Tumor containment: a more general mathematical analysis

Abstract

Clinical and pre-clinical data suggest that treating some tumors at a mild, patient-specific dose might delay resistance to treatment and increase survival time. A recent mathematical model with sensitive and resistant tumor cells identified conditions under which a treatment aiming at tumor containment rather than eradication is indeed optimal. This model however neglected mutations from sensitive to resistant cells, and assumed that the growth-rate of sensitive cells is non-increasing in the size of the resistant population. The latter is not true in standard models of chemotherapy. This article shows how to dispense with this assumption and allow for mutations from sensitive to resistant cells. This is achieved by a novel mathematical analysis comparing tumor sizes across treatments not as a function of time, but as a function of the resistant population size.

Appendices

Appendix A: Mutations from sensitive to resistant cells

The analysis in this appendix is related to the work of Martin et al. (1992); Hansen et al. (2017). Consider a basic Norton–Simon model with mutations, generalizing the Monro and Gaffney model [Norton and Simon (1977); Goldie and Coldman (1979); Monro and Gaffney (2009)]:

$$\begin{aligned} \begin{aligned} \frac{dS}{dt}&= g(N) \, (1 - L) S - \tau _1 g(N) S + \tau _2 g(N) R, \\ \frac{dR}{dt}&= g(N) \, R + \tau _1 g(N) S - \tau _2 g(N) R, \end{aligned} \end{aligned}$$

(NS with mutations)

where $\tau _1$ and $\tau _2$ are mutation and backmutation rates.

If the growth-rate function g is decreasing in N, an increase in the size of the sensitive population leads to two opposite effects: it slows down the development of existing resistant cells (the competition effect), but usually increases the number of mutations from sensitive to resistant cells (the mutation effect). This trade-off has been studied by Martin et al. (1992) and Hansen et al. (2017). They assume that: a) the sensitive population may me modified at will (eliminated and increased arbitrary quickly), and b) the initial frequency of resistant cells is arbitrary; in particular, it is not necessarily consistent with the assumed tumor growth model and mutations rates. Given these assumptions and a threshold tumour size $N^*$, they conclude that if the resistant population is low enough, then its growth is lower if sensitive cells are absent than if the tumor population is stabilized at $N^*$ (i.e. it is more important to minimize mutations than to maximise competition), while if the resistant population is large, its growth is lower if the tumour is stabilized at $N^*$ than in the absence of sensitive cells.

Here, we do not make these assumptions, but study whether Model (NS with mutations) or variants are compatible with our assumption that, during treatment, a larger sensitive population leads globally to a lower growth-rate of resistant cells. A key difference with previous analysis (Martin et al. 1992; Hansen et al. 2017) is that we try to estimate the initial resistant population.

Let $\phi _R$ denote the growth-rate function of resistant cells:

$$\begin{aligned} \phi _R(S, R) = g(N) R + \tau _1 g(N) S - \tau _2 g(N) R. \end{aligned}$$

Denoting by $x_R = R/N$ the resistant fraction, it is easily checked that $\partial \phi _R / \partial S \le 0$ if and only if:

$$\begin{aligned} \frac{x_R}{\tau _1} (1-\tau _1 - \tau _2) + 1 \ge - \frac{g(N)}{Ng'(N)}. \end{aligned}$$

(A1)

Since the resistant fraction increases during treatment, this condition is bound to be hardest to satisfy at treatment initiation.

The resistant fraction obtained from Model (NS with mutations) for the initial condition $S = 1$, $R=0$ is then (Goldie and Coldman 1979):

$$\begin{aligned} x_R = \frac{\tau _1}{\tau _1 +\tau _2} (1- N_0^{-\tau _1 - \tau _2}) \simeq \tau _1 \ln N_0, \end{aligned}$$

(A2)

where we used the approximation $N^{-\tau } \simeq 1 - \tau \ln N$ for $\tau$ small. Injecting (A2) into (A1) and using that $\tau _1$ and $\tau _2$ are much smaller than 1 leads to:

$$\begin{aligned} \ln N_0 + 1 \ge - \frac{g(N_0)}{N_0g'(N_0)}. \end{aligned}$$

(A3)

Note that this condition is independent of the mutation rates, as long as they are constant and small. This is because a higher mutation rate leads to a higher resistant fraction, and the ratio $x_R(0)/\tau _1$ turns out to be constant (up to the approximation $\tau _i<< 1$). Let us now consider various growth-models.

Case 1 (power-law model): $g(N) = \rho N^{- \gamma }$ with $0< \gamma < 1$. Equation (A3) becomes:

$$\begin{aligned} \ln N_0 + 1 \ge 1/\gamma . \end{aligned}$$

Typical choices for $\gamma$ are $\gamma = 1/3$ or $\gamma = 1/4$ (Gerlee (2013); Benzekry et al. (2014); our $\gamma$ corresponds to $1-\gamma$ in these references). The condition then holds by a huge margin for any detectable tumor size.

Case 2 (Gompertzian growth): $g(N) = \rho \ln (K/N)$. Equation (A3) becomes:

$$\begin{aligned} \ln N_0 + 1 \ge \ln (K/N_0), \end{aligned}$$

which is satisfied if $K \le e N^2$. Standard values of the carrying capacity in Gompertzian models are in the range $10^{12}-10^{13}$ (e.g., $K = 2 \times 10^{12}$ in Monro and Gaffney (2009)). Equation (A3) is then satisfied for any detectable tumor size.

Case 3 (logistic growth): $g(N) = \rho (1 - N/K)$. Equation (A3) becomes:

$$\begin{aligned} \ln N_0 + 1 \ge \frac{K}{N_0} - 1. \end{aligned}$$

This condition need not be satisfied, depending on the interpretation of the model and parameter choices. For instance, Monro and Gaffney (2009) take $N_0 = 10^{10}$. Then $\ln N_0 \simeq 23$ and the condition is roughly $K \le 2.5 \times 10^{11}$, which is not satisfied for standard values of the carrying capacity K. (The choice of carrying capacity may differ for a Gompertz or a logistic growth model. However, in Monro and Gaffney (2009), the lethal tumor size is taken to be $5 \times 10^{11}$ so in a logistic growth version of their model, K would have to be at least that large and Eq. (A3) would not be satisfied.)

The condition would however be satisfied for larger initial tumor sizes, modeling late-stage treatments. Actually, when logistic growth models are used in the adaptive therapy literature, the initial tumor size is often assumed to be a large fraction of the carrying capacity (e.g., Zhang et al. (2017); Strobl etal. (2020); Mistry (2020)). This may be interpreted as modeling late-stage treatments, or as a model of local growth. In the latter case, the carrying capacity should be seen as the maximal number of cells for the current tumor volume (or equivalently, the variables N, S, R, K should be interpreted as densities). Assuming $10^{10}$ tumor cells at tumor initiation, the estimate $x_R/\tau _1 \simeq 23$ would still be valid, and Eq. (A1) would become $K/N \le 25$, which is bound to be satisfied if N and K are densities.

Figure 6 represents the evolution of tumor size under containment and MTD for the three tumor growth models presented above. Containment fails after MTD for Gompertzian growth and the Power-Law model. For logistic growth, in the bottom-left panel, Condition (A3) is violated by a large margin, and Containment fails before MTD. In the bottom-right panel, Condition (A3) is only violated by a tiny margin, and Containment fails after MTD though our results do not guarantee it. Note that, strictly speaking, what our results predict when Condition (A3) is satisfied is that the resistant population is minimal under Containment, not necessarily the total tumor size, though this increasingly coincides as sensitive cells decay.

Fig. 6

Containment vs MTD for the three tumor growth models we consider. In all figures, $\tau _1=\tau _2=10^{-6}$, $K=2 \times 10^{12}$ and $\rho$ is chosen so that the initial growth rate of an untreated tumour is the same as for Gompertzian growth. Except in the Bottom-Right panel, $N_0=10^{10}$, $R_0=2.3 \times 10^5$ and $N^*= 7 \times 10^{10}$, as in Fig. 1. Top-left: Gompertzian growth. Top-right: Power-law model. Bottom-left: Logistic Growth. MTD fails slightly after Containment. Bottom-right: Logistic Growth with $N_0 = 6.9*10^{10}$, $R_0$ computed by means of (A2) and $N^*=2\times 10^{11}$. Containment fails after MTD though condition (A3) is not satisfied

Let us now consider three variants of Model (NS with mutations).

Variant 1: birth-death model. In Model (NS with mutations), the number of mutations is assumed proportional to the net growth-rate of the tumor. It would be natural to assume that the number of mutations is proportional to the net birth-rate, e.g.:

$$\begin{aligned} \begin{aligned} \frac{dS}{dt}&= S \left[ b(N) \, (1 - L) - d(N)\right] - \tau _1 b(N) S + \tau _2 b(N) R \\ \frac{dR}{dt}&= R \left[ b(N) - d(N)\right] + \tau _1 b(N) S - \tau _2 b(N) R \end{aligned} \end{aligned}$$

ss The main effect is to increase the effective mutation rate, that is, the average number of mutations relative to a given increase of tumor size. However, since the condition we found is insensitive to the mutation rates $\tau _1$ and $\tau _2$, this is unlikely to affect the previous analysis. For instance, if the turnover rate d(N)/b(N) is constant, effective mutation rates are multiplied by the factor $b(N)/(b(N)-d(N))$, but Condition (A3) is unchanged.

Variant 2: late-stage treatment. Condition (A3) and the previous analysis are better suited for a first line treatment than a second or third line treatment, especially if resistance to the first treatment may be associated with resistance to ulterior ones. However, in such a situation (late-stage treatment), the initial resistant population is likely to be larger than the one given by (A2), and so condition (A1) is more likely to be satisfied.

Variant 3: cost of resistance in the baseline growth-rate. Consider the following variant of Model (NS with mutations), with a different growth-rate parameter for sensitive cells and for resistant cells:

$$\begin{aligned} \begin{aligned} \frac{dS}{dt}&= \rho _S g(N) \, (1 - L) S - \tau _1 \rho _S g(N) S + \tau _2 \rho _R g(N) R, \qquad \qquad \\ \frac{dR}{dt}&= \rho _R g(N) \, R + \tau _1 \rho _S g(N) S - \tau _2 \rho _R g(N) R. \qquad \qquad \end{aligned} \end{aligned}$$

(Resistance cost)

(the terms g(N) in Model (NS with mutations) correspond here to terms of the form $\rho g(N)$, with $\rho = \rho _S$ or $\rho =\rho _R$.) The absolute growth-rate of resistant cells is now

$$\begin{aligned} \phi _R(S, R) = \rho _R g(N) R + \tau _1 \rho _S g(N) S - \tau _2 \rho _R g(N) R. \end{aligned}$$

The condition for $\partial \phi _R / \partial S$ to be nonpositive becomes:

$$\begin{aligned} \frac{x_R}{\tau _1} (\rho _R/\rho _S - \tau _2 \rho _R/\rho _S - \tau _1) + 1 \ge - \frac{g(N)}{N g'(N)}. \end{aligned}$$

(A4)

Moreover, if $\rho _R$ is substantially smaller than $\rho _S$, then the resistant fraction at treatment initiation is no longer given by (A2) but approximately by (see (Viossat and Noble 2020), Section 7):

$$\begin{aligned} x_R = \frac{\tau _1}{1- \rho _R/\rho _S}. \end{aligned}$$

(A5)

Injecting (A5) into (A4) and using that $\tau _1$ and $\tau _2$ are much smaller than 1 leads to the condition:

$$\begin{aligned} \frac{1}{1 - \rho _R/\rho _S} \ge - \frac{g(N)}{N g'(N)}. \end{aligned}$$

(A6)

For a Power-law model, the condition becomes $\rho _R/\rho _S \ge 1 - \gamma$. For $\gamma = 1/3$, this is satisfied if and only if $\rho _R/\rho _S \ge 2/3$. that is, if and only if the resistance cost is not too large.

For a Gompertzian model, the right-hand side is $\ln (K/N)$ and Eq. (A6) may be written: $\rho _R/\rho _S \ge 1--1/\ln (K/N)$. With Monro and Gaffney (2009) values: $N_0 = 10^{10}$, $K=2 \times 10^{12}$, this is satisfied if $\rho _R/\rho _S \ge 0.81$.

With logistic growth, the right-hand side is $K/N - 1$ and the condition may be written as $\rho _R/\rho _S \ge (K-2N)/(K-N)$.² Assuming for instance $\rho _R/\rho _S = 4/5$, this boils down to $K \le 6 N$. This would not be satisfied at treatment initiation if N and K represent total numbers of cells in the whole tumor (except possibly for a late-stage treatment), but seems likely to be satisfied in a model of local growth, where N and K are densities.

We conclude that in the absence of resistance costs, our analysis applies to several standard models of tumor growth with mutations, such as Power-law models or Gompertzian growth, and possibly to logistic growth, at least when it models local growth. However, if the baseline growth-rate of resistant cells is substantially smaller than the baseline growth-rate of sensitive cells, our assumptions become more restrictive and might fail even for Gompertzian growth. This is in line with Hansen et al. (2017)’s finding that, contrary to common wisdom, a resistance cost in the baseline growth rate may make it less likely that containment strategies outperform more aggressive treatments.

Note however that the fact that we can no longer prove that containment outperforms MTD does not mean that it would not do so. For instance, Fig. 7, right, compares containment and MTD for a Gompertzian growth model with a strong resistance cost: $\rho _R/\rho _S = 0.5$. Condition (A6) is then violated, but containment still fails long after MTD.

Fig. 7

Gomperzian growth with a resistance cost. Parameters $N_0$, K, $\rho _S$, $\tau _1$, $\tau _2$ as in Fig. 1. $R_0$ is obtained from (A5). Left: $\rho _R = 0.81 \rho _S$. Right: $\rho _R = 0.5 \rho _S$. The resistant population (unshown) still grows slower under Containment than under MTD, though our analysis does not guarantee it

Moreover, the analysis in the preprint version of Viossat and Noble (2020), Section 7 of the supplementary material, suggests that the mutation effect could only make MTD marginally superior to containment.

Appendix B: Proof of proposition 8

We first need a lemma.

Lemma 12

Assume that the sensitive population decreases when treated at $L_{max}$.

a) Let $\bar{N} \ge 0$. Consider a solution (N, R) of Model 3 under a treatment such that $L(t) = L_{max}$ whenever $N(t) > \bar{N}$. Let $\bar{t} \ge 0$ be such that $N(\bar{t}) \ge \bar{N}$. For all $t \ge \bar{t}$, $S(t) \le S(\bar{t})$. If moreover $N(t) \ge \bar{N}$ for all $t \ge \bar{t}$, then S is non-increasing on $[\bar{t}, +\infty )$.

b) Under containment (respectively, containment at $N_{min}$), once tumor size reaches $N^*$ for the first time (respectively, $N_{min}$), the sensitive population is non-increasing.

Proof

a) The idea is that when $N > \bar{N}$, S is non-increasing by assumption, and when $N(t) \le \bar{N}$ for $t > \bar{t}$, the sensitive population must have decreased since time $\bar{t}$ because the resistant population increased (by assumption) and total tumor size did not. Formally, let $t \ge \bar{t}$. If for all $\tau$ in $(\bar{t}, t)$, $N(\tau ) > \bar{N}$, hence $L(\tau ) = L_{max}$, then S is non-increasing on $[\bar{t}, t]$ by assumption, therefore $S(t) \le S(\bar{t})$. Otherwise, let $t_{max} = \max \{\tau \le t, N(\tau ) \le \bar{N} \}$. The previous argument implies that $S(t) \le S(t_{max})$. Moreover, since R is increasing,

$$\begin{aligned} S(t_{max}) = N(t_{max}) - R(t_{max}) \le \bar{N} - R(t_{max}) \le N(\bar{t}) - R(\bar{t}) = S(\bar{t}) \end{aligned}$$

where we used that $N(\bar{t}) \ge \bar{N}$ and $R(t_{max}) \le R(\bar{t})$ Therefore, $S(t) {\le S(t_{max})} \le S(\bar{t})$. Finally, if for any $t_1 \ge \bar{t}$, $N(t_1) \ge \bar{N}$, then the previous result applied from $t_1$ on shows that for any $t_2 \ge t_1$, $S(t_2) \le S(t_1)$, hence S is non-increasing on $[\bar{t}, \infty )$.

b) For containment, this follows from a) with $\bar{N} = N^*$ and the fact that once tumor size reaches $N^*$ under containment, it never becomes smaller. The proof for containment at $N_{min}$ is the same with $N_{min}$ replacing $N^*$. $\square$

We now prove Proposition 8. Proof of a): The inequality $S_{MTD}(t) \le S_{alt}(t)$ follows from Lemma 10, the fact that $\tilde{N}_{MTD}(r) \le \tilde{N}_{alt}(r)$ (see Eq. (2)), and the fact that $\tilde{S}_{MTD}(r)$ is non-increasing by Assumption (A2). The last inequality follows from Assumption (A1), or, independently of (A1), from Lemma 10, the fact that $\tilde{N}_{Cont}(r) \le \tilde{N}_{noTreat}(r)$ [see Eq. (2)], and that once Containment starts treating, $S_{Cont}$ is non-increasing [Lemma 12, item b)].

Let us now prove that $S_{alt}(t) \le S_{Cont}(t)$. Let $r_1 = \min \{r\ge R_0, \tilde{N}_{Cont}(r) = N^*\}$. For $t \le t_{Cont}(r_1)$, containment does not treat so $S_{Cont}(t) = S_{noTreat}(t) \ge S_{alt}(t)$ by Assumption (A1). Moreover, on $[r_1, R_{Cont}^{\infty })$, $\tilde{N}_{Cont}(r) \ge \tilde{N}_{alt}(r)$ [see Eq. (2)] and $\tilde{N}_{Cont}(r) \ge N^*$, therefore $\tilde{S}_{Cont}(r)$ is non-increasing by Lemma 12. Thus, by Lemma 10, $S_{Cont}(t) \ge S_{alt}(t)$ for any t in $[t_{Cont}(r_1), t_{alt}(R_{Cont}^{\infty }) \,)$.

Finally, let $t \ge \max (t_{Cont}(r_1), t_{alt}(R_{Cont}^{\infty }))$, that is, such that $R_{Cont}(t) \ge r_1$ and $R_{alt}(t) \ge R_{Cont}^{\infty }$. Since $N_{Cont}(t) \ge N^*$ for all $t \ge t(r_1)$, it follows from Lemma 12 that $S_{Cont}$ is non-increasing on $[t(r_1), +\infty [$, so $S_{Cont}(t) \ge S_{Cont}^{\infty }$. Thus, it suffices to show that $S_{alt}(t) \le S_{Cont}^{\infty }$. There are two cases.

Case 1: If $\tilde{N}_{alt}(R_{Cont}^{\infty }) \ge N^*$, then by Lemma 12, for all $t \ge t_{alt}(R_{Cont}^{\infty })$,

$$\begin{aligned} S_{alt}(t) \le S_{alt}(t_{alt}(R_{Cont}^{\infty }))= \tilde{S}_{alt} (R_{Cont}^{\infty }) \le S_{Cont}^{\infty } \end{aligned}$$

where the last inequality follows from the fact that for $r < R_{Cont}^{\infty }$, $\tilde{S}_{alt}(r) \le \tilde{S}_{Cont}(r)$ due to Eq. (1), so that

$$\begin{aligned} \tilde{S}_{alt}(R_{Cont}^{\infty }) = \lim _{r \rightarrow R_{Cont}^{\infty }} \tilde{S}_{alt}(r) \le \lim _{r \rightarrow R_{Cont}^{\infty }} \tilde{S}_{Cont}(r) = \lim _{t \rightarrow +\infty } S_{Cont}(t) = S_{Cont}^{\infty } \end{aligned}$$

Case 2: If $\tilde{N}_{alt}(R_{Cont}^{\infty }) < N^*$, then as long as $N_{alt}(t) \le N^*$,

$$\begin{aligned} S_{alt}(t) {=N_{alt}(t) - R_{alt}(t)} \le N^* - R_{alt}(t) \le N^* - R_{Cont}^{\infty } \le S_{Cont}^{\infty } \end{aligned}$$

where the second inequality uses that we consider the case where $R_{alt}(t) \ge R_{Cont}^{\infty }$. Moreover, if at some time $\bar{t}$, $N_{alt}(\bar{t}) = N^*$ (which must indeed happen), then $S_{alt}(\bar{t}) \le S_{Cont}^{\infty }$ by the previous argument, and for all $t \ge \bar{t}$, by Lemma 12, $S_{alt}(t) \le S_{alt}(\bar{t}) \le S_{Cont}^{\infty }$. This concludes the proof of a).

Proof of b): The inequality $S_{ContNmin}(t) \le S_{Int}(t)$ follows from Lemma 10, the fact that $\tilde{N}_{contNmin}(r) \le \tilde{N}_{Int}(r)$, and the fact that once tumor size reaches $N_{min}$, $S_{ContNmin}$ is non-increasing [Lemma 12, item b)]. The proof of $S_{Int}(t) \le S_{Cont}(t)$ is as the proof of $S_{alt}(t) \le S_{Cont}(t)$ (except that Assumption (A1) is not needed).

Proof of c): we first prove $S_{idContNmin} \le S _{idInt}$. Before tumor size reaches $N_{min}$, both treatments coincide, then until $t_{idContNmin}$, $N_{idContNmin} = N_{min} \le N_{idInt}$ while $R_{idContNmin} \ge R_{idInt}$, so $S_{idContNmin} \le S_{idInt}$. Finally, for $t \ge t_{idContNmin}$, $S_{idContNmin}(t)=0 \le S_{idInt}(t)$. The proof of $S_{idInt} \le S_{idCont}$ is similar.

Proof of d): the first inequality is trivial, as $S_{idMTD}(t)=0$ for all $t >0$. The last inequality follows from (A1) but also, independently of (A1), from the following argument: for $t \le t_{idCont}$, $N_{idCont} \le N_{noTreat}$ while $R_{idCont} \ge R_{noTreat}$ by Proposition 3, so $S_{idCont} \le S_{noTreat}$, and for $t \ge t_{idCont}$, $S_{idCont}=0$.

It remains to prove that $S_{idAlt}(t) \le S_{idCont}(t)$. Let $t_1 = \min \{t \ge 0, N_{idCont}(t)=N^*\}$. For $t \le t_1$, the result follows from Assumption (A1). For $t_1 \le t \le t_{alt}$, then, by Proposition 6, $N_{idAlt}(t)\le N^* = N_{idCont}(t)$ and $R_{idAlt}(t) \ge R_{idCont}(t)$, hence $S_{idAlt}(t) \le S_{idCont}(t)$. Finally, for $t \ge t_{idAlt}(t)$, $S_{idAlt}(t)=0$.

Appendix C: Comparison principles

The following comparison principle is standard:

Proposition 13

Let $\Omega$ be a nonempty open subset of $\mathbb {R}^{2}$. Let $f: \Omega \rightarrow \mathbb {R}$ be continuously differentiable. Consider the ordinary differential equation $x'(t) = f(t, x(t))$. Let $t_1 \ge t_0$. Let $u: [t_0, t_1] \rightarrow \mathbb {R}$ be solution of this ODE and let $v: [t_0, t_1] \rightarrow \mathbb {R}$ be a subsolution. That is, v is continuous, almost everywhere differentiable, $(t, v(t)) \in \Omega$ on $[t_0, t_1]$, and, almost everywhere, $v'(t) \le f(t, v(t))$. If furthermore $v(t_0) \le u(t_0)$, then $v(t) \le u(t)$ for all $t \in [t_0, t_1]$.

We want to apply a variant of this result to two equations: first (for Lemma 11),

$$\begin{aligned} \frac{d\tilde{N}}{dr} = G(r, \tilde{N}(r)) \text{ where } G(r, \tilde{N}) = \frac{f_N(\tilde{N}, r, \tilde{L}_2(r))}{f_R(\tilde{N}, r)}, \end{aligned}$$

(here $\tilde{N}(r)$ plays the role of u(t), and G the role of f in Proposition 13). Second (for Lemma 10),

$$\begin{aligned} \frac{dR}{dt} = F(R(t)) \text{ where } F(R) = f_R(\tilde{N}_2(R), R), \end{aligned}$$

where $\tilde{L}_2$ and $\tilde{N}_2$ are not continuously differentiable (otherwise Proposition 13 would directly apply), but piecewise $C^1$ in the strong sense we defined in Sect. 2. For instance, for $\tilde{L}_2$, which is defined on $[R_0, R_2^{\infty })$, there exist values $r_0 = R_0< r_1<... < r_n = R_2^{\infty }$ such that for each i in $\{1,.., n-1\}$, $\tilde{L}_2$ coincides on $[r_i, r_{i+1})$ with a continuously differentiable function $\tilde{L}_2^i$ defined on a neighborhood of $[r_i, r_{i+1})$.

We thus need variants of Proposition 13 where f is slightly less regular. The proof of these variants consists in repeated applications of Proposition 13.

Proposition 14

The conclusion $u(t) \le v(t)$ on $[t_0, t_1]$ of Proposition 13 still holds in the following cases:

a) if f takes the form $f(t, x) = \psi (t, x, \phi (t))$, where $\psi$ is continuously differentiable and $\phi$ is strongly piecewise $C^1$.

b) if f is a function of x only ($f: I \rightarrow \mathbb {R}$, where I is a nonempty interval) which is strongly piecewise $C^1$, and u and v are strictly increasing.

Proof

Proof of a): by assumption, there exists an integer n and real numbers $(\tau _k)_{0 \le k \le n}$ satisfying $t_0 = \tau _0< \tau _1<... < \tau _n = t_1$ such that on $A_k = [t_k, t_{k+1}) \times \mathbb {R}$, f coincides with a continuously differentiable function $f_k$ defined on a neighborhood of $A_k$. Assuming $u(\tau _k) \le v(\tau _k)$, Proposition 13 implies that $u(t) \le v(t)$ on $[t_k, t_{k+1})$ and this is also true at time $t_{k+1}$ by continuity of u and v. An induction argument then gives the result.

Proof of b): By assumption, there exist a finite number of values $x_0$,...,$x_n$ such that f coincides on $[x_k, x_{k+1})$ with a continuously differentiable function $\phi _k$ defined on a neighborhood of $[x_k, x_{k+1})$, and we may assume (up to adding an artificial point) that $x_0 \le u(t_0)$. Since u and v are strictly increasing, there exists an integer $q \le 2(n+1)$ and a sequence of times $\tau _0 = t_0< \tau _1<... < \tau _q = t_1$ such that on $(\tau _j, \tau _{j+1})$, u(t) and v(t) are never equal to one of the $x_k$. Assuming $u(\tau _j) \le v(\tau _j)$, then either: for all t in $[\tau _j, \tau _{j+1})$, u(t) and v(t) belong to the same interval $[x_k, x_{k+1})$ and Proposition 13 implies that $u(t) \le v(t)$ on $[\tau _j, \tau _{j+1})$; or there exists k such that for all t in $[\tau _j, \tau _{j+1})$, $u(t) \le x_k \le v(t)$ and the same inequality holds trivially. In both cases, $u(\tau _{k+1}) \le v(\tau _{k+1})$ by continuity of u and v. An induction argument then gives the result. $\square$