# Is there any hope to compute analytically $y=f(x)$ for which $g(x,y)=0$?

I have a two-variable function $$g(x,y)$$ with $$0 and $$\frac32. I am looking for $$y=f(x)$$ for which $$g(x,y)=0$$. Using ContourPlot

g[x_,y_]= -4 (-44 +
8 Abs[-2 (Sin[x/2 - (3 y)/2] -
3 Sin[x/2 + y/2]) (2 Sin[x/2 - (7 y)/2] +
Sin[(3 x)/2 - (5 y)/2] - 5 Sin[x/2 - (3 y)/2] -
3 Sin[(3 x)/2 - y/2] + 3 Sin[x/2 + y/2] +
6 Sin[x/2 + (5 y)/2])] +
4 Abs[6 + 9 Cos[2 x] + Cos[2 x - 4 y] - 4 Cos[x - 3 y] -
6 Cos[2 x - 2 y] + 8 Cos[x - y] - 10 Cos[2 y] +
12 Cos[x + y]] + -27 Cos[2 x] - Cos[2 x - 8 y] +
6 Cos[2 x - 6 y] - 11 Cos[2 x - 4 y] - 16 Cos[x - 3 y] +
12 Cos[2 x - 2 y] + 32 Cos[x - y] + 70 Cos[2 y] - 60 Cos[4 y] +
18 Cos[6 y] + 48 Cos[x + y] + 54 Cos[2 (x + y)] -
81 Cos[2 x + 4 y]);

p1 = ContourPlot[{g[x,y]== 0 }  , {x, 0, \[Pi]/2}, {y, 3/2, 5},
PlotPoints -> 90 , FrameLabel -> Automatic , AspectRatio -> 3];
p2 = ContourPlot[{y == \[Pi]/2, y == 3 \[Pi]/2}  , {x, 0, \[Pi]/
2}, {y, 3/2, 5}, ContourStyle -> Red, FrameLabel -> Automatic ,
AspectRatio -> 3];
Show[{p1, p2}]



I obtain this plot

My question. Is there any hope to find the explicit expressions for the functions describing those blue curves in the form of $$y=f(x)$$? those red lines are $$y=\frac{\pi}2$$ and $$3\frac{\pi}2$$. I use Solve[g[x,y]==0 ,y] and Reduce[g[x,y]==0 ,y] and after an hour it is still running.

P.S. More precisely, I want to compute the area enclosed between the red and blue curves analytically. (numerically, I can compute it using NIntegrate[Boole[...]])

• It does not seem to be a simple thing. Actually, there are no other ways than those you indicated. Further, even if these ways bring you to a solution after a long calculation, it will probably be cumbersome. What I would do in this case, is solve the problem numerically in several points. After that, I would try to fit the numerical solution with some simple analytical function. Mar 7, 2023 at 15:03
• Thank you for the comment. Dealing numerically sounds complicated too; that long blue curve is a bit scary to be approximated numerically! Mar 7, 2023 at 15:26
• Reduce[g[x, y] == 0 && 0 <= x <= \[Pi]/2, y, Reals] actually does return something, but it's so big that there's no way you can realistically do something with it. Mar 7, 2023 at 15:39

Here my numerical solution (inspired by @AlexeiBoulbitch useful comment):

upper contour

contUP=cont = ContourPlot[{g[x, y] == 0}, {x, 0, \[Pi]/2}, {y, 3 , 5},PlotPoints -> 90, FrameLabel -> Automatic, AspectRatio -> 3]


get the points

pi = contUP[[1, 1]][[1]];


points pi form a polygon, the enclosed area follows to

Area[Polygon[pi]]
(*0.76061*)


lower contour might be solved similarely (lower area gives 0.437233)

Hope it helps!

• Thanks, it confirms the result I had obtained by NIntegrate but it is still numeric' so, probably there is no hope to find analytic solutions. Mar 8, 2023 at 23:36