How to simplify the Euler-Lagrange equation of Brachistochrone in this way?
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I already know that in the Brachistochrone problem, we have Euler-Lagrange equation:
$$frac{1}{2y}sqrt{frac{1+y'^2}{y}}+frac{d}{dx}[frac{y'}{sqrt{y(1+y'^2)}}]=0$$
To solve this equation, we simplify the above equation and get:
$$frac{d}{dx}[frac{1}{sqrt{y(1+y'^2)}}]=0$$
How to get the second equation from the first equation?
variational-analysis
asked Nov 17 at 12:55
Perry_W
62
add a comment |
up vote
1
down vote
favorite
I already know that in the Brachistochrone problem, we have Euler-Lagrange equation:
$$frac{1}{2y}sqrt{frac{1+y'^2}{y}}+frac{d}{dx}[frac{y'}{sqrt{y(1+y'^2)}}]=0$$
To solve this equation, we simplify the above equation and get:
$$frac{d}{dx}[frac{1}{sqrt{y(1+y'^2)}}]=0$$
How to get the second equation from the first equation?
variational-analysis
asked Nov 17 at 12:55
Perry_W
62
For the record consider to include the Lagrangian.
– Qmechanic
Nov 17 at 20:56
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I already know that in the Brachistochrone problem, we have Euler-Lagrange equation:
$$frac{1}{2y}sqrt{frac{1+y'^2}{y}}+frac{d}{dx}[frac{y'}{sqrt{y(1+y'^2)}}]=0$$
To solve this equation, we simplify the above equation and get:
$$frac{d}{dx}[frac{1}{sqrt{y(1+y'^2)}}]=0$$
How to get the second equation from the first equation?
variational-analysis
asked Nov 17 at 12:55
Perry_W
62
I already know that in the Brachistochrone problem, we have Euler-Lagrange equation:
$$frac{1}{2y}sqrt{frac{1+y'^2}{y}}+frac{d}{dx}[frac{y'}{sqrt{y(1+y'^2)}}]=0$$
To solve this equation, we simplify the above equation and get:
$$frac{d}{dx}[frac{1}{sqrt{y(1+y'^2)}}]=0$$
How to get the second equation from the first equation?
variational-analysis
variational-analysis
asked Nov 17 at 12:55
Perry_W
62
asked Nov 17 at 12:55
Perry_W
62
asked Nov 17 at 12:55
Perry_W
62
asked Nov 17 at 12:55
Perry_W
62
asked Nov 17 at 12:55
Perry_W
62
62
For the record consider to include the Lagrangian.
– Qmechanic
Nov 17 at 20:56
add a comment |
For the record consider to include the Lagrangian.
– Qmechanic
Nov 17 at 20:56
For the record consider to include the Lagrangian.
– Qmechanic
Nov 17 at 20:56
For the record consider to include the Lagrangian.
– Qmechanic
Nov 17 at 20:56
add a comment |
1 Answer
1
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oldest
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up vote
0
down vote
I figured it out, it is just the application of the following equation:
$$F-y'frac{partial F}{partial y'}=C$$
To see the proof, please go to another questions:Euler-Lagrange formula
answered Nov 17 at 13:23
Perry_W
62
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I figured it out, it is just the application of the following equation:
$$F-y'frac{partial F}{partial y'}=C$$
To see the proof, please go to another questions:Euler-Lagrange formula
answered Nov 17 at 13:23
Perry_W
62
add a comment |
up vote
0
down vote
I figured it out, it is just the application of the following equation:
$$F-y'frac{partial F}{partial y'}=C$$
To see the proof, please go to another questions:Euler-Lagrange formula
answered Nov 17 at 13:23
Perry_W
62
add a comment |
up vote
0
down vote
up vote
0
down vote
I figured it out, it is just the application of the following equation:
$$F-y'frac{partial F}{partial y'}=C$$
To see the proof, please go to another questions:Euler-Lagrange formula
answered Nov 17 at 13:23
Perry_W
62
I figured it out, it is just the application of the following equation:
$$F-y'frac{partial F}{partial y'}=C$$
To see the proof, please go to another questions:Euler-Lagrange formula
answered Nov 17 at 13:23
Perry_W
62
answered Nov 17 at 13:23
Perry_W
62
answered Nov 17 at 13:23
Perry_W
62
answered Nov 17 at 13:23
Perry_W
62
62
add a comment |
add a comment |
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For the record consider to include the Lagrangian.
– Qmechanic
Nov 17 at 20:56