Existence and stability of periodic solutions of an impulsive differential equation and application to CD8 T-cell differentiation

Abstract

Unequal partitioning of the molecular content at cell division has been shown to be a source of heterogeneity in a cell population. We propose to model this phenomenon with the help of a scalar, nonlinear impulsive differential equation (IDE). To study the effect of molecular partitioning at cell division on the effector/memory cell-fate decision in a CD8 T-cell lineage, we study an IDE describing the concentration of the protein Tbet in a CD8 T-cell, where impulses are associated to cell division. We discuss how the degree of asymmetry of molecular partitioning can affect the process of cell differentiation and the phenotypical heterogeneity of a cell population. We show that a moderate degree of asymmetry is necessary and sufficient to observe irreversible differentiation. We consider, in a second part, a general autonomous IDE with fixed times of impulse and a specific form of impulse function. We establish properties of the solutions of that equation, most of them obtained under the hypothesis that impulses occur periodically. In particular, we show how to investigate the existence of periodic solutions and their stability by studying the flow of an autonomous differential equation. Then we apply those properties to prove the results presented in the first part.

Appendices

Proofs of Lemmas

1.1 Proof of Lemma 2

We first show (i). Let us assume that $X(\omega ):=(1+\alpha ) X(\omega ^-)>X(0)$. We then assume that there exists $k\in \mathbb {N}^*$ such that $X_{k+1}:=X((k+1)\omega )>X_k:=X(k\omega )$ and we show that $X((k+2)\omega )>X((k+1)\omega )$.

Thanks to the uniqueness of the solution of (1), and by integrating the solutions $Y_1=Y_1(\cdot ;X_{k+1},\alpha )$ and $Y_2=Y_2(\cdot ;X_k,\alpha )$ of (1) on $\left[ 0,\omega \right)$, the inequality $X((k+1)\omega )>X(k\omega )$ implies that $X((k+2)\omega ^-)>X((k+1)\omega ^-)$ and then

$$\begin{aligned} X((k+2)\omega ):=(1+\alpha )X((k+2)\omega ^-)>(1+\alpha )X((k+1)\omega ^-)=:X((k+1)\omega ). \end{aligned}$$

We showed that for all $k\in \mathbb {N}^*,\ X((k+1)\omega )>X(k\omega )$. Similarly, we can show that if $X(\omega )<X(0)$, then for all $k\in \mathbb {N}^*,\ X((k+1)\omega )<X(k\omega )$.

It remains to prove (ii). As done in the proof of (i) we show that if $X(\omega )=X(0)$, then for all $k\in \mathbb {N}^*,\ X(k\omega )=X(0)$. Because X is the solution of an autonomous equation on each interval $\left[ k\omega , (k+1)\omega \right)$, it follows that for all $k\in \mathbb {N}^*$, for all $t\in \left[ 0,\omega \right)$, $X(t)=X(k\omega +t)$ and then X is $\omega$-periodic. $\square$

1.2 Proof of Lemma 3

We first rewrite Eq. (2) in the form $X'(t)=\mathcal {H}\left( X(t)\right) -\delta X(t)$ with

$$\begin{aligned} \mathcal {H}: x\in \mathbb {R}^+ \mapsto \eta \dfrac{x^n}{\theta ^n+x^n}\in \left[ 0,\eta \right) . \end{aligned}$$

Hence, for all $X_0 \in \mathbb {R^+_*}$,

$$\begin{aligned} \varphi _\omega (X_0)=X(\omega ;X_0,0)=X_0e^{-\delta \omega }+e^{-\delta \omega }\int _0^\omega e^{\delta s}\mathcal {H}\left( X(s)\right) \mathrm {d} s, \end{aligned}$$

thus

$$\begin{aligned} \frac{X(\omega ;X_0,0)}{X_0}=e^{-\delta \omega }+e^{-\delta \omega }\int _0^\omega e^{\delta s}\frac{\mathcal {H}\left( X(s)\right) }{X_0} \mathrm {d}s. \end{aligned}$$

(14)

Since $\mathcal {H}<\eta$ on $\mathbb {R}^+$ and $e^{\delta s}\le e^{\delta \omega }$ for $s\in (0,\omega )$, one obtains

$$\begin{aligned} 0\le e^{-\delta \omega }\int _0^\omega e^{\delta s}\frac{\mathcal {H}\left( X(s)\right) }{X_0} \mathrm {d}s \le \omega \frac{\eta }{X_0} . \end{aligned}$$

Then it is clear from (14) that

$$\begin{aligned} \lim _{{X_0\rightarrow + \infty }} \frac{X(\omega ;X_0,0)}{X_0}=\exp (-\,\delta \omega ) \end{aligned}$$

and then, from (11), $\lim _{{X_0\rightarrow + \infty }}\varGamma _\omega (X_0)= \exp (\delta \omega )-1$.

We now study the limit of $\varGamma _\omega$ as $X_0$ goes to zero. Let us recall that if $X_0<X_u$, then $f(X_0)<0$ and $X(\ \cdot \ ;X_0,0)$ decreases on $[0,\omega ]$. Consequently, for $s\in [0,\omega ]$, $X(\omega )\le X(s) \le X_0$ and we have the inequality

$$\begin{aligned} \frac{\mathcal {H}\left( X(s)\right) }{X_0} = \eta \dfrac{X(s)^n}{\left( \theta ^n+X(s)^n\right) X_0 } \le \eta \dfrac{X_0^{n-1}}{ \theta ^n+X(\omega )^n } . \end{aligned}$$

Therefore

$$\begin{aligned} e^{-\delta \omega }\int _0^\omega e^{\delta s}\frac{\mathcal {H}\left( X(s)\right) }{X_0} \mathrm {d}s \le \omega \eta \dfrac{X_0^{n-1}}{ \theta ^n+X(\omega )^n } . \end{aligned}$$

Finally, from (14),

$$\begin{aligned} \lim _{{X_0\rightarrow 0^+}}\frac{X(\omega ;X_0,0)}{X_0}=\exp (-\,\delta \omega ), \end{aligned}$$

so $\lim _{{X_0\rightarrow 0^+}}\varGamma _\omega (X_0)=\exp (\delta \omega )-1$. This concludes the proof. $\square$

1.3 Proof of Lemma 4

Let us assume $\varTheta \in (X_u,X_s)$. The proof is similar if $\varTheta \in (0,X_u]$. In order to apply Theorem 2, we investigate the sign of $f'-f'\circ \varphi _\omega$ on $(0,+\,\infty )$.

As steady states of (2), $0,\ X_u$ and $X_s$ are fixed points of $\varphi _\omega$. Therefore, if $x\in \{0,X_u,X_s\}$, then $f'(x)-f'(\varphi _\omega (x))=0$.

If $X_0\in (0,X_u)\cup (X_s,+\,\infty )$, then $f(X_0)<0$ and the solution of (3) decreases on $[0,\omega )$, so $0<\varphi _\omega (X_0)<X_0$. According to Proposition 1, f is strictly convex on $(0,X_u)\subset (0,\varTheta )$ and strictly concave on $(X_s,+\,\infty )\subset (\varTheta ,+\,\infty )$. Consequently, for all $X_0\in (0,X_u)$, $f'(X_0)-f'(\varphi _\omega (X_0))>0$ and for all $X_0\in (X_s,+\,\infty )$, $f'(X_0)-f'(\varphi _\omega (X_0))<0$.

If $X_0\in (X_u,X_s)$, then $f(X_0)>0$, the solution of (3) increases on $[0,\omega )$ and $\varphi _\omega (X_0)>X_0$. Moreover, if $X_0\in (X_u,\varphi _\omega ^{-1}(\varTheta ))$, then $\varphi _\omega (X_0)\in (X_0,\varTheta )$, that is $X_u<X_0<\varphi _\omega (X_0)<\varTheta$. Since f is strictly convex on $(X_u,\varTheta )$ (Proposition 1), it follows $f'(X_0)-f'(\varphi _\omega (X_0))<0$ . Similarly, for all $X_0$ in $[\varTheta ,X_s)$, $f'(X_0)-f'(\varphi _\omega (X_0))>0$. It remains to prove that there exists a unique $X_c\in (\varphi _\omega ^{-1}(\varTheta ),\varTheta )$ such that $f'(X_c)-f'(\varphi _\omega (X_c))=0$.

Since $f, \varphi _\omega \in \mathcal {C}^{\infty }(\mathbb {R})$, it follows $f'-f'\circ \varphi _\omega \in \mathcal {C}^{\infty }(\mathbb {R})$ and then there exists $X_c \in (\varphi _\omega ^{-1}(\varTheta ),\varTheta )$ such that $f'(X_c)-f'(\varphi _\omega (X_c))=0$. To show the uniqueness of $X_c$, we suppose there exists $Y_c\in (\varphi _\omega ^{-1}(\varTheta ),\varTheta )$, $Y_c>X_c$, such that $f'(Y_c)-f'(\varphi _\omega (Y_c))=0$. Since $\varphi _\omega$ is increasing and $\varphi _\omega (X_s)=X_s$, it follows $X_c<Y_c<\varTheta<\varphi _\omega (X_c)< \varphi _\omega (Y_c)<\varphi _\omega (\varTheta )<X_s$. From Proposition 1, f is convex on $(\varphi _\omega ^{-1}(\varTheta ),\varTheta )$ so $f'(X_c)<f'(Y_c)$. On the other hand, from Proposition 1, f is concave on $(\varTheta ,X_s)$ so $f'(\varphi _\omega (X_c))>f'(\varphi _\omega (Y_c))$. There is a contradiction since $f'(X_c)=f'(\varphi _\omega (X_c))$ and $f'(Y_c)=f'(\varphi _\omega (Y_c))$.

Then $f'(\varphi _\omega (x))=f'(x)$ if and only if $x\in \{0,X_u,X_c,X_s\}$ and $f'(\varphi _\omega (x))-f'(x)<0$ (resp. $>0$) if $x\in (0,X_u)\cup (X_c,X_s)$ (resp. if $x\in (X_u,X_c)\cup [X_s,+\,\infty )$).

According to Theorem 2, since $f<0$ on $(0,X_u)\cup (X_s,+\,\infty )$, we conclude that $\varphi _\omega$ is convex on $(0,X_u)$ and on $(X_u,X_c)$ and concave on $(X_c,X_s)$ and on $(X_s,+\infty )$. Since $\varphi _\omega \in \mathcal {C}^\infty (\mathbb {R}^+)$, then $\varphi _\omega$ is convex on $(0,X_c)$ and concave on $(X_c,+\infty )$.

In the particular case $X_u=\varTheta$, f is strictly convex on $(0,X_u)$ and strictly concave on $(X_u,+\,\infty )$. Thus $f'(\varphi _\omega (x))-f'(x)<0$ if $x\in (0,X_u)\cup (X_u,X_s)$ and $f'(\varphi _\omega (x))-f'(x)<0$ if $x\in (X_s,+\infty )$. It follows that $\varphi _\omega$ is convex on $(0,X_u)$ and concave on $(X_u,+\infty )$, i.e. $X_c=X_u$. The converse is straightforward. $\square$

1.4 Proof of Lemma 5

The function $x\mapsto \varphi _\omega (x)/x$ is continuous on $(0,+\infty )$ and, in particular, on the closed interval $[X_u,X_s]$. Therefore there exists $\beta =\max _{x\in [X_u,X_s]}\varphi _\omega (x)/x$. Moreover, for all $x\in (X_u,X_s)$, $\varphi _\omega (x)/x>1$ and for all $x\in (0,X_u]\cup [X_s,+\,\infty )$, $\varphi _\omega (x)/x\le 1$ (cf. proof of Lemma 4). Consequently, $\beta =\max _{x>0}\varphi _\omega (x)/x$ and $\beta >1$. Hence there exists $x_\beta \in [X_u,X_s]$ such that $\varphi _\omega (x_\beta )=\beta x_\beta$ and for all $x>0$, $\varphi _\omega (x)\le \beta x$.

Let $x>0$ be such that $\varphi _\omega (x)/x=\beta$. Then $x\in \arg \max _{x>0}(\varphi _\omega /id)$, where id denotes the identity function on $\mathbb {R}^+$. As an extremum of a $\mathcal {C}^\infty$-function, x satisfies $(\varphi _\omega /id)'(x)=0$. That is $\varphi _\omega '(x)= \varphi _\omega (x)/x=\beta$. $\square$

1.5 Proof of Lemma 6 (i)

From Lemma 5 we know that I is non-empty and $I\subset (X_u,X_s)$. Moreover, $I=(\varphi _\omega /id)^{-1}(\{\beta \})$, where id is the identity function and $\varphi _\omega /id$ is continuous on $(0,+\,\infty )$, so I is a closed set. Let’s show that $I\subset (X_c,X_s)$.

By contradiction, let us suppose that there exists $x_0\in I$ such that $X_u<x_0 \le X_c$. According to Lemmas 4 and 5, for all $0<x<x_0$, $\varphi _\omega '(x)\le \varphi _\omega '(x_0)=\beta$. Then the function $x\mapsto \varphi _\omega (x)-\beta x$ is decreasing on $(0,x_0)$. Consequently, for all $x\in (0,x_0)$, $\varphi _\omega (x)-\beta x \ge \varphi _\omega (x_0)-\beta x_0=0$, by definition of $x_0$, and in particular $\varphi _\omega (X_u)-\beta X_u\ge 0$.

On the other hand, since $X_u$ is a steady state of (2), then $\varphi _\omega (X_u)=X_u$ and since $\beta >1$ (Lemma 5), then $\varphi _\omega (X_u)-\beta X_u=X_u(1-\beta )<0$. There is a contradiction, so $I\subset (X_c,X_s)$.

It remains to show that I is an interval. If $I=\{x_\beta \}$, given by Lemma 5, the proof is complete. Assume $x_0,y_0\in I$ with $x_0<y_0$ and let $z_0\in (x_0,y_0)$. Since $x_0>X_c$ and $\varphi _\omega$ is concave on $(X_c,+\infty )$, it follows from Lemma 5 and Definition 2 that

$$\begin{aligned} \beta =\varphi _\omega '(x_0)\ge \varphi _\omega '(z_0)\ge \varphi _\omega '(y_0)= \beta . \end{aligned}$$

Therefore $\varphi _\omega '(z_0)=\beta$ for all $z_0\in (x_0,y_0)$, so $\varphi _\omega$ is linear. Since $\varphi _\omega (x_0)=\beta x_0$, we deduce $\varphi _\omega (z_0)=\beta z_0$, that is $z_0\in I$ and I is an interval. $\square$

1.6 Proof of Lemma 6 (ii)

We first focus on the interval $(0,X_c)\subset (0,I_m)$. Since $\varphi _\omega$ is convex on $(0,X_c)$ (Lemma 4) and $\varphi _\omega (0)=0$, from the mean value theorem one has, for all $x\in (0,X_c)$, $\varphi _\omega '(x)\ge \varphi _\omega (x)/x$.

Let us suppose there exists $x_0\in (0,X_c)$ such that $\varphi _\omega '(x_0) =\varphi _\omega (x_0)/x_0$. Then we claim that for all $x\in (0,x_0)$, $\varphi _\omega (x)/x=\varphi _\omega (x_0)/x_0$. Indeed, from the convexity of $\varphi _\omega$,

$$\begin{aligned} \text { for all }x\in (0,x_0),\ \varphi _\omega (x)/x\le \varphi _\omega '(x)\le \varphi _\omega '(x_0)=\varphi _\omega (x_0)/x_0. \end{aligned}$$

(15)

Assume there exists $x\in (0,x_0)$ such that $\varphi _\omega (x)/x<\varphi _\omega (x_0)/x_0$. Then, from the mean value theorem, there exists $c\in (x,x_0)$ such that

$$\begin{aligned} \varphi _\omega '(c)=\frac{\varphi _\omega (x_0)-\varphi _\omega (x)}{x_0-x}>\frac{\varphi _\omega (x_0)-x\varphi _\omega (x_0)/x_0}{x_0-x}=\frac{\varphi _\omega (x_0)}{x_0}. \end{aligned}$$

There is a contradiction with (15), so for all $x\in (0,x_0)$, $\varphi _\omega (x)/x=\varphi _\omega (x_0)/x_0$.

On the other hand, according to Lemma 3,

$$\begin{aligned} \lim _{x\rightarrow 0^+}\varphi _\omega (x)/x=\lim _{x\rightarrow 0^+}(\varGamma _\omega (x)+1)^{-1}=\exp (-\delta \omega ). \end{aligned}$$

It follows that $\varphi _\omega (x_0)/x_0=\exp (-\,\delta \omega )$ and so, for all $x\in (0,x_0)$, $\varphi _\omega (x)=\exp (-\,\delta \omega )x$. Consequently, $\varphi _\omega$ coincides with the flow of the linear equation $Y'(t)=-\,\delta Y(t)$. That is impossible since for all $x>0$, $f(x)>-\,\delta x$, so the first assumption in the proof is false. We conclude that for all $x_0\in (0,X_c)$, $\varphi _\omega '(x_0)> \varphi _\omega (x_0)/x_0$.

Now, consider $x_0\in [X_c,I_m)$ and suppose that $\varphi _\omega '(x_0)\le \varphi _\omega (x_0)/x_0$. By concavity of $\varphi _\omega$ on $(X_c,+\infty )$ (Lemma 4), for all $x\ge x_0$, $\varphi _\omega '(x)\le \varphi _\omega '(x_0)\le \varphi _\omega (x_0)/x_0$. Consequently for any $y\in (x_0,+\infty ),$

$$\begin{aligned} \varphi _\omega (y)=\varphi _\omega (x_0)+\int _{x_0}^y \varphi _\omega ' (x) \mathrm {d}x \le \varphi _\omega (x_0)+\frac{\varphi _\omega (x_0)}{x_0}(y-x_0)=y\frac{\varphi _\omega (x_0)}{x_0}. \end{aligned}$$

In particular, $\varphi _\omega (I_m)\le I_m \frac{\varphi _\omega (x_0)}{x_0}$. However, by definition of I, $\varphi _\omega (I_m)= I_m\beta$ and, since $x_0\notin I=[I_m,I_M]$ , $\beta > \varphi _\omega (x_0)/x_0$ (Lemma 5). There is a contradiction. Therefore, for all $x_0\in [X_c,I_m)$, $\varphi _\omega '(x_0)> \varphi _\omega (x_0)/x_0$.

Similar arguments allow us to conclude that $\varphi _\omega '(x)<\varphi _\omega (x)/x$ for $x\in (I_M,+\infty )$.

Finally, if $x\in [I_m,I_M]$, from Lemmas 5 and 6 (i), $\varphi _\omega '(x)=\beta$ and $\varphi _\omega (x)=\beta x$. That achieves the proof. $\square$

Convexity of the flow of an autonomous equation

In this part, we aim at proving Theorem 2. In Sect. B.1 we first study a Cauchy problem where the right-hand side of the equation is a continuous piecewise linear function and we determine necessary and sufficient conditions to conclude on the strict convexity of the flow (Proposition B2). We generalize this result to a Cauchy problem with continuously differentiable right-hand side function in Sect. B.2 to prove Theorem 2.

1.1 Preliminary results

We set $a<b\in \mathbb {R}$, $n\in \mathbb {N^*}$ and $(c_i)_{0\le i \le n}$ a subdivision of [a, b] such that $c_0=a,\ c_n=b$ and, for all $i\in \{ 0,\dots , n-1 \}$, $c_i<c_{i+1}$. Then, we define the piecewise linear function $h_n$, on [a, b], by

$$\begin{aligned} \begin{array}{ll} h_n (x)= a_{i+1}x+b_{i+1}&\text { for } x\in [c_i,c_{i+1}),\ i\in \{0,\dots ,n-1\}, \end{array} \end{aligned}$$

where real numbers $(a_i)_{1\le i \le n}$, $(b_i)_{1\le i \le n}$ are such that $h_n$ is positive and continuous, that is

$$\begin{aligned} h_n(c_0)>0\text { and }\quad \forall \; 1\le i \le n-1,\ a_ic_i+b_i=a_{i+1}c_i+b_{i+1}=h_n(c_i)>0. \end{aligned}$$

(16)

Then, we introduce the autonomous Cauchy problem

$$\begin{aligned} \left\{ \begin{array}{ll} \begin{aligned} X'(t)&{}=h_n(X(t)), \quad t\ge 0,\\ X(0)&{}=X_0. \end{aligned} \end{array}\right. \end{aligned}$$

(17)

For the sake of simplicity, we set $h_n(c_n)=0$ such that for any $X_0\in [a,b)$, the solution of (17) is increasing and remains in [a, b), and $X=c_n=b$ is the unique steady state of (17).

Note that existence and uniqueness for the solution X of System (17) is guaranteed by Cauchy-Lipschitz theorem, most of the time this solution will be denoted by $X_n$. In particular, the restriction of $X_n$ on each interval $[c_{i-1},c_i]$, $i\in \{ 1,\ldots , n-1 \}$, is solution of a linear equation.

In the following, we first determine the values of t such that $X_n(t)=c_i$ for $i=0,\ldots ,n-1$ (Lemma B1) and then we provide an explicit expression for $X_n$ (Proposition B1).

Since $h_n(x)>0$ for all $x\in [c_0,c_n)$ and $h_n(c_n)=0$, it is clear that for all $X_0\in (c_0,c_n)$, $X_n$ is an increasing function and $\lim _{t\rightarrow +\infty } X_n(t)=c_n$. Straightforward calculations show that, if $X_0\in [c_{i},c_{i+1})$ for $i\in \{0,\ldots , n-2 \}$, then $X_n(t)=c_{i+1}$ if and only if $t=\lambda _{i+1}(X_0)$, where $\lambda _{i+1}$ is defined for all $x\in [c_i,c_{i+1})$ by

$$\begin{aligned} \lambda _{i+1}(x)= {\left\{ \begin{array}{ll} \frac{1}{a_{i+1}}\ln \left( \frac{h_n(c_{i+1})}{h_n(x)} \right) , &{}\quad \text{ if } \; a_{i+1} \ne 0, \\ \frac{c_{i+1}-x}{b_{i+1}}, &{}\quad \text{ if } \;a_{i+1}= 0. \end{array}\right. } \end{aligned}$$

By induction, we deduce the expression of $\lambda _k(X_0)$ for any $X_0<c_k$.

Lemma B1

Let $i\in \{ 0,\dots , n-2 \}$, $X_0\in [c_i,c_{i+1})$, and $X_n$ be the solution of (17). Let $k\in \{i+1,\ldots ,n-1 \}$, then $X_n(t)=c_k$ if and only if $t=\lambda _k(X_0)$, defined by

$$\begin{aligned} \lambda _k(X_0)=\lambda _{i+1}(X_0)+\sum _{p=i+1}^{k-1}\lambda _{p+1}(c_{p}). \end{aligned}$$

(18)

In other words, $\lambda _k(X_0)$ corresponds to the time for the solution $X_n$ of (17) to go from $X_0$ to $c_k$. Since $c_n=\lim _{t\rightarrow +\infty }X_n(t)$, we set $\lambda _n(X_0)=+\,\infty$. If $X_0\ge c_p$, for $p\in \{0,\dots , n-1\}$, we set $\lambda _p(X_0)=0$.

From Lemma B1, we can write a concise expression of $X_n(t)$ for any $t>0$,

Proposition B1

Let $i \in \{ 0,\ldots ,n-1 \}$, $X_0\in [c_i,c_{i+1})$ and $X_n$ be the solution of (17). Then, if $t\in [0,\lambda _{i+1}(X_0))$,

$$\begin{aligned} X_n(t)= {\left\{ \begin{array}{ll} \frac{h_n(X_0)}{a_{i+1}}e^{a_{i+1}t}-\frac{b_{i+1}}{a_{i+1}}, &{}\quad \text{ if } \; a_{i+1} \ne 0, \\ X_0+b_{i+1}t, &{}\quad \text{ if } \; a_{i+1}= 0. \\ \end{array}\right. } \end{aligned}$$

If $t\in \left[ \lambda _k(X_0),\lambda _{k+1}(X_0)\right)$ for $k\in \{i+1,\ldots ,n-1\}$, then

$$\begin{aligned} X_n(t)= {\left\{ \begin{array}{ll} \left( c_k-\frac{b_{k+1}}{a_{k+1}}\right) e^{a_{k+1}(t-\lambda _k(X_0))}-\frac{b_{k+1}}{a_{k+1}}, &{}\quad \text{ if } \; a_{k+1} \ne 0, \\ c_k+b_{k+1}\left( t-\lambda _k(X_0)\right) , &{}\quad \text{ if } \; a_{k+1} = 0. \end{array}\right. } \end{aligned}$$

(19)

We now consider $t>0$ fixed and $\varphi _{t}^n$ , the flow of (17) at time t, defined by

$$\begin{aligned} \begin{array}{lrlll} \varphi _{t}^n :&X_0\in [a,b]&\mapsto X_n(t)\in [a,b],\end{array} \end{aligned}$$

(20)

where $X_n$ is the solution of (17). Proposition B1 provides an explicit expression for $\varphi _{t}^n(X_0)$.

Note that, since $h_n$ is continuous, $\varphi _{t}^n$ is also continuous. In the following, we determine the convexity of $\varphi _t^n$ by studying its second derivative. However, if $X_0=c_k$ or $\varphi _t^n(X_0)=c_k$, the expression of $\varphi _t^n$ is not the same on both sides of $X_0$ and, in that case, we cannot directly compute the derivative of $\varphi _t^n$ . Consequently, we first study the convexity of $\varphi _t^n$ everywhere $X_0\ne c_k$ and $\varphi _t^n(X_0)\ne c_k$, and then we conclude on the convexity of $\varphi _t^n$ on the whole interval (a, b). For the sake of simplicity we define

$$\begin{aligned} J_n=\{c_k,\ k=1,\dots ,n-1 \} \cup \{(\varphi _t^n)^{-1}(c_k),\ k=1,\dots ,n-1 \}. \end{aligned}$$

(21)

Lemma B2

Let $X_0 \in (a,b)\backslash J_n$ and $i,k\in \{0,\dots , n-1\}$ such that $X_0\in (c_i,c_{i+1})$ and $\varphi _{t}^n(X_0)\in (c_k,c_{k+1})$. Then, on a neighbourhood of $X_0$, $\varphi _{t}^n$ is $\mathcal {C}^\infty$ and strictly convex (resp. strictly concave, resp. affine) if and only if $a_{k+1}>a_{i+1}$ (resp. $a_{k+1}<a_{i+1}$, resp. $a_{k+1}=a_{i+1}$).

Proof of Lemma B2

To prove Lemma B2, it suffices to study the sign of the second derivative of $\varphi _t^n$. Depending on whether $a_{i+1}$ and $a_{k+1}$ are null are not, $\varphi _t^n$ can be defined by four different expressions. Here we only provide a proof for the case $a_{i+1}\ne 0$ and $a_{k+1}\ne 0$. Other cases are analogous.

Let $X_0\in (c_i,c_{i+1})$ such that $\varphi _t^n(X_0)\in (c_k,c_{k+1})$, $k>i$. Then, according to Eqs. (18) and (19)

$$\begin{aligned} \begin{array}{ll} \varphi _t^n(X_0)&{}=\left( c_k-\frac{b_{k+1}}{a_{k+1}}\right) e^{a_{k+1}(t-\lambda _k(X_0))}-\frac{b_{k+1}}{a_{k+1}}, \\ &{}=\frac{h_n(c_k)}{a_{k+1}}e^{a_{k+1}t}\prod \limits _{p=i+1}^{k-1} e^{-a_{k+1}\lambda _{p+1}(c_p)}e^{-a_{k+1}\lambda _{i+1}(X_0)} \\ &{}=\frac{h_n(c_k)}{a_{k+1}}h_n(X_0)^\frac{a_{k+1}}{a_{i+1}}{M_{i,k}}^{a_{k+1}}-\frac{b_{k+1}}{a_{k+1}}, \end{array} \end{aligned}$$

(22)

where

$$\begin{aligned} M_{i,k}=e^{t}\left( \frac{1}{h_n(c_{i+1})}\right) ^\frac{1}{a_{i+1}}\prod _{p=i+1}^{k-1}\left( \frac{h_n(c_{p})}{h_n(c_{p+1})}\right) ^\frac{1}{a_{p+1}}>0. \end{aligned}$$

Consequently

$$\begin{aligned} \begin{array}{ll} \displaystyle {\frac{\mathrm {d}^2\varphi _t^{n} }{\mathrm {d}X_0^2}}(X_0)&{}=\frac{h_n(c_k){M_{i,k}}^{a_{k+1}}}{a_{k+1}}\left( \frac{a_{k+1}}{a_{i+1}}\right) \left( \frac{a_{k+1}}{a_{i+1}}-1\right) a_{i+1}^2\left( a_{i+1}X_0+b_{i+1}\right) ^{\left( \frac{a_{k+1}}{a_{i+1}}-2\right) }\\ &{}= h_n(c_k){M_{i,k}}^{a_{k+1}}(a_{k+1}-a_{i+1})\left( a_{i+1}X_0+b_{i+1}\right) ^{\left( \frac{a_{k+1}}{a_{i+1}}-2\right) }. \end{array} \end{aligned}$$

Moreover, on any neighbourhood of $X_0$ included in $(a,b)\backslash J_n$, $\varphi _t^n$ is linear and then its second derivative is constant. Finally, the convexity of $\varphi _t^n$ is determined by the sign of its second derivative, given by the sign of $(a_{k+1}-a_{i+1})$, as stated in Lemma B2. $\square$

Thanks to Lemma B2, we can determine the convexity of $\varphi _{t}^n$ on each interval included in $(a,b)\backslash J_n$. To extend this result to the whole interval (a, b) we need to introduce the following lemma.

Lemma B3

$\varphi _{t}^n$ is continuously differentiable on (a, b).

Proof

We know that $\varphi _t^n$ is $\mathcal {C}^\infty$ on $(a,b)\backslash J_n$. Let $X_0\in J_n$ and $i,k\in \{0,\dots ,n-1\}$ such that $X_0=c_i$ or $\varphi _t^n(X_0)=c_k$. We want to show that the derivative of $\varphi _t^n$ is continuous in $X_0$. For the sake of simplicity, we only show the proof for the case where $a_i,\ a_{i+1},a_k$ and $\ a_{k+1}$ are non null, $X_0=c_i$ and $\varphi _t^n(X_0)=c_k$. Other cases are analogous.

Let $X_1\in (c_{i-1},c_i)$ such that $\varphi _t^n(X_1)\in (c_{k-1},c_k)$ and $X_2\in (c_{i},c_{i+1})$ such that $\varphi _t^n(X_2)\in (c_{k},c_{k+1})$

According to (22),

$$\begin{aligned} \varphi _t^n(X_2)=\frac{h_n(c_k)}{a_{k+1}}h_n(X_2)^\frac{a_{k+1}}{a_{i+1}}{M_{i,k}}^{a_{k+1}}-\frac{b_{k+1}}{a_{k+1}} \end{aligned}$$

and straightforward calculations lead to

$$\begin{aligned} {\begin{array}{ll} \varphi _t^n(X_1)&{}=\frac{h_n(c_{k-1})}{a_{k}}h_n(X_1)^\frac{a_{k}}{a_{i}}{M_{i-1,k-1}}^{a_{k}}-\frac{b_{k}}{a_{k}}\\ &{}= \frac{h_n(c_{k})}{a_{k}}h_n(X_1)^\frac{a_{k}}{a_{i}} {M_{i,k}}^{a_{k}}\left( h_n(X_0)\right) ^{\frac{a_k}{a_{i+1}}-\frac{a_k}{a_i}}-\frac{b_{k}}{a_{k}}. \end{array}} \end{aligned}$$

Consequently

$$\begin{aligned} {(\varphi _t^n})'(X_2)=h_n(c_k)h_n(X_2)^{\frac{a_{k+1}}{a_{i+1}}-1}{M_{i,k}}^{a_{k+1}} \end{aligned}$$

(23)

and

$$\begin{aligned} {(\varphi _t^n})'(X_1)= h_n(c_{k})h_n(X_1)^{\frac{a_{k}}{a_{i}}-1} {M_{i,k}}^{a_{k}}\left( h_n(X_0)\right) ^{\frac{a_k}{a_{i+1}}-\frac{a_k}{a_i}}. \end{aligned}$$

(24)

On the other hand, since $\varphi _t^n$ is continuous and $\varphi _t^n(X_0)=c_k$, it follows that

$$\begin{aligned} \begin{array}{ll} c_k&=\varphi _t^n(X_0)=\underset{X_2\rightarrow X_0}{\lim }\varphi _t^n(X_2)=\frac{h_n(c_k)}{a_{k+1}}h_n(X_0)^\frac{a_{k+1}}{a_{i+1}}{M_{i,k}}^{a_{k+1}}-\frac{b_{k+1}}{a_{k+1}}. \end{array} \end{aligned}$$

That is

$$\begin{aligned} \begin{array}{ll} a_{k+1}c_k+b_{k+1}=h_n(c_k) \left( h_n(X_0)^\frac{1}{a_{i+1}}{M_{i,k}}\right) ^{a_{k+1}}. \end{array} \end{aligned}$$

(25)

According to (16), $h_n(c_k)=a_{k+1}c_k+b_{k+1}\ne 0$. It follows that

$$\begin{aligned} h_n(X_0)^\frac{1}{a_{i+1}}M_{i,k}=1. \end{aligned}$$

Applying (25)–(23) and (24), we get

$$\begin{aligned} \begin{array}{ll} \underset{X_2\rightarrow X_0^+}{\lim }{\varphi _t^n}'(X_2)=\frac{h_n(c_k)}{h_n(X_0)}\left( h_n(X_0)^{\frac{1}{a_{i+1}}}{M_{i,k}}\right) ^{a_{k+1}}=\frac{h_n(c_k)}{h_n(X_0)},\\ \underset{X_1\rightarrow X_0^-}{\lim }{\varphi _t^n}'(X_1)=\frac{h_n(c_k)}{h_n(X_0)}\left( h_n(X_0)^{\frac{1}{a_{i+1}}} {M_{i,k}}\right) ^{a_{k}}\left( h_n(X_0)\right) ^{\frac{a_k}{a_{i}}-\frac{a_k}{a_i}}= \frac{h_n(c_k)}{h_n(X_0)}, \end{array} \end{aligned}$$

and the proof is achieved. $\square$

Consequently, the convexity of $\varphi _{t}^n$ on (a, b) can be entirely determined, as stated in Proposition B2 hereafter.

Let us denote, for all $n\in \mathbb {N}^*$, by $h_n^*$ and ${{}^{*}}{h_{n}}$ the right and left derivatives of $h_n$ respectively. Then, $h_n^*$ and ${{}^{*}}{h_{n}}$ are defined everywhere in (a, b) and given by

$$\begin{aligned} \left\{ \begin{array}{ll} {{}^{*}}{h_{n}}(x)= h_n^*(x)=h_n'(x)=a_{i+1}, &{}\quad \text { if } \;x\in (c_{i},c_{i+1}),\\ {{}^{*}}{h_{n}}(x)=a_{i},\ h_n^*(x)=a_{i+1}, &{}\quad \text { if }\; x=c_i. \end{array} \right. \end{aligned}$$

(26)

Proposition B2

Let $X_0\in (a,b)$. Then $\varphi _{t}^n$ is strictly convex (resp. strictly concave, resp. affine) on a neighbourhood of $X_0$ if and only if $h_n^*(X_0)-h_n^*(\varphi _{t}^n(X_0))<0$ (resp. $>0$, resp. $=0$) and ${{}^{*}}{h_{n}}(X_0)-{{}^{*}}{h_{n}}(\varphi _{t}^n(X_0))<0$ (resp. $>0$, resp. $=0$).

Proposition B2 is an immediate consequence of Lemmas B2 and B3. Indeed, Lemma B2 provides necessary and sufficient conditions for $\varphi _t^n$ to be strictly convex (resp. concave) on any interval included in $(a,b)\backslash J_n$, with $J_n$ defined by (21), and Lemma B3 ensures that $\varphi _t^n$ in strictly convex (resp. concave) on a neighbourhood of $X_0\in J_n$ if and only if it is strictly convex (resp. concave) on both sides of $X_0$.

1.2 Generalisation to smooth functions

In this section, we consider the autonomous Cauchy problem

$$\begin{aligned} \left\{ \begin{array}{ll} \begin{aligned} X'(t)&{}=h(X(t)),\quad t\ge 0,\\ X(0)&{}=X_0\in [a,b], \end{aligned} \end{array}\right. \end{aligned}$$

(27)

where $h\in \mathcal {C}^1([a,b],\mathbb {R})$ is a Lipschitz continuous function, positive on (a, b) and satisfying $h(b)=0$.

We want to study the convexity of $\varphi _t$, the flow associated to System (27) at time $t>0$, defined by (10). Remark that the hypothesis $h(b)=0$ ensures that for any $X_0\in [a,b]$, X and $\varphi _ t$ remain in [a, b].

The main purpose of this section is to prove the following theorem, reproduced hereafter from Sect. 3.2 for the sake of clarity.

Theorem 2

Let $X_0\in (a,b)$. If $h'(\varphi _t(X_0))-h'(X_0)>0$ (resp. $<0$), then $\varphi _t$ is convex (resp. concave) on a neighbourhood of $X_0$.

Of course, this theorem can be directly adapted to the case where h is a negative function by studying $g=-\,h$. In particular, $g=-\,h$ is convex (resp. concave) if and only if h is concave (resp. convex).

In the following, we consider $h_n$ defined in B.1 to be the order n linear interpolation of h on [a, b] with uniform distribution of the nodes. That is, with the notations of B.1,

$$\begin{aligned} { \left\{ \begin{array}{ll} \forall \; i \in \{ 0,\dots , n \}, &{}\quad c_i=a+i\frac{b-a}{n},\\ \forall \;i \in \{ 0,\dots , n-1 \},&{}\quad a_{i+1}=\frac{h(c_{i+1})-h(c_i)}{c_{i+1}-c_i},\ b_{i+1}=h(c_i)-a_{i+1}c_i. \end{array}\right. } \end{aligned}$$

(28)

In particular, $h_n$ is continuous, for all $i \in \{ 0,\dots , n \},\ c_{i+1}-c_i=1/n$ and $h_n(c_i)=h(c_i)$. When there is ambiguity on the order n, we note $a_i=a^{(n)}_{i}$, $b_i=b_{i}^{(n)}$ and $c_i=c_{i}^{(n)}$. The flow $\varphi _t^n$ associated with $h_n$ is defined in (20).

Lemma B4

Let $X_0\in [a,b]$, then $\lim _{n\rightarrow +\infty }\varphi _{t}^n(X_0)=\varphi _t(X_0)$.

Proof of Lemma B4

Let $X_0\in [a,b]$ and $X_n$, X be the respective solutions of (17) and (27). It is easy to show that for any $\epsilon >0$, there exists $n_\epsilon \in \mathbb {N}^*$ such that for all $n\ge n_\epsilon$ and for all $x\in [a,b]$, $|h(x)-h_n(x)|<\epsilon$. Then for all $x\in [a,b]$, $h(x)-\epsilon \le h_n(x)\le h(x)+\epsilon$.

For any $\epsilon >0$, let us consider $X_\epsilon$ and $X_{-\epsilon }$ the respective solutions of $X_\epsilon '=h(X_\epsilon )+\epsilon$ and $X_{-\epsilon }'=h(X_{-\epsilon })-\epsilon$ satisfying $X_\epsilon (0)=X_{-\epsilon }(0)=X_0$. Then, for all $s>0$,

$$\begin{aligned} X_n(s)\in [X_\epsilon (s), X_{-\epsilon }(s)]. \end{aligned}$$

(29)

It remains to show $\lim _{\epsilon \rightarrow 0^+}X_\epsilon (t)=\lim _{\epsilon \rightarrow 0^+}X_{-\epsilon }(t)=X(t)$. We set $Z=X-X_\epsilon$. Since h is Lipschitz continuous, there exists $C_h>0$ such that, for all $s\ge 0$,

$$\begin{aligned} |Z'(s)|=|h(X(s))-h(X_\epsilon (s))-\epsilon |\le C_h|X(s)-X_\epsilon (s)|+\epsilon =|Z(s)|+\epsilon . \end{aligned}$$

Using the Gronwall lemma, one gets

$$\begin{aligned} \forall \; s\ge 0,\ |Z(s)| \le |Z(0)|e^{C_hs}+\frac{\epsilon }{C_h}\left( e^{C_hs}-1\right) =\frac{\epsilon }{C_h}\left( e^{C_hs}-1\right) . \end{aligned}$$

Consequently $\lim _{\epsilon \rightarrow 0^+}Z(t)=0$ and so $\lim _{\epsilon \rightarrow 0^+}X_\epsilon (t)=X(t)$. Similarly, we can show $\lim _{\epsilon \rightarrow 0^+}X_{-\epsilon }(t)=X(t)$. By using the squeeze theorem and (29), it follows that $\lim _{\epsilon \rightarrow 0^+}X_n(t)=X(t)$. $\square$

Using Lemma B4 and Proposition B2, we prove Theorem 2.

Proof of Theorem 2

Here we prove the convexity case. Proof for the concavity one is similar. Let $x \in (a,b)$ and $(x_n)_{n\in \mathbb {N}}$ be a sequence such that $\forall \;n\in \mathbb {N},\ x_n\in (a,b)$ and $\lim _{n\rightarrow +\infty } x_n =x$. We first show that

$$\begin{aligned} \lim _{n\rightarrow +\infty }h_n^*(x_n)=\lim _{n \rightarrow \infty } {{}^{*}}{h_{n}}(x_n)=h'(x), \end{aligned}$$

(30)

where $h_n^*$ and ${{}^{*}}{h_{n}}$ are given by (26).

For all $n\in \mathbb {N}^*$, there exists a unique $i \in \{ 0,\dots , n-1 \}$ such that $x_n\in [c_{i}^{(n)},c_{i+1}^{(n)})$, we note $i=i_n$. Since $\lim _{n\rightarrow +\infty } x_n =x$ and $\forall \; i\in \{ 0,\dots n-1\}$,

$$\begin{aligned} \lim _{n\rightarrow +\infty } \left| c_{i+1}^{(n)}-c_i^{(n)}\right| =\lim _{n\rightarrow +\infty } 1/n =0, \end{aligned}$$

it follows that

$$\begin{aligned} \lim _{n\rightarrow +\infty } c_{i_n-1}^{(n)}=\lim _{n\rightarrow +\infty } c_{i_n+1}^{(n)}=x. \end{aligned}$$

(31)

Using the mean value theorem and the definition of $a_i$ in (28), it follows that for all $n\in \mathbb {N}^*$, there exists $\alpha _n \in (c_{i_n-1}^{(n)},c_{i_n}^{(n)})$ and $\beta _n \in (c_{i_n}^{(n)},c_{i_n+1}^{(n)})$ such that

$$\begin{aligned} a_{i_n}=h'(\alpha _n) \text { and } a_{i_n+1}= h'(\beta _n). \end{aligned}$$

(32)

Since $h\in \mathcal {C}^1([a,b],\mathbb {R})$, using (31) and (32), it follows that

$$\begin{aligned} \lim _{n\rightarrow +\infty }a_{i_n}=\lim _{n\rightarrow +\infty }a_{i_n+1}=h'(x). \end{aligned}$$

(33)

On the other hand, according to (26), for all $n\in \mathbb {N}^*$,

$$\begin{aligned} h_n^*(x_n), {{}^{*}}{h_{n}}(x_n)\in \{a_{i_n},a_{i_n+1} \}. \end{aligned}$$

(34)

With (33) and (34) we conclude that (30) is true. In particular,

$$\begin{aligned} \left\{ \begin{array}{l} \lim _{n\rightarrow +\infty }h_n^*(x)=\lim _{n \rightarrow \infty }{{}^{*}}{h_{n}}(x)=h'(x),\\ \lim _{n\rightarrow +\infty }h_n^*(\varphi _{t}^n(x))=\lim _{n \rightarrow \infty }{{}^{*}}{h_{n}}(\varphi _{t}^n(x))=h'(\varphi (x)). \end{array}\right. \end{aligned}$$

(35)

Let $X_0\in (a,b)$. If we assume $h'(\varphi _t(X_0))>h'(X_0)$, since $h'$ and $\varphi _t$ are continuous there exists a neighbourhood $I_{X_0}$ of $X_0$ such that, for all $y\in I_{X_0}$, $h'(\varphi _t(y))>h'(x)$.

Moreover, (35) implies that there exists a neighbourhood $J_{X_0}\subset I_{X_0}$ of $X_0$ and $N_0\in \mathbb {N}^*$ such that for all $n\ge N_0$ and for all $y\in J_{X_0}$,

$$\begin{aligned} {{}^{*}}{h_{n}}(y)<{{}^{*}}{h_{n}}(\varphi _{t}^n(y))\quad \text { and } \quad h_n^*(y)<h_n^*(\varphi _{t}^n(y)). \end{aligned}$$

According to Proposition B2, for all $n\ge N_0$, $\varphi _{t}^n$ is strictly convex on $J_{X_0}$.

Finally, as the limit of a sequence of strictly convex functions (Lemma B2), $\varphi _{t}$ is convex on $J_{X_0}$. $\square$