Proof of the Fundamental Theorem of Curves
Theorem 1
Let’s say $a,b$ contains $0$ in an interval. And let’s say the following holds:
- (i): $\overline{\kappa}(s) > 0$ is $C^{1}$ in $(a,b)$
- (ii): $\overline{\tau}(s)$ is continuous in $(a,b)$
- (iii): $\mathbf{x}_{0}$ is a fixed point of $\mathbb{R}^{3}$
- (iV): $\left\{ D,E,F \right\}$ is the right-hand normal orthogonal basis of $\mathbb{R}^{3}$
Then, there uniquely exists a $C^{3}$ regular curve $\alpha : (a,b) \to \mathbb{R}^{3}$ whose parameter is the length of the arc from $\alpha (0)$, satisfying the following: $$ \begin{align*} \alpha (0) =& \left( \mathbf{x}_{0} \right) \\ T(0) =& D \\ N(0) =& E \\ B(0) =& F \\ \kappa (s) =& \overline{\kappa} (s) \\ \tau (s) =& \overline{\tau} (s) \end{align*} $$
- $\left\{ T, N, B , \kappa, \tau \right\}$ is a tool of Frenet-Serret.
- $C^{1}$ is the set of functions with continuous derivatives.
Explanation
The Fundamental Theorem of Curves is a powerful theorem that specifies a curve in $3$-dimensional space by its curvature and torsion, indicating that the following results are no coincidence:
- If $\kappa = 0$, it’s a straight line.
- If $\kappa \ne 0 , \tau = 0$, it’s a plane curve.
- If $\kappa / \tau$ is constant, it’s a helix.
- If $\tau = 0$ and $\kappa > 0$ are constant, it’s a circle.
- If $\tau \ne 0$ is constant and $\kappa > 0$ is constant, it’s a circular helix.
It’s aptly named the fundamental theorem in terms of guaranteeing uniqueness and existence.
Proof
Picard’s Theorem: For the initial value problem of a system of first-order ordinary differential equations, there exists a unique solution.
Frenet-Serret Formulas: If $\alpha$ is a unit speed curve with $\kappa (s) \ne 0$ then $$ \begin{align*} T^{\prime}(s) =& \kappa (s) N(s) \\ N^{\prime}(s) =& - \kappa (s) T(s) + \tau (s) B(s) \\ B^{\prime}(s) =& - \tau (s) N(s) \end{align*} $$
$$ \mathbf{u}_{j}^{\prime} = \sum_{i=1}^{3} a_{ij} (s) u_{i} \\ \left( a_{ij} \right) = \begin{bmatrix} 0 & \overline{\kappa} & 0 \\ -\overline{\kappa} & 0 & \overline{\tau} \\ 0 & \overline{\tau} & 0 \end{bmatrix} $$
Considering such an ODE system, by Picard’s Theorem, there exists a unique solution $\mathbf{u}_{j}(s)$ satisfying the following: $$ \begin{align*} \mathbf{u}_{1} (0) =& D \\ \mathbf{u}_{2} (0) =& E \\ \mathbf{u}_{3} (0) =& F \end{align*} $$ Now, we need to show that the solution satisfies the conditions we need.
Step 1. $\mathbf{u}_{i}(t)$ are orthogonal.
If we set $p_{ij} := \left< \mathbf{u}_{i}, \mathbf{u}_{j} \right>$ as $$ \begin{align*} p_{ij}^{\prime} =& \left< \mathbf{u}_{i}^{\prime}, \mathbf{u}_{j} \right> + \left< \mathbf{u}_{i}, \mathbf{u}_{j}^{\prime} \right> \\ =& \left< \sum_{k=1}^{3} a_{ki} \mathbf{u}_{k} , \mathbf{u}_{j} \right> + \left< \mathbf{u}_{i}, \sum_{k=1}^{3} a_{kj} \mathbf{u}_{k} \right> \\ =& \sum_{k=1}^{3} a_{ki} p_{kj} + \sum_{k=1}^{3} a_{kj} p_{ik} \end{align*} $$ then $p_{ij}$, by Picard’s Theorem, is the unique solution to the differential equation with initial value $$ p_{ij}^{\prime} = \sum_{k=1}^{3} \left( a_{ki} p_{kj} + a_{kj} p_{ik} \right) $$ and in $t = 0$, Kronecker delta function $p_{ij} (0) = \delta_{ij}$ is applied. Meanwhile, $$ \sum_{k=1}^{3} \left( a_{ki} \delta_{kj} + a_{kj} \delta_{} \right) = a_{ji} + a_{ij} = 0 = \delta_{ij}^{\prime} $$ thus $\delta_{ij} = p_{ij}$ itself exists as the unique solution to the given differential equation. Therefore, we obtain the following: $$ \left< \mathbf{u}_{i} , \mathbf{u}_{j} \right> = \delta_{ij} $$
Step 2. Regularity of the unit speed curve $\alpha$
$$ \alpha (s) := \mathbf{x}_{0} + \int_{0}^{s} \mathbf{u}_{1} (\sigma) d \sigma $$ For $s \in (a,b)$, let’s consider $\alpha (s)$ as above. First derivative, by the Fundamental Theorem of Calculus, $$ {{ d \alpha } \over { ds }} = \mathbf{u}_{1} (s) $$ Second derivative, by the original differential equation we considered, $$ {{ d^{2} \alpha } \over { ds^{2} }} = \mathbf{u}_{1}^{\prime} = \overline{\kappa} \mathbf{u}_{2} $$ Since $\overline{\kappa}$ and $\mathbf{u}_{2}$ are differentiable, one more differentiation $$ {{ d^{3} \alpha } \over { ds^{3} }} = \overline{\kappa}^{\prime} \mathbf{u}_{2} + \overline{\kappa} \mathbf{u}_{2}^{\prime} = \overline{\kappa}^{\prime} \mathbf{u}_{2} + \overline{\kappa} \left( -\overline{\kappa} \mathbf{u}_{1} + \overline{\tau} \mathbf{u}_{3} \right) $$ Since $\overline{\kappa}$ and $\overline{\tau}$ are continuous and $\mathbf{u}_{i}$ are differentiable, thus continuous, ${{ d^{3} \alpha } \over { ds^{3} }}$ is also continuous, and therefore $\alpha$ is $C^{3}$. As already seen in Step 1, $$ \left| {{ d \alpha } \over { ds }} \right| = \left| \mathbf{u}_{1} \right| = 1 $$ hence $\alpha$ is a unit speed curve.
Step 3. $\overline{\kappa} = \kappa, \overline{\tau} = \tau, \mathbf{u}_{1} = T, \mathbf{u}_{2} = N, \mathbf{u}_{3} = B$
Since $\alpha^{\prime} = \mathbf{u}_{1}$, naturally $\mathbf{u}_{1} = T$. According to Frenet-Serret Formulas, $$ \kappa N = T^{\prime} = \mathbf{u}_{1}^{\prime} = \overline{\kappa} \mathbf{u}_{2} $$ Since $N$ and $\mathbf{u}_{2}$ are unit vectors and $\overline{\kappa} > 0$, it must be $\overline{\kappa} = \kappa$, and therefore $N = \mathbf{u}_{2}$. $\left\{ \mathbf{u}_{1} , \mathbf{u}_{2}, \mathbf{u}_{3} \right\}$ is the orthogonal basis of $\mathbb{R}^{3}$, hence $\left[ \mathbf{u}_{1} , \mathbf{u}_{2}, \mathbf{u}_{3} \right] = \pm 1$ and in $s = 0$, $$ \left[ D, E, F \right] = \left[ \mathbf{u}_{1} , \mathbf{u}_{2}, \mathbf{u}_{3} \right] = \pm 1 $$ The scalar triple product $\left[ \mathbf{u}_{1} , \mathbf{u}_{2}, \mathbf{u}_{3} \right]$ is continuous2, so in fact, it must always be
$$ \left[ \mathbf{u}_{1} , \mathbf{u}_{2}, \mathbf{u}_{3} \right] = 1 $$ and therefore, $$ B = T \times N = \mathbf{u}_{1} \times \mathbf{u}_{2} = \mathbf{u}_{3} $$ Finally, once more according to Frenet-Serret Formulas, $$ -\tau N = B^{\prime} = \mathbf{u}_{3}^{\prime} = - \overline{\tau} \mathbf{u}_{2} $$ hence we obtain $N = \mathbf{u}_{2}$.
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