Show that $H(v, y) = frac{v^2}{2} - F(y)$ is a first integral
$begingroup$
$H(v, y) = frac{v^2}{2} - F(y)$ is a hamiltonian of $y' = v, v' = f(y)$
Linear first integrals are of the form $I(x) = b^Tx + c$ where $b in ℝ^d $ and $ c in ℝ$
Quadratic first integrals are of the form $I(x) = x^TMx+b^Tx+c$ where $M in ℝ^{d,d}$, $b in ℝ^d $ and $ c in ℝ$
I am unsure how to get the function $H(v,y)$ in either form or somehow else show it is indeed a first integral
Could someone please help
ordinary-differential-equations pde partial-derivative runge-kutta-methods hamilton-equations
asked Dec 2 '18 at 1:19
pablo_mathscobarpablo_mathscobar
996
$endgroup$
add a comment |
$begingroup$
$H(v, y) = frac{v^2}{2} - F(y)$ is a hamiltonian of $y' = v, v' = f(y)$
Linear first integrals are of the form $I(x) = b^Tx + c$ where $b in ℝ^d $ and $ c in ℝ$
Quadratic first integrals are of the form $I(x) = x^TMx+b^Tx+c$ where $M in ℝ^{d,d}$, $b in ℝ^d $ and $ c in ℝ$
I am unsure how to get the function $H(v,y)$ in either form or somehow else show it is indeed a first integral
Could someone please help
ordinary-differential-equations pde partial-derivative runge-kutta-methods hamilton-equations
asked Dec 2 '18 at 1:19
pablo_mathscobarpablo_mathscobar
996
$endgroup$
add a comment |
$begingroup$
$H(v, y) = frac{v^2}{2} - F(y)$ is a hamiltonian of $y' = v, v' = f(y)$
Linear first integrals are of the form $I(x) = b^Tx + c$ where $b in ℝ^d $ and $ c in ℝ$
Quadratic first integrals are of the form $I(x) = x^TMx+b^Tx+c$ where $M in ℝ^{d,d}$, $b in ℝ^d $ and $ c in ℝ$
I am unsure how to get the function $H(v,y)$ in either form or somehow else show it is indeed a first integral
Could someone please help
ordinary-differential-equations pde partial-derivative runge-kutta-methods hamilton-equations
asked Dec 2 '18 at 1:19
pablo_mathscobarpablo_mathscobar
996
$endgroup$
$H(v, y) = frac{v^2}{2} - F(y)$ is a hamiltonian of $y' = v, v' = f(y)$
Linear first integrals are of the form $I(x) = b^Tx + c$ where $b in ℝ^d $ and $ c in ℝ$
Quadratic first integrals are of the form $I(x) = x^TMx+b^Tx+c$ where $M in ℝ^{d,d}$, $b in ℝ^d $ and $ c in ℝ$
I am unsure how to get the function $H(v,y)$ in either form or somehow else show it is indeed a first integral
Could someone please help
ordinary-differential-equations pde partial-derivative runge-kutta-methods hamilton-equations
ordinary-differential-equations pde partial-derivative runge-kutta-methods hamilton-equations
asked Dec 2 '18 at 1:19
pablo_mathscobarpablo_mathscobar
996
asked Dec 2 '18 at 1:19
pablo_mathscobarpablo_mathscobar
996
asked Dec 2 '18 at 1:19
pablo_mathscobarpablo_mathscobar
996
asked Dec 2 '18 at 1:19
pablo_mathscobarpablo_mathscobar
996
asked Dec 2 '18 at 1:19
pablo_mathscobarpablo_mathscobar
996
996
add a comment |
add a comment |
1 Answer
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$begingroup$
Indeed your ODE is the Hamiltonian system
$$
dot y = H_v=v\
dot v = -H_y=F'(y)
$$
For any Hamiltonian system you have along solutions
$$
frac{d}{dt}H(v(t),y(t))=H_vdot v+H_ydot y=0
$$
so that the Hamiltonian function is a first integral.
answered Dec 2 '18 at 1:29
LutzLLutzL
58.7k42055
$endgroup$
$begingroup$
cheers mate, i appreciate the help
$endgroup$
– pablo_mathscobar
Dec 2 '18 at 1:50
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Indeed your ODE is the Hamiltonian system
$$
dot y = H_v=v\
dot v = -H_y=F'(y)
$$
For any Hamiltonian system you have along solutions
$$
frac{d}{dt}H(v(t),y(t))=H_vdot v+H_ydot y=0
$$
so that the Hamiltonian function is a first integral.
answered Dec 2 '18 at 1:29
LutzLLutzL
58.7k42055
$endgroup$
$begingroup$
cheers mate, i appreciate the help
$endgroup$
– pablo_mathscobar
Dec 2 '18 at 1:50
add a comment |
$begingroup$
Indeed your ODE is the Hamiltonian system
$$
dot y = H_v=v\
dot v = -H_y=F'(y)
$$
For any Hamiltonian system you have along solutions
$$
frac{d}{dt}H(v(t),y(t))=H_vdot v+H_ydot y=0
$$
so that the Hamiltonian function is a first integral.
answered Dec 2 '18 at 1:29
LutzLLutzL
58.7k42055
$endgroup$
$begingroup$
cheers mate, i appreciate the help
$endgroup$
– pablo_mathscobar
Dec 2 '18 at 1:50
add a comment |
$begingroup$
Indeed your ODE is the Hamiltonian system
$$
dot y = H_v=v\
dot v = -H_y=F'(y)
$$
For any Hamiltonian system you have along solutions
$$
frac{d}{dt}H(v(t),y(t))=H_vdot v+H_ydot y=0
$$
so that the Hamiltonian function is a first integral.
answered Dec 2 '18 at 1:29
LutzLLutzL
58.7k42055
$endgroup$
Indeed your ODE is the Hamiltonian system
$$
dot y = H_v=v\
dot v = -H_y=F'(y)
$$
For any Hamiltonian system you have along solutions
$$
frac{d}{dt}H(v(t),y(t))=H_vdot v+H_ydot y=0
$$
so that the Hamiltonian function is a first integral.
answered Dec 2 '18 at 1:29
LutzLLutzL
58.7k42055
answered Dec 2 '18 at 1:29
LutzLLutzL
58.7k42055
answered Dec 2 '18 at 1:29
LutzLLutzL
58.7k42055
answered Dec 2 '18 at 1:29
LutzLLutzL
58.7k42055
58.7k42055
$begingroup$
cheers mate, i appreciate the help
$endgroup$
– pablo_mathscobar
Dec 2 '18 at 1:50
add a comment |
$begingroup$
cheers mate, i appreciate the help
$endgroup$
– pablo_mathscobar
Dec 2 '18 at 1:50
$begingroup$
cheers mate, i appreciate the help
$endgroup$
– pablo_mathscobar
Dec 2 '18 at 1:50
$begingroup$
cheers mate, i appreciate the help
$endgroup$
– pablo_mathscobar
Dec 2 '18 at 1:50
add a comment |
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