Using the monotonicity

For a given box we will compute the jacobian matrix using interval analysis. Each row $j$ of this interval matrix enable to get some information of the corresponding function $F_j$.

• if the i-th column of the j-th row is an interval which is strictly negative or strictly positive, then $F_j$ is monotonic with respect to the unknowns $x_i$
• if the i-th column of the j-th row is equal to 0, then function $F_j$ does not depend on the variable $x_i$

In the first case the minimal and maximal value of $F_j$ will be obtained either for $x_i = \underline{x_i}$ or $x_i = \overline{x_i}$ and we are able to define the value of $x_i$ to get successively the minimal and maximal value as we know the sign of the gradient. But this procedure has to be implemented recursively. Indeed we have previously computed the jacobian matrix for $x_i = (\underline{x_i},\overline{x_i})$ but now $x_i$ have a fixed value: hence a component $J_{jk}$ of the j-th row which for $x_i = (\underline{x_i},\overline{x_i})$ was such that $\underline{J_{jk}}<0$ and $\overline{J_{jk}}>0$ may now be a strictly positive or negative intervals. Consequently the minimal and maximal value will be obtained for some combination of $x_i, x_k$ in the two sets $\{\underline{x_i},\overline{x_i}\}$ and $\{ \underline{x_k},\overline{x_k}\}$. Bus as $x_k$ has now a fixed value some other component of $J_k$ may become strictly negative or positive...

The algorithm for computing a sharper evaluation of $F_j$ is:

$(\underline{F_j},\overline{F_j})$=Evaluate$F_j$( $(\underline{x_1},\overline{x_1}),\ldots,(\underline{x_m},\overline{x_m})$)

1. compute $J_j((\underline{x_1},\overline{x_1}),\ldots,(\underline{x_m},\overline{x_m}))$
2. let $l_1$ be the number of components of $J_j$ such that $\underline{J_{jk}}>0$ or $\overline{J_{jk}}<0$ and let $x_l,\ldots,x_p$ be the variables for which this occur
3. if $l_1>0$
   loop: for all combination of $x_l,\ldots,x_p$ in the set $\{ \underline{x_l}, \overline{x_l},\ldots, \underline{x_p},\overline{x_p}\}$:
   • if $l_1< m$
     • compute $J_j((\underline{x_1},\overline{x_1}),\ldots, x_l, \ldots,x_p,\ldots,(\underline{x_m},\overline{x_m}))$
     • let $l_2$ be the number of components of $J_j$ such that $\underline{J_{jk}}>0$ or $\overline{J_{jk}}<0$
     • if $l_2 > l_1$, then $(\underline{F_u},\overline{F_u})$=Evaluate$F_j$( $(\underline{x_1},\overline{x_1}),\ldots,x_l,\ldots,x_m,\ldots,(\underline{x_m},\overline{x_m})$)
     • otherwise $(\underline{F_u},\overline{F_u})=F_j((\underline{x_1},\overline{x_1}),\ldots,x_l,\ldots,x_m,\ldots,(\underline{x_m},\overline{x_m})$)
     • if this is the first estimation of $F_j$ then $F_j=F_u$
     • otherwise
       • if $\underline{F_u}<\underline{F_j}$, then $\underline{F_j}=\underline{F_u}$
       • if $\overline{F_u}>\overline{F_j}$, then $\overline{F_j}=\overline{F_u}$
   • otherwise
     • $(\underline{F_u},\overline{F_u})=F_j((\underline{x_1},\overline{x_1}),\ldots,x_l,\ldots,x_m,\ldots,(\underline{x_m},\overline{x_m})$)
     • if this is the first estimation of $F_j$ then $F_j=F_u$
     • otherwise
       • if $\underline{F_u}<\underline{F_j}$, then $\underline{F_j}=\underline{F_u}$
       • if $\overline{F_u}>\overline{F_j}$, then $\overline{F_j}=\overline{F_u}$
4. end loop:
5. otherwise $(\underline{F_j},\overline{F_j})=F_j((\underline{x_1},\overline{x_1}),\ldots,(\underline{x_m},\overline{x_m}))$
6. return $(\underline{F_j},\overline{F_j})$

This procedure has to be repeated for each $F_j$.