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

$\displaystyle E(t)=\frac{1}{2}\int^{b}_{a}\frac{Q(\zeta,t)^{2}}{L(\zeta)}+\frac{Q(\zeta,t)^{2}}{C(\zeta)}d\zeta$



Did I differentiate that correctly? And then how can I remove the integral?

  • $\begingroup$ I have edited to provide more information. $\endgroup$
    – Jason Born
    May 1 '15 at 21:27

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}

  • $\begingroup$ This is consistent with the formula $ P = I V $ for electrical power. $\endgroup$ May 4 '15 at 5:01
  • $\begingroup$ Is $E'(t)$ actually useful for showing that the operator associated to the p.d.e. generates a contraction semigroup? $\endgroup$
    – Jason Born
    May 4 '15 at 17:19
  • $\begingroup$ @user3482534: I think that your question is answered in this set of presentation slides by Hans Zwart. I just found it online. $\endgroup$ May 4 '15 at 18:10
  • $\begingroup$ @user3482534: Hi user. I just wanted to follow up with you on this problem. Did you find the notes useful? Your given conditions and the Lumer-Phillips Theorem guarantee the existence of a contraction semigroup. $\endgroup$ May 6 '15 at 5:03
  • $\begingroup$ @Berrick_Caleb_Filimore Yes, thank you for your help. I wrote a solution which essentially mimicked what Hans Zwart did on page 29. $\endgroup$
    – Jason Born
    May 15 '15 at 14:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.