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.

For the solution of the ODE $dy/dt=f(t,y)$, $f$ has to be Lipschitz for all $t$. So, if $f$ is a function that is not differentiable with respect to $y$ but Lipschitz, what can I say about $f_y$? Can I estimate some norm of it? I can't say $||f_y||_C$ because $f$ is not $C^1$, however, I can say that $f_y$ is defined in a weak sense and then this weak derivative is bounded (by the Lipschitz property) and I can estimate $||f_y||_{\mathbb{L}^{\infty}}$ and have it bounded. Is that correct?

share|improve this question
    
In the ODE $dy/dt=f(t,y)$ the function $f$ is understood to be given, while $y$ is a solution we don't know yet. It seems that your question has nothing to do with the function $y(t)$, hence it's not really a question about the differential equation. Is this correct? –  user31373 Jul 11 '12 at 20:30
    
I just thought what does it mean to be Lipshitz in case of non differentiability. I understand that we don't know range for $y$ and thus can't say what the constant that bounds derivative is, but we just know it exists. –  Medan Jul 12 '12 at 1:00
add comment

1 Answer

up vote 1 down vote accepted

Yes, every Lipschitz function (on an open subset $\Omega\subset \mathbb R^n$) has a weak derivative which belongs to $L^\infty(\Omega)$. The $L^\infty$ norm of the derivative is bounded by the Lipschitz constant of the function.

Under appropriate assumptions on the geometry of $\Omega$ one can reverse this implication and obtain the global Lipschitz condition from the boundedness of the derivative.

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.