2

I know a version of Ladyzhenskaya's inequality, that is

Let $\Omega$ be bounded domain in $\mathbb R^2$, we have $$\|u\|^2_{L^4} \leq C\|u\|_{L^2}\|\nabla u\|_{L^2}, \quad \forall u\in H^1(\Omega).$$ So is it true when $\Omega$ is unbounded?

And where can i find an estimate for $\|u\|_{L^4}, u\in H^1(\Omega), \Omega$ is unbounded?

BTTD
  • 87

2 Answers2

1

In case $\Omega=\mathbb{R}^n$, $n\geqslant 1$, see theorem on page 125 of the paper by L.Nirenberg: On elliptic partial differential equations in Annali della Scuola Normale Superiore di Pisa, Cl.Sci., Sér. 3, t. 13, no. 2 (1959), p. 115-162 (http://booksc.org/book/11614874). What you are interested in, is just a particular case of the inequality now widely known as Gagliardo–Nirenberg inequality. On page 124 of his paper, L.Nirenberg shortly outlines the background of the theorem. Specifically, when he arrived to Edinburgh in August 1958 to present his short communicatition "Inequalities for derivatives" at the ICM, he learned that almost equivalent results had also been proved by E.Gagliardo.

Generally, without certain auxiliary conditions such as zero boundary conditions or linear functional relations, the Gagliardo–Nirenberg inequality cannot be valid for a domain $\Omega\neq\mathbb{R}^n$. This is quite obvious for bounded domains. For unbounded domains, the answer depends on the geometry of a particular domain, e.g., "valid" for a special Lipschitz domain while "not valid" for a layer. Nevertheless, when the Gagliardo–Nirenberg inequality is to be applied to some nonlinear problem, what really matters is just the ratio of the exponents raising each factor of the RHS. For the Gagliardo–Nirenberg inequality, the ratio is the best possible, staying exactly the same for a modified Gagliardo–Nirenberg inequality with norms of derivatives substituted by the same-order-Sobolev-space norms. For example, restricted to the OP context, given any bounded or unbounded uniformly Lipschitz domain $\Omega\in\mathbb{R}^2$, it is easy to construct an extension $\widetilde{u}\in H^1(\mathbb{R}^2)$   of a function $u\in H^1(\Omega)$ from $\Omega$ to the whole $\mathbb{R}^2$, such that $$ \|\widetilde{u}\|_{L^2(\mathbb{R}^2)}\leqslant C_1\|u\|_{L^2(\Omega)}\,,\quad \|\widetilde{u}\|_{H^1(\mathbb{R}^2)}\leqslant C_2\|u\|_{H^1(\Omega)} $$ with some positive constans $C_1\,,C_2$ depending only on $\Omega$. Now, by the Gagliardo–Nirenberg inequality for $\mathbb{R}^2$ readily follows the modified Gagliardo–Nirenberg inequality for domains: $$ \|u\|^2_{L^4(\Omega)}\leqslant \|\widetilde{u}\|^2_{L^4(\mathbb{R}^2)}\leqslant C_3\|\widetilde{u}\|_{L^2(\mathbb{R}^2)}\|\widetilde{u}\|_{H^1(\mathbb{R}^2)} \leqslant C\|u\|_{L^2(\Omega)}\|u\|_{H^1(\Omega)} $$ with constant $C=C_1\!\cdot\!C_2\!\cdot\!C_3\,$. In very many nonlinear problems, though not absolutely all, such modified version of the Gagliardo–Nirenberg inequality for domains proves equally effective as its original version for the whole space.

mkl314
  • 2,899
  • When $\Omega = \mathbb R^n$ then $H^1_0(\Omega) \equiv H^1(\Omega)$, so the Ladyzhenskaya's inequality is true for all functions $u\in H^1_0$. But here $\Omega$ is unbounded and it has boundary $\partial \Omega$. If $\Omega$ is bounded by one direction then the above inequality is true? – BTTD Apr 24 '14 at 07:49
  • As to unbounded domains $\Omega\neq\mathbb{R}^n$, the Gagliardo-Nirenberg inequality stays valid for the so-called special Lipschitz domain, i.e., a domain $\Omega={x=(x',x_n)\in\mathbb{R}^n,:;x_n>\psi(x')}$ with some uniformly Lipschitz on $\mathbb{R}^{n-1}$ function $\psi$. It is readily proved using an even extension wrt the boundary $x_n=\psi(x')$ to the whole $\mathbb{R}^n$. But this cannot be done for a strip $\Omega={x=(x_1,x_2)\in\mathbb{R}^2,:;0<x_2<1}$, in which case a hypothetical version of the Gagliardo-Nirenberg inequality for a strip can be easily disproved. – mkl314 Apr 24 '14 at 10:18
  • can you show me the proof for Gagliardo-Nirenberg inequality in special Lipschitz domain? – BTTD Apr 24 '14 at 10:32
  • For a special Lipshitz domain $\Omega= {x=(x_1,x_2)\in\mathbb{R}^2,:;x_2>\psi(x_1)}$, the following extension will do $$\widetilde{u}(x)=\begin{cases}u(x),;;x_2\geqslant\psi(x_1),,\ u\bigl(x_1,\psi(x_1)-x_2\bigr),,;;x_2<\psi(x_1),. \end{cases}$$ The rest part follows the argument employed in the edited answer above. – mkl314 Apr 24 '14 at 14:03
0

If $\Omega$ is the whole space, see e.g. the paper by Constantin and Seregin.

Umberto P.
  • 52,165