Generalized Lax-Milgram for Weak Formulation of 1D Linear Schrodinger I am interested in the variational formulation of the 1D Schrodinger equation:
$i u_t- \beta u_{xx} = 0 $ and $u(x,0)=u_0(x)$ which upon integration by parts yields:
$i(u_t,v) + \beta (u_x,v_x) = 0$ and the boundary terms vanish by appropriate testing with test functions $v$. We then have the continuous sesquilinear form $b(u,v) = i(u_t,v) + \beta (u_x,v_x)$. I would like to apply the generalized Lax-Milgram on $b(\cdot,\cdot)$, but I am having trouble showing the boundedness below (coercivity) of $b(\cdot,\cdot)$. Is there a slick way to show $b(u,u) > \alpha \|u\|_{H^1}^2$? 
 A: Your method is doomed to fail. For several reasons.

*

*Suppose that $u$ solves the linear Schrodinger equation. Using Fourier methods it is easy to see that $\| u(\cdot,t)\|_{L^2_x}$ is conserved and independent of $t$. This means that $u$ cannot be in $L^2(\mathbb{R}\times \mathbb{R}_+)$. So there is zero chance that Lax-Milgram can give you any indication on how to get a solution.


*Forgetting item 1 above. Observe that if Lax-Milgram were to work, you solution $u$ will satisfy $B(u,v) = \langle 0,v\rangle = 0$ for any $v$, since you are solving the homogeneous Schrodinger equation. This implies immediately that $B$ cannot be coercive.


*Forgetting items 1 and 2 above, suppose $u$ is a function in $L^2(\mathbb{R}\times\mathbb{R}_+)\cap C^2$, then necessarily $\liminf_{t\to\infty} \|u(\cdot,t)\|_{L^2_x} = 0$. But then unless $u_0 \equiv 0$, $B(u,u)$ must have a non-vanishing imaginary part.


*Lax-Milgram is intended to provide a weak solution to the linear partial differential equation $L u = f$ for $u$ belonging to some Hilbert space $H$. In the case $f = 0$ however the existence of a solution is trivial! Namely that $u = 0$ will solve the equation.
In your problem you are prescribing a boundary value $u_0$. This is not in the usual form of Lax-Milgram. Furthermore, by fixing the boundary value $u_0$ you destroy linearity; in other words, in makes no sense to look at the functional on a Hilbert space $H$, since you cannot add two elements with boundary value $u_0$ and obtain another element with boundary value $u_0$.
