4
$\begingroup$

I have a function $u \in L^2(0,T;H^{1}_{0} (U))$ with $u' \in L^2(0,T;L^2 (U))$ and $u''\in L^{2}(0,T;H^{-1} (U))$.

My professor wrote :

$$ u' \in W^{1,2}(0,T; H^{-1}(U)).$$

I don't know how to conclude this affirmation.

I was wondering about and if I define the function

$F\colon L^2(0,T;L^2 (U)) \rightarrow L^2(0,T;H^{-1}(U))$ given by $F(u) = \tilde{u}$, where $\langle \tilde{u}(t), f\rangle_{H^{-1}(U)H^{1}_{0}(U)} = \int_{0}^{T} f(x) u(t) \ dx$ . I can show that $F$ is linear and $\lVert F u \rVert\leq\lVert u \rVert$. I don't know if this help . Maybe this function help me how to see a element of $L^2(0,T;L^2 (U))$ in $L^2(0,T;H^{-1}(U))$. Someone can help me ?

Thanks in advance.

$\endgroup$
3
$\begingroup$

Your argument is right @Leandro. You have proved that every element $v\in L^2(0,T:L^2(U))$, is a element of $L^2(0,T:H^{-1}(U))$. By using the definition of $W^{1,2}$, you can conclude that $$u'\in W^{1,2}(0,T;H^{-1}(U))$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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