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.

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 Bx_A, x_B is an equilibrium solution of the rate equation, so is cx A,cx Bc x_A, c x_B for any c0c \geq 0.

Stoichiometry and Petri nets

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

H 2+OH 2O 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.

So if mass action kinetics holds true, the rate of the reaction forming H 2OH_2O above would be proportional to the product of the concentration of the reactants, [H 2][O][H_2][O]. These reactants each appear one time, if they appeared nn times for H 2H_2 and mm times for OO the reaction rate would be proportional to [H 2] n[O] m[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+2B3A A + 2 B \to 3 A
3AA+2B 3 A \to A + 2 B

We can represent these reaction rules using the following Petri net.

1-parameter family of equilibrium states

This was talked about in detail in several posts, starting in Part 2. Three reactants AA combine to produce one BB and one AA with proportionality constant r 2r_2. Likewise, two BB‘s combine with a single AA to produce three AA’s.

We can name these transformations as τ 1\tau_1 and τ 2\tau_2 so

τ 1:A+2B3A\tau_1: A + 2 B \to 3 A
τ 2:3AA+2B\tau_2: 3 A \to A + 2 B

which can alternatively be expressed in matrix notation as

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

τ 1:A+2B3A\tau_1: A + 2 B \to 3 A
τ 2:3AA+2B\tau_2: 3 A \to A + 2 B

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

In our case, we have two transitions τ 1\tau_1 and τ 2\tau_2. The input and output functions evaluate as follows.

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

and for τ 2\tau_2

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

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

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

and for τ 2\tau_2

(5)m(τ 2)=(3 0) m(\tau_2)= \left( \begin{array}{c} 3 \\ 0 \end{array} \right)
(6)n(τ 2)=(1 2) 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(τ 1)m(τ 1)=(2 2) n(\tau_1)-m(\tau_1)= \left( \begin{array}{c} 2 \\ -2 \end{array} \right)
(8)n(τ 2)m(τ 2)=(2 2) n(\tau_2)-m(\tau_2)= \left( \begin{array}{c} -2 \\ 2 \end{array} \right)

The Stochastic Petri Net

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

ddtX i= τTr(τ)X m(τ)(n i(τ)m i(τ)) \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 ii 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.

ddtX= τTr(τ)X m(τ)(n(τ)m(τ)) \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 iX_i are concentrations of chemical reactants. The ddt\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.

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

The population of species AA (BB) will be denoted X 1(t)X_1(t) (X 2(t)X_2(t)) and the rate of the reaction τ 1\tau_1 (τ 2\tau_2) as r 1r_1 (r 2r_2).

ddtX 1=2r 1X 1X 2 22r 2X 1 3 \frac{d}{d t}X_1 = 2 r_1 X_1 X_2^2 - 2 r_2 X_1^3


ddtX 2=2r 1X 1X 2 2+2r 2X 1 3 \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)ddt(X 1 X 2)=2(r 1 r 2 r 1 r 2)(X 1 3 X 1X 2 2) \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 1X_1 and X 2X_2 are given in the vector on the left. The proportionality of the reaction is given by the 2x22 x 2 matrix, containing r 1r_1 and r 2r_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 3X_1^3 and X 1X 2 2X_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

ddtΨ=HΨ \frac{d}{d t} \Psi = H \Psi

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

H= τTr(τ)(a n(τ)a m(τ)a m(τ)a m(τ)) 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 a 1 a 1 a 1a 2a 2a 1 a 2 a 2 a 1a 2a 2)+r 2(a 1 a 2 a 2 a 1a 1a 1a 1 a 1 a 1 a 1a 1a 1) 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

ddtX 1=0=2r 1X 1X 2 22r 2X 1 3 \frac{d}{d t}X_1 = 0 = 2 r_1 X_1 X_2^2 - 2 r_2 X_1^3


ddtX 2=0=2r 1X 1X 2 2+2r 2X 1 3 \frac{d}{d t} X_2 = 0 = - 2 r_1 X_1 X_2^2 + 2 r_2 X_1^3
f 1,...,f m f_1,...,f_m

be polynomials in

F[x 1,...,x n] F[x_1,...,x_n]

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

AF[y 1,...,y m] A \in F[y_1,...,y_m]

such that

A(f 1,...,f m)=0 A(f_1,...,f_m)=0

We now consider a wave function of the form

Ψ:=e bz 1e be cz 2e c \Psi := \frac{e^{b z_1}}{e^b}\frac{e^{c z_2}}{e^c}

We then calculate HΨH\Psi.

(10)ddtΨ=HΨ=(1 1)(r 1cb 2 r 1cb 2 r 2c 3 r 2c 3)(z 1 3 z 1z 2 2)Ψ \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 Ψ\Psi as

(r 1c 2br 2b 3)z 1 3+(r 2b 3r 1c 2b)z 1z 2 2=0 (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 1c 2br 2b 3=0 r_1 c^2 b - r_2 b^3 = 0
r 2b 3r 1c 2b=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

Ψ(r 1,r 2,c)=exp[r 1/r 2cz 1]exp[r 1/r 2c]exp[cz 2]exp[c] \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]}