When Will the Particle Reach Infinity with Given Initial Values?

[1670frank]

Homework Statement

At what time does the particle reach infinity given that H(p,x)=(1/2)p^2 -(1/2)x^4. And initial values are x(0)=1 and p(0)=1

Homework Equations

The hamiltonian equations i believe are given by the partial derivatives let d mean partial derivative so x'=dH/dp and p'=-dH/dx

The Attempt at a Solution

well so far i have that x'=p and p'=2x^3 but at this point, I am confused over how to impose the initial values... This is my attempt. By separation of variables, dx/dt=p so (1/p)x=t+ c by integration and now i impose initial value of x(0)=1which gives me c=1/p so i get t=(1/p)x-(1/p) and for dp/dt i got t=(1/2x^3)p - (1/2x^3) . At this point I am very confused what to do next or if I am doing this right in the first place... How can i tell when the particle reaches infinity with two times??

 

[Ray Vickson]

Homework Statement

At what time does the particle reach infinity given that H(p,x)=(1/2)x^2 -(1/2)x^4. And initial values are x(0)=1 and p(0)=1

Homework Equations

The hamiltonian equations i believe are given by the partial derivatives let d mean partial derivative so x'=dH/dp and p'=-dH/dx

The Attempt at a Solution

well so far i have that x'=p and p'=2x^2 but at this point, I am confused over how to impose the initial values... This is my attempt. By separation of variables, dx/dt=p so (1/p)x=t+ c by integration and now i impose initial value of x(0)=1which gives me c=1/p so i get t=(1/p)x-(1/p) and for dp/dt i got t=(1/2x^3)p - (1/2x^3) . At this point I am very confused what to do next or if I am doing this right in the first place... How can i tell when the particle reaches infinity with two times??

You wrote H(p,x)=(1/2)x^2 -(1/2)x^4. Did you mean H(p,x) = (1/2)p^2 - (1/2)x^4?

 

[1670frank]

Yes thanks for that :)

 

[haruspex]

i have that x'=p and p'=2x^2

p' = 2x³?

but at this point, I am confused over how to impose the initial values... This is my attempt. By separation of variables, dx/dt=p so (1/p)x=t+ c

Doesn't look valid to me. Try looking at x''.

 

[Ray Vickson]

Homework Statement

At what time does the particle reach infinity given that H(p,x)=(1/2)x^2 -(1/2)x^4. And initial values are x(0)=1 and p(0)=1

Homework Equations

The hamiltonian equations i believe are given by the partial derivatives let d mean partial derivative so x'=dH/dp and p'=-dH/dx

The Attempt at a Solution

well so far i have that x'=p and p'=2x^2 but at this point, I am confused over how to impose the initial values... This is my attempt. By separation of variables, dx/dt=p so (1/p)x=t+ c by integration and now i impose initial value of x(0)=1which gives me c=1/p so i get t=(1/p)x-(1/p) and for dp/dt i got t=(1/2x^3)p - (1/2x^3) . At this point I am very confused what to do next or if I am doing this right in the first place... How can i tell when the particle reaches infinity with two times??

If I were dong this question I would avoid the dynamical equations and use instead conservation of energy (i.e., constant H).

 

[1670frank]

Im just confused on how to use two initial conditions . Ray Vickson , what do you mean by the conservation of energy method?

 

[Ray Vickson]

Im just confused on how to use two initial conditions . Ray Vickson , what do you mean by the conservation of energy method?

Because of the form of the Hamiltonian, it is constant over time; that is, ${\cal{H}}(t) \equiv H(p(t),x(t))$ is constant. That means that for any t we have
p^2(t) - x^4(t) = p^2(0) - x^4(0) = 0. Since the mass is 1 (from the form of H) we have
p(t) = \frac{d x(t)}{dt},
so we get immediately a first-order DE for $x(t)$.