Getting independent equations by differentiation My question is related to the procedure used in the following paper: HEP paper. No need to even look into the physics, this question is strictly related to the solving of equation (27):
$$\lambda_{1}^2 \, e^{-m^2_{1}\tau}+
\lambda_{2}^2 \, e^{-m^2_{2}\tau}
=
f(\tau).  \;\;\;\; (1)$$
This $f(\tau)$ on the right side is some "well behaved" function of $\tau$ (infinitely differentiable, continuous), called $\Pi^{QCD}(\tau)$ in the paper. The authors are interested in finding $m_1$ and $m_2$ without knowing $\lambda_1$ and $\lambda_2$ (none of these four unknowns appears in $f(\tau)$).
They proceed to differentiate both sides of that equation (more than once) in relation to $\tau$ assuming all $m$ and $\lambda$ to be independent of $\tau$ and then divide the obtained equations by each other until the finally isolate $m_1$ and $m_2$ (see equations (27) to (39)). So, the first differentiation gives us:
$$-m^2_{1}\lambda_{1}^2 e^{-m^2_{1}\tau}-m^2_{2}\lambda_{2}^2 e^{-m^2_{2}\tau}
= \frac{d}{d\tau}f(\tau). \;\;\;\; (2)$$
Which together with (1), allow us to write ($Df$ means differentiation in relation to $\tau$):
$$\lambda_{1}^2 e^{-m^2_{1}\tau}=\frac{
Df(\tau) + f(\tau)\,m_{2}^2
}
{
m_{2}^2 -m_{1}^2  
},\;\;\;\; (3)$$
$$\lambda_{2}^2 e^{-m^2_{2}\tau}=
\frac{
Df(\tau) + f(\tau)\,m_{1}^2
}
{
m_{1}^2 -m_{2}^2
}.\;\;\;\; (4)$$
We can get rid of the $\lambda$'s by taking the derivative of any of the two previous equations and dividing the result by the original:
$$m_{1}
=
\sqrt{
-\frac{
Df(\tau)\,m_{2}^2 +D^2f(\tau)
}
{
Df(\tau) + f(\tau)\,m_{2}^2
}
},\;\;\;\; (5)$$
$$m_{2}=
\sqrt{
-\frac{
Df(\tau)\,m_{1}^2+ D^2f(\tau)
}
{
Df(\tau)+ f(\tau)\,m_{1}^2
}
}.\;\;\;\; (6)$$
If we take the derivative of equation (4) twice:
$$
m^4_{2}\lambda_{2}^2 e^{-m^2_{2}\tau}=
\frac{
D^3f(\tau) + D^2f(\tau)\,m_{1}^2
}
{
m_{1}^2 -m_{2}^2
},\;\;\;\; (7)$$
and divide (7) by (4), we get:
$$m^4_{2}\alpha(\tau)
+
m^2_{2}\beta(\tau)
+
\gamma(\tau)
=
0, \;\;\;\; (8)$$
where:
$$\alpha(\tau)=-Df(\tau)^2
+ f(\tau)\,D^2f(\tau)$$
$$\beta(\tau)=-D^2f(\tau)\,Df(\tau)
+D^3f(\tau)\,f(\tau)$$
$$\gamma(\tau)=D^3f(\tau)\,Df(\tau)
-D^2f(\tau)^2$$
Eq. (8) can easily be solved for $m_2$ and the same trick gets us the same equation for $m_1$. The answer for both $m_1$ and $m_2$ ends up being given as a function in $\tau$, but they look for intervals in $\tau$ where that function is flat (justifying the assumption that the $m$'s are $\tau$-independent).
Even assuming that (a) such interval with $\tau$ independence can be found, and (b) one takes care of avoiding divisions by zero along the way, this procedure still "smells bad". One starts from an equation with four unknowns and ends up with a determined solution for at least two of them. 
Under which conditions can such a procedure work (if at all)? 
 A: This is halfway between a comment and an answer.. If it does not satisfy you I'll just delete it.
First of all: It's not quite that easy to understand when systems of equations can be solved. Take for example the equation 
$$x_1^2 + \dots + x_n^2 = 0$$.
If $x_i \in \mathbb R$, then we conclude $x_i = 0 \ \ \forall i$. One equation, arbitrarily many unknown, we found them all.
But the important aspect is that yours is not an equation, it's an equivalence between two functions.
I mean an equation is something like $x^2 = 25$, it's true only for two values of $x$.
An equivalence is something like $$e^x = \sum \frac{x^n }{n!}$$
which is valid for every $x \in \mathbb R$.
With equivalence, you don't have only one condition, one constraint: you have infinitely many! it has to hold for every point!
So technically in your example you could get $4$ conditions (enough to find $m_1, m_2, \lambda_1, \lambda_2$) simply by plugging into $\tau = 1, 2, 3, 4$ or any other $4$ values.
On the other hand, there are too many conditions! Once you found those values, the relation should hold for $\tau = \sqrt 2$ or any other value! It may be that there are no $m_i$ such that it is true. You have "infinite" equations (constraints) and only $4$ unknowns (degrees of freedom). In general this has no solution, but it depends 
And in fact is what happens in this case; why didn't they just put $4$ values of $\tau$ and solved the system? Because the system has no solution, it would not hold for different values of $\tau$ than the $4$ one you choose. 
So you have to admit that $m_i$ depend on $\tau$, too. Now you have infinite constraints (like before) but also infinite degrees of freedom (every value that $m_1(\tau)$ can take, for example).
Still, there not need to be necessarily a solution, and this is getting into solving functional equations which are notoriously difficult to deal with, so what now?
The idea is to treat $m_i$ are constant, and justify that because we look into an interval when they really are constant. I have no idea how well this works, what are the errors you introduce, and in what sense what you found is a solution to the original equation, but those are probably different (and difficult) questions :-) 
