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.

1 Introduction

Let us consider polyconvex integrals of the Calculus of Variations. Partial regularity results (that is, the regularity of minimizers up to a subset of the set of definition and the study of the properties of the singular set; see for example Section 4.2 in [1] and Section 1 in [2]) are contained in [3,4,5,6,7,8,9,10]. Only few everywhere regularity results are available: [11] where the everywhere continuity is proved in the two-dimensional case, [12] where Hölder continuity for extremals is dealt with in dimension two, [13] where local boundedness is proved in the three-dimensional case. Global pointwise bounds are in [14,15,16,17,18,19]. Interesting results are contained in [20,21,22,23,24,25]; see also [26, 27]. Let us come back to [13]; in such a paper, the authors make an important step toward regularity: they prove boundedness of minimizers in the three-dimensional case; unfortunately, they make restrictions that rule out the most important polyconvex integral. In the present paper, we find a different set of assumptions, which allows us to deal with such a polyconvex integral. In the next section, we write assumptions and results; in Sect. 3 we collect some preliminaries and, in Sect. 4, we give the proof of the main theorem.

2 Assumptions and Results

In this paper we study the regularity of vectorial local minimizers of integral functionals

$$\begin{aligned} I(v,\varOmega ) =\int _\varOmega f(x,\,Dv(x)) \hbox {d}x, \end{aligned}$$

(1)

where $\varOmega \subset {\mathbb {R}}^3$ is an open, bounded set, $v:\varOmega \subset {\mathbb {R}}^3 \rightarrow {\mathbb {R}}^3$, $v=(v^1, v^2, v^3)$ and Dv is the Jacobian matrix of its partial derivatives

$$\begin{aligned} Dv= \left( v^\alpha _{x_i} \right) ^{\alpha =1,2,3} _{i=1,2,3} =\left( \begin{array}{llll} Dv^1\\ Dv^2\\ Dv^3 \end{array} \right) =\left( \begin{array}{llll} v^1_{x_1} &{} v^1_{x_2} &{} v^1_{x_3} \\ v^2_{x_1} &{} v^2_{x_2} &{} v^2_{x_3} \\ v^3_{x_1} &{} v^3_{x_2} &{} v^3_{x_3} \end{array} \right) , \end{aligned}$$

moreover, $f: \varOmega \times {\mathbb {R}}^{3\times 3} \rightarrow [0, +\infty [$ is a Carathéodory function such that for fixed x

$$\begin{aligned} \xi \rightarrow f(x, \xi )\; \mathrm { is \;polyconvex} \end{aligned}$$

that is

$$\begin{aligned} f(x,\xi ) = g(x, \xi , \text{ adj }_2 \xi , \text{ det }\xi ) \quad \text { with } \quad (\xi ,\lambda ,t) \rightarrow g(x, \xi , \lambda , t) \quad \text { convex}, \end{aligned}$$

(2)

see [28, 29]. When dealing with models in nonlinear elasticity, f is the stored-energy function; moreover, $\xi , \text{ adj }_2 \xi , \text{ det }\xi$ govern the deformation of line, surface and volume elements respectively. Our model is

$$\begin{aligned} f(x, Dv) = |Dv|^p + |\text{ adj }_2Dv|^q + |\text{ det }Dv|^r, \end{aligned}$$

(3)

where $\det Dv$ is the determinant of the matrix Dv, and $\text{ adj }_2 Dv$ denotes the adjugate matrix of order 2, whose components are

$$\begin{aligned}(\text{ adj }_2 Dv )_{ij} =(-1) ^{i +j} \det \left( \begin{array}{llll} v^\alpha _{x_k}, \ v^\alpha _{x_\ell }\\ v^\beta _{x_k}, \ v^\beta _{x_\ell } \end{array} \right) , \ \ i, j \in \{1,2,3\}, \end{aligned}$$

with $\alpha ,\beta \in \{1,2,3\}\setminus \{i\}$, $\alpha <\beta$, and $k,\ell \in \{1,2,3\} \setminus \{j\}$, $k<\ell$. Moreover, $(\text{ adj }_2 Dv)^\alpha$ denotes the $\alpha -$row of $\text{ adj }_2 Dv$, that is

$$\begin{aligned} (\text{ adj }_2 Dv)^\alpha = \left( (\text{ adj }_2 Dv)_{\alpha 1}, (\text{ adj }_2 Dv)_{\alpha 2},(\text{ adj }_2 Dv)_{\alpha 3}\right) . \end{aligned}$$

In paper [13], the authors consider densities f for which the following splitting holds true

$$\begin{aligned} f(x,Dv) = \sum \limits _{\alpha = 1}^{3} F^\alpha (x, Dv^\alpha ) + \sum \limits _{\beta = 1}^{3} G^\beta (x, (\text{ adj }_2 Dv)^\beta ) + H(x, \det Dv) \end{aligned}$$

(4)

for suitable nonnegative functions $F^\alpha , G^\beta , H$. Note that model (3), with $p \ne 2$, cannot be written as (4); see Lemma A.1 in “Appendix A”. In this paper, we succeed in dealing with model (3) and we prove the following

Theorem 2.1

Let $\varOmega$ be a bounded and open subset of ${\mathbb {R}}^3$. Assume that $1\le r< q < p \le 3$ with $2<p$ and

$$\begin{aligned} \frac{p}{p^*}< & {} \mathrm{min} \left\{ 1- \frac{q p^*}{p(p^* - q)}, 1- \frac{rp^*}{q(p^*-r)} \right\} , \qquad \mathrm{if } \quad \, 1< q \le 2, \nonumber \\ \frac{p}{p^*}< & {} \mathrm{min} \left\{ 1- \frac{2 p^*}{p(p^* - 2)} - \frac{(q-2)p^*}{q(p^* - 2)}, 1- \frac{rp^*}{q(p^*-r)} \right\} , \qquad \mathrm{if } \quad \, 2 < q;\qquad \end{aligned}$$

(5)

then all the local minimizers $u\in W_{loc} ^{1,p} (\varOmega ; {\mathbb {R}}^3)$ of

$$\begin{aligned} \int _\varOmega (|Du|^p + |\mathrm{adj}_2 Du|^q + |\mathrm{det} Du|^r) \end{aligned}$$

(6)

are locally bounded in $\varOmega$.

We recall that $p^*$ is the Sobolev exponent: $p^* = \frac{n p}{n-p}= \frac{3p}{3-p}$ when $p<n=3$; moreover, $p^*$ is any number greater than p when $p=n=3$, so it can be chosen large enough that (5) is satisfied by assuming only $1 \le r< q < p$. We notice that we have restricted ourselves to the case $p\le 3$ because, when $p>3$, every function in $W_{loc}^{1,p} (\varOmega )$ is trivially in $L_{loc}^\infty (\varOmega )$ by the Sobolev theorem. Note that we have existence of minimizers for (6) when $2 \le p$, $\frac{p}{p-1} \le q$ and $1<r$, provided a boundary datum ${\overline{u}} \in W^{1,p}(\varOmega ; {\mathbb {R}}^3)$, with finite energy, has been fixed; see Remark 8.32 (iii) in [29] and Theorem 3.1 in [13]. Condition (5) is satisfied, for example, when $p=\frac{14}{5}$, $q=2$, $r=\frac{3}{2}$ and this gives us the following.

Corollary 2.1

Let $\varOmega$ be a bounded and open subset of ${\mathbb {R}}^3$ and let $u\in W_{loc} ^{1,\frac{14}{5}} (\varOmega ; {\mathbb {R}}^3)$ be a local minimizer of

$$\begin{aligned} \int _\varOmega (|Du|^{\frac{14}{5}} + |\mathrm{adj}_2Du|^2 + |\mathrm{det}Du|^{\frac{3}{2}}); \end{aligned}$$

(7)

then u is locally bounded in $\varOmega$.

In the framework of Corollary 2.1, we have $\frac{p}{p-1}=\frac{14}{9} < 2 = q$, so the existence of minimizers is guaranteed as in the previous lines. Theorem 2.1 is a particular case of a more general result. Let us note that model (3) suggests we assume the following structure

$$\begin{aligned} f(x, \xi ) = F(x, |\xi |^2) + G(x, |\text{ adj }_2\xi |^2) + H(x, \text{ det }\xi ), \end{aligned}$$

(8)

where F, G and H are Carathéodory nonnegative functions. We assume p-growth with respect to $\xi$, q-growth with respect to $\text{ adj }_2\xi$ and r-growth with respect to $\text{ det }\xi$

$$\begin{aligned} k_1 t^{p/2} - k_2\le & {} F(x, t) \le k_3 t^{p/2} + a(x) \end{aligned}$$

(9)

$$\begin{aligned} k_1 t^{q/2} - k_2\le & {} G(x, t) \le k_3 t^{q/2} + b(x) \end{aligned}$$

(10)

$$\begin{aligned} 0\le & {} H(x, s) \le k_3 |s|^{r} + c(x), \end{aligned}$$

(11)

where $k_1$, $k_2$, $k_3$ are constants such that $k_1, k_3 \in ]0, +\infty [$ and $k_2 \in [0, +\infty [$ and $a, b, c:\varOmega \rightarrow [0, +\infty [$ are functions in $L^\sigma (\varOmega )$, $\sigma > 1$; as far as exponents p, q, r are concerned, we assume that $2 < p \le 3$ and $1 \le r< q < p$. Now we need to control the behavior of F with respect to the sum from below

$$\begin{aligned} F(x,t_1) + F(x,t_2) - k_2 \le F(x,t_1 + t_2). \end{aligned}$$

(12)

A weaker condition is needed for G:

$$\begin{aligned} G(x,t_1) - k_2 \le G(x,t_1 + t_2). \end{aligned}$$

(13)

We also need to control the behavior of F with respect to the sum from above:

$$\begin{aligned} F(x,t_1 + t_2) \le F(x,t_1) + F(x,t_2) + k_3 t_1 t_2^{\frac{p}{2} - 1} + a(x). \end{aligned}$$

(14)

Note that in (14) there is an extra term with the product between $t_1$ and $t_2$. When $q > 2$ we assume

$$\begin{aligned} G(x,t_1 + t_2) \le G(x,t_1) + G(x,t_2) + k_3 t_1 t_2^{\frac{q}{2} - 1} + b(x). \end{aligned}$$

(15)

When $q \le 2$ we do not need the product between $t_1$ and $t_2$ any longer; we require subadditivity

$$\begin{aligned} G(x,t_1 + t_2) \le G(x,t_1) + G(x,t_2) + b(x). \end{aligned}$$

(16)

Functions F verifying the previous assumptions are $F(x,t) = \gamma (x) t^{p/2}$ and $F(x,t) = \gamma (x) (1 + t^2)^{p/4}$, provided $\gamma (x)$ is positive and away from both 0 and $+\infty$; similar examples for G and H: see Remarks 3.2, $\ldots$, 3.7. Our main result is the following

Theorem 2.2

Let $\varOmega$ be a bounded and open subset of ${\mathbb {R}}^3$ and let f be as in (8); assume that conditions (9)–(16) hold with $1\le r< q < p \le 3$ such that $2<p$ and

$$\begin{aligned} \frac{p}{p^*}< & {} \mathrm{min} \left\{ 1- \frac{q p^*}{p(p^* - q)}, 1- \frac{rp^*}{q(p^*-r)}, 1 - \frac{1}{\sigma } \right\} , \quad \quad \mathrm{if } \, 1< q \le 2, \nonumber \\ \frac{p}{p^*}< & {} \mathrm{min} \left\{ 1- \frac{2 p^*}{p(p^* - 2)} - \frac{(q-2)p^*}{q(p^* - 2)}, 1- \frac{rp^*}{q(p^*-r)}, 1 - \frac{1}{\sigma } \right\} , \quad \mathrm{if } \, 2 < q.\nonumber \\ \end{aligned}$$

(17)

Then, all the local minimizers $u\in W_{loc} ^{1,p} (\varOmega ; {\mathbb {R}}^3)$ of I are locally bounded in $\varOmega$.

Note that $\frac{1}{\sigma } = 0$, if $\sigma = \infty$. In our Theorem 2.2, we assume (8); in [13] (4) was in force: in vectorial problems, some structure conditions are due to minimizers which can be unbounded: see De Giorgi’s counterexample [30]; see also [31], Section 3 in [1] and [32]. As far as exponents p, q, r are concerned, (17) is the same as (2.5) in [13] when $1<q\le 2$; if $2<q$ then (17) seems to require a bit more than (2.5) in [13]: see comparison (72).

The integrals we consider show a ${\tilde{p}}$ growth from below and a ${\tilde{q}}$ growth from above, so we are in the class of functionals with ${\tilde{p}},{\tilde{q}}$-growth. It is now well known, as in our result, that a restriction between ${\tilde{p}}$ and ${\tilde{q}}$ must be imposed due to counterexamples in [33,34,35,36,37]; see also [38, 39]; we refer to [1] for a detailed survey on the subject.

3 Preliminaries

In this section, we recall some standard definitions and collect several lemmas useful in our proofs.

First of all, we recall the following

Definition 3.1

A function $u\in W_{loc} ^{1,1} (\varOmega ; {\mathbb {R}}^3)$ is a local minimizer of (1) if $f(Du) \in L^1 _{loc} (\varOmega )$ and

$$\begin{aligned} I(u, \text{ supp } \, \varphi ) \le I(u+\varphi , \text{ supp } \, \varphi ), \end{aligned}$$

(18)

for all $\varphi \in W^{1,1} (\varOmega , {\mathbb {R}}^3)$ with $\text{ supp } \, \varphi \subset \subset \varOmega$.

All the norms we use on ${\mathbb {R}}^3$ and ${\mathbb {R}}^{3\times 3}$ will be the standard Euclidean ones and denoted by $| \cdot |$ in all cases. In particular, for matrices $\xi , \eta \in {\mathbb {R}}^{3\times 3}$ we write $\langle \xi , \eta \rangle : = \text {trace} (\xi ^T \eta )$ for the usual inner product of $\xi$ and $\eta$, and $| \xi | : = \langle \xi , \xi \rangle ^{\frac{1}{2}}$ for the corresponding Euclidean norm.

Lemma 3.1

For $a, b\ge 0$ we have that

$$\begin{aligned} a^m +b^m\le & {} (a+b)^m, \quad \, \mathrm{if} \quad m \ge 1, \end{aligned}$$

(19)

$$\begin{aligned} (a+b)^m\le & {} a ^m+b^m +mab^{m-1}, \quad \mathrm{if} \quad 1 \le m \le 2. \end{aligned}$$

(20)

Proof

When $m=1$, (19) and (20) are easy. We are left with the case $1<m$. It is obvious that (19) and (20) hold true both for $b=0$. We now assume $b>0$ and we let $t=a/b$. It suffices to show that for $t\ge 0$,

$$\begin{aligned} t^m +1\le & {} (t+1)^m, \quad \, \mathrm{if} \quad m > 1, \end{aligned}$$

(21)

$$\begin{aligned} (t+1)^m\le & {} t ^m+1 +mt, \quad \mathrm{if} \quad 1 < m \le 2. \end{aligned}$$

(22)

In order to prove (21), we let $h(t) =(t+1)^m- t^m-1$. Since

$$\begin{aligned} h(0)= 0 \end{aligned}$$

(23)

and, by $m>1$,

$$\begin{aligned} h'(t) =m[(t+1) ^{m-1} -t ^{m-1}]\ge 0, \end{aligned}$$

(24)

then $h(t)\ge 0$ and (21) follows.

Regarding (22), we let $g(t) =(t+1)^m- t^m-mt-1$. Since

$$\begin{aligned} g(0)= & {} 0, \end{aligned}$$

(25)

$$\begin{aligned} g'(t)= & {} m[(t+1) ^{m-1} -t ^{m-1} -1]\le m[ t^{m-1}+1 -t ^{m-1} -1]=0, \end{aligned}$$

(26)

where we used $1<m\le 2$ and Remark 3.1, then (22) follows. $\square$

Remark 3.1

We recall the well-known inequality: for $a, b \ge 0$ we have

$$\begin{aligned} (a + b)^m \le a^m + b^m, \qquad \text { if } \quad 0 < m \le 1. \end{aligned}$$

(27)

Lemma 3.2

Fix $m \in [-\frac{1}{2}, +\infty [$ and consider $V:{\mathbb {R}} \rightarrow {\mathbb {R}}$ as follows

$$\begin{aligned} V(s) = \left( 1 + s^2 \right) ^{m} s; \end{aligned}$$

(28)

then, $V:{\mathbb {R}} \rightarrow {\mathbb {R}}$ is strictly increasing.

Proof

We compute the first derivative

$$\begin{aligned} V^{\prime }(s) = \left( 1 + s^2 \right) ^{m-1} [(2m+1) s^2 + 1]; \end{aligned}$$

(29)

since $m \ge -\frac{1}{2}$, we have $V^{\prime }(s) > 0$ for every $s \in {\mathbb {R}}$. This ends the proof. $\square$

Lemma 3.3

Fix $p \in [2, +\infty [$; consider $w:{\mathbb {R}}^2 \rightarrow {\mathbb {R}}$ as follows

$$\begin{aligned} w(a,b) = [1 + (a + b)^2]^{p/4} - [1 + a^2]^{p/4} - [1 + b^2]^{p/4}. \end{aligned}$$

(30)

Then,

$$\begin{aligned} a \ge 0, \quad b \ge 0 \Longrightarrow w(0,0) \le w(a,b). \end{aligned}$$

(31)

Proof

We compute the first partial derivatives:

$$\begin{aligned} \frac{\partial w}{\partial a}(a,b)= & {} \frac{p}{4} [1 + (a + b)^2]^{(p/4) - 1} 2 (a + b) - \frac{p}{4} [1 + a^2]^{(p/4) - 1} 2 a\\= & {} \frac{p}{2} \{ V(a + b) - V(a) \} \end{aligned}$$

and

$$\begin{aligned} \frac{\partial w}{\partial b}(a,b)= & {} \frac{p}{4} [1 + (a + b)^2]^{(p/4) - 1} 2 (a + b) - \frac{p}{4} [1 + b^2]^{(p/4) - 1} 2 b\\= & {} \frac{p}{2} \{ V(a + b) - V(b) \} \end{aligned}$$

where V is given by (28) with $m = (p/4) - 1$. Note that $m \ge - 1/2$ since $p \ge 2$. Then, V is increasing so that, when $a\ge 0$ and $b \ge 0$, we have $V(a + b) - V(a) \ge 0$ and $V(a + b) - V(b) \ge 0$. This shows that

$$\begin{aligned} a \ge 0, \quad b \ge 0 \Longrightarrow \frac{\partial w}{\partial a}(a,b) \ge 0, \quad \frac{\partial w}{\partial b}(a,b) \ge 0. \end{aligned}$$

(32)

Then, $a \rightarrow w(a,b)$ increases and $b \rightarrow w(a,b)$ increases too, if we restrict ourselves to $a \ge 0$ and $b \ge 0$; thus,

$$\begin{aligned} w(0,0) \le w(0,b) \le w(a,b), \end{aligned}$$

(33)

provided $b \ge 0$ and $a \ge 0$. This ends the proof. $\square$

Corollary 3.1

Fix $p \in [2, +\infty [$; then,

$$\begin{aligned} a \ge 0, \quad b \ge 0 \Longrightarrow [1 + a^2]^{p/4} + [1 + b^2]^{p/4} - 1 \le [1 + (a + b)^2]^{p/4}. \end{aligned}$$

(34)

Proof

We write (31) explicitly and we get (34). $\square$

Lemma 3.4

Fix $p \in ]2, 3]$. If $a \ge 0$ and $b \ge 0$, then

$$\begin{aligned}{}[1 + (a + b)^2]^{p/4} \le [1 + a^2]^{p/4} + [1 + b^2]^{p/4} + \frac{p}{2} a b^{(p/2) - 1} + 1. \end{aligned}$$

(35)

Proof

Since $p \in ]2, 3]$ we have $\frac{p}{4} \in ]\frac{1}{2}, \frac{3}{4}]$ and we can use (27) with $m = \frac{p}{4}$:

$$\begin{aligned}{}[1 + (a + b)^2]^{p/4} \le [1]^{p/4} + [(a + b)^2]^{p/4} = 1 + (a + b)^{p/2}; \end{aligned}$$

(36)

now $\frac{p}{2} \in ]1, \frac{3}{2}]$ and we can use (20) with $m = \frac{p}{2}$:

$$\begin{aligned} 1 + (a + b)^{p/2}\le & {} 1 + a^{p/2} + b^{p/2} + \frac{p}{2} a b^{(p/2) - 1} \nonumber \\= & {} 1 + (a^2)^{p/4} + (b^2)^{p/4} + \frac{p}{2} a b^{(p/2) - 1} \nonumber \\\le & {} 1 + [1 + a^2]^{p/4} + [1 + b^2]^{p/4} + \frac{p}{2} a b^{(p/2) - 1}. \end{aligned}$$

(37)

This ends the proof. $\square$

Lemma 3.5

Fix $q \in ]1, 2]$. Then,

$$\begin{aligned} a \ge 0, \quad b \ge 0 \Longrightarrow [1 + (a + b)^2]^{q/4} \le [1 + a^2]^{q/4} + [1 + b^2]^{q/4} + 1. \end{aligned}$$

(38)

Proof

Since $q \in ]1, 2]$ we have $\frac{q}{4} \in ]\frac{1}{4}, \frac{1}{2}]$ and we can use (27) with $m = \frac{q}{4}$:

$$\begin{aligned}{}[1 + (a + b)^2]^{q/4} \le 1^{q/4} + [(a + b)^2]^{q/4} = 1 + (a + b)^{q/2}; \end{aligned}$$

(39)

now $\frac{q}{2} \in ]\frac{1}{2}, 1]$ and we can use (27) with $m = \frac{q}{2}$:

$$\begin{aligned} 1 + (a + b)^{q/2}\le & {} 1 + a^{q/2} + b^{q/2} \nonumber \\= & {} 1 + (a^2) ^{q/4} +(b^2) ^{q/4} \nonumber \\\le & {} 1 + [1 + a^2]^{q/4} + [1 + b^2]^{q/4}. \end{aligned}$$

(40)

This ends the proof. $\square$

Now we are able to give examples of functions F, G, H verifying conditions required in Theorem 2.2.

Remark 3.2

Fix $p \in ]2,3]$ and define

$$\begin{aligned} F(x,t) = \gamma (x) t^{p/2} \end{aligned}$$

(41)

for $t \in [0, +\infty [$, where $\gamma _1 \le \gamma (x) \le \gamma _2$ with $\gamma _1, \gamma _2 \in ]0, +\infty [$. Then (9), (12), (14) hold true with $k_1 = \gamma _1$, $k_2 = 0$, $k_3 = \frac{p}{2} \gamma _2$, $a(x) = 0$. Indeed, we use (19) and (20) with $m = p/2$ in Lemma 3.1 and we are done.

Remark 3.3

Fix $q \in ]1,3[$ and define

$$\begin{aligned} G(x,t) = \gamma (x) t^{q/2} \end{aligned}$$

(42)

for $t \in [0, +\infty [$, where $\gamma _1 \le \gamma (x) \le \gamma _2$ with $\gamma _1, \gamma _2 \in ]0, +\infty [$. Then, when $q>2$, (10), (13), (15) hold true with $k_1 = \gamma _1$, $k_2 = 0$, $k_3 = \frac{q}{2} \gamma _2$, $b(x) = 0$. Indeed, we use (20) with $m = q/2$ in Lemma 3.1 and we are done. Moreover, when $q \le 2$, (10), (13), (16) hold true with $k_1 = \gamma _1$, $k_2 = 0$, $k_3 = \gamma _2$, $b(x) = 0$. Indeed, when $q \le 2$, we use the well-known inequality (27) with $m = q/2$ and we are done.

Remark 3.4

Fix $r \in [1,3[$ and define

$$\begin{aligned} H(x,s) = \gamma (x) |s|^{r} \end{aligned}$$

(43)

for $s \in {\mathbb {R}}$, where $\gamma _1 \le \gamma (x) \le \gamma _2$ with $\gamma _1, \gamma _2 \in ]0, +\infty [$. Then, (11) holds true with $k_3 = \gamma _2$, $c(x) = 0$.

Remark 3.5

Fix $p \in ]2,3]$ and define

$$\begin{aligned} F(x,t) = \gamma (x) [1 + t^2]^{p/4} \end{aligned}$$

(44)

for $t \in [0, +\infty [$, where $\gamma _1 \le \gamma (x) \le \gamma _2$ with $\gamma _1, \gamma _2 \in ]0, +\infty [$. Then (9), (12), (14) hold true with $k_1 = \gamma _1$, $k_2 = \gamma _2$, $k_3 = \frac{p}{2} \gamma _2$, $a(x) = \gamma _2$. Indeed, we use (27) with $m = p/4$, (34) and (35).

Remark 3.6

Fix $q \in ]1,3[$ and define

$$\begin{aligned} G(x,t) = \gamma (x) [1 + t^2]^{q/4} \end{aligned}$$

(45)

for $t \in [0, +\infty [$, where $\gamma _1 \le \gamma (x) \le \gamma _2$ with $\gamma _1, \gamma _2 \in ]0, +\infty [$. Then, when $q>2$, (10), (13), (15) hold true with $k_1 = \gamma _1$, $k_2 = 0$, $k_3 = \frac{q}{2} \gamma _2$, $b(x) = \gamma _2$. Indeed, we use (27) with $m = q/4$, (35) and we are done. Moreover, when $q \le 2$, (10), (13), (16) hold true with $k_1 = \gamma _1$, $k_2 = 0$, $k_3 = \gamma _2$, $b(x) = \gamma _2$. Indeed, when $q \le 2$, we use (27) with $m = q/4$, (38) and we are done.

Remark 3.7

Fix $r \in [1,3[$ and define

$$\begin{aligned} H(x,s) = \gamma (x) [1 + |s|^2]^{r/2} \end{aligned}$$

(46)

for $s \in {\mathbb {R}}$, where $\gamma _1 \le \gamma (x) \le \gamma _2$ with $\gamma _1, \gamma _2 \in ]0, +\infty [$. Then, (11) holds true with $k_3 = 2^{r/2} \gamma _2$, $c(x) = 2^{r/2} \gamma _2$.

The following lemma can be found in [13] as Lemma 4.1.

Lemma 3.6

Consider the matrices $A,B \in {\mathbb {R}} ^{3\times 3}$:

$$\begin{aligned} A=\left( \begin{array}{llll} A^1\\ B^2\\ B^3 \end{array} \right) , \quad B=\left( \begin{array}{llll} B^1\\ B^2\\ B^3 \end{array} \right) . \end{aligned}$$

Then, the following estimates hold:

(a):

$|A|\le |A^1|+|B^2| +|B^3|$,

(b):

$|\det A| \le |A^1| |({\mathrm{adj}_2 B})^1|$,

(c):

$|(\mathrm{adj}_2 A)_{2j}| \le |A^1| |B^3|$ and $|(\mathrm{adj}_2 A)_{3j}| \le |A^1| |B^2|$, for all $j\in \{ 1,2,3\}$.

In order to get our main result, we have to prove a suitable Caccioppoli-type inequality for any component $u^\alpha$ of the local minimizer u of functional I (1) on every superlevel set $\{u^\alpha >k\}$. To this goal, we will use the following lemma (see [13] for a proof).

Lemma 3.7

Let $\varOmega$ be an open subset of ${\mathbb {R}}^3$. Consider a Carathéodory function $f: \varOmega \times {\mathbb {R}}^{3\times 3} \rightarrow [0,+\infty [$. Assume that there exist $c_1,c_3>0$ and $c_2\ge 0$ such that, for every $\xi \in {\mathbb {R}}^{3\times 3}$,

$$\begin{aligned} c_1 (|\xi |^p +|\mathrm{adj}_2\xi |^q) -c_2 \le f(x,\xi ) \le c _3 (|\xi |^p +|(\mathrm{adj}_2 \xi )| ^q +|\det \xi | ^r +1 +\omega (x)), \end{aligned}$$

with $1\le p, 1\le q, 1\le r, \omega (x) \ge 0$.

Let $u\in W_{loc}^{1,p} (\varOmega ; {\mathbb {R}}^3)$ be such that $x\rightarrow f(x,Du(x)) \in L_{loc} ^1 (\varOmega )$. Fix $\eta \in C_0 ^1 (\varOmega )$, $\eta \ge 0$ and $k\in {\mathbb {R}}$, and denote, for almost every $x\in \{u^1>k\} \cap \{\eta >0\}$,

$$\begin{aligned} A =\left( \begin{array}{cccc} \mu \eta ^{-1} (k-u^1) D\eta \\ Du^2\\ Du^3 \end{array} \right) . \end{aligned}$$

If

$$\begin{aligned} q< \frac{p^*p}{p^*+p} \ \ \text{ and } \ \ r <\frac{p^*q}{p^*+q} \end{aligned}$$

and $\omega \in L_{loc} ^1 (\varOmega )$, then

$$\begin{aligned} \eta ^\mu f(x,A) \in L^1 (\{u^1>k\} \cap \{\eta >0\}), \ \ \forall \mu \ge p^*. \end{aligned}$$

4 Proof of Theorem 2.2

We want to stress that the proof of our result follows the idea used in [13]: we provide the local boundedness of the minimizers by proving that each component is locally bounded. In the following lemma, we refer to the first component $u^1$: the core of the proof lies in the following Caccioppoli-type inequality, obtained on every superlevel set $\{u^1 > k \}$. We keep in mind that $p^* = \frac{np}{n-p}$ if $p<n=3$ and $p^*$ is any number $>p$ when $p=n=3$.

Proposition 4.1

(Caccioppoli-type estimate) Let f be as in (8) satisfying (9)–(16) with $1\le r< q < p \le 3$ such that

$$\begin{aligned} 2< p, \quad q< \frac{pp^*}{p+p^*}, \quad r<\frac{p^*q}{p^*+q}. \end{aligned}$$

(47)

Let $u\in W_{loc}^{1,p} (\varOmega ; {\mathbb {R}}^3)$ be a local minimizer of I. Let $B_R(x_0) \subset \subset \varOmega$ with $|B_R(x_0)|<1$; fixed $k\in {\mathbb {R}}$, denote

$$\begin{aligned} A_{k,\tau } ^1:= \{x\in B_\tau (x_0): u^1(x) >k\} \quad \ 0<\tau \le R. \end{aligned}$$

Then, there exists $C =C(k_1, k_2, k_3, p, q, r, p^*)>0$ such that, for every $0<s<t\le R$:

$$\begin{aligned} \begin{array}{lllll} &{}\displaystyle \int _{A_{k,s}^1 } |Du^1| ^p\hbox {d}x \le C \int _{A_{k,t}^1} \left( \frac{u^1-k}{t-s}\right) ^{p^*} \mathrm{d}x \displaystyle +C \biggm \{ 1+ \Vert a+b+c\Vert _{L^{\sigma }(B_R)}\\ &{} \quad + \displaystyle \left( \int _{B_R} (|D u^2|+|Du^3|) ^{p} \mathrm{d}x\right) ^{\frac{p^*(p-2)}{p(p^* - 2)} } \displaystyle +\left( \int _{B_R} \left( |Du^2| +|Du^3|\right) ^p \mathrm{d}x \right) ^{\frac{q p^*}{p(p^* - q)}}\\ &{} \quad \displaystyle + \left( \int _{B_R} |( \mathrm{adj}_2 Du) ^ 1 |^q \, \mathrm{d}x \right) ^{\frac{rp^*}{q(p^*-r)}} +1_{(2,+\infty )}(q) \left( \displaystyle \int _{B_R} (|D u^2| + |D u^3|)^p \mathrm{d}x\right) ^{\frac{2 p^*}{p(p^* - 2)}}\\ &{} \quad \times \left( \displaystyle \int _{B_R} |(\text{ adj }_2 Du)^1 |^q \mathrm{d}x \right) ^{\frac{(q-2)p^*}{q(p^* - 2)}} \biggm \} |A_{k,t}^1|^\theta , \end{array} \end{aligned}$$

(48)

where

$$\begin{aligned}&\theta := \mathrm{min} \left\{ 1- \frac{q p^*}{p(p^* - q)}, 1- \frac{rp^*}{q(p^*-r)}, 1 - \frac{1}{\sigma } \right\} , \quad \mathrm{if } \, 1< q \le 2, \\&\theta := \mathrm{min} \left\{ 1- \frac{2 p^*}{p(p^* - 2)} - \frac{(q-2)p^*}{q(p^* - 2)}, 1- \frac{rp^*}{q(p^*-r)}, 1 - \frac{1}{\sigma } \right\} , \quad \mathrm{if } \, 2 < q, \end{aligned}$$

with $\frac{1}{\sigma } = 0$ if $\sigma = \infty$.

Proof

The condition $|B_R(x_0)|=\frac{4\pi R^3}{3} <1$ ensures $R<1$. Let s, t be such that $0<s<t\le R$. Consider a cutoff function $\eta \in C_0^\infty (B_t(x_0))$ satisfying the following assumptions:

$$\begin{aligned} 0\le \eta \le 1, \ \eta \equiv 1 \text{ in } B_s(x_0), \ |D\eta | \le \frac{2 }{t-s}. \end{aligned}$$

Fixing $k\in {\mathbb {R}}$, define $w \in W_{loc} ^{1,p} (\varOmega ; {\mathbb {R}}^3)$,

$$\begin{aligned} w ^1 :=\max \{u^1-k, 0 \}, \ w ^2=0, \ w ^3 =0, \end{aligned}$$

and, for $\mu = p^*$,

$$\begin{aligned} \varphi := -\eta ^\mu w. \end{aligned}$$

For almost every $x \in \varOmega \setminus (\{\eta>0\} \cap \{u^1>k\})$ we have $\varphi =0$, thus

$$\begin{aligned} f(x,Du+D\varphi ) =f(x,Du) \end{aligned}$$

(49)

almost everywhere in $\varOmega \setminus ({\{\eta>0\} \cap \{u^1>k\}} )$.

For almost every $x\in \{\eta>0\} \cap \{u^1>k\}$ denote

$$\begin{aligned} A=\left( \begin{array}{cccc} \mu \eta ^{-1} (k-u^1) D\eta \\ Du^2\\ Du^3 \end{array} \right) . \end{aligned}$$

(50)

We notice that

$$\begin{aligned} Du+D\varphi =\left( \begin{array}{cccc} (1-\eta ^\mu ) Du^1 +\mu \eta ^{\mu -1} (k-u^1) D\eta \\ Du^2\\ Du^3 \end{array} \right) =(1-\eta ^\mu ) Du +\eta ^\mu A. \end{aligned}$$

Moreover, since for almost every $x\in \{\eta>0\} \cap \{u^1>k\}$,

$$\begin{aligned} \det (Du+D\varphi )=(1-\eta ^\mu ) \det Du +\eta ^\mu \det A \end{aligned}$$

and

$$\begin{aligned} \text{ adj }_2 (Du+D\varphi ) =(1-\eta ^\mu ) \text{ adj }_2 Du +\eta ^\mu \text{ adj }_2 A, \end{aligned}$$

then, since f is polyconvex, we get that

$$\begin{aligned} f(x,Du+D\varphi ) \le (1-\eta ^\mu ) f(x,Du) +\eta ^\mu f(x,A) \end{aligned}$$

(51)

almost everywhere in $\{\eta>0\} \cap \{u^1>k\}$.

By the minimality of u, $f(x,Du) \in L_{loc} ^1 (\varOmega )$; note that in our case we can use Lemma 3.7, deducing that

$$\begin{aligned} \eta ^\mu f(x,A) \in L^1 (\{\eta>0\} \cap \{u^1>k\}). \end{aligned}$$

Therefore, (49) and (51) imply $f(x,Du+D\varphi ) \in L_{loc}^1 (\varOmega )$.

By the local minimality of u, (49) and (51), recalling that $A_{k,t} ^1$ is the set $\{ x \in B_t (x_0): u^1(x) >k\}$, we have

$$\begin{aligned} \int _{A_{k,t}^1 \cap \{\eta>0\}} f(x,Du)\hbox {d}x\le & {} \int _{A_{k,t}^1 \cap \{\eta>0\}} f(x,Du+D\varphi )\hbox {d}x \\\le & {} \int _{A_{k,t}^1 \cap \{\eta >0\}} \{ (1-\eta ^\mu ) f(x,Du) +\eta ^\mu f(x,A) \} \hbox {d}x. \end{aligned}$$

The inequality above implies

$$\begin{aligned} \int _{A_{k,t}^1 \cap \{\eta>0\}}\eta ^\mu f(x,Du)\hbox {d}x \le \int _{A_{k,t}^1 \cap \{\eta >0\}}\eta ^\mu f(x,A)\hbox {d}x. \end{aligned}$$

Taking into account the expression of f (see (8)), we obtain from the above inequality that

$$\begin{aligned}&\displaystyle \int _{{A_{k,t}^1} \cap \{\eta>0\}}\eta ^\mu \left[ F(x,|Du|^2) +G(x,|\text{ adj }_2 Du|^2) +H(x,\det Du) \right] \hbox {d}x \nonumber \\&\quad \le \displaystyle \int _{{A_{k,t}^1} \cap \{\eta >0\}} \eta ^\mu \left[ F(x,|A|^2) +G(x,|\text{ adj }_2 A|^2) +H(x,\det A) \right] \hbox {d}x. \end{aligned}$$

(52)

Denote ${{\tilde{u}}} =(u^2, u^3)$ and

$$\begin{aligned} D{{\tilde{u}}} =\left( \begin{array}{llll} Du^2\\ Du^3 \end{array} \right) . \end{aligned}$$

We have

$$\begin{aligned} |Du|^2=|Du^1|^2+|D{\widetilde{u}}|^2; \end{aligned}$$

we use (12) with $t_1=|Du^1|^2$ and $t_2=|D{\widetilde{u}}|^2$, so that

$$\begin{aligned} \begin{array}{llll} &{}\displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu \left[ F(x,|Du^1|^2) +F(x,|D{{\tilde{u}}} | ^2) - k_2\right] \hbox {d}x\\ &{}\quad \le \displaystyle \int _{A_{k,t}^1\cap \{\eta >0\}} \eta ^\mu F(x,|Du|^2) \hbox {d}x. \end{array} \end{aligned}$$

(53)

Note that

$$\begin{aligned} |A|^2=|A^1|^2+|D{\widetilde{u}}|^2; \end{aligned}$$

by using (14) with $t_1=|A^1|^2$ and $t_2=|D{\widetilde{u}}|^2$, we obtain

$$\begin{aligned}&\int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu F(x,|A|^2)\hbox {d}x \nonumber \\&\quad \le \int _{A_{k,t}^1\cap \{\eta >0\}} \eta ^\mu \left[ F(x,|A^1|^2) +F(x,|D{{\tilde{u}}} | ^2)\right. \nonumber \\&\qquad \left. +\, k_3 |A^1|^2 (|D{{\tilde{u}}} | ^2)^{\frac{p}{2} -1}+a(x)\right] \hbox {d}x. \end{aligned}$$

(54)

Furthermore, setting

$$\begin{aligned} |{{\tilde{A}}} |^2 =|(\text{ adj }_2 Du)^2| ^2 +|(\text{ adj }_2 Du)^3|^2, \ \ \quad |\tilde{{{\tilde{A}}} } |^2 =|(\text{ adj }_2 A)^2| ^2 +|(\text{ adj }_2 A)^3|^2 \end{aligned}$$

(55)

and noticing

$$\begin{aligned} ( \text{ adj }_2 A)^1 =(\text{ adj }_2 Du)^1, \end{aligned}$$

we can write

$$\begin{aligned} |\text{ adj }_2 Du|^2= & {} |(\text{ adj }_2 Du) ^1|^2 +|(\text{ adj }_2 Du )^2| ^2 +|(\text{ adj }_2 Du)^3|^2= |(\text{ adj }_2 Du ) ^1|^2 +|{{\tilde{A}}}|^2,\\ |\text{ adj }_2 A|^2= & {} |(\text{ adj }_2 A) ^1|^2 +|(\text{ adj }_2 A )^2| ^2 +|(\text{ adj }_2 A)^3|^2= |(\text{ adj }_2 Du ) ^1|^2 +|\tilde{{{\tilde{A}}}} |^2. \end{aligned}$$

Applying (13) with $t_1=|(\text{ adj }_2 Du)^1|^2$ and $t_2=|{{\tilde{A}}}|^2$, we get

$$\begin{aligned}&\int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu \left( G(x, |(\text{ adj }_2 Du)^1 | ^2) -k_2\right) \hbox {d}x \nonumber \\&\quad \le \int _{A_{k,t}^1\cap \{\eta >0\}} \eta ^\mu G(x, |\text{ adj }_2 Du|^2 ) \hbox {d}x. \end{aligned}$$

(56)

Assumption (15) when $q> 2$ or (16) when $q\le 2$, with $t_1=|\tilde{{\tilde{A}}}|^2$ and $t_2=|(\text{ adj }_2 Du)^1|^2$, yields

$$\begin{aligned}&\displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu G(x, |\text{ adj }_2 A|^2) \hbox {d}x \le \displaystyle \int _{A_{k,t}^1\cap \{\eta >0\}} \eta ^\mu [G(x, |\tilde{{{\tilde{A}}}}|^2)\nonumber \\&\quad +\, G(x, |(\text{ adj }_2 Du)^1 | ^2)+ b(x)+1_{(2,+\infty )}(q) k_3|\tilde{{{\tilde{A}}}}|^2\left( |(\text{ adj }_2 Du)^1 | ^2\right) ^{\frac{q}{2}-1}] \hbox {d}x.\qquad \quad \end{aligned}$$

(57)

By virtue of (53),(54), (56) and (57), from (52), we get

$$\begin{aligned}&\int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu [F(x,|Du^1|^2) +F(x,|D{{\tilde{u}}} | ^2) - 2k_2 \\&\qquad + G(x,|(\text{ adj }_2 Du)^1|^2)+H(x,\text{ det }Du)] \hbox {d}x \\&\quad \le \int _{A_{k,t}^1\cap \{\eta >0\}} \eta ^\mu \Big [F(x,|A^1|^2) +F(x,|D{{\tilde{u}}} | ^2) +k_3|A^1|^2(|D{{\tilde{u}}}|^2 )^{\frac{p}{2} -1}\\&\qquad +\,a(x) + G(x,|\tilde{{{\tilde{A}}}}|^2) + G(x,|(\text{ adj }_2 Du)^1|^2)+ b(x)\\&\qquad +\,1_{(2,+\infty )}(q) k_3|\tilde{{{\tilde{A}}}}|^2\left( |(\text{ adj }_2 Du)^1 | ^2\right) ^{\frac{q}{2}-1}+H(x,\text{ det }A)\Big ] \hbox {d}x \end{aligned}$$

and then

$$\begin{aligned} \begin{array}{ll} &{}\displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu \left[ F(x,|Du^1|^2) - 2k_2 +H(x,\text{ det }Du)\right] \hbox {d}x\\ &{}\quad \le \displaystyle \int _{A_{k,t}^1\cap \{\eta >0\}} \eta ^\mu \Big [F(x,|A^1|^2) +k_3|A^1|^2(|D{{\tilde{u}}}|^2 )^{\frac{p}{2} -1}+a(x) + G(x,|\tilde{{{\tilde{A}}}}|^2)\\ &{}\qquad +\, b(x)+1_{(2,+\infty )}(q) k_3|\tilde{{{\tilde{A}}}}|^2\left( |(\text{ adj }_2 Du)^1 | ^2\right) ^{\frac{q}{2}-1}+H(x,\text{ det }A)\Big ] \hbox {d}x. \end{array} \end{aligned}$$

(58)

In order to estimate the first two terms on the right-hand side of (58), we recall that $\mu = p^*>p$ and

$$\begin{aligned} A^1 =\mu \eta ^{-1}(k-u^1)D\eta . \end{aligned}$$

By using the right-hand side of (9) and the fact $z^p \le 1+z^{p^*}$ if $z\ge 0$, we obtain

$$\begin{aligned}&\int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu F(x,|A^1|^2) \le \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu \left[ k_3|A^1|^p+a(x)\right] \hbox {d}x \nonumber \\&\quad \le \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu \left[ k_3\left\{ 1+|A^1| ^{p^*} \right\} +a(x)\right] \hbox {d}x \nonumber \\&\quad \le \int _{A_{k,t}^1\cap \{\eta>0\}} \left\{ \eta ^\mu ( k_3+a(x))+ k_3(2\mu )^{p^*} \eta ^{\mu -p^*} \left( \frac{u^1-k}{t-s}\right) ^{p^*} \right\} \hbox {d}x \nonumber \\&\quad \le \int _{A_{k,t}^1\cap \{\eta >0\}} \left\{ k_3+a(x)+ k_3 (2\mu )^{p^*} \left( \frac{u^1-k}{t-s}\right) ^{p^*} \right\} \hbox {d}x. \end{aligned}$$

(59)

We will write $d'$ to denote the Hölder conjugate of $d>1$: $d'=\frac{d}{d-1}$. Regarding the second term on the right-hand side in (58), notice that

$$\begin{aligned} (p-2) \left( \frac{p^*}{2}\right) ' <p; \end{aligned}$$

we use Young inequality with $\frac{p^*}{2}$, $\left( \frac{p^*}{2}\right) '$ and Hölder inequality with $\frac{p}{(p-2) \left( \frac{p^*}{2} \right) ' }$, $\frac{p}{p-(p-2) \left( \frac{p^*}{2} \right) ' }$:

$$\begin{aligned} \begin{array}{llll} &{}\displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} k_3\eta ^\mu |A^1|^2 |D{{\tilde{u}}} | ^{p-2} \hbox {d}x \\ &{}\quad \le \displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} k_3\eta ^\mu |A^1|^{p^*} \hbox {d}x +\int _{A_{k,t}^1\cap \{\eta>0\}}k_3\eta ^\mu |D{{\tilde{u}}} | ^{(p-2) \left( \frac{p^*}{2}\right) '} \hbox {d}x \\ &{}\quad \le \displaystyle k_3(2\mu )^{p^*} \int _{A_{k,t}^1\cap \{\eta >0\}} \left( \frac{u^1-k}{t-s}\right) ^{p^*} \hbox {d}x \\ &{}\qquad \displaystyle +\,k_3\left( \int _{B_R}|D{{\tilde{u}}} | ^p \hbox {d}x\right) ^{\left( 1-\frac{2}{p} \right) \left( \frac{p^*}{2}\right) ' } |A_{k,t}^1|^{1-\left( 1-\frac{2}{p} \right) \left( \frac{p^*}{2}\right) ' }. \end{array} \end{aligned}$$

(60)

Now we estimate the fourth term in (58). By using (10), (55), Lemma 3.6-(c) and Young inequality with exponents $\frac{p^*}{q}$ and $( \frac{p^*}{q})'$, we estimate

$$\begin{aligned}&\displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu G(x,|\tilde{{{\tilde{A}}}}|^2)\le \displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu k_3 ( (|\tilde{{{\tilde{A}}}}|^2)^{\frac{q}{2}}+b(x)) \hbox {d}x\\&\quad = \displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu (k_3 \left( |(\text{ adj }_2A)^2|^2 + |(\text{ adj }_2A)^3|^2 \right) ^{\frac{q}{2}}+b(x)) \hbox {d}x \\&\quad \le \displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu (3^{\frac{q}{2}}k_3 \left[ |A^1|^2 (|Du^2|^2 +|Du^3|^2) \right] ^{\frac{q}{2}} +b(x))\hbox {d}x \\&\quad \le \displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu 3^{\frac{q}{2}} k_3|A^1|^{p^*}\hbox {d}x + \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu 3^{\frac{q}{2}}k_3\left( |D {\tilde{u}}|\right) ^{q( \frac{p^*}{q})'} \hbox {d}x\\&\qquad + \displaystyle \int _{A_{k,t}^1\cap \{\eta >0\}} \eta ^\mu b(x) \hbox {d}x. \end{aligned}$$

Note that $q( \frac{p^*}{q})'<p$ when $q<\frac{pp^*}{p+p^*}$ and $\frac{pp^*}{p+p^*}>1$ if $p> \frac{2n}{n+1}$.

Moreover, Hölder inequality, with $\frac{p }{q} \Big / \left( \frac{p^*}{q}\right) '$ and $\left( \frac{p }{q} \Big / \left( \frac{p^*}{q}\right) '\right) '$, yields

$$\begin{aligned} \begin{array}{llll} \displaystyle \int _{A_{k,t}^1\cap \{\eta >0\}} \eta ^\mu |D {\tilde{u}}|^{q( \frac{p^*}{q})'} \hbox {d}x &{}\displaystyle \le \left( \int _{A_{k,t}^1} \left( |D \tilde{u}|\right) ^{ p } \hbox {d}x \right) ^{\frac{q}{p}(\frac{p^*}{q})'} |A_{k,t}^1| ^{1-\frac{q}{p}(\frac{p^*}{q})'}\\ &{}\displaystyle \le \left( \int _{B_R} \left( |D {{\tilde{u}}}|\right) ^{ p } \hbox {d}x \right) ^{\frac{q}{p}(\frac{p^*}{q})'} |A_{k,t}^1| ^{1-\frac{q}{p}(\frac{p^*}{q})'}; \end{array} \end{aligned}$$

(61)

therefore, if we note that $(\frac{p^*}{q})' = \frac{p^*}{p^* - q}$, we have

$$\begin{aligned} \begin{array}{llll} &{} \displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu G(x,|\tilde{{{\tilde{A}}}}|^2) \le \displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu 3^{\frac{q}{2}} k_3|A^1|^{p^*}\hbox {d}x\\ &{} \quad \displaystyle + 3^{\frac{q}{2}}k_3\left( \int _{B_R} |D {{\tilde{u}}}|^{ p } \hbox {d}x \right) ^{\frac{q p^*}{p(p^* - q)}} |A_{k,t}^1| ^{1-\frac{q p^*}{p(p^* - q)}}+ \int _{A_{k,t}^1\cap \{\eta >0\}} \eta ^\mu b(x) \hbox {d}x. \end{array} \end{aligned}$$

(62)

Eventually, if $q>2$, we have to estimate the sixth term in (58) and we use Lemma 3.6-(c) and Young inequality with exponents $\frac{p^*}{2}$ and $\left( \frac{p^*}{2} \right) '$, so having

$$\begin{aligned} \begin{array}{l} \displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu |\tilde{{{\tilde{A}}}}|^2\left( |(\text{ adj }_2 Du)^1 | ^2\right) ^{\frac{q}{2}-1} \hbox {d}x\\ \quad = \displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu \left( |(\text{ adj }_2 A)^2 | ^2 +|(\text{ adj }_2 A)^3 | ^2 \right) |(\text{ adj }_2 Du)^1 | ^{q-2} \hbox {d}x\\ \quad \le 3 \displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu |A^1|^2 (|Du^3|^2+|Du^2|^2)|(\text{ adj }_2 Du)^1 | ^{q-2} \hbox {d}x\\ \quad \le 3 \displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu |A^1|^{p^*} \hbox {d}x\\ \qquad + 3 \displaystyle \int _{A_{k,t}^1\cap \{\eta >0\}} \eta ^\mu |D{{\tilde{u}}} |^{2 \left( \frac{p^*}{2} \right) '}|(\text{ adj }_2 Du)^1 | ^{(q-2)\left( \frac{p^*}{2} \right) '} \hbox {d}x. \end{array} \end{aligned}$$

(63)

Observe that $2< q < \frac{p p^*}{p + p^*}$ implies $p > \frac{12}{5}$; so we have $2\left( \frac{p^*}{2} \right) '<p$ and we apply Hölder inequality with exponents $\frac{p}{2\left( \frac{p^*}{2} \right) '}$ and $\left( \frac{p}{2\left( \frac{p^*}{2} \right) '}\right) '$,

$$\begin{aligned} \begin{array}{llll} &{} \displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu |D\tilde{u} |^{2 \left( \frac{p^*}{2} \right) '}|(\text{ adj }_2 Du)^1 | ^{(q-2)\left( \frac{p^*}{2} \right) '} \hbox {d}x\\ &{}\quad \le \left( \displaystyle \int _{B_R} |D{{\tilde{u}}} |^p\right) ^{\frac{2}{p} \left( \frac{p^*}{2} \right) '}\\ &{} \qquad \times \left( \displaystyle \int _{A_{k,t}^1\cap \{\eta >0\}} \eta ^\mu |(\text{ adj }_2 Du)^1 | ^{(q-2)\left( \frac{p^*}{2} \right) '\left( \frac{p}{2\left( \frac{p^*}{2} \right) '}\right) '} \right) ^{1- \frac{2}{p}\left( \frac{p^*}{2}\right) '} \end{array} \end{aligned}$$

(64)

Furthermore, if ${(q-2)\left( \frac{p^*}{2} \right) '\left( \frac{p}{2\left( \frac{p^*}{2} \right) '}\right) '} <q$, we apply Hölder inequality again with exponents $\frac{q}{{(q-2)\left( \frac{p^*}{2} \right) '\left( \frac{p}{2\left( \frac{p^*}{2} \right) '}\right) '}}$ and its conjugate:

$$\begin{aligned}&\left( \int _{B_R} |D{{\tilde{u}}} |^p \right) ^{\frac{2}{p} \left( \frac{p^*}{2} \right) '} \left( \int _{A_{k,t}^1} |(\text{ adj }_2 Du)^1 | ^{(q-2)\left( \frac{p^*}{2} \right) '\left( \frac{p}{2\left( \frac{p^*}{2} \right) '}\right) '} \right) ^{1- \frac{2}{p}\left( \frac{p^*}{2}\right) '} \nonumber \\&\quad \le \left( \int _{B_R} |D{{\tilde{u}}} |^p \right) ^{\frac{2}{p} \left( \frac{p^*}{2} \right) '} \left[ \left( \int _{A_{k,t}^1} |(\text{ adj }_2 Du)^1 |^q \right) ^{\frac{(q-2)}{q}\left( \frac{p^*}{2} \right) '\left( \frac{p}{2\left( \frac{p^*}{2} \right) '}\right) '} \right. \nonumber \\&\qquad \qquad \qquad \left. \times |A^1_{k,t}|^{1-{\frac{(q-2)}{q}\left( \frac{p^*}{2} \right) '\left( \frac{p}{2\left( \frac{p^*}{2} \right) '}\right) '}} \right] ^{1- \frac{2}{p}\left( \frac{p^*}{2}\right) '}. \end{aligned}$$

(65)

Therefore, by (64) and (65), (63) becomes

$$\begin{aligned}&\int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu |\tilde{{{\tilde{A}}}}|^2\left( |(\text{ adj }_2 Du)^1 | ^2\right) ^{\frac{q}{2}-1} \hbox {d}x \le 3 \int _{A_{k,t}^1\cap \{\eta >0\}} \eta ^\mu |A^1|^{p^*} \hbox {d}x \qquad \nonumber \\&\quad +\, 3 \left( \int _{B_R} |D{{\tilde{u}}} |^p \hbox {d}x\right) ^{\frac{2}{p} \left( \frac{p^*}{2} \right) '} \left\{ \left( \int _{A_{k,t}^1} |(\text{ adj }_2 Du)^1 |^q \hbox {d}x \right) ^{\frac{(q-2)}{q}\left( \frac{p^*}{2} \right) '\left( \frac{p}{2\left( \frac{p^*}{2} \right) '}\right) '} \right. \nonumber \\&\qquad \qquad \qquad \left. \times |A^1_{k,t}|^{1-{\frac{(q-2)}{q}\left( \frac{p^*}{2} \right) '\left( \frac{p}{2\left( \frac{p^*}{2} \right) '}\right) '}} \right\} ^{1- \frac{2}{p}\left( \frac{p^*}{2}\right) '}. \end{aligned}$$

(66)

Please, note that the previous condition ${(q-2)\left( \frac{p^*}{2} \right) '\left( \frac{p}{2\left( \frac{p^*}{2} \right) '}\right) '} <q$ means $q < \frac{p p^*}{p + p^*}$. Finally, r-growth assumption (11) on H(x, .) yields

$$\begin{aligned} \displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu H(x, \det A) \hbox {d}x \le \displaystyle \int _{A_{k,t}^1\cap \{\eta >0\}} \eta ^\mu (k_3 |\det A|^r + c(x))\hbox {d}x. \end{aligned}$$

(67)

We compute $\det A$ with respect to the first row, see Lemma 3.6-(b),

$$\begin{aligned} \begin{array}{ll} \displaystyle \eta ^\mu |\det A|^r \le \eta ^\mu |A^1|^r |(\text{ adj }_2 Du)^1| ^r &{}\le \displaystyle (2\mu )^r \eta ^{\mu -r} \left( \frac{u^1-k}{t-s}\right) ^r |(\text{ adj }_2 Du)^1| ^r\\ &{}\le \displaystyle (2\mu )^r\left( \frac{u^1-k}{t-s}\right) ^r |(\text{ adj }_2 Du)^1| ^r. \end{array} \end{aligned}$$

Notice that $r<p< p^*$ and $\frac{rp^*}{p^*-r} <q$. By the Young inequality with exponents $\frac{p^*}{r}$ and $\frac{p^*}{p^*-r}$, one has

$$\begin{aligned} \left( \frac{u^1-k}{t-s}\right) ^r |(\text{ adj }_2 Du)^1| ^r \le \left( \frac{u^1-k}{t-s}\right) ^{p^*} +|( \text{ adj }_2 Du) ^ 1 |^{\frac{rp^*}{p^*-r}}. \end{aligned}$$

Hölder inequality with $\frac{q}{\frac{rp^*}{p^*-r} }$ and $\frac{q}{q- \frac{rp^*}{p^*-r} }$ leads to

$$\begin{aligned} \displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu |\det A|^r \hbox {d}x\le & {} \displaystyle (2\mu )^r \bigg [\int _{A_{k,t}^1\cap \{\eta>0\}} \left( \frac{u^1-k}{t-s}\right) ^{p^*}\hbox {d}x\nonumber \\&+ \displaystyle \int _{A_{k,t}^1\cap \{\eta>0\}}|( \text{ adj }_2 Du) ^ 1 |^{\frac{rp^*}{p^*-r}} \hbox {d}x \bigg ]\nonumber \\\le & {} \displaystyle (2\mu )^r \bigg [\int _{A_{k,t}^1\cap \{\eta >0\}} \left( \frac{u^1-k}{t-s}\right) ^{p^*}\hbox {d}x\nonumber \\&\displaystyle +\left( \int _{B_R} |( \text{ adj }_2 Du) ^ 1 |^q\hbox {d}x \right) ^{\frac{rp^*}{q(p^*-r)}} |A_{k,t}^1| ^{1-\frac{rp^*}{q(p^*-r)}}\bigg ].\nonumber \\ \end{aligned}$$

(68)

Therefore, (67) and (68) imply

$$\begin{aligned}&\int _{A_{k,t}^1\cap \{\eta>0\}} \eta ^\mu H(x, \det A) \hbox {d}x \le k_3 (2\mu )^r \bigg [\int _{A_{k,t}^1\cap \{\eta>0\}} \left( \frac{u^1-k}{t-s}\right) ^{p^*}\hbox {d}x \qquad \quad \nonumber \\&\quad + \left( \int _{B_R} |( \text{ adj }_2 Du) ^ 1 |^q\hbox {d}x \right) ^{\frac{rp^*}{q(p^*-r)}} |A_{k,t}^1| ^{1-\frac{rp^*}{q(p^*-r)}}\bigg ] + \int _{A_{k,t}^1\cap \{\eta >0\}} \eta ^\mu c(x)\hbox {d}x. \qquad \qquad \end{aligned}$$

(69)

By left-hand side inequalities in (9) and (11), using (58), (59), (60), (62), (66) and (69), we conclude

$$\begin{aligned} \begin{array}{lllll} &{}\displaystyle \int _{A_{k,s}^1 } |Du^1| ^p\hbox {d}x \le C \int _{A_{k,t}^1} \left( \frac{u^1-k}{t-s}\right) ^{p^*} \hbox {d}x \displaystyle +C \biggm \{ 1 + \Vert a+b+c\Vert _{L^{\sigma }(B_R)}\\ &{}\quad + \displaystyle \left( \int _{B_R} (|D u^2|+|Du^3|) ^{p} \hbox {d}x\right) ^{\left( 1-\frac{2}{p} \right) \left( \frac{p^*}{2}\right) ' }\\ &{} \quad \displaystyle +\left( \int _{B_R} \left( |Du^2| +|Du^3|\right) ^p \hbox {d}x \right) ^{\frac{q p^*}{p(p^* - q)}} \displaystyle + \left( \int _{B_R} |( \mathrm{adj}_2 Du) ^ 1 |^q \, \hbox {d}x \right) ^{\frac{rp^*}{q(p^*-r)}}\\ &{}\quad +1_{(2,+\infty )}(q) \left( \displaystyle \int _{B_R} (|D u^2| + |D u^3|)^p \hbox {d}x\right) ^{\frac{2}{p} \left( \frac{p^*}{2} \right) '}\\ &{}\quad \times \left( \displaystyle \int _{B_R} |(\text{ adj }_2 Du)^1 |^q \hbox {d}x \right) ^{\frac{(q-2)}{q}\left( \frac{p^*}{2} \right) '\left( \frac{p}{2\left( \frac{p^*}{2} \right) '}\right) '\left( 1- \frac{2}{p}\left( \frac{p^*}{2}\right) '\right) } \biggm \} |A_{k,t}^1|^\theta , \end{array} \end{aligned}$$

(70)

where

$$\begin{aligned}&\theta := \mathrm{min} \left\{ 1-\left( 1-\frac{2}{p} \right) \left( \frac{p^*}{2}\right) ', \, 1- \frac{q p^*}{p(p^* - q)}, \, 1- \frac{rp^*}{q(p^*-r)}, \, 1 - \frac{1}{\sigma }, \quad \right. \nonumber \\&\quad \left. 1_{[1,2]}(q) + 1_{(2,+\infty )}(q)\left( 1-\frac{(q-2)}{q}\left( \frac{p^*}{2} \right) '\left( \frac{p}{2\left( \frac{p^*}{2} \right) '}\right) '\right) \left( 1- \frac{2}{p}\left( \frac{p^*}{2}\right) '\right) \right\} \end{aligned}$$

and $C=C(k_1, k_2, k_3, p, q, r, p^*)>0$; moreover, $1_E(q) = 1$ if $q \in E$ and $1_E(q) = 0$ if $q \notin E$. Now we note that

$$\begin{aligned} 1-\left( 1-\frac{2}{p} \right) \left( \frac{p^*}{2}\right) ' \ge 1- \frac{ p^*}{p(p^* - 1)} \ge 1- \frac{q p^*}{p(p^* - q)}, \end{aligned}$$

where the last inequality is granted since $q \mapsto 1- \frac{q p^*}{p(p^* - q)}$ decreases. Then,

$$\begin{aligned}&\theta := \mathrm{min} \left\{ 1- \frac{q p^*}{p(p^* - q)}, \quad \, 1- \frac{rp^*}{q(p^*-r)}, \quad \, 1 - \frac{1}{\sigma }, \quad \qquad \qquad \qquad \qquad \right. \\&\quad \, \left. 1_{[1,2]}(q) + 1_{(2,+\infty )}(q)\left( 1-\frac{(q-2)}{q}\left( \frac{p^*}{2} \right) '\left( \frac{p}{2\left( \frac{p^*}{2} \right) '}\right) '\right) \left( 1- \frac{2}{p}\left( \frac{p^*}{2}\right) '\right) \right\} . \end{aligned}$$

Note that $\left( \frac{p^*}{2}\right) ' = \frac{p^*}{p^* - 2}$; then, the exponents in (70) can be written as follows

$$\begin{aligned} \left( 1 - \frac{2}{p} \right) \left( \frac{p^*}{2}\right) '= & {} \frac{p^*(p-2)}{p(p^* - 2)}, \quad \frac{2}{p} \left( \frac{p^*}{2}\right) ' \\= & {} \frac{2 p^*}{p(p^* - 2)},\left( \frac{p^*}{2} \right) '\left( \frac{p}{2\left( \frac{p^*}{2} \right) '}\right) ' \left( 1- \frac{2}{p}\left( \frac{p^*}{2}\right) '\right) \\= & {} \frac{p^*}{p^* - 2}, \end{aligned}$$

so that (70) turns out to be

$$\begin{aligned} \int _{A_{k,s}^1 } |Du^1| ^p\hbox {d}x\le & {} C \int _{A_{k,t}^1} \left( \frac{u^1-k}{t-s}\right) ^{p^*} \hbox {d}x +C \biggm \{ 1+ \Vert a+b+c\Vert _{L^{\sigma }(B_R)} \nonumber \\&+ \left( \int _{B_R} (|D u^2|+|Du^3|) ^{p} \hbox {d}x\right) ^{\frac{p^*(p-2)}{p(p^* - 2)} } \nonumber \\&+\left( \int _{B_R} \left( |Du^2| +|Du^3|\right) ^p \hbox {d}x \right) ^{\frac{q p^*}{p(p^* - q)}} \nonumber \\&+ \left( \int _{B_R} |( \mathrm{adj}_2 Du) ^ 1 |^q \, \hbox {d}x \right) ^{\frac{rp^*}{q(p^*-r)}} \nonumber \\&+1_{(2,+\infty )}(q) \left( \int _{B_R} (|D u^2| + |D u^3|)^p \hbox {d}x\right) ^{\frac{2 p^*}{p(p^* - 2)}} \nonumber \\&\times \left( \int _{B_R} |(\text{ adj }_2 Du)^1 |^q \hbox {d}x \right) ^{\frac{(q-2)p^*}{q(p^* - 2)}} \biggm \} |A_{k,t}^1|^\theta , \end{aligned}$$

(71)

where

$$\begin{aligned}&\theta := \mathrm{min} \left\{ 1- \frac{q p^*}{p(p^* - q)}, 1- \frac{rp^*}{q(p^*-r)}, 1 - \frac{1}{\sigma } \right\} , \quad \mathrm{if } \quad 1< q \le 2, \\&\theta := \mathrm{min} \left\{ 1- \frac{q p^*}{p(p^* - q)}, 1- \frac{rp^*}{q(p^*-r)}, 1 - \frac{1}{\sigma }, \right. \\&\qquad \qquad \qquad \left. \frac{p^*(p-2) -2p}{p(p^* - 2)} - \frac{(q-2)p^*}{q(p^* - 2)} \right\} , \quad \mathrm{if } \quad 2 < q. \end{aligned}$$

If $2< q < \frac{p p^*}{p + p^*}$, we have

$$\begin{aligned} 1- \frac{q p^*}{p(p^* - q)} > \frac{p^*(p-2) -2p}{p(p^* - 2)} - \frac{(q-2)p^*}{q(p^* - 2)} = 1- \frac{2 p^*}{p(p^* - 2)} - \frac{(q-2)p^*}{q(p^* - 2)} \end{aligned}$$

(72)

and

$$\begin{aligned} \theta = \mathrm{min} \left\{ 1- \frac{2 p^*}{p(p^* - 2)} - \frac{(q-2)p^*}{q(p^* - 2)}, 1- \frac{rp^*}{q(p^*-r)}, 1 - \frac{1}{\sigma } \right\} . \end{aligned}$$

This ends the proof of Proposition 4.1. $\square$

We now proceed with the proof of Theorem 2.2. We fix $x_0 \in \varOmega$ and $R_0 < \min \{ \text {dist}(x_0, \partial \varOmega ), \left( \frac{3}{4 \pi } \right) ^{1/3} \}$ such that

$$\begin{aligned} \int _{B_{R_0}} |u^1|^{p^*} \hbox {d}x < 1, \end{aligned}$$

(73)

where $B_{\rho }$ is the ball centered at $x_0$ with radius $\rho$. Note that $R_0 < 1$, $|B_{R_0}| < 1$ and $B_{R_0} \subset \subset \varOmega$. For every $R \in (0, R_0]$ we define the decreasing sequence of radii

$$\begin{aligned} \rho _h := \frac{R}{2} + \frac{R}{2^{h+1}}. \end{aligned}$$

Fix a positive constant $d \ge 1$ and define the increasing sequence of positive levels

$$\begin{aligned} k_h := d \left( 1 - \frac{1}{2^{h+1}} \right) . \end{aligned}$$

We define the “excess”

$$\begin{aligned} J_{h} := \int _{A^1_{k_h,\rho _h}} (u^1 - k_h)^{p^*} \hbox {d}x. \end{aligned}$$

We use our Caccioppoli inequality (48) and Proposition 2.4 of [13]: we get

$$\begin{aligned} J_{h+1} \le c \left( 2^{\frac{p^* p^*}{p}}\right) ^h (J_{h})^{\theta \frac{p^*}{p}}, \end{aligned}$$

where the positive constant c is independent of h. See also [40, 41]. Assumption (17) tells us that $\theta \frac{p^*}{p}>1$; then, we can use Lemma 2.5 of [13] with $\gamma : = \theta \frac{p^*}{p} - 1$, see also [42]:

$$\begin{aligned} J_h \le \left( 2^{\frac{p^* p^*}{p}}\right) ^{-\frac{h}{\gamma }} J_0, \end{aligned}$$

(74)

provided

$$\begin{aligned} J_0 \le c^{-\frac{1}{\gamma }} \left( 2^{\frac{p^* p^*}{p}}\right) ^{-\frac{1}{\gamma ^2}}. \end{aligned}$$

(75)

Note that

$$\begin{aligned} J_0 = \int _{A^1_{\frac{d}{2},R}} \left( u^1 - \frac{d}{2} \right) ^{p^*} \hbox {d}x \rightarrow 0 \qquad \text {as} \qquad d \rightarrow +\infty ; \end{aligned}$$

then, we can choose $d \ge 1$ large enough so that (75) holds true. Thus, we have (74) with $\gamma > 0$, so that $J_h \rightarrow 0$ as $h \rightarrow +\infty$; since $\frac{R}{2} < \rho _h$ and $k_h < d$, we also have

$$\begin{aligned} 0 \le \int _{A^1_{d,\frac{R}{2}}} \left( u^1 - d \right) ^{p^*} \hbox {d}x \le J_h; \end{aligned}$$

then,

$$\begin{aligned} \int _{A^1_{d,\frac{R}{2}}} \left( u^1 - d \right) ^{p^*} \hbox {d}x = 0 \end{aligned}$$

so that $u^1 \le d$ almost everywhere in $B_{\frac{R}{2}}$. We have proved that $u^1$ is locally bounded from above. In order to prove that $u^1$ is locally bounded from below, we note that $-u$ locally minimizes $\int _{\varOmega } {\tilde{f}}(x,Dz(x)) \hbox {d}x$, where ${\tilde{f}}(x,\xi ) = f(x,-\xi )$; then, we get that $-u^1$ is locally bounded from above, so $u^1$ is locally bounded from below. We have just shown that $u^1 \in L^{\infty }_{\text {loc}}(\varOmega )$.

Now we turn our attention to the second component $u^2$. We change the order of the two components $u^1$ and $u^2$: we get a new function v as follows:

$$\begin{aligned} v= \left( \begin{array}{c} u^2\\ u^1\\ u^3 \end{array} \right) ; \end{aligned}$$

then,

$$\begin{aligned} Dv= \left( \begin{array}{ccc} &{} Du^2 &{}\\ &{} Du^1 &{}\\ &{} Du^3 &{} \end{array} \right) \end{aligned}$$

and $\det Dv = - \det Du$; moreover $(\text {adj}_2 Dv)^1 = - (\text {adj}_2 Du)^2$, $(\text {adj}_2 Dv)^2 = - (\text {adj}_2 Du)^1$ and $(\text {adj}_2 Dv)^3 = - (\text {adj}_2 Du)^3$, so that

$$\begin{aligned} \text {adj}_2 Dv = - \left( \begin{array}{c} (\text {adj}_2 Du)^2\\ (\text {adj}_2 Du)^1\\ (\text {adj}_2 Du)^3 \end{array} \right) . \end{aligned}$$

If we write $C_{1,2}(\xi )$ to denote the matrix obtained from $\xi$ by inverting line 1 and line 2, we have $Dv = C_{1,2}(Du)$ and $\text {adj}_2 Dv = - C_{1,2} (\text {adj}_2 Du)$. Then, v is a local minimizer of $\int _{\varOmega } \tilde{{\tilde{f}}}(x,Dw(x)) dx$, where $\tilde{{\tilde{f}}}(x,\xi ) = f(x, C_{1,2}(\xi ))$. Thus, the first component $v^1$ is locally bounded: $u^2 = v^1 \in L^{\infty }_{\text {loc}}(\varOmega )$. In a similar way we deal with the third component $u^3$: we change the order of the two components $u^1$ and $u^3$; we get a new function w as follows:

$$\begin{aligned} w= \left( \begin{array}{c} u^3\\ u^2\\ u^1 \end{array} \right) ; \end{aligned}$$

then,

$$\begin{aligned} Dw= \left( \begin{array}{ccc} &{} Du^3 &{}\\ &{} Du^2 &{}\\ &{} Du^1 &{}\end{array} \right) \end{aligned}$$

and $\det Dw = - \det Du$; moreover $(\text {adj}_2 Dw)^1 = - (\text {adj}_2 Du)^3$, $(\text {adj}_2 Dw)^2 = - (\text {adj}_2 Du)^2$ and $(\text {adj}_2 Dw)^3 = - (\text {adj}_2 Du)^1$, so that

$$\begin{aligned} \text {adj}_2 Dw = - \left( \begin{array}{c} (\text {adj}_2 Du)^3\\ (\text {adj}_2 Du)^2\\ (\text {adj}_2 Du)^1 \end{array} \right) . \end{aligned}$$

If we write $C_{1,3}(\xi )$ to denote the matrix obtained from $\xi$ by inverting line 1 and line 3, we have $Dw = C_{1,3}(Du)$ and $\text {adj}_2 Dw = - C_{1,3} (\text {adj}_2 Du)$. Then, w is a local minimizer of $\int _{\varOmega } \tilde{\tilde{{\tilde{f}}}}(x,Dz(x)) dx$, where $\tilde{\tilde{{\tilde{f}}}}(x,\xi ) = f(x, C_{1,3}(\xi ))$. Thus, the first component $w^1$ is locally bounded: $u^3 = w^1 \in L^{\infty }_{\text {loc}}(\varOmega )$. This ends the proof of Theorem 2.2. $\square$

5 Conclusions

We have been able to prove boundedness for minimizers of the most important three-dimensional polyconvex integral, provided the growth exponents verify some restrictions. It would be interesting to understand what happens when such restrictions are not in force.

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.

$\square$