Michell trusses in two dimensions as a $\Gamma$-limit of optimal design problems in linear elasticity

Abstract

We reconsider the minimization of the compliance of a two dimensional elastic body with traction boundary conditions for a given weight. It is well known how to rewrite this optimal design problem as a nonlinear variational problem. We take the limit of vanishing weight by sending a suitable Lagrange multiplier to infinity in the variational formulation. We show that the limit, in the sense of $\Gamma$-convergence, is a certain Michell truss problem. This proves a conjecture by Kohn and Allaire.

Appendices

Appendix A: Proof of Theorem 3.7

In analogy to the proof of the relaxation leading to 1-quasiconvex integrands, we first need to prove an approximation lemma. This is slightly more complicated here than in the case of 1-quasiconvexity, since we cannot use the approximation by finite elements here, which is possible there (see [10]). Instead, we are going to need Whitney’s extension theorem, that we quote here in a version that can be found in Stein’s book [33]. Let $\Omega \subset \mathbb {R}^n$. Let the Greek letters $\alpha ,\beta ,\gamma \in \mathbb {N}_0^n$ denote multiindices. We will write $|\alpha |=\sum _i\alpha _i$, $\alpha !=\prod _i\alpha _i!$, and $\nabla ^\alpha =\partial _1^{\alpha _1}\dots \partial _n^{\alpha _n}$. Furthermore, for $x\in \mathbb {R}^n$, we write $x^\alpha =x_1^{\alpha _1}\dots x_n^{\alpha _n}$. We shall say that a function $f:\Omega \rightarrow \mathbb {R}$ belongs to $\mathrm {Lip}^{(k,1)}(\Omega )$ if there exists a collection of real-valued functions $\{f^{(\alpha )}:|\alpha |\le k\}$ and a constant $M>0$ such that

$$\begin{aligned} f^{(\alpha )}(y)=\sum _{\beta :|\alpha +\beta |\le k}\frac{f^{(\alpha +\beta )}(x)}{\beta !} (x-y)^{\beta } +R_\alpha (x,y) \end{aligned}$$

and

$$\begin{aligned} |f^{(\alpha )}(y)|\le M \text { and } |R_\alpha (x,y)|\le M|x-y|^{k+1-|\alpha |} \end{aligned}$$

for all $x,y\in \Omega$ and all multiindices $\alpha$ with $|\alpha |\le k$. The set $\mathrm {Lip}^{(k,1)}(\Omega )$ is a normed space, where the norm of f is given by the smallest constant M such that the above relations hold true.

Theorem A.1

(Theorem 4 in Chapter 6 of [33]) Let k be a non-negative integer and let $\Omega \subset \mathbb {R}^n$ be closed. Then there exists a continuous extension operator $\mathrm {Lip}^{(k,1)}(\Omega )\rightarrow \mathrm {Lip}^{(k,1)}(\mathbb {R}^n)$. The norm of this mapping has a bound that is independent from $\Omega$.

For a closed set $\Gamma \subset \mathbb {R}^n$, let $d_{\Gamma }\in C^\infty (\mathbb {R}^n;[0,\infty ))$ denote a regularized distance function, that is, a function with the property that there exists a constant $C>0$ such that

$$\begin{aligned} \begin{aligned} C^{-1}{\text {dist}}(x,\Gamma )&\le d_{\Gamma }(x)\le C{\text {dist}}(x,\Gamma )\quad \text { for all }x\in \mathbb {R}^N.\\ |\nabla ^\alpha d_{\Gamma }(x)|&\le C |d_{\Gamma }(x)|^{1-|\alpha |}\quad \text { for all multiindices } \alpha . \end{aligned} \end{aligned}$$

Such a regularized distance function exists by Theorem 2 of Chapter 6 in [33]. We will use it to construct suitable cutoff functions in Lemma A.2 below. This lemma is a preparation for the approximation lemma, Lemma A.3 below.

Let $\phi \in C^\infty ([0,\infty ))$ such that $\phi (t)=0$ for $t\le \frac{1}{2}$ and $\phi (t)=1$ for $t\ge 1$.

Lemma A.2

Let $p\ge 1$, $U\subset [0,1]^2$, $(u_\varepsilon )_{\varepsilon >0}$ a sequence in $W^{2,p}(U)$ that converges strongly to $u\in W^{2,p}(U)$ for $\varepsilon \rightarrow 0$, and let

$$\begin{aligned} \Gamma \subset \left( [0,1]\times \{0\}\cup \{0\}\times [0,1]\right) \cap \partial U \end{aligned}$$

be a closed set of positive $\mathcal {H}^1$ measure, and $u_0\in W^{2,p}(U)$ such that $u=u_0$ and $\nabla u=\nabla u_0$ on $\Gamma$. Furthermore let

$$\begin{aligned} \tilde{u}_\varepsilon (x)=\phi \left( \frac{d_{\Gamma }(x)}{\varepsilon }\right) u_\varepsilon (x)+\left( 1-\phi \left( \frac{d_{\Gamma }(x)}{\varepsilon }\right) \right) u_0(x). \end{aligned}$$

Then $\tilde{u}_\varepsilon =u_0$ and $\nabla \tilde{u}_\varepsilon =\nabla u_0$ on $\Gamma$, and

$$\begin{aligned} \begin{aligned} \Vert \tilde{u}_\varepsilon - u\Vert _{W^{2,p}(U)}&\rightarrow 0\\ \int _{U\cap \{x:d_\Gamma (x)\le \varepsilon \}}\left( 1+|\nabla ^2 u_\varepsilon |^p+|\nabla u|^p\right) \text {d}x&\rightarrow 0. \end{aligned} \end{aligned}$$

Proof

The first claim is obvious. To show the second claim, it suffices to show that $\Vert \nabla ^2(\tilde{u}_\varepsilon -u)\Vert _{L^p}\rightarrow 0$, since $\tilde{u}_\varepsilon -u$ and its gradient vanish on $\Gamma$. This is a straightforward computation, that estimates the integral over $|\nabla ^2(\tilde{u}_\varepsilon -u)|^p$ in the “bulk” $U\cap \{x:d_\Gamma (x)>\varepsilon \}$ and in boundary layer

$$\begin{aligned} U\cap \left\{ x:d_\Gamma (x)\le \varepsilon \right\} \subset (U\cap \{x:x_1<C\varepsilon \})\cup (U\cap \{x:x_2<C\varepsilon \}), \end{aligned}$$

where the constant $C>0$ is chosen appropriately. The contribution of the bulk vanishes in the limit $\varepsilon \rightarrow 0$ due to the assumption. Writing $\tilde{u}_\varepsilon -u$ and its gradient as integrals in $x_i$-direction in the set $U\cap \{x:x_i<C\varepsilon \}$ for $i=1,2$ and using Fubini’s theorem, the proof that the contribution of the boundary layer vanishes in the limit is a straightforward but lengthy computation that we omit here for the sake of brevity. The third claim is an immediate consequence of the $L^p$ integrability of $\nabla ^2 u$ and the strong convergence $\nabla ^2u_\varepsilon \rightarrow \nabla ^2 u$ in $L^p$. $\square$

Lemma A.3

Let $\Omega \subset \mathbb {R}^2$ satisfy Definition 2.1, and let $p\in [1,\infty )$. Furthermore let $u\in \{u_0\}+W^{2,p}_0(\Omega )$ and $\delta >0$. Then there exists $w\in \{u_0\}+W^{2,p}_0(\Omega )$ and $\Omega _w\subset \Omega$ such that $\Omega _w$ is the union of mutually disjoint closed cubes, w is piecewise a polynomial of degree k on $\Omega _w$, and furthermore

$$\begin{aligned} \begin{aligned} \Vert u-w\Vert _{W^{2,p}(\Omega )}&<\delta ,\\ \int _{\Omega \setminus \Omega _w}(1+|\nabla ^2u|^p+|\nabla ^2w|^p)\text {d}x&<\delta . \end{aligned} \end{aligned}$$

Proof

Let $v:=u-u_0\in W^{2,p}_0(\Omega )$. We may extend $u_0$ to $\mathbb {R}^2$ such that

$$\begin{aligned} \begin{aligned} \Vert u_0\Vert _{W^{2,p}(\mathbb {R}^2)}&\lesssim \Vert u_0\Vert _{W^{2,p}(\Omega )}. \end{aligned} \end{aligned}$$

Also, v may be understood as an element of $W^{2,p}(\mathbb {R}^2)$ by a trivial extension. Let $\{\Omega _i:i=1,\ldots ,P\}$ be an open cover of $\overline{\Omega }$, let $\{\psi _i:i=1,\ldots ,P\}$ be a subordinate partition of unity, and let $\xi _i:\Omega _i\rightarrow \xi _i(\Omega _i)=:\tilde{\Omega }_i$, $i=1,\ldots ,P$, be a set of $C^2$ diffeomorphisms such that $\xi _i(\Omega \cap \Omega _i)\subset (0,1)^2$ and $\Gamma _i:=\xi _i(\Omega _i\cap \partial \Omega )\subset [0,1]\times \{0\}\cup \{0\}\times [0,1]$. (Such $\xi _i$ exist by the assumption on $\Omega$.)

Let $\varphi \in C_c^\infty (\mathbb {R}^n)$ with $\int \varphi \text {d}x =1$, and $\varphi _\varepsilon =\varepsilon ^{-n}\varphi (\cdot /\varepsilon )$. For every i, we have that

$$\begin{aligned} \Vert (u- (\varphi _\varepsilon * u))\circ \xi _i^{-1}\Vert _{W^{2,p}(\tilde{\Omega }_i)}\rightarrow 0. \end{aligned}$$

By the previous lemma, there exists $\hat{u}^{(i)}_\varepsilon \in {W^{2,p}(\tilde{\Omega }_i)}$ such that $\hat{u}^{(i)}_\varepsilon =(\varphi _\varepsilon * u)\circ \xi _i^{-1}$ on $\tilde{\Omega }_i\cap \{x:d_{\Gamma _i(x)}\ge \varepsilon \}$,

$$\begin{aligned} \hat{u}^{(i)}_\varepsilon = u\circ \xi _i^{-1}, \quad \nabla \hat{u}^{(i)}_\varepsilon = \nabla u\circ \xi _i^{-1} \quad \text { on } [0,1]\times \{0\}\cap \tilde{\Omega }_i, \end{aligned}$$

and additionally,

$$\begin{aligned} \begin{aligned} \Vert \hat{u}^{(i)}_\varepsilon - u\circ \xi _i^{-1}\Vert _{W^{2,p}(U)}&\rightarrow 0\\ \int _{\tilde{\Omega }_i\cap \{x:d_{\Gamma _i(x)}<\varepsilon \}}1+|\nabla ^2 \hat{u}_\varepsilon ^{(i)}|^p+|\nabla u\circ \xi _i^{-1}|^p\text {d}x&\rightarrow 0. \end{aligned} \end{aligned}$$

Setting $u_\varepsilon ^{(i)}:=\hat{u}^{(i)}_\varepsilon \circ \xi _i$, we obviously have that $u_\varepsilon ^{(i)}=\varphi _\varepsilon \circ u$ on $\Omega _i\setminus \xi _i^{-1}(\tilde{\Omega }_i\cap \{x:d_{\Gamma _i(x)}< \varepsilon \})$, and

$$\begin{aligned} \begin{aligned} u^{(i)}_\varepsilon = u, \quad \nabla u^{(i)}_\varepsilon&= \nabla u \quad \text { on } \Omega _i\cap \partial \Omega ,\\ \Vert u^{(i)}_\varepsilon - u\Vert _{W^{2,p}(\Omega _i)}&\rightarrow 0\\ \int _{ \xi _i^{-1}(\tilde{\Omega }_i\cap \{x:d_{\Gamma _i(x)}<\varepsilon \})}1+|\nabla ^2 u_\varepsilon ^{(i)}|^p+|\nabla u|^p\text {d}x&\rightarrow 0. \end{aligned} \end{aligned}$$

Setting

$$\begin{aligned} \begin{aligned} \Omega _\varepsilon&:=\Omega \setminus \bigcup _{i=1}^N \xi _i^{-1}(\tilde{\Omega }_i\cap \{x:d_{\Gamma _i(x)}<\varepsilon \})\\ u_\varepsilon&:=\sum _i \psi _i u_\varepsilon ^{(i)}, \end{aligned} \end{aligned}$$

we have $u_\varepsilon =\varphi _\varepsilon * u$ on $\Omega _\varepsilon$ and

$$\begin{aligned} \begin{aligned} u_\varepsilon = u, \quad \nabla u_\varepsilon&= \nabla u \quad \text { on } \partial \Omega ,\\ \Vert u_\varepsilon - u\Vert _{W^{2,p}(\Omega )}&\rightarrow 0\\ \int _{\Omega \setminus \Omega _\varepsilon }1+|\nabla ^2 u_\varepsilon |^p+|\nabla u|^p\text {d}x&\rightarrow 0. \end{aligned} \end{aligned}$$

and hence we may fix $\varepsilon$ such that

$$\begin{aligned} \begin{aligned} \Vert u_\varepsilon -u\Vert _{W^{k,p}(\Omega )}&<\delta /2\\ \int _{\Omega \setminus \Omega _\varepsilon }(1+ |\nabla ^k u|^p+|\nabla ^k u_\varepsilon |^p)\text {d}x&<\delta /2. \end{aligned} \end{aligned}$$

Note that $\Omega _\varepsilon$ is closed and $u_\varepsilon \in C^\infty (\Omega _\varepsilon )$.

Let $0<h\ll 1$ to be chosen later, and let $Q_i$, $i=1,\ldots ,N$ be mutually disjoint closed cubes of sidelength h contained in $\Omega _\varepsilon$ such that

$$\begin{aligned} \begin{aligned} {\text {dist}}(Q_i,Q_j)&\ge h^{4/3} \quad \text { for all }i,j=1,\ldots ,N, i\ne j\\ {\text {dist}}(\partial \Omega _\varepsilon ,Q_i)&\ge h^{4/3} \quad \text { for all }i=1,\ldots ,N\\ {\mathcal L}^n\left( \Omega _\varepsilon \setminus \bigcup _i Q_i\right)&\le C h^{1/3}. \end{aligned} \end{aligned}$$

This is always possible assuming that h is small enough. Let $x_i$ denote the midpoint of $Q_i$. Then we define $\tilde{u}_\varepsilon$ to be the Taylor polynomial of degree two at $x_i$ on each $Q_i$,

$$\begin{aligned} \tilde{u}_\varepsilon (y)= \sum _{|\alpha |\le 2} \nabla ^\alpha u_\varepsilon (x_i) \frac{(y-x_i)^\alpha }{\alpha !}\quad \text { for }y\in Q_i. \end{aligned}$$

Let $V=(\cup _i Q_i)\cup \partial \Omega _\varepsilon$. We claim that there exists an extension of $\tilde{u}_\varepsilon$ from V to $\Omega _\varepsilon$ such that $\Vert \tilde{u}_\varepsilon \Vert _{W^{2,\infty }(\Omega _\varepsilon )}\lesssim \Vert u_\varepsilon \Vert _{W^{2,\infty }({\Omega _\varepsilon })}$. In order to prove our claim, we invoke Theorem A.1 with $k=1$. We verify that $\tilde{u}_\varepsilon \in \mathrm {Lip}^{(1,1)}(V)$: Firstly, we have for all $y\in V$ and all multiindices $\alpha$ with $|\alpha |\le 1$,

$$\begin{aligned} \begin{aligned} |\nabla ^\alpha \tilde{u}_\varepsilon (y)-\nabla ^\alpha u_\varepsilon (y)|\le C h^{2-|\alpha |}, \end{aligned} \end{aligned}$$

(50)

where the constant C depends on $\Vert u_\varepsilon \Vert _{C^{3}(\Omega _\varepsilon )}$. Furthermore, we have $u_\varepsilon \in \mathrm {Lip}^{(1,1)}(\Omega _\varepsilon )=C^{1,1}(\Omega _\varepsilon )$, namely there exists a constant $0<M_1\lesssim \Vert u_\varepsilon \Vert _{C^{1,1}(\Omega _\varepsilon )}$ such that for all $x,y\in {\Omega _\varepsilon }$ and all multiindices $\alpha$ with $|\alpha |\le 1$ we have

$$\begin{aligned} \begin{aligned} \nabla ^\alpha u_\varepsilon (y)&=\sum _{|\alpha +\beta |\le 1}\nabla ^{\alpha +\beta } u_\varepsilon (x)\frac{(y-x)^\beta }{\beta !}+R_\alpha (x,y) \end{aligned} \end{aligned}$$

(51)

with

$$\begin{aligned} |\nabla ^\alpha u_\varepsilon (x)|<M_1,\quad |R_\alpha (x,y)|<M_1|x-y|^{2-|\alpha |}. \end{aligned}$$

Now let $x,y\in Q_i$. Then we have for all multiindices $\alpha$ with $|\alpha |\le 1$,

$$\begin{aligned} \begin{aligned} \nabla ^\alpha \tilde{u}_\varepsilon (y)&=\sum _{|\alpha +\beta |\le 2}\nabla ^\alpha u_\varepsilon (x)\frac{(y-x)^{\beta }}{\beta !}. \end{aligned} \end{aligned}$$

(52)

Next let $x\in Q_i,y\in Q_j$ with $i\ne j$, or $x\in \partial \Omega$, $y\in Q_i$. In this case, we have $|x-y|\ge h^{4/3}$. By inserting (50) in (51), we obtain

$$\begin{aligned} \begin{aligned} \nabla ^\alpha \tilde{u}_\varepsilon (y)&= \sum _{|\alpha +\beta |\le 1}\nabla ^{\alpha +\beta }\tilde{u}_\varepsilon (x)\frac{(y-x)^{\beta }}{\beta !}+O\left( h^{3-|\alpha |}\right) + R_\alpha (x,y)\\&= \sum _{|\alpha +\beta |\le 1}\nabla ^{\alpha +\beta }\tilde{u}_\varepsilon (x)\frac{(y-x)^{\beta }}{\beta !}+ \tilde{R}_\alpha (x,y), \end{aligned} \end{aligned}$$

(53)

where we have introduced $\tilde{R}_\alpha =R_\alpha (x,y)+O(h^{3-|\alpha |})$, which by $|x-y|\ge h^{4/3}$ implies

$$\begin{aligned} |\tilde{R}_\alpha (x,y)-R_\alpha (x,y)|=O(h^{3-|\alpha |})= o(1)|x-y|^{2-|\alpha |}, \end{aligned}$$

where the last estimate holds since for all multiindices $\alpha$ with $|\alpha |\le 1$, we have

$$\begin{aligned} \frac{4}{3}<\frac{3-|\alpha |}{2-|\alpha |}. \end{aligned}$$

Summarizing (52) and (53), we have proved that $\tilde{u}_\varepsilon \in \mathrm {Lip}^{(1,1)}(V)$, with

$$\begin{aligned} |\nabla ^\alpha \tilde{u}_\varepsilon (x)|<M_2,\quad |\tilde{R}_\alpha (x,y)|<M_2|x-y|^{2-|\alpha |}, \end{aligned}$$

where

$$\begin{aligned} M_2\le M_1+o(1)\lesssim \Vert u_\varepsilon \Vert _{C^{1,1}({\Omega _\varepsilon })}\lesssim \Vert u_\varepsilon \Vert _{W^{2,\infty }({\Omega _\varepsilon })}. \end{aligned}$$

By Theorem A.1, there exists an extension of $\tilde{u}_\varepsilon$ to $\mathrm {Lip}^{(1,1)}({\Omega _\varepsilon })$, with

$$\begin{aligned} \Vert \tilde{u}_\varepsilon \Vert _{W^{2,\infty }({\Omega _\varepsilon })}\lesssim \Vert u_\varepsilon \Vert _{W^{2,\infty }({\Omega _\varepsilon })}. \end{aligned}$$

Comparing the extension with $u_\varepsilon$, we have the estimates

$$\begin{aligned} \begin{aligned} \int _{\Omega _\varepsilon \setminus V}|\nabla ^2 \tilde{u}_\varepsilon -\nabla ^2 u_\varepsilon |^p\text {d}x&\lesssim h^{1/3}\Vert u_\varepsilon \Vert _{W^{2,\infty }(\Omega _\varepsilon )}^p,\\ \int _{V}|\nabla ^2 \tilde{u}_\varepsilon -\nabla ^2 u_\varepsilon |^p\text {d}x&\lesssim {\mathcal L}^2(V)h^p. \end{aligned} \end{aligned}$$

(54)

Choosing h small enough, we have

$$\begin{aligned} \int _{{\Omega _\varepsilon }\setminus V}(1+|\nabla ^k u|^p)\text {d}x<\delta /3, \end{aligned}$$

and $\Vert \tilde{u}_\varepsilon -u_\varepsilon \Vert _{W^{2,p}(\Omega _\varepsilon )}<\delta /2$. We claim that the function $w\in W^{2,p}(\Omega )$, defined by

$$\begin{aligned} w(x)={\left\{ \begin{array}{ll}\tilde{u}_\varepsilon &{}\text { if }x\in \Omega _\varepsilon \\ u_\varepsilon &{}\text { else }\end{array}\right. } \end{aligned}$$

satisfies all the properties that are stated in the lemma. Indeed, we have $w=u_0$ and $\nabla w=\nabla u_0$ on $\partial \Omega$, w is a polynomial of degree 2 on on $\Omega _w:=\cup _i Q_i$, and

$$\begin{aligned} \begin{aligned} \Vert w-u\Vert _{W^{2,p}(\Omega )}&\le \Vert u-u_\varepsilon \Vert _{W^{2,p}(\Omega )}+\Vert u_\varepsilon -\tilde{u}_\varepsilon \Vert _{W^{2,p}(\Omega )}<\delta ,\\ \int _{\Omega \setminus \Omega _w}(1+|\nabla ^2 u|^p+|\nabla ^2 w|^p)\text {d}x&\le 2\delta /3+Ch^{1/3}\Vert u_\varepsilon \Vert _{W^{2,\infty }(\Omega _\varepsilon )}^p<\delta . \end{aligned} \end{aligned}$$

This proves the lemma. $\square$

Proof of Theorem 3.7

Let $u\in u_0+W^{2,p}_0({\Omega })$. By Lemma A.3, there exists $w\in u_0+W^{2,p}_0({\Omega })$ and ${\Omega }_w\subset {\Omega }$ such that ${\Omega }_w$ is a union of mutually disjoint closed cubes, ${\Omega }_w=\cup _{i=1}^NQ_i$, w is piecewise a polynomial of degree 2 on ${\Omega }_w$, and furthermore

$$\begin{aligned} \begin{aligned} \Vert u-w\Vert _{W^{2,p}({\Omega })}&<\varepsilon /2,\\ \int _{{\Omega }\setminus {\Omega }_w}1+|\nabla ^2 w|^p+|\nabla ^2 u|^p\text {d}x&<\varepsilon /2.\end{aligned} \end{aligned}$$

For $i \in \{1,\ldots ,N\}$, choose $\tilde{\xi }_i\in W_0^{2,\infty }([-1/2,1/2]^2;\mathbb {R}^m)$ such that

$$\begin{aligned} \int _{[-1/2,1/2]^2} f(\nabla ^2 w(x_i)+\nabla ^2\tilde{\xi }_i(x))\text {d}x< Q_2f(\nabla ^2 w(x_i))+\frac{\varepsilon }{N{\mathcal L}^2(Q_i)}, \end{aligned}$$

where $x_i$ denotes the center of the cube $Q_i$. We identify $\tilde{\xi }_i$ with its periodic extension to $\mathbb {R}^2$. Let $d_i$ denote the sidelength of the cube $Q_i$, and let $M_i\in \mathbb {N}$ to be chosen later. We define $\xi _i\in W^{2,p}_0(Q_i;\mathbb {R}^m)$ by

$$\begin{aligned} \xi _i(x)=\left( \frac{d_i}{M_i}\right) ^{2}\tilde{\xi }_i\left( \frac{M_i(x-x_i)}{d_i}\right) . \end{aligned}$$

Then we have

$$\begin{aligned} \begin{aligned} \int _{Q_i}f(\nabla ^2 w(x_i)+\nabla ^2\xi _i)\text {d}x&={\mathcal L}^2(Q_i)\int _{[0,1]^2} f(\nabla ^2 w(x_i)+\nabla ^2\tilde{\xi }_i(x))\text {d}x\\&< {\mathcal L}^2(Q_i)\,Q_2f(\nabla ^2 w(x_i))+\frac{\varepsilon }{N}\\&= \int _{Q_i}Q_2f(\nabla ^2 w(x))\text {d}x +\frac{\varepsilon }{N}. \end{aligned} \end{aligned}$$

Choosing $M_i$ large enough, we may assume

$$\begin{aligned} \Vert \xi _i\Vert _{W^{1,p}(Q_i;\mathbb {R}^m)}^p<\frac{(\varepsilon /2)^p}{N}. \end{aligned}$$

Now the function

$$\begin{aligned} v(x)={\left\{ \begin{array}{ll} w+\xi _i&{}\text { on }Q_i\\ w&{}\text { on }{\Omega }\setminus {\Omega }_w\end{array}\right. } \end{aligned}$$

has all the required properties. $\square$

Appendix B: Proof of Theorem 3.9

For the convenience of the reader, we repeat the statement. We set

$$\begin{aligned} \bar{G}_\lambda (\sigma ):={\left\{ \begin{array}{ll}2\sqrt{\lambda } \rho ^0(\sigma )-2|\det \sigma | &{} \text { if }\rho ^0(\sigma )\le \sqrt{\lambda }\\ |\sigma |^2+\lambda &{} \text { else,}\end{array}\right. } \end{aligned}$$

and the theorem we want to prove is

Theorem B.1

We have

$$\begin{aligned} Q_2F_\lambda (\sigma )=\bar{G}_\lambda (\sigma ). \end{aligned}$$

For the proof, we will need to carry out proofs of statements whose analogues for first gradients are well known. We closely follow the proofs in [11], adapting them to the current situation.

Definition B.2

Let $f:\mathbb {R}^{n\times n}_{\mathrm {sym}}\rightarrow \mathbb {R}$. We say that f is symmetric rank one convex if

$$\begin{aligned} f(t\xi _1+(1-t)\xi _2)\le t f(\xi _1)+(1-t) f(\xi _2) \end{aligned}$$

for all $t\in [0,1]$, and for all $\xi _1,\xi _2\in \mathbb {R}^{n\times n}_{\mathrm {sym}}$ such that $\xi _1-\xi _2=\alpha \eta \otimes \eta$ for some $\alpha \in \mathbb {R}$, $\eta \in \mathbb {R}^n$.

Furthermore, for $f:\mathbb {R}^{n\times n}_{\mathrm {sym}}\rightarrow \mathbb {R}$, we set

$$\begin{aligned} R^{\mathrm {sym}}f(\xi ):=\sup \{g(\xi ):g\le f \text { and } g \text { is symmetric rank one convex}\}. \end{aligned}$$

Lemma B.3

Let $\alpha ,\beta \in \mathbb {R}$, $t\in [0,1]$, $\varepsilon >0$ and

$$\begin{aligned} u_t:[a,b]\rightarrow \mathbb {R},\,\quad x\mapsto \frac{1}{2} (t\alpha +(1-t)\beta ) x^2. \end{aligned}$$

Then there exist $I,J\subset [0,1]$ and $u:[0,1]\rightarrow \mathbb {R}$ such that $\overline{I}\cup \overline{J}=[0,1]$, $I\cap J=\emptyset$, $|I|=t$, $|J|=(1-t)$ and

$$\begin{aligned}&u(0)=u_t(0),\quad u'(0)=u'_t(0),\\&u(1)=u_t(1),\quad u'(1)=u'_t(1),\\&\Vert u-u_t\Vert _{L^\infty }+\Vert u'-u_t'\Vert _{L^\infty }<\varepsilon \\&u''(x)= \alpha \quad \text { for }x\in I\\&u''(x)= \beta \quad \text { for }x\in J. \end{aligned}$$

Proof

For $q\in [0,1-t]$, let $\varphi _{t,q}:[0,1]\rightarrow \mathbb {R}$ be defined by

$$\begin{aligned} \varphi _{t,q}(x)={\left\{ \begin{array}{ll} (1-t)(\alpha -\beta )&{}\text { if } q\le x< q+t\\ -t(\alpha -\beta )&{}\text { else. }\end{array}\right. } \end{aligned}$$

Note that $\int _0^1\text {d}s\varphi _{t,q}(s)=0$ independently of q. In fact, we may choose q such that we also have

$$\begin{aligned} \int _0^1\text {d}s\int _0^s \text {d}\tilde{s} \varphi _{t,q}(\tilde{s})=0. \end{aligned}$$

This choice of q shall be fixed from now on. We extend $\varphi _{t,q}$ periodically on $\mathbb {R}$. For $k\in \mathbb {N}$, we set $\Phi _k(x):=\varphi _{t,q}(kx)$. Choosing $k\in \mathbb {N}$ large enough, we set

$$\begin{aligned} u(x):=u_t(x)+\int _0^x\text {d}s\int _0^s\text {d}\tilde{s}\, \Phi _k\left( \tilde{s}\right) . \end{aligned}$$

It is obvious that this function has all the desired properties (for large enough k). $\square$

Lemma B.4

Let $\varepsilon >0$, $t\in [0,1]$ and let $\xi _1,\xi _2\in \mathbb {R}^{n\times n}_{\mathrm {sym}}$ and $\alpha \in \mathbb {R}$, $\eta \in \mathbb {R}^n$ such that $\xi _1-\xi _2=\alpha \eta \otimes \eta$. Let $l:[0,1]^n\rightarrow \mathbb {R}$ be affine, and $u_t(x)=l(x)+\frac{1}{2} x^T(t\xi _1+(1-t)\xi _2)x$. Then there exists a function $u:[0,1]^n\rightarrow \mathbb {R}$ and open sets $\Omega _1,\Omega _2\subset [0,1]^n$ such that

$$\begin{aligned} \begin{aligned} \Vert u-u_t\Vert _{L^\infty }+\Vert u-u_t\Vert _{L^\infty }&<\varepsilon \\ \nabla ^2 u&= \xi _1\text { on }\Omega _1\\ \nabla ^2 u&= \xi _2\text { on }\Omega _2\\ \Vert \nabla ^2 u\Vert _{L^\infty }&\le C\\ |{\mathcal L}^n(\Omega _1)-t|&<\varepsilon \\ |{\mathcal L}^n(\Omega _2)-(1-t)|&<\varepsilon . \end{aligned} \end{aligned}$$

Proof

We may fill the cube $[0,1]^n$ by smaller cubes with one of the axes parallel to $\eta$, and set $u=u_t$ on the (small) remainder. In this way, we reduce the problem to the case where $\eta =e_1$. Now let $\Omega _\varepsilon :=[\varepsilon ,1-\varepsilon ]^n$, and $\eta \in C_0^\infty ((0,1)^n)$ such that

$$\begin{aligned} \begin{aligned} \eta&=1\text { on }\Omega _\varepsilon \\ \Vert \nabla \eta \Vert _{L^\infty }&\le \frac{L}{\varepsilon }\\ \Vert \nabla ^2\eta \Vert _{L^\infty }&\le \frac{L}{\varepsilon ^2}, \end{aligned} \end{aligned}$$

where L is some numerical constant that does not depend on $\varepsilon$. For $x\in [0,1]$, we write $\tilde{u}_t(s)=\frac{1}{2} (t(\xi _1)_{11}+(1-t)(\xi _2)_{11})s^2$. Let $\delta >0$ to be chosen later. According to Lemma B.3, we may choose $\tilde{u}:[0,1]\rightarrow \mathbb {R}$ and $I,J\subset [0,1]$ such that $\overline{I}\cup \overline{J}=[0,1]$, $I\cap J=\emptyset$, $|I|=t$, $|J|=(1-t)$, and

$$\begin{aligned}&\tilde{u}(0)=\tilde{u}_t(0),\quad \tilde{u}'(0)=\tilde{u}'_t(0),\\&\tilde{u}(1)=\tilde{u}_t(1),\quad \tilde{u}'(1)=\tilde{u}'_t(1),\\&\Vert \tilde{u}-\tilde{u}_t\Vert _{L^\infty }+\Vert \tilde{u}'-\tilde{u}_t'\Vert _{L^\infty }<\delta \\&\tilde{u}''(x)= \xi _1 \quad \text { for }x\in I\\&\tilde{u}''(x)= \xi _2 \quad \text { for }x\in J. \end{aligned}$$

We set $\psi (x_1,\ldots ,x_n)=\tilde{u}(x_1)$ and

$$\begin{aligned} u:=\eta (\psi +l)+(1-\eta ) u_t. \end{aligned}$$

Choosing $\delta$ small enough (e.g. $\delta <\min (\varepsilon ^3,\varepsilon ^3/L)$), this choice of u satisfies all the requirements. We leave it to the reader to carry out the straightforward computations that lead to this statement. $\square$

Lemma B.5

Assume that $f:\mathbb {R}^{n\times n}_{\mathrm {sym}}\rightarrow \mathbb {R}$ is bounded from above by a continuous function $\tilde{f}\in C^0(\mathbb {R}^{n\times n}_{\mathrm {sym}})$. Then we have

$$\begin{aligned} Q_2 f\le R^{\mathrm {sym}}f. \end{aligned}$$

Proof

Since $Q_2f$ is the largest 2-quasiconvex function that is less or equal to f, and $R^{\mathrm {sym}}f$ is the largest symmetric rank one convex function that is less or equal to f, it suffices to show that if $g:\mathbb {R}^{n\times n}\rightarrow \mathbb {R}$ is 2-quasiconvex, and $g\le f$, then it is symmetric-rank-one convex. So let us suppose $g:\mathbb {R}^{n\times n}\rightarrow \mathbb {R}$ is 2-quasiconvex, and let $\xi _1,\xi _2\in \mathbb {R}^{n\times n}$ and $\alpha \in \mathbb {R}$, $\eta \in \mathbb {R}^n$ such that $\xi _1-\xi _2=\alpha \eta \otimes \eta$. We need to show

$$\begin{aligned} g(t\xi _1+(1-t)\xi _2)\le t g(\xi _1)+(1-t) g(\xi _2) \end{aligned}$$

(55)

for all $t\in [0,1]$. Let $u_t(x):= \frac{1}{2} x^T(t\xi _1+(1-t)\xi _2)x$, and $\varepsilon >0$ to be chosen later. Let u be the approximating function of Lemma B.4, with the sets $\Omega _1,\Omega _2\subset [0,1]^n$ as in the statement of that lemma. Then we have $u-u_t\in W^{2,\infty }_0([0,1]^n)$ and

$$\begin{aligned} \nabla ^2 u= t\xi _1+(1-t)\xi _2+\nabla ^2(u-u_t). \end{aligned}$$

Hence

$$\begin{aligned} \begin{aligned} g(t\xi _1+(1-t)\xi _2)&=\int _{[0,1]^n}g(\nabla ^2 u_t)\text {d}x\\&\le \int _{[0,1]^n}g(\nabla ^2 u)\text {d}x\\&={\mathcal L}^n(\Omega _1)g(\xi _1)+{\mathcal L}^n(\Omega _2)g(\xi _2)+ \int _{[0,1]^n\setminus (\Omega _1\cup \Omega _2)} g(\nabla ^2 u)\text {d}x\\&\le {\mathcal L}^n(\Omega _1)g(\xi _1)+{\mathcal L}^n(\Omega _2)g(\xi _2)+ \int _{[0,1]^n\setminus (\Omega _1\cup \Omega _2)} \tilde{f}(\nabla ^2 u)\text {d}x \end{aligned} \end{aligned}$$

Choosing $\varepsilon$ small enough and using the properties of $u,\Omega _1,\Omega _2$ from the statement of Lemma B.4, we see that the right hand side is smaller than $tg(\xi _1)+(1-t)g(\xi _2)+\delta$ for any given $\delta >0$; here we also used the assumption that $\tilde{f}$ is continuous. This proves (55) and hence the lemma. $\square$

Definition B.6

Let $f:\mathbb {R}^{n\times n}_{\mathrm {sym}}\rightarrow \mathbb {R}$. We set $R^{\mathrm {sym}}_0 f=f$ and

$$\begin{aligned} R^{\mathrm {sym}}_{k+1}f(\xi )&:=\inf \{ t R^{\mathrm {sym}}_k(\xi _1)+(1- t )R^{\mathrm {sym}}_k(\xi _2):\\&\qquad t \xi _1+(1- t )\xi _2=\xi ,\xi _1-\xi _2=\alpha \eta \otimes \eta \text { for some }\alpha \in \mathbb {R},\,\eta \in \mathbb {R}^n\}. \end{aligned}$$

In complete analogy to Theorem 6.10, part 2 in [11], we show

Lemma B.7

Let $f:\mathbb {R}^{n\times n}_{\mathrm {sym}}\rightarrow \mathbb {R}$, and let $g:\mathbb {R}^{n\times n}_{\mathrm {sym}}\rightarrow \mathbb {R}$ be symmetric rank one convex with $g\le f$. Then we have

$$\begin{aligned} R^{\mathrm {sym}}f=\inf _{k\in \mathbb {N}}R^{\mathrm {sym}}_k f. \end{aligned}$$

Proof

First we observe that for any $k\in \mathbb {N}$, we have

$$\begin{aligned} g\le R^{\mathrm {sym}}_{k+1} f\le R^{\mathrm {sym}}_k f, \end{aligned}$$

and hence obtain that

$$\begin{aligned} R'f:=\inf _{k\in \mathbb {N}}R^{\mathrm {sym}}_k f=\lim _{k\rightarrow \infty }R^{\mathrm {sym}}_k f \end{aligned}$$

is well defined. For any symmetric rank one convex function g we have $R'g=g$, and hence $R'(R^{\mathrm {sym}}f)=R^{\mathrm {sym}}f$. Furthermore, if $g\le g'$, then $R'g\le R'g'$. Combining these observations with the fact $R^{\mathrm {sym}}f\le f$, we obtain

$$\begin{aligned} R^{\mathrm {sym}}f\le R'f\le f. \end{aligned}$$

It remains to show that $R'f$ is symmetric rank one convex. Assume that $\xi _1,\xi _2\in \mathbb {R}^{n\times n}_{\mathrm {sym}}$, and $\alpha \in \mathbb {R}$, $\eta \in \mathbb {R}^{n}$ such that $\xi _1-\xi _2=\alpha \eta \otimes \eta$. Let $\varepsilon >0$. By definition of $R'f$, there exist $i,j\in \mathbb {N}$ such that

$$\begin{aligned} R^{\mathrm {sym}}_if(\xi _1)\le R'f(\xi _1)+\varepsilon ,\quad R^{\mathrm {sym}}_jf(\xi _2)\le R'f(\xi _2)+\varepsilon . \end{aligned}$$

Without loss of generality, we may assume $i\le j$, which yields $R^{\mathrm {sym}}_jf(\xi _1)\le R^{\mathrm {sym}}_if(\xi _1)$. Thus we obtain for every $t\in [0,1]$,

$$\begin{aligned} \begin{aligned} R'f(t\xi _1+(1-t)\xi _2)&\le R^{\mathrm {sym}}_{j+1}(t\xi _1+(1-t)\xi _2)\\&\le tR^{\mathrm {sym}}_{j}(\xi _1)+(1-t)R^{\mathrm {sym}}_j(\xi _2)\\&\le tR'f(\xi _1)+(1-t)R'f(\xi _2)+\varepsilon . \end{aligned} \end{aligned}$$

Since $\varepsilon >0$ was arbitrary, we obtain that $R'f$ is symmetric rank one convex, which proves the lemma. $\square$

Lemma B.8

We have

$$\begin{aligned} R^{\mathrm {sym}}F_\lambda \le \bar{G}_\lambda . \end{aligned}$$

Proof

From the definition of $R^{\mathrm {sym}}F_\lambda$, we see that for $\xi \in \mathbb {R}^{2\times 2}_{\mathrm {sym}}$, $R\in SO(2)$, we have

$$\begin{aligned} R^{\mathrm {sym}}F_\lambda (R^T\xi R)=R^{\mathrm {sym}}F_\lambda (\xi ). \end{aligned}$$

Hence it suffices to consider $\xi$ of diagonal form,

$$\begin{aligned} \xi =\left( \begin{array}{cc} x&{}0\\ 0&{}y\end{array}\right) . \end{aligned}$$

We may assume $|x|+|y|< \lambda$, since otherwise we know $F_\lambda (\xi )=\bar{G}_\lambda (\xi )=\lambda +x^2+y^2$. Similarly, we may assume $0<|x|+|y|$, since otherwise $F_\lambda (\xi )=\bar{G}_\lambda (\xi )=0$. Let $\alpha ,\beta \in (0,1)$ to be chosen later, and set

$$\begin{aligned} \xi _1=\left( \begin{array}{cc} 0&{}0\\ 0&{}0\end{array}\right) ,\quad \xi _2=\left( \begin{array}{cc} x/\alpha &{}0\\ 0&{}0\end{array}\right) ,\quad \xi _3=\left( \begin{array}{cc} x&{}0\\ 0&{}y/\beta \end{array}\right) . \end{aligned}$$

Note that $\beta \xi _3+(1-\beta )(\alpha \xi _2+(1-\alpha ) \xi _1))=\xi$, and $\xi _3-(\alpha \xi _2+(1-\alpha )\xi _1),\xi _2-\xi _1$ are both symmetric-rank-one. By Lemma B.7, we have

$$\begin{aligned} \begin{aligned} R^{\mathrm {sym}}F_\lambda (\xi )\le&\beta F_\lambda (\xi _3)+(1-\beta )\left( \alpha F_\lambda (\xi _2)+(1-\alpha ) F_\lambda (\xi _1)\right) \\ =&\beta \left( \lambda +x^2+\frac{y^2}{\beta ^2}\right) +(1-\beta )\alpha \left( \lambda +\frac{x^2}{\alpha ^2}\right) . \end{aligned} \end{aligned}$$

Now we assume $|x|>0$. The right hand side in the last estimate is convex in $\alpha$; it attains its minimum at $\alpha = \frac{|x|}{\sqrt{\lambda }}$. Hence,

$$\begin{aligned} \begin{aligned} R^{\mathrm {sym}}F_\lambda (\xi )&\le \beta \left( \lambda +x^2+\frac{y^2}{\beta ^2}\right) +(1-\beta )2|x|\sqrt{\lambda }\\&= 2|x|\sqrt{\lambda }+\beta (\sqrt{\lambda }-|x|)^2+\frac{y^2}{\beta } \end{aligned} \end{aligned}$$

Choosing $\beta =|y|/(\sqrt{\lambda }-|x|)$, we obtain

$$\begin{aligned} R^{\mathrm {sym}}F_\lambda (\xi )\le 2\sqrt{\lambda }(|x|+|y|-|xy|)=\bar{G}_\lambda (\xi ). \end{aligned}$$

It remains to prove the claim for the case $|x|=0$. Then we have

$$\begin{aligned} \begin{aligned} R^{\mathrm {sym}}F_\lambda (\xi )\le&\beta F_\lambda (\xi _3)+(1-\beta ) F_\lambda (\xi _1)\\&= \beta \left( \lambda +x^2+\frac{y^2}{\beta ^2}\right) . \end{aligned} \end{aligned}$$

Again setting $\beta =|y|/(\sqrt{\lambda }-|x|)$, we obtain the same conclusion as before. This proves the lemma. $\square$

Proof of Theorem B.1

By Theorem 6.28 in [11], we have $\bar{G}_1=Q_1F_1$. We have $F_\lambda =\lambda F(\cdot /\sqrt{\lambda })$, and hence by the definition of the quasiconvex envelope (51) it is easily seen that $Q_1F_\lambda = \lambda Q_1F_1(\cdot /\sqrt{\lambda })$. It is also easily verified that $\bar{G}_\lambda =\lambda \bar{G}(\cdot /\sqrt{\lambda })$, and since $Q_1F_\lambda \le Q_2 F_\lambda$, we obtain $\bar{G}_\lambda \le Q_2F_\lambda$. By Lemma B.5 and Lemma B.8, we also have the opposite inequality $\bar{G}_\lambda \ge Q_2F_\lambda$. This proves the theorem. $\square$