Stability Analysis of a Model of Interaction Between the Immune System and Cancer Cells in Chronic Myelogenous Leukemia

Abstract

We describe here a simple model for the interaction between leukemic cells and the autologous immune response in chronic phase chronic myelogenous leukemia (CML). This model is a simplified version of the model we proposed in Clapp et al. (Cancer Res 75:4053–4062, 2015). Our simplification is based on the observation that certain key characteristics of the dynamics of CML can be captured with a three-compartment model: two for the leukemic cells (stem cells and mature cells) and one for the immune response. We characterize the existence of steady states and their stability for generic forms of immunosuppressive effects of leukemic cells. We provide a complete co-dimension one bifurcation analysis. Our results show how clinical response to tyrosine kinase inhibitors treatment is compatible with the existence of a stable low disease, treatment-free steady state.

Appendix

The following equations are satisfied by positive steady states of the model (4). They will be used in the technical lemmas.

$$\begin{aligned} {\bar{z}}&= \frac{s}{f({\bar{y}}_2)},\\ {\bar{z}}&= \frac{d_2}{\mu } \frac{M-{\bar{y}}_2}{M\frac{d_2}{r}+{\bar{y}}_2},\\ \frac{r}{K}{\bar{y}}_{1}&= \frac{(r+d_2) {\bar{y}}_2}{M\frac{d_2}{r}+{\bar{y}}_2}\\ a_1 \frac{{\bar{y}}_{1}}{{\bar{y}}_2}&= M\frac{d_2}{r}\frac{(r+d_2)}{M\frac{d_2}{r}+{\bar{y}}_2}. \end{aligned}$$

Lemma 1

Let ${\bar{x}}=({\bar{y}}_{1},{\bar{y}}_2,{\bar{z}})$ be a steady state of (4) such that $f'({\bar{y}}_2)>0$. The steady state ${\bar{x}}$ is asymptotically stable if and only if $\det (J({\bar{x}}))<0$.

Proof

The determinant of $J({\bar{x}})$ is the product of all its eigenvalues, so it is equal to $-\chi _J(0)$. We note that the polynomial $\chi _J$ is convex on ${\mathbb {R}}^+$ and that

$$\begin{aligned} \chi _J'(0) - \frac{\chi _J(0)}{r+d_2}= & {} \left( (r+d_2) f({\bar{y}}_2)-\mu {\bar{z}}{\bar{y}}_2 f'({\bar{y}}_2) + \frac{r}{K} {\bar{y}}_{1} a_1 \frac{{\bar{y}}_{1}}{{\bar{y}}_2}\right) \\&- \frac{1}{r+d_2} \left( \frac{r}{K} {\bar{y}}_{1} a_1 \frac{{\bar{y}}_{1}}{{\bar{y}}_2} f({\bar{y}}_2) - (r+d_2) \mu {\bar{z}} {\bar{y}}_2 f'({\bar{y}}_2) \right) \\= & {} (r+d_2) f({\bar{y}}_2) + \frac{r}{K} {\bar{y}}_{1} a_1 \frac{{\bar{y}}_{1}}{{\bar{y}}_2} - \frac{1}{r+d_2} \frac{r}{K} {\bar{y}}_{1} a_1 \frac{{\bar{y}}_{1}}{{\bar{y}}_2} f({\bar{y}}_2) \\= & {} \frac{r}{K} {\bar{y}}_{1} a_1 \frac{{\bar{y}}_{1}}{{\bar{y}}_2} +\frac{f({\bar{y}}_2)}{r+d_2} \left( (r+d_2)^2- \frac{r}{K} {\bar{y}}_{1} a_1 \frac{{\bar{y}}_{1}}{{\bar{y}}_2} \right) \\= & {} \frac{r}{K} {\bar{y}}_{1} a_1 \frac{{\bar{y}}_{1}}{{\bar{y}}_2} +\frac{f({\bar{y}}_2)}{r+d_2} \left( \left( \frac{r}{K} {\bar{y}}_{1} + a_1 \frac{{\bar{y}}_{1}}{{\bar{y}}_2}\right) ^2- \frac{r}{K} {\bar{y}}_{1} a_1 \frac{{\bar{y}}_{1}}{{\bar{y}}_2} \right) \\= & {} \frac{r}{K} {\bar{y}}_{1} a_1 \frac{{\bar{y}}_{1}}{{\bar{y}}_2} +\frac{f({\bar{y}}_2)}{r+d_2} \left( \left( \frac{r}{K} {\bar{y}}_{1}\right) ^2 + \left( a_1 \frac{{\bar{y}}_{1}}{{\bar{y}}_2}\right) ^2+ \frac{r}{K} {\bar{y}}_{1} a_1 \frac{{\bar{y}}_{1}}{{\bar{y}}_2} \right) >0. \end{aligned}$$

Here we used $f({\bar{y}}_2)>0$ given by Proposition 2. We consider two cases:

Case 1: If $\chi _J(0)<0$, then there exists a root in ${\mathbb {R}}_+^*$. In this case the steady state ${\bar{x}}$ is unstable.

Case 2: If $\chi _J(0)>0$, then $\chi _J'(0)>0$ and, by convexity, $\chi _J$ stays non-negative on ${\mathbb {R}}_+^*$. If $\chi _J$ admits three real roots, they are all negative, and the steady state is asymptotically stable. Otherwise, there exist two conjugate complex roots. In this case, we denote by x the negative real root and by z one of the two complex roots. First, note that since $f'({\bar{y}}_2)>0$, we have:

$$\begin{aligned} \chi _J\left( -\frac{r}{K} {\bar{y}}_{1}\right)= & {} -a_1 {\bar{y}}_{1} \mu {\bar{z}} f'({\bar{y}}_2) <0. \end{aligned}$$

As $\chi _J$ has only one sign change on ${\mathbb {R}}$, at x, we deduce that $-\frac{r}{K} {\bar{y}}_{1}<x$. Yet,

$$\begin{aligned} 2 {\text {Re}}(z) +x =-\frac{r}{K} {\bar{y}}_{1}-a_1 \frac{{\bar{y}}_{1}}{{\bar{y}}_2}-\frac{s}{f({\bar{y}}_2)} < x - a_1 \frac{{\bar{y}}_{1}}{{\bar{y}}_2} - \frac{s}{f({\bar{y}}_2)}. \end{aligned}$$

Hence

$$\begin{aligned} {\text {Re}}(z)< - \frac{1}{2} \left( a_1 \frac{{\bar{y}}_{1}}{{\bar{y}}_2} + \frac{s}{f({\bar{y}}_2)}\right) <0. \end{aligned}$$

We have shown that complex roots have negative real parts in the case where $f'({\bar{y}}_2)$ is positive. Therefore, the steady state is asymptotically stable. $\square$

Lemma 2

Let ${\bar{x}}=({\bar{y}}_{1},{\bar{y}}_2,{\bar{z}})$ be a steady state of (4). The determinant of the corresponding Jacobian matrix $\det (J({\bar{x}}))$ has a sign opposite to that of $A'B-AB'$.

Proof

We compute

$$\begin{aligned} A'B-AB'= & {} (f'(X)(M-X)-f(X))\left( X+\frac{d_2}{r}M\right) - f(X)(M-X) \\= & {} f'(X) (M-X)\left( X+\frac{d_2}{r}M\right) - f(X) M \left( 1+\frac{d_2}{r}\right) . \end{aligned}$$

Let ${\bar{x}}=({\bar{y}}_{1},{\bar{y}}_2,{\bar{z}})$ be a steady state for system (4). The determinant of the Jacobian matrix at ${\bar{x}}$ is

$$\begin{aligned} \det (J({\bar{x}}))= & {} \frac{r}{K} {\bar{y}}_{1} a_1 \frac{{\bar{y}}_{1}}{{\bar{y}}_2} f({\bar{y}}_2) - (r+d_2) \mu {\bar{z}} {\bar{y}}_2 f'({\bar{y}}_2) \\= & {} \frac{(r+d_2) {\bar{y}}_2}{M\frac{d_2}{r}+{\bar{y}}_2} M\frac{d_2}{r}\frac{(r+d_2)}{M\frac{d_2}{r}+{\bar{y}}_2} f({\bar{y}}_2) - (r+d_2) d_2 \frac{M-{\bar{y}}_2}{M\frac{d_2}{r}+{\bar{y}}_2} {\bar{y}}_2 f'({\bar{y}}_2) \\= & {} \frac{d_2(r+d_2){\bar{y}}_2}{ \left( M\frac{d_2}{r}+{\bar{y}}_2\right) ^2} \left( M \left( 1+\frac{d_2}{r}\right) f({\bar{y}}_2) - \left( M-{\bar{y}}_2\right) \left( M\frac{d_2}{r}+{\bar{y}}_2\right) f'({\bar{y}}_2) \right) \\= & {} -\frac{d_2(r+d_2){\bar{y}}_2}{ \left( M\frac{d_2}{r}+{\bar{y}}_2\right) ^2} (A'B-AB')({\bar{y}}_2) . \end{aligned}$$

Hence $sign(\det (J({\bar{x}})))=-sign((A'B-AB')({\bar{y}}_2))$.$\square$

Lemma 3

Let $x\ge 0$. The following conditions are equivalent:

• 1. $(A'B-AB')(x)=0$,

• 2. x is a double root of $P_\theta$, where $\theta :=\frac{A(x)}{B(x)}$.

Proof

Let x be a positive zero of $A'B-AB'$. We fix $\theta :=\frac{A(x)}{B(x)}$. Then,

$$\begin{aligned} P_\theta (x)= & {} -A(x)+\theta B(x) = 0, \\ P_\theta '(x)= & {} -A'(x)+\theta B'(x) =-\frac{A'B-A B'}{B}(x) = 0. \end{aligned}$$

Hence x is a double zero of $P_\theta$.

Reciprocally, let $(\theta ,x)$ be such that x is a double zero of $P_\theta$. Since $P_\theta (x)=0$, $\theta =\frac{A}{B}(x)$. Also, since $P_\theta '(x)=0$, $(A'B-A B')(x)=0$, which means that x is a zero of $A'B-A B'$. $\square$

Lemma 4

Consider the polynomial $P=X^3+a X^2 + b X+ c$, where $a,b,c\in {\mathbb {R}}_+^*$. If $a b >c$, then all roots of P have negative real part.

Proof

First, positivity of all coefficients ensures that the real roots of P cannot be positive. Second, as $P(0)=c>0$, P necessarily admits a real, negative root. It remains to characterize the two other roots. If real and negative, then the dominating root is negative and the lemma is proven. Therefore we may assume that the two remaining roots are complex, and we need to determine the sign of their real part.

Let $\lambda$ be the first negative root of P. We can factor

$$\begin{aligned} P=\big (X-\lambda \big )\big (X^2+(a+\lambda ) X+ (b+a \lambda + \lambda ^2)\big ). \end{aligned}$$

By our assumption, P admits a pair of complex roots. We want to compare $\big (X^2+(a+\lambda ) X+ (b+a \lambda + \lambda ^2)\big )$ and $(X-z)(X-{\bar{z}})$, where $z\in {\mathbb {C}}$. Separating the real and imaginary parts leads to

$$\begin{aligned} a+\lambda&=-2 {\text {Re}}(z), \\ b+a \lambda + \lambda ^2&=({\text {Re}}(z))^2+({\text {Im}}(z))^2. \end{aligned}$$

In order to find the sign of ${\text {Re}}(z)$, we need to compare a and $\lambda$. We have $P(-a)=-a b +c$. Since we assume that $a b >c$, we obtain $P(-a)<0$. As P has only one real root $\lambda$, we obtain $-a<\lambda$. This leads to ${\text {Re}}(z)<0$, and the dominating root of P has its real part negative. $\square$