Application of semigroup method We consider the following Cauchy problem with certain conditions on the boundary:
$$\eqalign{
  & {u_{tt}}(t,x) + {u_{tt}}(t,0) - {u_{xx}}(t,x) + u(t,L) = 0{\rm{  in ]0}}{\rm{,T[}} \times {\rm{]0}}{\rm{,L[}}  \cr 
  & {\rm{\{ u(0}}{\rm{,x)}}{\rm{,}}{{\rm{u}}_t}(0,x)\}  = \{ f(x),g(x)\}  \cr} $$
My problem is how to transfer the above equation to the Cauchy problem form and how to deal with the term $u(t,L)$ and ${u_{tt}}(t,L)$ :
$$\eqalign{
  & Y' = AY  \cr 
  & Y(0) = {Y_0} \cr} $$
How can I writing this equation under the Cauchy form problem?
Thanks.
 A: I'd say that the boundary conditions are relevant for the matter.
For example, let us assume that the boundary conditions are
$$u(t,0)=0,\qquad u_{tt}(t,L)+u_x(t,L)=-u_t(t,L).$$
If we write $v(t)=u(t,L)$, then the equation can be rewritten as
$$u_{tt}(t,x)=u_{xx}(t,x)-v(t)-v_{tt}(t).$$
But the boundary conditions imply
$$v_{tt}(t)=u_{tt}(t,L)=-u_t(t,L)-u_x(t,L)=-v_t(t)-u_x(t,L)$$
and thus the equation becomes
$$u_{tt}(t,x)=u_{xx}(t,x)-v(t)+v_t(t)+u_x(t,L).$$
Now, if we write
$$Y=\begin{pmatrix}
u\\ 
u_t\\ 
v\\
v_t
\end{pmatrix}$$
then
$$Y'=\begin{pmatrix}
u_t\\ 
u_{tt}\\
v_t\\
v_{tt}
\end{pmatrix}=
\begin{pmatrix}
u_t\\ 
u_{xx}-v+v_t+u_x(L)\\
v_t\\
-v_t-u_x(L)
\end{pmatrix}.
$$
So, the problem can be written in the desired form by taking
$$Y=\begin{pmatrix}
u\\ 
\tilde{u}\\ 
v\\
\tilde{v}
\end{pmatrix},\qquad A\begin{pmatrix}
u\\ 
\tilde{u}\\
v\\
\tilde{v}
\end{pmatrix}=
\begin{pmatrix}
\tilde{u}\\ 
u_{xx}-v+\tilde{v}+u_x(L)\\
\tilde{v}\\
-\tilde{v}-u_x(L)
\end{pmatrix},\qquad
Y_0=\begin{pmatrix}
f\\ 
g\\ 
f(L)\\
g(L)
\end{pmatrix}.
$$
Note that, in this setting, the phase space should have the form $H_1\times H_2\times \mathbb C\times\mathbb C$, where $H_1$ and $H_2$ are functional spaces (probably some Sobolev space), and the domain of the operator should include the condition $v=u(L)$.
These considerations are based in a similar problem that I studied and maybe it gives you some insights on how to deal with the boundary terms (but in my case there were no boundary terms in the equation).

Edit (in response to the comments).
Let me consider
$$\eqalign{(1)\qquad   & {u_{tt}} = {u_{txx}} + {u_{xx}}\cr    
(2)\qquad& {u_{tt}}(L) =  - {u_{tx}}(L) - {u_x}(L)  \cr    
(3)\qquad& u(0,t) = {u_t}(0,t) = 0  \cr
(4)\qquad    & u(L,t) = a,{u_t}(L,t) = b  \cr
(5)\qquad    & u(x,0) = v(x),{u_t}(x,0) = w(x) \cr}$$
Multiplying equation $(1)$ by $\overline{u}_t$, integrating over $[0,L]$, performing integration by parts and using the boundary conditions $(2)$-$(4)$ we get
$$\frac{1}{2}\frac{d}{dt}\left(\int_0^L\Big(|u_t(x,t)|^2+|u_x(x,t)|^2\Big)dx+|b(t)|^2\right)=-\int_0^L|u_{tx}(x,t)|^2\;dx.$$
So, the energy is given by
\begin{equation}
E(t)=\frac{1}{2}\int_0^L\Big(|u_t(x,t)|^2+|u_x(x,t)|^2\Big)dx+\frac{1}{2}|b(t)|^2.
\end{equation}
We consider the phase space
$$H=H_\star^1\times L^2\times\mathbb{C}\quad(\text{where } H_\star^1=\{f\in H^1\mid f(0)=0\})$$
equipped with the norm associated with the energy:
$$\|(u,U,b)\|_H^2=\int_0^\ell\Big(|U|^2+|u_x|^2\Big)dx+|b|^2.$$
Then, the problem can be written as
\begin{equation}
\left\{\begin{aligned}
&Y'=AY\\
&Y(0)=Y_0
\end{aligned}\right.
\end{equation}
where $Y=(u,U,b)$, $Y_0=(v,w,w(L))$ and $A:D(A)\subset H \to H$ is given by
$$AY=\begin{pmatrix}
U\\ 
U_{xx}+u_{xx}\\ 
U(L)
\end{pmatrix}$$
with domain
$$D(A)=\{Y\in H\mid u\in H^2,\;U\in H_\star^1\cap H^2,\; b=U(L) \}.$$
We have
$$\text{Re }\langle AY,Y\rangle_{H}=-\int_0^L|U|^2\;dx\leq 0$$
and thus $A$ is dissipative.
Here, of course, $\langle\cdot,\cdot\rangle_H$ is the inner product that induces the energy norm $\|\cdot\|_H$:
$$\langle(u^*,U^*,b^*),(u,U,b)\rangle_H=\int_0^L \Big(U^*\overline{U}+u^*_x\overline{u}_x\Big)dx+b^*\overline{b}.$$
