# Space $H^1([0, T], H^{-1}(U))$

My teacher uses the space $H^1([0, T], H^{-1}(U))$, where $H^1 = W^{1,2}$ (Sovolev space) and $H^{-1}(U)$ is the dual space of $H_0^1(U)$.

So we have $u \in H^1([0, T], H^{-1}(U))$ if $u(t)$ is in $H^{-1}(U)$ for all $t \in [0, T]$, $u \in L^2([0, T], H^{-1}(U))$ (which means that $\displaystyle\int_0^T||u(t)||_{H^{-1}(U)}^2dt < +\infty$) and $u' \in L^2([0, T], H^{-1}(U))$.

But I actually don't understand how $u'$, the weak derivative of $u$, is defined in this context. When we work on $H^1([0,T], \mathbb{R})$, we have that $u'$ is the weak derivative of $u$ if $$\int_0^T u(t) \phi'(t) dt = \int_0^T u'(t) \phi(t) dt$$ for all $\phi \in C_0^\infty([0, T], \mathbb{R})$

Here, I would like to say something like $$\int_0^T ||u(t)||_{H^{-1}(U)} ||\phi'(t)||_{H^{-1}(U)} dt = \int_0^T ||u'(t)||_{H^{-1}(U)} ||\phi(t)||_{H^{-1}(U)} dt$$ for all $\phi \in C_0^\infty([0, T], H^{-1}(U))$ but it seems to be a total non-sense since we only use the norm of $u'$ in $H^{-1}$... $u$ would then have plenty of derivatives!

So I suppose it can't be that, but I don't know what else it should be. I also think about something like : $$\int_0^T \langle u(t), \phi'(t) \rangle dt = \int_0^T \langle u'(t), \phi(t) \rangle dt$$ for all $\phi \in C_0^\infty([0, T], H_0^1(U)$) (where $\langle , \rangle$ means the action of the first element (which is in $H^{-1}$) on the second one (which is in $H_0^1$)), but I don't see why this should be the good definition... There aren't a lot of "analogies" with the definition for $H^1([0, T], \mathbb{R})$ that I recall before...

Nicolas

-
I believe you mean $H^1 = W^{1,2}$. –  al0 Jun 5 '13 at 15:47
Oh yeah sorry, I correct it. –  Nicolas Jun 5 '13 at 16:01
I believe your first definition is essentially correct. $u'$ is defined in the usual weak sense:
if $u \in L^1(0,T;X)$, then $v \in L^1(0,T;X)$ is the weak derivative, $u'=v$, if $$\int_0^T \varphi'(t)u(t) \, dt = -\int_0^T \varphi(t)v(t) \, dt$$ $\forall \varphi \in C_0^{\infty}(0,T)$. This is the definition Evan's gives.
But what would this product between $\varphi'$ and $u$ mean in this context where $X = H^{-1}(U)$? –  Nicolas Jun 5 '13 at 16:20
$\varphi$ is just a scalar test function, so usual multiplication. –  al0 Jun 5 '13 at 16:52