### Problem


I am trying to invert an equation of the form:


$R(l\_0)=(\int\_{0}^{l\_0} \rho(x) \, dx)(\int\_{l\_0}^{l} \rho(x) \, dx)$
where $0\leq l\_0 \leq l$


I.e. I want to find $\rho(x)$ given $R(l\_0)$ via some kind of integral transform, series solution, or discretized approximation. What I do not want to do is just blind curve fitting. Everything is real and positive. Is this possible, and if so, what should technique should I use for a solution? I suspect it may be helpful to convert this into some kind of product of two integral transforms where the kernels are Heaviside step functions. That might look like this:


$R(l\_0)=(\int\_{0}^{l} K\_1(x,l\_0)\rho(x) \, dx)(\int\_{0}^{l} K\_2(x,l\_0)\rho(x) \, dx)$


where $K\_1(x,l\_0) = H(l\_0-x)$ and $K\_2(x,l\_0) = H(x-l\_0)$.


### What I have done so far:


* I have looked in the [*Handbook of Integral Equations*](https://hupaa.com/book/HANDBOOK_OF_INTEGRAL_EQUATIONS.pdf) and think it has to do with difference kernels on finite intervals, but none of their integrals take the form of the product of two integral equations, even in the nonlinear sections as far as I can tell.
* I have tried to find double integrals with two kernels like in [this paper](https://www.jstor.org/stable/2037988?seq=1), but it seems like this is just proving the existence of something and not actually solving anything. Is this better to solve? $R(l\_0)=\int\_{0}^{l}\int\_{0}^{l}K\_1(x,l\_0)\rho(x)K\_2(y,l\_0)\rho(y) dx dy$
* Circular convolution looked promising, but I cannot see how I might apply it here.
* I have seem some sources that convert integral equations into differential equations, but the product rule seems to introduce additional problems here


### Motivation


The problem motivation is simple can be understood as follows: You have a loop of wire of length $l$ with resistivity along it $\rho(x)$. Without cutting the wire you want to figure out $\rho(x)$ using only a multimeter. Assume that $\int\_{0}^{l}\rho(x) = 1$ (i.e, the total resistance if you did cut the wire is 1). The only way you can take measurements is by probing it with an ohmmeter at $0$ and $l\_0$, which gives $R(l\_0)$.


I am quite new to functional analysis, so any other tips on even approximation techniques would be appreciated. I suspect but cannot prove a solution to this exists at all, so steps toward proof that there is no unique solution for a given $R(l\_0)$ would also be helpful.



How to find the inverse of a product of two integral equations
Let $F(y) = \int\_0^y \rho(x) dx$. Then your equation reduces to:


$$R(y) = F(y)(F(l) - F(y))$$


Note that for any value of $F(y)$, we get that $R(y) \leq \frac{F(l)^2}{4}$; this is easily proven by completing the square. As $F$ is an antiderivative, it is continuous, so by the intermediate value theorem (using $F(0) \leq \frac{F(l)}{2} \leq F(l)$) there is some $c$ such that $F(c) = \frac{F(l)}{2}$. Then $R(c) = \frac{F(l)^2}{4}$. Therefore $\max\_{y \in [0, l]} R(y) = \frac{F(l)^2}{4}$, so $F(l) = 2 \sqrt{\max\_{x \in [0, l]} R(x)}$


With $F(l)$ known, we can now solve for $F(y)$ easily: $F(y) = \frac{F(l) \pm \sqrt{F(l)^2 - 4 R(y)}}{2}$; we can easily check whether to use the positive or negative square root by whether $x$ is greater than or less than $c$. Finally, $\rho(x) = F'(x) = \frac{|R'(x)|}{2\sqrt{\left(\max\_{x \in [0, l]} R(x)\right) - R(x)}}$, as we have assumed that $\rho(x)$ is positive for all $x$.


