Optimization of radiation dosing schedules for proneural glioblastoma

Abstract

Glioblastomas are the most aggressive primary brain tumor. Despite treatment with surgery, radiation and chemotherapy, these tumors remain uncurable and few significant increases in survival have been observed over the last half-century. We recently employed a combined theoretical and experimental approach to predict the effectiveness of radiation administration schedules, identifying two schedules that led to superior survival in a mouse model of the disease (Leder et al., Cell 156(3):603–616, 2014). Here we extended this approach to consider fractionated schedules to best minimize toxicity arising in early- and late-responding tissues. To this end, we decomposed the problem into two separate solvable optimization tasks: (i) optimization of the amount of radiation per dose, and (ii) optimization of the amount of time that passes between radiation doses. To ensure clinical applicability, we then considered the impact of clinical operating hours by incorporating time constraints consistent with operational schedules of the radiology clinic. We found that there was no significant loss incurred by restricting dosage to an 8:00 a.m. to 5:00 p.m. window. Our flexible approach is also applicable to other tumor types treated with radiotherapy.

Appendix

1.1 Technical lemma

We prove here a technical lemma, which is quite standard but we provide a proof for completeness.

Lemma 3

For $a>0$ and f a bounded function on [0, a] and continuous at 0,

$$\begin{aligned} \lim _{\nu \rightarrow \infty }\nu \int _0^ae^{-\nu y}f(y)dy=f(0). \end{aligned}$$

Proof

First note that

$$\begin{aligned} \nu \int _0^ae^{-\nu y}f(y)dy-f(0)&=\nu \int _0^ae^{-\nu y}(f(y)-f(0))dy-f(0)e^{-a\nu }, \end{aligned}$$

and it thus suffices to establish that

$$\begin{aligned} \lim _{\nu \rightarrow \infty }\nu \int _0^ae^{-\nu y}(f(y)-f(0))dy=0. \end{aligned}$$

For $\nu >0$, define $\ell (\nu )=\log (\nu )/\nu$ and then consider the decomposition

$$\begin{aligned} \nu \int _0^ae^{-\nu y}(f(y)-f(0))dy&=\nu \int _0^{\ell (\nu )}e^{-\nu y}(f(y) -f(0))dy\\&\quad +\nu \int _{\ell (\nu )}^ae^{-\nu y}(f(y)-f(0))dy\\&\le \max _{y\le \ell (\nu )}|f(y)-f(0)|\nu \int _0^{\ell (\nu )}e^{-\nu y}dy\\&\quad +2\max _{y\le a}|f(y)|\nu \int _{\ell (\nu )}^ae^{-\nu y}dy\\&\le \max _{y\le \ell (\nu )}|f(y)-f(0)|+2\max _{y\le a}|f(y)|/\nu . \end{aligned}$$

Both terms on the final line in the previous display then go to 0 as $\nu \rightarrow \infty$ due to our assumptions on the function f. $\square$