I want to solve this parametric nonlinear system of equations in Mathematica:
$$x−1−(e^{a−x}+e^{b−y})/(c+e^{b−y})=0$$
$$y−1−(e^{b−y})/(c+e^{a−x})=0$$
and I used this code:
Solve[{x - 1 - (Exp[a - x] + Exp[b - y])/(c + Exp[b - y]) == 0, y - 1 - (Exp[b - y])/(c + Exp[a - x]) == 0}, {x, y}]
but I got this message: This system cannot be solved with the methods available to Solve
Does anyone know what's the problem?
I am pretty sure this can be done only numerically after evaluating the parameters a,b,c, for example, in such a way. We extract the LHSes of the equations by [[1]], square these by Map[#^2 &,...], sum the squares by Total, and evaluate the parameters by /. {a -> 1, b -> -2, c -> 3}. Then we apply NMinimize
NMinimize[Total[Map[#^2 &, {(x -
1 - (Exp[a - x] + Exp[b - y])/(c + Exp[b - y]) ==
0)[[1]], (y - 1 - (Exp[b - y])/(c + Exp[a - x]) ==
0)[[1]]}]] /. {a -> 1, b -> -2, c -> 3}, {x, y}]
{0., {x -> 1.26718, y -> 1.01305}}
If you are interested in the solutions over the complexes, then that approach should be modified.
plots:-implicitplot([-1 - (exp(1 - x) + exp(-2 - y))/(3 + exp(-2 - y)) + x = 0, -1 - exp(-2 - y)/(3 + exp(1 - x)) + y = 0], x = -5 .. 5, y = -5 .. 5, color = [blue, red]) and fsolve([-1 - (exp(1 - x) + exp(-2 - y))/(3 + exp(-2 - y)) + x = 0, -1 - exp(-2 - y)/(3 + exp(1 - x)) + y = 0])confirm my answer ( see here)
$\endgroup$
Commented
Jul 7, 2021 at 20:13
You can get solutions in terms of Root expressions for defined a,b,c.
Use the fact that f^2+g^2 has a minimum at intersection point to get additional equations.
You get two separated transzendental equations for x and y.
{f, g} = {x - 1 - (Exp[a - x] + Exp[b - y])/(c + Exp[b - y]),
y - 1 - (Exp[b - y])/(c + Exp[a - x])}
ContourPlot[
Evaluate[{f == 0, g == 0} /. {a -> 1, b -> 1, c -> 1}], {x, 0,
4}, {y, 0, 4}]
FindRoot[{f == 0, g == 0} /. {a -> 1, b -> 1, c -> 1}, {x, 2}, {y, 2}]
(* {x -> 1.69589, y -> 1.43284} *)
Plot3D[{0, f^2 + g^2 /. {a -> 1, b -> 1, c -> 1}}, {x, 0, 4}, {y, 0,
4}]
df = D[f^2 + g^2, x] // Together // Numerator // Simplify
dg = D[f^2 + g^2, y] // Together // Numerator // Simplify
elix = Eliminate[{f == 0, df == 0, g == 0, dg == 0}, x]
(* Log[-((E^b + c E^y - c E^y y)/(1 - y))] == -1 + a - E^b/(
E^b + c E^y) + y + (
E^b + c E^y - c E^y y)/((E^b + c E^y) (1 - y)) && E^b + c E^y != 0 *)
Solve[elix[[1]] /. {a -> 1, b -> 1, c -> 1}, y, Reals]
(* {{y -> Root[{-E^#1 - 2 E #1 + E #1^2 + E^#1 #1^2 +
Log[-((E + E^#1 - E^#1 #1)/(1 - #1))] (E + E^#1 (1 - #1) -
E #1) &, 1.43283890934916955086}]}} *)
eliy = Eliminate[{f == 0, df == 0, g == 0, dg == 0}, y]
Solve[eliy[[1]] /. {a -> 1, b -> 1, c -> 1}, x, Reals]
(* {{x -> Root[{-E - E^#1 - 2 E #1 - E^#1 #1 + E #1^2 + E^#1 #1^2 +
Log[-((E + E^#1 - E^#1 #1)/(2 - #1))] (2 E + E^#1 (2 - #1) -
E #1) &, 1.69589278283101356668}]}} *)
Solve[eliy[[1]] /. {a -> 1, b -> 1, c -> 1}, x, Reals] and Solve[elix[[1]] /. {a -> 1, b -> 1, c -> 1}, y, Reals] are numerical solutions camouflaged by Root[..]. Before solving, the parameters are evaluated.
$\endgroup$
Commented
Jul 8, 2021 at 4:16
Solve- it's too difficult. If a,b are known values please provide the numbers - in which case you could tryNSolvefor a numerical solution, otherwise if you need a symbolic solution in $a,b,c$ I'm afraid you're out of luck. $\endgroup$but I got this message:how long did it take you to get this message? its been running for 20 minutes on my 12.31. You should really mention the version number and the OS also. $\endgroup$