What is the minimum distance between a point on one torus and another point of the other torus?

Added 2019 June 2 0900Z:

I awoke in a nightmare about this problem.

A slice through the tori on the $z=0$ plane.

A similar slice exists on the $y=0$ plane.

The bright red circle, which is the innermost circle going around the red torus, surrounding the darker green circle within is at distance 1 from corresponding points on the darker green circle. Also the closest point on the lighter green circle is distance 1 from the darker red circle, which is the outermost circle going around the red torus.

Therefore, there are an uncounted infinite of points at which the distance is $1$ .

$\text{ParametricPlot}\left[\left( \begin{array}{cc} 2 \cos (u)+\frac{3}{2} & 2 \sin (u) \\ 4 \cos (u)+\frac{3}{2} & 4 \sin (u) \\ \cos (u)+\frac{3}{2} & -\sin (u) \\ -\cos (u)-\frac{9}{2} & -\sin (u) \\ \end{array} \right),\{u,0,2 \pi \},\text{PlotStyle}\to \{\text{Red},\text{Darker}[\text{Red}],\text{Darker}[\text{Green}],\text{Green}\}\right]$

A simple examination of the parametric equations reveals that they are all circles and the first three are concentric.

And, this is a far simpler solution to the problem.

End of 2019 June 2 0900Z addendum.

Added 2019 June 11 1745Z.

In fact, the entire two annuli regions, in blue, are of distance 1. The two annuli intersect in a line segment between the two tori, which is where my original solution is.

End of 2019 June 11 1745Z addendum.

The material below was edited 2019 June 4 1700Z to amplify and clarify several points.

First, I did a numerical minimization. The equivalent could be done by examining the plot given in the problem and then, using descending sized steps, search for the variables $u$ and $v$ to find values close to where the solution must be.

To do the numerical minimization, I changed the $u$ and $v$ in the second (green) parametric equations to $r$ and $s$ , respectively. This is necessary so that the positions on the two tori are independent. This is a four -dimensional optimization problem.

$\color{#EC7300}\text{The p1 formula has had }\pi \text{ added to } v \text{ to bring the solution to all zeroes.}$ Mea culpa: I forgot to mention this when writing this solution. Added: 2019 Jun 5 1600Z. This does not affect the minimum distance; it only affects the values of u, v, r and s at which this solution occurs. Please note well the addendum of 2019 June 2 0900Z above.

${\color{#D61F06}\text{p1}=\left\{\cos (u) (\cos (v)+3)+\frac{3}{2},\sin (u) (\cos (v)+3),\sin (v)\right\}}$

${\color{#20A900}\text{p2x}=\left\{\cos (r) (\cos (s)+3)-\frac{3}{2},-\sin (s),\sin (r) (\cos (s)+3)\right\}}$

$\Delta =\text{p1}-\text{p2x} \Rightarrow$

$\{-\cos (r) (\cos (s)+3)-\cos (u) (\cos (v)-3)+3,\,\sin (s)-\sin (u) (\cos (v)-3),\,$ $\sin (r) (-(\cos (s)+3))-\sin (v)\}$

Computing the Euclidean distance using the vector dot product:

$\sqrt{\Delta .\Delta } \Rightarrow$

$\sqrt{(-\cos (r) (\cos (s)+3)-\cos (u) (\cos (v)-3)+3)^2+(\sin (r) (-(\cos (s)+3))-\sin (v))^2+(\sin (s)-\sin (u) (\cos (v)-3))^2}$

The equation above is the same equation given by the Wolfram Mathematica 12 EuclideanDistance function with absolute function replaced by the identity function as the problem is using real numbers and the square of a real is always non-negative.

$\text{EuclideanDistance}[\text{p1},\text{p2x}]\text{/.}\, \text{Abs}\to \text{Identity} \Rightarrow \\ \sqrt{(-\cos (r) (\cos (s)+3)-\cos (u) (\cos (v)-3)+3)^2+(\sin (r) (-(\cos (s)+3))-\sin (v))^2+(\sin (s)-\sin (u) (\cos (v)-3))^2}$

$\text{NMinimize}[\{\text{EuclideanDistance}[\text{p1},\text{p2x}],0\leq u<2 \pi \land 0\leq v<2 \pi \land 0\leq r<2 \pi \land 0\leq s<2 \pi \}, \\ \{u,v,r,s\},\text{AccuracyGoal}\to 30]$

The values returned are all very close to $0$ or $2\pi$ ; hence, the simplification to $0$ .

Considering that trigonometric functions are being used, it seems that $u$ , $v$ , $r$ and $s$ are all $0$ . Now to show that that is a solution.

The plot below contains small patches of the red and green tori. The orange line is the solution indicated by the numerical minimization. The ends of the orange line are the points $\{\{\frac{7}{2}, \,0,\, 0\},\,\{\frac{5}{2} ,\,0 ,\, 0\}\})$ .

First, develop an equation for the Euclidean distance between arbitrary points on the two tori:

$d=\sqrt{(-\cos (r) (\cos (s)+3)-\cos (u) (\cos (v)-3)+3)^2+(\sin (r) (\cos (s)+3)+\sin (v))^2+(\sin (s)-\sin (u) (\cos (v)-3))^2}$

Now, compute the derivatives of the Euclidean distance with respect to each of the variables, respectively:

$\text{eu}=\frac{\partial d}{\partial u}=-\frac{(\cos (v)-3) (\sin (u) (\cos (r) (\cos (s)+3)-3)+\sin (s) \cos (u))}{\sqrt{(\cos (r) (\cos (s)+3)+\cos (u) (\cos (v)-3)-3)^2+(\sin (r) (\cos (s)+3)+\sin (v))^2+(\sin (s)-\sin (u) (\cos (v)-3))^2}}$

$\text{ev}=\frac{\partial d}{\partial v}=\frac{\sin (v) (\cos (u) (-(\cos (r) (\cos (s)+3)-3))+\sin (s) \sin (u)+3)+\sin (r) (\cos (s)+3) \cos (v)}{\sqrt{(\cos (r) (\cos (s)+3)+\cos (u) (\cos (v)-3)-3)^2+(\sin (r) (\cos (s)+3)+\sin (v))^2+(\sin (s)-\sin (u) (\cos (v)-3))^2}}$

$\text{er}=\frac{\partial d}{\partial r}=\frac{(\cos (s)+3) (\sin (r) (3-\cos (u) (\cos (v)-3))+\cos (r) \sin (v))}{\sqrt{(\cos (r) (\cos (s)+3)+\cos (u) (\cos (v)-3)-3)^2+(\sin (r) (\cos (s)+3)+\sin (v))^2+(\sin (s)-\sin (u) (\cos (v)-3))^2}}$

$\text{es}=\frac{\partial d}{\partial s}=\frac{-\sin (s) (\cos (r) (\cos (u) (\cos (v)-3)-3)+\sin (r) \sin (v)+3)-\cos (s) \sin (u) (\cos (v)-3)}{\sqrt{(\cos (r) (\cos (s)+3)+\cos (u) (\cos (v)-3)-3)^2+(\sin (r) (\cos (s)+3)+\sin (v))^2+(\sin (s)-\sin (u) (\cos (v)-3))^2}}$

Note that $\text{eu}$ , $\text{ev}$ , $\text{er}$ and $\text{es}$ are equations into which specific values can be substituted.

Then. check that the derivatives are all zero when the variables are all zero (the suspected answer).

$\text{eu}=0\land \text{ev}=0\land \text{er}=0\land \text{es}=0\text{/.}\, \{u\to 0,v\to 0,r\to 0,s\to 0\} \Rightarrow \text{True}$

That check is successful.

Then, evaluate the Euclidean distance formula developed above to get the answer to the problem.

$d\text{/.}\, \{u\to 0,v\to 0,r\to 0,s\to 0\} \Rightarrow 1$

The answer is $1$ .