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I have three positivity constraints (f1, f2, f3):

f1 = 1 - z[2, 3]^2 + z[1, 4]^2 (-1 + z[2, 3]^2) - z[2, 4]^2 + 
  z[1, 3]^2 (-1 + z[2, 4]^2) - 2 z[2, 3] z[2, 4] z[3, 4] - z[3, 4]^2 
- 2 z[1, 3] z[1, 4] (z[2, 3] z[2, 4] + z[3, 4]) + 
  z[1, 2]^2 (-1 + z[3, 4]^2) + 
  2 z[1, 2] (z[1, 4] (z[2, 4] + z[2, 3] z[3, 4]) + 
    z[1, 3] (z[2, 3] + z[2, 4] z[3, 4]))

and

f2 = 1 - z[1, 2]^2 - z[1, 3]^2 + 2 z[1, 2] z[1, 3] z[2, 3] - z[2, 3]^2

and

f3 = 1 - z[1, 2]^2

This six-dimensional constrained integration command

Integrate[Boole[f1 > 0 && f2 > 0 && f3 > 0], {z[1, 2], -1, 1}, {z[1, 3], -1, 1}, 
 {z[1, 4], -1, 1}, {z[2, 3], -1, 1}, {z[2, 4], -1, 1}, {z[3, 4], -1, 1}]

succeeds with the result 32 Pi^2/27.

I also have a fourth "separability" constraint (note the new nonnegative variable u)

f4 = -u^4 z[1, 4]^2 - z[2, 3]^2 + 
  2 u z[2, 3] (z[1, 2] z[2, 4] - z[1, 3] z[3, 4]) 
+ 2 u^3 z[1, 4] (z[1, 2] z[1, 3] - z[2, 4] z[3, 4]) 
- u^2 (-1 - z[1, 4]^2 z[2, 3]^2 + 2 z[1, 3] z[1, 4] z[2, 3] z[2, 4] + 
    z[2, 4]^2 - z[1, 3]^2 (-1 + z[2, 4]^2) - 
    2 z[1, 2] (z[1, 4] z[2, 3] + z[1, 3] z[2, 4]) z[3, 4] + 
    z[3, 4]^2 - z[1, 2]^2 (-1 + z[3, 4]^2))

I ideally would like to perform the enhanced integration

Integrate[Boole[f1 > 0 && f2 > 0 && f3 > 0 && f4 > 0 && 0 < u <1], 
 {z[1, 2], -1, 1}, {z[1, 3], -1, 1}, 
 {z[1, 4], -1, 1}, {z[2, 3], -1, 1}, 
 {z[2, 4], -1, 1}, {z[3, 4], -1, 1}]

but this does not seem doable, by any means. This being the case, I would then like to perform this integration, but after setting the variable u to certain nonnegative values, $u \in [0,1]$, say $u = 1$. But such exact integrations also do not seem doable.

So, now I would like to set $u = 1$, say, and perform the last integration numerically to as high a precision/accuracy as possible, since I conjecture (ref, sec. 9) the true values are simple and exact in nature (particularly so if divided by the aforementioned constant, 32 Pi^2/27).

I can obtain numerical results, such as $7.37592698928$, with $u = 1$, but the results seem not very trustworthy, getting messages such as:

"The global error of the strategy GlobalAdaptive has increased more \
than 10000 times. The global error is expected to decrease \
monotonically after a number of integrand evaluations. Suspect one of \
the following: the working precision is insufficient for the \
specified precision goal; the integrand is highly oscillatory or it \
is not a (piecewise) smooth function; or the true value of the \
integral is 0. Increasing the value of the GlobalAdaptive option \
MaxErrorIncreases might lead to a convergent numerical integration. \
NIntegrate obtained..."

So, how might one best formulate the integration command with the various options, having set $u = 1, 1/2, 1/4$ and other simple fractions $\in [0,1]$?

(A good/warm-up problem would be to try to obtain the three-constraints 32 Pi^2/27 result numerically.)

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    $\begingroup$ try "Method->LocalAdaptive" ( This is running a long time for me, maybe you have the patience to see if it converges ) $\endgroup$
    – george2079
    Feb 13, 2017 at 22:19
  • $\begingroup$ Thanks, george2079--will try tomorrow when I have the opportunity. $\endgroup$ Feb 14, 2017 at 3:02
  • $\begingroup$ In regard to the suggestion of george2079 to try "Method->LocalAdaptive", doing so seems to be discouraged, in light of the six-dimensional nature of the problem, since the Mathematica help information asserts: "For multidimensional integrals "GlobalAdaptive" is much faster because "LocalAdaptive" does partitioning along each axis, so the number of regions can explode combinatorically." $\endgroup$ Feb 14, 2017 at 23:23
  • $\begingroup$ @Paul Are you sure that the copied expressions are OK? I can't obtain your first result at all (MMA 11, Win7, seems to crash the kernel), and in the second case the choice of $u=1$ in inconsistent with your Boole expression, which contains $0<u<1$. $\endgroup$
    – MarcoB
    Feb 14, 2017 at 23:40
  • $\begingroup$ Thanks, MarcoB. The third and succeeding lines of the formula for f1, that is beginning with -2 z[1,3] need to be "linked" with the previous two. (So, maybe my copying was subpar.) Also, the 0 <u <1 constraint is replaceable by 0 <u <=1, and is not a critical issue, I think. In any case, I seem barely able to get 1% accuracy in the attempted numerical reproduction of 32 Pi^2/27, so hopes for high accuracy in the four-constraint problem seem quite unrealistic at this point. $\endgroup$ Feb 15, 2017 at 16:43

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