Can i tell numpy curve_fit to find the best parameters that meet some conditions?

Question

I have this set of experimental data:

x_data = np.array([0, 2, 5, 10, 15, 30, 60, 120])
y_data = np.array([1.00, 0.71, 0.41, 0.31, 0.29, 0.36, 0.26, 0.35])

t = np.linspace(min(x_data), max(x_data), 151)

scatter plot

I want to fit them with a curve that follows an exponential behaviour for t < t_lim and a linear behaviour for t > t_lim, where t_lim is a value that i can set as i want. I want to use curve_fit to find the best fit. I would like to find the best fit meeting these two conditions:

1. The end point of the first behaviour (exponential) must be the starting point of the second behaviour (linear): in other words, I don't want the jump discontinuity in the middle.
2. I would like the second behaviour (linear) to be descending.

I solved in this way:

t_lim = 15    

def y(t, k, m, q):
    return np.concatenate((np.exp(-k*t)[t<t_lim], (m*t + q)[t>=t_lim]))

popt, pcov = curve_fit(y, x_data, y_data, p0=[0.5, -0.005, 0.005])

y_model = y(t, k_opt, m_opt, q_opt)

I obtain this kind of curve: chart_plot

I don't know how to tell python to find the best values of m, k, q that meet the two conditions (no jump discontinuity, and m < 0)

Answer 1

Instead of trying to add these conditions as explicit constraints, I'd go about modifying the form of y so that these conditions are always satisfied.

For example, try replacing m with -m**2. That way, the coefficient in the linear part will always be negative.

For the continuity condition, how about this: For an exponential with a given decay factor and a linear curve with a given slope which are supposed to meet at a given t_lim there's only exactly one value for q that will satisfy that condition. You can explicitly compute that value and just plug that in.

Basically, q won't be a fit parameter anymore; instead, inside of y, you'd compute the correct q value based on k, m, t_lim.

Answer 2

This post is not a direct answer to the question. This is a preliminary study.

First : Fitting to a simple exponential function with only a constant (without decreasing or increasing linear part) :

The result is not bad considering the wide scatter on the right part.

Second : Fitting to an exponential function with a linear function (without taking account of the expected decreasing on the right).

The slope of the linear part is very low : 0.000361

But the slope is positive which is not as wanted.

Since the scatter is very large one suspects that the slope of the linear function might be governed mainly by the scatter. In order to check this hypothesis one make the same fitting calculus whitout one point. Taking only the seven first points (that is forgetting the eighth point) the result is :

Now the slope is negative as wanted. But this is an untruthful result.

Of course if some technical reason implies that the slope is necessarily negative one could use a picewise function made of an exponenlial and a linear function. But what is the credibility of such a model ?

This doesn't answer to the question. Neverthelss I hope that this inspection will be of interest.

For information :

The usual nonlinear regression methods are often non convergent in case of large scatter due to the difficulty to set initial values of the parameters sufficienly close to the unknown correct values. In order to avoid the difficulty the above fittings where made with a non usual method which doesn't requires "guessed" initial value. For the principle refer to : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales

In the referenced document the case of the function exponential and linear isn't fully treated. In order to overcome this deficiency the method is shown below with the numerical calculus (MathsCAD).

If more accuracy is needed use a nonlinear regression software with the values of p,a,b,c found above as initial values to start the iterative calculus.