I am currently trying to get into calculus of variation for a course of Image processing. In the lecture we learned that a smooth function u[a,b]->R that minimises: $$\int_a^b F(x,u,u') dx $$ satisfies the Euler-Lagrange equation $$F_{u}-\frac{d}{dx}F_{u'}=0$$
Okay, now i have an excericse in which the given F(x,u,u') equals $$ F(x,u,u')=\frac{1}{2}(u-f)^2+\alpha*\lambda^2*\sqrt{1+u_{x}^2/\lambda^2}$$
I then calculated $$F_{u}=u-f$$ and $$F_{ux} = \alpha *\lambda^2*\frac{1}{2\sqrt{1+u_x^2/\lambda^2}}*\frac{2u_x}{\lambda^2}=\frac{\alpha*u_x}{\sqrt{1+u_x^2/\lambda^2}}$$
Assuming that this is right I then get with the Euler-Lagrange Formula:
$$0=F_u -\frac{d}{dx}F_{ux}=u-f-\frac{d}{dx}(\frac{\alpha*u_x}{\sqrt{1+u_x^2/\lambda^2}})$$
And here my problems start by calculating the second part of this equation: $$\frac{d}{dx}(\frac{\alpha*u_x}{\sqrt{1+u_x^2/\lambda^2}})=\alpha*\frac{d}{dx}(\frac{u_x}{\sqrt{1+u_x^2/\lambda^2}}) = \alpha*\frac{\frac{d}{dx}(u_x)*\sqrt{1+u_x^2/\lambda^2}-u_x*\frac{d}{dx}(\sqrt{1+u_x^2/\lambda^2})}{1+\frac{u_x^2}{\lambda^2}}$$
Now I feel like i have problem understanding the basics. $u_x$ should be the partial derivative of u with respect to x. If i now take the "normal" derivative will i get the second order partial derivative? So is $\frac{d}{dx}(u_x) = u_{xx}$? And if i want to calculate $\frac{d}{dx}(\sqrt{1+u_x^2/\lambda^2})$ can I just assume that $u_x^2$is a normal variable or do I have to be carefull because basically $u_x$ is the partial derivative of a function?
Thanks for reading my question.
Max
Just assume $u_x$ is a normal variable and $\frac{\mathrm{d} }{\mathrm{d} x} (u_x) = u_{xx}$. Notice that u(x) is a function of one variable.