Consider for $r>0$ the Hilbert space $H=\mathbb{R} \times L^{2}[-r,0]$ with scalar product $$\langle x,y \rangle=x_0 y_0 + \int_{-r}^0 x_1(\theta) y_1(\theta) d\theta $$
for every $x=(x_0,x_1), y=(y_0,y_1) \in H$.
Consider $A \colon D(A) \subset H \to H$ defined by
$$Ax=[x_0+x_1(-r),x_1']$$
for every $x=(x_0,x_1) \in D(A)=\{x=(x_0,x_1) \in H: x_1 \in W^{1,2}[-T,0],x_0=x_1(0) \}$
Here $W^{1,2}$ is the Sobolev space and $x_1'$ is the derivative.
I want to derive the adjoint, i.e. by identifying $H=H^*$ we have $A^* \colon D(A^*) \subset H \to H$. In order to do this I compute:
for $x=(x_0,x_1) \in D(A), y=(y_0,y_1) \in H$, by integration by parts:
\begin{align}
\langle Ax,y \rangle & =[x_0+x_1(-r)]y_0+\langle x_1',y_1 \rangle_{L^2} \\
& = y_0x_0+x_1(-r)y_0 +x_1(0)y_1(0)-x_1(-r)y_1(-r) -\langle x_1,y_1' \rangle_{L^2} \\
& = y_0x_0+x_1(-r)y_0 +x_0y_1(0)-x_1(-r)y_1(-r) -\langle x_1,y_1' \rangle_{L^2}
\end{align}
which makes sense only for $y_1 \in W^{1,2}$.
Moreover since $D(A^*)=\{ y \in X^*: \langle Ax ,y\rangle\leq C |x|, x \in D(A) \}$ we must have that $y_0=0,y_1(-r)=0$,i.e. this suggests that $D(A^*)=\{y \in H:y_0=0,y_1 \in W^{1,2}[-T,0],y_1(-r)=0\}$ and
$$A^*y=[y_1(0),-y_1']$$ for $y=(y_0,y_1) \in D(A^*)$
Is it correct?