Are the Euler-Lagrange equations equivalent to the functional having a stationary point?
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 $q_0 in C^1([0,1], mathbb{R}^n)$ satisfies Euler-Lagrange, is $DS(q_0)=0$?
calculus-of-variations
asked Nov 25 at 19:12
Jannik Pitt
291316
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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 $q_0 in C^1([0,1], mathbb{R}^n)$ satisfies Euler-Lagrange, is $DS(q_0)=0$?
calculus-of-variations
asked Nov 25 at 19:12
Jannik Pitt
291316
add a comment |
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 $q_0 in C^1([0,1], mathbb{R}^n)$ satisfies Euler-Lagrange, is $DS(q_0)=0$?
calculus-of-variations
asked Nov 25 at 19:12
Jannik Pitt
291316
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 $q_0 in C^1([0,1], mathbb{R}^n)$ satisfies Euler-Lagrange, is $DS(q_0)=0$?
calculus-of-variations
calculus-of-variations
asked Nov 25 at 19:12
Jannik Pitt
291316
asked Nov 25 at 19:12
Jannik Pitt
291316
edited Dec 9 at 19:26
asked Nov 25 at 19:12
Jannik Pitt
291316
asked Nov 25 at 19:12
Jannik Pitt
291316
asked Nov 25 at 19:12
Jannik Pitt
291316
291316
add a comment |
add a comment |
1 Answer
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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 $q_0$,
$$
S(q_0 + varepsilon r_0)-S(q_0) = O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 partial_1{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 frac{d}{dt}partial_2{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilon frac{d}{dt}(r_0partial_2{cal{L}}(q_0,dot q_0, t))bigvert_0^1=O(varepsilon^2).
$$
answered Dec 5 at 6:40
mjqxxxx
31k23985
add a comment |
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1 Answer
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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 $q_0$,
$$
S(q_0 + varepsilon r_0)-S(q_0) = O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 partial_1{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 frac{d}{dt}partial_2{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilon frac{d}{dt}(r_0partial_2{cal{L}}(q_0,dot q_0, t))bigvert_0^1=O(varepsilon^2).
$$
answered Dec 5 at 6:40
mjqxxxx
31k23985
add a comment |
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 $q_0$,
$$
S(q_0 + varepsilon r_0)-S(q_0) = O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 partial_1{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 frac{d}{dt}partial_2{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilon frac{d}{dt}(r_0partial_2{cal{L}}(q_0,dot q_0, t))bigvert_0^1=O(varepsilon^2).
$$
answered Dec 5 at 6:40
mjqxxxx
31k23985
add a comment |
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 $q_0$,
$$
S(q_0 + varepsilon r_0)-S(q_0) = O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 partial_1{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 frac{d}{dt}partial_2{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilon frac{d}{dt}(r_0partial_2{cal{L}}(q_0,dot q_0, t))bigvert_0^1=O(varepsilon^2).
$$
answered Dec 5 at 6:40
mjqxxxx
31k23985
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 $q_0$,
$$
S(q_0 + varepsilon r_0)-S(q_0) = O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 partial_1{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilonint_{0}^{1}dt left(r_0 frac{d}{dt}partial_2{cal{L}}(q_0,dot q_0, t)+dot r_0partial_2{cal{L}}(q_0,dot q_0, t)right) \ =O(varepsilon^2)+varepsilon frac{d}{dt}(r_0partial_2{cal{L}}(q_0,dot q_0, t))bigvert_0^1=O(varepsilon^2).
$$
answered Dec 5 at 6:40
mjqxxxx
31k23985
answered Dec 5 at 6:40
mjqxxxx
31k23985
answered Dec 5 at 6:40
mjqxxxx
31k23985
answered Dec 5 at 6:40
mjqxxxx
31k23985
31k23985
add a comment |
add a comment |
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