gradient flow and what is, for example, $L^2$ gradient? Am I right that the gradient flow of a functional $E$ is
$$f_t = -\nabla E(f).$$
Solving this for $f$ gives you a minimiser of $E$ in some way?
Here the $\nabla$ denotes the gradient or the first variation or Gateaux derivative or whatever is appropriate. 
What is meant by "$L^2$ gradient (flow)" or "$H^{-1}$ gradient (flow)?"
Thanks
 A: If $E$ is a Frechet differentiable functional on a Hilbert space, the gradient of $E$ at $u$ is the element $w$ of the Hilbert space satisfying $E'(u)v = (w,v)$ for all $v$ in the Hilbert space.  The $E$ gradient flow starting of $u$ is the solution $\eta(t)$ of the diffential equation $\frac{d}{dt}I(\eta) = -\nabla E(\eta), \eta(0) = u$.  Sorry, I don't know how to make the accent mark in Frechet, nor do I know how this works more generally in Banach spaces.  If $\eta(t)$ converges to some $w$ as $t\to \infty$, then that $w$ is a critical point of $E$ (not necessarily a minimizer).  If someone says ''$L^2$ gradient flow'' or ''$H^{-1}$ gradient flow" they mean that the Hilbert space is $L^2$ or $H^{-1}$, but the gradient flow depends on the functional. 
A: Imagine that we start with a linear functional $f$ which takes $X$ to $\mathbb{R}$.  If we complete that to a space $\hat{X}$ which is obtained by equipping it with an inner product and taking the closure, then $f \in (\hat{X})^{*}$.  (Recall that Hilbert spaces are defined to be complete).
The Riesz representation theorem for the connection between a Hilbert space and its continuous dual space then then says $f(x)=\langle \nabla f, x\rangle$ for any $x$ in $\hat{X}$ and any $\nabla f$ in $(\hat{X})^{*}$.   Obviously, that inner product could be the $L^2$ norm or whichever you choose depending on the space.  $f$ can always be written $\langle \nabla f, x \rangle$ and $f$ having gradient just means $\nabla f \in X$.  
So a $L^2$ gradient flow is just a flow for a $L^2$ gradient, where the definition of a gradient flow is given by Stefan above.  As an example from differential geometry, take a map $\pi$ from $E$ to a compact, oriented manifold $M$ where there is a bundle metric $\langle \cdot, \cdot\rangle_E$.  In this case, the $L^2$ norm has to take $M$ to $\text{Sym}^2 (T^{*} M)$, so it takes the form
$\langle s, t \rangle_{L^{2}} = \int_M \langle s(x),  t(x) \rangle_E \: \text{d}v(x)$
where $s,t \in \Gamma(E)$.
