failure in initial user-supplied objective function evaluation
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sorry. I want to solve an equation with fsolve, but it just solve for n=0 and for others gives this error:
failure in initial user-supplied objective function evaluation. Fsolve cannot continue.
the code is:
c=3.8;
ncl=3; nc=3;
syms u w n
Jn=besselj(n,u);
Jnp=diff(Jn,u);
Kn=besselk(n,w);
Knp=diff(Kn,w);
J0=Jnp/(u*Jn);
K0=Knp/(w*Kn);
CH=(J0+K0)*(J0+(ncl/nc)^2*K0)-n^2*(1/u^2+1/w^2)*(1/u^2+(ncl/nc)^2*1/w^2);
CHL=limit(CH,w,0);
x0=(3);
for m=0:9
CHA=subs(CHL,n,m);
options = optimoptions('fsolve','Display','iter'); % Option to display output
fval= fsolve(matlabFunction(CHA),x0,options) % Call solver
end
can enyone help me.
2 个评论
Walter Roberson
2015-5-22
What is nc? You have ncl/nc twice, but you do not define nc. Is it n*c ? Should ncl/nc be read as (ncl/n)*c, or as ncl/(n*c) ?
Walter Roberson
2015-5-22
Instead of using
@(u)eval(CHA)
use
matlabFunction(CHA)
This is an efficiency change, but will not solve the problem itself.
回答（1 个）
Walter Roberson
2015-5-22
0 个投票
Your limit is signum(EXPRESSION)*infinity and so can only be 0 if EXPRESSION is 0 and if you give 0*infinity a definite meaning as being 0 instead of being undefined. Logically solving for signum(EXPRESSION)*infinity = 0 with 0*infinity being defined as 0 is the same as solving for EXPRESSION = 0, but is much more fragile at the best of times, since almost all of the results are going to be +/- infinity and when you finally get a different result it is the exact solution. And then there is the problem that MATLAB is going to return NaN for 0*inf when the signum is finally 0, so fsolve() is not going to be able to find that answer either.
The difficulty of finding the roots of EXPRESSION will depend upon what your "nc" means. Under at least one of the potential meanings, n=0 is a root, which is why you get a solution when n = 0; the solution is independent of the u, so if you are counting on any particular u result as being numerically meaningful, you should not be.
7 个评论
nima salyani
2015-5-22
Walter Roberson
2015-5-22
With the changed version of the code, the limit CHA is
-signum(n)^2*infinity
which is going to be -infinity for any n other than 0.
Notice in your code you have
Kn=besselk(n,w);
Knp=diff(Kn,u);
which differentiates a function of w with respect to u. The result is 0.
Walter Roberson
2015-5-22
Have another look at your
CH=(J0+K0)*(J0+(ncl/nc)^2*K0)-n^2*(1/u^2+1/w^2)*(1/u^2+(ncl/nc)^2*1/w^2);
Notice you have 1/w^2. Your CHL asks for the limit of that as w approaches 0. You are therefore asking for a value that approaches 1/0, and the only time that is not going to be infinite (or undefined) is if it is being multiplied by something that goes to 0 in w "faster" so that the limit is 0. But the only places you use w are in 1/w^2 expressions, so there is nothing that is going to 0 "faster". Your entire limit is therefor at the mercy of exactly which infinity or undefined behavior is going to take priority.
nima salyani
2015-5-23
Walter Roberson
2015-5-23
Which line did that error show up on? Did you somehow accidentally not execute the "syms" line?
nima salyani
2015-5-23
nima salyani
2015-5-24
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