Let $Q=(0,T)\times (0,1)$, and $u\in L^2(Q)$. We can define two notions of distribuitional derivative.
- The usual one: $D_tu,D_xu\in \mathcal{D}'(Q)$ defined by
$\langle\langle D_tu,\phi\rangle\rangle:=-((u,\phi_t)):=\iint_Qu(t,x)\phi_t(t,x)dxdt$ and $\langle\langle D_xu,\phi\rangle\rangle:=-((u,\phi_x))$ for all $\phi\in \mathcal{D}(Q)$.
- The vector one: Considering $u\in L^2(0,T;L^2(0,1))$ we can define $u_t\in \mathcal{D}(0,T;L^2(0,1)) $ as
$\langle u_t,\varphi\rangle:= -(u(\cdot,x),\varphi')_{L^2(0,T)}=\int_0^Tu(t,x)\varphi'(t)dt\in L^2(0,1)$ for all $\varphi\in \mathcal{D}(0,T).$
And, we can also define $u_x:[0,T]\to H^{-1}(0,1)$ as
$\langle u_x(t),\psi\rangle:=-(u(t,\cdot),\psi')_{L^2(0,1)}=\int_0^1 u(t,x)\psi'(x)dx$ for all $\psi \in \mathcal{D}(0,1).$
I would like to know if there is a context where those notions are the same, that is, $D_tu=u_t$ and $D_xu=u_x$. If possible, I would to like to have a good reference.
I have found the answer. That two notions coincide because the subspace of $C^\infty_c(Q)$ generated by the product of functions $\varphi\psi$, where $\varphi\in \mathcal{D}(0,T)$ and $\psi\in \mathcal{D}(0,1)$ is dense in $\mathcal{D}(Q)$. So, it easy to see that the two notions coincide when $\phi(t,x)=\varphi(t)\phi(x)$. By density we can define $u_t$ an $u_x$ as distribution in $\mathcal{D}(Q)$.
That result about density can be found in Theorem 4.3.1 of
.