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Experiments in trimolecular reactions (changes)

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A three concentration example

For the general terms and definitions used in this project see

Here we will consider a three trimolecular concentration reaction example from Feinberg page 2253.

Equation 6.7 gives the following chemical reaction network.

r 1:A 1→2A 1 r_1: A_1 \rightarrow 2 A_1
r 2:2A 1→A 1 r_2: 2 A_1 \rightarrow A_1
r 3:A 1+A 2→A 3 r_3: A_1 + A_2 \rightarrow A_3
r 4:A 3→A 1+A 2 r_4: A_3 \rightarrow A_1 + A_2
r 5:A 3→2A 2 r_5: A_3 \rightarrow 2 A_2
r 6:2A 2→A 3 r_6: 2 A_2 \rightarrow A_3

The stochastic Petri net

The rate equation

ddtX 1=r 1X 1−r 3X 1X 2−r 2X 1 2+r 4X 3 \frac{d}{d t} X_1 = r_1 X_1 - r_3 X_1 X_2 - r_2 X_1^2 + r_4 X_3
ddtX 2=−r 3X 1X 2+(r 4+2r 5)X 3−2r 6X 2 2 \frac{d}{d t} X_2 = -r_3 X_1 X_2 + (r_4+2 r_5) X_3 - 2 r_6 X_2^2
ddtX 3=r 3X 1X 2+r 6X 2 2−(r 4+r 5)X 3 \frac{d}{d t} X_3 = r_3 X_1 X_2 + r_6 X_2^2 - (r_4+r_5) X_3

The solutions of this equation are given as.

X 1=r 1r 2 X_1 = \frac{r_1}{r_2}
X 2=r 1r 3r 5r 2r 4r 6 X_2 = \frac{r_1 r_3 r_5}{r_2 r_4 r_6}
X 3=r 1 2r 3 2r 5r 2 2r 4 2r 6 X_3 = \frac{r_1^2 r_3^2 r_5}{r_2^2 r_4^2 r_6}

These concentration levels make the rate equations zero for all time.

The master equation

H=r 1(a 1 †a 1 †a 1−a 1 †a 1)+r 2(a 1 †a 1a 1−a 1 †a 1 †a 1a 1)+r 3(a 3 †a 1a 2−a 1 †a 2 †a 1a 2)+r 4(a 1 †a 2 †a 3−a 3 †a 3)+r 5(a 2 †a 2 †a 3−a 3 †a 3)+r 6(a 3 †a 2a 2−a 2 †a 2 †a 2a 2) H = r_1(a_1^\dagger a_1^\dagger a_1 - a_1^\dagger a_1) + r_2(a_1^\dagger a_1 a_1 - a_1^\dagger a_1^\dagger a_1 a_1)+ r_3(a_3^\dagger a_1 a_2 - a_1^\dagger a_2^\dagger a_1 a_2) + r_4(a_1^\dagger a_2^\dagger a_3 - a_3^\dagger a_3) + r_5(a_2^\dagger a_2^\dagger a_3 - a_3^\dagger a_3) + r_6(a_3^\dagger a_2 a_2 - a_2^\dagger a_2^\dagger a_2 a_2)

We let this Hamiltonian act on Ψ:=e c 1z 1e c 2z 2e c 3z 3\Psi := e^{c_1 z_1}e^{c_2 z_2}e^{c_3 z_3} and to satisfy HΨ=0H\Psi = 0 there must exist a choice of c 1c_1, c 2c_2 and c 3c_3 that will cause all the following equations to vanish together.

z 1;0=−r 1c 1+r 2c 1 2 z_1; 0 = -r_1c_1 + r_2 c_1^2
z 1 2;0=r 1c 1−r 2c 1 2 z_1^2; 0 = r_1 c_1 - r_2 c_1^2
z 3;0=r 3c 1c 2−(r 4+r 5)c 3+r 6c 2 2 z_3; 0 = r_3 c_1 c_2 - (r_4 +r_5) c_3 + r_6 c_2^2
z 2 2;0=−r 6c 2 2+r 5c 3 z_2^2; 0 = -r_6 c_2^2 + r_5 c_3
z 1z 2;0=r 4c 3−r 3c 1c 2 z_1 z_2; 0 = r_4 c_3 - r_3 c_1 c_2

By picking c 1=X 1c_1 = X_1, c 2=X 2c_2 = X_2 and c 3=X 3c_3 = X_3. This example has the following Petri net.

category: experiments, mathematical methods

Revised on September 20, 2012 13:01:50 by John Baez (137.132.3.9)
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