The following code works fine
eqs = {-1 + (3 Subscript[β, 1])/4 ==
0, -1 + (3 Subscript[β, 2])/4 ==
0, -1 + (
2 Subscript[β,
1] (-(11/2) + (15 Subscript[β, 2])/2))/(
9 Subscript[β, 2]) ==
0, -1 + ((1 + (15 Subscript[β, 1])/4) Subscript[β,
2])/(6 Subscript[β, 1]) == 0};
vB = {Subscript[β, 1], Subscript[β, 2]};
ContourPlot[
Evaluate[eqs], {Subscript[β, 1], 0, 2}, {Subscript[β, 2],
0, 2}, ContourStyle -> {Blue, Red, Green, Orange},
PlotLegends -> {"R?? = 1", "R?? = 1", "R?? = 1", "R?? = 1"},
AxesLabel -> {ToString[Subscript[β, 1]],
ToString[Subscript[β, 2]]},
PlotLabel -> "Analytical Boundaries"]
but was not working in original version where instead of Subscript[β, 1], Subscript[β, 2] I had vB[[1]], vB[[2]]:
ContourPlot[Evaluate[eqs], {vB[[1]], 0, 2}, {vB[[2]], 0, 2},
ContourStyle -> {Blue, Red, Green, Orange},
PlotLegends -> {"R?? = 1", "R?? = 1", "R?? = 1", "R?? = 1"},
AxesLabel -> {ToString[Subscript[β, 1]],
ToString[Subscript[β, 2]]},
PlotLabel -> "Analytical Boundaries"]
In my problem, vB may be any sublist (bifurcation parameters) of a bigger list, and it is disapointing that I cannot plot unless I copy the actual names of the parameters in each case into ContourPlot
ContourPlot[]
holds its arguments is to localize the variables. It cannot inspect the variable invB[[1]]
without evaluating, so something like @Bob's answer is probably the most convenient, unless one absolutely needs the variables localized. $\endgroup$