Algorithms: calculation of the maximums of a function subject to a restriction

The first figure below shows the graphs of a function $ f (x) $ plotted for different values ​​of a parameter $ lambda $that is to say $ f (x) = f (x, lambda = lambda_i); i = 1,2,3, ..; lambda $ It is maximum for the upper curve and decreases down. The second figure shows the values ​​of $ lambda $ That corresponds to the extremes.

Each curve has a maximum and a minimum. The dotted curve corresponds to the case that has no maximum or minimum, that is, the value of $ lambda $ corresponding to the dotted curve is the threshold for the existence of any end.

Note that Figure-1 is obtained from particle dynamics where we have an expression for $ f (x) $ and we can obtain figure 2 by calculating the extreme values. However, in hydrodynamics we can only obtain Figure 2 that exactly matches that of particle dynamics. In this sense, Figure 2 is the same for particle dynamics and hydrodynamics.

Figure 1
Figure 2

The following are my goals:

  • To find the threshold value of $ lambda $ (corresponding to the black dotted curve): This had already been calculated by finding the minimums of the $ lambda-x $ curve.
  • To find the value of $ lambda $ for which $ f (x) = 1 $ (corresponding to the black solid curve): Since all the points in the $ lambda-x $ the curve corresponds to the end, I have no idea how to use the constraint $ f (x) = 1 $ as the expression for $ f (x) $ not available.

Additional details and helpful suggestions:

  • In Figure 1, all curves are saturated in $ f (x) = 1 $ at large values ​​of
    $ x $. Then the restriction $ f (x) = 1 $ it is satisfied for the black (solid) curve as seen in the figure in $ x = 2.91 $ and also for all curves in general $ x $. So I think the asymptotic behavior of $ f (x) $ could be related to the values ​​of $ lambda $.

  • In figure 2, initially $ lambda $ abruptly decreases with $ x $ and after the minimums, $ lambda $ Varies according $ lambda = sqrt x $.

Any clue about the approach to solving the problem would be enough.