The case when curvature may be zero

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 Applet (\(k\geq 0\), 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 applet \((k\geq 0)\)