# Newton's laws and Uniqueness of Motion

1. Dec 20, 2011

### PhDorBust

How would you show mathematically that Newton's laws, when taken as given, always yield a motion and that this motion is always unique (given initial positions/velocities) for arbitrary systems?

Last edited: Dec 20, 2011
2. Dec 20, 2011

### Staff: Mentor

I don't understand the question.

3. Dec 20, 2011

### PhDorBust

Given an arbitrary system, show that when Newton's equations of motion are written out for the system as a whole and for the different subsystems, they will always have a unique solution.

Is that any better? Basically asking to show that when all the forces acting are known, that Newton's laws predict the motion.

4. Dec 20, 2011

### D H

Staff Emeritus
You can't. Imagine a point mass atop a frictionless inverted bowl. The bowl is continuous and everywhere differentiable. The gradient is downward except at the peak, where it is zero. The gradient in turn is everywhere differentiable except at the peak, where it has a discontinuity.

If you start the point mass at rest at the top of the bowl, one solution is that the point mass will just stay there forever. There are however an infinite number of other solutions. The point mass can stay at rest atop the bowl for an arbitrary amount of time and then start sliding down the bowl in any arbitrary direction.

To see that this is the case, instead of starting the point mass at rest at the top of the bowl, imagine starting it at the bottom. With just the right initial conditions, the point mass will come to rest at the peak and do so in finite time. (That discontinuity in the gradient is what allows the point mass to reach the unstable equilibrium point in finite time.) The only forces are gravity and the normal force, both of which are conservative. The system is time-reversible. Time reversal of this start-at-the-bottom problem yields a solution to the problem of the point mass starting at rest atop the bowl.

5. Dec 20, 2011

### micromass

Isn't this basically showing that a differential equation has a unique solution??

6. Dec 20, 2011

### D H

Staff Emeritus
Exactly. The concept of Lipschitz continuity is important here. Norton's dome is not Lipschitz continuous.

7. Dec 20, 2011

### olivermsun

Why will the point mass just randomly start moving?

8. Dec 20, 2011

### D H

Staff Emeritus
Because its a solution to the ODE. The equations of motion don't have a unique solution.

The problem is that the discontinuity is in the second derivative. This problem wouldn't exist if the inverted bowl was replaced by an inverted cone. In the case of an inverted cone, a point mass can be set into motion so that it will come to rest right at the peak of the cone, but this will only happen as t→∞.

9. Dec 20, 2011

### olivermsun

I must be confused about the geometry you are describing. Is this a bowl like the upper half of a sphere

Can you write the ODE?

Edit: okay, I see you were referring to a "bowl" with a singularity at the top. Cute.

Last edited: Dec 20, 2011
10. Dec 20, 2011

### cmb

Is the chaotic motion of a compound pendulum also an example of a dynamic system that is 'not Lipschitz continuous'? I'm thinking when it reaches the states at which the second pendulum is poised to go either way, you're actually looking at a series of potentially 'bifurcating' sequences underlying the behaviour. (Though, this example animated gif shows only one sequence of motions!)

(PS: Here's one you can play with; http://www.myphysicslab.com/dbl_pendulum.html )

Last edited: Dec 20, 2011
11. Dec 20, 2011

### D H

Staff Emeritus
$$\frac {d^2 r(t)}{dt^2} = \sqrt r$$

Given initial conditions r(0)=0, r'(0)=0, one solution is the trivial solution r(t)=0. It also has non-trivial solutions

$$r(t) = \begin{cases} 0 & t<t_0 \\ \frac{(t-t_0)^4}{144} & t\ge t_0 \end{cases}$$

This "bowl" is called Norton's dome.

Nope. If you knew the initial conditions to infinite precision you could predict the state at any point in the future.

12. Dec 21, 2011

### cmb

I'm not sure I understand any difference (excepting degrees of freedom) between a ball perched incipiently atop a spherical shell, to that of an inverted pendulum.

13. Dec 21, 2011

### olivermsun

At first glance, I'm not sure I agree that the non-trivial solutions satisfy Newton's First Law. Although it is an interesting point -- I'd never really given much thought to why the First Law might have been stated separately from the Second. But here might be one case where it might potentially see some use.

14. Dec 21, 2011

### espen180

Norton's dome is not spherical.

The difference is that in the second case, the future evolution of the system is uniquely determined, while in the former it is not.

A quick google search gave me this article, which may be interesting.

15. Dec 21, 2011

### D H

Staff Emeritus
Interesting take! It does indeed seem that Newton's first law rules out these non-trivial solutions. There is more to Newton's first than meets the eye.

Another way to look at it is that the indeterminate solutions to a classical mechanics problem represent a space of measure zero. While such solutions might exist, does it really matter? These solutions inevitably require perfect knowledge of position and momentum. (And we all know what quantum mechanics has to say about that.)