Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$\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 $x(0)=0$ .

share|improve this question
add comment

1 Answer 1

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 $|x(t)| \leq 0\cdot \exp t = 0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.