Marginal Material Stability

Abstract

Marginal stability plays an important role in nonlinear elasticity because the associated minimally stable states usually delineate failure thresholds. In this paper we study the local (material) aspect of marginal stability. The weak notion of marginal stability at a point, associated with the loss of strong ellipticity, is classical. States that are marginally stable in the strong sense are located at the boundary of the quasi-convexity domain and their characterization is the main goal of this paper. We formulate a set of bounds for such states in terms of solvability conditions for an auxiliary nucleation problem formulated in the whole space and present nontrivial examples where the obtained bounds are tight.

Appendices

Appendix A: Proof of Lemma 3.3

We note that ${\mathcal{S}}=W^{1,\infty}(\mathbb{R} ^{d};\mathbb{R} ^{m})\cap {\mathcal{S}} _{0}$. It will be important to use the following embedding theorem for the space ${\mathcal{S}}_{0}$.

Theorem A.1

Assume that $\phi \in{\mathcal{S}}_{0}$ and d≥3. Then there exists a unique constant $\boldsymbol{c}\in\mathbb{R} ^{m}$ such that $\phi - \boldsymbol{c}\in L^{\frac{2d}{d-2}}(\mathbb{R} ^{d};\mathbb{R} ^{m})$.

Proof

First, we remark that without loss of generality we may take m=1. Now recall the well-known potential theory operators. The Riesz transforms R _j are defined by

$${\mathcal{F}}(R_{j}f) (\xi )=i\frac{\xi_{j}}{|\xi |}\widehat{f}(\xi ), $$

where

$${\mathcal{F}}(f) (\xi )=\widehat{f}(\xi )=\int_{\mathbb{R} ^{d}}f( \boldsymbol{x} )\mathrm{e}^{2\pi i( \boldsymbol{x} ,\xi )}\,\mathrm{d}\boldsymbol{x} $$

is the Fourier transform. The operators R _j map $L^{2}(\mathbb{R} ^{d})$ into $L^{2}(\mathbb{R} ^{d})$. The Riesz potential I ₁ is defined by

$${\mathcal{F}}(I_{1}f) (\xi )=\frac{\widehat{f}(\xi )}{2\pi|\xi |}. $$

It maps $L^{2}(\mathbb{R} ^{d})$ into $L^{\frac{2d}{d-2}}(\mathbb{R} ^{d})$, (Stein 1970).

If ϕ were smooth and compactly supported, we would have

$$\phi=I_{1} \Biggl(\sum_{j=1}^{d}R_{j} \biggl(\frac{\partial\phi }{\partial x_{j}} \biggr) \Biggr). $$

Hence, we define

$$\psi( \boldsymbol{x})=I_{1} \Biggl(\sum_{j=1}^{d}R_{j}(g_{j}) \Biggr), $$

where $\boldsymbol{g}=\nabla\phi\in L^{2}(\mathbb{R} ^{d};\mathbb{R} ^{d})$.

Let η(x) be an arbitrary smooth compactly supported function. By definition of the distributional derivative we have

$$\int_{\mathbb{R} ^{d}} \biggl\{g_{k}\frac{\partial\eta}{\partial x_{j}}-g_{j} \frac{\partial\eta}{\partial x_{k}} \biggr\}\,\mathrm{d}\boldsymbol{x}= -\biggl\langle\phi,\frac{\partial^2 \eta}{\partial x_{k}\partial x_{j}} \biggr\rangle+\biggl\langle\phi,\frac{\partial^2 \eta}{\partial x_{j}\partial x_{k}} \biggr\rangle=0. $$

By Plancherel’s identity

$$\int_{\mathbb{R} ^{d}}(\widehat{g}_{k}\xi_{j}- \widehat{g}_{j}\xi _{k})\overline{\widehat{\eta}}\, \mathrm{d}\xi =0. $$

We conclude that

$$ \widehat{g}_{k}(\xi )\xi_{j}= \widehat{g}_{j}(\xi )\xi_{k} $$

(A.1)

for a.e. $\xi \in\mathbb{R} ^{d}$. Thus,

$$-2\pi i\xi_{k}\widehat{\psi}(\xi )=\sum_{j=1}^{d} \frac{\xi _{k}\xi _{j}\widehat{g}_{j}(\xi )}{|\xi |^{2}} =\widehat{g}_{k}(\xi ), $$

due to (A.1). By Plancherel’s identity

$$\int_{\mathbb{R} ^{d}}\psi( \boldsymbol{x})\frac{\partial\eta}{\partial x_{k}}\,\mathrm{d}\boldsymbol{x}= 2\pi i \int_{\mathbb{R} ^{d}}\xi_{k}\widehat{\psi}\,\overline{\widehat {\eta}}\,\mathrm{d}\xi = -\int_{\mathbb{R} ^{d}}\widehat{g}_{k}( \xi )\overline{\widehat{\eta }}\,\mathrm{d}\xi = -\int_{\mathbb{R} ^{d}}g_{k}( \boldsymbol{x})\eta( \boldsymbol{x})\,\mathrm{d}\boldsymbol{x}. $$

Therefore, ∇ψ=g=∇ϕ as distributions. The theorem is proved. □

The proof of Lemma 3.3 proceeds in different ways depending on the dimension d. If d=1, then

Let d=2. We estimate

$$\frac{1}{R}\int_{A_{R}}|\phi ||\nabla\phi |\, \mathrm{d}\boldsymbol{x}\le\|\phi \| _{\infty }\sqrt{3\pi} \biggl(\int _{A_{R}}|\nabla\phi |^{2}\,\mathrm{d}\boldsymbol{x} \biggr)^{1/2}\to0,\quad\text{as }R\to\infty. $$

By the Poincaré inequality

$$\int_{A_{R}}\big|\phi ( \boldsymbol{x})-\langle\phi \rangle_{A_{R}}\big|^{2} \,\mathrm{d}\boldsymbol{x}\le C_{0}R^{2}\int_{A_{R}}| \nabla\phi |^{2}\,\mathrm{d}\boldsymbol{x}, $$

where C ₀ is the Poincaré constant for A ₁. The boundedness of ϕ implies that there exists a sequence R=R _k such that

$$\lim_{k\to\infty}\langle\phi \rangle_{A_{R_{k}}}= \boldsymbol{c}. $$

Hence, by the triangle inequality

$$\biggl(\int_{A_{R}}|\phi - \boldsymbol{c}|^{2}\,\mathrm{d}\boldsymbol{x} \biggr)^{1/2}\le \biggl(\int_{A_{R}}\big|\phi -\langle\phi \rangle_{A_{R}}\big|^{2}\,\mathrm{d}\boldsymbol{x} \biggr)^{1/2}+ |A_{R}|^{1/2}\big|\langle\phi \rangle_{A_{R}}- \boldsymbol{c}\big|. $$

Then,

Now assume that d≥3. By Theorem A.1 there exists a unique constant c, such that $\phi - \boldsymbol{c}\in L^{\frac{2d}{d-2}}(\mathbb{R} ^{d};\mathbb{R} ^{m})$. Using the inequality ab≤(a ²+b ²)/2 we get

$$\frac{1}{R}\int_{A_{R}}|\phi - \boldsymbol{c}||\nabla\phi |\, \mathrm{d}\boldsymbol{x}+\frac {1}{R^{2}}\int_{A_{R}}|\phi - \boldsymbol{c}|^{2}\,\mathrm{d}\boldsymbol{x}\le \frac{1}{2}\int_{A_{R}}| \nabla\phi |^{2}\,\mathrm{d}\boldsymbol{x}+\frac{3}{2R^{2}}\int _{A_{R}}|\phi - \boldsymbol{c}|^{2}\,\mathrm{d}\boldsymbol{x}. $$

By Hölder inequality

$$\frac{1}{R^{2}}\int_{A_{R}}|\phi - \boldsymbol{c}|^{2}\, \mathrm{d}\boldsymbol{x}\le \biggl(\int_{A_{R}}|\phi - \boldsymbol{c}|^{\frac{2d}{d-2}}\, \mathrm{d}\boldsymbol{x} \biggr)^{\frac {d-2}{d}}\to 0,\quad\text{as }R\to\infty. $$

The lemma is proved.

Appendix B: Proof of Lemma 3.7

Let us begin with a technical lemma.

Lemma B.1

Suppose α(R)>0 is such that α(R)→0, as R→∞. Then there exists a monotonically increasing function h(R) with h(R)/R→0, as R→∞, such that

$$ \lim_{R\to\infty} \biggl(\frac{R}{h(R)} \biggr) \alpha(R)=0. $$

(B.1)

Proof

We define

$$h(R)=\max_{r<R} \bigl(r\sqrt{\alpha(r)} \bigr). $$

Then h(R) is monotonically increasing and h(R)/R→0, as R→∞. Indeed, for any ϵ>0

$$h(R)\le\max_{r<\epsilon R} \bigl(r\sqrt{\alpha(r)} \bigr)+ \max _{\epsilon R<r<R} \bigl(r\sqrt{\alpha(r)} \bigr)\le\epsilon R\sqrt{ \alpha(0)}+R\sqrt{\alpha(\epsilon R)}. $$

Therefore,

$$\mathop{\overline{\lim}}_{R\to\infty}\frac{h(R)}{R}\le\epsilon \sqrt{ \alpha(0)}. $$

Hence, h(R)/R→0, as R→∞. By definition of h(R) we have $h(R)\ge R\sqrt{\alpha(R)}$. Therefore,

$$\frac{R}{h(R)}\le\frac{1}{\sqrt{\alpha(R)}}. $$

Thus,

$$\biggl(\frac{R}{h(R)} \biggr)\alpha(R)\le\sqrt{\alpha(R)}\to 0,\quad\text{as }R\to\infty. $$

 □

Now let us prove Lemma 3.7. First observe that for any $\phi \in{\mathcal{S}}_{0}$

$$\lim_{R\to\infty}\int_{B_{R}}|\nabla\phi |^{2}\,\mathrm{d}\boldsymbol{x}=\|\nabla\phi \|_{2}^{2}, $$

while

$$\lim_{R\to\infty}\int_{A_{R}(h(R))}|\nabla\phi |^{2}\,\mathrm{d}\boldsymbol{x}=0. $$

Hence, we only need to prove that there exist $\boldsymbol{c}\in\mathbb{R} ^{m}$ and a monotonically increasing function h(R)=o(R) such that

$$ \lim_{R\to\infty}\frac{1}{h(R)^{2}}\int _{A_{R}(h(R))}|\phi - \boldsymbol{c} |^{2}\,\mathrm{d}\boldsymbol{x}=0. $$

(B.2)

Remark 3.6 implies that we need to prove (B.2) for d≥3. In that case, the constant $\boldsymbol{c}\in\mathbb{R} ^{m}$ is chosen so that $\phi - \boldsymbol{c}\in L^{\frac{2d}{d-2}}(\mathbb{R} ^{d};\mathbb{R} ^{m})$, which is possible by Theorem A.1. The Hölder inequality gives

$$\frac{1}{h(R)^{2}}\int_{A_{R}(h(R))}|\phi - \boldsymbol{c}|^{2}\, \mathrm{d}\boldsymbol{x}\le C \biggl(\frac{R}{h(R)} \biggr)^{\frac{2(d-1)}{d}} \biggl(\int _{A_{R}(h(R))}|\phi - \boldsymbol{c}|^{\frac{2d}{d-2}}\,\mathrm{d}\boldsymbol{x} \biggr)^{\frac{d-2}{d}}. $$

By Theorem A.1

$$\alpha(R)= \biggl(\int_{| \boldsymbol{x}|\ge R/2}|\phi - \boldsymbol{c}|^{\frac {2d}{d-2}}\, \mathrm{d}\boldsymbol{x} \biggr)^{\frac{d-2}{2(d-1)}}\to0,\quad\text{as }R\to\infty. $$

We see that in each of the three cases we have a function α(R)→0, as R→∞, which is independent of h(R). The application of Lemma B.1 concludes the proof of Lemma 3.7.

Appendix C: Proof of Theorem 3.9

Step 1: Asymptotics of $\int_{B_{R}}|\nabla\phi |^{2}\,\mathrm{d}\boldsymbol{x}$.

We write x=p+R ^T t and |x|²=|p|²+|t|². Therefore,

If we make the change of variables p=R u we obtain

$$\frac{1}{R^{d-k}}\int_{V_{R}}|\nabla\phi |^{2}\, \mathrm{d}\boldsymbol{x}=\int_{\{| \boldsymbol{t} |<R\} }\int_{\{| \boldsymbol{u}|<1\}} \bigl\{\big| \psi _{ \boldsymbol{t}}( \boldsymbol{t},R \boldsymbol{u})\big|^{2}+\big|\psi _{ \boldsymbol{p}}( \boldsymbol{t},R \boldsymbol{u})\big|^{2}\bigr\}\,\mathrm{d}\boldsymbol{u} \,\mathrm{d}\boldsymbol{t}. $$

Hence,

$$\frac{1}{R^{d-k}}\int_{V_{R}}|\nabla\phi |^{2}\, \mathrm{d}\boldsymbol{x}\le\int_{\{| \boldsymbol{u} |<1\} }\int_{{\mathbb{R} ^{k}}} \bigl \{\big|\psi _{ \boldsymbol{t}}( \boldsymbol{t},R \boldsymbol{u})\big|^{2}+\big|\psi _{ \boldsymbol{p}}( \boldsymbol{t},R \boldsymbol{u})\big|^{2}\bigr\}\,\mathrm{d}\boldsymbol{t} \,\mathrm{d}\boldsymbol{u}. $$

By the Riemann–Lebesgue lemma we get

$$\mathop{\overline{\lim}}_{R\to\infty}\frac{1}{R^{d-k}}\int _{V_{R}}|\nabla\phi |^{2}\,\mathrm{d}\boldsymbol{x} \le \omega_{d-k}\hspace{-8pt}\skew {33}-{\int _{Q_{d-k}}}\int_{{\mathbb{R} ^{k}}} \bigl\{\big| \psi _{ \boldsymbol{t}}( \boldsymbol{t}, \boldsymbol{p})\big|^{2}+\big|\psi _{ \boldsymbol{p}}( \boldsymbol{t}, \boldsymbol{p})\big|^{2}\bigr\}\,\mathrm{d}\boldsymbol{t} \,\mathrm{d}\boldsymbol{p}, $$

where ω _n is the volume of the n-dimensional unit ball. Thus,

$$\mathop{\overline{\lim}}_{R\to\infty}\frac{1}{R^{d-k}}\int _{B_{R}}|\nabla\phi |^{2}\,\mathrm{d}\boldsymbol{x} \le \omega_{d-k}\hspace{-8pt}\skew {33}-{\int _{Q_{d-k}}}\int_{{\mathbb{R} ^{k}}} \bigl\{\big| \psi _{ \boldsymbol{t}}( \boldsymbol{t}, \boldsymbol{p})\big|^{2}+\big|\psi _{ \boldsymbol{p}}( \boldsymbol{t}, \boldsymbol{p})\big|^{2}\bigr\}\,\mathrm{d}\boldsymbol{t} \,\mathrm{d}\boldsymbol{p}. $$

To get the reverse inequality we write

$$\int_{B_{R}}|\nabla\phi |^{2}\,\mathrm{d}\boldsymbol{x}=\int _{\{| \boldsymbol{t}|<R\}}\int_{\{| \boldsymbol{p} |<r(R, \boldsymbol{t})\}} \bigl\{|\psi _{ \boldsymbol{t}}|^{2}+|\psi _{ \boldsymbol{p}}|^{2}\bigr\}\, \mathrm{d} \boldsymbol{p}\,\mathrm{d} \boldsymbol{t}, $$

where $r(R, \boldsymbol{t})=\sqrt{R^{2}-| \boldsymbol{t}|^{2}}$. If we make a change of variables p=r(R,t)u we obtain

By the Riemann–Lebesgue lemma we get

for a.e. $\boldsymbol{t}\in\mathbb{R} ^{k}$. By Fatous’s lemma we get

$$\mathop{\underline{\lim}}_{R\to\infty}\frac{1}{R^{d-k}}\int _{B_{R}}|\nabla\phi |^{2}\,\mathrm{d}\boldsymbol{x} \ge \omega_{d-k}\int_{\mathbb{R} ^{k}} \hspace{-8pt}\skew {33}-{\int _{Q_{d-k}}}\bigl\{\big| \psi _{ \boldsymbol{t}}( \boldsymbol{t}, \boldsymbol{p})\big|^{2}+\big|\psi _{ \boldsymbol{p}}( \boldsymbol{t}, \boldsymbol{p} )\big|^{2}\bigr\}\,\mathrm{d} \boldsymbol{p}\,\mathrm{d} \boldsymbol{t}. $$

Hence, we obtain the asymptotics of $\int_{B_{R}}|\nabla\phi |^{2}\,\mathrm{d}\boldsymbol{x}$:

$$ \lim_{R\to\infty}\frac{1}{R^{d-k}}\int _{B_{R}}|\nabla\phi |^{2}\,\mathrm{d}\boldsymbol{x} = \omega_{d-k}\int_{\mathbb{R} ^{k}} \hspace{-8pt}\skew {33}-{\int _{Q_{d-k}}}\bigl\{\big| \psi _{ \boldsymbol{t}}( \boldsymbol{t}, \boldsymbol{p})\big|^{2}+\big|\psi _{ \boldsymbol{p}}( \boldsymbol{t}, \boldsymbol{p} )\big|^{2}\bigr\}\,\mathrm{d} \boldsymbol{p}\,\mathrm{d} \boldsymbol{t}. $$

(C.1)

In particular, we get

$$\mathop{\overline{\lim}}_{R\to\infty}\hspace{-8pt}\skew {33}-{\int _{B_{R}}}|\nabla\phi |^{2}\,\mathrm{d}\boldsymbol{x}\le\mathop{\overline{\lim}}_{R\to \infty} \frac{\omega_{d}}{R^{d}}\int_{V_{R}} |\nabla\phi |^{2}\, \mathrm{d}\boldsymbol{x}=0, $$

establishing (3.19).

Step 2: Proof of (3.17).

For any h(R)=o(R) we have, using (C.1),

$$\frac{\int_{A_{R}(h(R))}|\nabla\phi |^{2}\,\mathrm{d}\boldsymbol{x}}{\int_{B_{R}}|\nabla\phi |^{2}\,\mathrm{d}\boldsymbol{x}}= \frac{\int_{B_{R}}|\nabla\phi |^{2}\,\mathrm{d}\boldsymbol{x}-\int_{B_{R-h(R)}}|\nabla\phi |^{2}\,\mathrm{d}\boldsymbol{x}}{ \int_{B_{R}}|\nabla\phi |^{2}\,\mathrm{d}\boldsymbol{x}}=1- \biggl(1-\frac{h(R)}{R} \biggr)^{d-k}u_{R}, $$

where u _R →1, as R→∞. Thus, (3.17) is proved.

Step 3: Proof of (3.18). We have

$$\int_{A_{R}(h(R))}|\phi |^{2}\,\mathrm{d}\boldsymbol{x}\le\int _{| \boldsymbol{t}|<R}\int_{| \boldsymbol{p} |<R}\big|\psi ( \boldsymbol{t}, \boldsymbol{p})\big|^{2}\,\mathrm{d}\boldsymbol{p} \,\mathrm{d}\boldsymbol{t}. $$

The periodicity in the p variable implies that for any domain $\varOmega \subset\mathbb{R} ^{d-k}$

$$\int_{\varOmega}\big|\psi ( \boldsymbol{t}, \boldsymbol{p})\big|^{2}\,\mathrm{d}\boldsymbol{p}\le v( \varOmega)\hspace{-8pt}\skew {33}-{\int _{Q_{d-k}}}\big|\psi ( \boldsymbol{t}, \boldsymbol{p})\big|^{2}\,\mathrm{d}\boldsymbol{p}, $$

where v(Ω) is the (d−k)-volume of all period cells intersecting Ω. We further estimate that

$$v(\varOmega)\le|\varOmega+B_{M}|, $$

where M is the diameter of the period cell Q _d−k . When R>M we obtain

$$v\bigl(\bigl\{| \boldsymbol{p}|\le R+M\bigr\}\bigr)\le\omega_{d-k}(2R)^{d-k}. $$

Hence, we get the estimate

(C.2)

For convenience we introduce the truncated L ² norm

$$\| \boldsymbol{f}\|_{2,R}^{2}=\int_{B_{R}}\int _{Q_{d-k}}| \boldsymbol{f}( \boldsymbol{t}, \boldsymbol{p} )|^{2}\,\mathrm{d} \boldsymbol{p}\,\mathrm{d} \boldsymbol{t}. $$

Lemma C.1

For every $\psi \in{\mathcal{S}}_{k}(Q_{d-k})$ there exists a constant $\boldsymbol{c}\in \mathbb{R} ^{m}$ such that

$$\mathop{\underline{\lim}}_{R\to\infty}\frac{\|\psi - \boldsymbol{c}\| ^{2}_{2,R}}{R^{2}}=0. $$

Proof

The proof of the lemma is different depending on whether k=1, k=2 or k≥3.

If k=1 we can use the assumption of uniform boundedness of ψ and conclude that

$$\frac{\|\psi \|^{2}_{2,R}}{R^{2}}\le\frac{2\|\psi \|_{\infty }^{2}}{R}\to0,\quad\text{as }R\to\infty. $$

If k≥3, then, according to Theorem A.1, for a.e. p∈Q _d−k there exists a unique vector c(p) such that

$$\int_{\mathbb{R} ^{k}}\big|\psi ( \boldsymbol{t}, \boldsymbol{p})- \boldsymbol{c}( \boldsymbol{p})\big|^{\frac {2k}{k-2}}<\infty. $$

However, we need a sharper statement

Lemma C.2

There exists $\boldsymbol{c}\in\mathbb{R} ^{m}$ such that c(p)=c for a.e. p∈Q _d−k .

Proof

Let

$$\langle\psi \rangle_{Q_{d-k}}( \boldsymbol{t})=\frac{1}{|Q_{d-k}|}\int _{Q_{d-k}}\psi ( \boldsymbol{t}, \boldsymbol{p})\,\mathrm{d}\boldsymbol{p}. $$

The Poincaré inequality implies

$$\big\|\psi -\langle\psi \rangle_{Q_{d-k}}( \boldsymbol{t})\big\|_{2,R}^{2} \le C\|\psi _{ \boldsymbol{p}}\|_{2,R}^{2}. $$

Therefore, $\|\psi -\langle\psi \rangle_{Q_{d-k}}( \boldsymbol{t})\|_{2,R}$ is bounded as R→∞. Next observe that $\langle\psi \rangle_{Q_{d-k}}( \boldsymbol{t} )\in{\mathcal{S}}$ as a function of t. Hence, there exists $\boldsymbol{c}\in\mathbb{R} ^{m}$ such that $\langle\psi \rangle_{Q_{d-k}}( \boldsymbol{t})- \boldsymbol{c}\in L^{\frac{2k}{k-2}}(\mathbb{R} ^{k})$. It follows that

$$\big\| \boldsymbol{c}( \boldsymbol{p})- \boldsymbol{c}\big\|_{2,R}\le\big\| \boldsymbol{c}( \boldsymbol{p})-\psi \big\|_{2,R}+\big\|\psi -\langle \psi \rangle_{Q_{d-k}}( \boldsymbol{t})\big\|_{2,R} +\big\|\langle\psi \rangle_{Q_{d-k}}( \boldsymbol{t})- \boldsymbol{c}\big\|_{2,R}. $$

Let us apply the Hölder inequality to the first and third term on the left-hand side of the above inequality:

We conclude that

$$\mathop{\overline{\lim}}_{R\to\infty}\frac{1}{R^{2}}\big\| \boldsymbol{c}( \boldsymbol{p} )- \boldsymbol{c} \big\|_{2,R}^{2}<+\infty. $$

However, this would contradict

$$\big\| \boldsymbol{c}( \boldsymbol{p})- \boldsymbol{c}\big\|_{2,R}^{2}=|B_{R}|\int _{Q_{d-k}}\big| \boldsymbol{c}( \boldsymbol{p})- \boldsymbol{c} \big|^{2}\,\mathrm{d}\boldsymbol{p}, $$

unless c(p)=c for a.e. p∈Q _d−k . □

We will now establish Lemma C.1, in which the constant vector c is coming from Lemma C.2. For simplicity of notation ψ(t,p) will now stand for ψ−c. In order to prove Lemma C.1 we split the t-integral in the definition of ∥ψ∥_2,R into the integral over the ball {|t|<ϵR} and the annulus {ϵR<|t|<R}. Then we apply the same Hölder inequality to both integrals and obtain the estimate

Lemma C.2 then implies that for a.e. p∈Q _d−k

$$\mathop{\overline{\lim}}_{R\to\infty}\frac{1}{R^{2}}\int _{| \boldsymbol{t} |<R}\big|\psi ( \boldsymbol{t}, \boldsymbol{p} )\big|^{2}\,\mathrm{d}\boldsymbol{t}\le \omega_{k}^{\frac{k}{2}}\epsilon^{2} \biggl(\int _{\mathbb{R} ^{k}}\big|\psi ( \boldsymbol{t}, \boldsymbol{p})\big|^{\frac{2k}{k-2}}\,\mathrm{d}\boldsymbol{t} \biggr)^{\frac{k-2}{k}}. $$

Thus, for a.e. p∈Q _d−k

$$ \lim_{R\to\infty}\frac{1}{R^{2}}\int _{| \boldsymbol{t}|<R}\big|\psi ( \boldsymbol{t}, \boldsymbol{p} )\big|^{2}\,\mathrm{d}\boldsymbol{t}=0. $$

(C.3)

By Hölder inequality and Theorem A.1

$$\frac{1}{R^{2}}\int_{| \boldsymbol{t}|<R}\big|\psi ( \boldsymbol{t}, \boldsymbol{p})\big|^{2}\, \mathrm{d}\boldsymbol{t}\le \omega_{k}^{\frac{2}{k}} \biggl(\int _{| \boldsymbol{t}|<R}\big|\psi ( \boldsymbol{t}, \boldsymbol{p} )\big|^{\frac{2k}{k-2}}\,\mathrm{d}\boldsymbol{t} \biggr)^{\frac{k-2}{k}}\le C\int_{\mathbb{R} ^{k}}\big|\psi _{ \boldsymbol{t}}( \boldsymbol{t}, \boldsymbol{p})\big|^{2}\,\mathrm{d}\boldsymbol{t}. $$

By the Lebesgue dominated convergence theorem $\|\psi \|_{2,R}^{2}/R^{2}\to0$, as R→∞, since the function

$$\varPhi( \boldsymbol{p})=\int_{\mathbb{R} ^{k}}\big|\psi _{ \boldsymbol{t}}( \boldsymbol{t}, \boldsymbol{p})\big|^{2}\,\mathrm{d}\boldsymbol{t} $$

is integrable over Q _d−k .

The case k=2 is the most delicate. Let us define

$$\boldsymbol{c}_{R}( \boldsymbol{p})=\hspace{-8pt}\skew {33}-{\int _{| \boldsymbol{t}|<R}}\psi ( \boldsymbol{t}, \boldsymbol{p})\,\mathrm{d}\boldsymbol{t}. $$

Let R _n →∞ be a strictly monotonic sequence such that

$$\lim_{n\to\infty}\langle \boldsymbol{c}_{R_{n}} \rangle_{Q_{d-2}}= \boldsymbol{c} $$

for some vector $\boldsymbol{c}\in\mathbb{R} ^{m}$. We claim that

$$\lim_{R\to\infty}\big\| \boldsymbol{c}_{R}( \boldsymbol{p})-\langle \boldsymbol{c}_{R} \rangle _{Q_{d-2}}\big\|_{2}\,\mathrm{d}\boldsymbol{p}=0. $$

Indeed,

$$\big\| \boldsymbol{c}_{R}( \boldsymbol{p})-\langle \boldsymbol{c}_{R} \rangle_{Q_{d-2}} \big\|_{2}^{2}\le C\hspace{-8pt}\skew {33}-{\int _{| \boldsymbol{t}|<R}}\int_{Q_{d-2}}\big|\psi ( \boldsymbol{t}, \boldsymbol{p})-\langle\psi \rangle _{Q_{d-2}}( \boldsymbol{t})\big|^{2}\,\mathrm{d} \boldsymbol{p}\, \mathrm{d} \boldsymbol{t}. $$

Applying the Poincaré inequality for the inner integral we get

$$\big\| \boldsymbol{c}_{R}-\langle \boldsymbol{c}_{R} \rangle_{Q_{d-2}} \big\|_{2}^{2}\le C\hspace{-8pt}\skew {33}-{\int _{| \boldsymbol{t}|<R}}\int_{Q_{d-2}}\big|\psi _{ \boldsymbol{p}}( \boldsymbol{t}, \boldsymbol{p})\big|^{2}\,\mathrm{d} \boldsymbol{p}\,\mathrm{d} \boldsymbol{t} \to0,\quad \text{as } R\to\infty. $$

We now prove that

$$\lim_{n\to\infty}\frac{1}{R^{2}}\big\|\psi ( \boldsymbol{t}, \boldsymbol{p})- \boldsymbol{c} \big\|_{2,R}^{2}=0. $$

By triangle inequality we have

$$\big\|\psi ( \boldsymbol{t}, \boldsymbol{p})- \boldsymbol{c}\big\|_{2,R}\le\big\|\psi - \boldsymbol{c}_{R}( \boldsymbol{p}) \big\|_{2,R} +\big\| \boldsymbol{c}_{R}( \boldsymbol{p})-\langle \boldsymbol{c}_{R} \rangle_{Q_{d-2}}\big\|_{2,R} +\big\|\langle \boldsymbol{c}_{R} \rangle_{Q_{d-2}}- \boldsymbol{c}\big\|_{2,R}. $$

We compute

$$\big\|\langle \boldsymbol{c}_{R} \rangle_{Q_{d-2}}- \boldsymbol{c}\big\|_{2,R}^{2}=|Q_{d-2}| \pi R^{2}\big|\langle \boldsymbol{c}_{R} \rangle_{Q_{d-2}}- \boldsymbol{c}\big|^{2}. $$

Hence,

Therefore,

$$\lim_{R\to\infty}\frac{1}{R^{2}}\big\| \boldsymbol{c}_{R}( \boldsymbol{p})-\langle \boldsymbol{c}_{R} \rangle_{Q_{d-2}}\big\|_{2,R}^{2}=0. $$

Finally, we have

Using the uniform boundedness of ψ for the first term and the Poincaré inequality for the second term, we get

$$\big\|\psi - \boldsymbol{c}_{R}( \boldsymbol{p})\big\|_{2,R}^{2}\le C \epsilon^{2}R^{2}\|\psi - \boldsymbol{c}\| _{\infty}^{2} +R^{2}C_{\epsilon}\int_{Q_{d-2}}\int _{\{\epsilon R<| \boldsymbol{t}|<R\}}\big|\psi _{ \boldsymbol{t} }( \boldsymbol{t}, \boldsymbol{p})\big|^{2}\,\mathrm{d} \boldsymbol{t}\,\mathrm{d} \boldsymbol{p}. $$

Thus,

$$\mathop{\overline{\lim}}_{R\to\infty}\frac{1}{R^{2}}\big\|\psi - \boldsymbol{c} _{R}( \boldsymbol{p})\big\|_{2,R}^{2}\le C\epsilon^{2}\| \psi - \boldsymbol{c}\|_{\infty}^{2}. $$

The arbitrariness of ϵ>0 implies that

$$\lim_{R\to\infty}\frac{1}{R^{2}}\big\|\psi - \boldsymbol{c}_{R}( \boldsymbol{p}) \big\|_{2,R}^{2}=0. $$

Lemma C.1 is proved now.  □

Let R _n →∞ be the monotonically increasing sequence for which $\|\psi - \boldsymbol{c}\|_{2,R_{n}}/R_{n}\to0$, as n→∞. Let

$$\alpha(R)=\frac{\|\psi - \boldsymbol{c}\|_{2,R_{n}}}{R_{n}},\quad R_{n}\le R<R_{n+1}. $$

Then α(R)→0, as R→∞. By Lemma B.1 there exists a monotonically increasing function h(R) such that h(R)/R→0, as R→∞ and

$$\lim_{R\to\infty} \biggl(\frac{R}{h(R)} \biggr)\alpha(R)=0. $$

Hence,

$$\mathop{\underline{\lim}}_{R\to\infty} \biggl(\frac{\|\psi - \boldsymbol{c}\| _{2,R}}{h(R)} \biggr)^{2}\le \lim_{n\to\infty} \biggl(\frac{R_{n}}{h(R_{n})} \alpha(R_{n}) \biggr)^{2}=0. $$

The estimate (C.2) together with (3.17) now implies (3.18). Thus, we have proved that ${\mathcal{C}}_{k}\subset{\mathcal{S}}_{*}$ for any 1≤k≤d.

Step 4: Proof of the formula (3.24). We have

By the Riemann–Lebesgue lemma

Thus, in order to finish the proof of the theorem we need to show that

$$ \rho=\lim_{R\to\infty}\int_{| \boldsymbol{u}|\le1} \int_{| \boldsymbol{t}|\ge R\sqrt {1-| \boldsymbol{u}|^{2}}} W^{\circ}\bigl( \boldsymbol{F},\psi _{ \boldsymbol{t}}( \boldsymbol{t},R \boldsymbol{u}) \boldsymbol{R}+\psi _{ \boldsymbol{p}}( \boldsymbol{t},R \boldsymbol{u} ) \boldsymbol{Q}\bigr)\,\mathrm{d} \boldsymbol{t}\,\mathrm{d} \boldsymbol{u}=0. $$

(C.4)

Recall that $\phi \in W^{1,\infty}(\mathbb{R} ^{d};\mathbb{R} ^{m})$. Hence, there exists a number C>0 depending on ∥ϕ∥_1,∞, but independent of R such that

$$\rho\le C\mathop{\overline{\lim}}_{R\to\infty}\int_{| \boldsymbol{u}|\le 1} \int_{| \boldsymbol{t}|\ge R\sqrt{1-| \boldsymbol{u}|^{2}}} \bigl\{\big|\psi _{ \boldsymbol{t}}( \boldsymbol{t},R \boldsymbol{u})\big|^{2}+\big|\psi _{ \boldsymbol{p}}( \boldsymbol{t},R \boldsymbol{u})\big|^{2}\bigr\}\, \mathrm{d}\boldsymbol{t} \,\mathrm{d}\boldsymbol{u}. $$

For any ϵ∈(0,1) we have

$$\int_{| \boldsymbol{u}|\le1}\int_{| \boldsymbol{t}|\ge R\sqrt{1-| \boldsymbol{u}|^{2}}} \bigl\{\big|\psi _{ \boldsymbol{t}}( \boldsymbol{t},R \boldsymbol{u})\big|^{2}+\big|\psi _{ \boldsymbol{p}}( \boldsymbol{t},R \boldsymbol{u})\big|^{2}\bigr\}\,\mathrm{d}\boldsymbol{t} \,\mathrm{d}\boldsymbol{u}=T_{1}(R, \epsilon)+T_{2}(R,\epsilon), $$

where

If |u|≤1−ϵ and $| \boldsymbol{t}|\ge R\sqrt{1-| \boldsymbol{u}|^{2}}$ then $| \boldsymbol{t} |\ge R\sqrt{\epsilon(2-\epsilon)}$. In particular, $| \boldsymbol{t}|\ge\sqrt {(2-\epsilon)/\epsilon }$, if R>1/ϵ. Therefore, by the Riemann–Lebesgue lemma

Also, by the Riemann–Lebesgue lemma

We conclude that ρ=0, since

$$\lim_{\epsilon\to0}T_{1}^{\infty}(\epsilon)=\lim _{\epsilon\to 0}T_{2}^{\infty }(\epsilon)=0. $$

Appendix D: Proof of Theorem 3.14

We may assume, without loss of generality, that n=e ₁. By Lemma 3.2 in Müller and Šverák (2003), applied to the bounded domain Q=[0,1]^d, there exists a sequence of functions u _n (x) converging uniformly in Q to u ₀(x)=x ₁ a and such that ∥∇u _n ∥_∞ is a bounded sequence and for all 1≤j≤r

$$\lim_{n\to\infty}\big|\bigl\{ \boldsymbol{x}\in Q:\operatorname{dist}\bigl(\nabla \boldsymbol{u}_{n}( \boldsymbol{x}), \boldsymbol{H} _{j}\bigr)<1/n\bigr\}\big|=\lambda_{j}. $$

Let p _n (x) denote the function defined in the layer 0<x ₁<1, which is periodic with periods e ₂,…,e _d and equal to u _n (x) on Q. Finally, we let

$$\boldsymbol{v}_{n}( \boldsymbol{x})= \begin{cases} \boldsymbol{a},&\text{if }x_{1}\ge1,\\[4pt] \boldsymbol{0},&\text{if }x_{1}\le0,\\[4pt] \boldsymbol{p}_{n}( \boldsymbol{x}),&\text{if }0<x_{1}<1. \end{cases} $$

Clearly, v _n (x)→ϕ ₀(x) uniformly in $\mathbb{R} ^{d}$. However, the functions v _n (x) have jump discontinuities across the surfaces Γ _j,k ={x _j =k,0<x ₁<1}, j=2,…,d, $k\in\mathbb{Z}$, as well as the planes Π ₀={x ₁=0} and Π ₁={x ₁=1}. Let ϵ _n =∥v _n −ϕ ₀∥_∞. Then ϵ _n →0 as n→∞. Let

$$\varGamma=\varPi_{0}\cup\varPi_{1}\cup \Biggl(\bigcup _{j=1}^{d} \biggl(\bigcup _{k\in\mathbb{Z}}\varGamma_{j,k} \biggr) \Biggr) $$

be the entire singular set. When n is sufficiently large there is a $C^{\infty}(\mathbb{R} ^{d})$ function η _n (x) that is periodic with periods e ₂,…,e _d , which is equal to 0 on Γ and 1 on $\{ \boldsymbol{x}\in\mathbb{R} ^{d}:\operatorname{dist}( \boldsymbol{x},\varGamma)>\sqrt{\epsilon _{n}}\}$, and such that $\|\nabla\eta_{n}\|_{\infty}\le C/\sqrt{\epsilon_{n}}$. It follows that the function

$$\phi _{n}( \boldsymbol{x})=\bigl(1-\eta_{n}( \boldsymbol{x})\bigr)\phi _{0}( \boldsymbol{x})+\eta_{n}( \boldsymbol{x}) \boldsymbol{v} _{n}( \boldsymbol{x}) $$

is Lipschitz continuous with

$$\|\nabla\phi _{n}\|_{\infty}\le\|\nabla\phi _{0} \|_{\infty}+\| \nabla \boldsymbol{u}_{n}\|_{\infty}+C\sqrt{ \epsilon_{n}}. $$

Obviously, ϕ _n (x) converges uniformly to ϕ ₀(x). In addition ∇ϕ _n (x)=0 whenever $x_{1}<-\sqrt{\epsilon_{n}}$ or $x_{1}>1+\sqrt{\epsilon_{n}}$. It follows that (3.19) holds. Observe that ϕ _n (x) has periods e ₂,…,e _d , since both v _n (x) and η _n (x) do. Thus, $\psi _{n}\in{\mathcal{S}}_{1}^{0}$, where

$$\psi _{n}(t, \boldsymbol{p})=\phi _{n} \Biggl(t \boldsymbol{e}_{1}+\sum _{j=2}^{d}{p}_{j} \boldsymbol{e} _{j} \Biggr). $$

Hence, $\phi _{n}\in{\mathcal{C}}_{1}$. To finish the proof of the theorem we need to establish (3.29). This is a consequence of the formula (3.24) and the relation

Appendix E: Proof of Lemma 4.4

The lemma is best proved in the (t,p) variables, where instead of the [0,1]^d−k periodic field ψ we use a Q _d−k periodic field, which we denote ψ as well, so that

$$\phi ( \boldsymbol{x})=\psi ( \boldsymbol{R} \boldsymbol{x}, \boldsymbol{Q} \boldsymbol{x}),\qquad\psi ( \boldsymbol{t}, \boldsymbol{p})=\phi \bigl( \boldsymbol{R} ^{T} \boldsymbol{t}+ \boldsymbol{Q}^{T} \boldsymbol{p}\bigr). $$

In terms of ψ Eq. (4.12) becomes

$$ \begin{cases} \nabla_{ \boldsymbol{t}}\cdot( \boldsymbol{P}( \boldsymbol{F}+\psi _{ \boldsymbol{t}} \boldsymbol{R}+\psi _{ \boldsymbol{p}} \boldsymbol{Q}) \boldsymbol{R}^{T})+ \nabla_{ \boldsymbol{p}}\cdot( \boldsymbol{P}( \boldsymbol{F}+\psi _{ \boldsymbol{t}} \boldsymbol{R}+\psi _{ \boldsymbol{p}} \boldsymbol{Q}) \boldsymbol{Q} ^{T})= \boldsymbol{0},\\[4pt] \nabla_{ \boldsymbol{t}}\cdot( \boldsymbol{R} \boldsymbol{P}^{*}( \boldsymbol{F}+\psi _{ \boldsymbol{t}} \boldsymbol{R}+\psi _{ \boldsymbol{p}} \boldsymbol{Q} ) \boldsymbol{R}^{T})= \nabla_{ \boldsymbol{p}}\cdot(\psi _{ \boldsymbol{t}}^{T} \boldsymbol{P}( \boldsymbol{F}+\psi _{ \boldsymbol{t}} \boldsymbol{R}+\psi _{ \boldsymbol{p}} \boldsymbol{Q}) \boldsymbol{Q}^{T}),\\[4pt] \nabla_{ \boldsymbol{p}}\cdot( \boldsymbol{Q} \boldsymbol{P}^{*}( \boldsymbol{F}+\psi _{ \boldsymbol{t}} \boldsymbol{R}+\psi _{ \boldsymbol{p}} \boldsymbol{Q} ) \boldsymbol{Q}^{T})= \nabla_{ \boldsymbol{t}}\cdot(\psi _{ \boldsymbol{p}}^{T} \boldsymbol{P}( \boldsymbol{F}+\psi _{ \boldsymbol{t}} \boldsymbol{R}+\psi _{ \boldsymbol{p}} \boldsymbol{Q}) \boldsymbol{R}^{T}), \end{cases} $$

(E.1)

while relation (4.18) reads

$$ \begin{cases} \int_{\mathbb{R} ^{k}}\int_{Q_{d-k}} \boldsymbol{R}\widehat{ \boldsymbol{P} }^{*} \boldsymbol{R} ^{T}\,\mathrm{d} \boldsymbol{p}\,\mathrm{d} \boldsymbol{t}= \boldsymbol{0},\\[4pt] \int_{\mathbb{R} ^{k}}\int_{Q_{d-k}}\psi _{ \boldsymbol{p} }^{T}\widehat{ \boldsymbol{P}} \boldsymbol{R}^{T}\,\mathrm{d} \boldsymbol{p}\,\mathrm{d} \boldsymbol{t}= \boldsymbol{0}, \end{cases} $$

(E.2)

where

$$\widehat{ \boldsymbol{P}}=\widehat{ \boldsymbol{P}}(\psi _{ \boldsymbol{t}} \boldsymbol{R}+\psi _{ \boldsymbol{p}} \boldsymbol{Q})= \boldsymbol{P} ( \boldsymbol{F} +\psi _{ \boldsymbol{t}} \boldsymbol{R}+\psi _{ \boldsymbol{p}} \boldsymbol{Q})- \boldsymbol{P}( \boldsymbol{F}). $$

Replacing $\widehat{ \boldsymbol{P}}^{*}$ by its expression from (4.15) and using Eq. (E.1) we obtain

$$ \nabla_{ \boldsymbol{t}}\cdot\bigl( \boldsymbol{R}\widehat{ \boldsymbol{P}}^{*} \boldsymbol{R}^{T}\bigr)= \nabla_{ \boldsymbol{p}}\cdot\bigl(\psi _{ \boldsymbol{t}}^{T}\widehat{ \boldsymbol{P}} \boldsymbol{Q}^{T}\bigr) - \nabla_{ \boldsymbol{p}}\cdot\bigl( \boldsymbol{R} \boldsymbol{F}^{T} \boldsymbol{P}( \boldsymbol{F}+\psi _{ \boldsymbol{t}} \boldsymbol{R}+ \psi _{ \boldsymbol{p} } \boldsymbol{Q}) \boldsymbol{Q}^{T}\bigr), $$

(E.3)

since

$$\nabla_{ \boldsymbol{t}}\cdot\bigl( \boldsymbol{R} \boldsymbol{N}(\psi _{ \boldsymbol{t}} \boldsymbol{R}+\psi _{ \boldsymbol{p}} \boldsymbol{Q}) \boldsymbol{R}^{T}\bigr)= -\nabla_{ \boldsymbol{t}}\cdot\bigl( \psi _{ \boldsymbol{t}}^{T} \boldsymbol{P}( \boldsymbol{F})\bigr). $$

Let

Integrating (E.3) over Q _d−k and using the periodicity we obtain ∇ _t ⋅f ₁(t)=0. Similarly, integrating the third equation in (E.1) over Q _d−k we conclude that ∇ _t ⋅f ₂(t)=0. We estimate

$$\big|\widehat{ \boldsymbol{P}}^{*}\big|\le C\bigl(|\psi _{ \boldsymbol{t}}|^{2}+| \psi _{ \boldsymbol{p} }|^{2}\bigr),\qquad \big|\psi _{ \boldsymbol{p}}^{T} \widehat{ \boldsymbol{P}}\big|\le C\bigl(|\psi _{ \boldsymbol{t}}|^{2}+|\psi _{ \boldsymbol{p}}|^{2}\bigr), $$

since ψ _t and ψ _p are assumed to be uniformly bounded. Then $\phi \in{\mathcal{C}}_{k}$ implies that $\{ \boldsymbol{f}_{1}, \boldsymbol{f}_{2}\}\subset L^{1}(\mathbb{R} ^{k})$. The statement of the lemma follows from

$$\int_{\mathbb{R} ^{k}} \boldsymbol{f}_{1}( \boldsymbol{t})\,\mathrm{d}\boldsymbol{t}= \boldsymbol{0},\qquad\int _{\mathbb{R} ^{k}} \boldsymbol{f}_{2}( \boldsymbol{t})\,\mathrm{d}\boldsymbol{t}= \boldsymbol{0}, $$

which is a consequence of a simple observation that any L ¹ divergence-free vector field f(t) on $\mathbb{R} ^{k}$ must satisfy $\int_{\mathbb{R} ^{k}} \boldsymbol{f}\,\mathrm{d} \boldsymbol{t}= \boldsymbol{0}$. Indeed, f∈L ¹ implies that its Fourier transform $\widehat{ \boldsymbol{f}}(\omega )$ is continuous. ∇⋅f=0 implies that $\omega \cdot\widehat{ \boldsymbol{f}}(\omega )=0$ for any $\omega \in\mathbb{R} ^{k}$. Fixing ω≠0 we obtain

$$\frac{\omega \cdot\widehat{ \boldsymbol{f}}(\epsilon\omega )}{|\omega |}=\frac {\epsilon\omega \cdot\widehat{ \boldsymbol{f}}(\epsilon\omega )}{|\epsilon\omega |}=0. $$

Passing to the limit as ϵ→0 and using continuity of $\widehat{ \boldsymbol{f}}(\omega )$ we obtain $\omega \cdot\widehat{ \boldsymbol{f}}( \boldsymbol{0})=0$. Thus $\widehat{ \boldsymbol{f} }( \boldsymbol{0} )= \boldsymbol{0}$, since $\omega \in\mathbb{R} ^{k}\setminus\{ \boldsymbol{0}\}$ was arbitrary.

Appendix F: Proof of Theorem 4.6

When the subspace ${\mathcal{L}}$ described by R is fixed we can simplify our notation by regarding the first k components of x as t and the remaining components as p. Then ∇ϕ=[ψ _t ,ψ _p ]. We then have the corresponding splitting of P=[P ₁,P ₂] and

$$\boldsymbol{P}^{*}=W\bigl( \boldsymbol{F}+[\psi _{ \boldsymbol{t}},\psi _{ \boldsymbol{p}}]\bigr) \left [ \begin{array}{c@{\quad}c} { \boldsymbol{I}_{k}}&{ \boldsymbol{0}}\\[4pt] { \boldsymbol{0}}&{ \boldsymbol{I}_{d-k}} \end{array} \right ]- \left [ \begin{array}{c} \psi _{ \boldsymbol{t}}^{T}\\[4pt] \psi _{ \boldsymbol{p}}^{T} \end{array} \right ] [ \boldsymbol{P}_{1}, \boldsymbol{P}_{2}]=\left [ \begin{array}{c@{\quad}c} { \boldsymbol{P}_{1}^{*}}&{-\psi _{ \boldsymbol{t}}^{T} \boldsymbol{P}_{2}}\\[4pt] {-\psi _{ \boldsymbol{p}}^{T} \boldsymbol{P}_{1}}&{ \boldsymbol{P}_{2}^{*}} \end{array} \right ]. $$

Similarly, splitting the t and p components we have

where

$$\widehat{ \boldsymbol{P}}^{*}_{1}=W^{\circ}\bigl( \boldsymbol{F},[\psi _{ \boldsymbol{t}},\psi _{ \boldsymbol{p}}]\bigr) I _{k}-\psi _{ \boldsymbol{t}}^{T}\widehat{ \boldsymbol{P}}_{1}, \qquad \widehat{ \boldsymbol{P}}^{*}_{2}=W^{\circ}\bigl( \boldsymbol{F},[\psi _{ \boldsymbol{t}}, \psi _{ \boldsymbol{p}}]\bigr) \boldsymbol{I} _{d-k}-\psi _{ \boldsymbol{p}}^{T} \widehat{ \boldsymbol{P}}_{2}. $$

Next we use the generalized Clapeyron theorem (Grabovsky and Truskinovsky 2013a) for $\widehat{W}( \boldsymbol{H})$:

$$ \int_{| \boldsymbol{t}|\le R}\int_{Q_{d-k}} \widehat{W}(\nabla\phi )\,\mathrm{d} \boldsymbol{p}\,\mathrm{d} \boldsymbol{t} =\frac{1}{d}(T_{1}+T_{2}), $$

(F.1)

where

Next we observe that

$$\int_{\partial Q_{d-k}}\psi _{ \boldsymbol{t}}^{T}\widehat{ \boldsymbol{P}}_{2} \boldsymbol{n}_{ \boldsymbol{p} }\,\mathrm{d}S( \boldsymbol{p} )= \boldsymbol{0},\qquad \int _{\partial Q_{d-k}}(\widehat{ \boldsymbol{P}}_{2} \boldsymbol{n}_{ \boldsymbol{p}},\psi )\, \mathrm{d}S( \boldsymbol{p})=0, $$

since $\psi _{ \boldsymbol{t}}^{T}\widehat{ \boldsymbol{P}}_{2}$ and $(\widehat{ \boldsymbol{P} }_{2})^{T}\psi$ are Q _d−k -periodic. By the divergence theorem we obtain

$$\int_{\partial Q_{d-k}}\bigl(\widehat{ \boldsymbol{P}}_{2}^{*} \boldsymbol{n}_{ \boldsymbol{p}}, \boldsymbol{p}\bigr)\,\mathrm{d}S( \boldsymbol{p} )=\int_{Q_{d-k}}\bigl \{ \bigl(\nabla_{ \boldsymbol{p}}\cdot\widehat{ \boldsymbol{P}}_{2}^{*}, \boldsymbol{p} \bigr)+\operatorname {Tr}\widehat{ \boldsymbol{P}}_{2}^{*}\bigr\} \, \mathrm{d}\boldsymbol{p}. $$

The Noether–Eshelby equation gives

$$\nabla_{ \boldsymbol{p}}\cdot\widehat{ \boldsymbol{P}}_{2}^{*}= \nabla_{ \boldsymbol{t}}\cdot\bigl(\psi _{ \boldsymbol{p} }^{T}\widehat{ \boldsymbol{P}}_{1}\bigr). $$

We also compute

$$\operatorname{Tr}\widehat{ \boldsymbol{P}}_{2}^{*}=(d-k)\widehat {W}- \operatorname{Tr}\bigl(\psi _{ \boldsymbol{p}}^{T}\widehat{ \boldsymbol{P}}_{2}\bigr). $$

Hence, we obtain

$$T_{2}=(d-k)\int_{| \boldsymbol{t}|\le R}\int_{Q_{d-k}} \widehat{W}(\nabla\phi )\,\mathrm{d}\boldsymbol{p} \,\mathrm{d}\boldsymbol{t}+T_{2}', $$

where

$$T_{2}'=\int_{| \boldsymbol{t}|=R}\int _{Q_{d-k}}\bigl(\psi _{ \boldsymbol{p}}^{T}\widehat{ \boldsymbol{P} }_{1} \boldsymbol{n}_{ \boldsymbol{t}}, \boldsymbol{p}\bigr)\,\mathrm{d}\boldsymbol{p}\,\mathrm{d}S( \boldsymbol{t}) -\int _{| \boldsymbol{t}|\le R}\int_{Q_{d-k}}\operatorname{Tr}\bigl( \psi _{ \boldsymbol{p} }^{T}\widehat{ \boldsymbol{P}}_{2}\bigr)\,\mathrm{d} \boldsymbol{p} \,\mathrm{d} \boldsymbol{t}. $$

Substituting this back into (F.1) we obtain

$$\int_{| \boldsymbol{t}|\le R}\int_{Q_{d-k}}\widehat{W}(\nabla \phi )\,\mathrm{d}\boldsymbol{p} \,\mathrm{d}\boldsymbol{t}=\frac{1}{k}\bigl(\widehat{T}_{1}(R)+ \widehat{T}_{2}(R)\bigr), $$

where

We observe that due to (4.17)

$$\lim_{R\to\infty}\widehat{T}_{2}(R)=-\operatorname{Tr} \biggl(\int_{Y_{k}}\psi _{ \boldsymbol{p}}^{T}\widehat{ \boldsymbol{P}}_{2}\,\mathrm{d} \boldsymbol{p}\,\mathrm{d} \boldsymbol{t} \biggr)=0. $$

To finish the proof of the theorem we need to show that $\widehat {T}_{1}(R)\to0$, as R→∞.

We have $|\widehat{ \boldsymbol{P}}|\le C|\nabla\phi |$ and $|\widehat{ \boldsymbol{P} }^{*}|\le C|\nabla\phi |^{2}$, due to the uniform boundedness of ∇ϕ, where the constant C depends on ϕ, but is independent of R. Thus, $|\widehat {T}_{1}(R)|\le CK(R)$ for a.e. R>1, where

$$K(R)=\int_{| \boldsymbol{t}|=R}\int_{Q_{d-k}}\bigl\{R|\nabla \phi |^{2}+|\phi - \boldsymbol{c} ||\nabla\phi |\bigr\}\,\mathrm{d}\boldsymbol{p}\,\mathrm{d}S( \boldsymbol{t}), $$

where $\boldsymbol{c}\in\mathbb{R} ^{m}$ can be chosen arbitrarily. We have, after an application of the Cauchy–Schwartz inequality,

If k=1 or k=2 then the boundedness of ϕ implies that

$$ \lim_{R\to\infty}\frac{1}{R}\int _{R}^{2R}K(r)\,\mathrm{d}r=0. $$

(F.2)

If k≥3 then Lemma C.1 guarantees the choice of the constant $\boldsymbol{c}\in\mathbb{R} ^{m}$ such that (F.2) holds. Therefore,

$$\mathop{\underline{\lim}}_{R\to\infty}K(R)=0. $$

Hence,

$$\biggl \vert \int_{Y}\widehat{W}(\nabla\phi )\, \mathrm{d}\boldsymbol{x}\biggr \vert \le C\mathop{\underline{\lim}}_{R\to\infty}K(R)=0. $$

The theorem is proved.

Appendix G: Proof of Lemma 5.5

We will prove that ${\mathcal{G}}={\mathcal{G}}_{0}$, where

$${\mathcal{G}}_{0}=\bigl\{ \boldsymbol{A}=\operatorname{diag}(A_{1}, \ldots,A_{k}),\ A_{i}>0,\ i=1,\ldots,k,\ \operatorname{Tr} \boldsymbol{A}=1\bigr\}. $$

If we write 〈Γ(n)〉 _a in components,

$$\bigl(\bigl\langle\varGamma( \boldsymbol{n}) \bigr\rangle_{ \boldsymbol{a}} \bigr)_{ij}=\hspace{-8pt}\skew {33}-{\int _{\mathbb{S}^{k-1}}}\frac {{n}_{i}{n}_{j}}{ a_{i}a_{j}\sum_{s=1}^{k}a_{s}^{-2}{n}_{s}^{2}}\, \mathrm{d}S( \boldsymbol{n}), $$

we immediately see that ${\mathcal{G}}\subset{\mathcal{G}}_{0}$.

To each $\boldsymbol{a}=(a_{1},\ldots,a_{k})\in\mathbb{R} ^{k}$ we associate (without relabeling) the diagonal matrix $\boldsymbol{a}=\operatorname{diag}(a_{1},\ldots,a_{k})$. Let $\Delta=\{(a_{1},\ldots,a_{k}):a_{i}> 0,\ \sum_{i=1}^{k}a_{i}=1\}$. Then the smooth map F(a)=〈Γ(n)〉 _a maps Δ into itself. To prove the reverse inclusion ${\mathcal{G}}_{0}\subset{\mathcal{G}}$ we need to show that the map F:Δ→Δ is surjective. We first show that the differential of the map F is non-degenerate. This implies, via the inverse function theorem, that F(Δ) is an open subset of Δ.

In order to simplify the calculation we first change variables $\boldsymbol{b}= \boldsymbol{a}^{-1}/\operatorname{Tr}( \boldsymbol{a}^{-1})$. Then

$$F( \boldsymbol{a})=G \biggl(\frac{ \boldsymbol{a}^{-1}}{\operatorname{Tr} \boldsymbol{a}^{-1}} \biggr),\qquad G( \boldsymbol{b})=\hspace{-8pt}\skew {33}-{\int _{ \mathbb{S}^{k-1}}}\frac{ \boldsymbol{b} \boldsymbol{n}\otimes \boldsymbol{b} \boldsymbol{n}}{| \boldsymbol{b} \boldsymbol{n} |^{2}}\,\mathrm{d}S( \boldsymbol{n}). $$

We compute

$$dF( \boldsymbol{a})\eta =-dG \biggl(\frac{ \boldsymbol{a}^{-1}}{\operatorname{Tr} \boldsymbol{a} ^{-1}} \biggr) \frac{ \boldsymbol{a}^{-1}\operatorname{Tr}( \boldsymbol{a}^{-1}\eta \boldsymbol{a}^{-1})- \boldsymbol{a} ^{-1}\eta \boldsymbol{a} ^{-1}\operatorname{Tr} \boldsymbol{a}^{-1}}{(\operatorname{Tr} \boldsymbol{a}^{-1})^{2}}, $$

where η is a diagonal trace-free matrix. If

$$\boldsymbol{a}^{-1}\operatorname{Tr}\bigl( \boldsymbol{a}^{-1}\eta \boldsymbol{a}^{-1} \bigr)- \boldsymbol{a}^{-1}\eta \boldsymbol{a} ^{-1}\operatorname{Tr} \boldsymbol{a} ^{-1}= \boldsymbol{0} $$

then η=λ a for some scalar λ. Taking traces we conclude that λ=0. Hence, the map

$$\eta \mapsto\frac{ \boldsymbol{a}^{-1}\operatorname{Tr}( \boldsymbol{a}^{-1}\eta \boldsymbol{a} ^{-1})- \boldsymbol{a} ^{-1}\eta \boldsymbol{a}^{-1}\operatorname{Tr} \boldsymbol{a}^{-1}}{(\operatorname{Tr} \boldsymbol{a} ^{-1})^{2}} $$

is a non-degenerate linear transformation on the space of diagonal trace-free matrices. Hence, dF is non-degenerate if and only if dG is non-degenerate. We compute explicitly

$$dG\eta =2\hspace{-8pt}\skew {33}-{\int _{\mathbb{S}^{k-1}}} \biggl\{\frac{\eta \boldsymbol{n}\odot \boldsymbol{b} \boldsymbol{n} }{| \boldsymbol{b} \boldsymbol{n}|^{2}}- \frac{ \boldsymbol{b} \boldsymbol{n}\otimes \boldsymbol{b} \boldsymbol{n}}{| \boldsymbol{b} \boldsymbol{n}|^{4}}( \boldsymbol{b} \boldsymbol{n},\eta \boldsymbol{n} ) \biggr\} \,\mathrm{d}S( \boldsymbol{n}), $$

where η is diagonal and $\operatorname{Tr}\eta =0$. Suppose dGη=0 for some non-zero η. The lemma will be proved if we show that only for η=0. If this is not the case then we have $\operatorname{Tr}(\eta \boldsymbol{b}^{-1}dG\eta )=0$. We compute (using commutativity of the diagonal matrix multiplication)

$$\operatorname{Tr}\bigl(\eta \boldsymbol{b}^{-1}dG\eta \bigr)=2\hspace{-8pt}\skew {33}-{\int _{ \mathbb{S}^{k-1}}}\frac {|\eta \boldsymbol{n} |^{2}| \boldsymbol{b} \boldsymbol{n}|^{2}-( \boldsymbol{b} \boldsymbol{n},\eta \boldsymbol{n})^{2}}{| \boldsymbol{b} \boldsymbol{n}|^{4}}\,\mathrm{d}S( \boldsymbol{n}). $$

The Cauchy–Schwartz inequality implies that the integrand is non-negative. For it to be zero we would need η n=α(n)bn for almost all $\boldsymbol{n}\in\mathbb{S}^{k-1}$. The equivalent relation b ⁻¹ η n=α(n)n means that every unit vector is an eigenvector of b ⁻¹ η. Hence, there is a constant α ₀ such that η=α ₀ b. Taking the trace, we obtain α ₀=0 and the non-degeneracy of dG is proved.

The lemma will follow, if we show that if a _n →a ^∘∈∂Δ and F(a _n )→f ^∘, as n→∞ then f ^∘∈∂Δ. Let 1≤i,j≤k be a pair of indices such that $a^{\circ}_{i}=0$ and $a^{\circ}_{j}\neq0$. Such a pair exists, since a ^∘∈∂Δ. We claim that $f^{\circ }_{j}=0$, finishing the proof of the lemma. We estimate

$$F_{j}( \boldsymbol{a})\le\hspace{-8pt}\skew {33}-{\int _{\mathbb{S}^{k-1}}}\frac {a_{j}^{-2}n_{j}^{2}}{a_{j}^{-2}n_{j}^{2}+a_{i}^{-2}n_{i}^{2}}\, \mathrm{d}S( \boldsymbol{n})= \hspace{-8pt}\skew {33}-{\int _{\mathbb{S}^{k-1}}}\frac {a_{i}^{2}n_{j}^{2}}{a_{i}^{2}n_{j}^{2}+a_{j}^{2}n_{i}^{2}}\,\mathrm{d}S( \boldsymbol{n}). $$

Now, by the Lebesgue bounded convergence theorem,

$$f^{\circ}_{j}=\lim_{n\to\infty}F_{j}( \boldsymbol{a}_{n})=0. $$

Lemma 5.5 is now proved.