Uniqueness of solution to ODE

Question

The following ODEs came up in a physics problem: $$m\ddot r-mr\dot \theta^2+F=0$$ $$r\ddot \theta+2\dot r \dot \theta = 0 $$ I have been told that if the initial conditions are: $$F=mr(0) \dot \theta(0)^2$$ $$\dot r(0)=0$$ (We are given $r(0)$ and $\dot \theta(0)$ and suppose $\theta(0)=0$.)

Then the only solution is: $$r=r(0)$$ $$\dot \theta = \dot \theta(0)$$ which corresponds to uniform circular motion, but I haven't seen why this is true.

So my question is: Why must the above solution be the only one when the conditions are met?

Answer 1

$$ r\ddot \theta+2\dot r \dot \theta = 0 \implies 2\frac{\dot r}{r} + \frac{\ddot\theta}{\dot\theta} = 0 \implies \ln(r^2\dot\theta) = C \implies r^2\dot\theta = C \tag{1} $$ By the condition $F=mr(0) \dot \theta(0)^2$ and the equation $$ \ddot r-r\dot \theta^2+\frac{F}{m}=0 \tag{2} $$ at $t= 0$ you can see that $\ddot r(0) = 0$. Substituting $(1)$ in $(2)$ to eliminate $\dot \theta$ yields $$ \ddot r -C^2\frac{1}{r^3}+\frac{F}{m}=0 \tag{3} $$ Now taking all derivatives of $(3)$ at $t = 0$, knowing that $\ddot r(0) = \dot r(0) = 0$, should show recursively that $\frac{d^n}{dt^n}r(0) = 0, n\ge 1$, so the Taylor expansion of $r(t)$, (if it has one,) will have only one nonvanishing term, namely the constant one $r(0) \ne 0$, (it is essential for this method that this term to be nonzero).

This result and the same idea can be use to find $\frac{d^n}{dt^n}\theta(0) = 0, n \ge 2$ from the equation ($1$) rewritten as $$ \dot \theta = \frac{C}{r^2} $$

Uniqueness

Set $\dot r = u$ and $\dot \theta = v$, then \begin{align} \dot r &= u \\ \dot \theta &= v \\ \dot u &= rv^2-\frac{F}{m} \\ \dot v &= -2\frac{u}{r} v \\ \end{align} The function $$ f:\mathbb R^4\backslash\{(0,\theta,u,v)\} \to \mathbb R^4, \qquad (r,\theta,u,v) \mapsto \left( u,v,rv^2-\frac{F}{m},-2\frac{u}{r} v \right) $$ is locally Lipschitz in every neighborhood with $r \ne 0$ because it is differentiable, i.e., it has continuous partial derivatives. So one can use the on such neighborhoods. So the uniqueness of the solution follows from, as the "mathcounterexamples.ne" noted, the Picard-Lindelöf Theorem.

Answer 2

You can use at least two different approaches to prove the uniqueness of the solution.

Use Picard–Lindelöf theorem by proving that the map defining the ODE is Lipschitz continuous.

Solve directly the ODE, starting by noticing that $$r\ddot \theta+2\dot r \dot \theta = 0 $$ implies that $$r^2 \dot\theta = r^2(0) \dot\theta(0)$$ is constant. And then replacing $ \dot\theta$ in the first equation.