@MaPal93

*Let me be clear about my goal. I am looking for a (i) closed-form, (ii) real, and (iii) positive solution. Even just one solution satisfying all three constraints would be fantastic.*

*Now, it seems that sol[1] (2 equations in the lambdas) and sol[5] (3 equations in the lambdas) do not look weird and are the most promising for further analysis.*

sol[2..5] all involve solutions of high degree polynomials, which in general do not have solutions in terms of radicals. So when you put your parameters in, you will not likely get symbolic solutions. So sol[1], which is surprisingly just a quadratic is your best chance.

However, for the case with all parameters =1, an exhaustive analysis proves that there are no positive solutions for any of the 5 cases:

triade.mw

*My questions:*

*1) How did you find the unique and real but negative lambdas from sol[5]?*

In each case, you solve the third equation for lambda__1, substitute it into the second equation and solve it for lambda_2 and then substitute these solutions into the first equation and solve for lambda__3 (this is the backsubstitution on the triangularized system). See details in the file above.

*2) You showed me one example in which also sol[1] produces a real but negative solution. In the same comment you mentioned "but real and positive is harder". Why is it the case? If mathematically real and positive solutions are not ruled out a priori, is there a more systematic way to find them? (even just a loop which tries many substitutions and stops as soon as we find one would do, I guess?)*

I did it in detail on the attached. With parameters it will be harder, but some things can help in this sort of analysis, Descartes rule of signs etc.

*3) It's not clear to me the benefit of dividing out the common factor in the system with all parameters normalized to 1 (since we already found solutions and, factorized or not, these solutions would be the same). The benefit would be clear if such removable common factor existed in the uncalibrated equations ***EqN := ((numer@evala@:-Norm@numer)'tilde'@eval)(MyEqs)**. How to verify this possibility?

I agree. I assume factorizing before solving speeds up the solution process. Factorizing may not work with parameters but you can try the factor() command to see.

*4) Surely a naive question, but would a solution found as in 2) solve also the uncalibrated system? That is, would simplify(eval(EqN,sol)) still give me 0 for ***EqN := ((numer@evala@:-Norm@numer)'tilde'@eval)(MyEqs) **instead of **EqN := ((numer@evala@:-Norm@numer)'tilde'@eval)(MyEqs,P='tilde'1)**?

Seems very unlikely.

*5) Related to 4). *__My end goal is to study how lambda_1, lambda_2, and lambda_3 vary with my parameters__. Is it legit to pick a real and positive lambda_2 and lambda_3 (found as in 2)) and plug them back into uncalibrated EqN[1] (and solve for lambda_1), then pick lambda_1 and lambda_3 and plug them back into uncalibrated EqN[2] (and solve for lambda_2), finally, pick lambda_1 and lambda_2 and plug them back into uncalibrated EqN[3] (and solve for lambda_3)? See my original script 160523_stylized_triade.mw: the first equation is for lambda_1, second for lambda_2, and third for lambda_3.

I'm not sure I understand this, but in general the solution for all parameters=1 is not going to generalize to the case with parameters.

*Really thank you for helping me out with this.*

You're welcome. There are some interesting symmetries that might help you, For example EqN[2] and EqN[3] are the same if lambda__1 and lambda__2 are exchanged (and this was true for sol[1].) Is this expected?