FSolve - no solution found, last step ineffective

My function looks like this:

function F=Fn2(t)
global wn a g
F=(wn{1}^2)*a{1}(1,1)*sin(wn{1}*t)+(wn{2}^2)*a{2}(1,1)*sin(wn{2}*t)-g;
end

where wn, a, and g are defined in my main program. I call the function like this:

tlo=fsolve(@Fn2,1)

but fsolve can't seem to find a solution to the equation and is returning: "fsolve stopped because the last step was ineffective. However, the vector of function values is not near zero, as measured by the default value of the function tolerance."

"a" and "wn" are always real, and I've messed around with changing fsolve's x0 position. Any ideas why it's not working? Thanks in advance.

[Matt J]

Perhaps there is no solution, for example, as would occur when

   g >  wn{1}^2*abs(a{1}(1,1)) + wn{2}^2*abs(a{2}(1,1))

Have you tested a case where the solution is known?

[Matt J]

Incidentally, global variables are a discouraged method of passing fixed parameters to functions. You should really be doing this:

 tlo=fsolve(@(t) Fn2(t,wn,a,g) ,1)

with Fn2 appropriately rewritten to accept (t,wn,a,g) as input arguments. See more info at http://www.mathworks.com/help/optim/ug/passing-extra-parameters.html

[Eli]

Thanks Matt, I used Wolfram to find the numerical solution with the calculated values of wn and a, and came up with an answer, so the function should have some solution. I removed the global variables, but I still can't get it to work correctly.

[Matt J]

What are the values of wn,a, and g and what is the corresponding solution t?

[Eli]

wn{1}=4.628
a{1}(1,1)=-2.92
wn{2}=39.01
a{2}(1,1)=-.0049
g=32.17

This should yield tlo=0.766. I also plotted F exactly as it appears in the function Fn2 and can see that it clearly crosses the X axis around 0.766, so there's either a problem with how Fn2 is receiving input or with how I have fsolve set up

Answer 1

[Matt J]

FSOLVE can do it, but it has a rather narrow radius of capture around the desired tlo, roughly between 0.55 and .79

>> Fn2=@(t) (wn{1}^2)*a{1}(1,1)*sin(wn{1}*t)+(wn{2}^2)*a{2}(1,1)*sin(wn{2}*t)-g;

>> [tlo,fval] = fsolve(Fn2,.6,optimset('Display','none'))

tlo =

      0.7667

fval =

     7.1054e-15

[Matt J]

There are also other solutions, of course,

   >> [tlo,fval] = fsolve(Fn2,.51,optimset('Display','none'))

tlo =

      1.2644

fval =

    -3.5527e-14

[Matt J]

FZERO has a fair bit larger radius of capture, roughly between .4 and 1.01

 >> [tlo,fval] = fzero(Fn2,1)

tlo =

      0.7667

fval =

     7.1054e-15

[Eli]

Thanks again Matt, it seems like FZERO is the way to go because I won't always have an estimate about where tlo will be.