This problem is extremely ill conditioned as for the TestDomain
the functions intervals are:
The importance of normalizing the functions has been mentioned in
section 220.127.116.11. But in this example the use of the Jacobian matrix
enable to drastically reduce the computation time.
If we use epsilonf=0.1 and epsilon=0 and if we stop at the first solution we
find an exact solution using 73 boxes:
In the single bisection smear mode (i.e. only one variable is bisected in the process) the same root is obtained in 21080ms (about 50% less time than when using the full bisection mode) with only 40 boxes in the direct storage mode.
Note that for epsilonf=0.1 and epsilon=0 we find the only root with 73 boxes in 39760ms (41 boxes and 24650ms in the single bisection smear mode).
Note that we may improve the efficiency of the procedure by using simplification procedures such as the 2B (section 2.17) and the 3B method. An interesting point in this example is that the bisection mode 1 (bisecting along the variable whose interval has the largest diameter) is more effective than using the bisection mode 2 (using the smear function) with 53 boxes against 108 for the mode 2 for epsilonf=1e-6 and epsilon=1e-6. This can easily been explained by the complexity of the Jacobian matrix elements that leads to a large overestimation of their values when using interval: in that case the smear function is not very efficient to determine which variable has the most influence on the equations.