QSSA and Solvability Mark Sweeney University of Rochester July 18, 2017
Chemical Reaction Networks A CRN is described by three sets: ◮ species, S ◮ complexes, C ⊆ R S ≥ 0 (or Z S ≥ 0 ) ◮ reactions, R ⊆ C × C From these, we get a system of (first order) differential equations 2 / 37
CRN Example k 1 k 2 − − ⇀ E + S k –1 E · S − − → E + P ↽ − − S = { E , S , P , E · S } C = { E + S , E · S , E + P } R = { ( c 1 , c 2 ) , ( c 2 , c 1 ) , ( c 2 , c 3 ) } 3 / 37
CRN Example k 1 k 2 − − ⇀ E + S k –1 E · S − − → E + P ↽ − − d [ E ] = − k 1 [ E ][ S ] + k − 1 [ E · S ] + k 2 [ E · S ] dt d [ S ] = − k 1 [ E ][ S ] + k − 1 [ E · S ] dt d [ E · S ] = k 1 [ E ][ S ] − k − 1 [ E · S ] − k 2 [ E · S ] dt d [ P ] = k 2 [ E · S ] dt 4 / 37
QSSA Method ◮ Reduce to a model with fewer ODEs ◮ Quasi-steady-state-assumption (QSSA) simplifies the system by assuming some components do not accumulate ◮ Eliminates some intermediates by replacing ODEs with algebraic constraints 5 / 37
QSSA Example k 1 k 2 − − ⇀ E + S k –1 E · S − − → E + P ↽ − − d [ E ] = − k 1 [ E ][ S ] + k − 1 [ E · S ] + k 2 [ E · S ] dt d [ S ] = − k 1 [ E ][ S ] + k − 1 [ E · S ] dt d [ E · S ] = k 1 [ E ][ S ] − k − 1 [ E · S ] − k 2 [ E · S ] = 0 dt d [ P ] = k 2 [ E · S ] dt 6 / 37
QSSA Example 0 = k 1 [ E ][ S ] − k − 1 [ E · S ] − k 2 [ E · S ] ( k − 1 + k 2 )[ E · S ] = k 1 [ E ][ S ] [ E · S ] = k 1 [ E ][ S ] k − 1 + k 2 7 / 37
Galois Theory ◮ If L / k is a normal, separable extension of fields, the automorphisms of L over k form a group G (the Galois group) ◮ G is solvable if (and only if) each α ∈ L can be expressed in terms of elements of k , roots of unity, radicals, and + , − , × , ÷ ◮ Rules out a “quadratic formula” for polynomials with degree 5 or higher 8 / 37
Galois Theory Examples solvable: x 2 − 2 ← → Z / 2 Z x 4 − 5 ← → D 8 insolvable: x 5 − 3 x 2 + 1 ← → S 5 ( k = Q ) 9 / 37
QSSA & Galois Theory ◮ Work over k = Q ( k i , c j , ... ); adjoin all relevant constants QSSA ⇔ systems of polynomials ⇔ ideals in k [ x 1 , ..., x n ] ◮ Examples exist which reduce to insoluble univariate polynomials (over k ) 10 / 37
Main Questions Under what circumstances will QSSA work? When will it fail? 1. classes of networks 2. structural properties 3. small counterexamples 4. subnetworks/extensions 11 / 37
What does “possible” mean? Many different ways of framing QSSA: ◮ Finitely many solutions ◮ Solutions expressible in radicals 12 / 37
What does “possible” mean? Many different ways of framing QSSA: ◮ Finitely many solutions ◮ Solutions expressible in radicals ◆ Nondegenerate solutions ◆ Real solutions ◆ Positive solutions 12 / 37
Algebra Preliminaries Fix ideals I , J ⊆ k [ x 1 , ..., x n ] ◮ the variety , V ( I ) = { zeros of I in k n } ◮ similarly, V a ( I ) = { zeros of I in ( k a ) n } ◮ a Gr¨ obner basis of I : generalization of Gaussian Elimination ◮ the ideal quotient , I : J , which generalizes division 13 / 37
Reduction to Univariate Case Lemma Let I be an ideal in k [ x 1 , ..., x n ] . Then V a ( I ) is finite if and only if each intersection I ∩ k [ x i ] is nonzero. Almost always the case when using QSSA 14 / 37
Computing Intersections Lemma Let I be an ideal in k [ x 1 , ..., x n ] with Gr¨ obner basis G w.r.t. x 1 > x 2 > ... > x n Then G ∩ k [ x n ] generates I ∩ k [ x n ] . For reduced GBs, there is a unique generator 15 / 37
Checking Solvability ◮ Together, these suggest an algorithm: 1. Find the generators of I ∩ k [ x i ] 2. Compute their Galois groups 3. Check for solvability ◮ If all the generators are solvable, V ( I ) has solvable entries in every coordinate 16 / 37
A Simple Case Lemma Fix I ⊆ k [ x , y ] , k algebraically closed. If there exist f 1 , f 2 ∈ I such that f 1 is irreducible and f 2 �∈ � f 1 � , then V ( I ) is finite. Lemma Let I = � f 1 , ..., f n � and deg( f i ) = d i . If V ( I ) is finite, then deg( g ) ≤ d 1 d 2 ... d n , where I ∩ k [ x i ] = � g � . 17 / 37
A Simple Case ◮ S 4 is solvable ◮ if deg( f ) = n , Gal(f/k) embeds in S n Proposition (S.) If a CRN has at-most-bimolecular kinetics and we choose two “chemically reasonable” intermediates, QSSA is always possible. 18 / 37
Example k 2 k 1 − − ⇀ A − − → 2X k –2 2Y ↽ − − k 3 X + Y − − → B dx dt = 0 = − 2 k 2 x 2 − k 3 xy + 2 k − 2 y + ak 1 dy dt = 0 = − 2 k − 2 y 2 − k 3 xy + k 2 x 2 19 / 37
Example After computing a Gr¨ obner basis, we get 3 ) x 4 + (8 k − 2 k 2 k 3 ) x 3 f ( x ) =(8 k − 2 k 2 2 − 3 k 2 k 2 + ( − 8 ak − 2 k 1 k 2 + ak 1 k 2 3 − 4 k 2 − 2 k 2 ) x 2 − (2 k − 2 k 1 ak 3 ) x + (2 a 2 k − 2 k 2 1 ) ◮ Gal( f / k ) is isomorphic to D 8 ◮ For y , we obtain D 8 as well 20 / 37
Extending Solvability ◮ The proposition describes some common systems, but is limited ◮ In some circumstances solvability can be extended: 1. “treelike” mechanisms 2. nondegenerate and/or physically achievable 21 / 37
Oriented Species-Reaction Graph 2 A 1 X 2 2 2 -2 3 B 2 2 Y 22 / 37
QSSA OSR Graph 2 1 X 2 2 -2 2 3 2 2 Y 23 / 37
Extending Solvability Theorem (S.) Given a QOSR graph H and intermediates Q , QSSA is possible when there exists an equivalence relation ∼ on H such that H / ∼ has no directed cycles and QSSA is possible on each equivalence class in Q / ∼ under particular kinds of substitution 24 / 37
Extending Solvability Corollary (S.) If we use Proposition 1 to prove solvability for the previous theorem, QSSA is possible for the nondegenerate achievable steady states. 25 / 37
Pantea et al.: “Counterexample” -3 -5 k 1 − − ⇀ 2 Y k –1 2 B ↽ − − 2 2 k 2 3 5 X Z Y + B − − → Z + A 2 2 k 3 − − ⇀ Z + B k –3 2 X ↽ − − 4 2 k 4 A + X − − → Y + B 2 2 k 5 -1 1 − − ⇀ Y 2 Z k –5 2 A ↽ − − 26 / 37
Pantea et al.: “Counterexample” -3 -5 k 1 − − ⇀ 2 Y k –1 2 B ↽ − − 2 2 k 2 3 5 X Z Y + B − − → Z + A 2 2 k 3 − − ⇀ Z + B k –3 2 X ↽ − − 4 2 k 4 A + X − − → Y + B 2 2 k 5 − − ⇀ -1 Y 1 2 Z k –5 2 A ↽ − − Remove reaction − 5 as well 27 / 37
Modified Pantea Mechanism -3 2 X 3 Z 5 2 2 4 2 2 2 -1 Y 1
Modified Pantea Mechanism -3 2 Q 1 = { X , Z } X 3 Z 5 2 2 4 2 2 2 -1 Y 1
Modified Pantea Mechanism -3 2 Q 1 = { X , Z } X 3 Z 5 2 2 4 2 Q 2 = { Y } 2 2 -1 Y 1 28 / 37
Modified Pantea Mechanism Q 1 = { X , Z } and Q 2 = { Y } Φ x = − 2 k − 3 x 2 − k 4 ax + 2 k 3 bz Φ y = − 2 k 1 y 2 − k 2 by + 2 k − 1 b 2 + k 4 ax Φ z = − 2 k 5 z 2 − k 3 bz + k − 3 x 2 x ← → S 3 or { e } y ← → S 4 × Z 2 or Z 2 z ← → S 3 or { e } 29 / 37
Modified Pantea Mechanism ◮ Multiple Galois groups arise when a polynomial is reducible ◮ In this case, { e } and Z 2 correspond to degenerate solutions ( x = 0 or z = 0) ◮ These are irrelevant for actual chemistry, so we would like to remove them 30 / 37
Modified Pantea Mechanism ◮ If we want to remove the zeros of an ideal J from another ideal I , we take their saturation : ∞ I : J ∞ = � I : J m m =1 ◮ Similar to performing division 31 / 37
Modified Pantea Mechanism ◮ To encode nondegeneracy we want to cut out x = 0 or y = 0 or z = 0 ◮ Which is summarized by J = � xyz � ◮ The ideal we want: I ′ Q = I Q : J ∞ 32 / 37
Modified Pantea Mechanism ◮ After performing the same steps to find the Galois groups: x ← → S 3 y ← → S 4 × Z 2 z ← → S 3 33 / 37
Saturation ◮ Saturation is not immediately useful: it is easy to ignore a few solutions, but... Conjecture Corollary 1 only requires nondegeneracy (i.e. imaginary or negative concentrations are permissible) 34 / 37
Saturation ◮ Saturation removes the (infinitely many) degenerate solutions ahead of time ◮ This may (not) simplify computations ◮ Almost all “counterexamples” in CRNs lie at boundaries, so saturation may help generalize some of these results 35 / 37
Future Directions ◮ More (general) finiteness criteria ◮ More solvability criteria ◮ CRN structure ⇔ Galois group ◮ Weakening QSSA to nondegenerate and/or achievable concentrations 36 / 37
Acknowledgements Many thanks to: ◮ Dr. Anne Shiu, Ola Sobieska, Nida Obatake, Jonathan Tyler ◮ Texas A&M University ◮ The National Science Foundation 37 / 37
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