using lsqnonlin () for solution of an nonlinear equation

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KB am 26 Jul. 2014

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https://de.mathworks.com/matlabcentral/answers/143390-using-lsqnonlin-for-solution-of-an-nonlinear-equation

Bearbeitet: KB am 1 Aug. 2014

Please consider the following code:

function diff = myfun(D,Y) 
t = 0:30:600; 
l = length(t);
a = 0.5*0.75*10^(-4);
for i = 1:l
Yp(i)=0;
for n = 1:10
y(i,n) = exp(-D*t(i)*(pi^2)*(n^2)/(a^2))/n^2;
Yp(i) = Yp(i) + y(i,n);
end
end
diff = 1- (6*Yp/pi^2)-Y;
%function
Y = [0.0566 0.4432 0.5539 0.6783 0.7303 0.3569 0.4001 0.4278 0.4499 0.4720 ...
    0.4500 0.5157 0.5237 0.5492 0.5590 0.5799 0.5890 0.6000 0.6150 0.6300...
    0.6450];
D0 = 2.0*10^(-12);
options=optimset('LargeScale','on','Display','iter','MaxFunEvals',1e20,'TolFun',2e-50,'TolX',2e-50,'LevenbergMarquardt','on');
[D,resnorm,residual,exitflag,output,lambda]=lsqnonlin(@myfun,D0,[],[],options);

I am trying to get D value using the lsqnonlin() but I am getting the message:

Error using ==> feval
Undefined function or method 'diffusion' for input arguments of type 'double'.
Error in ==> lsqnonlin at 200
            initVals.F = feval(funfcn{3},xCurrent,varargin{:});
Caused by:
    Failure in initial user-supplied objective function evaluation. LSQNONLIN cannot continue.

Please help me with the code why I am getting this error message. What could be the better possible way to obtain the solution for D value.

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Answer 1

Matt J am 26 Jul. 2014

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https://de.mathworks.com/matlabcentral/answers/143390-using-lsqnonlin-for-solution-of-an-nonlinear-equation#answer_146502

Bearbeitet: Matt J am 26 Jul. 2014

I suspect that the code you are running is not the code you have shown. The posted myfun(D,Y) takes 2 arguments, but you are not passing a value for Y to myfun in any way. MATLAB should be complaining that Y is missing, unless your call to lsqnonlin looks something like this,

 [D,resnorm,residual,exitflag,output,lambda]=lsqnonlin(@(D) myfun(D,Y), D0,[],[],options);

Also, if D is a scalar, it would be probably be much simpler to use fminsearch instead of lsqnonlin to minimize norm(diff).

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KB am 26 Jul. 2014

Bearbeitet: Matt J am 26 Jul. 2014

Sorry to update the actual code:The actual one is as follows

function diff = diffusion(D,Y) 
t = 0:30:600;
l = length(t);
a = 0.5*0.75*10^(-4);
for i = 1:l
Yp(i)=0;
for n = 1:10
y(i,n) = exp(-D*t(i)*(pi^2)*(n^2)/(a^2))/n^2;
Yp(i) = Yp(i) + y(i,n);
end
end
y; 
Yp;
diff = 1- (6*Yp/pi^2)-Y;
%function
Y = [0.05698 0.1700 0.2398 0.2929 0.33510 0.3710 0.4101 0.4256 0.45016 0.47470 0.5010 0.51490 0.5303 0.54908 0.56506 0.5798 0.6012 0.6090 0.6190 0.63650 0.6514]; % experimental data
D0 = 1; 
options=optimset('LargeScale','on','Display','iter','MaxFunEvals',1e20,'TolFun',2e-50,'TolX',2e-50,'LevenbergMarquardt','on');
[D,Resnorm,residual,exitflag,output,lambda]=lsqnonlin(@(D)diffusion(D,Y),D0,[],[],options);
   D,Resnorm

which gives output:

Optimization terminated: first-order optimality less than OPTIONS.TolFun, and no negative/zero curvature detected in trust region model.

D =

1

Resnorm =

       5.8455
even if I change the D0 value, D value return same as D0 with similar Resnorm.
I have passed on the Y values as vector which is experimental data.

KB am 27 Jul. 2014

The term inside of exponential is unitless, D is in cm^2/s, a in micron or cm, t is s. However, when I am using the same code to solve for D value for another data set of Y.

function diff = diffusion(D,Y)

t = 0:30:600;

l = length(t);

a = 0.5*0.75*10^(-4);

for i = 1:l

Yp(i)=0;

for n = 1:10

y(i,n) = exp(-D*t(i)*(pi^2)*(n^2)/(a^2))/n^2;

Yp(i) = Yp(i) + y(i,n);

end

y;

Yp;

diff = 1- (6*Yp/pi^2)-Y;

%function

Y = [0.057854 0.24074 0.33023 0.39484 0.44657 0.49006 0.52771 0.56093 0.59065 0.61749 0.64194 0.66434 0.68496 0.70403 0.72172 0.73818 0.75352 0.76786 0.78127 0.79384 0.80563];

D0 = 0;

options=optimset('LargeScale','on','Display','iter','MaxFunEvals',1e20,'TolFun',2e-50,'TolX',2e-50,'LevenbergMarquardt','on');

[D,Resnorm,residual,exitflag,output,lambda]=lsqnonlin(@(D)diffusion(D,Y),D0,[],[],options);

   D,Resnorm
 it shows:
                                       Norm of      First-order 
 Iteration  Func-count     f(x)          step          optimality   CG-iterations
     0          2         6.66559                     7.02e+008
     1          4         3.50381   8.77753e-009              0            1
Optimization terminated: first-order optimality less than OPTIONS.TolFun,
 and no negative/zero curvature detected in trust region model.

D =

8.7775e-009

Resnorm =

3.5038

This D value supposed to give similar value in the range of 10^(-12). I tried to do it increasing/decreasing the Tolfun and TOlx, but could not solve the problem. The same code worked nice with previous data set but not for this one, although the new Y data set is little different than previous one.

KB am 30 Jul. 2014

hello, thanks, I have another problem. Recently, I have installed Matlab 2012a in my personal computer and trying to run the same code. Surprisingly, it is not running with 'LevenbergMarquardt' one, so adopted what you suggested. Still it is not running and shows following error:

Undefined function 'isfinite' for input arguments of type 'diffusion'.

Error in /Applications/MATLAB_R2012a.app/toolbox/shared/optimlib/finDiffEvalAndChkErr.p>finDiffEvalAndChkErr (line 35)

Error in /Applications/MATLAB_R2012a.app/toolbox/shared/optimlib/finitedifferences.p>finitedifferences (line 128)

Error in sfdnls (line 89) [J,~,~,numEvals] = finitedifferences(x,fun,[],lb,ub,valx,[],[], ...

Error in snls (line 178) [A,findiffevals] = sfdnls(xcurr,fvec,Jstr,group,[],funfcn{3},l,u,...

Error in lsqncommon (line 175) [xC,FVAL,LAMBDA,JACOB,EXITFLAG,OUTPUT,msgData]=...

Error in lsqnonlin (line 235) [xCurrent,Resnorm,FVAL,EXITFLAG,OUTPUT,LAMBDA,JACOB] = ...

Error in diffusion_call (line 4) [D,Resnorm,residual,exitflag,output]=lsqnonlin(@(D)diffusion(D,Y),D0,[],[],options);

The same code worked well in past but i have no idea it is not running in my computer.

KB am 30 Jul. 2014

Hello Matt, are you using the same code what I am using in R2011a

function diff = diffusion(D,Y)

t = 0:30:600;

l = length(t);

a = 0.5*0.76*10^(-4);

for i = 1:l

Yp(i)=0;

for n = 1:10

y(i,n) = exp(-D*t(i)*(pi^2)*(n^2)/(a^2))/n^2;

Yp(i) = Yp(i) + y(i,n);

end

y;

Yp;

diff = 1- (6*Yp/pi^2)-Y; %% function

Y = [0.0568 0.2376 0.3219 0.4102 0.4487 0.5023 0.5589 0.5787 0.6104 0.6119 0.6418 0.666 0.7015 0.7111 0.7301 0.74109 0.7621 0.7876 0.7901 0.8015 0.81826]; D0 = 0;

options=optimset('Display','iter','MaxFunEvals',1e20,'TolFun',2e-50,'TolX',2e-50);

[D,Resnorm,residual,exitflag,output]=lsqnonlin(@(D)diffusion(D,Y),D0,[],[],options);

D,Resnorm The output is:

                                         Norm of      First-order 
 Iteration  Func-count     f(x)          step          optimality   CG-iterations
     0          2         6.87754                     7.13e+008
     1          4         3.38556   8.91612e-009              0            0

Local minimum found.

Optimization completed because the size of the gradient is less than the selected value of the function tolerance.

criteria details

D =

8.9161e-009

Resnorm =

3.3856

Matt J am 30 Jul. 2014

Bearbeitet: Matt J am 30 Jul. 2014

Your code, and its output, are difficult to read. You should use the

button to format it distinctly from your text (like my posted code appears). In any case, the code you've shown is not what I ran. I'm still applying lsqnonlin to the scaled version of the problem:

fun=@(D)diffusion(D,Y);
[Dsc,Resnorm]=lsqnonlin(@(Dsc) fun((1e-10)*Dsc),D0,0,[],options);
 D=Dsc*1e-10;

KB am 1 Aug. 2014

Bearbeitet: KB am 1 Aug. 2014

Thanks Matt. You have been really awesome. Could you help me out with this data set:

and I am using similar code:

    function diff = kani(D,Y) 
    t = 0:120:7200;
    l=length(t);
    a = 0.5*0.75*10^(-4); 
    for i = 1:l
        Yp(i)=0;
        for n = 1:10
            y(i,n) = exp(-D*t(i)*(pi^2)*(n^2)/(a^2))/n^2;
            Yp(i) = Yp(i) + y(i,n);
        end
    end
    y; 
    Yp;
    diff = 1- (6*Yp/pi^2)-Y;
Y =    [ 0 0.0035 0.0066 0.0091 0.0121 0.0143 0.0150 0.0161 0.0167 0.0174 0.0188 0.0197 0.0199 0.0210 0.0218 0.0226 0.0229 0.0226 0.0237 0.0247 0.0244 0.0262 0.0249 0.0252 0.0248 0.0267 0.0285 0.0272 0.0279 0.0277 0.0292 0.0294 0.0289 0.0301 0.0282 0.0285 0.0299 0.0285 0.0304 0.0306 0.0309 0.0315 0.0317 0.0310 0.0310 0.0304 0.0339 0.0326 0.0320 0.0323 0.0335 0.0312 0.0332 0.0333 0.0316 0.0311 0.0315 0.0307 0.0313 0.0312 0.0328];
fun=@(D)kani(D,Y);
      D0 = 0;
      options = optimset('Display','iter','MaxFunEvals',1e20,'TolFun',2e-50,'TolX',2e-50);
      [Dsc,Resnorm]=lsqnonlin(@(Dsc) fun((1e-10)*Dsc),D0,0,[],options);
      D=Dsc*1e-10 
      Resnorm
      D_interval=linspace(0,2*D,1000);
      plot(D_interval, arrayfun(@(D)norm(fun(D))^2,D_interval));
      xlabel 'D'
      ylabel 'norm(diff)^2' }

and this is what I got:

I have tried to solve this by scaling the D value very low e^(-20), but still not getting it.

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using lsqnonlin () for solution of an nonlinear equation

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