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Calculus of variations with inverse of derivative composited within?
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Suppose $s in S subset mathbbR^+$ with c.d.f. $F(s)$. A real-valued function $beta(s) leq s$ is determined by a first order condition $gamma'(s - beta(s)) = g(F(s)) in mathbbR$, where $gamma$ is a real valued penalty function. I need to find the $gamma$ which
$$
min int_S (s - beta(s))^2 dF(s)
$$
subject to a constraint
$$
int_S gamma (s - beta(s)) dF(s) leq K in mathbbR.
$$
The Lagrangian with the first order condition plugged in, and change of variable, is
$$
mathcalL=int_0^1 [gamma'^-1 (g(x))]^2 + lambda gamma(gamma'^-1(g(x))) dx
$$
I am not looking for an analytical solution, but any guidance about which direction I should look into? I would like to obtain a differential equation about $gamma$ so that I can quantitatively analyze the function $gamma$.
calculus-of-variations optimal-control
asked Aug 3 at 21:19
Half Hogshead
84
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1
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Suppose $s in S subset mathbbR^+$ with c.d.f. $F(s)$. A real-valued function $beta(s) leq s$ is determined by a first order condition $gamma'(s - beta(s)) = g(F(s)) in mathbbR$, where $gamma$ is a real valued penalty function. I need to find the $gamma$ which
$$
min int_S (s - beta(s))^2 dF(s)
$$
subject to a constraint
$$
int_S gamma (s - beta(s)) dF(s) leq K in mathbbR.
$$
The Lagrangian with the first order condition plugged in, and change of variable, is
$$
mathcalL=int_0^1 [gamma'^-1 (g(x))]^2 + lambda gamma(gamma'^-1(g(x))) dx
$$
I am not looking for an analytical solution, but any guidance about which direction I should look into? I would like to obtain a differential equation about $gamma$ so that I can quantitatively analyze the function $gamma$.
calculus-of-variations optimal-control
asked Aug 3 at 21:19
Half Hogshead
84
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose $s in S subset mathbbR^+$ with c.d.f. $F(s)$. A real-valued function $beta(s) leq s$ is determined by a first order condition $gamma'(s - beta(s)) = g(F(s)) in mathbbR$, where $gamma$ is a real valued penalty function. I need to find the $gamma$ which
$$
min int_S (s - beta(s))^2 dF(s)
$$
subject to a constraint
$$
int_S gamma (s - beta(s)) dF(s) leq K in mathbbR.
$$
The Lagrangian with the first order condition plugged in, and change of variable, is
$$
mathcalL=int_0^1 [gamma'^-1 (g(x))]^2 + lambda gamma(gamma'^-1(g(x))) dx
$$
I am not looking for an analytical solution, but any guidance about which direction I should look into? I would like to obtain a differential equation about $gamma$ so that I can quantitatively analyze the function $gamma$.
calculus-of-variations optimal-control
asked Aug 3 at 21:19
Half Hogshead
84
Suppose $s in S subset mathbbR^+$ with c.d.f. $F(s)$. A real-valued function $beta(s) leq s$ is determined by a first order condition $gamma'(s - beta(s)) = g(F(s)) in mathbbR$, where $gamma$ is a real valued penalty function. I need to find the $gamma$ which
$$
min int_S (s - beta(s))^2 dF(s)
$$
subject to a constraint
$$
int_S gamma (s - beta(s)) dF(s) leq K in mathbbR.
$$
The Lagrangian with the first order condition plugged in, and change of variable, is
$$
mathcalL=int_0^1 [gamma'^-1 (g(x))]^2 + lambda gamma(gamma'^-1(g(x))) dx
$$
I am not looking for an analytical solution, but any guidance about which direction I should look into? I would like to obtain a differential equation about $gamma$ so that I can quantitatively analyze the function $gamma$.
calculus-of-variations optimal-control
asked Aug 3 at 21:19
Half Hogshead
84
edited Aug 3 at 22:08
asked Aug 3 at 21:19
Half Hogshead
84
asked Aug 3 at 21:19
Half Hogshead
84
asked Aug 3 at 21:19
Half Hogshead
84
84
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add a comment |Â
1 Answer
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I'll start with the final expression for your Langrangian, assuming that $g,gamma, gamma'$ are all invertible functions and well defined in the respective intervals that the compositions define.
By making the substitution $u=gamma'^-1(g(x))$ the Lagrangian takes the following form:
$mathcalL=int^gamma'^-1(g(1))_gamma'^-1(g(0))Big(fracgamma''(u)g'(g^-1(gamma'(u)))(lambdagamma(u)+u^2)Big)du$
which now has the form $mathcalL=intduF(gamma(u),gamma'(u), gamma''(u),u)$ and can be tackled using standard variational techniques. The Euler-Lagrange equation should read:
$fracd^2du^2(fracpartial Fpartialgamma'')-fracddu(fracpartial Fpartialgamma')+fracpartial Fpartial gamma=0$.
answered Aug 3 at 22:10
DinosaurEgg
1214
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I'll start with the final expression for your Langrangian, assuming that $g,gamma, gamma'$ are all invertible functions and well defined in the respective intervals that the compositions define.
By making the substitution $u=gamma'^-1(g(x))$ the Lagrangian takes the following form:
$mathcalL=int^gamma'^-1(g(1))_gamma'^-1(g(0))Big(fracgamma''(u)g'(g^-1(gamma'(u)))(lambdagamma(u)+u^2)Big)du$
which now has the form $mathcalL=intduF(gamma(u),gamma'(u), gamma''(u),u)$ and can be tackled using standard variational techniques. The Euler-Lagrange equation should read:
$fracd^2du^2(fracpartial Fpartialgamma'')-fracddu(fracpartial Fpartialgamma')+fracpartial Fpartial gamma=0$.
answered Aug 3 at 22:10
DinosaurEgg
1214
add a comment |Â
up vote
0
down vote
accepted
I'll start with the final expression for your Langrangian, assuming that $g,gamma, gamma'$ are all invertible functions and well defined in the respective intervals that the compositions define.
By making the substitution $u=gamma'^-1(g(x))$ the Lagrangian takes the following form:
$mathcalL=int^gamma'^-1(g(1))_gamma'^-1(g(0))Big(fracgamma''(u)g'(g^-1(gamma'(u)))(lambdagamma(u)+u^2)Big)du$
which now has the form $mathcalL=intduF(gamma(u),gamma'(u), gamma''(u),u)$ and can be tackled using standard variational techniques. The Euler-Lagrange equation should read:
$fracd^2du^2(fracpartial Fpartialgamma'')-fracddu(fracpartial Fpartialgamma')+fracpartial Fpartial gamma=0$.
answered Aug 3 at 22:10
DinosaurEgg
1214
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I'll start with the final expression for your Langrangian, assuming that $g,gamma, gamma'$ are all invertible functions and well defined in the respective intervals that the compositions define.
By making the substitution $u=gamma'^-1(g(x))$ the Lagrangian takes the following form:
$mathcalL=int^gamma'^-1(g(1))_gamma'^-1(g(0))Big(fracgamma''(u)g'(g^-1(gamma'(u)))(lambdagamma(u)+u^2)Big)du$
which now has the form $mathcalL=intduF(gamma(u),gamma'(u), gamma''(u),u)$ and can be tackled using standard variational techniques. The Euler-Lagrange equation should read:
$fracd^2du^2(fracpartial Fpartialgamma'')-fracddu(fracpartial Fpartialgamma')+fracpartial Fpartial gamma=0$.
answered Aug 3 at 22:10
DinosaurEgg
1214
I'll start with the final expression for your Langrangian, assuming that $g,gamma, gamma'$ are all invertible functions and well defined in the respective intervals that the compositions define.
By making the substitution $u=gamma'^-1(g(x))$ the Lagrangian takes the following form:
$mathcalL=int^gamma'^-1(g(1))_gamma'^-1(g(0))Big(fracgamma''(u)g'(g^-1(gamma'(u)))(lambdagamma(u)+u^2)Big)du$
which now has the form $mathcalL=intduF(gamma(u),gamma'(u), gamma''(u),u)$ and can be tackled using standard variational techniques. The Euler-Lagrange equation should read:
$fracd^2du^2(fracpartial Fpartialgamma'')-fracddu(fracpartial Fpartialgamma')+fracpartial Fpartial gamma=0$.
answered Aug 3 at 22:10
DinosaurEgg
1214
answered Aug 3 at 22:10
DinosaurEgg
1214
answered Aug 3 at 22:10
DinosaurEgg
1214
answered Aug 3 at 22:10
DinosaurEgg
1214
1214
add a comment |Â
add a comment |Â
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