TITLE: Abstract Wave Equation and Semigroups
QUESTION [5 upvotes]: If an operator $A$ on a Hilbert space $H$ generates a strongly continuous semigroup, does then the operator $B$ on $H \oplus H$ given by the matrix
$$ B := \begin{pmatrix} 0 & \mathrm{id} \\ A & 0\end{pmatrix}$$
generate a semigroup as well? This would then yield a solution to the wave equation $u^{\prime\prime} = A u$ on $H\oplus H$. 
If this is not generally true, I would be very interested in a counterexample. Also, if this is not generally true, what properties of $A$ are needed in order for this to be true?

REPLY [4 votes]: You need two additional assumptions: the operator $A$ has to be a so-called cosine function generator, and your product space has to be $V\times H$ with a space $V\subset H$. 
Cosine function generator is more than sectorial, it is more or less when the numerical range of $A$ is in a parabola (selfadjoint negative definite is ok, this is discussed in the references Liviu mentioned).
The space $V$ is difficult, but if $A$ is selfadjoint, then it is essentially $D(A^{1/2})$.
See Section VI.3 in

Klaus-Jochen Engel and
  Rainer Nagel, MR 1721989 One-parameter semigroups for linear
  evolution
  equations, ISBN:
  0-387-98463-1. (here a downloadable version)

or Section 7.4 in

Haase, Markus The functional calculus for sectorial operators.
  Operator Theory: Advances and Applications, 169. Birkhäuser Verlag,
  Basel, 2006. xiv+392 pp. ISBN: 978-3-7643-7697-0; 3-7643-7697-X