What is the formula to convert nonlinear bi-exponential decay function to linear by log transform

Question

My apologies if i have multiple questions in a single thread, but as they are all related, i want to ask them as one question. I have the following questions:

1) what will be the log transform (in terms of equation) of the following equation:

$$y = a_0\exp\left(-\frac x{a_1}\right) + (1-a_0)\exp\left(-\frac x{a_2}\right) + a_3\tag{1}$$ So the answer that i am looking for is what will be of the form $$\ln y=\ldots\tag{2}$$

I want to use this formula for fitting a bi-exponential decay model by converting nonlinear problem to linear.

for the following data \begin{array}{c|c} X&Y\\ \hline 3.3 &1\\ 8.5 &0.6898\\ 13.7 &0.3519\\ 18.9& 0.2593\\ 24.1& 0.2731\\ 29.3& 0.1435\\ 34.5& 0.1574\\ 39.7& 0.0972\\ 44.9& 0.0787\\ 50.1& 0.0741\\ 55.3& 0.0741\\ 60.6& 0.0278\\ 65.7& 0.0648\\ 70.9& 0.0556\\ 76.1& 0.0463\\ \end{array}

2) Also which function (lsqnonneg,lsqlin) should i use after conversion of nonlinear eq'n to linear with logarithmic values in MATLAB?

3) how does upper and lower bounds will change after logarithmic conversion ?

If this approach is not advisable then how can i choose upper and lower bounds for such an equation $(1)$.

Answer 1

You can't do the transform you are looking for. You can take the $a3$ term out by making $y'=y-a3$ but you still have a nonlinear fit. You can use a multidimensional function minimizer to estimate the parameters.

Answer 2

As said, you cannot linearize this model.

What you could do is to rewrite the model as $$y = a_0\left(\exp\left(-\frac x{a_1}\right)-\exp\left(-\frac x{a_2}\right)\right) + \exp\left(-\frac x{a_2}\right) + a_3$$ Now make a grid of values of $a_1$ and $a_2$; these being fixed, the model is linear with respect to $a_0$ and $a_3$.

Try to find a pair $(a_1,a_2)$ corresponding more or less to a minimum of the sum of squares. When you got it, you have all elements to start a full nonlinear regression.

Edit

Using your data, I built a grid $a_1=0.25 i$, $a_2=0.25 j$, $i$ varying from $1$ to $100$ and $j$ varying from $i+1$ to $100$ and, for each pair, I computed the sum of squares. Over that grid, the minimum value I obtained did correspond to $a_1=1$ and $a_2=11$ and, corresponding to these, the linear least-square regression was giving $a_0=-0.28$ and $a_3=0.06$. Using these four values as initial estimates, the non linear regression leads to $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} \\ a_0 & -0.2912 & 0.19288 \\ a_1 & +1.0575 & 2.93116 \\ a_2 & +10.904 & 1.64133 \\ a_3 & +0.0596 & 0.01726 \\ \end{array}$$ which is not very good.

The predicted values are $$\{1.0007,0.6516,0.4271,0.2877,0.2012,0.1475,0.11415,0.0934,0.0806, 0.0726,0.0677,0.0645,0.0627,0.0615,0.0608\}$$

Answer 3

$$y = a_0\exp\left(-\frac x{a_1}\right) + (1-a_0)\exp\left(-\frac x{a_2}\right) + a_3\tag{1}$$ An easy method (no iterative calculus, no initial guess) is explained in the paper

https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales

The case of functions made of a sum of exponentials is treated pages 71-73 with a numerical example in case of two exponentials. The notations are different from the notations in the above equation. This can be confusing. To make it clear the scheme of the method is re-edited below in the more general case of five parameters : $$y=a+b\exp(px)+c\exp(qx)\tag 2$$ with the related parameters : $$a=a_3\quad;\quad b=a_0\quad;\quad c=(1-a_0)\quad;\quad p=-\frac{1}{a_1}\quad;\quad q=-\frac{1}{a_2}$$

In fact, equation $(2)$ is not equivalent to equation $(1)$. They respectively involve 5 and 4 independent parameters since $c=1-b$. Equation $(1)$ is equivalent to equation $(3)$ : $$y=a+b\exp(px)+(1-b)\exp(qx)\tag 3$$

The above calculus have to be modified. After updating for equation $(3)$ the result appears not acceptable. The fitting is not good. This confirms the result from Claude Leibovici.

This draw to think that equation $(3)$ is not a good model to fit to the given data. Equation $(2)$ is recommended.