# The Azimuth Project Experiments in a one-parameter family of equilibrium states

For the general terms and definitions used in this project see

Here we will consider one of the eight 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.

The example is from Table 1 (8) and is said to have just one equilibrium state. 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 $c \geq 0$.

## Stoichiometry and Petri nets

Balanced chemical reaction equations are given as rules for how certain things combine. For instance,

$H_2 + O \to H_2O$

As we have seen, this method can be extended to systems outside of chemistry. The situation resides on mass action kinetics.

• (The law of mass action). The instantaneous rate of a reaction is taken to be directly proportional to the concentration of each reactant raised to the power of its stoichiometric coefficient.

So if mass action kinetics holds true, the rate of the reaction forming $H_2O$ above would be proportional to the product of the concentration of the reactants, $[H_2][O]$. These reactants each appear one time, if they appeared $n$ times for $H_2$ and $m$ times for $O$ the reaction rate would be proportional to $[H_2]^n[O]^m$. That’s it for the law of mass action. It works in practice.

The example we will consider here is given by the following stoichiometric equations.

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

We can represent these reaction rules using the following Petri net. This was talked about in detail in several posts, starting in Part 2. Three reactants $A$ combine to produce one $B$ and one $A$ with proportionality constant $r_2$. Likewise, two $B$‘s combine with a single $A$ to produce three $A$’s.

We can name these transformations as $\tau_1$ and $\tau_2$ so

$\tau_1: A + 2 B \to 3 A$
$\tau_2: 3 A \to A + 2 B$

which can alternatively be expressed in matrix notation as

(1)$\left( \begin{array}{c} A \\ 2B \end{array} \right) \to \left( \begin{array}{c} 3A \\ 0 \end{array} \right)$
(2)$\left( \begin{array}{c} 3A \\ 0 \end{array} \right) \to \left( \begin{array}{c} A \\ 2B \end{array} \right)$

## Connection to the familiar notation

We will name each of the transformation as follows.

$\tau_1: A + 2 B \to 3 A$
$\tau_2: 3 A \to A + 2 B$

As is typical, we introduce input and output functions. These functions are defined on transitions $\tau$.

• $m(\tau)$ gives the stoichiometric coefficients of the reactants (inputs).
• $n(\tau)$ gives the stoichiometric coefficients of the products (outputs).

In our case, we have two transitions $\tau_1$ and $\tau_2$. The input and output functions evaluate as follows.

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

and for $\tau_2$

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

Returning to matrix notation, the above can be equivalently expressed as.

(3)$m(\tau_1)= \left( \begin{array}{c} 1 \\ 2 \end{array} \right)$
(4)$n(\tau_1)= \left( \begin{array}{c} 3 \\ 0 \end{array} \right)$

and for $\tau_2$

(5)$m(\tau_2)= \left( \begin{array}{c} 3 \\ 0 \end{array} \right)$
(6)$n(\tau_2)= \left( \begin{array}{c} 1 \\ 2 \end{array} \right)$

It should be evident that each instance of the reaction should change the state of the system in a way that is related to the difference between the inputs and the outputs.

(7)$n(\tau_1)-m(\tau_1)= \left( \begin{array}{c} 2 \\ -2 \end{array} \right)$
(8)$n(\tau_2)-m(\tau_2)= \left( \begin{array}{c} -2 \\ 2 \end{array} \right)$

### The chemical rate equation

The chemical rate equation says how the expected number of quantities changes with time. The equation is deterministic. The equation is a good approximation to quantities in a chemical reaction when the numbers of the reactants is large and any fluctuations in these numbers is negligible.

The general form of the rate equation is

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

We can write this a vector equation, and include all $i$ such equations, by using index free notation. John did this in Part 7. Doing this, we arrive at the rate equation for the full system as.

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

Let’s take a closer look. The quantities $X_i$ are concentrations of chemical reactants. The $\frac{d}{d t}$ is the time derivative so the left hand side is saying that we want to calculate the rate of change of the reactants. That’s easy, we already knew that you’re thinking right? But what about the right hand side?

The law of mass action states that the rate of change of a concentration is proportional to the stoichiometric coefficient of the corresponding reactants.

• $r(\tau)$ is a real number that gives a strength of the reaction. This is only part of the proportionality we can expect from mass action kinetics.

• $X^{m(\tau)}$ gives precisely the power term that the law of mass action says we should expect.

• The quantity $[n_i(\tau)-m_i(\tau)]$ or $[n(\tau)-m(\tau)]$ gives the difference between the number of molecules or whatever going in ($m$) and the number going out ($n$). This combined with the $r(\tau)$ gives the full proportionality constant, as expected in the law of mass action.

Now lets write the rate equation for the example at hand.

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 $r_1$ ($r_2$).

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

and

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

Now we will combine these equations, into a matrix equation, written as

(9)$\frac{d}{d t}\left( \begin{array}{c} X_1 \\ X_2 \end{array} \right) = 2\left( \begin{array}{cc} r_1 & - r_2 \\ -r_1 & r_2 \end{array} \right)\left( \begin{array}{c} X_1^3 \\ X_1X_2^2 \end{array} \right)$

So the rate of change of the quantities $X_1$ and $X_2$ are given in the vector on the left. The proportionality of the reaction is given by the $2 x 2$ matrix, containing $r_1$ and $r_2$. This matrix is a linear operator (as you’d expect from a constant of proportionality!) and it acts on a vector containing $X_1^3$ and $X_1X_2^2$ on the right. This vector on the right contains the quantities raised to the power of their stoichiometric coefficients. That’s it.

proportional to the

### The master equation

The master equation has been reviewed previously in this series. The general form of the master equation is

$\frac{d}{d t} \Psi = H \Psi$

and the general for of the Hamiltonian operators $H$ we are considering in the Petri net field theory series are given as

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

In our case, the Hamiltonian operator becomes

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

This equation is not an approximation, its exact. It governs all information we can know about the process, at the level of the individual species or molecules or whatever interacting.

### Prof 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

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

and

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

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

We now consider a wave function of the form

$\Psi := \frac{e^{b z_1}}{e^b}\frac{e^{c z_2}}{e^c}$
• (Puzzle). A question about probability in relation to the distribution here.

We then calculate $H\Psi$.

(10)$\frac{d}{d t} \Psi = H\Psi = \left(\begin{array}{cc} 1 & 1 \end{array} \right) \left( \begin{array}{cc} r_1cb^2 & - r_1cb^2 \\ -r_2c^3 & r_2c^3 \end{array} \right)\left( \begin{array}{c} z_1^3 \\ z_1z_2^2 \end{array} \right) \Psi$

We find solutions where this vanishes, for non-zero $\Psi$ as

$(r_1 c^2 b - r_2 b^3)z_1^3 + (r_2 b^3 - r_1 c^2 b)z_1 z_2^2 = 0$

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

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

In this case, and in most but not all of the cases we have considered, these equation are not algebraically independent. Their sum vanishes, so a solution to one implies a solution to the other. We arrive at

$\Psi(r_1, r_2, c) = \frac{\text{exp}[\sqrt{r_1 /r_2}c z_1]}{\text{exp}[\sqrt{r_1 /r_2}c]}\frac{\text{exp}[c z_2]}{\text{exp}[c]}$