Evaluation of Shear-Induced Platelet Activation Models Under Constant and Dynamic Shear Stress Loading Conditions Relevant to Devices

Abstract

The advent of implantable blood-recirculating devices such as left ventricular assist devices and prosthetic heart valves provides a viable therapy for patients with end-stage heart failure and valvular disease. However, device-generated pathological flow patterns result in thromboembolic complications that require complex and lifelong anticoagulant therapy, which entails hemorrhagic risks and is not appropriate for certain patients. Optimizing the thrombogenic performance of such devices utilizing numerical simulations requires the development of predictive platelet activation models that account for variations in shear-loading rates characterizing blood flow through such devices. Platelets were exposed in vitro to both dynamic and constant shear stress conditions emulating those found in blood-recirculating devices in order to determine their shear-induced activation and sensitization response. Both these behaviors were found to be dependent on the shear loading rates, in addition to shear stress magnitude and exposure time. We then critically examined several current models and evaluated their predictive capabilities using these results. Shear loading rate terms were then included to account for dynamic aspects that are either ignored or partially considered by these models, and model parameters were optimized. Independent optimization for each of the two types of shear stress exposure conditions tested resulted in different sets of best-fit constants, indicating that universal optimization may not be possible. Inherent limitations of the current models require a paradigm shift from these integral-based discretized power law models to better address the dynamic conditions encountered in blood-recirculating devices.

Appendix

Derivation of the Cumulative Power Law Model (CPL) with Loading Rate

We wish to derive a power-law formulation that includes a shear loading term, $\dot{\tau }$, that accounts for the change in shear stress as opposed to assuming the shear stress is applied in a stepwise manner. The following approach utilizes the derivation method for a prior blood damage model.16 We start with the simple equation that accounts for shear stress and time (Eq. (A1)):

$$ {\text{PAS}}(\tau_{\text{const}} ,t) = C\tau^{\alpha } t_{\exp }^{\beta } $$

(A1)

This is the simple extrapolation of the Giersiepen expression for platelet damage to predict the instantaneous value of the platelet activation state (PAS) at a given time point (“original power law—OPL”). For the dynamic case where shear stress is a function of time, $\tau = \tau (t)$, we consider the integral form of the quantities $\tau$ and $t$. We first differentiate Eq. (A1) to obtain the loading rate term, yielding:

$$ \frac{d}{dt}\left( {C\tau (t)^{\alpha } t^{\beta } } \right) = C\beta \tau (t)^{\alpha } t^{\beta - 1} + C\alpha \tau (t)^{\alpha - 1} t^{\beta } \frac{d\tau (t)}{dt} $$

(A2)

The final term$\frac{d\tau (t)}{dt}$ is the shear loading rate term. We substitute this into Eq. (A1), and then integrate to obtain the form presented in Eq. (A2):

$$ \int\limits_{{t_{0} }}^{{t_{\text{total}} }} {\frac{d}{dt}\left( {C\tau (t)^{\alpha } t^{\beta } } \right)} dt = \int\limits_{{t_{0} }}^{{t_{\text{total}} }} {\left( {C\beta \tau (t)^{\alpha } t^{\beta - 1} + C\alpha \tau (t)^{\alpha - 1} t^{\beta } \frac{d\tau (t)}{dt}} \right)} dt $$

(A3)

Therefore, taking the integral yields the following without loss of generality:

$$ \begin{aligned} {\text{PAS}}(\tau (t),t)\left| {_{{t_{0} }}^{{t_{\text{total}} }} } \right. = & C\tau (t)^{\alpha } t^{\beta } \left| {_{{t_{0} }}^{{t_{\text{total}} }} } \right. \\ \, = & C\int\limits_{{t_{0} }}^{{t_{\text{total}} }} {\beta \tau (t)^{\alpha } t^{\beta - 1} dt + C\int\limits_{{t_{0} }}^{{t_{\text{total}} }} {\alpha \tau (t)^{\alpha - 1} t^{\beta } \frac{d\tau (t)}{dt}dt} + C_{0} } \\ \end{aligned} $$

(A4)

The constant of integration, C ₀, is defined as the non-zero initial platelet activation state at $t$ = 0, or PAS(t ₀). The absolute value of $d\tau (t)$ is taken to avoid violating the principle of causality. The cumulative power law PAS model (CPL) with loading rate then becomes:

$$ {\text{PAS}}_{\text{CPL}} (\tau (t),t)\left| {_{{t_{0} }}^{{t_{\text{total}} }} } \right. = C_{1} \int\limits_{{t_{0} }}^{{t_{\text{total}} }} {\beta \tau (t)^{\alpha } t^{\beta - 1} dt + C_{2} \int\limits_{{t_{0} }}^{{t_{\text{total}} }} {\alpha \tau (t)^{\alpha - 1} t^{\beta } \frac{{\left| {d\tau (t)} \right|}}{dt}dt} + {\text{PAS}}(t_{0} )} $$

(A5)

The constants C ₁ and C ₂ represent C for the constant and dynamic shear stress parts of the model, respectively. For constant shear stress conditions, where $\dot{\tau }$ = 0, Eq. (A5) can be rewritten:

$$ {\text{PAS}}_{\text{CPL}} (\tau (t),t)\left| {_{{t_{0} }}^{{t_{\text{total}} }} } \right. = C_{1} \int\limits_{{t_{0} }}^{{t_{\text{total}} }} {\beta \tau (t)^{\alpha } t^{\beta - 1} dt + {\text{PAS}}(t_{0} )} $$

(A6)

This is the original CPL model16 adapted to PAS. For utilization in a computational tool, such as MATLAB, Eq. (A5) is discretized:

$$ PAS_{CPL} (\tau,t) = C_{1} \beta \sum\limits_{i = 1}^{n} {t_{i}^{\beta - 1} \tau_{i}^{\alpha} \Updelta t} + C_{2} \alpha \sum\limits_{i = 1}^{n} {\tau_{i}^{\alpha - 1} t_{i}^{\beta} \frac{{\left| {\Updelta \tau_{i}} \right|}}{\diagup\!\!\!\!\!\!\Updelta t}} {\diagup\!\!\!\!\!\!\Updelta t}\, + PAS(t_{0}) $$

Rewriting $\Updelta \tau_{i}$, we get Eq. (A7):

$$ {\text{PAS}}_{\text{CPL}} ({{\uptau}},{\text{t}}) = C_{1} \beta \sum\limits_{i = 1}^{n} {t_{i}^{\beta - 1} \tau_{i}^{\alpha } \Updelta t} + C_{2} \alpha \sum\limits_{i = 1}^{n} {\tau_{i}^{\alpha - 1} t_{i}^{\beta } \left| {\tau_{i} - \tau_{i - 1} } \right|} + {\text{PAS}}(t_{0} ) $$

(A7)

Similarly, Eq. (A6) is discretized:

$$ {\text{PAS}}_{\text{CPL}} (\tau ,t) = C_{1} \beta \sum\limits_{i = 1}^{n} {t_{i}^{\beta - 1} \tau_{i}^{\alpha } \Updelta t} + {\text{PAS}}(t_{0} ) $$

(A8)

Derivation of the Modified Cumulative Power Law (MPL) Model with Loading Rate

As in the previous section, we want to derive a power-law formulation that includes a shear loading term, $\dot{\tau }$, that accounts for the change in shear stress as opposed to assuming the shear stress is applied in a stepwise manner. However, we want to truly account for loading history, where two groups of platelets exposed to different mechanical loadings are expected to show different responses in a subsequent loading, even if the latter is the same for both groups. That means that the PAS at each loading stage is dependent on the PAS of a previous loading scheme. Equations (A5) and (A6) do not satisfy these requirements, since only the initial activation, PAS(t ₀), is considered. Thus, we need to follow the Grigioni approach,17 with adaptation for PAS.25 This approach is different from that utilized for the earlier CPL formulation16 in that the latter does not consider the loading history of the platelet (i.e., a mechanical dose function). We start with the simple equation that accounts for shear stress and time (Eq. (A9)):

$$ {\text{PAS}}(\tau_{\text{const}} ,t) = C\tau^{\alpha } t^{\beta } $$

(A9)

Grouping the independent variables $\tau$ and t on one side, we get:

$$ \frac{{{\text{PAS}}(\tau ,t)}}{C} = \tau^{\alpha } t^{\beta } $$

(A10)

Here, we define a mechanical dose function, D:

$$ D = \root{\beta } \of {{\frac{{{\text{PAS}}(\tau ,t)}}{C}}} = \tau^{\alpha /\beta } t $$

(A11)

Thus, the function for PAS can be rewritten as:

$$ {\text{PAS}}(\tau ,t) = C \cdot D^{\beta } $$

(A12)

Applying the chain rule, we get the form:

$$ \frac{dD}{dt} = \frac{\partial D}{\partial t} + \frac{\partial D}{\partial \tau } \cdot \frac{d\tau }{dt} = \tau^{\alpha /\beta } + \frac{\alpha }{\beta }t \cdot \tau^{(\alpha /\beta ) - 1} \dot{\tau } $$

(A13)

Grigioni et al. neglected $\dot{\tau }$, claiming that it violates the principle of causality and causes a reduction in damage if shear stress is decreasing.17 We diverge from this approach by accounting for the shear loading rate and taking its absolute value so that its effect is always additive. In partial differential form, the above equation becomes:

$$ dD = \tau^{\alpha /\beta } dt + \frac{\alpha }{\beta }t \cdot \tau^{(\alpha /\beta ) - 1} d\tau $$

(A14)

Integrating with respect to time and shear stress, the above equation becomes:

$$ D(t) - D(t_{0} ) = \int\limits_{{t_{0} }}^{\phi } {\tau (\phi )^{\alpha /\beta } d\phi } + \frac{\alpha }{\beta }\int\limits_{{\tau (t_{0} )}}^{\tau (\phi )} {\phi \cdot \tau (\phi )^{(\alpha /\beta ) - 1} d\tau } $$

(A15)

We must rewrite D(t ₀) in terms of the initial platelet activation state, PAS(t ₀):

$$ \begin{gathered} {\text{PAS}}(t_{0} ) = C \cdot D(t_{0} )^{\beta } \hfill \\ D(t_{0} ) = \left( {\frac{{{\text{PAS}}(t_{0} )}}{C}} \right)^{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 \beta }}\right.\kern-0pt} \!\lower0.7ex\hbox{$\beta $}}}} \hfill \\ \end{gathered} $$

(A16)

Taking the derivative of the PAS function in terms of the dose D, we obtain:

$$ d({\text{PAS}}) = d(C \cdot D^{\beta } ) = Cd(D^{\beta } ) = C\beta D^{\beta - 1} dD $$

(A17)

Putting this equation in integral form yields:

$$ d({\text{PAS}}) = C\beta \left( {\int\limits_{{t_{0} }}^{\phi } {\tau (\xi )^{\alpha /\beta } d\xi } + \frac{\alpha }{\beta }\int\limits_{{\Upgamma (t_{0} )}}^{\Upgamma (\phi )} {\xi \cdot \tau (\xi )^{(\alpha /\beta ) - 1} d\tau } + D(t_{{^{0} }} )} \right)^{\beta - 1} dD $$

(A18)

The inner integrals represent the total mechanical stress loading dose D divided into constant and dynamic shear stress terms and applied over a time $\xi$, where $d\xi$ is the interval over which an elemental dose of shear stress is applied and $\Upgamma$ is the shear stress at time$\xi$. The outer integral is the summation of these doses over the duration of the experiment, with $\phi$ and $d\phi$ as the observation time points and the interval between them, respectively. Substituting Eq. (A14) for dD yields:

$$ d({\text{PAS}}) = C\beta \left( {\int\limits_{{t_{0} }}^{\phi } {\tau (\xi )^{\alpha /\beta } d\xi } + \frac{\alpha }{\beta }\int\limits_{{\tau (t_{0} )}}^{\tau (\phi )} {\xi \cdot \tau (\xi )^{(\alpha /\beta ) - 1} d\tau } + D(t_{{^{0} }} )} \right)^{\beta - 1} \times \left( {\tau^{\alpha /\beta } d\phi + \frac{\alpha }{\beta }t \cdot \tau^{(\alpha /\beta ) - 1} d\tau } \right) $$

(A19)

Expressing the platelet activation state (PAS) as the integral sum of the infinitesimal contributions represented by the above equation yields:

$$ \begin{gathered} {\text{PAS}} = C\beta \left[ {\int\limits_{{t_{0} }}^{{t_{\text{total}} }} {\left( {\int\limits_{{t_{0} }}^{\phi } {\tau (\xi )^{\alpha /\beta } d\xi } + \frac{\alpha }{\beta }\int\limits_{{\tau (t_{0} )}}^{\tau (\phi )} {\xi \cdot \tau (\xi )^{(\alpha /\beta ) - 1} d\tau } + D(t_{{^{0} }} )} \right)^{\beta - 1} \tau^{\alpha /\beta } d\phi } } \right. \hfill \\ \, \left. { + \int\limits_{{\tau (t_{0} )}}^{{\tau (t_{\text{total}} )}} {\frac{\alpha }{\beta }\left( {\int\limits_{{t_{0} }}^{\phi } {\tau (\xi )^{\alpha /\beta } d\xi } + \frac{\alpha }{\beta }\int\limits_{{\tau (t_{0} )}}^{\tau (\phi )} {\xi \cdot \tau (\xi )^{{^{(\alpha /\beta ) - 1} }} d\tau } + D(t_{{^{0} }} )} \right)^{\beta - 1} t \cdot \tau^{{^{(\alpha /\beta ) - 1} }} d\tau } } \right] \hfill \\ \end{gathered} $$

(A20)

PAS can be numerically computed by adding the mechanical doses acting on a platelet trajectory. The discrete elemental dose $\Updelta ({\text{PAS}})_{i}$ is sustained by a platelet in the i-th interval, from the instant $t_{i - 1}$ to $t_{i}$, and is expressed as:

$$ \begin{gathered} \Updelta ({\text{PAS}})_{i} = C\beta \left[ {\left( {\sum\limits_{j = 1}^{i} {\tau (t_{j} )^{\alpha /\beta } \Updelta t_{j} } + \frac{\alpha }{\beta }\sum\limits_{j = 1}^{i} {t_{j} \cdot \tau (t_{j} )^{(\alpha /\beta ) - 1} \left| {\Updelta \tau_{j} } \right|} + D(t_{{^{0} }} )} \right)^{\beta - 1} \tau (t_{i} )^{\alpha /\beta } \Updelta t_{i} } \right. \hfill \\ \, \left. { + \frac{\alpha }{\beta }\left( {\sum\limits_{j = 1}^{i} {\tau (t_{j} )^{\alpha /\beta } \Updelta t_{j} } + \frac{\alpha }{\beta }\sum\limits_{j = 1}^{i} {t_{j} \cdot \tau (t_{j} )^{(\alpha /\beta ) - 1} \left| {\Updelta \tau_{j} } \right|} + D(t_{{^{0} }} )} \right)^{\beta - 1} t_{i} \cdot \tau (t_{i} )^{(\alpha /\beta ) - 1} \left| {\Updelta \tau_{i} } \right|} \right] \hfill \\ \end{gathered} $$

(A21)

The time period $\Updelta t_{i}$ refers to the duration between experimental observations. Unlike previous derivations of this formula, we no longer assume the shear stress is constant in this interval, as we now have a shear loading term. The starting observation time, t ₀, is conventionally assumed to be equal to 0. The mechanical dose, due to dynamic behavior, is broken into multiple sub-intervals (referred to as the j-th interval in the above equation). Each sub-interval can be a constant shear stress dose if applied as such, or an approximation, where the dose is approximated as a constant shear stress part and a linearly ramped part, which includes the change in shear stress $\Updelta \tau_{j}$. Note that for the latter, the absolute value is taken in order not to violate the principle of causality.17 Thus, for each i-th interval, multiple j-th intervals may be present. From the integral form for the PAS function (Eq. (A20)), we derive the discrete form:

$$ \begin{gathered} {\text{PAS}} = C\beta \sum\limits_{i = 1}^{N} {\left[ {\left( {\sum\limits_{j = 1}^{i} {\tau (t_{j} )^{\alpha /\beta } \Updelta t_{j} } + \frac{\alpha }{\beta }\sum\limits_{j = 1}^{i} {t_{j} \cdot \tau (t_{j} )^{(\alpha /\beta ) - 1} \left| {\Updelta \tau_{j} } \right|} + D(t_{{^{0} }} )} \right)^{\beta - 1} \tau (t_{i} )^{\alpha /\beta } \Updelta t_{i} } \right.} \hfill \\ \, \left. { + \frac{\alpha }{\beta }\left( {\sum\limits_{j = 1}^{i} {\tau (t_{j} )^{\alpha /\beta } \Updelta t_{j} } + \frac{\alpha }{\beta }\sum\limits_{j = 1}^{i} {t_{j} \cdot \tau (t_{j} )^{{{\raise0.7ex\hbox{$\alpha $} \!\mathord{\left/ {\vphantom {\alpha \beta }}\right.\kern-0pt} \!\lower0.7ex\hbox{$\beta $}} - 1}} \left| {\Updelta \tau_{j} } \right|} + D(t_{{^{0} }} )} \right)^{\beta - 1} t_{i} \cdot \tau (t_{i} )^{(\alpha /\beta ) - 1} \left| {\Updelta \tau_{i} } \right|} \right] \hfill \\ \end{gathered} $$

(A22)

This represents the sum of PAS values over N intervals. In our experiments, PAS measurements are only taken during constant shear stress phases. Therefore, we can neglect the second part of the above equation. In addition, the term for D(t ₀) can also be substituted, yielding the discrete form of the modified cumulative power law model (MPL) with loading rate:

$$ {\text{PAS}}_{\text{MPL}} (\tau ,t) = C\beta \sum\limits_{i = 1}^{\rm N} {\left( {\sum\limits_{j = 1}^{i} {\tau (t_{j} )^{\alpha /\beta } \Updelta t_{j} } + \frac{\alpha }{\beta }\sum\limits_{j = 1}^{i} {t_{j} \cdot \tau (t_{j} )^{(\alpha /\beta ) - 1} \left| {\Updelta \tau_{j} } \right|} + \left( {\frac{{{\text{PAS}}(t_{0} )}}{C}} \right)^{1/\beta } } \right)^{\beta - 1} \tau (t_{i} )^{\alpha /\beta } \Updelta t_{i} } $$

(A23)

Making a similar simplification for Eq. (A20) yields:

$$ {\text{PAS}}_{\text{MPL}} (\tau ,t) = C\beta \left[ {\left. {\int\limits_{{t_{0} }}^{{t_{\text{total}} }} {\left( {\int\limits_{{t_{0} }}^{\phi } {\tau (\xi )^{\alpha /b} d\xi } + \frac{\alpha }{\beta }\int\limits_{{\tau (t_{0} )}}^{\tau (\phi )} {\xi \cdot \tau (\xi )^{(\alpha /\beta ) - 1} d\tau } + \left( {\frac{{{\text{PAS}}(t_{0} )}}{C}} \right)^{1/\beta } } \right)^{\beta - 1} \tau^{\alpha /b} d\phi } } \right]} \right. $$

(A24)

For constant shear stress conditions, where $\dot{\tau }$ = 0, Eqs. (A23) and (A24) can be rewritten as Eqs. (A25) and (A26), respectively. To allow comparison with other models in this study, we set C = C ₁. These 2 equations are the original MPL model for PAS:

$$ {\text{PAS}}_{\text{MPL}} (\tau ,t) = C_{1} \beta \sum\limits_{i = 1}^{\rm N} {\left( {\sum\limits_{j = 1}^{i} {\tau (t_{j} )^{\alpha /\beta } \Updelta t_{j} } + \left( {\frac{{{\text{PAS}}(t_{0} )}}{{C_{1} }}} \right)^{1/\beta} } \right)^{\beta - 1} \tau (t_{i} )^{\alpha /\beta } \Updelta t_{i} } $$

(A25)

$$ {\text{PAS}}_{\text{MPL}} (\tau ,t) = C_{1} \beta \left[ {\left. {\int\limits_{{t_{0} }}^{{t_{\text{total}} }} {\left( {\int\limits_{{t_{0} }}^{\phi } {\tau (\xi )^{\alpha /\beta } d\xi } + \left( {\frac{{{\text{PAS}}(t_{0} )}}{{C_{1} }}} \right)^{1/\beta } } \right)^{\beta - 1} \tau^{\alpha /\beta } d\phi } } \right]} \right. $$

(A26)

Equation (A24) provides a model that accounts for previous shear stress history and platelet activation, as well as the effect of the shear loading rate, the latter which was not addressed in prior models.8,25,27

Errors in Power Law Models Due to Discretization

The CPL model (Eq. (A8))16 is prone to accumulation of error resulting from the utilization of time steps in discrete form due to the use of powers on time. Consider the initial condition of PAS(t ₀) = PAS₀ at $t = t_{0}$, after which platelets are subjected to a constant shear stress, including at two subsequent times separated by $\Updelta t$. These time points are given at $t_{1} = t_{0} + \Updelta t$ and $t_{2} = t_{0} + 2\Updelta t$. Substituting into the OPL (Eq. (A1)) and rewriting yields:

$$ {\text{PAS}}_{1} = {\text{PAS}}(t_{1} ) = {\text{PAS}}_{0} + C\tau^{\alpha } (t_{1} - t_{0} )^{\beta } = {\text{PAS}}_{0} + C\tau^{\alpha } \Updelta t^{\beta } $$

(A27)

$$ {\text{PAS}}_{2} = {\text{PAS}}(t_{2} ) = {\text{PAS}}_{0} + C\tau^{\alpha } (t_{2} - t_{0} )^{\beta } = {\text{PAS}}_{0} + C\tau^{\alpha } (2\Updelta t)^{\beta } $$

(A28)

Alternatively, PAS₂ can be obtained by substituting the time of exposure from t ₁ to t ₂ into Eq. (A1) and adding PAS₁:

$$ {\text{PAS}}_{2} = {\text{PAS}}_{1} + C\tau^{\alpha } (t_{2} - t_{1} )^{\beta } = {\text{PAS}}_{0} + 2C\tau^{\alpha } \Updelta t^{\beta } $$

(A29)

Equations (A28) and (A29) are not equivalent if β ≠ 1, and therefore this form of the power law model16 is inconsistent.

While the mechanical dose term was redefined in the MPL17 to remove the power on time (Eq. (A11)), errors due to the discretization of the time intervals $\Updelta t_{i}$ and $\Updelta t_{j}$ are still present. Consider the case where shear stress, $\tau$, is constant over the time period $t_{0}$ to $t_{2}$. Thus, $\Updelta t = t_{2} - t_{1} = t_{1} - t_{0}$ and $2\Updelta t = t_{2} - t_{0}$, as described earlier. We set $\left| {\Updelta \tau_{j} } \right| = 0$ in Eq. (7), since the loading is instantaneous. For simplicity, we consider a single i-th interval, from $t_{0}$ to $t_{2}$, and set the i-th and j-th intervals equivalent in size. Substituting these values into Eq. (7), we obtain:

$$ \begin{gathered} {\text{PAS}}_{0 - 2} = \sum\limits_{i = 1}^{1} {C_{1} \beta \left[ {\sum\limits_{j = 1}^{1} {\left( {\tau_{j} } \right)^{\alpha /\beta } \Updelta t_{j} } + \left( {\frac{{{\text{PAS}}(t_{0} )}}{{C_{1} }}} \right)^{1/\beta } } \right]^{\beta - 1} \left( {\tau_{i} } \right)^{\alpha /\beta } \Updelta t_{i} } \hfill \\ \, = 2C_{1} \beta \left( \tau \right)^{\alpha /\beta } \Updelta t\left[ {2\left( \tau \right)^{\alpha /\beta } \Updelta t + \left( {\frac{{{\text{PAS}}(t_{0} )}}{{C_{1} }}} \right)^{1/\beta } } \right]^{\beta - 1} \hfill \\ \end{gathered} $$

(A30)

Alternatively, we consider 2 i-th and j-th intervals with the same sizes, from $t_{0}$ to $t_{1}$ and from $t_{1}$ to $t_{2}$. Substituting into Eq. (A21), and setting $\left| {\Updelta \tau_{i} } \right| = \left| {\Updelta \tau_{j} } \right| = 0$, we obtain:

$$ \begin{gathered} \Updelta ({\text{PAS}})_{0 - 1} = C_{1} \beta \left[ {\sum\limits_{j = 1}^{1} {\left( {\tau_{j} } \right)^{\alpha /\beta } } \Updelta t_{j} + \left( {\frac{{{\text{PAS}}(t_{0} )}}{{C_{1} }}} \right)^{1/\beta } } \right]^{\beta - 1} \left( {\tau_{i} } \right)^{\alpha /\beta } \Updelta t_{i} \hfill \\ \, = C_{1} \beta \left[ {\left( \tau \right)^{\alpha /\beta } \Updelta t + \left( {\frac{{{\text{PAS}}(t_{0} )}}{{C_{1} }}} \right)^{1/\beta } } \right]^{\beta - 1} \left( \tau \right)^{\alpha /\beta } \Updelta t \hfill \\ \end{gathered} $$

(A31)

$$ \begin{gathered} \Updelta ({\text{PAS}})_{1 - 2} = C_{1} \beta \left[ {\sum\limits_{j = 1}^{2} {\left( {\tau_{j} } \right)^{\alpha /\beta } } \Updelta t_{j} + \left( {\frac{{{\text{PAS}}(t_{0} )}}{{C_{1} }}} \right)^{1/\beta } } \right]^{a - 1} \left( {\tau_{i} } \right)^{\alpha /\beta } \Updelta t_{i} \hfill \\ \, = C_{1} \beta \left[ {\left( \tau \right)^{\alpha /\beta } \Updelta t + \left( \tau \right)^{\alpha /\beta } \Updelta t + \left( {\frac{{{\text{PAS}}(t_{0} )}}{{C_{1} }}} \right)^{1/\beta } } \right]^{a - 1} \left( \tau \right)^{\alpha /\beta } \Updelta t \, \hfill \\ \end{gathered} $$

(A32)

Adding Eqs. (A31) and (A32), we obtain:

$$ \begin{aligned} {\text{PAS}}_{0 - 2} = & \Updelta ({\text{PAS}})_{0 - 1} + \Updelta ({\text{PAS}})_{1 - 2} \\ \, = & C_{1} \beta \left( \tau \right)^{\alpha /\beta } \Updelta t\left( {\left[ {\left( \tau \right)^{\alpha /\beta } \Updelta t + \left( {\frac{{{\text{PAS}}(t_{0} )}}{{C_{1} }}} \right)^{1/\beta } } \right]^{\beta - 1} + \left[ {2\left( \tau \right)^{\alpha /\beta } \Updelta t + \left( {\frac{{{\text{PAS}}(t_{0} )}}{{C_{1} }}} \right)^{1/\beta } } \right]^{\beta - 1} } \right) \\ \end{aligned} $$

(A33)

We see Eqs. (A30) and (A33) are not equivalent since the first bracket of Eq. (A33) is missing an additional $\left( \tau \right)^{\alpha /\beta } \Updelta t$, and therefore, accuracy of the MPL model is still heavily dependent on discretization of time intervals.