부산대학교 도서관

When dealing with vectorial minimizers of functionals in the calculus of variations, a key concept for the semicontinuity is the notion of quasiconvexity, introduced by Morrey in 1952. However, the drawback of this definition lies in the difficulty to check it even for simpler Lagrangians. In some situations, such as for quadratic forms, quasiconvexity turns out to be equivalent to rank-one convexity. Still another condition, intermediate between convexity and quasiconvexity, is polyconvexity, introduced by J. Ball. \par The purpose of this article is to continue the analysis of quadratic forms using the link between different notions in the calculus of variations such as quasiconvexity, polyconvexity, rank-one convexity and corresponding interpretations in algebraic geometry. Quasiconvex quadratic forms in applied mathematics correspond to nonnegative biquadratic forms in algebraic geometry, while polyconvex ones are sums of squares in algebraic geometry. Let $\Cal{C}_{N, n}$ denote the convex cone of $N\times n$ quasiconvex quadratic forms. \par An important concept was introduced by Milton and explored in a series of papers; see [20,21,22,23]. \par Definition 1. A quasiconvex quadratic form $f(\xi)\: \Bbb{R}^{N\times n}\to \Bbb{R}$ is called \roster \item"$\bullet$" a weak (or Milton) extremal if one can not subtract a convex form from it, preserving the quasiconvexity; \item"$\bullet$" an extreme ray of $\Cal{C}_{N,n}$ (or a strong extremal) if one can not subtract a quasiconvex form from it, preserving the quasiconvexity of $f$. \endroster \par Extremal quasiconvex forms play an important role in applications such as bounds for effective tensors of electrical and magnetic properties of two-component two-dimensional composites. Another related problem concerns quadratic forms for which the associated tensor (acoustic tensor) has particular symmetry properties. In the case $n=N=3$, let $ f$ be a quadratic form on $\Bbb R^{3\times 3}$ corresponding to a linear elastic cubic tensor $T$ of rank four, $ f(\xi)=\xi T \xi^t$ where $\xi=\{\xi_{i,j}\}_{i,j=1}^3$. The first author and Milton showed that if $f$ is quasiconvex, then it can be represented as a sum of convex and null-Lagrangian form. Next, the authors turn to quadratic forms with cyclic and axis-reflexion symmetry. A particular such form is $$ q(\xi)=(\xi_{11}^2+\xi_{22}^2+\xi_{33}^2)-2(\xi_{11}\xi_{22}+\xi_{22}\xi_{33}+\xi_{33}\xi_{11})+(\xi_{12}^2+\xi_{23}^2+\xi_{31}^2). $$ They establish that this quadratic form is quasiconvex but not polyconvex; moreover, $q$ is shown to be strong extremal. In algebraic geometry, the problem of expressing a nonnegative homogeneous polynomial as a sum of squares was known as one of Hilbert's problems and was solved earlier by Choi and Lam in [17]. \par The first author and Milton found an interesting connection between extremal polynomials and extremal elasticity tensors that are at the boundary of being rank-one convex. They proved that an elasticity tensor with orthotropic symmetry is extreme if the determinant of its acoustic tensor is an extremal polynomial that is not a perfect square. In addition, they have some results concerning weak extremals. Namely, in the case $d=3$ they proved that if the determinant of the acoustic tensor of the form is identically zero, then the form is either weak extremal or polyconvex. Also, if the determinant of the acoustic tensor of the form is a perfect square, then the form is either weak extremal or polyconvex, or is a sum of a rank-1 form and an extremal whose acoustic tensor determinant is identically zero. \par In the present paper the authors are able to upgrade some of the previous results to the case of extreme rays (or strong extremal): \par Theorem 2. Let $f(\xi)= \xi C\xi^T \in \Cal{C}_3$, where $\xi \in \Bbb{R}^{3\times 3}$ and $C$ is a fourth-order tensor with the usual symmetries: $ C_{ijkl}=C_{kjil}=C_{ilkj}$. Assume that the determinant of the $y$-matrix of $f(x\otimes y)$ is an extremal polynomial. Then one has one of the following: \roster \item"$\bullet$" If the determinant of the acoustic tensor is not a perfect square, then $f$ must be an extreme ray of $\Cal{C}_3$. \item"$\bullet$" If the determinant of the acoustic tensor is a perfect square, then $f$ is either an extreme ray of $\Cal{C}_3$ or polyconvex. \endroster \par The main tool is the following result: \par Theorem 3. Let $f(\xi)= \xi C\xi^T \in \Cal{C}_3$, where $\xi \in \Bbb{R}^{3\times 3}$ and $C$ is a fourth-order tensor with the usual symmetries. Assume that the determinant of the $y$-matrix of $f(x\otimes y)$ is identically zero. Then $f$ must be a polyconvex form. \par Next, based on some recent results of Helton [24], the authors conjecture that any non-polyconvex weak extremal $f\in \Cal{C}_3$ is an extreme ray of $\Cal{C}_3$, and if $f\in \Cal{C}_3$ is a non-polyconvex extreme ray, then the determinant of the acoustic tensor is an extremal polynomial different from a perfect square.