Macroscopic and Microscopic Behavior of Narrow Elastic Ribbons

Abstract

A one-dimensional model for a narrow ribbon is derived from the plate theory of Kirchhoff by means of a power expansion in the width variable. The energy found coincides with the corrected Sadowsky’s energy. Furthermore, we derive the Euler-Lagrange equations and use them to study an equilibrium configuration of a twisted ribbon. Within this example we also describe how to construct the fine scale oscillations that develop in the deformed configuration.

Appendix

1.1 A.1 Convexification

The aim of this appendix is to determine the convex envelope of

$$ W(\mathbf{L})=\left \{ \textstyle\begin{array}{l@{\quad }l} Q(\mathbf{L}) & \mbox{if }\det \mathbf{L}=0 , \\ +\infty & \mbox{else}, \end{array}\displaystyle \right . $$

where $Q$ is a positive definite quadratic form defined on 2-by-2 symmetric matrices.

Consider the function

$$ Q_{\mathrm{det}}(\mathbf{L};\alpha )=Q(\mathbf{L})+\alpha \det \mathbf{L} $$

with $\alpha$ a real number. For fixed $\alpha$, $Q_{\mathrm{det}}( \cdot ;\alpha )$ is a quadratic function—recall that $\mathbf{L}$ may be represented by a $2\times 2$ matrix—and hence it is convex if and only if $Q_{\mathrm{det}}(\mathbf{L};\alpha )\ge 0$ for every $\mathbf{L}$. This inequality is satisfied for all $\mathbf{L}$ with determinant equal to zero, since $Q$ is positive definite. If $\det \mathbf{L}>0$, the condition $Q_{\mathrm{det}}(\mathbf{L};\alpha ) \ge 0$ for every $\mathbf{L}$ holds if and only if

$$ Q(\mathbf{L}/\sqrt{\det \mathbf{L}})+\alpha \ge 0\quad \mbox{for every } \mathbf{L}, $$

that is equivalent to

$$ \alpha \ge - Q(\mathbf{L})\quad \mbox{for every }\mathbf{L}\mbox{ with }\det \mathbf{L}=1, $$

or

$$ \alpha \ge - \min_{\det \mathbf{L}=1} Q(\mathbf{L})=:-\alpha ^{-}. $$

(60)

Similarly, if $\det \mathbf{L}<0$, the condition $Q_{\mathrm{det}}( \mathbf{L};\alpha )\ge 0$ for every $\mathbf{L}$ holds if and only if

$$ \alpha \le \min_{\det \mathbf{L}=-1} Q(\mathbf{L})=:\alpha ^{+}. $$

(61)

Note that $\alpha ^{-}, \alpha ^{+}>0$, since $Q$ is positive definite. Hence, $Q_{\mathrm{det}}(\cdot ;\alpha )$ is convex if and only if $-\alpha ^{-}\le \alpha \le \alpha ^{+}$, and

$$ W_{c}(\mathbf{L})=\sup_{-\alpha ^{-}\le \alpha \le \alpha ^{+}}Q_{\mathrm{det}}( \mathbf{L};\alpha ) $$

(62)

is also convex, for the point-wise supremum of convex functions is convex. It immediately follows that

$$ W_{c}(\mathbf{L})=Q(\mathbf{L})+\alpha ^{+}(\det \mathbf{L})^{+}+\alpha ^{-}( \det \mathbf{L})^{-}, $$

(63)

where $(\det \mathbf{L})^{+}$ is the positive part of $\det \mathbf{L}$, and $(\det \mathbf{L})^{-}$ is the negative part, i.e., $(\det \mathbf{L})^{-}=- \det \mathbf{L}$ if $\det \mathbf{L}<0$ and zero otherwise.

Clearly, $W(\mathbf{L})\ge Q_{\mathrm{det}}(\mathbf{L};\alpha )$ for any $\mathbf{L}$ and for any $\alpha$, hence $W(\mathbf{L})\ge W_{c}( \mathbf{L})$. Since $W_{c}$ is convex it also follows that

$$ W^{**}(\mathbf{L})\ge W_{c}(\mathbf{L}). $$

In the rest of the section we will show that the inequality above is indeed an equality, and that the lower bound can be achieved by a single lamination. More precisely, we show that given $\mathbf{L}$ we can find two symmetric tensors $\mathbf{A}$ and $\mathbf{B}$, and $0< t<1$ such that

$$ (1-t)\mathbf{A}+t\mathbf{B}=\mathbf{L}, \quad \mbox{and} \quad (1-t)W( \mathbf{A})+t W(\mathbf{B})=W_{c}(\mathbf{L}). $$

By the definition of $W$, the statement above is equivalent to: given $\mathbf{L}$ there exist two symmetric tensors $\mathbf{A}$ and $\mathbf{B}$ with $\det \mathbf{A}=\det \mathbf{B}=0$, and $0< t<1$ such that

$$ (1-t)\mathbf{A}+t\mathbf{B}=\mathbf{L}, \quad \mbox{and} \quad (1-t)Q( \mathbf{A})+t Q(\mathbf{B})=W_{c}(\mathbf{L}). $$

In this statement we may eliminate the tensor $\mathbf{A}$ by noticing that: given $\mathbf{L},\mathbf{B}$ and $t$, as above, there exists a symmetric tensor $\mathbf{A}$ with $\det \mathbf{A}=0$ and satisfying the equation

$$ (1-t)\mathbf{A}=\mathbf{L}-t\mathbf{B} $$

if and only if $\det (\mathbf{L}- t\mathbf{B})=0$. Since

$$\begin{aligned} (1-t)Q(\mathbf{A})+t Q(\mathbf{B}) &=\frac{1}{1-t} Q\bigl((1-t)\mathbf{A} \bigr)+t Q( \mathbf{B}) \\ &=\frac{1}{1-t} Q(\mathbf{L}-t\mathbf{B})+t Q(\mathbf{B}) \\ &=Q(\mathbf{L})+\frac{t}{1-t} Q(\mathbf{L}-\mathbf{B}), \end{aligned}$$

the statement to be proven is: given $\mathbf{L}$ there exist a symmetric tensor $\mathbf{B}$ with $\det \mathbf{B}=0$, and $0< t<1$ such that

$$ \det (\mathbf{L}- t\mathbf{B})=0, \quad \mbox{and} \quad Q( \mathbf{L})+\frac{t}{1-t} Q(\mathbf{L}-\mathbf{B})=W_{c}( \mathbf{L}). $$

(64)

Recalling (60) and (61), let $\mathbf{D}^{-}, \mathbf{D}^{+}$ be the two symmetric matrices with $\det \mathbf{D}^{-}=-1$ and $\det \mathbf{D}^{+}=1$ such that

$$ \alpha ^{-}=Q\bigl(\mathbf{D}^{+}\bigr), \qquad \alpha ^{+}=Q\bigl(\mathbf{D}^{-}\bigr). $$

(65)

Note that the tensors $\mathbf{D}^{-}$ and $\mathbf{D}^{+}$ only depend on the bending stiffness tensor $\mathbb{D}$. We now show that if $\det \mathbf{L}>0$ there exist a symmetric tensor $\mathbf{B}$ with null determinant and $0< t<1$ such that

$$ \det (\mathbf{L}- t\mathbf{B})=0, \quad \quad \det (\mathbf{L}- \mathbf{B})=- \frac{1-t}{t} \det \mathbf{L}, $$

(66)

and

$$ \mathbf{D}^{-}=\frac{\mathbf{L}-\mathbf{B}}{\sqrt{-\det (\mathbf{L}- \mathbf{B})}}. $$

(67)

Assuming, for the moment, the validity of this statement we deduce (64) in the case $\det \mathbf{L}>0$:

$$\begin{aligned} Q(\mathbf{L})+\frac{t}{1-t} Q(\mathbf{L}-\mathbf{B}) &=Q(\mathbf{L})+ \frac{t}{1-t} \bigl(-\det (\mathbf{L}-\mathbf{B})\bigr) Q\bigl( \mathbf{D}^{-}\bigr) \\ &=Q(\mathbf{L})+Q\bigl(\mathbf{D}^{-}\bigr)\det \mathbf{L} \\ &=Q(\mathbf{L})+\alpha ^{+}\det \mathbf{L}=W_{c}( \mathbf{L}), \end{aligned}$$

where we have used, in order, (67), (66), (65), and (63). Hence, assuming (66) and (67) we have shown that if $\det \mathbf{L}>0$ then $W^{**}(\mathbf{L})= W_{c}(\mathbf{L})$. A similar proof can be made for $\det \mathbf{L}<0$, while the result is trivial for $\det \mathbf{L}=0$.

We therefore only need to show the validity of (66) and (67). In doing it, we shall provide formulae that deliver $\mathbf{B}$ and $t$.

For $\det \mathbf{L}>0$, set $\mathbf{B}=\mathbf{L}-\eta \mathbf{D}^{-}$ where the real number $\eta$ is chosen by imposing that $\det \mathbf{B}=0$. With (1), this amount to solve the second order equation

$$ \det \mathbf{L}-\eta \mathbf{L}^{*}\cdot \mathbf{D}^{-} -\eta ^{2}=0, $$

(68)

where we used the fact that $\det \mathbf{D}^{-}=-1$. Of the two solutions we choose

$$ \eta =\frac{-\mathbf{L}^{*}\cdot \mathbf{D}^{-} + \sqrt{(\mathbf{L}^{*} \cdot \mathbf{D}^{-})^{2}+4\det \mathbf{L}}}{2}. $$

To keep the notation compact, we set

$$ \lambda =\frac{\mathbf{L}^{*}\cdot \mathbf{D}^{-}}{2 \sqrt{\det \mathbf{L}}} $$

so that

$$ \eta =\sqrt{\det \mathbf{L}}\bigl(\sqrt{\lambda ^{2}+1}-\lambda \bigr). $$

Now that $\mathbf{B}$ has been fixed, we find $t$ by solving (66)₁. We first write

$$ \mathbf{L}- t\mathbf{B}=\mathbf{L}- t\bigl(\mathbf{L}-\eta \mathbf{D}^{-} \bigr)=(1-t) \mathbf{L}+t \eta \mathbf{D}^{-}, $$

and by using (1), we write (66)₁ as

$$ (1-t)^{2}\det \mathbf{L}+(1-t)t \eta \mathbf{L}^{*}\cdot \mathbf{D}^{-}-t ^{2} \eta ^{2}=0, $$

which simplifies, thanks to (68), to

$$ \det \mathbf{L}+t\bigl(\eta \mathbf{L}^{*}\cdot \mathbf{D}^{-}-2 \det \mathbf{L}\bigr)=0. $$

Thus

$$ t=\frac{-\det \mathbf{L}}{\eta \mathbf{L}^{*}\cdot \mathbf{D}^{-}-2\det \mathbf{L}}=\frac{1}{2-2\lambda (\sqrt{\lambda ^{2}+1}-\lambda )}. $$

We may easily verify that $0< t<1$. Indeed, $t>0$ if and only if $2\lambda (\sqrt{\lambda ^{2}+1}-\lambda )<2$, which clearly holds if $\lambda \le 0$, while for $\lambda >0$ is equivalent to $\sqrt{ \lambda ^{2}+1}<\frac{1}{\lambda }+\lambda$ which is trivially verified. Similarly we may check that $t<1$.

From the definition of $\mathbf{B}$, we have that $\eta \mathbf{D}^{-}= \mathbf{L}-\mathbf{B}$. Taking the determinant, we find $-\eta ^{2}=\det ( \mathbf{L}-\mathbf{B})$ from which (67) follows. Finally, by applying (1) twice, we obtain that

$$\begin{aligned} \det (\mathbf{L}-\mathbf{B}) &=\det \mathbf{L}-\mathbf{L}^{*}\cdot \mathbf{B}= \det \mathbf{L}-\frac{\det \mathbf{L}}{t} + \frac{\det \mathbf{L}-\mathbf{L} ^{*}\cdot (t\mathbf{B})}{t} \\ &=\frac{t-1}{t}\det \mathbf{L}+\frac{\det (\mathbf{L}-t\mathbf{B})}{t}, \end{aligned}$$

and recalling (66)₁ we deduce (66)₂.

We close this appendix by summarizing the results found:

$$ W^{**}(\mathbf{L})=Q(\mathbf{L})+\alpha ^{+}(\det \mathbf{L})^{+}+\alpha ^{-}( \det \mathbf{L})^{-}, $$

where

$$ \alpha ^{-}=\min_{\det \mathbf{L}=1}Q(\mathbf{L})=Q\bigl( \mathbf{D}^{+}\bigr), \qquad \alpha ^{+}=\min _{\det \mathbf{L}=-1}Q(\mathbf{L})=Q\bigl(\mathbf{D}^{-}\bigr). $$

If $\det \mathbf{L}>0$, set

$$ \begin{aligned} \lambda &=\frac{\mathbf{L}^{*}\cdot \mathbf{D}^{-}}{2 \sqrt{\det \mathbf{L}}}, \quad \quad t=\frac{1}{2-2\lambda \bigl(\sqrt{\lambda ^{2}+1}-\lambda \bigr)}, \\ \eta &=\sqrt{\det \mathbf{L}}\bigl(\sqrt{\lambda ^{2}+1}-\lambda \bigr),\quad \quad \mathbf{B}=\mathbf{L}-\eta \mathbf{D}^{-}; \end{aligned} $$

while, if $\det \mathbf{L}<0$, set

$$ \begin{aligned} \lambda &=\frac{\mathbf{L}^{*}\cdot \mathbf{D}^{+}}{2\sqrt{-\det \mathbf{L}}}, \quad \quad t=\frac{1}{2+2\lambda \bigl(\lambda -\sqrt{\lambda ^{2}+1}\bigr)}, \\ \eta &=\sqrt{-\det \mathbf{L}}\bigl(\lambda -\sqrt{\lambda ^{2}+1}\bigr),\quad \quad \mathbf{B}=\mathbf{L}-\eta \mathbf{D}^{+}. \end{aligned} $$

Then, with

$$ \mathbf{A}=\frac{\mathbf{L}-t\mathbf{B}}{1-t}, $$

we have that

$$ (1-t)\mathbf{A}+t\mathbf{B}=\mathbf{L}, \quad \mbox{and} \quad (1-t)Q( \mathbf{A})+t Q(\mathbf{B})=W^{**}(\mathbf{L}). $$

1.2 A.2 Evaluation of $\alpha ^{\pm }$ for Various Symmetries

We first turn our attention to the computation of the constants $\alpha ^{+}$ and $\alpha ^{-}$ in the definition of the convexification $W^{**}$ in (24). According to (25) we must minimize the quadratic form $Q(\mathbf{L})$ over the manifolds $\{\mathbf{L}: \det \mathbf{L}=\pm 1\}$. We accomplish this task through the method of Lagrange multipliers by seeking stationary points of the augmented functional$(\mathbf{L},\beta )\mapsto Q(\mathbf{L})+\beta (\det \mathbf{L}\pm 1)$. We note that the determinant of a 2-by-2 matrix is a quadratic form. Thus, there exists a fourth-order tensor $\mathbb{E}$ such that

$$ \det \mathbf{L}=L_{11}L_{22}-L_{12}^{2}= \frac{1}{2}\mathbb{E}\mathbf{L} \cdot \mathbf{L}. $$

(69)

Thus, the augmented functional is

$$ (\mathbf{L},\beta )\mapsto \frac{1}{2} (\mathbb{D}+\beta \mathbb{E}) \mathbf{L}\cdot \mathbf{L}\mp \beta . $$

(70)

We now argue that the minima $\alpha ^{\pm }$ have the following property:

$$ \mathbb{D} \mp \alpha ^{\mp }\mathbb{E} \text{ is singular}, $$

(71)

that is $\alpha ^{\mp }$ are generalized eigenvalues of$\mathbb{D}$with respect to$\mathbb{E}$. To verify the last statement, let us recall the definition of $\mathbf{D}^{\pm }$, see (27), that is

$$ \mathbf{D}^{\pm }=\operatorname{argmin}_{\det \mathbf{L}=\pm 1} Q( \mathbf{L}). $$

(72)

Then there exist $\beta ^{\mp }$ such that the pair $(\mathbf{D}^{\pm }, \beta ^{\mp })$ is a stationary point of the augmented functional defined in (70). The stationarity conditions for such pair are

$$\begin{aligned} &\bigl(\mathbb{D}+\beta ^{\mp }\mathbb{E}\bigr) \mathbf{D}^{\pm }=\mathbf{0}, \end{aligned}$$

(73a)

$$\begin{aligned} &\frac{1}{2}\mathbb{E}\mathbf{D}^{\pm }\cdot \mathbf{D}^{\pm }\mp 1=0. \end{aligned}$$

(73b)

It follows from the second stationarity condition that $\mathbf{D}^{ \pm }\neq \mathbf{0}$. This implies that the tensor $\mathbb{D}+ \beta ^{\mp }\mathbb{E}$ is singular. Moreover, the minimum $\alpha ^{\mp }$ coincides with the Lagrange multiplier $\beta ^{\mp }$, up to sign:

$$ \alpha ^{\mp }=Q\bigl(\mathbf{D}^{\pm }\bigr)=\frac{1}{2} \mathbb{D}\mathbf{D}^{ \pm }\cdot \mathbf{D}^{\pm }\stackrel{ \text{(73a)}}{=}-\frac{ \beta ^{\mp }}{2}\mathbb{E}\mathbf{D}^{\pm } \cdot \mathbf{D}^{\pm }\stackrel{ \text{(73b)}}{=}\mp \beta ^{\mp }. $$

(74)

To carry our calculation further on, we observe that

$$ \mathbb{D}\mathbf{L}\cdot \mathbf{L}= \begin{pmatrix} \mathbb{D}_{1111} & \mathbb{D}_{1122} & \mathbb{D}_{1112} \\ \mathbb{D}_{1122} & \mathbb{D}_{2222} & \mathbb{D}_{1222} \\ \mathbb{D}_{1112} & \mathbb{D}_{1222} & \mathbb{D}_{1212} \end{pmatrix} \begin{pmatrix} L_{11} \\ L_{22} \\ 2L_{12} \end{pmatrix} \cdot \begin{pmatrix} L_{11} \\ L_{22} \\ 2L_{12} \end{pmatrix} $$

(75)

and that

$$ \mathbb{E}\mathbf{L}\cdot \mathbf{L}= \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1/2 \end{pmatrix} \begin{pmatrix} L_{11} \\ L_{22} \\ 2L_{12} \end{pmatrix} \cdot \begin{pmatrix} L_{11} \\ L_{22} \\ 2L_{12} \end{pmatrix} . $$

(76)

Thus, on denoting by $\mathbb{d}$ and $\mathbb{e}$ the matrices appearing in the definitions of the quadratic forms (75) and (76), we can write (71) as

$$det(\mathbb{d}\mp {\alpha }^{\mp }\mathbb{e})=0.$$

(77)

Orthotropic Response

In the special case, when the material is orthotropic with respect to the basis $(\mathbf{e}_{1},\mathbf{e}_{2})$, we have

$$ \mathbb{D}_{1112}=0, \quad \text{and} \quad \mathbb{D}_{1222}=0 $$

(the other vanishing components of $\mathbb{D}$ are obtained by application of the standard symmetries: $\mathbb{D}_{\alpha \beta \gamma \delta }=\mathbb{D}_{\beta \alpha \gamma \delta }=\mathbb{D} _{\gamma \delta \alpha \beta }$). In this case, the characteristic polynomial

$$p(\beta ):=det(\mathbb{d}+\beta \mathbb{e})=({\mathbb{D}}_{1212}-\frac{\beta }{2})({\mathbb{D}}_{1111}{\mathbb{D}}_{2222}-{({\mathbb{D}}_{1122}+\beta )}^2)$$

has the following set of roots:

$$ \{-\sqrt{\mathbb{D}_{1111}\mathbb{D}_{2222}}- \mathbb{D}_{1122},\sqrt{ \mathbb{D}_{1111}\mathbb{D}_{2222}}- \mathbb{D}_{1122}, 2\mathbb{D} _{1212}\}. $$

The positive definiteness of $\mathbb{D}$ implies that the first root is negative, and that the remaining roots are positive. The negative root must coincide with $\beta ^{-}=-\alpha ^{-}$, that is

$$ \alpha ^{-}=\sqrt{\mathbb{D}_{1111}\mathbb{D}_{2222}}+ \mathbb{D}_{1122}. $$

On the other hand, the minimality of $\alpha ^{+}$ yields

$$ \alpha ^{+}=\min \{\sqrt{\mathbb{D}_{1111} \mathbb{D}_{2222}}- \mathbb{D}_{1122}, 2\mathbb{D}_{1212} \}. $$

We note that the foregoing calculation can be used also when the plate is orthotropic with respect to a basis $(\mathbf{e}'_{1},\mathbf{e}'_{2})$, not necessarily coincident with $(\mathbf{e}_{1},\mathbf{e}_{2})$. In this case, the relevant strain energy is $W'(\mathbf{L})=W(\mathbf{R}\mathbf{L} \mathbf{R}^{T})$, where $\mathbf{R}=\mathbf{e}'_{\alpha }\otimes \mathbf{e} _{\alpha }$ is a rotation matrix. Then it can be easily shown that the convexification of $Q'$ is given by

$$ \bigl(W'\bigr)^{**}(\mathbf{L})=W^{**}\bigl( \mathbf{R}\mathbf{L}\mathbf{R}^{T}\bigr). $$

Isotropic Response

For the isotropic strain energy $Q(\mathbf{L})=d_{\mu }|\mathbf{L}|^{2}+\frac{d _{\lambda }}{2} (L_{11}+L_{22})^{2}$ the relevant components of $\mathbb{D}$ are $\mathbb{D}_{1111}=\mathbb{D}_{2222}=2d_{\mu }+d_{ \lambda }$, $\mathbb{D}_{1122}=d_{\lambda }$, $\mathbb{D}_{1212}=d _{\mu }$. Then

$$ \alpha ^{-}=2(d_{\mu }+d_{\lambda }), \qquad \alpha ^{+}=2d_{\mu }, $$

and so

$$\begin{aligned} W^{**}(\mathbf{L}) =& d_{\mu }|\mathbf{L}|^{2}+ \frac{d_{\lambda }}{2}(L_{11}+L_{22})^{2}+2d _{\mu }(\det \mathbf{L})^{+}+2(d_{\mu }+d_{\lambda }) (\det \mathbf{L})^{-} \\ =& d_{\mu }|\mathbf{L}|^{2}+\frac{d_{\lambda }}{2} | \mathbf{L}|^{2} +d_{ \lambda }\det \mathbf{L}+2d_{\mu }(\det \mathbf{L})^{+}+2(d_{\mu }+d_{ \lambda }) (\det \mathbf{L})^{-} \\ =& \biggl(d_{\mu }+\frac{d_{\lambda }}{2} \biggr) |\mathbf{L}|^{2} +2 d _{\mu }|\det \mathbf{L}| +d_{\lambda }\bigl(\det \mathbf{L}+2( \det \mathbf{L})^{-}\bigr) \\ =& \biggl(d_{\mu }+\frac{d_{\lambda }}{2} \biggr) \bigl(| \mathbf{L}|^{2} +2 |\det \mathbf{L}| \bigr). \end{aligned}$$

1.3 A.3 Compatibility Among Curvatures

The aim of this Appendix is to prove the following statement.

Let $\omega , \omega _{\mathbf{A}}$ and $\omega _{\mathbf{B}}$ be three two-dimensional open sets, with $\omega = \omega _{\mathbf{A}}\cup \omega _{\mathbf{B}}\cup \varGamma$, where $\varGamma =\overline{\omega }_{ \mathbf{A}}\cap \overline{\omega }_{\mathbf{B}}$ is a smooth curve and $\overline{\omega }_{\mathbf{A}}$ denotes the closure of $\omega _{ \mathbf{A}}$. Let $\mathbf{A}$ and $\mathbf{B}$, with $\mathbf{A}\ne \mathbf{B}$, be two 2-by-2 symmetric matrices with null determinant. If there exists a deformation $\mathbf{y}$ continuously differentiable on $\omega$ and twice differentiable on $\overline{\omega }_{\mathbf{A}}$ and on $\overline{\omega }_{\mathbf{B}}$ such that

$$ \mathbf{K}_{\mathbf{y}}=\mathbf{A}\mbox{ on }\overline{\omega }_{\mathbf{A}}, \quad \mbox{and}\quad \mathbf{K}_{\mathbf{y}}=\mathbf{B} \mbox{ on }\overline{ \omega }_{\mathbf{B}}, $$

where $\mathbf{K}_{\mathbf{y}}$ denotes the second fundamental form of $\mathbf{y}$, then the curve $\varGamma$ is straight and the matrices $\mathbf{A}$ and $\mathbf{B}$ are proportional, that is, there exists a constant $\sigma$ for which either $\mathbf{A}=\sigma \mathbf{B}$ or $\mathbf{B}=\sigma \mathbf{A}$.

We assume that a deformation $\mathbf{y}$ as described in the statement above exists and we study the consequences.

Since $\mathbf{A}\ne \mathbf{B}$, we may assume without loss of generality that $\mathbf{A}\neq \mathbf{0}$. Since $\mathbf{y}$ is continuously differentiable we have that $\nabla \mathbf{y}$ and the normal $\boldsymbol{\nu }= \partial _{1}\mathbf{y}\wedge \partial _{2}\mathbf{y}$ are continuous on $\omega$. Let $\mathbf{t}$ be the unit tangent to the curve $\varGamma$. Since $(\nabla \boldsymbol{\nu })\mathbf{t}= \partial _{\mathbf{t}}\boldsymbol{\nu }$ and since $\mathbf{K}=- (\nabla \mathbf{y})^{\top }\nabla \boldsymbol{\nu }$ we have that

$$ \mathbf{A}\mathbf{t}= \mathbf{B}\mathbf{t},\quad \mbox{on } \varGamma . $$

Thus $\det (\mathbf{A}-\mathbf{B})=0$, and the identity $\det (\mathbf{A}- \mathbf{B})=\det \mathbf{A}-\mathbf{A}^{*}\cdot \mathbf{B}+ \det \mathbf{B}$ implies that

$$ \mathbf{A}^{*}\cdot \mathbf{B}=0. $$

Let $(\lambda ,\mathbf{a}_{1})$ and $(0,\mathbf{a}_{2})$ be the pairs of eigenvalues and eigenvectors of $\mathbf{A}$, so that

$$ \mathbf{A}=\lambda \mathbf{a}_{1}\otimes \mathbf{a}_{1}, $$

with $\lambda \neq 0$. Then $\mathbf{A}^{*}= \lambda \mathbf{a}_{2}\otimes \mathbf{a}_{2}$ and $\mathbf{A}^{*}\cdot \mathbf{B}=0$ implies that $\mathbf{B}\mathbf{a}_{2}\cdot \mathbf{a}_{2}=0$. We may therefore write

$$ \mathbf{B}=\beta \mathbf{a}_{1}\otimes \mathbf{a}_{1}+ \tilde{\beta }( \mathbf{a}_{1}\otimes \mathbf{a}_{2}+ \mathbf{a}_{2}\otimes \mathbf{a}_{1}), $$

and, since $\det \mathbf{B}=0$, conclude that

$$ \mathbf{B}=\beta \mathbf{a}_{1}\otimes \mathbf{a}_{1}. $$

Setting $\sigma =\beta /\lambda$ we obtain $\mathbf{B}= \sigma \mathbf{A}$. Since $\mathbf{A}\ne \mathbf{B}$, from the identity $\mathbf{A}\mathbf{t}= \mathbf{B}\mathbf{t}$ we deduce that $\mathbf{a}_{1}\cdot \mathbf{t}=0$, which implies that $\mathbf{t}$ is constant and, in turn, that the curve $\varGamma$ is a straight segment.