Legendre-Hadamard Conditions for Two-Phase Configurations

Abstract

We generalize the classical Legendre-Hadamard conditions by using quadratic extensions of the energy around a set of two configurations and obtain new algebraic necessary conditions for nonsmooth strong local minimizers. The implied bounds of stability are easily accessible as we illustrate on a nontrivial example where quasiconvexification is unknown.

Appendices

Appendix A: Geometry of the Jump Set

1.1 A.1 Normal to the Jump Set

Suppose $\boldsymbol{F}_{-}(s)\in\mathfrak{J}_{-}$ is a smooth curve passing through $\boldsymbol{F}_{-}$ at $s=0$. We would like to show by the implicit function theorem that there are smooth curves $\boldsymbol{a}(s)$, $\boldsymbol{n}(s)$ passing through $\boldsymbol{a}$ and $\boldsymbol{n}$ at $s=0$ and satisfying

$$ \textstyle\begin{cases} (W_{\boldsymbol{F}}(\boldsymbol{F}_{-}(s)+\boldsymbol{a}(s)\otimes \boldsymbol{n}(s))-W_{\boldsymbol{F}}(\boldsymbol {F}_{-}(s)))\boldsymbol{n}(s)=\mathbf{0},\\ (W_{\boldsymbol{F}}(\boldsymbol{F}_{-}(s)+\boldsymbol{a}(s)\otimes \boldsymbol{n}(s))-W_{\boldsymbol{F}}(\boldsymbol {F}_{-}))^{T}\boldsymbol{a}(s)=\mathbf{0},\\ W(\boldsymbol{F}_{-}(s)+\boldsymbol{a}(s)\otimes\boldsymbol {n}(s))-W(\boldsymbol{F}_{-}(s))-W_{\boldsymbol{F}}(\boldsymbol {F}_{-}(s))\boldsymbol{n}(s)\cdot\boldsymbol{a}(s)=0. \end{cases} $$

(A.1)

Differentiating (A.1) with respect to $s$ at $s=0$ we obtain

$$ \textstyle\begin{cases} \boldsymbol{A}_{+}\dot{\boldsymbol{a}}+(\boldsymbol{B}_{+}+{}[ \!{}[\boldsymbol{P} ]\!])\dot{\boldsymbol {n}}+({}[\!{}[\mathsf{L} ]\!]\dot{\boldsymbol {F}}_{-})\boldsymbol{n}=\mathbf{0},\\ (\boldsymbol{B}_{+}+{}[\!{}[\boldsymbol{P} ]\! ])^{T}\dot{\boldsymbol{a}}+\boldsymbol{A}_{+}^{*}\dot {\boldsymbol{n}}+({}[\!{}[\mathsf{L} ]\!]\dot {\boldsymbol{F}}_{-})^{T}\boldsymbol{a}=\mathbf{0},\\ -\frac{1}{2}(\boldsymbol{A}_{+}\boldsymbol{a},\dot{\boldsymbol {a}})-\frac{1}{2}(\boldsymbol{A}_{+}^{*}\boldsymbol{n},\dot {\boldsymbol{n}})+({}[\!{}[\boldsymbol{P} ]\!] -\lbrace\!\!\lbrace\mathsf{L}\rbrace\!\!\rbrace{}[\!{}[ \boldsymbol{F} ]\!],\dot{\boldsymbol{F}}_{-})=0. \end{cases} $$

(A.2)

We now solve the first two equations in (A.2) for $\dot {\boldsymbol{a}}$, $\dot{\boldsymbol{n}}$. We obtain the system

$$ \mathbb{ C}_{+}\left [ \textstyle\begin{array}{c} {\dot{\boldsymbol{a}}}\\ {\dot{\boldsymbol{n}}} \end{array}\displaystyle \right ]=-\left [ \textstyle\begin{array}{c} {({}[\!{}[\mathsf{L} ]\!]\dot{\boldsymbol {F}}_{-})\boldsymbol{n}}\\ {({}[\!{}[\mathsf{L} ]\! ]\dot{\boldsymbol{F}}_{-})^{T}\boldsymbol{a}} \end{array}\displaystyle \right ], $$

(A.3)

where $\mathbb{ C}_{\pm}$ is given by (3.1). Observe that

$$\left [ \textstyle\begin{array}{c} {({}[\!{}[\mathsf{L} ]\!]\dot{\boldsymbol {F}})\boldsymbol{n}}\\ {({}[\!{}[\mathsf{L} ]\! ]\dot{\boldsymbol{F}}_{-})^{T}\boldsymbol{a}} \end{array}\displaystyle \right ]\cdot \left [ \textstyle\begin{array}{c} {\boldsymbol{a}}\\ {-\boldsymbol{n}} \end{array}\displaystyle \right ]=0. $$

By the non-degeneracy assumption the right-hand side of (A.3) is orthogonal to the null-space of $\mathbb{ C}_{+}$. Thus, (A.3) has a solution. This solution is determined uniquely by the property $\dot{\boldsymbol{n}}\cdot\boldsymbol{n}=0$, since the vector $[\mathbf{0},\boldsymbol{n}]$ does not belong to $(\mathbb{ R}[\boldsymbol{a},-\boldsymbol{n}])^{\perp}$. Thus, there is a well defined operator denoted $\mathbb{ C}_{+}^{-1}:(\mathbb{ R}[\boldsymbol{a},-\boldsymbol {n}])^{\perp}\to(\mathbb{ R}[\mathbf{0},\boldsymbol{n}])^{\perp}$ with the property $\mathbb{ C}_{+}(\boldsymbol{a},\boldsymbol{n})(\mathbb { C}_{+}^{-1}\boldsymbol{z})=\boldsymbol{z}$ for any $\boldsymbol{z}\in(\mathbb{ R}[\boldsymbol{a},-\boldsymbol {n}])^{\perp}$. We can write

$$\left [ \textstyle\begin{array}{c} {\dot{\boldsymbol{a}}}\\ {\dot{\boldsymbol{n}}} \end{array}\displaystyle \right ]=-\mathbb{ C}_{+}^{-1}\left [ \textstyle\begin{array}{c} {({}[\!{}[\mathsf{L} ]\!]\dot{\boldsymbol {F}}_{-})\boldsymbol{n}}\\ {({}[\!{}[\mathsf{L} ]\! ]\dot{\boldsymbol{F}}_{-})^{T}\boldsymbol{a}} \end{array}\displaystyle \right ]. $$

Substituting this into the third equation in (A.2) we obtain

$$ \frac{1}{2}\left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{+}(\boldsymbol{n})\boldsymbol{a}}\\ {\boldsymbol {A}_{+}^{*}(\boldsymbol{a})\boldsymbol{n}} \end{array}\displaystyle \right ]\cdot\mathbb{ C}_{+}^{-1} \left [ \textstyle\begin{array}{c} {({}[\!{}[\mathsf{L} ]\!]\dot{\boldsymbol {F}}_{-})\boldsymbol{n}}\\ {({}[\!{}[\mathsf{L} ]\! ]\dot{\boldsymbol{F}}_{-})^{T}\boldsymbol{a}} \end{array}\displaystyle \right ]+ \bigl({}\bigl[\!{}[\boldsymbol{P} ]\!\bigr]-\bigl\lbrace \! \! \lbrace\mathsf{L}\rbrace\!\!\bigr\rbrace {}\bigl[\!{}[\boldsymbol {F} ]\!\bigr], \dot{\boldsymbol{F}}_{-}\bigr)=0. $$

(A.4)

Observe that $\mathbb{ C}_{+}$ is a symmetric matrix and that

$$\left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{+}\boldsymbol{a}}\\ {\boldsymbol {A}_{+}^{*}\boldsymbol{n}} \end{array}\displaystyle \right ]\cdot \left [ \textstyle\begin{array}{c} {\boldsymbol{a}}\\ {-\boldsymbol{n}} \end{array}\displaystyle \right ]=0. $$

Therefore,

$$\left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{+}\boldsymbol{a}}\\ {\boldsymbol {A}_{+}^{*}\boldsymbol{n}} \end{array}\displaystyle \right ]\cdot\mathbb{ C}_{+}^{-1} \left [ \textstyle\begin{array}{c} {({}[\!{}[\mathsf{L} ]\!]\dot{\boldsymbol {F}}_{-})\boldsymbol{n}}\\ {({}[\!{}[\mathsf{L} ]\! ]\dot{\boldsymbol{F}}_{-})^{T}\boldsymbol{a}} \end{array}\displaystyle \right ]= \mathbb{ C}_{+}^{-1}\left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{+}\boldsymbol{a}}\\ {\boldsymbol {A}_{+}^{*}\boldsymbol{n}} \end{array}\displaystyle \right ]\cdot \left [ \textstyle\begin{array}{c} {({}[\!{}[\mathsf{L} ]\!]\dot{\boldsymbol {F}}_{-})\boldsymbol{n}}\\ {({}[\!{}[\mathsf{L} ]\! ]\dot{\boldsymbol{F}}_{-})^{T}\boldsymbol{a}} \end{array}\displaystyle \right ]. $$

We easily compute

$$\mathbb{ C}_{+}^{-1}\left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{+}\boldsymbol{a}}\\ {\boldsymbol {A}_{+}^{*}\boldsymbol{n}} \end{array}\displaystyle \right ]=\left [ \textstyle\begin{array}{c} {\boldsymbol{a}}\\ {\mathbf{0}} \end{array}\displaystyle \right ]. $$

Hence, (A.4) becomes

$$\biggl(\frac{1}{2}{}[\!{}[\mathsf{L} ]\!] {}[\!{}[ \boldsymbol{F} ]\!]+{}[\!{}[ \boldsymbol{P} ]\!]-\lbrace\!\! \lbrace\mathsf {L}\rbrace\!\!\rbrace{}[\!{}[\boldsymbol{F} ]\! ], \dot{\boldsymbol{F}}_{-} \biggr)=0. $$

In other words, $(\boldsymbol{N}_{-},\dot{\boldsymbol{F}}_{-})=0$, where

$$ \boldsymbol{N}_{-}=\mathsf{L}_{-}{}[ \!{}[\boldsymbol{F} ]\!]-{}[\!{}[\boldsymbol{P} ]\!]. $$

(A.5)

1.2 A.2 Foliation of the Simple Laminate Region

Now, suppose that $\boldsymbol{F}_{-}$ varies over an open subset $G$ of $\mathfrak{J}_{-}$ that is sufficiently small, so that due to the non-degeneracy assumption, the functions $\boldsymbol{a}(\boldsymbol{F}_{-})$ and $\boldsymbol {n}(\boldsymbol{F}_{-})$ are well-defined and smooth on $G$. Let us show that the line segments joining $\boldsymbol{F}_{-}$ and $\boldsymbol{F}_{+}=\boldsymbol{F}_{-}+\boldsymbol{a}(\boldsymbol {F}_{-})\otimes\boldsymbol{n}(\boldsymbol{F}_{-})$ foliate an open subset of $\mathbb{ M}$, as $\boldsymbol{F}_{-}$ ranges over $G$. We have a map $\boldsymbol{F}:G\times[0,1]\to \mathbb{ M}$ given by

$$ \boldsymbol{F}=\boldsymbol{F}_{-}+t\boldsymbol{a}( \boldsymbol {F}_{-})\otimes\boldsymbol{n}(\boldsymbol{F}_{-}). $$

(A.6)

If $G$ is sufficiently small then the only possibility that the map $\boldsymbol{F}$ is not injective is for $\boldsymbol{F}$ to be locally non-injective, i.e., for $d\boldsymbol{F}(\boldsymbol{F}_{-},t)$ to be singular for some $t\in[0,1]$. If this is the case, then there exists $\dot{\boldsymbol{F}}_{-}\in T_{\boldsymbol{F}_{-}}\mathfrak{J}_{-}$ and $t\in[0,1]$ such that

$$\dot{\boldsymbol{F}}_{-}+t(\dot{\boldsymbol{a}}\otimes\boldsymbol {n}+\boldsymbol{a}\otimes\dot{\boldsymbol{n}})=\boldsymbol {a}\otimes \boldsymbol{n}, $$

where $[\dot{\boldsymbol{a}},\dot{\boldsymbol{n}}]$ solves (A.3). If (A.6) is satisfied then there exist $\boldsymbol{u}\in\mathbb{ R}^{m}$ and $\boldsymbol{\eta}\in\mathbb{ R}^{d}$ such that

$$ \dot{\boldsymbol{F}}_{-}=\boldsymbol{u}\otimes \boldsymbol {n}+\boldsymbol{a}\otimes\boldsymbol{\eta},\qquad \boldsymbol{u}+t \dot{\boldsymbol{a}}=\lambda\boldsymbol{a},\qquad \boldsymbol{\eta}+t\dot{ \boldsymbol{n}}=(1-\lambda)\boldsymbol{n} $$

(A.7)

for some $\lambda\in\mathbb{ R}$. Substituting (A.7) into (A.3) we obtain

$$\widetilde{\mathbb{ C}}_{t}\left [ \textstyle\begin{array}{c} {\boldsymbol{u}}\\ {\boldsymbol{\eta}} \end{array}\displaystyle \right ]=\left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{+}(\boldsymbol{n})\boldsymbol{a}}\\ {\boldsymbol {A}_{+}^{*}(\boldsymbol{a})\boldsymbol{n}} \end{array}\displaystyle \right ],\quad \widetilde{ \mathbb{ C}}_{t}=(1-t)\mathbb{ C}_{+}+t\mathbb{ C}_{-}. $$

The solution $[\boldsymbol{u},\boldsymbol{\eta}]$ is determined up to a multiple of $[\boldsymbol{a},-\boldsymbol{n}]$. However, any choice of solution $[\boldsymbol{u},\boldsymbol{\eta}]$ gives one and the same value $\dot{\boldsymbol{F}}_{-}$ in (A.7). We therefore, write

$$\left [ \textstyle\begin{array}{c} {\boldsymbol{u}}\\ {\boldsymbol{\eta}} \end{array}\displaystyle \right ]=\widetilde{\mathbb{ C}}_{t}^{-1}\left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{+}(\boldsymbol{n})\boldsymbol{a}}\\ {\boldsymbol {A}_{+}^{*}(\boldsymbol{a})\boldsymbol{n}} \end{array}\displaystyle \right ], $$

where the inverse of the positive semidefinite matrix $\widetilde {\mathbb{ C}}_{t}$ is computed on $(\mathbb{ R}[\boldsymbol{a},-\boldsymbol{n}])^{\perp}$, where it is positive definite by assumption of local stability. The resulting $\dot{\boldsymbol{F}}_{-}$ has to belong to $T_{\boldsymbol{F}_{-}}\mathfrak{J}_{-}$, therefore $(\dot {\boldsymbol{F}}_{-},\boldsymbol{N}_{-})=0$, i.e.,

$$ q(t)=\left (\widetilde{\mathbb{ C}}_{t}^{-1} \left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{+}(\boldsymbol{n})\boldsymbol{a}}\\ {\boldsymbol {A}_{+}^{*}(\boldsymbol{a})\boldsymbol{n}} \end{array}\displaystyle \right ], \left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{-}(\boldsymbol{n})\boldsymbol{a}}\\ {\boldsymbol {A}_{-}^{*}(\boldsymbol{a})\boldsymbol{n}} \end{array}\displaystyle \right ] \right )=0. $$

(A.8)

Let us show that (A.8) is impossible. We observe that

$$q(0)=\bigl(\boldsymbol{A}_{-}(\boldsymbol{n})\boldsymbol{a}, \boldsymbol {a}\bigr)>0,\qquad q(1)=\bigl(\boldsymbol{A}_{+}( \boldsymbol{n})\boldsymbol {a},\boldsymbol{a}\bigr)>0. $$

For any $t\in(0,1)$ we have

$$\widetilde{\mathbb{ C}}_{t}>t\mathbb{ C_{-}}>0, $$

where the inequality is strict on $(\mathbb{ R}[\boldsymbol {a},-\boldsymbol{n}])^{\perp}$. Therefore,

$$ t\widetilde{\mathbb{ C}}_{t}^{-1}< \mathbb{ C}_{-}^{-1}. $$

(A.9)

We have

$$\widetilde{\mathbb{ C}}_{t}\left [ \textstyle\begin{array}{c} {\boldsymbol{a}}\\ {\boldsymbol{n}} \end{array}\displaystyle \right ]=2(1-t)\left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{+}(\boldsymbol{n})\boldsymbol{a}}\\ {\boldsymbol {A}_{+}^{*}(\boldsymbol{a})\boldsymbol{n}} \end{array}\displaystyle \right ] +2t\left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{-}(\boldsymbol{n})\boldsymbol{a}}\\ {\boldsymbol {A}_{-}^{*}(\boldsymbol{a})\boldsymbol{n}} \end{array}\displaystyle \right ]. $$

Therefore,

$$\left [ \textstyle\begin{array}{c} {\boldsymbol{a}}\\ {\boldsymbol{n}} \end{array}\displaystyle \right ]-\lambda \left [ \textstyle\begin{array}{c} {\boldsymbol{a}}\\ {-\boldsymbol{n}} \end{array}\displaystyle \right ]= 2(1-t)\widetilde{\mathbb{ C}}_{t}^{-1}\left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{+}(\boldsymbol{n})\boldsymbol{a}}\\ {\boldsymbol {A}_{+}^{*}(\boldsymbol{a})\boldsymbol{n}} \end{array}\displaystyle \right ] +2t\widetilde{\mathbb{ C}}_{t}^{-1}\left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{-}(\boldsymbol{n})\boldsymbol{a}}\\ {\boldsymbol {A}_{-}^{*}(\boldsymbol{a})\boldsymbol{n}} \end{array}\displaystyle \right ], $$

where $\lambda\in\mathbb{ R}$ is chosen such that the left-hand side is in $(\mathbb{ R}[\boldsymbol{a},-\boldsymbol{n}])^{\perp}$. Taking inner product with $[\boldsymbol{A}_{-}(\boldsymbol{n})\boldsymbol{a},\boldsymbol {A}_{-}^{*}(\boldsymbol{a})\boldsymbol{n}]$ we obtain

$$\bigl(\boldsymbol{A}_{-}(\boldsymbol{n})\boldsymbol{a},\boldsymbol{a} \bigr)=(1-t)q(t)+ t\left (\widetilde{\mathbb{ C}}_{t}^{-1} \left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{-}(\boldsymbol{n})\boldsymbol{a}}\\ {\boldsymbol {A}_{-}^{*}(\boldsymbol{a})\boldsymbol{n}} \end{array}\displaystyle \right ], \left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{-}(\boldsymbol{n})\boldsymbol{a}}\\ {\boldsymbol {A}_{-}^{*}(\boldsymbol{a})\boldsymbol{n}} \end{array}\displaystyle \right ] \right ). $$

By (A.9) we have

$$\bigl(\boldsymbol{A}_{-}(\boldsymbol{n})\boldsymbol{a},\boldsymbol{a} \bigr)< (1-t)q(t)+ \left (\mathbb{ C}_{-}^{-1}\left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{-}(\boldsymbol{n})\boldsymbol{a}}\\ {\boldsymbol {A}_{-}^{*}(\boldsymbol{a})\boldsymbol{n}} \end{array}\displaystyle \right ], \left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{-}(\boldsymbol{n})\boldsymbol{a}}\\ {\boldsymbol {A}_{-}^{*}(\boldsymbol{a})\boldsymbol{n}} \end{array}\displaystyle \right ] \right )=(1-t)q(t)+\bigl(\boldsymbol{A}_{-}( \boldsymbol {n})\boldsymbol{a},\boldsymbol{a}\bigr). $$

Thus, $q(t)>0$ for all $t\in[0,1]$. It follows that ${\mathcal {G}}=\boldsymbol{F}(G\times(0,1))$ is an open subset of $\mathbb{ M}$.

Appendix B: Formula for the Second Derivative of $\overline{W}(\boldsymbol{F})$

Let $\boldsymbol{F}_{0}\in\mathfrak{O}$ and let $\boldsymbol{F}(s)$ be any curve in $\mathfrak{O}$ passing through $\boldsymbol{F}_{0}$ when $s=0$. Then, for sufficiently small $s$ there exist unique $\boldsymbol{F}_{-}(s)\in \mathfrak{J}$, $\boldsymbol{a}(s)\otimes\boldsymbol{n}(s)$ and $t(s)\in(0,1)$, such that $\boldsymbol{F}(s)=t(s)\boldsymbol{F}_{+}(s)+(1-t(s))\boldsymbol {F}_{-}(s)$ and (2.5) holds, where $\boldsymbol{F}_{+}(s)=\boldsymbol{F}_{-}(s)+\boldsymbol {a}(s)\otimes\boldsymbol{n}(s)\in\mathfrak{J}$. Let $\boldsymbol{\xi}=\dot{\boldsymbol{F}}(0)$. In the calculations below dot over a symbol denotes derivative in $s$ at $s=0$, while a symbol without an argument $s$ refers to $s=0$. For example, $\boldsymbol{a}$ denotes $\boldsymbol{a}(0)$.

Differentiating (4.1) at $s=0$ we obtain

$$ \bigl\langle \overline{W}_{\boldsymbol{F}}(\boldsymbol {F}_{0}),\boldsymbol{\xi} \bigr\rangle =t\langle\boldsymbol{P}_{+}, \dot {\boldsymbol{F}}_{+} \rangle+(1-t)\langle\boldsymbol{P}_{-}, \dot {\boldsymbol{F}}_{-} \rangle- \dot{t} {}[\!{}[ W ]\!]. $$

(B.1)

To compute $\dot{\boldsymbol{F}}_{\pm}$ we need to express $\boldsymbol{F}_{\pm}(s)$ in terms of $\boldsymbol{F}(s)$ and $\boldsymbol{a}(s)\otimes\boldsymbol{n}(s)$:

$$ \boldsymbol{F}_{-}(s)=\boldsymbol{F}(s)-t(s) \boldsymbol{a}(s)\otimes \boldsymbol{n}(s),\qquad \boldsymbol{F}_{+}(s)= \boldsymbol{F}(s)+\bigl(1-t(s)\bigr)\boldsymbol {a}(s)\otimes\boldsymbol{n}(s). $$

(B.2)

Differentiating in $s$ at $s=0$ we get

$$ \dot{\boldsymbol{F}}_{+}=\boldsymbol{\xi}-\dot{t} {} [\!{}[ \boldsymbol{F} ]\!]+(1-t){}[\!{}[\dot {\boldsymbol{F}} ]\! ],\qquad \dot{\boldsymbol{F}}_{-}=\boldsymbol{\xi}-\dot{t} {} [\!{}[ \boldsymbol{F} ]\!]-t{}[\!{}[\dot{\boldsymbol {F}} ]\! ], $$

(B.3)

where ${}[\!{}[\dot{\boldsymbol{F}} ]\!]$ is just a compact form of

$$ {}[\!{}[\dot{\boldsymbol{F}} ]\!]=\boldsymbol {a} \otimes\dot{\boldsymbol{n}}+\dot{\boldsymbol{a}}\otimes \boldsymbol{n}. $$

(B.4)

Substituting (B.3) into (B.1) we obtain

$$\begin{aligned} \bigl\langle \overline{W}_{\boldsymbol{F}}(\boldsymbol {F}_{0}), \boldsymbol{\xi} \bigr\rangle =&\bigl\langle t\boldsymbol {P}_{+}+(1-t) \boldsymbol{P}_{-},\boldsymbol{\xi} \bigr\rangle +\dot{t}\bigl\{ {} [\!{}[ W ]\!]-\bigl\langle t\boldsymbol {P}_{+}+(1-t) \boldsymbol{P}_{-},{}[\!{}[\boldsymbol{F} ]\!] \bigr\rangle \bigr\} \\ &{}+t(1-t)\bigl\langle {}[\!{}[ \boldsymbol{P} ]\!],{}[\!{}[ \dot{\boldsymbol {F}} ]\!] \bigr\rangle . \end{aligned}$$

Using (2.5) we conclude that

$$ \overline{W}_{\boldsymbol{F}}\bigl(\boldsymbol{F}(s) \bigr)=t(s)W_{\boldsymbol {F}}\bigl(\boldsymbol{F}_{+}(s)\bigr)+ \bigl(1-t(s)\bigr)W_{\boldsymbol{F}}\bigl(\boldsymbol {F}_{-}(s)\bigr). $$

(B.5)

Differentiating (B.5) at $s=0$ and using (B.3) we see that

$$ \overline{W}_{\boldsymbol{F}\boldsymbol{F}}(\boldsymbol {F}_{0}) \boldsymbol{\xi}=\mathsf{L}_{t}\boldsymbol{\xi}-\dot {t} \boldsymbol{N}_{t}+t(1-t){}[\!{}[\mathsf{L} ]\! ]{}[ \!{}[\dot{\boldsymbol{F}} ]\!], $$

(B.6)

where we use the shorthand $X_{t}=tX_{+}+(1-t)X_{-}$.

In order to express $\dot{\boldsymbol{a}}$, $\dot{\boldsymbol{n}}$ and $\dot{t}$ in terms of $\boldsymbol{\xi}$ we differentiate the last two equations in (2.5) at $s=0$ and obtain

$$ \widetilde{\mathbb{ C}}_{t}\left [ \textstyle\begin{array}{c} {\dot{\boldsymbol{a}}}\\ {\dot{\boldsymbol{n}}} \end{array}\displaystyle \right ]-\dot{t}\left [ \textstyle\begin{array}{c} {{}[\!{}[\boldsymbol{A} ]\!]\boldsymbol{a}}\\ {{}[\!{}[\boldsymbol{A}^{*} ]\!]\boldsymbol{n}} \end{array}\displaystyle \right ]= -\left [ \textstyle\begin{array}{c} {({}[\!{}[\mathsf{L} ]\!]\boldsymbol{\xi })\boldsymbol{n}}\\ {({}[\!{}[\mathsf{L} ]\!] \boldsymbol{\xi})^{T}\boldsymbol{a}} \end{array}\displaystyle \right ]. $$

(B.7)

Differentiating the Maxwell relation we obtain $\langle\boldsymbol{N}_{\pm},\dot{\boldsymbol{F}}_{\pm} \rangle =0$ (which is expressing the fact that $\boldsymbol{N}_{\pm}$ are orthogonal to the jump set at $\boldsymbol {F}_{\pm}$). Using (B.3) to eliminate $\dot{\boldsymbol{F}}_{\pm}$ we obtain

$$\dot{t}\boldsymbol{A}_{-}(\boldsymbol{n})\boldsymbol{a}\cdot \boldsymbol{a}=\langle\boldsymbol{N}_{-},\boldsymbol{\xi} \rangle -t \bigl\langle \boldsymbol{N}_{-},{}[\!{}[\dot{\boldsymbol{F}} ]\! ] \bigr\rangle ,\qquad \dot{t}\boldsymbol{A}_{+}(\boldsymbol{n}) \boldsymbol{a}\cdot \boldsymbol{a}=\langle\boldsymbol{N}_{+}, \boldsymbol{\xi} \rangle +(1-t)\bigl\langle \boldsymbol{N}_{+},{}[ \!{}[\dot{\boldsymbol {F}} ]\!] \bigr\rangle . $$

We combine the two relations into a more symmetric form by multiplying the first equation by $1-t$, the second one by $t$ and adding:

$$\dot{t}= \frac{\langle\boldsymbol{N}_{t},\boldsymbol{\xi} \rangle +t(1-t)\langle{}[\!{}[\boldsymbol{N} ]\!] ,{}[\!{}[\dot{\boldsymbol{F}} ]\!] \rangle }{\boldsymbol{A}_{t}\boldsymbol{a}\cdot\boldsymbol{a}}. $$

Substituting this formula into (B.6) and (B.7) we obtain

$$ \langle\overline{W}_{\boldsymbol{F}\boldsymbol{F}}\boldsymbol{\xi },\boldsymbol{ \xi} \rangle=\langle\mathsf{L}_{t}\boldsymbol{\xi },\boldsymbol{\xi} \rangle-\frac{\langle\boldsymbol {N}_{t},\boldsymbol{\xi} \rangle^{2}}{\boldsymbol{A}_{t}\boldsymbol {a}\cdot\boldsymbol{a}}-t(1-t) \biggl(\bigl\langle {}[\!{}[ \mathsf{L} ]\! ]\boldsymbol{\xi},{}[\!{}[\dot {\boldsymbol{F}} ]\!] \bigr\rangle - \frac{\langle \boldsymbol{N}_{t},\boldsymbol{\xi} \rangle\langle{}[\! {}[\boldsymbol{N} ]\!],{}[\!{}[\dot {\boldsymbol{F}} ]\!] \rangle}{\boldsymbol {A}_{t}\boldsymbol{a}\cdot\boldsymbol{a}} \biggr), $$

(B.8)

and

$$ \mathbb{ D}(t)\left [ \textstyle\begin{array}{c} {\dot{\boldsymbol{a}}}\\ {\dot{\boldsymbol{n}}} \end{array}\displaystyle \right ]= -\left [ \textstyle\begin{array}{c} {({}[\!{}[\mathsf{L} ]\!]\boldsymbol{\xi })\boldsymbol{n}}\\ {({}[\!{}[\mathsf{L} ]\!] \boldsymbol{\xi})^{T}\boldsymbol{a}} \end{array}\displaystyle \right ]+\frac{\langle\boldsymbol{N}_{t},\boldsymbol{\xi} \rangle }{\boldsymbol{A}_{t}\boldsymbol{a}\cdot\boldsymbol{a}} \left [ \textstyle\begin{array}{c} {{}[\!{}[\boldsymbol{A} ]\!]\boldsymbol{a}}\\ {{}[\!{}[\boldsymbol{A}^{*} ]\!]\boldsymbol{n}} \end{array}\displaystyle \right ], $$

(B.9)

respectively, where

$$ \mathbb{ D}(t)=\widetilde{\mathbb{ C}}_{t}- \frac {t(1-t)}{\boldsymbol{A}_{t}\boldsymbol{a}\cdot\boldsymbol{a}}\left [ \textstyle\begin{array}{c} {{}[\!{}[\boldsymbol{A} ]\!]\boldsymbol{a}}\\ {{}[\!{}[\boldsymbol{A}^{*} ]\!]\boldsymbol{n}} \end{array}\displaystyle \right ]\otimes \left [ \textstyle\begin{array}{c} {{}[\!{}[\boldsymbol{A} ]\!]\boldsymbol{a}}\\ {{}[\!{}[\boldsymbol{A}^{*} ]\!]\boldsymbol{n}} \end{array}\displaystyle \right ]. $$

(B.10)

Hence,

$$ \bigl\langle \overline{W}_{\boldsymbol{F}\boldsymbol{F}}(\boldsymbol {F}_{0})\boldsymbol{\xi},\boldsymbol{\xi} \bigr\rangle =\langle\mathsf {L}_{t}\boldsymbol{\xi},\boldsymbol{\xi} \rangle-\frac{\langle \boldsymbol{N}_{t},\boldsymbol{\xi} \rangle^{2}}{\boldsymbol {A}_{t}\boldsymbol{a}\cdot\boldsymbol{a}}+t(1-t) \mathbb{ D}(t)^{-1}\boldsymbol{Z}(t)\cdot\boldsymbol{Z}(t), $$

(B.11)

where

$$\boldsymbol{Z}(t)=-\left [ \textstyle\begin{array}{c} {({}[\!{}[\mathsf{L} ]\!]\boldsymbol{\xi })\boldsymbol{n}}\\ {({}[\!{}[\mathsf{L} ]\!] \boldsymbol{\xi})^{T}\boldsymbol{a}} \end{array}\displaystyle \right ]+ \frac{\langle\boldsymbol{N}_{t},\boldsymbol{\xi} \rangle }{\boldsymbol{A}_{t}\boldsymbol{a}\cdot\boldsymbol{a}}\left [ \textstyle\begin{array}{c} {{}[\!{}[\boldsymbol{A} ]\!]\boldsymbol{a}}\\ {{}[\!{}[\boldsymbol{A}^{*} ]\!]\boldsymbol{n}} \end{array}\displaystyle \right ]. $$

In the last term of (B.11) $\mathbb{ D}(t)^{-1}\boldsymbol{Z}(t)$ is understood in the orthogonal complement to $[\boldsymbol{a},-\boldsymbol{n}]$. This is justified by the following lemma.

Lemma B.1

Assume that $\boldsymbol{F}_{\pm}$ is a non-degenerate pair in the sense of Definition  4.1. Assume that (3.2) holds. Then $\mathbb{ D}(t)$ is a positive-definite self-adjoint operator on $V=(\mathbb{ R}[\boldsymbol{a},-\boldsymbol{n}])^{\perp}$.

Proof

It is obvious that $\mathbb{ D}(t)$ is self-adjoint and that $\mathbb{ D}(t)[\boldsymbol{a},-\boldsymbol{n}]=0$. Thus, $\mathbb{ D}(t)$ is a self-adjoint operator on $V$. By assumption of non-degeneracy, $\widetilde{\mathbb{ C}}_{t}$ is a positive-definite self-adjoint operator on $V$. Thus, in order to prove the lemma we need to show that

$$ \widetilde{\mathbb{ C}}_{t}^{-1} \boldsymbol{X}\cdot\boldsymbol {X}< \frac{\boldsymbol{A}_{t}\boldsymbol{a}\cdot\boldsymbol {a}}{t(1-t)},\quad \boldsymbol{X}= \left [ \textstyle\begin{array}{c} {{}[\!{}[\boldsymbol{A} ]\!]\boldsymbol{a}}\\ {{}[\!{}[\boldsymbol{A}^{*} ]\!]\boldsymbol{n}} \end{array}\displaystyle \right ] $$

(B.12)

for every $t\in[0,1]$.

We observe that

$$\widetilde{\mathbb{ C}}_{t}\left [ \textstyle\begin{array}{c} {\boldsymbol{a}}\\ {\boldsymbol{n}} \end{array}\displaystyle \right ]=2\left [ \textstyle\begin{array}{c} {\widetilde{\boldsymbol{A}}_{t}\boldsymbol{a}}\\ {\widetilde {\boldsymbol{A}}_{t}^{*}\boldsymbol{n}} \end{array}\displaystyle \right ]. $$

We have

$$\boldsymbol{X}=\frac{1}{t}\left (\left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{+}\boldsymbol{a}}\\ {\boldsymbol {A}_{+}^{*}\boldsymbol{n}} \end{array}\displaystyle \right ]-\left [ \textstyle\begin{array}{c} {\widetilde{\boldsymbol{A}}_{t}\boldsymbol{a}}\\ {\widetilde {\boldsymbol{A}}_{t}^{*}\boldsymbol{n}} \end{array}\displaystyle \right ] \right ). $$

Hence,

$$\widetilde{\mathbb{ C}}_{t}^{-1}\boldsymbol{X}\cdot \boldsymbol {X}=\frac{\widetilde{\boldsymbol{A}}_{t}\boldsymbol{a}\cdot \boldsymbol{a}-2\boldsymbol{A}_{+}\boldsymbol{a}\cdot\boldsymbol {a}+\widetilde{\mathbb{ C}}_{t}^{-1}\boldsymbol{Y}_{+}\cdot \boldsymbol{Y}_{+}}{t^{2}}, $$

where

$$\boldsymbol{Y}_{+}=\left [ \textstyle\begin{array}{c} {\boldsymbol{A}_{+}\boldsymbol{a}}\\ {\boldsymbol {A}_{+}^{*}\boldsymbol{n}} \end{array}\displaystyle \right ]. $$

When $t=0$ or 1, the statement is obvious. For $t\in(0,1)$ we have

$$\widetilde{\mathbb{ C}}_{t}>(1-t)\mathbb{ C}_{+}, $$

in the sense of quadratic forms on $V$, so that

$$\widetilde{\mathbb{ C}}_{t}^{-1}\boldsymbol{Y}_{+} \cdot\boldsymbol {Y}_{+}< \frac{1}{1-t}\mathbb{ C}_{+}^{-1}\boldsymbol{Y}_{+}\cdot \boldsymbol{Y}_{+}=\frac{\boldsymbol{A}_{+}\boldsymbol{a}\cdot \boldsymbol{a}}{1-t}. $$

Thus, we obtain

$$\widetilde{\mathbb{ C}}_{t}^{-1}\boldsymbol{X}\cdot \boldsymbol {X}< \frac{\boldsymbol{A}_{t}\boldsymbol{a}\cdot\boldsymbol{a}}{t(1-t)}, $$

as required. □

Finally, let us show that the acoustic tensor of $\overline {W}(\boldsymbol{F})$ at $\boldsymbol{F}_{0}=t\boldsymbol{F}_{+}+(1-t)\boldsymbol{F}_{-}$ is nonnegative whenever it is nonnegative at both $\boldsymbol{F}_{+}$ and $\boldsymbol{F}_{-}$. Indeed, the last term in (B.11) is nonnegative, by Lemma B.1. It remains to observe that the function

$$\phi(t)=\langle\mathsf{L}_{t}\boldsymbol{\xi},\boldsymbol{\xi} \rangle-\frac{\langle\boldsymbol{N}_{t},\boldsymbol{\xi} \rangle ^{2}}{\boldsymbol{A}_{t}\boldsymbol{a}\cdot\boldsymbol{a}} $$

is concave in $t$, which we can see by differentiating $\phi(t)$ twice:

$$\phi''(t)=-\frac{2(\langle\boldsymbol{N}_{+},\boldsymbol{\xi} \rangle\boldsymbol{A}_{-}\boldsymbol{a}\cdot\boldsymbol{a}-\langle \boldsymbol{N}_{-},\boldsymbol{\xi} \rangle\boldsymbol {A}_{+}\boldsymbol{a}\cdot\boldsymbol{a})^{2}}{(\boldsymbol {A}_{t}\boldsymbol{a}\cdot\boldsymbol{a})^{3}}. $$

Thus, $\phi(t)$ attains its minimum at $t=0$ or $t=1$. Setting $t=0$ and 1 in (B.11) we obtain

$$ \overline{W}_{\boldsymbol{F}\boldsymbol{F}}(\boldsymbol{F}_{\pm })= \mathsf{L}_{\pm}-\frac{\boldsymbol{N}_{\pm}\otimes\boldsymbol {N}_{\pm}}{\boldsymbol{A}_{\pm}\boldsymbol{a}\cdot\boldsymbol{a}}. $$

(B.13)

Formula (4.4) now follows easily from (B.13).