Compartmental Modelling: Eigenvalues, Traps and Nonlinear Models

Transcription

1 Compartmental Modelling: Eigenvalues, Traps and Nonlinear Models Neil D. Evans School of Engineering, University of Warwick Systems Pharmacology - Pharmacokinetics Vacation School, University of Warwick, March 2014 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 1/32

2 More Compartmental Modelling 1 Linear (time-invariant) compartmental models 2 Some Examples Pseudo Steady State Approximation ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 2/32

3 Following definitions relate to compartmental model diagrams: Definition A compartmental model is strongly connected (SC) if, for any two compartments i and j, there exists a path from i to j, and a path from j to i. Definition A compartmental model is outflow connected (OC) if there is a path from any compartment to a compartment with an outflow. Definition A trap is a subset of compartments that is outflow-closed and there are no transfers/paths to compartments outside of the trap. ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 3/32

4 Example 1 (Drug distribution, 2 compartments). Widely used in pharmacokinetics (PK) to analyse distribution of drug after injection into blood: I a a 01 a 12 Example 2 (Drug distribution, 3 compartments). Also widely used in PK to analyse distribution of drug after injection into the blood: a 13 I a a 31 a 12 a 01 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 4/32

5 Linear (time-invariant) compartmental models ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 5/32

6 General n compartment linear model I i a ji I j to/from other compartments i j to/from other compartments a 0i a ij a 0j Since system linear: f ij (q(t), t) = a ij, and q(t) = Aq(t)+I(t) (1) where q 1 (t) a 11 a a 1n q 2 (t) q(t) =., A = a 21 a a 2n a ii = ( a 0i + ) a ji j i q n (t) a n1 a n2... a nn Constants a ij are rate constants, with units [T] 1 ; and for outflows, a 0i, called elimination rate constants. ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 6/32

7 General n compartment linear model I i k ij I j to/from other compartments i j to/from other compartments k ie k ji k je Since system linear: f ij (q(t), t) = a ij = k ji, and q(t) = Aq(t)+I(t) (1) where q 1 (t) k 11 k k n1 q 2 (t) q(t) =., A = k 12 k k n k ii = ( k ie + ) k ij j i q n (t) k 1n k 2n... k nn Constants k ji are rate constants, with units [T] 1 ; and for outflows, k ie, called elimination rate constants. ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 6/32

8 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 7/32

9 Spectral/modal analysis In the absence of inflows linear compartmental model is: q(t) = Aq(t), q(0) = q 0. (2) If A has n distinct eigenvalues λ 1, λ 2,..., λ n (Av i = λ i v i ), then any sum of exponentials q(t) = n c i v i e λ i t i=1 is a general solution of (2). The particular solution starts at q 0, namely: n q(0) = c i v i = q 0. i=1 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 8/32

10 Matrix exponential We can also write the particular solution in terms of the initial state q(0) and the matrix exponential of A: q(t) = n c i v i e λ i t = e At q 0, (3) i=1 where e At = I n + ta+ t2 2! A2 + + tr r! Ar +. Eigenvalues of compartmental models (i.e., of A) must satisfy: No eigenvalue can have a positive real part; There can be no purely imaginary eigenvalues; Model has a trap if and only if A has a zero eigenvalue. ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 9/32

11 Behaviour of exponentials For a fixed λ consider behaviour of exponential term: e λt λ = 0 gives constant term λ = 0 2 e λ t Time ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 10/32

12 Behaviour of exponentials For a fixed λ consider behaviour of exponential term: e λt λ = 0 gives constant term λ > 0 gives exponential rise (unstable) λ = 0 λ > 0 e λ t Time ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 10/32

13 Behaviour of exponentials For a fixed λ consider behaviour of exponential term: e λt λ = 0 gives constant term λ > 0 gives exponential rise (unstable) λ < 0 gives exponential decrease (stable) e λ t λ = 0 λ > 0 λ < 0 Half-life: τ e λτ = 1 2 τ = 1 λ ln Time ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 10/32

14 Behaviour of exponentials For a fixed λ consider behaviour of exponential term: e λt λ = 0 gives constant term λ > 0 gives exponential rise (unstable) λ < 0 gives exponential decrease (stable) e λ t λ = 0 λ 1 > 0 λ 2 < 0 2 λ 1 2 λ 2 Half-life: τ e λτ = 1 2 τ = 1 λ ln Time ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 10/32

15 Example: Topotecan Hydrolysis Reversible hydrolysis of anti-cancer agent topotecan (TPT) can be modelled via a two-compartment model: k o L H k c Rate constants correspond to first order processes for opening of lactone ring (L H) and ring-closing (H L). ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 11/32

16 Example: Topotecan Hydrolysis Reversible hydrolysis of anti-cancer agent topotecan (TPT) can be modelled via a two-compartment model: k o L H k c Rate constants correspond to first order processes for opening of lactone ring (L H) and ring-closing (H L). q L (t) = k o q L (t)+k c q H (t), q L (0) = d q H (t) = k o q L (t) k c q H (t), q L (0) = 0 and so ( ) ko k A = c k o k c ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 11/32

17 Example: Topotecan Hydrolysis Reversible hydrolysis of anti-cancer agent topotecan (TPT) can be modelled via a two-compartment model: k o L H k c Rate constants correspond to first order processes for opening of lactone ring (L H) and ring-closing (H L). q L (t) = k o q L (t)+k c q H (t), q L (0) = d q H (t) = k o q L (t) k c q H (t), q L (0) = 0 and so ( ) ko k A = c k o k c det(λi A) = 0 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 11/32

18 Example: Topotecan Hydrolysis Reversible hydrolysis of anti-cancer agent topotecan (TPT) can be modelled via a two-compartment model: k o L H k c Rate constants correspond to first order processes for opening of lactone ring (L H) and ring-closing (H L). and so ( ) ko k A = c k o k c q L (t) = k o q L (t)+k c q H (t), q L (0) = d q H (t) = k o q L (t) k c q H (t), q L (0) = 0 det(λi A) = (λ+k o )(λ+k c ) k o k c ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 11/32

19 Example: Topotecan Hydrolysis Reversible hydrolysis of anti-cancer agent topotecan (TPT) can be modelled via a two-compartment model: k o L H k c Rate constants correspond to first order processes for opening of lactone ring (L H) and ring-closing (H L). q L (t) = k o q L (t)+k c q H (t), q L (0) = d q H (t) = k o q L (t) k c q H (t), q L (0) = 0 and so ( ) ko k A = c k o k c det(λi A) = λ 2 +(k o + k c )λ ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 11/32

20 Eigenvalues: λ = 0, λ = (k o + k c ) Eigenvectors: (λi A) v = 0 ( ) ( ) λ+ko k c v1 = k o λ+k c v 2 ( ) 0 0 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 12/32

21 Eigenvalues: λ = 0, λ = (k o + k c ) Eigenvectors: (λi A) v = 0 λ = 0 ( ko k c k o k c )( v1 v 2 ) = ( ) 0 0 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 12/32

22 Eigenvalues: λ = 0, λ = (k o + k c ) Eigenvectors: (λi A) v = 0 λ = 0 ( ko k c k o k c )( v1 v 2 ) = ( ) 0 0 = k o v 1 = k c v 2 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 12/32

23 Eigenvalues: λ = 0, λ = (k o + k c ) Eigenvectors: (λi A) v = 0 λ = 0 ( ko k c k o k c )( v1 v 2 ) = ( ) 0 0 = E-vector: ( kc k o ) ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 12/32

24 Eigenvalues: λ = 0, λ = (k o + k c ) Eigenvectors: (λi A) v = 0 λ = 0 ( ko k c k o k c )( v1 v 2 ) = ( ) 0 0 = E-vector: ( kc k o ) λ = (k o + k c ) ( ) ( ) λ+ko k c v1 = k o λ+k c v 2 ( ) 0 0 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 12/32

25 Eigenvalues: λ = 0, λ = (k o + k c ) Eigenvectors: (λi A) v = 0 λ = 0 ( ko k c k o k c )( v1 v 2 ) = ( ) 0 0 = E-vector: ( kc k o ) λ = (k o + k c ) ( ) ( ) kc k c v1 = k o k o v 2 ( ) 0 0 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 12/32

26 Eigenvalues: λ = 0, λ = (k o + k c ) Eigenvectors: (λi A) v = 0 λ = 0 ( ko k c k o k c )( v1 v 2 ) = ( ) 0 0 = E-vector: ( kc k o ) λ = (k o + k c ) ( ) ( ) kc k c v1 = k o k o v 2 ( ) 0 0 = v 1 = v 2 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 12/32

27 Eigenvalues: λ = 0, λ = (k o + k c ) Eigenvectors: (λi A) v = 0 λ = 0 ( ko k c k o k c )( v1 v 2 ) = λ = (k o + k c ) ( ) ( ) kc k c v1 = k o k o v 2 ( ) 0 0 ( ) 0 0 = E-vector: ( kc k o ) ( ) 1 = E-vector: 1 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 12/32

28 Eigenvalues: λ = 0, λ = (k o + k c ) Eigenvectors: (λi A) v = 0 λ = 0 ( ko k c k o k c )( v1 v 2 ) = λ = (k o + k c ) ( ) ( ) kc k c v1 = k o k o v 2 ( ) 0 0 ( ) 0 0 = E-vector: ( kc k o ) ( ) 1 = E-vector: 1 Initial conditions: ( ) ( ) ( ) d kc 1 = c c k 2 o 1 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 12/32

29 Eigenvalues: λ = 0, λ = (k o + k c ) Eigenvectors: (λi A) v = 0 λ = 0 ( ko k c k o k c )( v1 v 2 ) = λ = (k o + k c ) ( ) ( ) kc k c v1 = k o k o v 2 ( ) 0 0 ( ) 0 0 = E-vector: ( kc k o ) ( ) 1 = E-vector: 1 Initial conditions: ( ) d = 0 d k c + k o ( kc k o ) + dk ( ) o 1 k c + k o 1 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 12/32

30 Solution is (ql ) ( ) ( ) (t) kc = c q H (t) 1 e 0t 1 + c k 2 e (ko+kc)t o 1 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 13/32

31 Solution is ( ) ql (t) = q H (t) d k c + k o ( kc k o ) + dk ( ) o 1 e (ko+kc)t k c + k o 1 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 13/32

32 Solution is ( ) ql (t) = q H (t) d k c + k o [( kc k o ) ( ) ] ko + e (ko+kc)t k c ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 13/32

33 Solution is ( ) ql (t) = q H (t) d k c + k o [( kc k o ) ( ) ] ko + e (ko+kc)t k c For example (k o = 2, k c = 3, d = 5) 5 q L 4 q H 3 Amount Time ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 13/32

34 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 14/32

35 General case In the presence of inflows linear compartmental model given by: and q(t) = Aq(t)+I(t), q(0) = q 0, (4) t q(t) = e At q 0 + e A(t s) I(s) ds. 0 Probably the most common forms for the inputs in pharmacokinetics: Impulsive input Constant infusion ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 15/32

36 Impulsive input Also called bolus injection in applications; corresponds to giving specific amount, d, in a negligible amount of time at t = τ. At t = τ = 0 easiest to include in initial conditions: and treat as homogeneous. q(0) = q 0 + d, If at t = τ 0 treat as homogeneous model for 0 t < τ such that q τ = q(τ); and then as homogeneous model for T = t τ 0 starting at q(τ) = q τ + d. Multiple impulsive inputs (at times τ i ) handled similarly: q(τ i+1 ) = q τi + d i. ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 16/32

37 Scalar example: Single input Consider a single compartment model subject to an impulsive input of size d at time t = τ. For 0 t < τ we have q(t) = aq(t), q(0) = q 0 = q(t) = q 0 e at. Then for t τ: q(t) = aq(t), q(τ) = q 0 e aτ + d = q(t) = ( q 0 e aτ + d ) e a(t τ). Hence the full general solution is given by: { q 0 e at 0 t < τ q(t) = q 0 e at + de a(t τ) t τ. ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 17/32

38 Scalar example: Multiple inputs For multiple impulsive inputs, of size d i at time τ i (i = 1...m), it can be shown that the general solution is given by: q 0 e at 0 t < τ 1 q 0 e at + d 1 e a(t τ 1) τ 1 t < τ 2 q(t) = q 0 e at + d 1 e a(t τ1) + d 2 e a(t τ 2) τ 2 t < τ 3. q 0 e at + m i=1 d ie a(t τ i) τ m t. i.e., sum of the homogeneous model starting from q 0 and the sum of the impulse responses for the inputs. ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 18/32

39 Constant input Solution with constant infusion at a certain rate, ie u(t) = d: q(t) = Aq(t)+d = q(t) = e At q 0 + t 0 e A(t s) d ds. Suppose eigenvalues λ i of A are distinct. Then there are constants c 11,..., c 1n and c 21,..., c 2n such that n n q 0 = c 1i v i and d = c 2i v i, i=1 where v i are eigenvectors corresponding to λ i. Then solution of compartmental model with constant input is given by: ( n ) ( t n ) q(t) = c 1i v i e λ i t + c 2i v i e λ i(t s) ds i=1 0 i=1 i=1 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 19/32

40 Constant input Solution with constant infusion at a certain rate, ie u(t) = d: q(t) = Aq(t)+d = q(t) = e At q 0 + t 0 e A(t s) d ds. Suppose eigenvalues λ i of A are distinct. Then there are constants c 11,..., c 1n and c 21,..., c 2n such that n n q 0 = c 1i v i and d = c 2i v i, i=1 where v i are eigenvectors corresponding to λ i. Then solution of compartmental model with constant input is given by: n (( q(t) = c 1i + c ) 2i v i e λ i t c ) 2i v i. λ i λ i i=1 i=1 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 19/32

41 Example: Topotecan Hydrolysis Now suppose active form of TPT k c added as continuous infusion: L H I(t) k o ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 20/32

42 Example: Topotecan Hydrolysis Now suppose active form of TPT k c added as continuous infusion: L H Solution given by: q(t) = c 11 v 1 e λ 1t +c 12 v 2 e λ 2t + t 0 I(t) k o ( c 21 v 1 e λ 1(t s) + c 22 v 2 e λ 2(t s) ) ds ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 20/32

43 Example: Topotecan Hydrolysis Now suppose active form of TPT k c added as continuous infusion: L H Solution given by: q(t) = t 0 I(t) ( c 21 v 1 e λ 1(t s) + c 22 v 2 e λ 2(t s) ) ds k o ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 20/32

44 Example: Topotecan Hydrolysis Now suppose active form of TPT k c added as continuous infusion: L H Solution given by: q(t) = t 0 I(t) ( c 21 v 1 + c 22 v 2 e λ 2(t s) ) ds k o ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 20/32

45 Example: Topotecan Hydrolysis Now suppose active form of TPT k c added as continuous infusion: L H Solution given by: ( ) ( )( kc 1 1 q(t) = c 21 t + c k 22 o 1 I(t) k o + k c k o ) 1 e (ko+kc)t k o + k c ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 20/32

46 Example: Topotecan Hydrolysis Now suppose active form of TPT k c added as continuous infusion: L H Solution given by: d q(t) = k c + k o [( kc k o ) t + 1 k o + k c ( ko k o I(t) k o ) ( 1 e (ko+kc)t)] ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 20/32

47 Example: Topotecan Hydrolysis Now suppose active form of TPT k c added as continuous infusion: L H Solution given by: d q(t) = k c + k o [( kc k o ) t + 1 k o + k c ( ko k o I(t) k o ) ( 1 e (ko+kc)t)] 3.5 q L 3 q H For example: k o = 2, k c = 3, d = 5 Amount Time ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 20/32

48 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 21/32

49 Linear compartmental models can be decomposed into isolated subsystems of four types: NIO(OC) Systems with no inflows, but at least one outflow (NIO); in addition, they are outflow-connected (OC). Asymptotic behaviour: All compartments decay to 0 exponentially: 0 only steady state (& asymptotically stable) IO(OC) Systems with inflows (IO) & outflow-connected (OC). For constant input u(t) d, system approaches non-zero steady state: q = A 1 d, and if eigenvalues of A are distinct, then: n (( q(t) = c 1i + c ) 2i v i e λ i t c ) 2i v i λ i λ i i=1 n i=1 ( ) c2i v i λ i ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 22/32

50 NINO(SC) Systems with no inflows & no outflows (NINO), but strongly connected (SC); they are themselves traps. The system (exponentially) approaches a steady state q = (q 1,...,q n) T, such that q = Aq = 0. Since there are no inflows or outflows the total amount in the system remains constant, A 0 = n i=1 q i say. Use this to replace equation for first compartment to get: Hq = M = (A 0, 0,...,0) T where H is nonsingular. The steady state is then given by q = H 1 M. ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 23/32

51 INO(SC) Systems with inflows & no outflows (INO), & strongly connected (SC). Unless inflow is bounded system continues to grow. If inflow is bounded system approaches steady state of NINO(SC) case, with total amount equal to initial amount in system plus total inflow. Example (Parent-Metabolite Model for Bromosulphthalein). Parent Metabolite I k 12 k 34 k k 21 k 43 k 2e k 3e ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 24/32

52 Some Examples Pseudo Steady State Approximation ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 25/32

53 Some Examples Pseudo Steady State Approximation Transfers between compartments can depend on time vector of rate constants k amounts in the compartments so that rate of change of amount in compartment i is where q i (t) = I i (t)+ n f ij (q(t), k, t) q j (t). (5) j=1 f ii (q(t), k, t) = f 0i (q(t), k, t)+ n f ji (q(t), k, t) j=1 j i ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 26/32

54 Example: Topotecan kinetics Some Examples Pseudo Steady State Approximation Only active form TPT L binds to nuclear DNA target & so important to monitor activity of drug from injection to target k i [B F (t) = B T L n (t)] L m TPT L L c TPT L k b B F (t) k e k cm k om k cc k oc k dl k dh L n TPT L H m TPT H H c TPT H Nucleus Cytoplasm Dose consists of bolus input of active drug administered into extracellular medium. ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 27/32

55 Example: Facilitated Diffusion Some Examples Pseudo Steady State Approximation Facilitated diffusion is one of simplest mechanisms for transporting material across cell membrane. Eg: consider entry of glucose into red blood cells: Extracellular Membrane Intracellular k 0 S+ C C +S k 0 K K K K CS k 1 k 1 CS State 2 State 1 Could assume reactions very fast compared with transfer of substrate across membrane; ie in equilibrium (a pseudo-steady state assumption/approximation). ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 28/32

56 Steady states Some Examples Pseudo Steady State Approximation Consider nonlinear compartmental model: n q i (t) = I i (t)+ f ij (q(t), k, t) q j (t). (6) j=1 ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 29/32

57 Steady states Some Examples Pseudo Steady State Approximation Consider nonlinear compartmental model: q(t) = F(q(t), k, t)q(t)+i(t) = f(q(t), k, t)+i(t) (6) where F is compartmental matrix of f ij ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 29/32

58 Steady states Some Examples Pseudo Steady State Approximation Consider nonlinear compartmental model: q(t) = F(q(t), k, t)q(t)+i(t) = f(q(t), k, t)+i(t) (6) where F is compartmental matrix of f ij Definition An equilibrium state or steady state ˆq for (6), with corresponding steady state inflow Î, occurs if and only if 0 = ˆq(t) = f(ˆq(t), k, t)+î(t). ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 29/32

59 Linearisation: Stability Some Examples Pseudo Steady State Approximation Consider small perturbations: q = ˆq + q and I = Î + I q(t) = f(q(t), k, t)+i(t) ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 30/32

60 Linearisation: Stability Some Examples Pseudo Steady State Approximation Consider small perturbations: q = ˆq + q and I = Î + I ˆq(t)+ q(t) = f(ˆq(t)+q(t), k, t)+î(t)+i(t) ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 30/32

61 Linearisation: Stability Some Examples Pseudo Steady State Approximation Consider small perturbations: q = ˆq + q and I = Î + I q(t) = f(ˆq + q(t), k, t)+î(t)+i(t) ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 30/32

62 Linearisation: Stability Some Examples Pseudo Steady State Approximation Consider small perturbations: q = ˆq + q and I = Î + I q(t) = f(ˆq + q(t), k, t)+î(t)+i(t) We can write f as: f(ˆq + q(t), k, t) = f(ˆq, k, t)+aq(t)+... ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 30/32

63 Linearisation: Stability Some Examples Pseudo Steady State Approximation Consider small perturbations: q = ˆq + q and I = Î + I We can write f as: where q(t) = f(ˆq + q(t), k, t)+î(t)+i(t) f(ˆq + q(t), k, t) = f(ˆq, k, t)+aq(t)+... A = f q =. f 1 f 1 q 1 f 2 f 2 q 1 f n q 1 q 2... f 1 q n f 2 q n q f n q 2... f n q n evaluated at ˆq ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 30/32

64 Linearisation: Stability Some Examples Pseudo Steady State Approximation Consider small perturbations: q = ˆq + q and I = Î + I We can write f as: where q(t) = f(ˆq, k, t)+aq(t)+î(t)+i(t) f(ˆq + q(t), k, t) = f(ˆq, k, t)+aq(t)+... A = f q =. f 1 f 1 q 1 f 2 f 2 q 1 f n q 1 q 2... f 1 q n f 2 q n q f n q 2... f n q n evaluated at ˆq ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 30/32

65 Linearisation: Stability Some Examples Pseudo Steady State Approximation Consider small perturbations: q = ˆq + q and I = Î + I We can write f as: where q(t) = Aq(t)+I(t) f(ˆq + q(t), k, t) = f(ˆq, k, t)+aq(t)+... A = f q =. f 1 f 1 q 1 f 2 f 2 q 1 f n q 1 q 2... f 1 q n f 2 q n q f n q 2... f n q n evaluated at ˆq ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 30/32

66 Linearisation: Stability Some Examples Pseudo Steady State Approximation Consider small perturbations: q = ˆq + q and I = Î + I We can write f as: where q(t) = Aq(t)+I(t) f(ˆq + q(t), k, t) = f(ˆq, k, t)+aq(t)+... A = f q =. f 1 f 1 q 1 f 2 f 2 q 1 f n q 1 q 2... f 1 q n f 2 q n q f n q 2... f n q n evaluated at ˆq Eigenvalues of A determine stability of steady state ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 30/32

67 Some Examples Pseudo Steady State Approximation Pseudo Steady State Approximation Consider an enzyme mediated reaction: S + E k 1 ES k 2 E + P. k 1 Let x S denote concentration of S, x P conc. of P, x E conc. of E & x C conc. of intermediate (ES). Applying law of mass action corresponding state-space model is: ẋ S = k 1 x S x E + k 1 x C ẋ P = k 2 x C ẋ E = k 1 x S x E +(k 1 + k 2 ) x C ẋ C = ẋ E = k 1 x S x E (k 1 + k 2 ) x C. Note that these equations are not in compartmental form. ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 31/32

68 Quasi Steady State Assumption Conservation Law: Some Examples Pseudo Steady State Approximation ẋ C + ẋ E = 0 = x E = e 0 x C so we can eliminate a variable (x E ). Suppose that enzyme bit of reaction fast compared to overall reaction: ẋ C = 0 = k 1 x S x E (k 1 + k 2 ) x C ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 32/32

69 Quasi Steady State Assumption Conservation Law: Some Examples Pseudo Steady State Approximation ẋ C + ẋ E = 0 = x E = e 0 x C so we can eliminate a variable (x E ). Suppose that enzyme bit of reaction fast compared to overall reaction: ẋ C = 0 = e 0 k 1 x S (k 1 + k 2 + k 1 x S ) x C ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 32/32

70 Quasi Steady State Assumption Conservation Law: Some Examples Pseudo Steady State Approximation ẋ C + ẋ E = 0 = x E = e 0 x C so we can eliminate a variable (x E ). Suppose that enzyme bit of reaction fast compared to overall reaction: ẋ C = 0 = e 0 k 1 x S (k 1 + k 2 + k 1 x S ) x C e 0 k 1 x = x C = S = e 0x S k 1 + k 2 + k 1 x S K M + x S ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 32/32

71 Quasi Steady State Assumption Conservation Law: Some Examples Pseudo Steady State Approximation ẋ C + ẋ E = 0 = x E = e 0 x C so we can eliminate a variable (x E ). Suppose that enzyme bit of reaction fast compared to overall reaction: ẋ C = 0 = e 0 k 1 x S (k 1 + k 2 + k 1 x S ) x C e 0 k 1 x = x C = S = e 0x S k 1 + k 2 + k 1 x S K M + x S where K M = (k 1 + k 2 )/k 1, let V M = e 0 k 2 : ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 32/32

72 Quasi Steady State Assumption Conservation Law: Some Examples Pseudo Steady State Approximation ẋ C + ẋ E = 0 = x E = e 0 x C so we can eliminate a variable (x E ). Suppose that enzyme bit of reaction fast compared to overall reaction: ẋ C = 0 = e 0 k 1 x S (k 1 + k 2 + k 1 x S ) x C e 0 k 1 x = x C = S = e 0x S k 1 + k 2 + k 1 x S K M + x S where K M = (k 1 + k 2 )/k 1, let V M = e 0 k 2 : ẋ P = ẋ S = V Mx S K M + x S ND Evans University of Warwick 24-Mar-14 Compartmental Modelling 32/32