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The question What equation can produce these curves? attracted a lot of good answers earlier today. I wanted to contribute my own which would be superior in some respect but then realized the OP might not require the additional criteria and would be happy with the answers available prior to that and preferring them for simplicity. Also, although I know some solutions to my extended specification of the problem, they are far from elegant. So I'm wondering what the community has to say.

The main idea of the problem stays the same: find a one-parametric family of functions whose graphs look like this:

curves

But with the following restrictions:

  • for each value of the parameter $a$ the function $f_a(x)$ is a $C^\infty$ bijection on $\mathbb{R}$ (so: no asymptotes and global minima / maxima),

  • for each $a$ it is also an involution, i.e., symmetric with respect to the line $y = x$,

  • the definition must be explicit (i.e., not in the form $F_a(x,y) = 0$).

  • explicit definitions are strongly preferred.

(Note that in the light of the second restriction the first could be reformulated so that $f_a(x)$ is defined (and smooth) for all $x \in \mathbb{R}$).

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The Vee
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  • An observation: changing coordinates to $s=x-y, t=x+y$, the graph looks like $t=a \cosh(s)$, or $t=a s^2$, or any such function. – lisyarus Jul 05 '16 at 17:02
  • Good observation. It looks like that but neither of the two functions could be used. The relation between t and s would need to satisfy that $\lim t/s = 1$ while $\lim s-t = ∞$, both as $s \to \infty$. Or something around these lines. – The Vee Jul 05 '16 at 17:07
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    For some smooth bijection $g$ from $\mathbb R$ to $\mathbb R$ and for every $a>0$, use $$f_a(x)=a^{-1}g^{-1}(g(0)+g(a)-g(ax)).$$ – Did Jul 05 '16 at 17:08
  • @Did: Wow, this looks fantastic with $g(x) = \sinh^{-1}(x)$. And super easy to verify. Answer? – The Vee Jul 05 '16 at 17:16
  • Yes the idea is to note that $g(ay)+g(ax)$ has the same value $g(0)+g(a)$ at $(x,y)=(1,0)$ and at $(x,y)=(0,1)$, and to choose the function $f_a$ whose graph is the set of points $(x,y)$ where $g(ay)+g(ax)=g(0)+g(a)$. Re an answer, you might want to write and post one yourself. – Did Jul 05 '16 at 17:22
  • @Did Why would I steal yours? If you post the basic idea I'm willing to supplement it with some graphs, special cases etc. but the reputation for the key idea belongs to you. – The Vee Jul 05 '16 at 17:30
  • You would steal nothing and reputation is basically a farce anyway... So, just go ahead! – Did Jul 05 '16 at 17:38

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