# why the solution to Hamilton's equations are unique

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Prove that ,in time independent case,the solution to Hamilton's equation(time independent) are unique for a given Hamiltonian;

that is ,if two trajectories in the gamma space pass through F(t+delta t) corresponding to the state of the system at time t+ delta t , then they both must have originated from the same F(t)

Question taken from "An introduction to the theory of the boltzmann equation"

Problems 1-4

I have been stuck on this for a long time

Tips I have:"The idea is to prove that no two trajectories can cross each other.
And the proof can be rather simple.
"

edited Aug 7, 2014

What did you try?

What is known about the Hamiltonian? (Without assumption the statement is not correct.)

the equation and the Hamiltonian of the system is time independent

Thank you for all of you. Prof.Moretti 's answer points me to a good theorem to answer my problem, which is really a simplified answer of  Ron Maimon 's

But I did not expect the answer is so mathematical. It is a problem from Statistical Physics,are there any more intuitive answer without too much math rigor ? the application of this problem is in fluid dynamics finally.

The intuitive answer is just Euler timestepping, or any other numerical integration algorithm for differential equations. The only issue is convergence to a unique continuum limit as the timestep is made small.

A differential equation may be discretized and solved, and the only obstacle to uniqueness happens when two solutions become coincident in the continuum limit, which only happens when the right hand side of the equation is too quickly varying. The essence of Euler (obvious) timestepping is better contained in the integral-equation iteration proof, because this only relies on Lebesgue integration, and can be generalized to all sorts of situations involving generalized functions and generalized pseudosolutions which obey the integral equation but only formally obey the differential equations.

The proof CANNOT be rather simple, as the statement is FALSE. The person who made up this problem is just not particularly competent.

I completely agree...

I finally figured out what the "easy proof" the person who posed this problem thought was supposed to be: it's a false argument in one dimension that goes like this: by Liouville's theorem, phase space area is preserved, so that if two trajectories collide in x, the p area of some rectangle must have become infinite. This argument is completely wrong, as shown by the explicit example, at best all it establishes is that these type of collisions are measure zero.

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You're stuck because it is false without further assumptions. For instance, consider $x(t) \propto t^4$, then $\ddot{x} \propto t^2 =\sqrt{x}$, so this is one solution to Hamilton's equations with

$$H = p^2 + x^{3 \over 2}$$

the equation of motion is $$\ddot{x} \propto \sqrt{x}$$ which this obeys, with x(0)=0 and v(0)=0 at t=0. The other solution is the trivial solution $x(t)=0$.

This is not in the spirit of the equations, it is just mathematical nonsense, due to the nondifferentiability of this particular H as a function of x. To get uniqueness results you just need H to not vary too much, and everywhere bounded differentiability is more than enough--- you can prove it just with a Lipschitz condition on $\nabla H$.

Then one standard existence/uniqueness proofs for ordinary differential equations $\dot x(t)=F(x(t),t)$ produce the solution by iterating the integral equation:

$$x_{n+1}(t) = \int_0^t F(x_n(t'),t') dt'$$

Considering this operation as a function on the space of all paths, it is a contraction mapping with a unique fixed point, at least when you choose t in a small enough interval and assuming F is differentiable (or Lipschitz), so that ||F(x)-F(y)||< C||x-y|| for a given constant C and a given norm. This method of proof is presented in any differential equations textbook, and it is a little bit easier than proving some integration algorithm converges in the continuum limit.

answered Aug 6, 2014 by (7,720 points)
edited Aug 6, 2014

Just to make clearer your (nice and correct) answer. The Lipshitz contintion on $H$ itself is not enough, it should be imposed on $\nabla H$. Indeed the Hamiltonian you wrote is Lipshitz, but  $\nabla_x H$ is not.

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There is a fundamental theorem in ODE stating that if $\vec{F} : I \times \Omega \to \Omega$, with $\Omega \subset \mathbb R^n$ is an open set and $I \subset \mathbb R$ is an interval, the system of differential equations written in normal form (i.e. the derivative of highest order separately appears in the left-hand side) :

$$\frac{d\vec{x}}{dt} = \vec{F}(t, \vec{x}(t))$$

admits a unique (maximal) solution with initial condition $\vec{x}(t_0)= \vec{x}_0$ and $(t_0, \vec{x}_0)\in I \times \Omega$ when $F$ is $(t, \vec{x})$-continuous and locally Lipschitz in the variable $\vec{x}$. The kernel of the proof of that theorem (proving the existence of a unique local solution around the initial condition) relies upon the famous Banach's fixed point theorem.

Hamilton equations already are written in normal form (differently from Euler-Lagrange ones which, however, can be re-written in that form).  Now $n=2m$, $\vec{x}= (q^1,\ldots, q^m, p_1,\ldots, p_m)$ and $\vec{F} = (\nabla_{\vec p} H, \nabla_{\vec {q}}H )$.

If the Hamiltonian function is sufficiently regular (jointly continuous and $C^2$ in $(q,p)$ is enough to fulfil the mentioned requirements), the mentioned theorem does the work.  It is worth stressing that for the existence of solutions, continuity of $\vec{F}$ is completely sufficient (there is a proof of existence due to Peano which is not based on Banach's theorem), but there are several counterexamples proving that it is by no sufficient for its uniqueness.

answered Aug 6, 2014 by (2,085 points)
edited Aug 6, 2014

Ha, this is nice !

I am wary of saying it depends on the Banach fixed point theorem, because unlike the contraction mapping thing, which is completely algorithmic and constructive, the general version of the Banach theorem is an axiom of choice monstrosity.

Indeed, by Banach theorem I meant the simplest algorithmic version of that class of theorems, the constructive one referred to contraction maps...

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An elementary direct strategy  to prove the result is the following. Let us make things simpler, at least initially, referring to the elementary equation in $\mathbb R$ (the procedure easily extends to $\mathbb R^n$ and to Hamilton equations in particular)

$$\frac{dx}{dt} = F(x(t))\:.$$

Let us assume that two solutions $x$ and $x'$  exist with the same initial condition $x(0)=x_0= x'(0)$. Finally suppose that $dF/dx$ is bounded by some positive real $L<+\infty$ (if $dF/dx$ is continuous, it is automatically true working in a bounded neighbourhood).

I go to prove that it must be $x(t)=x'(t)$ in a sufficiently small right neighbourhood of $t=0$.

The considered differential equation is equivalent to the integral one (as Ron said)

$$x(t) = x_0 + \int_0^t F(x(u)) du\:,$$

so that, we also have

$$x'(t) = x_0 + \int_0^t F(x'(u)) du\:,$$

and thus

$$x(t)-x'(t) = \int_0^t F(x(u)) -F(x'(u)) du \:.$$

Therefore, using Lagrange theorem (below, $y(u)$ is some unknown point between $x(u)$ and $x'(u)$),

$$|x(t)-x'(t)| \leq \int_0^t |F(x(u)) -F(x'(u))| du = \int_0^t \left|\frac{dF}{dx}|_{y(u)}(x'(u)-x(u)) \right|du$$

We can now use the bound $|dF/dx| \leq L$, obtaining

$$|x(t)-x'(t)| \leq L \int_0^t |x(u)-x'(u)| du \leq Lt \max_{u \in [0,t]} |x(u)-x'(u)|$$

Obviously, if $0\leq t\leq T$

$$Lt \max_{u \in [0,t]} |x(u)-x'(u)| \leq LT \max_{u \in [0,T]} |x(u)-x'(u)|$$

so that

$$|x(t)-x'(t)| \leq LT \max_{u \in [0,T]} |x(u)-x'(u)| \quad \forall t\in [0,T]$$

which eventually produces, since the bound must hold also for the value $t=t_0$ where the (continuous) function in the left-hand side above attains its maximum,

$$\max_{t \in [0,T]} |x(t)-x'(t)| \leq LT \max_{u \in [0,T]} |x(u)-x'(u)|\:,$$

i.e.

$$0 \leq \max_{t \in [0,T]} |x(t)-x'(t)| \leq LT \max_{t \in [0,T]} |x(t)-x'(t)|\:.$$

We can take $T>0$ so small that $0< LT<1$. In this case the found inequality can hold only if

$$\max_{t \in [0,T]} |x(t)-x'(t)| =0\:,$$

and thus $x(t)=x'(t)$ for $t\in [0,T]$.

We have found that, at least in a sufficiently small neighbourhood $[0,T]$ of the initial time (however the procedure can be implemented also for $t<0$), $x(t)=x'(t)$. This reasoning, gluing together several patches, establishes the uniqueness of the solution on its whole domain.

Let us pass to the true Hamilton equations. The proof is essentially identical for a $C^2$ Hamiltonian $H$ defined an open set of  $\mathbb R^{2n}$, with the following replacements.

$${\mathbb R} \to {\mathbb R}^{2n}$$

$$x \to \vec{x} = (\vec{q}, \vec{p})$$

$$F \to \vec{F}(\vec{x}) := S\nabla_{\vec{x}} H(\vec{x})$$

where $S$ is the symplectic $2n \times 2n$ matrix with $-I$, $I$ on the anti diagonal and $0$, $0$ on the diagonal when viewed as 2x2 block  matrix.

$$|\cdot | \to ||\cdot||$$

the Euclidean norm in $\mathbb R^{2n}$.

answered Aug 6, 2014 by (2,085 points)
edited Aug 7, 2014
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Intuitively, without any mathematical rigor, you can discretize time with a tiny step size $\delta t$ and replace the derivative by a finite difference quotient. Then it is easy to see that if the solutions agree at time $t+\delta t$ they also agree at time $t$ - just solve the equation in the opposite direction. The higher math is just needed to prove that the argument holds water when $\delta t\to 0$, and as the counterexample displayed by Ron Maimon shows, one indeed needs some care as the argument survives the limit only under some additional condition.

answered Aug 6, 2014 by (15,737 points)

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