Interpretation of transit compartments pharmacodynamic models as lifespan based indirect response models

Abstract

Transit compartments (TC) models are used to describe pharmacodynamic responses that involve drug action on cells undergoing differentiation and maturation. Such pharmacodynamic systems can also be described by lifespan based indirect response (LIDR) models. The purpose of this report is to investigate conditions under which the transit compartments models can be considered a special case of LIDR models. An integral representation of a solution to TC model has been used to determine the lifespan distribution for cell population described by this model. The distribution served as a basis for definition of new LIDRE (lifespan based indirect response with an effect on the lifespan distribution) models. Time courses of responses described by both types of models were simulated for a monoexponential pharmacokinetic function. The limit response was calculated as the number of transit compartments approached infinity. The difference between the limit response and TC responses were evaluated by computer simulations using MATLAB 7.7. TC models are a special case of LIDR models with the lifespan distribution described by the gamma function. If drug affects only the production of cells, then the cell lifespan distribution is time invariant. In this case an increase in the number of compartments results in a basic LIDR model with a point lifespan distribution. When the drug inhibits or stimulates cell aging, the cell lifespan distribution becomes time dependent revealing a new mechanism for drug effect on the gamma probability density function. The TC model with a large number of transit compartments converges to an LIDRE model. The limit LIDR models are approximated by the TC models when the number of compartments is at least 5. A moderate improvement in the approximation is observed if this number exceeds 20. The lifespan distribution for a cell population described by a TC model is described by the gamma probability density function. A drug affects this distribution only if it stimulates or inhibits the rate of cell maturation. If the number of transit compartments increases, then the TC model converges to a new type of LIDR model.

Appendices

Appendix 1

Derivation of Eq. 6

To simplify calculations we introduce the following variables

$$ {\text{k}}({\text{t}}) = {\text{k}}_{\text{in}} ({\text{t}}) - {\frac{{{\text{R}}_{0} }}{{{\text{T}}_{\text{R}} }}}{\text{E}}({\text{t}}) \quad{\text{and}}\quad{\text{Y}}_{\text{i}} = {\text{P}}_{\text{i}} - {\frac{{{\text{R}}_{0} }}{\text{n}}}\quad {\text{for i}} = 1, \ldots ,{\text{n}}. $$

(44)

Then the transit compartments model Eqs. 1–2 become

$$ {\frac{{{\text{dY}}_{1} }}{\text{dt}}} = {\text{k}}({\text{t}}) - {\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}\,{\text{E}}({\text{t}}){\text{Y}}_{1} ,\quad{\text{Y}}_{1} (0) = 0 $$

(45)

$$ {\frac{{{\text{dY}}_{\text{i}} }}{\text{dt}}} = {\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}{\text{E}}({\text{t}})\left( {{\text{Y}}_{{{\text{i}} - 1}} - {\text{Y}}_{\text{i}} } \right),\quad{\text{Y}}_{\text{i}} (0) = 0,\quad{\text{i}} = 2, \ldots ,{\text{n}} $$

(46)

Equation 45 can be solved using the integrating factor technique [24]:

$$ {\text{Y}}_{1} ({\text{t}}) = \int\limits_{0}^{\text{t}} {{\text{k}}(\tau )\exp \left( { - {\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}\int\limits_{\tau }^{t} {{\text{E}}({\text{z}}){\text{dz}}} } \right)} {\text{d}}\tau $$

(47)

We will show by mathematical induction that for ${\text{i}} = 1, \ldots ,{\text{n}}$

$$ {\text{Y}}_{\text{i}} ({\text{t}}) = {\frac{1}{{({\text{i}} - 1)!}}}\left( {{\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}} \right)^{{{\text{i}} - 1}} \int\limits_{0}^{\text{t}} {{\text{k}}(\tau )\left( {\int\limits_{\tau }^{\text{t}} {{\text{E}}({\text{z}}){\text{dz}}} } \right)^{{{\text{i}} - 1}} \exp \left( { - {\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}\int\limits_{\tau }^{\text{t}} {{\text{E}}({\text{z}}){\text{dz}}} } \right)} {\text{d}}\tau $$

(48)

Since the case i = 1 reduces to Eq. 47, it remains to show that Eq. 48 is true for i + 1 assuming it holds for i. Using the same integrating factor as for Eq. 47 one can solve Eq. 46 with i + 1 to obtain the following representation of Y_i+1:

$$ {\text{Y}}_{{{\text{i}} + 1}} ({\text{t}}) = {\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}\int\limits_{0}^{\text{t}} {{\text{E}}(\tau ){\text{Y}}_{\text{i}} (\tau )\exp \left( { - {\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}\int\limits_{\tau }^{\text{t}} {{\text{E}}({\text{z}}){\text{dz}}} } \right)} {\text{d}}\tau $$

(49)

One can substitute the right hand side of Eq. 48 for Y_i(τ) in Eq. 49:

$$ {\text{Y}}_{{\text{i}} + 1} ({\text{t}}) = {\frac{1}{({\text{i}} - 1)!}}\left( {{\frac{{\text{n}}}{{{\text{T}}_{{\text{R}}} }}}} \right)^{{\text{i}}} \int\limits_{0}^{{\text{t}}} {{\text{E}}(\tau )\int\limits_{0}^{\tau } {{\text{k}}({\text{s}})\left( {\int\limits_{{\text{s}}}^{\tau } {{\text{E}}({\text{z}}){\text{dz}}} } \right)^{{\text{i}} - 1} \exp \left( { - {\frac{{\text{n}}}{{{\text{T}}_{{\text{R}}} }}}\int\limits_{{\text{s}}}^{\tau } {{\text{E}}({\text{z}}){\text{dz}}} } \right)} {\text{ds}}\exp \left( { - {\frac{{\text{n}}}{{{\text{T}}_{{\text{R}}} }}}\int\limits_{\tau }^{{\text{t}}} {{\text{E}}({\text{z}}){\text{dz}}} } \right){\text{d}}\tau }$$

(50)

Equation 50 can be further reduced to

$$ {\text{Y}}_{{{\text{i}} + 1}} ({\text{t}}) = {\frac{1}{{({\text{i}} - 1)!}}}\left( {{\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}} \right)^{\text{i}} \int\limits_{0}^{\text{t}} {\int\limits_{0}^{\tau } {{\text{E}}(\tau ){\text{k(s)}}\left( {\int\limits_{\text{s}}^{\tau } {\text{E(z)dz}} } \right)^{{{\text{i}} - 1}} \exp \left( { - {\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}\int\limits_{s}^{\text{t}} {\text{E(z)dz}} } \right)} {\text{dsd}}\tau } $$

(51)

Changing the order of the first two integrals in Eq. 51 results in

$$ {\text{Y}}_{{{\text{i}} + 1}} ({\text{t}}) = {\frac{1}{{({\text{i}} - 1)!}}}\left( {{\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}} \right)^{\text{i}} \int\limits_{0}^{\text{t}} {\int\limits_{\text{s}}^{\text{t}} {{\text{E}}(\tau ){\text{k(s)}}\left( {\int\limits_{\text{s}}^{\tau } {\text{E(z)dz}} } \right)^{{{\text{i}} - 1}} \exp \left( { - {\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}\int\limits_{\text{s}}^{\text{t}} {\text{E(z)dz}} } \right)} {\text{d}}\tau {\text{ds}}} $$

(52)

The only part of the integrant that depends on τ is equal to

$$ \int\limits_{\text{s}}^{\text{t}} {{\text{E}}(\tau )\left( {\int\limits_{\text{s}}^{\tau } {\text{E(z)dz}} } \right)^{{{\text{i}} - 1}} {\text{d}}\tau } = {\frac{1}{\text{i}}}\int\limits_{\text{s}}^{\text{t}} {\left( {\int\limits_{\text{s}}^{\tau } {\text{E(z)dz}} } \right)^{\text{i}} {\text{d}}\tau } $$

(53)

Substituting Eq. 53 in the integrant of Eq. 52 gives

$$ {\text{Y}}_{{{\text{i}} + 1}} ({\text{t}}) = {\frac{1}{{{\text{i}}!}}}\left( {{\frac{\text{n}}{\text{T}}}} \right)^{\text{i}} \int\limits_{0}^{\text{t}} {{\text{k}}({\text{s}})\left( {\int\limits_{\text{s}}^{\text{t}} {\text{E(z)dz}} } \right)^{\text{i}} \exp \left( { - {\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}\int\limits_{\text{s}}^{\text{t}} {\text{E(z)dz}} } \right){\text{ds}}} $$

(54)

Equation 11 is equal to Eq. 48 with i + 1, which completes the recursive proof that Eq. 48 holds true for ${\text{i}} = 1, \ldots ,{\text{n}}$. Finally, to derive Eq. 6 one needs to replace Y_i(t) and k(τ) in Eq. 48 with the original P_i and k_in(t) using Eq. 44.

Appendix 2

Derivation of Eq. 16

Equation 6 with i = n implies that the solution for the n-th compartment is

$$ {\text{P}}_{\text{n}} ({\text{t}}) = {\frac{{{\text{R}}_{0} }}{\text{n}}} + {\frac{1}{{({\text{n}} - 1)!}}}\left( {{\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}} \right)^{{{\text{n}} - 1}} \int\limits_{0}^{\text{t}} {\left( {{\text{k}}_{\text{in}} (\tau ) - {\frac{{{\text{R}}_{0} }}{{{\text{T}}_{\text{R}} }}}{\text{E}}(\tau )} \right)\left( {\int\limits_{\tau }^{\text{t}} {\text{E(z)dz}} } \right)^{{{\text{n}} - 1}} \exp \left( { - {\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}\int\limits_{\tau }^{\text{t}} {\text{E(z)dz}} } \right)} {\text{d}}\tau $$

(55)

Changing the variables τ → t − τ in the first integral results in

$$ {\text{P}}_{\text{n}} ( {\text{t)}} = {\frac{{{\text{R}}_{ 0} }}{\text{n}}} + {\frac{1}{{({\text{n}} - 1)!}}}\left( {{\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}} \right)^{{{\text{n}} - 1}} \int\limits_{0}^{\text{t}} {\left( {{\text{k}}_{\text{in}} ({\text{t}} - \tau ) - {\frac{{{\text{R}}_{0} }}{{{\text{T}}_{\text{R}} }}}{\text{E}}({\text{t}} - \tau )} \right)\left( {\int\limits_{{{\text{t}} - \tau }}^{\text{t}} {\text{E(z)dz}} } \right)^{{{\text{n}} - 1}} \exp \left( { - {\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}\int\limits_{{{\text{t}} - \tau }}^{\text{t}} {\text{E(z)dz}} } \right)} {\text{d}}\tau $$

(56)

The difference in the integrand can be integrated separately as follows

$$ \begin{gathered} {\text{P}}_{\text{n}} ( {\text{t)}} = {\frac{{{\text{R}}_{0} }}{\text{n}}} - {\frac{1}{{({\text{n}} - 1)!}}}\,{\frac{{{\text{R}}_{0} }}{{{\text{T}}_{\text{R}} }}}\left( {{\frac{\text{n}}{\text{T}}}} \right)^{{{\text{n}} - 1}} \int\limits_{0}^{\text{t}} {{\text{E}}({\text{t}} - \tau )\left( {\int\limits_{{{\text{t}} - \tau }}^{\text{t}} {\text{E(z)dz}} } \right)^{{{\text{n}} - 1}} \exp \left( { - {\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}\int\limits_{{{\text{t}} - \tau }}^{\text{t}} {\text{E(z)dz}} } \right)} {\text{d}}\tau \hfill \\ + {\frac{1}{{({\text{n}} - 1)!}}}\left( {{\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}} \right)^{{{\text{n}} - 1}} \int\limits_{0}^{\text{t}} {{\text{k}}_{\text{in}} ({\text{t}} - \tau )\left( {\int\limits_{{{\text{t}} - \tau }}^{\text{t}} {\text{E(z)dz}} } \right)^{{{\text{n}} - 1}} \exp \left( { - {\frac{\text{n}}{{{\text{T}}_{\text{R}} }}}\int\limits_{{{\text{t}} - \tau }}^{\text{t}} {\text{E(z)dz}} } \right)} {\text{d}}\tau \hfill \\ \end{gathered} $$

(57)

The change of the variable

$$ {\text{x}} \to \int\limits_{{{\text{t}} - \tau }}^{\text{t}} {\text{E(z)dz}} $$

(58)

reduces the integral in the second term of Eq. 57 to the following form

$$ \begin{gathered} {\text{I}}_{\text{n}} = {\frac{1}{{({\text{n}} - 1)!}}}\,{\frac{{{\text{R}}_{0} }}{{{\text{T}}_{{\text{R}}} }}}\left( {{\frac{{\text{n}}}{{{\text{T}}_{{\text{R}}} }}}} \right)^{{\text{n}} - 1} \int\limits_{0}^{{\text{t}}} {{\text{E}}({\text{t}} - \tau )\left( {\int\limits_{{\text{t}} - \tau }^{{\text{t}}} {{\text{E}}({\text{z}}){\text{dz}}} } \right)^{{\text{n}} - 1} \exp \left( { - {\frac{{\text{n}}}{{{\text{T}}_{{\text{R}}} }}}\int\limits_{{\text{t}} - \tau }^{{\text{t}}} {{\text{E(z)dz}}} } \right)} {\text{d}}\tau = \hfill \\ {\frac{{{\text{R}}_{0} }}{{\text{n}}}}\,{\frac{1}{({\text{n}} - 1)!}}\left( {{\frac{{\text{n}}}{{{\text{T}}_{{\text{R}}} }}}} \right)^{{\text{n}}} \int\limits_{0}^{{\int\limits_{0}^{{\text{t}}} {{\text{E(z)dz}}} }} {{\text{x}}^{{\text{n}} - 1} \exp \left( { - {\frac{{\text{n}}}{{{\text{T}}_{{\text{R}}} }}}x} \right)} {\text{dx}} \hfill \\ \end{gathered} $$

(59)

The properties of the gamma distribution imply that [21]

$$ {\frac{1}{({\text{n}} - 1)!}}\left( {{\frac{{\text{n}}}{{{\text{T}}_{{\text{R}}} }}}} \right)^{{\text{n}}} \int\limits_{0}^{\infty } {{\text{x}}^{{\text{n}} - 1} \exp \left( { - {\frac{{\text{n}}}{{{\text{T}}_{{\text{R}}} }}}{\text{x}}} \right)} {\text{dx}} = 1 $$

(60)

Consequently,

$$ {\text{I}}_{{\text{n}}} = {\frac{{{\text{R}}_{0} }}{{\text{n}}}} - {\frac{1}{({\text{n}} - 1)!}}{\frac{{{\text{R}}_{0} }}{{{\text{T}}_{{\text{R}}} }}}\left( {{\frac{{\text{n}}}{{{\text{T}}_{{\text{R}}} }}}} \right)^{{\text{n}} - 1} \int\limits_{{\text{t}}}^{\infty } {{\text{E(t}} - \tau )\left( {\int\limits_{{\text{t}} - \tau }^{t} {{\text{E(z)dz}}} } \right)^{{\text{n}} - 1} \exp \left( { - {\frac{{\text{n}}}{{{\text{T}}_{{\text{R}}} }}}\int\limits_{{\text{t}} - \tau }^{{\text{t}}} {{\text{E(z)dz}}} } \right)} d\tau $$

(61)

where the original integration variable τ was restored in the integral in Eq. 60. The baseline conditions Eq. 3 imply that for τ > t

$$ {\frac{{{\text{R}}_{0} }}{{{\text{T}}_{{\text{R}}} }}} = {\text{k}}_{{\text{in}}0} = {\text{k}}_{{\text{in}}} ({\text{t}} - \tau ) \quad{\text{and}}\quad{\text{E}}({\text{t}} - \tau ) = 1 $$

(62)

Hence

$$ {\text{I}}_{{\text{n}}} = {\frac{{{\text{R}}_{0} }}{{\text{n}}}} - {\frac{1}{({\text{n}} - 1)!}}\left( {{\frac{{\text{n}}}{{{\text{T}}_{{\text{R}}} }}}} \right)^{{\text{n}} - 1} \int\limits_{{\text{t}}}^{\infty } {{\text{k}}_{{\text{in}}} ({\text{t}} - \tau )\left( {\int\limits_{{\text{t}} - \tau }^{{\text{t}}} {{\text{E(z)dz}}} } \right)^{{\text{n}} - 1} \exp \left( { - {\frac{{\text{n}}}{{{\text{T}}_{{\text{R}}} }}}\int\limits_{{\text{t}} - \tau }^{{\text{t}}} {{\text{E(z)dz}}} } \right)} d\tau $$

(63)

One can insert I_n back in Eq. 57 resulting in

$$ {\text{P}}_{{\text{n}}} = {\frac{1}{({\text{n}} - 1)!}}\left( {{\frac{{\text{n}}}{{{\text{T}}_{{\text{R}}} }}}} \right)^{{\text{n}} - 1} \int\limits_{0}^{\infty } {{\text{k}}_{{\text{in}}} ({\text{t}} - \tau )\left( {\int\limits_{{\text{t}} - \tau }^{{\text{t}}} {{\text{E(z)dz}}} } \right)^{{\text{n}} - 1} \exp \left( { - {\frac{{\text{n}}}{{{\text{T}}_{{\text{R}}} }}}\int\limits_{{\text{t}} - \tau }^{{\text{t}}} {{\text{E(z)dz}}} } \right)} {\text{d}}\tau $$

(64)

After multiplying both sides of Eq. 64 by nE(t)/T_R one arrives at Eq. 16.

Appendix 3

Derivation of Eq. 19

For given times t and τ let us define

$$ \hat{{\text{E}}}_{{\text{t}}} (\tau ) = \int\limits_{{\text{t}}}^{{\text{t}} + \tau } {{\text{E(z)dz}}} $$

(65)

Since the derivative

$$ {\frac{{{\text{d}}\hat{{\text{E}}}_{{\text{t}}} (\tau )}}{{\text{d}}\tau }} = {\text{E}}({\text{t}} + \tau ) $$

(66)

Equation 17 can be simplified to the following form

$$ \ell_{{\text{n}}} ({\text{t}},\tau ) = \left( {{\frac{{\text{n}}}{{{\text{T}}_{{\text{R}}} }}}} \right)^{{\text{n}}} {\frac{{\hat{{\text{E}}}_{{\text{t}}} (\tau )^{{\text{n}} - 1} }}{({\text{n}} - 1)!}}\exp \left( { - {\frac{{\text{n}}}{{{\text{T}}_{{\text{R}}} }}}\hat{{\text{E}}}_{{\text{t}}} (\tau )} \right){\frac{{{\text{d}}\hat{{\text{E}}}_{{\text{t}}} (\tau )}}{{\text{d}}\tau }} $$

(67)

Applying the change of variables $s \to \hat{{\text{E}}}_{{\text{t}}} (\tau )$ in the following integral, one can further reduce it to the integral of the gamma function γ_n(s) defined in Eq. 18:

$$ \int\limits_{0}^{\infty } {\ell_{{\text{n}}} ({\text{t}},\tau ){\text{d}}\tau = } \int\limits_{0}^{\infty } {\gamma_{{\text{n}}} {\text{(s)ds}} = 1} $$

(68)

The latter is a consequence of γ_n(s) as a probability density function. This proves Eq. 19.

Appendix 4

Derivation of Eq. 29

To prove Eq. 29 we use the definition of the Dirac delta function δ_τ0(τ) = δ(τ − τ₀) stating that for any test function ϕ(τ) (an indefinitely differentiable function with a compact support contained in the interval (0,∞)) the following relationship holds [19]:

$$ \int\limits_{0}^{\infty } {\delta (\tau - \tau_{0} )\phi (\tau ){\text{d}}\tau = \phi (\tau_{0} )} $$

(69)

Let ϕ(τ) be a test function, then according to Eq. 24

$$ \int\limits_{0}^{\infty } {\delta_{\tau 0} \bar{^ \circ }\Upphi_{{\text{t}}} (\tau )\phi (\tau ){\text{d}}\tau = } \int\limits_{0}^{\infty } {\delta \left( {\Upphi_{{\text{t}}} (\tau ) - \tau_{0} } \right)\Upphi_{{\text{t}}}^{\prime } (\tau )\phi (\tau ){\text{d}}\tau } $$

(70)

The substitution ${\text{s}} \to \Upphi_{{\text{t}}} (\tau )$ transforms the last integral in Eq. 70 to the following

$$ \int\limits_{0}^{\infty } {\delta \left( {{\text{s}} - \tau_{0} } \right)\phi \left( {\Upphi_{{\text{t}}}^{ - 1} ({\text{s}})} \right){\text{ds}}} = \phi \left( {\Upphi_{{\text{t}}}^{ - 1} (\tau_{0} )} \right) $$

(71)

where $\Upphi_{{\text{t}}}^{ - 1} ({\text{s}})$ is the inverse of $\Upphi_{{\text{t}}} (\tau )$. The integral in Eq. 71 is evaluated according to Eq. 69. Consequently, for any test function ϕ(τ)

$$ \int\limits_{0}^{\infty } {\delta_{\tau 0} \bar{^ \circ }\Upphi_{{\text{t}}} (\tau )\phi (\tau ){\text{d}}\tau } = \phi \left( {\Upphi_{{\text{t}}}^{ - 1} (\tau_{0} )} \right) $$

(72)

This proves Eq. 29.

Appendix 5

Derivation of Eqs. 30, 32, and 33

A general form for the basic LIDRE model for drug affecting the point distribution according to Eqs. 27 and 28 will be described as follows

$$ \frac{{\text{dR}}}{{\text{dt}}} = {\text{k}}_{{\text{in}}} ({\text{t}}) - \int\limits_{0}^{\infty } {{\text{k}}_{{\text{in}}} ({\text{t}} - \tau )\ell ({\text{t}} - \tau ,\tau )} {\text{d}}\tau $$

(73)

where after applying Eqs. 24, 65 and 66:

$$ \ell ({\text{t}},\tau ) = \delta_{{{\text{T}}_{{\text{R}}} }} \bar{^ \circ }\hat{{\text{E}}}_{{\text{t}}} (\tau ) = \delta \left( {\hat{{\text{E}}}_{{\text{t}}} (\tau ) - {\text{T}}_{{\text{R}}} } \right){\text{E}}({\text{t}} + \tau ) $$

(74)

Therefore Eq. 73 can be further simplified to

$$ \frac{{\text{dR}}}{{\text{dt}}} = {\text{k}}_{{\text{in}}} ({\text{t}}) - \int\limits_{0}^{\infty } {{\text{k}}_{{\text{in}}} ({\text{t}} - \tau )\delta } \left( {\hat{{\text{E}}}_{{\text{t}} - \tau } (\tau ) - {\text{T}}_{{\text{R}}} )} \right){\text{E(t)d}}\tau $$

(75)

To calculate the integral in Eq. 75 let us introduce a transformation Ψ_t(τ)

$$ \Uppsi_{{\text{t}}} (\tau ) = \hat{{\text{E}}}_{{\text{t}} - \tau } (\tau ) = \int\limits_{{\text{t}} - \tau }^{{\text{t}}} {\text{{E(z)dz}}} $$

(76)

One can verify that for any time t

$$ \Uppsi_{{\text{t}}} (0) = 0,\,\,\Uppsi_{{\text{t}}} (\infty ) = \infty ,\,\,{\text{and}}\,\,{\frac{{{\text{d}}\Uppsi_{{\text{t}}} }}{{\text{d}}\tau }}(\tau ) = {\text{E}}({\text{t}} - \tau ) > 0 $$

(77)

The change of variables s → Ψ_t(τ) transforms the integral in Eq. 74 to

$$ \int\limits_{0}^{\infty } {{\text{k}}_{{\text{in}}} \left( {{\text{t}} - \Uppsi_{{\text{t}}}^{ - 1} ({\text{s}})} \right)\delta ({\text{s}} - {\text{T}}_{{\text{R}}} ){\frac{{\text{E}}({\text{t}})}{{{\text{E}}\left( {{\text{t}} - \Uppsi_{{\text{t}}}^{ - 1} ({\text{s}})} \right)}}}{\text{ds}}} = {\text{k}}_{{\text{in}}} \left( {{\text{t}} - \Uppsi_{{\text{t}}}^{ - 1} ({\text{T}}_{{\text{R}}} )} \right){\frac{{\text{E(t)}}}{{{\text{E}}\left( {{\text{t}} - \Uppsi_{{\text{t}}}^{ - 1} ({\text{T}}_{{\text{R}}} )} \right)}}} $$

(78)

where the derivative of the inverse is

$$ {\frac{{{\text{d}}\Uppsi_{{\text{t}}}^{ - 1} }}{{\text{ds}}}} = {\frac{1}{{{\text{E}}\left( {{\text{t}} - \Uppsi_{{\text{t}}}^{ - 1} ({\text{s}})} \right)}}} $$

(79)

and the definition of the delta function δ(τ − T_R) has been utilized. To abbreviate the right hand side of Eq. 78 the following function of time t is introduced:

$$ {\text{T}}_{{\text{E}}} ({\text{t}}) = {\text{t}} - \Uppsi_{{\text{t}}}^{ - 1} ({\text{T}}_{{\text{R}}} ) $$

(80)

Then Eq. 73 simplifies to Eq. 30

$$ \frac{{\text{dR}}}{{\text{dt}}} = {\text{k}}_{{\text{in}}} ({\text{t}}) - {\text{k}}_{{\text{in}}} \left( {{\text{T}}_{{\text{E}}} ({\text{t}})} \right){\frac{{\text{E(t)}}}{{{\text{E}}\left( {{\text{T}}_{{\text{E}}} ({\text{t}})} \right)}}} $$

(81)

Notice that Eq. 81 was derived without an assumption of any specific form for k_in(t) like one specified by Eqs. 9 and 10. Equations 76 and 80 imply

$$ \int\limits_{{{\text{T}}_{{\text{E}}} ({\text{t}})}}^{{\text{t}}} {{\text{E(z)dz}} = {\text{T}}_{{\text{R}}} } $$

(82)

At t = 0, Eq. 82 becomes

$$ \int\limits_{{{\text{T}}_{{\text{E}}} (0)}}^{0} {{\text{E(z)dz}} = {\text{T}}_{{\text{R}}} } $$

(83)

Since T_R > 0 and E(z) > 0, then T_E(0) < 0. For z < 0, E(z) = 1 (see Eq. 3), which implies

$$ {\text{T}}_{\text{E}} \left( 0 \right) = - {\text{T}}_{\text{R}} $$

(84)

This proves Eq. 33. To derive a differential equation for T_E(t), calculate the derivative with respect to t of both sides of Eq. 82. The derivative of the right-hand side is zero, and the derivative of the integral in the left-hand side results in the following equation:

$$ {\text{E(t)}} - {\frac{{{\text{dT}}_{{\text{E}}} ({\text{t}})}}{{\text{dt}}}}{\text{E}}\left( {{\text{T}}_{{\text{E}}} ({\text{t}})} \right) = 0 $$

(85)

That results in Eq. 32.

Appendix 6

Proof of Eqs. 37–39

Integration of both sides of Eq. 26 leads to

$$ {\text{R}}_{{\text{n}}} ({\text{t}}) = {\text{R}}_{0} + \int\limits_{0}^{{\text{t}}} {{\text{k}}_{{\text{in}}} {\text{(s)ds}}} - \int\limits_{0}^{{\text{t}}} {\int\limits_{0}^{\infty } {{\text{k}}_{{\text{in}}} ({\text{s}} - \tau )\ell_{{\text{n}}} ({\text{t}} - \tau ,\tau ){\text{d}}\tau {\text{ds}}} } $$

(86)

Similarly, integration of Eq. 30 results in

$$ {\text{R}}_{\infty } ({\text{t}}) = R_{0} + \int\limits_{0}^{{\text{t}}} {{\text{k}}_{{\text{in}}} {\text{(s)ds}}} - \int\limits_{0}^{{\text{t}}} {{\text{k}}_{{\text{in}}} \left( {{\text{T}}_{{\text{E}}} ({\text{s}})} \right)} {\frac{{\text{E(s)}}}{{E\left( {{\text{T}}_{{\text{E}}} ({\text{s}})} \right)}}}{\text{ds}} $$

(87)

Upon subtraction of both sides of Eqs. 87 and 86:

$$ {\text{R}}_{\infty } ({\text{t}}) - {\text{R}}_{{\text{n}}} ({\text{t}}) = \int\limits_{0}^{{\text{t}}} {\int\limits_{0}^{\infty } {{\text{k}}_{{\text{in}}} ({\text{s}} - \tau )\ell_{{\text{n}}} ({\text{t}} - \tau ,\tau ){\text{d}}\tau {\text{ds}}} } - \int\limits_{0}^{{\text{t}}} {{\text{k}}_{{\text{in}}} \left( {{\text{T}}_{{\text{E}}} ({\text{s}})} \right)} {\frac{{\text{E(s)}}}{{{\text{E}}\left( {{\text{T}}_{{\text{E}}} ({\text{s}})} \right)}}}{\text{ds}} $$

(88)

Following the definition of Ψ_t(τ), Eq. 88 implies that

$$ \int\limits_{0}^{\infty } {{\text{k}}_{{\text{in}}} ({\text{s}} - \tau )\ell_{{\text{n}}} ({\text{t}} - \tau ,\tau ){\text{d}}\tau } = \int\limits_{0}^{\infty } {{\text{k}}_{{\text{in}}} \left( {{\text{s}} - \Uppsi_{{\text{s}}}^{ - 1} ({\text{z}})} \right){\frac{{\text{E(s)}}}{{{\text{E}}\left( {{\text{s}} - \Uppsi_{{\text{s}}}^{ - 1} ({\text{z}})} \right)}}}\gamma_{{\text{n}}} {\text{(z)dz}}} = \int\limits_{0}^{\infty } {\phi_{{\text{s}}} ({\text{z}})\gamma_{{\text{n}}} {\text{(z)dz}}} $$

(89)

Analogously, Eq. 80 implies

$$ {\text{k}}_{{\text{in}}} \left( {{\text{T}}_{{\text{E}}} ({\text{s}})} \right){\frac{{\text{E}}({\text{s}})}{{{\text{E}}\left( {{\text{T}}_{{\text{E}}} ({\text{s}})} \right)}}} = {\text{k}}_{{\text{in}}} \left( {{\text{t}} - \Uppsi_{{\text{s}}}^{ - 1} ({\text{T}}_{{\text{R}}} )} \right){\frac{{\text{E(s)}}}{{{\text{E}}\left( {{\text{s}} - \Uppsi_{{\text{s}}}^{ - 1} ({\text{T}}_{{\text{R}}} )} \right)}}} = \phi_{{\text{s}}} ({\text{T}}_{{\text{R}}} ) $$

(90)

Hence

$$ {\text{R}}_{\infty } ({\text{t}}) - {\text{R}}_{{\text{n}}} ({\text{t}}) = \int\limits_{0}^{{\text{t}}} {\int\limits_{0}^{\infty } {\phi_{{\text{s}}} ({\text{z}})\gamma_{{\text{n}}} {\text{(z)dzds}}} } - \int\limits_{0}^{t} {\phi_{{\text{s}}} ({\text{T}}_{{\text{R}}} ){\text{ds}}} $$

(91)

and Eq. 37 follows. To show that Eq. 37 holds true one can transform Eq. 91 further utilizing the fact that γ_n(τ) is a p.d.f.

$$ {\text{R}}_{\infty } ({\text{t}}) - {\text{R}}_{\text{{n}}} ({\text{t}}) = \int\limits_{0}^{{\text{t}}} {\int\limits_{0}^{\infty } {\left( {\phi_{{\text{s}}} {\text{(z)}} - \phi_{{\text{s}}} ({\text{T}}_{{\text{R}}} )} \right)\gamma_{{\text{n}}} {\text{(z)dzds}}} } $$

(92)

Since Eqs. 7–10 imply that ϕ_s(τ) is a bound function of s and τ, i.e. there is a constant M > 0 such that for any s and τ

$$ \phi_{{\text{s}}} (\tau ) \le {\text{M}} $$

(93)

Hence

$$ \left| {\int\limits_{0}^{\infty } {\left( {\phi_{{\text{s}}} ({\text{z}}) - \phi_{{\text{s}}} ({\text{T}}_{{\text{R}}} )} \right)\gamma_{{\text{n}}} {\text{(z)dz}}} } \right| \le 2{\text{M}}\int\limits_{0}^{\infty } {\gamma_{{\text{n}}} {\text{(z)dz}}} = 2{\text{M}} $$

(94)

The dominated convergence theorem [25] implies that for any t the right hand side of Eq. 92 approaches 0 as n → ∞, for any s

$$ \int\limits_{0}^{\infty } {\left( {\phi_{{\text{s}}} ({\text{z}}) - \phi_{{\text{s}}} ({\text{T}}_{{\text{R}}} )} \right)\gamma_{{\text{n}}} {\text{(z)dz}}} \to 0\,\,{\text{as n}} \to \infty $$

(95)

Equation 95 is true due to the approximate identity property of the gamma distribution p.d.f. γ_n(τ) cited in Eq. 36, which we will prove for an arbitrary continuous and bound function f(τ). Let ε > 0 be a small number and M > 0 be such |f(τ)| ≤ M for all τ. Then the continuity of f(τ) implies that there is a small ε > 0 such that for any T_R-ε < τ < T_R + ε

$$ |{\text{f}}(\tau ) -{\text{f}}({\text{T}}_{{\text{R}}} )| \le \varepsilon $$

(96)

The integral in Eq. 95 can be partitioned into a sum of three integrals:

$$ \begin{gathered} \int\limits_{0}^{\infty } {({\text{f}}(\tau ) - {\text{f(T}}_{{\text{R}}} ))\gamma_{{\text{n}}} (\tau ){\text{d}}\tau } = \hfill \\ \int\limits_{0}^{{{\text{T}}_{{\text{R}}} - \varepsilon }} {({\text{f}}(\tau ) - {\text{f(T}}_{{\text{R}}} ))\gamma_{{\text{n}}} (\tau ){\text{d}}\tau } + \int\limits_{{{\text{T}}_{{\text{R}}} - \varepsilon }}^{{{\text{T}}_{{\text{R}}} + \varepsilon }} {({\text{f}}(\tau ) - {\text{f}}({\text{T}}_{{\text{R}}} ))\gamma_{{\text{n}}} (\tau ){\text{d}}\tau } + \int\limits_{{{\text{T}}_{{\text{R}}} + \varepsilon }}^{\infty } {({\text{f}}(\tau ) - {\text{f(T}}_{{\text{R}}} ))\gamma_{{\text{n}}} (\tau ){\text{d}}\tau } \hfill \\ \end{gathered} $$

(97)

The first and third integral in Eq. 97 can be bound as follows:

$$ \left| {\int\limits_{0}^{{{\text{T}}_{{\text{R}}} - \varepsilon }} {\left( {{\text{f}}(\tau ) - {\text{f}}({\text{T}}_{{\text{R}}} )} \right)\gamma_{{\text{n}}} (\tau ){\text{d}}\tau } } \right| \le 2{\text{M}}\int\limits_{0}^{{{\text{T}}_{{\text{R}}} - \varepsilon }} {\gamma_{{\text{n}}} (\tau ){\text{d}}\tau } $$

(98)

and

$$ \left| {\int\limits_{{{\text{T}}_{{\text{R}}} - \varepsilon }}^{\infty } {\left( {{\text{f}}(\tau ) - {\text{f(T}}_{{\text{R}}} )} \right)\gamma_{{\text{n}}} (\tau ){\text{d}}\tau } } \right| \le 2{\text{M}}\int\limits_{{{\text{T}}_{{\text{R}}} - \varepsilon }}^{\infty } {\gamma_{{\text{n}}} (\tau ){\text{d}}\tau } $$

(99)

The bound for the second integral in Eq. 97 is

$$ \left| {\int\limits_{{{\text{T}}_{{\text{R}}} - \varepsilon }}^{{{\text{T}}_{{\text{R}}} + \varepsilon }} {\left( {{\text{f}}(\tau ) -{\text{f}}({\text{T}}_{{\text{R}}} )} \right)\gamma_{{\text{n}}} (\tau ){\text{d}}\tau } } \right| \le \varepsilon \int\limits_{{{\text{T}}_{{\text{R}}} - \varepsilon }}^{{{\text{T}}_{{\text{R}}} + \varepsilon }} {\gamma_{{\text{n}}} (\tau ){\text{d}}\tau } \le \varepsilon \int\limits_{0}^{\infty } {\gamma_{{\text{n}}} (\tau ){\text{d}}\tau } = \varepsilon $$

(100)

Since

$$ \int\limits_{0}^{{{\text{T}}_{{\text{R}}} - \varepsilon }} {\gamma_{{\text{n}}} (\tau ){\text{d}}\tau } \to 0 \quad {\text{and}}\,\int\limits_{{{\text{T}}_{{\text{R}}} - \varepsilon }}^{\infty } {\gamma_{{\text{n}}} (\tau ){\text{d}}\tau } \to 0 \quad {\text{as n}} \to \infty $$

(101)

the first and third integral in Eq. 97 can be bound by ε for large enough n, and consequently

$$ \left| {\int\limits_{0}^{\infty } {\left( {{\text{f}}(\tau ) - {\text{f}}({\text{T}}_{{\text{R}}} )} \right)\gamma_{{\text{n}}} (\tau ){\text{d}}\tau } } \right| \le 3\varepsilon $$

(102)

This completes the derivations of Eqs. 36, 37 and 38.