Suppose I have two dynamical passive systems with input-output denoted respectively with $u_1, y_1$ and $u_2 = y_1, y_2 $. Notice that $u_2=y_1$ means that the output of the first system serves as the input of the second system. The two systems are passive, in the sense that there exist certain storage functions $V_i$ such that:
$$ \int_0^\tau u_i(t)^T y_i(t) \mathrm{d} t \geq-V_i(x_i(0)), \quad \quad \text{ for } i=1,2 $$
Is it true that the overall system is also passive? More precisely, is it always true that one can find a storage function $V_c$ such that the following holds?
$$ \int_0^\tau u_1(t)^T y_2(t) \mathrm{d} t \geq-V_c(x_c(0)) $$
I know that this is true if the systems are connected in a feedback fashion but I was not able to prove that this is also true for the open loop configuration.
Pick $\dot{x}_i=-x_i+u_i,y_i=x_i$, $i=1,2$, and $u_2=y_1$. The system in series then maps $u_1$ to $y_2$ and can be expressed as
$$ \dot{x}=\underbrace{\begin{bmatrix}-1 & 0\\1 & -1\end{bmatrix}}_{\mbox{$A$}}x+\underbrace{\begin{bmatrix}1 \\ 0\end{bmatrix}}_{\mbox{$B$}}u,\ \ y=\underbrace{\begin{bmatrix}0 & 1\end{bmatrix}}_{\mbox{$C$}}x. $$
This system is passive if and only if there exists a symmetrix positive semidefinite matrix $P$ such that $A^TP+PA$ is negative semidefinite and $PB-C^T=0$.
The equality constraint imposes that $P$ is of the form $$ P=\begin{bmatrix}0 & 1\\1 & p_3\end{bmatrix} $$ and, therefore, cannot be positive semidefinite. This proves that the system is not passive.
One equivalent way for looking at it by considering the correspondence between passive systems and positive realness of the associated transfer function. In the present case, the transfer function of the overall system is
$$ H(s)=\dfrac{1}{(s+1)^2}, $$ and is obviously not positive real as the Nyquist plot does not stay in the right half plane of the complex plane.
On the other hand, this is true for systems in parallel. Define two passive systems $H_1$ and $H_2$ and let $y_1=H_1u_1$ and $y_2=H_2u_2$. We now define the parallel interconnection $u_1=u_2=u$ and $y=y_1+y_2$.
Since the systems are passive, then there exists a storage function $V_i$ such that $\dot{V}_i(x_i)\le u_i^Ty_i$, $i=1,2$.
Defining now $V=V_1+V_2$, we get
$$ \begin{align} \dot{V} &= \dot{V}_1+\dot{V}_2\\ &\le u_1^Ty_1+u_2^Ty_2\\ &= u^Ty_1+u^Ty_2\\ &= u^Ty \end{align} $$ which proves that the parallel interconnection is passive.