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$\cfrac{dx}{dt} =f(x) $ ,

where $f(x)=\begin{cases}x\sin\left(\cfrac{1}{x}\right) & x\not=0\\ 0&x=0 \end{cases}$

How to show that $x=0$ is the unique solution with $x(0)=0$ .

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up vote 4 down vote accepted

While $f(x)$ is not Lipschitz continuous, what you have is that it is Lipschitz at the origin. That is, for $x \neq 0$,

$$ |f(x) - f(0)| = | x\sin(1/x)| \leq |x| = |x - 0| $$

which means that we can repeat the uniqueness proof in Picard-Lindelof for this particular solution.

More precisely: suppose $x = x(t)$ solve the ordinary differential equation. Then in integral form

$$ x(t) - x(0) = \int_0^t f\circ x(s) \mathrm{d}s $$

which implies, using the fact that $|f(x)| \leq |x|$,

$$ |x(t) - x(0)| \leq \int_0^t |x(s)| \mathrm{d}s $$

If we satisfy the initial condition that $x(0) = 0$, the LHS is just $|x(t)|$. That is, we have

$$ |x(t)| \leq \int_0^t |x(s)| \mathrm{d}s $$

which implies by Gronwall's lemma that $|x(t)| \leq 0\cdot \exp t = 0$.

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