Energy Scaling Law for a Singularly Perturbed Four-Gradient Problem in Helimagnetism

Abstract

We study pattern formation in magnetic compounds near the helimagnetic/ferromagnetic transition point in case of Dirichlet boundary conditions on the spin field. The energy functional is a continuum approximation of a $J_1-J_3$ model and was recently derived in Cicalese et al. (SIAM J Math Anal 51: 4848–4893, 2019). It contains two parameters, one measuring the incompatibility of the boundary conditions and the other measuring the cost of changes between different chiralities. We prove the scaling law of the minimal energy in terms of these two parameters. The constructions from the upper bound indicate that in some regimes branching-type patterns form close to the boundary of the sample.

1 Introduction

We study a continuum variational problem that arises from a statistical mechanics description of magnetic compounds and describes pattern formation in case of incompatible boundary conditions. We prove a scaling law for the minimal energy in terms of the problem parameters. Such scaling results have proven useful in a huge variety of singularly perturbed non-(quasi-)convex models for pattern forming systems where explicit minimizers cannot be easily determined analytically or numerically, see, e.g., (Kohn 2007) for some examples. Pattern formation is then often related to competing terms in the energy functional, favoring rather uniform or highly oscillatory structures, respectively. The proofs of the scaling laws often involve branching-type constructions where structures oscillate on refining scales near the boundary, among many others see for instance the scaling laws in Kohn and Müller (1994), Conti (2000, 2006), Capella and Otto (2009, 2012), Chan and Conti (2015), Knüpfer et al. (2013), Bella and Goldman (2015), Conti and Zwicknagl (2016), Conti et al. (2020), Rüland and Tribuzio (2022)) for martensitic microstructure, (Kohn and Wirth 2014, 2016) for compliance minimization, (Choksi et al. 2004, 2008; Conti et al. 2016) for type-I-superconductors, (Ben Belgacem et al. 2002; Bella and Kohn 2014; Bourne et al. 2017; Conti et al. 2005) for compressed thin elastic films, (Conti and Ortiz 2016; Conti and Zwicknagl 2016) for dislocation patterns, and (Brancolini and Wirth 2017; Brancolini et al. 2018) for transport networks.

We point out that in particular a variety of magnetization patterns (including branching structures) have been successfully explained via scaling laws of continuum micromagnetic energies, see, e.g., Choksi and Kohn (1998), Choksi et al. (1998), Dabade et al. (2019), DeSimone et al. (2006a), DeSimone et al. (2006b), Knüpfer and Muratov (2011), Otto and Steiner (2010), Otto and Viehmann (2010), Venkatraman et al. (2020)). While these models typically contain local and non-local terms, we will focus on a purely local model that arises - at least heuristically - from a frustrated spin system, see Diep (2013), Diep (2015) for the general context and Cicalese and Solombrino (2015), Cicalese et al. (2019) for the specific setting considered here. Precisely, starting from a 2-dimensional square lattice $\varepsilon {\mathbb {Z}}^2$ with lattice width $\varepsilon >0$, we consider spin fields $v:\varepsilon {\mathbb {Z}}^2\rightarrow S^1$ and a configurational energy of the form (also called $J_1-J_3$-model)

$$\begin{aligned} E_\varepsilon (v):=-\alpha \sum _{\mid i-j\mid =1}v(\varepsilon i)\cdot v(j\varepsilon )+\sum _{\mid i-j\mid =2}v(\varepsilon i)\cdot v(j\varepsilon ) \end{aligned}$$

with some positive parameter $\alpha >0$, where the summation is taken over indices $i,j\in {\mathbb {Z}}^2\cap \frac{1}{\varepsilon } \Omega$. While the first term favors nearest neighbors to have aligned spins, the second term favors next-to-nearest neighbors (horizontally and vertically) to have opposite spins. Note that the model considered here does not take into account diagonal interactions, for a recent analysis in that case see (Cicalese et al. 2021) and the references therein. Depending on the size of the parameter $\alpha$, different minimizers arise. Precisely, the analysis in Cicalese and Solombrino (2015), Cicalese et al. (2019) shows that (at least locally) the energy is minimized by ferromagnetic configurations, i.e., constant spin fields, if $\alpha \ge 4$, while for small $\alpha <4$, the energy is minimized by helimagnetic configurations, i.e., spin fields in which spins rotate at a fixed angle $\phi =\pm \arccos (\alpha /4)$ between horizontal and vertical nearest neighbors, respectively. Such helical structures have recently been observed experimentally, see, e.g., Schoenherr et al. 2018; Uchida et al. 2006.

Of particular interest is the transition point $\alpha \nearrow 4$ where the ground state changes from a helimagnetic to a ferromagnetic structure. Mathematically, an asymptotic analysis in the sense of $\Gamma$-convergence in this transition regime in the limit of vanishing lattice spacing has been performed in Cicalese et al. (2019). In the sequel, we briefly sketch the idea as outlined there, for details and references see Cicalese et al. (2019). Heuristically, it can be shown that the appropriately rescaled normalized energy $E_\varepsilon -\min E_\varepsilon$ can be rewritten in terms of an appropriately rescaled version u of the angular lifting ${\tilde{u}}$, given by $v=(\cos {\tilde{u}},\sin {\tilde{u}})$, as an energy of the form

$$\begin{aligned} I_{\tau }(u)= \int \tau \left( \mid {\partial _{1} \partial _1 u}\mid ^2 + \mid {\partial _{2} \partial _2u}\mid ^2 \right) + \frac{1}{\tau } \left( ( 1 - \mid \partial _1u\mid ^2)^2 + ( 1 - \mid \partial _2u\mid ^2)^2 \right) \,dx \end{aligned}$$

(1)

with $\tau = \frac{\sqrt{2}\varepsilon }{\sqrt{4-\alpha }}$. Here, the preferred derivatives $\partial _1u=\pm 1$ and $\partial _2u=\pm 1$ are the order parameters describing the chiralities, i.e., they correspond to helical structures rotating clockwise ($-1$) and counterclockwise ($+1$) between horizontally and vertically adjacent spins, respectively. We note that this is a heuristic simplification where in particular discrete derivatives are approximated by continuous ones and we assume that there are no vortices in the spin field. However, the rigorous analysis of Cicalese et al. (2019), Cicalese and Solombrino (2015) supports such a perspective, at least in certain parameter regimes. Roughly speaking, by the classical Modica–Mortola result, one expects that the functional $I_\tau$ converges in the sense of convergence as $\tau \rightarrow 0$ to a functional that is finite only on fields $\nabla u\in BV$ satisfying the differential inclusion $\nabla u\in \{(\pm 1,\pm 1)\}$, see Cicalese et al. (2019) for the rigorous derivation. A respective rigorous result in terms of $\Gamma$-convergence in the regime $\tau \rightarrow \tau _0\in (0,\infty )$, relating the discrete $J_1-J_3$-spin model with boundary conditions to a continuum functional of the form (1) holds also true (see Cicalese and Solombrino 2015 for a one-dimensional local result and Ginster et al. (in preparation) for the two-dimensional setting considered here).

While the analysis in Cicalese et al. (2019) focuses on the local behavior, we study the system under Dirichlet boundary conditions on the spin field. More precisely, we start from the continuum model (1) on a square domain $\Omega =(0,1)^2$, and derive the scaling law of the minimal energy among configurations satisfying affine boundary conditions $u(0,y)=(1-2\theta )y$ at the left boundary. Here, the parameter $\theta \in (0,1/2)$ is a compatibility parameter, where for $\theta =0$, the boundary condition is compatible with the helical structures $(\pm 1,1)$, $\theta =1/2$ corresponds to a ferromagnetic configuration (at least in vertical direction), and $\theta \in (0,1/2)$ indicates that the spin field on the boundary rotates in vertical direction with an angle that is smaller than the optimal angle $\phi = \arccos (\alpha /4)$.

For the ease of notation, we present the proof of the scaling law for a slightly simplified functional, namely

$$\begin{aligned} J_\sigma (u):=\int _{(0,1)^2}{\text {dist}}^2\left( \nabla u,\left\{ (\pm 1,\pm 1)\right\} \right) d{\mathcal {L}}^2+\sigma \mid D^2u\mid ((0,1)^2) \end{aligned}$$

with affine boundary conditions on one boundary. Precisely, we show that there are two scaling regimes for the minimal energy,

$$\begin{aligned} \min _{u(0,y)=(1-2\theta )y}J_\sigma (u)\sim \min \left\{ \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid }+1\right) ,\theta ^2\right\} . \end{aligned}$$

The second scaling is attained, for example, by the affine functions $u(x,y) = (1-2\theta ) y \pm x$, while the first one, which is relevant for small $\sigma$, is attained by a branching-type construction. It turns out that in contrast to many other branching-type constructions, the length scale on which patterns form, depends only on the compatibility parameter $\theta$ but not on $\sigma$. Also, our upper bound construction does not show equi-partition of energy but indicates that the surface term plays a major role. We note that the second scaling implies that minimizers in this regime just fail to have gradients in BV, see Rüland et al. (2019).

Let us briefly comment on the differences of $J_\sigma$ compared to $I_\tau$. First, the double-well potential penalizing deviations from the preferred gradients $(\pm 1,\pm 1)$ is different. However, the main difference lies in the growth for large arguments which play no role in our estimates, and we can easily transfer our results to the original double-well potential, see Sect. 4.2. Next, the higher-order term in $I_\tau$ does not control the full Hessian but only the two diagonal components. This also does not influence the scaling properties, see Remark 3. Finally, we work in $J_\sigma$ with a BV-type regularization while $I_\tau$ contains a quadratic regularization term. As is well known for related problems (see, e.g., Schreiber 1994; Zwicknagl 2014) this usually does not qualitatively change the scaling regimes of the minimal energy, see also Remark 2.

Fig. 1

Left: The four preferred gradients can be combined so that the corresponding function is zero outside the rotated square. Right: Rescaled versions of the rotated square can be used to cover $(0,1)^2$ so that the resulting function u satisfies $u=0$ on $\partial (0,1)^2$ and $\nabla u \in \left\{ (\pm 1,\pm 1) \right\}$ a.e

The functional $J_\sigma$ is a four-gradient functional, and hence formally lies “in between” very well-studied problems, namely scalar models for martensitic microstructures (preferred gradients $(1,\pm 1)$), see, e.g., Kohn and Müller (1992), Kohn and Müller (1994), and the Aviles–Giga functional (preferred gradients in $S^1$), see, e.g., Aviles et al. (1987). While for the two-gradient problem, the minimal energy scales as $\min \{\sigma ^{2/3}\theta ^{2/3},\ \theta ^2\}$ (this follows by a change of variables¹ directly e.g. from Zwicknagl 2014), the minimal energy for the Aviles–Giga functional in our setting is 0.² Thus the scaling we prove here for the four-gradient setting indeed lies “in between.” However, while any test function for the two-gradient problem yields a test function for $J_\sigma$, this functional has much more flexibility which comes from the high compatibility of the four gradients, see Fig. 1. Roughly speaking, this allows to construct test functions with low energy by covering the domain with building blocks using only the preferred gradients in the spirit of a simplified convex integration for differential inclusions, see, e.g., Conti (2008), Müller and Šverák (1999), Pompe (2010), Rüland et al. (2018), Rüland et al. (2020), Rüland and Tribuzio (2022). This relation will be explored in Sect. 5. We remark that a similar functional with corresponding four preferred magnetizations, in which patterns form due to non-local terms, has been studied in Dabade et al. (2019), Venkatraman et al. (2020).

The rest of the article is structured as follows: After briefly collecting the notation in Sect. 2, the main result will be proven in Sect. 3. We state the energy scaling law, discuss the regimes and prove the upper bound in Sect. 3.1 and the lower bound in Sect. 3.2. In Sect. 4, several generalizations are considered, including p-growth, different double-well potentials, boundary conditions on the full boundary, and rectangles. Finally, in Sect. 5, consequences for solutions of the related differential inclusion as derived in Cicalese et al. (2019) are collected.

2 Notation and Preliminaries

We will write C or c for generic constants that may change from line to line but do not depend on the problem parameters. The notation $c_i$ with an index i indicates that these are fixed constants which do not change within a proof. We write $\log$ to denote the natural logarithm. For the ease of notation, we always identify vectors with their transposes.

For a measurable set $B\subset {\mathbb {R}}^n$ with $n=1,2$, we use the notation $\mid B\mid$ or ${\mathcal {L}}^n(B)$ to denote its n-dimensional Lebesgue measure.

For $\sigma >0$ and $\theta \in (0,1/2]$, we set

$$\begin{aligned} {\mathcal {A}}_\theta :=\left\{ u\in W^{1,2}((0,1)^2):\ \nabla u\in BV((0,1)^2),\ u(0,x_2)=(1-2\theta )x_2\right\} , \end{aligned}$$

and consider the functional $E_{\sigma ,\theta }:{\mathcal {A}}_\theta \rightarrow [0,\infty )$ by

$$\begin{aligned} E_{\sigma ,\theta }(u) = \int _{(0,1)^2} {\text {dist}}^2(\nabla u,K) \, dx + \sigma \mid D^2 u\mid (\Omega ) \end{aligned}$$

where

$$\begin{aligned} K:=\left\{ (\pm 1, \pm 1)\right\} . \end{aligned}$$

The expression $\mid D^2u\mid (\Omega )$ in the second term of the functional $E_{\sigma ,\theta }$ denotes the total variation of the vector measure $D^2u$. Note that $u\in {\mathcal {A}}_\theta$ in particular implies that $u\in W^{1,1}((0,1)^2)$ and $\nabla u\in BV$. Hence, u has a continuous representative on the closed square $[0,1]^2$, see, e.g., (Conti and Ortiz 2016, Lemma 9). We will always identify such functions with their continuous representatives.

For a Borel set $B\subset {\mathbb {R}}^2$ and $u\in W^{1,2}(B)$ with $\nabla u\in BV$, we use the notation $E_{\sigma ,\theta }(u;B)$ for the energy on B, i.e.,

$$\begin{aligned} E_{\sigma ,\theta }(u;B) = \int _{B} {\text {dist}}^2(\nabla u,K) \, dx + \sigma \mid D^2 u\mid (B) . \end{aligned}$$

(5)

3 Energy Scaling on the Square $(0,1)^2$

Our main result is the following scaling law for the minimal energy.

Theorem 1

There exists a constant $C_T>0$ such that for all $\sigma >0$ and all $\theta \in (0,\frac{1}{2}]$,

$$\begin{aligned} \frac{1}{C_T}\min \left\{ \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } + 1\right) , \theta ^2\right\}\le & {} \min _{u\in {\mathcal {A}}_\theta } E_{\sigma ,\theta }(u) \\\le & {} C_T \min \left\{ \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } + 1\right) , \theta ^2\right\} . \end{aligned}$$

We will prove the upper bound in Proposition 2 and the lower bound in Proposition 3.

Remark 1

We note some properties of the scaling regimes in Theorem 1.

• (i) If $\sigma \ge \theta ^2$ then

  $$\begin{aligned} \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } +1 \right) \ge \sigma \ge \theta ^2.\end{aligned}$$

• (ii) If $\sigma \in [\theta ^{k+1},\theta ^{k})$ for some $k\in {\mathbb {N}}$, then

  $$\begin{aligned} { \sigma k} \le \sigma (k+1) \le \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } + 1 \right) \le \sigma ( k+2 ) \le 3 \sigma {k}. \end{aligned}$$

3.1 The Upper Bound

In this section, we provide test functions to prove the upper bound in Theorem 1. Before we start the proof, let us briefly explain the heuristics of the construction of the test function in the regime $\sigma < \theta ^2$ in which the affine function does not yield the optimal scaling. Instead, we provide a branching construction which (up to a small interpolation layer) only uses the four preferred gradients. In particular, in the y-variable the function is a saw-tooth function with slope $\pm 1$, where the volume fraction of slope $+1$ is $1-\theta$ and of slope $-1$ is $\theta$ to match the slope $1-2\theta =(1-\theta )\cdot 1+\theta \cdot (-1)$ on the boundary, see Fig. 2. If we assume that $\partial _1 u = \pm 1$, we observe that from the boundary condition, we have $\mid u(x,y) - (1-2\theta )y \mid \le x$. On the other hand, assuming that $\partial _2 u (x,y) = \pm 1$, one obtains that the number of jumps of the y-derivative on the slice $\{x\} \times (0,1)$ is of order $\frac{\theta }{x}$, see Fig. 4. Following these estimates, we present a self-similar construction that refines in the k-th step from $x \approx \theta ^k$ approximately $\theta ^{-k+1}$ jumps of the y-derivative into approximately $\theta ^{-k}$ jumps at $x \approx \theta ^{k+1}$. If $\theta = \frac{1}{m}$ for some $m \in {\mathbb {N}}$ (i.e., $\theta ^{-k} \in {\mathbb {N}}$) this can be done in an exact manner, see Fig. 2, for other $\theta$ a modification is needed leading to slightly more complicated branching patterns, see Fig. 3. Moreover, we note that although other branching constructions are in principle possible, the construction presented below yields in every construction step an approximate balance between the occurring horizontal and vertical interfaces. The proof of the lower bound (see Sect. 3.2) indicates that this is essential for a function providing the optimal scaling.

Proposition 2

There exists a constant ${C_U}>0$ such that for all $\sigma >0$ and all $\theta \in (0,\frac{1}{2}]$ there exists $u\in {\mathcal {A}}_\theta$ such that

$$\begin{aligned} E_{\sigma ,\theta }(u) \le {C_U} \min \left\{ \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } + 1\right) , \theta ^2\right\} . \end{aligned}$$

Proof

Step 1: Preparation. We first note that the affine function $u_{\text {aff}}(x,y) = (1-2\theta ) y + x$ satisfies

$$\begin{aligned} E_{\sigma ,\theta }(u_{\text {aff}}) \le 4 \theta ^2. \end{aligned}$$

In view of Remark 1(i), it hence suffices to consider the case $\sigma < \theta ^2$ and to construct a function u such that $E_{\sigma ,\theta }(u) \le {C_U} \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } + 1 \right)$ with a constant $C_U$ chosen below. Let $k\in {\mathbb {N}}$ be such that $\theta ^{k+1}\le \sigma < \theta ^k$. To simplify notation, set

$$\begin{aligned} m := \lceil \frac{1}{\theta }\rceil <\frac{1}{\theta }+1 \qquad \text { and }\qquad \delta := \frac{1}{m} \le \theta . \end{aligned}$$

We note that we always have $1/\delta \in {\mathbb {N}}$ and $\delta =1/m>\theta /(\theta +1)$, and hence

$$\begin{aligned} \delta \theta - (\theta -\delta )=\delta (\theta +1)-\theta {>} 0. \end{aligned}$$

We point out that many of the expressions below simplify if $\theta ^{-1}\in {\mathbb {N}}$ since then $\delta =\theta$. For an illustration of the construction described in the next steps for $\theta = \frac{1}{3}$ see Fig. 2 (the case $\theta =1/2$ is sketched in Fig. 6).

Finally, we fix some $N\in {\mathbb {N}}$ to be chosen later (see (14)).

Step 2: Construction of the building block.

As in many branching constructions (see, e.g., Kohn and Müller 1994), we first construct an auxiliary function that acts as a building block for the construction of u. For the ease of notation, we describe an admissible function via its gradient field. By the boundary condition $u(0,x_2)=(1-2\theta )x_2$, this uniquely determines the function u.

Precisely, we define $V: (\delta \theta ,\theta ] \times {\mathbb {R}}\rightarrow {\mathbb {R}}^2$ as the function which is 1-periodic in y-direction and satisfies the following (see Fig. 3):

• (i) If $(x,y) \in {[}\delta \theta ,\theta {]} \times [{1-\delta },1)$ then

  $$\begin{aligned}{ V(x,y) = {\left\{ \begin{array}{ll} ( -1, -1) &{}\text { if } y\ge \ 1- \delta - (x-\theta ) \text { and } x\ge \delta \theta + \theta -\delta , \\ (1,-1) &{} \text { if } y \ge 1-\delta \theta \text { and } x \le \delta \theta + \theta -\delta , \\ ( 1, 1) &{}\text { else}.\end{array}\right. } } \end{aligned}$$

• (ii) If $(x,y) \in {[}\delta \theta ,\theta {]} \times [{1-2\delta ,1-\delta })$ then

  $$\begin{aligned} V(x,y) = {\left\{ \begin{array}{ll} ( 1 , -1 ) &{}\text { if } y\ge \max \{ 1-(1+\theta )\delta , 1-\theta + (x - \theta ) \}, \\ ( 1 , 1) &{}\text { if } y \le 1-(1+\theta )\delta \ {\text { and }\delta \theta } \le x\le 2\theta -\delta -{\delta \theta }, \\ ( -1 , 1 ) &{}\text { else.} \end{array}\right. } \end{aligned}$$

• (iii) If $(x,y) \in {[}\delta \theta ,\theta {]} \times {[(\ell -1)\delta ,\ell \delta )}$ for $1\le \ell \le \frac{1}{\delta }-2$ then

  $$\begin{aligned} V(x,y) = {\left\{ \begin{array}{ll}(1 ,-1 ) &{}\text { if } y\ge \max \{ (\ell -\theta )\delta ,\ {(\ell -\theta )}\delta + x - (\delta \theta + \theta - \delta + ({\ell }-1) \delta \theta )\}, \\ ( 1 , 1 ) &{}\text { if } y \le (\ell -{\theta }){\delta } {\text { and }} x \le \delta \theta + \theta - \delta + ({\ell }-1) \delta \theta , \\ ( -1 , 1 )&{}\text { else.} \end{array}\right. } \end{aligned}$$

Note that V is ${\text {curl}}$-free on $(\delta \theta ,\theta )\times {\mathbb {R}}$ as it is piecewise constant and $\nu \parallel (V^{-}-V^+)$ on its jump set $J_{V}$, where $\nu$ is the measure-theoretic normal to $J_{V}$, see also Fig. 3. Consequently, V is a gradient field on $(\delta \theta ,\theta )\times {\mathbb {R}}$, and additionally,

$V(x,y) \in K$ for almost all (x, y), and

$$\begin{aligned} \mid DV\mid {(}(\delta \theta ,\theta ) \times (0,1)) \le {2}(1-\delta )\theta m \, (\sqrt{2}+2) \le {8} m\theta \le {16}. \end{aligned}$$

We will use in the next step that for the second component $V^{(2)}$ of V, we have

$$\begin{aligned} V^{(2)}(\theta ,y)=V^{(2)}(\delta \theta ,\delta y) \qquad \text { for all }y\in {\mathbb {R}}. \end{aligned}$$

(8)

Step 3: Branching construction.

We now set $V_N: ( \delta ^N \theta , 1 ) \times (0,1) \rightarrow {\mathbb {R}}^2$ for the fixed number $N\in {\mathbb {N}}$ by

$$\begin{aligned} V_N(x,y) = {\left\{ \begin{array}{ll}( 1 , -1)&{}\text { if } x \ge \theta {\text { and }} y\ge 1-\theta , \\ (1, 1 ) &{}\text { if } x \ge \theta {\text { and }} y\le 1-\theta , \\ V(\delta ^{-k+1}x,\delta ^{-k+1}y) &{}\text { if } x \in {[}\delta ^k \theta ,\delta ^{k-1}\theta ) \text { for {some} } 1 \le k \le N. \end{array}\right. } \end{aligned}$$

We note that $V_N$ is ${\text {curl}}$-free as $\nu \parallel (V_N^- - V_N^+)$ on $J_{V_N}$, where $\nu$ is the measure-theoretic normal to $J_{V_N}$, see also (8) and Fig. 3. Moreover, $V_N(x,y) \in K$ for almost all $(x,y){\in (\delta ^N\theta ,1)}$, and

$$\begin{aligned} \mid DV_N\mid \left( (\delta ^N \theta ,1) \times (0,1)\right) \le ({16+2}) N + {2} \le {20} N. \end{aligned}$$

(9)

Let ${\tilde{u}}_N: ( \delta ^N \theta , 1 ) \times (0,1) \rightarrow {\mathbb {R}}$ be a potential, i.e., $\nabla {\tilde{u}}_N=V_N$, such that ${\tilde{u}}_N(\delta ^N\theta ,0) = 0$. Then notice that (see Fig. 4)

$$\begin{aligned} |{\tilde{u}}(\delta ^N\theta ,y) - (1-2\theta )y|\le \delta ^N 2 \theta . \end{aligned}$$

(10)

Finally, we interpolate linearly in x to satisfy the boundary condition, and eventually define $u_N: (0,1)^2 \rightarrow {\mathbb {R}}$ by

$$\begin{aligned} u_N(x,y) = {\left\{ \begin{array}{ll} {\tilde{u}}_N(x,y) &{}\text { if } x \ge \delta ^N \theta , \\ (1-2\theta )y + \delta ^{-N} \theta ^{-1} x \left( {\tilde{u}}_N(\delta ^N\theta ,y) - (1-2\theta )y\right) &{}\text { else.} \end{array}\right. } \end{aligned}$$

(11)

Step 4: Estimate for the energy.

By (11) and (10), we have for a.e. $(x,y) \in (0,\delta ^N\theta ) \times (0,1)$ that

$$\begin{aligned} \mid \partial _1 u_N(x,y)\mid&\le 2 \quad \text { and}\quad \mid \partial _2 u_N(x,y)\mid \\&\le (1-2\theta ) + \mid \partial _2 {\tilde{u}}_N(\delta ^N\theta ,y) - (1-2\theta )\mid \le 3. \end{aligned}$$

In particular, we have

$$\begin{aligned} \int _{(0,1)^2} {\text {dist}}(\nabla u_N,K)^2 \, dx dy \le {5} { \delta ^N \theta }. \end{aligned}$$

(12)

Next, by (9), and since ${\mathcal {L}}^1 \left( \left\{ y\in (0,1): \partial _2u_N(\delta ^N\theta ,y)=1\right\} \right) =1-\theta$ and ${\mathcal {L}}^1\left( \left\{ y\in (0,1): \partial _2u_N(\delta ^N\theta ,y)=-1\right\} \right) =\theta$, we have

$$\begin{aligned}&\mid D^2 u_N\mid ((0,1)^2) \nonumber \\&\quad \le {20 N} + \mid D^2 u_N\mid ((0,\delta ^N\theta ] \times (0,1)) \nonumber \\&\quad \le {20} N + \delta ^N \theta |\partial _2 \partial _2 u_N(\delta ^N\theta ,\cdot )|((0,1)) + 2 \int _0^1 \mid \partial _2 {\tilde{u}}_N(\delta ^N\theta ,y) - (1-2\theta )\mid \, dy \nonumber \\&\quad \le {20} N + \delta ^N \theta 2 \delta ^{-N} + 2 ((1-\theta ) 2\theta + (2-2\theta ) \theta ) \nonumber \\&\quad \le {20} N + {10}\theta \le {30} N. \end{aligned}$$

(13)

Now fix

$$\begin{aligned} N {:=} {\left\lceil \frac{\log \frac{\sigma }{\theta } }{\log \delta } \right\rceil }. \end{aligned}$$

(14)

Since $0<\sigma< \theta ^2< {1}$ we have $\frac{\log \frac{\sigma }{\theta } }{\log \delta } > 0$ and consequently $N \ge 1$. Combining (12) and (13) and using that $\mid \log \delta \mid \ge \mid \log \theta \mid$, we obtain

$$\begin{aligned} E_{\sigma ,\theta }(u_N) \le 5 \theta \delta ^{N} + 30 \sigma N&\le 5 \theta \delta ^{\frac{\log \frac{\sigma }{\theta }}{\log \delta }} + 30 \sigma \left( \frac{\log \frac{\sigma }{\theta }}{\log \delta } + 1 \right) \\&\le 5 \sigma + 30 \sigma \left( \frac{\log \sigma }{\log \delta } + 1 \right) \\&\le 5 \sigma + 30 \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } + 1 \right) . \end{aligned}$$

This concludes the proof of the upper bound with $C_U:=$35. $\square$

Fig. 2

Sketch of the construction for $\theta = 1/3$ and $N=4$. The regions of the four different gradients are color-coded as in Fig. 1

Fig. 3

Left: Gradient field V of the building block described in Step 2 for $\theta =2/5$ and $m=3$. Note that the left region of size $\theta -\delta$ is not needed in the construction but is rather added for the sake of an easier notation in the proof. On the other hand, due to Proposition 3 deleting this region from the construction cannot lead to an improved energy scaling. Right: The corresponding branching construction for $N=3$

Fig. 4

Sketch of the function $u(\delta ^k\theta ,\cdot )$ for $\delta = \theta =\frac{1}{3}$ and $k=0,1,2,3$

Remark 2

If we replace the BV-type regularization $\sigma \mid D^2u\mid$ by the smoother one $\sigma ^2\int _\Omega (D^2u)^2\,d{\mathcal {L}}^2$, we can use slight modifications of the above-constructed test functions to obtain the same upper bound on the energy scaling. Clearly, the function $u = (1-2\theta ) y \pm x$ produces again an energy of order $\theta ^2$. Hence, it remains to consider the branching regime $\sigma < \theta ^2$. Starting with the branching construction $u_N:(0,1)^2\rightarrow {\mathbb {R}}$ (N chosen as in (14)) which we can extend in y-direction so that $\nabla u_N$ is 1-periodic, we set

$$\begin{aligned}{\tilde{u}}(x,y):={\left\{ \begin{array}{ll} u_N(x-2\sigma ,y)&{} \quad \text { if } x \in (2\sigma , 1), \\ (1-2\theta )y&{}\quad \text { if }x\in (0,2\sigma ) \end{array}\right. } \end{aligned}$$

and smooth this function with a symmetric mollifier of support $B_{\sigma }(0)$. For the smoothed function a straightforward computation shows that one can estimate the term $\sigma ^2 \int _\Omega (D^2u)^2\,d{\mathcal {L}}^2$, up to a constant, by $\sigma N$. On the other hand, in addition to the region $(0,4\sigma ) \times (0,1)$ (outside of $(0,3\sigma ) \times (0,1)$ we have $\nabla {\tilde{u}} \in {\mathcal {K}}$) the gradient of the mollified function agrees with one of the preferred gradients except for a tube with width $2\sigma$ around the jump set of $J_{\nabla {\tilde{u}}}$. Thus, we can estimate the second term in the energy, up to a constant, by $\sigma N + 4\sigma$. Recalling the computation at the end of the proof of Proposition 2 leads to the claimed energy scaling.

3.2 The Lower Bound

In this section, we prove the ansatz-free lower bound in Theorem 1. Precisely, we show the following statement.

Proposition 3

There exists a constant ${C_L}>0$ such that for all $\sigma >0$, all $\theta \in (0,\frac{1}{2}]$, and all $u\in {\mathcal {A}}_\theta$

$$\begin{aligned} E_{\sigma ,\theta }(u)\ge {C_L}\min \left\{ \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } + 1\right) , \theta ^2\right\} . \end{aligned}$$

Remark 3

A careful inspection of the upcoming proof shows that the same lower bound holds true even if the term $\sigma \mid D^2u\mid (\Omega )$ is replaced by the term $\sigma (\mid \partial _1\partial _1 u\mid (\Omega ) + \mid \partial _2\partial _2 u\mid (\Omega ))$.

Proof

The proof is split in several steps. By Lemma 4, for fixed $\theta _0\in (0,1/2]$, there exists a constant ${c_A}>0$ such that for all $\theta \in [\theta _0,1/2]$, all $\sigma >0$ and all $u\in {\mathcal {A}}_\theta$,

$$\begin{aligned} E_{\sigma ,\theta }(u)\ge {c_A}\min \left\{ \sigma (\mid \log \sigma \mid + 1), 1\right\} . \end{aligned}$$

On the other hand, Corollary 7 shows that there exist $\theta _0\in (0,1/2]$, $k_0 \in {\mathbb {N}}$ with $k_0\ge 2$ and ${c_B}>0$ (depending only on $k_0$) such that for all $\sigma >0$ and all $\theta \in (0,\theta _0]$, we have

$$\begin{aligned} \min _{u\in {\mathcal {A}}_\theta } E_{\sigma ,\theta }(u) \ge {c_B}{\left\{ \begin{array}{ll}\min \{\sigma , \theta ^2\} &{}\text { if } \sigma \ge \theta ^{k_0}, \\ k \sigma &{}\text { if } \sigma \in [\theta ^{{k+1}},\theta ^{{k}}) \text {{for some} } k_0 \le k. \end{array}\right. } \end{aligned}$$

(15)

Note that by Remark 1, this indeed implies the assertion:

• (i) If $\sigma \ge 1$ then

  $$\begin{aligned} {\min _{u\in {\mathcal {A}}_\theta } E_{\sigma ,\theta }(u) \ge c_B\min \{ \sigma ,\theta ^2\} =c_B\theta ^2\ge c_B \min \left\{ \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } + 1\right) , \theta ^2 \right\} }. \end{aligned}$$

• (ii) If $\sigma \in [\theta ^{\ell +1},\theta ^\ell )$ for some $0\le \ell {<} k_0$ then by Remark 1(ii), we have

  $$\begin{aligned} (k_0+2)\sigma \ge (\ell +2)\sigma \ge \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } + 1\right) {,} \end{aligned}$$

  and consequently, by (15),

  $$\begin{aligned} \min _{u\in {\mathcal {A}}_\theta } E_{\sigma ,\theta }(u)\ge & {} c_B\frac{k_0+2}{k_0+2}\min \{\sigma , \theta ^2\} \ge \frac{c_B}{k_0+2} \min \left\{ \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } + 1 \right) , \theta ^2\right\} . \end{aligned}$$

• (iii) If $\sigma \in (\theta ^{k+1},\theta ^{k})$ for some $2\le k_0 \le k$ then again by Remark 1(ii), we have

  $$\begin{aligned} 2k{\sigma }\ge (k+2)\sigma \ge \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } + 1 \right) \ge \min \left\{ \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } + 1\right) , \theta ^2 \right\} \end{aligned}$$

  and consequently, by (15),

  $$\begin{aligned} \min _{u\in {\mathcal {A}}_\theta } E_{\sigma ,\theta }(u)\ge c_Bk\sigma \ge \frac{c_B}{2}\min \left\{ \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } + 1\right) , \theta ^2 \right\} . \end{aligned}$$

Hence, choosing $c_L = \min \{ c_A,\frac{c_B}{k_0+2}\}$ concludes the proof. $\square$

As outlined above, we will prove the lower bound (15) separately for large and small $\theta$, respectively. We build on some techniques that have been used for example in the derivation of scaling laws for martensitic microstructures, see, e.g., Conti (2006), Conti et al. (2017), Conti et al. (2020), Zwicknagl (2014).

Lemma 4

Let $\theta _0\in (0,1/2]$. There exists a constant ${c_A}>0$ (depending only on $\theta _0$) such that for all $\sigma >0$, all $\theta \in [\theta _0,1/2]$, and all $u\in {\mathcal {A}}_\theta$

$$\begin{aligned} E_{\sigma ,\theta }(u)\ge {c_A}\min \left\{ \sigma \left( |\log \sigma |+1\right) , 1\right\} . \end{aligned}$$

Proof

Let u be an admissible function such that

$$\begin{aligned} E_{\sigma ,\theta }(u) \le \min \{\sigma (\mid \log \sigma \mid + 1),1\}. \end{aligned}$$

(16)

(Otherwise we are done.) Note that there exist measurable functions ${\rho _x, \rho _y}: (0,1)^2 \rightarrow \{\pm 1\}$ such that almost everywhere in $(0,1)^2$ it holds

$$\begin{aligned}{} & {} \min \{\mid \partial _1 u - 1\mid ,\mid \partial _1 u + 1\mid \} = \mid \partial _1 u + {\rho _x}\mid \quad \text { and } \quad \\{} & {} \min \{\mid \partial _2 u - 1\mid ,\mid \partial _2 u + 1\mid \} = \mid \partial _2 u + {\rho _y}\mid . \end{aligned}$$

Step 1: Comparison of u on vertical slices to the boundary data

For almost every $x\in (0,1)$ and almost every $y\in (0,1)$ we have by the fundamental theorem of calculus that

$$\begin{aligned} u(x,y) -(1-2\theta )y&= u(x,y) - u(0,y) \\&= \int ^x_0 \partial _1 u(t,y) \,\textrm{d}t\\&= \int ^x_0 \left( \partial _1 u(t,y) + \rho _x(t,y)\right) \,\textrm{d}t- \int ^x_0\rho _x(t,y) \,\textrm{d}t. \end{aligned}$$

Consequently, it holds for almost every $x\in (0,1)$ that

$$\begin{aligned} \int _0^1 \mid u(x,y)-(1-2\theta )y\mid \,\textrm{d}y&\le \int _0^1 \int _0^x \mid \partial _1 u(t,y)+ {\rho _x}(t,y)\mid {\,\textrm{d}t\,\textrm{d}y}+ x \\&\le x^{\frac{1}{2}} \left( \int _0^1 \int _0^x \mid \partial _1 u(t,y)+ {\rho _x}(t,y)\mid ^2 \,\textrm{d}t\,\textrm{d}y\right) ^{\frac{1}{2}} + x \\&\le x^{\frac{1}{2}} E_{\sigma ,\theta }(u)^{\frac{1}{2}} + x. \end{aligned}$$

Step 2: A lower bound on the energy on many vertical slices.

We fix a constant

$$\begin{aligned} 0<c_1\le \frac{\theta _0^3}{4\cdot 64\cdot 72^2}. \end{aligned}$$

(17)

Let ${\tilde{x}} \in (0,1)$ be such that

• 1. ${\mid \partial _2\partial _2 u \mid (\{{\tilde{x}}\} \times (0,1))} \le \frac{c_1}{{\tilde{x}}}$, and

• 2. $\int _0^1 \min \{\mid \partial _2 u({\tilde{x}},s)-1\mid ,\mid \partial _2 u({\tilde{x}},s)+1\mid \}^2 \,\textrm{d}s\le \frac{\theta _0^2}{12}$.

We claim that this implies that ${\tilde{x}}\le \frac{c_1}{\theta _0}\min \left\{ \sigma \left( \mid \log \sigma \mid +1\right) ,1\right\}$ or ${\tilde{x}}\ge \frac{2c_1}{\theta _0}$, so that in particular for almost all ${\tilde{x}}\in (\frac{c_1}{\theta _0}\min \left\{ \sigma \left( \mid \log \sigma \mid +1\right) {, 1}\right\} ,\frac{2c_1}{\theta _0})$ at least one of the two properties fails. Note that by (17), this is an interval of length at least $\frac{c_1}{\theta _0}$ that is completely contained in (0, 1).

To see the claim, we proceed similarly to Conti et al. (2017) and subdivide the interval (0, 1) into the three subsets

$$\begin{aligned} M_1&:=\left\{ y\in (0,1): \partial _2 u({\tilde{x}},y) \ge 1-\theta _0 \right\} ,\\ M_2&:=\left\{ y\in (0,1): \partial _2 u({\tilde{x}},y) \le -1+\theta _0 \right\} , \\ \text { and } M_3&:=\left\{ y\in (0,1): -1+\theta _0< \partial _2 u({\tilde{x}},y) < 1-\theta _0 \right\} . \end{aligned}$$

Since the three sets form a partition of the interval (0, 1), one of them has measure at least $\frac{1}{3}$. From property 2. it follows immediately that $\mid M_3\mid \le \frac{1}{12}<\frac{1}{3}$.

We consider the remaining two cases separately.

• (a) Consider first the case that $\mid M_1\mid \ge \frac{1}{3}$. By the coarea formula, we have (using property 1.)

  $$\begin{aligned}&\int _{1-\frac{3\theta _0}{2}}^{1-\theta _0} {\mathcal {H}}^0(\partial \{ y{\in (0,1)}:\ \partial _2 u({\tilde{x}},y)> s\}) \,\textrm{d}s\\&\quad \le \int _{{\mathbb {R}}} {\mathcal {H}}^0(\partial \{ y{\in (0,1)}:\ \partial _2 u({\tilde{x}},y) > s\}) \,\textrm{d}s\\&\quad = {\mid \partial _2 \partial _2 u({\tilde{x}},\cdot )\mid (0,1)} \le \frac{c_1}{{\tilde{x}}}. \end{aligned}$$

  Therefore, there is some $s\in (1-\frac{3\theta _0}{2},1-\theta _0)$ such that ${\mathcal {H}}^0(\partial \{y \in (0,1): \partial _2 u({\tilde{x}},{y})>s \}) \le \frac{2c_1}{\theta _0{\tilde{x}}}$. Since $|\left\{ y:\ \partial _2 u({\tilde{x}},y) > s \right\} |\ge \mid M_1\mid \ge \frac{1}{3}$, there exists a family of disjoint open intervals $(I_k)_{k=1}^K$ such that

  $$\begin{aligned} K \le \left\lceil \frac{2c_1}{\theta _0{\tilde{x}}}\right\rceil ,\qquad \sum _{k=1}^K \mid I_k\mid \ge \frac{1}{3}\qquad \text { and }\qquad \bigcup _{k=1}^K I_k \subseteq \left\{ y \in (0,1): \partial _2 u({\tilde{x}},y) \ge s\right\} . \end{aligned}$$

  On each interval $I_k$, we have $\partial _2u({\tilde{x}},\cdot )\ge 1-\frac{3\theta _0}{2}$, which implies that for all $y\in I_k$ (recall that $1-\frac{3\theta _0}{2}=1-2\theta _0+\theta _0/2\ge 1-2\theta +\theta _0/2$)

  $$\begin{aligned} |u({\tilde{x}},y)-(1-2\theta )y|&= \bigg |u({\tilde{x}},0)+\int _0^y\left( \partial _2u({\tilde{x}},t)-(1-2\theta )\right) \,dt\bigg |\\&\ge \bigg ||u({\tilde{x}},0)|- \bigg |\int _0^y\left( \partial _2u({\tilde{x}},t)-(1-2\theta )\right) \,dt \bigg |\bigg |\\&\ge {\min _{\alpha \in {\mathbb {R}}}|\alpha -\frac{\theta _0}{2}y|,} \end{aligned}$$

  and hence

  $$\begin{aligned} \int _{I_k} \mid u({\tilde{x}},y)-(1-2\theta )y\mid \,\textrm{d}y&\ge \min _{\alpha \in {\mathbb {R}}}\int _{I_k}|\alpha -\frac{\theta _0}{2}y|\,\textrm{d}y\nonumber \\&= 2\frac{\theta _0}{2}\int _0^{\mid I_k\mid /2}y\,dy = \frac{\theta _0}{8} \mid I_k\mid ^2. \end{aligned}$$

  (18)

  Summing estimate (18) over k and using Step 1 yields by assumption (16)

  $$\begin{aligned} \sum _{k=1}^K \frac{\theta _0}{8} \, \mid I_k\mid ^2&\le \int _0^1 |u({\tilde{x}},y)-(1-2\theta )y|\,\textrm{d}y\le {\tilde{x}}^{\frac{1}{2}} E_{\sigma ,\theta }(u)^{\frac{1}{2}} + {\tilde{x}} \nonumber \\&\le {\tilde{x}}^{\frac{1}{2}} \min \{\sigma (\mid \log \sigma \mid +1), 1\}^{\frac{1}{2}} + {\tilde{x}}. \end{aligned}$$

  (19)

  There are two possibilities: If $K=1\ge \frac{2c_1}{\theta _0{\tilde{x}}}$ then ${\tilde{x}}\ge \frac{2c_1}{\theta _0}$. Otherwise, we have $1{<} K \le \left\lceil \frac{2c_1}{\theta _0{\tilde{x}}}\right\rceil \le \frac{4c_1}{\theta _0{\tilde{x}}}$ and from $\sum _{k=1}^K \mid I_k\mid \ge \frac{1}{3}$ we deduce by convexity and (19) that

  $$\begin{aligned} \frac{\theta _0^2}{72}{\cdot } \frac{{\tilde{x}}}{4c_1}&\le \frac{\theta _0}{8} \frac{{K}}{(3K)^2}\le \sum _{k=1}^K \frac{\theta _0}{8} \, \mid I_k\mid ^2 \le {\tilde{x}}^{\frac{1}{2}} \min \{\sigma (\mid \log \sigma \mid +1), 1\}^{\frac{1}{2}} + {\tilde{x}}. \end{aligned}$$

  (20)

  Note that for $t\ge \frac{4\cdot 64\cdot 72^2{\cdot }c_1^2}{\theta _0^4} \min \{\sigma (\mid \log \sigma \mid +1), 1\}$,

  $$\begin{aligned}t^{-1/2} \min \{\sigma (\mid \log \sigma \mid +1), 1\}^{1/2}+1\le \frac{\theta _0^2}{8\cdot 72{\cdot }c_1}, \end{aligned}$$

  which implies that (20) can only hold for ${\tilde{x}}{<} \frac{4\cdot 64\cdot 72^2c_1^2}{\theta _0^4} \min \{\sigma (\mid \log \sigma \mid +1), 1\}$. Note that $\frac{4\cdot 64\cdot 72^2{\cdot }c_1^2}{\theta _0^4}\le \frac{c_1}{\theta _0}$ which concludes the proof of the claim in this case.

• (b) Consider now the case $\mid M_2\mid \ge \frac{1}{3}$. We proceed along the lines of (a), and find that there is some $t\in (-1,-1+\theta _0)$ such that ${\mathcal {H}}^0(\partial \{y \in (0,1): \partial _2 u({\tilde{x}},\cdot )<t \}) \le \frac{c_1}{\theta _0{\tilde{x}}}$. Consequently, there exists a family of disjoint open intervals $(I_k)_{k=1}^K$ such that

  $$\begin{aligned} K \le \left\lceil \frac{c_1}{\theta _0{\tilde{x}}}\right\rceil ,\qquad \sum _{k=1}^K \mid I_k\mid \ge \frac{1}{3}\qquad \text { and }\qquad \bigcup _{k=1}^K I_k \subseteq \left\{ y \in (0,1): \partial _2 u({\tilde{x}},y) \le t\right\} . \end{aligned}$$

  In this case, we obtain that in each interval $I_k$,

  $$\begin{aligned} \int _{I_k} \mid u({\tilde{x}},y)-(1-2\theta )y\mid \,\textrm{d}y&\ge \min _{\alpha \in {\mathbb {R}}}\int _{I_k}|\alpha -(1-\theta _0)y|\,\textrm{d}y\\&= \frac{1-\theta _0}{4}\mid I_k\mid ^2\ge \frac{\theta _0}{8}\mid I_k\mid ^2, \end{aligned}$$

  and the rest follows as in case (a).

Step 3. Conclusion.

By Step 2, for almost all ${\tilde{x}}\in (\frac{c_1}{\theta _0}\min \left\{ \sigma \left( \mid \log \sigma \mid +1\right) ,1\right\} ,\frac{2c_1}{\theta _0})$ at least one of the properties 1. or 2. is not true. Consequently,

$$\begin{aligned} E_{\sigma ,\theta }(u) \ge \int _{\frac{c_1}{\theta _0}\min \left\{ \sigma \left( \mid \log \sigma \mid +1\right) ,1\right\} }^{ \frac{2c_1}{\theta _0}} \min \left\{ \frac{\sigma c_1}{x}, \frac{\theta _0^2}{12} \right\} \,\textrm{d}x. \end{aligned}$$

(21)

We now consider two cases separately.

• (i) If $\sigma \le \frac{1}{2}$ then we find for $x\ge \frac{c_1}{\theta _0}\min \{\sigma (\mid \log \sigma \mid +1), 1\}=\frac{c_1}{\theta _0}\sigma (\mid \log \sigma \mid +1)$ that

  $$\begin{aligned} \frac{\sigma c_1}{x}\le \frac{\sigma c_1\theta _0}{c_1\sigma (\mid \log \sigma \mid +1)}\le \theta _0. \end{aligned}$$

  Hence, $\min \{\frac{\sigma c_1}{x}, \frac{\theta _0^2}{12}\}\ge \frac{\theta _0}{12}\frac{\sigma c_1}{x}$ for all $x\in (\frac{c_1}{\theta _0}\min \left\{ \sigma \left( \mid \log \sigma \mid +1\right) ,1\right\} ,\frac{2c_1}{\theta _0})$, and since $\log (\mid \log \sigma \mid +1)\le \max \left\{ \frac{9}{10}\log (2),\ \frac{3}{4} \mid \log (\sigma )\mid \right\}$, we deduce from (21)

  $$\begin{aligned} E_{\sigma ,\theta }(u)&\ge {\frac{\theta _0}{12}} \int _{\frac{c_1}{\theta _0}\sigma (\mid \log \sigma \mid +1)}^{\frac{2c_1}{\theta _0}} \frac{\sigma c_1}{x} \,\textrm{d}x={\frac{\theta _0}{12}} c_1 \sigma \left( \log (2)-\log \left( \sigma (\mid \log \sigma \mid +1)\right) \right) \nonumber \\&\ge {\frac{\theta _0}{12}}c_1\sigma \left( \frac{1}{10}\log (2)+{\frac{1}{4}}\mid \log (\sigma )\mid \right) \ge {\frac{\theta _0c_1}{12}\frac{\log (2)}{10}}\sigma \left( \mid \log (\sigma )\mid +1\right) , \end{aligned}$$

  (23)

  which concludes the proof in this case.

• (ii) If $\sigma >1/2$, we have for all $x\le \frac{2c_1}{\theta _0}$ that $\frac{\sigma c_1}{x}\ge \frac{\sigma \theta _0}{2}\ge \frac{\theta _0}{4}\ge \frac{\theta _0^2}{12}$, and hence

  $$\begin{aligned} E_{\sigma ,\theta }(u) \ge \int _{\frac{c_1}{\theta _0}}^{ \frac{2c_1}{\theta _0}} \frac{\theta _0^2}{12}=\frac{c_1\theta _0}{12}. \end{aligned}$$

  (24)

If we choose $c_A:=\frac{\theta _0c_1}{12}\frac{\log (2)}{10}$, the assertion follows from (23) and (24). $\square$

The proof of the lower bound in the case $\theta \ll 1$ is split into two lemmata which combined lead to the estimate in Corollary 7. We start with a general estimate which will be relevant for the lower bound only if $\sigma \ge \theta ^{32}$.

Lemma 5

There exists $c_B^{(1)}>0$ such that for all $\theta \in (0,1/2]$, all $\sigma >0$, and all $u\in {\mathcal {A}}_\theta$, we have

$$\begin{aligned} E_{\sigma ,\theta }(u) \ge c_B^{(1)}\min \{ \theta ^2,\sigma \}. \end{aligned}$$

Proof

Let u be an admissible function such that $E_{\sigma ,\theta }(u) \le \frac{1}{2\cdot 24^2} \min \{\theta ^2,\sigma \}$ (otherwise there is nothing to prove).

Step 1: Choice of representative vertical and horizontal slices.

By Fubini and slicing, we find ${\tilde{x}}\in ( \frac{3}{4},1)$ satisfying

$$\begin{aligned}&\int _0^1 \min \left\{ \mid \partial _2 u({\tilde{x}},t) - 1\mid , \mid \partial _2 u({\tilde{x}},y) + 1\mid \right\} ^2 \,\textrm{d}y+ \sigma {|\partial _2 \partial _2 u({\tilde{x}},\cdot )|(0,1)}\nonumber \\&\quad \le 4\bigg (\int _{3/4}^1\int _0^1 \min \{\mid \partial _2 u(x,y) - 1\mid , \mid \partial _2 u(x,y) + 1\mid \}^2 \,\textrm{d}y\,\textrm{d}x\nonumber \\&\qquad + \sigma |D^2 u|((3/4,1) \times (0,1)) \bigg )\nonumber \\&\quad \le 4 E_{\sigma ,\theta }(u)\le \frac{{2}}{24^2}\min \{\theta ^2,\sigma \}. \end{aligned}$$

(25)

Similarly, there are $y_1\in (1/8,1/4)$ and $y_2\in (3/4,7/8)$ such that

$$\begin{aligned}&\int _0^1 \min \left\{ \mid \partial _1 u (x,y_1) - 1\mid ,\mid \partial _1 u(x,y_1) + 1\mid \right\} ^2 \,\textrm{d}x+ \sigma {|\partial _1 \partial _1 u(\cdot ,y_1)|(0,1) } \nonumber \\&\quad \le \frac{4}{24^2}\min \{\theta ^2,\sigma \} \quad \text { and }\nonumber \\&\int _0^1 \min \left\{ \mid \partial _1 u (x,y_2) - 1\mid ,\mid \partial _1 u(x,y_2) + 1\mid \right\} ^2 \,\textrm{d}x+ \sigma {|\partial _1 \partial _1 u(\cdot ,y_2)|(0,1)} \nonumber \\&\quad \le \frac{4}{24^2}\min \{\theta ^2,\sigma \} . \end{aligned}$$

(26)

Step 2: Partial derivatives do not jump between the wells on the chosen slices.

By the coarea formula,

$$\begin{aligned} \int _{-1/2}^{1/2}{\mathcal {H}}^0\left( \partial \left\{ {x\in (0,1)}:\,\partial _1u(x,{y_i})>s\right\} \right) \,\textrm{d}s\le {|\partial _1 \partial _1 u(\cdot ,y_i)|(0,1)} \quad {\text { for }i=1,2.} \end{aligned}$$

Hence, there exists ${\tilde{s}} \in (-\frac{1}{2},\frac{1}{2})$ such that

$$\begin{aligned}&{\mathcal {H}}^0(\partial \{{x}: \partial _1 u ({x},y_1)> {\tilde{s}}\}) + {\mathcal {H}}^0(\partial \{{x}: \partial _1 u ({x},y_2) > {\tilde{s}}\}) \\&\quad \le {|\partial _1 \partial _1 u(\cdot ,y_1)|\left( (0,1) \right) } + {|\partial _1 \partial _1 u(\cdot ,y_2)|(0,1) } \le \frac{8}{24^2}<1, \end{aligned}$$

and therefore

$$\begin{aligned} {\mathcal {H}}^0(\partial \{{x}: \partial _1 u ({x},y_1)> {\tilde{s}}\}) = {\mathcal {H}}^0(\partial \{{x}: \partial _1 u ({x},y_2) > {\tilde{s}}\}) = 0.\end{aligned}$$

In particular, we have for ${i=1,2}$ either $\partial _1 u({x},{y_i}) \ge {\tilde{s}}$ for almost every ${x} \in (0,1)$ or $\partial _1 u({x},{y_i}) \le {\tilde{s}}$ for almost every ${x}\in (0,1)$. Without loss of generality, we may assume that

$$\begin{aligned} \partial _1 u({x},y_1) \ge {\tilde{s}} \quad \text { for almost every }{x} \in (0,1), \end{aligned}$$

(27)

the other case can be treated analogously. Note that we will consider the two possibilities for $y_2$ separately in the sequel.

Proceeding analogously, we also find some ${\tilde{t}}\in (-\frac{3}{4},-\frac{1}{2})$ such that

$$\begin{aligned} {\mathcal {H}}^0\left( \partial \{y{\in (0,1)}:\partial _2u({\tilde{x}},y)>{\tilde{t}}\}\right) =0. \end{aligned}$$

(28)

Step 3: A lower bound for the energy.

By (27), for almost every ${x}\in (0,1)$ we have $\partial _1u({x},y_1)\ge {\tilde{s}}>-1/2$, which implies that

$$\begin{aligned} |\partial _1u({x},y_1)-1|<3|\partial _1u({x},y_1)+1|. \end{aligned}$$

Hence, by Hölder’s inequality, we have for every $x\in (0,1)$, using the choice of $y_1$ (see (26))

$$\begin{aligned} \bigg |\int _0^x \partial _1 u(t,y_1) \,\textrm{d}t- x \bigg |^2&\le \left( \int _0^x|\partial _1 u(t,y_1)-1|\,\textrm{d}t\right) ^2\le \int _0^{{x}} \mid \partial _1 u(t,y_1) - 1\mid ^2 \,\textrm{d}t\\&\le 9 \int _0^x \min \left\{ \mid \partial _1 u(t,y_1) - 1\mid ,\mid \partial _1 u (t,y_1) + 1\mid \right\} ^2 \,\textrm{d}t\\&\le \left( \frac{\theta }{4} \right) ^2. \end{aligned}$$

Consequently, by the fundamental theorem of calculus,

$$\begin{aligned} \left( u(x,y_1) - u(0,y_1)\right) -x =\int _0^x\partial _1u(t,y_1)\,\textrm{d}t-x\in \left( -\frac{\theta }{4},\frac{\theta }{4}\right) . \end{aligned}$$

(29)

We proceed similarly for $y_2$ where we consider the two cases from Step 2 separately. If $\partial _1u({x},y_2)\ge {\tilde{s}}$ for almost every ${x}\in (0,1)$ then as above for $y_1$,

$$\begin{aligned}&\left( u(x,y_2) - u(0,y_2)\right) -x \in \left( -\frac{\theta }{4},\frac{\theta }{4}\right) \qquad \text { for all }x\in (0,1). \end{aligned}$$

In the other case, i.e., if $\partial _1u({x},y_2)\le {\tilde{s}}$ for almost every ${x}\in (0,1)$ then we find similarly

$$\begin{aligned}&\left( u(x,y_2) - u(0,y_2)\right) +x \in \left( -\frac{\theta }{4},\frac{\theta }{4}\right) \qquad \text { for all }x\in (0,1). \end{aligned}$$

We consider the two cases separately.

Case 1: Suppose that $u(x,y_2) - u(0,y_2)-x \in (-\frac{\theta }{4},\frac{\theta }{4})$ for all $x\in (0,1)$.

Recalling (29), in this case, we have that $x-\frac{\theta }{4}<u(x,y_i)-u(0,y_i)<x+\frac{\theta }{4}$ for $i\in \{1,2\}$. Hence, using $y_2-y_1\ge \frac{1}{2}$ and the boundary condition at $x=0$, we obtain

$$\begin{aligned}&(1-3\theta )(y_2-y_1) \\&\quad \le (1-2\theta )(y_2-y_1) - \frac{1}{2} \theta =u(0,y_2)-u(0,y_1)-\frac{\theta }{2}\\&\quad =u(x,y_2)-\left( u(x,y_2)-u(0,y_2)\right) -\frac{\theta }{4}-u(x,y_1)+\left( u(x,y_1)-u(0,y_1)\right) -\frac{\theta }{4}\\&\quad \le u(x,y_2) - u(x,y_1) \\&\quad = u(0,y_2)+\left( u(x,y_2)-u(0,y_2)\right) -u(0,y_1)-\left( u(x,y_1)-u(0,y_1)\right) \\&\quad \le (1-2\theta )(y_2-y_1) + \frac{\theta }{2} \le (1-\theta ) (y_2-y_1), \end{aligned}$$

which implies that

$$\begin{aligned} -\frac{1}{2}\le 1-3\theta \le \frac{u(x,y_2)-u(x,y_1)}{y_2-y_1}\le 1-\theta \qquad \text { for all }x\in (0,1), \end{aligned}$$

(30)

and hence in particular

$$\begin{aligned} \bigg |\frac{u(x,y_2) - u(x,y_1)}{y_2-y_1} - 1 \bigg |\ge \theta \qquad \text { for all }x\in (0,1). \end{aligned}$$

We now consider ${\tilde{x}}$ as chosen in Step 1. Using the lower bound in (30) we deduce from (28) that for almost all $y \in (0,1)$ we have $\partial _2 u({\tilde{x}},y)> {\tilde{t}}>-\frac{3}{4}$, and hence for almost all $y\in (0,1)$

$$\begin{aligned} \mid \partial _2 u({\tilde{x}},y) - 1\mid \le 7 \min \{ \mid \partial _2 u({\tilde{x}},y) - 1\mid , \mid \partial _2 u({\tilde{x}},y) + 1\mid \} . \end{aligned}$$

Putting things together, we obtain by Hölder’s inequality (recall that $\mid y_2-y_1\mid \ge 1/2$), using (25) in the last step,

$$\begin{aligned} \theta ^2&\le \bigg |\frac{u({\tilde{x}},y_2) - u({\tilde{x}},y_1)}{y_2-y_1} - 1 \bigg |^2 {=} \bigg |\frac{1}{y_2-y_1}\int _{y_1}^{y_2} \partial _2 u ({\tilde{x}},y) \,\textrm{d}y- 1 \bigg |^2 \nonumber \\&\le \frac{1}{y_2-y_1} \int _{y_1}^{y_2} \mid \partial _2 u({\tilde{x}},y) - 1\mid ^2 \,\textrm{d}y\le 2 \int _0^1 \mid \partial _2 u({\tilde{x}},y) - 1\mid ^2 \,\textrm{d}y\nonumber \\&\le 98 \int _0^1 \min \{\mid \partial _2 u({\tilde{x}},y)-1\mid , \mid \partial _2 u({\tilde{x}},y)+1\mid \}^2 \,\textrm{d}y\le 392 E_{\sigma ,\theta }(u). \end{aligned}$$

(31)

This concludes the proof of the lower bound in this case if $c_B\le \frac{1}{392}$.

Case 2: Suppose that $u(x,y_2) - u(0,y_2) +x\in (-\frac{\theta }{4},\frac{\theta }{4})$ for all $x\in (0,1)$.

In this case, we have (recall (29))

$$\begin{aligned} x- & {} \frac{\theta }{4}<u(x,y_1)-u(0,y_1)<x+\frac{\theta }{4}\quad \text { and } \quad -x-\frac{\theta }{4}<u(x,y_2)\\- & {} u(0,y_2)<-x+\frac{\theta }{4}. \end{aligned}$$

Proceeding similarly to Case 1, we apply this with $x={\tilde{x}}$, and we obtain using that ${\tilde{x}}\in ( 3/4,1)$, ${\frac{3}{4} \ge y_2-y_1\ge 1/2}$, and the boundary condition at $x=0$,

$$\begin{aligned}&(-3-3\theta )(y_2-y_1) \\&\quad \le (1-2\theta )(y_2-y_1)-2-\frac{\theta }{2}\le (1-2\theta )(y_2-y_1)-2{\tilde{x}}-\frac{\theta }{2}\\&\quad \le u(0,y_2)-u(0,y_1)+\left( u({\tilde{x}},y_2)-u(0,y_2)\right) -\left( u({\tilde{x}},y_1)-u(0,y_1)\right) \\&\quad = u({\tilde{x}},y_2)-u({\tilde{x}},y_1)\\&\quad \le (1-2\theta )(y_2-y_1)-2{\tilde{x}}+\frac{\theta }{2}\le (1-2\theta )(y_2-y_1)-\frac{3}{2}+\frac{\theta }{2}\\&\quad \le (-1-\theta )(y_2-y_1). \end{aligned}$$

This yields in particular

$$\begin{aligned} -3-3\theta \le \frac{u({\tilde{x}},y_2) - u({\tilde{x}},y_1)}{y_2-y_1} \le -1-\theta \le {\tilde{t}}, \end{aligned}$$

and we deduce from (28) that $\partial _2u({\tilde{x}},y)\le {\tilde{t}}$ for almost all $y\in (0,1)$. Thus,

$$\begin{aligned} \mid \partial _2 u({\tilde{x}},y) + 1\mid = \min \{ \mid \partial _2 u({\tilde{x}},y) - 1\mid , \mid \partial _2 u({\tilde{x}},y) + 1\mid \} \qquad \text { for almost all }y\in (0,1). \end{aligned}$$

Since

$$\begin{aligned} \bigg |\frac{u({\tilde{x}},y_2) - u({\tilde{x}},y_1)}{y_2-y_1} + 1 \bigg |\ge \theta , \end{aligned}$$

the claimed lower bound on $E_{\sigma ,\theta }(u)$ follows as in (31) (with an even better constant).

This concludes the proof of the lemma. $\square$

Lemma 6

There exists a constant $c_B^{(2)}>0$ with the following property. For all $\theta \in (0,1/2]$ and all $\sigma \in (0,\theta ^k)$ for some $k\in {\mathbb {N}}$ with $k\ge 32$, we have

$$\begin{aligned} E_{\sigma ,\theta }(u)\ge c_B^{(2)}k\sigma . \end{aligned}$$

Proof

It suffices to consider $\sigma \in [\theta ^{k+1},\theta ^k)$ for some $k\ge 32$. Let $u\in {\mathcal {A}}_\theta$ be such that $E_{\sigma ,\theta }(u) \le k \sigma$ (otherwise there is nothing to prove). Let $K = \lfloor \frac{k}{8} \rfloor$. We fix ${c_2:=} \frac{1}{2000}$.

Step 1: Choice of representative vertical slices and reduction to auxiliary statement.

By Fubini’s theorem and standard slicing arguments we can find points $x_i \in (\frac{1}{2} \theta ^{2i},\frac{3}{2} \theta ^{2i})$, $i=1,\dots ,K$, such that $u(x_i,\cdot ) \in H^1(0,1)$, $\partial _2 u(x_i, \cdot ) \in BV(0,1)$ and

$$\begin{aligned}&\int _0^1 \min \{\mid \partial _2 u(x_i,y)-1\mid , \mid \partial _2 u(x_i,y)+1\mid \}^2 \, dy + \sigma \mid \partial _2 \partial _2 u(x_i,\cdot )\mid (0,1) \\&\quad \le \theta ^{-2i}E_{\sigma ,\theta }\left( u,(\frac{1}{2} \theta ^{2i},\frac{3}{2} \theta ^{2i}) \times (0,1)\right) . \end{aligned}$$

For an illustration, see Fig. 5. Note that by the assumption $\theta \le 1/2$, the intervals $(\frac{1}{2} \theta ^{2i},\frac{3}{2} \theta ^{2i})$ for $1=1,\dots ,K$ are pairwise disjoint and contained in (0, 1).

In the subsequent steps, we will prove the following result: There exists a constant $c_3>0$ (not depending on $\sigma$ and $\theta$) such that for each $1\le i\le K-1$ we have (recall that we chose $c_2$ above)

$$\begin{aligned}{} & {} \int _0^1 \min \{\mid \partial _2 u(x_i,y)-1\mid , \mid \partial _2 u(x_i,y)+1\mid \}^2 dy + \sigma \mid \partial _2 \partial _2 u(x_i,\cdot )\mid (0,1) \ge {c_2} \theta ^{-2i} \sigma \end{aligned}$$

(32)

or

$$\begin{aligned} E_{\sigma ,\theta }(u; (x_{i+1},x_i) \times (0,1)) \ge {c_3} \sigma . \end{aligned}$$

(33)

Note that this indeed implies the assertion since (using the choice of $x_i$)

$$\begin{aligned}&E_{\sigma ,\theta }(u)\ge \frac{1}{2}\sum _{i=1}^{K-1} \left( E_{\sigma ,\theta }(u,(\frac{1}{2} \theta ^{2i},\frac{3}{2} \theta ^{2i})\times (0,1)) +E_{\sigma ,\theta }(u,(x_{i+1},x_i)\times (0,1))\right) \\&\quad \ge \frac{1}{2}\sum _{i=1}^{K-1}\bigg ( \theta ^{2i} \int _0^1 \min \{\mid \partial _2 u(x_i,y)-1\mid , \mid \partial _2 u(x_i,y)+1\mid \}^2 \, dy \\&\qquad + \theta ^{2i} \sigma \mid \partial _2 \partial _2 u(x_i,\cdot )\mid (0,1) +E_{\sigma ,\theta }(u,(x_{i+1},x_i)\times (0,1))\bigg )\\&\quad \ge \frac{1}{2}\sum _{i=1}^{K-1}\min \{{c_2,c_3}\}\sigma \ge \frac{\min \{{c_2,c_3}\}}{{32}}k\sigma . \end{aligned}$$

Therefore, from now on, we fix some $1 \le i \le K-1$ and assume that (32) does not hold, i.e., we have

$$\begin{aligned}{} & {} \int _0^1 \min \{\mid \partial _2 u(x_i,y)-1\mid , \mid \partial _2 u(x_i,y)+1\mid \}^2 \, dy + \sigma \mid \partial _2 \partial _2 u(x_i,\cdot )\mid (0,1) \le {c_2}\theta ^{-2i} \sigma \end{aligned}$$

(34)

In the rest of the proof, we will show that (33) holds for a constant $c_3>0$ chosen below.

Step 2: Choice of a large representative portion with an almost constant derivative.

We note that inequality (34) implies that $\mid \partial _2 \partial _2 u(x_i,\cdot )\mid (0,1) \le {c_2} \theta ^{-2i}$ and

$$\begin{aligned}&\int _0^1 \min \{\mid \partial _2 u(x_i,y)-1\mid , \mid \partial _2 u(x_i,y)+1\mid \}^2 \, dy \nonumber \\&\quad \le {c_2} \theta ^{-2i} \sigma \nonumber \\&\quad \le {c_2} \theta ^{-2K+k} \le {c_2} \theta ^3. \end{aligned}$$

(35)

We proceed similarly to the proof of Lemma 5. By the choice of $c_2$, the set

$$\begin{aligned} {P_1:=}\{y\in (0,1):&\partial _2u(x_i,y)\in (-1+\theta ,1-\theta ) \text { or }\partial _2u(x_i,y)\le -1-\theta \\&\text { or }\partial _2u(x_i,y)\ge 1+\theta \} \end{aligned}$$

has measure (much) less than 1/3. Next we show that also the set $P_1:=\{y\in (0,1):\ -1-\theta \le \partial _2u(x_i,y)\le -1+\theta \}$ has measure less than 1/3. For a contradiction, let us assume that ${\mathcal {L}}^1(P_1) \ge 1/3$. We find $y_2,y_1 \in (0,1)$ such that $y_2-y_1 \ge 1-\frac{1}{12}$ and for $j=1,2$

$$\begin{aligned} \int _0^1 \min \{\mid \partial _1 u(x,y_j)-1\mid , \mid \partial _1 u(x,y_j)+1\mid \}^2 \, dx \le 24 E_{\sigma ,\theta }(u). \end{aligned}$$

Then we estimate, using that $E_{\sigma ,\theta }(u)\le k\sigma \le \theta ^2$,

$$\begin{aligned} \int _{y_1}^{y_2} \partial _2 u(x_i,t) \, dt&= u(x_i,y_2) - u(x_i,y_1) \\&\ge u(0,y_2) - u(0,y_1) - {2}x_i -{2} x_i^{\frac{1}{2}} \left( 24 E_{\sigma ,\theta }(u) \right) ^{\frac{1}{2}} \\&\ge (1-2\theta )(1-\frac{1}{12}) - {3} \theta ^{2} - {2}\left( 36 \theta ^{4} \right) ^{\frac{1}{2}}. \end{aligned}$$

Since ${\mathcal {L}}^1\left( (y_2,y_1)\cap P_1\right) \ge {\mathcal {L}}^1(P_1)-{\mathcal {L}}^1\left( (0,1)\setminus (y_1,y_2)\right) \ge \frac{1}{3}-\frac{1}{12}$, we have for $\theta _0\le \frac{1}{16}$

$$\begin{aligned}&\int _{(y_2,y_1)\cap \{\partial _2 u{(x_i,\cdot )}\ge 0\}} \partial _2 u(x_i,t) \, dt\nonumber \\&\quad \ge (1-2\theta )(1-\frac{1}{12}) - {3} \theta ^{2} - {12} \theta ^2 - (-1+\theta )(\frac{1}{3}-\frac{1}{12}) \nonumber \\&\quad \ge \frac{14}{12} - \frac{1}{16} \left( 2 + \frac{3}{16} + \frac{3}{4} + \frac{1}{4}\right) \nonumber \\&\quad \ge {\frac{9}{10}}. \end{aligned}$$

(36)

On the other hand, we have by (35) and the choice of $c_2$

$$\begin{aligned}&\int _{(y_2,y_1)\cap \{\partial _2 u(x_i,\cdot )\ge 0\}} \partial _2 u(x_i,t) \, dt \nonumber \\&\quad \le \left( \int _{(y_2,y_1)\cap \{\partial _2 u(x_i,\cdot )\ge 0\}} (\partial _2 u(x_i,t)-1)^2 \, dt\right) ^{\frac{1}{2}} + \mid \{\partial _2 u(x_i,\cdot )\ge 0\}\mid \nonumber \\&\quad \le {c_2}^{\frac{1}{2}}\theta + \frac{2}{3}{<0.67}. \end{aligned}$$

(37)

Combining (36) and (37) yields a contradiction. Consequently, the set

$$\begin{aligned} P_2:=\{y\in (0,1):\ 1-\theta \le \partial _2u(x_i,y)\le 1+\theta \} \end{aligned}$$

has measure at least 1/3. Hence, using the coarea formula, we derive that there exists ${\tilde{\theta }} \in (\theta ,3/2 \theta )$ such that $A_i= \{ \partial _2 u(x_i,\cdot ) \in (1-{{\tilde{\theta }}},1+{{\tilde{\theta }}})\}$ is of finite perimeter and such that

$$\begin{aligned} {\mathcal {L}}^1(A_i) \ge \frac{1}{3} \text { and } \quad {\mathcal {H}}^0(\partial ^* A_i) \le {2} {c_2} \theta ^{-2i-1}. \end{aligned}$$

Let us now consider the disjoint intervals $I_l = (\frac{l}{9{c_2}} \theta ^{2i+1}, \frac{ (l+1)}{9{c_2}} \theta ^{2i+1})$ for $l\in {\mathbb {N}}_0$. For $\theta _0$ so small that ${{c_2}\theta _0^{-3} \ge 2}$ it follows that at least ${\frac{{c_2}}{2}} \theta ^{-2i-1}$ of those intervals are (up to a set of measure 0) contained in $A_i$. Indeed, by volume considerations the number of intervals $I_l$ intersecting $A_i$ is larger than $3{c_2} \theta ^{-2i-1}$. In addition, the number of intervals that contain a point of $\partial ^* A_i$ is bounded by the number of points in $\partial ^* A_i$. Eventually, there might be an interval intersecting $A_i$, which does not contain a point from $\partial ^* A_i$ but is not fully contained in (0, 1). Hence, the number of intervals that are fully contained in $A_i$ is at least $3{c_2} \theta ^{-2i-1} - {c_2} {2} \theta ^{-2i-1} - 1 \ge {\frac{{c_2}}{2}} \theta ^{-2i-1}$. In particular, we can find an interval $I_{{\bar{l}}} \subseteq A_i$ such that

$$\begin{aligned}&E_{\sigma ,\theta }(u; (0,1) \times I_{{\bar{l}}}) \le {40} \, {\mathcal {L}}^1(I_{{\bar{l}}}) \, E_{\sigma ,\theta }(u) \quad \nonumber \\ \text { and } \quad&E_{\sigma ,\theta }(u; (x_{i+1},x_i) \times I_{{\bar{l}}}) \le {40} \, {\mathcal {L}}^1(I_{{\bar{l}}}) \, E_{\sigma ,\theta }(u; (x_{i+1},x_i) \times (0,1)) \end{aligned}$$

(38)

Step 3: Estimate on horizontal difference quotients between $\{x_i\} \times (0,1)$ and $\{x_{i+1}\} \times (0,1)$.

Let us write $I_{{\bar{l}}} = (a_{{\bar{l}}},b_{{\bar{l}}})$ and estimate for $t \in (a_{{\bar{l}}}, a_{{\bar{l}}} + \mid I_{{\bar{l}}}\mid /2)$

$$\begin{aligned}&|u(x_i,t+ \mid I_{{\bar{l}}}\mid /2) - u(x_i,t) - u(x_{i+1},t+ \mid I_{{\bar{l}}}\mid /2) + u(x_{i+1},t) |\nonumber \\&\quad \ge |u(x_i,t+ \mid I_{{\bar{l}}}\mid /2) - u(x_i,t) - (1-2\theta ) \mid I_{{\bar{l}}}\mid /2 |-|u(x_{i+1},t) - (1-2\theta )t|\nonumber \\&\qquad - |u(x_{i+1},t+\mid I_{{\bar{l}}}\mid /2) - (1-2\theta )(t+\mid I_{{\bar{l}}}\mid /2)|. \end{aligned}$$

(39)

By the definition of $A_i$ and $I_{{\bar{l}}}$, we find for the first term of the right-hand side (using ${\tilde{\theta }}\in (\theta ,3/2 \theta )$)

$$\begin{aligned} |u(x_i,t+ \mid I_{{\bar{l}}}\mid /2) - u(x_i,t) - (1-2\theta ) \mid I_{{\bar{l}}}\mid /2 |\, dt \ge \frac{\mid I_{{\bar{l}}}\mid \theta }{4}. \end{aligned}$$

For the second term, we follow the argument of Step 1 in the proof of Lemma 4 and find for almost every $t \in (a_{{\bar{l}}}, a_{{\bar{l}}} + \mid I_{{\bar{l}}}\mid /2)$

$$\begin{aligned}&|u(x_{i+1},t) - (1-2\theta )t|\\&\quad = |u(x_{i+1},t) - u(0,t)|\\&\quad \le x_{i+1}^{\frac{1}{2}} \, \left( \int _{0}^1 \min \{\mid \partial _1 u(s,t) -1\mid , \mid \partial _1 u(s,t) +1\mid \}^2 \, ds \right) ^{\frac{1}{2}} \, ds + x_{i+1}. \end{aligned}$$

The third term in (39) can be treated similarly. Putting things together, (39) yields for almost every t

$$\begin{aligned}&|u(x_i,t+ \mid I_{{\bar{l}}}\mid /2) - u(x_i,t) - u(x_{i+1},t+ \mid I_{{\bar{l}}}\mid /2) + u(x_{i+1},t) |\nonumber \\&\quad \ge \frac{\mid I_{{\bar{l}}}\mid \theta }{4} - 2 x_{i+1} - x_{i+1}^{\frac{1}{2}} \, \left( \int _{0}^1 \min \{\mid \partial _1 u(s,t) -1\mid , \mid \partial _1 u(s,t) +1\mid \}^2 \, ds \right) ^{\frac{1}{2}} \nonumber \\&\qquad -x_{i+1}^{\frac{1}{2}} \, \left( \int _{0}^1 \min \{\mid \partial _1 u(s,t+{\frac{\mid I_{{\bar{l}}}\mid }{2}}) -1\mid , \mid \partial _1 u(s,t+{\frac{\mid I_{{\bar{l}}}\mid }{2}}) +1\mid \}^2 \, ds \right) ^{\frac{1}{2}}. \end{aligned}$$

(40)

On the other hand, we may estimate similarly to above for almost every t

$$\begin{aligned}&|u(x_i,t+ \mid I_{{\bar{l}}}\mid /2) - u(x_i,t) - u(x_{i+1},t+ \mid I_{{\bar{l}}}\mid /2) + u(x_{i+1},t) |\nonumber \\&\quad \le |u(x_i,t+ \mid I_{{\bar{l}}}\mid /2) - u(x_i,t) - (1-2\theta ) \mid I_{{\bar{l}}}\mid /2|\nonumber \\&\qquad + |u(x_{i+1},t+ \mid I_{{\bar{l}}}\mid /2) - u(x_{i+1},t) - (1-2\theta ) \mid I_{{\bar{l}}}\mid /2|\nonumber \\&\quad \le 2\theta |I_{{\bar{l}}}|+ 2x_{i+1} + x_{i+1}^{\frac{1}{2}} \, \left( \int _{0}^1 \min \{\mid \partial _1 u(s,t) -1 \mid , \mid \partial _1 u(s,t) +1\mid \}^2 \, ds \right) ^{\frac{1}{2}} \nonumber \\&\qquad + x_{i+1}^{\frac{1}{2}} \, \left( \int _{0}^1 \min \{\mid \partial _1 u(s,t+{\frac{\mid I_l\mid }{2}}) -1\mid , \mid \partial _1 u(s,t+{\frac{\mid I_l\mid }{2}}) +1\mid \}^2 \, ds \right) ^{\frac{1}{2}} . \end{aligned}$$

(41)

Next, we notice that by the choice of $I_{{\bar{l}}}$ (see (38))

$$\begin{aligned}&\int _{a_{{\bar{l}}}}^{a_{{\bar{l}}} + \mid I_{{\bar{l}}}\mid /2} \int _{0}^1 \min \{\mid \partial _1 u(s,t) -1\mid , \mid \partial _1 u(s,t) +1\mid \}^2 \, ds \, dt \\&\quad \le E_{\sigma ,\theta }(u; (0,1) \times I_{{\bar{l}}}) \le {40} \mid I_{{\bar{l}}}\mid \, E_{\sigma ,\theta }(u) \quad \text { and }\\&\int _{a_{{\bar{l}}}}^{a_{{\bar{l}}} + \mid I_{{\bar{l}}}\mid /2} \int _{0}^1 \min \{\mid \partial _1 u(s,t+\mid I_{{\bar{l}}}\mid /2) -1\mid , \mid \partial _1 u(s,t+\mid I_{{\bar{l}}}\mid /2) +1\mid \}^2 \, ds \\&\quad \le E_{\sigma ,\theta }(u; (0,1) \times I_{{\bar{l}}}) \le {40} \mid I_{{\bar{l}}}\mid \, E_{\sigma ,\theta }(u) . \end{aligned}$$

Hence, there exists a subset of $(a_{{\bar{l}}},a_{{\bar{l}}}+\mid I_{{\bar{l}}}\mid /2)$ whose measure is at least $\frac{1}{4} \mid I_{{\bar{l}}}\mid$ such that for all its elements t it holds

$$\begin{aligned}&\int _{0}^1 \min \{\mid \partial _1 u(s,t) -1\mid , \mid \partial _1 u(s,t) +1\mid \}^2 \, ds \le {320} E_{\sigma ,\theta }(u) \text { and } \nonumber \\&\int _{0}^1 \min \{\mid \partial _1 u(s,t+{\frac{\mid I_{{\bar{l}}}\mid }{2}}) -1\mid , \mid \partial _1 u(s,t+{\frac{\mid I_{{\bar{l}}}\mid }{2}}) +1\mid \}^2 \, ds \le {320} \, E_{\sigma ,\theta }(u). \end{aligned}$$

(42)

Next, we note that it holds by assumption that (recall that $\sigma <\theta ^k$, $k\ge 32$, $i\le \frac{k}{8}$, and $\theta \le \frac{1}{2}$)

$$\begin{aligned} E_{\sigma ,\theta }(u)\le k\sigma \le k \theta ^k \le \theta ^{\frac{k}{2}} \le \theta ^{2i+2}. \end{aligned}$$

Hence, for all the t from above we obtain from (40) and (42)

$$\begin{aligned}&\mid u(x_i,t+ \mid I_{{\bar{l}}}\mid /2) - u(x_i,t) - u(x_{i+1},t+ \mid I_{{\bar{l}}}\mid /2) + u(x_{i+1},t) \mid \nonumber \\&\quad \ge \frac{\mid I_{{\bar{l}}}\mid \theta }{4} - 2 x_{i+1} - 2 x_{i+1}^{\frac{1}{2}} \sqrt{{320}} \sqrt{E_{\sigma ,\theta }(u)} \nonumber \\&\quad \ge \frac{1}{36 {c_2}} \theta ^{2i+2} - 3 \theta ^{2i+2} - 2 \sqrt{{480}} \theta ^{2i+2}\nonumber \\&\quad \ge \theta ^{2i+2} (50 - {3 - 22}) \nonumber \\&\quad \ge \theta ^{2i+2}. \end{aligned}$$

(43)

On the other hand, we obtain similarly for the same $t \in I_{{\bar{l}}}$ from (41) and (42)

$$\begin{aligned}&|u(x_i,t+ \mid I_{{\bar{l}}}\mid /2) - u(x_i,t) - u(x_{i+1},t+ \mid I_{{\bar{l}}}\mid /2) + u(x_{i+1},t) |\nonumber \\&\quad \le 2\theta \frac{1}{9{c_2}} \theta ^{2i+1} + 2x_{i+1} + {2} x_{i+1}^{\frac{1}{2}} \sqrt{{320} E_{\sigma ,\theta }(u)} \nonumber \\&\quad \le (500+3+ {2 \sqrt{480}}) \theta ^{2i+2}. \end{aligned}$$

(44)

By definition of $x_{i}$ and $x_{i+1}$ we have $\frac{1}{8} \theta ^{2i} \le x_{i} - x_{i+1} \le \frac{3}{2}\theta ^{2i}$. Together with (43) and (44) this yields for the $t \in I_{{\bar{l}}}$ from above that

$$\begin{aligned} \frac{2}{3} \theta ^2&\le \bigg |\frac{u(x_i,t+ \mid I_{{\bar{l}}}\mid /2)- u(x_{i+1},t+ \mid I_{{\bar{l}}}\mid /2) }{x_{i} - x_{i+1}}- \frac{u(x_i,t) - u(x_{i+1},t)}{x_{i} - x_{i+1}} \bigg |\nonumber \\&\le 8(500+3+ {2 \sqrt{480}}) \theta ^2. \end{aligned}$$

(45)

We now choose $\theta _0\in (0,1/2]$ small enough such that $8(500+3+ {2 \sqrt{480}})\theta _0^2\le 1/2$. Hence, roughly speaking, at most one of the difference quotients occurring in (45) can be close (at the order of $\theta ^2$) to $\{\pm 1\}$. Precisely, summarizing the results of this step, there is a universal constant ${c_4}>0$ and a subset of $I_{{\bar{l}}}$ whose measure is at least $\frac{1}{4} \mid I_{{\bar{l}}}\mid$ such that for all t in this subset it holds

$$\begin{aligned} \bigg |\, \bigg |\frac{u(x_i,t) - u(x_{i+1},t)}{x_{i}-x_{i+1}}\bigg |- 1 \bigg |\ge {c_4} \theta ^2. \end{aligned}$$

Step 4: Conclusion of estimate (33).

Let us now assume that for a point t from Step 2 it holds $\mid \partial _1 \partial _1 u(\cdot ,t)\mid (x_{i+1},x_i) < \frac{1}{2}$. Then we may assume without loss of generality for almost all $s \in (x_{i+1},x_i)$ that $\mid \partial _1 u (s,t) - 1\mid \le 3 \min \{\mid \partial _1 u (s,t) - 1\mid , \mid \partial _1 u (s,t) + 1\mid \}$ and thus

$$\begin{aligned}&\int _{x_{i+1}}^{x_i} \min \{\mid \partial _1 u (s,t) - 1\mid , \mid \partial _1 u (s,t) + 1\mid \}^2 \, ds \\&\quad \ge \frac{1}{9} \int _{x_{i+1}}^{x_i} \mid \partial _1 u (s,t) - 1\mid ^2 \, ds \\&\quad \ge \frac{1}{9(x_{i}-x_{i+1})} \left( \int _{x_{i+1}}^{x_i} \partial _1 u(s,t) - 1 \, ds \right) ^2 \\&\quad = \frac{1}{9} (x_{i}-x_{i+1}) \left( \frac{u(x_i,t) - u(x_{i+1},t)}{x_i - x_{i+1}} - 1 \right) ^2 \\&\quad \ge \frac{{c_4}^2}{72} \theta ^{2i} \theta ^4 \\&\quad \ge \frac{{c_4}^2}{72}\sigma . \end{aligned}$$

For the last estimate, we used again that $2i+4 \le \frac{k}{4} + 4 \le k$ as $k\ge 32$. Consequently,

$$\begin{aligned} E_{\sigma ,\theta }(u, (x_{i+1},x_i) \times I_{{\bar{l}}})&\ge \sigma \mid I_{{\bar{l}}}\mid \, \min \{ \frac{1}{72} {c_4}^2, \frac{1}{4} \}. \end{aligned}$$

This concludes the proof of (33) for $\theta _0 \le \min \left\{ \root 3 \of {c_2/2}, 1/16, \sqrt{\frac{1}{16(500+3+2\sqrt{480})}} \right\}$ with ${c_3} \le \min \{ \frac{1}{72} {c_4}^2, \frac{1}{4}\}$, and hence the assertion is proven. $\square$

Fig. 5

In the proof of the lower bound representative vertical slices (dashed lines) at $x_i$ ($x_{i+1}$) are chosen in a neighborhood around $\theta ^{2i}$ (green) ($\theta ^{2i+2}$ (red)). On a representative vertical slice $\{x_i\} \times (0,1)$ intervals $\{x_i\} \times I_l$ (vertical blue line) are identified in which u is almost affine. Then difference quotients of u between $x_{i+1}$ and $x_i$ are estimated along horizontal lines (horizontal blue line)

Combining the estimates of Lemma 5 and 6, we obtain the following lower bound.

Corollary 7

There exist ${c_B>0}$, $k_0\in {\mathbb {N}}$, and $\theta _0\in (0,1/2]$ such that for all $\theta \in (0,\theta _0]$, all $\sigma >0$ and all $u\in {\mathcal {A}}_\theta$,

$$\begin{aligned} { E_{\sigma ,\theta }(u)} \ge {c_B}{\left\{ \begin{array}{ll}\min \{ \theta ^2,\sigma \} &{}\text { if } \sigma \ge \theta ^{k_0}, \\ k \sigma &{}\text { if } \sigma \in [\theta ^{k+1},\theta ^{k}) \text { {for some} } k_0 \le k. \end{array}\right. } \end{aligned}$$

4 Generalizations of the Scaling Result

4.1 General p

We consider for $1\le p<\infty$ the energy $E_{\sigma ,\theta }: {\mathcal {A}}_{\theta } \rightarrow [0,\infty )$

$$\begin{aligned} E_{\sigma ,\theta }^p(u) = \int _{(0,1)^2} {\text {dist}}^p(\nabla u,K) \, dx + \sigma \mid D^2 u\mid (\Omega ). \end{aligned}$$

Corollary 8

Let $p\in (1,\infty )$. There exists a constant $C>0$ such that for all $\sigma >0$ and all $\theta \in (0,\frac{1}{2}]$,

$$\begin{aligned} \frac{1}{C}\min \left\{ \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } +1\right) , \theta ^p\right\} \le \min _{u\in {\mathcal {A}}_\theta } E^p_{\sigma ,\theta }(u) \le C \min \left\{ \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } +1\right) , \theta ^p\right\} . \end{aligned}$$

Proof

Fix $p \in (1,\infty )$. For an upper bound one can use the constructions for $p=2$ from Proposition 2. Clearly, the function $u(x,y) = (1-2\theta )y \pm x$ produces an energy of order $\theta ^p$. On the other hand, it can be seen from the proof of Proposition 2 that the function constructed via branching $u_N$ satisfies $\nabla u_N \in K$ except for an interpolation region of size $\theta ^N$ on which it holds $\mid \nabla u_N\mid \le 5$. Hence, one obtains again

$$\begin{aligned} \int _{(0,1)^2} {\text {dist}}(\nabla u_N,K)^p \, dx dy \le C \theta ^N. \end{aligned}$$

and $\mid D^2u_N\mid ((0,1)^2) \le CN$. Setting $N=\lceil \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } \rceil$ yields as in the proof of Proposition 2, $E_{\sigma ,\theta }^p(u_N) \le C \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } +1\right)$.

The lower bounds can be shown analogously to the case $p=2$. $\square$

4.2 Classical Double-Well Potential

We define the function $W: {\mathbb {R}}^2 \rightarrow {\mathbb {R}}$, $W(x,y) = (1-x^2)^2 + (1-y^2)^2$, and the energy $F_{\sigma ,\theta }: {\mathcal {A}}_{\theta } \rightarrow [0,\infty )$,

$$\begin{aligned} F_{\sigma ,\theta }(u) = \int _{(0,1)^2} W(\partial _1 u,\partial _2 u) \, dx + \sigma \mid D^2 u\mid (\Omega ). \end{aligned}$$

The following corollary shows that the scaling law for $\min E_{\sigma ,\theta }$ and $\min F_{\sigma ,\theta }$ are the same.

Corollary 9

There exists a constant $C>0$ such that for all $\sigma >0$ and all $\theta \in (0,\frac{1}{2}]$ it holds

$$\begin{aligned} \frac{1}{C}\min \left\{ \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } +1\right) , \theta ^2\right\} \le \min _{u\in {\mathcal {A}}_\theta } F_{\sigma ,\theta }(u) \le C \min \left\{ \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid } +1\right) , \theta ^2\right\} . \end{aligned}$$

Proof

First note that

$$\begin{aligned} W(x,y) = (1-x)^2(1+x)^2 + (1-y)^2(1+y)^2 \ge {\text {dist}}((x,y),K)^2. \end{aligned}$$

Consequently, $\inf E_{\sigma ,\theta }\le \inf F_{\sigma ,\theta }$ and the lower bound follows from Theorem 1.

Again, it can be easily checked that the competitors from the proof of Proposition 2 produce a corresponding upper bound for ${\min F_{\sigma ,\theta }}$. $\square$

Fig. 6

Left: Sketch of the function $u_N$ as constructed in Proposition 2 for $\theta =\frac{1}{2}$ and $N=4$. The regions of constant gradients are color-coded, the ruled region indicates the necessary interpolation to meet the boundary values. For larger N the gradient of $u_N$ is only changed in the interpolation region. Right: Sketch of the function $u = \lim _{N\rightarrow \infty } u_N$ as discussed in the proof of Corollary 10

Fig. 7

Sketch of a competitor such that $u=0$ on $\partial (0,1)^2$. The construction is obtained by subdividing the square $(0,1)^2$ through diagonals into four triangles. On each of those triangles, a version of the construction sketched in Fig. 6 is used

4.3 Boundary Conditions on Whole Boundary

In this section we show that in the symmetric case $\theta = \frac{1}{2}$, we can replace the boundary condition $u(0,\cdot ) = 0$ by the more restrictive boundary condition $u = 0$ on $\partial (0,1)^2$.

Corollary 10

Let $\sigma > 0$. Then there exists $u \in W^{1,\infty }((0,1)^2)$ such that $u = 0$ on $\partial (0,1)^2$ such that

$$\begin{aligned} E_{\sigma ,1/2} \le C \min \{1, \sigma (\mid \log \sigma \mid + 1)\}. \end{aligned}$$

Proof

Clearly $u=0$ meets the more restrictive boundary conditions and satisfies $E_{\sigma ,1/2}(u) = 2$.

In the branching regime, $\sigma < 1/4$, we recall that in the proof of Proposition 2 we constructed functions $u_N$ such that $u_N(0,\cdot ) = 0$ and $E_{\sigma ,1/2}(u_N) \le C (\sigma N + 2^{-N})$, see Fig. 6. It is easy to see that the limit $u = \lim u_N$ exists in $L^1$, c.f. the proof of Proposition 12. The function u belongs to $W^{1,\infty }((0,1)^2)$ and satisfies $\nabla u \in K$ almost everywhere and $u(0,\cdot ) = 0$. Moreover, $u(x,x) = u(x,1-x) = 0$ for all $0 \le x \le \frac{1}{2}$. In addition, $\nabla u \in BV_{loc}((0,1)^2)$ with $\mid D \nabla u\mid ((2^{-N},1-2^{-N})^2) \le CN$. Then define ${\tilde{u}}: (0,1)^2 \rightarrow {\mathbb {R}}$ in the following way: For $(x,y) \in (2^{-N},1-2^{-N})^2$ we set

$$\begin{aligned} {\tilde{u}}(x,y) = {\left\{ \begin{array}{ll} u(x,y) &{}\text { if } x \le y \le 1-x, \\ u(1-x,y) &{}\text { if } 1-x \le y \le x, \\ u(y,x) &{}\text { if } y \le x \le 1-y, \\ u(1-y,1-x) &{}\text { if } 1-y \le x \le y. \end{array}\right. } \end{aligned}$$

One checks that $\mid {\tilde{u}}(x,y)\mid \le 2^{-N}$ for all $(x,y) \in {\partial (2^{-N},1-2^{-N})^2}$. Then one interpolates on $(0,1)^2 \setminus {(2^{-N},1-2^{-N})^2}$ so that ${\tilde{u}} = 0$ on $\partial (0,1)^2$. See Fig. 7 for an illustration of ${\tilde{u}}$. The energy estimates on the interpolation layer are analogous to the computations in the proof of Proposition 2. Choosing as in Proposition 2$N \approx \mid \log \sigma \mid$ leads to $E_{\sigma ,1/2}({\tilde{u}}) \le C \sigma (\mid \log \sigma \mid +1)$. $\square$

4.4 Scaling Law on Rectangles

Proposition 11

There is a constant $C>0$ with the following property: For $L>0$ consider the rectangle $\Omega _L:=(0,L)\times (0,1)$ and set

$$\begin{aligned} {\mathcal {A}}_\theta ^{(L)}:=\left\{ u\in W^{1,2}({\Omega _L}):\ \nabla u\in BV({\Omega _L}),\ u(0,y)=(1-2\theta )y\right\} . \end{aligned}$$

We define

$$\begin{aligned} s(L,\theta ,\sigma ):={\left\{ \begin{array}{ll} \min \left\{ L\theta ^2,\ \sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid }+1\right) \right\} &{}\quad \text { if }L\ge \theta \\ \min \left\{ L\theta ^2,\ \sigma \left( 1+\frac{\log (L/\sigma )}{\mid \log \theta \mid }\right) \right\} &{}\quad \text { if }\sigma \le L<\theta \\ L\theta ^2&{}\quad \text { if }L\le \min \{\theta ,\ \sigma \}. \end{array}\right. } \end{aligned}$$

Then for all $\sigma$, $L>0$ and $\theta \in (0,1/2)$,

$$\begin{aligned}&\frac{1}{C}s(L,\theta ,\sigma ) \le \min _{u\in {\mathcal {A}}_\theta ^{(L)}}E_{\sigma ,\theta }(u;\Omega _L)\le Cs(L,\theta ,\sigma ), \end{aligned}$$

where we use the notation $E_{\sigma ,\theta }(u;\Omega _L)$ as defined in (5).

Proof

Upper Bound. The affine function $u(x,y)=(1-2\theta )y$ satisfies $E_{\sigma ,\theta }(u;((0,L)\times (0,1))\le L\theta ^2$. This in particular concludes the proof of the upper bound if $L\le \min \{\theta ,\sigma \}$.

For the other two regimes, we use the test function ${u_N}:(0,1)^2\rightarrow {\mathbb {R}}$ constructed in the proof of Proposition 2 and note that

$$\begin{aligned}{u_N(\theta ,y)={\left\{ \begin{array}{ll} \alpha +y&{}\quad \text { if }y\in (0,1-\theta )\\ \alpha + (2-2\theta )-y&{}\quad \text { if }y\in (1-\theta , 1) \end{array}\right. }} \end{aligned}$$

for some value $\alpha \in {\mathbb {R}}$. We now define the auxiliary function $u:(0,\infty )\times (0,1)\rightarrow {\mathbb {R}}$ via

$$\begin{aligned}u(x,y):={\left\{ \begin{array}{ll} u_N(x,y)&{}\text { if }x\in (0,\theta ),\ y\in (0,1)\\ \alpha +y-(x-\theta )&{}\text { if }x\in (\theta ,\infty ),\ y\in (0,\min \{1,\ x+1-2\theta \})\\ \alpha +(2-2\theta )-y+(x-\theta )&{}\text { if }x\in (\theta ,2\theta ),\ y\in (x+1-2\theta ,1). \end{array}\right. } \end{aligned}$$

Note that this construction resembles the truncated branching construction used for martensitic microstructures (Conti and Zwicknagl 2016; Conti et al. 2020; Zwicknagl 2014). We claim that restrictions of the so-defined function u yield the respective energy scalings. We consider the cases from the definition of $s(L,\theta ,\sigma )$ separately.

• (i) If $L\ge \theta$ then by Proposition  2 we have

  $$\begin{aligned} E_{\sigma ,\theta }(u;\Omega _L)\le E_{\sigma ,\theta }(u_N;\Omega _\theta )+\sigma +\sigma \theta \sqrt{2} \le c\sigma \left( \frac{\mid \log \sigma \mid }{\mid \log \theta \mid }+1\right) . \end{aligned}$$

• (ii) If $\sigma \le L<\theta$ then there exists $m\in {\mathbb {N}}$ such that $\delta ^{m+1}\le L/\theta <\delta ^m$, where $\delta = \frac{1}{\lceil \theta ^{-1} \rceil } \le \theta$ as in the proof of Proposition 2. We note that $m<\frac{\mid \log L/\theta \mid }{\mid \log \delta \mid } \le \frac{\mid \log \sigma /\theta \mid }{\mid \log \delta \mid }\le N$ with $N\in {\mathbb {N}}$ as defined in the proof of Proposition 2, see (14).We estimate using the computations from the proof of Proposition 2

  $$\begin{aligned}&E_{\sigma ,\theta }(u;(0,L)\times (0,1)) \\&\quad \le E_{\sigma ,\theta }(u_N;(0,\theta \delta ^m)\times (0,1))\le c\left( \theta \delta ^N+\sigma (N-m)\right) \\&\quad \le c\sigma \left( 1+\frac{\mid \log \sigma / \theta \mid }{\mid \log \delta \mid }-\frac{\mid \log L/\theta \mid }{\mid \log \delta \mid }\right) \le c\sigma \left( 1+\frac{\log (L/\sigma )}{\mid \log \theta \mid }\right) . \end{aligned}$$

Lower bound. Step 1: $L\ge 1$. In this case, the proof follows from the lower bound for $E_{\sigma ,\theta }$ on $(0,1)^2$. We will from now on focus on the case $L<1$.

Step 2: $\sigma \ge \theta _0^{k_0}$. An argument along the lines of the proof of Lemma 5 shows that for all admissible functions u,

$$\begin{aligned} E_{\sigma ,\theta }(u,\Omega _L)\ge c\min \{L\theta ^2,\sigma \}. \end{aligned}$$

This in particular concludes the proof of the lower bound in the case $L<\min \{\theta ,\sigma \}$.

Step 3: $L\in [\theta ,1)$. In this case, the proof of the lower bound follows as for $L=1$. More precisely, for large $\theta \ge \theta _0$ both, $\theta$ and L are of order one, and we can proceed similarly to the proof of Lemma 4 to obtain a lower bound $E_{\sigma ,\theta }(u;\Omega _L)\ge c\min \{\sigma (\mid \log \sigma \mid +1),1\}$. For $\sigma \ge \theta ^{k}$ for some $k\ge 33$, we can directly use Lemma 6 since this proof uses only the energy on a domain that is contained in $(0,\theta )\times (0,1)\subseteq \Omega _L$.

Step 4: $L\in [\sigma ,\theta )$. First note that by the argument in step 2 it always holds $E_{\sigma ,\theta }(u,\Omega _L) \ge c \min \{L\theta ^2,\sigma \}$. This shows the desired lower bound as long as $\frac{\log (L/\sigma )}{\mid \log \theta \mid } \le 33$.

If $\sigma \in [\theta ^{k+1},\theta ^k)$ for some k and $L\in [\theta ^{m+1},\theta ^m)$ for some $m\in {\mathbb {N}}$ such that $k-m \ge 32$ define $K:= \left\lfloor \frac{k-m}{8} \right\rfloor$. Then one can find for $i=1,\dots ,K$ points $x_i \in (\frac{1}{2} L \theta ^{2i},\frac{3}{2} L \theta ^{2i})$ such that $u(x_i,\cdot ) \in H^1(0,1)$, $\partial _2 u(x_i,\cdot ) \in BV(0,1)$ and

$$\begin{aligned}&\int _0^1 \min \{ \mid \partial _2 u(x_i,y)-1\mid ,\mid \partial _2u(x_i,y)+1\mid \}^2 \, dy + \sigma \mid \partial _2 \partial _2 u(x_i,\cdot )\mid (0,1) \\&\quad \le L^{-1} \theta ^{-2i} E_{\sigma ,\theta }\left( u,(\frac{1}{2} L \theta ^{2i}, \frac{3}{2} L \theta ^{2i} \times (0,1) \right) . \end{aligned}$$

With this notation a lower bound can then be proven with minor modifications along the lines of part (B) in the proof of Proposition 3. $\square$

5 Regularity of Solutions to the Differential Inclusion

Refining the construction in the proof of Proposition 2, yields a solution to the differential inclusion problem derived in Cicalese et al. (2019) subject to boundary conditions. While our scaling shows that the resulting gradient is not in BV (cf. also Rüland et al. 2019), we can in the spirit of Rüland et al. (2018), Rüland et al. (2019), Rüland et al. (2020) exploit regularity properties of it.

Proposition 12

There exists a function $u\in W^{1,\infty }((0,1)^2)$ such that $u(0,y)=(1-2\theta )y$, $\nabla u\in K$ almost everywhere, and $\nabla u\in W^{s,q}$ for all $0<s<1$ and $q \in (1,\infty )$ such that $\frac{1}{q} > s$. Moreover, $\nabla u \in BV_{loc}((0,1)^2)$ and consequently ${\text {dim}}_{{\mathcal {H}}}(J_u) = 1$.

Proof

Fix $s \in (0,1)$ and $q \in (1,\infty )$ such that $\frac{1}{q} > s$. Then there exists $p \in (1,\infty )$ such that $\frac{1}{q} = \frac{1-s}{p} + s$.

Now, recall the branching construction of the function $u_N: (0,1)^2 \rightarrow {\mathbb {R}}$ in the proof of Proposition 2, see Fig. 6. In particular, we note that for all $N\in {\mathbb {N}}$ we have

• (i) $u_N(0,y) = (1-2\theta ) y$ in $L^1(0,1)$,

• (ii) $\nabla u_N \in K$ for almost all $x \in (\theta ^N,1) \times (0,1)$,

• (iii) if $M > N$ then $\nabla u_M = \nabla u_N$ for almost every $x \in (\theta ^N,1) \times (0,1)$,

• (iv) $\Vert u_{N+1} - u_N \Vert _{L^1} \le C\theta ^N$,

• (v) $\Vert \nabla u_N\Vert _{L^{\infty }} \le C$,

• (vi) $\Vert \nabla u_N \Vert _{BV} \le CN$,

• (vii) $\Vert \nabla u_N - \nabla u_{N+1} \Vert _{L^p} \le C \theta ^N$,

• (viii) ${\Vert \nabla u_N - \nabla u_{N+1} \Vert _{BV}} \le C$.

First note for $M>N$ that by (iv)

$$\begin{aligned} \Vert u_{M} - u_N \Vert _{L^1} \le \sum _{k=N}^{M-1} \Vert u_{k+1} - u_k \Vert _{L^1}\le C\sum _{k=N}^{\infty } \theta ^N. \end{aligned}$$

Hence, $(u_N)_N$ forms a Cauchy sequence in $L^1$. Similarly, one shows using (vii) that $(\nabla u_N)_N$ is Cauchy in $L^p$. Consequently, $(u_N)_N$ is a Cauchy sequence in $W^{1,1}$ and converges strongly in $W^{1,1}$ to some $u \in W^{1,1}((0,1)^2)$. As the trace is continuous with respect to strong convergence in $W^{1,1}$ we obtain from (i) that $u(0,y) = (1-2\theta ) y$ in $L^1(0,1)$. Moreover, it follows from (ii) and (iii) that $\nabla u \in K$. In particular, $u \in W^{1,\infty }$.

Next, we apply an interpolation inequality between $L^p$ and BV, see (Rüland et al. 2018, Corollary 2.1) to $\nabla u_{N+1} - \nabla u_N$ which yields using (v), (vii) and (viii)

$$\begin{aligned} \Vert \nabla u_{N+1} - \nabla u_N \Vert _{W^{s,q}} \le C \Vert \nabla u_{N+1} - \nabla u_N \Vert _{L^p}^{1-s} \, \Vert \nabla u_{N+1} - \nabla u_N \Vert _{BV}^s \le C \theta ^{(1-s)N}. \end{aligned}$$

Hence, we obtain for $M > N$ that

$$\begin{aligned} \Vert \nabla u_{M} - \nabla u_N \Vert _{W^{s,q}}&\le \sum _{k=N}^{M-1} \Vert \nabla u_{k+1} - \nabla u_k \Vert _{W^{s,q}} \le C \sum _{k=N}^{\infty } \left( \theta ^{1-s}\right) ^k \end{aligned}$$

In particular, $(\nabla u_N)_N$ is a Cauchy sequence in $W^{s,q}$. Its limit is already identified to be $\nabla u$. Consequently, $\nabla u \in W^{s,q}$. Eventually, we remark that (iii) and (vi) imply that $(\nabla u_N)_N$ is bounded in $BV_{loc}((0,1)^2)$. By BV-compactness, it follows $\nabla u \in BV_{loc}((0,1)^2)$. It remains to show that ${\text {dim}}_{{\mathcal {H}}}(J_{\nabla u}) =1$. Since $\nabla u \in BV_{loc}((0,1)^2)$, we have for $s>1$ that ${\mathcal {H}}^s(J_{{\nabla u}}) \le \sum _k {\mathcal {H}}^s(J_{{\nabla u}} \cap (1/k,1-1/k)^2) = 0$. On the other hand, it follows from the energy scaling result Theorem 1 for that ${\mathcal {H}}^1(J_{{\nabla u}})=+\infty$ which implies ${\text {dim}}_{{\mathcal {H}}}(J_{\nabla u}) \ge 1$. $\square$