# Nonlinear fitting: how do I split the linear and the nonlinear problems?

7 views (last 30 days)

Show older comments

Samuele Bolotta
on 25 Mar 2021

Commented: Alan Weiss
on 25 Aug 2021

I am fitting a function to some data I simulated. I managed to get intelligent constraints that help the fit quite a bit, even with a lot of noise.

This is the function and as you can see, c(1) and c(2) are linear, while lam(1), lam(2), lam(3) and lam(4) are nonlinear. I am following the procedure explained here (https://it.mathworks.com/help/optim/ug/nonlinear-data-fitting-problem-based-example.html#NonlinearDataFittingProblemBasedExample-4) to split linear and nonlinear parameters.

% Create a function that computes the value of the response at times t when the parameters are c and lam

diffun = ((c(1)) .* ((1 - exp(-t / lam(1))) .* exp(-t / lam(2))) * (Vm - (-70)) + ...

((c(2)) .* ((1 - exp(-t / lam(3))) .* exp(-t / lam(4))) * -30));

This is the code that I came up with, but for some reason it's not working. To generate the data:

function [EPSC, IPSC, CPSC, t] = generate_current(G_max_chl, G_max_glu, EGlu, EChl, Vm, tau_rise_In, tau_decay_In, tau_rise_Ex, tau_decay_Ex,tmax)

dt = 0.1; % time step duration (ms)

t = 0:dt:tmax-dt;

% Compute compound current

IPSC = ((G_max_chl) .* ((1 - exp(-t / tau_rise_In)) .* exp(-t / tau_decay_In)) * (Vm - EChl));

EPSC = ((G_max_glu) .* ((1 - exp(-t / tau_rise_Ex)) .* exp(-t / tau_decay_Ex)) * (Vm - EGlu));

CPSC = IPSC + EPSC;

end

To fit the function:

% Simulated data

[EPSC,IPSC,CPSC,t] = generate_current(80,15,0,-70,-30,0.44,15,0.73,3,120);

ydata = awgn(CPSC,25,'measured'); % Add white noise

% Values

Vm = -30;

% Initial values for fitting

gmc = 40; gmg = 20; tde = 0.2; tdi = 8; tre = 1.56; tri = 3;

% Objective function

c = optimvar('c',2); % Linear parameters

lam = optimvar('lam',4); % Nonlinear parameters

% Bounds

c.LowerBound = [0, 0];

c.UpperBound = [200, 200];

lam.LowerBound = [0.16,7.4,1.1,2.6];

lam.UpperBound = [0.29,8.4,2.3,3.3];

x0.c = [gmc,gmg]; % Starting values

x0.lam = [tri,tdi,tre,tde]; % Starting values

% Create a function that computes the value of the response at times t when the parameters are c and lam

diffun = ((c(1)) .* ((1 - exp(-t / lam(1))) .* exp(-t / lam(2))) * (Vm - (-70)) + ...

((c(2)) .* ((1 - exp(-t / lam(3))) .* exp(-t / lam(4))) * -30));

%Solve the problem using solve starting from initial point x02

x02.lam = x0.lam;

%To do so, first convert the fitvector function to an optimization expression using fcn2optimexpr.

F2 = fcn2optimexpr(@(x) fitvector(x,t,ydata),lam,'OutputSize',[length(t),1]);

% Create a new optimization problem with objective as the sum of squared differences between the converted fitvector function and the data y

ssqprob2 = optimproblem('Objective',sum((F2' - ydata).^2));

[sol2,fval2,exitflag2,output2] = solve(ssqprob2,x02)

% Plot

resp = evaluate(diffun,sol2);

hold on

plot(t,resp)

hold off

The error is:

Solving problem using lsqnonlin.

Error using optim.problemdef.OptimizationProblem/solve

Matrix dimensions must agree.

Error in SplittFit (line 37)

[sol2,fval2,exitflag2,output2] = solve(ssqprob2,x02)

Caused by:

Failure in initial objective function evaluation. LSQNONLIN cannot continue.

Not sure why.

##### 0 Comments

### Accepted Answer

Alan Weiss
on 25 Mar 2021

You should follow the example more closely. In the example the lambda variables only are declared to be optimization variables; the c variables are not, and are computed by backslash for given values of the lambda variables in the fitvector function.

For your case you will need to update the fitvector function from the example to handle a 4-D lambda vector that differs from your 4-D function because you have some (1-exp(-t/lambda)) terms, not just exp(-t/lambda). You need to write out the linear equations in c and solve those equations in the fitvector function.

Good luck,

Alan Weiss

MATLAB mathematical toolbox documentation

##### 4 Comments

Yanxin Liu
on 25 Aug 2021

Alan Weiss
on 25 Aug 2021

The response function can have any form. For complicated examples, see Fit ODE, Problem-Based and Fit an Ordinary Differential Equation (ODE), which have objective functions given by the solution of ODEs.

By no means are you required to split the response into a linear and a nonlinear part; when you can, then you can use the trick of fitting the linear part after fitting the nonlinear part.

Alan Weiss

MATLAB mathematical toolbox documentation

### More Answers (0)

### See Also

### Categories

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!