Is a function with values in $\mathcal{D}'(\Omega)$ an element of $\mathcal{D}'(\Omega\times(0,T))$?.

Question

The question is in the end of the post. First, I need some preliminary considerations.

In the Temam's book (as well as in the Dautray's book), the author proves that there exists $u\in C([0,T];H)\cap L^2(0,T;V)$ with $u'\in L^2(0,T;V')$ such that $$u_t(t)-Au(t)=f(t)\quad\text{in}\quad V',\qquad\forall\ t\in(0,T),\tag{1}$$ where $f\in L^2(0,T;V')$ is given and $A:V\to V'$ is the operator defined by $$\langle Av,w\rangle_{V',V}=(v,w)_{H_0^1(\Omega)},\quad\forall \ v,w\in V.$$ The spaces are defined by $V={\overline{\mathcal{V}}}^{H^1(\Omega)\times H^1(\Omega)}$ and $H={\overline{\mathcal{V}}}^{L^2(\Omega)\times L^2(\Omega)}$, where $\Omega$ is a bounded open subset of $\mathbb{R}^2$ with smooth boundary and $$\mathcal{V}=\{v\in\mathcal{D}(\Omega)\times\mathcal{D}(\Omega)\mid\operatorname{div} v=0\}.$$

After proving existence the author considers the functions $$U(t)=\int_0^tu(s)\;ds,\qquad F(t)=\int_0^Tf(s)\;ds.$$ From the regularity of $u$ and $f$ we have $U\in C([0,T];V)$ and $F\in C([0,T];V')$. And integrating $(1)$ we get $$u(t)-u(0)-AU(t)=F(t)\quad\text{in}\quad V',\qquad\forall \ t\in(0,T).$$ This implies $$\langle u(t)-u(0)-\Delta U(t)-F(t),v\rangle_{V',V}=0,\quad\forall\ v\in V,\;t\in(0,T).$$ Identifynig (by extension) $S(t):=u(t)-u(0)-\Delta U(t)-F(t)$ with a element of $H^{-1}(\Omega)$, it follows as consequence of the "Rham's Theorem" that, for each $t\in (0,T)$, there exists $P(t)\in\mathcal{D}'(\Omega)$ such that $$u(t)-u(0)-\Delta U(t)-F(t)=\nabla P(t)\quad\text{in}\quad\mathcal{D}'(\Omega).\tag{2}$$

After justifying that $\nabla P\in C([0,T]; [H^{-1}(\Omega)]^2)$ and $P\in C([0,T];L^2(\Omega))$, the author says that we can "differentiate $(2)$ in the $t$ variable, in the distribution sense in $Q:=\Omega\times (0,T)$", which yields $$u_t-\Delta u-f=\nabla p\quad\text{in}\quad\mathcal{D}'(Q),\tag{3}$$ where $p=\partial_t P$.

Question: It is not clear for me how to get $(3)$ from $(2)$. To differenciate $(2)$ with respect to $t$ in the sense of $\mathcal{D}'(Q)$ we need to indentify the terms of $(2)$ with elements of $\mathcal{D}'(Q)$ , right? How is this identification made?

To prove $(3)$ we have to show that, for each $\varphi\in \mathcal{D}(Q)$, $$\langle u_t-\Delta u-f,\varphi\rangle=\langle \nabla (\partial_t P),\varphi\rangle.$$ Assuming that $P\in \mathcal{D}'(Q)$, we have $$\langle \nabla (\partial_t P),\varphi\rangle =\langle \partial_t (\nabla P), \varphi\rangle \overset{(2)}{=}\langle \partial_t (u(t)-u(0)-\Delta U(t)-F(t)), \varphi\rangle,$$ but why is $P$ in $\mathcal{D}'(Q)$ and why $$\langle \partial_t (u(t)-u(0)-\Delta U(t)-F(t)), \varphi\rangle=\langle u_t(t)-\Delta u(t)-f(t), \varphi\rangle?$$

Answer 1

I will focus on the main question of whether a continuous mapping from $(0,T)$ to $\mathcal{D}'(\Omega)$ is a distribution in $\mathcal{D}'((0,T)\times\Omega)$.

In analogy to the definition $$ \mathcal(D)'(\Omega) = \mathcal{L}(\mathcal{D}(\Omega), \mathbb{C}) $$ is is possible to define vector-valued distributions, i.e., distributions with values in a locally convex space $E$ by $$ \mathcal{D}'(\Omega, E) = \mathcal{L}(\mathcal{D}(\Omega), E). $$ Then is possible to show that the space $\mathcal{C}(\Omega; E)$ of $E$-valued continuous functions can be embedded into $\mathcal{D}'(\Omega,E)$. I think what you need here is this for the case $E=\mathcal{D}'(\Omega)$ then you get in particular that $$ \mathcal{C}((0,T),\mathcal{D}'(\Omega)) \hookrightarrow \mathcal{D}'((0,T),\mathcal{D}'(\Omega)). $$ The other ingredient you need is L. Schwartz kernel theorem, which states that $$ \mathcal{D}'(\Omega_1,\mathcal{D}'(\Omega_2))= \mathcal{D}'(\Omega_1\times\Omega_2). $$ Combining these results yields $$ \mathcal{C}((0,T),\mathcal{D}'(\Omega)) \hookrightarrow \mathcal{D}'((0,T),\mathcal{D}'(\Omega)) = \mathcal{D}'((0,t)\times \Omega) $$ as you need. Providing all the details for the above arguments would need more space than an answer here allows, but you can find them for example in the book "Topological Vector Spaces, Distributions and Kernels" by F. Trèves or in the book "Lectures on Mixed Problems in Partial Differential Equations and Representations of Semi-groups" by L. Schwartz.

Answer 2

$P$ belongs to $\mathcal{D}'(Q)$ because $C([0,T]; L^2(\Omega))\subset L^2(0,T; L^2(\Omega))\cong L^2(\Omega\times (0,T))\subset\mathcal{D}'(Q)$ and $\langle \partial_t (u(t)-u(0)-\Delta U(t)-F(t)), \varphi\rangle=\langle u_t(t)-\Delta u(t)-f(t), \varphi\rangle$ by the explanation in this post.