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The theory as I know it

Let $\mathcal{H}$ be a Hilbert space and $(A, D(A))$ a self-adjoint operator acting on it. The Spectral Theorem (cfr. Reed & Simon Methods of modern mathematical physics, vol.I, §VIII.6) asserts the existence of a unique projection-valued measure (PVM) $P$ s.t.

$$A= \int_{-\infty}^\infty \lambda dP_{\lambda},$$

thus allowing us to define functions of $A$ and especially, for real $t$,

$$e^{i t A}=\int_{-\infty}^{\infty}e^{i t \lambda} dP_{\lambda}.$$

Turns out (Reed & Simon, §VIII.7) that $(e^{itA})_{t \in \mathbb{R}}$ is a strongly continuous unitary group with generator $(A, D(A))$.

In particular, since the operator $-\Delta$ is self-adjoint on $H^2(\mathbb{R}^d)$, we can define solution of the free Schrödinger equation with inital datum $u_0 \in H^2(\mathbb{R}^d)$

$$\text{(S)}\ \begin{cases} i u_t= \Delta u \\ u(0, x)=u_0(x) \end{cases}$$

the function $u(t,x)=e^{i t \Delta}u_0(x)$.

Problem

Is the requirement $u_0 \in H^2(\mathbb{R}^d)$ really necessary? Would $u_0 \in H^1(\mathbb{R}^d)$ suffice? In fact I read in some course notes: "since $-\Delta$ is self-adjoint on $L^2(\mathbb{R}^d)$ with form domain $H^1(\mathbb{R}^d)$, we can define the Schrödinger group $e^{i t \Delta}$ [...] and so $u(t, x)=e^{i t \Delta}u_0(x)$ is solution of (S) for $u_0 \in H^1(\mathbb{R}^d)$."

The author then stops explaining. How could he weaken the regularity request on $u_0$ so much?

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1 Answer 1

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Well, this is sloppy language. You can solve the equation only for $H^2$. However, you can define the exponential function and the unitary group as you do, and call $u(t,\cdot) = e^{it\Delta}u_0(\cdot)$ the generalized (mild) solution for all $u_0\in L^2$. Why not?

Now if multiply the original equation by a nice function with compact support, then you can integrate by part once, and what you get you can write down also for $H^1$ functions. They remain invariant under the group, and are usually called the weak solutions.

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Let me see if I got it. In an abstract framework, we are trying to solve $$\begin{cases} u_t=Au \\ u(0)=u_0 \end{cases},$$ where $(A, D(A))$ is a self-adjoint operator and $u_0 \in \mathcal{H}$. What the author did is switching to the quadratic form $q_A(u, v)=(Au, v)$ which is defined in a bigger domain than $D(A)$ and consider the equation $$\begin{cases}(u_t, v)=q_A(u, v), \quad v \in D(q) \\ u(0)=u_0 \end{cases}.$$ [continue...] –  Giuseppe Negro Apr 21 '11 at 15:19
    
[...] When $u_0 \in D(A)$, $e^{itA}u_0$ is a solution of the first problem, ok. When $u_0 \in D(q)$, $e^{itA}u_0$ must be a solution of the second problem. Correct? Can you explain me a little bit how to better justify this, if you don't mind? Thank you. –  Giuseppe Negro Apr 21 '11 at 15:23

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