Slow–Fast Model and Therapy Optimization for Oncolytic Treatment of Tumors

Abstract

The present work studies models of oncolytic virotherapy without space variable in which virus replication occurs at a faster time scale than tumor growth. We address the questions of the modeling of virus injection in this slow–fast system and of the optimal timing for different treatment strategies. To this aim, we first derive the asymptotic of a three-species slow–fast model and obtain a two-species dynamical system, where the variables are tumor cells and infected tumor cells. We fully characterize the behavior of this system depending on the various biological parameters. In the second part, we address the modeling of virus injection and its expression in the two-species system, where the amount of virus does not appear explicitly. We prove that the injection can be described by an instantaneous jump in the phase plane, where a certain amount of tumors cells are transformed instantly into infected tumor cells. This description allows discussing qualitatively the timing of different injections in the frame of successive treatment strategies. This work is illustrated by numerical simulations. The timing and amount of injected virus may have counterintuitive optimal values; nevertheless, the understanding is clear from the phase space analysis.

A Proofs

1.1 A.1 Proof of Proposition 2 (Stability of Equilibria)

Proof

The fact that the system admits only the three equilibria $(0,0), (1,0), (x^*,y^*)$ follows from a simple computation. We now study the stability of these equilibria.

Equilibrium (0, 0) The Jacobian matrix is given by:

$$\begin{aligned} \begin{pmatrix} \mu &{}0\\ 0 &{}-\lambda \end{pmatrix}, \end{aligned}$$

it is a saddle equilibrium. The only possibility to converge toward this equilibrium is to follow its stable manifold, which is the line $x=0$. In other words, all cancer cells are infected.

Equilibrium (1, 0) The Jacobian matrix is given by:

$$\begin{aligned} \begin{pmatrix} -\mu &{} \dfrac{- N\gamma }{{1} +\beta }\\ 0&{}\dfrac{N\gamma }{{1} +\beta }-\gamma \end{pmatrix}. \end{aligned}$$

This equilibrium is a saddle if $N-{1} > \beta$, and for $N-{1} < \beta$ it is stable.

Endemic equilibrium The Jacobian matrix is given by:

$$\begin{aligned} \begin{pmatrix} -\mu \lambda x^*(1+\dfrac{1}{N})+\lambda \mu \frac{1}{N} &{} -\gamma \\ (1-x^*)(1-\dfrac{1}{N})\mu &{}0 \end{pmatrix}. \end{aligned}$$

The determinant is positive for $N>1$ since $x^*<1$. Moreover, the trace is negative if and only if

$$\begin{aligned} \frac{1}{N}<x^*(1+\frac{1}{N}) \Leftrightarrow \beta >\frac{N-1}{N+1}. \end{aligned}$$

Therefore, the endemic equilibrium is stable if $\beta >\beta _c=\dfrac{N-1}{N+1}$ and unstable if $\beta <\beta _c$. $\square$

1.2 A.2 Proof of Proposition 3 (Hopf Bifurcation Analysis)

Proof

We follow the approach proposed in Kuznetsov (1998). Let us study the following orbitally equivalent system obtained by changing the time parametrization:

$$\begin{aligned} \left\{ \begin{aligned}&x_0'(t)=-x_0y_0N\gamma +\mu x_0(1-x_0)(\beta +x_0),\\&y_0'(t)=x_0y_0\gamma (N-1)-\gamma \eta y_0. \end{aligned}\right. \end{aligned}$$

The equilibrium of interest is:

$$\begin{aligned} x^*(\beta )=\frac{\beta }{N-1},\quad y^*(\beta )= \dfrac{\mu }{N\gamma }x^*(1-x^*). \end{aligned}$$

The Jacobian matrix of the system in $(x^*,y^*)$ for $\beta =\beta _c$ is:

$$\begin{aligned} A(\beta _c)=\begin{pmatrix} 0&{}-N\gamma x^*_c\\ (N-1)\gamma y^*_c&{}0 \end{pmatrix}. \end{aligned}$$

Its eigenvalues are $i\omega$ and $-i\omega$ with

$$\begin{aligned} \omega ^2=N\gamma ^2(N-1)x^*_cy^*_c. \end{aligned}$$

The following two eigenvectors p and q are such that $Aq=i\omega q$, $A^Tp=-i\omega p$ and $\langle p,q\rangle =1$:

$$\begin{aligned} q=\begin{pmatrix}N\gamma x^*_c\\ -i\omega \end{pmatrix},\quad p=\frac{1}{2\omega N\gamma x^*_c}\begin{pmatrix} \omega \\ -iN\gamma x^*_c\end{pmatrix}. \end{aligned}$$

The system in the appropriate basis is approximated by:

$$\begin{aligned} \dot{\xi }=A\xi +\frac{1}{2}B(\xi ,\xi )+\frac{1}{6}C(\xi ,\xi ,\xi )+\dots \end{aligned}$$

where

$$\begin{aligned} B(\xi ,\eta )=2\begin{pmatrix}-\frac{N\gamma }{2}(\xi _1\eta _2+\xi _2\eta _1)+\xi _1\eta _1(\mu (1-\beta )-3\mu x^*_c)\\ \frac{1}{2}\gamma (N-1)(\xi _1\eta _2+\xi _2\eta _1) \end{pmatrix} \end{aligned}$$

and

$$\begin{aligned} C(\xi ,\eta ,\zeta )=6\begin{pmatrix} -\xi _1\eta _1\zeta _1\\ 0 \end{pmatrix} . \end{aligned}$$

Finally, it remains to compute

$$\begin{aligned} g_{20}=\langle p,B(q,q)\rangle , \quad g_{11}=\langle p, B(q,\bar{q})\rangle , \quad g_{2,1}=\langle p, C(q,q,\bar{q})\rangle . \end{aligned}$$

The computations lead to

$$\begin{aligned} g_{20}= & {} N\gamma x^*_c\left( \gamma (N-1)+\mu (1-\beta )-3\mu x^*_c\right) +i\omega N\gamma , \\ g_{11}= & {} N\gamma x^*_c \left( \mu (1-\beta )-3\mu x^*_c\right) ,\qquad \qquad g_{21}=-3\mu N^2\gamma ^2(x^*_c)^2. \end{aligned}$$

The first Lyapunov exponent is then given by:

$$\begin{aligned} \begin{aligned} l_1&=\frac{1}{2\omega ^2}\Re (ig_{20}g_{11}+\omega g_{12}) =\frac{-N^2\gamma ^2\mu (x^*)^2}{\omega (N+1)}.\end{aligned} \end{aligned}$$

The first Lyapunov exponent is thus negative; therefore, the unique and stable limit cycle bifurcates from the equilibrium via a supercritical Hopf bifurcation for $\beta <\beta _c$. $\square$

1.3 A.3: Treatment Approximation

Proof of Proposition 4

We consider the solutions $(X_0, Y_0, Z_0)$ of the first-order system (8):

$$\begin{aligned} \begin{pmatrix} X_0'(u)\\ Y_0'(u)\\ Z_0'(u) \end{pmatrix} = \begin{pmatrix} -\lambda x_0(t_0)Z_0(u)-\lambda X_0(u)Z_0(u)\\ \lambda x_0(t_0)Z_0(u)+\lambda X_0(u)Z_0(u)\\ (-\lambda x_0(t_0)-\delta )Z_0(u)-\lambda X_0(u)Z_0(u) \end{pmatrix} \end{aligned}$$

with initial conditions $X_0(0)=0=Y_0(0)$ and $Z_0(0)=V>0$.

Since $Y_0'(u)= -X_0'(u)$, it suffices to study the couple $(X_0,Z_0)$. From the last equation, we deduce that $Z_0(u)>0$ for all $u\ge 0$. Similarly, we obtain that $X_0(u)+x_0(t_0)>0$ for all $u\ge 0$. Thus, we deduce that

$$\begin{aligned} Z_0'(u) < -\delta Z_0(u). \end{aligned}$$

This leads to

$$\begin{aligned} Z_0(u)\le V\exp (-\delta u). \end{aligned}$$

(11)

As a consequence $Z_0(u)\rightarrow 0$ as $u\rightarrow \infty$. Let us now remark that

$$\begin{aligned} X_0'(u)=Z_0'(u)+\delta Z_0(u) \end{aligned}$$

and therefore by integration

$$\begin{aligned} X_0(u)={Z_0(u)}-V+{\delta }\int _0^uZ_0(s){\mathrm{d}}s. \end{aligned}$$

From (11), we deduce that $X_0$ converges as $u\rightarrow \infty$ toward a negative limit

$$\begin{aligned} -Q=\lim _{u\rightarrow \infty } X_0(u) = -V+ \delta \int _0^\infty Z_0(s){\mathrm{d}}s <0. \end{aligned}$$

Moreover, it follows from $X_0'(u)=-\lambda (x_0(t_0)+X_0(u))Z_0(u)$ and Duhamel’s formula that

$$\begin{aligned} X_0(u)+x_0(t_0) = x_0(t_0)\exp (-\lambda \int _0^uZ_0(s)\,{\mathrm{d}}s). \end{aligned}$$

By letting $u\rightarrow \infty$ and using the fact that $\displaystyle \int _0^\infty Z_0(s){\mathrm{d}}s=\dfrac{V-Q}{\delta }$, one sees that Q satifies the following implicit equation:

$$\begin{aligned} Q+x_0(t_0)\exp (\beta (Q-V))=x_0(t_0). \end{aligned}$$

(12)

$\square$

Proof of Theorem 1

${\hbox {Step 1: Prove that the solutions of}}$ (8) $\hbox {are close approximations of} (X, Y,Z).$

To do so, we first study the regularity of the solutions of (7) with respect to the parameter $\epsilon$ as a parameter. Let us denote by G the function ${\mathbf {R}}\times {\mathbf {R}}^3\times {\mathbf {R}}\rightarrow {\mathbf {R}}^3$ such that the system (7) reads

$$\begin{aligned} \begin{pmatrix}X'(u)\\ Y'(u)\\ Z'(u)\end{pmatrix}=G(u, (X(u), Y(u), Z(u)), \epsilon ). \end{aligned}$$

To study the dependency in $\epsilon$, we consider the extended system for a fixed value $\epsilon _0>0$

$$\begin{aligned} \left\{ \begin{aligned}&(X'(u), Y'(u), Z'(u),\epsilon '(u))=(G(u, (X(u), Y(u), Z(u), \epsilon (u)),0),\\&(X(0), Y(0), Z(0))=(0,0,V),\qquad \epsilon (0)=\epsilon _0.\end{aligned}\right. \end{aligned}$$

Let us denote by $u\mapsto \psi (u,(0,0,V,\epsilon _0))$ the flow associated with this differential equation, and then, our aim is to control

$$\begin{aligned} \psi (u,(0,0,V,\epsilon _0))-\psi (u,(0,0,V,0)= & {} \int _0^u \widetilde{G}\bigl (s, \psi (s,(0,0,V,\epsilon _0)\bigr )\\&-\widetilde{G}\bigl (s,\psi (s,(0,0,V,0)\bigr ) {\mathrm{d}}s, \end{aligned}$$

where $\widetilde{G}(s,(X,Y,Z,\epsilon ))=(G(s,X,Y,Z,\epsilon ),0)$. If $\widetilde{G}$ has bounded derivatives with respect to its second variable, then there exists a positive constant L such that

$$\begin{aligned} |\psi (u,(0,0,V,\epsilon _0))-\psi (u,(0,0,V,0)|\le L \int _0^u|\psi (s,(0,0,V,\epsilon _0))-\psi (s,(0,0,V,0)|{\mathrm{d}}s, \end{aligned}$$

which leads using the Grönwall lemma to

$$\begin{aligned} \psi (u,(0,0,V,\epsilon _0))-\psi (u,(0,0,V,0)\le \epsilon _0 e^{Lu}. \end{aligned}$$

(13)

${\mathrm{Step 2: Regularity of} \widetilde{G}.}$

We recall that:

$$\begin{aligned}&\widetilde{G}(u,X,Y,Z,\epsilon )\\&\quad =\begin{pmatrix} -\lambda x_\epsilon (t_0+\epsilon u)Z-\lambda XZ+\epsilon \left[ -\lambda Xz_\epsilon (t_0+\epsilon u) +\mu X(1-2x_\epsilon (t_0+\epsilon u)-X)\right] \\ \lambda x_\epsilon (t_0+\epsilon u)Z+\lambda XZ+\epsilon \left[ \lambda Xz_\epsilon (t_0+\epsilon u)-\gamma Y \right] \\ (-\lambda x_\epsilon (t_0+\epsilon u)-\delta )Z(u)-\lambda XZ+\epsilon \left[ -\lambda Xz_\epsilon (t_0+\epsilon u)+N\gamma Y \right] \\ 0 \end{pmatrix} \end{aligned}$$

Clearly, $\widetilde{G}$ is smooth, but its derivative is not uniformly bounded. We now prove that the trajectories remain in a compact set, where $\widetilde{G}$ has thus bounded derivatives.

Let us first prove that the solutions $(x_\epsilon , y_\epsilon ,z_\epsilon )$ of (2) are bounded uniformly in time, by a constant that depends only on the initial condition (and not on $\epsilon$). Clearly, all coordinates remain positive. The upper bound can be obtain since $x_\epsilon$ has a logistic behavior, and $x_\epsilon +y_\epsilon$ has negative derivative if $y_\epsilon \ge \frac{\mu }{\lambda }x_\epsilon (1-x_\epsilon )$ and $z_\epsilon$ has negative derivative as soon as $z_\epsilon \ge \frac{N\gamma y_\epsilon }{\delta +\lambda x_\epsilon }$. A similar argument shows that the solutions (X, Y, Z) of (8) are uniformly bounded through times and for $\epsilon$ in some compact $[0,\epsilon _{max}]$. This allows to obtain (13).

${\mathrm{Step 3: Conclusion}}$ Finally combining (13) and (9), if $(X_\epsilon ,Y_\epsilon ,Z_\epsilon )$ denote the solution of (7) for a given value of $\epsilon$, we obtain:

$$\begin{aligned}|X_\epsilon (u)+Q|&\le |X_\epsilon (u)-X_0(u)|+ |X_0(u)+Q|\\&\le \epsilon e^{Lu} +\int _{u}^\infty Z_0(s) {\mathrm{d}}s \\&\le \epsilon e^{Lu} + \frac{V}{\delta } e^{-\delta u} \end{aligned}$$

Let us fix $A>0$, and then, for a time $u_\epsilon =-\frac{1}{\delta }ln(\epsilon ^A\delta /V)$, we have that

$$\begin{aligned} |X_\epsilon (u_\epsilon )+Q|&\le \epsilon e^{Lu_\epsilon } + \epsilon ^A \end{aligned}$$

In particular, if A satisfies that $1-AL/\delta >0$, then it comes

$$\begin{aligned} \lim _{\epsilon \rightarrow 0}|X_\epsilon (u_\epsilon )+Q|=0 \end{aligned}$$

which leads as $\epsilon u_\epsilon \rightarrow 0$ to

$$\begin{aligned} \lim _{\epsilon \rightarrow 0}\bar{x}_\epsilon \left( t_0+\epsilon u_\epsilon \right) = x_0-Q. \end{aligned}$$

Similarly, we have that

$$\begin{aligned} \lim _{\epsilon \rightarrow 0}\bar{y}_\epsilon \left( t_0+\epsilon u_\epsilon \right) =y_0+Q. \end{aligned}$$

To conclude, let us prove that $\lim _{\epsilon \rightarrow 0}\bar{z}_\epsilon (t_0+\epsilon u_\epsilon )$ is finite. The naive bound given by (13) is not sufficient to conclude since $u_\epsilon \rightarrow \infty$ and

$$\begin{aligned} \frac{1}{\epsilon } Z_\epsilon (u_\epsilon ) \le e^{Lu_\epsilon } \rightarrow \infty \end{aligned}$$

Therefore, let us come back to the exact ODE satisfied by $Z_\epsilon$:

$$\begin{aligned} Z_\epsilon '(u)=\left( -\lambda \left( x_\epsilon \left( t_0+\epsilon u\right) +X_\epsilon (u)\right) -\delta \right) Z_\epsilon (u)+\epsilon \left[ -\lambda X_\epsilon z_\epsilon (t_0+\epsilon u)+N\gamma Y_\epsilon \right] \end{aligned}$$

From the previous computations, we have that on the time interval $[0, u_\epsilon ]$, there exists a positive constant M independent of $\epsilon$ such that

$$\begin{aligned} Z_\epsilon '(u)\le -\delta Z_\epsilon (u)+\epsilon M. \end{aligned}$$

As a consequence for all $u\in [0, u_\epsilon ]$

$$\begin{aligned} Z_\epsilon (u)\le (V-\frac{\epsilon M}{\delta })e^{-\frac{\delta }{\epsilon M}u}+\frac{\epsilon M}{\delta } \end{aligned}$$

Using this upper bound and the fact that $\bar{z}_\epsilon$ remains positive for all times, we deduce that

$$\begin{aligned} \bar{z}_\epsilon (t_0+\epsilon u_\epsilon )=z_\epsilon (t_0+\epsilon u_\epsilon )+\frac{1}{\epsilon }Z_\epsilon ( u_\epsilon ) \end{aligned}$$

remains bounded as $\epsilon \rightarrow 0$. We finally use Theorem 1.2 from Levin and Levinson (1954) on the solutions of singularly perturbed differential equations to conclude that for any $T>t>t_0$, the solution $(\bar{x}_\epsilon , \bar{y}_\epsilon )$ are asymptotically equal to the solution of the system (5) starting from $(x_0-Q,y_0+Q)$ on the interval [t, T]. $\square$