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Experiments in multiple equilibrium states (Rev #5)

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 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 \ge 0.

The chemical reaction network

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

2A+B3A 2 A + B \to 3 A
3A2A+B 3 A \to 2 A + B

The input and output functions then give.

m(τ 1)={2,1} m(\tau_1)=\{2,1\}
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)={2,1} n(\tau_2)=\{2,1\}

The Stochastic Petri Net

multiple equilibrium states

The chemical rate equation

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))

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


ddtX 2=r 1X 1 2X 2+r 2X 1 3 \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= 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, this becomes

H=r 1(a 1 a 1 a 1 a 1a 1a 2a 1 a 1 a 2 a 1a 1a 2)+r 2(a 1 a 1 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_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

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


ddtX 2=0=r 1X 1 2X 2+r 2X 1 3 \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 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

then its called the annihilating polynomial.

We consider a wave function of the form

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

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

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