Euler-Lagrange equations without lower boundary conditions

Question

Could someone help me with the following problem?

Let be the functional $F(y)=\int_a^bf(t,y,y')$, ($f\in{C^2[a,b]\times{\mathbb{R}}\times{\mathbb{R}}}$) with $y\in{C^2[a,b]}$ and $y(b)=0$, prove that if we have a extreme, $y^*$, of that functional with these conditions then $\delta_{y'}f(a,y^*(a),y'^*(a))=0$.

Answer 1

Hints:

1. Define momentum $$p~:=~\frac{\partial f}{\partial y^{\prime}}.\tag{1}$$ If one varies infinitesimally the functional $$F[y]~:=~ \int_a^b\! dt~ f(t,y(t),y^{\prime}(t)) \tag{2}$$ without discarding boundary contributions, one finds $$ \delta F[y]~=~ \int_a^b\! dt~\left( \frac{\partial f(t)}{\partial y(t)}\delta y(t)+p(t) ~\delta y^{\prime}(t)\right)$$ $$~\stackrel{\text{int. by parts}}{=}~ p(b)~\underbrace{\delta y(b)}_{=0}- p(a)~\delta y(a)+ \int_a^b\! dt~\left( \frac{\partial f(t)}{\partial y(t)}-\frac{d p(t)}{dt}\right) \delta y(t) .\tag{3}$$

2. Besides the given boundary condition $y(b)=0$, one concludes from formula (3) that a stationary configuration must obey $$ p(a)~=~0\quad\text{and} \quad\forall t\in [a,b]: \frac{\partial f(t)}{\partial y(t)}-\frac{d p(t)}{dt}~=~0.\tag{4}$$