fsolve + fminbnd combination... Problem !
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
Good evening to everybody. My problem is the following:
I want to solve a non-linear equation G(r)-(t-texp)=0 for the variable r with the following complications:
- r is in the upper limit of a definite integral of the integrand F(x).
-"texp" is a 1x10 array experimental points.
- t, which is another variable, are the minimums of another function H defined itself by G, H[t, G(r)].
I am not sure about the combination of algorithms that I must use. I have tried this scheme:
x0=ones(1,10)
rteor=@(r,t) fsolve(integral(G(x),1,r)-(t-texp),x0);
t = fminbnd(@(r,t) H,lb,up);
, but it does not work. I am almost sure fsolve is not well written. I have tried to make combinations with arrayfun inside fsolve, but with no success. I think it is a problem of dimensions and/or declaring variables. Can anybody help me please?
Thanks !!!
2 Kommentare
Matt J
am 5 Okt. 2018
Bearbeitet: Matt J
am 5 Okt. 2018
So r is a scalar and t is a function of r through the equation
t(r) = argmin_z H(z,G(r) )
That means you have 10 equations in one unknown, r. How do you expect to satisfy 10 different equations with only 1 degree of freedom?
Or, do you mean you are looking for 10 different r(i), corresponding to each texp(i) separately?
Antworten (1)
Matt J
am 5 Okt. 2018
Bearbeitet: Matt J
am 5 Okt. 2018
This might be what you want.
G=@(r) integral( @F,1,r);
t=@(r) fminbnd(@(z) H(z,G(r)),lb,up);
for i=1:10
r_solution(i) = fzero(@(r) G(r)-(t(r)-texp(i)) , r0 );
end
11 Kommentare
Torsten
am 9 Okt. 2018
And you claim that
2 - The equation I have is not I(t), but I[r_teor(t)], where r_teor(t) is in the form t(r_teor), being "r_teor" in an upper integration limit.
is not muddled ? Nobody is able to understand what you write here.
Siehe auch
Kategorien
Mehr zu Calculus finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!