Energy norm. Why is it called that way?

Let $\Omega$ be an open part of $\mathbb{R}^n$. The adhering to

$$\lVert u \rVert_{1, 2}^2=\int_{\Omega} \lvert u(x)\rvert^2\, dx + \int_{\Omega} \lvert \nabla u(x)\rvert^2\, dx$$

specifies a standard on $H^1(\Omega)$ room, that is occasionally called power standard .

I do not really feel very easy with the physical definition this name recommends. Specifically, I see 2 non - uniform amounts, $\lvert u(x)\rvert^2$ and also $\lvert \nabla u(x)\rvert^2$, being summed with each other. Just how can this be literally regular?

Possibly some instance can aid me below. Thanks.

19
2022-06-07 14:32:27
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Answers: 2

When u is the rate area of a thick, incompressible uniform liquid with thickness readied to 1, after that the first indispensable is symmetrical to the kinetic power of the liquid circulation. The 2nd indispensable is called enstrophy (see Wikipedia).

Enstrophy remains in this scenario something various than power, consequently the term "energy norm" is a little deceptive. For a viscuous circulation defined by the Navier - Stokes formulas, the enstrophy gauges just how quickly the liquid circulation dissipates power as a result of rubbing, it is feasible to show under some light presumptions that $$ \frac{d}{dt} \text{energy} = - \text{viscosity} * \text{enstrophy} $$ So, literally talking, liquid moves with limited power and also limited enstrophy are specifically those whose rate areas are components of $H^1$. And also literally intriguing liquid moves need to have limited enstropy to make sure that the procedure of power loss as a result of rubbing is defined by the version.

4
2022-06-07 14:59:23
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To expand on my comment (see as an example http://online.itp.ucsb.edu/online/lnotes/balents/node10.html):

The expression which you offer can be taken the power of a $n$ - dimensional flexible manifold being extended in the $n+1$ measurement (as an example for $n=2$, membrane layer in 3 measurement) ; $u$ is the variation area.

Allow me return the devices $$E[u]= \frac{a}{2}\int_{\Omega} \lvert u(x)\rvert^2\, dx + \frac{b}{2} \int_{\Omega} \lvert \nabla u(x)\rvert^2\, dx.$$ The first term attempts to bring the manifold back to stability (with $u=0$), the 2nd term punishes rapid adjustments in the variation. The power is not uniform and also entails a particular size range $$\ell_\text{char} = \sqrt{\frac{b}{a}}.$$ This is the range over which the manifold returns back to stability (precede) if extended at some time. With $b=0$, the manifold would certainly return quickly, you extend it at some time and also definitely close the manifold is back at $u=0$. With $a=0$ the manifold would certainly never ever go back to $u=0$. Just the competitors in between $a$ and also $b$ brings about the physics which we anticipate for flexible manifold. This competitors is thoroughly pertaining to the reality that there is a particular size range showing up.

It is necessary that physical regulations are not uniform, in order to have particular size ranges (like $\ell$ in your instance, the Bohr distance for the hydrogen trouble, $\sqrt{\hbar/m\omega}$ for the quantum harmonic oscillator, ). The power of systems just come to be range stable at 2nd order stage changes. This is a solid problem on power functionals to the expand that individuals identify all feasible 2nd order stage changes.

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2022-06-07 14:59:12
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