fsolve + fminbnd combination... Problem !

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 !!!

[Matt J]

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]

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

Answer 1

[Matt J]

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

[Sergio Quesada]

Thank you so much, Matt...

I have tried your lines. This is my complete code:

%Previous definition of scalars
F=@(x) FC.*exp(-(V.*a)./(30.*x.*log(x)));  % integrand

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));
%Trapezoid integration to reduce calculus time

inpts=(1+(5.*((texp-t0).^0.25))).*ones(1,10);
for i=1:10
      r_teor(i) =  fzero(@(r)  Int(r)-(t(r)-texp(i)), inpts);
   end

   Itotal=@(i) (FD./FC).*(rteor(i).*exp((V.*a)./(30.*rteor(i).*log(rteor(i)))))+Io./((rteor(i)).^4);
  SumError=@(i) sum((Itotal(i)-Iexp).^2); %Iexp are also 1x10
%SumError is H, the function to find minimums

   t=@(i)  fminbnd(@(i) SumError(i),6e-3,t1-5e-5,options);

, and it gives the following error:

Error using fzero (line 433)
Second argument must be a scalar or vector of length 2.

Error in ampliando_integral_definitivo_6 (line 12)
    r_teor(i) =  fzero(@(r)  Int(r)-(t(r)-texp(i)), inpts);

Hope you know how to avoid it... Thank you again !!

[Matt J]

Yes, the second argument of fzero must be a vector of length 2 or a scalar. However, inpts is length 10. What is inpts and what is your rationale for passing it to fzero?

[Sergio Quesada]

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]

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.

[Torsten]

r_teor=@(i)  fsolve(@(r)Int(r)-(t(i)-texp(i)), inpts(i));

But where is t(i) ?

[Sergio Quesada]

A few lines below

t= fminbnd(SumError(i),6e-3,t1-5e-5,options);

[Torsten]

Below is too late for fsolve ...

[Sergio Quesada]

Yes, that's exactly my problem. I can change i(t) to be defined at the beggining of the for-end, but in that case SumError(i) has not been defined before... And if I take SumError to the beggining, then r_teor(i) has not been defined...

Plase, tell me that you know how to solve it !!

[Matt J]

You need to restate your problem (in equation form). In the beginning, you said that t was a function of r (not r_teor). Now, you've changed your mind and the thing you are trying to solve is very muddled.

In any case, you will not be able to use fsolve for this. Your functions is clearly non-differentiable.

[Sergio Quesada]

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

[Torsten]

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.