Designing Drug Regimens that Mitigate Nonadherence

Abstract

Medication adherence is a well-known problem for pharmaceutical treatment of chronic diseases. Understanding how nonadherence affects treatment efficacy is made difficult by the ethics of clinical trials that force patients to skip doses of the medication being tested, the unpredictable timing of missed doses by actual patients, and the many competing variables that can either mitigate or magnify the deleterious effects of nonadherence, such as pharmacokinetic absorption and elimination rates, dosing intervals, dose sizes, and adherence rates. In this paper, we formulate and analyze a mathematical model of the drug concentration in an imperfectly adherent patient. Our model takes the form of the standard single compartment pharmacokinetic model with first-order absorption and elimination, except that the patient takes medication only at a given proportion of the prescribed dosing times. Doses are missed randomly, and we use stochastic analysis to study the resulting random drug level in the body. We then use our mathematical results to propose principles for designing drug regimens that are robust to nonadherence. In particular, we quantify the resilience of extended release drugs to nonadherence, which is quite significant in some circumstances, and we show the benefit of taking a double dose following a missed dose if the drug absorption or elimination rate is slow compared to the dosing interval. We further use our results to compare some antiepileptic and antipsychotic drug regimens.

Appendix

In this appendix, we prove the theorems of Sect. 3.

Proof of Theorem 1

For any dosing protocol f and any integers $N\ge M$, define

$$\begin{aligned} A_{M,N}&:= \sum _{n=M}^{N}\alpha ^{N-n}f(X_{n}),\quad B_{M,N} := \sum _{n=M}^{N}\beta ^{N-n}f(X_{n}). \end{aligned}$$

(55)

For $t\in [0,\tau ]$, define the random variable

$$\begin{aligned} C_{M,N}(t) :=\kappa (\alpha ^{t/\tau }A_{M,N}-\beta ^{t/\tau }B_{M,N}). \end{aligned}$$

(56)

Further, for any $N\in \mathbb {Z}$, define the almost sure limits,

$$\begin{aligned} \begin{aligned} A_{-\infty ,N}&:=\lim _{M\rightarrow -\infty }A_{M,N},\quad B_{-\infty ,N} :=\lim _{M\rightarrow -\infty }B_{M,N},\\ C_{-\infty ,N}(t)&:=\lim _{M\rightarrow -\infty }C_{M,N}(t), \end{aligned} \end{aligned}$$

(57)

and notice that $A_{-\infty ,0}=A$, $B_{-\infty ,0}=B$, and $C_{-\infty ,0}(t)=C(t)$ where A and B are defined in (24) and C(t) is defined in (26). To see why the almost sure convergence in (57) is guaranteed, note first that the function f must be bounded since its domain $\{0,1\}^{m+1}$ is finite. Hence, the terms in the sums in (55) are bounded by the terms in a geometric series, and so the Weierstrass M-test ensures the almost sure convergence in (57).

Notice that the drug concentration in (10) with $f_{n}=f(X_{n})$ can be written as

$$\begin{aligned} c({t})&=\kappa \Big (\alpha ^{{t}/\tau -N({t})}\sum _{n=0}^{N({t})}\alpha ^{N(t)-n}f_{n}-\beta ^{{t}/\tau -N({t})}\sum _{n=0}^{N({t})}\beta ^{N(t)-n}f_{n}\Big ),\quad t\ge 0. \end{aligned}$$

Therefore,

$$\begin{aligned} c(N\tau +t) =C_{0,N}(t),\quad \text {for any }t\in [0,\tau ]\,\, \text {and integer } N\ge 0. \end{aligned}$$

(58)

Since $\{X_{n}\}_{n\in \mathbb {Z}}$ is stationary, it follows that

$$\begin{aligned} C_{0,N}(t) =_{d }C_{-N,0}(t),\quad \text {for any integer }N\ge 0, \end{aligned}$$

(59)

where $=_{d }$ denotes equality in distribution. Now, as in (57), we have that

$$\begin{aligned} \begin{aligned} \lim _{N\rightarrow \infty }C_{-N,0}(t) =C(t) =\kappa (\alpha ^{t/\tau }A-\beta ^{t/\tau }B),\quad \text {for }t\in [0,\tau ]. \end{aligned} \end{aligned}$$

(60)

Equations (58), (59), and (60) yield (26). We note that random variables akin to (60) are sometimes called random pullback attractors because they take an initial condition and pull it back to the infinite past (Crauel 2001; Mattingly 1999; Schmalfuß 1996; Lawley et al. 2015; Lawley and Keener 2019).

Since f is bounded, $C_{0,N}(t)$ can be bounded by a deterministic constant independent of N, and thus (59), (60), and the bounded convergence theorem yield

$$\begin{aligned} \mathbb {E}[(C_{0,N}(t))^{j}] \rightarrow \mathbb {E}[({C(t)})^{j}],\quad \text {as }N\rightarrow \infty \text { for all }j>0. \end{aligned}$$

(61)

Combining (61) with (58) and (59) yields the second equality in (27). Combining the second equality in (27) with the bounded convergence theorem yields the second equality in (28).

Finally, the same argument that gave the second equality in (27) gives the second equality in (29) upon noticing that (58), (59), and (60) all still hold when integrated from $t=0$ to $t=\tau$.

To obtain the first equalities in (27)-(29) for the time averages, we first note that Theorem 7.1.3 in Durrett (2019) ensures that $\{C_{-\infty ,n}(t)\}_{n\in \mathbb {Z}}$ is ergodic for any fixed $t\ge 0$ since $\{X_{n}\}_{n\in \mathbb {Z}}$ is ergodic and stationary. We thus have that

$$\begin{aligned} \lim _{N\rightarrow \infty }\frac{1}{N}\sum _{n=0}^{N-1}(C_{-\infty ,n}(t))^{j} =\mathbb {E}[(C_{-\infty ,0}(t))^{j}] =\mathbb {E}[(C(t))^{j}]\quad \text {with probability one}, \end{aligned}$$

by Birkhoff’s ergodic theorem (see, for example, Theorem 7.2.1 in Durrett (2019)). By (58), we have that for any $t\in [0,\tau ]$,

$$\begin{aligned} \frac{1}{N}\sum _{n=0}^{N-1}(c(n\tau +t))^{j} =\frac{1}{N}\sum _{n=0}^{N-1}(C_{0,n}(t))^{j}. \end{aligned}$$

Hence, in order to prove the first equality in (27), it remains to prove

$$\begin{aligned} \lim _{N\rightarrow \infty }\frac{1}{N}\sum _{n=0}^{N-1}(C_{-\infty ,n}(t))^{j} =\lim _{N\rightarrow \infty }\frac{1}{N}\sum _{n=0}^{N-1}(C_{0,n}(t))^{j}\quad \text {with probability one}. \end{aligned}$$

(62)

To prove this, we first note the bound,

$$\begin{aligned} C_{-\infty ,n}(t)-C_{0,n}(t) \!=\!\kappa \sum _{i=-\infty }^{-1}(\alpha ^{t/\tau +n-i}-\beta ^{t/\tau +n-i})f(X_{i})&\le \frac{2\kappa f_{+}(\max \{\alpha ,\beta \})^{n+1}}{1-\max \{\alpha ,\beta \}}\\&=:K\gamma ^{n}, \end{aligned}$$

where $f_{+}:=\sup _{x\in \{0,1\}^{m+1}}f(x)<\infty$ since $\{0,1\}^{m+1}$ is finite.

If $j\ge 1$ is an integer, then the binomial theorem implies the general identity for $a,b\in \mathbb {R}$,

$$\begin{aligned} (a+b)^{j}=a^{j}+b\sum _{k=1}^{j}{j\atopwithdelims ()k}b^{k-1}a^{j-k}. \end{aligned}$$

Therefore,

$$\begin{aligned} (C_{-\infty ,n}(t))^{j} =(C_{0,n}(t)+C_{-\infty ,n}(t)-C_{0,n}(t))^{j} \le (C_{0,n}(t))^{j}+K_{0}\gamma ^{n}, \end{aligned}$$

(63)

for a suitably chosen deterministic constant $K_{0}$ independent of n. Since f is nonnegative, it follows that

$$\begin{aligned} (C_{-\infty ,n}(t))^{j} -K_{0}\gamma ^{n} \le (C_{0,n}(t))^{j} \le (C_{-\infty ,n}(t))^{j}, \end{aligned}$$

(64)

which then yields (62) for the case that $j\ge 1$ is an integer.

If $j>0$ is not an integer, then notice that the function $g(a)=a^{j-\lfloor j\rfloor }$ for $a>0$ is concave and therefore subadditive, and thus

$$\begin{aligned} \begin{aligned} (C_{-\infty ,n}(t))^{j-\lfloor j\rfloor }&\le (C_{0,n}(t))^{j-\lfloor j\rfloor } +(C_{-\infty ,n}(t)-C_{0,n}(t))^{j-\lfloor j\rfloor }\\&\le (C_{0,n}(t))^{j-\lfloor j\rfloor } +(K\gamma ^{n})^{j-\lfloor j\rfloor }. \end{aligned} \end{aligned}$$

(65)

Therefore, (63) and (65) imply

$$\begin{aligned} (C_{-\infty ,n}(t))^{j}&\le \big [(C_{0,n}(t))^{j-\lfloor j\rfloor } +(K\gamma ^{n})^{j-\lfloor j\rfloor }\big ]\big [(C_{0,n}(t))^{{\lfloor j\rfloor }}+K_{0}\gamma ^{n}\big ]\\&\le (C_{0,n}(t))^{j}+K_{1}\gamma _{1}^{n}, \end{aligned}$$

for suitably chosen deterministic constants $K_{1}>0$ and $\gamma _{1}\in (0,1)$ which are independent of n. Hence, (64) holds with $K_{0}$ replaced by $K_{1}$ and $\gamma$ replaced by $\gamma _{1}$ and (62) follows.

Having proven the first equality in (27), the first equality in (28) then follows from the bounded convergence theorem upon noting that

$$\begin{aligned} \lim _{T\rightarrow \infty }\frac{1}{T}\int _{0}^{T}(c(t))^{j}\,d t&=\lim _{N\rightarrow \infty }\frac{1}{N\tau }\int _{0}^{N\tau }(c(t))^{j}\,d t\\&=\lim _{N\rightarrow \infty }\frac{1}{N}\sum _{n=0}^{N-1}\frac{1}{\tau }\int _{n\tau }^{(n+1)\tau }(c(t))^{j}\,d t. \end{aligned}$$

Finally, the first equality in (29) follows from applying the same argument that gave the first equality in (27) to the integrals $\int _{0}^{\tau }C_{-\infty ,n}(t)\,d t$ and $\int _{0}^{\tau }C_{0,n}(t)\,d t$. $\square$

Proof of Theorem 2

Equations (32) and (35) follow immediately from the definition of C(t) in (60). It thus remains to prove (33) and (36), which generalizes the proof of Theorem 1 in [20]. Recalling the definitions in (55) and (57), notice that

$$\begin{aligned} {A} =_{d }A_{-\infty ,1} =\alpha {A}+f(X_{1}),\quad {B} =_{d }B_{-\infty ,1} =\beta {B}+f(X_{1}), \end{aligned}$$

(66)

where $=_{d }$ denotes equality in distribution. Taking the expectation of (66) and rearranging yields (33).

To obtain $\mathbb {E}[A^{2}]$, we first square (66), take expectation, and rearrange to obtain

$$\begin{aligned} \mathbb {E}[A^{2}] =\frac{1}{1-\alpha ^{2}}\Big (2\alpha \mathbb {E}\big [{A}f(X_{1})\big ]+\mathbb {E}\big [(f(X_{1}))^{2}\big ]\Big ). \end{aligned}$$

(67)

By definition of expectation, we have that $\mathbb {E}\big [(f(X_{1}))^{2}\big ] =\sum _{x}(f(x))^{2}\pi (x)$. To compute $\mathbb {E}[{A}f(X_{1})]$, let $1_{E}\in \{0,1\}$ denote the indicator function on an event E, which means $1_{E}=1$ if E occurs and $1_{E}=0$ otherwise. Thus,

$$\begin{aligned} \mathbb {E}[{A}f(X_{1})] =\sum _{x}f(x)\mathbb {E}[{A}1_{X_{1}=x}]. \end{aligned}$$

(68)

Multiplying (66) by $1_{X_{1}=x}$, taking expectation, and using that $({A},X_{0})$ is equal in distribution to $(A_{-\infty ,1},X_{1})$ yields

$$\begin{aligned} \mathbb {E}[{A}1_{X_{0}=x}] =\mathbb {E}[A_{-\infty ,1}1_{X_{1}=x}] =\alpha \mathbb {E}[{A}1_{X_{1}=x}] +f(x)\pi (x),\quad x\in \{0,1\}^{m+1}. \end{aligned}$$

(69)

The conditional expectation tower property (Theorem 5.1.6 in Durrett (2019)) implies

$$\begin{aligned} \mathbb {E}[{A}1_{X_{1}=x}] =\sum _{y}\mathbb {E}[{A}1_{X_{1}=x}1_{X_{0}=y}] =\sum _{y}\mathbb {E}[{A}1_{X_{0}=y}]P(y,x), \end{aligned}$$

(70)

where P is in (18). Combining (69) and (70) yields the following system of linear algebraic equations for $\mathbb {E}[{A}1_{X_{0}=x}]$,

$$\begin{aligned} \mathbb {E}[{A}1_{X_{0}=x}] =\alpha \sum _{y}\mathbb {E}[{A}1_{X_{0}=y}]P(y,x)+f(x)\pi (x),\quad x\in \{0,1\}^{m+1}. \end{aligned}$$

(71)

If we define the vectors ${{u}}_{\alpha }\in \mathbb {R}^{2^{m+1}}$ and $v\in \mathbb {R}^{2^{m+1}}$ by

$$\begin{aligned} {{u}}_{\alpha }(x)&:=\mathbb {E}[{A}1_{X_{0}=x}],\quad v(x) :=f(x)\pi (x), \end{aligned}$$

then (34) solves (71). We note that the Perron–Frobenius theorem guarantees the invertibility of $I-\alpha P^{\top }$ since $I-\alpha P^{\top }=\alpha (\alpha ^{-1}I-P^{\top })$ and $\alpha \in (0,1)$. Therefore, (69) implies

$$\begin{aligned} \mathbb {E}[Af(X_{1})] =\sum _{x}f(x)\mathbb {E}[A1_{X_{1}=x}] =\sum _{x}f(x)\frac{1}{\alpha }(u_{\alpha }(x)-f(x)\pi (x)). \end{aligned}$$

(72)

Combining (72) with (67) yields the formula for $\mathbb {E}[A^{2}]$ given by (36) upon replacing $\beta$ by $\alpha$. The analogous argument yields $\mathbb {E}[B^{2}]$ (given by (36) upon replacing $\alpha$ by $\beta$). To obtain $\mathbb {E}[AB]$, we first observe that

$$\begin{aligned} AB =_{d }A_{-\infty ,1}B_{-\infty ,1} =(\alpha A+f(X_{1}))(\beta B+f(X_{1})). \end{aligned}$$

(73)

Taking the expectation of (73) and rearranging yields

$$\begin{aligned} \mathbb {E}[AB] =\frac{1}{1-\alpha \beta }\Big (\alpha \mathbb {E}\big [{A}f(X_{1})\big ]+\beta \mathbb {E}\big [{B}f(X_{1})\big ]+\mathbb {E}\big [(f(X_{1}))^{2}\big ]\Big ). \end{aligned}$$

Using (72) and the analogous equation for $\mathbb {E}\big [{B}f(X_{1})\big ]$ yields the formula for $\mathbb {E}[AB]$ in (36) and completes the proof. $\square$

Proof of Theorem 5

Since missing doses can only decrease the concentration for the single dose protocol, we have the following pair of inequalities,

$$\begin{aligned} \sup _{t\in [0,\tau ],N\ge 0,\xi }c^{single }(N\tau +t)&\le \sup _{t\in [0,\tau ],N\ge 0}c^{perf }(N\tau +t), \end{aligned}$$

(74)

$$\begin{aligned} \sup _{t\in [0,\tau ],\xi }C^{single }(t)&\le \sup _{t\in [0,\tau ]}C^{perf }(t). \end{aligned}$$

(75)

But, setting $\xi _{n}=1$ for all n yields $c^{single }(N\tau +t)=c^{perf }(N\tau +t)$ and $C^{single }(t)=C^{perf }(t)$, and thus the inequalities in (74) and (75) can be replaced by equalities. Hence, we have obtained the first and third equalities in (40).

The second equality in (40) and the first equality in (43) follow from the convergence in distribution in (26) in Theorem 1. To see this, note first that for any dosing protocol, we have by (58) and (59) that

$$\begin{aligned} c(N\tau +t) =C_{0,N}(t) =_{d }C_{-N,0}(t) \le C(t),\quad \text {for any }t\in [0,\tau ], N\ge 0, \end{aligned}$$

(76)

where $C_{M,N}(t)$ is defined in (56) and $C(t):=C_{-\infty ,0}(t)$ is defined in (57). The inequality in (76) holds because f is nonnegative. The $=_{d }$ in (76) denotes equality in distribution, where the probability measure $\mathbb {P}$ on the set of sequences $\xi =\{\xi _{n}\}_{n\in \mathbb {Z}}$ can be any measure as described in Sect. 3.1. In particular, we can take $\mathbb {P}$ to be the probability measure for the case that $\{\xi _{n}\}_{n\in \mathbb {Z}}$ are iid with $\mathbb {P}(\xi _{n}=1)=p\in (0,1)$ as in Sect. 3.3. The important point is that for this choice of $\mathbb {P}$, a supremum over sequences $\xi$ is the same as an essential supremum over sequences $\xi$ (since for any finite sequence $\{\zeta _{i}\}_{i=1}^{M}\in \{0,1\}^{M}$ and any $n\in \mathbb {Z}$, we have $\mathbb {P}(\xi _{n}=\zeta _{1},\dots ,\xi _{n+M}=\zeta _{M})>0$). Therefore, (76) implies that

$$\begin{aligned} \sup _{t\in [0,\tau ],N\ge 0,\xi }c(N\tau +t) \le \sup _{t\in [0,\tau ],\xi }C(t). \end{aligned}$$

(77)

To obtain equality in (77), fix $t\in [0,\tau ]$ and let $\delta >0$. By definition of supremum,

$$\begin{aligned} \mathbb {P}\big (C(t)>\sup _{\xi }C(t)-\delta \big )>0, \end{aligned}$$

where we again take $\mathbb {P}$ to be as in Sect. 3.3 so that a supremum over $\xi$ and an essential supremum over $\xi$ are equivalent. Hence, the convergence in distribution in (26) in Theorem 1 implies that we can take N sufficiently large so that

$$\begin{aligned} \mathbb {P}\big (c(N\tau +t)>\sup _{\xi }C(t)-\delta \big )>0. \end{aligned}$$

Since $\delta >0$ is arbitrary, we thus have that

$$\begin{aligned} \sup _{N\ge 0,\xi }c(N\tau +t) \ge \sup _{\xi }C(t). \end{aligned}$$

(78)

Since $t\in [0,\tau ]$ is arbitrary, combining (78) and (77) implies that the inequality in (77) can be replaced by equality. Since this holds for any dosing protocol, we have obtained the second equality in (40) and the first equality in (43).

The final equality in (40) and the maximizing time $t^{*}$ in (41) follow from a simple calculus exercise. The formula in (42) follows from combining (40)-(41) with the definition of $\theta$ in (16).

We now prove the inequality in (43). Fix $t\in [0,\tau ]$ and recall that $C^{double }(t)$ is

$$\begin{aligned} C^{double }(t) =\kappa (\alpha ^{t/\tau }A^{double }-\beta ^{t/\tau }B^{double }) =\sum _{n=0}^{\infty }K_{n}(t)g_{n}, \end{aligned}$$

(79)

where $K_{n}(t)$ is the coefficient,

$$\begin{aligned} K_{n}(t) :=\kappa (\alpha ^{n+t/\tau }-\beta ^{n+t/\tau })\ge 0,\quad n\ge 0, \end{aligned}$$

and

$$\begin{aligned} g_{n}:=f^{double }(X_{-n}),\quad n\ge 0. \end{aligned}$$

(80)

By definition of the double dose protocol, we have that for all $n\ge 0$,

$$\begin{aligned} g_{n}\in \{0,1,2\} and if g_{n}=2, then g_{n+1}=0. \end{aligned}$$

(81)

We claim that there is a finite nonnegative integer $n^{*}\ge 0$ such that

$$\begin{aligned} \begin{aligned} K_{n}(t)&< K_{n+1}(t)\quad \text {if }0\le n\le n^{*}-1,\\ K_{n}(t)&\ge K_{n+1}(t)\quad \text {if }n\ge n^{*}. \end{aligned} \end{aligned}$$

(82)

That is, the sequence $\{K_{n}(t)\}_{n\ge 0}$ is strictly increasing in n for $n<n^{*}$ and nonincreasing in n for $n>n^{*}$. To prove this claim, we momentarily treat n as a continuous variable and differentiate $K_{n}(t)$ with respect to n,

$$\begin{aligned} \frac{\partial }{\partial n}K_{n}(t) =\kappa (\alpha ^{n+t/\tau }\ln \alpha -\beta ^{n+t/\tau }\ln \beta ). \end{aligned}$$

(83)

Rearranging (83) shows that $\frac{\partial }{\partial n}K_{n}(t)=0$ if and only if

$$\begin{aligned} n= n_{0} := \frac{\ln (\ln \beta /\ln \alpha )}{\ln (\alpha /\beta )}-t/\tau \in \mathbb {R}. \end{aligned}$$

Note further that the second derivative is negative at $n_{0}$,

$$\begin{aligned} \frac{\partial ^{2}}{\partial n^{2}}K_{n}(t)\Big |_{n=n_{0}}&=\kappa (\alpha ^{n+t/\tau }(\ln \alpha )^{2} -\beta ^{n+t/\tau }(\ln \beta )^{2})\Big |_{n=n_{0}}\\&=\kappa \beta ^{n_{0}+t/\tau }(\ln \beta )(\ln \alpha -\ln \beta )<0. \end{aligned}$$

Hence, $K_{n}(t)\le K_{n_{0}}(t)$ for all $n\in \mathbb {R}$. Therefore, if $n_{0}\le 0$, then the claim is satisfied with $n^{*}=0$. If $n_{0}>0$, then the claim is satisfied by either $n^{*}=\lfloor n_{0}\rfloor \ge 0$ or $n^{*}=\lceil n_{0}\rceil \ge 1$, where $\lfloor \cdot \rfloor$ and $\lceil \cdot \rceil$ denote the floor and ceiling functions, respectively. To distinguish between the case $n^{*}=\lfloor n_{0}\rfloor \ge 0$ or $n^{*}=\lceil n_{0}\rceil \ge 1$ for $n_{0}>0$, one merely checks if $K_{\lfloor n_{0}\rfloor }(t)<K_{\lceil n_{0}\rceil }(t)$ or $K_{\lfloor n_{0}\rfloor }(t)>K_{\lceil n_{0}\rceil }(t)$. We note that if $K_{\lfloor n_{0}\rfloor }(t)=K_{\lceil n_{0}\rceil }(t)$, then we can simply take $n^{*}=\lfloor n_{0}\rfloor$. Therefore, we have verified the claim.

We now claim that if $C^{double }(t)$ is to be maximized, then we must have that

$$\begin{aligned} g_{n}=1\quad \text {for all }0\le n\le n^{*}-1. \end{aligned}$$

(84)

The claim in (84) is vacuously true if $n^{*}=0$, so suppose $n^{*}\ge 1$. To prove the claim in (84), we start with $n=0$. It is immediate that the maximizing value must either be $g_{0}=1$ or $g_{0}=2$, since setting $g_{0}=0$ only makes the first term in (79) smaller compared to if $g_{0}=1$ or $g_{0}=2$, and it does not allow any other term to be larger than if $g_{0}=1$. If $g_{0}=2$, then we must set $g_{1}=0$ by (81). However, we claim that $C^{double }(t)$ is certainly larger if $g_{0}=g_{1}=1$ compared to if $g_{0}=2$ and $g_{1}=0$. To see this, note first that the values of $g_{i}$ for $i\ge 2$ are unconstrained by either choice. Further, since $n^{*}\ge 1$, (82) implies that $K_{0}(t)<K_{1}(t)$ and thus

$$\begin{aligned} K_{0}(t)\cdot 2+K_{1}(t)\cdot 0 < K_{0}(t)\cdot 1+K_{1}(t)\cdot 1. \end{aligned}$$

Therefore, $C^{double }(t)$ is larger if $g_{0}=g_{1}=1$ rather than $g_{0}=2$ and $g_{1}=0$. At this point in the argument, it is still not determined if $C^{double }(t)$ is larger by taking $g_{0}=g_{1}=1$ or $g_{0}=1$ and $g_{1}=2$. Nevertheless, we conclude that $C^{double }(t)$ is maximized by setting $g_{0}=1$ in this case that $n^{*}\ge 1$. Repeating this argument shows that we must take $g_{n}=1$ for all $n<n^{*}$ in order to maximize $C^{double }(t)$, and thus, we have verified the claim in (84).

We further claim that if $C^{double }(t)$ is to be maximized, then

$$\begin{aligned} if g_{n}=0, then g_{n-1}=2,\quad \text {for all }n\ge 1. \end{aligned}$$

(85)

To see why (85) holds, observe that if $g_{n}=0$ and $g_{n-1}\ne 2$ for some $n\ge 1$, then one could change the value of $g_{n}$ to be $g_{n}=1$ without changing the value of $g_{i}$ for any $i\ne n$, and this would make the value of $C^{double }(t)$ larger. Hence, (85) holds for any sequence $\{g_{n}\}_{n\ge 0}$ which maximizes $C^{double }(t)$.

Now, let $\{g_{n}\}_{n\ge 0}$ be any sequence as in (80)-(81) that satisfies (84) and (85). We claim that the corresponding value of $C^{double }(t)$ in (79) satisfies

$$\begin{aligned} C^{double }(t) \le K_{n'}(t)+\sum _{n=0}^{\infty }K_{n}(t) =K_{n'}(t)+C^{perf }(t), \end{aligned}$$

(86)

where $n'\ge n^{*}$ is the smallest integer such that $g_{n}=2$,

$$\begin{aligned} n' :=\inf \{n\ge 0:g_{n}=2\}\ge n^{*}, \end{aligned}$$

(87)

where we set $K_{n'}(t)=0$ if $n'=\infty$ in the case that $g_{n}=1$ for all $n\ge 0$ (note that (86) is trivially satisfied in this case). In words, the claim in (86) means that the concentration for the double dose protocol is always less than the concentration for perfect adherence plus the concentration from a single dose taken $n'$ dosing times in the past. Note that (85) and (87) imply that

$$\begin{aligned} g_{n}=1,\quad \text {for all }0\le n\le n'-1. \end{aligned}$$

(88)

Using the definition of $C^{double }(t)$ in (79), (88), and the definition of $n'$ in (87), the claim in (86) is equivalent to

$$\begin{aligned} K_{n'}(t)+\sum _{n=0}^{\infty }K_{n}(t)(1-g_{n}) =\sum _{n=n'+1}^{\infty }K_{n}(t)(1-g_{n}) \ge 0. \end{aligned}$$

(89)

Define the sequence of $\{S_{n}\}_{n\ge n'+1}$ by

$$\begin{aligned} S_{n} :=\sum _{i=n'+1}^{n-1}(1-g_{i}),\quad n\ge n'+1, \end{aligned}$$

where $S_{n'+1}=0$. Note that since $g_{n}\in \{0,1,2\}$, we are assured that

$$\begin{aligned} S_{n}-S_{n-1} =1-g_{n-1} \in \{-1,0,1\}. \end{aligned}$$

(90)

Further, (81) implies that

$$\begin{aligned} if S_{n}-S_{n-1}=-1, then S_{n+1}-S_{n}=1. \end{aligned}$$

(91)

In addition, since (87) implies that $g_{n'}=2$, (81) implies

$$\begin{aligned} S_{n'+2}=1-g_{n'+1}=1. \end{aligned}$$

(92)

In words, (90) means that successive terms in the sequence $\{S_{n}\}_{n\ge n'+1}$ can change by at most $\pm 1$, and (91) means that if successive terms decrease by 1, then the next term increases by 1. Since $S_{n'+1}=0$ and $S_{n'+2}=1$ by (92), we conclude that

$$\begin{aligned} S_{n}\ge 0\quad \text {for all }n\ge n'+1. \end{aligned}$$

(93)

Using summation by parts, we have that

$$\begin{aligned} \begin{aligned} \sum _{n=n'+1}^{\infty }K_{n}(t)(1-g_{n})&=\sum _{n=n'+1}^{\infty }K_{n}(t)(S_{n+1}-S_{n})\\&=\sum _{n=n'+2}^{\infty }S_{n}(K_{n-1}(t)-K_{n}(t))\ge 0, \end{aligned} \end{aligned}$$

(94)

by (93) and the fact that $n'\ge n^{*}$ and $K_{n}(t)$ is nonincreasing in n for $n\ge n^{*}$ as in (82). In (94), we also used that $\lim _{n\rightarrow \infty }K_{n}(t)=0$ and $S_{n'+1}=0$. Hence, we have verified (89) and thus (86).

Therefore, (86) implies that

$$\begin{aligned} \sup _{t\in [0,\tau ],\xi }C^{double }(t)&\le \sup _{t\in [0,\tau ],n'\ge 0}\Big (K_{n'}(t)+\sum _{n=0}^{\infty }K_{n}(t)\Big )\\&= \sup _{t\in [0,\tau ],n'\ge 0}\Big (K_{n'}(t)+C^{perf }(t)\Big )\\&\le \sup _{t\in [0,\tau ],n'\ge 0}K_{n'}(t)+\sup _{t\in [0,\tau ]}C^{perf }(t). \end{aligned}$$

We obtained $\sup _{t\in [0,\tau ]}C^{perf }(t)$ in (40)-(41), and a straightforward calculus exercise yields

$$\begin{aligned} \sup _{t\in [0,\tau ],n'\ge 0}K_{n'}(t) =\sup _{s\ge 0}\kappa (\alpha ^{s/\tau }-\beta ^{s/\tau }) =\kappa (\alpha ^{s^{*}/\tau }-\beta ^{s^{*}/\tau }), \end{aligned}$$

where $s^{*}$ is given in (44). The bound in (45) follows from combining (43) and (44) with the definition of $\theta$ in (16). $\square$