# Derivative of Banach-space valued functions, and weak solutions

1. Mar 26, 2012

### zmhl0910

Hello everyone,

I'm new to this forum and this is my first post. I hope this is the right place to ask the following.

I've been reading up a little on PDE theory lately, and the following point has been causing me a bit of confusion.

In the definition of the weak solution of a parabolic PDE, one considers Banach-space valued functions

$u\in L^2\left(0,T; H_0^1(\Omega)\right)$​

such that

$u' \in L^2\left(0,T; H^{-1}(\Omega)\right)$​

Here $T>0$ and $\Omega\subset\mathbb{R}^n$ ($n\ge 3$) is bounded open.

Does $u'$ here refer to the weak derivative of $u$, i.e. the function defined by

$\int_0^T u'(t)\phi(t) \;\mathrm{d}t := -\int_0^T u(t) \phi'(t) \;\mathrm{d}t \quad \mbox{for all \phi\in C^\infty_c(0,T)}$​

(Here the integrals are Bochner integrals.)

If so, how does one make sense of the above expression? Specifically, my gripe is that the left-hand side technically belongs to $H^{-1}(\Omega)$ while the right-hand side belongs to $H_0^1(\Omega)$. Is one supposed to identify both sides via the self-duality of $H_0^1(\Omega)$ as a Hilbert space?

In which case, what is the rationale for distinguishing between $L^2\left(0,T; H_0^1(\Omega)\right)$ and $L^2\left(0,T; H^{-1}(\Omega)\right)$ (for example, in the definition of the weak solution to a parabolic PDE) in the first place?