Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

We say $u \in L^2(0,T;V)$ has weak derivative $u' \in L^2(0,T;V')$ if $$\int_0^T \langle u',v\rangle_{V',V} = -\int_0^T (u,v')_{H}$$ holds for all $v \in C_0^\infty(0,T; V)$. This relation can be written as $$\int_0^T \frac{d}{dt}(u,v)_H = 0 \tag{1}$$ in some sense.

I am interested in defining a weak derivative for more abstract spaces. My question is, does the form (1) make sense to take as a motivation for the definition? Does it have some physical or mathematical meaning?

Thanks. Clarification: Just to make sure, I am not asking about why we define weak derivative in the way of my first formula. I am just asking whether such a motivation that I gave after that makes mathematical sense. I think this might be a stupid question anyway.

share|improve this question
    
The order change of $d/dt$ and the integration does not make sense. –  i707107 Mar 12 '13 at 18:24
    
Thanks, I changed the question. –  Chris Mar 12 '13 at 18:39
2  
If you define the weak derivative using the first equation, then the second equation holds. If you want the second equation to hold for $v$ smooth and $u$ not, then you have no choice but to define the weak derivative using the first equation. –  Deane Yang Mar 13 '13 at 2:05

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.