Can anyone explain how the professor goes from line 4 to 5 of the derivation? In particular, how is:
$$\frac{\delta L}{\delta u}h'=-\frac{d}{dx}\frac{\delta L}{\delta u'}h$$
The professor states that this can be derived from integration by parts but I still don't see how it's possible. I wrote out the integration by parts:
$$\int \frac{\delta L}{\delta u'}h' =\frac{\delta L}{\delta u'}h - \int \frac{d}{dx}\frac{\delta L}{\delta u'}h$$
How can I relate this to the above?
On
Regarding why $h$ is zero on the boundary: The slide doesn't mention, but in Calculus of Variations your problems usually have boundary conditions, otherwise the minimum often does not exist. That is, canonical problems are of the form $$\min E(u) = \int_0^1 L(u'(x), u(x), x) dx $$ $$\text{s.t. } \ u(0) = a, \ \ u(1) = b, \ u \in C^1([0, 1])$$ If $u^*$ is a local min, then we can say that $\phi(t) = E(u^* + th)$ has a local min at $0$ for functions h which vanish at 0 and at 1. If the $h$ doesn't vanish, then $u^* + th$ will not satisfy the boundary conditions, so we can't say that $\phi(t)$ has a local min at $0$.
So that's why we pick $h$ in the first place to be a function which vanishes at the boundary.
It's just integration by part :
$$\int \left(\frac{\partial L}{\partial u'}h'\right)=\left[h\frac{\partial L}{\partial u'}\right]_{boundary}-\int \frac{d}{dx}\frac{\partial L}{\partial u'}h$$
But since $h$ has compact support, $h$ is $0$ on the boundary. Your equality follow.