since \mu is scalar, the max problem could be solved (locally) simply by a kind of search method, for example you could use grid search combined with quadratic interpolation (there is ready to use software for this: ''fmin'', simply take the negative to make minimization to maximization. Now what about ''x''? Make x a function of \mu by

solving the minimization problem for \mu as given value. That means, you must write

for this ''fmin'' a code which calls the minimizer (for x) with \mu as a given parameter and returns its optimal value as input value for one function value in the maximization phase. Since all this can be based on existing public code, it remains as your job to write this function routine only.   whereas ''fmin'' uses function values only, most unconstrained codes for multidimensional problems require (or work at least better) if you can provide the gradient for x. Simply for x, no chain rule involved, since \mu is fixed and a given number in this minimization. But most codes are user friendly and provide some means like finite differences if you cannot provide the gradient.

for codes look here:

http://plato.asu.edu/guide.html

in the section problems/ unconstrained

good luck

peter

In this problem x is in fact the "main" variable, while mu is there merely to determine the objective value in the space of x. Perhaps it is possible to code a descent method in x-space. If mu can be determined easily then you can code a descent algorithm with respect to x alone. 

Fully agree with Michael Patriksson, I think that's what you actually need to do

A silly search could serve.

I recommend you use any algorithm to find the minimum of the function respect to x, and modify the algorithm so that every step you find the maximum with respect to /mu.

For example, if you follow the negative gradient of the function respect to x, after each movement you call a routine that maximizes the function with respect to /mu. You do this until the function value is constant.

I can mention algorithms popular among physicists and chemical physicists. I attach two papers as an example.These methods are especially good for higher dimensions. You can find another analogous papers of these authors.

I hope, it will be useful

Transitions states in chemical reactions are saddle points.  We made some interesting progress treating these as multi-objective optimization problems instead -- the Pareto front becomes the reaction path, and the extreme points on the Pareto front are the initial reactants and the transition state.

Perhaps the formulation as a MOO problem might be useful in your domain too.

You can find some easy-to-implement min-max algorithms in the book Numerical Optimization Techniques by Yurij G. Evtushenko (Springer-Verlag New York, Inc. New York, NY, USA ©1985 ISBN:0-387-90949-4 )

I would replace the "max_{\mu} f(x,\mu)" by a variable "\theta" with the following constraints:

\theta >= f(x,\mu)    \forall u \in R, x \in R^{n}.  

Then, the problem  is the following:

min_{x, \mu, \theta} \theta

st: \theta >= f(x,\mu) 

If your original problem have more constraints, they can be added to the transformed problem.

Finally, yo can use a non-linear programming solver in order to get numerical solutions.

Regards, 

Hans

Here is how I, the lover of interval calculus, would attack such a problem:

1. Establish n+1 dimensional box as a search space.  Each edge is an interval [xmin, xmax], including the coordinate/unknown \mu.  This is the first and only element on the list L.

2. repeat:  take a box from list L and evaluate the limits of grad f within this box gradF(box) = [gradmin, gradmax].  Remove current box from list L. If gradF contains the number 0 then bisect current box perpendicularly to its longest edge and append two new boxes to the list L, otherwise do nothing (current box disappears from further considerations).

until: either the list L becomes empty (no solution at all) or contains only small boxes.

3. Examine the remaining boxes, one by one, and decide which one is the best for you.  This means evaluate now the range [fmin, fmax] (not gradF) for every box still on the list L.

Here, however, we have only one real problem: how to verify that the box of interest indeed contains the saddle point, not the local extremum?  Global extrema are rather obvious.