Problem using lscurvefit - 'Initial point is a local minimum'

I am attempting to fit a complicated function using lscurvefit. The function I am attempting to fit looks like this:

xdata./( 
(3*x(1)./(1.28*xdata.^0.46).^2) 
+ (((4.178*(x(2)-0.418*x(1)))./(1.28*xdata.^0.46)).*(1-xdata).^(1/3)) 
+ (6.806 * (x(2) - 1.588 * x(1)) * (1-xdata).^(2/3)) 
)

where xdata is on the order of 1e-3 and the output of the function is on the order of 1e-7 to 1e-6, always increasing. The constants I am attempting to solve for, x(1) and x(2), will be between 0 and 60, so my initial vector x0 is [30 30].

The code I am using looks like this:

options = optimset('TolX',1e-6, 'TolFun',1e-6);
lb = [0 0];
ub = [99 99];

fcn = @(x,xdata) [the above function]

x0 = [30 30];

xsolve = lsqcurvefit(fcn,x0,eps,kdim,lb,ub,options);

I am told no matter what my initial guess is that my function is already at an optimum. I am assuming that there is something happening with the scaling in this problem, but I am having a hard time seeing where it is, as the solution vector does appear to be well scaled for a solution. I have tried changing TolX and TolFun to see if I could find a meaningful solution, but to no avail.

If anyone can offer some advice, it would be much appreciated.

Thank you

Answer 1

[Torsten]

Try to fit 1/kdim against the inverse of your function f.

After rearranging you'll see that 1/f takes the form

1/f = f1(eps)*x(1)+f2(eps)*x(2).

To get optimum x(1) and x(2), just do

xsol = [f1(eps) f2(eps)]\(1./kdim)

At least xsol can be used as a good starting guess for a nonlinear fit.

Best wishes

Torsten.

[henry kagey]

Thank you for the reply, Torsten. I followed your advice to invert and separate, and was able to find an optimum for the constants x(1) and x(2).

I appreciate that you took the time to offer up a potential solution. I am curious about what you saw that made you think to reframe the problem: was it the nature of inputs to lscurvefit? Or was it something else?

Thanks again!