我将尝试不同的方法,看看是否能找到解决方案
Ef[J2_, D2_, θA1_, θC1_, φA1_, φC1_] := J2 (Sin[θA1] Sin[θC1]*
(Cos[φA1] Cos[φC1] + Sin[φA1] Sin[φC1]) + Cos[θA1] Cos[θC1]) -
D2 (Sin[θA1] Cos[φA1] Cos[θC1] - Cos[θA1] Sin[θC1] Cos[φC1]);
J2=1/10;D2=1/10;
NMinimize[{Ef[J2, D2, θA1, θC1, φA1, φC1], 0<=θA1<=Pi&&0<=θC1<=Pi&&
0<=φA1<=2 Pi&&0<=φC1<=2 Pi}, {θA1, θC1, φA1, φC1},WorkingPrecision->32]
这立即给了我
{-0.1414213562373095048801688724209560058569711627807200358098`32.,
{θA1 -> 1.3609367058891979125401402005460735210021598955174397243367`32.,
θC1 -> 2.5660541110980434566496343521125515860033321979399030055079`32.,
φA1 -> 3.1415926535897927498306628362843651999778557816889010918926`32.,
φC1 -> 6.2831853071795854493988148705854136770428612595496860217646`32.}}
我注意到φA1 几乎可以肯定是 Pi,φC1 几乎可以肯定是 2 Pi,所以
s=Simplify[Ef[J2, D2, θA1, θC1, φA1, φC1]/.{φA1->Pi,φC1->2 Pi}]
给予
(Cos[θA1 + θC1] + Sin[θA1 + θC1])/10
现在我试图找到解决方案
FullSimplify[Minimize[{s, 0<=θA1<=Pi&&0<=θC1<=Pi}, {θA1, θC1}]]
给予
{-1/(5*Sqrt[2]), {θA1 -> Pi, θC1 -> Pi/4}}
仔细检查以确保没有错误。
看看 -Sqrt[2]/10 有多少个局部最小值,哪个满足你找到的约束
Length[Select[Partition[Flatten[Table[
{Ef[J2, D2, θA1, θC1, φA1, φC1], θA1, θC1, φA1, φC1},
{θA1,0,Pi,Pi/4},{θC1,0,Pi,Pi/4},{φA1,0,2Pi,Pi/64},{φC1,0,2 Pi,Pi/64}]],
5],#[[1]]==-Sqrt[2]/10&]]
可能还有更多。