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Let $\mathcal{L} \in C^{1}(\mathbb{R}^n \times \mathbb{R}^n \times \mathbb{R}, \mathbb{R}$) and $S:= \ C^1([0,1], \mathbb{R}^n) \ni \gamma \mapsto \int_0^1dt \ \mathcal{L}(\gamma(t), \dot{\gamma}(t),t) \in \mathbb{R}$.

If the Fréchet-derivative of $S$ vanishes at a point $q_0 \in C^1([0,1], \mathbb{R}^n)$, then the Euler-Lagrange equations $$\partial_1 \mathcal{L}(q_0(t), \dot{q_0}(t), t)=\frac{d}{dt}\partial_2\mathcal{L}(q_0(t), \dot{q_0}(t),t)$$ are satisfied. Does the converse hold? That is, if $q_0 \in C^1([0,1], \mathbb{R}^n)$ satisfies Euler-Lagrange, is $DS(q_0)=0$?

Jannik Pitt
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1 Answers1

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Intuitively, yes... if $DS(q_0)$ were nonzero, there would be some nice function $r_0$, vanishing at $t=0$ and $t=1$, such that $S(q_0 + \varepsilon r_0) - S(q_0)$ was $\Theta(\varepsilon)$. But since the Euler-Lagrange equations are satisfied at $q_0$, $$ S(q_0 + \varepsilon r_0)-S(q_0) = O(\varepsilon^2)+\varepsilon\int_{0}^{1}dt \left(r_0 \partial_1{\cal{L}}(q_0,\dot q_0, t)+\dot r_0\partial_2{\cal{L}}(q_0,\dot q_0, t)\right) \\ =O(\varepsilon^2)+\varepsilon\int_{0}^{1}dt \left(r_0 \frac{d}{dt}\partial_2{\cal{L}}(q_0,\dot q_0, t)+\dot r_0\partial_2{\cal{L}}(q_0,\dot q_0, t)\right) \\ =O(\varepsilon^2)+\varepsilon \frac{d}{dt}(r_0\partial_2{\cal{L}}(q_0,\dot q_0, t))\big\vert_0^1=O(\varepsilon^2). $$

mjqxxxx
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