Fitting exponential using derivatives

Hello everyone!

I have a question regarding fitting data. I have an example data set

time=linspace(0,200,20);
r=linspace(0,1,20);

where r is a function of time.

And i have to fit the data set to the model equation

y=a*1.5^(c+d)*(5-r)^c*(3-r)^d

however, i do not have the y values, so I have no idea how to solve it.

I started with calculating derivatives and then linearizing it with logarithms

dt = diff(time);               
dr = diff(r);             
dr_dt = dr ./ dt; 

ln(dr_dt) = ln(a) + (c+d)*ln(1.5) + c*ln(5−r) + d*ln(3−r)

However, I have no how to calculate a, c and d values. Any ideas?

[Walter Roberson]

where t is a function of x.

Before and after that statement, you never refer to x or t again, so it is difficult to understand the relevance of that statement.

[Walter Roberson]

What exactly do you have? You have time, and what else?

[CherryLady]

Yes, you are right, I have changed the notation already, thanks!

[CherryLady]

r is the concentration of a reagent, y would be reaction kinetic expression

[Matt J]

ln(dr_dt) = ln(a) + (c+d)*ln(1.5) + c*ln(5−r) + d*ln(3−r)

This only follows mathematically if y=dr_dt, which you haven't expressly mentioned.

[Walter Roberson]

y would be reaction kinetic expression

But you do not have measured reaction kinetic expressions values ? What do you have measured? You have measured time; you have implied r values based on time... what else?

[Sam Chak]

Hi @CherryLady

Actually, I'm scared of seeing the power operator in differential equations. If one is not careful, improper computations can yield complex values. Even for attentive individuals, if the exponent parameters 'c' or 'd' are not integers but floating-point numbers, a negative real base can also produce complex values.

I also noticed that you intend to perform numerical differentiation using the operation diff(r)./diff(t) and then attempt to fit the model to the ​ data. Please note that there may be a discrepancy between the analytical and numerical results at high slopes.

You can also use this toy problem to test whether you can estimate the parameters using the fit() or lsqcurvefit() commands.

%% differential equation with "unknown" parameters
a       = 1;    % parameter 1
c       = 2;    % parameter 2
d       = 3;    % parameter 3
dr      = @(t, r) a*1.5^(c + d)*(5 - r).^c.*(3 - r).^d;

%% artificially generate data (about 50 points)
numpts  = 50;
tspan   = linspace(0, 10, numpts+1);
r0      = 4;    % initial value
[t, r]  = ode45(dr, tspan, r0);

%% create finer sampling points via Akima-styled interpolation
tq      = linspace(0, 10, 10*numpts+1);     % 10 times finer
rq      = interp1(t, r, tq, 'makima');

%% compare actual data and interpolated data
figure
plot(t, r, ':o', tq, rq), grid on
legend('Actual data', 'Interpolated data')
xlabel('t'), ylabel('r(t)')

%% perform numerical differentiation
drq = gradient(rq)./gradient(tq);
figure
hold on
plot(t, dr(t, r), ':o')
plot(tq, drq), grid on
hold off
legend('analytical dr', 'numerical dr', 'location', 'east')
xlabel('t'), ylabel('dr(t)')
axis([-2, 10, -8, 1])

Answer 1

[Matt J]

however, i do not have the y values

If y=dr_dt, then you can approximate y numerically,

y=gradient(r,t)

However, it be advisable to instead fit r(t) with a smoothing spline, and differentiate it analytically, e.g., with fnder.

Answer 2

[Torsten]

As far as I understand, your model is a differential equation

dr/dt = a*1.5^(c+d)*(5-r)^c*(3-r)^d, r(0) = r0

with unknown parameters a, c and d.

Further, you have measurements for r over time: (t0,r0),(t1,r1),...,(tn,rn).

In order to determine a, c and d, you will have to couple a parameter fitting tool (e.g. lsqcurvefit) with an integrator for ordinary differential equations (e.g. ode15s).

Take a look at the example

https://uk.mathworks.com/matlabcentral/answers/43439-monod-kinetics-and-curve-fitting

and StarStrider's answer code to learn how to proceed.