Optimal Therapy Scheduling Based on a Pair of Collaterally Sensitive Drugs

Abstract

Despite major strides in the treatment of cancer, the development of drug resistance remains a major hurdle. One strategy which has been proposed to address this is the sequential application of drug therapies where resistance to one drug induces sensitivity to another drug, a concept called collateral sensitivity. The optimal timing of drug switching in these situations, however, remains unknown. To study this, we developed a dynamical model of sequential therapy on heterogeneous tumors comprised of resistant and sensitive cells. A pair of drugs (DrugA, DrugB) are utilized and are periodically switched during therapy. Assuming resistant cells to one drug are collaterally sensitive to the opposing drug, we classified cancer cells into two groups, $A_\mathrm{R}$ and $B_\mathrm{R}$, each of which is a subpopulation of cells resistant to the indicated drug and concurrently sensitive to the other, and we subsequently explored the resulting population dynamics. Specifically, based on a system of ordinary differential equations for $A_\mathrm{R}$ and $B_\mathrm{R}$, we determined that the optimal treatment strategy consists of two stages: an initial stage in which a chosen effective drug is utilized until a specific time point, T, and a second stage in which drugs are switched repeatedly, during which each drug is used for a relative duration (i.e., $f \Delta t$-long for DrugA and $(1-f) \Delta t$-long for DrugB with $0 \le f \le 1$ and $\Delta t \ge 0$). We prove that the optimal duration of the initial stage, in which the first drug is administered, T, is shorter than the period in which it remains effective in decreasing the total population, contrary to current clinical intuition. We further analyzed the relationship between population makeup, $\mathcal {A/B} = A_\mathrm{R}/B_\mathrm{R}$, and the effect of each drug. We determine a critical ratio, which we term $\mathcal {(A/B)}^{*}$, at which the two drugs are equally effective. As the first stage of the optimal strategy is applied, $\mathcal {A/B}$ changes monotonically to $\mathcal {(A/B)}^{*}$ and then, during the second stage, remains at $\mathcal {(A/B)}^{*}$ thereafter. Beyond our analytic results, we explored an individual-based stochastic model and presented the distribution of extinction times for the classes of solutions found. Taken together, our results suggest opportunities to improve therapy scheduling in clinical oncology.

Appendices

Appendix A: Derivations of explicit expressions

1.1 A.1: Details of Equations (4), (5), (6), (7), (9), (10) and (12)

• 1. $T_\mathrm{max}$: Equation (4)

$T_\mathrm{max}$ is a minimum point of $C_p(t)$ (from (3)). Therefore,

$$\begin{aligned}&\ C_p'(T_\mathrm{max})=0 \\&\quad \Leftrightarrow \ -(g-s)\left( \displaystyle \frac{r-s}{g+r-s}C_\mathrm{S}^0 \right) e^{-(g-s) \ T_\mathrm{max}} \\&\quad + r \left( \displaystyle \frac{g \ (C_\mathrm{S}^0 + C_\mathrm{R}^0) + (r - s) \ C_\mathrm{R}^0}{g+r-s} \right) e^{r \ T_\mathrm{max}} = 0 \\&\quad \Leftrightarrow \ (g-s)\left( \displaystyle \frac{r-s}{g+r-s}C_\mathrm{S}^0 \right) e^{-(g-s) \ T_\mathrm{max}} \\&\quad = r \left( \displaystyle \frac{g \ (C_\mathrm{S}^0 + C_\mathrm{R}^0) + (r - s) \ C_\mathrm{R}^0}{g+r-s} \right) e^{r \ T_\mathrm{max}} \\&\quad \Leftrightarrow \ (g-s)(r-s)C_\mathrm{S}^0 \ e^{-(g-s) \ T_\mathrm{max}} = r \left( (g \ (C_\mathrm{S}^0 + C_\mathrm{R}^0) + (r - s) \ C_\mathrm{R}^0) \right) e^{r \ T_\mathrm{max}} \\&\quad \Leftrightarrow \ e^{(g+r-s)T_\mathrm{max}}=\displaystyle \frac{(g-s)(r-s)C_\mathrm{S}^0}{r \left( g \ (C_\mathrm{S}^0 + C_\mathrm{R}^0) + (r - s) \ C_\mathrm{R}^0 \right) } \\&\quad \Leftrightarrow \ e^{(g+r-s)T_\mathrm{max}}=\displaystyle \frac{(g-s)(r-s)}{r (g(\mathcal {(R/S)}_0+1) + (r-s)\mathcal {(R/S)}_0)} \\&\quad \Leftrightarrow \ T_\mathrm{max}=\frac{\ln \left[ \displaystyle \frac{(g-s)(r-s)}{r (g(\mathcal {(R/S)}_0+1) + (r-s)\mathcal {(R/S)}_0)} \right] }{g+r-s} \end{aligned}$$

• 2. $T_\mathrm{min}$: Equation (5)

Let us consider the case of drug switch with DrugA being the “pre-switch” drug and DrugB being the “post-switch” drug. If, at a specific time point $t_1$, cell population is decreasing faster by continuing DrugA-therapy than by changing drug to DrugB,

$$\begin{aligned} C_\mathrm{P}'(0 \ |p_A,\{B_\mathrm{R}(t_1),A_\mathrm{R}(t_1)\}) > C_\mathrm{P}'(0 \ |p_B,\{A_\mathrm{R}(t_1),B_\mathrm{R}(t_1)\}), \end{aligned}$$

from Eq. (3) where $\left( \begin{array}{l} B_\mathrm{R}(t_1) \\ A_\mathrm{R}(t_1) \end{array} \right) = \left( \begin{array}{ll} e^{-(g_A - s_A) \ t_1} &{}\quad 0 \\ \displaystyle \frac{g_A \ (e^{r_A \ t_1} - e^{-(g_A - s_A) \ t_1})}{g_A+r_A-s_A} &{}\quad e^{r_A \ t_1} \end{array} \right) \left( \begin{array}{l} B_\mathrm{R}(0) \\ A_\mathrm{R}(0) \end{array} \right)$ evaluated from Eq. (2). Then,

$$\begin{aligned}&\ C_\mathrm{P}'(0 \ |p_A,\{B_\mathrm{R}(t_1),A_\mathrm{R}(t_1)\})> C_\mathrm{P}'(0 \ |p_B,\{A_\mathrm{R}(t_1),B_\mathrm{R}(t_1)\}) \\&\quad \Leftrightarrow \ -(g_A-s_A)\left( \displaystyle \frac{r_A-s_A}{g_A+r_A-s_A}B_\mathrm{R}(t_1) \right) \\&\quad + r_A \left( \displaystyle \frac{g_A \ (A_\mathrm{R}(t_1) + B_\mathrm{R}(t_1)) + (r_A - s_A) \ A_\mathrm{R}(t_1)}{g_A+r_A-s_A} \right) \\&\quad> -(g_B-s_B)\left( \displaystyle \frac{r_B-s_B}{g_B+r_B-s_B}A_\mathrm{R}(t_1) \right) \\&\quad + r_B \left( \displaystyle \frac{g_B \ (A_\mathrm{R}(t_1) + B_\mathrm{R}(t_1)) + (r_B - s_B) \ B_\mathrm{R}(t_1)}{g_B+r_B-s_B} \right) \\&\quad \Leftrightarrow \ r_A \ A_\mathrm{R}(t_1) + s_A \ B_\mathrm{R}(t_1)> r_B \ B_\mathrm{R}(t_1) + s_B \ A_\mathrm{R}(t_1) \\&\quad \Leftrightarrow \ \displaystyle \frac{A_\mathrm{R}(t_1)}{B_\mathrm{R}(t_1)}> \displaystyle \frac{r_B - s_A}{r_A - s_B} \\&\quad \Leftrightarrow \ \displaystyle \frac{\displaystyle \frac{g_A \ (e^{r_A \ t_1} - e^{-(g_A - s_A) \ t_1})}{g_A+r_A-s_A} \ B_\mathrm{R}(0)+ e^{r_A \ t_1} \ A_\mathrm{R}(0)}{e^{-(g_A - s_A) \ t_1} \ B_\mathrm{R}(0)}> \displaystyle \frac{r_B - s_A}{r_A - s_B} \\&\quad \Leftrightarrow \ \displaystyle \frac{g_A \ (e^{(g_A+r_A-s_A) \ t_1} - 1)}{g_A+r_A-s_A} + e^{(g_A+r_A-s_A) \ t_1} \ \mathcal {(A/B)}_0> \displaystyle \frac{r_B - s_A}{r_A - s_B} \\&\quad \Leftrightarrow \ e^{(g_A+r_A-s_A) \ t_1} \left( \displaystyle \frac{g_A}{g_A+r_A-s_A} + \mathcal {(A/B)}_0 \right) > \displaystyle \frac{r_B - s_A}{r_A - s_B} + \displaystyle \frac{g_A}{g_A+r_A-s_A} \\&\quad \Leftrightarrow \ t_1 < \frac{\ln \left[ \displaystyle \frac{(r_A-s_A)(r_B-s_A)+g_A(r_A+r_B-s_A-s_B)}{(r_A-s_B) (g_A + (g_A+r_A-s_A)\mathcal {(A/B)}_0)} \right] }{g_A+r_A-s_A} \\&\quad \left( = T_\mathrm{min}(\{s_A,r_A,g_A\},\{s_B,r_B\},\mathcal {(A/B)}_0) \right) . \end{aligned}$$

Similarly,

$$\begin{aligned} t_1>T_\mathrm{min}(\{s_A,r_A,g_A\},\{s_B,r_B\},\mathcal {(A/B)}_0), \end{aligned}$$

if and only if the population is dropping faster using DrugB than by continuing to use DrugA, and

$$\begin{aligned} t_1=T_\mathrm{min}(\{s_A,r_A,g_A\},\{s_B,r_B\},\mathcal {(A/B)}_0), \end{aligned}$$

if and only if the population is dropping at an equal rate with either drug.

The general form of $T_\mathrm{min}$ is

$$\begin{aligned}&T_\mathrm{min}(\{s_1,r_1,g_1\},\{s_2,r_2\},\mathcal {(R/S)}_0)\\&\quad =\displaystyle \frac{\displaystyle \ln \left[ \displaystyle \frac{(r_1-s_1)(r_2-s_1)+g_1(r_1+r_2-s_1-s_2)}{(r_1-s_2) (g_1 + (g_1+r_1-s_1)\mathcal {(R/S)}_0)} \right] }{g_1+r_1-s_1}, \end{aligned}$$

where the parameters of “pre-switch” and “post-switch” drugs are $\{s_1,r_1,g_1\}$ and $\{s_2,r_2,g_2\}$, respectively, and initial population makeup, $\mathcal {(R/S)}_0$, is the resistant cell population divided by the sensitive cell population for the “pre-switch” drug.

• 3. $T_\mathrm{gap}$: Equations (6)–(7)

$$\begin{aligned}&T_\mathrm{gap}(\{s_1,r_1,g_1\},\{s_2,r_2\}) = T_\mathrm{max}(\{s_1,r_1,g_1\},\mathcal {(R/S)}_0)\\&\qquad -T_\mathrm{min}(\{s_1,r_1,g_1\},\{s_2,r_2\},\mathcal {(R/S)}_0) \\&\quad = \frac{\ln \left[ \displaystyle \frac{(g_1-s_1)(r_1-s_1)}{r_1 (g(\mathcal {(R/S)}_0+1) + (r_1-s_1)\mathcal {(R/S)}_0)} \right] }{g_1+r_1-s_1} \\&\qquad - \frac{\ln \left[ \displaystyle \frac{(r_1-s_1)(r_2-s_1)+g_1(r_1+r_2-s_1-s_2)}{(r_1-s_2) (g_1 + (g_1+r_1-s_1)\mathcal {(R/S)}_0)} \right] }{g_1+r_1-s_1} \\&\quad = \frac{\ln \left[ \displaystyle \frac{(g_1-s_1)(r_1-s_1)}{r_1 (g(\mathcal {(R/S)}_0+1) + (r_1-s_1)\mathcal {(R/S)}_0)} \ \displaystyle \frac{(r_1-s_2) (g_1 + (g_1+r_1-s_1)\mathcal {(R/S)}_0)}{(r_1-s_1)(r_2-s_1)+g_1(r_1+r_2-s_1-s_2)} \right] }{g_1+r_1-s_1} \\&\quad =\frac{\ln \left[ \displaystyle \frac{(g_1-s_1)(r_1-s_1)(r_1-s_2)}{r_1((r_1-s_1)(r_2-s_1)+g_1(r_1+r_2-s_1-s_2))} \right] }{g_1+r_1-s_1} \end{aligned}$$

And,

$$\begin{aligned}&T_\mathrm{min}(\{s_1,r_1,g_1\},\{s_2,r_2\},\mathcal {(R/S)}_0)< T_\mathrm{max}(\{s_1,r_1,g_1\},\mathcal {(R/S)}_0) \\&\quad \Leftrightarrow T_\mathrm{gap}(\{s_1,r_1,g_1\},\{s_2,r_2\})> 0 \\&\quad \Leftrightarrow \frac{\ln \left[ \displaystyle \frac{(g_1-s_1)(r_1-s_1)(r_1-s_2)}{r_1((r_1-s_1)(r_2-s_1)+g_1(r_1+r_2-s_1-s_2))} \right] }{g_1+r_1-s_1}> 0 \\&\quad \Leftrightarrow \ln \left[ \displaystyle \frac{(g_1-s_1)(r_1-s_1)(r_1-s_2)}{r_1((r_1-s_1)(r_2-s_1)+g_1(r_1+r_2-s_1-s_2))} \right]> 0 \\&\quad \Leftrightarrow \displaystyle \frac{(g_1-s_1)(r_1-s_1)(r_1-s_2)}{r_1((r_1-s_1)(r_2-s_1)+g_1(r_1+r_2-s_1-s_2))}> 1 \\&\quad \Leftrightarrow (g_1-s_1)(r_1-s_1)(r_1-s_2)> r_1((r_1-s_1)(r_2-s_1)+g_1(r_1+r_2-s_1-s_2)) \\&\quad \Leftrightarrow g_1 \ s_1 \ s_2 - s_1^2 \ s_2 + r_1 \ s_1 \ s_2> g_1 \ r_1 \ r_2 - s_1 \ r_1 \ r_2 + r_1^2 \ r_2 \\&\quad \Leftrightarrow (g_1 + r_1 - s_1)(s_1 \ s_2 - r_1 \ r_2)> 0 \\&\quad \Leftrightarrow r_1 \ r_2 - s_1 \ s_2 >0 \quad \Leftrightarrow \quad r_1 \ r_2 < s_1 \ s_2 \end{aligned}$$

Similarly, $T_\mathrm{gap}=0$ if and only if $r_1 \ r_2 = s_1 \ s_2$, and $T_\mathrm{gap}<0$ if and only if $r_1 \ r_2 > s_1 \ s_2$.

• 4. $\mathcal {A/B}$ at $T_\mathrm{max}$ and $T_\mathrm{min}$: Equations (9)–(10).

It is clear that

$$\begin{aligned} \mathcal {A/B}(T_\mathrm{min}^A)=\mathcal {A/B}(T_\mathrm{min}^B)=\displaystyle \frac{r_B-s_A}{r_A-s_B}, \end{aligned}$$

and

$$\begin{aligned} \mathcal {A/B}(T_\mathrm{max}^A)=\displaystyle \frac{-s_A}{r_A} \quad \text {and} \quad \mathcal {A/B}(T_\mathrm{max}^B)=\displaystyle \frac{r_B}{-s_B} \end{aligned}$$

by the expressions of $A_\mathrm{R}(t)$, $B_\mathrm{R}(t)$, $T_\mathrm{max}$ and $T_\mathrm{min}$ from Eqs. (2), (5) and (4).

Otherwise, it can be proved more simply using the concept of $T_\mathrm{min}$ and $T_\mathrm{max}$. Since $C_\mathrm{S}'(t)+C_\mathrm{R}'(t)=s \ C_\mathrm{S}(t) + r \ C_\mathrm{R}(t)$, from the differential system (1), the derivatives of $A_\mathrm{R}(t)+B_\mathrm{R}(t)$ are $s_A \ B_\mathrm{R}(t) + r_A \ A_\mathrm{R}(t)$ and $s_B \ A_\mathrm{R}(t) + r_B \ B_\mathrm{R}(t)$ under DrugA and DrugB, respectively. At $T_\mathrm{min}$ (whether it is $T_\mathrm{min}^A$ or $T_\mathrm{min}^B$), the derivatives of total populations are equivalent either under DrugA or under DrugB. Then,

$$\begin{aligned}&s_A \ B_R(T_\mathrm{min}) + r_A \ A_R(T_\mathrm{min}) = s_B \ A_R(T_\mathrm{min}) + r_B \ B_R(T_\mathrm{min}) \\&\displaystyle \frac{A_R(T_\mathrm{min})}{B_R(T_\mathrm{min})}=\displaystyle \frac{r_B-s_A}{r_A-s_B} \\&\mathcal {A/B}(T_\mathrm{min})=\displaystyle \frac{r_B-s_A}{r_A-s_B} \end{aligned}$$

Therefore,

$$\begin{aligned} \mathcal {A/B}(T_\mathrm{min}^A)=\mathcal {A/B}(T_\mathrm{min}^B)=\displaystyle \frac{r_B-s_A}{r_A-s_B}, \end{aligned}$$

Under DrugA at $T_\mathrm{max}^A$, $A_R'(t)+B_R'(t)=0$. Therefore,

$$\begin{aligned}&s_A \ B_R(T_\mathrm{max}^A) + r_A \ A_R(T_\mathrm{max}^A)=0 \\&\mathcal {A/B}(T_\mathrm{max}^A)=\displaystyle \frac{-s_A}{r_A}. \end{aligned}$$

Similarly, $\mathcal {A/B}(T_\mathrm{max}^B)=\displaystyle \frac{r_B}{-s_B}$.

• 5. $k^*$: Equation (12)

The sizes of the subpopulations after $\Delta t$-long therapy with DrugA started from initial population makeup of $\mathcal {A/B}(0)=\mathcal {(A/B)}^*$ are

$$\begin{aligned} \left( \begin{array}{l} B_\mathrm{R}(\Delta t) \\ A_\mathrm{R}(\Delta t) \end{array} \right) = \left( \begin{array}{ll} e^{-(g_A - s_A) \ \Delta t} &{}\quad 0 \\ \displaystyle \frac{g_A \ (e^{r_A \ \Delta t} - e^{-(g_A - s_A) \ \Delta t})}{g_A+r_A-s_A} &{}\quad e^{r_A \ \Delta t} \end{array} \right) \left( \begin{array}{l} K \\ K \ \mathcal {(A/B)}^* \end{array} \right) \end{aligned}$$

derived from Eq. (2), with some constant K scaling population size. Then, the population makeup at the $\Delta t$ and its derivative in terms of $\Delta t$ are

$$\begin{aligned} \mathcal {(A/B)}_{\Delta t}:=\displaystyle \frac{A_\mathrm{R}(\Delta t)}{B_\mathrm{R}(\Delta t)}=\displaystyle \frac{g_A \ (e^{(g_A+r_A-s_A) \ \Delta t} - 1)}{g_A+r_A-s_A} + e^{(g_A+r_A-s_A) \ \Delta t} \mathcal {(A/B)}^*\\ \displaystyle \frac{d \ (\mathcal {(A/B)}_{\Delta t})}{d \ (\Delta t)}= g_A \ e^{(g_A+r_A-s_A) \ \Delta t} + (g_A+r_A-s_A) e^{(g_A+r_A-s_A) \ \Delta t} \mathcal {(A/B)}^* \end{aligned}$$

The time taken from $t=\Delta t$ to reach back to the time of $\mathcal {A/B}(t)=\mathcal {(A/B)}^*$ given DrugB is

$$\begin{aligned}&T_\mathrm{min}(\{s_B,r_B,g_B\},\{s_A,r_A\},1/\mathcal {(A/B)}_{\Delta t}) \\&\quad = \displaystyle \frac{\displaystyle \ln \left[ \displaystyle \frac{(r_B-s_B)(r_A-s_B)+g_B(r_B+r_A-s_B-s_A)}{(r_B-s_A) (g_B + (g_B+r_B-s_B)/\mathcal {(A/B)}_{\Delta t})} \right] }{g_B+r_B-s_B} \end{aligned}$$

from Eq. (5).

Then, the relative ratio between the periods of DrugA and DrugB, $k'$, illustrated in Fig. 9, and its limit, $k^*$, can be derived using:

$$\begin{aligned} k'&= \displaystyle \frac{\Delta t}{T_\mathrm{min}(\{s_B,r_B,g_B\},\{s_A,r_A\},1/\mathcal {(A/B)}_{\Delta t})} \\ k^*&= \displaystyle \lim _{\Delta t \rightarrow 0} k' = \displaystyle \lim _{\Delta t \rightarrow 0} \displaystyle \frac{(g_B+r_B-s_B) \ \Delta t}{\displaystyle \ln \left[ \displaystyle \frac{(r_B-s_B)(r_A-s_B)+g_B(r_B+r_A-s_B-s_A)}{(r_B-s_A) (g_B + (g_B+r_B-s_B)/\mathcal {(A/B)}_{\Delta t})} \right] } \\&= \displaystyle \lim _{\Delta t \rightarrow 0} \displaystyle \frac{(g_B+r_B-s_B) \ \Delta t}{-\displaystyle \ln \left[ \displaystyle g_B + (g_B+r_B-s_B)/\mathcal {(A/B)}_{\Delta t} \right] + \displaystyle \ln K} \\&\quad \text { with } K= \displaystyle \frac{(r_B-s_B)(r_A-s_B)+g_B(r_B+r_A-s_B-s_A)}{r_B-s_A} \\&= \displaystyle \lim _{\Delta t \rightarrow 0} \displaystyle \frac{g_B+r_B-s_B}{-\displaystyle \frac{d}{d \ (\Delta t)} \displaystyle \ln \left[ g_B + (g_B+r_B-s_B)/\mathcal {(A/B)}_{\Delta t} \right] } \\&\quad \text {by L'Hospital's rule} \\&= \displaystyle \lim _{\Delta t \rightarrow 0} \displaystyle \frac{g_B+r_B-s_B}{-\displaystyle \frac{(g_B+r_B-s_B)(-(\mathcal {(A/B)}_{\Delta t})^{-2})}{g_B + (g_B+r_B-s_B)/\mathcal {(A/B)}_{\Delta t}} \displaystyle \frac{d \ (\mathcal {(A/B)}_{\Delta t})}{d \ (\Delta t)}} \\&= \displaystyle \frac{g_B+r_B-s_B}{\displaystyle \frac{(g_B+r_B-s_B)/(\mathcal {(A/B)}^*)^2}{g_B + (g_B+r_B-s_B)/\mathcal {(A/B)}^*} (g_A+(g_A+r_A-s_A)\mathcal {(A/B)}^*)} \\&\quad \text { since, } \displaystyle \lim _{\Delta t \rightarrow 0} \mathcal {(A/B)}_{\Delta t} = \mathcal {(A/B)}^* \\&\quad \text { and } \displaystyle \lim _{\Delta t \rightarrow 0} \displaystyle \frac{d \ (\mathcal {(A/B)}_{\Delta t})}{d \ (\Delta t)}= g_A + (g_A+r_A-s_A) \mathcal {(A/B)}^* \\&= \displaystyle \frac{g_B \ \mathcal {(A/B)}^* + (g_B+r_B-s_B)}{g_A / \mathcal {(A/B)}^* + (g_A+r_A-s_A)} \\&= \displaystyle \frac{(r_A-s_B)((r_A-s_A)(r_B-s_A)+g_A(r_A+r_B-s_A-s_B))}{(r_B-s_A)((r_B-s_B)(r_A-s_B)+g_B(r_A+r_B-s_A-s_B))} \\&\quad \text { since, } \mathcal {(A/B)}^*=\displaystyle \frac{r_B-s_A}{r_A-s_B} \end{aligned}$$

1.2 A.2: Differential system of instantaneous drug switch

The goal of this section is to derive the simple differential equations of $V=\{A_\mathrm{R},B_\mathrm{R}\}$ under instantaneous drug switch (Theorem A.8). For the sake of convenience, we want to use matrix operations and equations based on the vectors and matrices defined below.

Definition

$\mathbb {D}_A:=\left( \begin{array}{ll} r_A &{} g_A \\ 0 &{} s_A - g_A \end{array} \right)$ , $\mathbb {D}_B:=\left( \begin{array}{ll} s_B - g_B &{} 0 \\ g_B &{} r_B \end{array} \right)$, $V(t):=\left( \begin{array}{l} A_\mathrm{R}(t) \\ B_\mathrm{R}(t) \end{array} \right)$,

$$\begin{aligned}&\mathbb {M}_A(t):=\left( \begin{array}{ll} e^{r_A \ t} &{} \displaystyle \frac{g_A \ (e^{r_A \ t} - e^{-(g_A-s_A) \ t})}{g_A+r_A-s_A} \\ 0 &{} e^{-(g_A - s_A) \ t} \end{array} \right) ,\\&\mathbb {M}_B(t):=\left( \begin{array}{ll} e^{-(g_B - s_B) \ t} &{} 0 \\ \displaystyle \frac{g_B \ (e^{r_B \ t} - e^{-(g_B-s_B) \ t})}{g_B+r_B-s_B} &{} e^{r_B \ t} \end{array} \right) , \\&\mathbb {A}_\epsilon :=\mathbb {M}_A(f \ \epsilon ), \mathbb {B}_\epsilon :=\mathbb {M}_B((1-f)\epsilon ), \\&\min \left[ V(t_1), V(t_2), \cdots , V(t_n)\right] :=\left( \begin{array}{l} \min \left[ A_\mathrm{R}(t_1), A_\mathrm{R}(t_2), \cdots , A_\mathrm{R}(t_n)\right] \\ \min \left[ B_\mathrm{R}(t_1), B_\mathrm{R}(t_2), \cdots , B_\mathrm{R}(t_n)\right] \end{array} \right) , \\&\max \left[ V(t_1), V(t_2), \cdots , V(t_n)\right] :=\left( \begin{array}{l} \max \left[ A_\mathrm{R}(t_1), A_\mathrm{R}(t_2), \cdots , A_\mathrm{R}(t_n)\right] \\ \max \left[ B_\mathrm{R}(t_1), B_\mathrm{R}(t_2), \cdots , B_\mathrm{R}(t_n)\right] \end{array} \right) . \end{aligned}$$

Proposition A.1

Using Drug A therapy:

$$\begin{aligned} V'(t) = \mathbb {D}_A \ V(t), \ V(t_0+\Delta t) = \mathbb {M}_A(\Delta t) \ V(t_0). \end{aligned}$$

Using Drug B therapy:

$$\begin{aligned} V'(t) = \mathbb {D}_B \ V(t), \ V(t_0+\Delta t) = \mathbb {M}_B(\Delta t) \ V(t_0). \end{aligned}$$

Proposition A.2

Both $A_\mathrm{R}$ and $B_\mathrm{R}$ are monotonic functions under either therapy. In the presence of Drug A, $A_\mathrm{R}$ is increasing, and $B_\mathrm{R}$ is decreasing. And, in the presence of Drug B, $A_\mathrm{R}$ is decreasing, and $B_\mathrm{R}$ is increasing.

Proposition A.3

$\left. {\mathbb {A}_\epsilon } \right| _{\epsilon =0} = \left. {\mathbb {B}_\epsilon } \right| _{\epsilon =0} = I_2$ for all $0 \le f \le 1$

Proposition A.4

$\left. {\displaystyle \frac{d}{d\epsilon } \mathbb {A}_\epsilon } \right| _{\epsilon =0} = f \ \mathbb {D}_A$, $\left. {\displaystyle \frac{d}{d\epsilon } \mathbb {B}_\epsilon } \right| _{\epsilon =0} = (1-f) \mathbb {D}_B$ for all $0 \le f \le 1$

Lemma A.5

$\displaystyle \lim _{\epsilon \rightarrow 0} \displaystyle \frac{\mathbb {B}_\epsilon \mathbb {A}_\epsilon -I_2}{\epsilon }= f \ \mathbb {D}_A + (1-f) \mathbb {D}_B$ for all $0 \le f \le 1$

Proof

$$\begin{aligned} \displaystyle \lim _{\epsilon \rightarrow 0} \displaystyle \frac{\mathbb {B}_\epsilon \mathbb {A}_\epsilon -I_2}{\epsilon }&=\displaystyle \lim _{\epsilon \rightarrow 0} \displaystyle \frac{\frac{d}{d\epsilon } (\mathbb {B}_\epsilon \mathbb {A}_\epsilon -I_2)}{\frac{d}{d\epsilon } \epsilon }&\text {(by L'Hospital's Rule)} \\&= \displaystyle \lim _{\epsilon \rightarrow 0} \displaystyle \frac{\frac{d\mathbb {B}_\epsilon }{d\epsilon } \mathbb {A}_\epsilon + \mathbb {B}_\epsilon \frac{d\mathbb {A}_\epsilon }{d\epsilon }}{1}&\\&=f \ \mathbb {D}_A + (1-f) \mathbb {D}_B&(\text {by Propositions}~ \hbox {A}.3-\hbox {A}.4) \end{aligned}$$

$\square$

Lemma A.6

$\displaystyle \lim _{\epsilon \rightarrow 0} \displaystyle \frac{(\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^n-I_2}{n \ \epsilon }= f \ \mathbb {D}_A + (1-f) \mathbb {D}_B$ for any positive integer, n, and for all $0 \le f \le 1$

Proof

Let $F(n) :=\displaystyle \lim _{\epsilon \rightarrow 0} \displaystyle \frac{(\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^n-I_2}{n \ \epsilon }$ and $L :=f \ \mathbb {D}_A + (1-f) \mathbb {D}_B$.

Then, we need to prove that $F(n) = L$ for $n = 1,2,3,\ldots$

If $n=1$,

$$\begin{aligned} F(n)&= F(1) = L&(\text {by Lemma}~ \hbox {A}.5) \end{aligned}$$

Otherwise, if $n \ge 2$ and $F(m) = L$ for all $1 \le m \le n-1$,

$$\begin{aligned} F(n)&= \displaystyle \lim _{\epsilon \rightarrow 0} \displaystyle \frac{(\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^n-I_2}{n \ \epsilon } \\&=\displaystyle \lim _{\epsilon \rightarrow 0} \displaystyle \frac{((\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^{n-1}-I_2) (\mathbb {B}_\epsilon \mathbb {A}_\epsilon )+(\mathbb {B}_\epsilon \mathbb {A}_\epsilon -I_2)}{n \ \epsilon } \\&= \displaystyle \frac{n-1}{n} \lim _{\epsilon \rightarrow 0} \displaystyle \frac{((\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^{n-1}-I_2) (\mathbb {B}_\epsilon \mathbb {A}_\epsilon )}{(n-1) \ \epsilon } + \frac{1}{n}\displaystyle \lim _{\epsilon \rightarrow 0} \displaystyle \frac{\mathbb {B}_\epsilon \mathbb {A}_\epsilon -I_2}{\epsilon } \\&= \frac{n-1}{n}F(n-1) + \frac{1}{n}F(1) \\&= \frac{n-1}{n}L + \frac{1}{n}L \quad \qquad \text {(by the inductive assumption)} \\&= L \end{aligned}$$

Therefore, proved. $\square$

Lemma A.7

$\displaystyle \lim _{\epsilon \rightarrow 0} \displaystyle \frac{\mathbb {A}_\epsilon (\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^n-I_2}{(n+f) \ \epsilon }= \frac{(n+1)f}{n+f} \ \mathbb {D}_A + \frac{n(1-f)}{n+f} \mathbb {D}_B$ for any positive integer, n, and for all $0 \le f \le 1$

Proof

Using mathematical induction, if $n=1$,

$$\begin{aligned}&\displaystyle \lim _{\epsilon \rightarrow 0} \displaystyle \frac{\mathbb {A}_\epsilon (\mathbb {B}_\epsilon \mathbb {A}_\epsilon )-I_2}{(1+f) \ \epsilon } \\&\quad = \displaystyle \frac{1}{1+f} \lim _{\epsilon \rightarrow 0} \displaystyle \frac{\mathbb {A}_\epsilon (\mathbb {B}_\epsilon \mathbb {A}_\epsilon -I_2)+(\mathbb {A}_\epsilon -I_2)}{\epsilon } \\&\quad = \displaystyle \frac{1}{1+f} \left[ \lim _{\epsilon \rightarrow 0} \mathbb {A}_\epsilon \lim _{\epsilon \rightarrow 0} \displaystyle \frac{\mathbb {B}_\epsilon \mathbb {A}_\epsilon -I_2}{\epsilon } + \lim _{\epsilon \rightarrow 0} \displaystyle \frac{\mathbb {A}_\epsilon -I_2}{\epsilon } \right] \\&\quad = \displaystyle \frac{1}{1+f} \left[ I_2 (f \ \mathbb {D}_A + (1-f) \mathbb {D}_B) + \left. {\displaystyle \frac{d}{d\epsilon } \mathbb {A}_\epsilon } \right| _{\epsilon =0} \right] ~ (\text {by Proposition}~\hbox {A}.3~\text {and Lemma}~\hbox {A}.5) \\&\quad = \displaystyle \frac{1}{1+f} \left[ (f \ \mathbb {D}_A + (1-f) \mathbb {D}_B) + f \ \mathbb {D}_A \right] \qquad (\text {by Proposition}~ \hbox {A}.4) \\&\quad = \displaystyle \frac{2 \ f}{1+f} \mathbb {D}_A + \displaystyle \frac{1-f}{1+f} \mathbb {D}_B \quad \qquad \text {The equality is true for} \ n=1 \end{aligned}$$

If $n \ge 2$, and the equality works for all integers $1 \le m \le n-1$,

$$\begin{aligned}&\displaystyle \lim _{\epsilon \rightarrow 0} \displaystyle \frac{\mathbb {A}_\epsilon (\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^n-I_2}{(n+f) \ \epsilon }&\\&\quad = \frac{1}{n+f} \left[ \displaystyle \lim _{\epsilon \rightarrow 0} \displaystyle \frac{(\mathbb {A}_\epsilon (\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^{n-1}-I_2)(\mathbb {B}_\epsilon \mathbb {A}_\epsilon )+(\mathbb {B}_\epsilon \mathbb {A}_\epsilon -I_2)}{\epsilon } \right]&\\&\quad = \frac{1}{n+f} \left[ ((n-1)+f) \displaystyle \lim _{\epsilon \rightarrow 0} \displaystyle \frac{(\mathbb {A}_\epsilon (\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^{n-1}-I_2)}{((n-1)+f)\epsilon } \displaystyle \lim _{\epsilon \rightarrow 0} (\mathbb {B}_\epsilon \mathbb {A}_\epsilon ) + \displaystyle \lim _{\epsilon \rightarrow 0} \displaystyle \frac{\mathbb {B}_\epsilon \mathbb {A}_\epsilon -I_2}{\epsilon } \right]&\\&\quad = \frac{1}{n+f} \left[ ((n-1)+f) \left( \frac{n \ f}{(n-1)+f} \ \mathbb {D}_A + \frac{(n-1)(1-f)}{(n-1)+f} \mathbb {D}_B \right) (I_2 \ I_2) \right.&\\&\quad \left. + (f \ \mathbb {D}_A + (1-f) \mathbb {D}_B) \right]&\\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text {by the inductive assumption and Proposition}~\hbox {A}.3~\text {and Lemma}~\hbox {A}.5) \\&\quad = \frac{(n+1)f}{n+f} \ \mathbb {D}_A + \frac{n(1-f)}{n+f} \mathbb {D}_B \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text {(The equality is true for} \ n \ge 2 \text {)} \end{aligned}$$

Therefore, proved. $\square$

Theorem A.8

If Drug A and Drug B are prescribed in turn with a relative intensity of f and $1-f$, and are switched instantaneously, V obeys

$$\begin{aligned} \displaystyle \frac{\mathrm{d}V}{\mathrm{d}t} = (f \ \mathbb {D}_A + (1-f)\mathbb {D}_B)V \end{aligned}$$

Proof

For any time point $t_0$, let us define $V_\epsilon (t)$ as a vector-valued function of $A_\mathrm{R}(t)$ and $B_\mathrm{R}(t)$ describing the cell population dynamics under a periodic therapy starting at $t_0$ with DrugA assigned at $t_0+m \ \epsilon \le t < t_0+(m+f)\epsilon$ and DrugB at $t_0+(m+f)\epsilon \le t < t_0+(m+1)\epsilon$ for $m=0,1,2,3,...$. Then, by Proposition A.1 and the definitions of $\mathbb {A}$ and $\mathbb {B}$,

$$\begin{aligned} V_\epsilon (t_0+m \ \epsilon )&= (\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^m \ V(t_0),\ \ \ \ \ V_\epsilon (t_0+(m+f)\epsilon ) = \mathbb {A}_\epsilon (\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^m \ V(t_0) \end{aligned}$$

(*1)

where $V(t_0)=\left( \begin{array}{l} A_\mathrm{R}(t_0) \\ B_\mathrm{R}(t_0) \end{array} \right)$. And, $V_0(t)$ represents instantaneous drug switching.

For any $\Delta t>0$ and any positive integer n, there exists $\epsilon =\epsilon (n,\Delta t)$ such that

$$\begin{aligned} \frac{\Delta t}{n+1}< \epsilon \le \frac{\Delta t}{n} \ \ \ \ \ \ \ \ \ \ \text {or} \ \ \ \ \ \ \ \ \ \ 1 \le \displaystyle \frac{\Delta t}{n \ \epsilon } < 1 + \displaystyle \frac{1}{n}. \end{aligned}$$

Then by the squeeze theorem,

$$\begin{aligned} \lim _{\Delta t \rightarrow 0} \epsilon (n, \Delta t) = 0 \text { for any positive integer}~ n,~\text {and } \lim _{n \rightarrow \infty } \displaystyle \frac{\Delta t}{n \ \epsilon (n, \Delta t)} = 1 \text { for any}~ \Delta t > 0. \end{aligned}$$

(*2)

For such $\Delta t$, n and $\epsilon (n,\Delta t)$, $V_\epsilon (t_0+\Delta t)$ is bounded, since local extrema can occur only when drugs are switched by Proposition A.2. That is,

$$\begin{aligned}&\min \left[ V_\epsilon (t_0+n \ \epsilon ), V_\epsilon (t_0+(n+f)\epsilon ), V_\epsilon (t_0+(n+1)\epsilon )\right] \ \le \ V_\epsilon (t_0+\Delta t)&\\&\quad \le \ \max \left[ V_\epsilon (t_0+n \ \epsilon ), V_\epsilon (t_0+(n+f)\epsilon ), V_\epsilon (t_0+(n+1)\epsilon )\right] , \end{aligned}$$

(*3)

Also,

$$\begin{aligned}&\displaystyle \lim _{\Delta t \rightarrow 0} \displaystyle \frac{\lim _{n \rightarrow \infty }V_{\epsilon (n,\Delta t)}(t_0+n \ \epsilon (n,\Delta t))-V(t_0)}{\Delta t} \\&\quad = \displaystyle \lim _{\Delta t \rightarrow 0} \ \displaystyle \lim _{n \rightarrow \infty } \displaystyle \frac{(\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^n-I_2}{\Delta t} V(t_0)\displaystyle&(\text {by} (*1)) \\&\quad = \displaystyle \frac{\lim _{\Delta t \rightarrow 0} \lim _{n \rightarrow \infty } \left[ (\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^n-I_2 \right] /(n \ \epsilon )}{\lim _{\Delta t \rightarrow 0} \lim _{n \rightarrow \infty } \Delta t / (n \ \epsilon )} V(t_0)&\\&\quad = \displaystyle \frac{\lim _{n \rightarrow \infty } \left[ \lim _{\Delta t \rightarrow 0} \left[ (\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^n-I_2 \right] /(n \ \epsilon ) \right] }{\lim _{\Delta t \rightarrow 0} \left[ \lim _{n \rightarrow \infty } \Delta t / (n \ \epsilon ) \right] } V(t_0)&\\&\quad = \displaystyle \frac{\lim _{n \rightarrow \infty } \left[ \lim _{\epsilon \rightarrow 0} \left[ (\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^n-I_2 \right] /(n \ \epsilon ) \right] }{\lim _{\Delta t \rightarrow 0} 1} V(t_0)&(\text {by}~ (*2)) \\&\quad = \lim _{n \rightarrow \infty } \left[ f \ \mathbb {D}_A + (1-f) \mathbb {D}_B \right] V(t_0)&(\text {by Lemma}~\hbox {A}.6) \\&\quad = (f \ \mathbb {D}_A + (1-f) \mathbb {D}_B) V(t_0). \end{aligned}$$

(*4)

And,

$$\begin{aligned}&\displaystyle \lim _{\Delta t \rightarrow 0} \displaystyle \frac{\lim _{n \rightarrow \infty }V_{\epsilon (n,\Delta t)}(t_0 + (n+f) \ \epsilon (n,\Delta t))-V(t_0)}{\Delta t} \\&\quad = \displaystyle \lim _{\Delta t \rightarrow 0} \ \displaystyle \lim _{n \rightarrow \infty } \displaystyle \frac{\mathbb {A}_\epsilon (\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^n-I_2}{\Delta t} V(t_0)\displaystyle&(\text {by}~ (*1)) \\&\quad = \displaystyle \frac{\lim _{\Delta t \rightarrow 0} \lim _{n \rightarrow \infty } \left[ \mathbb {A}_\epsilon (\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^n-I_2 \right] /((n+f) \ \epsilon )}{\lim _{\Delta t \rightarrow 0} \lim _{n \rightarrow \infty } \Delta t / ((n+f) \ \epsilon )} V(t_0)&\\&\quad = \displaystyle \frac{\lim _{n \rightarrow \infty } \left[ \lim _{\Delta t \rightarrow 0} \left[ (\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^n-I_2 \right] /((n+f) \ \epsilon ) \right] }{\lim _{\Delta t \rightarrow 0} \left[ \lim _{n \rightarrow \infty } (\Delta t / (n \ \epsilon ))(n/(n+f)) \right] } V(t_0)&\\&\quad = \displaystyle \frac{\lim _{n \rightarrow \infty } \left[ \lim _{\epsilon \rightarrow 0} \left[ (\mathbb {B}_\epsilon \mathbb {A}_\epsilon )^n-I_2 \right] /((n+f) \ \epsilon ) \right] }{\lim _{\Delta t \rightarrow 0} 1} V(t_0)&(\text {by}~ (*2)) \\&\quad = \lim _{n \rightarrow \infty } \left[ \frac{(n+1)f}{n+f} \ \mathbb {D}_A + \frac{n(1-f)}{n+f} \mathbb {D}_B \right] V(t_0)&(\text {by Lemma}~ \hbox {A}.7) \\&\quad = (f \ \mathbb {D}_A + (1-f) \mathbb {D}_B) V(t_0) \end{aligned}$$

(*5)

Similar to (*4),

$$\begin{aligned} \displaystyle \lim _{\Delta t \rightarrow 0} \displaystyle \frac{\lim _{n \rightarrow \infty }V_{\epsilon (n,\Delta t)}(t_0+(n+1) \ \epsilon (n,\Delta t))-V(t_0)}{\Delta t}&= (f \ \mathbb {D}_A + (1-f) \mathbb {D}_B) V(t_0) \end{aligned}$$

(*6)

By (*4)–(*6),

$$\begin{aligned}&\min \left[ \displaystyle \lim _{\Delta t \rightarrow 0} \displaystyle \frac{\lim _{n \rightarrow \infty }V_\epsilon (t_0+n \ \epsilon )-V(t_0)}{\Delta t}, \displaystyle \lim _{\Delta t \rightarrow 0} \displaystyle \frac{\lim _{n \rightarrow \infty }V_\epsilon (t_0+(n+f) \ \epsilon )-V(t_0)}{\Delta t} \right. , \\&\quad \left. \displaystyle \lim _{\Delta t \rightarrow 0} \displaystyle \frac{\lim _{n \rightarrow \infty }V_\epsilon (t_0+(n+1) \ \epsilon )-V(t_0)}{\Delta t} \right] = \max \left[ \displaystyle \lim _{\Delta t \rightarrow 0} \displaystyle \frac{\lim _{n \rightarrow \infty }V_\epsilon (t_0+n \ \epsilon )-V(t_0)}{\Delta t}, \right.&\\&\quad \left. \displaystyle \lim _{\Delta t \rightarrow 0} \displaystyle \lim _{\Delta t \rightarrow 0} \displaystyle \frac{\lim _{n \rightarrow \infty }V_\epsilon (t_0+(n+f) \ \epsilon )-V(t_0)}{\Delta t}, \displaystyle \lim _{\Delta t \rightarrow 0} \displaystyle \frac{\lim _{n \rightarrow \infty }V_\epsilon (t_0+(n+1) \ \epsilon )-V(t_0)}{\Delta t} \right]&\\&\quad = (f \ \mathbb {D}_A + (1-f) \mathbb {D}_B) V(t_0) \end{aligned}$$

(*7)

Then, by (*3), (*7) and the squeeze theorem,

$$\begin{aligned} \displaystyle \left. \frac{\mathrm{d}}{\mathrm{d}t} V_0 \right| _{t=t_0}= \displaystyle \lim _{\Delta t \rightarrow 0} \displaystyle \frac{\lim _{n \rightarrow \infty }V_\epsilon (t_0 + \Delta t)-V(t_0)}{\Delta t} = (f \ \mathbb {D}_A + (1-f) \mathbb {D}_B) V(t_0) \end{aligned}$$

Therefore,

$$\begin{aligned} \displaystyle \frac{\mathrm{d}V}{\mathrm{d}t} = (f \ \mathbb {D}_A + (1-f)\mathbb {D}_B)V \end{aligned}$$

$\square$

1.3 A.3: Population dynamics with the optimal regimen

In this section, we want to write the differential equations of $V=\{A_\mathrm{R},B_\mathrm{R}\}$ under the optimal control strategy described in Section 3.3. Based on “Appendix A.2” and a couple of lemma/theorem, we will reach to a concise form of a differential system described at Theorem A.11.

Lemma A.9

$\left\{ \displaystyle \frac{r_A r_B - s_A s_B}{r_A+r_B-s_A-s_B},\left( \begin{array}{l} \mathcal {(A/B)}^* \\ 1 \end{array} \right) \right\}$ is an eigenpair of $f^* \ \mathbb {D}_A + (1-f^*)\mathbb {D}_B$ with $\mathcal {(A/B)}^*$ and $f^*=k^*/(1+k^*)$ defined by Eqs. (9) and (12).

Proof

Let $\mathbb {D}^*:=f^* \ \mathbb {D}_A + (1-f^*)\mathbb {D}_B$, and $\lambda =\displaystyle \frac{r_A r_B - s_A s_B}{r_A+r_B-s_A-s_B}$. Then,

$$\begin{aligned} \mathbb {D}^*-\lambda \ I_2 = C_1 \left( \begin{array}{l} C_2 \ U^T \\ C_3 \ U^T \end{array} \right) , \end{aligned}$$

where $U=\left( \begin{array}{l} 1 \\ -\mathcal {(A/B)}^* \end{array} \right)$ along with

$$\begin{aligned}&C_1=-(g_A(r_A-s_B)+g_B(r_B-s_A)\\&\quad +(r_B-s_A)(r_A-s_B))(r_A+r_B-s_A-s_B)/(r_A-s_B), \\&C_2=g_A((r_A-s_B)(r_B-s_B)+g_B(r_A+r_B-s_A-s_B), \\&C_3=-g_B((r_B-s_A)(r_A-s_A)+g_A(r_A+r_B-s_A-s_B)). \end{aligned}$$

Since $U^T \ V = 0$ where $V=((r_B-s_A)/(r_A-s_B),1)^T$, $(\lambda ,V)$ is an eigenpair of $\mathbb {D}^*$. $\square$

Theorem A.10

In Stage 2 of the optimal strategy, both $A_\mathrm{R}$ and $B_\mathrm{R}$ change with a constant net proliferation rate,

$$\begin{aligned}\lambda = \displaystyle \frac{r_A r_B - s_A s_B}{r_A + r_B - s_A - s_B}. \end{aligned}$$

Proof

Without a loss of generality, let us prove it only when $\mathcal {A/B}(0)<\mathcal {(A/B)}^*$.

If $\mathcal {A/B}(0)<\mathcal {(A/B)}^*$, DrugA has a better effect initially. So following the optimal therapy scheduling, DrugA is assigned alone at the beginning as long as $T_\mathrm{min}^A=T_\mathrm{min}(p_A,p_B,\mathcal {A/B}(0))$ (Stage 1), and then, Stage 2 starts at $T_\mathrm{min}^A$ with initial condition

$$\begin{aligned} V(T_\mathrm{min}^A)=\mathbb {M}_A(T_\mathrm{min}^A)V(0)=C \left( \begin{array}{l} \mathcal {(A/B)}^* \\ 1 \end{array} \right) \quad \quad \qquad \qquad \quad \quad \quad \quad \text {(**1)} \end{aligned}$$

where $C=\displaystyle \frac{P(0)}{1+\mathcal {A/B}(0)} \left( \displaystyle \frac{(r_A-s_A)(r_B-s_A)+g_A(r_A+r_B-s_A-s_B)}{(r_A-s_B)(g_A+\mathcal {A/B}(0)(g_A+r_A-s_A))}\right) ^{-\frac{g_A-s_A}{g_A+r_A-s_A}}$.

By Theorem A.8, in Stage 2, V(t) obeys

$$\begin{aligned} \displaystyle \frac{\mathrm{d}V}{\mathrm{d}t} = \mathbb {D}^* V,~\hbox {where}~\mathbb {D}^* = f^* \mathbb {D}_A + (1-f^*)\mathbb {D}_B \quad \quad \qquad \qquad \quad \quad \quad \quad \text {(**2)} \end{aligned}$$

By Lemma A.9, $V(T_\mathrm{min}^A)$ is an eigenvector of $\mathbb {D}^*$ with the corresponding eigenvalue, $\lambda$. Then, the solution of (**2) with the initial value (**1) is

$$\begin{aligned} V(t+T_\mathrm{min}^A)=e^{\lambda \ t}V(T_\mathrm{min}^A). \end{aligned}$$

$\square$

Theorem A.11

With optimal therapy utilizing DrugA and DrugB, V obeys the following equations and solutions.

If $\mathcal {A/B}(0)<\mathcal {(A/B)}^*$,

$$\begin{aligned} \displaystyle \frac{\mathrm{d}V}{\mathrm{d}t} = \left\{ \begin{array}{ll} \mathbb {D}_A V &{} \text {if }\quad 0 \le t \le T_\mathrm{min}^A\\ \lambda \ V &{} \text {if }\quad t> T_\mathrm{min}^A \end{array} \right. ~\hbox {and}~ V(t) = \left\{ \begin{array}{ll} \mathbb {M}_A(t) V(0) &{} \text {if }\quad 0 \le t \le T_\mathrm{min}^A\\ e^{\lambda \ (t-T_\mathrm{min}^A)} V(T_\mathrm{min}^A) &{} \text {if }\quad t > T_\mathrm{min}^A \end{array} \right. \end{aligned}$$

Similarly, if $\mathcal {A/B}(0) \ge \mathcal {(A/B)}^*$,

$$\begin{aligned} \displaystyle \frac{\mathrm{d}V}{{rm d}t} = \left\{ \begin{array}{ll} \mathbb {D}_B V &{} \text {if }\quad 0 \le t \le T_\mathrm{min}^B\\ \lambda \ V &{} \text {if }\quad t> T_\mathrm{min}^B \end{array} \right. ~\hbox {and}~ V(t) = \left\{ \begin{array}{ll} \mathbb {M}_B(t) V(0) &{} \text {if }\quad 0 \le t \le T_\mathrm{min}^B\\ e^{\lambda \ (t-T_\mathrm{min}^B)} V(T_\mathrm{min}^B) &{} \text {if }\quad t > T_\mathrm{min}^B \end{array} \right. \end{aligned}$$

Proof

Straightforward, by Theorem A.10 $\square$

Appendix B: Sensitivity analysis on optimal scheduling

The two determinant quantities of optimal control scheduling are (i) the duration of the first stage ($T_\mathrm{min}^1$), and (ii) the relative intensity between two drugs in the second stage ($k^*$). Here, we show sensitivity analysis on the quantities related to them, $T_\mathrm{gap}$ and $f^*$, over a range of (scaled) model parameters. Additionally over the same range, we studied how much our $T_\mathrm{min}$-based optimal scheme is better than the $T_\mathrm{max}$-based scheme evaluated by the integral in Eq. (13).

• 1. Sensitivity analysis of $T_\mathrm{gap}$

  Using $g_1$, we non-dimentionalize all the values, like

$$\begin{aligned} \{\overline{s_1},\overline{r_1}|\overline{s_2},\overline{r_2}\} := \displaystyle \frac{1}{g_1} \{s_1,r_1|s_2,r_2\} \quad \hbox {and} \quad \overline{T_\mathrm{gap}}:= g_1 \ T_\mathrm{gap} \end{aligned}$$

then,

$$\begin{aligned} \overline{T_\mathrm{gap}}(\{\overline{s_1},\overline{r_1}\},\{\overline{s_2},\overline{r_2}\})&:= \frac{\ln \left[ \displaystyle \frac{(1-\overline{s_1})(\overline{r_1}-\overline{s_1})(\overline{r_1}-\overline{s_2})}{\overline{r_1}((\overline{r_1}-\overline{s_1})(\overline{r_2}-\overline{s_1})+(\overline{r_1}+\overline{r_2}-\overline{s_1}-\overline{s_2}))} \right] }{1+\overline{r_1}-\overline{s_1}} \end{aligned}$$

In general, cells mutate slower than they proliferate , so we ran sensitivity analysis on $T_\mathrm{gap}$ for all $a \gg 1$ for ${a \in \{-\overline{s_1}, -\overline{s_2}, \overline{r_1}, \overline{r_2}\}}$. Figure 12 shows $T_\mathrm{gap}$ over the range of ${20 \le -\overline{s_1}, -\overline{s_2}, \overline{r_1}, \overline{r_2} \le 100}$. So, under the assumption that $g_1 \ll \min \{-s_1,-s_2,r_1,r_2\}$,

$$\begin{aligned} T_\mathrm{gap}(\{s_1,r_1\},\{s_2,r_2\}) \approx \displaystyle \frac{\ln \left[ \displaystyle \frac{-s_1(r_1-s_2)}{r_1(r_2-s_1)} \right] }{r_1-s_1}, \end{aligned}$$

which approximates the contour curves in Fig. 12.

Fig. 12

(Colour figure Online) Contour maps of $T_\mathrm{gap}$ over ranges of $10 \le a \le 100$ for ${a \in \{-\overline{s_1},-\overline{s_2},\overline{r_1},\overline{r_2}\}=\{-s_1,-s_2,r_1,r_2\}/g_1}$ and $r_1 r_2 < s_1 s_2$ (Condition (6)). As $-s_2$ decreases and/or $r_2$ increases, the optimal switching timing to the second drug is delayed ($T_\mathrm{min}\uparrow$ and $T_\mathrm{gap}\downarrow$). As $r_1$ increases, $T_\mathrm{gap}$ decreases. Also, $T_\mathrm{gap}$ and $s_1$ have a non-monotonic relationship as shown on the graphs

• 2. Sensitivity analysis of $f^*$

Regarding the regulated intensities among the two drugs, $k^*$, we assumed that $g_1 \approx g_2 := g$, similarly assuming that they are both much smaller than $\{-s_1,-s_2,r_1,r_2\}$. Then, we normalized all the parameters with the unit of g, like

$$\begin{aligned} {\{\overline{s_1},\overline{r_1}|\overline{s_2},\overline{r_2}\} := \displaystyle \frac{1}{g} \{s_1,r_1|s_2,r_2\}.} \end{aligned}$$

$k^*$ can be rewritten in terms of the dimensionless parameters.

$$\begin{aligned} {k^*(\{\overline{s_1},\overline{r_1}\},\{\overline{s_2},\overline{r_2}\}) = \displaystyle \frac{(\overline{r_1}-\overline{s_2})((\overline{r_1}-\overline{s_1})(\overline{r_2}-\overline{s_1})+(\overline{r_1}+\overline{r_2}-\overline{s_1}-\overline{s_2}))}{(\overline{r_2}-\overline{s_1})((\overline{r_2}-\overline{s_2})(\overline{r_1}-\overline{s_2})+(\overline{r_1}+\overline{r_2}-\overline{s_1}-\overline{s_2}))}} \end{aligned}$$

In this sensitivity analysis, we use

$$\begin{aligned} f^*:=\displaystyle \frac{k^*}{1+k^*}, \end{aligned}$$

which represents intensity fraction of the initially better drug out of the total therapy. We evaluated $f^*$ over the same ranges of $\{s_1,s_2,r_1,r_2\}$, like the previous exercise (see Fig. 13) over the range $\max \{g_1,g_2\} \ll \min \{-s_1,-s_2,r_1,r_2\}$, so $k^*$ and $f^*$ can be approximated by the simpler forms:

$$\begin{aligned} k^* \approx \displaystyle \frac{r_1-s_1}{r_2-s_2} \quad \hbox {and} \quad f^* \approx \displaystyle \frac{r_1-s_1}{r_1+r_2-s_1-s_2} \end{aligned}$$

Fig. 13

(Colour figure Online) Contour maps of $f^*$ over ranges of $10 \le a \le 100$ for ${a \in \{-\overline{s_1},-\overline{s_2},\overline{r_1},\overline{r_2}\}=\{-s_1,-s_2,r_1,r_2\}/g}$ and $r_1 r_2 < s_1 s_2$ (Condition 6). $k^*$ (or $f^*$) increases, as $r_1$ and/or $-s_1$ decreases and/or as $r_2$ and/or $-s_2$ increases

• 3. Sensitivity analysis of Integral (13)

To study the sensitivity of the advantage of using the optimal control defined by Integral (13), we assumed that $g_1 \approx g_2 \approx g = 0.001$. Then, similar to the previous studies, we explored the sensitivity of the normalized parameters in terms of g, that is (Fig. 14):

$$\begin{aligned} {\{\overline{s_1},\overline{r_1}|\overline{s_2},\overline{r_2}\} := \displaystyle \frac{1}{g} \{s_1,r_1|s_2,r_2\}.} \end{aligned}$$

Fig. 14

(Colour figure Online) Contour maps of the measured advantageous effect of the optimal therapy defined by the integration (13) over ranges of $10 \le a \le 100$ for ${a \in \{-\overline{s_1},-\overline{s_2},\overline{r_1},\overline{r_2}\}}$ and $r_1 r_2 < s_1 s_2$ (Condition (6)) Here, ${\{-\overline{s_1},-\overline{s_2},\overline{r_1},\overline{r_2}\}=\{-s_1,-s_2,r_1,r_2\}/g}$ and $g=0.001$. The measured effect increases as $r_1$, $r_2$ decreases and/or $-s_1$ increases

Appendix C: Clinical implementation of instantaneous switch in the optimal strategy

In clinical practice, the instantaneous drug switch which we suggest in the second stage of the optimal treatment scheduling is not implementable. Therefore, we compared similar schedules to the optimal case. In the “similar” schedules, the first stage, using an initial drug, remained the same as the optimal schedule. However the second part, where we previously used an instantaneous switch (with $\Delta t = 0$), was modified to use a fast switch ($\Delta t \gtrsim 0$). Figure 15a, b shows how instantaneous switching ($\Delta t=0$) and fast switching (multiple choices of $\Delta t \gtrsim 0$) compare in terms of population size using different drug parameters. As expected, the smaller $\Delta t$ is , the closer to the ideal case. And, a choice of a reasonably small $\Delta t$ (like 1 day or 3 days) results in an outcome quite close to the optimal scenario.

We repeated this exercise with $k^*$ (from Eq. 12) instead of $k(\Delta t)$ modulated by $\Delta t$ (Fig. 15c, d). Only small differences are observed between Fig. 15a, b and c, d which justifies the general usefulness of $k^*$ independent of $\Delta t$.

Fig. 15

Graphs showing regular drug switching in Stage 2 with different $\{\Delta t, k(\Delta t,p_A,p_B)\}$: $\Delta t=1$ day (blue), $\Delta t=4$ days (red), $\Delta t=7$ days (green), and $\Delta t=10$ days (magenta). Parameters/conditions: $p_A=\{-0.18,0.008,0.00075\}$/day, $p_B=\{-0.9,0.016,0.00125\}$/day and $\{A_\mathrm{R}^0,B_\mathrm{R}^0\}=\{0.1,0.9\}$, a Total population histories, $C_\mathrm{P}^n$ for $n \in \{1,4,7,10\}$ days, b Differences between the optimal population history $C_\mathrm{P}^*$ (i.e., when $\Delta t=0$) and each case with positive $\Delta t$ (i.e., $C_\mathrm{P}^n-C_\mathrm{P}^*$). The inserts interesting ranges. c and d are equivalent with (a) and (b) except that $k^*(p_A,p_B)\}$ has been used instead of $k(\Delta t,p_A,p_B)\}$ (Color figure online)

Appendix D: Stochastic simulation codes

The computational code written in Python will be provided at Github (https://github.com/nryoon12/Optimal-Therapy-Scheduling-Based-on-a-Pair-of-Collaterally-Sensitive-Drugs).