Bone metastasis treatment modeling via optimal control

Abstract

Metastatic disease is a lethal stage of cancer progression. It is characterized by the spread of aberrant cells from a primary tumor to distant tissues like the bone. Several treatments are used to deal with bone metastases formation, but they are palliative since the disease is considered incurable. Computational and mathematical models are used to understand the underlying mechanisms of how bone metastasis evolves. In this way, new therapies aiming to reduce or eliminate the metastatic burden in the bone tissue may be proposed. We present an optimal control approach to analyze some common treatments for bone metastasis. In particular, we focus on denosumab treatment, an anti-resorptive therapy, and radiotherapy treatment which has a cell killing action. We base our work in a variant of an existing model introduced by Komarova. The new model incorporates a logistic equation in order to describe the bone metastasis evolution. We provide proofs of existence and uniqueness of solutions to the corresponding optimal control problems for each treatment. Moreover, we present some numerical simulations to analyze the effectiveness of both treatments when different interactions between cancer and bone cells occur. A discussion of the obtained results is provided.

Appendix A: Proof of Theorem 2

We follow (Fister et al. 1998) to show uniqueness of the optimal solution for the model (4) under certain conditions over the final time. First, we state some basic results.

Lemma 6

Let $a,b,c, \bar{a}, \bar{b}, \bar{c}$ be real positive numbers such that they are bounded by some positive constant M. Then

• i) $ab - \bar{a}\bar{b} \le M (|a-\bar{a}| + |b-\bar{b}|).$

• ii) $(ab - \bar{a}\bar{b}) (c - \bar{c}) \le M ( (a-\bar{a})^2 + (b-\bar{b})^2 + (c-\bar{c})^2).$

$\square$

Now, we proceed to prove Theorem 3.

Proof

Let suppose that there are two optimal pairs $(x, \lambda , u)$ and $(\bar{x}, \bar{\lambda }, \bar{u})$ that solve the problem (4) and the adjoint system (12), where $u=u_D$, $x=(x_1, x_2, x_3)$, $x_1=C$, $x_2=B$, $x_3=T$, and $\lambda = (\lambda _1, \lambda _2, \lambda _3)$. Let $m > 0$ be fixed. Then there exist functions $y_1, y_2, y_3$ and $\mu _1, \mu _2, \mu _3$ (also with bar) such that $x_i = y_i e^{mt},\ \bar{x}_i = {y}_i e^{mt},\ \lambda _i = \mu _i e^{-mt},\ \bar{\lambda }_i = \bar{\mu }_i e^{-mt}$. Then:

$$\begin{aligned} u&= \max \left\{ 0, \min \left\{ 1, \frac{\alpha _1 e^{m g_1 t} y_1 y_2^{g_1}\mu _1}{2 B}\right\} \right\} , \\ \bar{u}&= \max \left\{ 0, \min \left\{ 1, \frac{\alpha _1 e^{m g_1 t} \bar{y}_1 \bar{y}_2^{g_1}\bar{\mu }_1}{2 B}\right\} \right\} . \end{aligned}$$

Substituting into the optimality system (4b)–(4d) and (12) we get:

$$\begin{aligned} y_1' e^{mt} + m y_1 e^{mt} =&\alpha _1 e^{mt} e^{m g_1 t} y_1 y_2^{g_1} (1 - u) - \beta _1 e^{mt} y_1 + \sigma _1 e^{2mt}y_1 y_3,\\ y_2' e^{mt} + m y_2 e^{mt} =&\alpha _2 e^{m g_2 t} e^{m t} y_1^{g_2} y_2 - \beta _2 e^{mt} y_2 + \sigma _2 e^{2mt} y_2 y_3,\\ y_3' e^{mt} + m y_3 e^{mt} =&\alpha _3 e^{mt} y_3\left( 1 - e^{mt} y_3/K\right) - \beta _3 e^{mt} y_3 + \sigma _3 e^{m g_2 t} e^{m t}y_1^{g_2} y_3 \\&+ \sigma _4 e^{m g_1 t} e^{mt} y_2^{g_1} y_3,\\ \mu _1' e^{-mt} - m \mu _1 e^{-mt} =&- e^{-mt} \mu _1(\alpha _1 e^{m g_1 t} y_2^{g_1}(1-u) - \beta _1 + \sigma _1 e^{mt} y_3) \\&- e^{-mt} \mu _2(\alpha _2 g_2 e^{m(g_2-1)t} y_1^{g_2-1} e^{mt} y_2) \\&- e^{-mt} \mu _3(\sigma _3 g_2 e^{m(g_2-1)t} y_1^{g_2-1} e^{mt} y_3), \\ \mu _2' e^{-mt} - m \mu _2 e^{-mt} =&- e^{-mt} \mu _1 (\alpha _1 g_1 e^{mt} e^{m(g_1-1)t} y_1 y_2^{g_1-1}(1-u)) \\&- e^{-mt} \mu _2 (\alpha _2 e^{m g_2 t} y_1^{g_2} - \beta _2 + \sigma _2 e^{mt} y_3) \\&- e^{-mt} \mu _3(\sigma _4 g_1 e^{m(g_1-1)t} y_2^{g_1-1} e^{mt} y_3), \\ \mu _3' e^{-mt} - m \mu _3 e^{-mt} =&- e^{-mt} \mu _1(\sigma _1 e^{mt} y_1) - e^{-mt} \mu _2(\sigma _2 e^{mt} y_2) \\&- e^{-mt} \mu _3(\alpha _3 (1 - 2 e^{mt} y_3/K) - \beta _3 \\&+ \sigma _3 e^{m g_2 t} y_1^{g_2} + \sigma _4 e^{m g_1 t} y_2^{g_1} ) \\&- 2 e^{mt} y_3. \end{aligned}$$

We can divide the first three equations by $e^{mt}$ and the other three by $e^{-mt}$. Simplifying:

$$\begin{aligned} y_1' + m y_1 =&\alpha _1 e^{m g_1 t} y_1 y_2^{g_1} (1 - u) - \beta _1 y_1 + \sigma _1 e^{mt} y_1 y_3,\\ y_2' + m y_2 =&\alpha _2 e^{m g_2 t} y_1^{g_2} y_2 - \beta _2 y_2 + \sigma _2 e^{mt} y_2 y_3,\\ y_3' + m y_3 =&\alpha _3 y_3 \left( 1 - e^{mt} y_3/K\right) - \beta _3 y_3 + \sigma _3 e^{m g_2 t} y_1^{g_2} y_3 + \sigma _4 e^{m g_1 t} y_2^{g_1} y_3,\\ \mu _1' - m \mu _1 =&- \mu _1(\alpha _1 e^{m g_1 t} y_2^{g_1}(1-u) - \beta _1 + \sigma _1 e^{mt} y_3) - \mu _2(\alpha _2 g_2 e^{m g_2 t} y_1^{g_2-1} y_2)\\&- \mu _3(\sigma _3 g_2 e^{m g_2 t} y_1^{g_2-1} y_3),\\ \mu _2' - m \mu _2 =&- \mu _1 (\alpha _1 g_1 e^{m g_1 t} y_1 y_2^{g_1-1}(1-u)) - \mu _2 (\alpha _2 e^{m g_2 t} y_1^{g_2} - \beta _2 + \sigma _2 e^{mt} y_3) \\&- \mu _3(\sigma _4 g_1 e^{m g_1 t} y_2^{g_1-1} y_3),\\ \mu _3' - m \mu _3 =&- \mu _1(\sigma _1 e^{mt} y_1) - \mu _2(\sigma _2 e^{mt} y_2) \\&- \mu _3(\alpha _3 (1 - 2 e^{mt} y_3/K) - \beta _3 + \sigma _3 e^{m g_2 t} y_1^{g_2} + \sigma _4 e^{m g_1 t} y_2^{g_1}) - 2 e^{2mt} y_3. \end{aligned}$$

The system related to the other optimal solution $(\bar{x},\bar{\lambda },\bar{u})$ is analogous. Subtracting the corresponding equations related to $(x,\lambda ,u)$ and $(\bar{x},\bar{\lambda },\bar{u})$ we get:

$$\begin{aligned} (y_1 - \bar{y_1})' + m (y_1 - \bar{y_1}) =&\alpha _1 e^{m g_1 t} (y_1 y_2^{g_1} (1 - u) - \bar{y_1} \bar{y_2}^{g_1} (1 - \bar{u})) \\&- \beta _1 (y_1 - \bar{y_1}) + \sigma _1 e^{mt} (y_1 y_3 - \bar{y_1} \bar{y_3}), \\ (y_2 - \bar{y_2})' + m (y_2 - \bar{y_2}) =&\alpha _2 e^{m g_2 t} (y_1^{g_2} y_2 - \bar{y_1}^{g_2} \bar{y_2}) - \beta _2 (y_2 - \bar{y_2}) \\&+ \sigma _2 e^{mt} (y_2 y_3 - \bar{y_2} \bar{y_3}), \\ (y_3 - \bar{y_3})' + m (y_3 - \bar{y_3}) =&\alpha _3 \left( (y_3 - \bar{y_3}) - e^{mt} (y_3^2 - \bar{y_3}^2)/K\right) \\&- \beta _3 (y_3 - \bar{y_3}) + \sigma _3 e^{m g_2 t} (y_1^{g_2} y_3 - \bar{y_1}^{g_2} \bar{y_3}) \\&+ \sigma _4 e^{m g_1 t} (y_2^{g_1} y_3 - \bar{y_2}^{g_1} \bar{y_3}), \\ (\mu _1 - \bar{\mu _1})' - m (\mu _1 - \bar{\mu _1}) =&- \alpha _1 e^{m g_1 t} (\mu _1 y_2^{g_1} - \bar{\mu _1} \bar{y_2}^{g_1}) \\&+ \alpha _1 e^{m g_1 t} (\mu _1 y_2^{g_1} u - \bar{\mu _1} \bar{y_2}^{g_1} \bar{u})\\&+ \beta _1 (\mu _1 - \bar{\mu _1}) - \sigma _1 e^{mt} (\mu _1 y_3 - \bar{\mu _1} \bar{y_3}) \\&- \alpha _2 g_2 e^{m g_2 t} (\mu _2 y_1^{g_2-1} y_2 - \bar{\mu _2} \bar{y_1}^{g_2-1} \bar{y_2}) \\&- \sigma _3 g_2 e^{m g_2 t} (\mu _3 y_1^{g_2-1} y_3 - \bar{\mu _3} \bar{y_1}^{g_2-1} \bar{y_3}), \\ (\mu _2 - \bar{\mu _2})' - m (\mu _2 - \bar{\mu _2}) =&- \alpha _1 g_1 e^{m g_1 t} ( \mu _1 y_1 y_2^{g_1-1} - \bar{\mu _1} \bar{y_1} \bar{y_2}^{g_1-1}) \\&+ \alpha _1 g_1 e^{m g_1 t} ( \mu _1 y_1 y_2^{g_1-1} u - \bar{\mu _1} \bar{y_1} \bar{y_2}^{g_1-1} \bar{u}) \\&- \alpha _2 e^{m g_2 t} (\mu _2 y_1^{g_2} - \bar{\mu _2} \bar{y_1}^{g_2}) \\&+ \beta _2 (\mu _2 - \bar{\mu _2}) - \sigma _2 e^{mt} (\mu _2 y_3 - \bar{\mu _2} \bar{y_3}) \\&- \sigma _4 g_1 e^{m g_1 t} (\mu _3 y_2^{g_1-1} y_3 - \bar{\mu _3} \bar{y_2}^{g_1-1} \bar{y_3}), \\ (\mu _3 - \bar{\mu _3})' - m (\mu _3 - \bar{\mu _3}) =&- \sigma _1 e^{mt} (\mu _1 y_1 - \bar{\mu _1} \bar{y_1}) - \sigma _2 e^{mt} (\mu _2 y_2 - \bar{\mu _2} \bar{y_2}) \\&- \alpha _3 (\mu _3 - \bar{\mu _3}) + \frac{2 \alpha _3 e^{mt}}{K} (\mu _3 y_3 - \bar{\mu _3} \bar{y_3}) \\&+ \beta _3 (\mu _3 - \bar{\mu _3}) \\&- \sigma _3 e^{m g_2 t} (y_1^{g_2} - \bar{y_1}^{g_2}) - \sigma _4 e^{m g_1 t} (y_2^{g_1} - \bar{y_2}^{g_1}) \\&- 2 e^{2mt} (y_3 - \bar{y_3}). \end{aligned}$$

Now, we multiply each equation by the left-hand side without the derivative and then integrate from 0 to a time T. We present next the result of doing this just for $y_1$ and $\mu _1$ since the other variables have similar expressions:

$$\begin{aligned}&\left. \frac{1}{2}(y_1 - \bar{y_1})^2\right| _0^T + m \int _{0}^{T} (y_1 - \bar{y_1})^2 dt \\&\quad =\alpha _1 \int _{0}^{T} (y_1 - \bar{y_1}) e^{m g_1 t} (y_1 y_2^{g_1} - \bar{y_1} \bar{y_2}^{g_1}) dt\\&\qquad - \alpha _1 \int _{0}^{T} (y_1 - \bar{y_1}) e^{m g_1 t} (y_1 y_2^{g_1} u - \bar{y_1} \bar{y_2}^{g_1} \bar{u}) dt \\&\qquad - \beta _1 \int _{0}^{T} (y_1 - \bar{y_1})^2 dt + \sigma _1 \int _{0}^{T} (y_1 - \bar{y_1}) e^{mt} (y_1 y_3 - \bar{y_1} \bar{y_3}) dt,\\&\qquad -\left. \frac{1}{2}(\mu _1 - \bar{\mu _1})^2\right| _0^T + m \int _{0}^{T} (\mu _1 - \bar{\mu _1})^2 dt = \\&\qquad \alpha _1 \int _{0}^{T} (\mu _1 - \bar{\mu _1}) e^{m g_1 t} (\mu _1 y_2^{g_1} - \bar{\mu _1} \bar{y_2}^{g_1}) dt \\&\qquad - \alpha _1 \int _{0}^{T} (\mu _1 - \bar{\mu _1}) e^{m g_1 t} (\mu _1 y_2^{g_1} u - \bar{\mu _1} \bar{y_2}^{g_1} \bar{u}) dt \\&\qquad - \beta _1 \int _{0}^{T} (\mu _1 - \bar{\mu _1})^2 dt + \sigma _1 \int _{0}^{T} (\mu _1 - \bar{\mu _1}) e^{mt} (\mu _1 y_3 - \bar{\mu _1} \bar{y_3}) dt \\&\qquad + \alpha _2 g_2 \int _{0}^{T} (\mu _1 - \bar{\mu _1}) e^{m g_2 t} (\mu _2 y_1^{g_2-1} y_2 - \bar{\mu _2} \bar{y_1}^{g_2-1} \bar{y_2}) dt \\&\qquad + \sigma _3 g_2 \int _{0}^{T} (\mu _1 - \bar{\mu _1}) e^{m g_2 t} (\mu _3 y_1^{g_2-1} y_3 - \bar{\mu _3} \bar{y_1}^{g_2-1} \bar{y_3}) dt. \\ \end{aligned}$$

On the other hand, we also have:

$$\begin{aligned} \int _0^T (u-\bar{u})^2 dt =&\int _0^T \left( \max \left\{ 0, \min \left\{ 1, \frac{\alpha _1 e^{m g_1 t} y_1 y_2^{g_1}\mu _1}{2 B}\right\} \right\} \right. \\&\left. - \max \left\{ 0, \min \left\{ 1, \frac{\alpha _1 e^{m g_1 t} \bar{y}_1 \bar{y}_2^{g_1}\bar{\mu }_1}{2 B}\right\} \right\} \right) ^2 dt\\ \le&\int _0^T \left( \frac{\alpha _1 e^{m g_1 t} y_1 y_2^{g_1}\mu _1}{2 B} - \frac{\alpha _1 e^{m g_1 t} \bar{y}_1 \bar{y}_2^{g_1}\bar{\mu }_1}{2 B} \right) ^2 dt \\ \le&\frac{\alpha _1}{2B} \int _0^T \left( y_1 y_2^{g_1} \mu _1 - \bar{y_1} \bar{y_2}^{g_1} \bar{\mu _1}\right) ^2 dt. \end{aligned}$$

Now, using that the function $f(y_1, y_2, \mu _1) = y_1 y_2^{g_1} \mu _1$ is locally Lipschitz, we can conclude that there exists a positive constant L such that:

$$\begin{aligned}&\frac{\alpha _1}{2B} \int _0^T \left( y_1 y_2^{g_1} \mu _1 - \bar{y_1} \bar{y_2}^{g_1} \bar{\mu _1}\right) ^2 dt \\&\quad \le \frac{\alpha _1 L}{2B} \int _0^T \left( (y_1 - \bar{y_1})^2 + (y_2 - \bar{y_2})^2 + (\mu _1 - \bar{\mu _1})^2 \right) dt. \end{aligned}$$

Another useful inequality is derived from using Lemma 6 two times successively and the locally Lipschitz condition for $f(y_2) = y_2^{g_1}$. Hence we have:

$$\begin{aligned} |y_1 - \bar{y_1}| (y_1 y_2^{g_1} u - \bar{y_1} \bar{y_2}^{g_1} \bar{u}) \le M_5 ((y_1-\bar{y_1})^2 + (u-\bar{u})^2 + (y_2 - \bar{y_2})^2 ) \end{aligned}$$

for some constant $M_5 > 0$. Now, using the previous results and summing up the expression for the six variables we can get: Summing the above six equations and grouping terms we get:

$$\begin{aligned}&\left( m - L_{11} - L_{12} e^{mT} - L_{13} - L_{21} e^{m g_2 T} - L_{22} e^{mT} - L_{31} - L_{32} e^{mT} \right. \\&\quad \left. - L_{33} e^{m g_2 T} - L_{34} - L_{41} - L_{42} - L_{43} e^{mT} - L_{44} e^{m g_2 T} - L_{45} e^{m g_2 T} - L_{51} - L_{52} \right. \\&\quad - L_{53} e^{m g_2 T} - L_{54} e^{mT} - L_{55} - L_{61} e^{mT} - L_{62} e^{mT} - L_{63} - L_{64} e^{mT} - L_{65} e^{m g_2 T} \\&\quad \left. - L_{66} e^{2mT} - L_{67} \right) \int _0^T \left( (y_1-\bar{y_1})^2 \right. \\&\quad \left. +(y_2-\bar{y_2})^2 +(y_3-\bar{y_3})^2+(\mu _1-\bar{\mu _1})^2 +(\mu _2-\bar{\mu _2})^2 +(\mu _3-\bar{\mu _3})^2 \right) dt \le 0. \end{aligned}$$

This can be rewritten as:

$$\begin{aligned}&\left( m - C_1 - C_2 e^{mT} - C_3 e^{m g_2 T} - C_4 e^{2mT} \right) \\&\quad \int _0^T \left( (y_1-\bar{y_1})^2 +(y_2-\bar{y_2})^2 +(y_3-\bar{y_3})^2 \right. \\&\quad \left. +(\mu _1-\bar{\mu _1})^2 +(\mu _2-\bar{\mu _2})^2 +(\mu _3-\bar{\mu _3})^2 \right) dt \le 0. \end{aligned}$$

So if $m - C_1 - C_2 e^{mT} - C_3 e^{m g_2 T} - C_4 e^{2mT} > 0$ then $y_1 = \bar{y_1}$, $y_2 = \bar{y_2}$, $y_3 = \bar{y_3}$, $\mu _1 = \bar{\mu _1}$, $\mu _2 = \bar{\mu _2}$ and $\mu _3 = \bar{\mu _3}$, and therefore the OC solutions u and $\bar{u}$ are the same. $\square$