optical flow Euler-Langrange equation

Question

I have a problem understanding how optical flow functional is plugged into Euler-Lagrange equation. The functional is:

$\iint[(I_xu+I_yv+I_t)^2+\alpha^2(||\nabla u||^2 +||\nabla v ||^2)]dxdy$

Euler_Langrange equation is: $\frac{\partial L}{\partial f} - \frac {d}{dt} \frac {\partial L} {\partial f'} = 0$ and for two variables we have $\frac{\partial L}{\partial u} - \frac {d}{dx}\frac {\partial L}{\partial u_x} - \frac {d}{dy}\frac {\partial L}{\partial u_y}=0$ and the same for v. This becomes then

$I_x(I_x u+I_y v+I_t)-\alpha^2 \Delta u = 0$ where $\Delta u = \frac{\partial ^2}{\partial x^2} + \frac {\partial^2}{\partial y^2}$

I just can't understand how the terms in the last eqution have been generated, I would appreciate it if anyone can walk me throughout the steps.

Thanks in advance

Answer 1

Hint:

$$|| \nabla u ||^{2} = u_{x}^{2}+u_{y}^{2}$$ Where subscript denotes partial differetiation. Thus, taking the last two terms of the Euler-Lagrange equation \begin{eqnarray} -\frac{d}{dx}\frac{\partial L}{\partial u_{x}}-\frac{d}{dy}\frac{\partial L}{\partial u_{y}} &=& -\frac{d}{dx}(2u_{x})--\frac{d}{dy}(2u_{y}) \\ &=& -2u_{xx}-2u_{yy} \\ &=& -2\Delta u \end{eqnarray} Obviously I omitted the factor of $\alpha^{2}$ in my calculation. Carrying on throughon with the first term in the Euler-Lagrange equation; $$\frac{\partial L}{\partial u} = 2I_{x}(I_{x}u+I_{y}v+I_{t})$$ and then the Euler-Lagrange equation beomces the final line in your question.