I don't get the relevance of u bar here, and how is $$2(u_{kl} - \bar{u}_{kl})$$
the derivative of the expanded square terms above? For the first part of the expression above, I'm confused why it's not $4u_{kl}$ since first, we seem to be adding 2 $u_{ij}^2$ terms together, and second I don't know what happened to the $-2(u_{i+1 j} + u_{i j+1})$ term.
motoole2
It is a little confusing---I agree. Let me try to clarify.
So first of all, let's consider taking the derivative with respect to $u_{k,l}$, and do this in two steps. First, let's expand the summation to only include terms that include $u_{k,l}$, since all other terms will be zeroed out anyways. We get the following:
And just to clarify, $\bar{u}_{kl}$ is defined as the local average (see bottom of slide).
nssampat
The derivation you have here makes sense, but the first expression you have written does not seem to match up with the expression at the top of the slide, right? I only see the second and fourth terms you wrote in the original expression from the top of the slide.
motoole2
You have to consider all terms with respect to all values $i$ and $j$ (i.e., given $N$ values for $i$ and $j$, the equation in the slide has a total of $4N$ smoothness terms, or $4$ for every value of $i$ and $j$). The first equation that I wrote here contains only the terms that include $u_{k,l}$. The second and fourth terms appear for $i = k$ and $j = l$. The first term appears for $i = k-1$ and $j = l$. The third term appears for $i = k$ and $j = l-1$.
I don't get the relevance of u bar here, and how is $$2(u_{kl} - \bar{u}_{kl})$$
the derivative of the expanded square terms above? For the first part of the expression above, I'm confused why it's not $4u_{kl}$ since first, we seem to be adding 2 $u_{ij}^2$ terms together, and second I don't know what happened to the $-2(u_{i+1 j} + u_{i j+1})$ term.
It is a little confusing---I agree. Let me try to clarify.
So first of all, let's consider taking the derivative with respect to $u_{k,l}$, and do this in two steps. First, let's expand the summation to only include terms that include $u_{k,l}$, since all other terms will be zeroed out anyways. We get the following:
$\frac{1}{4}(u_{k,l} - u_{k-1,l})^2 + \frac{1}{4}(u_{k,l} - u_{k+1,l})^2 + \frac{1}{4}(u_{k,l} - u_{k,l-1})^2 + \frac{1}{4}(u_{k,l} - u_{k,l+1})^2 + \cdots$
$\quad\quad\lambda(I_x u_{k,l} + I_y u_{k,l} + I_t)^2$
Second, let's take the derivative with respect to $u_{k,l}$:
$\frac{1}{2}(u_{k,l} - u_{k-1,l}) + \frac{1}{2}(u_{k,l} - u_{k+1,l}) + \frac{1}{2}(u_{k,l} - u_{k,l-1}) + \frac{1}{2}(u_{k,l} - u_{k,l+1}) + \cdots$
$\quad\quad2\lambda(I_x u_{k,l} + I_y u_{k,l} + I_t) I_x$
Simplifying, we get our solution:
$2(u_{k,l} - \frac{1}{4}(u_{k-1,l} + u_{k+1,l} + u_{k,l-1} + u_{k,l+1})) + 2\lambda(I_x u_{k,l} + I_y u_{k,l} + I_t) I_x$
And just to clarify, $\bar{u}_{kl}$ is defined as the local average (see bottom of slide).
The derivation you have here makes sense, but the first expression you have written does not seem to match up with the expression at the top of the slide, right? I only see the second and fourth terms you wrote in the original expression from the top of the slide.
You have to consider all terms with respect to all values $i$ and $j$ (i.e., given $N$ values for $i$ and $j$, the equation in the slide has a total of $4N$ smoothness terms, or $4$ for every value of $i$ and $j$). The first equation that I wrote here contains only the terms that include $u_{k,l}$. The second and fourth terms appear for $i = k$ and $j = l$. The first term appears for $i = k-1$ and $j = l$. The third term appears for $i = k$ and $j = l-1$.