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Experiments in multiple equilibrium states (Rev #5, changes)

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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 )

Prof 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