb{R}^K

describes how each transition causes a change in complexes:

=ts \partial = t - s

As we saw last time, there is a Hamiltonian

H: K K H : \mathbb{R}^K \to \mathbb{R}^K

describing a Markov processes on the set of complexes, given by

H=s H = \partial s^\dagger

The star of the show is the rate equation. This describes how the number of things of each species changes with time. We write these numbers in a list and get a vector x Sx \in \mathbb{R}^S with nonnegative components. The rate equation says:

dxdt=YHx Y \displaystyle{ \frac{d x}{d t} = Y H x^Y }

We are looking for equilibrium solutions of the rate equation, where the number of things of each species doesn’t change with time at all:

YHx Y=0 Y H x^Y = 0

In fact we will find complex balanced equilibrium solutions, where even the number of complexes of each kind don’t change:

Hx Y=0 H x^Y = 0

More precisely, we have:

Deficiency Zero Theorem (Child’s Version). Suppose we have a reaction network obeying these two conditions:

  1. It is weakly reversible, meaning that whenever there’s a transition from one complex κ\kappa to another κ\kappa', there’s a sequence of transitions going back from κ\kappa' to κ\kappa.

  2. It has deficiency zero, meaning imdkerY={0} \mathrm{im} d \cap \mathrm{ker} Y = \{ 0 \} .

Then for any choice of rate constants there exists an complex balanced equilibrium solution of the rate equation where all species are present in nonzero amounts. In other words, there exists x(0,) Sx \in (0,\infty)^S with

Hx Y=0 H x^Y = 0

Proof of the zero deficiency theorem

category: blog