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Is this system of equations unsolvable?

Wang Jiapeng
Posted 3 days ago

POSTED BY: Wang Jiapeng
5 Replies
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It does not mean that real analytical solutions do not exist. The problem is that their domain is probably very very complicated, as you have a lot of parameters.
POSTED BY: Gianluca Gorni

Remove Reals and you get solutions.

sol = Solve[equations, vars];

Table[Simplify[equations /. s], {s, sol}]
(*
{{True, True, True, True, True, True, True},
 {True, True, True, True, True, True, True},
 {True, True, True, True, True, True, True},
 {True, True, True, True, True, True, True},
 {True, True, True, True, True, True, True},
 {True, True, True, True, True, True, True},
 {True, True, True, True, True, True, True},
 {True, True, True, True, True, True, True},
 {True, True, True, True, True, True, True},
 {True, True, True, True, True, True, True}}
*)

Determining the constraints on the parameters to get real-only solutions may not be practical.
POSTED BY: Michael Rogers
POSTED BY: Michael Rogers

Two real trivial solutions are guaranteed:

equations /. {{X -> 0, Y -> 0, Z -> -1, M -> 0, A -> 0, 
   B -> 0}, {X -> 0, Y -> 0, Z -> 1, M -> 0, A -> 0, B -> 0}}

In general the situation is very complicated:

Solve[equations /. {\[Omega]a -> 2, \[Omega]c -> 1, \[Kappa]1 -> 
    2, \[Lambda]x -> 3, \[Lambda]y -> 1}, vars, Reals]
POSTED BY: Gianluca Gorni
Wang Jiapeng
Wang Jiapeng
Posted 2 days ago

Hello Professor, I was wondering why we can immediately obtain an analytical solution when the real-number restriction is removed, even though all my variables are indeed not complex. Does adding the real-number constraint definitely mean that no analytical solution exists
POSTED BY: Wang Jiapeng
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