When do equations represent the same curve?

Question

Suppose we have two sets of parametric equations $\mathbf c_1(u) = (x_1(u), y_1(u))$ and $\mathbf c_2(v) = (x_2(v), y_2(v))$ representing two 2D planar curves. When I say "2D planar curves" I mean that $\mathbf c_1(u)$ and $\mathbf c_2(u)$ are mappings from compact intervals in $\mathbb R$ to $\mathbb R^2$. We can assume (without loss of generality, I think) that $\mathbf c_1:I \to \mathbb R^2$ and $\mathbf c_2:I \to \mathbb R^2$, where $I=[0,1]$. You can assume some continuity or differentiability of $\mathbf c_1(u)$ and $\mathbf c_2(u)$, if that helps.

I'm interested to know how we can determine that these two sets of equations represent the same curve. In other words, how can I determine that $\mathbf c_1(I)$ and $\mathbf c_2(I)$ are the same point set.

An interesting special case: what if the parametric equations are all rational functions? In this case, it's often possible to implicitize -- i.e. convert to equations of the form $f_1(x,y)=0$ and $f_2(x,y)=0$. Then, if the two curves are the same point set, I would guess that something can be said about $f_1$ and $f_2$? Maybe one is a multiple of the other, or something like that??

Even simpler (but still interesting): what if all the functions involved are polynomials.

The implicitization doesn't necessarily solve the original problem, though. It's clear that $\mathbf c_1(I)$ is a subset of the zero set $Z_1 = \{(x,y) \in \mathbb R^2 : f_1(x,y) = 0\}$, but it might be a proper subset. So, even if we know how to relate $Z_1$ and $Z_2$, this might not tell us much about how $\mathbf c_1(I)$ is related to $\mathbf c_2(I)$. Can we say anything about when the implicitization approach will work and when it won't?

My question was inspired by this one.

There might be some connection with this question, but both the question and the answer are written in jargon that's not familiar to me.

This has practical applications -- curves in engineering and manufacturing are often described by using rational or polynomial parameterizations, and it would be nice if we had some way to identify when two curves are the same. In engineering & manufacturing, we only care about the shapes of curves (i.e. sets like $\mathbf c_1(I)$ and $\mathbf c_2(I)$), not their parameterization. For example, a circular wheel is still circular, regardless of how the circle curve is parameterized. The parameterization is artificial, in some sense, and I want to be able to ignore its effects when comparing two curves.

In case it matters to anyone, this isn't homework :-).

Example (for the rational case)

$$\mathbf c_1(t) = \left( \frac{1 - (2 - \sqrt2)t - (\sqrt2 - 1)t^2} {1 - (2 - \sqrt2)t + (2 - \sqrt2)t^2}, \frac{\sqrt2 t - (\sqrt2 - 1)t^2} {1 - (2 - \sqrt2)t + (2 - \sqrt2)t^2} \right)$$

$$\mathbf c_2(t) = \left( \frac{1 -t^2}{1 + t^2}, \frac{2t} {1 + t^2} \right)$$

Here $\mathbf c_1(I) = \mathbf c_2(I)$. They are both the first quadrant of the unit circle, actually.

Progress (December 2017)
Apparently, if two two implicit equations $f_1(x,y)=0$ and $f_2(x,y)=0$ represent the same curve, and $f_1$ and $f_2$ are both irreducible polynomials, then one must be a constant multiple of the other. This result is mentioned (without proof) in this paper by Sendra, so I suppose it must be well-known.

Answer 1

Consider the simpler problem of two parametrized curves $(x_i(t),y_i(t))$ that start at $(0,0)$ at time $t=0$, and equality of trajectories up to reparametrization, which is stronger than equality of point sets and a more natural condition as it is local (up to matching of starting points). Heuristically, and to some extent rigorously, there is a usable criterion.

We want that when $x_1 = x_2$, then $y_1 = y_2$ so that the bivariate function $x_1(t)-x_2(s)$ divides $y_1(t)-y_2(s)$ (and vice versa) in a suitable ring of functions. Their ratio is an invertible function with positive values, at least for nonzero $s,t$ near $0$. In fact we need it to be positive only for nonzero $(s,t)$ at which $x_1(s)=x_2(t)$ or the same for $y$.

Example: parabola and half-parabola.

Curve A is $(t,t^2)$.

Curve B is $(s^2,s^4)$.

$x_1(t)-x_2(s) = t - s^2$

$y_1(t)-y_2(s) = t^2 - s^4$

Ratio is $(t + s^2)$

This is positive near the locus where $x_1(t)=x_2(s)$ (namely $t=s^2$). On the locus where $y_1(t)=y_2(s)$ (namely $t^2=s^4$), this is positive for $t>0$ and negative for $t < 0$. The positivity condition knows which half of the parabola is curve B! That is a good sign that this is either the complete answer to the simplified problem, or on the right track.

Finding an intersection point of two parametric curves OR detecting a difference between the curves is simpler than the general problem of curve intersection. Take a point on one curve, solve for the parameter values that would place its $x$ coordinate on the other curve, and test whether the $y$ coordinates are the same. For algebraic parameterizations this calculation can be done exactly.

For the point-set equality problem, locate the zeros of the $st$ ratio. These parameter values segment the two curves into arcs. Then there is a combinatorial problem of orienting and matching (by the procedure given above) identical arcs of the two curves, and testing whether both curves are covered by the matched arcs.

Answer 2

Disclaimer. This answer is more like a question. Since the (+100) bounty has been raised by the same author, it will not be attached to this answer. But if someone can prove my parametrization conjecture (see below), that would be great. (Update. The bounty hasn't been awarded to anyone)

Because we are going to calculate definite integrals, it is important to have closed intervals for the parameters $t$ . For the example as presented in the question: $$ \mathbf c_1(0) = (1,0) \quad ; \quad \mathbf c_1(1) = (0,1) \\ \mathbf c_2(0) = (1,0) \quad ; \quad \mathbf c_2(1) = (0,1) $$ So we are lucky: if both parameters $t$ are allowed to be restricted to the interval $[0,1]$ then the two curves at least have the same end-points.
Consider the expression $\left[x(t)y'(t)-y(t)x'(t)\right] dt$ . It is twice the area of an infinitesimal triangle with vertices at $(0,0)$ , $(x(t),y(t))\,$ and $\,(x(t+dt),y(t+dt))$ : $$ 2 \times \mbox{area}\,\Delta = det\begin{bmatrix} x(t) & x(t+dt) \\ y(t) & y(t+dt) \end{bmatrix} = \left[x(t)\frac{y(t+dt)-y(t)}{dt} - y(t)\frac{x(t+dt)-x(t)}{dt}\right] dt $$ Now we are going to calculate the following quantities, with the end-points $\,(a,b) = (0,1)\,$ and $\,m,n\,$ positive or zero integers. Integration over an area instead of an arc length is to be preferred for some good reasons (: e.g. suppose that the curve is traversed back and forth at some places). $$ M_{m,n} = \int_a^b x(t)^m y(t)^n \left[x(t)y'(t)-y(t)x'(t)\right] dt $$ The quantities $M_{m,n}$ , not at all by coincidence, are similar to variances (in statistics terms) or moments of inertia (in physics terms). It is conjectured that the outcome of these integrals is independent of any parametrization. This may be called a Parametrization Conjecture.
It is assumed to be true in the sequel. Can someone prove or disprove?

MAPLE has been invoked to save time and effort. Definitions for the example as given in the OP's question and first few steps:

x1(t) := (1 - (2 - sqrt(2))*t - (sqrt(2) - 1)*t^2)/
         (1 - (2 - sqrt(2))*t + (2 - sqrt(2))*t^2);
y1(t) := (sqrt(2)*t - (sqrt(2) - 1)*t^2)/
         (1 - (2 - sqrt(2))*t + (2 - sqrt(2))*t^2);
x2(t) := (1-t^2)/(1+t^2); y2(t) := 2*t/(1+t^2);
x1d(t) := simplify(diff(x1(t),t)); y1d(t) := simplify(diff(y1(t),t));
x2d(t) := simplify(diff(x2(t),t)); y2d(t) := simplify(diff(y2(t),t));
M00 := int(x1(t)*y1d(t)-x1d(t)*y1(t),t=0..1);
N00 := int(x2(t)*y2d(t)-x2d(t)*y2(t),t=0..1);
verify(M00,N00,equal);
                           true

For the example at hand, the lowest order (area) moments are indeed exactly the same for the two parametrizations: $$ M_{0,0} = \frac{\pi}{2} \\ M_{1,0} = 1 \quad ; \quad M_{0,1} = 1 \\ M_{2,0} = \frac{\pi}{4} \quad ; \quad M_{1,1} = \frac{1}{2} \quad ; \quad M_{0,2} = \frac{\pi}{4} \\ M_{3,0} = \frac{2}{3} \quad ; \quad M_{2,1} = \frac{1}{3} \quad ; \quad M_{1,2} = \frac{1}{3} \quad ; \quad M_{0,3} = \frac{2}{3} \\ M_{4,0} = \frac{3\pi}{16} \quad ; \quad M_{3,1} = \frac{1}{4} \quad ; \quad M_{2,2} = \frac{\pi}{16} \quad ; \quad M_{1,3} = \frac{1}{4} \quad ; \quad M_{0,4} = \frac{3\pi}{16} \\ M_{5,0} = \frac{8}{15} \; ; \; M_{4,1} = \frac{1}{5} \; ; \; M_{3,2} = \frac{2}{15} \; ; \; M_{2,3} = \frac{2}{15} \; ; \; M_{1,4} = \frac{1}{5} \; ; \; M_{0,5} = \frac{8}{15} $$ And so on and so forth. In this way it may be confirmed, step by step, that the two parametrizations represent one and the same curve. It remains unsatisfactory that we cannot establish everything in one step, though: MAPLE could not calculate for the general expression $\,M_{m,n}$ .

Answer 3

It would be easier for you if you have a nice set of definitions to understand your problem. So here I drop some of them hoping they will let you arrive at the correct answer. If you still have problems just ask in the comments.

By a $C^1$ path in $\Re^n$ is meant a continuously differentiable function $\gamma:[a,b] \rightarrow \Re^n$.

The $C^1$ path $\gamma:[a,b] \rightarrow \Re^n$ is said to be smooth if $\gamma' (t) \not= 0$ for all $t \in [a,b]$.

Having defined that, suppose that $\alpha:[a,b] \rightarrow \Re^n$ and $\beta:[c,d] \rightarrow \Re^n$ are two $C^1$ paths that are "geometrically equivalent" in the sense that they have the same initial point and the same terminal point, i.e., $\alpha(a)= \beta(c)$ and $\alpha(b)= \beta(d)$ respectively. Then we say that the path $\alpha:[a,b] \rightarrow \Re^n$ is equivalent to the path $\beta:[c,d] \rightarrow \Re^n$ if and only if there exists a $C^1$ function $$\psi:[a,b] \rightarrow [c,d]$$ such that $\psi([a,b])=[c,d], \alpha = \beta \circ \psi$ and $\psi'(t)> 0$ for all $t\in[a,b].$

The set $C \subseteq \Re^n$ is called a curve if and only if it is the image of a smooth path $\gamma$ which is one-to-one. And any one-to-one smooth path which is equivalent to $\gamma$ is then called a parametrization of $C$.

Finally if $\vec{x}=\gamma(t) \in C$ then $\vec{T}(\vec{x}) =\frac{\gamma'(t)}{|\gamma'(t)|}$ is a unit tangent vector to $C$ at $\vec{x}$, and $\vec{T}(\vec{x})$ is independent of the chosen parametrization $\gamma$ of $C$. Such a continuous mapping $T:C \rightarrow \Re^n$ is called an orientation for $C$. So an oriented curve is then a pair $(C,T)$ or just $C$. And $-C$ is the same geometric curve with the opposite orientation $(C,-T)$.

Answer 4

The first representation of the OP's example repeated: $$ \mathbf c_1(t) = \left( \frac{1 - (2 - \sqrt2)t - (\sqrt2 - 1)t^2} {1 - (2 - \sqrt2)t + (2 - \sqrt2)t^2}, \frac{\sqrt2 t - (\sqrt2 - 1)t^2}{1 - (2 - \sqrt2)t + (2 - \sqrt2)t^2} \right) \quad \mbox{with} \quad 0 \le t \le 1 $$ The second representation of the OP's example repeated: $$ \mathbf c_2(u) = \left( \frac{1 -u^2}{1 + u^2}, \frac{2u}{1 + u^2} \right) \quad \mbox{with} \quad 0 \le u \le 1 $$ The equality problem for the two representations can be formulated as solving $u$ and $t$ from the equations $\mathbf c_1(t) = \mathbf c_2(u)$ in such a way that the solutions are a bijection.
Let a computer algebra system (MAPLE) do the hard work:

c1x(t) := (1-(2-sqrt(2))*t-(sqrt(2)-1)*t^2)/(1-(2-sqrt(2))*t+(2-sqrt(2))*t^2);
c1y(t) := (sqrt(2)*t-(sqrt(2)-1)*t^2)/(1-(2-sqrt(2))*t+(2-sqrt(2))*t^2);
c2x(u) := (1-u^2)/(1+u^2); c2y(u) := 2*u/(1+u^2);
solve(c1x(t)=c2x(u),u);
solve(c1y(t)=c2y(u),u);
solve(c1x(t)=c2x(u),t);
solve(c1x(t)=c2x(u),t);

Throwing away redundant outcomes gives the following end-result. $$ u = -\frac{(\sqrt2+1)\,t}{t-2-\sqrt2} \quad ; \quad t = \frac{(2+\sqrt2)\,u}{u+1+\sqrt2} \\ \mathbf c_1(t) = \mathbf c_2\left(-\frac{(\sqrt2+1)\,t}{t-2-\sqrt2}\right) \quad ; \quad \mathbf c_2(u) = \mathbf c_1\left(\frac{(2+\sqrt2)\,u}{u+1+\sqrt2}\right) $$