how to introduce time into calculus of variations for image processing? I'm studying some topics about calculus of variation applied to image processing.
I'd like to understand how to introduce time parameter to evolve an image in an iterative way.
For example, let's consider the isotropic diffusion example, which should be one of the simplest functionals and that is defined by:
$I(x,y)$ = the target image (single function of two variables)
$F(x,y,I,\nabla I) = F(x,y,I,I_x,I_y) = \|\nabla I\|^2$ = function of $I$ and $\nabla I$
The functional to extremize is
$$
E(I) = \int_\Omega \|\nabla I\|^2 d\Omega
$$
We want to satisfy the E-L equation, then:
$$
\begin{align*}
\frac{\partial F}{\partial I} &= div \left( \frac{\partial F}{\partial \nabla I} \right)\\\\\\\\
\frac{\partial \| \nabla I \|^2}{\partial I} &= \frac{\partial \| \nabla I \|^2}{\partial I_{xx}} + \frac{\partial \| \nabla I \|^2}{\partial I_{yy}}
\end{align*}
$$
Now, I'm stucked. I'd like to understand how to derive the following relation, that can be found in many papers:
$$
\frac{\partial I}{\partial t} = div \left( \nabla I \right) = \Delta I
$$
In particular, I can't understand when and how the time variable $t$ is introduced to derive that formula from the E-L equation.
Thank you in advance
 A: The idea is to interprete the gradient algorithm as time-discretization of a time-dependent equation:
If $E$ is given by $E(I)=\frac12\int\|\nabla I\|^2dx$, then the first derivative at $I\in H^1(\Omega)$ in direction $\delta I\in H^1(\Omega)$ is given by
$$
E'(I)\delta I = \int_\Omega \nabla I \cdot \nabla \delta I\ dx.
$$
Integration by-parts and assuming $\delta I=0$ on the boundary of $\Omega$ gives (for $I\in H^2(\Omega)$)
$$
E'(I) = - \int_\Omega div(\nabla I)\delta I\ dx.
$$
Thus, the gradient of $E$ with respect to the $L^2$-inner product is given by $\nabla E(I)= -\Delta I$.
Now suppose an iterate $I_k$ is given, and you want to apply one gradient step to obtain $I_{k+1}$:
$$
I_{k+1} = I_k + h \nabla E(I_k) = I_k + t_k (-\Delta I_k).
$$
Now you interprete $I_k$ to be approximations of the function value at $t=k\cdot h$ of a function $I:\mathbb R\to H^1(\Omega)$. Rewriting the equation above as difference quotient gives
$$
\frac1h(I_{k+1}-I_k) = -\Delta I_k,
$$
$h\searrow 0$ yields
$$
\frac{\partial I}{\partial t} = -\Delta I.
$$
