Generating a contraction semigroup on an energy space Consider the system of partial differential equations
$\displaystyle\frac{\partial Q}{\partial t}(\zeta, t)=-\frac{\partial}{\partial\zeta}\frac{\phi(\zeta, t)}{L(\zeta)}$
$\displaystyle\frac{\partial \phi}{\partial t}(\zeta, t)=-\frac{\partial}{\partial \zeta}\frac{Q(\zeta, t)}{C(\zeta)}$
$Q(\zeta,t)$ is the charge at position $\zeta\in[a,b]$ and time $t>0$, and $\phi(\zeta,t)$ is the magnetic flux at position $\zeta$ and time $t$. $C$ is the distributed capacity and $L$ is the distributed inductance.
Let the voltage and the current be given by $V=Q/C$ and $I=\phi/L$ respectively.
Set boundary conditions: $V(a,t)=0$ and $V(b,t)=RI(b,t)$ with $R>0$.
I want to show that the differential operator associated to the given system of partial differential equations with the given boundary conditions generates a contraction semigroup on the energy space.
We know that the energy of this space is
$\displaystyle E(t)=\frac{1}{2}\int^{b}_{a}\frac{Q(\zeta,t)^{2}}{L(\zeta)}+\frac{Q(\zeta,t)^{2}}{C(\zeta)}d\zeta$
Then
$\displaystyle\frac{dE(t)}{dt}=2\int^{b}_{a}\frac{\phi(\zeta,t)\cdot\partial_{t}\phi(\zeta,t)}{L(\zeta)}+\frac{Q(\zeta,t)\cdot\partial_{t}Q(\zeta,t)}{C(\zeta)}d\zeta$
Did I differentiate that correctly? And then how can I remove the integral?
 A: Your definition of the energy is incorrect. It should be
$$
\forall t \in \Bbb{R}_{> 0}: \quad
E(t) \stackrel{\text{df}}{=}
\frac{1}{2}
\int_{a}^{b}
\left\{
\frac{[\phi(\zeta,t)]^{2}}{L(\zeta)} + \frac{[Q(\zeta,t)]^{2}}{C(\zeta)}
\right\} \mathrm{d}{\zeta}.
$$
Taking the time derivative, we get
\begin{align}
    E'(t)
& = \frac{1}{2}
    \int_{a}^{b}
    \left[
    \frac{2 \cdot \phi(\zeta,t) {\phi_{t}}(\zeta,t)}{L(\zeta)} +
    \frac{2 \cdot Q(\zeta,t) {Q_{t}}(\zeta,t)}{C(\zeta)}
    \right] \mathrm{d}{\zeta} \\
& = \int_{a}^{b}
    \left[
    \frac{\phi(\zeta,t) {\phi_{t}}(\zeta,t)}{L(\zeta)} +
    \frac{Q(\zeta,t) {Q_{t}}(\zeta,t)}{C(\zeta)}
    \right] \mathrm{d}{\zeta} \\
& = \int_{a}^{b}
    \left[
    \frac{\phi(\zeta,t)}{L(\zeta)} \cdot {\phi_{t}}(\zeta,t) +
    \frac{Q(\zeta,t)}{C(\zeta)} \cdot {Q_{t}}(\zeta,t)
    \right] \mathrm{d}{\zeta} \\
& = - \int_{a}^{b}
      \left\{
      \frac{\phi(\zeta,t)}{L(\zeta)} \cdot
      \frac{\partial}{\partial \zeta} \left[ \frac{Q(\zeta,t)}{C(\zeta)} \right] +
      \frac{Q(\zeta,t)}{C(\zeta)} \cdot
      \frac{\partial}{\partial \zeta}
      \left[ \frac{\phi(\zeta,t)}{L(\zeta)} \right]
      \right\} \mathrm{d}{\zeta} \\
& = - \int_{a}^{b}
      \frac{\partial}{\partial \zeta}
      \left[ \frac{\phi(\zeta,t)}{L(\zeta)} \frac{Q(\zeta,t)}{C(\zeta)} \right]
      \mathrm{d}{\zeta} \qquad (\text{By the Product Rule.}) \\
& = - \left[
      \frac{\phi(\zeta,t)}{L(\zeta)} \frac{Q(\zeta,t)}{C(\zeta)}
      \right]_{\zeta = a}^{\zeta = b} \\
& = - \left[
      \frac{\phi(b,t)}{L(b)} \frac{Q(b,t)}{C(b)} -
      \frac{\phi(a,t)}{L(a)} \frac{Q(a,t)}{C(a)}
      \right] \\
& = - [I(b,t) V(b,t) - I(a,t) V(a,t)] \qquad (\text{By definition.}) \\
& = I(a,t) V(a,t) - I(b,t) V(b,t).
\end{align}
