Catenary curved string with differing linear densities - Linear density distribution Do you guys know if there's some kind of weight distribution that fits a problem like this one?
Let there be a string with two different sections with different linear densities, like the following image, where the concentrated loads may be taken as being positioned in the center of each string section. So I can then calculate the equivalent catenary and be able to use the catenary equations.
I need to know if there's a way to find an equivalent linear density that can be applied in the middle of the complete string.
So far I have tried the obvious assumption that there's a linear distribution but the following equation yields a somewhat not credible result, it seems that, because a large part of the system's total mass is concentrated near x = 0 coordinate.

I'm wondering that there may be some kind of exponential behaviour to the linear density distribution, as exemplified by the m(x) curve, but I would like to be more sure of it.
Thanks for your attention,
Regards.

The first image depicts the problem, the second one is the linear density distribution along the x axis and the third one is the concentrated weight distribution along the x axis.
 A: The catenary equation can be derived by considering the minimization of the functionals,
\begin{align}
f[y]&=\int_0^{l_2}\mu gy\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx \tag{pot. e.}\\
g[y]&=\int_0^{l_2}\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx\tag{length}
\end{align}
where $l_2$ is the point we're hanging a chain of length $l_1$.
Next, we can define the functional,
$$
S[y]=f[y]+\lambda g[y]=\int_0^{l_2}\mathcal{L}(y,\,y',\,x)\,\mathrm{d}x
$$
where $\lambda$ is a parameter that helps us determine the correct chain length and $y'=\mathrm{d}y/\mathrm{d}x$. From this definition, we have that our Lagrangian is,
$$\mathcal{L}\triangleq\left(\mu gy+\lambda\right)\sqrt{1+(y')^2}.$$
So by applying the Legendre transform, we get our Hamiltonian,
\begin{align}
\mathcal{H}&=y'\frac{\partial\mathcal{L}}{\partial y'}-\mathcal{L} \\
&=y'\left(\mu gy+\lambda\right)\cdot\frac{y'}{\sqrt{1+(y')^2}}-\left(\mu gy+\lambda\right)\sqrt{1+(y')^2}.
\end{align}
If we then let the Hamiltonian be the constant total energy (density), $\varepsilon$, then we end up with,
$$\left(\frac{\mu gy+\lambda}{\varepsilon}\right)^2=1+(y')^2$$
Under the constant mass density case, we can trivially solve this to get that common $\propto\cosh(x)$ function.
In the non-constant density, the ODE is a little different,
$$\frac{\mathrm dy}{\mathrm dx}=\left(\left(\frac{g\mu(x)y(x)+\lambda}{\varepsilon}\right)^2-1\right)^{1/2}\tag{1}$$
and may be a bit more complicated to solve, possibly requiring some numerical solution to fit the two boundary conditions ($y(0)=y_0$ and $y(l_2)=y_2$) along with the condition that $g[y]=l_1$ (i.e., length is fixed).
Since you are interested in an effective mass density when given two different-but-uniform ropes, your mass density is,
$$ 
\mu(x)=\begin{cases} \mu_1 & x<x_p \\ \mu_2 & \text{otherwise} \end{cases}
$$
where $x_p$ is the joint. This seems to point you towards having two separate linear cases and an additional constraint that $y(x_p^-)=y(x_p^+)$ (i.e., the heights at the joint as measured from above and from below must match). This likely will require some numerical minimization methods to satisfy.
Once you have that solution, it may be possible to add an additional parameter of the equivalent mass density, $\mu_\text{equiv}$, by fitting the constant-density case with the additional condition that $y(x_p)=y_p$ (i.e., the curve must also pass through the joint), though I don't know if this would result in a solution or not.
