Energy norm. Why is it called that way?

Solution 1:

To expand on my comment (see e.g. http://online.itp.ucsb.edu/online/lnotes/balents/node10.html):

The expression which you give can be interpreted as the energy of a $n$-dimensional elastic manifold being elongated in the $n+1$ dimension (e.g. for $n=2$, membrane in three dimension); $u$ is the displacement field.

Let me put back the units $$E[u]= \frac{a}{2}\int_{\Omega} \lvert u(x)\rvert^2\, dx + \frac{b}{2} \int_{\Omega} \lvert \nabla u(x)\rvert^2\, dx.$$ The first term tries to bring the manifold back to equilibrium (with $u=0$), the second term penalizes fast changes in the displacement. The energy is not homogenous and involves a characteristic length scale $$\ell_\text{char} = \sqrt{\frac{b}{a}}.$$ This is the scale over which the manifold returns back to equilibrium (in space) if elongated at some point. With $b=0$, the manifold would return immediately, you elongate it at some point and infinitely close the manifold is back at $u=0$. With $a=0$ the manifold would never return to $u=0$. Only the competition between $a$ and $b$ leads to the physics which we expect for elastic manifold. This competition is intimately related to the fact that there is a characteristic length scale appearing.

It is important that physical laws are not homogenous, in order to have characteristic length scales (like $\ell$ in your example, the Bohr radius for the hydrogen problem, $\sqrt{\hbar/m\omega}$ for the quantum harmonic oscillator, ...). The energy of systems only become scale invariant in the vicinity of second order phase transitions. This is a strong condition on energy functionals to the extend that people classify all possible second order phase transitions.

Solution 2:

When u is the velocity field of an viscous, incompressible homogeneous fluid with density set to 1, then the first integral is proportional to the kinetic energy of the fluid flow. The second integral is called enstrophy (see Wikipedia).

Enstrophy is in this situation something different than energy, therefore the term "energy norm" is slightly misleading. For a viscuous flow described by the Navier-Stokes equations, the enstrophy measures how fast the fluid flow dissipates energy due to friction, it is possible to show under some mild assumptions that $$ \frac{d}{dt} \text{energy} = - \text{viscosity} * \text{enstrophy} $$ So, physically speaking, fluid flows with finite energy and finite enstrophy are precisely those whose velocity fields are elements of $H^1$. And physically interesting fluid flows should have finite enstropy so that the process of energy loss due to friction is described by the model.