A Boundedness Result for Minimizers of Some Polyconvex Integrals

Abstract

We consider polyconvex functionals of the Calculus of Variations defined on maps from the three-dimensional Euclidean space into itself. Counterexamples show that minimizers need not to be bounded. We find conditions on the structure of the functional, which force minimizers to be locally bounded.

Appendix: Comparison Between Two Structures

Lemma A.1

We assume that $F^{\alpha }, G^{\alpha }: {\mathbb {R}}^3 \mapsto [0, +\infty [$ and $H:{\mathbb {R}} \mapsto [0, +\infty [$; let $p, q, r \in ]0, +\infty [$ with $p \ne 2$. Then, it is false that

$$\begin{aligned} \sum \limits _{\alpha = 1}^{3} F^{\alpha }(\xi ^{\alpha }) + \sum \limits _{\alpha = 1}^{3} G^{\alpha }((\mathrm {adj}_2 \xi )^{\alpha }) + H( \det \xi ) = |\xi |^p + |\mathrm {adj}_2 \xi |^q + | \det \xi |^r \end{aligned}$$

(76)

for every $\xi \in {\mathbb {R}}^{3 \times 3}$.

Proof

We argue by contradiction: if (76) holds true, then we can use (76) with

$$\begin{aligned} \xi = \left( \begin{array}{lll} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 \end{array} \right) \end{aligned}$$

(77)

and we get

$$\begin{aligned} \text{ adj }_2\xi = \left( \begin{array}{lll} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 \end{array} \right) , \end{aligned}$$

(78)

with $\det \xi = 0$, so that

$$\begin{aligned} \sum \limits _{\alpha = 1}^{3} F^{\alpha }((0, 0, 0)) + \sum \limits _{\alpha = 1}^{3} G^{\alpha }((0, 0, 0)) + H( 0 ) = 0; \end{aligned}$$

(79)

we keep in mind that $F^{\alpha }, G^{\alpha }, H \ge 0$ and we get

$$\begin{aligned} F^{\alpha }((0, 0, 0)) = G^{\alpha }((0, 0, 0)) = H( 0 ) = 0, \end{aligned}$$

(80)

for every $\alpha = 1, 2, 3$. Now we use (76) with

$$\begin{aligned} \xi = \left( \begin{array}{lll} t &{} 0 &{} 0\\ 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 \end{array} \right) \end{aligned}$$

(81)

and we get

$$\begin{aligned} \text{ adj }_2\xi = \left( \begin{array}{lll} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 \end{array} \right) , \end{aligned}$$

(82)

with $\det \xi = 0$, so that

$$\begin{aligned} F^{1}((t, 0, 0)) + F^{2}((0, 0, 0)) + F^{3}((0, 0, 0)) + \sum \limits _{\alpha = 1}^{3} G^{\alpha }((0, 0, 0)) + H( 0 ) = |t|^p;\quad \end{aligned}$$

(83)

we keep in mind (80) and we get

$$\begin{aligned} F^{1}((t, 0, 0)) = |t|^p, \end{aligned}$$

(84)

for every $t \in {\mathbb {R}}$. In a similar manner, taking

$$\begin{aligned} \xi = \left( \begin{array}{lll} 0 &{} 0 &{} 0\\ t &{} 0 &{} 0\\ 0 &{} 0 &{} 0 \end{array} \right) , \end{aligned}$$

(85)

we get

$$\begin{aligned} F^{2}((t, 0, 0)) = |t|^p, \end{aligned}$$

(86)

for every $t \in {\mathbb {R}}$. In the same way, taking

$$\begin{aligned} \xi = \left( \begin{array}{lll} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0\\ t &{} 0 &{} 0 \end{array} \right) , \end{aligned}$$

(87)

we get

$$\begin{aligned} F^{3}((t, 0, 0)) = |t|^p, \end{aligned}$$

(88)

for every $t \in {\mathbb {R}}$. Eventually, we take

$$\begin{aligned} \xi = \left( \begin{array}{lll} 1 &{} 0 &{} 0\\ 1 &{} 0 &{} 0\\ 1 &{} 0 &{} 0 \end{array} \right) \end{aligned}$$

(89)

and (76) implies

$$\begin{aligned} \sum \limits _{\alpha = 1}^{3} F^{\alpha }((1, 0, 0)) + \sum \limits _{\alpha = 1}^{3} G^{\alpha }((0, 0, 0)) + H( 0 ) = 3^{p/2}; \end{aligned}$$

(90)

we use (80), (84), (86), (88) and we get

$$\begin{aligned} 3 = 3^{p/2}: \end{aligned}$$

(91)

such an equality is a contradiction, since $p \ne 2$. This ends the proof of Lemma A.1.

$\square$