Are the Euler-Lagrange equations equivalent to the functional having a stationary point?











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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 $$partial_1 mathcal{L}(q_0(t), dot{q_0}(t), t)=frac{d}{dt}partial_2mathcal{L}(q_0(t), dot{q_0}(t),t)$$ are satisfied. Does the converse hold? That is, if $q_0$ satisfies Euler-Lagrange, is $DS(q_0)=0$?










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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 $$partial_1 mathcal{L}(q_0(t), dot{q_0}(t), t)=frac{d}{dt}partial_2mathcal{L}(q_0(t), dot{q_0}(t),t)$$ are satisfied. Does the converse hold? That is, if $q_0$ satisfies Euler-Lagrange, is $DS(q_0)=0$?










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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 $$partial_1 mathcal{L}(q_0(t), dot{q_0}(t), t)=frac{d}{dt}partial_2mathcal{L}(q_0(t), dot{q_0}(t),t)$$ are satisfied. Does the converse hold? That is, if $q_0$ satisfies Euler-Lagrange, is $DS(q_0)=0$?










      share|cite|improve this question















      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 $$partial_1 mathcal{L}(q_0(t), dot{q_0}(t), t)=frac{d}{dt}partial_2mathcal{L}(q_0(t), dot{q_0}(t),t)$$ are satisfied. Does the converse hold? That is, if $q_0$ satisfies Euler-Lagrange, is $DS(q_0)=0$?







      calculus-of-variations






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      edited Nov 30 at 19:01

























      asked Nov 25 at 19:12









      Jannik Pitt

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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 $q_0$,
          $$
          S(q_0 + varepsilon r_0)-S(q_0) = O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 partial_1{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 frac{d}{dt}partial_2{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilon frac{d}{dt}(r_0partial_2{cal{L}}(q_0,dot q_0, t))bigvert_0^1=O(varepsilon^2).
          $$






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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 $q_0$,
            $$
            S(q_0 + varepsilon r_0)-S(q_0) = O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 partial_1{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 frac{d}{dt}partial_2{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilon frac{d}{dt}(r_0partial_2{cal{L}}(q_0,dot q_0, t))bigvert_0^1=O(varepsilon^2).
            $$






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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 $q_0$,
              $$
              S(q_0 + varepsilon r_0)-S(q_0) = O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 partial_1{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 frac{d}{dt}partial_2{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilon frac{d}{dt}(r_0partial_2{cal{L}}(q_0,dot q_0, t))bigvert_0^1=O(varepsilon^2).
              $$






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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 $q_0$,
                $$
                S(q_0 + varepsilon r_0)-S(q_0) = O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 partial_1{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 frac{d}{dt}partial_2{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilon frac{d}{dt}(r_0partial_2{cal{L}}(q_0,dot q_0, t))bigvert_0^1=O(varepsilon^2).
                $$






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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 $q_0$,
                $$
                S(q_0 + varepsilon r_0)-S(q_0) = O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 partial_1{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 frac{d}{dt}partial_2{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilon frac{d}{dt}(r_0partial_2{cal{L}}(q_0,dot q_0, t))bigvert_0^1=O(varepsilon^2).
                $$







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                answered 2 days ago









                mjqxxxx

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