I am trying to understand mathematically what happens with the following system: A uniform, inextensible string of length $2s$ hanging at its ends from two poles of equal height $h.$ We know that the curve assumed by the string at any one instant is a catenary. What happens, however, as the poles are brought together along a straight line in a uniform manner?
In precise terms, what I want to find are the single-variable real functions $a(t),\,b(t)$ such that the quantity $$\int_0^{x_0}\sqrt{1+{y'}^2}\mathrm dx$$ is fixed and equal to $s$ as $x_0=x_0(t)$ is varied, where $x_0$ satisfies the equation $h=y(x_0,t)$ for all $t,$ and $$y(x,t)=a(t)+b(t)\cosh\frac{x}{b(t)}.$$
Of course, this should satisfy some natural conditions. We must, for example, have that as $t\to\beta,$ the function $y$ becomes constant. In particular, we must have then that $y=h$ for all $x,$ as $t\to \beta,$ where the range of values of $t$ is given as $[\alpha,\beta]$ with $\alpha<0$ and $\beta>0.$ These may be finite or infinite. Now, the initial configuration is also given. That is, we must have that for all $x$ and $t=0,$ the curve is given by $$y=b(0)\cosh\frac{x}{b(0)}.$$ Clearly, we must have that $b(t)>0$ for all $t.$ I presently can't think of any other necessary condition I might have skipped. I'll provide details as needed. Well, another thing is that as $t\to\beta,$ we must have that the function becomes singular -- that is, it fails to define $y$ as a function of $x,$ and is rather some ray coincident with the line $x=0.$
Now the integral is easy to evaluate, and that gave me the equation $(h-a)^2=b^2+s^2.$ However, this doesn't determine $a,b$ uniquely.
Could someone offer more help in order to determine the functions $a,\,b$ so that they satisfy all the required functions? Thanks.