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Given a self-intersecting parametric surface how to find the curve of its self-intersection (mathematically)? For example given this surface (in Wolfram Language):

surface[u_,v_]:={Cos[u],Sin[u]+Cos[v],Sin[v]};

ParametricPlot3D[surface[u,v],{u,0,2π},{v,-π,π}]

enter image description here

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Let the parametric equations of the surface be

$$\begin{cases}x=x(u,v),\\ y=y(u,v),\\ z=z(u,v).\end{cases}$$

The double points (self-intersections) are the solutions of a system of three equations in four unknowns.

$$\begin{cases}x(u,v)=x(u',v'),\\ y(u,v)=y(u',v'),\\ z(u,v)=z(u',v').\end{cases}$$

If you also have the implicit equation $f(x,y,z)=0$, you need to solve a single equation in two unknowns.

$$f(x(u,v),y(u,v),z(u,v))=0.$$


With your example,

$$\begin{cases}\cos u=\cos u',\\ \sin u+\cos v=\sin u'+\cos v',\\ \sin v=\sin v'.\end{cases}$$

giving

$$\begin{cases}u=2k\pi\pm u',\\ v=2k\pi+v'\lor v=(2k+1)\pi-v',\\ \sin u+\cos v=\pm\sin u\pm\cos v.\end{cases} $$

There are four solutions, corresponding to (from the third equation)

$$0=0,\\ \cos u=0,\\ \sin v=0,\\ \sin u+\cos v=0. $$

The first identity represents the whole surface while the next three are curves.

  • $(0,\pm1+\cos v,\sin v)$ is a pair of tangent circles,

  • $(\cos u,\sin u\pm1,0)$ is also a pair of tangent circles,

  • $(\cos u, 0, \pm\cos u)$ is a pair of orthogonal line segments.

You can see the line segments on the left picture and a pair of circles on the right.

enter image description here

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  • $\begingroup$ Nice exposition! It looks like there's a second self-intersection curve in the $xz$-plane for the given example, though; there may be an error in your derivation. $\endgroup$ Nov 14 '17 at 20:27
  • $\begingroup$ Note that (e.g.) for $\cos u = 0$, the $y$-component should be $\sin u + \cos v = \pm 1 + \cos v$, so you have a pair of circles that intersect at the origin. The second component has a similar error. I'm not sure how to fix the third component. $\endgroup$ Nov 14 '17 at 21:12
  • $\begingroup$ @MichaelSeifert: you are right, I plugged $0$ where $\pm1$ was required. Thanks for the notice. $\endgroup$
    – user65203
    Nov 14 '17 at 22:33

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