"Ramesh Kudenatti" <ramesh@bub.ernet.in> wrote in message <jcggph$abc$1@newscl01ah.mathworks.com>...
> My interesting problem is given below:
> (C_mn(W)  T*I)X=0, where T is a known eigenvalue (say 1.53). I need to find the value of W. Now how to find the value of W ?
       
I'm a little puzzled by your expression "(C_mn(W)T*I)X=0". The C_mn you defined earlier appeared to be a scalar quantity for each individual pair of values, m and n, and would become a square matrix only with some set of these values.
In any case, the problem of finding a W which produces a given eigenvalue, T, looks like the same kind of problem as for producing a zero eigenvalue, requiring some kind of searching algorithm which successively narrows down on a solution. The hard part would be generating the matrix of C_mn values for a large number of possible values of W as part of a searching algorithm. In this case the objective function to be used would presumably be, for the eigenvalue closest to T, the difference between the eigenvalue and T, thereby allowing both positive and negative values, and you could perhaps use 'fzero' (provided W is a real quantity.) (Note that such an objective function might be discontinuous in some areas but hopefully continuous at a crossing over the T value where continuity is important.)
I see that the Symbolic Toolbox has available the function airyAi(z,1) which gives values of the first derivative of the Airy function. You defined C_mn in terms of an integral taken with respect to k from infinity to +infinity of an integrand involving that first derivative. Is this the form you would have to use to evaluate each C_mn element each time a different value of W is selected? This is the part that I would think would be very time consuming, numerically computing such an integral for each element of C(W) every time W is altered.
I would recommend first, computing the whole set of eigenvalues against a wide range of varying W and making a simultaneous plot of them, (using dot or asterisk symbols rather than connecting lines.) This would give you a feeling for the nature of the problem you face. You could possibly observe in a rough way one or more of them crossing the T value.
Roger Stafford
