Finite and Infinitesimal Flexibility of Semidiscrete Surfaces

The functions $|\dot{f}_{1}|$, $|\Delta f_{0}|$, and $|\Delta f_{1}|$ are infinitesimally preserved by infinitesimal flexions, hence $G_{11}$, $G_{22}$, and $G_{33}$ vanish.

The invariance of angles between $\dot{f}_{1}$ and $\Delta f_{0}$, and $\dot{f}_{1}$ and $\Delta f_{1}$ yield the equations $G_{12}+G_{21}=0$ and $G_{13}+G_{31}=0$, respectively. $\square$

We start with Proposition 5$(i)$. Consider three vectors

We introduce the notation

By Proposition 3 there exists a unique solution $(G_{11},G_{12},\ldots,G_{33})$ satisfying the initial conditions $G_{ij}(a)=c_{ij}$. For every point $t\in[a,b]$ the values $\mathcal{D}\dot{f}_{1}$, $\mathcal{D}\Delta f_{0}$, and $\mathcal{D}\Delta f_{1}$ of the tangent vector $\mathcal{D}f$ are uniquely defined in the basis $(\dot{f}_{1},\Delta f_{0},\Delta f_{1})$ by Eq. (10): here we substitute the solution of System A with the initial conditions $G_{ij}(a)=c_{ij}$ to the right hand side of Eq. (10). Hence, there exists a unique tangent vector $\mathcal{D}f$ of $f$ satisfying System A and the boundary conditions

This concludes the proof of the first item of the proposition. $\square$

Consider a point $h\in U$. The element $(G_{11},G_{12},\ldots,G_{33})$ itself satisfies a system of linear differential equations (System A). The coefficients of this system depend only on a point of $\tilde{\Omega}_{3{\times}3}^{1}$. Since the point $h$ is in general position, there exists a positive real constant $K$ such that for a sufficiently small neighborhood $V_{h}$ of $h$ the dependence is $K$-Lipschitz, i.e., for $p$ and $q$ from $V_{h}$ every coefficient $c$ of System A satisfies the inequality

Hence the solutions for $t\in[a,b]$ satisfy the Lipschitz condition for a fixed initial data on $V_{h}$. (This is clear from the fact that the solution of the system with small coefficients $c(p)-c(q)$ will be almost constant, the difference in each coordinate will not be greater than $9(b-a)K\|p-q\|$.) Finally the solution for $t\in[a,b]$ satisfies the Lipschitz condition for a fixed parameter and different initial data on $V_{h}$ (See Proposition 1.10.1 in [Cartan1967]). Therefore, for some constants $\overline{K}_{l}$ we have

From System A we know that the $\dot{G}_{i,j}$ linearly depend on $G_{11},G_{12},\ldots,G_{33}$, therefore, we get the Lipschitz condition for the derivatives: for some constants $\tilde{K}_{l}$ we have

Thus there exists a real number $\hat{K}$ such that for all points $p$ and $q$ in $V_{h}$,

Thus ${\mathcal{G}}$ satisfies a Lipschitz condition on $V_{h}$. Therefore, ${\mathcal{G}}$ satisfies a Lipschitz condition on $[0,\Lambda]\times V_{h}$ (since ${\mathcal{G}}$ is autonomous). $\square$

The proof of the Lemma is straightforward. $\square$

Let $\gamma=\{f^{\lambda}\}$, $\lambda\in[-\varepsilon,\varepsilon]$, be a normalized isometric deformation in $C^{1,2,1}_{0}([a,b],\mathbb{R}^{3})$ such that all 2-ribbon surfaces of the family are weakly regular. Every normalized deformation satisfies Condition (13) at every point $\lambda\in[-\varepsilon,\varepsilon]$ with the positive choice of the sign.

Since $\gamma$ is an isometric deformation, $\mathcal{D}_{\gamma}f^{\lambda}$ is an infinitesimal flexion at every point $\lambda\in[-\varepsilon,\varepsilon]$. Hence the corresponding functions $G_{ij}^{\lambda}$ satisfy system A (by Proposition 4). Let us now write ${\mathcal{D}_{\gamma}}\dot{f}_{1}^{\lambda}$, ${\mathcal{D}_{\gamma}}\Delta f_{0}^{\lambda}$, and ${\mathcal{D}_{\gamma}}\Delta f_{1}^{\lambda}$ in the basis $e_{1}$, $e_{2}$, $e_{3}$ using functions $G_{ij}^{\lambda}$. Recall that

Hence in the basis $(e_{1},e_{2},e_{3})$ we have

Similarly we have:

Hence by Definition 17 the corresponding derivatives in the space $\tilde{\Omega}^{1}_{3{\times}3}$ satisfy:

for every $\lambda\in[-\varepsilon,\varepsilon]$.

Conversely, let $Z^{-1}(f^{\lambda})$ satisfy

for every $\lambda\in[-\varepsilon,\varepsilon]$. Then the corresponding ${\mathcal{D}_{\gamma}}\dot{f}_{1}^{\lambda}$, ${\mathcal{D}_{\gamma}}\Delta f_{0}^{\lambda}$, and ${\mathcal{D}_{\gamma}}\Delta f_{1}^{\lambda}$ are defined as in (16) and (3.4.3). Hence the correspondent scalar products $G_{ij}^{\lambda}$ satisfy System A for $\lambda\in[-\varepsilon,\varepsilon]$. Thus ${\mathcal{D}_{\gamma}}f^{\lambda}$ is an infinitesimal flexion for every $\lambda\in[-\varepsilon,\varepsilon]$. Hence by $\{f^{\lambda}\}$ is an isometric deformation on $[-\varepsilon,\varepsilon]$. From construction it follows that $\{f^{\lambda}\}$ is a normalized deformation. $\square$

On the one hand, by Lemma 4 normalized isometric deformations of $f$ with a fixed initial position are in one-to-one correspondence with Solution of Eq. (15) satisfying $\gamma(0)=Z^{-1}(f)$. On the other hand, by Proposition 9 for sufficiently small positive $\varepsilon$ there exists a unique solution $\gamma$ of Eq. (15) satisfying $\gamma(0)=Z^{-1}(f)$. Hence, there exists a unique normalized isometric deformations of $f$ (with the parameter in $[-\varepsilon,\varepsilon]$ for sufficiently small positive $\varepsilon$). Therefore, $f$ has one degree of finite flexibility. $\square$

Note that

Let us prove the statement of the lemma for an arbitrary point $t_{0}$. Without loss of generality we fix $\mathcal{D}\dot{f}_{1}(t_{0})=0$ and $\mathcal{D}\Delta f_{1}(t_{0})=0$ (this is possible since every flexion is isometric to a flexion with such properties and isometries of flexions do not change the functions in the formula of the lemma). Then $\mathcal{D}\Delta f_{0}(t_{0})$ is proportional to $\dot{f}_{1}(t_{0}){\times}\Delta f_{0}(t_{0})$, and hence there exists some real number $\alpha$ with

Thus we immediately get

Let us express the summands for $\mathcal{D}\dot{\Phi}(t_{0})$. We start with $\big\langle\mathcal{D}\Delta\dot{f}_{0}(t_{0}),\Delta f_{1}(t_{0})\big\rangle$. First we note that

We do calculations at a point $t_{0}$ again assuming that $\mathcal{D}\dot{f}_{1}(t_{0})=0$ and $\mathcal{D}\Delta f_{1}(t_{0})=0$ (by choosing an appropriate isometric representative of the deformation). Let us show that $\mathcal{D}\dot{f}_{2}(t_{0})=0$. First, note that

Secondly we show that the inner products of $\mathcal{D}\Delta\dot{f}_{1}(t_{0})$ and the vectors $\dot{f}_{1}(t_{0})$, $\Delta f_{1}(t_{0})$, and $\dot{f}_{1}(t_{0}){\times}\Delta f_{1}(t_{0})$ are all zero (this would imply that $\mathcal{D}\Delta\dot{f}_{1}(t_{0})=0$).

From Eq. (7) we have

Further, from Eq. (5), we get

Finally, from the equation $\mathcal{D}(\dot{f}_{1},\Delta f_{1},\Delta\dot{f}_{1})=0$ we obtain

Therefore, $\mathcal{D}\Delta\dot{f}_{1}(t_{0})=0$, and hence $\mathcal{D}\dot{f}_{2}(t_{0})=0$.

From Eqs. (1) and (9) we get

Therefore, for some real number $\beta_{1}$ we have

By a similar reasoning (since we have shown that $\mathcal{D}\dot{f}_{2}(t_{0})=0$) we get

Since $\frac{\partial}{\partial t}\big(\mathcal{D}(\dot{f}_{1},\Delta f_{1},\dot{f}_{ 2})\big)=0$, at point $t_{0}$ we have

Hence,

and, therefore $\beta_{1}=\beta_{2}$. This implies

and

The last two formulas imply the statement of Proposition 11. $\square$

We restrict ourselves to the case of a point. Without loss of generality we assume that $\mathcal{D}\dot{f}_{1}(t_{0})=0$ and $\mathcal{D}\Delta f_{1}(t_{0})=0$. So as we have seen before, there exists $\alpha$ such that

and hence

Let us calculate $\mathcal{D}\langle\ddot{f}_{1},\ddot{f}_{1}\rangle=2\langle\mathcal{D}\ddot{f} _{1},\ddot{f}_{1}\rangle$. Decompose

Since

we get

By Eq. (8) we have

Hence after the substitution of $\mathcal{D}\Delta f_{0}(t_{0})$ in the first summand one gets

Therefore, we obtain

Since the statement does not depend on the choice of the basis and invariant under isometries, we get the statement for all the points. $\square$

The statement follows directly from Propositions 11 and 12. $\square$

Let $\mathcal{D}^{1}f^{1}$ and $\mathcal{D}^{2}f^{2}$ be isometrically nontrivial flexions of $f^{1}$ and $f^{2}$ respectively. Since $f^{1}$ and $f^{2}$ are strongly regular $(n{-}1)$-ribbon surfaces for $n\geq 4$, Theorem 13 can be applied. By Theorem 13(i) the surfaces $f^{1}$ and $f^{2}$ are strongly isometrically nontrivial. Hence by Theorem 13(i) in case $n>4$ or by Theorem 1 (see Remark 12 above) in case $n=4$ the induced flexions $\tilde{\mathcal{D}}^{1}f^{12}$ and $\tilde{\mathcal{D}}^{2}f^{12}$ are proportional, i.e, there exists $\alpha$ such that

Thus the surface $f$ has the combined infinitesimal flexion $\mathcal{D}f$ that induces $\mathcal{D}^{1}f^{1}$ and $\alpha\mathcal{D}^{2}f^{2}$. This flexion is infinitesimally nontrivial, since the induced ones are infinitesimally nontrivial. Hence $f$ has at least one degree of infinitesimal flexibility.

On the other hand by Theorem 13(i) the surface $f$ has at most one degree of infinitesimal flexibility. Hence $f$ is infinitesimally flexible and has one degree of infinitesimal flexibility. $\square$

By Proposition 13(ii) there exists monotonous functions $\xi_{1}$ and $\xi_{2}$ such that $\gamma_{1}\circ\xi_{1}$ and $\gamma_{2}\circ\xi_{2}$ are normalized isometric deformations of $f$. Hence by Theorem 4 in case $n\geq 3$ and Theorem 2 (see Remark 12 above) in case $n=2$ we have

in some neighborhood of 0. Set $\xi=\xi_{2}\circ\xi_{1}^{-1}$. The function $\xi$ is the monotonous function such that $\gamma_{1}(\lambda)=\gamma_{2}(\xi(\lambda))$ in some small neighborhood of 0. $\square$

By Theorem 4(ii) the surfaces $f^{1}$ and $f^{2}$ have one degree of finite flexibility. Therefore, there exist unique normalized isometric deformations $\gamma_{1}$ and $\gamma_{2}$ of $f^{1}$ and $f^{2}$ respectively. They induce two deformations $\tilde{\gamma}_{1}$ and $\tilde{\gamma}_{2}$ of the $(n{-}2)$-ribbon surface $f^{12}$. By Proposition 14, since $n-2\geq 2$, these two deformations locally parameterize the same curve in $C_{0}^{1,2,2,\ldots,2,1}([a,b],\mathbb{R}^{3})$, i.e., in the segment $[-\varepsilon,\varepsilon]$ for some $\varepsilon>0$ there exists a locally increasing function $\xi$ such that $\tilde{\gamma}_{1}(\lambda)=\tilde{\gamma}_{2}(\xi(\lambda))$.

Now consider the deformation $\gamma$ of the surface $f$ inducing both isometric deformations $\gamma_{1}$ for $f_{1}$ and $\gamma_{2}\circ\xi$ for $f_{2}$ in the segment $[-\varepsilon,\varepsilon]$ for some $\varepsilon>0$. The deformation $\gamma$ is isometric, since the induced ones are isometric. In addition, $\gamma$ is normalized, since its restriction to $f^{1}$ is a normalized isometric deformation. Hence $f$ is finitely flexible (and not finitely regularly rigid). Therefore, by Theorem 13(ii) $f$ has one degree of finite flexibility. $\square$