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Relaxation of convex functionals: the gap problem. (English) Zbl 1025.49012
The paper paper deals with integral representation properties of relaxed functionals of integral energies, and with the corresponding Lavrentiev phenomenon.
Let $$\Omega$$ be a bounded domain in $${\mathbb R}^N$$, let $$f\colon\Omega\times{\mathbb R}^{d\times N}\to[0,+\infty[$$ be Carathéodory, satisfy the following growth conditions $z ^\alpha\leq f(x,z)\leq C(1+ z ^\beta)\text{ for all }z\in{\mathbb R}^{d\times N},\;{\mathcal L}^N\hbox{-a.e. }x\in\Omega$ for some $$C>0$$ and $$1<\alpha\leq\beta<{N\alpha\over N-1}$$, and let $$f$$ be convex in the $$z$$ variable for $${\mathcal L}^N$$-a.e. $$x\in\Omega$$. For every open subset $$A$$ of $$\Omega$$ and $$u\in L^1(A;{\mathbb R}^d)$$ let ${\mathcal F}(u,A)=\inf\left\{\liminf_{n\to+\infty}\int_A f(x,\nabla u(x))dx : \{u_n\}\subseteq W^{1,\beta}_{\text{loc}}(A;{\mathbb R}^d),\;u_n\to u\text{ in }L^1(A;{\mathbb R}^d)\right\}.$ In the paper it is first proved that if $$A$$ is an open subset of $$\Omega$$, and if $$u\in L^1(A;{\mathbb R}^d)$$ satisfies $${\mathcal F}(u,A)<+\infty$$, then $$u\in W^{1,\alpha}(A;{\mathbb R}^d)$$ and for every open subset $$B$$ of $$A$$ ${\mathcal F}(u,A)=\int_B f(x,\nabla u(x))dx+\mu_s(u,B),$ where $$\mu_s(u,\cdot)$$ is a nonnegative Radon measure singular with respect to $${\mathcal L}^N$$.
Then, some conditions ensuring that $$\mu_s(u,\cdot)=0$$, i.e., that no Lavrentiev phenomenon occurs, are proposed.
Finally, an example is discussed showing that, if $$f(x,\cdot)$$ is no more convex but only quasiconvex for $${\mathcal L}^N$$-a.e. $$x\in\Omega$$ and $$\alpha<\beta$$, then the above representation result can be no more true.

##### MSC:
 49J45 Methods involving semicontinuity and convergence; relaxation 74B20 Nonlinear elasticity
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##### References:
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