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How can I check if the initial value problem $\begin{cases} u'(t)=f(u(t)), \\ u(0) = 3 \end{cases}$ where $f:\mathbb R \to \mathbb R$, $f(u):= -u^2 \cos(u)^2 $ is uniquely solvable on $[0, \infty)$?

A little help would be much appreciated

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    Is $f$ a locally Lipschitz continuous function? If so then you have uniqueness. (You may not have global existence, but you will have uniqueness where you have existence.) – 2017-01-04
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    No not necessarily – 2017-01-04
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    ...? Why do you say that? Note that I said *locally*. I realize $f$ is not globally Lipschitz. – 2017-01-04
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    see http://math.stackexchange.com/questions/963612/uniqueness-of-ivp-solution-with-a-condition-weaker-than-lipschitz – 2017-01-04

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Note that, in $[0,\pi]$, one has $$ |f'(u)|=|2u\cos^2u-u^2\sin(2u)|\le2|u|+u^2\le\pi^2+2\pi$$ and hence $f(u)$ is Lipschitz in $[0,\pi]$. Therefore by the existence and uniqueness theorem, there is a unique solution $u(t)$ to the equation. Extend $u(t)$ to $[0,t_1)$, where $[0,t_1)$ is the largest interval such that $u(t)$ exists. Clearly $t_1>\pi$. Now show $t_1=\infty$. Otherwise $t_1<\infty$ and $u(t_1)=\infty$. From the equation, one has $$ -\frac{u'}{u^2}=\cos^2u$$ and hence $$ 0\le-\frac{u'}{u^2}\le 1.$$ Integrating from $0$ to $t$ gives $$ 0\le\frac{1}{u(t)}-\frac{1}{u(0)}\le t$$ or $$ \frac13\le\frac{1}{u(t)}\le t+\frac13.$$ So $$ \frac{3}{3t+1}\le u(t)\le3$$ from which one derives $|u(t_1)|<\infty$, contradictory to the assuppumtion $u(t_1)=\infty$.

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    Well explained, thanks a lot brother! – 2017-01-06
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    Why do we need to assume that $u(t_1)=\infty$? Also, how does this show uniqueness of the solution? – 2018-11-02
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    @sequence, it means $u(t_1)$ is brew up. – 2018-11-02
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    @xpaul Why does it need to blow up? – 2018-11-02
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    @sequence, see the definition of $t_1$. If |$u(t_1)|<\infty$, then $[0,t_1)$ will not be the largest interval. – 2018-11-02
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    @xpaul For an ODE to be uniquely solvable does one need to have its solution to be defined on $\mathbb{R}$? – 2018-11-02
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    @sequence, not really. For example $y'=y^3,y(0)=1$, its solution is $y=\frac{1}{\sqrt{1-2t}}$. Here $t_1=\frac12$. – 2018-11-02
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    @xpaul Does this equation have a unique solution because its partial derivative with respect to $y$ is bounded on the domain of $y$? I'm wondering how we can apply the Existence-Uniqueness Theorem in this case because $y$ is not bounded on $\left(-\infty, -\frac12\right]$. So, as much as I understand, we can only apply the EUT only on a subset of the domain of $y(t)$. Which will tell us that there are locally unique solutions in the domain of $y$. – 2018-11-02
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Let $\phi$ and $F(r)$ be measurable functions on $(0,a)$, $\phi(t) \geq 0 $, $\int_{a}^a \phi(t) \, \mathrm{d}t < \infty$ and $F(r) >0 $. Also assume that $$\int_{0}^\delta \frac{1}{F(r)} \, \mathrm{d}r = \infty$$

With these assumptions if for small $t$ and small $|x|$ we have

$$ |f(t,x)-f(t,y)| \leq \phi(t) F(|x-y|) $$

then the following initial value problem has unique solution.

$$ x'=f(t,x), \quad x(0=0, $$

Locally Lipschitz functions satisfy the above assumptions.

The proof of this theorem is essentially same as the uniqueness in the case which function is lipschitz.