a question related to calculus of variations
$begingroup$
Consider a particle with coordinates $(x(t),y(t))$ on a smooth curve $phi(x,y)=0$. If the particle moves from $(x(0),y(0))$ to $(x(tau),y(tau))$ for $tau >0$ such that its kinetic energy is minimized, then
$(a)$ $frac{ddot{x}}{phi_x}=frac{ddot{y}}{phi_y}$.
$(b)$ $dot{x}^2(0)+dot{y}^2(0)=dot{x}^2(tau)+dot{y}^2(tau)$.
$(c)$ $dot{x}phi_x+dot{y}phi_y=0$.
$(d)$ $dot{x}^2(0)=dot{x}^2(tau)$.
Now, if we consider this problem as minimizing a functional $J[x(t),y(t)]=int_0^tau F(x,dot{x},y,dot{y},t)dt$ in two dependent variables representing coordinates and one independent variable representing time, then Euler-Lagrange equation will give a family of extremals from which we can conclude the answer. But I am unable to find a way to relate the K.E. as a functional as written above, and include the curve $phi(x,y)=0$ in the same. So is there another method for this problem or am I on the right track? Any help will be appreciated.
calculus-of-variations
asked Dec 13 '18 at 18:11
am_11235...am_11235...
1747
$endgroup$
add a comment |
$begingroup$
Consider a particle with coordinates $(x(t),y(t))$ on a smooth curve $phi(x,y)=0$. If the particle moves from $(x(0),y(0))$ to $(x(tau),y(tau))$ for $tau >0$ such that its kinetic energy is minimized, then
$(a)$ $frac{ddot{x}}{phi_x}=frac{ddot{y}}{phi_y}$.
$(b)$ $dot{x}^2(0)+dot{y}^2(0)=dot{x}^2(tau)+dot{y}^2(tau)$.
$(c)$ $dot{x}phi_x+dot{y}phi_y=0$.
$(d)$ $dot{x}^2(0)=dot{x}^2(tau)$.
Now, if we consider this problem as minimizing a functional $J[x(t),y(t)]=int_0^tau F(x,dot{x},y,dot{y},t)dt$ in two dependent variables representing coordinates and one independent variable representing time, then Euler-Lagrange equation will give a family of extremals from which we can conclude the answer. But I am unable to find a way to relate the K.E. as a functional as written above, and include the curve $phi(x,y)=0$ in the same. So is there another method for this problem or am I on the right track? Any help will be appreciated.
calculus-of-variations
asked Dec 13 '18 at 18:11
am_11235...am_11235...
1747
$endgroup$
add a comment |
$begingroup$
Consider a particle with coordinates $(x(t),y(t))$ on a smooth curve $phi(x,y)=0$. If the particle moves from $(x(0),y(0))$ to $(x(tau),y(tau))$ for $tau >0$ such that its kinetic energy is minimized, then
$(a)$ $frac{ddot{x}}{phi_x}=frac{ddot{y}}{phi_y}$.
$(b)$ $dot{x}^2(0)+dot{y}^2(0)=dot{x}^2(tau)+dot{y}^2(tau)$.
$(c)$ $dot{x}phi_x+dot{y}phi_y=0$.
$(d)$ $dot{x}^2(0)=dot{x}^2(tau)$.
Now, if we consider this problem as minimizing a functional $J[x(t),y(t)]=int_0^tau F(x,dot{x},y,dot{y},t)dt$ in two dependent variables representing coordinates and one independent variable representing time, then Euler-Lagrange equation will give a family of extremals from which we can conclude the answer. But I am unable to find a way to relate the K.E. as a functional as written above, and include the curve $phi(x,y)=0$ in the same. So is there another method for this problem or am I on the right track? Any help will be appreciated.
calculus-of-variations
asked Dec 13 '18 at 18:11
am_11235...am_11235...
1747
$endgroup$
Consider a particle with coordinates $(x(t),y(t))$ on a smooth curve $phi(x,y)=0$. If the particle moves from $(x(0),y(0))$ to $(x(tau),y(tau))$ for $tau >0$ such that its kinetic energy is minimized, then
$(a)$ $frac{ddot{x}}{phi_x}=frac{ddot{y}}{phi_y}$.
$(b)$ $dot{x}^2(0)+dot{y}^2(0)=dot{x}^2(tau)+dot{y}^2(tau)$.
$(c)$ $dot{x}phi_x+dot{y}phi_y=0$.
$(d)$ $dot{x}^2(0)=dot{x}^2(tau)$.
Now, if we consider this problem as minimizing a functional $J[x(t),y(t)]=int_0^tau F(x,dot{x},y,dot{y},t)dt$ in two dependent variables representing coordinates and one independent variable representing time, then Euler-Lagrange equation will give a family of extremals from which we can conclude the answer. But I am unable to find a way to relate the K.E. as a functional as written above, and include the curve $phi(x,y)=0$ in the same. So is there another method for this problem or am I on the right track? Any help will be appreciated.
calculus-of-variations
calculus-of-variations
asked Dec 13 '18 at 18:11
am_11235...am_11235...
1747
asked Dec 13 '18 at 18:11
am_11235...am_11235...
1747
asked Dec 13 '18 at 18:11
am_11235...am_11235...
1747
asked Dec 13 '18 at 18:11
am_11235...am_11235...
1747
asked Dec 13 '18 at 18:11
am_11235...am_11235...
1747
1747
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint.
Defining the Lagrangian with $X = (x(t),y(t))$
$$
L(X,dot X,lambda) = frac m2left(dot x(t)^2+dot y(t)^2right)+lambda phi(x(t),y(t))
$$
we have the Euler-Lagrange movement equations
$$
L_X-frac{d}{dt}left(L_{X'}right) = left{begin{array}{rcl}m ddot x(t)-lambdaphi_x(x(t),y(t)) & = & 0\ m ddot y(t)-lambdaphi_y(x(t),y(t)) & = & 0end{array}right.
$$
answered Dec 13 '18 at 18:43
CesareoCesareo
8,7193516
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1 Answer
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1 Answer
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$begingroup$
Hint.
Defining the Lagrangian with $X = (x(t),y(t))$
$$
L(X,dot X,lambda) = frac m2left(dot x(t)^2+dot y(t)^2right)+lambda phi(x(t),y(t))
$$
we have the Euler-Lagrange movement equations
$$
L_X-frac{d}{dt}left(L_{X'}right) = left{begin{array}{rcl}m ddot x(t)-lambdaphi_x(x(t),y(t)) & = & 0\ m ddot y(t)-lambdaphi_y(x(t),y(t)) & = & 0end{array}right.
$$
answered Dec 13 '18 at 18:43
CesareoCesareo
8,7193516
$endgroup$
add a comment |
$begingroup$
Hint.
Defining the Lagrangian with $X = (x(t),y(t))$
$$
L(X,dot X,lambda) = frac m2left(dot x(t)^2+dot y(t)^2right)+lambda phi(x(t),y(t))
$$
we have the Euler-Lagrange movement equations
$$
L_X-frac{d}{dt}left(L_{X'}right) = left{begin{array}{rcl}m ddot x(t)-lambdaphi_x(x(t),y(t)) & = & 0\ m ddot y(t)-lambdaphi_y(x(t),y(t)) & = & 0end{array}right.
$$
answered Dec 13 '18 at 18:43
CesareoCesareo
8,7193516
$endgroup$
add a comment |
$begingroup$
Hint.
Defining the Lagrangian with $X = (x(t),y(t))$
$$
L(X,dot X,lambda) = frac m2left(dot x(t)^2+dot y(t)^2right)+lambda phi(x(t),y(t))
$$
we have the Euler-Lagrange movement equations
$$
L_X-frac{d}{dt}left(L_{X'}right) = left{begin{array}{rcl}m ddot x(t)-lambdaphi_x(x(t),y(t)) & = & 0\ m ddot y(t)-lambdaphi_y(x(t),y(t)) & = & 0end{array}right.
$$
answered Dec 13 '18 at 18:43
CesareoCesareo
8,7193516
$endgroup$
Hint.
Defining the Lagrangian with $X = (x(t),y(t))$
$$
L(X,dot X,lambda) = frac m2left(dot x(t)^2+dot y(t)^2right)+lambda phi(x(t),y(t))
$$
we have the Euler-Lagrange movement equations
$$
L_X-frac{d}{dt}left(L_{X'}right) = left{begin{array}{rcl}m ddot x(t)-lambdaphi_x(x(t),y(t)) & = & 0\ m ddot y(t)-lambdaphi_y(x(t),y(t)) & = & 0end{array}right.
$$
answered Dec 13 '18 at 18:43
CesareoCesareo
8,7193516
edited Dec 13 '18 at 22:08
answered Dec 13 '18 at 18:43
CesareoCesareo
8,7193516
answered Dec 13 '18 at 18:43
CesareoCesareo
8,7193516
answered Dec 13 '18 at 18:43
CesareoCesareo
8,7193516
8,7193516
add a comment |
add a comment |
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