Differential equation for Hamiltonian

Question

Consider the differential equation $$\dot q=\frac{\partial H}{\partial v},\ \dot v=-\frac{\partial H}{\partial q}$$ where $H(q,v)=\frac{1}{2}v^2+\frac{1}{q^4}$. Show that for every initial value there exist $c,C>0$ such that $|q(t)|\geq c$ and $|v(t)|\leq C$ for all $t$ in the interval of existence of the solution and thus that the flow is complete (every IVP has a unique solution that is defined on the whole real axis $\mathbb R$).

First, I observed that $H$ is conserved, meaning $\frac{d}{dt}H(\varphi(t))=0$ for every solution $\varphi(t)$. Also, we see that $$\dot q=\frac{\partial H}{\partial v}=v\\\dot v=- \frac{\partial H}{\partial q}=\frac{4}{q^5}$$All I can think of right now is deriving the first one and plugging it into the second one, obtaining $$\ddot q-4q^{-5}=0$$ but I can't possibly solve this equation so there's got to be something more general to do here. My problem is that I can't make any kind of statement about how monotonuous the solution is to derive some information about the limits of $q(t)$ and $v(t)$.

Answer 1

You can at least come to a first order differential equation by multiplying with $\dot q$ and then integration: $$\dot q \ddot q = 4 q^{-5} \dot q$$ $$\frac{d}{dt}(\frac12 \dot q^2) = \frac{d}{dt}(-q^{-4})$$ $$\dot q = \pm 2 (C-q^{-4})$$

Answer 2

Since $H$ is constant (and hence bounded) on a trajectory, you know that for any initial condition that there is some $K$ such that ${1 \over 2} v^2(t) + {1 \over q^4(t)} \le K$ for all $t$.

In particular ${1 \over 2} v^2(t) \le K$ and $ {1 \over q^4(t)} \le K$ for all $t$, or $|v(t)| \le \sqrt{2K}$ and $|q(t)| \ge {1 \over \sqrt[4]{K}}$ for all $t$.

The existence & uniqueness follow from the usual theorems for ODEs. For example, Theorem 1 in Kantorovich & Akilov, "Functional Analysis", XVI.4.2, p.487. The proof is straightforward, but technically tedious.

Pick some initial condition $(q(0),v(0))$, which gives a value $K= H(q(0),v(0)) >0$. Let $D=\{(q,v)| {1 \over 2} v^2 \le 2K, {1 \over q^4} \le 2K \}$. Let $f((q,v)) = (v,{4 \over q^5})$. Let $M= \sup_{(q,v) \in D} f((q,v))$, and $L=\sup_{(q,v) \in D} \|Df((q,v)) \|$.

Let $D'=\{(q,v)| {1 \over 2} v^2 \le K, {1 \over q^4} \le K \} \subset D$ and choose $\delta>0$ such that $B_\infty((q,v)), \delta) \subset D$ for any $(q,v) \in D'$ (this is what will allow us to extend the solution indefinitely.)

Suppose $(q(t_0),v(t_0))\in D'$. Then the above theorem shows that there is a unique solution defined on the interval $[t_0, t_0+\eta]$ where $\eta < \min ({\delta \over M}, {1 \over L})$. Since $(q(t_0+\eta),v(t_0+\eta))\in D'$, we see that there is a unique solution for all $t \ge t_0$.

In particular, there is a solution defined for $t \in [0,\infty)$.

The same analysis applies to the reverse time system $\dot{(q,v)} = - f((q,v))$, hence we see the solution is defined on all of $\mathbb{R}$.

Note: Some versions of the above theorem show that there is a unique solution defined on $[t_0-\eta, t_0+\eta]$ in which case there is no need for to look at the reverse time dynamics. In fact, the proof in K&A can be easily modified to obtain this result (I presume they were concerned with $t \ge t_0$ which is why they did not include the $t<t_0$ situation).

Note: Note that $D'$ consists of two connected slabs, one with $q>0$ and one with $q<0$. In particular, if $q(0)>0$, then $\dot{v(t)}>0$ for all $t$ and so $v$ is monotonically increasing. One can continue this qualitative analysis to derive more characteristics of the solution.