An Approximation of the $M_2$ Closure: Application to Radiotherapy Dose Simulation

Abstract

Particle transport in radiation therapy can be modelled by a kinetic equation which must be solved numerically. Unfortunately, the numerical solution of such equations is generally too expensive for applications in medical centers. Moment methods provide a hierarchy of models used to reduce the numerical cost of these simulations while preserving basic properties of the solutions. Moment models require a closure because they have more unknowns than equations. The entropy-based closure is based on the physical description of the particle interactions and provides desirable properties. However, computing this closure is expensive. We propose an approximation of the closure for the first two models in the hierarchy, the $M_1$ and $M_2$ models valid in one, two or three dimensions of space. Compared to other approximate closures, our method works in multiple dimensions. We obtain the approximation by a careful study of the domain of realizability and by invariance properties of the entropy minimizer. The $M_2$ model is shown to provide significantly better accuracy than the $M_1$ model for the numerical simulation of a dose computation in radiotherapy. We propose a numerical solver using those approximated closures. Numerical experiments in dose computation test cases show that the new method is more efficient compared to numerical solution of the minimum entropy problem using standard software tools.

Appendix: Computation of Moments in \(H^i\)

1.1 Computation of Moments in $H^1$

Consider $\bar{\lambda }\in \mathscr {L}_1$, and the associated ansatz $\psi _1$ (defined in (21a)) and its moments $(\psi ^1,\psi ^2)$ given by

$$\begin{aligned} \psi _1(\varOmega )= & {} \exp \left( \lambda _1\varOmega _1+\lambda _4\varOmega _1^2+\lambda _5\varOmega _2^2+\lambda _6\varOmega _3^2\right) , \\ (\psi ^1, \psi ^2)= & {} \left\langle (\varOmega ,\ \varOmega \otimes \varOmega ) \psi _1 \right\rangle \in H^1. \end{aligned}$$

One can remark that $\psi _1$ is an even function of $\varOmega _2$ or $\varOmega _3$, and therefore the moment of $\psi _1$ according to any odd polynomial is zero, in particular

$$\begin{aligned} \psi _2^1 = \psi _3^1 = \psi _{1,2}^2 = \psi _{1,3}^2 = \psi _{2,3}^2 = 0. \end{aligned}$$

With those computations, the moments $\psi ^1$ and $\psi ^2$ actually reads

$$\begin{aligned} \psi ^1 = \psi _1^1 e_1, \quad \psi ^2 = Diag\left( \psi _{1,1}^2,\psi _{2,2}^2,\psi _{3,3}^2\right) , \end{aligned}$$

and $\psi ^1$ is therefore an eigenvector of $\psi ^2$. Using Notations 1 leads to write $N^1$ and $N^2$ under the form (23a), and one may observe that the eigenvectors of $N^2-N^1\otimes N^1$ are along the cartesian axis $e_i$.

Using again evenness of $\psi _1$, one obtains

$$\begin{aligned} \psi _{1,1,2}^2 = \psi _{1,1,3}^2 =\psi _{1,2,3}^2 =\psi _{2,2,2}^2 =\psi _{2,2,3}^2 =\psi _{2,2,3}^2 =\psi _{3,3,3}^2 = 0. \end{aligned}$$

Using the fact that $tr(\varOmega \otimes \varOmega ) = 1$, one obtains that

$$\begin{aligned} \sum \limits _{j=1}^3 \psi _{i,j,j}^3 = \int _{S^2} \varOmega _i tr(\varOmega \otimes \varOmega ) \psi _1(\varOmega ) d\varOmega = \psi _{i}^1. \end{aligned}$$

This leads to write $N^3$ under the form (24a).

Proposition 4

Consider realizable moments $(\psi ^0,\psi ^1,\psi ^2)\in \mathscr {R}_2$ such that $\psi ^1$ is an eigenvector of $\psi ^2$.

Then the rotated normalized moments $(N^1,N^2)$ given by (17) are in $H^1$.

Proof

Under those hypothesis, the decomposition (17) can be simplified. Indeed, since $\psi ^1$ is an eigenvector of $\psi ^2$, then $N^1$ is an eigenvector of $N^2$. So a rotation R diagonalizing $N^2$ will send $N^1$ onto one of the cartesian axis (chose R such that $N^1$ is along $e_1$).

Then this rotation also diagonalizes $N^2-N^1\otimes N^1$ since it diagonalizes both $N^2$ and $N^1\otimes N^1$ and one can write $N^1$ and $N^2$ under the form (21a).

Finally one can prove that the unique exponential representation for moments $(N^1,N^2)$ satisfying (23a) is (21a) by using Theorem 1 with $\bar{m}(\varOmega ) = (\varOmega _1, \ \varOmega _1^2, \ \varOmega _2^2,\ \varOmega _3^2).$ Indeed this theorem provide the existence of a unique function $\psi$ of the form (21a) satisfying

$$\begin{aligned} (N_1^1,N_{1,1}^2,N_{2,2}^2,N_{3,3}^2) = \left\langle \bar{m}\psi \right\rangle . \end{aligned}$$

Computing the other moments of such a function (21a) read

$$\begin{aligned} \psi _2^1 = \psi _3^1 = \psi _{1,2}^2 = \psi _{1,3}^2 = \psi _{2,3}^2 = 0, \end{aligned}$$

i.e. it satisfies the other moment constraints. Then the unique function (10) satisfying all the moment constraints has the form (21a).$\square$

1.2 Computation of Moments in $H^2$

Consider $\bar{\lambda }\in \mathscr {L}_2$, and the associated ansatz $\psi _2$ (defined in (21b)) and its moments $(\psi ^1,\psi ^2)$ given by

$$\begin{aligned} \psi _2(\varOmega )= & {} \exp \left( \lambda _5+\lambda _1\varOmega _1+(\lambda _4-\lambda _5)\varOmega _1^2\right) , \\ (\psi ^1, \psi ^2)= & {} \left\langle (\varOmega ,\ \varOmega \otimes \varOmega ) \psi _2 \right\rangle \in H^2. \end{aligned}$$

The computations of the previous subsection hold. Since $\psi _2$ does not depend on $\varOmega _2$ nor $\varOmega _3$, one deduces that

$$\begin{aligned} \psi _{2,2}^2= & {} \int _{S^2} \varOmega _2^2 \exp \left( \lambda _5+\lambda _1\varOmega _1+(\lambda _4-\lambda _5)\varOmega _1^2\right) d\varOmega = \frac{\psi ^0}{2},\\ \psi _{3,3}^2= & {} \int _{S^2} \varOmega _3^2 \exp \left( \lambda _5+\lambda _1\varOmega _1+(\lambda _4-\lambda _5)\varOmega _1^2\right) d\varOmega = \frac{\psi ^0}{2}, \end{aligned}$$

in particular, $\psi _{2,2}^2 = \psi _{3,3}^2$. Using Notations 1 leads to write $N^1$ and $N^2$ under the form (23b).

Similarily, one has

$$\begin{aligned} \psi _{1,2,2}^2= & {} \int _{S^2} \varOmega _1 \varOmega _2^2 \exp \left( \lambda _4+\lambda _1\varOmega _1\right) d\varOmega = \frac{\psi _1^1}{2},\\ \psi _{1,3,3}^2= & {} \int _{S^2} \varOmega _1 \varOmega _3^2 \exp \left( \lambda _4+\lambda _1\varOmega _1\right) d\varOmega = \frac{\psi _1^1}{2}, \end{aligned}$$

This leads to write $N^3$ under the form (24b).

Proposition 5

Consider realizable moments $(\psi ^0,\psi ^1,\psi ^2)\in \mathscr {R}_2$ such that $\psi ^1$ is an eigenvector of $\psi ^2$ and $\psi _{2,2}^2=\psi _{3,3}^2$.

Then the rotated normalized moments $(N^1,N^2)$ given by (17) are in $H^2$.

The proof is identical to the one of Proposition 4 with $\bar{m}(\varOmega ) = (1,\varOmega _1,\varOmega _1^2)$.

1.3 Computation of Moments in $H^3$

Consider $\bar{\lambda }\in \mathscr {L}_3$, and the associated ansatz $\psi _3$ (defined in (21c)) and its moments $(\psi ^1,\psi ^2)$ given by

$$\begin{aligned} \psi _3(\varOmega )= & {} \exp \left( \lambda _4+\lambda _1\varOmega _1\right) , \\ (\psi ^1, \psi ^2)= & {} \left\langle (\varOmega ,\ \varOmega \otimes \varOmega ) \psi _3 \right\rangle \in H^2. \end{aligned}$$

The computations of the previous subsections hold. In this case, the ansatz $\psi _3$ is the $M_1$ ansatz defined in (13), and therefore $\psi _{1,1}^2$ is the Eddington factor $\chi _2$ defined in Sect. 2.3. Using Notations 1 leads to write $N^1$ and $N^2$ under the form (23c).

The form of $N^3$ is not simplified compared to the previous case.

Proposition 6

Consider realizable moments $(\psi ^0,\psi ^1,\psi ^2)\in \mathscr {R}_2$ such that $\psi ^1$ is an eigenvector of $\psi ^2$, $\psi _{2,2}^2=\psi _{3,3}^2$ and $\psi _{1,1}^2 = \psi ^0 \chi _2\left( |\psi ^1|/\psi ^0\right)$.

Then the rotated normalized moments $(N^1,N^2)$ given by (17) are in $H^2$.

The proof is identical to the one of Proposition 4 with $\bar{m}(\varOmega ) = (1,\ \varOmega _1)$.