3.1 Differential inclusions in the orthogonal group

In this section, we will consider some background material concerning differential inclusions into the orthogonal group. Consider the relaxation of the orthogonal group $O({\mathbb {R}}^n)$, i.e. its convex hull given by

\[ \text{co}\,O({\mathbb{R}}^n)=\{X\in \mathcal{L}({\mathbb{R}}^n): X^\ast X\leq I\}. \]

Theorem 3.1 Let $\Omega \subset {\mathbb {R}}^n$ be a Lipschitz domain and let $\phi \in W^{1,\infty }(\Omega,{\mathbb {R}}^n)$ be a Lipschitz map such that $D\phi (x)\in int co O({\mathbb {R}}^n)$ for a.e. $x\in \Omega$. Then the differential inclusion

(3.1)\begin{equation} \left\{\begin{array}{@{}rl} Du(x)\in O({\mathbb{R}}^n) & \text{for a.e. $x\in \Omega$ },\\ u(x)=\phi(x) & \text{for a.e. $x\in \partial\Omega$ } \end{array} \right. \end{equation}

for $u\in W^{1,\infty }(\Omega,{\mathbb {R}}^n)$ possesses infinitely many solutions that are nowhere $C^1$.

The key to this result is that $O({\mathbb {R}}^n)$ possesses many rank-one connections, i.e. there exit matrices $X,Y\in O({\mathbb {R}}^n)$ such that

\[ X-Y=u\otimes v \]

for some $u,v\in {\mathbb {R}}^n$.

A proof of theorem 3.1 can be found in [Reference Székelyhidi30, § 5.1].

Solutions of differential inclusions need not necessarily be very irregular.

Definition 3.2 Let $u\in C^{0,1}(\Omega,{\mathbb {R}}^m)$ be a Lipschitz map. Let

(3.2)\begin{equation} \Sigma(u)=\{x\in \Omega: u \text{is not}\ C^1\ \text{in a neighbourhood of}\ x\} \end{equation}

denote the singular set of $u$.

In [Reference Dacorogna, Marcellini and Paolini10], the authors consider solutions of (3.1) with affine boundary values generated by origami maps. These maps are piecewise $C^1$ and the Hausdorff measure $\mathscr {H}^{n-1}(\Sigma (u))$ is locally finite in the interior of the domain. However, to satisfy the boundary conditions, $\Sigma (u)$ will become fractal-like as we approach the boundary and $\mathscr {H}^{n-1}(\Sigma (u))=+\infty$ in the whole domain. Similar types of solutions are also considered in [Reference Iwaniec, Verchota and Vogel19] and also for the energy–momentum equations for the Dirichlet energy in [Reference Iwaniec and Onninen18, § 3.6]. In both cases $Du(x)\in K$, where $K$ is a finite subset of $O({\mathbb {R}}^n)$.

3.2 Generic ill-posedness

As we have seen in proposition 2.5, the mapping $T: \mathcal {L}({\mathbb {R}}^n)\to \mathcal {L}({\mathbb {R}}^n)$ is not invertible for a frame-indifferent Lagrangian $W$. In particular, the level sets of

\[ T(X)=Y \]

for some fixed $Y\in T(\mathcal {L}({\mathbb {R}}^n))$ are $O({\mathbb {R}}^n)$-invariant.

Proof. If $Du(x)\in O({\mathbb {R}}^n)A$ a.e., then

\[ T(Du(x))=T(A)=Y \]

is constant a.e. and hence a weak solution of $\text {div}\,T(Du(x))=0$.

This leads to differential inclusions of the form

\[ Du(x)\in T^{{-}1}(Y) \]

for some fixed $Y$ such that $T^{-1}(Y)\neq \varnothing$.

There are similar results for the lack of partial regularity for the Euler–Lagrange equations for elliptic systems. In [Reference Müller and S̆verák26, theorem 4.1], the authors show that there exists a smooth strongly quasiconvex function $W:\mathcal {L}({\mathbb {R}}^2)\to {\mathbb {R}}$ such that there exists a Lipschitz continuous solution of $\text {div}DW(Du(x))=0$ that is nowhere $C^1$. This is done by rewriting the Euler–Lagrange equation as a differential inclusion and using methods from convex integration theory. The result was extended in [Reference Székelyhidi31] to apply also to smooth strongly polyconvex functions. Moreover, the solutions in [Reference Székelyhidi31] are also weak local minimizers. This also applies to the example in [Reference Müller and S̆verák26] by the work of [Reference Kristensen and Taheri22]. It is however important to note that weak local minimizers $u_0\in W^{1,\infty }$ need not be weak solutions of the energy–momentum equations. Indeed, let $\Omega \subset {\mathbb {R}}^n$ be a domain and let $\lambda \in C^\infty _0(\Omega,{\mathbb {R}}^n)$. Let $u_0\in W^{1,\infty }(\Omega,{\mathbb {R}}^n)$ and let $u_\varepsilon (x)=u_0(x+\varepsilon \lambda (x))$ be an inner variation. Then

\begin{align*} \vert Du_0(x)-Du_\varepsilon(x)\vert& =\vert Du_0(x)-Du_0(x+\varepsilon \lambda(x))(I+\varepsilon D\lambda (x))\vert\\ & \geq \left\vert \vert Du_0(x)-Du_0(x+\varepsilon \lambda(x))\vert -\varepsilon \vert Du(x+\varepsilon\lambda(x))D\lambda (x)\vert\right\vert\\ & \geq \left\vert \vert Du_0(x)-Du_0(x+\varepsilon \lambda(x))\vert -\varepsilon \Vert \vert Du\vert \Vert_{L^\infty(\Omega)}\Vert \vert D\lambda\vert \Vert_{L^\infty(\Omega)}\right\vert \end{align*}

Since $Du_0$ is not continuous, it may happen that $\vert Du_0(x)-Du_0(x+\varepsilon \lambda (x))\vert \geq 1+\varepsilon \Vert \vert Du\vert \Vert _{L^\infty (\Omega )}\Vert \vert D\lambda \vert \Vert _{L^\infty (\Omega )}$ for every $\varepsilon >0$ and thus that $\Vert u_0-u_\varepsilon \Vert _{W^{1,\infty }(\Omega )}>1$ for all $\varepsilon >0$. Thus, inner variations need not be close in $W^{1,\infty }(\Omega )$-norm. Therefore, $u_0$ being a weak local minimizer need not imply that $u_0$ solves the energy–momentum equations.

It is therefore a natural question if the differential inclusions giving solutions to the energy–momentum equations are also weak solutions of the Euler–Lagrange equations? For this purpose, we first consider a special class of solutions given by laminations, the reason being that any solution in, for example, [Reference Dacorogna, Marcellini and Paolini10, Reference Iwaniec, Verchota and Vogel19] is locally a lamination outside a small closed set. Given a first-order partial differential operator $\mathscr {A}$ with constant coefficients and its associated symbol $\mathbb {A}(\xi )$ with $\xi \in \mathbb {S}^{n-1}$, consider functions of the form

(3.3)\begin{equation} u(x)=\lambda h(\langle x,\xi\rangle)+\mu(1-h(\langle x,\xi\rangle)) \end{equation}

where $h:{\mathbb {R}}\to \{0,1\}$ is measurable and $\lambda -\mu \in \text {ker}\,\mathbb {A}(\xi )$. These are solutions to two-state rigidity problem

\[ \left\{\begin{array}{@{}l} \mathscr{A}u(x)=0\ \text{in the sense of distributions},\\ u(x)\in \{\mu,\lambda\}.\end{array} \right. \]

In the case when $\mu,\lambda \in O({\mathbb {R}}^n)$ are rank-one connected, then the laminate solution (3.3) gives a solution to the differential inclusion $Du(x)\in O({\mathbb {R}}^n)$ a.e. with $\mathcal {A}=\text {curl}$, where $\text {curl}$ denotes the matrix curl operator. In this case, the boundary values for non-trivial measurable functions $h$ are, however, not smooth. Furthermore, we want the rank-one connected laminate to be such that $v(x)=DW(D(x))$ is two-state laminate solution for $\mathscr {A}=\text {div}$, where $\text {div}$ is the matrix divergence. By [Reference Kristensen and Raita21, p. 8], $\mathbb {A}(\xi )X=X\xi$ for an $n\times n$-matrix $X$ and hence we must have that $(DW(\mu )-DW(\lambda ))\xi =0$ as well. This leads us to the following proposition.

Proposition 3.4 Let $W$ satisfy the assumptions in theorem 1.1. Let $A,B\in O({\mathbb {R}}^n)$ be rank-one connected and consider the laminate

(3.4)\begin{equation} Du(x)=A h(\langle x,\xi\rangle)+B(1-h(\langle x,\xi\rangle)) \end{equation}

for some measurable $h:{\mathbb {R}}\to \{0,1\}$ and such that $B-A=a\otimes \xi$ for some $a\in {\mathbb {R}}^n$ and $\xi \in \mathbb {S}^{n-1}$. Then the laminate is a distributional solution to the Euler–Lagrange equation

\[ {\rm div} DW(Du(x))=0 \]

if and only if $\langle \xi, T(I)\xi \rangle =-W(I)$ or equivalently if and only if $\langle \xi, DW(I)\xi \rangle =0$.

Proof. Let $\lambda =DW(A)$ and $\mu =DW(B)$. Then for the matrix field $M(x)=DW(Du(x))$ to be divergence free in the sense of distributions, we must have $(DW(B)-DW(A))n=0$. Since $O({\mathbb {R}}^n)$ acts transitively on itself, there exists a $U\in O({\mathbb {R}}^n)$ such that

\[ B=UA. \]

On the other hand, we have

\[ B-A=a\otimes \xi \]

which implies

\[ (U-I)A=a\otimes \xi \quad \Longrightarrow \quad U-I= a\otimes \xi\circ A^\ast \]

Using that $DW$ is $O({\mathbb {R}}^n)$-equivariant by remark 2.5, we find that

\begin{align*} (DW(B)-DW(A))\xi& =(UDW(A)-DW(A))\xi=(U-I)DW(A)\xi\\ & = a\otimes \xi\circ A^\ast{\circ} DW(A)\xi =a\langle \xi, A^\ast DW(A)\xi\rangle\\ & =a\langle \xi, (T(A)+W(A)I)\xi\rangle=a(\langle \xi, T(A)\xi\rangle+W(A)\vert \xi\vert^2)\\ & =a(\langle \xi, T(A)\xi\rangle+W(A)) \end{align*}

Thus, $\langle \xi, T(A)\xi \rangle +W(A)=0$. Since $A\in O({\mathbb {R}}^n)$ and $T$ and $W$ are $O({\mathbb {R}}^n)$-invariant $\langle \xi, T(A)\xi \rangle +W(A)=\langle \xi, T(I)\xi \rangle +W(I)=0$.

Remark 3.5 If $W$ is the energy density of a hyperelastic material, it is physically reasonable that scalings $x\mapsto tx$ for $t>0$ cost energy. Consequently, the function

\[ j(t)=W(Du(x))=W(tI) \]

should have a minimum at $t=1$. Since

\[ j'(t)=\langle DW(tI),tI\rangle \]

we find that the condition $j'(1)=0$ implies that $0={\rm tr} (DW(I))={\rm tr} (T(I)+W(I)I)={\rm tr} (T(I))+nW(I)$. Since we have for an ON-basis $\{e_j\}_{j=1}^n$

\[ {\rm tr} (T(I))+nW(I)=\sum_{j=1}^n(\langle e_j,T(I)e_j\rangle +W(I)), \]

we see that the condition $j'(1)=0$ can be seen as an averaged condition of the previous proposition 3.4.

It is easy to produce frame-indifferent smooth functions $W$ which satisfy $DW(I)=0$, $W(X)=(\vert X\vert ^2-4)^2$ for example will do. However, $W$ is not rank-one convex and therefore not polyconvex either. In example 4.5, we give a less trivial example coming from geometric function theory and in particular the study of mappings of finite distortion. This functional is not globally polyconvex but polyconvex when restricted to $GL_+({\mathbb {R}}^n)$.

Proof. By a previous remark, it is sufficient to consider solutions with $Du(x)\in \{I,R\}$ and $R\in O({\mathbb {R}}^n)$ rank-one connected to $I$. Let $R-I=a\otimes \xi$ for some $\xi \in S^{n-1}$ and some $a\in {\mathbb {R}}^n$. Furthermore, since $R^\ast R=I$, we get the equation

\[ \xi \otimes a+a\otimes \xi={-}\vert a\vert^2\xi\otimes \xi \]

which implies that $a=-2\xi$.

For $u$ to solve the Euler–Lagrange equations, we must use the $\text {O}({\mathbb {R}}^n)$-equivariance of $DW$

\begin{align*} 0=(DW(R)-DW(I))\xi& =(R-I)DW(I)\xi={-}2(\xi\otimes \xi)\circ (DW(I)\xi)\\& ={-}2\xi\langle \xi,DW(I)\xi\rangle \end{align*}

and so $\langle \xi,DW(I)\xi \rangle =0$. Now consider the function

\[ \phi(t)=W(I-2t\xi \otimes \xi) \]

By frame indifference, $\phi (0)=\phi (1)$ and

\[ \phi'(t)=\langle DW(I-2t\xi \otimes \xi),R-I\rangle={-}2\langle DW(I-2t\xi \otimes \xi),\xi \otimes \xi \rangle. \]

By assumption, $\phi (t)$ is strictly convex. Hence, $\phi '(0)\neq 0$ and $\phi '(1)\neq 0$. On the other hand

\[ \phi'(0)={-}2\langle DW(I),\xi \otimes \xi \rangle={-}2\langle \xi,DW(I)\xi\rangle=0 \]

contradicting the strict convexity of $\phi$.

Proof of theorem 1.2 We note that by the frame indifference of $W$ (remark 2.6) $DW(I)^\ast =DW(I)$. Furthermore, since any rank-one connected matrix $R\in \text {O}({\mathbb {R}}^n)$ to $I$ is of the form $I-2\xi \otimes \xi$ for some $\xi \in S^{n-1}$, we have for $\phi _\xi (t)=W(I-2t\xi \otimes \xi )$

\[ \phi_\xi'(t)={-}2\langle DW(I-2t\xi \otimes \xi), \xi\otimes \xi\rangle \]

and the strict rank-one convexity assumption implies that $\phi _\xi '(0)=\phi '_\xi (1)\neq 0$ for all $\xi \in S^{n-1}$ we find that the quadratic form $Q(\xi )=\langle DW(I), \xi \otimes \xi \rangle =\langle \xi,DW(I)\xi \rangle \neq 0$ for all $\xi \in S^{n-1}$. Thus, either $Q(\xi )>0$ or $Q(\xi )<0$. We may assume the former case. Hence, $DW(I)$ is positive definite. Set $A=DW(I)$ and define the linear map $L\in \mathcal {L}(\mathcal {L}({\mathbb {R}}^n))$ by

\[ L(X)=XA. \]

Then $L$ is symmetric and positive definite. Indeed,

\begin{align*} \langle Y,L(X)\rangle& ={\rm tr} (Y^\ast XA)={\rm tr} (AY^\ast X)={\rm tr} ((YA^\ast)^\ast X)={\rm tr} ((YA)^\ast X)\\ & =\langle L(Y),X\rangle. \end{align*}

Furthermore, using that $A$ is diagonalizable with diagonal matrix $D$ and eigenvalues $0<\lambda _1\leq \lambda _2\leq...\leq \lambda _n$ such that $A=RDR^\ast$ and setting $Y=XR$, we find

\begin{align*} \langle X,L(X)\rangle& ={\rm tr} (X^\ast XRDR^\ast)={\rm tr} (R^\ast X^\ast XRD)={\rm tr} ((X^\ast R)^\ast XRD)\\ & =\langle Y,YD\rangle=\sum_{j=1}^n\lambda_j \langle Y,Y(e_j\otimes e_j)\rangle=\sum_{j=1}^n\lambda_j \langle Y(e_j),Y(e_j)\rangle\\ & \geq \lambda_1 \sum_{j=1}^n\langle Y(e_j),Y(e_j)\rangle=\lambda_1 \vert Y\vert^2=\lambda_1\vert X\vert^2. \end{align*}

Now assume that $u$ is a solution of the differential inclusion $Du(x)\in O({\mathbb {R}}^n)$ that is not $C^1$ and that in addition $u$ is a weak solution of the Euler–Lagrange equations on some domain $\Omega \subset {\mathbb {R}}^n$. Then for any $\varphi \in C^\infty _0(\Omega, {\mathbb {R}}^n)$ and using the $\text {O}({\mathbb {R}}^n)$-equivariance of $W$, we find

\begin{align*} 0=\int_{\Omega}\langle DW(Du(x)),D\varphi(x)\rangle {\rm d}x& =\int_{\Omega}\langle Du(x)DW(I),D\varphi(x)\rangle {\rm d}x\\& =\int_{\Omega}\langle L(Du(x)),D\varphi(x)\rangle {\rm d}x. \end{align*}

Thus, $u$ is a weak solution of the very strongly elliptic constant coefficient equation (in the sense of [Reference Giaquinta and Martinazzi14, definition 3.36 (3.16), p.53])

\[ \text{div}\,L(Du(x))=0. \]

However, by elliptic regularity theory ([Reference Giaquinta and Martinazzi14, theorem 4.11]) $u\in C^\infty$, a contradiction.

4.1 Invertibility of the reduced energy–momentum tensor in a number of interesting cases

In this section, we will consider a number of important functionals that occur in non-linear elasticity and geometric functions theory. We will show that in all these cases, the reduced energy–momentum mapping $\mathcal {T}$ is typically not injective, yet the conditions of theorem 4.1 are not satisfied. In addition, they all have the feature that their Lagrangian $W$ in addition to being frame-indifferent is also isotropic, i.e.

\[ W(R^\ast XR)=W(X) \]

for all $R\in O({\mathbb {R}}^n)$ and all $X\in \mathcal {L}({\mathbb {R}}^n)$. We begin with the Dirichlet $p$-energy.

Example 4.4 Let $1< p<+\infty$ and let for $u\in W^{1,p}(\Omega,{\mathbb {R}}^n)$

\[ \mathscr{D}_p[u]=\int_{\Omega}\vert Du(x)\vert^p\,{\rm d}x. \]

Since $W(X)=\vert X\vert ^p$ we find that $DW(X)=p\vert X\vert ^{p-2}X$ and the energy–momentum mapping becomes

\[ T(X)=p\vert X\vert^{p-2}X^\ast X-\vert X\vert^pI=p\text{tr}(X^\ast X)^{(p-2)/2}X^\ast X-\text{tr}(X^\ast X)^{p/2}I. \]

The reduced energy–momentum mapping becomes with $Y=X^\ast X$

\[ \mathcal{T}(Y)=p\text{tr}(Y)^{(p-2)/2}Y-\text{tr}(Y)^{p/2}I. \]

Note that

\[ \text{tr}(\mathcal{T}(Y))=p\text{tr}(Y)^{(p-2)/2}\text{tr}(Y)-\text{tr}(Y)^{p/2}\text{tr}(I)=(p-n)\text{tr}(Y)^{p/2}, \]

which is different from $0$ if and only if $p\neq n$. Let $Z\in \mathcal {T}(\text {Sym}_+(n))$. Consider the equation for $p\neq n$

\[ p\text{tr}(Y)^{(p-2)/2}Y-\text{tr}(Y)^{p/2}I=Z. \]

Taking traces of both sides gives us

\[ (p-n)\text{tr}(Y)^{p/2}=\text{tr}(Z)\ \Longrightarrow\ \text{tr}(Y)=((p-n)^{{-}1}\text{tr}(Z))^{2/p}. \]

Thus,

(4.2)\begin{equation} Y=\frac{Z+\text{tr}(Y)^{p/2}I}{p\text{tr}(Y)^{p/2-1}}=\frac{Z+\frac{1}{p-n}\text{tr}(Z)I}{p(\frac{1}{p-n}\text{tr}(Z))^{1-2/p}}. \end{equation}

Since $Y$ is positive semi-definite, it has a unique positive semi-definite square root $\sqrt {Y}$. In particular, all solutions of $T(X)=Z$ are given by

\[ T^{{-}1}(Z)=O({\mathbb{R}}^n)\sqrt{\frac{Z+\frac{1}{p-n}\text{tr}(Z)I}{p(\frac{1}{p-n}\text{tr}(Z))^{1-2/p}}} \]

and the situation (4.1) cannot occur. Furthermore, by theorem 1.2 and in view of Uhlenbeck's regularity result [Reference Uhlenbeck34], weak solutions $u\in W^{1,p}(\Omega,{\mathbb {R}}^n)$ of the Euler–Lagrange equations of the Dirichlet $p$-energy are always $C^{1,\alpha }$ for some $0<\alpha <1$. Hence, the weak solutions of the energy–momentum equations which are not $C^{1,\alpha }$ are not weak solutions of the Euler–Lagrange equations. In the conformally invariant case $p=n$, the formula (4.2) does not hold and in fact we now show that $\mathcal {T}(Y)$ is not injective. It follows from the equation $\mathcal {T}(Y)=Z$ that $[Y,Z]=0$, so if $Z$ is diagonalizable so is $Y$. Thus, we restrict to considering only diagonal matrices $Y$ and only consider the case $n=2$. We find that if

\[ Y=\begin{bmatrix} \alpha & 0\\ 0 & \beta \end{bmatrix},\quad Z=\begin{bmatrix} c & 0\\ 0 & -c \end{bmatrix} \]

where $c\geq 0$ we find the system of equations

\[ \left\{\begin{array}{@{}l} \alpha-\beta=c \\ \beta-\alpha={-}c \end{array} \right. \]

Hence, if $\alpha =t$, $\beta =t-c$ and $t\geq c$ parametrizes the solutions. Thus, $\mathcal {T}$ is not an injective map. On the other hand, we can find no two $t_1,t_2$ such that the condition of theorem 4.1 is satisfied.

Example 4.5 $q$-mean distortion

Let $u:\Omega \subset {\mathbb {R}}^n \to {\mathbb {R}}^n$ be a map in $W^{1,n}(\Omega,{\mathbb {R}}^n)$ and consider the $q$-mean distortion functional

\[ \mathscr{K}_q[u]=\int_{\Omega}\mathbb{K}(x,u)^q\,{\rm d}x=\int_{\Omega}\left(\frac{\vert Du(x)\vert^n}{J(x,u)}\right)^q\,{\rm d}x \]

where $q\geq 1$. $W$ is a priori only well-defined when $\det (X)>0$, however we can extend $W$ to $\widetilde {W}\in C^\infty (\text {GL}_+({\mathbb {R}}^n)\cup \text {GL}_-({\mathbb {R}}^n))$ as a frame-indifferent function by defining

\[ \widetilde{W}(X)=\left(\frac{\vert X\vert^n}{\vert \det(X)\vert}\right)^q. \]

This extension is however not polyconvex due to the blow up when $\det (X)=0$ (except in the case $X=tI$ and $t\to 0$). The only polyconvex extension is to define $\mathbb {K}(x,u)^q=+\infty$ whenever $\det (X)\leq 0$. For more on this functional, we refer the reader to [Reference Iwaniec, Martin and Onninen17] and references there in. We will however use $\widetilde {W}$ and by abuse of notation also write $W$ for its frame-indifferent extension. Set $W(X)=(f(X)g(X))^q$ where $f(X)=\vert X\vert ^n$ and $g(X)=\vert \det (X)\vert ^{-1}$. Then

\[ DW(X)=q\left(\frac{\vert X\vert^n}{\vert \det(X)\vert}\right)^{q-1}[g(X)Df(X)+f(X)Dg(X)]. \]

By (A.1)

\begin{align*} Df(X)& =n\vert X\vert^{n-2}X,\\ Dg(X)& ={-}\frac{\text{sgn}(\det(X))}{\vert \det(X)\vert^2}\text{adj}(X)^\ast \end{align*}

Thus,

\begin{align*} DW(X)=q\left(\frac{\vert X\vert^n}{\vert \det(X)\vert}\right)^{q-1}\left[\frac{n\vert X\vert^{n-2}X}{\vert \det(X)\vert}-\frac{\text{sgn}(\det(X))\vert X\vert^n}{\vert \det(X)\vert^2}\text{adj}(X)^\ast\right] \end{align*}

and

\begin{align*} T(X)& =X^\ast DW(X)-W(X)I\\ & =q\left(\frac{\vert X\vert^n}{\vert \det(X)\vert}\right)^{q-1}\left[\frac{n\vert X\vert^{n-2}X^\ast X}{\vert \det(X)\vert}-\frac{\text{sgn}(\det(X))\vert X\vert^n}{\vert \det(X)\vert^2}X^\ast \text{adj}(X)^\ast\right]\\& \quad-\left(\frac{\vert X\vert^n}{\vert \det(X)\vert}\right)^qI\\ & =q\left(\frac{\vert X\vert^n}{\vert \det(X)\vert}\right)^{q-1}\left[\frac{n\vert X\vert^{n-2}X^\ast X}{\vert \det(X)\vert}-\frac{\vert X\vert^n}{\vert \det(X)\vert}I\right]-\left(\frac{\vert X\vert^n}{\vert \det(X)\vert}\right)^qI \end{align*}

Thus,

\begin{align*} T(I)& =q\left(\frac{\vert I\vert^n}{\vert \det(I)\vert}\right)^{q-1}\left[\frac{n\vert I\vert^{n-2}I}{\vert \det(I)\vert}-\frac{\vert I\vert^n}{\vert \det(I)\vert}I\right]-\left(\frac{\vert I\vert^n}{\vert \det(I)\vert}\right)^qI\\ & ={-}W(I)I \end{align*}

and $\langle T(I)\xi,\xi \rangle =-W(I)$.

Hence, $T$ does satisfy the assumptions of proposition 3.4 and there are stationary points of the functional which are nowhere $C^1$. Indeed, as an explicit example, take $n=2$ and $q=1$ and let

\[ R=\begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}. \]

Then $I-R=2e_2\otimes e_2$ and so $R$ is rank-one connected to $I$. Furthermore,

\[ DW(I)=\frac{2I}{\vert \det(I)\vert}-\frac{\text{sgn}(\det(I))\vert I\vert^2}{\vert \det(I)\vert^2}\text{adj}(I)^\ast{=}0, \]

and

\[ \text{adj}(R)^\ast{=}\begin{bmatrix} -1 & 0\\ 0 & 1 \end{bmatrix}, \]

and

\[ DW(R)=\frac{2R}{\vert \det(R)\vert}-\frac{\text{sgn}(\det(R))\vert R\vert^2}{\vert \det(R)\vert^2}\text{adj}(R)^\ast{=}2[R+\text{adj}(R)^\ast]=0. \]

Thus, $u$ is also a solution to the Euler–Lagrange equations. The reduced energy–momentum tensor becomes

\begin{align*} \mathcal{T}(Y)& =q\left(\frac{\text{tr}(Y)^{n/2}}{ \sqrt{\det(Y)}}\right)^{q-1}\left[\frac{n(\text{tr}(Y))^{(n-2)/2}Y}{\sqrt{\det(Y)}}-\frac{\text{tr}(Y)^{n/2}}{\sqrt{\det(Y)}}I\right]-\left(\frac{\text{tr}(Y)^{n/2}}{\sqrt{\det(Y)}}\right)^qI\\ & =q\left(\frac{\text{tr}(Y)^{n/2}}{ \sqrt{\det(Y)}}\right)^{q}\left[\frac{nY}{{\rm tr} (Y)}-I\right]-\left(\frac{\text{tr}(Y)^{n/2}}{\sqrt{\det(Y)}}\right)^qI\\ & =\left(\frac{\text{tr}(Y)^{n/2}}{ \sqrt{\det(Y)}}\right)^{q}\left[\frac{qnY-(q+1)\text{tr}(Y)I}{{\rm tr} (Y)}\right], \end{align*}

provided $\det (Y)\neq 0$ and ${\rm tr} (Y)\neq 0$. We now specialize to the case when $n=2$ and $q=1$. Then

\begin{align*} \text{tr}(\mathcal{T}(Y))& ={-}2\left(\frac{\text{tr}(Y)}{\sqrt{\det(Y)}}\right)\\ \det(\mathcal{T}(Y))& =\det\left[\frac{2}{\sqrt{\det(Y)}}(Y-{\rm tr} (Y)I)\right]=\frac{4}{\det(Y)}\det(Y-{\rm tr} (Y)I)\\ & =\frac{2}{\det(Y)}(({\rm tr} (Y-{\rm tr} (Y)I))^2-{\rm tr} ((Y-{\rm tr} (Y)I)^2))\\ & =\frac{2}{\det(Y)}(({\rm tr} (Y))^2-{\rm tr} (Y^2-2{\rm tr} (Y)Y+({\rm tr} (Y))^2I))\\ & =\frac{2}{\det(Y)}(({\rm tr} (Y))^2-{\rm tr} (Y^2))=\frac{4}{\det(Y)}\det(Y)=4 \end{align*}

Assume that $Z\in \text {im}(\mathcal {T})$ and that furthermore $Y$ is a diagonal matrix. Then we recall that $Y-{\rm tr} (Y)I=-\text{adj} (Y)$. This gives us the equation

\[ \frac{2}{\sqrt{\det(Y)}}[Y-{\rm tr} (Y)I]=Z,\ \Longleftrightarrow -2\frac{\text{adj} (Y)}{\sqrt{\det(Y)}}=Z \]

or upon multiplying by $Y$,

(4.3)\begin{equation} -2\sqrt{\det(Y)}I=YZ. \end{equation}

Set $t=\sqrt {\det (Y)}$. This gives the one-parameter family of solutions

(4.4)\begin{equation} Y(t)={-}2tZ^{{-}1} \end{equation}

for $t>0$ given a solution $Y_0$ such that $\mathcal {T}(Y_0)=Z$. For example, if $Y_0=I$, then $Z=-2I$, and $Y(t)=tI$. Again one may verify that no two solutions of (4.4) satisfy the assumptions of theorem 4.1.

Example 4.6 Let $W:\mathcal {L}({\mathbb {R}}^n)\to {\mathbb {R}}$ be given by

\[ W(X)=\vert X\vert^p+\vert X^{{-}1}\vert^p\vert \det(X)\vert. \]

This is an example from [Reference Iwaniec, Martin and Onninen17] and comes from geometric function theory and non-linear elasticity. $W$ is polyconvex when restricted to $\text {GL}_+({\mathbb {R}}^n)$ but not on the entire $\mathcal {L}({\mathbb {R}}^n)$. Using (A.2), we find that

\begin{align*} DW(X)& =p\vert X\vert^{p-2}X-p\vert \det(X)\vert\vert X^{{-}1}\vert^{p-2}(X^\ast X X^\ast)^{{-}1}\nonumber\\& \quad +\text{sgn}(\det(X))\vert X^{{-}1}\vert^{p}\text{adj}(X)^\ast,\nonumber \end{align*}

\begin{align*} X^\ast DW(X)& =p\vert X\vert^{p-2}X^\ast X-p\vert \det(X)\vert\vert X^{{-}1}\vert^{p-2}X^\ast(X^\ast X X^\ast)^{{-}1}\nonumber\\& \quad +\text{sgn}(\det(X))\vert X^{{-}1}\vert^{p}X^\ast\text{adj}(X)^\ast\nonumber\\ & =p\vert X\vert^{p-2}X^\ast X-p\vert \det(X)\vert\vert X^{{-}1}\vert^{p-2}(X^\ast X)^{{-}1}\nonumber\\& \quad +\text{sgn}(\det(X))\vert X^{{-}1}\vert^{p}\det(X)I\nonumber\\ & =p\vert X\vert^{p-2}X^\ast X-p\vert \det(X)\vert\vert X^{{-}1}\vert^{p-2}(X^\ast X)^{{-}1}+\vert \det(X)\vert\vert X^{{-}1}\vert^{p} I \nonumber\end{align*}

Thus,

\[ T(X)=p\vert X\vert^{p-2}X^\ast X-p\vert \det(X)\vert\vert X^{{-}1}\vert^{p-2}(X^\ast X)^{{-}1}-\vert X\vert^{p} I. \]

The reduced energy–momentum tensor becomes

\[ \mathcal{T}(Y)=p(\text{tr}(Y))^{(p-2)/2}Y-p\sqrt{\det(Y)}\text{tr}(Y^{{-}1})^{(p-2)/2}Y^{{-}1}-\text{tr}(Y)^{p/2} I. \]

We now assume that $p=n=2$. By the Cayley–Hamilton theorem, we have

\[ Y^{{-}1}=\frac{1}{\det(Y)}[\text{tr}(Y)I-Y]. \]

This gives

\begin{align*} \mathcal{T}(Y)& =2Y-2\sqrt{\det(Y)}Y^{{-}1}-\text{tr}(Y)I\\& =2Y-2\sqrt{\det(Y)}\frac{1}{\det(Y)}[\text{tr}(Y)I-Y]-\text{tr}(Y)I\\ & =2Y-2\frac{\text{tr}(Y)I}{\sqrt{\det(Y)}}+\frac{2}{\sqrt{\det(Y)}}Y-\text{tr}(Y)I\\ & =2\left(1+\frac{1}{\sqrt{\det(Y)}}\right)Y-\left(1+\frac{2}{\sqrt{\det(Y)}}\right)\text{tr}(Y)I \end{align*}

Let $Y=\alpha I$, $\alpha >0$. Then

\begin{align*} \mathcal{T}(\alpha I)& =2\left(1+\frac{1}{\alpha}\right)\alpha I-\left(1+\frac{2}{\alpha}\right)\text{tr}(\alpha I)I\\ & =2\left(1+\frac{1}{\alpha}\right)\alpha I-\left(1+\frac{2}{\alpha}\right)2\alpha I\\ & =2\alpha\left(1+\frac{1}{\alpha}-1-\frac{2}{\alpha}\right)I={-}2I. \end{align*}

Thus, $\mathcal {T}$ is not injective, but there exists no two $\alpha _1\neq \alpha _2$ so that $\alpha _1I$ and $\alpha _2I$ satisfy the assumptions of theorem 4.1.

Example 4.7 Ball class

Let $\Omega \subset {\mathbb {R}}^n$ be a Lipschitz domain and define the class of mappings

\[ \mathscr{A}_{p,q}:=\{u:\Omega \to {\mathbb{R}}^n: Du\in L^p,\ \text{adj} (Du)\in L^q\}, \]

where $p\geq n-1$ and $q\geq p/(p-1)$. This is studied in [Reference Ball1, Reference Ball2, Reference Šverák29], see also [Reference Fusco and Hutchinson12] for a similar class of polyconvex functionals. The associated energy to this function class is

\[ I[u]=\int_{\Omega}\vert Du(x)\vert^p+\vert \text{adj}(Du)\vert^q\,{\rm d}x \]

and

\[ W(X)=\vert X\vert^p+\vert \text{adj}X\vert^q \]

which is a frame-indifferent strictly polyconvex function. Using that $W(X)= \vert {\rm tr} (X^\ast X)\vert ^{p/2}+ \vert {\rm tr} (\text{adj} (X)^\ast X)\vert ^{q/2}$ and lemma A.11, we have

\begin{align*} DW(X)& =p\vert X\vert^{p-2}X+q\vert \text{adj} (X)\vert^{q-2}(-\det(X)^{{-}1}\text{adj} (X^\ast X X^\ast)\\& \quad+\det(X)^{{-}1}\vert \text{adj} (X)\vert^2\text{adj} (X)^\ast)\\ T(X)& =p\vert X\vert^{p-2}X^\ast X+q\vert \text{adj} (X)\vert^{q-2}({-}\text{adj} (X^\ast X)+\vert \text{adj} (X)\vert^2I) \end{align*}

and thus,

\begin{align*} T(I)& =p\vert I\vert^{p-2}I^\ast I+q\vert \text{adj} (I)\vert^{q-2}({-}\text{adj} (I^\ast I)+\vert \text{adj} (I)\vert^2I)\\ & =pn^{(p-2)/2}I+q(n-1)n^{(q-2)/2}I. \end{align*}

Thus, $T$ does not satisfy the assumptions of proposition 3.4.

The reduced energy–momentum tensor becomes

\[ \mathcal{T}(Y)=p\vert {\rm tr} (Y)\vert^{(p-2)/2}Y+q\vert {\rm tr} (\text{adj} (Y))\vert^{(q-2)/2}({-}\text{adj} (Y)+ {\rm tr} (\text{adj} (Y)I) \]

We now specialize to dimension $n=2$ and choose $p=q=2$. We get

\[ \mathcal{T}(Y)=2Y+2({-}\text{adj} (Y)+{\rm tr} (\text{adj} (Y))I). \]

If $Y$ is a diagonal matrix, then ${\rm tr} (\text{adj} (Y))={\rm tr} (Y)$ and we get

\[ \mathcal{T}(Y)=2Y-2\text{adj} (Y)+2{\rm tr} (Y)I=4Y \]

We now consider the equation $\mathcal {T}(Y)=Z$ for some positive definite diagonal matrix $Z$, which implies that $Y=Z/4$. Thus, we have a unique solution and the assumptions of theorem 4.1 are not satisfied.