A mathematical analysis of rebound in a target-mediated drug disposition model: I.Without feedback

Abstract

We consider the possibility of free receptor (antigen/cytokine) levels rebounding to higher than the baseline level after one or more applications of an antibody drug using a target-mediated drug disposition model. Using geometry and dynamical systems analysis, we show that rebound will occur if and only if the elimination rate of the drug–receptor product is slower than the elimination rates of the drug and of the receptor. We also analyse the magnitude of rebound through approximations and simulations and demonstrate that it increases if the drug dose increases or if the difference between the elimination rate of the drug–receptor product and the minimum of the elimination rates of the drug and of the receptor increases.

Appendix: Proofs of invariance and stability results

Proof of Lemma 2.1

The phase space is bounded by the three planes $x=0, y=0$ and $z=0$. If the vector field given by the Eqs. (4)–(6) at these planes is directed inside or is tangent to the octant, then the given region is invariant (Smith 1995).

Taking the inner product of the vector field with the inward pointing normal at the plane $x=0$, we see from (4) that

$$\begin{aligned} {\dot{x}}|_{x=0}=\mu k_2z \end{aligned}$$

and so ${\dot{x}}|_{x=0}\ge 0$ when $z\ge 0$. Hence the vector field is directed inside the octant at the plane $x=0$.

Similarly, from (5), we obtain

$$\begin{aligned} {\dot{y}}|_{y=0}=k_3+k_2z \end{aligned}$$

and so ${\dot{y}}|_{y=0}>0$ when $z\ge 0$. In the same vein, from (6), we obtain

$$\begin{aligned} {\dot{z}}|_{z=0}=\frac{xy}{\mu } \end{aligned}$$

and so ${\dot{z}}|_{z=0}\ge 0$ when $x,y\ge 0$.

Note that at the $x$-axis ($y=z=0$), the vector field is tangent to the plane $z=0$, and is pointing inwards as ${\dot{y}}|_{y=z=0}=k_3$. At the $y$-axis ($x=z=0$), we have ${\dot{x}}=0$ and ${\dot{z}}=0$, hence $x$ and $z$ will stay zero for all time and the $y$-axis is invariant. On the $y$-axis, ${\dot{y}}>0$ if $y<1$ and ${\dot{y}}<0$ is $y>1$. Hence all orbits on the $y$-axis tend to the stable steady state at $y=1$.

Thus, we can conclude that any trajectory with $x(0), y(0), z(0)\ge 0$ must satisfy $x(t), y(t), z(t)\ge 0$ for all $t>0$. $\square$

Proof of Lemma 2.2

There are two steady state solutions of Eqs. (4)–(6), one of which is given by (7). The other solution is not in our phase space, and hence is not physically relevant, as the values of $x$ and $y$ are negative. $\square$

Proof of Corollary 2.4

By negating Lemma 2.3, it follows immediately that if the maximal $v(\tau )$ value is above 1, then $k_3>k_4$. And Lemma 2.3 implies that if $k_3>k_4$, then $v(\tau )\ge 1$ for all $\tau$. We only have to show that $v(\tau )$ cannot be equal to 1 for all time. This follows immediately with a contradiction argument. Assume that $k_3>k_4$ and $v(\tau )=1$ for all $\tau \ge 1$. Then the $v$-dynamics (22) gives for all $\tau \ge 0$ that $0=(k_3-k_4)z$, hence $z(\tau )=0$. Then (6) implies that $0=\frac{xy}{\mu }$ for all $\tau \ge 0$. However at $\tau =0$, we have $\frac{xy}{\mu }=\frac{1}{\mu }$. This is a contradiction, thus the assumption that $v\equiv 1$ is false. $\square$

Proof of Theorem 2.5

In Lemma 2.3 we have shown that $\lim _{\tau \rightarrow \infty }u(\tau )=0$. Since $u=x+\mu z$ and $x$ and $z$ are non-negative functions, we conclude that $\lim _{\tau \rightarrow \infty }x(\tau )= \lim _{t\rightarrow \infty }z(t)=0$ also. Hence, every trajectory will converge to a neighbourhood of the $y$-axis. We have also seen in Lemma 2.1 that the $y$-axis is invariant and that on this axis, all orbits tend to the stable steady state at $y=1$. So intuitively it is clear that all trajectories will converge to the steady state (7) as $\tau \rightarrow \infty$. To make this formal, we consider the $y$-dynamics in a neighbourhood of the $y$-axis.

Let $\varepsilon >0$. Since $\lim _{\tau \rightarrow \infty }x(\tau )= \lim _{\tau \rightarrow \infty }z(\tau )=0$, then there is some $T>0$ such that for $\tau >T$, we have

$$\begin{aligned} 0\le x(\tau )\le \varepsilon \mu k_3 \quad \text{ and } \quad 0\le z(\tau )\le \varepsilon k_3/k_2. \end{aligned}$$

Thus with (5), we get for $\tau >T$,

$$\begin{aligned} k_3 [1-(1+\varepsilon ) y]\le {\dot{y}} \le k_3 [1 + \varepsilon - y]. \end{aligned}$$

In other words, $\frac{d}{d\tau }[e^{k_3(1+\varepsilon )\tau } (y-\frac{1}{1+\varepsilon })] \ge 0$ and $\frac{d}{d\tau }[e^{k_3\tau }(y-(1+\varepsilon ))] \le 0$ and we can conclude that for $\tau \ge T$

$$\begin{aligned} - \frac{\varepsilon }{1+\varepsilon } \!+\! e^{-k_3(1\!+\!\varepsilon )(\tau -T)}\left( y(T)\!-\!\frac{1}{1\!+\!\varepsilon } \right) \le y(\tau )\!-\!1 \le \varepsilon \!+\! e^{\!-\!k_3(\tau \!-\!T)} (y(T)-(1\!+\!\varepsilon )) \end{aligned}$$

Hence by choosing $\varepsilon$ sufficiently small and for all $\tau >T$ sufficiently large, we can make $|y(\tau )-1|$ arbitrarily small. So we conclude that $y(\tau )\rightarrow 1$ for $\tau \rightarrow \infty$. $\square$

Proof of Lemma 2.7

This observation follows quickly from the expressions for the eigenvalues $\lambda _1$ and $\lambda _2$. Indeed, from (10) we find that

$$\begin{aligned} k_2+k_4+\lambda _1=\frac{1}{2}\left( -(1+k_1-k_2-k_4)+ \sqrt{(1+k_1-k_2-k_4)^2+4k_2}\right) >0, \end{aligned}$$

since $k_2>0$ and hence the square root term is larger than the magnitude of the first term. Similarly, using (12) we get

$$\begin{aligned} k_2+k_4+\lambda _2=\frac{1}{2}\left( -(1+k_1-k_2-k_4)- \sqrt{(1+k_1-k_2-k_4)^2+4k_2}\right) <0. \end{aligned}$$

$\square$

Proof of Lemma 2.8

Let $k_2,k_4>0$ be fixed. Differentiating the expression for $\lambda _1$, (10), with respect to $k_1$ gives

$$\begin{aligned} \frac{\partial \lambda _1}{\partial k_1} = -\frac{1}{2} \left( 1 - \frac{1+k_1-k_2-k_4}{\sqrt{(1+k_1-k_2-k_4)^2+4k_2}}\right) . \end{aligned}$$

The second term in the brackets is less than one in modulus (since $k_2>0$), and so this derivative is always negative. Thus, $\lambda _1$ is a monotonically decreasing function of $k_1$.

To derive the stated bounds on $-\lambda _1$ that involve $k_4$, we note that from (9) we have

$$\begin{aligned} \lambda _1\!+\!k_4\!=\!\frac{1}{2} \left( \!-\!(1\!+\!k_1\!+\!k_2\!-\!k_4)\!+\!\sqrt{(1\!+\!k_1\!+\!k_2\!-\!k_4)^2 \!+\!4(k_4\!-\!k_1)k_2}\right) .\quad \end{aligned}$$

(29)

When $k_1<k_4$, the square root term in (29) always dominates the first term, and so $\lambda _1+k_4>0$. When $k_1=k_4$, we have

$$\begin{aligned} \lambda _1+k_4=\frac{1}{2}(-(1+k_2)+|1+k_2|)=0. \end{aligned}$$

Finally, when $k_1>k_4$, we note that $1+k_1+k_2-k_4>0$ and in this case, the first term in (29) dominates the square root term and so we find that $\lambda _1+k_4<0$.

Similarly, for the bounds on $-\lambda _1$ involving $k_1$, from (9) we obtain

$$\begin{aligned} \lambda _1+k_1=\frac{1}{2}\left( -(1-k_1+k_2+k_4)+\sqrt{(1-k_1+k_2+k_4)^2 +4(k_1-k_4)}\right) . \end{aligned}$$

The sign of $\lambda +k_1$ is then obtained in the same way as for the sign of $\lambda +k_4$ above. $\square$