In a practical experiment, we have two independent variables, $x>0$ and $y \ge 0$, and one dependent variable, $z \ge 0$.
Theoretically, we know that at any fixed $x$, we have the relation between $y$ and $z$ is linear without intercept, and the slope changes as $x$ changes. But practically, due to experimental errors of measurements, we have intercept that is not identically zero.
Also theoretically, we know that the slope, $A$, as a function of $x$ is given by $A(x)=a_1+a_2e^{a_3x}$.
So, we can conclude that $z(x,y)=(a_1+a_2e^{a_3x})y$.
After implementing the experiment, we found that the intercept, $B$, as a function of $x$ is given by $B(x)=b_1+b_2e^{b_3x}$
My thinking to evaluate $z$ at any given $x,y$ should be of the form:
$z(x,y)=(a_1+a_2e^{a_3x})y+(b_1+b_2e^{b_3x})$
However, that will not satisfy the theoretical condition that for any fixed $x$, we have a linear relation between $y$ and $z$.
So to summarize my problem:
From the experiment, having $x$-Slope relation being of the form $A(x)=a_1+a_2e^{a_3x}$, and having $x$-Intercept relation being of the form $B(x)=b_1+b_2e^{b_3x}$,
what is the best model for $z$ in terms of $x$ and $y$ so that, forcefully, $z(x,0)=0$?
Considering $z(x,y)=(a_1+a_2e^{a_3x})y$ only will have big error for extreme values of $y$.
See the following data, as an example ONLY:
EDIT: Adding Charts:
This shows that for a fixed $x$, we have a linear relation between $y$ and $z$. This relation should be without an intercept. But due to some reasons (not mathematical reasons but experimental issues of the devices not well calibrated, and so on), then unfortunately getting intercept for each $x$.
The blue points, represent the slope at different values of $x$, and that is very fine as the theory of the experiment say. It may look decreasing little for higher values of $x$, but that is also due to measurement error and it should be:
$$\text{slope}=k_1+k_2e^{k_3x}$$
The red points, represent the intercept at different values of $x$, and that is not fine as the theory of the experiment say, it should be always at $0$. But as I mentioned earlier that this intercept we are getting only because of calibration needed. However, the intercept values are nicely decreasing, and not randomly scattered. And following the same form of the slope, lets say the form is:
$$\text{intercept}=j_1+j_2e^{j_3x}$$
Now the required is to follow these $4$ things:
$1.$ For any fixed $x$, we have a linear relation between $y$ and $z$ (that theoretically should be without intercept.
$2.$ In the $y-z$ relation, the slope is increasing only and has an upper bound as $x$ increases. Its form is given above.
$3.$ In the $y-z$ relation, the intercept (which should not be there theoretically, but practically) is decreasing only and has a lower bound as $x$ increases. Its form is given above.
$4.$ Even though we, theoretically, have no intercept, we should keep it in consideration as we are getting it practically. But the consideration should not make $z(x,0)$ to be nonzero. So $z(x,0)$ must be identically (not approximately) $0$.
Your help would be appreciated. THANKS!
As already observed the condition $z(x,0)=0$ cannot be satisfied with an equation on the form : $$z(x,y)=(a_1+a_2e^{a_3x})y+(b_1+b_2e^{b_3x})$$ except if $b_2=0$ which isn't acceptable to fit to the data.
This means that the function $z(x,y)$ is not limear wrt $y$. So one look for an equation not linear wrt $y$ but which is almost linear on a limited range ( on the range $0.9<y<4$ in the present case). A possible way is to replace $(b_1+b_2e^{b_3x})$ by $(b_1+b_2e^{b_3x})f(y)$ with a function $f(y)\simeq 1$ on the useful $y$ range and $f(y)\to 0$ for $y\to 0$.
Moreover one observes that $z(x,y)$ is slightly decreasing with increasing large $x$ at constant $y$. This draw to replace $(b_1+b_2e^{b_3x})$ by $(b_1+b_2x+b_3e^{b_4x})$ with $b_2<0$. The proposed combination becomes even more complicated : $$z(x,y)=(a_1+a_2e^{a_3x})y+(b_1+b_2x+b_3e^{b_4x})f(y)$$
With several exponentials and so many parameters the nonlinear regression will be problematic.
A simplified version is shown below (in setting $b_4=a_3$ which is rather arbitrary nevertheless leading to a preliminaty result).