Euler-Lagrange equation for the Brachistochrone problem with friction
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The WolframMathworld website has a page containing the derivation for the brachistochrone with and without friction.
I am able to follow the derivation for the with-friction case up to the point of writing out the Euler-Lagrange equation from the functional derived from the physics of the situation. This is done in step (29) of the demonstration in the linked website.
I am asking for an algebraic demonstration of how applying the Euler-Lagrange equation to the integrand in step (28) results in the expression in step (29).
calculus-of-variations euler-lagrange-equation
asked Jan 10 at 5:57
AkaashAkaash
285
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add a comment |
$begingroup$
The WolframMathworld website has a page containing the derivation for the brachistochrone with and without friction.
I am able to follow the derivation for the with-friction case up to the point of writing out the Euler-Lagrange equation from the functional derived from the physics of the situation. This is done in step (29) of the demonstration in the linked website.
I am asking for an algebraic demonstration of how applying the Euler-Lagrange equation to the integrand in step (28) results in the expression in step (29).
calculus-of-variations euler-lagrange-equation
asked Jan 10 at 5:57
AkaashAkaash
285
$endgroup$
add a comment |
$begingroup$
The WolframMathworld website has a page containing the derivation for the brachistochrone with and without friction.
I am able to follow the derivation for the with-friction case up to the point of writing out the Euler-Lagrange equation from the functional derived from the physics of the situation. This is done in step (29) of the demonstration in the linked website.
I am asking for an algebraic demonstration of how applying the Euler-Lagrange equation to the integrand in step (28) results in the expression in step (29).
calculus-of-variations euler-lagrange-equation
asked Jan 10 at 5:57
AkaashAkaash
285
$endgroup$
The WolframMathworld website has a page containing the derivation for the brachistochrone with and without friction.
I am able to follow the derivation for the with-friction case up to the point of writing out the Euler-Lagrange equation from the functional derived from the physics of the situation. This is done in step (29) of the demonstration in the linked website.
I am asking for an algebraic demonstration of how applying the Euler-Lagrange equation to the integrand in step (28) results in the expression in step (29).
calculus-of-variations euler-lagrange-equation
calculus-of-variations euler-lagrange-equation
asked Jan 10 at 5:57
AkaashAkaash
285
asked Jan 10 at 5:57
AkaashAkaash
285
asked Jan 10 at 5:57
AkaashAkaash
285
asked Jan 10 at 5:57
AkaashAkaash
285
asked Jan 10 at 5:57
AkaashAkaash
285
285
add a comment |
add a comment |
1 Answer
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Let $f(x,y,xi) := sqrt{frac{1 + xi^2}{2g(y - mu x)}}$. Then $$t = int f(x,y,y'),dx$$ and the Euler-Lagrange equation is given by $$frac{partial f}{partial y} - frac{d}{dx}frac{partial f}{partial xi} = 0.$$
We now compute these derivatives $$frac{partial f}{partial y} = -sqrt{frac{1 + xi^2}{2g}}frac{1}{2(y - mu x)^{3/2}},$$ $$frac{partial f}{partial xi} = frac{xi}{sqrt{(1 + xi^2)2g(y - mu x)}},$$
and evaluate them at $(x,y(x),y'(x))$.
The Euler-Lagrange equation can then be rewritten as $$frac{d}{dx}left(frac{y'(x)}{sqrt{(1 + y'(x)^2)2g(y(x) - mu x)}}right) = sqrt{frac{1 + y'(x)^2}{2g}}frac{1}{2(y(x) - mu x)^{3/2}}.$$
Equation (29) follows upon expanding the derivative on the left hand side in the previous equation. The computations are a little lengthy, but I'm sure you can fill in the details.
answered Jan 10 at 7:11
GiovanniGiovanni
5,25351326
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1 Answer
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1 Answer
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active
oldest
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active
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$begingroup$
Let $f(x,y,xi) := sqrt{frac{1 + xi^2}{2g(y - mu x)}}$. Then $$t = int f(x,y,y'),dx$$ and the Euler-Lagrange equation is given by $$frac{partial f}{partial y} - frac{d}{dx}frac{partial f}{partial xi} = 0.$$
We now compute these derivatives $$frac{partial f}{partial y} = -sqrt{frac{1 + xi^2}{2g}}frac{1}{2(y - mu x)^{3/2}},$$ $$frac{partial f}{partial xi} = frac{xi}{sqrt{(1 + xi^2)2g(y - mu x)}},$$
and evaluate them at $(x,y(x),y'(x))$.
The Euler-Lagrange equation can then be rewritten as $$frac{d}{dx}left(frac{y'(x)}{sqrt{(1 + y'(x)^2)2g(y(x) - mu x)}}right) = sqrt{frac{1 + y'(x)^2}{2g}}frac{1}{2(y(x) - mu x)^{3/2}}.$$
Equation (29) follows upon expanding the derivative on the left hand side in the previous equation. The computations are a little lengthy, but I'm sure you can fill in the details.
answered Jan 10 at 7:11
GiovanniGiovanni
5,25351326
$endgroup$
add a comment |
$begingroup$
Let $f(x,y,xi) := sqrt{frac{1 + xi^2}{2g(y - mu x)}}$. Then $$t = int f(x,y,y'),dx$$ and the Euler-Lagrange equation is given by $$frac{partial f}{partial y} - frac{d}{dx}frac{partial f}{partial xi} = 0.$$
We now compute these derivatives $$frac{partial f}{partial y} = -sqrt{frac{1 + xi^2}{2g}}frac{1}{2(y - mu x)^{3/2}},$$ $$frac{partial f}{partial xi} = frac{xi}{sqrt{(1 + xi^2)2g(y - mu x)}},$$
and evaluate them at $(x,y(x),y'(x))$.
The Euler-Lagrange equation can then be rewritten as $$frac{d}{dx}left(frac{y'(x)}{sqrt{(1 + y'(x)^2)2g(y(x) - mu x)}}right) = sqrt{frac{1 + y'(x)^2}{2g}}frac{1}{2(y(x) - mu x)^{3/2}}.$$
Equation (29) follows upon expanding the derivative on the left hand side in the previous equation. The computations are a little lengthy, but I'm sure you can fill in the details.
answered Jan 10 at 7:11
GiovanniGiovanni
5,25351326
$endgroup$
add a comment |
$begingroup$
Let $f(x,y,xi) := sqrt{frac{1 + xi^2}{2g(y - mu x)}}$. Then $$t = int f(x,y,y'),dx$$ and the Euler-Lagrange equation is given by $$frac{partial f}{partial y} - frac{d}{dx}frac{partial f}{partial xi} = 0.$$
We now compute these derivatives $$frac{partial f}{partial y} = -sqrt{frac{1 + xi^2}{2g}}frac{1}{2(y - mu x)^{3/2}},$$ $$frac{partial f}{partial xi} = frac{xi}{sqrt{(1 + xi^2)2g(y - mu x)}},$$
and evaluate them at $(x,y(x),y'(x))$.
The Euler-Lagrange equation can then be rewritten as $$frac{d}{dx}left(frac{y'(x)}{sqrt{(1 + y'(x)^2)2g(y(x) - mu x)}}right) = sqrt{frac{1 + y'(x)^2}{2g}}frac{1}{2(y(x) - mu x)^{3/2}}.$$
Equation (29) follows upon expanding the derivative on the left hand side in the previous equation. The computations are a little lengthy, but I'm sure you can fill in the details.
answered Jan 10 at 7:11
GiovanniGiovanni
5,25351326
$endgroup$
Let $f(x,y,xi) := sqrt{frac{1 + xi^2}{2g(y - mu x)}}$. Then $$t = int f(x,y,y'),dx$$ and the Euler-Lagrange equation is given by $$frac{partial f}{partial y} - frac{d}{dx}frac{partial f}{partial xi} = 0.$$
We now compute these derivatives $$frac{partial f}{partial y} = -sqrt{frac{1 + xi^2}{2g}}frac{1}{2(y - mu x)^{3/2}},$$ $$frac{partial f}{partial xi} = frac{xi}{sqrt{(1 + xi^2)2g(y - mu x)}},$$
and evaluate them at $(x,y(x),y'(x))$.
The Euler-Lagrange equation can then be rewritten as $$frac{d}{dx}left(frac{y'(x)}{sqrt{(1 + y'(x)^2)2g(y(x) - mu x)}}right) = sqrt{frac{1 + y'(x)^2}{2g}}frac{1}{2(y(x) - mu x)^{3/2}}.$$
Equation (29) follows upon expanding the derivative on the left hand side in the previous equation. The computations are a little lengthy, but I'm sure you can fill in the details.
answered Jan 10 at 7:11
GiovanniGiovanni
5,25351326
answered Jan 10 at 7:11
GiovanniGiovanni
5,25351326
answered Jan 10 at 7:11
GiovanniGiovanni
5,25351326
answered Jan 10 at 7:11
GiovanniGiovanni
5,25351326
5,25351326
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