# The Azimuth Project Experiments in multiple equilibrium states (changes)

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We have so far always first considered the rate equation and then talked about the master equation and given heuristic arguments on how they are different. I wanted to start from the master equation and show how the rate equation arises from this. That’s exactly what we are going to do in this post.

## The master equation

We can also arrive at a master equation. Now say we have a probability,

$\psi_{i_1...i_k}$ of having $i_1$ things in state 1, $i_k$ things in state $k$ etc. We write

$\psi = \sum \psi_{i_1...i_k}z_1^{i_1}\cdots z_k^{i_k}$

and $a_i^\dagger \psi = z_i \psi$, $a_i\psi = \frac{\partial}{\partial z_i} \psi$ then the master equation says

$\frac{d}{dt}\psi = H\psi$

Each transition corresponds to an operator. There is a term $a_1^{\dagger n_1}\cdots a_k^{\dagger n_1} a_1^{m_1}\cdots a_k^{m_k}$

In general, the master equation is written as

$H = \sum_{\tau \in T} r(\tau) (a_1^{\dagger n_1(\tau)}\cdots a_k^{\dagger n_k(\tau)}a_1^{m_1(\tau)}\cdots a_k^{m_k(\tau)}- N_1^{\underline m_1}\cdots N_k^{\underline m_k})$

We will attempt to solve $H\Psi = 0$, by trying

$\Psi = \prod_{i=1}^k \sum_{n_i = 0}^\infty \frac{\alpha_i^{n_i}}{n_i!} z_i^{n_i} = \prod_{i=1}^k e^{\alpha_i z_i}$
$(a_1^{\dagger n_1(\tau)}\cdots a_k^{\dagger n_k(\tau)}) \Psi = (z_1^{n_1(\tau)}\cdots z_k^{n_k(\tau)})\Psi$
$(a_1^{m_1(\tau)}\cdots a_k^{m_k(\tau)}) \Psi = (\alpha_1^{m_1(\tau)}\cdots \alpha_k^{m_k(\tau)}) \Psi$
$(N_1^{\underline m_1}\cdots N_k^{\underline m_k})\Psi = (a_1^{\dagger m_1(\tau)} a_1^{m_1(\tau)})\cdots (a_k^{\dagger m_k(\tau)} a_k^{m_k(\tau)})\Psi = (\alpha_1^{m_1(\tau)}\cdots \alpha_k^{m_k(\tau)})(z_1^{m_1(\tau)}\cdots z_k^{m_k(\tau)})\Psi$

So the master equation acting on $\Psi$ becomes

$H\Psi = \sum_{\tau \in T} r(\tau)(\alpha_1^{m_1(\tau)}\cdots \alpha_k^{m_k(\tau)})\left(z_1^{n_1(\tau)}\cdots z_k^{n_k(\tau)} - z_1^{m_1(\tau)}\cdots z_k^{m_k(\tau)}\right) \Psi$

We will assume that

$H\Psi = 0$

From which it follows that

$\frac{d}{d z_i}H\Psi = 0$

We will make use of the compact notation for the Hamiltonian as

$H = \sum_{\tau} r(\tau)\alpha^{m(\tau)}(z^{n(\tau)}-z^{m(\tau)})$
$\frac{d}{d z_i}H\Psi = \frac{d H}{d z_i}\Psi + \frac{d\Psi}{dz_i}H$

This becomes

$\sum_{\tau} r(\tau)\alpha^{m(\tau)}(n_i z^{n'(\tau)}-m_i z^{m'(\tau)}) \Psi + \alpha_i H \Psi=0$

but from the assumption, the second term vanishes. Hence,

$\sum_{\tau} r(\tau)\alpha^{m(\tau)}(n_i z^{n'(\tau)}-m_i z^{m'(\tau)}) \Psi=0$

and we let

$z_1=z_2=z_3=...=z_k=1$

which yields a solution for non-zero $\Psi$ as

$\sum_{\tau} r(\tau)\alpha^{m(\tau)}(n_i -m_i)=0$

which gives a solution to the rate equation, as asserted. Creation ex nihilo, something from nothing.

## Some notation

Consider the nth derivative of the product of $\Psi$ and $H$.

$\left(\frac{d}{d z_i}\right)^n H\Psi = \sum_{l+m=n}H^{(l)}\Psi^{(m)}$

and so $H^{(l)}$ and $\Psi^{(m)}$ become short hand for the lth and mth derivative respectively. In the case we consider here, this takes the simpler form as

$\sum_{l+m=n}H^{(l)}\Psi^{(m)} = \left(\sum_{l+m=n} \alpha_i^{(m)} H^{(l)}\right)\Psi$

And so we are left with the scalar $\alpha_i$ to the mth power times the lth derivative of the Hamiltonian operator. We will evaluate this at

$z_1=z_2=z_3=...=z_k=1$

Yielding

$H^{(l)} = \sum_{\tau}r(\tau)\alpha^{m(\tau)}\left(\Pi_{q=0}^l(n_i-q) - \Pi_{q=0}^l(m_i-q)\right)$

Note that the term $\Pi_{q=0}^l(n_i-q)$ will vanish iff we count over a value where $n_i=q$. The same is true for the term $\Pi_{q=0}^l(m_i-q)$.

## This is the moment

We will consider the number operator $N$ acting $l$ times on $H\Psi$

$N^l H \Psi=$

From above we know that

$H\Psi = \sum_{\tau} r(\tau)\alpha^{m(\tau)}(z^{n(\tau)}-z^{m(\tau)})\Psi$

and also that

$N^q (H \Psi) = \sum_{l+m=q}(N^l H)(N^m \Psi)= \sum_{l+m=q}\alpha_i^m (N^l H)\Psi$

For now we will consider each term $(N^l H)\Psi$ separately.

$(N^l H)\Psi = \sum_{\tau}r(\tau)\alpha^{m(\tau)} (n_i^l z^{n(\tau)} - m_i^l z^{m(\tau)})\Psi$

When we set $z_1=...=z_k=1$ we arrive at

$\sum_{\tau}r(\tau)\alpha^{m(\tau)} (n_i^l(\tau) - m_i^l(\tau))$
• (Result) For $l=1$ we recover that the mean vanishes for appropriate choice of $\alpha^{m(\tau)}$, given that the rate equation vanishes with appropriate choice of $X^{m(\tau)}$.

It is a straight forward application of the factor theorem to show that $(n_i-m_i)$ is a factor of $(n_i^l(\tau) - m_i^l(\tau))$ so we have

$\sum_{\tau}r(\tau)\alpha^{m(\tau)} (n_i(\tau) - m_i(\tau))\text{Poly}((n_i(\tau), m_i(\tau))$

We know from the rate equation that

$\sum_{\tau}r(\tau)\alpha^{m(\tau)} (n_i(\tau) - m_i(\tau)) = 0$

## Example of multiple equilibrium states

For the general terms and definitions used in this project see

Here we consider examples from Table 1 in the following paper.

P. M. Schlosser and M. Feinberg, A theory of multiple steady states in isothermal homogeneous CFSTRs with many reactions, Chemical Engineering Science 49 (1994), 1749–1767.

• By ‘just one equilibrium state’, I presume they mean just a 1-parameter family of equilibrium states. After all, if $x_A, x_B$ is an equilibrium solution of the rate equation, so is $c x_A, c x_B$ for any $c \ge 0$.

### The chemical reaction network

The example we consider here is from Table 1 (7), giving the chemical reaction network defined by

$2 A + B \to 3 A$
$3 A \to 2 A + B$

The input and output functions then give.

$m(\tau_1)=\{2,1\}$
$n(\tau_1)=\{3,0\}$

and for $\tau_2$

$m(\tau_2)=\{3,0\}$
$n(\tau_2)=\{2,1\}$

### The Stochastic Petri Net ### The chemical rate equation

The general form of the rate equation is

$\frac{d}{d t}X_i=\sum_{\tau\in T}r(\tau)X^{m(\tau)}(n_i(\tau)-m_i(\tau))$

The population of species $A$ ($B$) will be denoted $X_1(t)$ ($X_2(t)$) and the rate of the reaction $\tau_1$ ($\tau_2$) as $r_1$ ($r_2$).

$\frac{d}{d t}X_1 = r_1 X_1^2 X_2 - r_2 X_1^3$

and

$\frac{d}{d t} X_2 = r_1 X_1^2 X_2 + r_2 X_1^3$

### The master equation

The general form of the master equation is

$H = \sum_{\tau\in T} r(\tau)(a^\dagger^{n(\tau)}a^{m(\tau)} - a^\dagger^{m(\tau)} a^{m(\tau)})$

In our case, this becomes

$H = r_1 (a_1^\dagger a_1^\dagger a_1^\dagger a_1 a_1 a_2 - a_1^\dagger a_1^\dagger a_2^\dagger a_1 a_1 a_2) + r_2 (a_1^\dagger a_1^\dagger a_2^\dagger a_1 a_1 a_1 - a_1^\dagger a_1^\dagger a_1^\dagger a_1 a_1 a_1 )$

### Proof that if the rate equation vanishes then so does the mater equation

Since the rate equation is known to vanish from the Deficiency Zero Theorem, we have that

$\frac{d}{d t}X_1 = 0 = r_1 X_1^2 X_2 - r_2 X_1^3$

and

$\frac{d}{d t} X_2 = 0 = r_1 X_1^2 X_2 + r_2 X_1^3$
• (Algebraic independence) Let
$f_1,...,f_m$

be polynomials in

$F[x_1,...,x_n]$

are called algebraically independent if there is no non-zero polynomial

$A \in F[y_1,...,y_m]$

such that

$A(f_1,...,f_m)=0$

then its called the annihilating polynomial.

We consider a wave function of the form

$\Psi := e^{b z_1} e^{c z_2}$

We then calculate $H\Psi$. We find solutions where this vanishes, for non-zero $\Psi$ as

$(r_1 b^2 c - r_2 b^3)z_1^3 + (r_2 b^3 - r_1 b^2 c)z_1^2 z_2$

For this to vanish, each of the terms must vanish separately. Hence we arrive at a system of equations

$r_1 b^2 c - r_2 b^3 = 0$
$r_2 b^3 - r_1 b^2 c = 0$

In this case, and in every case I’ve considered so far, a solution of one of these equations, gives a solution of the second.

category: experiments