If I have the data points for the Morse Potential '$V(r)$' at given distance '$r$' of the Morse Potential described by: $V(r) = D[exp(-2\alpha(r-r_{eq}))-2exp(-\alpha(r-r_{eq}))]$, is it possible to find the parameters D,$\alpha$ and $r_{eq}$, using the least squares method?
The sample data looks like this
$V(r)$=[-1360.12,-1368.53,-1374.21,-1378.09,-1380.65,-1382.22,-1383.09,-1383.48,-1383.56,-1383.45,-1383.22,-1382.93,-1382.61,-1382.27,-1381.94,-1381.62,-1381.32,-1381.04,-1380.77,-1380.53,-1380.31,-1380.1,-1379.91,-1379.73,-1379.58,-1379.43,-1379.3,-1379.18,-1379.08,-1378.98,-1378.90,-1378.83,-1378.77,-1378.71,-1378.67,-1378.63,-1378.6,-1378.57,-1378.55,-1378.54,-1378.54,-1378.54,-1378.55,-1378.57,-1378.6,-1378.63,-1378.67,-1378.71,-1378.77,-1378.83]
r = [1.0 ,1.1 ,1.2 ,1.3 ,1.4 ,1.5 ,1.6 ,1.7 ,1.8 ,1.9 ,2.0 ,2.1 ,2.2 ,2.3 ,2.4 ,2.5 ,2.6 ,2.7 ,2.8 ,2.9 ,3.0 ,3.1 ,3.2 ,3.3 ,3.4 ,3.5 ,3.6 ,3.7 ,3.8 ,3.9 ,4.0 ,4.1 ,4.2 ,4.3 ,4.4 ,4.5 ,4.6 ,4.7 ,4.8 ,4.9 ,5.0 ,5.1 ,5.2 ,5.3 ,5.4 ,5.5 ,5.6 ,5.7 ,5.8 ,5.9]
You cannot have a good fitting with the function $$V(r) = D[e^{-2\alpha(r-r_{eq})}-2e^{-\alpha(r-r_{eq})}]$$ because a constant term is missing.
Better use the function :
$$V(r) = a+D[e^{-2\alpha(r-r_{eq})}-2e^{-\alpha(r-r_{eq})}] \tag 1$$ where $a$ is a constant.
The fitting for the parameters $D,\alpha,r_{eq},a$ can be done thanks to non-linear regression. In fact this is least squares method but iterative. The principle can be found in many papers on the web, for example https://mathworld.wolfram.com/NonlinearLeastSquaresFitting.html
Of course non-linear robust fitting is not easy to implement. In practice one uses available softwares.
Note that the above function is a particular case of the more general function made of two exponentials : $$V(r)=a+b\,e^{p\,r}+c\,e^{q\,r} \tag 2$$
Eq.$(2)$ is related to $(1)$ in particular case of $\quad p=-2\alpha \quad,\quad q=-\alpha\quad,\quad b=De^{2\alpha r_{eq}} \quad,\quad c=-2De^{\alpha r_{eq}}$
Thus the above numerical values of the parameters convenient for Eq.$(2)$ are not valid for Eq.$(1)$. If we require $p=2q$ the fitting will be slightly different (See at end).
FOR INFORMATION.
The above numerical values where computed thanks to the method explained in : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales
This method transforms the non-linear equation to a linear equation (wrt the parameters) thanks to an integral equation. In the present case the non-iterative calculus is shown below :
CASE OF EQUATION $(1)$ which is equivalent to :
$$V(r)=a+b\,e^{-2\alpha\,r}+c\,e^{-\alpha\,r} \tag 2$$
$b=De^{2\alpha r_{eq}} \quad,\quad c=-2De^{\alpha r_{eq}}\quad\implies\quad D=\frac{b^2}{4c}\quad,\quad r_{eq}=-\frac{1}{\alpha}\ln(-\frac{b}{2c}).$
A non-linear least squares regression is necessary only due to one parameter $\alpha$. The iterative calculus is simple and very fast starting from a rather good initial value $\alpha_{init}=-\frac{2p+q}{4}\simeq 1.4\quad$ from the values of $p$ and $q$ already known from the above calculus. The result is : $$\alpha\simeq 1.433$$ Finally the linear least squares regression for $a,b,c$ , then the calculus of $D$ and $r_{eq}$ gives :