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



Evolution equation with inverse operator
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.


