Theorem 6.2.7 in the book Nonlinear Analisis of Gasisnski and Papageorgiou states: If $u\in W^{1,p}(\Omega)\cap L^\infty(\Omega)$ and $\Delta_pu \in L^\infty(\Omega)$ then we have $u\in C^{1,\beta}_0(\overline\Omega)$ with $\beta\in(0,1)$. Now, I know that $u\in W^{1,p}_0(\Omega)\cap L^\infty(\Omega)$ is a weak solution to the problem $\Delta_pu=f(u)$ in $\Omega$; $u=0$ on $\partial\Omega$. Where $f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous bounded function. The point is that $\Delta_pu=f(u)$ is not a pointwise equation but rather a distributional one. Does the fact that $f\circ u\in L^\infty(\Omega)$ imply $\Delta_pu \in L^\infty(\Omega)$?. (I want to conclude that $u\in C^{1,\beta}_0(\overline\Omega)$)
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1$\begingroup$ Just a remark about terminology. It seems that your $u$ solves a semilinear equation, presuming that $f(u)$ is non-linear, rather than a quasilinear one. The latter condition would imply that the highest derivative term (which for you is $\Delta_p u$) would have $u$-dependent coefficients. $\endgroup$– Igor KhavkineCommented Feb 24, 2014 at 23:22
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