I'm having trouble understanding how the Euler Lagrange equations are derived for the optical flow equation in the paper: https://scihub.wikicn.top/10.1007/978-3-540-24673-2_3 Eq 7 on page 4 of the pdf.
$$ E(\mathit{u,v}) = \int \psi (|I(\mathbf{x+w}) - I(\mathbf{x})|^2 + \gamma |\nabla I(\mathbf{x+w}) - \nabla I(\mathbf{x})|^2) + \alpha \int \psi (|\nabla u|^2 + |\nabla v|^2) $$
Euler lagrange equations will be :
$$ \dfrac{\partial L}{\partial u} - \dfrac{d}{dx} \dfrac{\partial L}{\partial u_{x}} - \dfrac{d}{dy} \dfrac{\partial L}{\partial u_{y}} = 0 $$
$$ \dfrac{\partial L}{\partial v} - \dfrac{d}{dx} \dfrac{\partial L}{\partial v_{x}} - \dfrac{d}{dy} \dfrac{\partial L}{\partial v_{y}} = 0 $$
I would really appreciate it if someone could walk me through the u partial derivative part. Apologies if this is straightforward. I'm new to both optical flow and variational calculus.
Thanks in advance.
Given that $\mathbf{w}=(u, v, 1)^T$, so you follow the definition to compute the first term through the chain rule: $$\frac{\partial L}{\partial u}=\psi'(|I(\mathbf{x}+\mathbf{w})-I(\mathbf{x})|^2+\gamma|\nabla I(\mathbf{x}+\mathbf{w})-\nabla I(\mathbf{x})|^2)\ 2[(I(\mathbf{x}+\mathbf{w})-I(\mathbf{x})\frac{\partial I}{\partial x}(\mathbf{x+w})+\gamma (\nabla I(\mathbf{x+w})-\nabla I(\mathbf{x}))\cdot \nabla \frac{\partial I}{\partial x}(\mathbf{x+w})].$$
Next you can calculate the dependence on $\nabla u$: $$\frac{\partial L}{\partial \nabla u}=\alpha\psi'(|\nabla u|^2+|\nabla v|^2)\ 2\nabla u.$$
You can assemble the last two term (it probably should have been three terms with the one on $\frac{\partial}{\partial t}$) by $$\nabla\cdot\frac{\partial L}{\partial \nabla u}=\nabla\cdot(\alpha\psi'(|\nabla u|^2+|\nabla v|^2)\ 2\nabla u).$$
Note that in that paper, they put divergence instead $\rm{div}=\nabla\cdot$