I am trying to use the version of the Lumer-Phillips, which is stated as the following
Let $A$ be a linear operator defined on a dense linear subspace $D(A)$ of the reflexive Banach space $X$. Then $A$ generates a contraction semigroup if and only if $A$ is closed and both $A$ and its adjoint operator $A^∗$ are dissipative.
My operator $\mathcal{L}: D(\mathcal{L}) \subseteq X \longrightarrow X$ is defined by \begin{equation} D(\mathcal{L})=H^2(\mathbb{R}) \end{equation} \begin{equation} X = L^2 (\mathbb{R}) \end{equation} \begin{equation} \mathcal{L} = - \nu \partial_x -\frac{\mu_2}{2} x^2 + \gamma \partial_xx . \end{equation}
My question is how do I show that $D(\mathcal{L})$ is dense in $X$? (Is it dense?) I do not know where to start!
Furthermore, I was wondering how one went about defining the spaces if the operator was to act on a complex variable? (So it made up part of the complex Ginzburg Landau equation).
Kindest regards,
Catherine
This is an standard result in the theory of functional spaces, that you can find in any text book (for instance, I recommend the one by Brezis on functional analysis).
Indeed, smooth (i.e. $C^\infty$ functions) with compact support are dense in $L^2(\mathbb R)$, and this is a subspace of $H^2(\mathbb R)$.
There are two main ideas involved in the proof i) truncation ii) convolution with an approximation of the delta function ("standard molifier")