euler-lagrange equation expansion
$begingroup$
Euler-Lagrange equation
$$frac{partial f}{partial y}-frac{d}{dx}frac{partial f}{partial y'} = 0$$
Can also be written as
$$f'_y-f''_{xy'}-f''_{yy'}y'-f''_{y'y'}y''=0$$
In my book it is provided as a self-evident fact. It is not evident to me at all. Please help me with the derivation, I have no idea where to even start
euler-lagrange-equation
asked Jan 17 at 15:45
user3600124user3600124
287
$endgroup$
add a comment |
$begingroup$
Euler-Lagrange equation
$$frac{partial f}{partial y}-frac{d}{dx}frac{partial f}{partial y'} = 0$$
Can also be written as
$$f'_y-f''_{xy'}-f''_{yy'}y'-f''_{y'y'}y''=0$$
In my book it is provided as a self-evident fact. It is not evident to me at all. Please help me with the derivation, I have no idea where to even start
euler-lagrange-equation
asked Jan 17 at 15:45
user3600124user3600124
287
$endgroup$
add a comment |
$begingroup$
Euler-Lagrange equation
$$frac{partial f}{partial y}-frac{d}{dx}frac{partial f}{partial y'} = 0$$
Can also be written as
$$f'_y-f''_{xy'}-f''_{yy'}y'-f''_{y'y'}y''=0$$
In my book it is provided as a self-evident fact. It is not evident to me at all. Please help me with the derivation, I have no idea where to even start
euler-lagrange-equation
asked Jan 17 at 15:45
user3600124user3600124
287
$endgroup$
Euler-Lagrange equation
$$frac{partial f}{partial y}-frac{d}{dx}frac{partial f}{partial y'} = 0$$
Can also be written as
$$f'_y-f''_{xy'}-f''_{yy'}y'-f''_{y'y'}y''=0$$
In my book it is provided as a self-evident fact. It is not evident to me at all. Please help me with the derivation, I have no idea where to even start
euler-lagrange-equation
euler-lagrange-equation
asked Jan 17 at 15:45
user3600124user3600124
287
asked Jan 17 at 15:45
user3600124user3600124
287
asked Jan 17 at 15:45
user3600124user3600124
287
asked Jan 17 at 15:45
user3600124user3600124
287
asked Jan 17 at 15:45
user3600124user3600124
287
287
add a comment |
add a comment |
1 Answer
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$begingroup$
$y=y(x)$, $f=f(x,y,y')$. Then
$$frac{d}{dx}frac{partial f}{partial y'}=frac{d}{dx}y'_{y'}(x,y,y')=
f''_{xy'}+f''_{yy'}y'+f''_{y'y'}y''$$
answered Jan 17 at 16:11
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
add a comment |
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1 Answer
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$begingroup$
$y=y(x)$, $f=f(x,y,y')$. Then
$$frac{d}{dx}frac{partial f}{partial y'}=frac{d}{dx}y'_{y'}(x,y,y')=
f''_{xy'}+f''_{yy'}y'+f''_{y'y'}y''$$
answered Jan 17 at 16:11
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
add a comment |
$begingroup$
$y=y(x)$, $f=f(x,y,y')$. Then
$$frac{d}{dx}frac{partial f}{partial y'}=frac{d}{dx}y'_{y'}(x,y,y')=
f''_{xy'}+f''_{yy'}y'+f''_{y'y'}y''$$
answered Jan 17 at 16:11
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
add a comment |
$begingroup$
$y=y(x)$, $f=f(x,y,y')$. Then
$$frac{d}{dx}frac{partial f}{partial y'}=frac{d}{dx}y'_{y'}(x,y,y')=
f''_{xy'}+f''_{yy'}y'+f''_{y'y'}y''$$
answered Jan 17 at 16:11
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
$y=y(x)$, $f=f(x,y,y')$. Then
$$frac{d}{dx}frac{partial f}{partial y'}=frac{d}{dx}y'_{y'}(x,y,y')=
f''_{xy'}+f''_{yy'}y'+f''_{y'y'}y''$$
answered Jan 17 at 16:11
Aleksas DomarkasAleksas Domarkas
1,62317
answered Jan 17 at 16:11
Aleksas DomarkasAleksas Domarkas
1,62317
answered Jan 17 at 16:11
Aleksas DomarkasAleksas Domarkas
1,62317
answered Jan 17 at 16:11
Aleksas DomarkasAleksas Domarkas
1,62317
1,62317
add a comment |
add a comment |
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