Asymptotic analysis of a TMDD model: when a reaction contributes to the destruction of its product

Abstract

The multi-scale dynamics of a two-compartment with first order absorption Target-Mediated Drug Disposition (TMDD) pharmacokinetics model is analysed, using the Computational Singular Perturbation (CSP) algorithm. It is shown that the process evolves along two Slow Invariant Manifolds (SIMs), on which the most intense components of the model are equilibrated, so that the less intensive are the driving ones. The CSP tools allow for the identification of the components of the TMDD model that (i) constrain the evolution of the process on the SIMs, (ii) drive the system along the SIMs and (iii) generate the fast time scales. Among others, such diagnostics identify (i) the factors that determine the start and the duration of the period in which the ligand-receptor complex acts and (ii) the processes that determine its degradation rate. The counterintuitive influence of the process that transfers the ligand from the tissue to the main compartment, as it is manifested during the final stage of the process, is studied in detail.

Appendices

A Appendix: The CSP basis vectors for the $\hbox {M}=2$ case

Considering the state vector defined as $\mathbf {y}=[L_d,~L,~L_t,~R,~RL]^T$, the CSP vectors in the case where $M=2$ are computed as follows. Since the fast variables are R and RL, a proper initial guess is:

$$\begin{aligned} \mathbf {a}_r= & {} \begin{bmatrix} 0&\quad 0\\0&\quad 0\\0&\quad 0\\1&\quad 0\\0&\quad 1 \end{bmatrix} \quad \mathbf {a}_s =\begin{bmatrix} 1&\quad 0&\quad 0 \\ 0&\quad 1&\quad 0 \\ 0&\quad 0&\quad 1 \\ 0&\quad 0&\quad 0 \\ 0&\quad 0&\quad 0 \end{bmatrix}\\ \mathbf {b}^r= & {} \begin{bmatrix} 0&\quad 0&\quad 0&\quad 1&\quad 0\\0&\quad 0&\quad 0&\quad 0&\quad 1 \end{bmatrix} \quad \mathbf {b}^s =\begin{bmatrix} 1&\quad 0&\quad 0&\quad 0&\quad 0 \\ 0&\quad 1&\quad 0&\quad 0&\quad 0 \\0&\quad 0&\quad 1&\quad 0&\quad 0 \end{bmatrix} \end{aligned}$$

With such a choice it is guaranteed that $\mathbf {a}_r$ has a component in the fast subspace of the tangent space along $SIM_2$ (Lam and Goussis 1994); i.e., a component along the axis of the fast variables R and RL. After one $\mathbf {b}^r$ and one $\mathbf {a}_r$-refinements, the following CSP vectors are obtained:

$$\begin{aligned} \mathbf {a}_r= & {} \left[ \mathbf {a}_1 ~ \mathbf {a}_2 \right] =\begin{bmatrix} 0&\quad 0 \\ k_{int}k_{on}L XD_{en}/X_1&\quad -k_{deg}k_{off}XD_{en}/X_1\\ 0&\quad 0\\ D_{en}(X^2D_{en}+Zk_{deg}k_{off})/X_1&\quad k_{int}k_{off}ZD_{en}/{X_1}\\ k_{int}k_{on}LZD_{en}/{X_1}&\quad D_{en}(XD_{en}+Yk_{on}L)/X_1 \end{bmatrix}\nonumber \\ \end{aligned}$$

(1)

$$\begin{aligned} \mathbf {a}_s= & {} \left[ \mathbf {a}_3 ~ \mathbf {a}_4 ~ \mathbf {a}_5\right] =\begin{bmatrix} 1&\quad 0&\quad 0 \\ 0&\quad 1&\quad 0\\0&\quad 0&\quad 1\\0&\quad -{YD_{en}}/{X_1}&\quad 0\\0&\quad {ZD_{en}}/{X_1}&\quad 0 \end{bmatrix} \end{aligned}$$

(2)

$$\begin{aligned} \mathbf {b}^r= & {} \begin{bmatrix} 0&\quad Y/X&\quad 0&\quad 1&\quad 0\\0&\quad -Z/X&\quad 0&\quad 0&\quad 1 \end{bmatrix} \end{aligned}$$

(3)

$$\begin{aligned} \mathbf {b}^s= & {} \begin{bmatrix} 1&0&0&0&0 \\ 0&\quad (XD_{en})^2/X_1&\quad 0&\quad -k_{int}k_{on}LXD_{en}/X_1&\quad k_{deg}k_{off}XD_{en}/X_1 \\ 0&\quad 0&\quad 1&\quad 0&\quad 0 \end{bmatrix} \end{aligned}$$

(4)

where as mentioned in Sect. 6:

$$\begin{aligned} X= & {} \frac{k_{deg}(k_{int}+k_{off})+k_{int}k_{on}L}{Den}\\ Y= & {} \frac{k_{int}k_{on}R}{Den} \\ Z= & {} \frac{k_{deg}k_{on}R}{Den} \\ Den= & {} \left( k_{el}+k_{pt}\right) \left[ k_{int}k_{on}L +k_{deg}(k_{int}+k_{off})\right] +k_{deg}k_{int}k_{on}R\\ and~X_1= & {} D_{en}(X^2D_{en}+Zk_{deg}k_{off}+Yk_{int}k_{on}L) \end{aligned}$$

B Appendix: The CSP basis vectors for the $\hbox {M}=3$ case

Considering the state vector as in “Appendix A”, the CSP vectors in the case where $M=3$ are computed as follows. Since the variables associated the most to the fast time scales are L, RL and R, a proper initial guess is:

$$\begin{aligned} \mathbf {a}_r= & {} \begin{bmatrix} 0&\quad 0&\quad 0\\1&\quad 0&\quad 0\\0&\quad 0&\quad 0\\0&\quad 0&\quad 1\\0&\quad 1&\quad 0 \end{bmatrix} \quad \mathbf {a}_s=\begin{bmatrix} 1&\quad 0\\ 0&\quad 0\\ 0&\quad 1 \\ 0&\quad 0 \\ 0&\quad 0 \end{bmatrix}\\ \mathbf {b}^r= & {} \begin{bmatrix} 0&\quad 1&\quad 0&\quad 0&\quad 0\\0&\quad 0&\quad 0&\quad 0&\quad 1\\0&\quad 0&\quad 0&\quad 1&\quad 0 \end{bmatrix} \quad \mathbf {b}^s=\begin{bmatrix} 1&\quad 0&\quad 0&\quad 0&\quad 0 \\ 0&\quad 0&\quad 1&\quad 0&\quad 0 \end{bmatrix} \end{aligned}$$

With such a choice it is guaranteed that $\mathbf {a}_r$ has a component in the fast subspace of the tangent space along $SIM_3$ (Lam and Goussis 1994); i.e., a component along the axis of the fast variables L, RL and R. After one $\mathbf {b}^r$ and one $\mathbf {a}_r$-refinements, the following CSP vectors are obtained:

$$\begin{aligned} \mathbf {a}_r= & {} \begin{bmatrix} 0&\quad 0&\quad 0\\&\quad&\quad \\ \dfrac{S_1}{S}&\quad -\dfrac{S_2}{S}&\quad \dfrac{S_3}{S}\\&\quad&\quad \\dfrac{S_4}{S}&\quad -\dfrac{S_5}{S}&\quad \dfrac{S_6}{S}\\&\quad&\quad \\ \dfrac{S_7}{S}&\quad \dfrac{S_8}{S}&\quad \dfrac{S_9}{S}\\&\quad&\quad \\dfrac{S_{10}}{S}&\quad \dfrac{S_{11}}{S}&\quad \dfrac{S_{12}}{S} \end{bmatrix} \end{aligned}$$

(1)

$$\begin{aligned} \mathbf {a}_s= & {} \begin{bmatrix} 1&\quad 0\\ \dfrac{k_aX}{X(k_{el}+k_{pt})+k_{deg}Y}&\quad \dfrac{k_{tp}X}{X(k_{el} +k_{pt})+k_{deg}Y}\\ 0&\quad 1 \\ -\dfrac{k_aY}{X(k_{el}+k_{pt})+k_{deg}Y}&\quad -\dfrac{k_{tp}Y}{X(k_{el} +k_{pt})+k_{deg}Y} \\ \dfrac{k_aZ}{X(k_{el}+k_{pt})+k_{deg}Y}&\quad \dfrac{k_{tp}Z}{X(k_{el} +k_{pt})+k_{deg}Y} \end{bmatrix} \end{aligned}$$

(2)

where

• $S_1=D_{en}(X(k_{el}+k_{pt})+Yk_{deg})^2+k_{pt}k_{tp}(2k_{deg} k_{off}Z k_{int}k_{on}LY)$

• $S_2=k_{deg}k_{off}k_{pt}k_{tp}X$

• $S_3=k_{int}k_{on}k_{pt}k_{tp}LX$

• $S_4=D_{en}Xk_{pt}(X(k_{el}+k_{pt})+Yk_{deg})$

• $S_5=k_{deg}k_{off}k_{pt}(X(k_{el}+k_{pt})+Yk_{deg})$

• $S_6=k_{int}k_{on}k_{pt}L(X(k_{el}+k_{pt})+Yk_{deg})$

• $S_7=D_{en}XYk_{pt}k_{tp}$

• $S_8=k_{deg}k_{off}k_{pt}k_{tp}Y$

• $S_9=D_{en}(X(k_{el}+k_{pt})+k_{deg}Y)^2+k_{pt}k_{tp}(X^2D_{en} +k_{deg}k_{off}Z)$

• $S_{10}=D_{en}XZk_{pt}k_{tp}$

• $S_{11}=D_{en}((k_{el}+k_{pt})X+Yk_{deg})^2+k_{pt}k_{tp}(D_{en}X^2 +k_{int}k_{on}LY)\qquad +2k_{deg}^2(k_{el}+k_{pt})(k_{off}+k_{int})Y$

• $S_{12}=k_{pt}k_{tp}k_{deg}k_{on}LY$

• $S=D_{en}(X(k_{el}+k_{pt}+Yk_{deg})^2+k_{pt}k_{tp}(X^2D_{en} +k_{deg}k_{off}Z)$

$$\begin{aligned} \mathbf {b}^r= & {} \begin{bmatrix} \mathbf {b}^1\\ \mathbf {b}^2 \\ \mathbf {b}^3 \end{bmatrix} =\begin{bmatrix} -k_aX&\quad 1&\quad -k_{tp}X&\quad 0&\quad 0 \\ -k_aZ&\quad 0&\quad -k_{tp}Z&\quad 0&\quad 1\\ k_aY&\quad 0&\quad k_{tp}Y&\quad 1&\quad 0 \end{bmatrix} \end{aligned}$$

(3)

$$\begin{aligned} \mathbf {b}^s= & {} \begin{bmatrix} \mathbf {b}^4\\ \mathbf {b}^5 \end{bmatrix} =\begin{bmatrix} 1&\quad 0&\quad 0&\quad 0&\quad 0 \\ -\dfrac{H_2}{H_1}&\quad \dfrac{H_3}{H_1}&\quad \dfrac{H_4}{H_1}&\quad -\dfrac{H_5}{H_1}&\quad \dfrac{H_6}{H_1} \end{bmatrix} \end{aligned}$$

(4)

where

• $H_1=D_{en}^2+k_{pt}k_{tp}X_1$

• $H_2=k_ak_{pt}X_1$

• $H_3=k_{pt}XD_{en}^2(X(k_{el}+k_{pt}+k_{int}Z)$

• $H_4=((k_{el}+k_{pt})XD_{en}+k_{int}ZD_{en})^2$

• $H_5=k_{pt}k_{int} k_{on} L ((k_{el}+k_{pt}) XD_{en}+k_{int}ZD_{en})$

• $H_6=k_{pt}k_{deg} k_{off} ((k_{el}+k_{pt})XD_{en} +k_{int}ZD_{en})$