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How can we solve a pair of equations that have to be solved self-consistently (even by plotting them simultaneously) .

y = 1/2 + 1/\[Pi] ArcTan[c (0.5 - x)];
x = 1/2 + 1/\[Pi] ArcTan[c (0.5 - y)];

For c=0.5 we must have a plot as below one in which the vertical axis is y and horizontal one is devoted to x.

enter image description here

For c=Pi we must have

enter image description here

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  • $\begingroup$ FindRoot for solving and ContourPlot for plotting $\endgroup$ – Lotus Mar 27 at 10:06
  • $\begingroup$ I wanted to plot the first function y vs x ordinary and by InverseFunction[f][y] I wanted to plot the second one (x vs y). For the second one the output plot is not symmetric relative to bisectrix of the first quarter of the coordination plate. $\endgroup$ – Unbelievable Mar 27 at 10:38
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With[{c = 0.5},
 ContourPlot[{
   y == 1/2 + 1/π ArcTan[c (0.5 - x)],
   x == 1/2 + 1/π ArcTan[c (0.5 - y)]
   }, 
  {x, 0, 1}, {y, 0, 1}]
 ]

Mathematica graphics

With[{c = π},
 ContourPlot[{
   y == 1/2 + 1/π ArcTan[c (0.5 - x)],
   x == 1/2 + 1/π ArcTan[c (0.5 - y)]
   }, 
  {x, 0, 1}, {y, 0, 1}]
 ]

Mathematica graphics

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  • 1
    $\begingroup$ One can even use MeshFunctions to mark the intersection point. $\endgroup$ – J. M. will be back soon Mar 27 at 13:36

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