The first figure below shows the plots of a function $f(x)$ plotted for different values of a parameter $\lambda$, i.e., $f(x)=f(x,\lambda=\lambda_i);i=1,2,3,..;\lambda$ is maximum for the topmost curve and decreases downwards. The second figure shows the values of $\lambda$ that corresponds to the extrema.
Each curve has one maxima and one minima. The dotted curve corresponds to the case that has no maxima or minima, i.e., the value of $\lambda$ corresponding to the dotted curve is the threshold for existence of any extrema.
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 which matches exactly with that of particle dynamics. In this sense, figure-2 is the same for particle dynamics as well as hydrodynamics.
The following are my goals:
- To find the threshold value of $\lambda$ (corresponding to the black dotted curve): This I had already calculated by finding the minima 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 on the $\lambda-x$ curve corresponds to extremum, I do not have any idea about how to use the constraint $f(x)=1$ as the expression for $f(x)$ is not available.
Additional details and useful hints:
In figure-1, all the curves saturates at $f(x)=1$ at large values of $x$. So the constraint $f(x)=1$ is satisfied for the black(solid) curve as seen in the figure at $x=2.91$ and also for all the curves at large $x$. So I think that the asymptotic behaviour of $f(x)$ might be related to the values of $\lambda$.
In figure-2, initially $\lambda$ decreases steeply with $x$ and after the minima, $\lambda$ varies as $\lambda=\sqrt x$.
Any hint about the approach to solve the problem would be sufficient.