[ Energy scaling and branched microstructures in a model for shape-memory alloys with $\operatorname{SO}(2)$ invariance\
]{} [October 20, 2014]{}\
Allan Chan and Sergio Conti\
[*Institut für Angewandte Mathematik, Universität Bonn\
53115 Bonn, Germany* ]{}\

Domain branching near the boundary appears in many singularly perturbed models for microstructure in materials and was first demonstrated mathematically by Kohn and Müller for a scalar problem representing the elastic behavior of shape-memory alloys. We study here a model for shape-memory alloys based on the full vectorial problem of nonlinear elasticity, including invariance under rotations, in the case of two wells in two dimensions. We show that, for two wells with two rank-one connections, the energy is proportional to the power $2/3$ of the surface energy, in agreement with the scalar model. In a case where only one rank-one connection is present, we show that the energy exhibits a different behavior, proportional to the power $4/5$ of the surface energy. This lower energy is achieved by a suitable interaction of the two components of the deformations and hence cannot be reproduced by the scalar model. Both scalings are proven by explicit constructions and matching lower bounds.

Introduction
============

The peculiar elastic properties of shape-memory alloys have attracted a large interest in the mechanics literature, as discussed for example in [@B9_Bhatta; @OtsukaWayman1999], starting with the works of Ball and James [@B9_BallJames87; @B9_BallJames92] also in the vectorial calculus of variations, see also [@ChipotKinderlehrer88; @Dacorogna1989; @MuellerLectureNotes] and the references therein. The typical model, based on nonlinear elasticity, takes the form $$\label{eqfctnonlineaelast}
\int_\Omega W(Du) dx\,,$$ where $u:\Omega\to\R^n$ is the elastic deformation. The energy density $W:\R^{n\times n}\to\R$ is minimized by several copies of the set of proper rotations $\operatorname{SO}(n)$, which correspond to the several phases of the material. The macroscopic material behavior can be analyzed using the theory of relaxation, which – under appropriate continuity and growth conditions – leads to the study of the quasiconvex envelope of $W$ [@Dacorogna1989; @MuellerLectureNotes]. Partial results in this direction have been obtained for example in [@B9_BallJames92; @Sverak1993; @DolzmannLN; @DolzmannKirchheim2003; @ContiDolzmannKirchheim2007; @ContiDolzmanntwowell]. The theory of relaxation does not, however, make any prediction on the length scale of the microstructure, as the functional (\[eqfctnonlineaelast\]) is scale invariant.

A more precise analysis of the microstructure is possible if one includes a singular perturbation, typically of the form $\varepsilon|D^2u|(\Omega)$ (in the sense of the total variation of the gradient of the $BV$ function $Du$) or $\varepsilon^2 \|D^2u\|_{L^2}^2$, where $\varepsilon$ represents a (scaled) surface energy. A scalar simplification of this type of model was proposed by Kohn and Müller in 1992 [@B9_KohnMuller92; @B9_KohnMuller94]. They have shown that the minimum energy scales as $\varepsilon^{2/3}$, and that this scaling is achieved by a self-similar microstructure, which refines close to the boundary. Finer results were later obtained in [@B9_Conti00; @B9_Conti2006; @B9_Zwicknagl], including in particular the statement of asymptotic self-similarity for the minimizers of the Kohn-Müller functional. Their approach was extended to many other physical problems, including for example micromagnetism [@CK98; @CKO99; @Viehmanndiss], magnetic structures in superconductors [@B9_CKO2004; @B9_CCKO2008], dislocation structures in plasticity [@B9_ContiOrtiz05], coarsening in thin film growth [@ContiOrtiz2008].

We consider in this work the full vectorial problem of nonlinear elasticity with surface energy, namely, the functional $$\label{intro:1}
  \int_{\Omega}W(Du)\,dx + \varepsilon |D^2u|(\Omega)\,.$$ Here, $u:\Omega\subset\R^n\rightarrow\R^n$ represents the elastic deformation, $W:\R^{n\times n}\rightarrow\R$ the energy density, which is minimized on a set $K\subset\R^{n\times n}$ and grows quadratically. For definiteness, we shall focus on $$\label{intro:2}
 W(F) = \operatorname{dist}^2(F,K)= \inf\{|F-G|^2: \ G\in K\}\,,$$ although all results hold for any energy with the same set of minimizers and the same growth. The small positive parameter $\varepsilon$ in (\[intro:1\]) represents the surface energy, and the admissible deformations are those $u\in
W^{1,2}(\Omega;\R^n)$ with $Du\in BV(\Omega;\R^{n\times n})$ which obey specific boundary conditions. By $|D^2u|(\Omega)$ we denote the total variation of the measure $D^2u$ over the set $\Omega$. The set $K\subset\R^{n\times n}$ of energy-minimizing deformation gradients, as used in (\[intro:2\]), depends on the crystallography of the phase transformation considered.

In this paper we focus on the two-dimensional case, $n=2$, and study two different crystallographic settings. The first situation we address, denoted case $K_1$, is $$\label{assumption:K1}
 A_1= {\ensuremath{\begin{pmatrix}1&-\alpha \\ 0&1 \end{pmatrix}}}, \quad B_1={\ensuremath{\begin{pmatrix}1&\alpha \\ 0&1 \end{pmatrix}}}, \quad K_1=\operatorname{SO}(2)A_1\cup\operatorname{SO}(2)B_1$$ for some $\alpha>0$. These matrices have the same determinant and are representative of volume-preserving phase transformations in which the two wells have two rank-one connections, in the sense that $\det(A_1-QB_1)=0$ has two solutions $Q\in
\operatorname{SO}(2)$. This is the vectorial situation that was represented by the scalar model of Kohn and Müller; indeed, one can obtain a model similar to the one they used if $K_1$ is reduced to the two matrices $\{A_1, B_1\}$, see [@B9_Zwicknagl]. The two components of $u$ are in this case decoupled, only the first one gives a nontrivial energy contribution.

We show that for this problem the energy is proportional to $\varepsilon^{2/3}$ if the aspect ratio of the domain is of order 1 (Theorem \[Theorem:K1\]), thereby extending the Kohn-Müller result to the present vectorial model. The presence of a second rank-one connection becomes, however, important in the case of long and thin domains, in which microstructure is generated along vertical oscillations instead of horizontal ones, as we discuss after Theorem \[Theorem:K1\] and in Section \[secubtheo1\] below.

The second situation we consider, denoted case $K_2$, is $$\label{assumption:K2}
A_2={\ensuremath{\begin{pmatrix}1&0 \\ 0&1-\alpha \end{pmatrix}}}, \quad B_2={\ensuremath{\begin{pmatrix}1&0 \\ 0&1+\alpha \end{pmatrix}}}, \quad K_2=\operatorname{SO}(2)A_2\cup\operatorname{SO}(2)B_2$$ for some $\alpha\in(0,1)$. For $\alpha=1$, the matrix $A_2$ becomes singular; treating this unphysical case would generate some complications in the proofs, for clarity we restrict in the relevant parts to $\alpha<1/2$. Clearly, any other condition $\alpha<\alpha_*$, for any fixed $\alpha_*<1$, would lead to the same results. Physically relevant values of the spontaneous strain are typically of a few percent, up to about $10\%$ [@B9_Bhatta; @OtsukaWayman1999], and therefore well within the range $(0,1/2)$.

The matrices in (\[assumption:K2\]) have a different determinant and are representative of phase transformations with one rank-one connection, in the sense that $\det(A_2-QB_2)=0$ has a unique solution $Q\in 
\operatorname{SO}(2)$. In this case, the invariance under rotations and the vectorial nature of the deformation can interact to reduce the energy cost of domain branching. The resulting scaling, proportional to $\varepsilon^{4/5}$, differs from the one derived for the scalar problem. This difference can be understood as a consequence of the degeneracy of the rank-one connection (in the same sense as for double roots of polynomials). To see this, let us consider a typical branching pattern, as illustrated in the top right part of Figure \[fig:PD2\], which consists of a fine-scale mixture of regions with $Du$ close to $\operatorname{SO}(2)A_j$ and regions with $Du$ close to $\operatorname{SO}(2)B_j$, with interfaces approximately horizontal. Since $u$ does not jump across the interface, if $u$ is sufficiently regular, the tangent component of the gradient does not jump either. In other words, given a vector $v\in S^1$ tangent to the interface, then the directional derivative $Duv$ must be the same on the two sides of the interface. If $Du\in \operatorname{SO}(2)A_j$ on one side, and $Du\in \operatorname{SO}(2)B_j$ on the other side, then necessarily $|A_jv|=|B_jv|$. For $j=1$, this equation is equivalent to $$(v_1-\alpha v_2)^2+v_2^2 = 
  (v_1+\alpha v_2)^2+v_2^2 \,,$$ which simplifies to $\alpha v_1v_2=0$. For $j=2$ the condition $|A_jv|=|B_jv|$ is equivalent to $$v_1^2 + (1-\alpha)^2 v_2^2
=  v_1^2 + (1+\alpha)^2 v_2^2\,,$$ which simplifies to $\alpha v_2^2=0$. It is easy to see that $v_2=0$ (corresponding to $v=e_1$) is a simple solution in the first case, but a degenerate solution in the second case. Correspondingly, $$\text{$|A_1v|-|B_1v|=O(\alpha|v-e_1|)$ but $|A_2v|-|B_2v|=O(\alpha|v-e_1|^2)$.  }$$ Therefore, small deviations of the interfacial direction from $v=e_1$ generate less elastic energy in the case $K_2$ than in the case $K_1$, leading to a lower global energy. If rotations are neglected, which is the case if one treats the scalar case, then the two terms are of the same order, $$A_1v-B_1v=-2\alpha v_2 e_1=O(\alpha |v-e_1|)$$ and $$A_2v-B_2v=-2\alpha v_2 e_2=O(\alpha |v-e_1|)\,.$$ Therefore, the difference between $K_1$ and $K_2$ is not present in scalar models. The detailed exploitation of this effect in the construction requires an appropriate interaction between the two components and will be discussed in Section \[secUBtheo2\], see in particular Lemma \[eqconstronerecttheo2\].

From a mechanical viewpoint, in most materials the phases are symmetry-related and in particular one expects the eigenstrains to have the same determinant [@B9_BallJames87; @B9_Bhatta; @OtsukaWayman1999], which corresponds to our case $j=1$. The two-dimensional geometry we discuss corresponds to a three-dimensional setting in which both eigendeformations coincide in the third dimension. This would be, for example, the case for the three-dimensional eigenstrains $$A_1'= 
 \begin{pmatrix}
   1 & -\alpha & 0\\
 0 & 1 & 0 \\
0 & 0 &\gamma
 \end{pmatrix} \text{ and }
 \quad B_1'=
 \begin{pmatrix}
   1 & \alpha & 0\\
 0 & 1 & 0 \\
0 & 0& \gamma
 \end{pmatrix}$$ for some $\gamma>0$, the three-dimensional deformation would then be of the form $\varphi(x_1,x_2,x_3)=(u_1(x_1,x_2),u_2(x_1,x_2),\gamma x_3)$.

In thin films, however, a different relation between the two- and the three-dimensional problem is possible. Indeed, if the film is sufficiently thin, then incompatibility in the transverse direction leads to minor energy contributions. It is therefore reasonable to expect microstructure which is only compatible in-plane [@ChaudhuriMueller2006; @Hornung2008]. In this case, the two eigenstrains have the same determinant as $3\times 3$ matrices, but there is no reason for the $2\times 2$ blocks which determine the in-plane microstructure to have the same determinant, as for example with the three-dimensional eigenstrains $$A_2'= 
 \begin{pmatrix}
   1 & 0 & 0\\
 0 & 1-\alpha & 0 \\
0 & 0 &1/(1-\alpha)
 \end{pmatrix} \text{ and }
 \quad B_2'=
 \begin{pmatrix}
   1 & 0 & 0\\
 0 & 1+\alpha & 0 \\
0 & 0& 1/(1+\alpha)
 \end{pmatrix}\,.$$ It would be interesting to see if the difference between $K_1$ and $K_2$ results in a different scaling in the thin-film geometry studied in [@ChaudhuriMueller2006; @Hornung2008].

If no rank-one connection is present, then the energy is much larger, and typically no microstructure is formed.

Before stating the main results we introduce the key parameters and the set of admissibile functions. We focus on a rectangle, and assume throughout $$\label{assumption:1}
\Omega= (0,L)\times (0,H), \quad H,L>0, \quad \alpha\in(0,1),\quad \varepsilon>0\,,$$ in some parts additionally $\alpha< 1/2$. We shall impose Dirichlet boundary conditions corresponding to the average of $A_j$ and $B_j$, which, with the present choices, is in both cases the identity, $\frac12(A_j+B_j)=\Id$. Precisely, the admissible deformations are in the set $$\label{assumption:2}
\calM=\{u\in W^{1,2}(\Omega;\R^2): \ Du\in BV(\Omega;\R^{2\times 2}), \ u(x)=x \ \forall x\in \partial\Omega \}.$$ For $j\in\{1,2\}$, we are interested in the scaling of the minimum of $$\label{assumption:E}
E_j^\varepsilon[u,\Omega] = \int_{\Omega}\operatorname{dist}^2(Du,K_j)dx + \varepsilon
|D^2u|(\Omega)$$ over all $u\in\calM$. Existence of minimizers follows immediately from the direct method of the calculus of variations and will not be further addressed here. In the following, $c$ will denote a generic positive constant, which may change its value from line to line but in particular does not depend on the four parameters $\alpha$, $\varepsilon$, $H$ and $L$. We shall denote by $E_j^\eps[v,\omega]$ the integral of the expression in (\[assumption:E\]) over a subset $\omega\subset \Omega$.

\[Theorem:K1\] There is $c>0$ such that, under the assumptions (\[assumption:K1\]), (\[assumption:1\]), (\[assumption:2\]) and (\[assumption:E\]), one has $$\frac{1}{c}f(\alpha,\varepsilon,L,H) \leq \min_{u\in\calM} E_1^\eps[u,\Omega] \leq c
f(\alpha,\varepsilon,L,H)\,,$$ where $$\begin{aligned}
1\nonumber
    f(\alpha,\varepsilon,L,H)=\min&\Bigl\{
\alpha^{4/3}\varepsilon^{2/3}L^{1/3}H+\alpha\varepsilon L,\\
\nonumber
&\phantom{\Bigl\{}
\alpha^{4/3}\varepsilon^{2/3}LH^{1/3}+\alpha^4LH+\alpha\varepsilon H,\\
&\phantom{\Bigl\{}
\alpha^2 LH\Bigr\}\,.
\label{eqdefFtheo1}
  \end{aligned}$$ The constant $c$ does not depend on $\alpha$, $\varepsilon$, $L$ and $H$.

Follows from Lemma \[lemmaUBK1\] and Lemma \[lemmaLBK1\].

![Sketches of the deformation $u$ in the different phases appearing in Theorem \[Theorem:K1\].[]{data-label="fig:PD2"}](figPD2-crop){width="11cm"}

Theorem \[Theorem:K1\] shows that, depending on the values of the four parameters, different types of microstructure give the optimal energy scaling. In Figure \[fig:PD2\] we sketch the different microstructures (using the constructions from the upper bound) and in Figure \[fig:PD1\] the corresponding phase diagram. We now briefly discuss the different phases.

![Schematic phase diagram in case $K_1$, in the $L/\eps$, $H/\eps$ plane.[]{data-label="fig:PD1"}](figPD1-crop){width="8cm"}

1.  If the domain is very small, then the boundary data and the regularization term dominate the picture and the transformation to martensite is not energetically convenient. The energy scales as $\alpha^2 LH$, which is easily attained by the deformation $u(x)=x$. We call this austenitic (A) phase. From (\[eqdefFtheo1\]), one easily sees that this phase is the optimal one if $L/\eps<\alpha^{-1}$ or $H/\eps<\alpha^{-1}$.

2.  In the branching regime (BR) the deformation gradient $Du$ oscillates between a value close to $A_1$ and a value close to $B_1$ in the interior of the sample; the boundaries are approximately horizontal and the oscillations become finer approaching the left and right boundary. The total energy, if $L$ and $H$ are sufficiently large compared to $\eps$, is proportional to $\alpha^{4/3}\eps^{2/3}L^{1/3}H$, as predicted by Kohn and Müller.

3.  Horizontal laminate. If the sample is long and thin, then it is convenient to have a single oscillation in most of the sample, which then branches only in small regions close to the left and right boundary. The dominant energy contribution then originates from the single long interface and is proportional to $\alpha\eps 
      L$.

4.  Vertical branching. If $H$ is larger than $L$, then it may be more convenient to use the second rank-one connection and have a vertical branching, qualitatively similar to a 90-degrees rotation of regime BR. The rotation, since we are dealing with a fully nonlinear model, does not bring the matrices $A_1$, $B_1$ to the matrices $\Id\pm \alpha
      e_1\otimes e_2$, but instead leaves a second-order perturbation, which generates an additional energy cost proportional to $\alpha^4LH$ (see discussion in Lemma \[lemmaUBK1\] for details). Therefore, this regime can be subdivided into two parts: for small $H$, the $\alpha^{4/3}\eps^{2/3}LH^{1/3}$ term dominates (regime VB$_1$), for large $H$, the $\alpha^4LH$ term dominates (regime VB$_2$).

5.  As in the case of phase HL, if $L$ is very small, then a long vertical interface will dominate.

The boundaries between the different regions can be easily obtained from (\[eqdefFtheo1\]), and are sketched in Figure \[fig:PD1\].

![Schematic phase diagram in case $K_2$, in the $L/\eps$, $H/\eps$ plane.[]{data-label="fig:PD3"}](figPD3-crop){width="11cm"}

In the case of a single rank-one connection the energy scaling is different, and the phase diagram simpler, see Figure \[fig:PD3\]. Here, only the three regimes A, HL and BR play a role.

\[Theorem:K2\] There is $c>0$ such that, under the assumptions (\[assumption:K2\]), (\[assumption:1\]), (\[assumption:2\]) and (\[assumption:E\]), and for $\alpha<1/2$, one has $$\frac{1}{c}g(\alpha,\eps,L,H) \leq \min_{u\in\calM} E_2^\eps[u,\Omega]  \leq c g(\alpha,\eps,L,H) \,,$$ where $$g(\alpha,\eps,L,H)=\min\Bigl\{\alpha^{6/5}\varepsilon^{4/5}L^{1/5}H+\alpha\eps L
,\alpha^2 LH\Bigr\}\,.$$ The constant $c$ does not depend on $\alpha$, $\varepsilon$, $H$ and $L$.

Follows from Lemma \[lemmaUBK2\] and Lemma \[lemmaLBK2\].

The situation in which the dependence on $\alpha$ is not analyzed is substantially simpler, in particular for the lower bounds. Indeed, a result similar to Theorem \[Theorem:K1\] without dependence on $\alpha$ and with $H=L$ was first proven in [@B9_Chan; @B9_ContiOWJun2007]. These results are discussed in the PhD thesis [@B9_ChanDiss] and, in the simplified case $\alpha\sim 1$ and $H=L$, are presented in [@ChanContiSFBBericht]. Here we focus for simplicity on the case that $\Omega$ is a rectangle, keeping the two lengths separated to emphasize, in the case of Theorem \[Theorem:K1\], the competition between the two orientations of the microstructure. Although the detailed comparison between the different patterns is geometry-dependent, the leading-order scaling in $\eps$ in the small-$\eps$ limit is not expected to depend on the shape of $\Omega$. This is addressed (for similar models) in [@B9_KohnMuller94] for parallelograms and in [@B9_Diermeier] for general polygons. The case of piecewise smooth domains is treated [@B9_BCDM2002] for a model of thin-film blistering, without obtaining significant differences with the case of a rectangle.

In closing this introduction, we briefly review the literature on energy scaling in martensitic phase transformations. The scalar model $$E_\mathrm{KM}[u]=
  \begin{cases}\displaystyle
    \int_{\Omega} (\partial_1 u)^2 dx + \eps |\partial_2\partial_2 u|(\Omega) & \text{ if } \partial_2 u \in\{-1,1\}\text{ a.e.,}\\
\infty & \text{ otherwise,}
  \end{cases}$$ corresponding to a rigid version of our $K_1$ case, was proposed by Kohn and Müller in 1992. They have shown that, if the Dirichlet boundary data $u(0,\cdot)=0$ are imposed, for small $\eps$, the energy scales as $\eps^{2/3}L^{1/3}H$ [@B9_KohnMuller92; @B9_KohnMuller94], the same result as in Theorem \[Theorem:K1\]. It was later shown that the minimizers of their functional are asymptotically self similar [@B9_Conti00]. Kohn and Müller also studied a softer problem, in which the Dirichlet boundary condition is eliminated and instead the term $\beta \|u(0,\cdot)\|_{H^{1/2}((0,H))}^2$ is added to the energy $E_\mathrm{KM}$, where $u(0,\cdot)$ is the trace of the deformation on the $x_1=0$ boundary. Physically, this corresponds to the elastic energy of the material “outside the domain”. Their results, which were completed in [@B9_Conti2006; @B9_Zwicknagl], state that, for small $\eps$ and $\beta$, the optimal energy scales as $$\label{eqscalingKMbeta}
  \min\{\eps^{2/3}L^{1/3}H, \eps^{1/2}\beta^{1/2}L^{1/2}H\}\,.$$ The first option corresponds to a branched pattern, the second one to a laminar, one-dimensional pattern, in which the interfaces are exactly parallel to $e_1$ and the incompatibility is accommodated by the boundary term. We remark that the minimum in the assertion of Theorem 1 in [@B9_Conti2006] misses a term proportional to $\eps^{1/2}$. This term, which was correctly identified in [@B9_Zwicknagl], arises since, at the end of the proof, one can only choose $l'=\min\{1/\beta, \eps^{-1/2}\}$ and is only relevant for large values of $\eps$. The correct bound is $J_{\eps,\beta}\ge c \min\{\eps^{1/2}\beta^{1/2}L^{1/2}H,
\eps^{2/3}L^{1/3}H, \eps^{1/2}H^{3/2}, \beta H^2\}$.

Variants, in which the sharp condition $\partial_2 u\in\{-1,1\}$ is replaced by a penalization and using different norms for the singular perturbation, were studied by Zwicknagl [@B9_Zwicknagl], leading for small $\eps$ to a result similar to (\[eqscalingKMbeta\]). The geometrically nonlinear version was addressed by Dolzmann and Müller [@B9_DolzmannMueller1995], who proved a rigidity result that implies that the energy scaling of the present problem is superlinear in $\eps$. Lorent [@Lorent2006] has shown that the energy scaling of the singularly perturbed problem and the one of a finite-element approximation are the same, $\Gamma$-convergence to a sharp-interface model (after dividing the energy by $\eps$) was derived in [@B9_ContiSchweizer2006b], quantitative rigidity estimates for low-energy states of singularly perturbed multiwell problems were obtained in [@B9_Lorent2005; @B9_ContiSchweizer2006b; @B9_ChermisiConti2010; @JerrardLorent2013]. The energy scaling for the geometrically nonlinear model was obtained in [@B9_Chan; @B9_ContiOWJun2007], a version of Theorem \[Theorem:K1\] with a geometrically linear model and without the explicit dependence on $H$ and $L$ was given in [@B9_Diermeier]. A somewhat different approach, in which not only the deformation but also the shape of the inclusion was optimized, was followed in [@KnuepferKohnOtto2013] in a geometrically linear setting, mimicking nucleation of martensite and leading to different scalings.

The rest of this paper discusses the proofs of Theorem \[Theorem:K1\] and Theorem \[Theorem:K2\]. The upper bounds are proven in Section \[sectUB\], the lower bounds in Section \[seclowerbound\].

Upper bounds {#sectUB}
============

The upper bounds are proven by explicit constructions. Whereas the basic construction in the proof of Theorem \[Theorem:K1\] is a direct generalization of the one by Kohn and Müller, the different scaling in Theorem \[Theorem:K2\] arises from a somewhat different construction, which exploits the vectorial nature of the problem and the invariance under rotations of the energy density. Therefore, we start with a discussion of the latter, and postpone the proof of Theorem \[Theorem:K1\] to Section \[secubtheo1\] below. In what follows, $c$ denotes a positive constant which does not depend on $\alpha$, $\varepsilon$, $L$ and $H$ and may be different from line to line. In this section we write $(x,y)$ for generic points in $\R^2$.

Proof of the upper bound in Theorem \[Theorem:K2\] {#secUBtheo2}
--------------------------------------------------

The key idea is to use a number of period-doubling steps which join fine-scale oscillations close to the boundary with coarse-scale oscillations in the interior of the sample. The difference between the two cases considered and the main novelty in the present construction reside in the treatment of the period-doubling step, which we now present.

In order to fix the boundary conditions on the internal boundaries, we fix a continuous $1$-periodic function $\vartheta:\R\to\R$ such that $\vartheta(0)=0$, $\vartheta'=1$ on $(-1/4,1/4)$, $\vartheta'=-1$ on $(1/4,3/4)$ (equivalently, $\vartheta(t)=\operatorname{dist}(t+\frac14,\Z)-\frac14$) and scale it according to $\vartheta_h(t)=h\vartheta(t/h)$. Then $\vartheta_h$ is $h$-periodic and has derivative $\pm1$ almost everywhere. We observe that the function $$w_h(x,y)=
  \begin{pmatrix}
    x\\y +\alpha\vartheta_h(y)
  \end{pmatrix}$$ fulfills $Dw_h\in\{A_2,B_2\}$ almost everywhere. Further, on each line $\{(x,y): y\in h\Z\}$ we have $w_h(x,y)=(x,y)$, corresponding to the prescribed boundary conditions. The construction will be based on using $w_h(x,y)$ for selected values of $x$, with different values of $h$, larger in the center of the domain and smaller close to $x=0$ and $x=L$. The key ingredient in the proof is the following period-doubling construction, which permits to join different values of $h$ with small energetic cost.

\[eqconstronerecttheo2\] Let $0<h\le\ell$, $\omega=(x_0,x_0+\ell)\times(y_0,y_0+h)$, $\alpha\in(0,1)$, $\eps>0$. Then there is a function $v:\omega\to\R^2$ such that $$E_2^\eps[v, \omega]\le 
    c \left(\alpha^2\frac{h^5}{\ell^3} +  \alpha \eps \ell\right)\,,$$ with $\|Dv-\Id\|_{L^\infty}\le c\alpha$ and fulfilling the boundary conditions $v(x,y)=(x,y)$ for $y\in\{y_0,y_0+h\}$, $v(x_0,y)=(x_0,y+\alpha \vartheta_{h/2}(y-y_0))$, $v(x_0+\ell,y)=(x_0+\ell,y+\alpha\vartheta_{h}(y-y_0))$. The constant $c$ is universal.

We assume without loss of generality $x_0=y_0=0$. We fix a smooth interpolation function $\gamma:[0,1]\rightarrow [0,1]$ with $\gamma(0)=0$, $\gamma(1)=1$ and the first two derivatives vanishing at both points, say $\gamma(x)=10x^3-15x^4+6x^5$. We subdivide $\omega$ into the five sets $$ \begin{split}
  \omega_1 &= \left\{ (x,y)\in\omega: \ 0< y <
    \frac{h}{8}+\frac{h}{8}\gamma\left(\frac x\ell\right)\right\}, \\
  \omega_2 &= \left\{ (x,y)\in\omega: \
    \frac{h}{8}+\frac{h}{8}\gamma\left(\frac
      x\ell\right)<y<\frac{3h}{8}+\frac{h}{8}\gamma\left(\frac x\ell\right)\right\}, \\ 
  \omega_3 &= \left\{ (x,y)\in\omega: \
    \frac{3h}{8}+\frac{h}{8}\gamma\left(\frac
      x\ell\right)<y<\frac{5h}{8}-\frac{h}{8}\gamma\left(\frac x\ell\right)\right\}, \\ 
  \omega_4 &= \left\{ (x,y)\in\omega:  \ \frac{5h}{8}-\frac{h}{8}\gamma\left(\frac x\ell\right)<y<\frac{7h}{8}-\frac{h}{8}\gamma\left(\frac x\ell\right)\right\}, \\
  \omega_5 &= \left\{ (x,y)\in\omega: \ \frac{7h}{8}-\frac{h}{8}\gamma\left(\frac x\ell\right)<y<h\right\},
 \end{split}$$ see Figure \[fig:2\]. In the domains $\omega_1$, $\omega_3$ and $\omega_5$ we shall construct $v$ so that $Dv$ is close to $B_2=\diag(1,1+\alpha)$, in the other two so that it is close to $A_2=\diag(1,1-\alpha)$. Since the boundaries are curved, there will necessarily be deviations from $A_2$ and $B_2$, and in particular the $x$-derivative will not be identically $e_1$.

![Left panel: subdivision of the domain $\omega$ used in Lemma \[eqconstronerecttheo2\]. Right panel: subdivision of the domain used in Lemma \[eqconstronerecttheo2bl\].[]{data-label="fig:2"}](fig2-crop){width="10cm"}

The basic idea is to set $v_2(\cdot,0)=0$, according to the boundary conditions, and then to construct $v_2(x,y)$ by integrating $\partial_2 v_2(x,\cdot)$ from $0$ to $y$, with $\partial_2
v_2=1+\alpha$ in $\omega_1$, $\omega_3$, $\omega_5$, and $\partial_2
v_2=1-\alpha$ on the other two sets. This leads to $$v_2(x,y)=
\begin{cases}
{(1+\alpha)y} & \text{on $\omega_1$}, \\
{(1-\alpha)y+\alpha h(\frac{1}{4}+\frac{1}{4}\gamma\left(\frac x\ell\right))} &
\text{on $\omega_2$}, \\ 
{(1+\alpha)y-\frac{\alpha h}{2}} & \text{on $\omega_3$}, \\
{(1-\alpha)y+\alpha h (\frac{3}{4}-\frac{1}{4}\gamma\left(\frac x\ell\right))} & \text{on $\omega_4$}, \\
{(1+\alpha)y-\alpha h} & \text{on $\omega_5$}.
\end{cases}$$ Computing $\partial_1 v_2$, we easily see that it is of order $\alpha
h/\ell$ in $\omega_2$ and $\omega_4$. In particular, if we would simply set $v_1(x,y)=x$, we would obtain $$\label{eqdval}
  |Dv-A_2|\sim \frac{\alpha h}{\ell} \text{  in  $\omega_2$ and $\omega_4$}\,,$$ and therefore an energy at least of order $\alpha^2h^3/\ell$ in $\omega$.

It is, however, possible, via a more careful construction of $v_1$, to reduce the energy further. The idea is to use rotations, in a way similar to what was done in the study of folding pattern in blistered thin films [@B9_BCDM2000; @B9_BCDM2002]. To understand the key idea it is helpful to first consider the linearized model, in which $\operatorname{dist}(Dv, \operatorname{SO}(2) A_2)$ is replaced by $|(Dv+Dv^T)/2-A_2|$. The easier choice for $v_1$, namely, $v_1(x,y)=x$, would make the first diagonal entry in the gradient, $\partial_1v_1$, vanish. The other natural option, namely, to make the off-diagonal entries (which are equal) vanish, leads to the possibility of defining $v_1$ by solving $\partial_2 v_1+\partial_1v_2=0$. By the same estimate as in (\[eqdval\]), the term $\partial_1v_2$ is of order $\alpha h/\ell$, hence integrating in $y$ we would obtain a component $v_1$ of order $\alpha
h^2/\ell$. The diagonal entry in the gradient $\partial_1 v_1$ would then be of order $\alpha (h/\ell)^2$. The additional factor $h/\ell$ in $\partial_1v_1$ with respect to $\partial_2v_1$ arises from the fact that we took an integral in the $y$ direction and a derivative in the $x$-direction. On most of our domain, $h$ will be substantially smaller than $\ell$. Therefore, this construction leads to a substantial reduction in energy.

Since we are dealing with a geometrically nonlinear model, there will be an additional error term of order $(\partial_1 v_2)^2\sim \alpha^2 (h/\ell)^2$, arising from the linearization. We also need to take care of the fact that we are not linearizing around the identity, but instead around the matrix $A_2$. Therefore, we solve $\partial_2 v_1+(1-\alpha)\partial_1v_2=0$ instead of $\partial_2
v_1+\partial_1v_2=0$, see (\[eqlinearize\]) below.

A detailed computation based on this idea motivates the definition $$v_1(x,y)=
\begin{cases}
{x} & \text{on $\omega_1$}, \\
{x+\alpha(1-\alpha)\frac{h}{4\ell}\gamma'\left(\frac
    x\ell\right)(\frac{h}{8}+\frac{h}{8}\gamma\left(\frac x\ell\right)-y)} & \text{on $\omega_2$},
\\ 
{x-\alpha(1-\alpha)\gamma'\left(\frac x\ell\right)\frac{h^2}{16\ell}}& \text{on $\omega_3$}, \\
{x-\alpha(1-\alpha)\frac{h}{4\ell}\gamma'\left(\frac
    x\ell\right)(\frac{7h}{8}-\frac{h}{8}\gamma\left(\frac x\ell\right)-y)} &
\text{on $\omega_4$}, \\ 
{x} & \text{on $\omega_5$}.
\end{cases}$$ This concludes the definition of $v$. We remark that $v_1$ satisfies the boundary conditions at $x=0$ and $x=\ell$ in the regions $\omega_2$, $\omega_3$ and $\omega_4$ only if $\gamma'(0)=\gamma'(1)=0$. Therefore, in this case (at variance with the $K_1$ case, see below) the simple choice $\gamma(s)=s$ would not have been possible. The boundary conditions on $\gamma''$, which give continuity of $Dv$ across vertical boundaries, are only a matter of convenience.

In the rest of the proof, we estimate the energy of the function we constructed. First, we easily check that $v$ is continuous and obeys the boundary conditions. By construction, $Dv=B_2\in K_2$ in the sets $\omega_1$ and $\omega_5$. A short computation shows that $|Dv-B_2|=|\partial_1v_1-1|\le c \alpha h^2/\ell^2$ in $\omega_3$. In order to estimate the energy in $\omega_2$ and $\omega_4$, we compute $$\label{eqlinearize}
  \begin{split}
 \min_{Q\in \operatorname{SO}(2)} |Dv-QA_2|
\le& \min_{\varphi\in\mathbb{R}}
|\partial_1v_1-\cos\varphi|+|\partial_2v_2-(1-\alpha)\cos\varphi|\\
&
+|\partial_1v_2-\sin\varphi|+|\partial_2v_1+(1-\alpha)\sin\varphi|\\
  \le& |\partial_1 v_1-1|+|\partial_2 v_2-(1-\alpha)| \\
&+ 
|\partial_2 v_1+(1-\alpha)\partial_1  v_2| + c |\partial_1 v_2|^2    \,,
  \end{split}$$ where we have chosen $\varphi$ such that $\sin\varphi=\partial_1v_2$ if $\partial_1v_2\in[-1,1]$, $\varphi=0$ otherwise. Inserting the definition of $v$, one sees that the second and third term vanish, the remainder can be estimated by $$\begin{aligned}
1
\mathrm{dist}(Dv,K_2)\le& 
c\alpha \frac{h^2}{\ell^2} \|\gamma''\|_{L^\infty}
+ c\alpha\frac{h^2}{\ell^2} \|\gamma'\|_{L^\infty}^2
+ c \Bigl(\alpha \frac{h}{\ell} \|\gamma'\|_{L^\infty}\Bigr)^2\\
\le&
c\alpha \frac{h^2}{\ell^2}\quad \text{ in }\omega_2\cup\omega_4\,,\end{aligned}$$ and therefore, $$ \int_{\omega_2\cup\omega_4}\operatorname{dist}^2(Dv(x,y),K_2)\, dxdy  \leq
 c\alpha^2\frac{h^5}{\ell^3}.$$ A simple computation shows that $|Dv-\Id|\le c\alpha$ on $\omega$.

We finally turn to the interfacial part of the energy. In each of the $\omega_i$ we see by inspection that $|D^2v|\le c \alpha
h/\ell^2$ (we use repeatedly that $h\le\ell$ to simplify expressions). The jump contributions can be estimated by $2\|Dv-\Id\|_{L^\infty}$ times the length of the jump set; since $\gamma$ is monotone, this is bounded by $4(h+\ell)$. We conclude that $$\varepsilon |D^2v|(\omega)\leq c\varepsilon\left(\alpha (\ell+h) + \alpha
  h\ell \frac{h}{\ell^2}\right)\leq c\varepsilon \alpha \ell\,.$$ This completes the proof.

We now turn to the boundary layer. One could use a linear interpolation as in [@ChanContiSFBBericht]. For the sake of variety, we present an alternative construction, which is based on a variant of the construction of Lemma \[eqconstronerecttheo2\] and does not use gradients far from $A$, $B$ and $\Id$.

\[eqconstronerecttheo2bl\] Let $0<h\le\ell$, $\omega=(x_0,x_0+\ell)\times(y_0,y_0+h)$, $\alpha\in(0,1)$, $\eps>0$. Then there is a function $v:\omega\to\R^2$ such that $$E_2^\eps[v, \omega]\le 
 c \left(\alpha^2h\ell +  \alpha \eps \ell\right)\,,$$ with the boundary conditions $v(x,y)=(x,y)$ for $y\in\{y_0,y_0+h\}$ or $x=x_0$, $v(x_0+\ell,y)=(x_0+\ell,y+\alpha\vartheta_{h}(y-y_0))$ and $\|Dv-\Id\|_{L^\infty}\le c\alpha$. The constant $c$ is universal.

The construction is similar to the one in Lemma \[eqconstronerecttheo2\]. Again, it suffices to treat the case $x_0=y_0=0$. With $\gamma$ as in Lemma \[eqconstronerecttheo2\], we set $$\begin{split}
 \omega_B' &= \left\{ x\in (0,\ell), \ 0<y<\frac{h}{4}\gamma
   \left(\frac{x}{\ell}\right)\right\}, \\
 \omega_M' &= \left\{ x\in(0,\ell), \
   \frac{h}{4}\gamma\left(\frac{x}{\ell}\right)<y<\frac{h}{2}-\frac{h}{4}\gamma\left(\frac{x}{\ell}\right)
\right\}, 
 \\ 
 \omega_A &= \left\{x\in (0,\ell), \
   \frac{h}{2}-\frac{h}{4}\gamma\left(\frac{x}{\ell}\right)<y<
   \frac{h}{2}+\frac{h}{4}\gamma\left(\frac{x}{\ell}\right)
\right\}\,,\\
 \omega_M'' &= \left\{ x\in(0,\ell), \
\frac{h}2+\frac{h}{4}\gamma\left(\frac{x}{\ell}\right)<y<h-\frac{h}{4}\gamma\left(\frac{x}{\ell}\right)
\right\}, 
 \\ 
 \omega_B'' &= \left\{ x\in (0,\ell), \ h-\frac{h}{4}\gamma \left(\frac{x}{\ell}\right)<y<h\right\}, 
 \end{split}$$ see Figure \[fig:2\]. We set $v_1(x,y)=x$ and $$  v_2(x,y)=
 \begin{cases}
 y+\alpha y & \text{on $\omega_B'$}, \\
 y+\frac{1}{4} \alpha h\gamma\left(\frac{x}{\ell}\right) & \text{on
   $\omega_M'$}, \\ 
 y+\alpha(\frac{h}{2}-y) & \text{on $\omega_A$},\\
 y-\frac{1}{4}\alpha h\gamma\left(\frac{x}{\ell}\right) & \text{on
   $\omega_M''$}, \\ 
 y+\alpha (y-h) & \text{on $\omega_B''$}.
 \end{cases}$$ One then checks that $v_2$ is continuous, $|D^2v_2|\le c\alpha h/\ell^2$ inside each of the five sets, $Dv=B_2$ on $\omega_B'\cup\omega_B''$, $Dv=A_2$ in $\omega_A$, and $|Dv-\Id|\le 
\alpha h \|\gamma'\|_{L^\infty}/\ell$ in $\omega_M'\cup\omega_M''$. Therefore, $\|Dv-\Id\|_{L^\infty}\le c\alpha$ and, treating the jump terms as above, $$E_2^{\varepsilon}[v,(0,\ell)\times (0,h)] \leq c\alpha^2
h\ell+c\eps\alpha \ell \,.$$

![Global pattern used in Lemma \[lemmaUBK2\] for the construction of the upper bound of Theorem \[Theorem:K2\].[]{data-label="fig:1"}](fig1c-crop){width="12cm"}

We are now ready to present the full construction.

\[lemmaUBK2\] Under the assumptions of Theorem \[Theorem:K2\], there is a deformation $u\in \calM$ such that $$E_2^\eps[u,\Omega]\le  c \min\Bigl\{ \alpha^{6/5}\eps^{4/5} L^{1/5}H + \alpha
  \eps L, \alpha^2 LH\Bigr\}\,.$$ The constant $c$ is universal.

Inserting $u(x)=x$, we easily obtain a construction with energy $\alpha^2
LH$. Therefore, we only need to consider the case in which the first term in the minimum is smaller than the second. Then, necessarily, $\eps^{4/5}\alpha^{6/5} L^{1/5}H \le \alpha^2 LH$, which is equivalent to $$\label{eqepsalphaL}
  \eps \le \alpha L\,.$$

We now start the construction. We will focus on $\Omega'=(0,L/2)\times (0,H)$, the other part is analogous (with minor sign changes in the formulas). The basic idea in the construction is to have coarse folds in the central part, which then refine in a self-similar way approaching to the boundary, up to a point where a different construction for the boundary layer is needed, see Figure \[fig:1\].

Let $N\in\N\setminus\{0\}$ be the number of oscillations in the center, $\theta\in(1/4,1/2)$ a geometric factor, both will be chosen below. We cut the domain in vertical stripes, so that in each of them, say $(x_{i+1},x_i)\times(0,H)$, the construction will be $h_i$-periodic in the $y$ direction. We define $$x_i = \frac{L}{2}\theta^{i}\quad \text{ and } \quad  h_i= \frac{H}{2^{i}N}$$ and denote the width of the $i$-th stripe by $\ell_i=\theta^{i}(1-\theta)\frac{L}{2}$.

Since the construction in Lemma \[eqconstronerecttheo2\] requires $h_i \leq \ell_i$, we need to stop this procedure at a finite $\tau$, defined as the largest integer $i$ with $h_i\leq \ell_i$. Further, we need to choose $N$ such that $H/N\le (1-\theta)L/2$, to be able to start.

For $i\in [0,\tau]\cap \N$ and $y\in (0,H)$, we set $$\label{eqvperiodiclines}
  v(x_i,y)={\ensuremath{\begin{pmatrix}x_i \\ y+\alpha \vartheta_{h_i}(y) \end{pmatrix}}}\,.$$ This permits to treat each refinement step separately. In each stripe of the form $(x_{i+1},x_i)\times(0,H)$, $i=0, \dots, \tau-1$, we use $N2^i$ times the construction of Lemma \[eqconstronerecttheo2\]. In the set $(0,x_\tau)\times (0,L)$ we use the same procedure with Lemma \[eqconstronerecttheo2bl\]. The result is continuous thanks to (\[eqvperiodiclines\]).

It remains to estimate the energy. This is given by the sum over the stripes of $N2^i$ times the energy in each rectangle plus the boundary energy, which on the boundary of each rectangle is controlled by $\eps\|Dv-\Id\|_{L^\infty}
(h_i+\ell_i)\le c \alpha \eps \ell_i$. We obtain $$E_2^\eps[u,\Omega'] \leq c\sum_{i=0}^{\tau-1} 2^iN \left( 
\alpha^2 \frac{h_i^5}{\ell_i^3} + \alpha \eps \ell_i\right) + c N2^\tau
\left(\alpha^2 h_\tau\ell_\tau + \alpha \eps \ell_\tau\right) \,.$$ By definition of $\tau$, we have $h_\tau\le\ell_\tau\le 2 h_\tau$, therefore, the boundary term is comparable to the $i=\tau$ term in the series and can be included in the series. Using the definitions of $h_i$ and $\ell_i$, the estimate becomes $$E_2^\eps[u,\Omega'] \leq c\sum_{i=0}^{\tau}\left(
  \frac{\alpha^2H^5}{2^{4i}N^4\theta^{3i}L^3(1-\theta)^3} +
  \alpha\varepsilon2^{i}N\theta^{i}(1-\theta)L \right)\,.$$ If $\theta\in
(2^{-4/3},2^{-1})$, both $\sum_i 1/(2^4\theta^3)^{i}$ and $\sum_i (2\theta)^i$ are converging geometric series. The precise choice of $\theta$ in this interval does not influence the scaling, but only the prefactor, which we did not compute explicitly. In principle, one could choose the value of $\theta$ that makes the energy smallest, as was discussed in [@B9_ChanDiss]. However, this is not the only parameter that one could minimize. For example, it is not clear that the optimal aspect ratios $\ell_i/h_i$ should take the form $\ell_i/h_i=\theta^i$. Further, in converging geometric series, the initial terms play a dominant role, and it is reasonable to expect the first ones to be special. The energy of the $i=0$ step, for example, could be lowered by using a laminate instead of a branching pattern (branching helps to reduce the energy of the “next” step, but for $i=0$ there is no “next” step). Since we are only interested in the scaling, we ignore the optimization of $\theta$ and, for definiteness, choose an arbitrary value.

Setting $\theta = 2^{-5/4}$, we obtain $$ E_2^\eps[u,\Omega']\leq c \left(\frac{\alpha^2H^5}{N^4L^3} + \alpha\varepsilon NL\right) .$$ We choose $N$ so that the sum is smallest, subject to the constraints $N\ge 1$, $N\in\N$ and $N\ge 4H/L$ (as above, we only need to obtain the optimal exponents, not the prefactors). Precisely, we set $N=\lceil \alpha^{1/5}H/(\varepsilon^{1/5}L^{4/5})+4H/L\rceil$, where $\lceil t\rceil=\min\{z\in\Z: z\ge t\}$. Estimating $N\ge \alpha^{1/5}H/(\varepsilon^{1/5}L^{4/5})$ in the first term and $N\le \alpha^{1/5}H/(\varepsilon^{1/5}L^{4/5})+4H/L+1$ in the second one, we conclude $$E_2^{\varepsilon}[u,\Omega'] \leq c \alpha^{6/5}\varepsilon^{4/5}HL^{1/5} +c
 \alpha\eps L+ c\alpha\eps H\,.$$ Including the set $\Omega\setminus\Omega'$ only doubles the energy. By (\[eqepsalphaL\]), we see that the term $\alpha \eps H$ can be absorbed into the first one, and the proof is concluded.

Proof of the upper bound in Theorem \[Theorem:K1\] {#secubtheo1}
--------------------------------------------------

The proof is similar to the one of Theorem \[Theorem:K2\], but the construction of the period-doubling step in Lemma \[eqconstronerecttheo2\] becomes simpler. We only discuss the differences. The presence of two rank-one connections, however, leads in this case to the presence of two competing constructions, which are essentially 90-degrees rotated copies of each other. The relation is discussed in the proof of Lemma \[lemmaUBK1\] below.

\[eqconstronerecttheo1\] Let $0<h\le\ell$, $\omega=(x_0,x_0+\ell)\times(y_0,y_0+h)$, $\alpha\in(0,1)$, $\eps>0$. Then there is a function $v:\omega\to\R^2$ such that $$E_1^\eps[v, \omega]\le 
    c \left(\alpha^2\frac{h^3}{\ell} +  \alpha \eps \ell\right)\,,$$ with $\|Dv-\Id\|_{L^\infty}\le c\alpha$ and with the boundary conditions $v(x,y)=(x,y)$ for $y\in\{y_0,y_0+h\}$, $v(x_0,y)=(x_0+\alpha \vartheta_{h/2}(y-y_0),y)$, $v(x_0+\ell,y)=(x_0+\ell+\alpha\vartheta_{h}(y-y_0)),y)$. The constant $c$ is universal.

Under the same assumptions as in Lemma \[eqconstronerecttheo2\], and with the same domain subdivision, we define $$v_1(x,y)=
\begin{cases}
 x+\alpha y & \text{ on }\omega_1, \\
 x-\alpha y +\alpha \frac{h}{4}+\alpha\frac h4\gamma(\frac x\ell) & \text{ on }\omega_2, \\
 x+\alpha y - \frac{\alpha h}{2} & \text{ on }\omega_3, \\
 x-\alpha y + \frac{3\alpha h}{4} -2 \alpha\frac h4 \gamma(\frac x\ell) & \text{ on }\omega_4, \\
 x+\alpha y -\alpha h & \text{ on }\omega_5.
\end{cases}$$ In this case, the strain terms arising from the derivatives of $\gamma$ lie in the diagonal entry $\partial_1v_1$ of $Dv$. Therefore their energy contribution cannot be reduced using the other component and the rotations, hence we can simply take $v_2(x,y)=y$. The estimate (\[eqdval\]) is optimal, and correspondingly we obtain $\operatorname{dist}(Dv,K)\le c\alpha h/\ell$ everywhere. The strain gradient term can be estimated in the same way as in Lemma \[eqconstronerecttheo2\].

The boundary layer is also similar.

\[eqconstronerecttheo1bl\] Let $0<h\le\ell$, $\omega=(x_0,x_0+\ell)\times(y_0,y_0+h)$, $\alpha\in(0,1)$, $\eps>0$. Then there is a function $v:\omega\to\R^2$ such that $$E_1^\eps[v, \omega]\le 
    c \left(\alpha^2h\ell +  \alpha \eps \ell\right)\,,$$ with $\|Dv-\Id\|_{L^\infty}\le c\alpha$ and with the boundary conditions $v(x,y)=(x,y)$ for $y\in\{y_0,y_0+h\}$ or $x=x_0$, $v(x_0+\ell,y)=(x_0+\ell+\alpha\vartheta_{h}(y-y_0)),y)$. The constant $c$ is universal.

We use the domain subdivision from Lemma \[eqconstronerecttheo2bl\] with $$  v_1(x,y)=
 \begin{cases}
   x+\alpha y & \text{on $\omega_B'$}, \\
 x+\frac{1}{4} \alpha h\gamma\left(\frac{x}{\ell}\right) & \text{on
   $\omega_M'$}, \\ 
 x+\alpha(\frac{h}{2}-y) & \text{on $\omega_A$},\\
 x-\frac{1}{4}\alpha h\gamma\left(\frac{x}{\ell}\right) & \text{on
   $\omega_M''$}, \\ 
 x+\alpha (y-h) & \text{on $\omega_B''$}, 
 \end{cases}$$ and estimate the various energy terms as in Lemma \[eqconstronerecttheo1\].

We remark that in both Lemma \[eqconstronerecttheo1bl\] and Lemma \[eqconstronerecttheo1\] a simpler choice of $\gamma$ (for example, $\gamma(x)=x$) would have been possible, since no derivatives of $\gamma$ enter the definition of $v$.

\[lemmaUBK1\] Under the assumptions of Theorem \[Theorem:K1\], there are two functions $u,v\in\calM$ such that $$E_1^\eps[u,\Omega]\le c \min\Bigl\{\alpha^{4/3} \eps^{2/3} H L^{1/3}+\alpha\eps
  L, \alpha^2 LH\Bigr\}$$ and $$E_1^\eps[v,\Omega]\le c \min\Bigl\{\alpha^4 LH +  \alpha^{4/3} \eps^{2/3} L
  H^{1/3}+\alpha\eps H,
\alpha^2 L H\Bigr\}\,.$$ The constant $c$ is universal.

The construction of $u$ is identical to the one discussed in Lemma \[lemmaUBK2\], with Lemma \[eqconstronerecttheo1bl\] instead of Lemma \[eqconstronerecttheo2bl\] and Lemma \[eqconstronerecttheo1\] instead of Lemma \[eqconstronerecttheo2\], only the choices of $\theta$ and $N$ differ. In particular, one obtains $$E_1^\eps[u,\Omega'] \leq c\sum_{i=0}^{\tau}\left(
  \frac{\alpha^2H^3}{2^{2i}N^2\theta^{i}L(1-\theta)} +
  \alpha\varepsilon2^{i}N\theta^{i}(1-\theta)L \right)\,.$$ In this case, the series converges for all $\theta\in (1/4,1/2)$. We set, arbitrarily, $\theta=1/3$ and obtain $$E_1^\eps[u,\Omega']\leq c \left(\frac{\alpha^2H^3}{N^2L} + \alpha\varepsilon NL\right) .$$ We choose $N=\lceil \alpha^{1/3}H/(\varepsilon^{1/3}L^{2/3})+4H/L\rceil$ and conclude $$E_1^{\varepsilon}[u,\Omega] \leq c \alpha^{4/3}\varepsilon^{2/3}HL^{1/3}+c\alpha\eps (L+H)\,.$$ The treatment of the bound $\alpha^2LH$ and the elimination of the term $\alpha \eps H$ are analogous to the other case.

We now turn to $v$. In the case $K_1$ there are two rank-one connections, and if $H$ is much larger than $L$, it is convenient to use a rotated pattern with long thin stripes along $e_2$. This construction can be obtained by rotating the one just derived. We set $$Z={\ensuremath{\begin{pmatrix}0&1 \\ 1&0 \end{pmatrix}}}$$ and identify it with a linear map from $\R^2$ to $\R^2$. Let $u$ be the function constructed above, using $\tilde
H=L$, $\tilde L=H$. Then $u$ is the identical map on the boundary of $\tilde\Omega=(0,H)\times (0,L)$. We set $$v(x,y)=Z u (Z(x,y))\,.$$ Then $v$ obeys the required boundary conditions on $\partial\Omega$ and $|D^2v|(\Omega)=|D^2u|(\tilde\Omega)$. In order to estimate the strain energy, we observe that $Dv = Z Du (Z(x,y))Z$. We assert that for all $F\in
\R^{2\times 2}$ one has $$\label{eqdistzfza}
  \operatorname{dist}(ZFZ,K)\le \operatorname{dist}(F,K)+\alpha^2\,.$$ This then implies $$E_1^\eps[v,\Omega]\le 2 E_1^\eps[u,\tilde\Omega] + 2\alpha^4 LH\,,$$ as required.

It remains to prove (\[eqdistzfza\]). We first show that there is a rotation $R_a\in \operatorname{SO}(2)$ such that $|R_aZA_1Z-A_1|\le\alpha^2$, and correspondingly for $B_1$. We set $$R_a=\frac{1}{\sqrt{1+\alpha^2}}{\ensuremath{\begin{pmatrix}1&-\alpha \\ \alpha&1 \end{pmatrix}}}\in\operatorname{SO}(2)$$ and compute $$R_aZA_1Z={\ensuremath{\begin{pmatrix}\sqrt{1+\alpha^2}&-\alpha/\sqrt{1+\alpha^2} \\ 0&1/\sqrt{1+\alpha^2} \end{pmatrix}}}\,,$$ which gives $|R_aZA_1Z-A_1|\le\alpha^2$ for all $\alpha\in (-1,1)$. Let now $Q\in \operatorname{SO}(2)$, $F\in \R^{2\times 2}$ be such that $\operatorname{dist}(F,K)=|F-QA_1|$ (the case with $B_1$ is the same). We compute $$\begin{aligned}
1
\operatorname{dist}(ZFZ,K)&\le |ZFZ-ZQA_1Z| + \operatorname{dist}(ZQA_1Z, K)\\
&= |F-QA_1| + \operatorname{dist}(ZA_1Z,K)
\le \operatorname{dist}(F,K)+\alpha^2\,,\end{aligned}$$ where we used that $Z\in \operatorname{O}(2)\setminus \operatorname{SO}(2)$ implies $ZQZ\in \operatorname{SO}(2)$, and obtain (\[eqdistzfza\]).

Lower bounds {#seclowerbound}
============

General Strategy
----------------

As in many singularly perturbed nonconvex problems, the lower bound arises by the interaction of the following effects: the second-gradient term limits the oscillations of the gradient in the interior of the domain, and therefore brings the deformation $u$ away from the interpolation of the boundary data. The strain term then makes a deviation of the deformation from the interpolation of the boundary data expensive. The main point in the proof is to make these effects quantitative; in the present case, the main difficulty resides in the gradient term, which does not control individual derivatives but only the deviation of the deformation gradient from the set of matrices $K_j$.

In order to make the strategy quantitative, it is convenient to localize at the appropriate length scale. We shall denote by $\lambda$ a length scale chosen at the end of the proof. Qualitatively, it can be understood as a typical length scale of the microstructure in the interior of the domain. Since energy can concentrate, we shall choose (in Lemma \[Lemma:Epart\]\[Lemma:Epart1\]) a square of side length $\lambda$ as well as a horizontal and a vertical stripe of width $\lambda$ on which the energy is no higher than on average. Then we shall show (Lemma \[Lemma:Poincare\]\[Lemma:Poincare1\]) that on this square any low-energy deformation is approximately affine, with gradient in $K_j$. In order to relate this to the boundary data, exploiting the elastic energy on the stripes, we need to eliminate rotations from the picture. This is done in Lemma \[lemmaaverage0\], which shows that if a vector has length approximately unity, and one component averages (because of the boundary data) to 1, then the vector is approximately constant. We then separate the two different settings we consider. In the case $K_1$, addressed in Section \[seccaseK1\], the key remaining difficulty is the treatment of the two possible orientations for the microstructure, which requires a separation of two cases. In the case $K_2$, addressed in Section \[seccaseK2\], the key difficulty is in the treatment of the rotational invariance of the elastic energy and the degeneracy of the rank-one connection. As we know from the upper bound, see the discussion in (\[eqlinearize\]), the two components can interact to substantially reduce the energy. Therefore, an optimal lower bound will require a careful treatment of this interaction. This will be done using a suitable test function to integrate twice by parts, see Lemma \[lemmaLBK2\] below. In this section, we use $x=(x_1,x_2)$ for an element of $\R^2$.

\[Lemma:Epart\]\[Lemma:Poincare\] Assume $H,L>0$, $j\in\{1,2\}$, $u\in \calM$, $\alpha\in (0,1)$, $\eps>0$. For any $\lambda\in (0,\min\{L,H\}]$, there are stripes $S=(0,L)\times (s,s+\lambda )\subset\Omega$, $S' = (s',s'+\lambda)\times
(0,H)\subset\Omega$ such that, writing for brevity $E_j=E_j^\eps[u,\Omega]$, the following holds:

1.  \[Lemma:Epart1\] $$\begin{aligned}
    1
    \label{epart1}
    E_j^{\varepsilon}[u,S]&\leq c\frac{\lambda}{H}  E_j, \\
    \label{epart2}
    E_j^{\varepsilon}[u,S']&\leq c \frac{\lambda}{L}  E_j, \\
    \label{epart3}
    E_j^{\varepsilon}[u,S\cap S']&\leq c  \frac{\lambda^2}{LH}  E_j,\\
    \label{epart4}
    \|\partial_1 u_2\|_{L^1(S\cap S')} &\leq c\frac{\lambda^2}{LH}  
    \|\partial_1 u_2\|_{L^1(\Omega)}\,.\end{aligned}$$

2.  \[Lemma:Poincare1\] There are $ F\in
    K_j$ and $a\in\R^2$ such that $$\label{Poincare}
    \int_{S\cap S'}|Du-F|\,dx \leq
    c\frac{\lambda^3}{\varepsilon LH } E_j
    + c \frac{\lambda^2}{L^{1/2}H^{1/2}} E_j^{1/2}$$ and $$\label{Poincare2}
    \int_{S\cap S'}|u(x)-Fx-a|\,dx \leq
    c\frac{\lambda^4}{\varepsilon LH } E_j
    + c \frac{\lambda^3}{L^{1/2}H^{1/2}} E_j^{1/2}.$$

The constant $c$ is universal.

![Geometry in Lemma \[Lemma:Epart\].[]{data-label="Bild:Epart"}](figL1-crop){width="5cm"}

Before starting the proof, we recall the Poincaré inequality for $BV$ functions. We shall use it only on squares, but exploit the explicit dependence of the constant on the size of the square. Precisely, there is a constant $c_P>0$ such that for any square $q_r=x_*+(0,r)^2$ and any function $v\in BV(q_r;\R^m)$ one has $$\label{eqpoincare}
  \int_{q_r} |v-\overline v| dx \le c_P \,r\, |Dv|(q_r)\,,$$ where $\overline v\in \R^m$ is the average of $v$ over $q_r$. For $r=1$, this assertion is a special case, for example, of [@AFP Th. 3.44]; the same estimate for generic $r$ follows by applying the $r=1$ bound to $v_r(x)=v(rx)$.

\[Lemma:Epart1\]: We subdivide $\Omega$ into $M=\lfloor H/\lambda\rfloor\ge H/(2\lambda) $ disjoint horizontal stripes $S_k=(0,L)\times(k\lambda, (k+1)\lambda)$, where $\lfloor t\rfloor =
\max\{z\in\Z: z\le t\}$. Since $$\sum_{k=0}^{M-1} E_j^\varepsilon[u, S_k]\le E_j\,,$$ we have $$\# \Bigl\{ k\in [0,M)\cap \N: E_j^\varepsilon[u, S_k]\ge \frac5M E_j \Bigr\}\le \frac{M}{5}$$ (in particular, for $M<5$, the set is empty). Analogously, we have $N=\lfloor L/\lambda\rfloor\ge L/(2\lambda) $ disjoint vertical stripes $S_i'=(i\lambda, (i+1)\lambda)\times(0,H)$, and at most $N/5$ of them do not satisfy the corresponding bound. Since for any of the “bad” choices of $k$ all choices of $h$ need to be eliminated, the bound (\[epart1\]) is satisfied by all choices of the pair $(k,i)$ except for $NM/5$. Analogously, (\[epart2\]) is violated for at most $N/5$ choices of $i$, which corresponds to at most $NM/5$ choices of the pair $(k,i)$. Therefore, the total number of choices of $(k,i)$ which do not satisfy (\[epart1\]) and (\[epart2\]), with $c=10$, is no larger than $2NM/5$.

A similar estimate leads to $$\# \Bigl\{ (k,i)\in [0,M)\times[0,N)\cap \N^2: E_j^\varepsilon[u, S_k\cap S_i']\ge \frac5{MN} E_j \Bigr\}\le \frac{MN}{5}\,.$$ Hence, for all but $MN/5$ of the possible choices (\[epart3\]) holds, with $c=20$, and the same for $\|\partial_1 u_2\|_{L^1(S_k\cap S_i')}$. Since we have four conditions, and each of them excludes at most $\lfloor MN/5\rfloor $ choices, there are necessarily some left, for which all four conditions are valid. The resulting geometry is illustrated in Figure \[Bild:Epart\].

\[Lemma:Poincare1\]: Let $F'\in\R^{2\times2}$ be the average of $Du$ over $Q=S\cap S'$. Writing $F'=F'-Du+Du$, we obtain $$\begin{aligned}
1
  \lambda^2 \operatorname{dist}(F',K)&\le \|Du-F'\|_{L^1(Q)}+ \|\operatorname{dist}(Du,K)\|_{L^1(Q)}\,.\end{aligned}$$ We define $F$ as the matrix in $K$ closest to $F'$ and write $|Du-F|\le |Du-F'|+\operatorname{dist}(F',K)$, so that $$\begin{aligned}
1
\|Du-F\|_{L^1(Q)}&\le   \|Du-F'\|_{L^1(Q)}+\lambda^2 \operatorname{dist}(F',K)\\
&\le 2\|Du-F'\|_{L^1(Q)}+ \|\operatorname{dist}(Du,K)\|_{L^1(Q)}\,.\end{aligned}$$ Using Poincaré’s inequality (\[eqpoincare\]) on the first term and Hölder’s inequality on the second, and then recalling (\[epart3\]), we obtain $$\begin{aligned}
1
\|Du-F\|_{L^1(Q)}
&  \le   c\lambda |D^2u|(Q)
+\lambda\|\operatorname{dist}(Du,K)\|_{L^2(Q)}\\
& \le c \frac{\lambda^3}{LH\varepsilon}  E_j +
c  \lambda^2 \frac{E_j^{1/2}}{L^{1/2}H^{1/2}}\,,\end{aligned}$$ which concludes the proof of (\[Poincare\]). To prove (\[Poincare2\]), it suffices to define $a\in\R^2$ as the average of $u(x)-Fx$ over $Q$ and apply Poincaré’s inequality once more.

The above estimates immediately give the lower bound in the case of very thin domains.

\[lemmathin\] Let $L,H>0$, $\alpha\in (0,1)$, $\eps>0$. Then for every $u\in \calM$ we have $E_j\ge c \min\{\alpha \eps(L+H), \alpha^2 LH\}$. The constant $c$ is universal.

Before starting, we recall that by the trace theorem in $W^{1,1}$, there is a universal constant $c_T>0$ such that for any square $q_r=(0,r)^2$, $$\label{eqtrace}
  \int_{\partial q_r} |w| d\calH^1 \le c_T \int_{q_r}\left( \frac1r |w|+|Dw|\right) dx\,.$$ The case $r=1$ is given, for example, in [@EvansGariepy Th. 1, Sect. 4.3], the same estimate for generic $r$ follows by applying the $r=1$ bound to $w_r(x)=w(rx)$.

We use Lemma \[Lemma:Epart\] with $\lambda=\min\{L,H\}$ and observe that $Q=S\cap S'$ is a square with at least two sides on $\partial\Omega$. By the trace estimate (\[eqtrace\]) applied to $w(x)=u(x)-Fx-a$, (\[Poincare\])–(\[Poincare2\]) imply $$\int_{\partial Q} |u(x)-Fx-a|\, d\calH^1(x)\le 
c\frac{\lambda^3}{\varepsilon LH } E_j
+ c \frac{\lambda^2}{L^{1/2}H^{1/2}} E_j^{1/2}
\,.$$ On the other hand, we show below that there is a universal constant $c_*>0$ such that $$\label{eqFIdrand}
    c_*\alpha\lambda^2 \le         \int_{\partial Q\cap\partial\Omega}
    |x-Fx-a|\, d\calH^1(x) \text{ for all $a\in \R^2$, $F\in K_j$}\,.$$ Since $u(x)=x$ on $\partial\Omega$, we obtain $$c_*\alpha\lambda^2 \le c \frac{\lambda^3}{\varepsilon LH } E_j
+ c \frac{\lambda^2}{L^{1/2}H^{1/2}} E_j^{1/2}\,.$$ At least one of the two addends must be larger than $    c_*\alpha\lambda^2/2$, therefore, $$E_j\ge c \min\Bigl\{\frac{\alpha\varepsilon LH}{\lambda}, \alpha^2 LH\Bigr\}\,,$$ inserting $\lambda$ this implies the assertion.

It remains to prove (\[eqFIdrand\]). Let one of the sides of $Q$ contained in $\partial\Omega$ be of the form $q+(0,\lambda)e_i$, $i=1$ or $2$. Since for any $v\in \R^2$, we have $$\min_{a'\in \R^2} \int_0^\lambda |tv+a'| dt =\frac14  \lambda^2|v|\,,$$ we obtain (with $a'=q-Fq-a$, $v=e_i-Fe_i$) $$\frac14\lambda^2 |Fe_i-e_i|\le\int_0^\lambda | (\Id-F)(q+te_i)-a| dt\,.$$ If there are two orthogonal sides of $\partial Q$ contained in $\partial\Omega$, this gives control of the norm of $F-\Id$. If instead only two parallel sides of $\partial Q$ are contained in $\partial\Omega$, we need one more step. We assume for notational simplicity that they are both horizontal, so that the previous equation gives control of $|Fe_1-e_1|$. We write $$\begin{aligned}
1
 \lambda^2 |Fe_2-e_2|&=
  \int_0^\lambda |(\Id-F)\lambda e_2|dt\\
&\le
  \int_0^\lambda |(\Id-F)te_1-a|dt+\int_0^\lambda|(\Id-F)(te_1+\lambda e_2)-a| dt \,.\end{aligned}$$ The first integral corresponds to the bottom side, the last one to the top side. In both cases, we conclude $$\frac14\lambda^2 (|Fe_1-e_1|+|Fe_2-e_2|)   \le      2 
  \int_{\partial Q\cap\partial\Omega}     |x-Fx-a|\, d\calH^1(x)\,.$$ Finally, if $F\in K_2$, we have $|Fe_2|=1\pm \alpha$, which implies $|Fe_2-e_2|\ge \alpha$ and concludes the proof of (\[eqFIdrand\]) with $c_*=1/8$. If $F\in K_1$, then $$\begin{aligned}
1
  \alpha&=|Fe_1\cdot Fe_2| = | (Fe_1-e_1)\cdot Fe_2 + e_1\cdot (Fe_2-e_2)|
\\ &  \le 2 |Fe_1-e_1|+|Fe_2-e_2|\,,\end{aligned}$$ and the proof of (\[eqFIdrand\]) is concluded with $c_*=1/16$.

Before closing this general part, we present a lemma which permits to deal with the boundary data.

\[lemmaaverage0\] Let $\omega\subset\R^n$ be bounded and measurable, $v\in L^1(\omega;\R^2)$, $d\in L^2(\omega;[0,\infty))$ and $e\in 
  S^1$ be such that $v\cdot e-1$ has average 0 and $|v|\le 1+d$ almost everywhere. Then $$\|v\cdot e-1\|_{L^1}\le 2 |\omega|^{1/2}\|d\|_{L^2}$$ and $$\|v\cdot e^\perp\|_{L^1}\le
     3 |\omega|^{3/4}\|d\|_{L^2}^{1/2}+|\omega|^{1/2}\|d\|_{L^2}\,.$$ All norms are taken over $\omega$.

We assume $e=e_1$ without loss of generality. Since $v_1-1$ has average 0, we have (writing $f_\pm=\max\{\pm f,0\}$) $$\begin{aligned}
1\label{eqv1m1}
    \|v_1-1\|_{L^1}&=\int_\omega \left[(v_1-1)_++(v_1-1)_-\right] dx\\
& = 2\int_\omega(v_1-1)_+\, dx\le 2 \int_\omega (|v|-1)_+ dx
    \le 2 \|d\|_{L^1}\,.  \nonumber
  \end{aligned}$$ Using Hölder’s inequality, this leads directly to the first bound given in the statement. To address the second one, we write $v_2^2=|v|^2-v_1^2$ which implies, recalling $|v|\le 1+d$, $$\label{eqv2dv}
  \|v_2\|_{L^2}^2 = \|v\|_{L^2}^2-\|v_1\|_{L^2}^2
\le \|(1+d)^2\|_{L^1} -\|v_1\|_{L^2}^2\,.$$ In order to estimate the last term, we write $1=v_1+(1-v_1)$ and use a triangular inequality and (\[eqv1m1\]), $$|\omega|\le \|v_1\|_{L^1}+\|1-v_1\|_{L^1} \le \|v_1\|_{L^1} + 2 \|d\|_{L^1}\,.$$ Hölder’s inequality then gives $$|\omega|-2\|d\|_{L^1} \le \|v_1\|_{L^1}\le |\omega|^{1/2}\|v_1\|_{L^2}\,.$$ If $2\|d\|_{L^1}\le |\omega|$, squaring and inserting in (\[eqv2dv\]), we obtain $$\begin{aligned}
1
  \|v_2\|_{L^2}^2 \le& \|(1+d)^2\|_{L^1} -(|\omega|^{1/2}-2
  |\omega|^{-1/2}\|d\|_{L^1})^2 \\
\le& |\omega|+2\|d\|_{L^1}+\|d\|_{L^2}^2 -|\omega|+4\|d\|_{L^1}-
4|\omega|^{-1}\|d\|_{L^1}^2 \\
\le & 6\|d\|_{L^1}+\|d\|_{L^2}^2 \,.\end{aligned}$$ In the case $2\|d\|_{L^1}\ge |\omega|$, we instead write, again from (\[eqv2dv\]), $$\|v_2\|_{L^2}^2 \le \|(1+d)^2\|_{L^1} \le
    |\omega|+2\|d\|_{L^1}+\|d\|_{L^2}^2
    \le 4\|d\|_{L^1}+\|d\|_{L^2}^2\,.$$ In both cases, Hölder’s inequality gives $$\|v_2\|_{L^2}^2\le  6\|d\|_{L^1}+\|d\|_{L^2}^2 
    \le 6|\omega|^{1/2}\|d\|_{L^2}+\|d\|_{L^2}^2    \,.$$ Another application of Hölder’s inequality and of the estimate $(x+y)^{1/2}\le x^{1/2}+y^{1/2}$ gives, rounding for brevity $\sqrt 6\le 3$, $$\begin{aligned}
1
    \|v_2\|_{L^1}&\le |\omega|^{1/2} \|v_2\|_{L^2}\\
&\le
    |\omega|^{1/2}( 6|\omega|^{1/2}\|d\|_{L^2}+\|d\|_{L^2}^2 )^{1/2}
    \le 3 |\omega|^{3/4}\|d\|_{L^2}^{1/2}+|\omega|^{1/2}\|d\|_{L^2}\,.\end{aligned}$$

Proof of the lower bound in Theorem \[Theorem:K1\] {#seccaseK1}
--------------------------------------------------

At this point, we specialize to the first case. We write for brevity $E=E_1=E_1^\eps[u,\Omega]$ and $K=K_1$.

\[Lemma:dux\] Assume $H,L>0$, $\alpha\in (0,1)$, $\eps>0$, $u\in \calM$. Let $S,S'$ be as in Lemma \[Lemma:Poincare\]. Then $$\label{u1x1bA}
\int_{S\cap S'}|u_1(x)-x_1|\,dx \leq c 
\lambda^2 \frac{L^{1/2}}{H^{1/2}} E^{1/2}$$ and $$\label{u2x2bA}
\int_{S\cap S'}|u_2(x)-x_2|\,dx \leq c 
\lambda^2  \frac{H^{1/2}}{L^{1/2}} (E+\alpha^4LH)^{1/2}\,.$$ Further, $$\label{eqpartial1u2}
  \|\partial_1 u_2\|_{L^1(\Omega)} \le
  c (HL)^{3/4} E^{1/4}+ c
(HL)^{1/2} E^{1/2}\,.$$ The constant $c$ is universal.

This follows from repeated application of Lemma \[lemmaaverage0\]. We first use $v=\partial_1 u$, $e=e_1$, $d=\operatorname{dist}(Du,K)$, $\omega=S$. By the boundary conditions, $\int_S (\partial_1u_1-1)dx=0$, and $|v|\le 1+d$ because $|Fe_1|=1$ for all $F\in K$ and $|(Du)e_1|\le \min_{F\in K}|Du-F|+|Fe_1|= 1+\operatorname{dist}(Du,K)$. We obtain, using (\[epart1\]), $$\|\partial_1 u_1-1\|_{L^1(S)} \le
 c (\lambda L)^{1/2} 
\left(\frac{\lambda E}{H}\right)^{1/2}
  = c \lambda \frac{L^{1/2}E^{1/2}}{H^{1/2}}\,.$$ Integrating, $$\|u_1-x_1\|_{L^1(S\cap S')}\le \lambda \|\partial_1u_1-1\|_{L^1(S)}$$ proves (\[u1x1bA\]). Using $\omega=\Omega$ with the same $v$, $e$, and $d$, we get $$\|\partial_1 u_2\|_{L^1(\Omega)} \le
  3 (H L)^{3/4}  E^{1/4}+ (H
  L)^{1/2} 
E^{1/2}\,,$$ which proves (\[eqpartial1u2\]). We now turn to the other component, and set $v=\partial_2 u$, $e=e_2$, $d=\operatorname{dist}(Du,K)+\alpha^2$, $\omega=S'$. Since for all $F\in K$, we have $|Fe_2|=\sqrt{1+\alpha^2}\leq
 1+\alpha^2/2$, we obtain $|v|=|(Du)e_2|\leq 1+ \frac12\alpha^2+ \operatorname{dist}(Du,K)$. In this case, recalling (\[epart2\]), $$\|d\|_{L^2(S')}^2 \le c \frac{\lambda E}L +2\lambda H\alpha^4
  \le c \frac{\lambda (E+\alpha^4LH)}{L}\,.$$ Proceeding as above, Lemma \[lemmaaverage0\] gives $$\|\partial_2 u_2-1\|_{L^1(S')} \le
 c (\lambda H)^{1/2} 
\left(\frac{\lambda (E+\alpha^4LH)}{L}\right)^{1/2}\,,$$ and therefore (\[u2x2bA\]).

\[lemmaLBK1\] Under the assumptions of Theorem \[Theorem:K1\] one has, for any $u\in \calM$, $$\begin{aligned}
1
E_1^\eps[u,\Omega]\ge c \min\Bigl\{ & \alpha^{4/3}\varepsilon^{2/3} L^{1/3}H +\alpha \eps L, \\
&\alpha^{4/3}\varepsilon^{2/3} LH^{1/3} +\alpha \eps H 
+ \alpha^4LH ,\alpha^2 LH\Bigr\}\,.\end{aligned}$$ The positive constant $c$ is universal.

We fix $\lambda\in(0,\min\{L,H\}]$, the precise value will be chosen below, and choose the stripes $S$ and $S'$ as in Lemma \[Lemma:Epart\].

We first show that there is $c_*>0$ such that for every $F\in K$ and $a\in \R^2$, one has $$\label{K1cases}
\begin{split}
&\text{if $|F_{21}|<\alpha/2$ (case 1):} \quad c_*\alpha\lambda^3 \leq {\ensuremath{|| (x-Fx-a)\cdot e_1 ||}}_{L^1(S\cap S')}, \\
&\text{if $|F_{21}|\geq\alpha/2$ (case 2):} \quad c_*\alpha\lambda^3  \leq {\ensuremath{|| (x-Fx-a)\cdot e_2 ||}}_{L^1(S\cap S')}.
\end{split}$$ Both inequalities follow from the fact that for all $\xi\in\R$ one has (as in the proof of (\[eqFIdrand\]) above) $$  \min_{\eta\in\R} \int_0^\lambda |\xi t + \eta| dt
  = \int_0^\lambda |\xi t - \frac12 \xi \lambda| dt = \frac14 \lambda^2 |\xi|\,.$$ To prove the first inequality in (\[K1cases\]), we estimate $$\begin{aligned}
1
 {\ensuremath{|| (x-Fx-a)\cdot e_1 ||}}_{L^1(S\cap S')}
& = \int_s^{s+\lambda} \int_{s'}^{s'+\lambda} |(x_1-F_{11}x_1-a_1)-F_{12}x_2| dx_2 dx_1\\
&\ge \int_s^{s+\lambda} \min_{\eta\in\R} \int_{s'}^{s'+\lambda} |\eta-F_{12}x_2| dx_2 dx_1
\ge \frac14 \lambda^3 |F_{12}|\,.\end{aligned}$$ Since $F\in K$ implies $F_{21}=\sin\varphi$, $F_{12}=\pm\alpha\cos\varphi-\sin\varphi$ for some $\varphi\in\R$, in the first case, we have $|F_{12}|\ge \alpha
(\sqrt{3/4}-1/2)\ge c\alpha$, and the proof is concluded. The second case is simpler, $$\begin{aligned}
1
 {\ensuremath{|| (x-Fx-a)\cdot e_2 ||}}_{L^1(S\cap S')}
& =  \int_{s'}^{s'+\lambda} \int_s^{s+\lambda} |(x_2-F_{22}x_2-a_2)-F_{21}x_1| dx_1 dx_2\\
&\ge \int_{s'}^{s'+\lambda}  \min_{\eta\in\R} \int_{s}^{s+\lambda} |\eta-F_{21}x_1| dx_1 dx_2
\ge \frac14 \lambda^3 |F_{21}|\,.\end{aligned}$$ This concludes the proof of (\[K1cases\]).

For $i\in\{1,2\}$, we write $$|(x-Fx-a)\cdot e_i| \leq |(u(x)-x)\cdot e_i| + |u(x)-Fx-a|\,.$$ In case $i$, we obtain $$\label{K1cases2}
c_*\alpha \lambda^3  \leq \|u_i(x)-x_i\|_{L^1(S\cap S')} + \|u(x)-Fx-a\|_{L^1(S\cap S')}\,.$$ At this point, we distinguish the two cases.

[**Case 1 ($|F_{21}|<\alpha/2$).**]{} We estimate the first term in (\[K1cases2\]) by (\[u1x1bA\]) and the second by (\[Poincare2\]), $$\label{eqcase1ineq12}
c_*\alpha\lambda^3\leq 
c\lambda^2\frac{L^{1/2}}{H^{1/2}} E^{1/2}
+c\frac{\lambda^4}{\varepsilon LH } E
+ c \frac{\lambda^3}{L^{1/2}H^{1/2}} E^{1/2}
 \,.$$ Since $\lambda\le L$, the last term can be absorbed into the first one, and $$\begin{aligned}
1\label{eqcase1ineq1}
c_*\alpha\lambda^3\leq &
c\lambda^2 \frac{L^{1/2}}{H^{1/2}} E^{1/2}
+c\frac{\lambda^4}{\varepsilon LH } E\,,\end{aligned}$$ which can be rewritten as $$E\ge c \min\Bigl\{ \frac{\alpha^2 H\lambda^2} {L} , \frac{\alpha\varepsilon LH}{\lambda}   \Bigr\}\,.$$ The right-hand side is maximized by choosing $\lambda=(\varepsilon/\alpha)^{1/3}L^{2/3}$. Since we can only choose $\lambda\in (0,\min\{L,H\}]$, we set $$\lambda=\min\left\{(\varepsilon/\alpha)^{1/3}L^{2/3}, L, H\right\}$$ and obtain $$E\ge c \min\Bigl\{ \alpha^{4/3}\varepsilon^{2/3} L^{1/3}H   ,
\alpha^2 LH,
\frac{\alpha^2 H^3}{L}\Bigr\}\,.$$ We recall that by Lemma \[lemmathin\], we have $E\ge c \min\{\alpha \eps(L+H), \alpha^2 LH\}$. Either $E\ge\alpha^2 LH$ or $E$ is larger than the first expression. Therefore, we can combine the two estimates to produce $$E\ge c \min\Bigl\{ \alpha^{4/3}\varepsilon^{2/3} L^{1/3}H +\alpha \eps(L+H)  ,
\frac{\alpha^2 H^3}{L}+\alpha \eps(L+H), \alpha^2 LH\Bigr\}\,.$$ We show that the second entry in the minimum can be dropped. Indeed, by the arithmetic-geometric mean inequality, $$\alpha^{4/3}\varepsilon^{2/3} L^{1/3}H =
\left(\frac{\alpha^2 H^3}{L}\right)^{1/3}\left(\alpha \eps L\right)^{2/3}
\le \frac13 \, \frac{\alpha^2 H^3}{L} + \frac23\, \alpha \eps L\,.$$ Therefore, the second term is larger than the first one and is never the minimum. Analogously, if $\alpha\eps H\ge \alpha^{4/3}\varepsilon^{2/3} L^{1/3}H$, then $\eps\ge \alpha L$, but in this case, the inimum equals $\alpha^2
LH$. Therefore, we can drop the addend $\alpha\eps H$ in the first term. We conclude that in case 1, $$\label{eqlastboundcase1}
E\ge c \min\Bigl\{ \alpha^{4/3}\varepsilon^{2/3} L^{1/3}H +\alpha \eps L, \alpha^2 LH\Bigr\}\,.$$

[**Case 2 ($|F_{21}|\ge\alpha/2$).**]{} If $\|\partial_1 u_2\|_{L^1(Q)} \le\lambda^2\alpha/4$, then (\[Poincare\]) gives $$\frac14\lambda^2\alpha\le 
c\frac{\lambda^3}{\varepsilon LH } E
+ c \frac{\lambda^2}{L^{1/2}H^{1/2}} E^{1/2}\,,$$ which implies (\[eqcase1ineq12\]), hence this case has already been treated. Otherwise, from (\[epart4\]) and (\[eqpartial1u2\]) we immediately obtain $$\frac14\alpha\lambda^2 \le c\frac{\lambda^2}{LH} 
  \|\partial_1 u_2\|_{L^1(\Omega)} \le
  c  \lambda^2   \left(\frac{ E}{LH}\right)^{1/4}+ c
\lambda^2
\left(\frac{E}{LH}\right)^{1/2}\,.$$ If $E\ge LH$, there is nothing to prove, hence we can ignore the second term. Otherwise, $$\label{eqcase2alpha4}
  E\ge c\alpha^4 LH\,,$$ which in particular permits to estimate $$\label{eqcase2alpha4b}
  E+\alpha^4LH \le c E\,.$$ Using (\[u2x2bA\]) and (\[Poincare2\]) in (\[K1cases2\]) yields $$\begin{aligned}
1
c_*\alpha\lambda^3\leq &
c\lambda^2
  \frac{H^{1/2}}{L^{1/2}} (E+\alpha^4 LH)^{1/2}
+c\frac{\lambda^4}{\varepsilon LH } E
+ c \frac{\lambda^3}{L^{1/2}H^{1/2}} E^{1/2}\,.\end{aligned}$$ The last term can be absorbed into the first one, using (\[eqcase2alpha4b\]) we obtain $$\begin{aligned}
1
c_*\alpha\lambda^3\leq &
c\lambda^2  \frac{H^{1/2}}{L^{1/2}} E^{1/2}
+c\frac{\lambda^4}{\varepsilon LH } E\end{aligned}$$ for all $\lambda\in(0,\min\{L,H\}]$. This is the same as (\[eqcase1ineq1\]) up to swapping $L$ and $H$, and gives $$E\ge c \min\Bigl\{ \alpha^{4/3}\varepsilon^{2/3} H^{1/3}L  ,
\frac{\alpha^2 L^3}{H}, \alpha^2 LH\Bigr\}\,.$$ Recalling Lemma \[lemmathin\], $$E\ge c \min\Bigl\{ \alpha^{4/3}\varepsilon^{2/3} H^{1/3}L +\alpha \eps(L+H)  ,
\frac{\alpha^2 L^3}{H}+\alpha \eps(L+H), \alpha^2 LH\Bigr\}\,,$$ and we drop the same irrelevant terms as in the previous case. Recalling (\[eqcase2alpha4\]), we conclude $$E\ge c \min\Bigl\{ \alpha^{4/3}\varepsilon^{2/3} H^{1/3}L +\alpha \eps H 
+ \alpha^4LH , \alpha^2 LH\Bigr\}\,,$$ which together with (\[eqlastboundcase1\]) concludes the proof.

Proof of the lower bound in Theorem \[Theorem:K2\] {#seccaseK2}
--------------------------------------------------

We now come to the proof of the lower bound in Theorem \[Theorem:K2\]. In this case the local structure leads to oscillations in $\partial_2u_2$, and the branching can only be horizontal.

There are two ways of producing a lower bound on the energy. The first one is to estimate $\partial_1u_2$ using Lemma \[lemmaaverage0\], as was done in [@ChanContiSFBBericht]. This gives the optimal $\eps^{4/5}$ scaling, but not the optimal scaling in the parameter $\alpha$. Indeed, this argument is based on a purely nonlinear effect, and would not produce any bound in the linear setting.

We show here that the bound from [@ChanContiSFBBericht] can be improved. The key idea is to use [*two*]{} partial integrations, to pass from the $\partial_2u_2$ term to the $\partial_1 u_1$ one, and to use for the leading-order term the estimate for $\partial_1 u_1$ in Lemma \[lemmaaverage0\], which scales as $E$ instead of $E^{1/2}$. Swapping the indices between the partial derivative and the component of $u$ is done using the invariance under rotations. In order to be able to do the repeated partial integration, it is helpful to test the derivatives with a smooth test function, instead of simply integrating over a suitable domain, as done in Section \[seccaseK1\].

The key result in this section, which proves the lower bound of Theorem \[Theorem:K2\], is the following.

\[lemmaLBK2\] Under the assumptions of Theorem \[Theorem:K2\], one has $$\inf \{ E_2^\eps[u,\Omega]:u\in\calM\} \ge c\min\Bigl\{\alpha^{6/5}\varepsilon^{4/5}L^{1/5}H +
    \alpha \eps
    L, \alpha^2 LH\Bigr\}\,.$$ The positive constant $c$ is universal.

Since the proof involves a treatment of the value of $Du$ on segments, to avoid a notationally complex treatment of traces we first use density to show that it suffices to treat smooth functions. Let $u\in\calM$, $\delta>0$. We extend $u$ by $u(x)=x$ on $\R^2\setminus\Omega$. Since $\Omega$ is convex, by scaling we can find $\tilde u\in\calM$ such that $u(x)=x$ in a neighbourhood of $\partial\Omega$ and $E_2^\eps[\tilde u,\Omega]\le 
E_2^\eps[ u,\Omega]+\delta$. For $\rho>0$, we let $u_\rho\in C^\infty(\R^2;\R^2)$ be a mollification of $\tilde u$ on a scale $\rho$. For sufficiently small $\rho$, one has $u_\rho\in\calM$. At the same time, for $\rho\to0$, one has $|D^2u_\rho|(\Omega)\to |D^2\tilde u|(\Omega)$ and $Du_\rho\to D\tilde u$ in $L^2(\Omega;\R^{2\times2})$. Therefore, there is $\rho_\delta>0$ such that $$E_2^\eps[u_{\rho_\delta},\Omega] \le  E_2^\eps[\tilde u,\Omega]+\delta\le 
E_2^\eps[ u,\Omega]+2\delta\,.$$ We conclude that $$\inf \{ E_2^\eps[u,\Omega]:u\in\calM\} =
    \inf \{ E_2^\eps[u,\Omega]:u\in\calM\cap C^\infty(\overline\Omega;\R^2)\} \,.$$

Let now $u\in\calM\cap C^\infty(\overline\Omega;\R^2)$, this function will be fixed for the rest of the proof. Let $\lambda\in(0,\min\{L,H\}]$, the precise value will be chosen below. We choose the stripes $S$ and $S'$ with Lemma \[Lemma:Epart\] and set $Q=S\cap S'$, $I=(s,s+\lambda)$. Let $F$ be as in Lemma \[Lemma:Poincare\]\[Lemma:Poincare1\]. By (\[Poincare\]) there is $x_1^*$ such that the segment $I^*=\{x_1^*\}\times I\subset Q$ has the property $$\label{eqDuFI}
  \int_{I} |Du-F|(x_1^*,t)dt \le \frac{1}{\lambda} \int_Q |Du-F|dx \le
  c\frac{\lambda^2}{\eps LH} E + c\frac{\lambda E^{1/2}}{L^{1/2}H^{1/2}}\,.$$

We exploit the boundary conditions, using again Lemma \[lemmaaverage0\]. In particular, we set $v=\partial_1u$, $e=e_1$, $d=\operatorname{dist}(Du,K)$, $\omega=S$ and obtain, recalling (\[epart1\]), $$\label{eqK2d1u1}
  \|\partial_1 u_1-1\|_{L^1(S)} \le
 c \lambda \frac{L^{1/2}E^{1/2}}{H^{1/2}}$$ and $$\|\partial_1u_2\|_{L^1(S)}\le 
  c \lambda \frac{L^{3/4} E^{1/4}}{H^{1/4}}+ c \lambda 
\frac{L^{1/2} E^{1/2}}{H^{1/2}}\,.$$ Since, in the case $E\ge HL$, there is nothing to prove, we can replace the last estimate with the simpler one $$\label{eqK2d1u2}
    \|\partial_1u_2\|_{L^1(S)}\le 
  c \lambda \frac{L^{3/4} E^{1/4}}{H^{1/4}}\,.$$ This will only be used in the estimate of the error term arising from the linearization of $\operatorname{SO}(2)$, in the second line of (\[eqcase1k2a\]). The leading-order terms will be estimated with (\[eqK2d1u1\]).

At this point, the proof starts to differ significantly from the one for $K_1$. We fix one test function $\varphi\in C^\infty_c([0,1];[0,1])$ with $\varphi=
1$ on $(1/4,3/4)$, and define $f\in C^\infty_c(I)$ by $f(\lambda t+s)=\varphi(t)$, so that $${\ensuremath{|| f ||}}_{L^1(I)} \ge\frac\lambda2 , \quad {\ensuremath{|| f ||}}_{L^{\infty}}=1, \quad
 {\ensuremath{|| f' ||}}_{L^{\infty}}\le\frac{c}{\lambda},  \quad {\ensuremath{|| f'' ||}}_{L^\infty}\le\frac{c}{\lambda^2}\,.$$ We distinguish two cases, depending on the value of $F_{22}$.

[**Case 1: $|F_{22}-1|\geq\alpha/2$**]{}. With a triangular inequality and a partial integration we write $$\begin{aligned}
1 \nonumber
  \frac\alpha4\lambda\le \frac\alpha2\|f\|_{L^1(I)}
&\le \left| \int_I (F_{22}-1) f dx_2\right|\\  \nonumber
&\le \left| \int_I (F_{22}-\partial_2u_2) f dx_2\right|
+\left| \int_I (\partial_2u_2-1) f dx_2\right|\\ \nonumber
&\le  \int_I |Du-F| dx_2 
+\left| \int_I (u_2-x_2) f'(x_2) dx_2\right|\\
&\le  \int_I |Du-F| dx_2 
+\left| \int_{(0,x_1^*)\times I}\partial_1u_2(x) f'(x_2) dx\right|\,,
\label{eqcase1k2a}\end{aligned}$$ where we used $\|f\|_{L^\infty}=1$ and in the last step the boundary data $u_2(0,x_2)=x_2$. In all integrals over $I$, the functions $u$ and $Du$ are evaluated at $(x_1^*,x_2)$. At this point we trade $\partial_1 u_2$ with $\partial_2 u_1$ in the last term. To do this, we use the structure of rotations. First, we choose $G\in L^\infty(S;K)$ with $|Du-G|=\operatorname{dist}(Du,K)$ almost everywhere, then choose $\sigma \in L^\infty(S;\{-1,1\})$ such that $G\in\operatorname{SO}(2)\text{diag}(1,1+\sigma\alpha)$ almost everywhere. A direct computation shows that $G_{12}+(1+\sigma\alpha)G_{21}=0$, and therefore $$ (1+\sigma\alpha)\partial_1u_2+\partial_2u_1 =
 (1+\sigma\alpha)(\partial_1u_2-G_{21})+ (\partial_2u_1-G_{12}) \,.$$ We multiply this expression with $f'(x_2)$ and integrate it over $(0,x_1^*)\times
I$ to obtain $$\begin{aligned}
1 \nonumber
 \left|\int_{(0,x_1^*)\times I}\partial_1u_2(x)f'(x_2)\,dx \right|
\le&  
  \left|\int_{(0,x_1^*)\times I}\partial_2u_1(x)f'(x_2)\,dx \right|\\
&+ \alpha  \nonumber
\|f'\|_{L^\infty}\int_{(0,x_1^*)\times I} |\partial_1u_2(x)| dx \\
&+ 3\int_S |f'|(x_2)\, |Du-G|(x)\, dx\,.
\label{eqk2lonif1}\end{aligned}$$ We treat the different terms separately. The term with $\partial_2 u_1$ is integrated by parts and then estimated using (\[eqK2d1u1\]), $$\begin{aligned}
1
\left|\int_{(0,x_1^*)\times I}\partial_2u_1(x)f'(x_2)\,dx \right|&=
\left|\int_{(0,x_1^*)\times I}(u_1(x)-x_1)f''(x_2)\,dx \right|\\
&\le \|f''\|_{L^\infty} \|u_1-x_1\|_{L^1(S)} \\
&\le L \|f''\|_{L^\infty} \|\partial_1u_1-1\|_{L^1(S)} \\
&\le L \frac{c}{\lambda^2} 
 c \lambda \frac{L^{1/2}E^{1/2}}{H^{1/2}}
=  \frac{c}{\lambda} 
  \frac{L^{3/2}E^{1/2}}{H^{1/2}}\,.\end{aligned}$$ For the second term, we use $ \|f'\|_{L^\infty}\le c/\lambda$ and (\[eqK2d1u2\]) to obtain $$\begin{aligned}
1
  \alpha \|f'\|_{L^\infty} \int_{(0,x_1^*)\times I} |\partial_1u_2|(s,x_2)\,dx
&\le
c\alpha \frac{L^{3/4} E^{1/4}}{H^{1/4}}\,.\end{aligned}$$ Finally, the last term is bounded, using (\[epart1\]), $$\begin{aligned}
1
  \int_S |f'| |Du-G| dx\le \|f'\|_{L^\infty} \int_S \operatorname{dist}(Du,K)dx \le c \frac{L^{1/2} E^{1/2}}{H^{1/2}}\,. \end{aligned}$$ The estimate (\[eqcase1k2a\]) becomes, using (\[eqDuFI\]), (\[eqk2lonif1\]) and the three previous estimates, $$\label{eqk2alletermsad}
  \alpha\lambda \le
  c\frac{\lambda^2}{\eps LH} E + c\frac{\lambda E^{1/2}}{L^{1/2}H^{1/2}}
+ \frac{c}{\lambda}    \frac{L^{3/2}E^{1/2}}{H^{1/2}}
+ c\alpha \frac{L^{3/4} E^{1/4}}{H^{1/4}}
+ c \frac{L^{1/2} E^{1/2}}{H^{1/2}}  
\,.$$ Since $\lambda\le L$, the second term and the last one can be absorbed into the third one, and we have $$\alpha\lambda \le
  c\frac{\lambda^2}{\eps LH} E 
+ \frac{c}{\lambda}    \frac{L^{3/2}E^{1/2}}{H^{1/2}}
+ c\alpha \frac{L^{3/4} E^{1/4}}{H^{1/4}}$$ for all $\lambda\in (0,\min\{L,H\}]$. Therefore, $$E\ge c \min\Bigl\{ \frac{\alpha\eps LH}{\lambda}, \frac{\alpha^2 \lambda^4 H}{L^3}, 
  \frac{\lambda^4 H}{L^3}\Bigr\}\,.$$ Since $\alpha\le 1$, the last term can be ignored. Finally, we choose $\lambda$ as $$\lambda=\min\Bigl\{ \frac{\eps^{1/5} L^{4/5} }{\alpha^{1/5}}, L,H\Bigr\}$$ and obtain $$E\ge c\min\Bigl\{  \alpha^{6/5} \eps^{4/5} L^{1/5} H, \alpha^2 LH, \frac{\alpha^2 H^5}{L^3}\Bigr\}\,.$$ By Lemma \[lemmathin\], we have $E\ge c \min\{\alpha \eps(L+H), \alpha^2 LH\}$. Therefore, $$E\ge c\min\Bigl\{  \alpha^{6/5}\eps^{4/5} L^{1/5} H+\alpha\eps(L+H), \alpha^2 LH,
  \frac{\alpha^2 H^5}{L^3}+\alpha\eps(L+H)\Bigr\} \,.$$ Since, by the arithmetic-geometric mean inequality, $$\eps^{4/5}
\alpha^{6/5} L^{1/5} H=
\left(\frac{\alpha^2 H^5}{L^3}\right)^{1/5}\left(\alpha\eps L\right)^{4/5}
\le\frac15\,\frac{\alpha^2 H^5}{L^3} +\frac45\,\alpha\eps L\,,$$ the last term is always larger than the first one and can be dropped. Further, if $\eps\alpha H\ge\alpha^{6/5} \eps^{4/5} L^{1/5} H$, then $\eps\ge \alpha L$, and therefore the relevant energy bound is $\alpha^2 LH$. We conclude $$E\ge c\min\Bigl\{\alpha^{6/5} \eps^{4/5}  L^{1/5} H+\alpha\eps L, \alpha^2 LH\Bigr\}\,,$$ and therefore the proof in this case.

[**Case 2: $|F_{22}-1|<\alpha/2$**]{}. Since $|Fe_2|=1\pm \alpha$, in this case necessarily $|F_{12}|\ge \alpha/2$. Therefore, proceeding as above, $$\begin{aligned}
1
  \frac14\alpha\lambda\le \frac12\alpha \|f\|_{L^1(I)}
&= \left| \int_I F_{12} f(x_2) dx_2\right|\\
&\le \left| \int_I (F_{12}-\partial_2u_1) f(x_2) dx_2\right|
+\left| \int_I \partial_2u_1 f(x_2) dx_2\right|\\
&\le  \int_I |Du-F| dx_2 
+\left| \int_I (u_1-x_1) f'(x_2) dx_2\right|\,.\end{aligned}$$ In turn, recalling (\[eqK2d1u1\]), $$\left| \int_I (u_1-x_1) f'(x_2) dx_2\right|\le
\|f'\|_{L^\infty}  \int_{(0,x_1^*)\times I}|\partial_1 u_1-1|\, dx
\le c \frac{L^{1/2}E^{1/2}}{H^{1/2}}\,.$$ Using (\[eqDuFI\]) we obtain also in this case (\[eqk2alletermsad\]), which concludes the proof.

Acknowledgements {#acknowledgements .unnumbered}
================

We thank Barbara Zwicknagl for interesting discussions and comments on this work. Most of the present results are part of the PhD thesis [@B9_ChanDiss]. The first author would like to thank Ruth Joachimi for her encouraging support. This work was partially supported by the Deutsche Forschungsgemeinschaft through SFB 611, Project B9. \#1[0=]{}

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