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Reduction of the Stoichiometric Submodel

4. Transformation and Reduction 51

4.2. Model Reduction

4.2.1. Reduction of the Stoichiometric Submodel

Conservation relations and stoichiometric cycles are related to the left and right null space of the stoichiometric matrices, respectively. Consider the mole balances of a thermodynamic model

˙

c=N J+NeJe.

Letb ∈null([N, Ne]T), then for all flux vectorsJ and Je the conservation relation bT c= const holds. Conservation relations define an invariant manifold in the concentration space, and thus the system can be reduced to this manifold. The reduced system contains less components.

Letb ∈null(N), then the effect of a certain flux distributionJ on the change of concentrations

˙

c is indistinguishable from the effect of a flux distribution J +b. Thus, with respect to the concentrations c, the flux vector is redundant and the dynamics of the original system can be described by a reduced system with less fluxes.

4.2.1.1. Reduction of Conservation Relations

We transform the concentrations c by a partitioned transformation such that cˆ2 is a constant quantity (c˙ˆ2 = 0) and thuscˆ2(ˆµ1,µˆ2,µˆe) = ˆc2,0. We solve this algebraic equation in order to gain a function µˆ2(ˆµ1,µˆe). Finally, after applying this function to all occurrences of µˆ2 in the first subsystem, we consider the first subsystem independently from the second subsystem. Thus, we gain a reduced description of the original system. The following corollary discusses the details of this procedure.

Corollary 4.21 (Reduction of conservation relations). Let M be a thermodynamic model (see Definition 3.1, p. 40). LetTc = [Tc,1T , Tc,2T ]T be a square and invertible matrix withTc,2N = 0and Tc,2Ne = 0. Assume that the matrix Tc,2∂c/∂µ Tc,2T is invertible for all µ∈Ri0 and µe ∈Ri0,e. Let M˜ be a thermodynamic model with

N˜ =Tc,1N, ˜c(˜µ, µe) =Tc,1c(Tc,1T µ˜+Tc,2T µˆ2(˜µ, µe), µe), S˜=S, R(˜˜ µ, µe) =R(Tc,1T µ˜+Tc,2T µˆ2(˜µ, µe), µe), N˜e =Tc,1Ne, µ˜0 = invTΛ−1

c (Tc,10 where µˆ2(˜µ, µe) is a solution of

Tc,2c(Tc,1T µ˜+Tc,2T µˆ2, µe) = Tc,2c(µ0, µe,0)

and Λc is a symmetric, invertible matrix with Tc,1ΛcTc,2T = 0. Then, a solution of M can be reconstructed from a solution of the reduced system M˜ by

c=Tc−1

c˜ Tc,2c(µ0, µe,0)

, µ=Tc,1T µ˜+Tc,2T µˆ2(˜µ, µe),

J = ˜J , ∆µ= ∆˜µ.

4. Transformation and Reduction

Proof. TransformM withˆc=Tcc(cf. §4.7, p. 53). The first subsystem (subscript 1) corresponds to the reduced systemM˜. Observe that c˙ˆ2 = 0 and thus cˆ2 =Tc,2c(µ0, µe,0). From this we can compute a functionµˆ2(ˆµ1, µe)by solving the equationTc,2c(Tc,1T µˆ1+Tc,2T µˆ2, µe) =Tc,2c(µ0, µe,0). If the solution exists, it is locally unique because the matrix Tc,2∂c/∂µ Tc,2T is invertible. This follows from the Inverse Function Theorem. If Tc,2∂c/∂µ Tc,2T would be singular for all µ and µe, the model equations would not determine a locally unique trajectory forµˆ2. If the function ˆ

µ2(ˆµ1, µe) does not exist or is not globally unique, the trajectory ofM does not exist or is not unique, respectively. The possible non-uniqueness of the chemical potentials will be discussed in §4.44 (p. 66). With the function µˆ2(ˆµ1, µe) we get µ= Tc,1T µˆ1+Tc,2T µˆ2(ˆµ1, µe). We use this relation to derive the functions R(ˆˆ µ1, µe) and ˆc1(ˆµ1, µe). Observe that the first part of the equations (cˆ1, µˆ1) can now be considered independently of the second (ˆc2,µˆ2) and that cˆ2 and ˆ

µ2 can be reconstructed from ˆc1 and µˆ1. Thus, the subsystem Mˆ1 is a reduced version of Mˆ. Replacing the subscript1 with a tilde (M˜ = ˆM1) yields the expressions given in the corollary.

The initial value for the reduced chemical potentialsµˆ1,0 = ˜µ1,0 follows from the relations given in §4.8 (p. 53). The state variables of the original system can be reconstructed by using the inverse transformation matrixTc−1, ˆc2 =Tc,2c(µ0, µe,0)and the function µˆ2(˜µ, µe).

§ 4.22 (Symmetry and positive semi-definiteness of R˜ and ∂˜c/∂µ˜). The matricesR˜ and ∂˜c/∂µ˜ of the reduced model are symmetric and positive semi-definite.

Proof. The resistance matrices of the reduced system and of the original system are equal at corresponding state and input vectors: R(˜˜ µ, µe) = R(Tc,1T µ˜ +Tc,2T µˆ2(˜µ, µe)). Thus, R˜ is symmetric and positive semi-definite.

Differentiating the expression for˜c(˜µ, µe)in Corollary 4.21 yields for the derivative ofc(˜˜µ, µe) that ∂˜c/∂µ˜=Tc,1∂c/∂µ Tc,1T +Tc,1∂c/∂µ Tc,2T ∂µˆ2/∂µ˜. The total differential of the conservation relations in Corollary 4.21 isTc,2∂c/∂µ Tc,1T dµ˜+Tc,2∂c/∂µ Tc,2T dˆµ2 = 0. This leads to∂µˆ2/∂µ˜ =

−(Tc,2∂c/∂µ Tc,2T )−1Tc,2∂c/∂µ Tc,1T . Combining these results yields:

∂˜c/∂µ˜=Tc,1 (∂c/∂µ−∂c/∂µ Tc,2T (Tc,2∂c/∂µ Tc,2T )−1Tc,2∂c/∂µ)

| {z }

L

Tc,1T . Because the matrixL is symmetric, the derivative ∂˜c/∂µ˜ is symmetric.

IfLis positive semi-definite, the derivative∂c/∂˜ µ˜is positive semi-definite. However, from the expression forL given above it is not obvious if L is positive semi-definite. For this reason, an alternative expression forLthat allows to prove its positive semi-definiteness is derived. Assume initially that ∂c/∂µ is positive definite. Then, the matrices W = Tc,1Λc(∂c/∂µ)−1 and Z = [WT, Tc,2T ] can be defined. LetL0 be the matrix L0 = ΛcTc,1T (Tc,1Λc(∂c/∂µ)−1ΛcTc,1T )−1Tc,1Λc. The matricesL0 andL are equal becauseZ is a invertible matrix andL Z =L0Z = (ΛcTc,1T ,0). In this representation, it can be seen easily now that the matrix L0 = L is symmetric and positive definite. In the limit case where ∂c/∂µ is only positive semi-definite, the limit of L may also be only positive semi-definite.

4. Transformation and Reduction

§ 4.23 (Computation of Tc,1,Tc,2 and Λc). Let Y be a matrix with full rank and with span(Y) = \

µ∈Ri0eRi0,e

span ∂c

∂µ(µ, µe)

.

In all cases considered in this thesis, the matrix ∂c/∂µ is diagonal (e. g. in an ideal dilute solution) or the result of a linear transformation of a system with a diagonal matrix ∂c/∂µ. Then, the column space span(∂c/∂µ) is independent of µ and µe, and one may choose Y =

∂c/∂µ(µ, µe) with an arbitrary state vector µ and input vector µe. If ∂c/∂µ is invertible, then Y can be chosen to be the identity matrix (Y = I). If ∂c/∂µ is diagonal, then Y can be chosen to be a diagonal matrix with diagonal elements Yii = 1 if ∂ci/∂µi 6= 0 and Yii = 0 if ∂ci/∂µi = 0 for i = 1. . . i0. With the matrix Y defined above, the matrix YT ∂c/∂µ Y has always full rank. Using the matrix Y, the matrices Tc,1, Tc,2 and Λc with the properties demanded in Corollary 4.21 can be gained easily by computing kernel matrices: (1) Compute a kernel matrix X with X YT N = 0and X YTNe = 0. The choice Tc,2 =X YT guarantees that Tc,2N = 0, Tc,2Ne = 0 and that Tc,2∂c/∂µ Tc,2T =X YT ∂c/∂µ Y XT has full rank. (2) Choose an arbitrary symmetric and invertibleΛc, e. g.Λc=I. (3) ComputeTc,1as a right kernel matrix of Tc,2T Λc with Tc,2T ΛcTc,1 = 0.

§ 4.24 (Non-linear equation). The reduction involves the solution of the non-linear equation Tc,2c(µ0, µe,0) = Tc,2c(Tc,1T µˆ1+Tc,2T µˆ2, µe). Thus, the computational complexity of the reduction of conservation relations may be large.

§ 4.25 (Invariance of the entropy production). The entropy production of the internal fluxes J is invariant under the above described reduction method: σ[˜s] = ∆˜µTJ˜= ∆µT J =σ[s].

§ 4.26 (Invariance of the Gibbs energy). For dˆc2 = Tc,2dc = 0, the differential of the Gibbs energy is invariant under the above described reduction method: dg = µT dc = (dµ˜T Tc,1 + dˆµ2Tc,2)dc =dµ˜T d˜c. The information on the dependency of the Gibbs energy on the change of the concentration of the conserved moietydˆc2 =Tc,2dc is lost during the reduction process.

Example 4.27 (Reduction of a conservation relation). Consider a reactionAB with mass-action kinetics in an ideal dilute solution in a closed system:

N = −1

1

, c(µ) =

c exp((µA−µA)/(RT)) c exp((µB−µB)/(RT))

, R(µ) =ρ RA/(RT), µB/(RT)).

The system contains the conservation relationcA+cB = const. We choose the transformation matrices for the reduction as

Tc,1 = −1 1

, Tc,2 = 1 1

withΛc=I. The variablecˆ2 =cA+cBis the concentration of the conserved moiety. The variable ˆ

c1 = cB−cA describes the difference of the concentrations of B and A. Although ˆc1 may be

4. Transformation and Reduction

negative, it is a concentration in the sense of Definition 3.1 (p. 40). We have µA = −ˆµ1+ ˆµ2 and µB = ˆµ1 + ˆµ2. From the conservation relation we get the following dependency of µˆ1 and ˆ

µ2:

cA,0+cB,0 =c exp((−µˆ1+ ˆµ2−µA)/(RT)) +c exp((ˆµ1+ ˆµ2−µB)/(RT)).

Solving this equation forµˆ2 and replacing µ˜= ˆµ1 yields ˆ

µ2(˜µ) = RT log

cA,0 +cB,0

c · 1

exp((−µ˜−µA)/(RT)) + exp((˜µ−µB)/(RT))

. Substituting this expression into˜c= ˆc1 =−cA+cB with ci =c exp((µi−µi)/(RT))and into R=ρ RA/(RT), µB/(RT)) yields the transformed functions

˜

c(˜µ) = (cA+cB)·−exp((−˜µ−µA)/(RT)) + exp((+˜µ−µB)/(RT)) + exp((−˜µ−µA)/(RT)) + exp((+˜µ−µB)/(RT)) and

R(˜ˆ µ) = ρ R(−˜µ+ ˆµ2(˜µ),µ˜+ ˆµ2(˜µ)).

Thus, the original system with two compounds and simple functions c(µ) and R(µ) can be reduced to one with one compound but more complex functions ˜c(˜µ)and R(˜˜ µ).

4.2.1.2. Reduction of Stoichiometric Cycles

The reduction of stoichiometric cycles reduces the number of fluxes. It proceeds analogously to the reduction of conservation relations. While the reduction of conservation relations is based on the left null space of the stoichiometric matrices, the reduction of stoichiometric cycles uses the right null space.

We transform the fluxesJ by a partitioning transformation such that∆ˆµ2 = 0 and such that Jˆ2 vanishes from the mole balances.

Corollary 4.28 (Reduction of cycles). Let M be a thermodynamic model (see Definition 3.1, p. 40) and let TJ = [TJ,1, TJ,2] be a square and invertible matrix with N TJ,2 = 0 and S TJ,2 = 0.

Assume that the matrixTJ,2T R(µ, µe)TJ,2T is invertible for allµ∈Ri0 and µe ∈Ri0,e. Let ΛJ be a symmetric, invertible matrix with TJ,2T ΛJTJ,1 = 0. Then, a trajectory ofM can be reconstructed from a trajectory of the reduced system M˜ with

N˜ =N TJ,1, S˜=S TJ,1, N˜e =Ne, c(˜˜µ, µe) = c(˜µ, µe), µ˜00, R(˜˜ µ, µe) = (TJ,1T R(˜µ, µe)TJ,1)−(TJ,1T R(˜µ, µe)TJ,2) (TJ,2T R(˜µ, µe)TJ,2)−1 (TJ,2T R(˜µ, µe)TJ,1) where

c= ˜c, µ= ˜µ,

J = (TJ,1−TJ,2(TJ,2T R TJ,2)−1(TJ,2T R TJ,1)) ˜J , ∆µ= invΛ−1

J (TJ,1T ) ∆˜µ.

4. Transformation and Reduction

Proof. TransformM with J =TJJˆ(see §4.7, p. 53). Observe that∆ˆµ2 = 0. BecauseTJ,2T R TJ,2 is invertible, we can solve the equation TJ,2T R TJ,11 + TJ,2T R TJ,22 = ∆ˆµ2 = 0 for Jˆ2 and get Jˆ2 = −(TJ,2T R TJ,2)−1(TJ,2T R TJ,1) ˆJ1. If TJ,2T R TJ,2 would be singular, the model equation would not uniquely determine the fluxes Jˆ2. The possible non-uniqueness of the fluxes will be discussed in Corollary 4.71 (p. 76). Entering the Jˆ2 computed above into the equation TJ,1T R TJ,11+TJ,1T R TJ,22 = ∆ˆµ1 yields

(TJ,1T R(˜µ, µe)TJ,1)−(TJ,1T R(˜µ, µe)TJ,2) (TJ,2T R(˜µ, µe)TJ,2)−1 (TJ,2T R(˜µ, µe)TJ,1)

| {z }

R˜

1

|{z}J˜

= ∆ˆµ1

|{z}∆˜µ

. The formulas for the reconstruction of the original variables from the reduced ones follow from the expressions given in §4.8 (p. 53).

§ 4.29 (Symmetry and positive semi-definiteness of R˜ and ∂˜c/∂µ˜). The matricesR˜ and ∂˜c/∂µ˜ of the reduced model are symmetric and positive semi-definite.

Proof. The derivative∂˜c/∂µ˜is symmetric and positive semi-definite because ˜c(˜µ, µe) =c(˜µ, µe) and ∂c/∂µ is symmetric and positive semi-definite.

The symmetry of R˜ is obvious from the expression given in Corollary 4.28.

The resistance matrix R˜ resulting from the reduction of stoichiometric cycles and the deriva-tive∂˜c/∂µ˜resulting from the reduction of conservation relations (see §4.22, p. 59) have the same structure. For this reason, the proof of the positive semi-definiteness of R˜ is completely analog to the proof of the positive semi-definiteness of∂˜c/∂µ˜ in §4.22 (p. 59).

§ 4.30 (Computation of TJ,1, TJ,2 and ΛJ). Let Y be a matrix with full rank and with span(Y) = \

µ∈Ri0eRi0,e

span (R(µ, µe)).

In all cases considered in this thesis, the matrixR is diagonal or the result of a linear transfor-mation of a system with a diagonal matrixR. Then, the column space span(R)is independent of µ and µe, and one may choose Y = R(µ, µe) with an arbitrary state vector µ and input vector µe. If R is invertible, then Y can be chosen to be the identity matrix (Y = I). If R is diagonal, then Y can be chosen to be a diagonal matrix with diagonal elements Yii = 1 if Rii 6= 0 and Yii = 0 if Rii = 0 for i = 1. . . i0. With the matrix Y defined above, the matrix YT R Y has always full rank. Using the matrix Y, the matrices TJ,1, TJ,2 and ΛJ with the properties demanded in Corollary 4.28 can be gained easily by computing kernel matrices: (1) Compute a kernel matrixXwithN Y X = 0andS Y X = 0. The choiceTJ,2 =Y X guarantees that N TJ,2 = 0, S TJ,2 = 0 and that TJ,2T R TJ,2 =XT YT R Y X has full rank. (2) Choose an arbitrary invertible, symmetric ΛJ, e. g. ΛJ =I. (3) Compute TJ,1 as a right kernel matrix of TJ,2T ΛJ with TJ,2T ΛJTJ,1 = 0.

§ 4.31(Invariance of entropy production). The entropy production is invariant under the above described reduction method: σ[˜s] =σ[s].

4. Transformation and Reduction

Proof. Using the equations for the reconstruction of the original states given in Corollary 4.28 we get for σ[s] = ∆µT JT:

σ[s] = ∆˜µT (TJ,1T ΛJTJ,1)−1TJ,1T ΛJ(TJ,1−TJ,2(TJ,2T R TJ,2)−1(TJ,2T R TJ,1))

| {z }

I

J˜=σ[˜s].

§ 4.32(Invariance of the Gibbs energy). The differential of the Gibbs energy is invariant under the above described reduction method: dg=µT dc= ˜µT d˜c=d˜g.

Example 4.33. The network in Example 3.11 (Equation 3.2, p. 42) contains a stoichiometric cycle. The matrices

TJ,1 =

 1 0 0 1 0 0

, TJ,2 =

 1 1 1

, ΛJ =

+1 −1 0

−1 0 1

0 1 1

fulfill the conditions TJ,2T ΛJTJ,1 = 0 and N TJ,2 = 0. Here,TJ,1 is chosen such that J˜1 =J1 and J˜2 =J2. This leads to the given matrix ΛJ. A reduction with these matrices yields a system with

N˜ =

−1 0 1 −1

0 1

, R˜ = 1 R1 +R2+R3

R1(R2+R3) −R1R2

−R1R2 R2(R1+R3)

.

This system contains a minimal number of fluxes and corresponds to the system given in Equa-tion 3.3 (p. 42).