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Let's consider the following Schrodinger equation $$iu_t+\Delta u+F(u)=0$$ in $\mathbb{R}^n$. In Cazenave's book, "Semilinear Schrodinger equation", he defines $H^1$-weak solution as $u\in L^\infty(0,T;H^1(\mathbb{R}^n))\cap W^{1,\infty}(0,T;H^{-1}(\mathbb{R}^n))$ such that the equation $iu_t+\Delta u+F(u)=0$ is satisfied in $H^{-1}(\mathbb{R}^n)$. Does it imply that the equation is satisfied in the space of distribution $D'(\mathbb{R}^n)$ in the sense $$\int u[i\varphi_t+\Delta\varphi]+<F(u),\varphi>_{D'(\mathbb{R}^n),D(\mathbb{R}^n)}=0$$

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    $\begingroup$ If the equation is satisfied in $H^{-1}$, it means that is is satisfied when integrated against any function in $H^1$ and formally integrated by parts to remove the Laplacian. Since $D\subset H^1$, the answer is yes. $\endgroup$ Nov 27, 2014 at 8:59
  • $\begingroup$ My doubt is the following: let's consider for instance the term $u_t$; in $H^{-1}$ it is the distribution $<u_t,\varphi>_{H^{-1}(\mathbb{R}^n),H^1(\mathbb{R}^n)}$, with $\varphi\in H^1(\mathbb{R}^n)$. The restriction of this distribution to $D(\mathbb{R}^n)$ coincides with the distribution $<u_t,\varphi>_{D'(\mathbb{R}^n),D(\mathbb{R}^n)}=-\int u\varphi_t$? $\endgroup$
    – Sue
    Nov 27, 2014 at 9:08
  • $\begingroup$ Can you be more specific about the underlying spaces? I thought originally that $\phi$ is independent of $t$, and you test the solution on each time slice separately (since you mentioned $H^{-1}(\mathbb R^n)$ and $D'(\mathbb R^n)$ as test function spaces), but it seems that this is perhaps not what you meant. Did you mean that the Schrödinger equation is in $\mathbb R^{n+1}$ with one dimension for time? $\endgroup$ Nov 27, 2014 at 13:11
  • $\begingroup$ I mean that the equation is satisfied in the mentioned spaces for almost every $t\in (0,T)$. $\endgroup$
    – Sue
    Nov 27, 2014 at 13:24
  • $\begingroup$ But then the test function only lives on $\mathbb R^n$ (since you test on one time slice at a time) and you cannot integrate by parts with respect to time. In that formulation $\phi$ is independent of time. The statement is true if you replace $u\phi_t$ with $u_t\phi$ in the last equation (I originally read it this way) as I stated in my first comment. $\endgroup$ Nov 27, 2014 at 13:34

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