# Difference between weak and distributional derivatives

I'm studing weak and distributional derivatives and solutions and I have a few questions about it.

1. From my understanding, one defines a weak derivative of $$u \in L^{1}_{loc}(\Omega)$$ such that

$$\int_{\Omega} D^{\alpha} u \varphi dx = (-1)^{|a|}\int_{\Omega} v D^{\alpha}\varphi dx$$

for all test functions $$\varphi \in C^{\infty}_{c}$$ and $$v \in L^{1}_{loc}(\Omega)$$. So $$v$$ is a weak derivative of u.

My question is: what is the difference between this definition and the definition of distributional derivatives? From what I understood, the integral of ($$\int_{\Omega} u \varphi dx$$) defines a distribution so the distributional derivative is defined as seen in equation above.

I understand that while working with weak derivatives, I'm differentiating functions and when one is working with distributional derivatives it's obviously differentiating distributions. But what is the difference between these two definitions?

1. Is it necessary, for the definition of a weak derivative, that the function $$u \in L^{1}_{loc}$$?

Thanks.

Concerning your second question: Yes, it's necessary to give meaning to the integral $$\int_a^b u\varphi'\,dx$$ for all test functions.
As you guessed correctly, the link between the definitions is the integral $$T_u(\varphi) := \int_a^bu\varphi\,dx.$$ And indeed, $$T_u$$ is a distribution. Such distributions are called regular. For a general distribution $$T$$ the derivative $$T'$$ is again a distribution and is defined by $$T'(\varphi) := -T(\varphi')$$. If $$u$$ has a weak derivative, then $$T_u'(\varphi) = -T_u(\varphi') = -\int_a^bu\varphi'\,dx = \int_a^b u'\varphi\,dx = T_{u'}(\varphi),$$ such that $$T_u' = T_{u'}$$. However, $$u$$ might not have a weak derivative. In this case, one just has $$T_u'(\varphi) = -T_u(\varphi') = -\int_a^bu\varphi'\,dx,$$ which is not a regular distribution.
You can always differentiate distributions (an infinite number of times). When you see something like $$\Delta u\in L^2$$ somewhere (possibly with the remark "in the distributional sense"), it means that $$\Delta T_u$$ is a regular distribution $$T_v$$ with $$v\in L^2$$.
• So it's a weak derivative when I can "go back" to a function (i used $v$ for represent $u'$). On the other hand, sometimes I can't really "go back" to a function (the derivative is a singular distribution)? So the two concepts are really linked to that definition but the result can be fundamentally different... Commented Aug 1, 2021 at 3:20
• If by "go back" you mean that $T_u'$ is incidently regular ("a function"), then yes. Commented Aug 1, 2021 at 3:29