# Solving a non linear system

I'm trying to solve this set of equations (getting A, x and y) for a given set of k and p (i'm actually trying to solve for $k\in [-\pi, \pi]$)

F1[A_, x_, y_, k_, p_] := A * (1 - Exp[-x/2])*(1 - Exp[-y/2]) - (1 + Exp[-x-y])^2
F2[A_, x_, y_, k_, p_] := A - 1 - p*(Cosh[x] + Cosh[y])^2
F3[A_, x_, y_, k_, p_] := Cosh[y] - Cosh[x] - Cos[k]


I know that i should find a set of two solution so i play on the initial guess to find the two. I easily find the first set by doing:

g[k_?NumericQ, p_?NumericQ] :=
FindRoot[{F1[A, x, y, k, p], F2[A, x, y, k, p],
F3[A, x, y, k, p]}, {{A, 1}, {x, 1}, {y, 1}},
MaxIterations -> \[Infinity]]


And I find almost what i want. However, If I want to find the other roots, with a A smaller but still positive, impossible to get something working. I try:

g[k_?NumericQ, p_?NumericQ] :=
FindRoot[{F1[A, x, y, k, p], F2[A, x, y, k, p],
F3[A, x, y, k, p]}, {{A, 1/10}, {x, 1/10}, {y, -1/10}},
MaxIterations -> \[Infinity]]


and always end up with:

FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.

I would like to know how i can go over this error. And if someone know how to automate the research for several roots

• I checked and If p->0 then A ->1. for -Pi<k<Pi.FindRoot says true. It's impossible for 0<A<1 ? – Mariusz Iwaniuk Sep 8 '18 at 19:08
• i have the two limit function K in the cases where p->0 and p-> infinity. I'm interested in the intermediate for p between 1 and 10. I have a criteria that says $A>4/p^2$ when p-> infinty – sailx Sep 8 '18 at 19:39
• As @MariuszIwaniuk suggested, there are no solutions for 1 > A > 0. This can be verified by solving the first expression for A and plotting it as a function of {x, y}. – bbgodfrey Sep 9 '18 at 1:07