# Global solution of nonlinear Schrödinger equation via blow-up argument

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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• 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 $t$. – fedja May 14 at 17:10
• Sorry, I dont understand that. Is that related to lyapunov criteria? – R. N. Marley May 14 at 17:16
• 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. – fedja May 14 at 17:28
• 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 $S=\Vert \nabla u\Vert_{L^2}=\Vert u\Vert_{H^1}$? – R. N. Marley May 14 at 17:36
• Erm... Have you read my stupid rants about the bounded component at all? – fedja May 14 at 18:35