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Although I read all the similar posts here, I still can't find the answer in my question so I 'll try to pose it as clear as I can.

DEFINITION $1$: Let $f\in L^1_{loc}(I, X)$. Then $\;\;\langle T_f, \phi \rangle=\int_{I} f\phi$ for $\phi \in D(I)$, defines a $X-$valued distribution on $I$, i.e $T_f \in D'(I,X)$

DEFINITION $2$: Let $T_f \in D'(I, X)$. We define the distributional derivative of $T_f$, $T_f' \in D'(I,X)$, by $\langle T'_f,\phi \rangle=-\langle T_f,\phi' \rangle$, for $\phi \in D(I)$

DEFINITION $3$: Let $u \in L^1(0,T:;X)$. We say that $v \in L^1(0,T:;X)$ is the weak derivative of $u$ written $u'=v\;$ provided $\int_{0}^T \phi'(t)u(t)\; dt=-\int_{0}^T \phi(t)v(t) \;dt$ for all scalar test functions $\phi \in C^{\infty}_c(0,T)$

So, it is clear to me that the weak time derivative of $u$ is in fact the distributional derivative of $u$. I am a bit familiar to distribution theory but not in the $X-$valued one. However an analog should exist between these two and this is what I 'm trying to understand here.

Suppose now that $u \in L^2(0,T;H^1_0(U))$ with $u_t \in L^2(0,T;H^{-1}(U))$. If $\phi \in C^{\infty}_c(0,T;H^1_0)$ then what can we say about

$(*)\int_{0}^T \int_U u_t(t,x) \phi (t,x) \;dxdt=$?

After some studying on this topic, I found that most calculations use the inner product as well as the dual pairing. More specifically, definition $3$ is equivalent to:

$(**)\int_{0}^T (u(t),\psi'(t))_{L^2}=-\int_{0}^T \langle u_t(t),\psi(t) \rangle_{H^{-1},H^1}$ for $\psi \in C^{\infty}_c(0,T;H^1_0)$

If I understood right, $(*)$ using $(**)$ can be written as:

$\int_{0}^T \int_U u_t(t,x) \phi (t,x) \;dxdt=-\int_{0}^T \int_U u(t,x) \phi_t (t,x) \;dxdt \Leftrightarrow \\ \int_{0}^T \langle u_t(t,\cdot),\phi(t,\cdot) \rangle_{H^{-1},H^1}=-\int_{0}^T (u(t,\cdot),\phi'(t,\cdot))_{L^2}$

Is the above correct?

I feel that what I 'm asking is quite elementary or silly but I've been stuck to definitions and expressions while all I want is to understand the explicit calculation of the distributional derivative of $u$.

Any help is much appreciated. Thanks in advance!

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Is the above correct?

I'd say yes, provided that $u\in L^2(0,T;L^2(U))$ instead of $L^2(0,T;H^{-1}(U))$. To see this, we could use the integration by parts formula:

Theorem (Dautray, p. 473 and p. 477). Let $V$ and $H$ be Hilbert spaces such that

  • $V$ is continuously and densely embedded into $H$
  • $H$ is identified with $H'$
  • $H'$ is continuously and densely embedded into $V'$

(A) If $u\in L^2(a,b;V)$ and $u_t\in L^2(a,b;V')$, then $u\in C([a,b]; H)$.

(B) If $u,v\in L^2(a,b;V)$ and $u_t,v_t\in L^2(a,b;V')$, then $$\int_a^b\langle u_t(t),v(t)\rangle_{V'\times V}\;dt=(u(b),v(b))_H-(u(a),v(a))_H-\int_a^b\langle u(t),v_t(t)\rangle_{V\times V'}\;dt\tag{1}$$ which has a sense by (A).

Therefore:

  • If $u,v\in L^2(0,T;H_0^1(U))$ and $u_t,v_t\in L^2(0,T;H^{-1}(U))$, then \begin{align*} \int_0^T\langle u_t(t),v(t)\rangle_{H^{-1}\times H_0^1}\;dt=(u(T),v(T))_{L^2}-(u(0),v(0))_{L^2}\\ -\int_0^T\langle u(t),v_t(t)\rangle_{H_0^1\times H^{-1}}\;dt\tag{2} \end{align*}

  • If we have the additional regularity $v\in C_c^\infty(0,T;H_0^1(U))\subset L^2(0,T;H_0^1(U))$, then $v_t\in C_c^\infty(0,T;H_0^1(U))\subset C_c^\infty(0,T;H^{-1}(U))\subset L^2(0,T;H^{-1}(U))$. In this case, the last equality becomes $$\int_0^T\langle u_t(t),v(t)\rangle_{H^{-1}\times H_0^1}\;dt=-\int_0^T( u(t),v_t(t))_{L^2}\;dt.\tag{3}$$

  • If we have the additional regularity $u_t\in L^2(0,T;L^2(U)) \subset L^2(0,T;H^{-1}(U))$, then the last equality becomes $$\int_0^T(u_t(t),v(t))_{L^2}\;dt=-\int_0^T( u(t),v_t(t))_{L^2}\;dt,\tag{4}$$ which can be rewritten as $$\int_0^T\int_U u_t(t,x)v(t,x)\;dx\;dt=-\int_0^T\int_U u(t,x)v_t(t,x)\;dx\;dt.\tag{5}$$

Thus $(5)$ is the same as $(4)$, which is a particular case of $(3)$, which is a particular case of $(2)$, which is a particular case of $(1)$.

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  • $\begingroup$ This is a great help! Thanks a lot for your detailed answer!! $\endgroup$ – kaithkolesidou Jan 28 at 10:11

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