Evolution equation with inverse operator I want to prove the existence and uniquness of the following problem by the semigroup method 
$$
\eqalign{
  & {u_{tt}} + {u_{xxtt}} + {u_t} - {u_{xx}} = 0  \cr 
  & u(t,0) = u(t,l) = 0  \cr 
  & u(0) = {u_{0{\rm{   }}}}  ,u'(0) = {u_1} \cr} 
$$ 
i wrote it as a Cauchy problem :
 $$U' = AU$$ with : $U = ({v^1},{v^2})$ and  $A$ is defined as : $$
A = \begin{pmatrix}
   0 & I  \\[3pt]
   \Big(I + \frac{\partial ^2}{\partial x^2}\Big)^{ - 1}\!\!\frac{\partial ^2}{\partial x^2} &  - \Big(I + \frac{\partial ^2}{\partial  x^2}\Big)^{ - 1}
 \end{pmatrix} 
$$
My question is how can i treat the operator ${(I + {{{\partial ^2}} \over {\partial {x^2}}})^{ - 1}}$ ? 
Thank you
 A: There are two cases.

Case 1. $l\notin \pi\mathbb N$. $\,$ In this case, the operator $A$ is bounded on $L^2(0,l)\oplus L^2(0,l)$ and generates a uniformly
  continuous group $\{e^{At}\}_{t\in\mathbb R}$ (i.e., one can solve
  forward and backward in time).

To see that the operator $(I+\partial_x^2)^{-1}\partial_x^2$ initially
defined on $H^2(0,l)\cap H_0^1(0,l)$ extends to a bounded operator on
$L^2(0,l)$ it is easiest to identify $L^2(0,l)$ with $l^2$ via the map
$\sum_{k=1}^\infty \alpha_k \sin\left(\lambda_k x\right)\mapsto
\{\alpha_k\}_{k=1}^\infty$, where $ \lambda_k = k\pi/l$. The action of
$(I+\partial_x^2)^{-1}\partial_x^2$ is then given by
$\{\alpha_k\}_k\mapsto \left\{
\frac{\lambda_k^2}{\lambda_k^2-1}\,\alpha_k\right\}_k$. Boundnedness
on $L^2(0,l)$ follows from
$\left\{\frac{\lambda_k^2}{\lambda_k^2-1}\right\}_k\in l^\infty$.

Case 2. $k_0 = l/\pi\in\mathbb N$. $\,$ In this case, $\lambda_{k_0}=1$ and one needs to impose the condition 
  $\langle u_0+u_1,\sin x\rangle=0$, where $\langle\;,\,\rangle$ is the scalar
  product in $L^2(0,l)$ (see below). So, one has to work on the
  space 
  $X = \{(u_0,u_1)\in L^2(0,l)\oplus L^2(0,l)\colon \langle u_0+u_1,\sin x\rangle=0\}$. Again, $A$ can be realized as a bounded
  operator on $X$ and generates a uniformly continuous group.

To see the latter, one has to take care of how to define
$(I+\partial_x^2)^{-1}$ on $L^2(0,l)$. Write $L^2(0,l)= Y\oplus
Y^\perp$, where $Y = \{u\in L^2(0,l)\colon \langle u,\sin
x\rangle=0\}$, and $Y^\perp$ is spanned by $\sin x$. Then $Y$ reduces
$I+\partial_x^2$ and $(I+\partial_x^2)\bigr|_{Y}\colon H^2(0,l)\cap
H_0^1(0,l)\cap Y\subset Y\to Y$ is an isomorphism. On $Y^\perp$, one sets
$(I+\partial_x^2)^{-1}\bigr|_{Y^\perp} = -\left(1/2\right)I$.

In case 2, let $w(t) = \int_0^l u(t,x)\sin x\,dx$ (i.e., $\sqrt{2/l}\,w$ is the Fourier coefficient $\alpha_{k_0}$). Then $w(t) = -\,\int_0^l u_{xx}(t,x)\sin x\,dx$ by an integration by parts and
\begin{multline*}
  w'(t) + w(t) = w''(t) - w''(t) + w'(t) + w(t) \\
  = \int_0^l \left( u_{tt}(t,x) - u_{ttxx}(t,x) + u_t(t,x) - u_{xx}(t,x)\right)dx = 0.
\end{multline*}
With $w_j = \int_0^l u_j(x)\sin x\,dx$ for $j=0,1$ one obtains $w(t) =
w_0\, e^{-t}$ and $w_1 = w'(0) =-\,w_0$.
Hence, $\langle u_0+u_1,\sin x\rangle=0$ is a necessary condition for solvability in this case.
