# Question on vector valued distributions: weak time derivative calculation

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!

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 $$(1)$$.

• This is a great help! Thanks a lot for your detailed answer!! – kaithkolesidou Jan 28 '19 at 10:11