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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?

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

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  • $\begingroup$ 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

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