fsolve + fminbnd combination... Problem !

Question

Sergio Quesada 2018-10-5

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/422453-fsolve-fminbnd-combination-problem

评论： Torsten 2018-10-9

在 MATLAB Online 中打开

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 个评论
显示无隐藏无

Matt J 2018-10-5

编辑：Matt J 2018-10-5

在 MATLAB Online 中打开

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?

Sergio Quesada 2018-10-6

编辑：Sergio Quesada 2018-10-6

在 MATLAB Online 中打开

Thank you for your answer, Matt. Yes, r is an scalar, and for each value of r, I want 10 different solutions G(r) - (texp-t) = 0, to be solve, each one for each value of texp. I call these solutions "rteor". I am trying this now:

rteor = @(r,t) fsolve(@(r) arrayfun (@(T) integral(F,1,r)-(T-t),texp), x0)
t = fminbnd(H(r,t),6e-3,t1-5e-5,options);

, with no results

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Matt J 2018-10-5

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/422453-fsolve-fminbnd-combination-problem#answer_340006

编辑：Matt J 2018-10-5

在 MATLAB Online 中打开

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 个评论
显示 9更早的评论隐藏 9更早的评论

Sergio Quesada 2018-10-6

编辑：Sergio Quesada 2018-10-6

在 MATLAB Online 中打开

Inpts are my initial points for fzero. I want them to be given by a formula. This is now my script:

nombre='C2';t0=6.00E-3;FC=4.24E-1;a=6.05;FD=8.17E-4;Io=15.53;V=25; % Parámetros de inicio
fID=fopen('Para_Leer_C2_2219_ptos.txt','r');% Nombre de archivo de datos experimentales
Columnexpi=1;Columnexpf=10; % # de filas inicial y final de datos experimentales
formatSpec='%f %f'; sizeSpec=[2,Inf]; % 2 columnas separados por un espacio y con '.' como separador
datos=fscanf(fID,formatSpec,sizeSpec); close('all');N=Columnexpf-Columnexpi+1; 
texp=datos(1, Columnexpi:Columnexpf);Iexp=datos(2, Columnexpi:Columnexpf); t1=texp(1);
F=@(x) FC.*exp(-(V.*a)./(30.*x.*log(x))); part=1e-3;  
L=@(r) 1:part:r;
me=@(r) arrayfun(F,L(r));ue=@(r) F(max(L(r)));
Int=@(r) part.*(sum(me(r))-(ue(r)./2));
options = optimset('Display','iter','TolX',1e-4,'PlotFcns','@optimplotx');
for i=1:N
inpts=(1+(5.*((texp(i)-t0).^0.25)));
r_teor=@(i)  fsolve(Int(r)-(t(i)-texp(i)), inpts(i));
Ifar=@(i)(FD./FC).*(r_teor(i).*exp((V.*a)./(30.*r_teor(i).*log(r_teor(i)))));
Itrans=@(i) Io./((r_teor(i)).^4);
Itotal=@(i) Ifar(i)+Itrans(i);
SumError=@(i) sum((Itotal(i)-Iexp).^2);
t= fminbnd(SumError,6e-3,t1-5e-5,options);
end

, giving:

Undefined function 'functions' for input arguments of type 'inline'.

Thank you for your help !!

Sergio Quesada 2018-10-8

编辑：Sergio Quesada 2018-10-8

在 MATLAB Online 中打开

Now I am on this point:

nombre='C2';t0=6.00E-3;FC=4.24E-1;a=6.05;FD=8.17E-4;Io=15.53;V=25; 
fID=fopen('Para_Leer_C2_2219_ptos.txt','r');
Columnexpi=1;Columnexpf=10; 
formatSpec='%f %f'; sizeSpec=[2,Inf]; 
datos=fscanf(fID,formatSpec,sizeSpec); close('all');N=Columnexpf-Columnexpi+1; 
texp=datos(1, Columnexpi:Columnexpf);Iexp=datos(2, Columnexpi:Columnexpf); t1=texp(1);
F=@(x) FC.*exp(-(V.*a)./(30.*x.*log(x))); part=1e-3;  
L=@(r) 1:part:r;
me=@(r) arrayfun(F,L(r));ue=@(r) F(max(L(r)));
Int=@(r) part.*(sum(me(r))-(ue(r)./2)); 
for i=1:N
inpts=@(i) (1+(5.*((texp(i)-t0).^0.25))); 
options = optimoptions('fsolve','Display','none'); 
r_teor=@(i)  fsolve(Int(r)-(t(i)-texp(i)), inpts(i)); 
Ifar=@(i)(FD./FC).*(r_teor(i).*exp((V.*a)./(30.*r_teor(i).*log(r_teor(i)))));
Itrans=@(i) Io./((r_teor(i)).^4);
Itotal=@(i) Ifar(i)+Itrans(i);
SumError=@(i) sum((Itotal(i)-Iexp).^2); 
options = optimset('Display','iter','TolX',1e-4,'PlotFcns','@optimplotx');
t= fminbnd(SumError(i),6e-3,t1-5e-5,options); 
end

, and it says:

Undefined function or variable 'r'.
Error in program>@(i)fsolve(Int(r)-(t(i)-texp(i)),inpts(i))

I have tried to write syms r above, and then it says:

Error using  :  (line 38)
Unable to compute number of steps from 1 to r by 1/1000.

Sergio Quesada 2018-10-9

编辑：Sergio Quesada 2018-10-9

在 MATLAB Online 中打开

Yes, Matt. I am continous changing the program because I am trying to solve it. The problem is not muddled at all:

1- I have a collection of experimental points (texp, Iexp) that I want to fit.

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.

As simple as that. This is my last code tried:

   nombre='C2';t0=6.00E-3;FC=4.24E-1;a=6.05;FD=8.17E-4;Io=15.53;V=25; 
  fID=fopen('Para_Leer_C2_2219_ptos.txt','r');
  Columnexpi=1;Columnexpf=10; 
  formatSpec='%f %f'; sizeSpec=[2,Inf]; 
  datos=fscanf(fID,formatSpec,sizeSpec); close('all');N=Columnexpf-Columnexpi+1; 
  texp=datos(1, Columnexpi:Columnexpf);Iexp=datos(2, Columnexpi:Columnexpf); t1=texp(1);
  F=@(x) FC.*exp(-(V.*a)./(30.*x.*log(x))); part=1e-3;
for i=1:N
t=@(i) (1+(5.*((texp(i)-t0).^0.25)))    
r_teor=@(i)  fsolve(Int(r)-(t(i)-texp(i)), inpts(i),options);
L=@(r) 1:part:rteor(i);
me=@(r) arrayfun(F,L(r)); ue=@(r) F(max(L(r)));
Int=@(r) part.*(sum(me(r))-(ue(r)./2)); 
options = optimoptions('fsolve','Display','none'); 
Ifar=@(i)(FD./FC).*(r_teor(i).*exp((V.*a)./(30.*r_teor(i).*log(r_teor(i)))));
Itrans=@(i) Io./((r_teor(i)).^4);
Itotal=@(i) Ifar(i)+Itrans(i);
SumError=@(i) sum((Itotal(i)-Iexp).^2); 
options = optimset('Display','iter','TolX',1e-4,'PlotFcns','@optimplotx');
t= fminbnd(SumError(i),6e-3,t1-5e-5,options); 
end