Principle of minimum action and application to real problems

[cianfa72]

TL;DR Summary

How the principle of minimum action help us to find out the actual path taken from a body

I've a doubt regarding the application of the principle of minimum action to real cases.

Pick an inertial frame with a potential $V$ defined on it. The principle (aka Hamilton's principle) claims that the actual path taken from a body gives rise to a "stationary" action when calculated from a starting location A to the final location B (such a path might be a local minimum, maximum or a saddle for the action's integral). So far so good.

Now the point is that actually we know only the starting position and velocity's vector of the body. Therefore, how such principle help us to find out the actual path taken from the body ? Thanks.

 

[renormalize]

Now the point is that actually we know only the starting position and velocity's vector of the body. Therefore, how such principle help us to find out the actual path taken from the body ? Thanks.

Hamilton's principle of stationary action yields only the Euler-Lagrange differential equations (equations-of-motion) that an arbitrary stationary-path must satisfy. To solve for a specific stationary-path, these equations-of-motion must be supplemented by freely-specifiable initial conditions.

 

[Dale]

the actual path taken from a body gives rise to a "stationary" action when calculated from a starting location A to the final location B (such a path might be a local minimum, maximum or a saddle for the action's integral). So far so good.

Now the point is that actually we know only the starting position and velocity's vector of the body. Therefore, how such principle help us to find out the actual path taken from the body ?

They are equivalent. Sometimes to distinguish you call them boundary conditions instead of initial conditions. But either way it is the same amount of information required to pick a specific solution out of the same family of equations.

 

[cianfa72]

They are equivalent. Sometimes to distinguish you call them boundary conditions instead of initial conditions. But either way it is the same amount of information required to pick a specific solution out of the same family of equations.

Basically you are saying that the Euler-Lagrange (or Hamilton) Ordinary Differential Equation (ODE) Dirichlet problem which consist in assigning initial and final spatial locations (boundary conditions) at given times $t_0$ and $t$, actually gives rise to an unique solution. Therefore, given the problem's specific ODE, there is a bijective map between the initial value problem (IVP) solution set and the boundary value problem's solution set.

 

[martinbn]

Basically you are saying that the Euler-Lagrange (or Hamilton) Ordinary Differential Equation (ODE) Dirichlet problem which consist in assigning initial and final spatial locations (boundary conditions) at given times $t_0$ and $t$, actually gives rise to an unique solution. Therefore, given the problem's specific ODE, there is a bijective map between the initial value problem (IVP) solution set and the boundary value problem's solution set.

Think about it this way. If you have a second order ODE the general solution will have two constants in it. To find the particular solution you need two pieces of data, for example initial position and velocity or initial position and final position.

 

[cianfa72]

Think about it this way. If you have a second order ODE the general solution will have two constants in it. To find the particular solution you need two pieces of data, for example initial position and velocity or initial position and final position.

Ok yes. From a theoretical point of view by Picard–Lindelöf theorem our second order ODE has got an IVP's unique solution. I believe that, under the same theorem's conditions, such a result also extends to boundary value problem in which initial and final positions are given rather than initial position and velocity.

 

[Dale]

Basically you are saying that the Euler-Lagrange (or Hamilton) Ordinary Differential Equation (ODE) Dirichlet problem which consist in assigning initial and final spatial locations (boundary conditions) at given times $t_0$ and $t$, actually gives rise to an unique solution. Therefore, given the problem's specific ODE, there is a bijective map between the initial value problem (IVP) solution set and the boundary value problem's solution set.

Basically, yes, but it isn’t an bijection.

Suppose you have a differential equation whose solutions are the set of all geodesics on a sphere. There is not a one to one map between specific elements of the set of solutions and initial value pairs. Any point on the geodesic can serve as an initial value and there are two initial directions that select the same geodesic. Similarly, any pair of non-antipodal points on the geodesic can serve as the boundary values to select the same geodesic.

So the mapping between initial value pairs and solutions is many to one. And the mapping between boundary value pairs and solutions is also many to one. Which means that the mapping between initial value pairs and boundary value pairs is many to many.

 

[cianfa72]

Basically, yes, but it isn’t an bijection.

Maybe I was unclear. I meant the set of solution paths of a second order ODE IVP problem between the starting time $t_0$ and the ending $t$ vs. the set of solution paths of the ODE boundary value problem between the same start and end times.

Searching about it, I believe the topic I'm facing to is known as two point ODE boundary value problem.

Suppose you have a differential equation whose solutions are the set of all geodesics on a sphere. There is not a one to one map between specific elements of the set of solutions and initial value pairs. Any point on the geodesic can serve as an initial value and there are two initial directions that select the same geodesic. Similarly, any pair of non-antipodal points on the geodesic can serve as the boundary values to select the same geodesic.

So the mapping between initial value pairs and solutions is many to one. And the mapping between boundary value pairs and solutions is also many to one. Which means that the mapping between initial value pairs and boundary value pairs is many to many.

Even in this case (the geodesic equation is actually a second order ODE) we can see that the two points ODE boundary value problem between the given starting and ending times ($t_0$ and $t$ respectively) may not have an unique solution (think of the two geodesic paths connecting antipodal points on the sphere between given start and end times).

Of course whether the given "two point ODE boundary value problem between $t_0$ and $t$" has an unique solution than it is the same (unique) solution of the "corresponding" IVP problem.

 

[cianfa72]

Think about it this way. If you have a second order ODE the general solution will have two constants in it. To find the particular solution you need two pieces of data, for example initial position and velocity or initial position and final position.

Thinking again about it, I believe is not exactly the same. Take for example the solutions of second order geodesic ODE between two antipodal points on the 2-sphere. The IVP problem between given $t_0$ and $t$ has always a unique solution while the two point boundary value problem has not.

 

[Dale]

Even in this case (the geodesic equation is actually a second order ODE) we can see that the two points ODE boundary value problem between the given starting and ending times (t0 and t respectively) may not have an unique solution (think of the two geodesic paths connecting antipodal points on the sphere between given start and end times).

Yes. That is precisely why I used this specific example. It is simple and clear and contains all of the usual pitfalls.