0
votes

Below is a code, which works fine without function/function handles. However, I have to write a function so that I can use optimization function of "lsqcurvefit" in MATLAB and optimize parameters constant(1) and constant(2).

L=zeros(n,length(S)); % Height of elements
L(:,1)=Linitial; % Height of elements for day zero
for jj=1:length(S)
   if jj==1
       continue
   end
    for j=1:n % n is 101 (previously defined...)
    
     L(j,jj) = @(constant, i_dt) L(j,jj-1) - (i_dt(j,jj).*constant(1).*((L(j,jj-1)./hs)-1)^constant(2)); %*****
     
    end
end

height_pred = @(constant,i_dt) sum(L); %heigh_pred(1) is known and previously defined
options = optimoptions('lsqcurvefit','FinDiffType','central',...
    'TolFun',1e-10);
constant = lsqcurvefit(height_pred,constant0,i_dt,height_meas,lb,ub,options);

When I run above code, I receive the error "Conversion to double from function_handle is not possible" for line with 5 stars (*****).

If I remove function handle and instead write a function at the end of my code like below:

L=zeros(n,length(S)); % Height of elements
L(:,1)=Linitial; % Height of elements for day zero
for jj=1:length(S)

      for j=1:n % n is 101 (previously defined...)
         L(j,jj) = Nelfun(j,jj);
      end
end

height_pred = @(constant,i_dt) sum(L); %heigh_pred(1) is known and previously defined
options = optimoptions('lsqcurvefit','FinDiffType','central',...
    'TolFun',1e-10);
constant = lsqcurvefit(height_pred,constant0,i_dt,height_meas,lb,ub,options);

function L = Nelfun(constant,i_dt)
for jj=1:145
   if jj==1
       L(:,1)=0.0187600000000000;
       continue
   end
      for j=1:101
         L(j,jj-1);
         L(j,jj) = L(j,jj-1) - (i_dt(j,jj).*constant(1).*((L(j,jj-1)./hs)-1)^constant(2)); %*****
      end
end
end

I receive the error

"Index exceeds matrix dimensions" for line with (*****). 

Size of matrix i_dt is (101,145), L is (101,145), height_meas(1,145), and height_pred(1,145).

L(j,jj) for jj==1 is known but then should be calculated and optimized for L(:,2) to L(:,145).

              ***

Thanks to Praveen, "Index exceeds matrix dimensions" issue is resolved, but now I receive the error

"Error using snls (line 183)
Finite difference Jacobian at initial point contains Inf or NaN values. lsqcurvefit cannot continue." 

for below code:

lb = [0.000000000001,0.05]; ub = [1,50]; constant0 = [1.06e-09,15.2];

height_pred = @(constant,i_dt) sum(Nelfun(constant,i_dt)); 

options = optimoptions('lsqcurvefit','FinDiffType','central',...
    'TolFun',1e-10);
constant = lsqcurvefit(height_pred,constant0,i_dt,height_meas,lb,ub,options);

function L = Nelfun(constant,i_dt)
L=zeros(size(i_dt));
n=101; % number of nodes
ei=2.05613504572765; 
e0=ei.*ones(n,1);
Hsolid=0.613847219422639; 
hs = Hsolid./(n-1);
for JJ=1:size(i_dt,2)
   if JJ==1
     for j=1:n
         Linitial0=hs.*(1+e0);
         Linitial0(1)=0;
     end
     L(:,JJ) = Linitial0;
     continue
   end
     for J=1:size(i_dt,1)
         L(J,JJ) = L(J,JJ-1) - (i_dt(J,JJ).*constant(1).*((L(J,JJ-1)./hs)-1)^constant(2));
     end
end
end
1

1 Answers

0
votes

In the first case, you can't make an array of function handles. you can use cell arrays but I can't understand why you need it.

Let's work your second case.

%I don't know what is hs, I assume it a scalar
hs=myhs;
%call the Nelfun in your function handle
height_pred = @(constant,i_dt) Nelfun(constant,i_dt,hs);


options = optimoptions('lsqcurvefit','FinDiffType','central',...
    'TolFun',1e-10);
% Constant0 should be 2 element array
constant0=[0,0];
constant = lsqcurvefit(height_pred,constant0,i_dt,height_meas,lb,ub,options);

function L = Nelfun(constant,i_dt,hs)% why hs is not passed
L=zeros(size(i_dt)); % Always initialize before filling it up
for jj=1:size(i_dt,2) %use adaptive indexing
   if jj==1
       L(:,jj)=0.0187600000000000;
       continue
   end
      for j=1:size(i_dt,1) %use adaptive indexing
         L(j,jj) = L(j,jj-1) - (i_dt(j,jj).*constant(1).*((L(j,jj-1)./hs)-1)^constant(2));
      end
   
end