ON THE EXISTENCE OF POSITIVE STEADY STATES
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1 ON THE EXISTENCE OF POSITIVE STEADY STATES FOR DEFICIENCY-ONE MASS ACTION SYSTEMS WITH TWO LINKAGE CLASSES Balázs Boros Institute of Mathematics Eötvös Loránd University Budapest, Hungary AMS Central Fall Sectional Meeting Recent Developments in the Theory and Applications of Reaction Network Models Chicago, IL, October 2-4, 2015 BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
2 CHEMICAL REACTION NETWORKS X = {X 1, X 2,..., X n } - set of species C - set of complexes Y ({0} [1, )) n C - matrix of complexes R - set of reactions (X, C, R) - chemical reaction network l - number of linkage classes t - number of terminal strong linkage classes BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
3 CHEMICAL REACTION NETWORKS X = {X 1, X 2,..., X n } - set of species C - set of complexes Y ({0} [1, )) n C - matrix of complexes R - set of reactions (X, C, R) - chemical reaction network l - number of linkage classes t - number of terminal strong linkage classes BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
4 CHEMICAL REACTION NETWORKS X = {X 1, X 2,..., X n } - set of species C - set of complexes Y ({0} [1, )) n C - matrix of complexes R - set of reactions (X, C, R) - chemical reaction network l - number of linkage classes t - number of terminal strong linkage classes BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
5 CHEMICAL REACTION NETWORKS X = {X 1, X 2,..., X n } - set of species C - set of complexes Y ({0} [1, )) n C - matrix of complexes R - set of reactions (X, C, R) - chemical reaction network l - number of linkage classes t - number of terminal strong linkage classes BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
6 CHEMICAL REACTION NETWORKS X = {X 1, X 2,..., X n } - set of species C - set of complexes Y ({0} [1, )) n C - matrix of complexes R - set of reactions (X, C, R) - chemical reaction network l - number of linkage classes t - number of terminal strong linkage classes BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
7 CHEMICAL REACTION NETWORKS X = {X 1, X 2,..., X n } - set of species C - set of complexes Y ({0} [1, )) n C - matrix of complexes R - set of reactions (X, C, R) - chemical reaction network l - number of linkage classes t - number of terminal strong linkage classes BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
8 CHEMICAL REACTION NETWORKS X = {X 1, X 2,..., X n } - set of species C - set of complexes Y ({0} [1, )) n C - matrix of complexes R - set of reactions (X, C, R) - chemical reaction network l - number of linkage classes t - number of terminal strong linkage classes BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
9 MASS ACTION SYSTEMS κ = (κ ij ) (i,j) R R R + - rate coefficients (X, C, R, κ) - mass action system A κ = κ 1,1 κ C, κ 1, C κ C, C C i=1 κ 1,i C i=1 κ C,i x Y = n s=1 x y s,1 s. n s=1 x y s, C s ẋ(τ) = Y A κ x(τ) Y, state space: R n 0 BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
10 MASS ACTION SYSTEMS κ = (κ ij ) (i,j) R R R + - rate coefficients (X, C, R, κ) - mass action system A κ = κ 1,1 κ C, κ 1, C κ C, C C i=1 κ 1,i C i=1 κ C,i x Y = n s=1 x y s,1 s. n s=1 x y s, C s ẋ(τ) = Y A κ x(τ) Y, state space: R n 0 BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
11 MASS ACTION SYSTEMS κ = (κ ij ) (i,j) R R R + - rate coefficients (X, C, R, κ) - mass action system A κ = κ 1,1 κ C, κ 1, C κ C, C C i=1 κ 1,i C i=1 κ C,i x Y = n s=1 x y s,1 s. n s=1 x y s, C s ẋ(τ) = Y A κ x(τ) Y, state space: R n 0 BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
12 MASS ACTION SYSTEMS κ = (κ ij ) (i,j) R R R + - rate coefficients (X, C, R, κ) - mass action system A κ = κ 1,1 κ C, κ 1, C κ C, C C i=1 κ 1,i C i=1 κ C,i x Y = n s=1 x y s,1 s. n s=1 x y s, C s ẋ(τ) = Y A κ x(τ) Y, state space: R n 0 BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
13 MASS ACTION SYSTEMS κ = (κ ij ) (i,j) R R R + - rate coefficients (X, C, R, κ) - mass action system A κ = κ 1,1 κ C, κ 1, C κ C, C C i=1 κ 1,i C i=1 κ C,i x Y = n s=1 x y s,1 s. n s=1 x y s, C s ẋ(τ) = Y A κ x(τ) Y, state space: R n 0 BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
14 δ, E κ +, AND S ẋ(τ) = Y A κ x(τ) Y DEFINITION (DEFICIENCY) δ = dim(ker Y ran A κ ) (provided that l = t) E κ + = {x R n + Y A κ x Y = 0} - positive steady states S = span{y j Y i (i, j) R} R n - stoichiometric subspace (p + S) R n 0 for p Rn + - positive stoichiometric classes BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
15 δ, E κ +, AND S ẋ(τ) = Y A κ x(τ) Y DEFINITION (DEFICIENCY) δ = dim(ker Y ran A κ ) (provided that l = t) E κ + = {x R n + Y A κ x Y = 0} - positive steady states S = span{y j Y i (i, j) R} R n - stoichiometric subspace (p + S) R n 0 for p Rn + - positive stoichiometric classes BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
16 δ, E κ +, AND S ẋ(τ) = Y A κ x(τ) Y DEFINITION (DEFICIENCY) δ = dim(ker Y ran A κ ) (provided that l = t) E κ + = {x R n + Y A κ x Y = 0} - positive steady states S = span{y j Y i (i, j) R} R n - stoichiometric subspace (p + S) R n 0 for p Rn + - positive stoichiometric classes BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
17 NECESSARY CONDITION FOR THE EXISTENCE OF A POSITIVE STEADY STATE Recall Thus, implies E κ + = {x R n + Y A κ x Y = 0}. E κ + there exists y R C + with y ker(y A κ). BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
18 WEAKLY REVERSIBLE (WR) NETWORKS THEOREM WR implies the existence of y R C + with y ker A κ. CONJECTURE (DENG, FEINBERG, JONES, NACHMAN) WR implies the non-emptiness of (p + S) E κ + for all p R n +. THEOREM (BB, 2013) WR and δ = 1 together imply the non-emptiness of (p + S) E κ + for all p R n +. BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
19 NON-WR NETWORKS THEOREM Assume δ = 1, t = l = 1, and non-wr. Then (Feinberg, 1979, 1987, 1995) E κ + implies (p + S) E κ + = 1 for all p R n + and (BB, 2010, 2012) E κ + if and only if there exists y R C + with y ker(y A κ). BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
20 DEFICIENCY-ONE THEOREM THEOREM Assume (I) δ r = 0 or 1 ( r 1, l), (II) δ 1 + δ δ l = δ, and (III) l = t. Then (Feinberg, 1979, 1987, 1995) WR implies E κ +, (Feinberg, 1979, 1987, 1995) E κ + implies (p + S) E κ + = 1 for all p R n +, and (BB, 2010, 2012) E κ + if and only if there exists y R C + with y ker(y A κ). BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
21 EXAMPLE: E κ +, BUT THERE EXISTS A p R n + WITH (p + S) E κ + = X 1 κ 12 X 2 2X 1 + X 2 κ 34 3X 1 } E+ κ = {x R 2 + x 1 x 2 = κ 12 /κ 34 } and E0 {x κ = bd(r 2 +) x 1 = 0 BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
22 QUESTIONS Assume l = t, not all the linkage classes are WR, δ 1 = = δ l = 0, δ = 1, and there exists y R C + with y ker(y A κ). Do we have E κ +? Do we have (p + S) E κ + for all p R n +? (Under some extra assumption.) BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
23 EXAMPLE: AN ENVZ-OMPR MODEL IN WHICH ATP (RESP. ADP) IS THE COFACTOR IN PHOSPHO-OMPR DEPHOSPHORYLATION BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
24 EXAMPLE: IDHKP-IDH SYSTEM (PHOSPHORYLATION OF THE ENZYME ISOCITRATE DEHYDROGENASE BY A BIFUNCTIONAL KINASE-PHOSPHATASE) BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
25 C (1) C (1), C (2) C (2) Assume for the rest of this talk l = t = 2 and both of the two linkage classes are non-wr, δ 1 = δ 2 = 0 and δ = 1, and there exists y R C + with y ker(y A κ). Let 0 h ker Y ran A κ with h(c (1)) > 0 and h(c (2)) > 0. [ ] [ y(1) y Let y = and y y(2) = ] (1) y be such that (2) A κ y = 0 and A κ y = h and y and y have appropriate sign structure. Let Ŷ be the augmented matrix of complexes, i.e., Y (1) Y (2) Ŷ = R (n+2) ( C(1) + C(2) ) BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
26 C (1) C (1), C (2) C (2) Assume for the rest of this talk l = t = 2 and both of the two linkage classes are non-wr, δ 1 = δ 2 = 0 and δ = 1, and there exists y R C + with y ker(y A κ). Let 0 h ker Y ran A κ with h(c (1)) > 0 and h(c (2)) > 0. [ ] [ y(1) y Let y = and y y(2) = ] (1) y be such that (2) A κ y = 0 and A κ y = h and y and y have appropriate sign structure. Let Ŷ be the augmented matrix of complexes, i.e., Y (1) Y (2) Ŷ = R (n+2) ( C(1) + C(2) ) BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
27 C (1) C (1), C (2) C (2) Assume for the rest of this talk l = t = 2 and both of the two linkage classes are non-wr, δ 1 = δ 2 = 0 and δ = 1, and there exists y R C + with y ker(y A κ). Let 0 h ker Y ran A κ with h(c (1)) > 0 and h(c (2)) > 0. [ ] [ y(1) y Let y = and y y(2) = ] (1) y be such that (2) A κ y = 0 and A κ y = h and y and y have appropriate sign structure. Let Ŷ be the augmented matrix of complexes, i.e., Y (1) Y (2) Ŷ = R (n+2) ( C(1) + C(2) ) BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
28 C (1) C (1), C (2) C (2) Assume for the rest of this talk l = t = 2 and both of the two linkage classes are non-wr, δ 1 = δ 2 = 0 and δ = 1, and there exists y R C + with y ker(y A κ). Let 0 h ker Y ran A κ with h(c (1)) > 0 and h(c (2)) > 0. [ ] [ y(1) y Let y = and y y(2) = ] (1) y be such that (2) A κ y = 0 and A κ y = h and y and y have appropriate sign structure. Let Ŷ be the augmented matrix of complexes, i.e., Y (1) Y (2) Ŷ = R (n+2) ( C(1) + C(2) ) BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
29 C (1) C (1), C (2) C (2), SUFFICIENT CONDITION FOR E κ + Then x R n + is in E+ κ (i.e., Y A κ x Y = 0) if and only if there exist β 1, β 2, γ 1, and γ 2 > 0 with log(x) [ Ŷ log(γ 1 ) log(β1 y = ] (1) + y(1)) log(β log(γ 2 ) 2 y and (2) + y(2)) β 1 γ 1 = β 2 γ 2. THEOREM (BB) Assume further that [0,..., 0, 1, 1] / ran Ŷ. Then E κ +. BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
30 C (1) C (1), C (2) C (2), SUFFICIENT CONDITION FOR E κ + Then x R n + is in E+ κ (i.e., Y A κ x Y = 0) if and only if there exist β 1, β 2, γ 1, and γ 2 > 0 with log(x) [ Ŷ log(γ 1 ) log(β1 y = ] (1) + y(1)) log(β log(γ 2 ) 2 y and (2) + y(2)) β 1 γ 1 = β 2 γ 2. THEOREM (BB) Assume further that [0,..., 0, 1, 1] / ran Ŷ. Then E κ +. BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
31 C (1) C (1), C (2) C (2), SUFFICIENT CONDITION FOR E κ +, TWO APPLICATIONS The theorem on the previous slide applies both for X 1 κ 12 X 2 2X 1 + X 2 κ 34 3X 1 and BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
32 C (1) C (1), C (2) C (2), SUFFICIENT CONDITION FOR (p + S) E κ + FOR ALL p R n + THEOREM (BB) Assume that Ŷ ε = [0,..., 0, 1, 1] and ε(c(2), y > 0) h(c(1), y > 0) ε(c(1), y = 0) h(c(2), y = 0), ε(c(2), y = 0) h(c(1), y = 0) ε(c(1), y > 0) h(c(2), y > 0). (1) Then (p + S) E κ + for all p R n +. BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
33 ε 5 h 1 >(ε 2 + ε 3 ) h 4 BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28 AN EXAMPLE WITH (1) NOT SATISFIED κ κ X 1 κ 45 3X 1 + 3X 2 3X 1 + X 2 2X 1 + 2X 2 X 2 κ / Ŷ = , h = 0 1 1, ε = 3/4 5/4 0, κ κ 12 κ 23 κ 32 1 κ A κ = 0 κ 23 κ 32 κ 45 0, y = 23 1 κ 320, y = κ κ 21 1 κ κ 450 ε 5 (h 1 + h 2 )<ε 3 h 4
34 AN EXAMPLE WITH (1) NOT SATISFIED, CASE E κ + = 0 Analysis of the ODE gives: for κ 32 κ 4 45 > 4 27 κ3 23 κ2 12 we have E κ + = 0. FIGURE : κ 12 = 1, κ 23 = 1, κ 32 = 1, κ 45 = 1 BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
35 AN EXAMPLE WITH (1) NOT SATISFIED, CASE E κ + = 1 Analysis of the ODE gives: for κ 32 κ 4 45 = 4 27 κ3 23 κ2 12 we have E κ + = 1. FIGURE : κ 12 = 3 3 2, κ 23 = 1, κ 32 = 1, κ 45 = 1 BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
36 AN EXAMPLE WITH (1) NOT SATISFIED, CASE E κ + = 2 Analysis of the ODE gives: for κ 32 κ 4 45 < 4 27 κ3 23 κ2 12 we have E κ + = 2. FIGURE : κ 12 = 3, κ 23 = 1, κ 32 = 1, κ 45 = 1 BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
37 C (1) C (1), C (2) C (2), SUFFICIENT CONDITION FOR (p + S) E κ + FOR ALL p R n + THEOREM (BB, REPEATED FROM SLIDE 16) Assume that Ŷ ε = [0,..., 0, 1, 1] and ε(c(2), y > 0) h(c(1), y > 0) ε(c(1), y = 0) h(c(2), y = 0), ε(c(2), y = 0) h(c(1), y = 0) ε(c(1), y > 0) h(c(2), y > 0). Then (p + S) E κ + for all p R n +. COROLLARY (BB) Assume that C = {i C y i > 0} {i C y i > 0}, Ŷ ε = [0,..., 0, 1, 1], and ε(c (2)) h(c (1)) ε(c (1)) h(c (2)). Then (p + S) E κ + for all p R n +. BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
38 C (1) C (1), C (2) C (2), C (1) = C (2) = 1, SUFFICIENT CONDITION FOR (p + S) E κ + FOR ALL p R n + COROLLARY (REPEATED FROM SLIDE 21) Assume that C = {i C y i > 0} {i C y i > 0}, Ŷ ε = [0,..., 0, 1, 1], and ε(c (2)) h(c (1)) ε(c (1)) h(c (2)). Then (p + S) E κ + for all p R n +. COROLLARY (BB) Assume that C (1) = C (2) = 1, Ŷ ε = [0,..., 0, 1, 1], and ε(c (2)) h(c (1)) ε(c (1)) h(c (2)). (2) Then (p + S) E κ + for all p R n +. BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
39 A C (1) = C (2) = 1 EXAMPLE WITH (2) NOT SATISFIED κ 12 κ X 1 + 2X 2 23 X 1 + 3X 2 κ 45 2X 1 X 1 + X 2 2X Ŷ = , h = 1 1 1, ε = 1 0 1, κ κ 12 κ κ 23 0 A κ = κ 45 0 κ 45 0, y = , y = 2 κ 12 1 κ κ 450 ε 5 (h 1 + h 2 )=ε 3 h 4, BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
40 A C (1) = C (2) = 1 EXAMPLE WITH (2) NOT SATISFIED, CASE 1 Analysis of the ODE gives: for 4 κ 23 κ 45 <κ 2 12 we have E κ + = 0 and the trajectories are diverging. FIGURE : κ 12 = 4, κ 23 = 1, κ 45 = 1 BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
41 A C (1) = C (2) = 1 EXAMPLE WITH (2) NOT SATISFIED, CASE 2 Analysis of the ODE gives: for 4 κ 23 κ 45 >κ 2 12 we have E κ + = 0 and the trajectories are converging to the boundary. FIGURE : κ 12 = 1, κ 23 = 1, κ 45 = 1 BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
42 A C (1) = C (2) = 1 EXAMPLE WITH (2) NOT SATISFIED, CASE 3 Analysis of the ODE gives: for 4 κ 23 κ 45 =κ 2 12 we have E+ κ = {x R 2 + x 2 = κ 45 /κ 23 } and the trajectories are converging either to the positive steady states or to the boundary. FIGURE : κ 12 = 2, κ 23 = 1, κ 45 = 1 BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
43 A C (1) = C (2) = 1 EXAMPLE WITH (2) NOT SATISFIED, BUT STILL HAVE (p + S) E κ + FOR ALL p R n + 3X 1 κ 12 2X 1 + X 2 3X 2 κ 34 X 1 + 2X 2 } E+ κ = {x R 2 + x 2 = (κ 12 /κ 34 ) 1 3 x1 = half line in the positive orthant ([ 1 S = span 1 ]) (p + S) E κ + = 1 for all p R 2 + BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
44 THESE SLIDES ARE AVAILABLE AT BALÁZS BOROS (EÖTVÖS UNIV., BUDAPEST) EXISTENCE OF POSITIVE STEADY STATES CHICAGO, OCTOBER 2-4, / 28
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