ATRACTOR

Now let us see what additional assumptions are needed to ensure the uniqueness of the curve, in the case where the curvature may have zero values.

Let \(f:\, I\rightarrow\mathbb{R}^{3}\). Denote by \(Z_{f}\) the set of zeros of \(f\). \[Z_{f}=\left\{ s\in I,f(s)=0\right\} \]

Theorem 1: Let \(P_{0}\) be a point in \(\mathbb{R}^{3}\) and let \(TF_{0}\) be a orthonormal triple with positive orientation. Furthermore, let \(k:\, I\rightarrow\mathbb{R}^{3}\) be a nonnegative continuous function, with a finite number of zeros, that is, such that \[Z_{k}=\left\{ s_{1},s_{2},s_{3},s_{4},...,s_{n}\right\} \] Let \(\tau:\, I\setminus Z_{k}\rightarrow\mathbb{R}^{3}\) be a continuous function.

Let \(TF_{s}=\left(T_{s},N_{s},B_{s}\right)\) be the Frenet-Serret frame at instant \(s\in I\setminus Z_{k}\).

Let \(TF_{s_{i}}=\left(T_{s_{i}},N_{s_{i}},B_{s_{i}}\right)\) be an orthonormal triple with positive orientation, corresponding to instant \(s_{i} \in Z_{k}\), which satisfies \[lim_{s\rightarrow s_{i^{-}}}T_{s}=T_{s_{i}}\]

Then, there exists a unique \(C^{2}\) curve \(c:\, I\rightarrow\mathbb{R}^{3}\), parameterized by arc length, with initial point \(P_{0}\) and initial Frenet-Serret frame \(TF_{0}\), whose curvature function coincides with \(k\), whose torsion function coincides with \(\tau\) and whose Frenet-Serret frames at instants \(s_{i}\) are precisely \(TF_{s_{i}}\).

Proof. Since \(f\) has a finite number of zeros, they are isolated points and therefore it suffices to prove the result for a function with a single zero.

Let \(s_{1}\in I=\left[I_{0},I_{1}\right]\) be that single zero.

The curve in \(\left[I_{0},s_{1}\right[\) is unique and \(C^{2}\) (by the Fundamental Theorem of Curves).

Similarly, the curve in \(\left]s_{1},I_{1}\right]\) is also unique and \(C^{2}\) - note that now \(TF_{0}=TF_{s_{1}}\) and \[lim_{s\rightarrow s_{1^{-}}}c(s)=P_{0}.\]

It remains to examine what happens at \(s_{1}\).

The curve defined in \(I\) is \(C^{1}\) by the assumption on the choice of the associated frame at \(s_{1}\) (that forces the tangent to change "smoothly" at \(s_{1}\)).

Since the curvature is continuous and tends to zero at \(s_{1}\), we have \[lim_{s\rightarrow s_{1^{-}}}f''=(0,0,0)\] and \[lim_{s\rightarrow s_{1^{+}}}f''=(0,0,0).\]

Moreover \(k(s_{1})=0\), that is, \(f''(s_{1})=(0,0,0)\) and thus the curve is \(C^{2}\).      QED

3D app (\(k\geq 0\), finite number of zeros)

Theorem 2: Let \(P_{0}\) be a point in \(\mathbb{R}^{3}\) and let \(TF_{0}\) be a orthonormal triple with positive orientation. Furthermore, let \(k:\, I\rightarrow\mathbb{R}^{3}\) be a nonnegative continuous function, with zero set \[Z_{k}=\{s_{i}\}_{i=1,...,N}\cup\{[a_{j},b_{j}]\}_{j=1,...,M},\] where \(b_{j-1}<a_{j}<b_{j}<a_{j+1}\) and \(s_{i}\) do not belong to any of intervals \([a_{j},b_{j}]\) (for every \(i\) and \(j\)).

Let \(\tau:\, I\setminus Z_{k}\rightarrow\mathbb{R}^{3}\) be a continuous function.

Let \(TF_{s}=\left(T_{s},N_{s},B_{s}\right)\) be the Frenet-Serret frame at instant \(s\in I\setminus Z_{k}\).

Let \(TF_{s_{i}}=\left(T_{s_{i}},N_{s_{i}},B_{s_{i}}\right)\) be an orthonormal triple with positive orientation, corresponding to instant \(s_{i}\in Z_{k}\), which satisfies \[lim_{s\rightarrow s_{i^{-}}}T_{s}=T_{s_{i}}\]

Let \(TF_{b_{i}}=\left(T_{b_{i}},N_{b_{i}},B_{b_{i}}\right)\) be an orthonormal triple with positive orientation, corresponding to instant \(b_{i}\), which satisfies \[lim_{s\rightarrow b_{i^{-}}}T_{s}=T_{b_{i}}.\]

Then, there exists a unique \(C^{2}\) curve \(c:\, I\rightarrow\mathbb{R}^{3}\), parameterized by arc length, with initial point \(P_{0}\) and initial Frenet-Serret frame \(TF_{0}\), whose curvature function coincides with \(k\), whose torsion function coincides with \(\tau\), whose Frenet-Serret frames at instants \(s_{i}\) are precisely \(TF_{s_{i}}\) and whose Frenet-Serret frames at instants \(b_{i}\) are \(TF_{b_{i}}\).

Proof. Since the case of isolated points has already been studied in the previous theorem and the number of intervals that make \(Z_{f}\) is finite, it remains to prove the case \(Z_{f}=[a_{1},b_{1}]\).

Let \(I=\left[I_{0},I_{1}\right]\).

The curve in \(\left[I_{0},a_{1}\right[\) is unique and \(C^{2}\) (by the Fundamental Theorem of Curves).

The curve in \(\left]a_{1},b_{1}\right[\) is a straight line (it has zero curvature) and therefore \(C^{\infty}\) (Fundamental Theorem of Curves). In order to be \(C^{1}\), it must start at \[lim_{s\rightarrow a_{1^{-}}}c(s)\] and must have the direction of \[lim_{s\rightarrow a_{1^{-}}}T_{s}.\

Hence, the straight line is unique.

The curve in \(\left]b_{1},I_{1}\right]\) is unique and \(C^{2}\) (Fundamental Theorem of Curves) - note that now \(TF_{0}=TF_{b_{1}}\) and \[P_{0}=lim_{s\rightarrow b_{1^{-}}}c(s)=\mbox{"end" point of the straight line}.\]

The curve defined in \(I\) is \(C^{1}\) by the choice of the straight line and by the assumption on the choice of the Frenet-Serret frame at pela escolha do segmento de recta e pela restrição imposta à instant \(b_{1}\).

Since curvature is continuous and tends to zero at \(a_{1}\) and \(b_{1}\), we have \[lim_{s\rightarrow a_{1^{-}}}f''=(0,0,0)\] and \[lim_{s\rightarrow b_{1^{+}}}f''=(0,0,0).\]

Moreover \(k\left(\left[a_{1},b_{1}\right]\right)=0\), that is, \(f''\left(\left[a_{1},b_{1}\right]\right)=(0,0,0)\) and so the curve is \(C^{2}\).      QED

3D app \((k\geq 0)\)