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Set $u_0\in H^1 (\mathbb{R} ^N)$ and $1<\alpha < \frac{N+2}{N-2}$. I want to show that there exists $\varepsilon > 0$ s.t. if $\Vert u _0 \Vert _ {H^1} < \varepsilon$, then there is global solution (defined for all $t\in\mathbb{R})$ of $$ u_t = i (\Delta u + \vert u \vert ^{\alpha -1} u) $$ $$ u(\cdot,0)=u_0 $$

I have proved the existence of solution (local). Now I want to prove that is in fact global. In order to do that, I thought to use the blow-up alternative, that is, I just have to show that $\Vert u \Vert _{H^1(\mathbb{R}^N)}$ is bounded, where $u$ is a $H^1-$solution. It is clear that $ \Vert u \Vert _{H^1(\mathbb{R}^N)} = \Vert u \Vert _{L^2(\mathbb{R}^N)} + \Vert \nabla u \Vert _{L^2\mathbb{R}^N)}$.

By the mass conservation, $\Vert u \Vert _{L^2(\mathbb{R}^N)} = \Vert u _0\Vert _{L^2(\mathbb{R}^N)} $. By the energy conservation, $ \frac{1}{2}\Vert \nabla u\Vert_{L^2} ^2 - \frac{1}{\alpha + 1} \Vert u \Vert_{L^{\alpha+1}} ^{\alpha +1} \equiv E\in\mathbb{R} $ therefore, by Sobolev embedding, $$ \Vert \nabla u\Vert_{L^2} ^2\leq 2E + \frac{2}{\alpha + 1} \Vert u \Vert_{L^{\alpha+1}} ^{\alpha +1}\leq 2E + \frac{2C}{\alpha + 1} \Vert u \Vert_{H^1} ^{\alpha +1}$$

If I show that $\Vert \nabla u\Vert_{L^2} ^2 \leq 2E + \frac{2C}{\alpha + 1} \Vert u \Vert_{H^1} ^{\alpha +1} \leq \text{constant}$ I finish. I have the assumption that $\Vert u _ 0 \Vert _{H^1}$ is small, does it implies that $\Vert u \Vert _{H^1}$ is also small? As I am reading on Internet, I found something called Lyapunov stability. Maybe is useful here, can anyone explain how could be used in this situation? Thanks in advance. Any idea is welcome!

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    $\begingroup$ Yes, because the set $\{t\le \varepsilon+Ct^{1+\alpha}\}$ has a bounded connected component near $0$ for sufficiently small $\varepsilon$ and you cannot go out of it and jump to the unbounded component because $\|u\|_{H_1}$ is a continuous function of $t$. $\endgroup$ – fedja May 14 at 17:10
  • $\begingroup$ Sorry, I dont understand that. Is that related to lyapunov criteria? $\endgroup$ – R. N. Marley May 14 at 17:16
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    $\begingroup$ Depends on what you mean by "Lyapunov criteria". What I'm saying is that the $H^1$-norm $S$ of your solution always satisfies the inequality $S^2\le\varepsilon+CS^{1+\alpha}$ (sorry for the missing square in the previous remark) due to the conservation laws where $\varepsilon$ (norm squared plus energy of $u_0$) is as small as you want. The set of non-negative $S$ satisfying this inequality has 2 connected components: one bounded, in which you initially are, and one unbounded. Due to the continuity of $S$, you cannot jump from the bounded component to the unbounded, so $S$ is globally bounded. $\endgroup$ – fedja May 14 at 17:28
  • $\begingroup$ By conservation laws, I have $$ \Vert \nabla u\Vert_{L^2} ^2\leq 2E + \frac{2}{\alpha + 1} \Vert u \Vert_{L^{\alpha+1}} ^{\alpha +1}\leq 2E + \frac{2C}{\alpha + 1} \Vert u \Vert_{H^1} ^{\alpha +1}$$ it seems that you are calling $S=\Vert \nabla u\Vert_{L^2}=\Vert u\Vert_{H^1}$? $\endgroup$ – R. N. Marley May 14 at 17:36
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    $\begingroup$ Erm... Have you read my stupid rants about the bounded component at all? $\endgroup$ – fedja May 14 at 18:35

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